Functional Analysis Kalton

Universitext Series Editors Sheldon Axler Department of Mathematics, San Francisco State University San Francisco, California, USA Vincenzo Capasso Dipartimento di Matematica, Università degli Studi di Milano Milano, Italy Carles Casacuberta Depto. Àlgebra i Geometria, Universitat de Barcelona Barcelona, Spain Angus Mcintyre Queen Mary University of London, London, United Kingdom Kenneth Ribet Department of Mathematics, University of California Berkeley, California, USA Claude Sabbah CNRS, Ecole polytechnique Centre de mathématiques Palaiseau, France Endre Süli Worcester College, University of Oxford, Oxford, United Kingdom Wojbor A. Woyczynski Department of Mathematics, Case Western Reserve University Cleveland, Ohio, USA Universitext is a series of textbooks that presents material from a wide variety of mathematical disciplines at master’s level and beyond. The books, often well class- tested by their author, may have an informal, personal even experimental approach to their subject matter. Some of the most successful and established books in the series have evolved through several editions, always following the evolution of teaching curricula, to very polished texts. Thus as research topics trickle down into graduate-level teaching, first textbooks written for new, cutting-edge courses may make their way into Universitext. More information about this series at http://www.springer.com/series/223 Adam Bowers • Nigel J. Kalton An Introductory Course in Functional Analysis 2123 Neither the publisher nor the authors or the editors give a warranty. in this publication does not imply. the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. LLC 2014 This work is subject to copyright. or by similar or dissimilar methodology now known or hereafter developed.Adam Bowers Nigel J. that such names are exempt from the relevant protective laws and regulations and therefore free for general use. with respect to the material contained herein or for any errors or omissions that may have been made. express or implied. reprinting. service marks. San Diego University of Missouri. The use of general descriptive names. even in the absence of a specific statement. reuse of illustrations. trademarks.com) . computer software. specifically the rights of translation. The publisher. recitation.springer. Columbia La Jolla. registered names. reproduction on microfilms or in any other physical way.1007/978-1-4939-1945-1 Library of Congress Control Number: 2014955345 Springer New York Heidelberg Dordrecht London © Springer Science+Business Media. CA Columbia. and transmission or information storage and retrieval. electronic adaptation. All rights are reserved by the Publisher. Printed on acid-free paper Springer is part of Springer Science+Business Media (www. broadcasting. etc. whether the whole or part of the material is concerned. MO USA USA ISSN 0172-5939 ISSN 2191-6675(electronic) Universitext ISBN 978-1-4939-1944-4 ISBN 978-1-4939-1945-1 (eBook) DOI 10. Kalton (deceased) Department of Mathematics Department of Mathematics University of California. To the memory of Nigel J. Kalton (1946–2010) . All functional analysts should be grateful to Adam for his kind en- deavour. Fate had in store that he completed this work alone. and find it rewarding to master a few pages of its daunting and ciphered chapters. a post-doctoral student during that year.Foreword Mathematicians are peculiar people who spend their life struggling to understand the great book of Mathematics. Adam Bowers. It will therefore be a perfect base for teaching a one-semester (or two) graduate course in functional analysis. and for the splendid textbook he provides. no matter what. Fortunately. his mathematical legacy is accessible. That book is now closed and we are left with the grief. and share with his colleagues and students his enlightening vision. Paris. gathered very careful notes of all classes and agreed with Nigel that these notes would eventually result in a textbook. and the duty to follow Nigel’s example and to keep working. But this book was wide open in front of Nigel Kalton. which includes the last graduate course he taught in Columbia during the Academic Year 2009–2010. and enjoy. Indeed this book is a smooth and well-balanced introduction to functional analysis. Please open this book. A cascade falling from so high is a powerful force. Adam Bowers decided that this tribute to Nigel should be co-authored by the master himself. constantly motivated by applications which make clear not only how but why the field developed. and a beautiful sight. who could browse through it with no apparent effort. France Gilles Godefroy January 2014 vii . and all of the relevant theory is provided. The current text seeks to give an introduction to functional analysis that will not overwhelm the beginner. we begin with a discussion of normed spaces and define a Banach space. As such. In the cases when a complete treatment would be more of a distraction than a help. but rather as a first glimpse at a vast and ever-deepening subject. The material is meant to be covered from beginning to end. Over the course of the semester. Nigel Kalton could always show something new. Without him. and allows us to work in a setting which is more intuitive than is necessary ix . While some experience with measure theory and complex analysis is expected. Nigel Kalton taught what would be his final course. The additional structure in a Banach space simplifies many proofs. Sadly. At the time. one need not be an expert. and all of the advanced theory used throughout the text can be found in an appendix. if I had the good fortune to teach a functional analysis course of my own. and watching him present a subject he loved was a joy in itself. About this book This book. No matter how well one knew a subject. It is not meant to function as a reference book. is meant as an introduction to the topic of functional analysis. this work—and indeed mathematics itself—suffers from a terrible loss. Nigel was unexpectedly taken from us before the text was complete. because I simply enjoyed watching him lecture. The book is designed so that a graduate student with a minimal amount of advanced mathematics can follow the course. and should fit comfortably into a one-semester course. I sat in on the course. I took notes diligently. usually as needed. The text is essentially self-contained. Nigel Kalton’s notes would make the best foundation. the necessary information has been moved to an appendix. I was a postdoctoral fellow at the University of Missouri- Columbia and Nigel was my mentor.Preface During the Spring Semester of 2010. he suggested we turn the notes into a textbook. It had occurred to me that someday. which was an introduction to functional analysis. as the title suggests. When I happened to mention to Nigel what I was doing. The origins of functional analysis lie in attempts to solve differential equations using the ideas of linear algebra. It is hoped that the reader will find the material intriguing and seek to learn more. About Nigel Kalton Nigel Kalton was born on 20 June 1946 in Bromley. we have sacrificed some generality for the sake of the reader’s comfort. Conway [8] or (moving in another direction) Topics in Banach Space Theory by Albiac and Kalton [2]. 4.D. we introduce the celebrated Hahn–Banach Theorem and explore its many consequences. For further study. While these spaces will lack some of the advantages of Banach spaces. The exercises come in varying degrees of diffculty. we relax our requirements and consider a broader class of objects known as locally convex spaces. 2. Some are very straightforward. 3. 7. considerable and interesting things can and will be said about them. The inquisitive mind would do well with the classic text Functional Analysis by Walter Rudin [34]. In Chap. As a final flourish. we meet the key examples of Banach spaces—examples which will appear again and again throughout the text. Consequently. and see how the Hahn–Banach Theorem appears in this context.x Preface for the development of the theory. In Chap. In Chap. He held positions at Lehigh University. where we see an example of how techniques from functional analysis can be used to solve a system of differential equations. At the end of each chapter. Banach spaces enjoy many interesting properties as a result of having a complete norm. we will prove the Wiener Inversion Theorem. and we will encounter results which allow us to do unexpected things. 6. We will meet the spectrum of an operator and see how it relates to the seemingly unrelated concept of maximal ideals of an algebra. but some introduce new concepts and ideas and are meant to expose the reader to a broader selection of topics. in 1970. Many of the exercises are directly related to topics in the chapter and are meant to complement the discussion in the textbook. such as sum the series ∞ 1 n=1 n 4 . His thesis was awarded the Rayleigh Prize for research excellence. and more. which provides a nontrivial result about Fourier series. Warwick University. He studied mathematics at Trinity College Cambridge. 8. In Chap. and (hopefully) understanding. but some are quite challenging. We conclude the course in Chap. University College of Swansea. We will continue our discussion of compact operators in Chap. After a general discussion of topological preliminaries. the University of . with a discussion of Banach algebras. extreme points. where we first meet compact operators. We will glimpse these ideas in Chap. including the Baire Category Theorem. England. we investigate some of the consequences of completeness. where he took his Ph. which covers the material of this text. the reader will find a collection of exercises. we consider topics such as Haar measure. the reader might wish to peruse A Course in Operator Theory by John B. the Open Mapping Theorem. and the Closed Graph Theorem. 4. D. he was appointed a Curators’ Professor. Among the many honors Nigel Kalton received. Casazza. The article (which is [6] in the references) is a good starting point to learn about the life and work of Nigel Kalton. Kalton was appointed the Luther Marion Defoe Distinguished Professor of Mathematics and then he was appointed to the Houchins Chair of Mathematics in 1985. Notices of the AMS. and Roman Vershynin: A Tribute to Nigel J. and the Banach Medal from the Polish Academy of Sciences in 2005 (the highest honor in his field). During his career. and inspired countless mathematicians. please visit the the Nigel Kalton Memorial Website developed by Fritz Gesztesy and hosted by the University of Missouri: http://kaltonmemorial. Joe Diestel.edu/ A very nice tribute to Nigel Kalton appeared in the Notices of the American Mathematical Society with contributions from Peter Casazza. he wrote over 270 articles and books. the Weldon Springs Presidential Award for outstanding research (at the University of Missouri) in 1987. 942–951. mentored 14 Ph. Coordinating Editor. No. In 1984. served on many editorial boards. and Michigan State University before taking a permanent position in the mathematics department at the University of Missouri in 1979.Preface xi Illinois. Peter G.missouri. 7 (2012). Kalton (1946–2010). he was awarded the Chancellor’s Award for outstanding research (at the University of Missouri) in 1984. Vol. Gilles Godefroy. . the highest recognition bestowed by the University of Missouri. 59. For more information about Nigel Kalton. In 1995. students. pp. Aleksander Pełczyński. I owe my deepest gratitude to Professor Nigel Kalton. family. I assure the reader that if there are any errors in this book. Nadia Gal. they belong to me. who thoroughly read the first draft and provided invaluable suggestions and corrections. the staff at Springer. I am very grateful to Professor Gilles Pisier for sharing his knowledge of the Approximation Problem and providing insight into a source of great confusion. Simon Cowell. including friends. In particular. La Jolla. I offer my sincere thanks to Greg Piepmeyer. and Daniel Fresen for many helpful comments. and the anonymous reviewers. Brian Tuomanen. I offer this text as my humble tribute to his memory. I also wish to thank Minerva Catral. who was both an inspiration and mentor to me.Acknowledgements I owe much to those who have helped me during the creation of this text. Since I cannot show him the gratitude and admiration I feel. without which this book would not exist. Of course. California Adam Bowers January 2014 xiii . and I am deeply grateful to Professor Godefroy for all of his advice and generosity. I am indebted to Jennifer Kalton and Gilles Godefroy for their kind support. . . .8 Duals of Quotients and Subspaces . . . . . . . . . . . . .4 Applications of the Open Mapping Theorem . . 64 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Biduals. . . . . . . . . . . . . . . . . . . 24 Exercises . . . . .3 Completeness in Function Spaces . . . . . . . . . . . . . . . 48 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2. . . . . . . .1 Sequence Spaces . . . . .4 Haar Measure for Compact Abelian Groups . . . .2 Sublinear Functionals and the Extension Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 3. . . 1 Exercises . . . . . . . . . . . . 77 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 11 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7 New Banach Spaces From Old . . . 11 2. . . . . . . . . . . . . . . . . . . . . . . . . . . 61 4. . . . . . 39 3. . . . . . . . . .2 Applications of Category . . . . . . . . . . . . . . . . 58 4 Consequences of Completeness . . . . . . . . . . . . 50 3. . . . 61 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Contents 1 Introduction . . . . . . . . . . . . . . . . . . . . .6 The Adjoint of an Operator . .1 General Topology . . . . . 71 4. . . . . . . . . . . . . . . . .1 The Axiom of Choice . . . . . . . . . . .2 Function Spaces . . . . . . . . . . . . . . 81 5 Consequences of Convexity . . . . . . . . . .1 The Baire Category Theorem . . . . . . . . . . .3 The Open Mapping and Closed Graph Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3. . . . . . . . . . . . . . . . . . . . . . and More . .3 Banach Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 5. . . . . . . . . . . . . . . . . . . . . . .4 The Geometric Hahn–Banach Theorem . . . .2 Topological Vector Spaces . . . . . . . . . . . . . . . . . . . . 26 3 The Hahn–Banach Theorems . . . . . 44 3.5 Duals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 5. . 86 5. . . . . . . . . . . . . . . . . . . . . . . . . . 32 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Exercises . 7 2 Classical Banach Spaces and Their Duals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 5. . . . . . . . . . . . . . . . 93 xv . . . . . . . . . . . . . . . . . . .3 Some Metrizable Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 6 Compact Operators and Fredholm Theory . . . . . . . . . . . . . . . . . . . . . . . . . 108 5.4 Sturm–Liouville Systems . . . . . 118 5. . . . . . 205 Appendix A Basics of Measure Theory . . . . 181 8. . . . 148 7 Hilbert Space Theory . . . . . . . . . . . . . . . . . . . . . . .2 Commutative Algebras . . . . . . . .3 Hilbert–Schmidt Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 A Rank-Nullity Theorem for Compact Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 5. . . . . . . . . . . . . . .1 Basics of Hilbert Spaces . . 196 8. . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 7. . .9 Haar Measure on Compact Groups . . . . . . . . . . . . . . . 141 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 8. . . . . . . . . . . . . . . . . 170 Exercises . . . . . . . . . . . . 106 5. . . . . . . . . . . . . . . . . . . . . . . 151 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 Operators on Hilbert Space . . . . . . . . .10 The Banach–Stone Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 6. . . . . . . . . . . . . . . . . . . . . . . . . .1 Compact Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6 Mazur’s Theorem . . . . . . . . . . . . . . .xvi Contents 5. . . . . . . . . 202 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 Exercises . . . . . . . 227 . 207 Appendix B Results From Other Areas of Mathematics . . . . . . . . . . . 219 References . . . . . . . . . . . . . . . . . . . . . . 111 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 7. . . . . . . . . . . . . . . . . . 129 6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 8 Banach Algebras . . . . .8 Milman’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . .1 The Spectral Radius . . . . . . . . . . . . . . .7 Extreme Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3 The Wiener Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5 Goldstine’s Theorem . . . . . . . . . . . . . . . . . . . z) + d(z. The goal of functional analysis is to generalize the well-known and very successful results of linear algebra on finite-dimensional vector spaces to the more complicated and subtle infinite-dimensional spaces. that is.1 A normed space is a real or complex vector space X together with a real-valued function x → x defined for all x ∈ X. the norm of x is given by the absolute value (or modulus) of x. Property (N2) is called subadditivity or the triangle inequality. DOI 10. (N2) x + y ≤ x + y for all x and y in X. y. such that (N1) x ≥ 0 for all x ∈ X. y) = d(y. and (M3) d(x. and x = 0 if and only if x = 0. Any normed space is metrizable. As was the case with a norm. Universitext. N. (x. certain issues (such as summability) become much more delicate. we can always introduce a metric on a normed space by d(x.1007/978-1-4939-1945-1_1 . LLC 2014 1 A. (M3). A metric is a measure of distance on the set X. x) for all x and y in X. in infinite dimensions. called a norm. and Property (N3) is called homogeneity. An Introductory Course in Functional Analysis. J. © Springer Science+Business Media. is called symmetry. and d(x. y) ≤ d(x.2 A metric space is a set X together with a map d : X × X → R that satisfies the following properties: (M1) d(x. which is denoted (in either case) by |x|. The function d is said to be a metric on X. Definition 1. Definition 1. y) = x − y. y) ∈ X × X. and (N3) λx = |λ| x for all x ∈ X and scalars λ. Property (N1) is known as non-negativity or positive-definiteness. Bowers. (M1) is known as non-negativity and (M2) is called the triangle inequality. Perhaps the most natural structure imposed upon a vector space is that of a norm. y) for all x. Of course. (M2) d(x. and accordingly additional structure is imposed. A norm on a vector space X is essentially a way of measuring the size of an element in X. y) = 0 if and only if x = y. If x is in R (or C).Chapter 1 Introduction Functional analysis is at its foundation the study of infinite-dimensional vector spaces. The final property. Kalton. y) ≥ 0 for all x and y in X. and z in X. then for every n ∈ N. A sequence (xn )∞ n=1 in X is said to converge to a point x in X if for any > 0 there exists an N ∈ N such that d(x. If T is unbounded. spaces in which every Cauchy sequence does converge. Such a space is called incomplete. 5. A sequence (xn )∞ n=1 in X is called a Cauchy sequence if for any > 0 there exists an N ∈ N such that d(xn . A linear map T : X → Y is called bounded if there exists a constant M ≥ 0 such that T x ≤ M for all x ≤ 1.. A convergent sequence will always be a Cauchy sequence. much study is made of linear maps on Banach spaces. A basic goal of functional analysis is to solve equations of the form T x = y. By the choice of xn . n ∈ N. y) = x − y for all (x.3 Let X be a metric space with metric d. For these spaces we have a special name. An early and important result in functional analysis is the following proposition. if T is a bounded linear map.4 A normed space X is called a Banach space if it is a complete metric space in the metric given by d(x. and T : X → Y is a given linear map (e. a differential or integral operator). where x ∈ X. then T = sup{ T x : x ≤ 1}.) Definition 1. we say the sequence is convergent and write limn→∞ xn = x. Proposition 1. Suppose X and Y are Banach spaces (or simply normed spaces) over a scalar field K (which is R or C).5 Let X and Y be normed spaces. we are generally interested in normed spaces that are complete. (ii) T is continuous at zero. This is not always true in infinite-dimensional normed spaces. While every norm determines a metric. xm ) < whenever n ≥ N and m ≥ N.3. It is well-known that a scalar-valued sequence converges if and only if it is a Cauchy sequence. because we imagine it lacks certain desirable points. For this reason. Definition 1. (We will see some examples of this in Sect. In such a case. that is. n n . Consequently. not every metric space can have a norm. and (iii) T is bounded. Of particular interest are bounded linear maps. but there may be Cauchy sequences that do not converge to an element of the normed space. we have that x 1 n ≤ . there exists an element xn ∈ X such that xn ≤ 1 and T xn ≥ n.6 Let X and Y be normed spaces and suppose T : X → Y is a linear map. xn ) < whenever n ≥ N . A linear map that is not bounded is called unbounded. Proof The implication (i) ⇒ (ii) is clear.g. assume T is continuous at zero. y ∈ Y . y) ∈ X × X. To show (ii) ⇒ (iii). Definition 1. The following are equivalent: (i) T is continuous. but is unbounded. We denote the smallest such M by T .2 1 Introduction This metric measures the size of the displacement between x and y. that is. n This is a contradiction. n→∞ n However. because x x has norm 1. Y ). Furthermore. Now we show (iii) ⇒ (i). n n By continuity at zero. x ∈ X}. Therefore. x T (x) = T · x ≤ T x. x ∈ X. Y ) is a vector space over the scalar field K with vector space operations given by: (αS + βT )(x) = αS(x) + βT (x). This allowed us to conclude that T was continuous. The proof of Proposition 1. x n T xn T = n ≥ 1. however. then L(X. X) of bounded linear maps from X to itself is often denoted by L(X).6 provides an alternative method for computing T . Definition 1. A function f : X → Y that satisfies the condition f (x) − f (y) ≤ Kx − y for a constant K > 0 is called Lipschitz continuous with Lipschitz constant K. where α and β are in K. we say T ∈ L(X. In fact. n ∈ N. it must be that x n lim T = T (0) = 0. we showed that T x − T y ≤ T x − y for all x and y in X. and x is in X. Assume that T is bounded and let x ∈ X. then T = inf{K : T x ≤ K x.6. Then T ( x x ) ≤ T . By the linearity of T .8 Let X and Y be normed spaces. T must be bounded. If T : X → Y is a linear map. (x. The set L(X. T x ≤ T x for all x ∈ X. The word operator implies both boundedness and linearity.Introduction 3 x n xn It follows that → 0 as n → ∞. Corollary 1. even though this is redundant. This completes the proof. 2 Remark 1. y) ∈ X × X. this property is stronger than continuity. Frequently. but a Lipschitz continuous map between Banach spaces need not be linear. S and T are in L(X. In the proof of Proposition 1. Any bounded linear map T between Banach spaces is Lipschitz continuous (with Lipschitz constant T ). The set L(X. . we use the above inequality together with linearity: T x − T y = T (x − y) ≤ T x − y. and consequently → 0 in X as n → ∞. when showing (iii) ⇒ (i). Y ) denotes the set of all bounded linear maps (or operators) from X to Y . x To prove the continuity of T . Y ) is a bounded linear operator.9 If X and Y are normed spaces.7. so that T x = lim Tn x. x ∈ X. sup Tn < ∞. observe that for all x and y in X. the sequence (Tn x)∞ n=1 is a Cauchy sequence in Y because Tn x − Tm x ≤ Tn − Tm x. (See Exercise 1. ∞ Thus. The other conditions are also straightforward and are left to the reader. It follows that lim Tn x exists for each x ∈ X.11 If X is a normed space and Y is a Banach space. T (x + y) = lim Tn (x + y) = lim Tn (x) + lim Tn (y) = T x + T y. Consequently. T is bounded and T ≤ supn∈N Tn . It remains to show that T = lim Tn in the norm on L(X. Y ).10 If X and Y are normed spaces. then L(X. Proof We verify the triangle inequality.5. The linearity of T follows from the continuity of the vector space operations. For each m and n in N. Y ).3) |Tm − Tn | ≤ Tm − Tn . S + T = sup{(S + T )(x) : x ≤ 1} ≤ S + T . m≥n m≥n . n→∞ We will show that T is linear and bounded. Y ). For n ∈ N. Y ) is a Banach space. since (Tn )∞ n=1 is a Cauchy sequence in L(X. n∈N T x ≤ sup Tn x ≤ sup Tn x. T ∈ L(X. defines a norm on L(X. it follows that (Tn )n=1 is a Cauchy sequence of scalars. n∈N n∈N Therefore. n→∞ n→∞ n→∞ We now show that T is bounded. Let x ∈ X be such that x ≤ 1. Proof Suppose (Tn )∞ n=1 is a Cauchy sequence in L(X. and so. (T − Tn )(x) ≤ sup (Tm − Tn )(x) ≤ sup Tm − Tn . for all m and n in N. Y ). for each x ∈ X. we have (from Exercise 1. For each x ∈ X. Denote this limit n→∞ by T x. then T = sup{ T x : x ≤ 1}. Let x ∈ X be such n→∞ that x ≤ 1.) 2 Proposition 1. and so T ∈ L(X. To see this. Y ). Then (S + T )(x) = Sx + T x ≤ Sx + T x ≤ S + T .4 1 Introduction Proposition 1. Y ). Therefore. 8.) The next definitions are central to all that follows.12 Let X and Y be normed spaces. we say f preserves the norm. 2 When X and Y are normed spaces. Definition 1. We remind the reader that a one-to-one map is called an injection and an onto map is called a surjection. The dual space of X is the space L(X. In such a case. (See Exercise 1. If · α and · β are two norms on X. then they are called equivalent norms provided there are constants c1 > 0 and c2 > 0 such that c1 xα ≤ xβ ≤ c2 xα . . · α ) and (X. m≥n The sequence (Tn )∞ n=1 is a Cauchy sequence.13. If in addition T x = x for all x ∈ X.) We will see that much can be learned about a Banach space by studying the properties of its dual space. A bijective linear map T : X → Y is called an isomorphism if both T and T −1 are continuous.6. then T is called an isometric isomorphism. Let X be a vector space. y) ≤ δ}.) If there exists an isometry between the spaces X and Y . An injective surjection is also called a bijection. Definition 1. δ) = {y ∈ M : d(x. they are said to be isometric. which is denoted X ∗ . and hence the result. The closed ball of radius δ about x is the set B(x. Elements of X∗ are called (bounded) linear functionals. x ∈ X. Definition 1. Taking the supremum over all x ∈ X with x ≤ 1. 5. (See Exercise 1. we have T − Tn ≤ sup Tm − Tn . We will encounter these concepts in greater generality in Chap.) When f is linear. but for now we will restrict our attention to metric spaces. · β ). y) < δ}. the open ball of radius δ about x is the set B(x. and even indicate the origins of the subject we are studying.Introduction 5 by Corollary 1. We close this section by recalling some topological notions that will appear frequently throughout the remainder of this text. the dual space X ∗ is a Banach space with the norm given by x ∗ = sup{|x ∗ (x)| : x ≤ 1} for x ∗ ∈ X∗ . In such an event.13 Let X be a normed space over the scalar field K (which is either R or C). (In this case. For x ∈ M. We remark that for any normed space X. there is a linear isomorphism between the spaces (X. this condition is equivalent to stating that f (x) = x for all x ∈ X. (In which case. a (not necessarily linear) function f : X → Y is called an isometry if f (x) − f (y) = x − y for all x and y in X. δ) = {y ∈ M : d(x. K). we say f preserves distances.14 Let M be a metric space with metric d. we say X and Y are isomorphic. Compact sets have many desirable properties. Later. Also. if K is a compact set and f : K → R is a continuous function. Definition 1. (See Exercise 1. we say that M1 and M2 are homeomorphic. then f is bounded and attains its maximum and minimum values. is the union of all open sets that are subsets of E. (This is the content of the Extreme Value Theorem. Observe that UX = int(BX ). . δ) = x + int(δBX ). . we will show that any bounded linear bijection between Banach spaces is necessarily an isomorphism. A set is called closed if its complement is open. where xn ∈ E for all n ∈ N. If there exists a countable dense subset of M. The closure of E is denoted E and is the intersection of all closed sets which contain E as a subset. we can write the closed and open balls in X (respectively) as: B(x. converges to some x ∈ E.) Digression: Historical Comments Functional analysis has its beginnings in the work of Fourier (1768–1830) and the study of differential equations. We remark that a set E in a metric space is closed if and only if any convergent sequence (xn )∞n=1 . then M is called separable. δ) = x + δBX and B(x. Uαn } such that K ⊆ Uα1 ∪ · · · ∪ Uαn . then the closed unit ball of X is the set BX = {x ∈ X : x ≤ 1} and the open unit ball of X is the set UX = {x ∈ X : x < 1}. in order to emphasize the underlying linear structure. the interior of E. and consequently has addition and scalar multiplication. Fourier’s goal was to find solutions to differential equations such as d 2y y + 2 = g(x). Notice that an isomorphism is a linear homeomorphism. . For any subset E of M. For example. denoted int(E). In particular.6 1 Introduction A subset U of M is called open if for every x ∈ U there exists a δ > 0 such that B(x. we have special terminology and notation for the open ball B(0. then E is called dense in M. We will use both notations for the open unit ball in X. A set K in a metric space M is called compact if any collection {Uα }α∈A of open sets for which K ⊆ α∈A Uα contains a finite subcollection {Uα1 . Homeomorphic spaces are considered identical from a topological point of view.15 If X is a normed space. dx . δ) ⊆ U . 1). If such a homeomorphism exists. then a map f : M1 → M2 is called a homeomorphism if it is a continuous bijection with continuous inverse.) If M1 and M2 are metric spaces.30. any compact set in a metric space can be covered by finitely many open balls of radius δ for any δ > 0. This is the notation we will generally use when working in a Banach space (or normed vector space). (See Corollary 4.10. .) For a normed space. 1) and the closed ball B(0. if X is a Banach space (or simply a normed vector space). If E = M. If K is any compact Hausdorff space. Banach himself died shortly after the war in 1945. To this end. ∞). With the advent of Lebesgue’s thesis in 1903. 1). y) f (y) dy = g(x). the space C(K) denotes the collection of all scalar-valued continuous functions on K.1] Note that this maximum is attained.36). called the supremum norm: f = max |f (t)|.2 Let X be a normed space. 1) for p ∈ [1. Exercises Exercise 1. When equipped with the norm f = max|f (x)|. 0 where K and g were given. Fredholm was trying to find continuous functions f on the unit interval [0. 1]. Much of the work of the Lwów school survived in what is now called the Scottish Book [24]. (However. f ∈ C(K). 1) for all p ∈ [1. 1] is denoted C[0.) The notion of a measurable function was critical in the understanding of these spaces—when restricted to continuous functions they are not complete.1 Show that in a metric space a convergent sequence is a Cauchy sequence. by Lusin’s Theorem (Theorem A. 1] is compact. 1]. and the subsequent development of measure theory. Exercise 1.Exercises 7 where g is some prescribed function. 1] can be generalized. Banach and his school at Lwów made many advances in the area.) Norbert Wiener (in 1919) and Stefan Banach (in his thesis in 1920) independently introduced the axioms of what are now called Banach spaces. t∈[0. ·) is a Banach space. ∞). Show that in a normed . In 1908. (See Appendix A for the relevant definitions.) The space (C[0. and was the first Banach space considered (although the term was not applied until much later). The space of continuous functions on [0. because f is continuous and [0. 1]. There is a natural norm on this space. x∈K this space is a Banach space. Hilbert and Schmidt studied L2 (0. x ∈ [0. the continuous functions are dense in Lp (0. In 1902. The space C[0. 1] that satisfied equations such as 1 f (x) + K(x. A sequence (xn )∞ n=1 in X is said to be bounded if there exists a M > 0 such that xn ≤ M for all n ∈ N. he applied to this setting the techniques of the already well-developed theory of linear algebra. Fredholm applied similar techniques to solve integral equations. but their work was cut short by the Second World War. Riesz studied the more general spaces Lp (0. and F. more examples of Banach spaces were discovered. 1] instead of the supremum. 1]. (It is for this reason we compute the maximum over t in [0. f ∈ C[0. ) Define a norm · on C (1) [0. Prove that X∗ is a Banach space. Exercise 1.8 1 Introduction space a convergent sequence is bounded.10. 1] that have continuous derivative. Exercise 1. 1] by f = maxt∈[0. Exercise 1. · ). there does not exist a c > 0 such that T (x) ≥ cx for all x ∈ X.13 Suppose · α and · β are equivalent norms on a vector space X. 1] be the space of all continuous functions on [0. .9 Let X be a normed vector space and suppose T and S are in X ∗ . Exercise 1.4 Let X and Y be normed vector spaces. Show that the map 1 A(f ) = f (x) dx 0 defines a bounded linear functional on the normed vector space (C[0. must a bounded sequence necessarily converge? Exercise 1.12 Let X and Y be Banach spaces and let T ∈ L(X. Show that the map B(f ) = f (0) defines a linear functional on the normed vector space (C (1) [0. Y ). Exercise 1. Show that there is an isomorphism from (X. · ) that is not bounded.5 Finish the proof of Proposition 1. x ∈ X. (This is sometimes called the reverse triangle inequality. 1] and define a norm · on C[0.7 Let C[0.3 Let X be a normed space. show that there is a constant c such that S(x) = c T (x) for all x ∈ X. A function f : X → Y is said to be bounded on X if there exists a M > 0 such that f (x) ≤ M for all x ∈ X.10 Show that a subset of a metric space M is closed if and only if it is sequentially closed.6 Let X be a normed space. That is.1] |f (t)|. 1]. 1] be the space of continuous functions on [0. Exercise 1.) Exercise 1. 1] by f = maxt∈[0.8 Let C (1) [0.1] |f (t)|. If T (x) = 0 implies that S(x) = 0. Exercise 1. · α ) onto (X. Exercise 1. · β ). Show there exists a sequence (xn )∞ n=1 of norm one elements in X such that lim T (xn ) = 0. (This space is called the space of continuously differentiable functions. Show that |x − y| ≤ x − y for all x and y in X. Show that a nonzero bounded linear operator T : X → Y is not bounded on X. show that E is a closed subset of M if and only if every convergent sequence in E has its limit in E. 1]. Show that T is an isomorphism if and only if there are constants c1 > 0 and c2 > 0 such that c1 x ≤ T x ≤ c2 x.11 Let X and Y be normed vector spaces and suppose T : X → Y is a linear surjection. that is. n→∞ Exercise 1. Suppose T is not bounded below. Is the converse true? That is. Define a scalar-valued function · T on X by xT = T (x)Y . x ∈ X.Exercises 9 Exercise 1. Use the preceding exercises to conclude that X is a Banach space if and only if Y is a Banach space.15 Suppose · α and · β are two norms on a vector space X. · X ) and (Y . Show that (X. · α ) and (X. Show that the two norms are equivalent if and only if the normed spaces (X. Exercise 1. · β ) is a complete normed space. . · α ) is a complete normed space if (X.16 Suppose · α and · β are two equivalent norms on a vector space X. Y ) is an isomorphism.17 Let X and Y be isomorphic normed vector spaces. · Y ) be normed vector spaces and assume the map T ∈ L(X. Prove that · T is a norm on X and show that it is equivalent to the original norm · X . Exercise 1. · β ) have the same open sets. Exercise 1.14 Let (X. 1. 2. ∞) is the collection of sequences . .718).Chapter 2 Classical Banach Spaces and Their Duals In the next two sections. . . . . . we denote by en the sequence with 1 in the nth coordinate. 2. we will consider the classical sequence and function spaces. We will therefore take for granted that the various Banach spaces are indeed Banach spaces—putting off until Sect.1 The set p of p-summable sequences for p ∈ [1.3 the proofs that they are complete in the given norms. 0. The main purpose of these sections is to make the necessary definitions and to identify the dual spaces for these classical spaces. Also. . Definition 2. we let e = (1.) be the constant sequence with 1 in every coordinate (not to be confused with the base of the natural logarithm e ≈ 2. 1. 1. 0. so that en = (0. . and 0 elsewhere.1 Sequence Spaces In the context of sequence spaces. . .) for all n ∈ N. . ξ = (ξn )∞ n=1 ∈ p .) : |ξn |p < ∞ . . We leave the proof of this fact to the exercises. J. N. . ξn . the space ∗p can be identified with q . DOI 10. Kalton. LLC 2014 11 A. p is a Banach space when given the p-norm (for 1 ≤ p < ∞). An Introductory Course in Functional Analysis.1007/978-1-4939-1945-1_2 .27. Lemma 2. ξ2 . . ∞ p = (ξ1 . where 1 p + q1 = 1. ∞). n=1 The set p is a vector space under component-wise addition and scalar multiplication. (See Exercise 2. © Springer Science+Business Media. n=1 Define the p-norm on p by ∞ 1/p ξ p = |ξn | p . . ∞). .2 For p ∈ (1.]) Furthermore. Bowers.) The next lemma will identify the dual space of p for p ∈ (1. [See Theorem A. .7. (This is a nontrivial fact which we will take as given. Universitext. .1) n=1 where ξ = (ξn )∞ n=1 is any sequence in p . ζ p q/p ηq Therefore. n=1 n=1 n=1 Let ξ = ζ ζ p . We begin by constructing a sequence ζ = (ζn )∞ n=1 so that ζn = |ηn | q−1 (sign ηn ) for each n ∈ N. for any η = (ηn )∞ n=1 in q .29). n=1 n=1 It follows that φη is a bounded linear functional on p and φη ≤ ηq . we start by assuming the scalars are real. By Hölder’s Inequality (Theorem A. ∞) as the sequence space q . there is a linear functional φη on p such that ηq = φη and such that φη (ξ ) = ∞ ∞ n=1 ξn ηn for all ξ = (ξn )n=1 in p .12 2 Classical Banach Spaces and Their Duals Proof For simplicity. where p1 + q1 = 1. q/p the sequence ζ is in p and ζ p = ηq . Then ξ is a sequence in p such that ξ p = 1 and such that q φη (ζ ) ηq q− q φη (ξ ) = = = ηq p = ηq . First. We claim that φη is in fact equal to ηq . (2. we will demonstrate how elements in q determine linear functionals on p . Then ∞ ∞ ∞ |ζn |p = |ηn |(q−1)p = |ηn |q . Consequently. Next. this series is absolutely convergent (whence φη is linear) and ∞ 1/p ∞ 1/q |φη (ξ )| ≤ |ξn | p |ηn | q = ξ p ηq . n=1 n=1 n=1 where (q − 1)p = q follows from the assumption that 1 p + 1 q = 1. it will suffice to find a sequence ξ in p such that ξ p = 1 and φη (ξ ) = ηq . We will show thatthe sequence η is an element of q such that ηq = ψ and such that ψ(ξ ) = ∞ ∞ i=1 ξi ηi for all ξ = (ξi )i=1 in p . Observe that ∞ ∞ ∞ φη (ζ ) = ζn · η n = |ηn |q−1 (sign ηn ) · ηn = |ηn |q = ηqq . In order to show this. Let η = (ηn )∞ n=1 be an element in q and define a scalar-valued function φη on p by ∞ φη (ξ ) = ξn η n .) We wish to identify the dual space of p for p ∈ (1. we will show that all linear functionals on p can be obtained in this way. We have demonstrated that any sequence in q determines a bounded linear functional on p . Let ψ ∈ ∗p and define a sequence η = (ηi )∞ i=1 by letting ηi = ψ(ei ) for each i ∈ N. (We will consider the complex case at the end of the proof. 4) n→∞ n→∞ i=n+1 . and consequently ∞ 1/p lim ξ − ξ (n) p = lim |ξi |p = 0. To see this. computing directly. we conclude that n n p1 n p1 ζi e i = |ζi |p = |ηi |q . it suffices to show that ψ = φη . . observe that i=1 |ξ i | p < ∞. For each i ∈ N. .2. (2. 0. it follows that p1 n n |ηi | ≤ ψ q |ηi | q . . define ζi = |ηi |q−1 (sign ηi ). Suppose ξ = (ξi )∞ i=1 ∈ p and let ξ (n) = (ξ1 . (2. i=1 i=1 Dividing. ξn . It remains to show that ψ ≤ ηq and that ψ = φη . and so η ∈ q and ψ ≥ ηq . we have n n n |ζi | = p |ηi | (q−1)p = |ηi |q . . where φη is defined by (2.3) i=1 i=1 i=1 i=1 From (2.1). and consequently n n n p1 ψ ζi ei ≤ ψ ζi ei = ψ |ηi | q .2) and (2. the linear functional ψ is bounded on p . . we will show that η is in fact an element of q . i=1 p i=1 i=1 By assumption. ) for each n ∈ N. Since we have already demonstrated that φη ≤ ηq . i=1 i=1 i=1 n Computing the p -norm of the finite sequence i=1 ζi ei . Then. (2.3). by assumption. . for any n ∈ N. . i=1 i=1 This inequality holds for all n ∈ N. (n) We ∞ claim that ξ converges to ξ in the norm on p .1 Sequence Spaces 13 First. we obtain n n n n ψ ζi ei = ζi ψ(ei ) = ζi η i = |ηi |q . we see that n 1− p1 n q1 ψ ≥ |ηi | q = |ηi | q .2) i=1 i=1 p i=1 However. ∞) as the space q . which is the desired result. The proof in this case is essentially the same.4 The space ∗1 can be identified with ∞ . for n ∈ N. In order to do this. by the continuity of ψ. but we write η(ξ ) because we are viewing η as a linear functional in ∗p . we wish to identify the dual space of 1 . In the above equation. Proof The proof is similar to the proof of Lemma 2. where φη is the linear functional corresponding to η that appears in (2. n→∞ i=1 It follows that ψ = φη .14 2 Classical Banach Spaces and Their Duals Now.2 and is left to the reader. when defining ζn . observe that n n n ψ ξ (n) =ψ ξi ei = ξi ψ(ei ) = ξi ηi . Lemma 2. i=1 i=1 i=1 Therefore.2.1). the dual space action is ∞ η(ξ ) = ξ n ηn . The object η is a sequence in the space q .3 The set ∞ of bounded sequences is the collection of sequences ∞ ∞ = (ξn )n=1 : sup |ξn | < ∞ . 2 The previous theorem identified the dual space of p for p ∈ (1. n=1 where now ξ = (ξn )∞ ∞ n=1 is in 1 and η = (ηn )n=1 is in ∞ . (See Exercise 2. we write η(ξ ) as shorthand for φη (ξ ). ξ = (ξn )∞ n=1 ∈ ∞ . n∈N The set ∞ is a vector space under component-wise addition and scalar multiplication and is a Banach space when given the supremum norm. where p1 + q1 = 1. Now assume the scalar field is C.) As in Lemma 2. let ζn = |ηn |q−1 ρn . n=1 where ξ = (ξn )∞ ∞ n=1 is in p and η = (ηn )n=1 is in q . via the dual space action ∞ η(ξ ) = ξ n ηn . where ρn is a complex number such that |ρn | = 1 and ηn ρn = |ηn |. ∞ ψ(ξ ) = lim ψ(ξ (n) )= ξi ηi = φη (ξ ). we must introduce a new space of sequences.2. Next. n∈N Define the supremum norm on ∞ by ξ ∞ = sup |ξn |. 2 . Definition 2. However. The argument proceeds as it did in the real case. 1. . . let ξ = e.2.7 Let c0 be the space of all sequences converging to 0: c0 = (ξn )∞n=1 : lim ξ n = 0 . ).5 Let p ∈ [1. we can find ξ ∈ ∞ such that lim ξ − ξ ∞ = lim sup |ξk | = 0.5 motivates the next definition. with the convention that q = ∞ when p = 1. n→∞ The space c is also a Banach space with the supremum norm. however.2 and is left as an exercise for the reader. While 1 ⊆ ∗∞ . the spaces do not coincide. ∞) and suppose q is such that 1 p + 1 q = 1. (See .1 Sequence Spaces 15 Since we identify the dual space of p with q .2. . the constant (n) sequence having every term equal to 1. Definition 2. (n) (2. If p and q are conjugate exponents that are both finite. then ∗p = q and ∗q = p . Then ∗p = q . but e − e(n) ∞ = 1 for all n ∈ N.) The last sequence space we discuss is the space c.6 If p ∈ [1. The relationship between the exponents p and q in Theorem 2. As a simple example. and so e is an element of ∞ . then p and q are called conjugate exponents. Since ∗1 = ∞ .5) is 0.8 c0∗ = 1 . (See Exercise 2. ξn . Perhaps surprisingly. We summarize our results in the following theorem. That is. Let us now consider the space of sequences for which the limit in (2. it is understood that we mean there is a way of identifying the linear functionals in ∗p with the sequences in q and the identification is shown explicitly by the dual space action η(ξ ) = ∞ ∞ n=1 ξn ηn . Proof See Lemmas 2. 0. Definition 2.2 and 2. Theorem 2.4. n∈N Theorem 2. it is natural to ask if 1 is the dual space of ∞ . Definition 2. .4) for the case p = ∞. it is standard to write ∗p = q . . the dual space of c is also 1 . The proof is similar to that of Lemma 2. We have that e∞ = 1. The argument used in the proof of Lemma 2. When we write this.9 Let c be the space of all convergent sequences: c = (ξn )∞ n=1 : lim ξn exists . ∞) fails in the case p = ∞ because there exist bounded sequences (ξn )∞ n=1 that do not satisfy the equation that corresponds to (2. n→∞ The space c0 is a Banach space with the supremum norm ξ ∞ = sup |ξn |. ∞) and if q is such that p1 + q1 = 1. . where ξ = (ξn )n=1 is in p ∞ and η = (ηn )n=1 is in q .2 for p ∈ (1.5) n→∞ n→∞ k>n where ξ = (ξ1 . with the convention that q = ∞ when p = 1. albeit with a slightly different dual space action. . ξ = (ξn )∞ n=1 ∈ c0 . let (Ω. which is either R or C.) 2. Σ.16 2 Classical Banach Spaces and Their Duals Example 2. K denotes the underlying scalar field.) Definition 2.) It is straightforward to show that c0 is a closed subspace of c. and in turn c is a closed subspace of ∞ . (See Appendix A for a more detailed discussion. but for simplicity we will assume μ(Ω) < ∞. μ) be a measure space. then the characteristic function of A is the function . The theorems in this section are true for positive σ -finite measure spaces.1. As before.2 Function Spaces In this section. (See Exercise 2.10 If A is a subset of Ω. We begin by recalling some definitions from measure theory.22. where μ is a positive measure. 11 For p ∈ [1. μ) be a measure space. (See Theorem 2. ∞). μ) = f : Ω → K a measurable function : |f | dμ < ∞ . The set Lp (μ) is actually a collections of equivalence classes of measurable functions. Two functions in Lp (μ) are considered equivalent if they differ only on a set of μ-measure zero. Definition 2. f ∈ Lp (μ). Definition 2. Despite this. this is a nontrivial result. The essential supremum norm of a measurable function f is defined to be f ∞ = inf {K : μ(|f | > K) = 0} . the set of p-integrable functions (or Lp -functions) on (Ω. μ) is the collection Lp (Ω. (As with p in the previous section. Σ.) As in the case of sequence spaces. rather than equivalence classes of functions. μ) or Lp (μ) when there is no risk of confusion. p Ω We often write this space as Lp (Ω. 1 if ω ∈ A. The characteristic function of A is a measurable function if and only if A is a measurable subset of Ω.]) Define the p-norm on Lp (μ) by p1 f p = |f |p dμ . Σ. we will usually speak of the elements in Lp (μ) as functions. χA (ω) = 0 if ω ∈ A. we must consider the case p = ∞ separately. . Ω We will show that Lp (μ) is a Banach space in the p-norm (for 1 ≤ p < ∞) in the next section.12 Let (Ω. Σ. We remark that the set Lp (μ) is a vector space under pointwise addition and scalar multiplication.25. [See Theorem A.27. First. The set L∞ (μ) is a vector space under pointwise addition and scalar multiplication. We note that φg is linear (by the linearity of the integral) and |φg (f )| ≤ f p gq (by Hölder’s Inequality). and then we will show that all linear functionals on Lp (μ) can be achieved in this way. We will see that. If p and q are conjugate exponents. μ) is the collection L∞ (Ω. Theorem 2. because the quantity f ∞ is the smallest number K such that |f | ≤ K a. the dual action of Lq (μ) on Lp (μ) is given by integration. then Lp (μ)∗ = Lq (μ).(μ). We often write L∞ (Ω. ∞). where p ∈ [1. where p ∈ [1. analogous to the case of sequence spaces.2 Function Spaces 17 The set of essentially bounded functions (or L∞ -functions) on (Ω. Σ. In order to show equality of the norms. then the dual action of g on f is given by g(f ) = f gdμ. φg is a bounded linear functional and φg ≤ gq . In this case. (See Theorem 2. where p1 + q1 = 1. As is the case for Lp (μ) when p is finite. the set L∞ (μ) is a collection of equivalence classes of measurable functions and (as before) we consider two functions to be equal in L∞ (μ) if they differ only on a set of μ-measure zero. however. As before. μ) be a positive finite measure space. Proof Start by assuming the scalars are real. if f ∈ Lp (μ) and g ∈ Lq (μ). (2. Ω Notice that g is a function on Ω.2. f ∈ Lp (μ). and define a scalar-valued function φg on Lp (μ) by φg (f ) = f gdμ. where p and q are conjugate exponents.6) Ω We will show first that φg is a linear functional on Lp (μ) such that φg = gq . μ) = {f : Ω → K a measurable function : f ∞ < ∞} . We begin with the case p ∈ (1. Σ. we write Lp (μ)∗ = Lq (μ) to indicate the identification of linear functionals in Lp (μ)∗ with elements of the function space Lq (μ). we write g(f ) because we view g as an element of the dual space Lp (μ)∗ .e. We wish to identify the dual space of Lp (μ). That is. Here.26. it suffices to find a function f in Lp (μ) such that f p = 1 and such that φg (f ) = gq . ∞). The essential supremum defines a norm on L∞ (μ) and L∞ (μ) is a Banach space when given this norm. Σ. Thus. Then .) We use the terminology “essentially bounded” to describe the functions in L∞ (μ) and call f ∞ the “essential supremum” of |f | in L∞ (μ). define a scalar-valued function h on Ω by letting h(x) = |g(x)|q−1 (sign g(x)) for each x ∈ Ω.13 Let (Ω. ∞). the dual space of Lp (μ) is Lq (μ). however. Let g ∈ Lq (μ). μ) or L∞ (μ) when there is no risk of confusion. let f + = f χ{x:f (x)≥0} and f − = −f χ{x:f (x)<0} . Σ. To see this. It is routine to show that ν is finitely additive (by the linearity of ψ). Because ψ is bounded. By the Radon–Nikodým Theorem (Theorem A. Define a measure ν on (Ω. by Lebesgue’s Dominated Convergence Theorem (Theorem A. To that end. for every A ∈ Σ. It follows that h q is in Lp (μ). let ψ be a bounded linear functional in the dual space Lp (μ)∗ .6) for some g in Lq (μ). Next. and it is countably additive by the continuity of ψ. let f = h h p . observe that φg (h) = gq . n→∞ n→∞ Ω Therefore. and observe that f = f + − f − . We now wish to show that any bounded linear functional in Lp (μ)∗ can be written as in (2. . That is.24). there exists a measurable function g ∈ L1 (μ) such that ν(A) = A g dμ for all A ∈ Σ.17). Therefore. Now. there exists a sequence of simple measurable functions (fn )∞ n=1 such that fn ≥ fn−1 for all n ∈ N. and such that f − fn p → 0 as n → ∞. hp gq q/p Therefore. Since (Ω. |ν(A)| = |ψ(χA )| ≤ ψ χA Lp (μ) = ψ μ(A)1/p = 0. f ≥ 0. By the continuity of ψ. it follows that f ∈ Lp (μ). A Ω By linearity. Let f ∈ L∞ (μ) be a real nonnegative essentially bounded measurable function. f ∈ L∞ (μ) ∩ Lp (μ). φg is a bounded linear functional on Lp (μ) and φg = gq . it follows that ψ(f ) = Ω f gdμ whenever f is a simple measurable function. ψ(f ) = f gdμ. μ) is a finite measure space. suppose A ∈ Σ is such that μ(A) = 0. ψ(f ) = lim ψ(fn ) = lim fn gdμ. Thus. ψ(χA ) = ν(A) = gdμ = χA gdμ. Ω To extend this to an arbitrary real function in L∞ (μ) ∩ Lp (μ). ν is absolutely continuous with respect to μ. Σ) by ν(A) = ψ(χA ) for all A ∈ Σ. Then f p = 1 and q φg (h) gq φg (f ) = = = gq . Ω Ω where (q − 1)p = q follows from the assumption that p1 + q1 = 1. We also claim that ν μ.18 2 Classical Banach Spaces and Their Duals p1 p1 hp = |g| (q−1)p dμ = |g| dμ q = gq/p q . As before. p1 |g| dμ = |ψ(hn )| ≤ ψ hn p = ψ q |g| dμ q . We have already established that ψ(fn ) = Ω fn g dμ for all n ∈ N. we obtain 1− p1 q1 ψ ≥ |g| dμq = |g| dμ q . {|g|≤n} {|g|≤n} Thus. for each n ∈ N. ∞) when the scalar field is R. by Hölder’s Inequality. the linear functional ψ is bounded on Lp (μ). we let hn = χ{|g|≤n} |g|q−1 ρ for each n ∈ N. For each n ∈ N. and once again observing that (q − 1)p = q. g is in Lq (μ) and gq ≤ ψ. Therefore. It remains to show that ψ = φg . and so it follows that |ψ(hn )| ≤ φ hn p . In order to extend this result to C. It also proves that for any bounded linear functional ψ on Lp (μ). We have now proven the theorem for p ∈ (1. it follows that f g ∈ L1 (μ). {|g|≤n} {|g|≤n} Therefore. 1/q 1/q gq = lim inf χ{|g|≤n} |g|q dμ ≤ lim inf χ{|g|≤n} |g|q dμ ≤ ψ. This argument proves that φg is a bounded linear functional on Lp (μ) for all g ∈ Lq (μ). by Fatou’s Lemma (Theorem A. We also know that ψ(fn ) → ψ(f ) as n → ∞. {|g|≤n} By assumption. we argue as above.2 Function Spaces 19 We claim that g ∈ Lq (μ). Thus. Computing the Lp -norm of hn . Ω n→∞ n→∞ Ω Therefore. lim fn g dμ = f g dμ. {|g|≤n} {|g|≤n} Dividing. ψ(f ) = φg (f ) for all nonnegative functions f in Lp (μ). we may extend this to all real functions in Lp (μ) by writing f = f + − f − . Then.2. we see that p1 p1 hn p = |g| (q−1)p dμ = |g| dμ q . Since f ∈ Lp (μ) and g ∈ Lq (μ).16). then we may choose a sequence (fn )∞n=1 of simple measurable functions such that fn increases to f almost everywhere and such that fn → f in the Lp -norm as n → ∞. If f is a real nonnegative function in Lp (μ). we have hn ∈ L∞ (μ) ∩ Lp (μ) and ψ(hn ) = |g|q dμ. Similarly. n→∞ Ω Ω by Lebesgue’s Dominated Convergence Theorem. but define the function h by the rule h = |g|q−1 ρ. where ρ : Ω → C is a function such that |ρ| = 1 and g ρ = |g|. because ψ is a continuous linear functional on Lp (μ). define a function hn on Ω by letting hn = χ{|g|≤n} |g|q−1 (sign g). there exists a function g ∈ Lq (μ) such that . Suppose that f ∈ Lp (N. 1]: 1 (a) (Integration) f → f (t) dt. 1/p ∞ 1/p f p = |f |p dm = |f (n)|p . 0 (b) (Point evaluation) f → f (s) for s ∈ K. 1].14 Consider the measure space (N.17 Consider the following linear functionals on C[0. 2N . respectively. and m is counting measure on N (i. To extend this result to complex functions f in Lp (μ). f ∈ C(K). m). Then. If there is any risk of confusion. where 2N denotes the power set of N (the collection of all subsets of N). we consider the following example. We denote the collection of scalar-valued continuous functions on K by C(K). Define the supremum norm on C(K) by f ∞ = sup |f (t)|. m) corresponds to the sequence (f (n))∞ n=1 in p . ∞) and is left to the reader. For p = 1.16 We use the notation · ∞ to represent both the supremum norm on C(K) (for a compact metric space K) and the essential supremum norm on L∞ (μ) (for a measure space (Ω. respectively. Example 2. Therefore..15 Let K be a compact metric space. the proof is similar to the case when p ∈ (1. Such a case can be seen in the following example. Theorem 2. the set function for which m(A) is the cardinality of the set A ⊆ N). Let us now consider spaces of continuous functions. 1). In this case.) 2 As is the case with sequence spaces. the dual of L∞ (μ) need not be L1 (μ). N n=1 We see that f in Lp (N. The same conclusion holds for p = ∞.) If we wish to emphasize the underlying scalar field.20 2 Classical Banach Spaces and Their Duals ψ(f ) = φg (f ) for all real functions f in Lp (μ). since a continuous function attains its supremum on compact sets. where (f ) and (f ) are the real and imaginary parts of f . 2N . (See Exercise 2. A. m) = p . Definition 2. write f = (f ) + i (f ). we will write CR (K) or CC (K). we write C(K) = C[0. Observe that. 0 . (See Theorem 2. and use linearity. we will write · C(K) and · L∞ (μ) to denote the norm on C(K) and L∞ (μ). where K = [0. for f ∈ C(K).27. Lp (N. 2N . 1 ≤ p ≤ ∞. ∞). To that end.3.e. t∈K The set C(K) is a vector space under pointwise addition and scalar multiplication and is a Banach space when given the supremum norm. m). 2N . 1]. Example 2.13 remains true when μ is a positive σ -finite measure. where p ∈ [1. the quantity f ∞ is actually the maximum of |f |. μ)). We wish to identify the dual space of C(K). 1 (c) (Integration against L1 functions) f → f (t) g(t) dt for g ∈ L1 (0. Remark 2. by means of a Dirac measure: .2 Function Spaces 21 We use L1 (0.17(b)) can be thought of as integration. 1) to denote the Banach space of L1 -functions on [0. Note that point evaluation (Example 2.2. 1] with Lebesgue measure. 17(c) provides a bounded linear functional on C[0. 1]. A where B denotes the collection of Borel measurable subsets of [0. This. 1] (as will be seen in Theorem 2. Definition 2. and f (s) = f (t) δs (dt). A We leave the verification of this as an exercise. the set function δs is a measure on [0. A ∈ B. 1 if s ∈ A. Example 2. (See Exercise 2. below).1] 0 for f ∈ C(K) and g ∈ L1 (0. [0. 1). let us recall several definitions from measure theory.18 Let (K. can be realized as integration against a measure. For any s ∈ [0. too. The total variation of μ is defined for all A ∈ B by ⎧ ⎫ ⎨ ⎬ |μ|(A) = sup |μ(Aj )| : (Aj )j ∈F is a finite measurable partition of A .4.17 are not all of the linear functionals on C[0. 1] because 1 1 f (t) g(t) dt ≤ sup |f (t)| · |g(t)| dt = f ∞ g1 . 0 t∈[0. B) be a measurable space and let μ be a measure on K. or a measure that is concentrated on a set of Lebesgue measure zero. A ⊆ [0. Then 1 f (t) g(t) dt = f dν.1] The linear functionals given in Example 2. 1])∗ . Before we identify this space. Let ν(A) = g(t) dt. but they do give a hint to the true nature of (C[0.20. 1].7) 0 ifs ∈ A.17(c) has total variation |ν|(A) = |g(t)| dt. 0 [0. 1] is a Borel set.) . δs (A) = (2. ⎩ ⎭ j ∈F We remark here that the measure ν defined in Example 2. A Dirac measure is an example of a singular measure.1] The measure δs is also known as the Dirac mass at s. 1]. 8). 1]. 1/3 = 1/3 + 0/32 + 0/33 + · · · . It represented a milestone in analysis and identified the dual space of C(K) as a space of measures on K. then M(K) consists of all bounded signed Borel measures on K. and in this case M(K) = {μ + iν : μ. We denote by M(K) the space of all Borel measures on K having finite total variation. f ∈ C(K). called the Riesz Representation Theorem. If x ∈ (0. 1]. then. was proved by F. For example. This is a Banach space with the total variation norm. although not in this generality. 1] is an infinite series x = j =1 δj (x)/3 . then there exists a Borel measure ν on K such that φ(f ) = f dν. 1]. When the scalar field is C. The Riesz Representation Theorem is a classical result in measure theory and. The next theorem. 1] → [0. but also ∞ 1 2 0 2 2 = j = + 2 + 3 + ··· . i. If x = 0. all Borel measures on K determine bounded linear functionals on C(K) according to (2.17 turned out to be given by a measure on [0.19 Suppose K is a compact metric space.20 (Riesz Representation Theorem) Suppose K is a compact metric space. A given number may have two ternary expansions. the measures are bounded complex measures. The dual of C(K) can be identified with the space M(K). 1] may have multiple ternary expansions. 2} for all j ∈ N. 1]. Example 2.e. Theorem 2. Riesz in 1909 [31]. We now give another example of a bounded linear functional on C[0. (2. that each of the bounded linear functionals in Example 2. which is given by νM = |ν|(K) for all ν ∈ M(K). where δj (x) ∈ {0. 1. We will define a map G : [0.22 2 Classical Banach Spaces and Their Duals Definition 2. and x has nonterminating ternary expansion x = ∞ j j =1 δj (x)/3 . In particular. if φ ∈ C(K)∗ . If the underlying scaler field is R. then let G(x) = 0. It is no surprise. 1])∗ as the space M[0. consequently. The Riesz Representation Theorem identifies (C[0. νare signed (real-valued) measures}. The ternary expansion is said to terminate if the sequence (δj (x))∞ j =1 has only finitely many nonzero terms. then let .8) K and φ = |ν|(K). it can be shown that x has a unique ternary expansion x = ∞ j j =1 δj (x)/3 that does not terminate. Furthermore.) Despite the fact that x ∈ (0.21 ∞(the Cantorj function) A ternary expansion for x ∈ [0. 1] of Borel measures on the unit interval.. there exists an N ∈ N such that δj (x) = 0 for all j ≥ N . we will not prove it here. 3 j =2 3 3 3 3 (This equality is easily verified with a geometric series argument. Interpret the result as a number in base 2.) Since G : [0.2 Function Spaces 23 ⎧∞ ⎪ ⎪ δj (x)/2 ⎪ ⎪ if δj (x) = 1 for any j ∈ N. (See Exercise 2. 0 ≤ a < b ≤ 1. In other words.16. 1]. Next. Observe that G(0) = 0 and G(1) = 1. We call G the Cantor function. The Cantor function G has the remarkable property that it is continuous (see Exercise 2. and nondecreasing. replace each 2 with a 1. every linear functional on c corresponds to an element of 1 (K). It can be shown that C has Lebesgue measure zero. An arbitrary element in 1 (K) is of the . and this is the value of G(x). Let C = {x : δj (x) = 1 for any j ∈ N}.17) and nondecreasing on the interval [0. but G (x) = 0 for almost every x. (See Sect. If x contains a 1. or a measure that is concentrated on a set of Lebesgue measure zero. 1]. 1]. ⎪ ⎪ 2j ⎪ ⎨ j =1 G(x) = (2. b]) = G(b) − G(a). which is a singular measure that has no atoms. and so the map 1 f → f dμG . which we call Cantor measure. then replace each digit after the first 1 by 0. 1] and express x in base 3. where K = N ∪ {∞} is the one-point compactification of the natural numbers. 1] such that μG ([a.1 for the relevant definitions. viewed as a subspace of R. c = C(K). the dual space of c is c∗ = 1 (N ∪ {∞}). f ∈ C[0. ⎪ ⎩ j j =1 2 2 An alternate way of describing this procedure is the following: Let x ∈ [0.15.) The space K is topologically equivalent to the compact set {1/n : n ∈ N} ∪ {0}. The measure μG . 1]. which is precisely the set on which G is not locally constant. the Cantor measure is an example of a nonatomic (or diffuse) singular measure.2. It is also known as Lebesgue’s Devil Staircase. there exists a measure μG on [0. (See Parts (d) and (e) of Exercise 2. 1] is continuous.) In fact. The Cantor measure is an example of a singular measure.20). 0 determines a bounded linear functional on C[0.16.22 The sequence space c can be viewed as a space of continuous functions on a compact metric space. B. In particular.9) ⎪ ⎪ ⎪ ⎪ δj (x)/2 N−1 1 ⎪ ⎪ + N if δN (x) = 1 and δj (x) = 1 for j < N. (See Exercise 2. This set. is a Borel probability measure on [0. Example 2. is known as the Cantor set. A measurable set E is called an atom for a measure μ if (i) μ(E) > 0 and (ii) μ(F ) = 0 for any measurable subset F of E for which μ(E) > μ(F ).) By the Riesz Representation Theorem (Theorem 2. 1] → [0. We can exhibit this correspondence explicitly. positive. suppose X is complete in the norm · and suppose n=1 xn < ∞. ∞ Proof First. 2 Let yn0 = 0. The total variation of μξ is ξ 1 (K) . K j →∞ n=1 It is natural to wonder if there are any Banach spaces with trivial dual space.3 Completeness in Function Spaces In this section. which we prove in Sect. The Hahn–Banach Theorem implies there are no Banach spaces X with X∗ = {0}.23 (Hahn–Banach Theorem) Suppose X is a real Banach space and let E be a linear subspace of X. Then Sm − Sn ≤ j =n+1 xk for each m and n in N. (See Theorem 3. let Sn = x1 + · · · + xn .24 (Cauchy Summability Criterion) A normedspace X is complete if and only if every series ∞ n=1 x n converges in X whenever ∞ n=1 xn < ∞.9. Any one-dimensional subspace of X will have non-trivial linear functionals. ∞ Now suppose ∞ n=1 xn converges whenever ∞ n=1 xn < ∞. m For n ∈ N. We begin by proving a very useful lemma. If φ ∈ E ∗ . In particular. Then k ynk = (ynj − ynj −1 ). Lemma 2. we will prove that the function spaces of the previous section are complete in their given norms. by assumption.) Theorem 2. then there exists a bounded linear functional ψ ∈ X∗ such that ψ = φ and ψ(x) = φ(x) for all x ∈ E. f ∈ c = C(K). The dual action of μξ on c is given by integration: ∞ f dμξ = f (n) ξn + lim f (j ) ξ∞ . ξ∞ . The linear functional that corresponds to ξ is a measure μξ on K defined by μξ (n) = ξn for n ∈ N ∪ {∞}. The sequence spaces are left for the exercises. Pick an increasing sequence (nk )k∈N of natural numbers such that 1 yp − yq < k . 3.24 2 Classical Banach Spaces and Their Duals form ξ = (ξn )∞ ∞ n=1 . ∞ |μξ |(K) = |ξn | + |ξ∞ |. Therefore. j =1 . Let (yn )n=1 be a Cauchy sequence in X. whenever p > q ≥ nk . k ∈ N. The Hahn–Banach Theorem. where (ξn )n=1 ∈ 1 (N) and ξ∞ ∈ K. since X is complete. n=1 This quantity is finite. that is. we will prove the following. and the Hahn–Banach Theorem states that these can be extended to the entire space X.and so (Sn )∞n=1 is a Cauchy sequence. 2. guarantees there are no such Banach spaces. the series ∞ x j =1 j converges. and consequently. then ym − ynK ≤ 1/2K .25 Let (Ω. It follows that the subsequence (ynk )k=1 converges to some element in X. because |f (ω)| ≤ g(ω) for all ω ∈ Ω. observe that · p is a norm on Lp (Ω. we will use the Cauchy Summability Criterion (Lemma 2.(μ). Let > 0 be given. μ) be a positive measure space. Theorem 2. Ω Proof We will make liberal use of the theorems in Appendix A.24). and so we see that ∞ ∞ j =1 ynj − ynj −1 < ∞. we have ynj − ynj −1 < 2j1−1 for all j ≥ 2. ∞ ∞ Assume (f k )k=1 is a sequence of functions in Lp (Ω. Observe also that n fk (ω) ≤ g(ω). define gn on Ω by g n (ω) = n k=1 |fk (ω)| ∞ for all ω ∈ Ω. μ). μ) is a Banach space when given the norm 1/p f p = |f |p dμ . Σ. n→∞ n→∞ n→∞ It follows that lim inf |gn |p < ∞ a. μ) be a positive measure space. Σ.e. The space L∞ (Ω. For each n ∈ N. μ).e. Let y be the limit of this subsequence. Observe that (gn )n=1 is a sequence of nonnegative measurable functions and that gn ≤ gn+1 for each n ∈ N. 2 2 whenever m > nK . (ym )∞ m=1 is a convergent sequence. μ). by Minkowski’s Inequality.(μ) and gp ≤ M. as required.e. We may choose K ∈ N large enough so that ynK − y < /2 and 1/2K < /2. By definition.2. Let M = ∞ k=1 f k p . k=1 fk converges to f in Lp (Ω.(μ).(μ). It follows that f exists a. By Fatou’s Lemma. then Lp (Ω. thefunction g = ∞ k=1 |fk | exists a.26 Let (Ω. if m > nK . lim inf |gn |p dμ ≤ lim inf |gn |p dμ = lim inf gn pp ≤ M p < ∞. 2 Theorem 2. If 1 ≤ p < ∞. To show that · p is complete. μ) is a Banach space when given the essential supremum norm · ∞ . 1 ym − y ≤ ym − ynK + ynK − y < K + < . n→∞ ∞ p p n |fk | = lim |fk | = lim inf |gn |p < ∞ a.3 Completeness in Function Spaces 25 By construction. by the Lebesgue Dominated Convergence Theorem. as required. f ∈ Lp (Ω. k=1 n Therefore. Thus. First. . ω ∈ Ω. ∞ Now let f = k=1 fk .e. n→∞ n→∞ k=1 k=1 Consequently. Therefore. μ) such that k=1 fk p < ∞. We have determined that f ∈ L∞ (Ω. k=1 k=n+1 provided n ≥ n0 − 1. 2 Theorem 2.27 Let K be a compact metric space. C0 (K) = C(K) and all finite Borel measures on K are regular.26 2 Classical Banach Spaces and Their Duals ∞ Proof Again we use Lemma ∞ 2. f = ∞ k=1 fk exists a.e. Note that M(K) is necessarily a Banach space as the dual space of a Banach space. measurable Let N = ∞ k=1 Nk . and that the dual of 1 can be identified with ∞ . Therefore.5.(μ) and ∞ ∞ |f (ω)| ≤ |fk (ω)| ≤ fk ∞ = M. there exists some n0 ∈ N such that k=m fk ∞ < whenever m ≥ n0 . of By ∞ assumption. for each k ∈ N. (See Sects. (See Proposition 1.26. This is not inconsistent notation because. but now we must show that f is the limit n ∞ k=1 fk in the essential supremum norm.27 remains true if we replace K with a locally compact Hausdorff space and consider the space C0 (K) of continuous functions on K vanishing at infinity. for each k ∈ N. but in this case it denotes the Banach space of regular Borel measures on K with the total variation norm. there exists a set Nk such that μ(Nk ) = 0 and |fk (ω)| ≤ fk ∞ for all ω ∈ Ω\Nk . Then n ∞ f − f k ∞ ≤ fk ∞ < . Let > 0 be given.24. Exercise 2. Proof The proof is similar to that of Theorem 2. and is left to the reader. (See Exercise 2.11. k=1 f k ∞ < ∞.1 Show that c0 is a closed subspace of c and that c is a closed subspace of ∞ . Suppose M < ∞.(μ). This completes the proof. . with modifications related to continuity. Consequently.) 2 Theorem 2. Let (fk )k=1 be a sequence of functions in L∞ (Ω. when K is a compact metric space.6 for the relevant definitions.2 Show that the dual of c0 can be identified with 1 . 5.1 and A. μ).) When K is a locally compact Hausdorff space.) Exercises Exercise 2.e. Then μ(N ) = 0 (as a countable union of measure-zero sets) and fk (ω) < ∞ for all ω ∈ Ω\N and all k ∈ N. The space C(K) of scalar-valued continuous functions on K is a Banach space when given the supremum norm · ∞ . By assumption. we have |fk | ≤ fk ∞ a. k=1 k=1 for all ω ∈ Ω\N. Therefore. the dual space C0 (K)∗ is still M(K). μ) and let M = k=1 fk ∞ . 14 Let 2p be the finite-dimensional vector space R2 equipped with the · p -norm for 1 ≤ p ≤ ∞.7 Use the theorems of Sect. Exercise 2. Σ. 1] when equipped with the · ∞ -norm. . 1]. and define a measure on [0. Show that L1 (Ω. μ) is a positive measure space and let 1 ≤ p < q ≤ ∞. then show f p ≤ f q for all measurable functions f . Conclude that c0 is a Banach space. 1] by ν(A) = A g(t) dt.13. (b) Show that the assumption μ(Ω) < ∞ cannot be omitted in (a).3 Let (Ω. Prove that C(K) is a Banach space when given the supremum norm. μ). (a) Denote by np the finite-dimensional vector space Rn equipped with the norm 1/p (x1 . Exercise 2.12 Suppose (Ω.) Exercise 2. Show that x ∈ p for all p ≥ r and prove that xp → x∞ as p → ∞. . .8 Verify that any Cauchy sequence in c0 (equipped with the supremum norm) converges to a limit in c0 . 1]. Show that |ν|(A) = A |g(t)| dt for all Borel subsets A of [0. where A is a Borel subset of [0. (b) Show that p ⊆ q .) Exercise 2. (This completes the proof of Theorem 2. but not when equipped with the · 1 -norm. 2.3 remain true for σ -finite measure spaces. (a) If 1 ≤ p < q ≤ ∞.10 Let 1 ≤ p < q ≤ ∞.4 Let g ∈ L1 (0. δ0 (f ) = f (0) for all f ∈ C[0.6 Let δ0 denote the linear functional on C[0. 1] given by evaluation at 0.12. (That is. . (a) Prove that if μ(Ω) < ∞. . (You may assume the theorems of Sect. where Cp.17(c) and the comments following it. Exercise 2. but q is not a subset of p . (See Example 2.9 Prove that c0 is not a Banach space in the · 2 -norm.) Exercise 2.3 to prove that p is complete in the p-norm for 1 ≤ p ≤ ∞. then f p ≤ Cp. Exercise 2. Exercise 2. the Banach space of L1 -functions on [0.q f q for all measurable functions f .5 Let K be a compact metric space. prove Theorem 2. That is.11 Let x ∈ r for some r < ∞.27. μ) is a positive measure space such that μ(Ω) = 1. (c) Find a real-valued function f on [0. Exercise 2. Show that δ0 is bounded on C[0. Exercise 2. (See Exercise 2. 1). μ)∗ can be identified with L∞ (Ω.) Exercise 2.q is a constant that depends on p and q.) (b) Assume that f is an essentially bounded measurable function and prove that f p → f ∞ as p → ∞. xn )p = |x1 |p + · · · + |xn |p . 2. 1] with Lebesgue measure.Exercises 27 Exercise 2. 1]. μ) be a positive finite measure space. Show that the norms · p and · q are equivalent on Rn . 1] such that f p < ∞ but f q = ∞.13 Suppose (Ω. (a) Show that a closed subset of a complete metric space is a complete metric space. can be written explicitly as follows: ∞ 3 −1 n−1 3k + 1 3k + 2 K = [0. from each remaining set remove the middle third. Show that μG (K) = 1. From each of these. (x. where k ∈ N. From this set.15 Let M be a metric space with subset E.21.2 . and B2∞ represent geometrically? (b) Let a and b be nonzero real numbers and define a function on R2 by 1/2 x2 y2 (x. (b) Show that N.) (b) Evidently K is not empty because it contains the endpoints of the middle third sets.22. the so-called middle third. Continue this process indefinitely to create Cantor’s Middle Thirds Set. Conclude that the one-point compactification N ∪ {∞} of the natural numbers is a compact metric space. 1] \ . B22 .) Exercise 2. This leaves the union of two closed intervals: [0. y) ∈ R2 . Show that K contains ∞ other numbers by showing that 41 ∈ K. (See Example 2. 13 ] ∪ [ 23 .) (d) Let m be Lebesgue measure on [0. (See Appendix B. where the latter set is given the subspace topology inherited from R.16.17 Let K be Cantor’s Middle Thirds Set from Exercise 2. (e) Let μG be the Cantor measure from Example 2. (Hint: You may wish to use the fact that every nonzero number has a unique nonterminating ternary expansion.1 and E1.) (c) Show that the one-point compactification of the natural numbers N ∪ {∞} (from part (b)) is homeomorphic to {1/n : n ∈ N} ∪ {0}. is a locally compact metric space with the metric d(x. This set. A set V is said to be open in the subspace topology on E if there exists a set U that is open in M and V = U ∩ E. . 19 ] ∪ [ 29 . y) = |x − y| for all x and y in N. (d) Conclude that c is a Banach space. again remove the middle third.16 Consider the interval [0. (Hint: Consider −kj the geometric series j =1 3 . What remains is the union of four closed intervals: [0. Once again. two closed intervals remain. the set of natural numbers.1 and E1. y)E = 2 + 2 . Exercise 2. 1]. 1]. n=1 k=0 3n 3n (a) Show that K coincides with the Cantor set C of Example 2. Exercise 2. Call these two sets E1. a b Prove that · E is a norm on R2 and identify geometrically the closed unit ball in (R2 . After the first middle third is removed.1 for the definition of the one-point compactification. 13 ] ∪ [ 23 . 23 ). 79 ] ∪ [ 89 .2 . (Hint: Use diagonalization and K = C. which we denote K. After the middle thirds are removed from E1.28 2 Classical Banach Spaces and Their Duals (a) What do the closed unit balls B21 . 1]. 1]. remove the open subinterval ( 13 .) (c) Show that K is uncountable. Show that m(K) = 0. there will remain .21. · E ). 2n − 1}.k . Observe that ∞ 2 n K= En.k < x < an. E2. .2n .k . E2.1 . . .Exercises 29 four closed intervals.k ≤ x ≤ bn.1 . 2n }. . as they appear on the unit interval.3 . . . ordering them from left to right. After the process has been repeated n times. . . Label these sets En. . Label these sets E2. . show that (Gn )n=1∞ converges uniformly to G. .k = [an.21). .k k ⎨ 3−n 2n + 3−n 2n if an.4 . and deduce that G is continuous on [0. n=1 k=1 For each n ∈ N and each k ∈ {1. there will remain 2n closed intervals. again from left to right. .k −x k−1 x−an. each of length 31n .k1 and x2 ∈ En. so that x1 ∈ En.2 . 2n }. the cantor function (from Exam- ple 2. . 1]. bn. .k2 implies that x1 < x2 whenever k1 < k2 . 1] → [0. . E2. En. . Gn (x) = ⎪ ⎩k 2n if bn.k ] and define a function Gn : [0.k for k ∈ {1. let En. 1] as follows: ⎧ ⎪ bn.k+1 for k ∈ {1. as they appear on the unit interval. and is usually accepted by mathematicians.1 The Axiom of Choice The Zermelo–Fraenkel Axioms (ZF) is a list of accepted statements upon which mathematics can be built. Kalton. below. DOI 10. Indeed. defined on X .1007/978-1-4939-1945-1_3 . LLC 2014 31 A. J.4. When the system includes the Axiom of Choice. given in Theorem 3. © Springer Science+Business Media. it follows that f (xn ) → f (x0 ) as n → ∞. it is often called (ZFC).Chapter 3 The Hahn–Banach Theorems Several theorems in functional analysis have been labeled as “the Hahn–Banach Theorem. Bowers. For example. By the choice of xn . An Introductory Course in Functional Analysis. 3. The Axiom of Choice was formulated by Ernst Zermelo in 1904. Universitext. and so xn → x0 as n → ∞. N. We are using the Axiom of Choice when we choose xn ∈ Un for each n ∈ N. While the Axiom of Choice is easy to believe. we choose a sequence (xn )∞n=1 such that x n ∈ Un . but the Axiom of Choice is even stronger. and the continuity of f . it is often used without realizing it. accepting it results in some unexpected consequences. A choice function on a collection X of nonempty sets is a function c. there exists a choice function defined on X .” At the heart of all of them is what we call here the Hahn–Banach Exten- sion Theorem. This theorem is at the foundation of modern functional analysis. and its use is so pervasive that its importance cannot be overstated. They form what is possibly the most common foundation of mathematics. and so on. consider the sequential characterization of continuity: Suppose f is a continuous function. From a countable collection of open sets (Un )∞ n=1 that decrease to a point x0 . allowing us to choose uncountably many points simultaneously. Axiom of Choice For any collection X of nonempty sets. such that c(A) ∈ A for every set A in X . each one from a different set. and so difficult to imagine.1 Let P be a set. 3. and those pieces can be reassembled to form two solid spheres of radius 1. If A ⊆ P . then P contains a maximal element. Naturally. then b is called an upper bound for A if a ≤ b for all a ∈ A. we can define a reverse partial order ≥ in the obvious way. A detailed discussion of the history of the Axiom of Choice and its equivalent formulations can be found in [26].2 Sublinear Functionals and the Extension Theorem We briefly turn our attention from linear functionals to sublinear functionals. The Hausdorff Maximality Principle was formulated by Felix Hausdorff in 1914. An- other (equivalent) formulation of the Axiom of Choice is the Hausdorff Maximality Principle. An element a ∈ P is called maximal if a ≤ b implies b = a. b.2 Suppose V is a vector space. every chain is contained in a maximal chain. and as a result has led some to question the Axiom of Choice. ≤ ) is a partially ordered set. by Max Zorn. in 1935. and reinterpret Zorn’s Lemma in terms of lower bounds and minimal elements. Of interest to us is an alternate (but equivalent) formulation of the Axiom of Choice known as Zorn’s Lemma. The Banach–Tarski Paradox violates standard geometric intuition. Despite these reservations. The disjoint pieces will by necessity be non-measurable. Zorn’s lemma was proposed later. A map p : V → R is called a sublinear functional if . b} ⊆ C ⇒ (a ≤ b or b ≤ a). Definition 3. Hausdorff Maximality Principle In a partially ordered set. Both are equivalent to the Axiom of Choice. Definition 3. (iii) a ≤ a. The formulation we will usually use is Zorn’s Lemma. A relation ≤ on P is said to be a partial order if for every a. Let us recall the relevant definitions.32 3 The Hahn–Banach Theorems Banach–Tarski Paradox A three-dimensional solid sphere of radius 1 can be split into a finite number of disjoint pieces. (ii) (a ≤ b and b ≤ a) ⇒ a = b. and c in P : (i) (a ≤ b and b ≤ c) ⇒ a ≤ c. Zorn’s Lemma Suppose (P . A subset C of P is a chain in P if (C. ≤ ) is totally ordered: {a. If every chain in P has an upper bound. we will take the Axiom of Choice for granted. and (ii) p(x + y) ≤ p(x) + p(y) for all x and y in V (subadditivity). n∈N defines a sublinear functional. x ∈ E.5 A sublinear functional p on a real vector space E is linear if and only if p(−x) = −p(x) for all x ∈ E. then the norm function defined by p(x) = x for x ∈ E is a sublinear functional. 2 Let PE be the collection of all sublinear functionals on a real vector space E. Condition (ii) is also called the triangle inequality.3 The following are examples of sublinear functionals: (a) Any linear functional is a sublinear functional. s∈K defines a sublinear functional. then there is an extension fˆ : E → R of f that is linear and satisfies fˆ(x) ≤ p(x).4 (Hahn–Banach Extension Theorem) Let E be a real vector space and let p be a sublinear functional on E. . Lemma 3. Then. then the conclusion follows. (c) If C(K) is the Banach space of real-valued continuous functions on a compact metric space K. Example 3. (d) If ∞ is the Banach space of bounded sequences of real numbers. f ∈ C(K). x ∈ V. we will prove several preliminary lemmas. If V is a subspace of E and if f : V → R is a linear functional such that f (x) ≤ p(x). Before proving the Hahn–Banach Extension Theorem. Theorem 3. Suppose now that p(−x) = −p(x) for all x ∈ E. then p(ξ ) = sup ξn . Proof Certainly. then p(f ) = maxf (s).3. Note that condition (i) implies that p(0) = 0. ξ = (ξn )∞ n=1 ∈ ∞ . p(x + y) = −p(−x − y) ≥ − (p(−x) + p(−y)) = p(x) + p(y).2 Sublinear Functionals and the Extension Theorem 33 (i) p(α x) = α p(x) for all α ≥ 0 and x ∈ V (positive homogeneity). We define an order on the set PE by saying p ≤ q whenever p(x) ≤ q(x) for all x ∈ E. (b) If E is a normed space. The reverse inequality is simply subadditivity of p. if p is linear. by the triangle inequality. we have −p(−x) ≤ −p(−v) + p(x − v) ≤ q(v) + p(x − v). and so r(x) = q(x). r(x) = inf {q(v) + p(x − v) : v ∈ V } ≥ q(x). and so the set {q(v) + p(x − v) : v ∈ V } has a lower bound. Proof Define r(x) = inf {q(v) + p(x − v) : v ∈ V } . we have p(−v) ≤ p(−x) + p(x − v). To see this. by the subadditivity of q. It remains only to show that r is subadditive on E. Next. using the subadditivity of p.7 If p ∈ PE . Taking the infimum over elements in V . It is clear that r is positively homogeneous and that r(x) ≤ p(x) for all x ∈ E (by taking ν = 0). 2 2 By the sublinearity of p and q.6 Suppose p ∈ PE and V is a linear subspace of a real vector space E. It follows that −p(−v) ≤ q(v). and so. For x ∈ E. Let x and y be in E and suppose > 0. i∈I . we show the set {q(v) + p(x − v) : v ∈ V } has a lower bound for all x ∈ E. we have r(x + y) < r(x) + r(y) + . q(v) + p(x − v) ≥ q(v) + q(x − v) ≥ q (v + (x − v)) = q(x). suppose that x ∈ V . Rearranging this inequality. We claim that also r(x) = q(x) for each x ∈ V . Thus. 0 ≤ q(v) + q(−v) ≤ q(v) + p(−v). If q is a sublinear functional on V such that q ≤ p|V . Pick v and w in V so that q(v) + p(x − v) < r(x) + and q(w) + p(y − w) < r(y) + . then there exists a minimal q ∈ PE such that q ≤ p. x ∈ E. To show that r is well defined. 2 Lemma 3. q(v + w) + p (x + y − (v + w)) < r(x) + r(y) + . the quantity r(x) is well-defined for each x ∈ E. Since > 0 was arbitrary. and also using the fact that −p( − v) ≤ q(v). then there exists a sublinear functional r ∈ PE such that r|V = q and r ≤ p. Then for all v ∈ V . q(−v) ≤ p(−v). Thus. Fix x ∈ E and let v ∈ V . let r(x) = inf ri (x). r is subadditive and the proof is complete.34 3 The Hahn–Banach Theorems Lemma 3. This is true for all v ∈ V . This infimum is achieved (when v = x). By assumption. Proof Let P = {r ∈ PE : r ≤ p} and let C = (ri )i∈I be a chain in P . Thus. Without loss of generality.4). and so q0 ∈ P . 2 Lemma 3. and so q0 = q by the minimality of q in P . we have that 0 ≤ q(αx) + q(−αx). by the positive homogeneity of q. Let i ∈ I and x ∈ E. r is a lower bound of the chain C. 2 2 Since C is a chain. and so f = q|V .8 If q ∈ PE is minimal. Proof By Lemma 3. Suppose q0 is an element of PE such that q0 ≤ q. It follows that q0 is an element of P such that q0 ≤ q. First we must show that r is well- defined. Proof of the Hahn–Banach Extension Theorem By Lemma 3. By the subadditivity of q. Then. By subadditivity. This is true for all > 0. the set {ri (x) : i ∈ I } has a lower bound. Thus. Then q0 ≤ p. there exists . We claim that q is actually minimal in PE . by Zorn’s Lemma. This completes the proof. and so we have the desired result. Then r = q by the minimality of q. then q is linear. If α < 0. For any > 0.7. there exists a sublinear functional r on E such that r ≤ q and r|V = f .6. assume rj ≤ ri . By Lemma 3. then f (αx) = q(αx). and so f (αx) = −α q( − x) = −q(−αx) ≤ q(αx). 0 ≤ ri (x) + ri ( − x). Thus. by Lemma 3.3. Therefore (taking α = 1). ri and rj are comparable. there exists a sublinear functional q on E such that q|V = f and q ≤ p. and so r is subadditive. It follows that f ≤ q on the subspace V . Fix an x ∈ E and let V = {αx : α ∈ R}. P contains a minimal element q. Let x and y be elements of E. we have q(x) = f (x) = −q(−x). Suppose α ≥ 0. Positive homogeneity is clear. since ri ∈ P . Now we show subaditivity. and thus r is well-defined. it suffices to show q(−x) = −q(x) for all x ∈ E. Then r(x + y) ≤ rj (x + y) ≤ rj (x) + rj (y) ≤ ri (x) + rj (y). Therefore r(x + y) < r(x) + r(y) + . Define a linear functional on V by f (αx) = −α q(−x). ri (x) ≥ −ri (−x) ≥ −p(−x). The choice of x was arbitrary.6.2 Sublinear Functionals and the Extension Theorem 35 We claim that r is a sublinear functional on E. 2 We are now ready to prove the Hahn–Banach Extension Theorem (Theorem 3. By construction.5. there exist indices i and j in I such that ri (x) < r(x) + and rj (y) < r(y) + . α ∈ R. the space of all real bounded sequences. 2 Theorem 3. By construction. fˆ ∈ (∞ )∗ . we have em ∈ c for every m ∈ N.1) n→∞ The map f is bounded and f = 1. there is a linear functional fˆ ∈ X ∗ such that fˆ ≤ p and fˆ|V = f . fˆ( − x) ≤ f ( − x). where ξ = (ξn )∞ n=1 . This implies that fˆ = 0.36 3 The Hahn–Banach Theorems a minimal sublinear functional q0 on E such that q0 ≤ q. Then f ≤ p on V .9 (Hahn–Banach Theorem for real normed spaces) Suppose X is a real normed vector space and let V be a linear subspace of X. ∞ Then there exists a sequence of scalars (αn )∞ n=1 such that n=1 |αn | = 1 and such that ∞ fˆ(ξ ) = α n ξn . ξ = (ξn )∞ n=1 ∈ c. . n=1 Consequently. We will now show that fˆ ∈ 1 . ξ = (ξn )∞ n=1 ∈ ∞ . Let fˆ = q0 . so that em (m) = 1 and em (n) = 0 if m = n. by the Hahn–Banach Extension Theorem (Theorem 3. αm = 0 for all m ∈ N. (Note that. (3. We know that c is a subspace of ∞ . 2 Example 3.) Suppose tothe contrary that fˆ ∈ 1 . by (3. We must show fˆ|V = f . If f ∈ V ∗ . Since fˆ ≤ q and q|V = f . Then fˆ(x) ≤ f (x). Proof Define p(x) = f x for x ∈ X. we then have fˆ(x) ≥ f (x).2). among other things. Furthermore. Certainly.) Define a linear functional f : c → R by f (ξ ) = lim ξn .1. the map q0 is linear on E. then denote the nth coordinate of x by x(n). We conclude that fˆ ∈ 1 . ∞ fˆ(em ) = αn em (n) = αm . there exists a linear extension fˆ : ∞ → R such that fˆ|c = f and fˆ = 1.4).10 Consider the space c of real convergent sequences. Let x ∈ V .) Therefore. By Lemma 3. By Theorem 3.9 (the Hahn–Banach Theorem for real normed spaces). Then fˆ is linear and fˆ ≤ p. as required. this shows that 1 is not reflexive.2) n=1 If x is a scalar-valued sequence. Let em be the sequence with a 1 in the mth coordinate and zeros elsewhere. The norm on this Banach space is ξ = supn∈N |ξn |.8. It follows directly that fˆ = f . by (3. Thus. fˆ(em ) = f (em ) = lim em (n) = 0. By linearity. n→∞ On the other hand. which is a contradiction (because fˆ = 1). It follows that fˆ(x) = f (x) for all x ∈ V . we have fˆ ≤ f on V . (3. (See Exercise 2. then there exists an extension fˆ ∈ X∗ such that fˆ|V = f and fˆ = f . (Note that f is defined only on V .1). It is worth noting that f˜ has an additional property: If ξ is a bounded sequence of nonnegative real numbers. No formula exists for constructing fˆ or f˜ and. 0. we have f (ξ ) ≥ f (η) whenever ξn ≥ ηn for all n ∈ N. . Hahn–Banach extensions are generally not unique. f˜(ξ ) ≥ 0. and so f˜(−ξ ) ≤ 0. in fact. When an extension of a bounded linear functional is found using the Hahn–Banach Theorem. for one thing. Observe that fE (ξ ) = f (ξ ) = fO (ξ ) for all sequences ξ ∈ c. By the Hahn–Banach Theorem. it can be shown that f˜ ∈ 1 . but this relies on the Axiom of Choice. there exists a linear functional f˜ : ∞ → R such that f˜|c = f and f˜ ≤ p. Now let 1E = (0. these bounded linear functionals can be extended to .10. The extensions fˆ and f˜ in Examples 3.3. Therefore. (That is. . 1. As a simple extension.12 The existence of fˆ and f˜ in the previous examples is guaranteed by the Hahn–Banach Theorem. f˜ ≤ p on ∞ . this time with the sublinear functional p(ξ ) = supn∈N ξn .) This means. ξ = (ξn )∞ n=1 ∈ c. then f˜(ξ ) ≥ 0.13 Let c be the space of convergent sequences and let f (ξ ) = lim ξn n→∞ for all ξ = (ξn )∞ n=1 in c. it is sometimes called a Hahn–Banach extension of the functional. ) be the sequence having 0 in each odd coordinate and 1 in each even coordinate. As was the case in Example 3. Then f is a linear functional on c with f = 1. as the following example illustrates. for sequences ξ = (ξn )∞ ∞ n=1 and η = (ηn )n=1 ˜ ˜ in ∞ . by Theorem 3. let X be the smallest subspace of ∞ that contains all sequences in c and the sequence 1E . no formula can exist. (See [36]. Thus. n→∞ n→∞ where x = (xn )∞n=1 is a sequence in X. 1. by the linearity of f˜. As in Example 3. n→∞ Then f is a norm one linear functional on c. suppose that ξ = (ξn )∞n=1 is a bounded sequence such that ξn ≥ 0 for all n ∈ N. Each of the linear functionals fE and fO are bounded on X and have norm one. observe that.2 Sublinear Functionals and the Extension Theorem 37 Example 3.) We will extend the linear functional f to X in two ways: fE (x) = lim x2n and fO (x) = lim x2n+1 . that any linear functional on ∞ which can be written explicitly must belong to 1 . Then p(−ξ ) = sup (−ξn ) ≤ 0.10 and 3. It is clear that f ≤ p on c. . where ξ = (ξn )∞ n=1 .11 Consider again the space c of real convergent sequences. then ∗∞ = 1 . Example 3. Let X denote the subspace of ∞ generated by c and 1E . To see this. If the Axiom of Choice is replaced by a weaker assumption. n∈N By construction. let f (ξ ) = lim ξn .11 (respectively) are both Hahn–Banach extensions of the same bounded linear functional f .10. Remark 3.4 (the Hahn–Banach Extension Theorem). By Theorem 3. Observe that {χ[0. and so f0 ≤ f . namely the set of even numbers and the set of odd numbers. ∞). which is certainly contained in ∞ . the space Lp (0. and consequently ∗∞ is a “monster. If f ∈ V ∗ . Proof Let XR be the underlying real Banach space (i.14 (Separability and classical spaces) The comments above show that the space ∞ of bounded sequences is not a separable space. we found two distinct Hahn–Banach extensions for the bounded linear functional f by partitioning N into two sets. The space ∞ . consider the set S of all sequences of zeros and ones. Define fˆ(x) = fˆ0 (x) − i fˆ0 (ix). ∞). they are both Hahn–Banach extensions of f to ∞ . Consequently. The space L∞ (0. (3. the size of the continuum (also denoted c). 1].3) . because the set {ek : k ∈ N} is a countable dense subset of p . Furthermore.36). and so ξ − η∞ = 1.9 (the Hahn–Banach Theorem for real normed spaces). We can find further distinct Hahn–Banach extensions for f by repeating the same argument using different partitions of N. Remark 3. however. because the set t : k ∈ N ∪ {0} is dense in C[0. by Lusin’s Theorem (Theorem A. the space p of p-summable sequences is separable when p ∈ [1. the linear functional f0 has an extension fˆ0 : XR → R that is linear and such that fˆ0 = f0 ≤ f . where p ∈ [1. Any two distinct elements ξ and η in S must differ in at least one coordinate.38 3 The Hahn–Banach Theorems norm one linear functionals fE and fO (respectively) on ∞ . Define f0 : V → R by f0 (v) = (f (v)) for all v ∈ V . 1) of (equivalence classes of) p-integrable measurable functions on [0.. however. where ek is the sequence with 1 in the k th coordinate and zero in every other coordinate. When p ∈ [1. On the other hand. The size of S is |S| = 2ℵ0 . Because fE |c = fO |c = f . x ∈ X. is vast. ∞)." 1] of continuous functions on [0. by the Weierstrass Approximation Theorem. 1) of essentially bounded measurable functions on [0. It is easily seen. 1] is not separable. and so ∞ is far from separable. because fE (1E ) = 1 and fO (1E ) = 0. |f0 (v)| ≤ |f (v)| ≤ f v. Then f0 is a real linear functional on V . In the preceding example.e.s] − χ[0. forget you can use complex scalars). and both fE and fO have norm 1.” To get some perspective on the size of ∞ . ∞). ! kThe space C[0. Ultimately. however. 1] is dense in this space. 1].15 (Hahn–Banach Theorem for complex normed spaces) Suppose X is a complex normed vector space and let V be a linear subspace of X. Theorem 3. then there exists an extension fˆ ∈ X∗ such that fˆ|V = f and fˆ = f . 1] is separable.x] : 0 < x ≤ 1} is an uncountable collection of functions in L∞ (0. this is because p is a separable space when p ∈ [1. because C[0. 1) and χ[0.t] ∞ = 1 whenever s = t. for all v ∈ V . that fE and fO are distinct linear functionals on ∞ . is also separable. we can actually write down an explicit formula for the linear functionals on p . 4).2. Notice that (3.3 Banach Limits In Example 3.11. called shift-invariance: L (ξn )∞ ∞ n=1 = L (ξn+1 )n=1 . fˆ ≤ f . (We called the linear functional f˜ in Example 3. We know fˆ ≥ f . (3.) In this section. It follows that.5) Shift-invariance.3): For x ∈ X. fˆ(v) = f0 (v) − if0 (iv) = (f (v)) + i (f (v)) = f (v).6) n→∞ n→∞ for all (ξn )∞ n=1 in ∞ . (3. together with the other properties mentioned above.6) implies (3. Therefore. Consequently. and so |fˆ(x)| = |eiθ fˆ(x)| = |fˆ0 (eiθ x)| ≤ fˆ0 eiθ x ≤ f x. fˆ(eiθ x) ∈ R. . (3. by linearity. we will show the existence of bounded linear functionals L that satisfy an additional property.4) n→∞ whenever this limit exists. To see that it is also C-linear. leads to the following inequality: lim inf ξn ≤ L (ξn )∞ n=1 ≤ lim sup ξn .6) are of interest because they generalize the notion of limits. We know that fˆ(x) ∈ C. fˆ(ix) = fˆ0 (ix) + i fˆ0 (x) = i fˆ(x). 2 3. and the proof is complete.3 Banach Limits 39 Observe that if v ∈ V . Fix θ ∈ R so that eiθ fˆ(x) ∈ R. then (since f is complex-linear) f0 (iv) = (f (iv)) = (if (v)) = − (f (v)) . fˆ is R-linear. fˆ is an extension of f . Therefore. Then. It remains to show that fˆ = f . Suppose x ∈ X with x ≤ 1. since fˆ is an extension of f .11.3. and such that L (ξn )∞ n=1 = lim ξn . By construction. for all v ∈ V . fˆ(eiθ x) = fˆ0 (eiθ x). Linear functionals that satisfy (3. put ix into (3. we showed the existence of a bounded linear functional L : ∞ → R on the real Banach space ∞ such that L = 1 and L (ξn )∞n=1 ≤ supn∈N ξn . Then.e. (3. Proof of Theorem 3. x ∈ V. . n∈N where ξ = (ξn )∞n=1 ∈ ∞ . we say that L is invariant under the shift operator.16. or is shift-invariant. x ∈ V.) Let T ∈ T and x ∈ V .3). we will have a set I containing only one map. then there exists a linear functional f on V such that f ≤ p and f (T x) = f (x). For example. ξ ∈ ∞ ..16 (Invariant Hahn–Banach Theorem) Let V be a real vector space and suppose that T is a commutative collection of linear maps on V (i. The following functions are sublinear functionals on ∞ that satisfy the hypotheses of Theorem 3. we can restate the shift-invariance property in terms of the shift operator T : L(T ξ ) = L(ξ ). Then q is a sublinear functional i∈I on V . Example 3.7. Before proving Theorem 3. by the definition of q. let us consider some situations where it can be used. Given this definition. consider the real Banach space ∞ and let T : ∞ → ∞ be the shift operator defined in (3. and so the commutativity assumption will be satisfied trivially. By Theorem 3. Suppose (qi )i∈I is a chain in C and let q = inf qi . Theorem 3. We now prove a form of the Hahn–Banach Theorem for sublinear functionals which are invariant under some collection of linear maps. T ∈ T . and n→∞ (iii) p(ξ ) = sup |ξn |.16. n∈N (ii) p(ξ ) = lim sup ξn . If p is a sublinear functional on V such that p(T x) ≤ p(x). ST = T S for all S and T in T ). We will use Zorn’s Lemma to show there exists a minimal element q ∈ C.17 In many cases.16 Consider the collection C of sublinear functionals q such that q ≤ p and such that q(T x) ≤ q(x) for all x ∈ V and T ∈ T .16 with T = {T }: (i) p(ξ ) = sup ξn . T ∈ T . (ξn )∞ n=1 ∈ ∞ .7) When L satisfies this equation.40 3 The Hahn–Banach Theorems We define the shift operator T : ∞ → ∞ by T (ξn )∞ ∞ n=1 = (ξn+1 )n=1 . each one of these sublinear functionals will lead to a shift-invariant bounded linear functional on ∞ . (See the proof of Lemma 3. i∈I We conclude that q ∈ C and q is a lower bound for the chain (qi )i∈I . q(T x) ≤ inf qi (x) = q(x). let x ∈ V and T ∈ T . Therefore. for each i ∈ I . Consequently. n Thus. any chain in C has a lower bound. T ∈ T . and hence qi (T x) ≤ qi (x). We assumed qi ∈ C. C contains a minimal element. since q ∈ C. Consequently.3. we have q(T n−1 x) ≤ q(T n−2 x) ≤ · · · ≤ q(T 2 x) ≤ q(T x) ≤ q(x). Thus. it follows that qn = q for all n ∈ N. S and T commute. by Zorn’s Lemma. We wish to show that qn ∈ C. It therefore follows that q(T x) ≤ qi (x) for all i ∈ I . Now. and hence x + T x + · · · + T n−1 x q(x) = q . and so Sx + T Sx + · · · + T n−1 Sx x + T x + · · · + T n−1 x qn (Sx) = q =q S n n x + T x + · · · + T n−1 x ≤q = qn (x). T ∈ T . n for all x ∈ V . x ∈ V. and n ∈ N. for all x ∈ V .3 Banach Limits 41 q(T x) ≤ qi (T x). (x − T x) + T (x − T x) + · · · + T n−1 (x − T x) q(x − T x) = q n . Observe that. By the minimality of q in C. n n n for all x ∈ V . Suppose S ∈ T . n Note that qn is a sublinear functional. qn ∈ C for each n ∈ N. because q is sublinear and T is linear. i ∈ I. For all n ∈ N. By assumption. Let n ∈ N and define the nth Cesàro mean by x + T x + · · · + T n−1 x qn (x) = q . x x x + T x + · · · + T n−1 x qn (x) = q ≤q + ··· + q = q(x). x ∈ V. Now let q be a minimal element of C and let T ∈ T . n n Since this is true for all n ∈ N. Observe that (i) implies both (ii) and (iii). n∈N n≥k This is true for all k ∈ N. Shortly. ·) : H × H → R. we conclude that q(x − T x) ≤ 0 for all x ∈ V and T ∈ T . n→∞ Motivated by the preceding example. we have f (x) = f (T x) for all x ∈ V and T ∈ T . Let T : ∞ → ∞ be the shift operator on the real Banach space ∞ and consider the sublinear functional p on ∞ defined by p(ξ ) = supn∈N ξn for all ξ = (ξn )∞ n=1 in ∞ . (ii) L(ξ ) = lim ξn whenever the limit exists. Observe that p(T ξ ) ≤ p(ξ ) for all ξ ∈ ∞ . we use the invariance of L under the shift operator. we make a definition. By a similar argument. for any sequence ξ = (ξn )∞ n=1 in ∞ . Definition 3. L(ξ ) = L(T ξ ) for all ξ ∈ ∞ . We claim that the map L has the following properties. which proves (i).8. where ξ = (ξn )∞ n=1 ∈ ∞ : (i) lim inf ξn ≤ L(ξ ) ≤ lim sup ξn . 2 Example 3. we recall some definitions. that satisfies the following properties: . that is. we also deduce that q( − x − T ( − x)) ≤ 0 for all x ∈ V and T ∈ T . and n→∞ (iii) L(ξ ) ≥ 0 whenever ξn ≥ 0 for all n ∈ N.16 with T = {T } to conclude that there exists a linear functional L on ∞ such that L ≤ p that is shift-invariant. By Lemmas 3. (i) L(T ξ ) = L(ξ ).19 A real vector space H is called a real inner product space if there is a map (·.18 A linear functional L on ∞ is called a Banach limit if. by the linearity of f .7 and 3. and so we conclude that L(ξ ) ≤ lim sup ξn . then (by shifting k times) L(ξ ) = L (ξn+k )∞ n=1 ≤ sup ξn+k = sup ξn . and n→∞ (iii) L(ξ ) ≥ 0 whenever ξn ≥ 0 for all n ∈ N. where T is the shift operator. called an inner product.42 3 The Hahn–Banach Theorems x − T nx 1 1 =q ≤ q(x) + q( − T n x) n n n 1 1 ≤ q(x) + q( − x). Definition 3. ξ ∈ ∞ . n→∞ n→∞ (ii) L(ξ ) = lim ξn whenever the limit exists. We now invoke Theorem 3. and so the proof is complete. If k ∈ N. It follows that both f (x − T x) ≤ 0 and f ( − x − T ( − x)) ≤ 0 for all x ∈ V and T ∈ T . Therefore. This is the desired result.17 (revisited). we will present an application of a Banach limit. there exists a linear functional f such that f ≤ q. First. To show that (i) is true. A similar n→∞ argument shows lim inf ξn ≤ L(ξ ). |(S n x. y) for all x and y in H .19.21 Let H be a real Hilbert space and suppose · is the complete norm induced by the inner product on H . and so S is necessarily bounded. (ii) (x. and (iii) bilinear. ·). The concept of an inner product space exists also when the underlying scalar field is C. y) + β(x . (See Exercise 3. S n y)| ≤ S n x S n y ≤ C 2 x y. A bounded linear map S : H → H is said to be an orthogonal operator if (Sx. (See Theorem 7. Note that S is an orthogonal operator if and only if S n x = x for all n ∈ N and x ∈ H . A bounded linear map S : H → H is said to be similar to an orthogonal operator if it is orthogonal with respect to an equivalent inner product on H . This inequality. ·) on a vector space H can always be used to define a norm by the formula # x = (x. αy + βy ) = α(x. y) = (y. We will study the topic of Hilbert spaces in greater depth in Chapter 7. β} ⊆ R. This norm on H is said to be induced by the inner product. (See Definition 7. respectively.3. (We will verify that this formula defines a norm in Section 7. for n ∈ N and {x. Proof Denote the inner product on H by (·.1. y} ⊆ H . x).20 A real inner product space H is called a real Hilbert space if it is a complete normed space when given the norm induced by the inner product. y) + β(x. then S is similar to an orthogonal operator. it is called (i) positive definite. x . y)| ≤ xy for all x and y in H .) In particular. then H is a Hilbert space precisely when (H .) Proposition 3. is a fundamental tool. x). If H is an inner product space with induced norm · . Sy) = (x. then Sx = x.) An inner product (·. but will not prove it until later. Suppose H is a real Hilbert space with inner product (·. where {x. x) ≥ 0 for all x ∈ H . When a map possesses the three properties listed in Definition 3. (Two inner products on H are equivalent if they induce equivalent norms.1.8) for all n ∈ N and x ∈ H . x) = 0 if and only if x = 0.3 Banach Limits 43 (i) (x. y) = α(x. y } ⊆ H and {α. and (x. · ) is a Banach space. If S : H → H is an invertible bounded linear map and there exist positive constants c and C such that cx ≤ S n x ≤ Cx. but the properties defining an inner product must be modified in this setting. Let L be a Banach limit on ∞ . S n y))n∈N is a sequence in ∞ for each x and y in H . y. x ∈ H. It follows that ((S n x. Define a new inner product on . y ). (3.) Definition 3. By the Cauchy–Schwarz Inequality. We will make use of it now.2.) We will make use of the following significant fact: If H is an inner product space. and (iii) (αx + βx . ·). (ii) symmetric. known as the Cauchy– Schwarz Inequality.3. if n = 1. then |(x. y) and (x. we will apply the Hahn–Banach Theorem to the setting of compact groups. Therefore. We call G a metric group if it is both a group and a metric space. 2 3.8) and property (iii) of a Banach limit. for a given x ∈ G. x ∈ H. Properties (i)–(iv) are known as the group axioms. S is orthogonal with respect to ·. ·. where ·. ·) are equivalent inner products on H . Hence. that is. y) → xy and x → x −1 are continuous for x and y in G. cx ≤ |||x||| ≤ Cx. {x.) If x ∈ H and (See Exercise |||x||| = x.44 3 The Hahn–Banach Theorems H by x. group multiplication is denoted by juxtaposition. y} ⊆ G. x = L S n x2 n∈N . ||| · ||| and · are equivalent norms on H . S n+1 y) n∈N = x. the group (G. so that x · y is written xy. By the shift-invariance of a Banach limit (property (i)). there exists x −1 ∈ G such that x · x −1 = e = x −1 · x. Therefore. (ii) (associativity) (x · y) · z = x · (y · z) for all {x. S n y) n∈N . then ||| · ||| is a norm on H and |||x|||2 = x. Sy = L (S n+1 x. x.4 Haar Measure for Compact Abelian Groups In this section. By (3. (iii) (identity) There exists an element e ∈ G such that x · e = x = e · x for all x ∈ G. . then G is called a compact metric group. and so ·. called multiplication. √ 3. y} ⊆ H. · is an inner product equivalent to the original. z} ⊆ G. the map x → x −1 is a well-defined operation on G (called inversion). if both the maps (x. A group is a pair (G. ·). where G is a set and · is a binary operation on G.15 to show that this defines a real inner product on H . S is similar to an orthogonal operator. that satisfies the following properties: (i) (closure) x · y ∈ G for all {x. If a metric group G is also a compact topological space. the inverse x −1 is necessarily unique. Sx. When the multiplication is understood. y. {x. · and (·. Consequently. A simple calculation shows that. (iv) (inverses) For x ∈ G. y} ⊆ H. We will adopt this convention when there is no risk of confusion. Frequently. and if the group operations of multiplication and inversion are continuous on G. y = L (S n x. ·) is often abbreviated to G. y. if the group multiplication is commutative. 1} . then there is a unique translation-invariant Borel probability measure on G.) The Cantor group can be given a metric using the formula ∞ 1 |xk − yk | d(x. but it is not compact. often called the circle group or the torus. (In other words.24 If (G. +) be an abelian metric group. When a group is abelian. Of particular interest in this section are abelian groups. The group operations are given by matrix multiplication and matrix inversion. Definition 3. $ (iii) The Cantor group is the countable product ∞ N n=1 Z2 = {0. +) is a compact abelian metric group. called the orthogonal group. That is. because C\{0} is an open subset of C. the group multiplication is often denoted by addition (+) and the inverse x −1 is then written as −x (provided this causes no confusion). d(X.4 in the appendix. and so is the metric. Proof Let C(G) be the space of real-valued continuous functions on G equipped with the norm f ∞ = max|f (x)|. . Multiplication in T is taken from C. The punctured complex plane C\{0} with the standard multiplication and metric is itself a metric group. f ∈ C(G).) Both the torus and the Cantor group are abelian. A Borel measure λ on G is translation-invariant if λ(B) = λ(x + B) for all x ∈ G and Borel subsets B ⊆ G. 1} . Then On is a compact metric group. (Nor is it complete. Let T = {Tx : x ∈ G}. Theorem 3. define an operator Tx : C(G) → C(G) by Tx f (y) = f (x +y) for all f ∈ C(G) and y ∈ G. For each x ∈ G. written T = {eiθ : 0 ≤ θ < 2π }. the identity in T is 1 and the inverse of eiθ is e−iθ . Y ) = |||Y − X||| for all X and Y in On .4 Haar Measure for Compact Abelian Groups 45 Example 3. Consequently.) (ii) Let On denote the collection of n × n orthogonal matrices (n < ∞). is a compact group. (See Theorem B. The set T is a commuting family of operators on C(G) because G is abelian.3. x∈G Note that this maximum is attained because f is continuous on the compact set G. We call the elements of T rotations. but the orthogonal group is not if n ≥ 2. A group G is called abelian if xy = yx for all x and y in G. Elements of the Cantor group are sequences of zeros and ones. Note that the Cantor group is a compact space by Tychonoff’s Theorem.23 Let (G. The metric on On is induced by the operator norm ||| · ||| on On . k=1 2k 1 + |xk − yk | where x = (xk )∞ ∞ N k=1 and y = (yk )k=1 are elements of the set {0. and the group operation is given by component-wise addition (mod 2). y) = .22 The following are examples of compact metric groups: (i) The unit circle in C. (3. It follows that λx = λ. we make the following claim: f dλx = T−x f dλ.4.46 3 The Hahn–Banach Theorems Define a sublinear functional p on C(G) by p(f ) = maxf (x). Furthermore. f ∈ C(G). Therefore.4.1). Thus. we have that T−x χB (y) = χB (y − x).16 (the Invariant Hahn–Banach Theorem).4. This is true for all x ∈ G. . using (3. there exists a Borel measure λ on G such that φ(f ) = f dλ. we have that φ(1) = 1. Thus. x∈G Certainly. G G G for all f ∈ C(G). p(Tx f ) = p(f ) for all x ∈ G and f ∈ C(G). φ(f ) ≤ maxx∈G f (x) and φ( − f ) ≤ maxx∈G (−f (x)) = −minx∈G f (x). Let x ∈ G and define a measure λx on G by λx (B) = λ(x + B) for all Borel sets B in G. f ∈ C(G).2) holds for f = χB . f ∈ C(G).4.2) holds for simple functions.20 (the Riesz Representation Theorem).2) holds for continuous functions. By the definition of the map T−x . To that end. G Since φ(f ) ≥ 0 whenever f ≥ 0. and so p is invariant under rotations.9) x∈G In particular. because of (3. we have that λ is a positive measure. The linear functional φ was chosen (via the Invariant Hahn–Banach Theorem) so that φ(T−x f ) = φ(f ) for all f ∈ C(G). Our goal is to show that λ = λx . and by the density of simple functions in C(G). (3. By construction. By linearity. It follows that λ(G) = 1. Thus. T−x χB (y) λ(dy) = χB (y−x) λ(dy) = λ(B +x) = λx (B) = χB (y) λx (dy). minf (x) ≤ φ(f ) ≤ maxx∈G f (x). by Theorem 2. G G G Therefore. by Theorem 3. It follows that φ is a bounded linear functional on C(G). where B is a Borel subset of G. f dλx = T−x f dλ = φ(T−x f ) = φ(f ) = f dλ. we see |φ(f )| ≤ f ∞ for all f ∈ C(G). Consequently. as well. Let f ∈ C(G). We now show λ is translation-invariant.10) G G To prove this. (3. there exists a linear functional φ on C(G) such that φ ≤ p and φ(Tx f ) = φ(f ) for all x ∈ G and f ∈ C(G). and so λ is translation-invariant. (3.10). (3. let B be a Borel set in G and let y ∈ G. and consequently λ is a probability measure on G. 28 Let G be a compact abelian metric group with Haar measure λ. however. Haar measure on T is given by m/(2π). f (x + y) λ(dx) = f (x) λ(dx). Since Haar measure is the unique measure with these properties. Thus. Proof Define a probability measure μ on G by μ(B) = λ( − B) for all Borel subsets B in G. x ∈ G. 2 Definition 3. If B is any Borel subset of G. T 2π 0 For this reason.24.3. Let f ∈ C(G). f (x)λ(dx) = f (y) μ(dy). (See Section 5. we have restricted our attention to topologies arising from a metric. Indeed. G G It follows that λ = μ. we conclude that μ = λ. some authors write T = [0. μ(x + B) = λ( − x + ( − B)) = λ( − B) = μ(B). f ∈ C(G). our proof requires only that G is a compact abelian topological group. To be more precise. since λ and μ are probability measures (and so λ(G) = μ(G) = 1). 2 .) Example 3.) Corollary 3. then 2π 1 f dλ = f eiθ m(dθ ). G G G G By translation-invariance. By the translation invariance of λ. μ is a translation invariant Borel probability measure on G. then λ(−B) = λ(B).9. We will consider more general topological spaces in Chapter 5 and we will revisit the topic of Haar measure at that time. At this time. G G Therefore.25 The unique translation-invariant Borel probability measure on a compact abelian metric group is called Haar measure. G G and f (x + y) μ(dy) = f (y) μ(dy). 2π ) and λ = m/(2π ). y ∈ G. we do not actually use the metric on G. and if −B is the set {−x : x ∈ B} of inverses of elements in B. f (x + y) λ(dx) μ(dy) = f (x + y) μ(dy) λ(dx). Assume μ is a translation-invariant probability measure on G.4 Haar Measure for Compact Abelian Groups 47 It remains to show that λ is unique. 2π). Remark 3. By Fubini’s Theorem. (The factor 2π is needed in the denominator so that λ is a probability measure. and so the proof is complete.26 In the proof of Theorem 3. where m is Lebesgue measure on [0. if f ∈ C(T) and λ is Haar measure on T.22 (i).27 Let T = {eiθ : 0 ≤ θ < 2π } be the torus from Example 3. . a3 . define a closed linear subspace Ex of X by Ex = {αx : α ∈ K}. 2 For any x in a Banach space X. . and so X ∗ = ∞ . .5 Duals. ) with |aj | ≤ 1 for all j ≥ 2 will determine a norming functional for ξ . (See Proposition 1. Any linear functional φ on ∞ will be a norming element for ξ provided both φ(ξ ) = 1 and φ = 1. Let ξ = (ξ1 . There are many norming functionals in this case. . any element in ∞ of the form (1. For x ∈ X. . 0. . Therefore. the map xf∗ is a bounded linear functional on X such that xf∗ (x) = x and xf∗ = 1. there exists a bounded linear functional xf∗ ∈ X∗ that extends f . 1. Proposition 3. .29 Let X be a Banach space with dual space X ∗ . an element x ∗ of norm 1 such that x ∗ (x) = x.9 for real spaces. then ! " x = sup |x ∗ (x)| : x ∗ ∈ BX∗ . for all α ∈ K. and More Let X be a Banach space. Define a map f : Ex → K by f (αx) = αx. 1. ). let X = 1 . Biduals. and there exists an x ∗ ∈ X∗ such that x ∗ = 1 and x ∗ (x) = x. and that element is the constant sequence (1. Indeed. In this case. . . That is. Any Banach limit will satisfy these criteria.48 3 The Hahn–Banach Theorems 3. where BX is the closed unit ball of X. the linear functional xf∗ is an element in BX∗ with the property that xf∗ (x) = x. In this case. (Recall that ∗2 = 2 (Theorem 2. This element may or may not be unique. 1. ξ2 . 16 . Theorem 3.11. ). as well as other linear functionals. . . (iii) In 2 . there is a unique norming element in ∞ .) (iv) Consider the space ∞ and let ξ = (1. the space X∗ is a Banach space. By the Hahn–Banach Theorem for normed spaces (Theorem 3. . ) be an ∞ 2 1/2 element of 2 . . Example 3. Proof Let K denote the scalar field. If x ∈ X. a2 . that is. ! " x ≥ sup |x ∗ (x)| : x ∗ ∈ BX∗ . n12 . ξ ∈ 1 and ξ = 1.5). . 4 . 0. This time. . ).) Proposition 3. . Recall that the dual space X∗ of X is the space of bounded linear functionals on X. let ξ = (1. 1. . Then f is linear and f = 1. 9 . Consider the summable sequence ξ = 1 1 1 ∞ 1 π2 1. On the other hand. a4 . . Then ξ ∈ 1 and ξ = n=1 n2 = 6 . 0. When X∗ is equipped with the operator norm x ∗ = sup{|x ∗ (x)| : x ∈ BX }.29 guarantees the existence of a so-called norming element in the dual space X ∗ . (i) Let X = 1 . .15 for complex spaces). . the norming functional is always unique.30 We consider the existence of norming elements in a few real sequence spaces. . Then ξ = n=1 ξn and the norming element is ξ/ξ . . . Observe that |x ∗ (x)| ≤ x for all x ∗ ∈ BX∗ . . The result follows. (ii) Once again. B.) It follows that j is an isometry. From this. X∗ is reflexive. Proof Let X be a Banach space and suppose j is the natural embedding of X into its bidual X∗∗ . (3. This is possibly the only case where we can explicitly see the proper inclusion of a Banach space in its bidual.32 The natural embedding of a Banach space into its bidual is an isometric isomorphism onto a closed subspace of the bidual. x ∗ . i.5 Duals. Proof First. Let X be a Banach space. Then (X∗ )∗ = X. for all x ∈ X. one can show that j is a linear injection onto its image. the bidual of X∗ is ∗ (X∗ )∗∗ = (X∗ )∗ = X∗ .35 A Banach space X is reflexive if and only if its dual X ∗ is reflexive. We call j the natural embedding of X into its bidual and ·. Proposition 3. Thus.11) is sometimes written x ∗ . x ∗ ≤1 x ∗ ≤1 (The last equality follows from Proposition 3.e. it is common to suppress the map j and simply view X as a closed subspace of X∗∗ . Example 3. (iii) The sequence space c0 is not reflexive.31 Let X be a Banach space. Definition 3. The function spaces Lp (Ω. If x ∈ X. In light of this. The spaces L1 (μ) and L∞ (μ) will not be reflexive unless the support of μ is a finite set. Theorem 3.34 (i) The sequence spaces p are reflexive whenever 1 < p < ∞. By definition.5.33 A Banach space X is called reflexive if the natural embedding j of X into its bidual is a surjection. j (x) = sup |j (x)(x ∗ )| = sup |x ∗ (x)| = x.29.32 suggests that an exact copy of X sits inside of X ∗∗ . and More 49 Definition 3. The equation in (3. The last equality follows from the assumption that X is reflexive. we can conclude that j (X) is closed in X∗∗ and that j is an isomorphism onto j (X). The bidual of X is the space X ∗∗ = (X∗ )∗ . Define a map j : X → X∗∗ by letting j (x) ∈ X∗∗ be the linear functional on X ∗ defined by j (x)(x ∗ ) = x ∗ (x). · the dual space action between a Banach space (written on the left) and its dual (written on the right). By direct computation. c0∗∗ = ∗1 = ∞ .) 2 Theorem 3. x ∗ ∈ X∗ . c0 ⊂ ∞ . The spaces 1 and ∞ are not reflexive. Therefore. Biduals. j (x) = x. if j (X) = X∗∗ . In fact. μ) be a measure space.3.) We now show that j is an isometry.5. having vast and mysterious biduals.11) for all x ∈ X. μ) are reflexive whenever 1 < p < ∞. that is. (See Exercise 3. . (ii) Let (Ω. assume X is reflexive. (See Exercise 3. then j (x) ∈ X∗∗ . B.. X ∗ is reflexive. . In this section we will generalize the notion of a matrix adjoint to infinite-dimensional Banach spaces. y) = (x. This map is well-defined because j is an injection onto j (X). a contradiction.k=1 with entries bj k = akj for each j and k in the set {1. x ∈ X. and so x ∗∗∗ corresponds to some x ∗ ∈ X∗ . By assumption. λ ∈ K. j (X)) = δ |λ|. x ∈ X. (See Theorem 3. Therefore.6 The Adjoint of an Operator Suppose n is a natural number. ·) denotes the inner product on Cn . y} ⊆ Cn . For λ ∈ K and x ∈ X.k=1 is an n × n complex matrix. n}. Let δ = d(x ∗∗ . This implies both x ∗ > 0 and x ∗ = 0. One of the important properties of the matrix adjoint is (Ax.36 Let X and Y be Banach spaces and let T : X → Y be a bounded linear operator.50 3 The Hahn–Banach Theorems Now assume X ∗ is reflexive. 2 3. there exists an element x ∗ ∈ X∗ such that x ∗∗ (x ∗ ) = 1 and x ∗ |X = 0. Then there exists an x ∗∗ ∈ X∗∗ such that x ∗∗ ∈ j (X). We wish to show X is reflexive. where (·.) Let E = span{x ∗∗ . δ We conclude that φ is bounded on E ⊆ X ∗∗ and φ ≤ 1/δ. . Definition 3. . λx ∗∗ + j (x) = |λ| x ∗∗ − j ( − λ−1 x) ≥ |λ| d(x ∗∗ . by assumption.4). Then δ > 0. . A∗ y). by the Hahn–Banach Extension Theorem (Theorem 3. X is reflexive. Consequently.32. because j (X) is closed in X∗∗ . Let j : X → X∗∗ be the natural embedding of X into its bidual. j (x) : x ∈ X}. If A = (aj k )nj. j (X)) = inf{x ∗∗ − j (x) : x ∈ X}. {x. The map T ∗ : Y ∗ → X∗ defined by (T ∗ y ∗ )(x) = (y ∗ ◦ T )(x). . is called the adjoint of T . It follows that 1 |φ(λx ∗∗ + j (x))| = |λ| ≤ λx ∗∗ + j (x). y ∗ ∈ Y ∗ . there exists an element of X ∗∗∗ that extends φ. then the matrix adjoint (or conjugate transpose) of A is the n × n matrix A∗ = (bj k )nj. Define a linear functional φ : E → K (where K denotes the scalar field) by φ λx ∗∗ + j (x) = λ. Assume that j is not a surjection. In particular. Therefore. there exists x ∗∗∗ ∈ X∗∗∗ such that x ∗∗∗ (x ∗∗ ) = 1 and x ∗∗∗ (j (x)) = 0 for all x ∈ X. we conclude that T ≤ T ∗ . Proof It is not hard to show T ∗ is linear.36. Suppose x ∈ X. Since the choice of was arbitrary. T ∗∗ (x) = T (x) for all x ∈ X. Then for any y ∗ ∈ Y ∗ . y ∗ . T xY = (T ∗ y ∗ )(x) ≤ T ∗ y ∗ X∗ xX ≤ T ∗ y ∗ X∗ ≤ T ∗ . and hence the result. To show T ∗ is bounded. T < T ∗ + . ·. the adjoint T ∗ : Y ∗ → X∗ also has an adjoint T ∗∗ : X∗∗ → Y ∗∗ . Proposition 3. let y ∗ ∈ Y ∗ . j (x) = x. Consequently. That is. T ∗ y ∗ = T x. There exists x ∈ X such that xX ≤ 1 and T xY > T −.6 The Adjoint of an Operator 51 Owing to the abundance of Banach spaces. T ∗∗ x = T ∗ y ∗ .3. T ∗∗ x ∈ Y ∗∗ . We have T ∗ y ∗ ∈ X∗ . then T ∗∗ |X = T .29. Y ) denote the Banach space of bounded linear operators from X to Y . It follows that y ∗ . Proof Let j be the natural embedding of X into its dual X ∗∗ (which we think of as the inclusion map). there exists y ∗ ∈ Y ∗ such that y ∗ Y ∗ = 1 and y ∗ (T x) = T xY .11) and Definition 3. T ∗ y ∗ .39 Let X and Y be Banach spaces. If T : X → Y is a bounded linear operator. Let the action of Y ∗∗ on Y ∗ be represented by ·. T is bounded and T ≤ T . The proof of the following proposition will afford one such occasion.37. Then T ∗ y ∗ X∗ = sup |(T ∗ y ∗ )(x)| = sup |y ∗ (T x)|. The map taking T to T ∗ is a linear isometry from L(X. To prove the reverse inequality. 2 For an operator T between Banach spaces X and Y . Proof This follows from Proposition 3. If T : X → Y is a bounded linear operator. by (3. By definition. we think of X as a subspace of its bidual X ∗∗ . x≤1 ∗ Hence. T ∗ y ∗ X∗ ≤ sup y ∗ Y ∗ T xY = T y ∗ Y ∗ . we will sometimes find it convenient to denote the norm on a Banach space X by · X . we begin by letting > 0.38 Let X and Y be Banach spaces and let L(X. and so. In the following proposition. Proposition 3.37 Let X and Y be two Banach spaces. T ∗∗ x = T x. then the adjoint T ∗ is a bounded linear operator and T ∗ = T . whence. By Proposition 3. X∗ ). 2 Corollary 3. (T ∗∗ x)(y ∗ ) = y ∗ . j (x). y ∗ for all y ∗ ∈ Y ∗ . x≤1 x≤1 Therefore. Y ) to L(Y ∗ . 2 . Observe that the adjoint operator V ∗ : Lq (0. t We can define the Volterra operator V for p = 1. as well.41 is a special case of the next example. A similar argument will yield the same adjoint V ∗ . 0 0 0 t for all f ∈ Lp (0. 0 0 Therefore. If T S = T ◦ S : X → Z. . 1).41 (The Volterra operator). TheVolterra operator defined in Example 3. 1) and g ∈ Lq (0. for all t ∈ [0. where 1 < p < ∞.) 2 Example 3. 1) denote the real Banach space of p-integrable real-valued functions on [0.8. (See Exercise 3. 1) satisfies the equation 1 1 1 s ∗ f (s) V g(s) ds = Vf (s) g(s) ds = f (t) dt g(s) ds. however. Let Lp (0. 0 0 0 0 for all f ∈ Lp (0. then V ∗ is a map from (L∞ (0. 1) → Lp (0. 1))∗ to (L∞ (0. 1). t 1/q t 1/p |Vf (t)| ≤ 1q ds |f (s)|p ds ≤ t 1/q f p . 1]. We therefore conclude that 1 V ∗ g(t) = g(s) ds. 1). 1) → Lq (0. t ∈ [0. 1). f ∈ Lp (0. then the adjoint map (T S)∗ : Z ∗ → X∗ is given by (T S)∗ = S ∗ T ∗ . 0 p It follows that V is bounded and V ≤ (1/p)1/p . 1/p 1/p 1 p 1 Vf p ≤ t 1/q dt f p = f p . 1 s 1 1 f (t) dt g(s) ds = g(s) ds f (t) dt. We must show that V is well-defined. By Hölder’s Inequality. If p = ∞. and this map is not so easy to compute.52 3 The Hahn–Banach Theorems Proposition 3. 1] with respect to Lebesgue measure. By Fubini’s Theorem. 1) and g ∈ Lq (0. Y . We now compute the adjoint of the operator V . g ∈ Lq (0. 0 We call V the Volterra operator on Lp (0. and Z be Banach spaces and suppose both S : X → Y and T : Y → Z are bounded linear operators. 1]. 1). Let p and q be conjugate exponents (so that 1/p + 1/q = 1). Define a map V : Lp (0. 1) by t Vf (t) = f (s) ds.40 Let X. Proof The proof is left to the reader. 1))∗ . 3. y2 ) = (x1 + x2 . 1]. the direct sum is denoted by X ⊕ Y . Proposition 3. 1) → Lp (0. This demonstrates the original goal of functional analysis: To generalize linear algebra to an infinite-dimensional setting. Suppose V and W are closed subspaces of X. For p ∈ [1. and so TK f p ≤ K∞ f p . while the second is comparable to subtraction. y) ∈ X × Y. y) = xX + yY . (In this case. TK ≤ K∞ . and • λ · (x. we will show two common ways to construct new Banach spaces from given ones. (x. ∞). Proposition 3.44 Let X and Y be Banach spaces. and (x2 .7 New Banach Spaces From Old 53 Example 3. 1]).3. 1) by 1 TK f (s) = K(s. 2 In some cases. Thus. 0 This example can be thought of as a “continuous” analog of a matrix adjoint. The direct sum X ⊕Y is a Banach space under the norm (x. y1 ). y2 ) are in X × Y and λ is a scalar. a Banach space can be decomposed into a direct sum of closed subspaces. then X is isomorphic (as a vector space) to V ⊕ W . 1). When given this vector space structure. g ∈ Lq (0. define a linear map TK : Lp (0. 1] × [0. Definition 3. t) f (t) dt. λ y).) .41 reveals 1 TK∗ g(s) = K(t. The direct sum of X and Y is the set X × Y equipped with component-wise addition and scalar multiplication: • (x1 . where (x. 0 Then |TK f (s)| ≤ K∞ f p for all s ∈ [0. f ∈ Lp (0.43 Let X an Y be Banach spaces. y). 1).45 Let X be a Banach space. A calculation similar to that in Example 3. If X = V + W and V ∩ W = {0}.42 Suppose K ∈ L∞ ([0. (x1 . y1 + y2 ). y1 ) + (x2 . The first method is essentially a means of summing two spaces. we write X = V ⊕ W .7 New Banach Spaces From Old In this section. s) g(t) dt. Proof The proof that this norm is complete follows directly from the fact that · X and · Y are complete norms. y) = (λ x. let x + Y = inf y∈Y x + y. respectively. A priori. which we shall meet in Section 4. Definition 3.30). The quotient space X/Y is the set of all cosets of Y in X. . λw) = λ φ(x). then φ(x + x ) = φ (v + v ) + (w + w ) = (v + v . Consequently. That is. Note that the zero vector in X/Y is Y = 0 + Y . By construction.44). in fact.45 need not be an isometry between the given norm on X and the norm on V ⊕ W (as given in Proposition 3. φ is linear. These two norms will always be equivalent.3 as a consequence of the Open Mapping Theorem. w ) be elements of V × W . We now consider a second operation used to create new Banach spaces. and hence a vector space isomorphism. For each x ∈ X. Define a map φ : X → V ⊕W by φ(v + w) = (v. φ is onto. That is. it may not be clear that this map is well-defined. Suppose that x ∈ X can be written in two ways as the sum of elements from V and W . φ(x + x ) = φ(x) + φ(x ). w ) are in V × W . we wish to find a linear bijection (which need not be a homeomorphism). It follows that each x ∈ X has a unique representation of the form x = v + w. where (v. X = {v+w : v ∈ V . The map Q : X → X/Y defined by Qx = x + Y for x ∈ X is called the quotient map. X/Y = {x +Y : x ∈ X}. let (v. Therefore. Furthermore.e. where v ∈ V and w ∈ W . where x and x are in X and α is a scalar. and so φ is well-defined. If x = v + w and x = v + w .. Therefore.54 3 The Hahn–Banach Theorems Proof We wish to establish a vector space isomorphism between the spaces X and V ⊕ W . and consequently v − v = w − w ∈ V ∩ W = {0}. Then v + w = v + w . w). w ). v = v and w = w . Furthermore. Then · defines a complete norm on X/Y (called the quotient norm). w + w ) = (v. Next. because x = v + w if φ(x) = (v. 2 It is perhaps worth mentioning that the map φ in the proof of Proposition 3. This will follow from the Bounded Inverse Theorem (Corollary 4. Proposition 3. i.47 Let X be a Banach space with closed subspace Y . w) + (v . w ∈ W }. w) and (v .46 Let X be a Banach space and let Y be a closed subspace of X. φ is one-to-one. The quotient space X/Y is a vector space with addition and scalar multiplication given by (x + Y ) + (x + Y ) = (x + x ) + Y and α(x + Y ) = (αx) + Y . however. We leave it to the reader to verify this fact. (v. w) ∈ V × W. w) and (v . a homeomorphism (a continuous bijection with continuous inverse). because φ is. By assumption. w). if λ is a scalar. then φ(λ x) = φ(λ v + λ w) = (λv. suppose that x = v + w and x = v + w . respectively.46 has norm 1. (Recall Definition 1.3. A map T : X → Z is called an open map if T (U ) is an open set in Z whenever U is an open set in X. Suppose ∞ n n=1 x converges to x ∈ X. Let x1 and x2 be in X and suppose > 0. Now let α be a nonzero scalar. It is clear that · is nonnegative and 0 + Y = 0. (3. (See Exercise 3. Therefore. we used the fact that x + Y ≤ x for each x ∈ X. We will show that x + Y = Y .7. it follows that x ∈ Y . and hence n=1 xn converges in X (by completeness of the norm on X). then αx + Y = inf αx + y = inf αx + αz = |α| inf x + z. n=1 (xn + Y ) converges to x + Y in X/Y . Then x1 + x2 + y1 + y2 ≤ x1 + y1 + x2 + y2 < x1 + Y + x2 + Y + . If X and Z are Banach spaces and T is linear. Suppose (xn )∞ n=1 is a sequence in X such that x n=1 n +Y < ∞.) Therefore. There exist elements y1 and y2 in Y such that x1 + y1 < x1 + Y + /2 and x2 + y2 < x2 + Y + /2. This follows from the definition of the norm on X/Y . it suffices to show that the open unit ball of X is mapped to an open set in Z. x − ( − yn ) < 2−n . Then. Therefore. we conclude α(x + Y ) = |α|x + Y . If x ∈ X. we conclude that · satisfies the triangle inequality.7 New Banach Spaces From Old 55 Proof We first show that · is a norm on the quotient X/Y . n ≥ N. for any > 0.) We will prove the quotient map Q : X → X/Y is an open map by showing that Q(UX ) = UX/Y . Next.24). For each n ∈ N.15). ∞ Then n=1 xn < ∞. because 0 ∈ Y . Since Y is a closed subspace of X. Suppose x + Y = 0. for all n ∈ N. n n (x + Y ) − (xk + Y ) ≤ x − xk < . It follows that the quotient norm is complete. y∈Y z∈Y z∈Y (Observe that z = y/α.12) k=1 k=1 ∞ Consequently. there exists yn ∈ Y such that x + yn < 2−n . . pick ∞ n x ∈ x n +Y such that xn ≤ 2xn +Y . then in order to show that T is an open map. For this we use the Cauchy Summability ∞ Criterion (Lemma 2. and hence is a norm on X/Y . there exists an N ∈ N such that x − nk=1 xk < for all n ≥ N . It remains to show the norm · is complete on X/Y . Since the choice of was arbitrary. and consequently x + Y = Y .5. and so · is homogeneous. we show the triangle inequality. Equivalently. By the definition of the norm. The quotient map Q : X → X/Y has the additional property that it is also an open map. the quotient map Q : X → X/Y of Definition 3. 2 In (3. and so the sequence ( − y n )∞ n=1 converges to x. as required.) Let UX and UX/Y be the open unit balls in X and X/Y . 2 There is a close connection between direct sums and quotient spaces. and so Q(UX ) ⊆ UX/Y . A simple calculation shows that E is a linear subspace of X. Since the choice of was arbitrary. Proof Let E = ker(T ).4. T0 makes the following diagram commute. as required. and so T (x − x ) = 0. then T0 is a continuous linear bijection. If T : X → Y is a bounded linear operator.56 3 The Hahn–Banach Theorems We already know that Q ≤ 1. If z = z + y . and hence Q is an open map. then z < 1 and Qz = z + Y . (Thus. Consequently. T X Y Q T0 X/ ker(T ) In particular.48 Let T : X → Y be a bounded linear operator between Banach spaces. This proves that UX/Y ⊆ Q(UX ). We have established that the image of UX is UX/Y . we may identify X/W with V .) . let > 0 and choose x ∈ x + E such that x < x + E + . Then T x = T x and T0 (x + E) = T x ≤ T x ≤ T x + E + T . Thus. by Proposition 3. it follows that T0 ≤ T . For any x ∈ X. as claimed. By assumption. In Sec- tion 4. Definition 3. To show the reverse inequality. we will show that X = V ⊕ W if and only if there exists a continuous projection P : X → V such that W = ker(P ) and V = P (X). We now derive an important proposition which is analogous to a well-known fact of linear algebra. and hence T x = T x . and so T ≤ T0 . The set {x : T x = 0} is called the kernel of T and is denoted ker(T ). This result will prove valuable to us on several occasions. let x + E = x + E for x and x in X. Therefore. there is some y ∈ Y such that z + y < 1.49. we have that T x − T x = T (x − x ) = 0. Proposition 3. To that end. The rest of the proposition follows directly.49 Let X and Y be Banach spaces. Furthermore. where Q : X → X/ker(T ) is the quotient map. the map T0 is well-defined. if T is a surjection. Define the map T0 : X/E → Y by T0 (x + E) = T x for all x ∈ X. then ker(T ) is a closed subspace of X and there exists an injective linear operator T0 : X/ker(T ) → Y such that T0 = T . we have T x = T0 (x + E) ≤ T0 x + E ≤ T0 x. Then E = T −1 ({0}) is a closed set in X because T is continuous and {0} is a closed set in Y . z + Y = inf{z + y : y ∈ Y } < 1. It follows that x − x ∈ E. Now let z + Y be any element of UX/Y . We therefore conclude that T0 = T . We must verify that the map T0 is well-defined. because E is the kernel of T . 3.8 Duals of Quotients and Subspaces 57 3.8 Duals of Quotients and Subspaces Definition 3.50 Let X be a Banach space. If E is a closed subspace of X, then the annihilator of E is the set E ⊥ = {x ∗ ∈ X∗ : x ∗ (x) = 0 for all x ∈ E} ⊆ X∗ . If F is a closed subspace of X ∗ , then the pre-annihilator of F is the set F⊥ = {x ∈ X : x ∗ (x) = 0 for all x ∗ ∈ F } ⊆ X. It is easy to check that E ⊥ and F⊥ are always closed in X ∗ and X, respectively. Proposition 3.51 If X is a Banach space and E is a closed subspace of X, then: (i) E ∗ can be naturally identified with X ∗ /E ⊥ , and (ii) (X/E)∗ can be naturally identified with E ⊥ . Proof (i) Define a map ρ : X∗ /E ⊥ → E ∗ by ρ(x ∗ + E ⊥ ) = x ∗ |E , x ∗ ∈ X∗ . We claim this map is well-defined. To see this, suppose x1∗ and x2∗ are two elements of X ∗ such that x1∗ + E ⊥ = x2∗ + E ⊥ (i.e., they are in the same coset). It follows that x1∗ − x2∗ ∈ E ⊥ , and consequently (x1∗ − x2∗ )(e) = 0 for all e ∈ E. Thus x1∗ |E = x2∗ |E , and so ρ(x1∗ + E ⊥ ) = ρ(x2∗ + E ⊥ ). We have established that the definition of ρ(x ∗ + E ⊥ ) does not depend on the choice of representative in the coset x ∗ + E ⊥ , and hence the map ρ is well-defined. We can show by direct computation that ρ is linear. We also have that ρ is injective, because x ∗ |E = 0 if and only if x ∗ ∈ E ⊥ . We now define a map ψ : E ∗ → X∗ /E ⊥ . For any φ ∈ E ∗ , let ψ(φ) = xφ∗ + E ⊥ , where xφ∗ is any Hahn–Banach extension of φ to an element of X ∗ . In order to show this map is well-defined, we must demonstrate that ψ(φ) is independent of choice of Hahn–Banach extension of φ. To that end, let φ ∈ E ∗ and suppose φ has Hahn– Banach extensions x1∗ and x2∗ in X∗ . Since these are both extensions of φ, it follows that x1∗ |E = x2∗ |E = φ. Thus, (x1∗ − x2∗ )|E = 0, and so x1∗ − x2∗ ∈ E ⊥ . We conclude that x1∗ and x2∗ are in the same coset, and therefore x1∗ + E ⊥ = x2∗ + E ⊥ . Once again, it is easy to see that ψ is an injective linear map. It is also easy to see that ψ ◦ ρ = IdX∗ /E ⊥ and ρ ◦ ψ = IdE ∗ . (Here, we use IdX∗ /E ⊥ and IdE ∗ to denote the identity maps on X ∗ /E ⊥ and E ∗ , respectively.) Consequently, ρ is a linear bijection with inverse ψ. 58 3 The Hahn–Banach Theorems We now show ρ is an isometry by showing that φ = xφ∗ + E ⊥ , where φ ∈ E ∗ and xφ∗ is any Hahn–Banach extension of φ. Certainly, xφ∗ + E ⊥ ≤ xφ∗ = φ. Suppose xφ∗ +E ⊥ < xφ∗ . Then there exists some z∗ ∈ xφ∗ +E ⊥ with z∗ < xφ∗ . Since z∗ and xφ∗ are in the same coset, it must be the case that z∗ |E = xφ∗ |E = φ, but z∗ |E ≤ z∗ < φ, a contradiction. Therefore, xφ∗ + E ⊥ = φ. (ii) Denote the quotient map by Q : X → X/E. Let φ ∈ (X/E)∗ . For any e ∈ E, Qe = E. Consequently, φ ◦ Q(e) = φ(E) = 0, and so φ determines an element of E ⊥ through the identification φ → φ ◦ Q. This identification preserves norms because Q(UX ) = UX/E (see the comments following the proof of Proposition 3.47), and so φ ◦ Q(UX ) = φ(UX/E ). To see that an element of E ⊥ determines an element of (X/E)∗ , let x ∗ ∈ E ⊥ . Define φ ∈ (X/E)∗ by φ(x + E) = x ∗ (x) for all x ∈ X. We must show φ is well- defined. Let x and x be in the same coset, so that x − x ∈ E. Then, x ∗ (x − x ) = 0, and hence x ∗ (x) = x ∗ (x ). Thus, φ(x + E) = φ(x + E), and so φ is well-defined. To complete the proof, observe that φ ◦Q = x ∗ , and refer to the previous paragraph. 2 The identifications in the preceding proposition lead to a remarkable corollary. Corollary 3.52 Let X be a Banach space. For any closed set E in X, we have the identification E ∗∗ = E ⊥⊥ . Proof First apply (i), and then (ii), of Proposition 3.51. 2 Exercises Exercise 3.1 Let X be a real inner product space with inner product (·, ·) and associated norm · . Prove the Parallelogram Law: If x and y are elements of X, then x + y2 + x − y2 = 2 (x2 + y2 ). Exercise 3.2 Let X be a real inner product space with inner product (·, ·) and associated norm · . Verify the polarization formula: If x and y are in X, then 1 (x, y) = x + y2 − x − y2 . 4 Exercise 3.3 Let H be a Hilbert space with norm · . Show that a bounded linear map S : H → H is an orthogonal operator if and only if S n x = x for all n ∈ N and x ∈ H . (Hint: Use Exercise 3.2.) Exercise 3.4 Let X and Y be normed vector spaces. If x is a nonzero vector in X, and y ∈ Y , show there exists a bounded linear map T : X → Y such that T (x) = y. Exercise 3.5 In this exercise, we complete the proof of Theorem 3.32. Let X be a Banach space with bidual X∗∗ and let j : X → X∗∗ be the natural embedding. Exercises 59 (a) Show that j is an injective bounded linear map. (b) Show that j (X) is a closed subset of X ∗∗ . (You may wish to use Exercise 1.10.) Exercise 3.6 Let X and Y be normed spaces. Show that if L(X, Y ) is a Banach space, then Y must be a Banach space. (This is the converse to Proposition 1.11.) Exercise 3.7 Let X and Z be Banach spaces and let T ∈ L(X, Z). Show that T is an open map if and only if T maps the open unit ball of X to an open set in Z. Exercise 3.8 Prove Proposition 3.40. Exercise 3.9 Let p be a sublinear functional on a real vector space V . Show that p(x) = max{f (x) : f ≤ p, f linear}, x ∈ V. Conversely, show that a functional q of the form q(x) = sup f (x), x ∈ V, f ∈A where A is some collection of linear functionals, is necessarily sublinear. Exercise 3.10 (Fekete’s Lemma [10]) Let (an )∞ n=1 be a sequence of real numbers such that am+n ≤ am + an , {m, n} ⊆ N. Show that if the sequence (an /n)∞ n=1 is bounded below, then an an lim = inf . n→∞ n n∈N n an am (Hint: For any m ∈ N, show that lim sup ≤ by writing n = k m + r, where n→∞ n m {k, r} ⊆ N and 0 ≤ r ≤ m − 1.) Exercise 3.11 Let V be a real vector space and suppose p is a sublinear functional on V . Suppose T : V → V is a linear map such that p(T x) = p(x). Show, using Exercise 3.10, that 1 q(x) = lim p(x + T x + · · · + T n−1 x), x ∈ V, n→∞ n defines a sublinear functional with q ≤ p. Show further that if f is a linear functional with f ≤ p, then f is T -invariant (i.e., f (T x) = f (x) for all x ∈ V ) if and only if f ≤ q. Exercise 3.12. Show that a linear functional L on ∞ is a Banach limit if and only if ξk+1 + · · · + ξk+n L(ξ ) ≤ lim sup , ξ = (ξj )∞ j =1 ∈ ∞ . n→∞ k∈N n Exercise 3.13 A sequence ξ ∈ ∞ is called almost convergent to α if L(ξ ) = α for all Banach limits L. 60 3 The Hahn–Banach Theorems (a) Show that ξ is almost convergent to α if and only if ξk+1 + · · · + ξk+n lim sup − α = 0. n→∞ k∈N n (b) Show that for any θ , the sequence (sin (nθ ))∞ n=1 is almost convergent to 0. Exercise 3.14 If x = (xj )∞ ∞ j =1 and y = (yj )j =1 are sequences in ∞ , let xy be the ∞ sequence (xj yj )j =1 . Show for any Banach limit L, there are sequences x and y in ∞ such that L(xy) = L(x) L(y). (Notice that L(xy) = L(x) L(y) if x and y are in c.) Exercise 3.15 Prove that ·, · is an inner product on H in the proof of Proposi- tion 3.21. Exercise 3.16 Show that a bounded linear functional f : c0 → R has a unique Hahn–Banach extension f˜ : ∞ → R. Exercise 3.17 Let E = {(xn )∞ n=1 ∈ 1 : x2k−1 = 0 for all k ∈ N}. Show that E is a closed subspace of 1 . Prove that any nonzero bounded linear functional on E has more than one Hahn–Banach extension to 1 . 1 Exercise 3.18 Let E = f ∈ L2 (0, 1) : xf (x) dx = 0 . Define a bounded 0 linear functional Λ on E by 1 Λ(f ) = x 2 f (x) dx, f ∈ E. 0 Find the (unique) Hahn–Banach extension of Λ to L2 (0, 1) and determine Λ. 1 1 (Hint: Use the fact that x 2 f (x) dx = (x 2 + ax)f (x) dx on E, for all a ∈ R.) 0 0 we will state and prove the Baire Category Theorem and see some of its applications.. 1] can be equipped with many metrics. Completeness of a metric is a very profitable property. If (Un )∞ n=1 is a sequence of dense open subsets of M. while (− π2 . but not to (− π2 . Two important examples are the metrics arising from the norms 1 1/2 f ∞ = max |f (s)| and f 2 = |f (s)| ds 2 . Correspondingly. J. Universitext. then n=1 Un is dense in M.1 The Baire Category Theorem In this section. s∈[0.e. Certainly. the © Springer Science+Business Media. Naturally. 1] of continuous functions on the interval [0. Bowers. all closed sets are Fσ -sets. however. However. d) is a%complete metric ∞ space. all open sets are Gδ -sets. We remind the reader that a Gδ -set is a countable intersection of open sets. Observe that. whereas the metric induced by the second norm is not (i. π2 ) is not. Recall that a set D is dense in a topological space M if and only if D ∩ U = ∅ for all nonempty open sets U in M. A notion that will prove fruitful is that of a Gδ -set. Therefore. although metric spaces were not formally defined until later. but not all Fσ -sets are closed.1. as we shall see in this chapter. π2 ). The metric arising from the first norm is complete. the hypothesis is not. LLC 2014 61 A. N. It originated in Baire’s 1899 doctoral thesis. DOI 10. The two spaces R and (− π2 . The first theorem we shall meet is a classical result about metric spaces called the Baire Category Theorem. 1]. there exist Cauchy sequences that fail to converge). 4. Theorem 4. while the conclusion of the Baire Category Theorem is topological in nature.1 (Baire Category Theorem) Suppose (M. π2 ) are homeomorphic (via the mapping x → arctan x). R is complete.1] 0 where f ∈ C[0.1007/978-1-4939-1945-1_4 . Kalton. An Introductory Course in Functional Analysis. an Fσ -set is the countable union of closed sets.Chapter 4 Consequences of Completeness The space C[0.1 applies directly to R.) Theorem 4. (See Exercise 4. but not all Gδ -sets are open. n=1 is countable. Since Gn is assumed to be dense. R\Q is the intersection of countably many sets.3. where x = p/q is written in lowest terms. %∞ Corollary 4. The same statement holds for all% x ∈ I . The set n=1 Un is precisely the set of all points of continuity of f . let us consider some consequences of it. Consequently. The preceding example shows that the set of irrational numbers R\Q is a Gδ -set. Suppose x ∈ Un . By definition. let the oscillation of f over I be given by oscI f = sup {|f (x) − f (y)| : {x. y} ⊆ I } . Before proving Theorem 4.1 holds for every metric space which is homeomorphic to a complete metric space.1. this can happen .3). and so I ⊆ Un . say Gn = ∞ m=1 Umn . there exists an open interval I such that x ∈ I and oscI f < n . let Un = {x ∈ R : ∃ an open interval I in R such that x ∈ I and oscI f < n }.%the set Gn can be written as an intersection of countably many open sets. then n=1 Gn is also a dense Gδ -set. each of which is open in R.1 (The Baire Category Theorem). The function f is continuous precisely on the set R\Q. Proof By assumption.2 If (Gn )∞ n=1 is a sequence of dense Gδ -sets. the set ∞ ∞ ∞ Gn = Umn n=1 n=1 m=1 is %∞ a dense set. but we could have shown that directly.3 If f : R → R is a function. the set of irrational numbers is a Gδ -set.4 does lead one to ask a natural question: Does there exist a function g : R → R which is continuous precisely on Q? By Proposition 4.4 Define a function f : R → R by f (x) = 0 for all x ∈ Q. for each n ∈ N. Since the sequence of open sets (Umn )∞ m. it must be the case that Umn is dense for each m and n in N. Note that R\Q = R\{q}. By Theorem 4. Let (n )∞ n=1 be a sequence of positive real numbers such that n → 0 as n → ∞. without the aid of Proposition 4. 2 Proposition 4. and consequently (by Proposition 4. then the set of points of continuity of f is a Gδ -set. q∈Q Thus. Example 4. and f (x) = 1/q for all x ∈ Q\{0}. the intersection ∞ n=1 Un is a Gδ -set.3.62 4 Consequences of Completeness conclusion of Theorem 4. f (0) = 1. %∞ Un is an open set. the set G n=1 n is a G δ -set. and hence the result. Therefore. Proof For any open interval I in R. For each n ∈ N. 2 Example 4. n ). 2 The name Baire Category Theorem is derived from a complementary formulation of Theorem 4. For any z ∈ M and > 0. there exists a point x2 ∈ U2 and an 2 ≤ 21 such that B(x2 . If U is a dense open set in M. Again arguing as before. ∩ Un ⊆ V ∩ (U1 ∩ · · · ∩ Un ). by construction. If Q is also a Gδ -set. It follows that. xn ) ≤ + ··· + < m + · · · + n+1 = n − m < n .4. xm−1 ) + · · · + d(xn+1 . Continuing inductively. xn ) ≤ n 2 . 21 ∩ U2 ⊆ V ∩ U1 ∩ U2 . n−1 ). xn ) ≤ d(xm . there exists a point x ∈ M such that x = lim xn . 2 for all n ∈ N. 2 ) ⊆ B x1 . as before. Hence. Consequently. 20 ∩ U1 ⊆ V ∩ U1 . 2 % N. n ) ⊆ B xn−1 . and hence d(x. n−1 x ∈ B(xn . % Let V be a nonempty open set in M. 20 ∩ U1 = ∅. This is a clear contradiction. d(xm . d) be a complete metric space and % suppose (Un )∞ n=1 is a sequence of dense open subsets of M. 0 ) ⊆ V . 2 2 2 2 2 2 2 It follows that (xn )∞ n=0 is a Cauchy sequence. Observe that n < 1 2n for all n ∈ N. If m > n.1 The Baire Category Theorem 63 only if Q is a Gδ -set. x ∈ V ∩ ∞ and so x ∈ V ∩ (U1 ∩ · · · ∩ Un−1 ) for all n ∈ % n=1 Un . by completeness. Proof of Theorem 4. n−1 and with the further property that n−1 B(xn . xn ). This fact motivates the following definitions. Thus. ⊆ B(xn−1 . let B(z. Both Q and R\Q are dense in R. n ) ⊆ B xn−1 . n→∞ We conclude that d(xm . 1) such that B(x0 . We conclude there exists no function that has Q as the set of points of continuity. m−1 n 1 1 1 1 1 d(xm . 1 By assumption. xn ) ≤ n for all n ∈ N. then xm ∈ B(xn . and so B x1 .1 (the Baire Category Theorem) Let (M.1. Therefore. Pick any x0 ∈ V and 0 ∈ (0. ) = {x ∈ M : d(x. the set n=1 Un is dense in M. By assumption. Suppose m > n. then the intersection Q ∩ (R\Q) = ∅ would have to be dense. the set U2 is dense in M. z) < } be the open ball about z of radius . We have shown that the intersection %∞ of ∞ n=1 Un and any open set V in M is nonempty. . and R\Q is a Gδ -set. The sequence (xn )∞ n=0 was chosen so that d(xn+1 . and so B x0 . we construct sequences (xn )∞ ∞ n=0 and (n )n=0 such that n ≤ 2 and xn ∈ V ∩ (U1 ∩ · · · ∩ Un ). the set U1 is dense in M. xn ) < n for all m > n. By the triangle inequality. then its complement M\U is closed with empty interior. 1 ) ⊆ B x0 . 2 ∩ U2 = ∅. by Corollary 4. Our goal is to show that ∞ n=1 Un is also dense in M.2 (a consequence of the Baire Category Theorem). We will show that ∞ n=1 Un and V have nonempty intersection. there exists a point x1 ∈ U1 and an 1 ≤ 20 such that B(x1 . Thus. we will learn the Uniform Boundedness Principle and see the important role it plays in the study of Fourier series. that is to say. This notion of size applies to any metric space. it is disjoint from ∞ n=1 Vn . 2 Theorem 4. Let G = n=1 Un . Since G is a subset of n=1 En .5 Let M be a topological space. This contradicts Corollary 4. In this case. any dense Gδ -set is second category. 4. Example 4. let Vn be the complement of En . In particular. if int(E) = ∅. A second category set is also known as non-meager. ∞ Then there exists a sequence (En )∞ n=1 of nowhere dense sets such that G ⊆ n=1 En .64 4 Consequences of Completeness Definition 4. Proposition 4. If λ represents Lebesgue measure on R. we have disjoint dense Gδ -sets.9 (Uniform Boundedness Principle) Let X and Y be Banach spaces and let Ti : X → Y be a bounded linear operator for each i ∈ I . Thus.8 Consider the real line R.7 (Baire Category Theorem. Then 1 λ(G) = 0. A set G ⊆ M is called first category (also known as meager) ∞ if there exists a sequence (En )∞n=1 of nowhere dense sets such that G ⊆ n=1 E n . 2 The intuition behind the proof of Theorem 4. we will investigate some implications of category in Banach spaces. by small we mean meager. the complement of G is first category. ∞ Therefore. and so G is not first category.2. Consider the set of rational numbers Q ⊆ R.1.Suppose that G is first category. for each n ∈ N. complementary version) In a complete metric space. It is natural to wonder if there is any relationship between the two. we may assume En is closed for each n ∈ N. then λ(Q) = 0 (because the set of ∞ rational numbers is countable). Proof Let G be a dense Gδ -set. there exists a sequence (U %n∞)n=1 of open sets such that Q ⊆ Un and λ(Un ) < n for each n ∈ N. but G cannot be first category. Therefore. Proof A countable union of countably many sets is a countable union of sets. The complement of a meager set is called a residual set. On the other hand.2 Applications of Category In this section. by Theorem 4. We have now for the set of real numbers two natural notions of smallness: first category and zero measure. Now. Then Vn is%both open and dense for each n ∈ N. ∞the intersection n=1 V% n is a dense Gδ -set. but must be of infinite measure. A set is called second category if it is not first category. where I is an index . Theorem 4. Without loss of generality. For further reading on the analogies between topological spaces and measure spaces. since Q ⊆ G and Q is dense in R.6 A countable union of first category sets is first category. A set E ⊆ M is said to be nowhere dense if the closure of E in M has empty interior.7 is that a countable union of small sets is still small. the curious reader might consider Measure and Category by John Oxtoby [29]. x ∗ ∈BX∗ x ∗ ∈BX∗ . it is necessary and sufficient that for each x ∈ X. By assumption. it follows that An is a closed set. To show sufficiency. Theorem 4. it must be that Ti (δu)Y ≤ 2n0 for all i ∈ I . then sup Ti (x)Y < ∞ for each x ∈ X if and only if sup Ti < ∞. Certainly. for a given a ∈ A. For each a ∈ A. the function φa is bounded and linear. since δ > 0 is a constant. Now let u ∈ BX . Thus.2 Applications of Category 65 set. Indeed. The set A is called weakly bounded if supa∈A |x ∗ (a)| < ∞ for all x ∗ ∈ X∗ . and the proof is complete. every x ∈ X is contained in An for some n ∈ N. Since Ti is continuous for each i ∈ I . define a scalar-valued function φa on X ∗ by φa (x ∗ ) = x ∗ (a). there exists some x0 ∈ An0 and δ > 0 such that x0 + δBX = {x : x − x0 X ≤ δ} ⊆ An0 . If {Ti : i ∈ I } is a collection of bounded linear operators from X to Y . For each a ∈ A. Consequently. we conclude that Ti (u)Y ≤ 2n0 /δ for all u ∈ BX . We chose x0 to be in An0 .4.6. Proof Let A be a subset of X. and so it follows that Ti (x0 + δu)Y ≤ n0 for all i ∈ I .12 Let X be a Banach space. Proof Necessity is immediate. not every An can be nowhere dense. Then x0 + δu ∈ An0 . Ti ≤ 2n0 /δ for each i ∈ I . φa = sup |φa (x ∗ )| = sup |x ∗ (a)| = a. Therefore. let An = {x : Ti (x)Y ≤ n for all i ∈ I } = {x : Ti (x)Y ≤ n} = Ti−1 (nBY ) . there exists a constant Cx > 0 such that Ti (x)Y ≤ Cx for all i ∈ I . By Proposition 4. if A is bounded. The set A is said to be bounded if supa∈A a < ∞. and so there exists some n0 ∈ N such that the closed set An0 has nonempty interior.11 Let X be a Banach space and let A be a subset of X. x ∗ ∈ X∗ . 2 The following theorem is a more concise statement of Theorem 4.9. and thus Ti (x0 )Y ≤ n0 for all i ∈ I . assume that for each x ∈ X there exists a constant Cx > 0 such that Ti (x)Y ≤ Cx for all i ∈ I .10 (Uniform Boundedness Principle) Let X and Y be Banach spaces. i∈I i∈I The set nBY is the closed ball centered at 0 ∈ Y with radius n. then it is weakly bounded. and therefore we must have X = ∞ n=1 An . Assume now that A is weakly bounded. Theorem 4. A subset of X is bounded if and only if it is weakly bounded. For each n ∈ N. By linearity. In order that there exists a uniform bound M such that Ti ≤ M for all i ∈ I . i∈I i∈I We now provide some definitions that lead to a straightforward application of the Uniform Boundedness Principle. Definition 4. 2 The next result is in some sense a converse to Theorem 4. If lim Sn x n→∞ exists for each x ∈ X.15. Sup- pose (Sn )∞ n=1 is a sequence of bounded linear operators from X to Y .12 and is left to the reader. n∈N n∈N Taking the supremum over x ∈ BX provides the desired result. and so there exists some natural number N such that Sm y − Sn y < /3 whenever m ≥ N and n ≥ N.66 4 Consequences of Completeness By the weakly bounded assumption on A. 2 Definition 4. Then. Since x is in the closure of E.14 Let X be a Banach space. . If (Sn )∞ n=1 is a sequence of uniformly bounded linear operators from X to Y . we conclude that supa∈A φa < ∞. then (i) sup Sn < ∞. Proof We need only show that E is closed. there exists some M > 0 such that Sn ≤ M for all n ∈ N. the sequence (Sn y)∞ n=1 is a Cauchy sequence in Y . A subset of X∗ is bounded if and only if it is weak∗ -bounded. we have shown that A is bounded. the closure of E. By assumption.16 Let X and Y be Banach spaces. Theorem 4. Let > 0.13 Let X be a Banach space and let A be a subset of X ∗ . sup |φa (x ∗ )| = sup |x ∗ (a)| < ∞. Theorem 4. observe that T x ≤ sup Sn x ≤ sup Sn x. n→∞ Proof (i) If (Sn x)∞ n=1 converges. Proof The argument parallels the proof of Theorem 4. Let m and n be natural numbers such that m ≥ N and n ≥ N . then T is a bounded linear operator. To show that T is bounded. The Uniform Bounded- ness Principle then implies that supn∈N Sn < ∞. by Theorem 4. By assumption. for each x ∗ ∈ X∗ . and n∈N (ii) if T x = lim Sn x for all x ∈ X.15 (Banach-Steinhaus Theorem) Let X and Y be Banach spaces. The set A is called weak∗ -bounded if supa ∗ ∈A |a ∗ (x)| < ∞ for all x ∈ X. a∈A a∈A Thus.10 (the Uniform Boundedness Principle). (ii) A simple check reveals that T is linear. Theorem 4. Since φa = a. there exists some y ∈ E such that x − y < /(3M). then the set E = {x : limn→∞ Sn xexists} is a closed linear subspace of X. 2 The following significant theorem is a consequence of the Uniform Boundedness Principle. then supn∈N Sn x < ∞. Suppose that x ∈ E. as required. We will consider the space C(T) of complex-valued continuous functions on the compact Hausdorf space T. lim Sn x exists.2 Applications of Category 67 Sm x − Sn x ≤ Sm x − Sm y + Sm y − Sn y + Sn y − Sn x <M + +M = . 3M 3 3M Therefore. it is natural to ask the following: Does the Fourier series of a general continuous function f converge to f ? To make this question more precise. By an abuse of notation. and so x ∈ E. the sequence (Sn x)∞ n=1 is a Cauchy sequence in Y . 2π). To that end. and then showing that these norms are not uniformly bounded. (4.2) n∈Z If f is a trigonometric polynomial. does SN f always converge uniformly to f ?) If the answer to this question is “yes. For each n ∈ Z. we have for each θ ∈ [0. 2π ).17 (Convergence of Fourier Series) Let T = {eiθ : 0 ≤ θ < 2π } be the one-dimensional torus. In k∈Z this case.3).” then the partial sum operators SN must be uniformly bounded. Fix N ∈ N and f ∈ C(T). by substituting (4. by the Uniform Boundedness Principle. we define the Fourier series of f by fˆ(n) einθ .3) n=−N The question now becomes: Is it the case that SN f − f ∞ → 0 as N → ∞ for all f in C(T)? (In other words.) Since a trigonometric polynomial is equal to its Fourier series.4.1) 0 2π We wish to determine if it is possible to reconstruct f from its Fourier coefficients. N 2π 2π N dφ dφ SN f (θ ) = f (φ) e−inφ einθ = f (φ) ein(θ −φ) .1) into (4. Let f ∈ C(T). define for each N ∈ N the N th partial sum operator SN : C(T) → C(T) as follows: N SN f (θ) = fˆ(n) einθ .9. θ ∈ [0. We will therefore show the answer is “no” by computing the operator norm SN for each N ∈ N. (4. it is easy to see that fˆ(n) = an for each n ∈ Z. 2π ). define the nth Fourier coefficient by 2π ˆ dθ f (n) = f (θ) e−inθ . Computing directly. (4. Since Y is complete. we write f (θ ) = f (eiθ ) for all θ ∈ [0. 2 n→∞ Example 4. (See Exercise 4. a subset of the complex plane C. n=−N 0 2π 0 n=−N 2π . then there exists a sequence of scalars (ak )k∈Z with only finitely many nonzero terms such that f (θ) = ak eikθ for all θ ∈ T. Applying the definitions. (4. 0 sin (φ/2) 2π This equality is valid for all f ∈ C(T).5) Let f ∈ C(T). because SN∗ δ0 is in M(T). 1 − ei(θ−φ) ei(θ −φ) − 1 We reduce the final fraction by dividing the numerator and denominator by e 2 (θ −φ) . 2π sin (N + 1 )φ dφ SN∗ δ0 M = 2 . SN∗ δ0 M ≤ SN∗ ≤ M. (4.37.5). By Proposition 3. and so dψ = (N + 21 ) dφ. i and so the above equation becomes N 1 ei(N+ 2 )(θ−φ) − e−i(N+ 2 )(θ −φ) 1 sin (N + 21 )(θ − φ) e in(θ−φ) = = .(2N+1)π ) dψ ≤ M. It is easy to see that δ0 M = 1. SN∗ ≤ M for all N ∈ N. We calculate the sum of this geometric series as follows: N 2N 1 − ei(2N +1)(θ −φ) ein(θ−φ) = e−iN(θ −φ) ein(θ−φ) = e−iN (θ −φ) n=−N n=0 1 − ei(θ −φ) −iN(θ−φ) e − ei(N+1)(θ−φ) ei(N +1)(θ −φ) − e−iN(θ −φ) = = . Then (2N+1)π ∗ sin ψ dψ SN δ0 M = ψ (2N + 1)π . each adjoint operator SN∗ : M(T) → M(T) must also be bounded by M. 0 sin 2N+1 Combining this with (4.4).68 4 Consequences of Completeness The sum appearing in the rightmost integral above is a geometric series with constant ratio ei(θ−φ) . Therefore.8)). e 2 (θ−φ) − e− 2 (θ−φ) sin θ −φ i i n=−N 2 Substituting this into the formula for SN f (θ). Let δ0 denote the Dirac measure at 0 (which was defined in (2. we conclude 2π sin (N + 21 )(θ − φ) dφ SN f (θ) = f (φ) . and as a consequence. we conclude that ∞ sin ψ 1 1 χ(0. π 0 2N + 1 sin 2Nψ+1 . and using (4. 0 sin (φ/2) 2π We make a change of variables: Let ψ = (N + 21 ) φ. we see 2π ∗ sin (N + 21 )φ dφ f . SN δ0 = SN f (0) = f (φ) . that is.4) 0 sin θ −φ 2 2π Assume there exists a constant M > 0 such that SN ≤ M for all N ∈ N. 6) 0 2π where SN is the N th partial sum operator. since 0 | sinψψ | dψ = ∞. it cannot be true that supN∈N SN ≤ M for some M ≥ 0. To see this.18 Before the advent of functional analysis. however. by Fatou’s Lemma.17 that DN ∗ f does not converge uniformly to f for each f ∈ C(T). n=−N sin (α/2) Observe that 2π dφ SN f (θ ) = f (φ) DN (θ − φ) = (DN ∗ f )(θ ). and in particular the Uniform Boundedness Principle. N→∞ 2N + 1 sin 2N+1 ψ ψ Consequently. Thus. k=0 (k + 1)π 0 π k=0 k + 1 Therefore. and so it is not possible that the Fourier series of f converges uniformly to f for all f ∈ C(T). we actually proved that (SN f (0))∞N =1 fails to converge to f (0) for all f in a dense Gδ -subset of C(T).2 Applications of Category 69 for all N ∈ N. π 0 ψ ∞ This is. however.19 For any natural number N . A quick application of l’Hôpital’s Rule reveals 1 sin ψ sin ψ lim = . a contradiction.4. 1 ∞ sin ψ dψ ≤ M.17. in Example 4. we can find a related kernel for which uniform convergence does hold for all continuous functions on T. In fact. (4. Definition 4. We know from Example 4. observe that ∞ (k+1)π ∞ sin ψ sin ψ dψ = ψ ψ dψ 0 k=0 kπ ∞ ∞ 1 π 2 1 ≥ sin ψ dψ = = ∞. α ∈ R. the Dirichlet kernel of degree N is the function N sin (N + 21 )α DN (α) = e = inα . there are many functions for which the Fourier series of f does not even converge pointwise to f . Remark 4. the only way to demonstrate that SN f did not converge uniformly to f for all f ∈ C(T) was to construct an explicit function for which the desired convergence failed. . Proof Let f ∈ C(T). and using the translation-invariance of Lebesgue measure. N Using (4. where A and B are real numbers. 2π dt TN f (θ) = f (t) KN (θ − t) = (KN ∗ f )(θ ). t ∈ R. 2 Lemma 4.6).70 4 Consequences of Completeness Definition 4. 2π ). 0 2π It follows that TN f C(T) ≤ f C(T) KN L1 (T) . θ ∈ [0. Lemma 4.22 If TN is the N th Cesàro mean. coupled with the periodicity of functions on T. we can derive a relationship between KN and TN . For t ∈ R. N k=0 The N th Cesàro mean of f ∈ C(T) is the function given by the formula 1 TN f = (S0 f + · · · + SN −1 f ). 2N k=0 sin2 (t/2) The first equality in the conclusion of the lemma follows directly from the above series. 1 sin (k + 21 )t 1 sin (k + 21 )t sin (t/2) N−1 N−1 KN (t) = = N k=0 sin (t/2) N k=0 sin2 (t/2) 1 cos (kt) − cos ((k + 1)t) N−1 = . we have 2π dt TN f (θ ) = f (θ − t) KN (t) . then TN = 1. .20 For any natural number N .21 If KN is the Fejér kernel of degree N . because it is a telescoping series. For any N ∈ N and θ ∈ [0. 0 2π We now find a closed-form formula for the Fejér kernel of degree N . 2N sin (t/2) N sin2 (t/2) Proof Recall the identity sin A sin B = 1 2 ( cos (A − B) − cos (A + B)). By a straightforward change of variables. The second equality in the conclusion of the lemma follows from the application of a half-angle identity. the Fejér kernel of degree N is the function 1 N−1 KN (t) = Dk (t). then 1 1 − cos (N t) sin2 (N t/2) KN (t) = 2 = . 2π). 2π 0 2π 0 N k=0 2π N k=0 0 n=−k Observe that . To find the value of KN L1 (T) .4.21. we compute it directly: N −1 2π 1 1 2π 2π N−1 k 1 1 KN (t) dt = Dk (t) dt = int e dt . we know that KN ≥ 0. and so KN L1 (T) = 1 2π 0 KN (t) dt.3 The Open Mapping and Closed Graph Theorems 71 2π By Lemma 4. then it is easy to see that TN f → f uniformly. By the linearity of T .24 Let X and Y be topological spaces.23 If f ∈ C(T). Before stating the next proposition. BX is a closed set and int(BX ) is an open set (in X). By the Weierstrass Approximation Theorem. 2 4. By Theorem 4. the result follows. 2π 0 if n = 0. A map T : X → Y is called open (or an open map) if T (U ) is open in Y whenever U is open in X. we recall that BX = {x : x ≤ 1} and int(BX ) = {x : x < 1}. The map T is an open map if and only if there exists a δ > 0 such that δBY ⊆ T (BX ). Naturally. Since TN f = KN ∗ f . and so T (intBX ) is open in Y . Definition 4. N−1 k 2π N −1 1 2π 1 1 KN (t) dt = eint dt = 2π = 1. Proposition 4. Proof If f is a trigonometric polynomial. we have 0 ∈ T (intBX ). then KN ∗ f → f uniformly. where X is a normed space. Proposition 4. 2 The next proposition provides a good example of how the Uniform Boundedness Principle is actually used in practice (in the guise of the Banach-Steinhaus Theorem). the set of functions for which this limit exists is closed. Since TN 1 = KN (t) dt = 1. It follows that TN f → f in C(T) as N → ∞ for all f ∈ C(T). Proof Assume T is an open map. eint dt = 0 2π if n = 0.25 Let X and Y be Banach spaces and suppose T : X → Y is a bounded linear operator. 2π 0 2πN k=0 n=−k 0 2π N k=0 2π 1 It follows that TN ≤ KN L1 (T) = 1. It follows that intBX .16. we 2π 0 conclude that TN = 1. The set intBX is open in X. Consequently.3 The Open Mapping and Closed Graph Theorems We begin this section with a topological definition. the trigonometric polynomials are dense in C(T). then T maps X onto Y .72 4 Consequences of Completeness contains a basic neighborhood of 0. We wish to show that T (U ) is open in Y whenever U is an open set in X. If U is an open set in X. then T is invertible and T −1 is continuous.26 Let X and Y be Banach spaces and suppose T : X → Y is a bounded linear operator. Consequently. there exists some x ∈ BX such that δy − T x < ν. and so the set T (U ) is open in Y . It must be that y ∈ yBY . Conversely. (i) If T is open.7) ŷ δ δ .25. T is an open map. The preimage of an open set is open. then δy ∈ δBY ⊆ T (BX ). there exists an element x ∈ BX such that ŷ x ν −T < . By our assumption. 2 Definition 4. and so for any ν > 0. T is almost open. If ŷ is an element of Y such that 0 < ŷ < δ. (ii) If T is one-to-one and open. The map T is said to be almost open if there exists a δ > 0 such that δBY ⊆ T (BX ). Proposition 4. To that end. Let y ∈ Y . there exists a δ > 0 such that δBY ⊆ T (intBX ) ⊆ T (BX ). Since T is also assumed to be one-to-one. and so we conclude that T −1 is continuous. (ii) If T is open. There exists some x ∈ U such that T x = y. and U is open in X. 1) such that δBY ⊆ T (BX ). Therefore. then T is open. y ∈ intT (U ).28 Let X and Y be Banach spaces and suppose T : X → Y is a bounded linear operator. then T is onto. and so y y y∈ · δBY ⊆ T (BX ). Since x ∈ U . assume there exists a δ > 0 such that δBY ⊆ T (BX ). Therefore.27 Let X and Y be Banach spaces and suppose T : X → Y is a bounded linear operator. by (i). as required. there exists a δ > 0 such that δBY ⊆ T (BX ). there is some ν > 0 such that x + νBX ⊆ U . there exists an x ∈ BX such that y − T (x/δ) < ν/δ. (4. It follows that y + νT (BX ) ⊆ T (U ). let U be open in X and let y ∈ T (U ). then (T −1 )−1 (U ) = T (U ) is open in Y (since T is an open map). and so T is onto. δ δ It follows that y = T y δ x for some x ∈ BX . If y ∈ BY . then ŷ/ŷ ∈ BY . It remains to show that T −1 : Y → X is continuous. By Proposition 4. Consequently. the inverse function T −1 is well-defined. Proof (i) Suppose T is an open map. If T is almost open. 2 Corollary 4. Proof By assumption. for any y ∈ BY and ν > 0. and therefore there exists a δ ∈ (0. for any ν > 0. Therefore. this implies that y + νδBY ⊆ T (U ). x2 ≤ 2 β. we construct a sequence (xn )∞ n=1 such that n ν n ν n−1 y − T (xk ) ≤ β. (4. δ δ Continuing inductively. By (4. as required. Again we use (2.9) and the continuity of T . By Proposition 4. it will suffice to show that int(δBY ) ⊆ T (BX ). Then y < δ. n→∞ k=1 Consequently. If ŷ = 0. Let β be any real number such that y < β < δ and choose a real number ν such that 0 < ν < δ − β. x̂ ≤ .8). We assumed y < δ. This implies that y − T (x1 ) is an element of Y with the property that y − T (x1 ) < β < δ. let y ∈ int(δBY ). δ δ Now observe that ν < δ. Theorem 4. n y = lim T (xk ) = T (x). and so β ν n−1 ∞ ∞ β 1 β xk ≤ = = < 1. and so. By the triangle inequality. by (4. we can find an x̂ ∈ X such that ν ŷ ŷ − T (x̂) ≤ ŷ.29 (Open Mapping Theorem) Suppose X and Y are Banach spaces. and so x ∈ BX .8) δ δ If ŷ = 0. xn ≤ β. then we choose x̂ = xδŷ . Denote this limit in X by x.7). and we find an x2 ∈ X such that ν 2 ν y − T (x1 ) − T (x2 ) ≤ β. there exists an x1 ∈ X such that ν β y − T (x1 ) ≤ β. To that end.4. Therefore. by the Cauchy Summability Criterion (Lemma cauchy-criterion). x ≤ ∞ k=1 xk < 1. for any ŷ ∈ int(δBY ) and ν > 0. then T is an open map. we choose x̂ = 0. (4. x1 ≤ . . k=1 k=1 δ δ δ 1− ν δ δ−ν Since X is a Banach space. and so T is an open map. int(δBY ) ⊆ T (BX ). we have y ∈ T (BX ). Our goal is to show that T is an open map. only this time with the element y − T (x1 ). 2 The next theorem is one of the cornerstones of functional analysis. where x is the element of BX chosen to satisfy (4.3 The Open Mapping and Closed Graph Theorems 73 Consequently.25. we have β < δ − ν.9) k=1 δ δn for each n ∈ N.24). Since ν < δ − β. If T : X → Y is a bounded surjective operator. ∞ k=1 xk converges to an element in X. 28. Thus.30).49. where Q is the quotient map onto X/ker(T ). Consider the following commuting diagram: T X Y Q T0 X / ker(T ) In this diagram. say (xn )∞ n=1 . up to an isomorphism. Up to an equivalence. Consequently.30 (Bounded Inverse Theorem) Let X and Y be Banach spaces. Therefore. T0 is an isomorphism. Proof By assumption. the converse is also true. T = T0 ◦ Q. there exists an element y ∈ Y and a number δ > 0 such that y + δBY ⊆ T (BX ).7. any quotient map is an open map. (See Exercise 4. by the Bounded Inverse Theorem (Corollary 4. let X be a separable Banach space. T is a continuous bijection with continuous inverse.26. by Corollary 4. but they are in fact equivalent. Since T is surjective. If T : X → Y is a bounded linear bijection. by Corollary 4.31 By Theorem 4.74 4 Consequences of Completeness Proof Observe that Y = T (X) = ∞ n=1 n T (BX ).21. the map T0 is a continuous linear bijection. as well. T is almost open. 2 We have stated the Bounded Inverse Theorem as a corollary to the Open Mapping Theorem. It follows that T is an open map. The closed unit ball BX has a countable dense subset. by Proposition 3. By assumption.29. A simple calculation reveals that −y + δBY ⊆ T (BX ). This means that any open map is a quotient map.) Example 4. because T0 need not be an isometry.26. by Theorem 4. Suppose X and Y are Banach spaces and let T : X → Y be a bounded linear operator that is an open mapping. Consequently. and so is an isomorphism between the Banach spaces X and Y . the set T (BX ) has non-empty interior.) Example 4.29. Therefore. by Theorem 4. we know that T is a surjection. Consequently. (Note that the norms of T and Q might not be equal. any continuous linear bijection between Banach spaces is an isomorphism. by Proposition 4. it is an open map. Therefore. then T −1 is a bounded linear operator. where Q is a quotient map and T0 is an isomorphism. Therefore. To see this. the bounded linear operator T is an open map. 2 Corollary 4. and so it must be the case that δBY ⊆ T (BX ). Define a bounded . the map T is a continuous bijection.32 Banach-Mazur Characterization of Separable Spaces) A rather remarkable result (dating back to 1933) is that every separable Banach space can be realized as a quotient of 1 . Because T is an injective open map. it follows that the inverse T −1 is bounded. This demonstrates that the open map T can be written as T0 ◦ Q. the separable Banach space X is isometrically isomorphic to 1 /ker(T ). there exists a bounded linear functional f : Y → K. (xn )n=1 is dense in BX . n=1 n=1 n=1 (Recall xn ∈ BX for all n ∈ N. so that en = (0. T ≤ 1. the map T is a surjection. where K is the field of scalars. . . (See the diagram below.49 (noting that T is a surjection).26.) Consequently. If x ∗ ∈ (kerT )⊥ . The above argument shows that T is an almost open map (with δ = 1). It was assumed that x ∗ ∈ (kerT )⊥ . By Proposition 4.30). we conclude that (xn )∞ ∞ n=1 ⊆ T (B1 ).28. n=1 The linearity of T follows from the summability of the terms in the sequence ξ ∈ 1 . Certainly en ∈ B1 for each n ∈ N. the linear functional x ∗ determines an element fˆ of (X/kerT )∗ via the identification fˆ(x + kerT ) = x ∗ (x).3 The Open Mapping and Closed Graph Theorems 75 linear operator T : 1 → X by ∞ T (ξ ) = ξn x n . by Corollary 4. observe: ∞ ∞ ∞ T (ξ ) = ξn x n ≤ |ξn | xn ≤ |ξn | = ξ . Recall that en denotes the sequence with 1 in the nth coordinate and 0 elsewhere. . .33 Let X and Y be Banach spaces and suppose T : X → Y is a surjective bounded linear operator. By assumption. Therefore. The following factorization result is a consequence of the Bounded Inverse Theorem. then there exists some f ∈ Y ∗ such that x∗ = f ◦ T .31. by Proposition 3. ξ = (ξn )∞ n=1 ∈ 1 . By the Bounded Inverse Theorem (Corollary 4. Since T (en ) = xn for each n ∈ N. we conclude that T0 is an isomorphism of Banach spaces.51. and so T (B1 ) ⊆ BX . there is a continuous linear bijection T0 : X/ker(T ) → Y such that T = T0 ◦ Q. 0. . Proof Let Q : X → X/ker(T ) be the quotient map. an almost open map is an open map. By Proposition 3. Thus. but in our current context: T 1 X Q T0 1 / ker(T ) As we saw in Example 4. Hence. T0 is an isometry because BX = T (B1 ). Since X/ker(T ) is isomorphic to Y . and one we will use on several occasions. Lemma 4. 1. ). To show that T is bounded. 0. and so BX = T (B1 ). such that fˆ = f ◦ T0 . Consider the diagram from Example 4. the map T0 is an isomorphism.31.) . . Furthermore. for all x ∈ X.4. . for all x ∈ X. First. Under this norm.) Example 4. Thus.34 Let X and Y be Banach spaces and suppose T : X → Y is a linear map. Theorem 4. By definition. the linearity assumption in the Closed Graph Theorem is necessary to prove continuity. Consider the map f : R → R given by . (See Proposition 3. The graph of T is the subset of X × Y given by G(T ) = {(x. S −1 x = (x. T x) = x. It is also true that the Closed Graph Theorem implies the Open Mapping Theorem.35). If T has a closed graph.20. by the Bounded Inverse Theorem (Corollary 4.30). and hence continuous. y) = x + y.36 In general. let (x. and consequently T ≤ S −1 . G(T ) is closed. we conclude that S −1 is bounded. and also that the Bounded Inverse Theorem implies the Closed Graph Theorem (Theorem 4. 2 We have now shown that the Open Mapping Theorem implies the Bounded Inverse Theorem (Corollary 4.76 4 Consequences of Completeness T X Y Q f T0 X / ker(T) K fˆ Therefore. then T is continuous.44. X × Y is a Banach space. we need a definition. x ∗ (x) = fˆ(x + kerT ) = fˆ(Qx) = (f ◦ T0 ◦ Q)(x) = (f ◦ T )(x). We call G(T ) a closed graph if it is closed as a subset of X × Y . Clearly. Therefore. Define a map S : G(T ) → X by S(x. Definition 4.30). Proof For x ∈ X and y ∈ Y . and consequently all three are equivalent. T x) : x ∈ X}. x ∈ X. T x) = S −1 x ≤ S −1 x.35 (Closed Graph Theorem) Let X and Y be Banach spaces and suppose T : X → Y is a linear map.) By assumption. (See Exercise 4. T x). and so x + T x = (x. and hence also a Banach space. as required. 2 We now turn our attention to an alternate formulation of the Open Mapping Theorem. S is a bijection and S ≤ 1. It follows that T is bounded. T x ≤ S −1 x for all x ∈ X. 1/x if x = 0. The graph of f is closed. but f is certainly not continuous. . f (x) = 0 if x = 0. 4. 1] → [0. and so (f (xn ))n=1 has a convergent subsequence. We have already established that φ(f ) = 0 whenever f is a trigonometric polynomial. A simple computation shows that φn ≤ 1 for all n ∈ Z. define a linear functional on L1 (T) by φn (f ) = fˆ(n) for all f ∈ L1 (T). then a closed graph does indeed suffice for continuity. which contradicts (4.4 Applications of the Open Mapping Theorem dθ Consider the torus T. The next proposition.17. 2π as L1 (T). we will abbreviate L1 T. for any f ∈ L1 (T). we define the Fourier coefficients of f by 2π dθ fˆ(n) = f (θ) e−inθ . For any f ∈ L1 T. where K is an arbitrary compact set.10) ∞ The interval ∞ [0. By definition. Suppose y is the limit of this subsequence. the set {f : lim|n|→∞ φn (f ) exists} is a closed linear subspace of L1 (T).4 Applications of the Open Mapping Theorem 77 If we restrict our attention to functions f : [0. In the preceding paragraph. 1] → [0. since φ is continuous.16. the doubly infinite The significance ˆ sequence f (n) is always an element of c0 (Z). n ∈ Z. 1]. By Theorem 4. but this implies y = f (x). 1] → [0. |f (xn ) − f (x)| > . Assume to the contrary that there is a function f : [0. 1] and a sequence (xn )∞ n=1 in [0. then lim fˆ(n) = 0. but this argument works equally well for a function f : K → K. then there exists some N ∈ N such that φn (f ) = 0 for all |n| ≥ N . n ∈ N. |n|→∞ Proof For each n ∈ Z.) dθ For ease of notation.37 (Riemann-Lebesgue Lemma) If f ∈ L1 (T). and so the sequence of linear functionals is uniformly bounded. Therefore. the map defined by φ(f ) = lim|n|→∞ φn (f ) for f ∈ L1 (T) is a bounded linear functional on L1 (T).4). Then there exists some x ∈ [0. for some > 0. 2π . Since the graph of f is closed. and hence the limit exists for all elements of L1 (T). we considered a function f : [0. This may lead one to ask if n∈Z the converse is true. 1] with closed graph that is not continuous. y) ∈ G(f ). Theorem 4. which makes use of the Open Mapping . and since the trigonometric polynomials are dense in the space of integrable functions. In particular. By Theorem 4. 2 of Theorem 4.15. say f (xnk ) k=1 . (4. it follows that (x. lim|n|→∞ φn exists on a dense subset of L1 (T). the point (xnk .4. 1] is compact. we have that φ(f ) = 0 for all f ∈ L1 (T).3. 0 2π (See Example 4. This proves the theorem. If f is a trigonometric polynomial.37 is that. f (xnk )) is in G(f ) for all k ∈ N. 1] converging to x such that. 1]. ψj is a bounded linear functional on p for each j ∈ N. suppose the sequence ηξ = (ηj )∞ j =1 is an element of p .k=1 is an infinite matrix. let n ψj (ξ ) = lim aj k ξk = ηj . . In particular.38 There exists a sequence ξ ∈ c0 (Z) which is not the Fourier transform of a function in L1 (T).30 (the Bounded Inverse Theorem). is impossible. We suppose (aj k )∞ j . any bounds can be established for the rate at which the se- 1 functions quence fˆ(n) decays. however. ξ ∈ p . F −1 is also a bounded linear bijection. where 1 ≤ p ≤ ∞. (See Section I. Proof For each j ∈ N.k=1 is a scalar array such that the series ∞ ηj = a j k ξk k=1 converges for all ξ = (ξk )∞ k=1 ∈ p and j ∈ N. If the map A : p → p is defined by A(ξ ) = ηξ for each ξ ∈ p .) Now consider the sequence space p = p (N).39 Suppose (aj k )∞ j . n∈Z It suffices to show that F is not a surjection.38. however. Proof Define a map F : L1 (T) → c0 (Z) by F(f ) = fˆ(n) . Proposition 4. 2 In light of Theorem 4. because 1 (Z) is separable and L∞ (T) is not. then A is a bounded linear operator on p .15. which implies that (F −1 )∗ : L∞ (T) → 1 (Z) is also a bijection. Our assumption that F is a surjection has led us to conclude that the inverse F −1 : c0 (Z) → L1 (T) is a bijection. f ∈ L1 (T).37 and Proposition 4. Furthermore. Then F is a bounded linear bijection. n→∞ k=1 By Theorem 4. it is tempting to wonder if.78 4 Consequences of Completeness Theorem (in the form of the Bounded Inverse Theorem). Proposition 4. It turns out. shows that the converse is not true. (The proofs of separability and nonseparability are similar to those given in Remark 3. and therefore must conclude that F does not map L1 (T) onto c0 (Z). (See Exercise 4. that Fourier coefficients can decay n∈Z arbitrarily slowly. by Corollary 4. We use the bounded linear functionals (ψj )∞ j =1 to define the map A : p → p by ∞ Aξ = ψj (ξ ) j =1 .) We have arrived at a contradiction. for f ∈ L (T).12. where aj k ∈ R for each j and k in N. The map F is a bounded linear operator with F = 1.14. Suppose F does map L1 (T) onto c0 (Z).4 of [20].) This. and hence an isomorphism. and F is injective because Fourier coefficients are unique. Definition 4. there exist positive constants c and C such that c xα ≤ xβ ≤ C xα . A is linear because ψj is linear for each j ∈ N. In such a case. Provided the map H is well-defined. by the Closed Graph Theorem (Theorem 4. For each j ∈ N.4. · β ) are isomorphic as Banach spaces. We begin with a definition. Example 4.49. f ∈ Lp (R). We pause for a moment to recall that a map T : X → Y is called an isomorphism of Banach spaces if it is a continuous linear bijection with a continuous inverse. and V and X/W are isomorphic as Banach spaces. the same calculation that shows that H is well-defined also shows that H is bounded. −∞ x − y where 1 < p < ∞. It remains only to show that A is bounded. By Proposition 3. then the two norms are equivalent.13.7: direct sums and quotient spaces. the graph of A is closed.) We now turn our attention back to the topics of Section 3. A map P : X → V is called a projection if P x = x for all x ∈ V . Note that if (X. where Q : X → X/W is the quotient map. this map is well defined. and so ψj (ξ (n) ) → ψj (ξ ). Therefore. Proof of Theorem 4. that is. and suppose ∞ ζ = (ζj )j =1 ∈ p is such that Aξ → ζ in p . it will follow directly (by the Closed Graph Theorem) that it is bounded (and hence continuous).42 Suppose that P : X → V is a continuous projection and W = ker(P ). it must be that ζj = ψj (ξ ) for each j ∈ N. we say X and Y are isomorphic.41 Let X be a Banach space with closed subspace V . then X and V ⊕ W are isomorphic as Banach spaces.42 Let X be a Banach space with closed subspace V . the map ψj is a (n) continuous linear functional on p . we will show that the graph of A is closed and apply the Closed Graph Theorem. and hence ζ = Aξ . The next example provides a further illustration. Suppose (ξ (n) )∞ n=1 is a sequence in p such that ξ (n) → ξ in p .39 demonstrates the general principle that if a linear map is properly defined. Conversely. it will tend to be bounded. x ∈ X. · α ) and (X. In order to do this. If P : X → V is a continuous projection and W = ker(P ). Furthermore. then there exists a continuous projection P : X → V with W = ker(P ). 2 Proposition 4. Consequently.35). there exists a linear map P0 : X/W → V such that the diagram below commutes. . if X = V ⊕W . (See Exercise 4. As a practical matter. and so A is continuous.4 Applications of the Open Mapping Theorem 79 By assumption.) Theorem 4.40 The Hilbert transform is the map H : Lp (R) → Lp (R) defined by the formula ∞ 1 Hf (x) = f (y) dy. (Note that this is equivalent to the condition P 2 = P . however. 49. we have that P x = 0. let (v. φ(v + w) = (P (v + w). Define φ : X → V ⊕ W by φ(x) = (P x.45. we show that φ is a bijection. Then P (v) = v and P (w) = 0. (v + w) − P (v + w)) = (v. x −P x) = (0. We know φ is well-defined by our earlier remarks. w) ∈ V × W . · ) is isomorphic to V ⊕ W equipped with the norm from Proposition 3. Consequently. and hence φ is injective. From this we conclude that x = P x = 0. Thus. where v ∈ V and w ∈ W . We next show that φ is continuous by showing that it is bounded: φ(x)V ⊕W = (P x. φ is a surjection.44: (v. by Proposition 3. To show that φ is surjective. It follows that x = 0. Then x has a unique representation of the form v + w. Now suppose X = V ⊕ W . by the Bounded Inverse Theorem (Corollary 4. It remains only to show that P is continuous. Any continuous linear bijection will have a continuous inverse by the Bounded Inverse Theorem (Corollary 4. and so x ∈ V implies that P x = x. Thus. x − P x). we wish to show (X. and so P x = 0 and x −P x = 0. Then (P x. 0). The next step is to show that X is the vector space direct sum of its subspaces V and W . Suppose φ(x) = (0. P is a projection onto V . v + w − v) = (v. We conclude that X = V ⊕ W as vector spaces. 0). Since x ∈ W . Define P (x) = v. φ is a continuous linear bijection.80 4 Consequences of Completeness P X V Q P0 X/W (Recall that W is closed by Proposition 3. φ is linear because P is linear and because of the way the vector space operations are defined in V ⊕ W . w) = v + w. every x ∈ X can be written as the sum of elements from V and W as follows: x = P x + (x − P x). and consequently an isomorphism. Therefore. w). V ∩ W = {0}. x ∈ X. Suppose (xn )∞ n=1 is a convergent sequence in X such . We will show continuity by means of the Closed Graph Theorem. w) ∈ V × W. However. Next. We now show that X is isomorphic to V ⊕ W . x − P x)V ⊕W = P x + x − P x ≤ (2 P + 1)x. Specifically. (v.30). Furthermore. Then P is a projection and W = ker(P ). Furthermore.) The continuity of P0 follows from the continuity of P and the fact that Q is an open map.30). Suppose that x ∈ V ∩ W . and so P0 is an isomorphism. 2 Show that the space C[0. We need to show that P x = v. we call V a complemented subspace of X. 1] of continuous functions on the closed interval [0. . P v = v. 1]. 1]. Show that the closed set [0. 1] and let Y = C[0. 1] in the norm · 2 is L2 (0. 1] be the space of continuously differentiable functions on [0. 1] is a Gδ -set. by Lusin’s Theorem. Suppose h : M → E is a homeomorphism onto its image (i. Equip both spaces with the supremum norm · ∞ .7 Let X = C (1) [0.e. 0 (The completion of C[0. 1]. f ∈ C[0. 1) is an Fσ -set. and h−1 |h(M) is continuous). h is a continuous one-to-one map..) Exercise 4. Exercise 4. Exercises Exercise 4. (Hint: Consider Y as a dense subspace in its completion. n→∞ Therefore. Exercise 4. For this reason. Show that h(M) is a Gδ -set. we have xn − P xn ∈ ker(P ) = W .5 Show that if Y is a normed space which is homeomorphic to a complete metric space. Then there exists some x ∈ X and v ∈ V such that xn → x and P xn → v as n → ∞. Show that E = X. · ∞ is not a Banach space. If M is a closed subspace of X. 1). Conclude that C (1) [0. 1]. 2 We have seen that whenever a closed subspace V is the image of a continuous projection in X. and hence P x = v. when V is the image of a projection. By assumption. Show that T has closed graph. as required. P (x − v) = 0. For each n ∈ N.6 Let X and Y be Banach spaces and let T : X → Y be a bounded linear operator. but T is not continuous. Exercise 4. show that either T (M) is first category in Y or T (M) = Y . lim xn − P xn = x − v ∈ W. Exercise 4.Exercises 81 that (P xn )∞ n=1 converges in V . and so P x = P v.) Exercise 4.1 Let R be given the standard topology. Since W is closed. then Y is a Banach space.3 Let M and E be complete metric spaces. Show that the open set (0.4 Let X be a Banach space and suppose E is a dense linear subspace which is a Gδ -set. Define a linear map T : X → Y by T (f ) = f for all functions f ∈ C (1) [0. there is a closed subspace W such that X = V ⊕ W . 1] is not complete in the norm 1 1/2 f 2 = |f (s)| ds 2 . 1). Exercise 4. or T −1 will be bounded by Corollary 4.14 Let f : [1. n→∞ ξn n→∞ x→∞ Exercise 4.19 Let 1 ≤ p ≤ ∞ and let V = {(xj )∞ j =1 ∈ p : x2k = 0 for all k ∈ N}.11 Show that the Cesàro means N1 (S1 f + · · · + SN f ) converge to f in the L1 -norm for every f ∈ L1 (T). If T : X → Y is a bijection. Show that Lp (Ω. lim ξn = ∞. and n→∞ ξn+1 lim = 1. (Hint: X and Y cannot both be Banach spaces.21 Show that the Closed Graph Theorem (Theorem 4. μ) be a probability space and suppose there exists a sequence of disjoint sets (En )∞ n=1 such that μ(En ) > 0 for all n ∈ N.40 is well-defined. Show the set M = {φ f : f ∈ C[0.13 Show that the Hilbert transform of Example 4.9 Let (ak )k∈Z be a sequence of complex scalars with only finitely many nonzero terms. then prove that lim f (x) = 0. then so is T ∗ . Exercise 4. 1].12 Let X and Y be Banach spaces. μ) = Lq (Ω.35) implies the Open Mapping Theorem (Theorem 4. Exercise 4. Exercise 4. such that T is a bounded linear bijection. Exercise 4.30. 1] be a function which is not identically 0. If both V and X/V are separable. Define a trigonometric polynomial f : T → C by f (θ ) = ak eikθ .20 Find an example of a map T : X → Y .10 Show that there exists a function f ∈ L1 (T) whose Fourier series fails to converge to f in the L1 -norm.18 Identify the quotient space c/c0 . then show X is separable. Exercise 4.15 Show that L2 (0. Exercise 4. . ∞) → R be a continuous function.) Exercise 4.82 4 Consequences of Completeness Exercise 4. show that the adjoint map T ∗ : Y ∗ → X∗ is also a bijection.29). where X and Y are normed spaces. 1] if and only if φ(x) = 0 for some x ∈ [0. Show that p /V is isometrically isomorphic to p . k=−N then there exists an f ∈ L1 (T) such that f − SN f L1 (T) does not tend to 0.8 Let φ ∈ C[0.17 Let X be an infinite-dimensional Banach space and suppose V is a closed subspace of X. μ) if 1 ≤ p < q < ∞. If lim f (ξn x) = 0 for all x ≥ 1. 1]} is of the first category in C[0. Exercise 4. but T −1 is not bounded. Suppose (ξn )∞ n=1 is a strictly increasing sequence of real numbers with ξ1 ≥ 1. Exercise 4. 1) is of the first category in L1 (0. k∈Z Show that fˆ(n) = an for all n ∈ Z. Conclude that if T is an isomorphism of Banach spaces. Exercise 4. Precisely.16 Let (Ω. show that if N SN f = fˆ(k) eikθ . Exercise 4. we wish to explore the geometric aspects of the Hahn–Banach The- orem. ∅}. τ ). as follows: W ⊆ E × F is open if © Springer Science+Business Media.1 General Topology Let E be a set. DOI 10. These spaces. τ ) a topological space.1007/978-1-4939-1945-1_5 . Example 5. If {Ui }i∈I is a (possibly uncountable) family of sets in τ . Example 5. denoted τ × τ . A topology τ on E is a collection of subsets called open sets satisfying the following three criteria: 1. Universitext. J. which include Banach spaces. 3. then τ is called the discrete topology. 5. a neighborhood of x is any subset N of E for which there exists an open set U such that x ∈ U and U ⊂ N . 2. When E is equipped with a topology τ . The collection τ contains both E and the empty set ∅. N.1 Let (E. An Introductory Course in Functional Analysis. is local convexity. then U ∩ V is in τ . If every set in E is open. τ ) and (F . Kalton. are sufficiently complex that we can say something interesting about their structure. then i∈I Ui is in τ . and in this chapter we will get some idea of how they fit into a more general topological framework. We will first recall some notions from general topology and then introduce the concept of a topological vector space. For x a point in E. If U and V are in τ . τ ) be a topological space. When there is no ambiguity. τ ) and (F .Chapter 5 Consequences of Convexity In this chapter. τ ) be two topological spaces. If (E. and say U is open in E when U ∈ τ . Banach spaces are topological vector spaces where the topology is determined by a complete norm. then τ is called the indiscrete topology. then a function f : E → F is called continuous if f −1 (U ) is open in E whenever U is open in F . Bowers. The crucial property. we will suppress the τ and simply write E for the topological space (E.2 Let (E. If τ = {E. we call the pair (E. LLC 2014 83 A. τ ) are topological spaces. it turns out. The product E × F can be given the product topology. A local base at x is a collection η of open sets. such that any neighborhood U of x contains an element of η. we see that any metric . For $ i ∈ I . τi ) is a topology with a base consisting of sets of the form & U i1 × · · · × U in × Ei . there exists a number rx > 0 such that B(x. Thus. Example 5.in } where Uij is open in Eij for ij ∈ I . rx ). or just the metric topology on M. τ ) be a topological space. . because the product in that example is finite. let (Ei . Let E be a topological space and let x ∈ E.4 (and the comments following it).2 as a special case. The topology τ is said to have a base of open sets {U i }i∈I if for each open set V ∈ τ . if the metric d is understood. the collection of open balls B(x.. n ∈ N. For each x ∈ M and r > 0. any point in the metric space M has a local base. rx ) ⊆ V . . When τ has a base. In fact. . where I is a (possibly uncountable) index set.4. there exists an index set J ⊆ I such that V = i∈J Ui . 1/n) : n ∈ N}. (E. all of which contain x. From Example 5. For x ∈ M. we say the base generates the topology τ . This set is the open ball about x of radius r. V = x∈V B(x. i∈I where Ui ∈ τ and Vi ∈ τ for each i ∈ I . V ∈ τ }. We declare a subset V of M to be open if for each x ∈ V . n}. z) < r}. This example contains Example 5. Example 5.2.4 Suppose M is a set with a metric d. r) = {z ∈ M : d(x. . τ ) with a countable local base at every point x ∈ E is called first countable. In Example 5. We will show that the metric d determines a topology on M. Observe that all but finitely many elements of the product are the entire space. there is an r > 0 such that B(x. A topological space (E.3 (Product topology). if we consider the collection of sets η = {B(x. For each x ∈ V . Let I be a (possibly uncountable) index set.. .84 5 Consequences of Convexity W = (Ui × Vi ). Further. τi ) be a topological space. The product topology on the product each i∈I (Ei . τ ) is called second countable if τ has a countable base. and j ∈ {1. then η is a countable local base at x. In Example 5. the product topology on E × F is generated by the base {U × V : U ∈ τ . r) ⊆ V .. The collection of all such open sets forms a topology on M called the metric topology on M generated by the metric d. and so the collection of open balls forms a base for the metric topology. r) for all r > 0 forms a local base at x. let B(x. Let (E. i∈I \{i1 . Suppose V is open in the metric topology. let U = B(x. Since x = y. Define a metric on E by . (See Exercise 5. M will not be second countable unless M is separable. if the open balls in (E. and U ∩ V = ∅. and U ∩ V = ∅. d) is a metric space and let x and y be two distinct points in M.6 Any nonempty set E with the discrete topology (see Example 5. τ ) a Hausdorff space if for any distinct points x and y in E there exist open sets U and V in τ such that x ∈ U . τ ) is metrizable if there exists a metric d such that d generates the topology τ .9. We say (E. τ ) be a topological space. y ∈ V . Furthermore. δ/2).) Let (E. however. y ∈ V . We call (E.5 Any metrizable space is a Hausdorff space.1 General Topology 85 space M is first countable. d) form a base for the topology τ . Let δ = d(x. it follows that d(x. δ/2) and V = B(y.1) is metrizable. y) > 0. That is. x ∈ U . Example 5. y). Example 5. To see this. suppose (M. Then U and V are open in the metric topology on M.5. 1 if x = y. y) ∈ E × E. A topological space is said to be compact if any open cover contains a finite open subcover. the real line R with its standard topology is locally compact. A notion of fundamental importance in topology is that of a convergent sequence. For example. It is easy to see that d is. the limit of a sequence need not be unique. The spaces we consider. Of particular interest to us is the notion of compactness. If X is a topological space.8. all compact spaces are locally compact. . For a (not necessarily compact) topological space. (Note that this notion of a limit agrees with the standard definition of a limit in a metric space.) . let X be a topological open sets such that X ⊆ U ∈U U there exists a finite collection {U1 . we say x is the limit of the sequence (xn )∞n=1 and we write x = n→∞ lim xn . Naturally. in fact. . This metric is called the discrete metric on E and it is not hard to show that d generates the discrete topology on E. d(x. Some well-known properties of compact sets are treated in the exercises at the end of this chapter. In such a case. (See Exercise 5.) A topological space is said to be locally compact if every point has a compact neighborhood. but the converse need not be true. a metric. . we define a compact subset in a similar way: A subset E of a topological space X is compact if any cover of E by sets open in X admits a finite subcover of E. y) = for (x. and (xn )∞ n=1 is a sequence of elements from X. Un } of elements in U such that X ⊆ U1 ∪ · · · ∪ Un .) In general. then (xn )∞ n=1 is said to converge to a point x ∈ X if for every open neighborhood U of x there exists an N ∈ N such that xn ∈ U for all n ≥ N . are Hausdorff spaces.2. (See Exercise 5. but not compact. Then X is compact if for any collection U of precise. however. 0 if x = y. and limits are necessarily unique in a Hausdorff space. . To be more space. We will denote the origin by 0. In this case. Let X be a vector space over the field K (which is either R or C). x ∈ X. Scalar multiplication is continuous. and so the map λ → λx is continuous. Any first countable topological space is a sequential space. By assumption. Thus. Proof Let X be a topological vector space. Lemma 5. namely an underlying linear structure. τ ) is called a topological vector space. x2 ) ∈ X × X. A vector topology is determined by a base of neighborhoods at the origin.8 In a topological vector space. V is a neighborhood of 0. since sets can be translated and scaled continuously. 5.7 Any normed vector space X is a topological vector space. x2 ) → x1 + x2 . Suppose V is an open neighborhood of 0 and let x ∈ X.2 Topological Vector Spaces We now consider topological spaces with additional structure. We call X a sequential space if every sequentially open set is open. That is. V is absorbent. x ∈ X. 2 . A topology τ on X is called a vector topology if the maps (λ. In particular. τ ). and so 0 ∈ {λ : λx ∈ V }. any metric space is a sequential space. Therefore. λ > 0. (X. Equivalently. where the topology is given by the base of open balls: x + λ(intBX ). are both continuous. the set {λ : λx ∈ V } is open in K. λ ∈ K. Consequently.86 5 Consequences of Convexity A subset U of a topological space X is called sequentially open if every sequence (xn )∞ n=1 that converges to a point in U is eventually in U . Example 5. We have established that the set {λ : λx ∈ V } is open in K and contains 0. A set V ∈ η is called balanced if λV ⊆ V for all scalars λ such that |λ| ≤ 1. if both scalar multiplication and addition are continuous in the topology on X. if there exists some N ∈ N such that xn ∈ U for all n ≥ N . any open neighborhood of the origin is absorbent. A set V ∈ η is called absorbent if X = ∞ n=1 n V . it must contain n1 for a sufficiently large n ∈ N. and (x1 . x) → λx. the topology on X is generated by the metric d given by the formula d(x. (x1 . y) ∈ X × X. Let η be a base of neighborhoods of the origin in a topological vector space (X. y) = x − y for (x. We conclude that xn ∈ V . and consequently x ∈ nV . That is. and this W is the required set. Proof Let (X.2 Topological Vector Spaces 87 Proposition 5. Certainly 0 ∈ V ∈η V . a key notion was that of the dual space. and hence αW ⊆ U for all |α| ≤ δ. Let V ∈ η. Let s : K × X → X be scalar multiplication. It follows that x = y. By assumption. For each open neighborhood of 0. % there exists a set It follows that x % ∈ V0 . such a V can be constructed. and consequently X is a Hausdorff space. By the continuity of addition. The set W is balanced. In the more general context of topological vector spaces. Therefore. This is true for every V ∈ η. and so there exists some δ > 0 and an open neighborhood W of 0 in X such that δBK × W ⊆ s −1 (U ). % Assume X is%a Hausdorff space. 2 In our discussions of normed spaces. It remains to verify (iii). s(δBK × W ) ⊆ U . Since X is a Hausdorff space. there exists a set W ∈ η such that W + W ⊆ V . By assumption. x) = λx for all λ ∈ K and x ∈ X. x − y = w2 − w1 ∈ W − W . As demonstrated earlier in this proof. and so x ∈ V ∈η V . we may assume that η satisfies the conclusions of Proposition 5. Let η be the collection of all such balanced sets. Let V ∈ η. x + W and y + W are not disjoint.9. Certainly. By Proposition 5. (ii) V is absorbent. there are open sets U and W such that 0 ∈ U and x ∈ W and U ∩ W = ∅. Then X is a Hausdorff space if and only if V ∈η V = {0}. We will show that x ∈ V ∈η V . Then (i) follows from the construction and (ii) follows from Lemma 5. U contains a subset W ∈ η. Suppose x = 0. there exist two open neighborhoods U1 and U2 of 0 ∈ X such that U1 + U2 ⊆ V . and (iii) there exists W ∈ η such that W + W ⊆ V . Let V = αW. 2 Proposition 5. balanced. . since U is open in X. By assumption. The dual space X∗ consists of all continuous linear scalar-valued functionals on X. this will remain true. and so x − y ∈ V ∈η V = {0}. α∈δBK Then V is open.11 Let X be a topological vector space. We will show that X is a Hausdorff space. Therefore. Then there exist elements w1 and w2 in W such that x + w 1 = y + w2 .9 Any topological vector space has a base of neighborhoods η of the origin such that for all V ∈ η: (i) V is balanced. (0. Thus. Let U = U1 ∩ U2 . Then U is an open neighborhood of 0 such that U + U ⊆ V . and contained in U . Therefore. and % so we conclude x − y ∈ W + W ⊆ V . τ ) be a topological vector space and let U be a neighborhood of the origin. and consequently %V0 ∈ η such that V0 ⊆ U . so that s(λ. Proof Without loss of generality.8.5. η is a base of open sets about the origin.9. 0) ∈ s −1 (U ). s is continuous. Definition 5.10 Let X be a topological vector space with%η a base of open sets about the origin. Let x and y be elements of X that cannot be separated by disjoint open sets. Now assume V ∈η V = {0}. V ∈η V = {0}. the preimage s −1 (U ) is open in K × X. The result follows. we have f (t) = pt p−1 − p(1 − t)p−1 . The metric d is even complete. 2 p p p From the above proposition. 1) is a Banach space. 1) when 0 < p < 1. then it is true that f + gpp ≤ f pp + gpp . we conclude that f +gp ≤ f p +gp whenever 0 < p < 1. The details of the proof are left to the reader.88 5 Consequences of Convexity 5. but do not have a norm structure. The proof of this fact is similar to the case when 1 ≤ p < ∞. This fact follows from the proposition below. and so f has a local maximum at t = 1/2. Differentiating a second time. 1). On the other hand. b} ⊆ R and 0 < p < 1. (See Exercise 5. Since f (0) = f (1) = 1. f (1/2) = 23−p p(p − 1) < 0. we consider some examples of real topological vector spaces which are metrizable. if not a normed space. because it is no longer subadditive. Proof Without loss of generality. g) = f − gpp determines a metric on Lp (0.24) p requires only subadditivity of the norm. It follows that if 0 < p < 1. it will suffice to show that f is a concave function.3 Some Metrizable Examples In this section. 1) is a metrizable space. a property which is shared with · p when 0 < p < 1. Consequently. we can use Lemma 2. and so f has one critical point. if 0 < p < 1. 0 If p ≥ 1. 1) denotes the collection of all (equivalence classes of) Lebesgue measurable real-valued functions f on [0. then Lp (0. Calculating the first derivative of f .) . 1]. We require only techniques of elementary differential calculus. If 0 < p < 1. Example A: Lp (0.12 If {a. Let a = t and b = 1 − t.12. we have f (t) = p(p − 1)t p−2 + p(p − 1)(1 − t)p−2 .24 to prove that Lp (0. however. the symbol Lp (0. Proposition 5. and let f (t) = t p + (1 − t)p . 1) is complete when 0 < p < 1. 0 < p < 1 If 0 < p < ∞. We will show that f (t) ≥ 1 for all t ∈ [0. Observe that the proof of the Cauchy Summability Criterion (Lemma 2. Therefore. then · p does not determine a norm. d(f . Consequently. which is at t = 1/2. 1] such that 1 1/p f p = |f (t)|p dt < ∞. then Lp (0. then |a + b|p ≤ |a|p + |b|p . assume that a and b are nonnegative real numbers such that a + b = 1. simply observe that 2−N B + 2−N B ⊆ B whenever N > 1/p. 0 2 and similarly. 1) is a complete metric space. Then φ is bounded on ∂B.3 Some Metrizable Examples 89 Let B = {f : f p < 1}. Thus. since Lp (0. even when 0 < p < 1. 1 |φ(f )| ≤ 2 · φ(1/2)1/p = φ 21− p . 1) be such that f p = 1.29) and the Closed Graph Theorem (Theorem 4. To see property (iii).a) and h = f χ(a. we have the two bounds |φ(g)| ≤ φ(1/2)1/p and |φ(h)| ≤ φ(1/2)1/p . and consequently the notation φ has not yet been given a meaning. . 1). f p =1 This should be taken as the definition of φ in this context.35) for these spaces. However. The function φ is a linear functional. 1) with f p = 1. 0 is continuous with range [0. 1)∗ for 0 < p < 1. 1) does not satisfy a Hahn–Banach Theorem if 0 < p < 1. Taking the supremum over all functions f ∈ Lp (0.1). we have 1 φ ≤ φ 21− p . Let f ∈ Lp (0. this can happen only if φ = 0.9. The map t t → |f (s)|p ds.1) . hp = (1/2)1/p . The preceding remark guarantees that Lp (0. By the linearity of φ. so that φ = sup |φ(f )| < ∞. On the other hand. together with the definition of φ. t ∈ [0. there exists some a ∈ [0. 1]. but not on a normed space.5. 1) by g = f χ(0. The first two properties are clear. Therefore. It is also possible to prove a version of the Open Mapping Theorem (Theorem 4. 1)∗ = {0}. 1]. Define two functions g and h in Lp (0. We will now compute Lp (0. again using the linearity of φ. we can apply the Baire Category Theorem (Theorem 4. Suppose φ is a continuous linear functional on Lp (0. By the choice of a. Then the collection of open sets (2−n B)∞ n=1 determines a countable base at 0 which satisfies the conclusions of Proposition 5. a 1/p 1 1/p gp = |f (s)|p ds = . 1] such that 0 |f (s)|p ds = 1/2. This implies that Lp (0. a by the Intermediate Value Theorem. 1) is determined by convergence in Lebesgue measure. as required. and a sequence converges when it converges in Lebesgue measure. To see that φ is subadditive on [0. we identify functions if they agree almost everywhere. 1]. In order to verify this. ∞). 1) is metrizable and is induced by the metric 1 |f (t) − g(t)| d(f . it suffices to show that the same sequences converge in each topology (since both spaces are sequential spaces). ∞). 1) coincides with that induced by the metric d.90 5 Consequences of Convexity Example B: L0 (0. To see that the topology on L0 (0. A simple application of the quotient rule reveals that φ (x) = 1/(1 + x)2 . and so φ is strictly increasing for all x ≥ 0. fn ) → 0 |f −fn | as n → ∞. but it does satisfy the triangle . (As usual. Then d(f . The only property of a metric that is not immediate is the triangle inequality.) The topology on L0 (0. but it remains valid for sequences that converge in measure on a σ -finite measure space. Let 1 |f (t)| f 0 = dt. 1+x+y 1+x+y 1+x+y 1+x 1+y because x ≥ 0 and y ≥ 0. 1]. 0 1 + |f (t)| Observe that d(f . We claim the topology on L0 (0. and hence in measure.17 for almost everywhere convergence. Therefore.) It remains to show that the metric d is complete. It follows that fn → f in measure. the triangle inequality follows 1 readily from the fact that d(f . g) = f − g0 for all measurable functions f and g on [0. The reverse implication. lim m{t : |f (t) − fn (t)| ≥ } = 0. n→∞ where m is Lebesgue measure on [0. is true by the Lebesgue Dominated Convergence Theorem. observe that x+y x y x y φ(x +y) = = + ≤ + = φ(x)+φ(y). it suffices to show that the function φ(x) = x/(1 + x) is a nondecreasing subadditive function on [0. we define a set to be open when it is sequentially open. More precisely. 1) the set of all (equivalence classes of) scalar-valued Lebesgue measurable functions on [0. Suppose the sequence (fn )∞ n=1 of measurable functions converges in the metric d to a measurable function f . Recall that a sequence (fn )∞ n=1 of measurable functions converges in Lebesgue measure to a measurable function f if for every > 0. 1]. · 0 is not a norm (it is not homogeneous). 1) We denote by L0 (0. g) = dt. 0 1 + |f (t) − g(t)| where f and g are measurable functions. f ∈ L0 (0. Given these properties. Certainly. (We state the Lebesgue Dominated Convergence Theorem in Theorem A. g) = 0 φ(|f (t) − g(t)|) dt. 1+|f −fn | → 0 in the L1 -norm. that convergence in measure implies convergence in d. 1). 5.3 Some Metrizable Examples 91 1 inequality, because f 0 = 0 φ(|f (t)|) dt and φ is subadditive on [0, ∞). Conse- quently, we may use Lemma 2.24 (the Cauchy Summability Criterion) to prove that d is a complete metric space (because the proof of Lemma 2.24 does not require homogeneity of the norm). ∞ Suppose (fn )∞n=1 is a sequence of measurable functions such that n=1 fn 0 < ∞. Then, by Fubini’s Theorem, ∞ ∞ 1 1 ∞ |fn (t)| |fn (t)| fn 0 = dt = dt < ∞. n=1 n=1 0 1 + |fn (t)| 0 n=1 1 + |fn (t)| It follows that, for almost every t ∈ [0, 1], there exists some Mt > 0 such that ∞ |fn (t)| n=1 1+|fn (t)| ≤ Mt . Consequently, by the subadditivity of φ, for every N ∈ N, N N φ |fn (t)| ≤ φ(|f (t)|) ≤ Mt < ∞ a.e.(t). n=1 n=1 Since φ is a strictly increasingfunction on the interval [0, ∞), we conclude Nthat, for∞al- ∞ most every t, the sequence ( N n=1 |fn (t)|)N=1 converges. Therefore, ( n=1 fn )N =1 converges almost everywhere, and hence in measure. Therefore, L0 (0, 1) is a complete metric space. As was the case in Example A (where 0 < p < 1), the dual space of L0 (0, 1) is trivial; that is, L0 (0, 1)∗ = {0}. We leave the verification of this fact as an exercise. (See Exercise 5.14.) Example C: ω = RN $ J be a (possibly uncountable) index set. Let R denote J Let the product space R j ∈J j , where R j = R for each j ∈ J . An element x in R J is a function x : J → R, where x(j ) ∈ R(= Rj ) for each j ∈ J . When the space RJ is equipped with the product topology, it becomes a topological vector space. The vector space operations are done pointwise; that is, if x and y are elements in RJ , then (x + y)(j ) = x(j ) + y(j ) for each j ∈ J . Convergence, too, is pointwise: xn → x in RJ as n → ∞ if xn (j ) → x(j ) in Rj as n → ∞ for each j ∈ J. If J is an uncountable index set, then RJ is not metrizable, since RJ with the product topology is not first countable; i.e., it does not have a countable local base at 0. (See Example 5.4.) In this example, we are interested in countable index sets, and so we let J = N. We denote RN by the Greek letter ω. Generally, we think of ω as the collection of all sequences in R. If ξ ∈ ω, we let ξk = ξ (k) for each k ∈ N, and we write ξ = (ξk )∞ k=1 . In this context, the vector space operations are done coordinate-wise. Convergence is also now viewed coordinate-wise, so that ξ (n) → ξ in ω as n → ∞ if ξk(n) → ξk in R as n → ∞ for each k ∈ N. 92 5 Consequences of Convexity Unlike RJ when J is uncountable, the space ω is first countable. A base of neighborhoods at the origin is formed by sets of the type (−1 , 1 ) × · · · × (−n , n ) × R × R × · · · , (5.1) where n ∈ N and i > 0 for each i ∈ {1, . . . , n}. If we denote elements of ω by ξ = (ξk )∞ k=1 , then the set in (5.3.1) can be written {ξ : |ξ1 | < 1 , · · · , |ξn | < n }. To identify a countable base, consider the sets with i = k1 , where k ∈ N, for all i ∈ {1, . . . , n} and n ∈ N. Not$only is ω first countable, but it is also metrizable. Recall that ω was defined to be ∞ k=1 Rk . Denote the metric on Rk by dk . We define a metric d on ω by ∞ 1 dk (ξk , ηk ) d(ξ , η) = k 1 + d (ξ , η ) , k=1 2 k k k where ξ = (ξk )∞ ∞ k=1 and η = (ηk )k=1 . We now wish to identify the space dual to ω. To that end, we prove the following proposition. Proposition 5.13 Let X be a topological vector space and let K be the field of scalars. A linear functional f : X → K is continuous if and only if there exists a neighborhood V of 0 such that the set f (V ) is bounded in K. Proof Let UK be the open unit ball in K. If f is continuous, then f −1 (UK ) is an open neighborhood of 0, and f (f −1 (UK )) ⊆ BK is bounded in K. Now suppose V is a neighborhood of 0 such that f (V ) is bounded in K. By definition, there is some M > 0 such that f (V ) ⊆ MUK . Let > 0 be given. Then f ( M V ) ⊆ UK . Therefore, |f (x)| < whenever x ∈ M V . In other words, f is continuous at zero. Continuity then follows from the linearity of f . 2 We can now use the preceding proposition to identify the continuous linear functionals on ω. Let f ∈ ω∗ . By Proposition 5.13, there must be some neighborhood V of 0 such that f (V ) is bounded in R. Without loss of generality, we may assume V is a basic set, say V = {ξ : |ξ1 | < 1 , · · · , |ξn | < n } for some n ∈ N. The set f (V ) is bounded, and so there exists some M > 0 such that |f (ξ )| ≤ M for any ξ ∈ V . Let ξ = (0, . . . , 0, ξn+1 , . . . ). Then ξ ∈ V , and so too is any constant multiple of ξ . Therefore, for any K > 0, we have that |f (Kξ )| ≤ M, and hence |f (ξ )| ≤ M/K, by the linearity of f . Since this inequality holds for all K > 0, it must be that f (ξ ) = 0. This is true for any ξ ∈ V having ξi = 0 for all i ∈ {1, . . . , n}. Thus, because f is linear, it follows that f (ξ ) = f (ξ ) for any ξ and ξ that agree on the first n coordinates. Define a function g : Rn → R by g(ξ1 , . . . , ξn ) = f (ξ1 , . . . , ξn , 0, . . . ). Since f is linear and continuous, it follows that g ∈ (Rn )∗ . Consequently, there exists αk ∈ R 5.4 The Geometric Hahn–Banach Theorem 93 for each k ∈ {1, . . . , n} such that n g(ξ1 , . . . , ξn ) = αk ξk , (ξk )nk=1 ∈ Rn . k=1 Since the value of f (ξ ) depends only on the first n coordinates of ξ , we conclude that n f (ξ ) = αk ξk , ξ = (ξk )∞k=1 ∈ ω. k=1 5.4 The Geometric Hahn–Banach Theorem In this section, we will meet the Hahn–Banach Theorem without the advantages of a norm structure. The key property a space must have, we shall see, is local convexity. Definition 5.14 Let X be a real or complex vector space. A subset V of X is called convex if given any x and y in V , we have (1 − t)x + ty ∈ V for all t ∈ [0, 1]. That is, if two points are in V , then the line segment joining them is also in V . A balanced convex set is called absolutely convex. Lemma 5.15 Let X be a real or complex vector space. A subset V of X is absolutely convex if and only if αx + βy ∈ V whenever x and y are in V and α and β are scalars such that |α| + |β| ≤ 1. Proof We first observe that V is absolutely convex if the latter condition holds: to show balance, take β = 0; to show convexity, let α = t − 1 and β = t. Now suppose V is absolutely convex. Let x and y be in V and suppose α and β are scalars such that |α| + |β| ≤ 1. We wish to show αx + βy ∈ V . Observe that α β αx + βy = (α + β)x + (α + β)y. α+β α+β Since V is balanced, x = (α + β)x and y = (α + β)y are both elements of V . Thus, by convexity, α β αx + βy = x + y ∈ V . α+β α+β This completes the proof. 2 Definition 5.16 A topological vector space is locally convex if there is a base of neighborhoods of 0 consisting of convex sets. By Proposition 5.9, we can always take the elements of a base in a locally convex topological vector space to be balanced, and hence absolutely convex. Example 5.17 Any normed space is locally convex. It is easy to see that balls with center at the origin are convex. 94 5 Consequences of Convexity p=∞ p=2 0 < p <1 p=1 Fig. 5.1 Closed unit balls in 2p for various values of p Example 5.18 Consider the space 2p of ordered pairs in the · p norm for p > 0. (We use the term “norm” here even though it is not a norm when 0 < p < 1.) If p ≥ 1, then the unit ball is convex and balanced; however, if 0 < p < 1, then the unit ball is balanced, but not convex. (See Fig. 5.1.) Example 5.19 Let X = Lp (0, 1) for 0 < p < 1. (See Example A in Sect. 5.3.) We claim that the only nonempty open convex subset of X is X. To show this, let V be a nonempty open convex subset of X. Without loss of generality, assume 0 ∈ V . Then there exists some δ > 0 such that δB ⊆ V , where B = {f : f p < 1}. (We remind the reader that · p is not a norm in this case.) p Choose any f ∈ X. Because p < 1, there is some n ∈ N such that np−1 f p < δ. Pick real numbers {t0 , t1 , . . . , tn } so that 0 = t0 < t1 < · · · < tn = 1 and such that tk 1 |f (s)|p ds = f pp , k ∈ {1, . . . , n}. tk−1 n p p For each k ∈ {1, . . . , n}, let gk = n f χ(tk−1 ,tk ] . Then gk p = np−1 f p < δ, and therefore gk ∈ δB ⊆ V . This is true for each k ∈ {1, . . . , n}, and so {g1 , . . . , gn } ⊆ V . Observe that f = n1 (g1 + · · · + gn ). Since V is convex, it follows that f ∈ V . The choice of f ∈ X was arbitrary, and so V = X, as required. Note that this argument implies that Lp (0, 1)∗ = {0}, a fact we first observed in Example A in Sect. 5.3. The next theorem is a geometric version of the Hahn–Banach Theorem. This version of the theorem is not set in the context of a complete normed space, but in that of a locally convex topological vector space. Theorem 5.20 (Hahn–Banach Separation Theorem) Let E be a real locally convex topological vector space. Let K be a closed nonempty convex subset of E. If x0 ∈ K, then there exists a continuous linear functional f on E such that f (x0 ) > sup f (y). y∈K x ∈ E. From this we conclude that x0 ∈ K + W . (Here we use the fact that W = −W . λ λ p(αx) = inf{λ > 0 : αx ∈ λV } = α inf > 0 : x ∈ V = αp(x). Because p(x) and p(y) are infima. for otherwise there would exist some k ∈ K and w ∈ W such that x0 − w = k. contradicting the fact that the intersection of x0 + W with K is empty. It follows that xλ0 ∈ V for all λ ≥ 1. We will demonstrate this by showing that f is bounded on some neighborhood of zero and applying Proposition 5.13. Thus. Suppose to the contrary that p(x0 ) ≤ 1.5. Let x and y be in E and let > 0. Because K ⊆ V . This is a contradiction. In particular V is absorbent. and so there exists an element k ∈ K and elements w1 and w2 in W such that x0 + 21 w1 = k + 21 w2 . We claim p is sublinear.) Since K is closed and x0 ∈ K. 1 1 1 1 x0 = k + w2 − w1 ∈ K + W − W. α α This proves positive homogeneity. Since xλ0 → x0 as λ → 1. Since . and so p(x) < ∞ for all x ∈ E. We now make use of Exercise 3. and so p(x + y) ≤ p(x) + p(y). By the definition of p. Recall that V = K + 21 W .9. because W is balanced. Therefore p is sublinear. The choice of was arbitrary. we have that 21 W − 21 W ⊆ W . (x0 + 21 W ) ∩ (K + 21 W ) = ∅. x+y λ x μ y = + ∈ V. For any x ∈ E and α ≥ 0. By definition. we have that xλ ∈ V and μy ∈ V . and so it must be that p(x0 ) > 1. It remains to show that p is subadditive. There exists a linear functional f on E such that f ≤ p and f (x0 ) > 1. and because p(x) ≤ 1 for all x ∈ V . use a translation. We now show p(x0 ) > 1. Recall that every neighborhood of 0 is absorbent. we may assume 0 ∈ K. 2 2 2 2 Because W is absolutely convex. as required. p(x) ≤ 1 for all x ∈ V . (If not. λ+μ λ+μ λ λ+μ μ Therefore.) Let V = K + 21 W . p(x + y) ≤ λ + μ < p(x) + p(y) + . Then V is a convex neighborhood of 0. This implies that x0 ∈ K + W . and consequently (x0 + 21 W ) ∩ V = ∅. It follows that there exists an absolutely convex open neighborhood W of 0 such that (x0 + W ) ∩ K = ∅. there exists some open neighborhood N of x0 such that N ∩ K = ∅. y∈K It remains to show that f is continuous. we have sup f (y) ≤ 1 < f (x0 ). Hence. By the convexity of V . there exist real numbers λ > 0 and μ > 0 such that p(x) < λ < p(x)+ 2 and p(y) < μ < p(y)+ 2 .4 The Geometric Hahn–Banach Theorem 95 Proof Without loss of generality. Define a function p : E → R by p(x) = inf{λ > 0 : x ∈ λV }. we conclude that x0 ∈ V . where α ∈ A and n ∈ N. What distinguishes a semi-norm from a norm is that a semi-norm p may satisfy p(x) = 0 even when x = 0.96 5 Consequences of Convexity 0 ∈ K. A function p : X → R is called a semi-norm if the following three conditions are satisfied: (i) p(x) ≥ 0 for all x ∈ X.20. n) = {x : pα (x) < 1/n}. by Proposition 5. 2 Definition 5. Theorem 5. The set W is balanced. Let V (α. and thus |f (x)| ≤ 1 for all x ∈ 21 W . y} ⊆ X. (ii) p(x + y) ≤ p(x) + p(y) for all {x.13. and (iii) p(α x) = |α| p(x) for all α ∈ K and x ∈ X. If x0 ∈ K. we have demonstrated that f ( 21 W ) ⊆ [−1. by Theorem 5. x∈K Proof Ignoring multiplication by complex scalars.21 Suppose E is a real locally convex topological vector space and K is a closed linear subspace of E. 1]. there exists a continuous linear functional f on E such that f (K) = 0 and f (x0 ) > 0. If η is the collection of all finite intersections of the sets V (α. α ∈ A. By construction. Therefore. then there exists a continuous linear functional f on E such that (f (x0 )) > sup (f (x)). It remains to show that addition and scalar multiplication are continuous.) There is also a version of Theorem 5. we may assume U is .24 Suppose {pα }α∈A is a family of semi-norms on a vector space X. Without loss of generality. we have that 21 W ⊆ K + 21 W = V . there exists a real linear functional g on E such that g(x0 ) > supx∈K g(x). n ∈ N. Therefore. we call the property in (ii) subadditivity (or the triangle inequality) and we call the property in (iii) homogeneity. As in the case of a norm. This defines a topology for which all members of η are absolutely convex (that is. Proof We define a topology on X by declaring a set E ⊆ X to be open if and only if E is a (possibly empty) union of translates of elements in η. The functional f is the desired continuous linear functional on E.23 Let X be a vector space and let K denote the scalar field. 2 Example 5.22 Let E be a complex locally convex topological vector space. convex and balanced). the linear functional f is continuous. Let U be an open neighborhood of 0 in X. If x0 ∈ K. Now. n). Consequently. we may treat E as a vector space over R. then.20.20 for complex topological vector spaces. Let K be a closed nonempty convex subset of E. Theorem 5.20. then η determines a locally convex vector topology on X in which the elements of η form an absolutely convex base of neighborhoods at 0. by Theorem 5. and hence f (x) ≤ 1 for all x ∈ 21 W . define a complex linear functional on E by f (x) = g(x) − ig(ix) for all x ∈ E. (See Exercise 5. f (x) ≤ p(x) ≤ 1 for all x ∈ V . there exists some n ∈ N such that x ∈ nV . Observe that x = nv1 and w − x = n v 1+|κ|n 2 for some choice of v1 and v2 in V . 2n1 )∩· · ·∩V (αk . . then V + V ⊆ U (because pα is subadditive for every α ∈ A). n1 ) ∩ · · · ∩ V (αk . The Minkowski functional of U on X is the function pU : X → R defined by pU (x) = inf{λ > 0 : x ∈ λU }. Now.4 The Geometric Hahn–Banach Theorem 97 an element of η. nk } ⊆ N. Then κx − λw = (κ − λ)x + λ(x − w). αk } ⊆ A and {n1 . To show the subadditivity of pU .25 Suppose X is a topological vector space and let U be an absorbent subset of X. Let x and y be elements in X and let > 0. as above. addition is continuous. Since V is an open neighborhood of 0. . where K is the scalar field. Suppose that X is a locally convex topological vector space.2) for {α1 . Such sets are absorbent. 2 Definition 5. We will show there exists an open neighborhood W of x and a δ > 0 such that λW ⊆ κx + U for all |κ − λ| < δ. 2nk ). by the definition of pU . Thus. δ). Hence. By the definition of pU . .5. U = V (α1 .2). 1 + |κ|n Therefore. . Then X has a base of neighborhoods of 0 that are absolutely convex. . Proposition 5. there exist numbers λ1 > 0 . nk ) (5. 2n1 ) ∩ · · · ∩ V (αk . . |λ|n κx − λw ∈ |κ − λ| nV + V ⊆ V + V ⊆ U. where U is written as in (5. The Minkowski functional pU is a semi-norm on X. and so each such set will give rise to a well-defined Minkowski functional. it is absorbent. Note that pU (x) < ∞ for all x ∈ X.26 Let X be a topological vector space and let U be an absorbent absolutely convex subset of X. . If V = V (α1 . Thus. Proof Certainly pU (x) ≥ 0 for each x ∈ X. because V is balanced. Therefore. . let x ∈ X and κ ∈ K. n 1 + |κ|n Suppose w ∈ W and λ ∈ B(κ. A basic open neighborhood of κx can be written as κx + U . and so the proof is complete. λn κx − λw = (κ − λ)nv1 − v2 . x ∈ X. because U is absorbent. we will use the convexity of U . 2nk ). 1 + |κ|n It follows that scalar multiplication is continuous. Let V = V (α1 . Let 1 n δ= and W = x + V. . ! " pU (αx) = |α| inf λ > 0 : x ∈ λ U = |α| pU (x). (See Exercise 5. where > 0 and {x1∗ . pU is a semi-norm on X. in fact.) So far. where K is the field of scalars. . and so pU (x + y) ≤ pU (x) + pU (y). say η. the so-called weak topology. To distinguish between the norm and weak topologies on X. then there exists a base of absolutely convex neigborhoods of 0.24. . Any weakly open set is necessarily open in the norm topology (the basic sets are intersections of . The weak and norm topologies are generally quite different. We have already said much about the norm topology of a Banach space X. The weak topology on X is the topology it inherits as a subspace of the space ∗ ∗ KX with the product topology. . The weak topology on X (or the w-topology) is defined by a base of neighborhoods at 0 of the form W (x1∗ . pU (x + y) ≤ λ1 + λ2 < pU (x) + pU (y) + . Definition 5. Letting λ = λ/|α|. we will frequently denote X with the norm topology by (X. sign(α)U = U . and we identify X with a subspace of KX by identifying ∗ x ∈ X with x̂ ∈ KX via the relationship x̂(x ∗ ) = x ∗ (x) for all x ∗ ∈ X∗ . the family of semi-norms {pU }U ∈η generates a locally convex vector topology on X.) Finally. By the convexity of U . By Proposition 5. (Compare to the proof of Theorem 5. It is necessarily the case that x/λ1 and y/λ2 are both in U . as claimed. Let α ∈ K. . we have considered general topological vector spaces.17. 2 If X is a locally convex topological vector space. Banach spaces. Therefore. xn∗ . . we show homogeneity. . We now consider a new topology on X.20. · ) and X with the weak topology by (X.26. 1 ≤ i ≤ n}. Computing directly. We leave it as an exercise to show that the topology generated by {pU }U ∈η is. By Theorem 5. The choice of was arbitrary. The space KX is the collection of all functions from ∗ X∗ into the scalar field K. w). we have λ λ pU (αx) = inf{λ > 0 : αx ∈ λU } = |α| inf >0 : x∈ · sign(α)U . We now focus our attention on topological vector spaces that have a complete norm structure—that is. the Minkowski functional pU is a semi-norm on X for each U ∈ η. . |α| |α| Since U is balanced.27 Let X be a topological vector space. x+y λ1 x λ2 y = + ∈ U. ) = {x : |xi∗ (x)| < . xn∗ } ⊆ X∗ for n ∈ N. λ1 + λ 2 λ1 + λ 2 λ1 λ1 + λ 2 λ2 Therefore. the original topology.98 5 Consequences of Convexity and λ2 > 0 such that pU (x) < λ1 < pU (x) + 2 and pU (y) < λ2 < pU (y) + 2 . Without loss of generality. there exists some g ∈ Lq (T). and so en → 0 in the weak topology in this case. then em − en p = 21/p . ) be the constant sequence with all terms equal to 1. .29 Consider the Banach space Lp (T) of p-integrable complex-valued functions on the torus T = [0.28 Consider p for 1 ≤ p < ∞. and then X must be finite-dimensional. 1. That is to say.5. and so it is “harder” for a function on (X. if both maps are continuous. where p > 1 and q is the exponent conjugate to p.30. let en be the sequence with 1 in the nth coordinate. define a function fn : T → C by fn (θ ) = einθ . n→∞ Since this is true for all x ∗ ∈ q . consider the identity map IdX on X. In this case. w) to be continuous than a function on (X. Example 5. but IdX : (X. If m and n are elements of N such that m = n. This sequence is bounded. A sequence converges to 0 in the norm topology if and only if every “strong” neighborhood of the origin eventually contains the sequence. · ) → (X.4 The Geometric Hahn–Banach Theorem 99 preimages of open sets under continuous maps).) The weak topology on X generally has fewer open sets. we may consider only those sequences converging to 0. The above conclusion does not remain true when p = 1. However. we have that e(en ) = 1. n→∞ In other words. we say that (xn )n=1 ∞ ∞ converges weakly to 0 (or xn → 0 weakly). . Let e = (1. such that dθ Λ(f ) = f (θ) g(θ ) . T 2π . (We will demonstrate this shortly. if x ∗ = (xn∗ )∞ n=1 is a sequence in (p )∗ = q . For each n ∈ N. . 1. Consequently. · ) need not be. For example. it is more “difficult” for a sequence to converge in the norm topology than to converge in the weak topology. xn → 0 weakly if and only if lim x ∗ (xn ) = 0. By duality. Indeed. Example 5. the norm topology has more open neighborhoods about 0 than the weak topology. f ∈ Lp (T). then lim x ∗ (en ) = xn∗ = 0. Conse- quently. For each n ∈ N. then the topologies must coincide. The sequence (xn )n=1 converges weakly ∗ to 0 precisely when it converges coordinate-wise to 0 in KX . and 0 elsewhere. w) is always continuous. ·). where 1 ≤ p < ∞. On the other hand. where 1/p + 1/q = 1. For each n ∈ N. Let Λ ∈ Lp (T)∗ . but not every set open in the norm topology will be weakly open.) Let us consider which sequences converge in X with the weak topology. we conclude that en → 0 weakly. the sequence (en )∞n=1 does not converge in the norm topology. and so e ∈ ∞ = (1 )∗ . xn converges to 0 in the weak topology if and only if every weak neighborhood of the origin eventually contains the sequence (xn )∞ n=1 . (See Proposition 5. q = ∞. The map IdX : (X. w) → (X. for all x ∗ ∈ X∗ . 2π ). If a sequence (xn )∞n=1 converges to 0 in the weak topology on X. where θ ∈ T. j (λy) = 0 for all j ∈ {1. Fix some x ∗ ∈ X∗ . 1 ≤ j ≤ Nn }. define ∗ En = span{xn. and (iii) the weak topology on X is metrizable. there exists a finite sequence (aj )N n j =1 . w) is first countable. Consequently. It remains to show (iii) ⇒ (i). We claim x ∗ ∈ (kerT )⊥ . there then exists some f ∈ (KNn )∗ such that x ∗ (x) = (f ◦ T )(x) for all x ∈ X. . Wn ⊆ {x : |x ∗ (x)| ≤ 1}. and so |x ∗ (λy)| ≤ 1 for all λ ∈ K. . and so fn → 0 weakly. This can occur only if |x ∗ (y)| ≤ 1/λ for all λ ∈ K. . Proof The implications (i) ⇒ (ii) ⇒ (iii) are clear. fn Lp (T) = 1 for all n ∈ N. for all n ∈ N and all j ∈ {1.33. (Recall Definition 3. and thus x ∗ (y) = 0. The following are equivalent: (i) dim(X) < ∞. .) To verify this. ∗ By the definition of T . Nn }. . Then (X. x ∈ X. then it follows that xn. For this fixed n. Since KNn is finite-dimensional. . This remains true for any y ∈ ker(T ). we have that xn. . However. n→∞ n→∞ T 2π n→∞ This last equality follows from the Riemann–Lebesgue Lemma (Theorem 4. then all linear functionals are continuous.1 (x). define a linear map T : X → KNn .j ∈ X∗ . .j (x)| ≤ n . and so we have that x ∗ ∈ (kerT )⊥ . where K is the scalar field. by ∗ ∗ T (x) = (xn. .50. and Nn ∈ N. Naturally. Therefore. suppose y ∈ ker(T ). and consequently must contain Wn for some n ∈ N. Nn }. Proposition 5. ∗ where xn. and so there exists a weak base of neighborhoods (Wn )∞ n=1 at the origin of the form ∗ Wn = {x : |xn. .j : 1 ≤ j ≤ Nn }. dθ lim Λ(fn ) = lim einθ g(θ ) = lim ĝ(−n) = 0. Nn }.37). n > 0. (ii) the weak topology on X coincides with the norm topology on X. .N n (x)). . xn. limn→∞ Λ(fn ) = 0 for all Λ ∈ Lq (T). By design. . λy ∈ Wn for all λ ∈ K. If X is a finite-dimensional Banach space. By Lemma 4.30 Let X be a Banach space. . and so fn → 0 in the norm topology. For each n ∈ N.j (y) = 0 for all j ∈ {1. . ∗ if λ ∈ K. The set {x : |x ∗ (x)| ≤ 1} is a weak neighborhood of 0 in X. . Assume the weak topology on X is metrizable.100 5 Consequences of Convexity Therefore. ) Now.N n (x)) = ∗ aj xn. Weakly open sets are open in the norm topology. Hence. ∗ ∗ We have shown that each x ∈ X is in En for some n ∈ N. ξNn ) = aj ξ j . the set {x : |x ∗ (x)−x ∗ (x2 )| < /2} is a weak neighborhood of x2 . the easier it is for a function to be continuous. .31 Let X be a Banach space. Therefore. j =1 and so x ∗ ∈ En . When we say a topology is weaker.20). the space X is finite-dimensional. and so X is finite-dimensional. w) is a Hausdorff topological space. (The idea is that it is “easier” to be continuous in the norm topology. . (ξj )N j =1 ∈ K . there exists an x ∗ ∈ X ∗ such that = x ∗ (x2 − x1 ) > 0. 2 Proposition 5. by Theorem 4. . We therefore conclude that X∗ = ∞ E n=1 n . X ∗ = ∞ ∗ k=1 kEn = En .7 (the complementary version of the Baire Category Theorem). By the Hahn– Banach Separation Theorem (Theorem 5. it is more likely that a given preimage is open.4 The Geometric Hahn–Banach Theorem 101 such that Nn f (ξ1 . We conclude that En is an open neighborhood of the origin in X ∗ . xn. But the norm topology contains all of the weakly open sets. A function is continuous if the preimage of any open set is open. . because with more open sets. (ii) A linear functional is continuous in the weak topology if and only if it is continuous in the norm topology. f is continuous in the weak topology on X. f is continuous in the norm topology on X. (X. Thus. but the converse need not be true. . For each n ∈ N. n Nn j =1 Therefore. there exists some n ∈ N such that int(En ) = ∅. the space En is finite-dimensional. suppose f is a norm continuous linear functional. and consequently is absorbent. because there are more open sets. and so is closed.1 ∗ ∗ (x). Therefore. 2 The weak topology on X is the weakest topology on X such that all norm continuous linear functionals remain continuous. The norm topology on X is stronger than the weak topology on X. The stronger the topology on the domain. Therefore. so f −1 (V ) is open in the norm topology. Proof (i) Assume x1 and x2 are elements in X such that x1 = x2 . Then: (i) The weak topology on X is a Hausdorff topology. Thus. (ii) If a linear functional f is continuous in the weak topology on X. Therefore. . . we mean that it contains fewer open sets. and these two neighborhoods are disjoint. Then f ∈ X ∗ . x ∈ X. . the set {x : |x ∗ (x) − x ∗ (x1 )| < /2} is a weak neighborhood of x1 .5. because it contains more open sets. and so the set {x : |f (x)| < } is a weak neighborhood of 0 (by the definition of the weak topology). as well. Nn x ∗ (x) = f (xn. .j (x). then f −1 (V ) is a weakly open set whenever V is an open set in the scalar field. In this example we have found a sequence which converges in the weak∗ topology on 1 . .28. This happens because the weak∗ topology has fewer open sets than the weak topology. 1 ≤ i ≤ n}. then the sets {x ∗ : |x ∗ (x) − x1∗ (x)| < /2} and {x ∗ : |x ∗ (x) − x2∗ (x)| < /2} are disjoint weak∗ -open sets containing x1∗ and x2∗ . where φx (x ∗ ) = x ∗ (x) for all x ∗ ∈ X∗ . and so the sequence (en )∞n=1 converges to 0 in the weak topology on 1 . xn . (ii) Certainly. The Banach space X ∗ has also a weak topology that is induced by it’s dual space (X∗ )∗ = X∗∗ (the bidual of X). . The weak∗ topology on X ∗ is weaker than the weak topology on X∗ . (X ∗ .34 Let X be a Banach space. where e = (1.102 5 Consequences of Convexity Definition 5. We use (X ∗ .) Proof (i) Let x1∗ and x2∗ be elements in X∗ such that x1∗ = x2∗ . 1. The weak∗ topology on X∗ is the weakest topology on X∗ for which the linear functionals φx are continuous for all x ∈ X. ) = {x ∗ : |x ∗ (xi )| < .) Example 5. (ii) A linear functional f on X∗ is weak∗ -continuous if and only if there exists some x ∈ X such that f (x ∗ ) = x ∗ (x) for all x ∗ ∈ X∗ . (In other words. . This is true for every ∗ ξ ∈ c0 . . It remains only to show that any weak∗ -continuous linear functional on X ∗ can be achieved in this way. w∗ )∗ = X. the space of all scalar-valued functions on X. Then there exists some x ∈ X such that x ∗ (x) = x2∗ (x). . the weak∗ topology is weaker than the weak topology). The weak∗ topology on X∗ is the topology inherited from viewing X ∗ as a subspace of KX . The weak∗ topology on X ∗ (or the w∗ -topology) is defined by a base of neighborhoods at 0 of the form W ∗ (x1 . .) If = |(x1∗ − x2∗ )(x)|. . (That is to say. if f (x ∗ ) = x ∗ (x) for all x ∗ ∈ X∗ . we endow KX with the product topology.33 Consider the sequence space 1 .) is the constant sequence with all terms equal to 1). respectively. but not in the weak topology on 1 . . (Only those coming from X. As before. it follows that en (ξ ) = ξn → 0 as n → ∞. . . Then: (i) The weak∗ topology on X ∗ is a Hausdorff topology. then f is continuous in the weak∗ topology. because it requires fewer members in X ∗∗ to be continuous. Suppose ξ = (ξk )∞ k=1 is an element of c0 . . xn } ⊆ X for n ∈ N. In this weak topology. Proposition 5. w∗ ) to denote X ∗ with the weak∗ topology. Since c0 consists of sequences that converge to 0. The space 1 can also be given a weak∗ topology as the dual space of c0 . where > 0 and {x1 .32 Let X be a topological vector space. we saw that the sequence (en )∞ n=1 did not converge to 0 (because e(en ) = 1 for all n ∈ N. (Otherwise they would be the same as linear functionals on X. In Example 5. we saw that 1 had a weak topology induced upon it by ∗1 = ∞ . . Observe that any x ∈ X can be thought of as a linear functional on X ∗ via the mapping x → φx . Choose x1 . . xn }. an } ⊆ K such that n f (x ∗ ) = φ(T x) = φ(x ∗ (x1 ).33). From this we conclude that f (x ∗ ) = 0. there exists a basic neighborhood of (X∗ . . . Now. . . . we need a lemma. and so (λx ∗ ) ∈ W ∗ (x1 . there exists a finite collection of scalars {a1 . there exists a real number > 0 and a finite set {x1 . ... we have that x ∗ (λxj ) = 0 for j ∈ {1. .. We recall that each element of X has n a unique representation of the form αi xi .. By the triangle inequality. by T (x ∗ ) = (x ∗ (x1 ). then. . i i i=1 i=1 i=1 i=1 (5. . xn . however. . . It follows that |f (λx ∗ )| ≤ 1. . however. Before we state and prove this proposition. Then x ∗ (xj ) = 0 for each j ∈ {1. we will prove Proposition 5. . . x ∗ (xn )) = aj x ∗ (xj ). Thus. Define a map T : X∗ → Kn . By Lemma 4.33.37 which (in some sense) demonstrates that it is “hard” to be compact in a normed space. If X is a reflexive space (recall Definition 3. . . xn } ⊆ X such that |f (x ∗ )| ≤ 1 for all x ∗ ∈ W ∗ (x1 . Thus. ). where K is the scalar field. 2 j =1 Remark 5. xn in X so that X = span{x1 . where αi ∈ K for each i ∈ {1.36 All norms on a finite-dimensional vector space are equ ivalent.. ).33 we saw that the weak∗ topology may be strictly weaker than the weak topology. n}. . .3 In Example 5. . n n n n αi x i ≤ |αi | · xi ≤ (maxxi ) |αi | = (maxxi ) αi xi . Therefore. . n}. .. for any λ ∈ K. Shortly. xn .5... We will find positive constants c and C such that c|||x||| ≤ x ≤ C|||x||| for all x ∈ X. . .4 The Geometric Hahn–Banach Theorem 103 Assume f is a continuous linear functional on (X ∗ . and consequently |f (x ∗ )| ≤ 1/|λ| for all λ = 0. . i=1 Define a norm |||·||| on X as follows: n n αi x i = |αi |.. w∗ ) on which f is bounded. there exists a bounded linear functional φ : Kn → K such that f = φ ◦T . . j =1 n The desired element of X is x = aj xj . let · be another norm on X. . ..13. x ∗ (xn )). which is of independent interest.. x ∗ ∈ X∗ . . By Proposition 5. Lemma 5. . then the weak and weak∗ topologies coincide. . and hence f ∈ (kerT )⊥ . .3) . n}. Suppose x ∗ ∈ ker(T ). x ∗ ∈ X∗ . w∗ ). i=1 i=1 It is straightforward to show that this does indeed define a norm on X. Proof Let X be a finite-dimensional vector space over the scalar field K. . . To see this. By the Extreme Value Theorem. αn ) ≥ c for all (α1 .104 5 Consequences of Convexity Thus. i=1 i=1 i=1 i=1 The last inequality follows from the Cauchy–Schwarz Inequality. .4). From this. . for any (α1 . i=1 n n n αi x i ≥ c |αi | = c αi xi . .. .3) and (5. Since the norm · was arbitrary. observe that n n n n f (α1 .. Then BX is compact in the norm topology on X if and only if dim(X) < ∞. . .. since f is continuous on a compact set. .. αn ) = αi xi .36. . . . . we conclude that the two norms are equivalent. . . . . By Lemma 5. Proof Suppse X is a finite-dimensional normed vector space. . βn ) = α i xi − βi xi ≤ α i xi − βi xi i=1 i=1 i=1 i=1 n n n 1/2 n 1/2 = (αi − βi )xi ≤ |αi − βi |xi ≤ |αi − βi |2 xi 2 ...4) i=1 i=1 i=1 Combining (5. Let c be that minimum value. Next. αn ) in Kn . Then n f (α1 . . it follows that f is continuous. . (5. This means that αi xi ≥ c for all i=1 n (α1 . . Alternately. S is compact by the Heine–Borel Theorem. . αn ) in S. αn ) : |αi | = 1 . . Therefore. the function f attains a minimum value on the set S. X is homeomorphic to Kn with the Euclidean norm. i=1 Observe that S is a closed and bounded subset of Kn . we may choose C = maxi xi . define a set ! n " S = (α1 . αn )−f (β1 . . 2 Proposition 5. BX is compact by the Heine–Borel Theorem.. it follows that all norms on X are equivalent. Therefore. αn ) in Kn such that |αi | = 1. .37 Suppose X is a Banach space (or just a normed linear space). i=1 We claim that the function f is continuous. Define a function f : S → R+ by n f (α1 .. . .38 (Tychonoff’s Theorem) Let I be$ an arbitrary index set. we have |x ∗ (x)| ≤ x. n=1 2 ∞ and so BX ⊆ F . If x ∗ ∈ BX∗ .5. Therefore. Since BX is absorbent. Since BX is compact. . x∈X . we have BX ⊆ F + 1 B 2n X for all n ∈ N. . let us recall a theorem from general topology. as required. ∞ 1 F + n BX = F . but we do wish to point out it relies on the Axiom of Choice. = xj + BX . known as the Banach-Alaoglu Theorem. ∗ ∗ Proof Let X be a Banach space over the scalar field K. X = F . Recall $ that X in the weak ∗ topology is achieved by viewing X as a subspace of K = x∈X K in the product X topology. If {Ki }i∈I is a collection of compact topological spaces. This is a recursive statement.36 (because F is finite- dimensional). xk } of elements in X. Then (5. as a consequence of Lemma 5. 2 2 2 4 4 Continuing recursively. then i∈I Ki is compact in the product topology. We are now ready to state and prove the Banach-Alaoglu Theorem. and so we apply it to itself to get 1 1 1 1 1 BX ⊆ F + F + BX = F + F + BX = F + BX . & φ(BX∗ ) ⊆ xBK . Theorem 5. 2 While the unit ball in a Banach space can be compact in the norm topology only if the space is finite-dimensional.5) j =1 2 j =1 2 Let F = span{x1 . suppose BX is compact in the norm topology on X. ∞ 1 BX ⊆ F+ B X .. . Consequently.5) implies that BX ⊆ F + 21 BX .39 (Banach-Alaoglu Theorem) If X is a Banach space. F is closed. But F is a vector space. Denote by B(x. the unit ball in the weak∗ topology will always be compact. Theorem 5. (5. Thus. We will not prove this theorem.. there exists a finite sequence {x1 .4 The Geometric Hahn–Banach Theorem 105 Next. we have X = ∞ n=1 nBX ⊆ n=1 nF . Therefore. x ∗ ∈ X∗ . such that k 1 k 1 BX ⊆ B xj . r) the open ball of radius r centered at x ∈ X. then for each x ∈ X. xk }. . and so X ⊆ F . We make this explicit by defining the map φ : X∗ → KX by φ(x ∗ ) = (x ∗ (x))x∈X . . n=1 2n However. Before proving this statement. then BX∗ is compact in the weak∗ topology on X ∗ . and as such can be given a weak∗ topology.20). since X ⊆ X ∗∗ .) Denote the closure of BX in the weak∗ topology (w∗ ) (w∗ ) on X∗∗ by BX . there exists a weak∗ -continuous linear functional f on X∗∗ such that (w∗ ) f (x0∗∗ ) > sup {f (u∗∗ ) : u∗∗ ∈ BX }.39). w∗ )|X = (X. Our goal is to show the equality BX = BX∗∗ . Therefore. If we restrict to the subspace X. w).106 5 Consequences of Convexity where BK is the closed unit ball in the scalar field K and $ xBK is the closed ball of radius x centered at the origin. (If X is complex. we see (w∗ ) that BX ⊆ BX∗∗ . the set BX∗∗ is a compact (and hence closed) set in the weak∗ topology on X ∗∗ . The weak∗ topology on X ∗∗ is the weakest topology under which elements of X∗ define continuous functions on X ∗∗ . 5. 2 The Banach–Alaoglu Theorem as given here is due to Leonidas Alaoglu [1]. then the weakest topology under which elements of X ∗ are continuous is the weak topology on X. the argument is similar. the restriction of the weak∗ topology on X∗∗ to X is the weak topology on X. By the Hahn–Banach Separation Theorem (The- orem 5. Recall that X can be thought of as a subspace of it’s bidual X ∗∗ . (w∗ ) Suppose x0∗∗ ∈ BX∗∗ \BX . while the second ensures it is bounded. {α1 . There is no reason the image of BX∗ would be all of A. and hence φ(BX∗ ) is compact in the product topology on KX . the image is precisely the collection of elements in the following set: & f : f (α1 x1 + α2 x2 ) = α1 f (x1 ) + α2 f (x2 ) xBK . Banach did not have the notions of general topology available to him. By the Banach–Alaoglu Theorem (Theorem 5.5 Goldstine’s Theorem Let X be a Banach space. (5.α2 }⊆R x∈X {x1 . although the result was known to Banach. In other words. For simplicity. The product A = x∈X xBK is compact. we will assume X is real. by Tychonoff’s Theorem. but it is a closed subspace. Theorem 5. φ(BX∗ ) is a closed subset of the compact set A. as required. The space X ∗∗ is the dual space for X ∗ . Proof Let X be a Banach space. Therefore (X ∗∗ . It follows that BX∗ is compact in the weak∗ topology on X ∗ .6) . then BX is weak∗ - dense in BX∗∗ . Indeed. and so he could not formulate it in this way.40 (Goldstine’s Theorem) If X is a Banach space.x2 }⊆X (The first set of relations ensures f ∈ KX is linear.) Therefore. Since X is reflexive. . and so BX is compact (and hence closed) in the weak∗ topology on X ∗∗ . the set BX∗∗ is compact in the weak∗ topology on X∗∗ . yn∗ . by Theorem 4. we use Theorem 5. we assumed that X was a real Banach space for the sake of simplicity. The result follows.40). but instead of Theorem 5. Therefore. If T : X → Y is a linear map. norm-to-norm continuous). Proof Assume first that X is reflexive.1) becomes (w∗ ) x0∗∗ (x ∗ ) > sup {u∗∗ (x ∗ ) : u∗∗ ∈ BX } ≥ sup{x ∗ (x) : x ∈ BX } = x ∗ . 2 Proposition 5.5.e. 1 ≤ j ≤ n}.20.7) x≤1 x≤1 Since (5. since f is continuous in the weak∗ topology on X∗∗ . X is reflexive. 2 In the proof of Goldstine’s Theorem.42 Suppose X and Y are Banach spaces (or simply normed linear spaces).. w) continuous. Theorem 5. Now. . w) continuous.12. we conclude that the set T (BX ) is weakly bounded in Y . Then y ∗ is continuous in the weak topology on Y . contradicting the assumption that x0∗∗ ∈ BX∗∗ . we conclude that BX = BX∗∗ . By Goldstine’s Theorem (Theorem 5. which is the Hahn–Banach Separation Theorem for complex spaces. Let y ∗ ∈ Y ∗ .39). We will show that (ii) implies (i). Therefore. By the Banach–Alaoglu Theorem (Theorem 5. assume instead that BX is weakly compact.41 A Banach space X is reflexive if and only if the closed unit ball BX is weakly compact. there exists an x ∗ ∈ X∗ such that f (x ∗∗ ) = x ∗∗ (x ∗ ) for all x ∗∗ ∈ X∗∗ . y ∗ ◦ T ∈ X∗ . ) = {y : |yj∗ (y)| < . (iii) T is (X. Assume (ii). The weak topology on X is the restriction of the weak∗ topology on X ∗∗ .5 Goldstine’s Theorem 107 By Proposition 5. which is the Hahn–Banach Separation Theorem for real spaces. Consequently. Proof Certainly (iii) implies (ii). BX is compact in the weak topology on X. (5. since T is norm-to-weak continuous. and so sup |y ∗ (T x)| = sup |(y ∗ ◦ T )(x)| < ∞.22. This implies x0∗∗ > 1. The argument is similar when the Banach space is complex. and then (i) implies (iii).5. Therefore. . Therefore. · ) to (Y . Thus. . the functional y ∗ ◦ T is continuous in the norm topology on X. since BX is closed and dense in BX∗∗ . the weak∗ topology on X ∗∗ coincides with the weak topology on X.34.2) holds for each y ∗ ∈ Y ∗ . (ii) T is (X. and we replace f with (f ). Now assume (i). (5. say WY = WY (y1∗ .. T (BX ) is bounded in the norm topology. Then BX = BX∗∗ . Consider a weak neighborhood in Y . then the following are equivalent: (i) T is bounded (i. We wish to show that T (BX ) is bounded in the norm topology on Y . w) to (Y .5. . . Now suppose x = (xn )n=1 is any element in B1 . Then y ∗ ∈ WY ∗ = WY ∗ (T x1 . 2 5. and hence norm-closed in Y . and so it follows that x ∈ WX = WX (T ∗ y1∗ .44 If T : X → Y is a bounded linear map between Banach spaces. If X is reflexive..42. |T ∗ yj∗ (x)| < for all j ∈ {1. . . so that n=1 |xn | ≤ 1. as required. . Recall that the adjoint operator T ∗ was defined so that T ∗ ◦ y ∗ = y ∗ ◦ T . . then X is reflexive [19]. n}. . Then for each j ∈ {1.. . yn∗ } ⊆ Y ∗ and > 0.. T xn . n=1 n=1 Since ξ ∞ = 1. .. 1) or [−1. A convex subset of X is closed if and only if it is weakly closed.108 5 Consequences of Convexity for {y1∗ .. Therefore.. . say WX∗ = WX∗ (x1 . then T ∗ : Y ∗ → X∗ is weak∗ -to-weak∗ continuous. ). n}. This is not true in general (i. Similarly. . Then x ∗ (BX ) could be either (−1.C. Example 5. We conclude that T −1 (WY ) ⊆ WX . and so T ∗ is weak∗ -to-weak∗ continuous. then T (BX ) is weakly compact. . We conclude this section with a result about the adjoint operator. n}. for {x1 . and so T is weak-to-weak continuous. which is a weak∗ neighborhood of Y ∗ . . Theorem 5. . We see that a linear functional on a reflexive Banach space attains its maximum value on the closed unit ball. It was a long standing question whether or not this property characterized reflexive spaces. James showed that it did when he proved the statement: If every bounded linear functional on X attains its maximum value on the closed unit ball. Proposition 5. Let ξ in ∞ be the bounded sequence ∞ ξ = (1 − 1/n)∞ ∞ n=1 . 1). . . and so |y ∗ (T xj )| < for all j ∈ {1. . Then ∞ ∞ |ξ (x)| = ξn xn ≤ ξn |xn | < 1. . . If X is reflexive. . . . Suppose y ∗ ∈ Y ∗ is such that T ∗ y ∗ ∈ WX∗ . . WY ∗ ⊆ (T ∗ )−1 (WX∗ ). R. as required.. 2 Suppose that T : X → Y is a bounded linear mapping between real Banach spaces. 1]. 1 ≤ j ≤ n}. xn } ⊆ X and > 0. Equality is obtained by running through the same argument in reverse. .. Consider a weak∗ neighborhood in X ∗ . T ∗ yn∗ . Proof The proof is very similar to the proof that (i) implies (iii) in Proposition 5. In 1964. . we have a norm-one element ξ ∈ ∗1 such that ξ (B1 ) = (−1.e. Consider any x ∗ ∈ X∗ with x ∗ = 1. ). for non-reflexive spaces X). Suppose x ∈ X is such that T x ∈ WY . This implies that (T ∗ )−1 (WX∗ ) ⊆ WY ∗ . .6 Mazur’s Theorem In this section we explore the consequences of convexity in weak topologies. a weak neighborhood of X. . .. Recall that ∗1 = ∞ . Then |T ∗ y ∗ (xj )| < for all j ∈ {1. n}. xn .43 Consider the real Banach space X = 1 . ) = {x ∗ : |x ∗ (xj )| < . But T ∗ y ∗ (x) = y ∗ (T x) for all x ∈ X.45 (Mazur’s Theorem) Let X be a locally convex topological vector space. we have |yj∗ (T x)| < . then the second interval (the closed one) is the only option.. 20). We know that en → 0 weakly as n → ∞ (see Example 5. there exists no convex combination of elements in {en : n ∈ N} that will approximate 0. if 1 < p < ∞. m ∈ N . This set is convex and closed in the norm topology. then 1 1 p 1/p n 1 (e 1 + · · · + e n ) = 1 = n p −1 −−−→ 0. j =1 In the above example. and hence weakly closed by Mazur’s Theorem (Theorem 5. let en be the sequence with 1 in the nth coordinate. As usual. . Let co(E) denote the set of convex linear combinations of elements in E: m m co(E) = λj ej : λj ≥ 0.46 Consider the real sequence space p . · · · . m m co(A) = λj aj : aj ∈ A. and so we give the following definition. assume X is a real topological vector space. by the Hahn–Banach Separation Theorem (Theorem 5. λn . for each n ∈ N. (w) This contradicts the assumption x0 is in the weak closure of K. 2 Example 5. there is an x ∗ ∈ X∗ such that x ∗ (x0 ) > sup{x ∗ (x) : x ∈ K}. j =1 j =1 The closed convex hull of A is the closure of the convex hull and is denoted co(A). λ2 . A weakly closed set is always strongly closed. We conclude that 0 can be approximated (in norm) by convex linear combinations of elements in E = {en : n ∈ N}. Then n λ1 e1 + · · · + λn en 1 = (λ1 . Assume x0 ∈ K \K. The convex hull of A is denoted by co(A) and consists of all convex linear combinations of elements in A. we introduced the notation co(E) to denote the set of convex linear combinations of elements in the set E. 0.45).5. since 0 is a weak limit point of E. and so 0 is in the weak closure of the set E = {en : n ∈ N}. Then. let λ1 e1 + · · · + λn en be any convex combination of elements from {en : n ∈ N}. Suppose K is (w) closed in the original topology.47 Let X be a topological vector space and let A be any subset of X. . It follows that co(E) contains 0. and let K denote the closure of K in the weak (w) topology. n p n j =1 n→∞ The same cannot be said of 1 —in this case. and 0 elsewhere. regardless of convexity. To see this. This idea will prove important later. where 1 < p < ∞. Definition 5. and so K = K. λj > 0.6 Mazur’s Theorem 109 Proof Without loss of generality. )1 = λj = 1. λj = 1. m ∈ N . . λj = 1. j =1 j =1 We denote the closure of co(E) in the norm topology by co(E). that is. The convex hull of A is the smallest convex subset of X that contains A. . In fact.28). 19. Let W be a subset of a topological vector space X. we saw that the only nonempty open convex subset of X is X itself.3. Our goal is to show that α > −∞ and that f (x0 ) = α for some x0 ∈ K. for all {x. α > −∞. and so n=1 nK = ∅. 1). Consequently. we claim that if X is a reflexive Banach space and if f is a convex function. K is weakly closed. As before. y} ⊆ X and t ∈ [0. (See Example A in Sect. We now give an example of how convexity can help to solve optimization problems. By the Nested Interval Property (Corollary B. where 0 < p < 1. Let α = inf{f (x) : x ∈ K}. there are no nonzero linear functionals bounded on any nonempty open subsets of Lp (0. . Despite this. If f is a linear functional on X such that |f (x)| ≤ M for all x ∈ W . convex. 5. 1]. it follows that there are no nonzero linear functionals bounded on the unit ball of Lp (0. 1) if 0 < p < 1. Indeed. if BX = {f : f p ≤ 1}. If x0 ∈ n=1 nK . 1). Therefore. define Kn = {x ∈ K : f (x) ≤ −n}.48 Let X = Lp (0. it must be that |f (x)| ≤ M for all x ∈ co(W ). as required. Suppose α = −∞ and. Suppose for a moment that X is reflexive. But this implies that there is some x0 ∈ K such that f (x0 ) ≤ −n for all n ∈ N.3). it must be that ∞ K n=1 n = ∅. Since no nonzero linear functionals are bounded on Lp (0.7). Then BX is weakly compact. We summarize in the following proposition. but not in the same topology: f is continuous in the norm topology. then x 0 ∈ K and f (x0 ) = α. by The- orem 5. and nonempty (since α = −∞). 1) (see Example A in Sect. Since K is closed in norm. an impossibility. 5.110 5 Consequences of Convexity Example 5. Consequently. We begin by making a definition. Define for n ∈ N a sequence of sets Kn = {x ∈ K : f (x) ≤ α + 1/n}.) We now have continuity and compactness. %∞ (Kn )∞ n=1 is a nested %∞ sequence of weakly compact nonempty sets. (Kn )∞ n=1 forms a nested sequence of weakly compact % nonempty sets. the set Kn is closed (and hence weakly closed by Mazur’s Theorem). as there is no compactness assumption made on K. Our problem is as follows: Suppose K is a closed bounded convex set in a Banach space X and let f : X → R be a continuous convex function. For each n ∈ N. Definition 5. 1).) In Example 5. It follows that K is weakly compact. for each n ∈ N. by Mazur’s Theorem. In the case of Lp (0. then co(BX ) = X.49 Let X be a vector space. we use the fact that BX is weakly compact and absorbent. Does there exist some x0 ∈ K such that f (x0 ) = min{f (x) : x ∈ K}? There is no reason we should assume this minimum exists. then. from the example above. then f does attain its minimum value on K. where 0 < p < 1. A function f : X → R is called convex if f (tx + (1 − t)y) ≤ tf (x) + (1 − t)f (y). (Here.41. by the definition of the convex hull. the convex hull of the unit ball is the entire space. and K is compact in the weak topology. 2 A special case of the above is f (x) = u − x.7 Extreme Points In this section. ·. the function representing the distance between x ∈ X and a fixed point u ∈ X. we have 1 = x. there exists some point x0 ∈ K which is closest to u. A point x ∈ K is an extreme point of K if it does not lie on a line segment in K. If X is a reflexive Banach space. a triangle has an extreme point at each vertex. then there exists an x0 ∈ K such that f (x0 ) = min{f (x) : x ∈ K}. then there exists a x0 ∈ K such that u − x0 = minu − x. Proof See the discussion preceding the statement of the proposition. If X is reflexive. (5.50 Suppose K is a nonempty closed bounded convex set in a Banach space X and let f : X → R be a continuous convex function. x∈K That is. (See Fig.2 Some elementary convex objects Proposition 5. For example. u = v = 1 (otherwise x < 1).) 5. x is an extreme point of K provided that the following is true: If u and v are elements of K such that x = (1 − t)u + tv for some t ∈ (0. Now let {u.7 Extreme Points 111 Fig.8) . x. and so we must consider only points on the boundary ∂BX .2. Definition 5. Note that no point of the interior of BX can be extreme.) Example 5. x + tv. Denote the inner product on 2 by ·. x = (1 − t)u. (i) X = 2 . then x = u = v. 1). while any boundary point of a circle is an extreme point. 1). and if K is a closed and bounded convex set in X (not containing u). By the triangle inequality. Since x = 1. 5. 5. we consider sets K that are convex in some vector space X. the boundedness assumption on K is not needed for this statement to be true. That is. Suppose x ∈ 2 is such that x = 1.52 We now determine the extreme points for the unit ball BX in several cases where X is a real Banach space. (Actually.51 Let X be a vector space and suppose K is a convex subset of X.5. v} ⊆ B2 and suppose x = (1 − t)u + tv for some t ∈ (0. The set of extreme points of K is denoted ex(K). and therefore any f on the boundary of the unit ball is an extreme point of the unit ball BLp (0. Observe that u. where 1 < p < ∞.1) implies that u. which is equivalent to p a 1/p |f | = b1/p |g|. The choice of f was arbitrary in ∂BLp (0. 1) such that f 1 = 0 |f (s)| ds = 1. Let f be a function in L1 (0. 1). 1).112 5 Consequences of Convexity By assumption. Again using the triangle inequality. x is an extreme point of B2 whenever x is on the boundary ∂B2 . This happens only if there are positive constants a and b such that a(|f |p−1 )q = b|g|p (as members of Lp (0. x = 1 (again using the triangle inequality). 0 < t < 1. Suppose f ∈ Lp (0. Define a new function F on [0. A similar argument shows |f | = |h| in Lp (0. this equality (which is valid almost everywhere) becomes a|f | = b|g|p . If f is an extreme point of BL1 (0. 1 < p < ∞. 1/q |φ(g)| ≤ |f (x)| p−1 |g(x)| dx ≤ |f (x)|(p−1)q dx gp = f p/q p gp . and thus we have equality in the Cauchy–Schwarz Inequality.1) . 1] by t F (t) = |f (s)| ds. 1). v = x. 1). Since f p = gp . From these equalities. we conclude that f = g and f = h. there exists a linear functional φ in Lp (0. By the Intermediate Value Theorem. and so (5. This can only happen if u = x. By Hölder’s Inequality. h} ⊆ Lp (0. 1)). 1). It follows that f is an extreme point of BLp (0. 1]. 0 The function F is continuous with F (0) = 0 and F (1) = 1.1) . (ii) X = Lp (0. 0 p Since φ(f ) = f p . 1). Let {g. Because q = p−1 p . Similarly. t ∈ [0. are the elements of the boundary ∂Bp . 1) is such that f p = 1. . x = v. Therefore. this can only happen if |f | = |g| in Lp (0. by the triangle inequality (as in (i)). together with the assumption that f is a convex combination of g and h.1) . 1)∗ such that φ = 1 and φ(f ) = 1. in this case we can write φ explicitly: 1 φ(k) = |f (x)|p−1 (signf (x)) k(x) dx. x = 1 = ux. 1) and suppose that f = (1 − t)g + th for some t ∈ (0. there exists a τ ∈ (0. We will take our cue from the preceding example.1) . A similar argument shows that the extreme points of the unit ball in p . By the Hahn–Banach Theorem.1) . In fact. we have that φ(g) = φ(h) = 1. Then gp = hp = 1. where p1 + q1 = 1 (and so q = p−1 p ). we have 1 = φ(f ) = (1 − t)φ(g) + tφ(h).7. then f is on the boundary 1 of BL1 (0. we have equality in Hölder’s Inequality. (iii) X = L1 (0. Since the left and right sides of the above inequality are both equal to 1. k ∈ Lp (0. 1) such that F (τ ) = 1/2. Since f was an arbitrary element of ∂BL1 (0. we conclude that the unit ball in L1 (0.1) . Suppose x = (xk )∞ k=1 is a sequence in Bc0 . then define y and z as follows: 1 1 . 1) has no extreme points.1) . We have found distinct functions g and h in BL1 (0. 1). Then lim xk = 0. We will define sequences y = (yk )∞ ∞ k=1 and z = (zk )k=1 in c0 so that x = 2 y + 2 z. We will show there are no extreme points in Bc0 .1) such that f = 21 g + 21 h. (iv) X = c0 . Therefore. By the choice of τ . f is not an extreme point of the unit ball of L1 (0. we have g1 = 1 and h1 = 1.τ ) and h = 2f χ(τ .7 Extreme Points 113 Let g = 2f χ(0.5. and so there exists some n ∈ N such k→∞ that |xn | < 1/2. If xn = 0. . xk if k = n. The previous sequences work only if xn = 0. If instead xn = 0. yk = and zk = 0 if k = n 2xn if k = n. We have x = 21 y + 21 z and. because |xn | < 1/2. then define y and z so that: . the sequences y and z are in Bc0 . xk if k = n. . The set of extreme points in B1 is {±en : n ∈ N}. the constant functions 1 and −1. we can put a small “wiggle” in the function. xk if k = n. and so yn = a1 − ab zn and yk = − ab zk for k = n. xk if k = n.1] are the two functions χ[0. (That is. yk = 1 and zk = 2 if k = n − 1 2 if k = n. then we may use an argument similar to the perturbation argument used in (iv). (That is. (v) X = C[0. if K is a compact Hausdorff space. Let y = (yk )∞ ∞ k=1 and z = (zk )k=1 be elements of B1 such that en = ay + bz. the extreme points of BC(K) are all functions f ∈ C(K) for which |f (s)| = 1 for all s ∈ K. (Note that these conditions imply yn > 0 and zn > 0.1] and −χ[0. Therefore. let us show that each of these points is indeed extreme in B1 . 1].1] is a continuous function such that |f (t)| < 1 for some t ∈ (0. In the case that C(K) is a complex Banach space. Once again making use of the triangle inequality.1] . k =n k =n a a a a k=1 a a . where the sequences y and z are in Bc0 . we have x = 21 y + 21 z. (vi) X = 1 . we have b 1 b b ∞ 1 2b |yk | + yn = |zk | + − zn = |zk | + − zn . we see that y1 = 1 and z1 = 1.) If f ∈ BC[0. In this case the extreme points of BC[0. where a and b are positive numbers such that a + b = 1. 1). First. Once again. By assumption.) Similarly.) We have that ayn + bzn = 1 and ayk + bzk = 0 for all k = n. Computing y1 directly. which are the constant functions 1 and −1 on K. then the extreme points of BC(K) are the two functions χK and −χK . we know that a = 0. Fix some n ∈ N. x is not an extreme point of the set Bc0 . 55 Let E be a topological vector space with nonempty subset K. Therefore. m2 }.39. Definition 5. m2 }. Define y = (yk )∞ ∞ k=1 and z = (zk )k=1 in 1 as follows: ⎧ ⎧ ⎪ ⎨xk if k ∈ {m1 . z} ⊆ B1 . where {y. Now we show that no other element of ∂B1 is an extreme point. The result then follows from the Krein–Milman Theorem. yk = xm1 + if k = m1 . and so z must in fact be en .53. the set BX∗ is compact in the w∗ -topology whenever X is a Banach space. Proof By Theorem 5. b 1 2b y1 = z1 + − zn . Without loss of generality. courtesy of Example 5. Theorem 5.53 (Krein–Milman Theorem) Suppose E is a locally convex Haus- dorff topological vector space. Therefore. we now introduce a definition and a lemma. a a a a A little arithmetic (and the fact that a + b = 1) reveals that zn = 1. conclude that neither c0 nor L1 (0. Then there must be at least two non-zero entries. ⎪ ⎨xk if k ∈ {m1 . en is an extreme point. since the unit balls in these spaces have no extreme points. It follows that x = 21 y + 21 z. xm2 . then K = co(exK). 1 − xm2 }. Choose some constant > 0 such that < min{xm1 . 2 What makes this result so interesting is that we can now. 1) is a dual space of a Banach space. let us observe a consequence. In order to proceed with the proof of Theorem 5. and zk = xm1 − if k = m1 . 1 − xm1 .54 If X is a Banach space. we may assume both terms are positive. x is not an extreme point. C[0. it does not have enough (only two!) to construct the entire unit ball using only convex linear combinations. a a a But y1 = 1 and z1 = 1. ex(K) = ∅.114 5 Consequences of Convexity Thus. The next theorem describes a condition under which the set of extreme points is never empty. either.53. but x = ±en for any n ∈ N. ⎪ ⎩ ⎪ ⎩ xm2 − if k = m2 xm2 + if k = m2 . In particular. While the unit ball of C[0. A subset F of K is called extremal (in K) if F is a nonempty compact convex set such . and so b 1 2b b + 1 − 2bz n 1 = (1) + − zn = . say xm1 and xm2 .52. 1] cannot be the dual space of any Banach space. A similar argument shows that −en is an extreme point for each n ∈ N. then BX∗ = co(w ) (exBX∗ ). If K is a nonempty compact convex subset of E. 1] does have extreme points. Therefore. Suppose x ∈ B1 with x1 = 1. Before proving Theorem 5. ∗ Corollary 5. Without loss of generality. Then G is a compact and % convex subset of F . Suppose {u. in in I .. The space E is Hausdorff. Proof Let K be a nonempty compact convex subset of the locally convex topological vector space E and suppose F is an extremal set in K. Let ! " G0 = x ∈ F0 : φ(x) = maxφ(ξ ) . By the Hahn–Banach Separation Theorem (Theorem 5.7 Extreme Points 115 that the following holds: If {u. By Zorn’s Lemma. We will show that F0 consists of only one element. and therefore attains its maximum on the compact set F0 . there exists a maximal element in the partially ordered set of subsets of F that are extremal in K. and we know that F0 is extremal. G is extremal in K. and so it must be the case that F0 contains only one element. . For each i ∈ I . . v} ⊆ F . Hence. v} ⊆ Gi . Suppose {u. 1). ξ ∈F0 Let M denote the maximum value of φ on F0 . because C is a chain of subsets. The set F0 is a single-point set and an extremal set. and so {u} and {v} are closed sets. Denote this minimal extremal set by F0 . there exists a continuous linear functional φ on E such that φ(u) = φ(v). the one element of F0 is an extreme point. Every extremal subset of K contains an extreme point. so G = i∈I Gi . v} ⊆ F0 .56 Suppose K is a nonempty compact convex subset of a locally convex topological vector space. We claim that G0 is extremal in K. 1).20 for real spaces and Theorem 5. Therefore. we have that (1 − t)u + tv ∈ G0 . but this violates the minimality of F0 . This implies G0 is extremal. It is also a proper subset of F0 . 1). . The set G0 is nonempty (by the continuity of φ). and so M = φ((1 − t)u + tv) = (1 − t)φ(u) + tφ(v). v} ⊆ K and (1 − t)u + tv ∈ G for all t ∈ (0. Suppose C = (Gi )i∈I is a chain of%subsets of F that are extremal in K. by the Finite Intersection Property.5. and convex. We know that G0 ⊆ F0 . It follows that for all t % {u. φ is not constant on F0 . for all t ∈ (0. We have derived a contradiction. Assume to the contrary that there exist distinct elements u and v in F0 . 1). v} ⊆ G0 . . where G ≥ H whenever G ⊆ H . and consequently {u. We claim that G is extremal in K. Let G be the intersection of all sets in C. G is nonempty. Therefore. we have nk=1 Gik = Gij for some j ∈ {1. Consider the partially ordered set of all subsets of F that are extremal in K. In particular. For each t ∈ (0. then {u. n}. and hence {u. 1). v} ⊆ K and (1 − t)u + tv ∈ G0 for all t ∈ (0. v} ⊆ i∈I Gi = G. 1). For any finite collection of indices i1 . We wish to find a maximal element of this partially ordered set (which in turn will be a minimal extremal subset of F ). compact. But Gi is extremal in K. .22 for complex spaces). Thus. so that φ(x) = M for all x ∈ G0 . because φ is not constant. and so (1 − t)u + tv ∈ Gi ∈ (0. Lemma 5. The functional φ is continuous on E.. we may assume φ is real-valued. v} ⊆ K and (1 − t)u + tv ∈ F for all t ∈ (0. . there is a minimal subset of F that is extremal in K. . From this we conclude that φ(u) = φ(v) = M. we have G ⊆ Gi . hence {u. ∗ This set is convex.56 assures us that P(K) must have at least one extreme point. (See Exercise 5. Local convexity is a necessary condition.20).8 Milman’s Theorem Suppose K is a compact Hausdorff space. however. that is. (5. K = K0 . y∈K The set G0 is nonempty because f is continuous and K is compact. The Riesz Representation Theorem identifies the dual space of the space of continuous functions on K as the space of regular Borel measures on K. and consequently contains an extreme point of K. see Exercise 5. A simple computation shows that P(K) is an extremal set in the unit ball of M(K). Suppose x ∈ K\K0 . we may assume E is a real topological vector space. by Lemma 5. Therefore.57 Let K be a compact Hausdorff space. (Again. and so must contain an extreme point. (See [4]. The Krein–Milman Theorem originally appeared in a work by Krein and Mil- man in 1940 [21]. Lemma 5. Let K0 = co(exK). there exists a continuous linear functional f on E such that f (x) > maxf (y). A probability measure in M(K) is an extreme point of P(K) if and only if it is a Dirac measure. The local convexity assumption on E was needed to invoke the Hahn–Banach Separation Theorem (Theorem 5. .56. We define the probability measures on K to be elements in the set P(K) = {μ ∈ M(K) : μ ≥ 0. Proof of Theorem 5.) 5.) By the Banach–Alaoglu Theorem (Theorem 5.27. which can be seen from the equality P(K) = {μ ∈ BM(K) : K 1 dμ = 1}.) Since P(K) is a w∗ -compact convex extremal set.53 Without loss of generality. and hence P(K) is w∗ -compact as a w∗ -closed subset. is a contradiction.56. a fact which was not shown until the 1970s [32].9) y∈K0 Let ! " G0 = z ∈ K : f (z) = maxf (y) . because K0 contains all of the extreme points of K. The Krein–Milman Theorem has a deep relationship with the Axiom of Choice.35. and K0 and G0 are disjoint.39). μM = 1}.116 5 Consequences of Convexity We are now prepared to prove the Krein–Milman Theorem. the closed convex hull of the set of extreme points in K.56).27. the unit ball BM(K) is compact in the w∗ -topology. by Lemma 5. C(K)∗ = M(K). By the Hahn–Banach Separation Theorem (Theorem 5. it is also disjoint from K0 because of (5.20). The set G0 is extremal (see the argument in the proof of Lemma 5. The set K is extremal in itself. Proposition 5. It is also closed in the w -topology.) We recall that the norm on M(K) is the total variation norm: μM = |μ|(K) for all μ ∈ M(K). as required.9). This. (See Theorem A. The entire μ-mass of K is contained in K\V . (They “live” on different sets. μ(W1 ) μ(K\W1 ) where B is any measurable subset of K. If V is the maximal open set of μ-measure zero. then K\V is called the support of μ.57.58 Suppose μ is a positive nonzero measure on K. Define measures μ1 and μ2 on K as follows: μ(W1 ∩ B) μ((K\W1 ) ∩ B) μ1 (B) = and μ2 (B) = . they have non-zero μ-measure. Let F = K\V . μ(V ) = sup{μ(E) : E is a compact subset of V } = 0. This is not possible. Since W1 and W2 are not subsets of V . and so (by subadditivity) μ(E) ≤ μ(U1 ) + · · · + μ(Un ) = 0. Now suppose μ is an extreme point of P(K). In the proof of Proposition 5. μ is an extreme point of P(K). so to speak. We wish to show that F contains only one point. We note that μ(K\W1 ) = 0 since μ ≥ 0 and W2 ⊆ K\W1 . t ∈ W2 . the support of μ is defined to be the support of |μ|. In the above proof. Suppose there exist probability measures μ and ν such that δs = (1 − t)μ + tν for some t ∈ (0. for every x ∈ K. say E ⊆ U1 ∪ · · · ∪ Un . Definition 5. μ = ν = δs . it follows that μ = μ1 = μ2 . Let {s. Suppose E is a compact subset of V . Assume to the contrary that F contains more than one point. say s.59 (Milman’s Theorem) Suppose E is a locally convex Hausdorff topological vector space and let K be a compact subset of E.) Theorem 5. since μ1 (W1 ) = 1 and μ2 (W1 ) = 0. Then μ(F ) = 1. By the regularity of μ. We will show that μ = δs for some s ∈ K. Let U = {U open : μ(U ) = 0} and let V = U ∈U U . then ex(K) ⊆ D. as required. there can be no more than one point in the set F . t} ⊆ F . We have derived a contradiction. As a result.8 Milman’s Theorem 117 Proof We first show that δs is an extreme point of P(K) for s ∈ K. We claim μ(V ) = 0. as required.5. it was certainly the case that μ1 = μ2 . If μ is a signed (or complex) measure. 1). Furthermore. The measure μ is nonnegative. it follows that μ = δs . By assumption. and μ(W1 ) μ1 + μ(K\W1 ) μ2 = μ. Thus. If D is a closed subset of K such that K = co(D). μ(E) = 0 for all compact subsets of V . The collection of sets U forms an open cover of E. Therefore. and W1 ∩ W2 = ∅. Thus. observe that the measures μ1 and μ2 were defined in such a way that they had disjoint supports. Therefore. and so by compactness there exists a finite subcover. Both μ1 and μ2 are probability measures. there . since μ(W1 ) + μ(K\W1 ) = 1. there are open sets W1 and W2 in K such that s ∈ W1 . By the Hausdorff property. Since μ(F ) = 1. the set V is the maximal open set of μ-measure zero. however. It follows that μ({s}) = ν({s}) = 1. This motivates the next definition. By Lemma 5. (5. {μ. where the superspace is equipped with the product topology. This proves the second part of the theorem. Now.53 (the Krein–Milman Theorem) and Proposition 5. and so T −1 (x) is extremal. We noted at the time that the metrizability assumption was not needed.118 5 Consequences of Convexity exists a μx ∈ P(D) such that f (x) = D f (y) μx (dy) for all linear functionals f ∈ E∗.57. the restriction T |P(D) : P(D) → K is a surjection.56. we will extend this . the collection of all scalar-valued functions on E ∗ . and so x = s ∈ D. The inclusion is made ∗ explicit by the embedding ρ : E → KE defined by ρ(e) = (f (e))f ∈E ∗ . (See the comments after Definition 5. we make ∗ the identification co(w ) ({δs }s∈D ) = P(D). using the tools of the previous sections. and consequently T (δs ) = x. 1). 3. Suppose x ∈ ex(K). For each s ∈ D. defined by T (μ) = f (y) μ(dy) . Therefore. that is. 1). It remains to prove the first part of the theorem. (We often make this identification implicitly. 2 5. It follows that (1 − t)T (μ) + tT (ν) = x for all t ∈ (0.4 that if G is a compact abelian metrizable group. ν} ⊆ P(D) and (1 − t)μ + tν ∈ T −1 (x) for all t ∈ (0. there exists some s ∈ D such that δs ∈ T −1 (x). T (δs ) = s. suppressing the letter ρ.10). By assumption. Suppose {μ.9 Haar Measure on Compact Groups We now turn our attention to topological groups. We claim that the set T −1 (x) ⊆ P(D) is an extremal set. e ∈ E. T −1 (x) contains an extreme point of P(D). By assumption. D f ∈E ∗ ∗ Observe that T is continuous in the w -topology on M(D). x is an extreme point in K. however. that the extreme points of K are in D. then there exists a unique translation-invariant probability measure on G. ∗ Proof Observe that E ⊆ KE . By (5. ν} ⊆ T −1 (x). we have that K = co(w) (D).45). Therefore.) ∗ Introduce a map T : M(D) → KE . The result follows. and so by Mazur’s Theorem (Theorem 5.) By Theorem 5. We saw in Sect. μ ∈ M(D). Therefore.27. and consequently T μ = T ν = x. K = co(D). we have T (δs ) = (f (s))f ∈E ∗ = s. it follows that T maps co(w ) ({δs }s∈D ) onto co(w) (D).10) ∗ By the w∗ -continuity of T . then. We note that we have the weak closure of co(D) in E because the topology E inherits ∗ from KE is the weak topology. Substituting into (5. Let us review some definitions. The σ -algebra on G is implicitly taken to be the Borel σ -algebra generated by the open sets in G. (Notice that this will imply uniqueness. If G is compact in the given topology.11) G G G G By the left-invariance of λ. we assume we can find a left-invariant probability measure λ and a right-invariant probability measure μ. By Fubini’s Theorem. we obtain f (t) λ(dt) = f (s) μ(ds). A measure μ on G is called left-invariant if μ(gB) = μ(B) for all B ∈ B and g ∈ G. Definition 5. (s. Classical examples of topological groups include Rn (where the group multiplication is given by addition) and the set of orthogonal n × n matrices On (where the group multiplication is matrix multiplication). this measure is also the unique right-invariant probability measure on the Borel sets of G. .11). (5. Theorem 5. Similarly. G G G G G since μ(G) = 1. f (s · t) λ(dt) μ(ds) = f (s · t) μ(ds) λ(dt). however. it is traditional to use a plus symbol (+). f (s · t) μ(ds) λ(dt) = f (s) μ(ds) λ(dt) = f (s) μ(ds). Multiplication in a group is usually denoted either with a dot (·) or by juxtaposition.9 Haar Measure on Compact Groups 119 result to include compact topological groups that are not abelian. When G is a compact group. We will show that λ = μ. f (s · t) λ(dt) μ(ds) = f (t) λ(dt) μ(ds) = f (t) λ(dt).60 A group G is called a topological group if the set G is endowed with a topology for which the group operations (multiplication and inversion) (s. are continuous. provided it will not result in any confusion. t) ∈ G × G.5. Definition 5. G G G G G since λ(G) = 1.61 Let G be a compact group with Borel algebra B. Furthermore. G G The choice of f ∈ C(G) was arbitrary. There exists a unique left-invariant probability measure on the Borel sets of G. When the group is abelian. and so by duality we conclude λ = μ. t) → s · t and s → s −1 . then G is called a compact group. Proof First. Correspondingly. by the right-invariance of μ.62 (Existence of Haar Measure) Suppose G is a compact group. the measure μ is called right-invariant if μ(Bg) = μ(B) for all B ∈ B and g ∈ G. we denote the space of continuous functions on G by C(G).) Let f ∈ C(G). We denote the Borel σ -algebra on G by B. Then. because P(G) ∈ K. Claim 1 Let f ∈ C(G). If (Ci )i∈I is a chain in K. P(G) is left-invariant. (A similar argument will produce a right-invariant measure. We know that K is nonempty. If s and s are in G. t∈G This proves Claim 1. . whenever μ ≥ 0.) We begin by defining for each s ∈ G a left-multiplication operator Ls : C(G) → C(G) by (Ls f )(t) = f (s · t). for f ∈ C(G). Furthermore. and so the map (s. then f dL∗s μ− f dL∗s μ = Ls f dμ− Ls f dμ ≤ Ls f −Ls f ∞ μM . Define a partial order ≤ on K so that A ≤ B when A ⊆ B. by the definition of the adjoint. for any given > 0. let K be the collection of all weak∗ -compact convex subsets of P(G) that are left-invariant. t ∈ G.e. then L∗s (P(G)) ⊆ P(G). and Lu ◦ Ls = Lu·s . Claim 2 Let μ ∈ C(G). Therefore.. To that end. The gist of Claim 3 is that the set P(G) is invariant under multiplication on the left. all weak∗ -compact convex subsets K such that L∗s K ⊆ K for all s ∈ G. we conclude the map (s. If f ≥ 0. We wish to estimate. that is. i. f dL∗s μ = Ls f dμ. The map s → Ls f is continuous from G into C(G). L−1 s = Ls −1 . whenever u and s are in G. Thus. t∈G Multiplication in the group is continuous. ∗ ∗ Ls μ(G) = 1 Ls μ(dt) = Ls (1) μ(dt) = μ(G) = 1. G G G G The rest follows from Claim 1. the quantity Ls f − Ls f ∞ = sup |f (s · t) − f (s · t)|. Claim 3 If s ∈ G. This proves Claim 3. In this case.120 5 Consequences of Convexity It remains to prove the existence of a left-invariant measure on G. and so if s · s −1 ∈ V . t) → f (s · t) is continuous on G × G. we have t = t. The map s → L∗s μ is w∗ -continuous from G into M(G). G G Thus. L∗s μ is in P(G) whenever μ is a probability measure. then Ls f − Ls f ∞ = sup |f (s · t) − f (s · t)| < . Observethat L∗s : M(G) → M(G). Observe that Ls = 1. there exists an open neighborhood V of the identity such that |f (s · t) − f (s · t )| < whenever s · s −1 ∈ V and t · t −1 ∈ V . We wish to find a set with this property that contains only one element. L∗s μ ≥ 0 for any μ ∈ P(G). then f dL∗s μ = Ls f dμ ≥ 0. for s and s in G. Since the group G is compact. t) → f (s · t) is in fact uniformly continuous. Recalling the definition of ν. if there exists some linear bijection T : X → Y such that T = T −1 = 1. Therefore. Theorem 5.10 The Banach–Stone Theorem 121 % then C = i∈I Ci is nonempty.53). Furthermore. and L∗s ν is extreme in K. then K1 and K2 are homeomorphic.10 The Banach–Stone Theorem In this section. there is some extreme point in K. and so K0 ∈ K. there exists a minimal element of K. We also have that K0 is weak∗ -compact. by convexity. Let ν = 21 (μ1 + μ2 ). we see that 1 ∗ L∗s ν = (L μ1 + L∗s μ2 ). every extreme point of K is in E. Thus L∗u (E) ⊆ E. By the Krein–Milman Theorem (Theorem 5. Therefore. and so E is left-invariant. Define a new set E = {L∗s ν : s ∈ G}. we discover that ν = μ1 = μ2 . We wish to show that K is a single-point set. K0 is left-invariant. apply L∗s −1 to all sides). L∗u (L∗s ν) = L∗u L∗s ν = L∗s·u ν ∈ E. L∗s ν = L∗s μ1 = L∗s μ2 . Therefore. By construction. If we multiply all sides of this equation by s −1 on the left (that is. For all u and s in G. If C(K1 ) and C(K2 ) are isometrically isomorphic. E ⊆ K. But K0 ⊆ K and K is minimal in K.64 (Banach–Stone Theorem) Suppose K1 and K2 are compact Haus- dorff spaces. and by Milman’s Theorem (Theorem 5. if T : C(K1 ) → C(K2 ) is an isometric isomorphism. This violates the assumption that μ1 and μ2 are distinct. Assume to the contrary that μ1 and μ2 are distinct elements in K.63 Let G be a compact group. Definition 5. the set E is weak∗ -compact (as the image of the compact set G under a weak∗ -continuous mapping). The unique left-invariant probability measure on the Borel subsets of G is called Haar measure on G. we prove a classical theorem about the structure of spaces of continuous functions. by the left-invariance of K. That is.59). ∗ Let K0 = co(w ) (E). K = K0 . Furthermore. by the Finite Intersection Property. there exists some s ∈ G such that L∗s ν is extreme in K. Then ν ∈ K. The measure λ is the desired left-invariant probability measure on G. We recall that two Banach spaces X and Y are called isometrically isomorphic if there exists a continuous linear bijection that preserves norms. say K. 5. because it is weak∗ -closed in the weak∗ -compact set K. By Claim 2. K contains only one element. C is a lower bound for the chain (Ci )i∈I .5. Thus. say λ. by left-invariance. by Zorn’s Lemma. . 2 s But L∗s μ1 and L∗s μ2 are in K. The set K0 is convex by construction. Furthermore. Therefore. In order to prove this lemma. Now let F be a closed set in K. ψ is continuous in the w∗ topology on Δ. We will not prove Urysohn’s Lemma. Define a map ψ: K→Δ by ψ(s) = δs for every s ∈ K. and so ψ is an injection as well as a surjection. Theorem 5. too.66 Let K be a compact Hausdorff space. Therefore. This means that δs (f ) = δt (f ) for all f ∈ C(K). s ∈ K2 . since for every f ∈ C(K). Proof We remindthe reader that for each s ∈ K.122 5 Consequences of Convexity then there exists some u ∈ C(K2 ) such that |u(s)| = 1 for all s ∈ K2 . then K is homeomorphic to Δ with the subspace topology inherited from (M(K). Then F is compact. Therefore. s ∈ K. we will observe that. We recall that a topological space X is normal if for disjoint closed sets E and F . we will provide a simple lemma that will not only help us now.64 Let K1 and K2 be compact Hausdorff spaces and suppose T : C(K1 ) → C(K2 ) is an isometric isomorphism. we need to make use of another result from general topology. K and because f is continuous (by assumption). That is. (See Exercise 5. Suppose that s and t are two distinct points in K such that ψ(s) = ψ(t). Proof of Theorem 5. then C(K) separates the points of K.) We are now prepared to prove the Banach–Stone Theorem. then there is a function f ∈ C(K) such that f (a) = 0 and f (b) = 1.4. however.65 (Urysohn’s Lemma) A topological space X is normal if and only if any two disjoint closed subsets A and B can be separated by a continuous function.7. Since ψ is continuous for Δ with the w∗ topology. it must be closed (in that topology). f (s) = f (t) for all f ∈ C(K). ψ(s)(f ) = f dδs = f (s). however. (See Exercise 5. if a and b are distinct points in K. if K is a compact Hausdorff space. the Dirac measure at s is a measure δs defined so that K f dδs = f (s) for all f ∈ C(K). Thus. We wish to show that K1 and K2 . (See the comments before the statement of Lemma 5. if there exists a continuous function f : X → [0. Since ψ(F ) is a compact set in a Hausdorff topology.66. Certainly. Then δs = δt . 1] such that f |A = 0 and f |B = 1. ψ(s) = ψ(t) only if s = t. This contradicts the fact that C(K) separates the points of K. Before proving the Banach–Stone Theorem. but will come in handy later. We next show that ψ is a homeomorphism by showing that it is a continuous closed map (so that it maps closed sets to closed sets). it follows that ψ(F ) is w∗ -compact in Δ. where φ : K2 → K1 is a homeomorphism. ψ is a surjection.) Therefore. w∗ ). The set Δ is closed in the w∗ topology and Δ ⊆ BM(K) . Clearly. That is. there exist disjoint open sets U and V such that E ⊆ U and F ⊆ V .) Lemma 5. If Δ = {δs : s ∈ K}. ψ is a closed map. as a consequence. and it follows that ψ is a homeomorphism. and such that Tf (s) = u(s) f (φ(s)). because it is a closed subset of the compact set K. Δ is w∗ -compact. we have Tf (s) = (Tf ) dδs = f d(T ∗ δs ) = f dδφ(s) = f (φ(s)). φ is continuous. Thus.66. and from this we deduce that T −1 g = T −1 h = f (because f is an extreme point).28. Let ψ : K1 → {δt : t ∈ K1 } and ϕ : K2 → {δs : s ∈ K2 } be defined by ψ(t) = δt and ϕ(s) = δs for all t ∈ K1 and s ∈ K2 . and so T ∗ (BM(K2 ) ) = BM(K1 ) .) As an isometry. We wish to show that φ is a homeomorphism. K1 K2 K2 (Here we use the fact that χK1 = 1 on K1 and χK2 = 1 on K2 . the extreme points in M(K1 ) and M(K2 ) are the Dirac masses. s ∈ K2 . T ∗ will map extreme points to extreme points. The adjoint T ∗ : M(K2 ) → M(K1 ) is also an isometry because (T ∗ )−1 = (T −1 )∗ . Observe that. φ(s) = ψ −1 (δφ(s) ) = ψ −1 (T ∗ δs ) = (ψ −1 ◦ T ∗ ◦ ϕ)(s).10 The Banach–Stone Theorem 123 are homeomorphic. It can be shown that these two facts imply that T ∗ μ ∈ P(K1 ). we can show that φ −1 = ϕ −1 ◦ (T −1 )∗ ◦ ψ. and as such we must show that both φ and φ −1 are continuous. and hence T ∗ : M(K2 ) → M(K1 ) is weak∗ -to- weak∗ continuous. there must be some ts ∈ K1 (depending on s) such that T ∗ δs = δts .5. let f be an extreme point in BC(K1 ) and suppose that g and h are functions in BC(K2 ) such that Tf = 21 (g + h). by Proposition 5. T ∗ δs = δφ(s) . If we define an operator S : C(K1 ) → C(K2 ) by Sf = (Tf )/T (χK1 ) for all functions f ∈ C(K1 ). Consequently. Then. which follows from the fact that T ∗ is a bijection from M(K2 ) onto M(K1 ). Define a map φ : K2 → K1 by φ(s) = ts for each s ∈ K2 . Then f = 21 (T −1 g + T −1 h). Finally. since χK1 (which is identically equal to 1 on K1 ) is extreme in BC(K1 ) . for all s ∈ K2 .57. Therefore. K2 K1 K1 The factor u appearing in the statement of the theorem does not appear now because of the normalization we made at the beginning of the proof. The map φ is a bijection from K2 onto K1 . We assumed T was continuous. To see this. the maps ψ and ϕ are homeomorphisms. and so φ −1 is continuous. and so for each s ∈ K2 . Observe that T maps extreme points of BC(K1 ) to extreme points of BC(K2 ) . By Lemma 5. In particular. Then ∗ ∗ (T μ)(K1 ) = 1d T μ = T (1) dμ = 1 dμ = μ(K2 ) = 1. and so Tf is an extreme point in BC(K2 ) whenever f is an extreme point in BC(K1 ) .) The measure T ∗ μ is then an element of BM(K1 ) such that T ∗ μ(K1 ) = 1. then S is an isometry such that S(χK1 ) = χK2 . (See Exercise 5. its image T (χK1 ) is extreme in BC(K2 ) . Suppose μ ∈ P(K2 ).44. φ is a homeomorphism. Had we not assumed . Therefore. By Proposi- tion 5. Similarly. we have that |T (χK1 )(s)| = 1 for all s ∈ K2 . It follows that g = h = Tf . we may assume without loss of generality that T (χK1 ) = χK2 . show that a sequence (xn )∞ n=1 in X cannot converge to two distinct limits x and x̃. . That is. (c) Every compact subset of a Hausdorff space is closed. then show there exist disjoint open sets U and V such that A ⊆ U and B ⊆ V .) Exercise 5. For example.1 Let a and b be real numbers such that a < b. If f : X → Y is continuous. (b) The image of a compact space under a continuous map is compact.) Exercise 5.) Exercise 5.) Exercises Exercise 5. Use Urysohn’s Lemma (Theorem 5.124 5 Consequences of Convexity T (χK1 ) = χK2 . That is. That is.6 to show that C(X) separates the points of X.2 Prove the following theorems: (a) Every closed subset of a compact space is compact. show there is a function f ∈ C(X) such that f (a) = 0 and f (b) = 1. (That is. which is a function with the property that |T (χK1 )(s)| = 1 for all s ∈ K2 . (d) Let X be a compact space and Y be a Hausdorff space. b) is not a compact set in R by finding an open cover with no finite subcover.7 Suppose X is a compact Hausdorff space.3 Let K be a compact topological space and suppose φ : K → E is a continuous one-to-one map. Exercise 5.8 Show that limits in a Hausdorff space are unique. (See [15] for more. so that no norm structure is required in the proof. (Use the standard topology on R.6 Suppose X is a compact Hausdorff space. Remark T here are other ways to prove that K1 and K2 are homeomorphic using properties of C(K1 ) and C(K2 ).9 Prove a metric space is second countable if and only if it is separable. f (C) is closed in Y whenever C is closed in X. Exercise 5. it is possible to prove that K1 and K2 are homeomorphic using only the fact that C(K1 ) and C(K2 ) are isomorphic as rings. then f is a closed map.4 Let X and Y be topological spaces. (Hint: Use Exercise 5. Exercise 5. where E is a Hausdorff topological space. show there exist disjoint open sets U and V such that E ⊆ U and F ⊆ V .65) and Exercise 5. Exercise 5. Show that φ is a homeomorphism onto its image φ(K).2. Show explicitly that (a. if E and F are disjoint closed subsets of X. If φ : X → Y is a continuous closed bijection. Show that X is a normal space. (Hint: See Exercise 5. then we would have u = T (χK1 ). If A and B are disjoint compact subsets of X. if X is a Hausdorff space.) Exercise 5.5 Let X be a Hausdorff space. if a and b are distinct points in X.5. show that φ is a homeomorphism. Verify that the topology on X is generated by the family of Minkowski functionals {pU }U ∈η . . where φ(x) = x/(1 + |x|).16 Suppose (Ω.) Exercise 5. define f 0 = exp log |f (ω)| μ(dω) . Furthermore. y) ∈ A × A defines a metric on A. then show f p ≤ f q for all measurable functions f . Deduce that xn → 0 in X if and only if pU (xn ) → 0 for all U ∈ η. g) = min 1. (b) Show that ρ(x. show that d(fn . Exercise 5. Prove that lim+ f p = exp log |f (ω)| μ(dω) . g) = f − gp . 0 Prove that d is a metric on L0 (0.) If d(f . Exercise 5. Define 1 d(f .17 Let X be a locally convex topological vector space with η a base of absolutely convex neighborhoods of 0. f (y)) for (x. (Hint: Use p Lemma 2.13. where 0 < p < 1. μ) is a positive measure space such that μ(Ω) = 1. Exercise 5. For any measurable function f . despite the fact that · p is not a norm.e. 1) is a topological vector space (i. ∞).15 Suppose (Ω. which still applies in this case. d) is a metric space and A is a set. does d define a metric on L0 (μ)? Exercise 5.Exercises 125 Exercise 5. y) = d(f (x).10 (a) Suppose (M. Show that d is a metric on R that is not complete. Ω (See Exercise 5. Let L0 (μ) denote the space of all (equivalence classes of) μ-measurable functions on Ω. |f (s) − g(s)| ds.) Exercise 5. g} ⊆ L0 (0. 1].11 Let d(x.4. p→0 Ω −∞ where we adopt the convention that e = 0.13 Let L0 (0.15. Conclude that L0 (0. 1) denote the space of all (equivalence classes of) Lebesgue measurable functions on [0. μ) is a positive measure space such that μ(Ω) = 1. 1). 1). Exercise 5. 1) is identically zero. (Compare to Exercise 2.. show that dA (x. show that addition and scalar multiplication are continuous). where f p = 0 |f (t)|p dt. (b) Assume that f is a measurable function such that f r < ∞ for some r ≤ 1.. {f . f ) → 0 if and only if f → 0 in measure. y) = |φ(x) − φ(y)|. y) = | log (y/x)| defines a metric on the set R+ = (0. g) = f − g0 for all measurable functions f and g in L0 (μ). If f : A → M is an injective function. (a) If 0 < p < q ≤ 1. 1).12 Show that the space Lp (0. is a complete metric p p 1 space with metric d(f .14 Show that any continuous linear functional on L0 (0. 126 5 Consequences of Convexity Exercise 5.18 Consider the set ∂B2 = {x ∈ 2 : x2 = 1}. Show that ∂B2 is closed in the norm topology, but not the weak topology on 2 . (This example shows that the convexity assumption cannot be omitted from Mazur’s Theorem.) Exercise 5.19 Let K be a compact subset of a Hausdorff topological vector space E, and suppose C is a closed subset of E. Show that C − K = {x − y : x ∈ C, y ∈ K} is a closed subset of E. Exercise 5.20 Let E be a locally convex topological vector space and suppose K is a closed linear subspace of E. If x0 ∈ K, show that there exists a continuous linear functional f ∈ E ∗ such that f (x0 ) = 1, but f (x) = 0 for all x ∈ K. Exercise 5.21 Let E be a real locally convex topological vector space. Suppose K is a nonempty compact convex subset of E, and C is a nonempty closed convex subset of E, and that K ∩ C = ∅. Show there is a continuous linear functional φ on E such that inf φ(x) > sup φ(y). x∈C y∈K (We say φ separates K and C.) Exercise 5.22 Let X be a real Banach space and let E be a weak∗ -closed subspace of X∗ . If φ is a weak∗ continuous linear functional on E with φ = 1, show for any > 0 there exists an x ∈ X with x < 1 + such that φ(e∗ ) = e∗ (x) for all e∗ ∈ E. (Hint: Consider the sets C = {e∗ ∈ E : φ(e∗ ) = 1} and K = (1 + )−1 BX∗ .) Exercise 5.23 Let (X, · ) be a real reflexive Banach space and let φ ∈ X∗ . Define a map f : X → R by f (x) = 21 x2 − φ(x) for all x ∈ X. Show that f attains a minimum value. Exercise 5.24 Let (fn )∞ n=1 be a bounded sequence in C[0, 1]. Show that fn (s) → 0 for every s ∈ [0, 1] if and only if fn → 0 weakly. Exercise 5.25 Let p ∈ [1, ∞) and for each n ∈ N let en be the sequence with 1 in the nth coordinate, and 0 elsewhere. Show that the sequence (nen )∞ n=1 does not converge weakly to 0 in p . (Compare to Example 5.28.) Exercise 5.26 Let p ∈ (1, ∞) and for each n ∈ N define a function fn : [0, 1] → R by fn (x) = n1/p χ[0,1/n] (x) for all x ∈ [0, 1]. Show that fn → 0 weakly in Lp (0, 1), but not in norm. (Recall that χA is the characteristic function of the measurable set A.) Exercise 5.27 Suppose K is a real compact Hausdorff space. Show that the set P(K) of regular Borel probability measures on K is a convex and w∗ -closed subset of M(K), the set of regular Borel measures on K. Show that P(K) is an extremal set in the unit ball of M(K). (See Sect. 5.8.) Exercise 5.28 Let K be a compact Hausdorff space and let ν be a Borel measure on K so that νM(K) ≤ 1 and ν(K) = 1. Show that ν is a probability measure. Exercises 127 Exercise 5.29 Let G be a group that is also a topological space. Show that G is a topological group if and only if the map g : G × G → G defined by g(x, y) = x −1 y is continuous. Exercise 5.30 Let X be a real separable Banach space. Show that BX∗ is metrizable in the weak∗ topology. (Hint: Let (xn )∞ n=1 be a countable dense subset in X and define φ(x ∗ ) = (x ∗ (xn ))∞ N n=1 ∈ R .) Exercise 5.31 Let X be a Banach space. If x ∈ X, use the Banach–Alaoglu Theorem to prove that there exists an element x ∗ ∈ X∗ such that x ∗ = 1 and x ∗ (x) = x. (Note: We proved this in Proposition 3.29 using the Hahn–Banach Theorem.) Exercise 5.32 A subset E of a topological vector space X is called bounded if for every open neighborhood V of 0, there exists an n ∈ N such that E ⊆ nV . Show that any compact subset of a topological vector space is bounded. Exercise 5.33 A topological vector space X has the Heine–Borel property if every closed and bounded subset of X is compact. (See Exercise 5.32 for the definition of a bounded set in a topological vector space.) (a) Show that a Banach space has the Heine–Borel property if and only if it is finite-dimensional. (Hint: Use Lemma 5.36.) (b) Show that (X∗ , w∗ ) has the Heine–Borel property if X is a Banach space. Exercise 5.34 Show that C[0, 1] is not reflexive by showing that BC[0,1] is not compact in the weak topology. (Hint: Find a Λ ∈ C[0, 1]∗ such that Λ(BC[0,1] ) is open.) Exercise 5.35 Let X be an infinite-dimensional Banach space. Show that (X∗ , w∗ ) is of the first category in itself. Chapter 6 Compact Operators and Fredholm Theory 6.1 Compact Operators Suppose X is a vector space (over R or C) and let T : X → X be a linear operator. Let us recall some basic definitions from linear algebra. The kernel (or nullspace) of T is the subspace of X given by ker(T ) = {x : T x = 0}. The range (or image) of T is given by ran(T ) = {T x : x ∈ X}. We say that x ∈ X is an eigenvector of T if x = 0 and there exists some scalar λ (called an eigenvalue) such that T x = λx. The behavior of linear operators on finite-dimensional vector spaces has been studied for a long time and is well understood. Some of the most basic and important theorems of linear algebra rely heavily on the dimension of the underlying vector space X. Consider the following well-known theorems. Theorem 6.1 Let X be a finite-dimensional vector space and suppose T : X → X is a linear operator. Then: (i) The map T is one-to-one if and only if T maps X onto X. (ii) dim(X) = dim(kerT ) + dim(ranT ) (Rank-Nullity Theorem). (iii) If X is a nontrivial complex vector space, then T has at least one eigenvector. Notice that (ii) implies (i), and certainly (ii) is dependent upon the underlying dimension of X. To see how (iii) depends on the finite-dimensionality of X, recall that eigenvalues can be calculated by solving the equation det (λI − A) = 0 for λ, where I is the identity matrix and A is any matrix representation of T . Because X is finite-dimensional, the expression det (λI − A) is a polynomial, and as such must have a root because C is algebraically closed (by the Fundamental Theorem of Algebra). These theorems rely on the finite-dimensionality of X, and so any attempt to generalize them to infinite-dimensional vector spaces requires careful consideration. In fact, as stated, Theorem 6.1 is false in a general Banach space. To see this explicitly, we now consider some examples, where the Banach spaces can be either real or complex. © Springer Science+Business Media, LLC 2014 129 A. Bowers, N. J. Kalton, An Introductory Course in Functional Analysis, Universitext, DOI 10.1007/978-1-4939-1945-1_6 130 6 Compact Operators and Fredholm Theory Example 6.2 Let {p, q} ⊆ [1, ∞] and consider the shift operators T : p → p given by T (ξ1 , ξ2 , ξ3 , . . .) = (0, ξ1 , ξ2 , ξ3 , . . .), (ξk )∞ k=1 ∈ p , and S : q → q given by S(η1 , η2 , η3 , . . .) = (η2 , η3 , η4 , . . .), (ηk )∞ k=1 ∈ q . The map T is clearly one-to-one, but certainly is not onto. On the other hand, the map S is onto, but clearly not one-to-one. We therefore cannot extend Theorem 6.1(i) to infinite dimensions. Since the spaces p and q have infinite dimensions, it is not obvious if the statement of Theorem 6.1(ii) has any significant meaning in its current form. To see that Theorem 6.1(iii) does not extend to infinite-dimensional vector spaces, suppose (ξk )∞k=1 is an eigenvector for T . Then there exists some scalar λ such that T (ξ1 , ξ2 , ξ3 , . . .) = λ(ξ1 , ξ2 , ξ3 , . . .), or (0, ξ1 , ξ2 , ξ3 , . . .) = (λξ1 , λξ2 , λξ3 , . . .). The only way this can happen is if ξk = 0 for all k ∈ N, contradicting the assumption that (ξk )∞ k=1 is an eigenvector. It follows that T does not have any eigenvectors. The map S, however, has many eigenvectors. (See Exercise 6.1.) In the case p = q, notice that S ◦ T is the identity map on p , but T ◦ S is not. It is also worth mentioning that if p and q are conjugate exponents (1/p + 1/q = 1), then S = T ∗ . Example 6.3 Let T : C[0, 1] → C[0, 1] be defined by Tf (x) = xf (x). It is clear that T is a bounded linear operator and that T ≤ 1. If λ is an eigenvalue for T , then there exists some function f ∈ C[0, 1] such that xf (x) = λf (x) for all x ∈ [0, 1]. It follows that (x − λ)f (x) = 0 for all x ∈ [0, 1]. This can happen only if f = 0. Therefore, T has no eigenvectors. We see that our intuition from linear algebra can fail us in infinite dimensions. Indeed, for general linear operators on Banach spaces, much of what we know for finite-dimensional vector spaces fails to remain true. For this reason, we impose additional conditions on our operators. We now define one class of operators for which our intuition can serve as a guide. Definition 6.4 Let X and Y be Banach spaces. A bounded linear operator T : X → Y is said to be a finite rank operator if dim(ranT ) < ∞. In such a case, the number dim(ranT ) is called the rank of T and is denoted rank(T ). For finite rank operators, we can still use some tools from linear algebra. This class of operators is too small, however, and so we wish to introduce a less restrictive condition on our operators. To that end, we introduce a definition. Definition 6.5 Let X be a topological space. A subset E of X is called relatively compact if it has compact closure; that is, if E is compact in X. n=1 n∈N n=1 Thus. It is important to note that the image of the unit ball under a compact operator need not be a compact set. the weak and weak∗ topologies on X coincide. assume λn > 0 for each n ∈ N. The proof of this fact relies on the assumption that X is reflexive. By the Banach–Alaoglu Theorem (Theorem 5.) Example 6. ∞ ∞ |T (ξ )| = λn ξn ≤ sup |ξn | λn = ξ c0 λ1 .1) ξ ∈Bc0 n=1 for all N ∈ N. however. and so T (BX ) is weakly compact in Y .6 Let X be a Banach space. the unit ball BX is weakly compact. (6. the set T (BX ) is compact if and only if T is a compact operator.10. (That is. by the Heine–Borel Theorem. A bounded linear map T : X → Y is called a compact operator if T (BX ) is a relatively compact set in Y .9 Suppose that X and Y are Banach spaces such that X is reflexive. and zero elsewhere.39). by Mazur’s Theorem (Theorem 5.42. In Example 6. and so T (BX ) is closed in the norm topology on Y . and so is relatively compact. T is continuous in the weak topologies on X and Y . The set T (Bc0 ) is not closed. Definition 6. BX is dense in BX∗∗ in the weak∗ topology on X ∗∗ . T = λ1 because N T = sup |T (ξ )| ≥ |T (e1 + · · · + eN )| = λn .39).1 Compact Operators 131 Example 6. For each ξ ∈ c0 . and let T : X → Y be a bounded linear operator. ξ = (ξn )∞n=1 ∈ c0 . (See Example 6.7 Let X and Y be Banach spaces. it is also compact because it is a rank 1 operator. the operator T is weakly continuous.9. if T is a compact operator. then T (BX ) may not be closed (and so not compact) even if T is a compact operator.45).6.10 Let λ = (λn )∞n=1 be a sequence in 1 and define a function T : c0 → R by ∞ T (ξ ) = λn ξn . Therefore.) We have shown that T is bounded and linear (and hence continuous). Example 6. Example 6. By Proposition 5. w∗ ). BX has weak∗ -compact closure. We know that T is weakly continuous. then T (BX ) is compact in the norm topology on Y . Therefore.8 Finite rank operators are compact.40). BX∗∗ is compact in the weak∗ topology on X∗∗ . We illustrate this in the next example. Observe that T is linear because the series defining T (ξ ) is absolutely convergent for all ξ ∈ c0 . by the Banach–Alaoglu Theorem (Theorem 5. If X is not reflexive. (Recall that ek is the sequence with a 1 in the k th coordinate. Therefore. because . n=1 For simplicity. in (X ∗∗ . It follows that T (BX ) is weakly closed in Y . In fact.) Since X is reflexive. T is bounded and T ≤ λ1 . By Goldstine’s Theorem (Theorem 5. a notion closely related to compactness is that of total boundedness.1 ) ∪ · · · ∪ B 1 (x2.n1 ). (See also Exercise 5. K1 can be covered by a finite collection of closed balls of radius 1/2.n2 }. Let K2 = K1 ∩ B 1 (x2 ).) In the case of a metric space. Definition 6. .43. . see Example 5. Thus. but there is no ξ ∈ c0 for which this maximum is ξ ∈Bc0 attained. Since K is totally bounded. y) ≤ ε}. say K ⊆ B 1 (x1. then any subset of E is also totally bounded.34. K is totally bounded. . we will denote the open ball of radius ε about x in M by Bε (x) and the closed ball of radius ε about x in M by B ε (x). Let K1 = K ∩ B 1 (x1 ). there is a finite set {x1 . The set K1 is totally bounded (because it is a subset of K). y) < ε} and B ε (x) = {y ∈ M : d(x. The collection of sets {Bε (x) : x ∈ K} forms an open cover of K. and so (by compactness) admits a finite subcover. such that K1 ∩ B 1 (x2 ) cannot be covered by 2 finitely many sets in V. .n2 ). . call it x1 . .n1 }. x2.132 6 Compact Operators and Fredholm Theory sup |T (ξ )| = λ1 . .1). Certainly. λ1 ). Thus. x1. For ease of notation. there is some member of the collection {x2.1 . if a subset E of a metric space is totally bounded. We will assume K is not compact and derive a contradiction. For another example of a compact operator T : X → Y such that T (BX ) is not closed in Y . T (Bc0 ) = ( − λ1 . Theorem 6. which is not closed. say K1 ⊆ B 1 (x2. by (6. 2 . . call it x2 . such that K ∩ B 1 (x1 ) cannot be covered by finitely many sets in V. Now suppose that K is a totally bounded closed subset of M. 2 2 Since K1 cannot be covered by finitely many sets in V. That is.12 A closed subset of a complete metric space is compact if and only if it is totally bounded. . . In fact. . We wish to show that K is compact. there is some member of the collection {x1. Let V be a collection of open sets that cover K and suppose V does not contain a finite subcover of K. Suppose K is a compact (and hence closed) subset of M. . Let ε > 0 be given.1 ) ∪ · · · ∪ B 1 (x1.1 . but has a compact closure.11 A subset of a metric space is called totally bounded if it can be covered by finitely many closed balls of radius ε for any ε > 0. it can be covered by a finite collection of closed balls of radius 1. Thus. xN } in K such that K ⊆ Bε (x1 ) ∪ · · · ∪ Bε (xN ) ⊆ B ε (x1 ) ∪ · · · ∪ B ε (xN ). Bε (x) = {y ∈ M : d(x. Proof Let M be a complete metric space with metric d. Since K cannot be covered by finitely many sets in V. n (ii) K ⊇ K1 ⊇ K2 ⊇ K3 ⊇ · · · . X). We wish to show that the operator αS + βT is compact. and (iii) Kn cannot be covered by finitely many sets in V for any n ∈ N. v) ∈ Y × Y. Y ) such that Tn − T → 0 as n → ∞. The next theorem shows that the collection of compact operators is well-behaved. For each n ∈ N. Theorem 6. Define a map φ : Y × Y → Y by φ(u. Y ) is a bounded linear operator and let (Tn )∞ n=1 be a sequence of compact operators in K(X. yN } in Y such that N ε Tn (BX ) ⊆ yj + B Y . Y ) of compact operators from X to Y forms a closed linear subspace of L(X. Y ). By assumption. Y ) is a closed subspace. This will follow if we show. Y ). there exists some N ∈ N such that B 1 (xn ) ⊆ V for all n ≥ N . there is a point y ∈ K (because K is closed) such that yn → y as n → ∞. the cover V contains a finite subcover of K. and H is compact. It remains to show that K(X. We chose yn ∈ Kn for each n ∈ N. By assumption. choose yn ∈ Kn . and so K is compact. We use K(X) to denote K(X. there is an open set V ∈ V such that y ∈ V . because Kn ⊆ B 1 (xn ) for n n all n ∈ N. Definition 6. Hence. Since (αS + βT )(BX ) ⊆ H . for any ε > 0. . and so αS + βT is a compact operator. We wish to show that T (BX ) is relatively compact. (u. it follows that (αS + βT )(BX ) is compact. Y ) is a linear subspace. as the continuous image of a compact set.6. . Therefore. it suffices to show that its closure is totally bounded. Since M is assumed to be a complete metric space. . the set S(BX )×T (BX ) is compact in Y ×Y . Because Tn (BX ) is relatively compact. Then the sequence (yn )∞ n=1 is a Cauchy sequence.1 Compact Operators 133 Continuing inductively. By Theorem 6. The map φ is continuous because Y is a topological vector space. There exists an n ∈ N such that Tn − T < ε/2. there exists a finite set of points {y1 . by the properties in (i) and (ii). Fix ε > 0. we find a sequence of points (xn )∞ n=1 in M and construct a sequence of subsets (Kn )∞ n=1 of K such that (i) Kn ⊆ B 1 (xn ) for all n ∈ N. the set H = φ(S(BX )×T (BX )) is compact in Y . Suppose that S : X → Y and T : X → Y are compact operators and let α and β be scalars. Proof First.12. The set K(X. V is an open cover of K and y ∈ K. Suppose T ∈ L(X. the set T (BX ) is contained in a finite union of balls of radius ε in Y . we will show that K(X. v) = αu + βv. and so it follows from (i) that xn → y as n → ∞. j =1 2 . and so we have obtained a contradiction. . Therefore. Since xn → y as n → ∞. above.14 Let X and Y be Banach spaces. This violates property (iii).13 We denote the collection of compact operators from X to Y by the symbol K(X. Denote the integers in N1 by (n1j )∞ j =1 .. we can cover K using a finite number of open balls of radius 21 . where n1j < n1k whenever j < k. 2 Remark 6. all the terms of which are in the ball B 1 (x2 ). Observe that (an1j )j =1 is a ∞ ∞ subsequence of (an )n=1 . Suppose K is a compact set and let (an )∞ n=1 be a sequence in K. Y ) is a subspace of L(X.15 There is a more straightforward proof that K(X. x) < δ}. where 2 n2j < n2k whenever j < k. Proof Let d be a complete metric on M. At least one of these balls contains an1j for infinitely many j ∈ N. in a metric space. The theorem remains valid even without the completeness assumption. Therefore. and so we opt for a theorem with a simpler proof. We define a subsequence (ani )∞ ∞ i=1 of (an )n=1 by letting ani = anii for each i ∈ N. there is a x1 ∈ K such that an ∈ B1 (x1 ) for infinitely many n ∈ N. that is. as required. anj ) < + = ε. T (BX ) is contained in a finite union of ε-balls. and hence compact. and consequently are in the open ball B 1 (xN ). because K is compact. all the terms of which are in B1 (x1 ). for each i ∈ N we find xi ∈ K and a subsequence (anij )∞ j =1 of (an(i−1)j )∞ j =1 such that anij ∈ B 1i (xi ). every sequence has a convergent subsequence). and we obtain the desired result. anj ) ≤ d(ani . then ani and anj are members of the N th subsequence.e. We will prove this shortly (for complete metric spaces). ε ε d(ani . let ε > 0. we will denote by Bδ (s) the open ball of radius δ about s ∈ M. We will show that (an )∞ n=1 has a convergent subsequence. That is. xN ) + d(xN . For notational simplicity. To see this. A closed subset K of M is compact if and only if it is sequentially compact. There exists a subsequence (xnk )k=1 such that ∞ (Sxnk )k=1 converges (not necessarily in S(BX )). Choose a positive integer N ∈ N such that N1 < 2ε . Observe that (an2j )∞ j =1 is a subsequence of (a ∞ n1j )j =1 . and so T x ∈ N j =1 (yj + εBY ). Since K is compact. Then ∞ ((αS + βT )(xnkj ))j =1 converges. Y ) that uses the fact that. Let (xn )∞ ∞ n=1 be a sequence in BX . then Tn x − T x < ε/2. At least one of these must contain an for infinitely many values of n ∈ N. we can cover K with a finite number of open balls of radius 1. Bδ (s) = {x ∈ M : d(s. Let N1 = {n : n ∈ N and an ∈ B1 (x1 )}. T is a compact operator.16 Let M be a complete metric space. not necessarily in T (BX )). Thus. If i > N and j > N. There exists a further subsequence (xnkj )∞ ∞ j =1 such that (T xnkj )j =1 converges (again. compactness is equivalent to sequential compactness (i. We conclude that T (BX ) is totally bounded. We will now show that compactness is equivalent to sequential compactness in a complete metric space.134 6 Compact Operators and Fredholm Theory If x ∈ BX . ∞ We claim that (ani )i=1 is a Cauchy sequence. By N the triangle inequality. there exists a x2 ∈ K such that an1j ∈ B 1 (x2 ) for infinitely many j ∈ N. 2 Continuing inductively. Theorem 6. but for now let us take this fact for granted. Let 2 N2 = {n1j : j ∈ N and an1j ∈ B 1 (x2 )}. Denote the integers in N2 by (n2j )∞ j =1 . As before. That is. but we need it only for Banach spaces. 2 2 . 6.1 Compact Operators 135 We have established that the subsequence (ani )∞ i=1 is a Cauchy sequence. By assumption, M is a complete metric space, and so the Cauchy sequence (ani )∞ i=1 converges to some point in the closed set K. Since any sequence in K has a convergent subsequence, K is sequentially compact. Now suppose K is sequentially compact. We will show that K is totally bounded. (Then K will be compact by Theorem 6.12.) Suppose to the contrary that K is not totally bounded. Then there exists some ε > 0 such that K cannot be covered by finitely many closed balls with radius ε. Pick any a1 ∈ K. Since K cannot be covered by finitely many ε-balls, the set K\Bε (a1 ) is nonempty. Choose some a2 ∈ K\Bε (a1 ). Similarly, K\(Bε (a1 ) ∪ Bε (a2 )) cannot be empty, and must contain some element, say a3 . Proceeding inductively, we construct a sequence of points (an )∞ n=1 in K such that an+1 ∈ Bε (a1 ) ∪ · · · ∪ Bε (an ) for all n ∈ N. It follows that d(an , am ) ≥ ε for all m = n, and so (an )∞ n=1 cannot have a convergent subsequence. This contradicts the assumption that K is sequentially compact. Therefore, K is totally bounded, and so is compact (by Theorem 6.12). We return now to the space of compact operators between Banach spaces X and Y . In addition to being a closed subspace of L(X, Y ), the collection of compact operators K(X, Y ) also possesses an ideal structure. Theorem 6.17 Let W , X, Y , and Z be Banach spaces. If T : X → Y is a compact operator and the maps A : W → X and B : Y → Z are bounded linear operators, then the composition B ◦ T ◦ A : W → Z is compact. In particular, the space K(X) is a two-sided ideal in L(X). Proof The map A is bounded, and consequently A(BW ) ⊆ A BX . We thus conclude that T A(BW ) ⊆ AT (BX ), the latter set being compact, since T is a compact operator. Finally, BT A(BW ) ⊆ A B T (BX ) . The set on the right of the previous inclusion is compact as the continuous image (under B) of a compact set. Therefore, BT A(BW ) is a closed subset of a compact set, and so is compact. Example 6.18 Suppose X is a Banach space. The identity map Id : X → X is a compact operator if and only if dim(X) < ∞. Indeed, if A : X → X is any invertible bounded linear operator, then A is compact if and only if dim(X) < ∞. This follows from Proposition 5.37. Example 6.19 Let p ∈ [1, ∞] be given and let a = (aj )∞ j =1 be a bounded sequence (i.e., an element of ∞ ). Define a map Ta : p → p by Ta (ξ1 , ξ2 , ξ3 , . . . ) = (a1 ξ1 , a2 ξ2 , a3 ξ3 , . . . ), (ξj )∞ j =1 ∈ p . We claim this map is a bounded linear operator. Linearity is clear. To see it is bounded, let ξ = (ξj )∞ j =1 ∈ p . Then ⎛ ⎞1/p ∞ Ta ξ p = ⎝ |aj ξj |p ⎠ ≤ a∞ ξ p . j =1 136 6 Compact Operators and Fredholm Theory This implies that Ta ≤ a∞ . In fact, Ta (ei ) = ai ei for all i ∈ N, and so it follows that Ta = a∞ . (Note that ei is an eigenvector with corresponding eigenvalue ai for each i ∈ N.) Proposition 6.20 Ta is compact if and only if a ∈ c0 . Proof If Ta is a compact operator, then Ta (Bp ) is a compact set, and consequently any sequence in Ta (Bp ) must have a convergent subsequence. Suppose that a ∈ c0 . Then there exists some δ > 0 and a subsequence (anj )∞ j =1 such that |anj | > δ for all j ∈ N. Then, for all natural numbers j and k, 1/p Ta enj − Ta enk p = anj enj − ank enk p = |anj |p + |ank |p > δ. It follows that (Ta enj )∞ j =1 has no convergent subsequence, and so Ta is not a compact operator. Now suppose a = (aj )∞ j =1 ∈ c0 . For each n ∈ N, define a sequence in c0 by the rule a (n) = (a1 , . . . , an , 0, . . . ). The sequence a (n) has only finitely many nonzero terms, and so Ta (n) is a finite rank operator, and hence is compact. For each n ∈ N, Ta − Ta (n) = Ta−a (n) = sup |ak |. k>n Since a ∈ c0 , this quantity tends to 0 as n → ∞. Thus, Ta is the uniform limit of a sequence of finite rank operators. Therefore, Ta is compact, because K(p ) is a closed subspace of L(p ), by Theorem 6.14. 2 Example 6.21 Let K be a continuous scalar-valued function on the compact space [0, 1] × [0, 1]. Define TK : C[0, 1] → C[0, 1] by 1 TK f (s) = K(s, t) f (t) dt, f ∈ C[0, 1], s ∈ [0, 1]. 0 We must show that TK is well-defined; that is, we must show that TK f is a continuous function on [0, 1]. If sn → s as n → ∞, then 1 1 K(sn , t) f (t) dt −→ K(s, t) f (t) dt, 0 0 as n → ∞, by the Dominated Convergence Theorem. Consequently, TK f is continuous on [0, 1], by the sequential characterization of continuity. It is clear that TK is linear. To compute the bound, let f ∈ C[0, 1]. Then 1 |TK f (s)| ≤ |K(s, t) f (t)| dt ≤ K∞ f ∞ . 0 This bound is uniform in s ∈ [0, 1], and so TK f ∞ ≤ K∞ f ∞ . Therefore, TK is bounded and TK ≤ K∞ . Proposition 6.22 If K is a continuous scalar-valued function on [0, 1]×[0, 1], then TK is a compact operator. 6.1 Compact Operators 137 Proof Suppose K is a polynomial on [0, 1] × [0, 1], say K(s, t) = nj=1 aj s mj t nj , where (mj )nj=1 and (nj )nj=1 are finite sequences in N, and where (aj )nj=1 is a finite sequence of scalars. Then n 1 TK f (s) = aj s m j t nj f (t) dt. j =1 0 Thus, TK f is a polynomial in the linear span of {s m1 , . . . , s mn }. This holds for all f ∈ C[0, 1], and so TK is a finite rank operator, and hence compact. Now suppose K ∈ C([0, 1] × [0, 1]) is a continuous function. By the Weierstrass Approximation Theorem, there exists a sequence of polynomials (Kn )∞ n=1 such that K − Kn ∞ → 0 as n → ∞. Then TK − TKn = TK−Kn ≤ K − Kn ∞ → 0, as n → ∞. Therefore, T is the limit of a sequence of finite rank operators, and so is compact (by Theorem 6.14). Digression: Historical Comments The linear operator TK from Example 6.21 was considered by Fredholm in 1903, in what is considered by some to be the first paper on functional analysis [12]. The function K is called the Fredholm kernel (or simply the kernel) of the operator TK (not to be confused with the nullspace kerTK ). Fredholm was interested in solving integral equations of the form 1 g(s) = K(s, t) f (t) dt, 0 where f , g, and K had specified properties. This integral equation is an extension of the matrix equation n yj = aj k xk , k=1 where A = is an n × n matrix (for some n ∈ N), x and y are n-dimensional (aj k )nj,k=1 vectors, and y = Ax. Fredholm’s work was before the advent of measure theory, and so the focus was on the space of continuous functions on [0, 1]. The space C[0, 1] is an infinite- dimensional vector space, and integral operators determine linear operators on this vector space. Fredholm set about trying to apply techniques of linear algebra to these kinds of operators. With the advent of measure theory and Riesz’s work on Lp -spaces, the focus on integral operators broadened to include these larger spaces of functions. For any p in the interval [1, ∞], we define a Fredholm operator on Lp (0, 1) to be any operator TK : Lp (0, 1) → Lp (0, 1) of the form 1 TK f (s) = K(s, t) f (t) dt, f ∈ Lp (0, 1), s ∈ [0, 1], (6.2) 0 138 6 Compact Operators and Fredholm Theory where K is a prescribed measurable function on [0, 1] × [0, 1]. The function K is known as the kernel of TK . (It is also variously known as the nucleus or the Green’s function for TK .) In order to insure that TK is well-defined, the kernel K generally has some additional assumptions imposed upon it. For example, in the following proposition, which can be seen as an extension of Proposition 6.22, the kernel K is assumed to be an essentially bounded measurable function. Proposition 6.23 If K ∈ L∞ ([0, 1] × [0, 1]), then TK is a compact operator on Lp (0, 1) for 1 < p < ∞. Proof Let p ∈ (1, ∞) and suppose f ∈ Lp (0, 1). Then 1 1/p 1 TK f p = |TK f (s)|p ds ≤ TK f ∞ = sup K(s, t)f (t) dt . 0 s∈[0,1] 0 Thus, TK f p ≤ K∞ f 1 ≤ K∞ f p . (Note that · 1 ≤ · p on a probability space, by Hölder’s Inequality.) Therefore, TK f ∈ Lp (0, 1) whenever f ∈ Lp (0, 1) and TK f p ≤ K∞ f p . We have established that TK is well-defined and TK ≤ K∞ . To prove that TK is compact, we use an argument similar to the one we used in Proposition 6.22. If K = χA×B , where A and B are measurable subsets of [0, 1], then for all f ∈ Lp (0, 1) and s ∈ [0, 1], 1 TK f (s) = χA (s) χB (t) f (t) dt = f (t) dt χA (s). 0 B Thus, TK f ∈ span{χA }, and so TK is a rank-one operator. Therefore, TK is compact. Now suppose n K= ci χAi ×Bi , (6.3) i=1 where n ∈ N and, for each i ∈ {1, . . . , n}, the sets Ai and Bi are measurable and ci is a scalar. Then for all f ∈ Lp (0, 1) and s ∈ [0, 1], 1 n n TK f (s) = ci χAi (s) χBi (t) f (t) dt = ci f (t) dt χAi (s). 0 i=1 i=1 Bi Consequently, TK f ∈ span{χA1 , . . . , χAn }, and so TK is an operator of rank at most n. Thus, TK is compact. If K is a bounded measurable function on [0, 1] × [0, 1], then the compactness of TK follows from a density argument. Since K is a bounded function, it is also true the inequality becomes 1 1 1/2 TK f 2 ≤ |K(s. 1] × [0. In the proof of Proposition 6. where r = max{p. t)|q dt ds f p . (6. 1]) for all r ∈ (1. it follows that TK f = lim TKn f for all f ∈ Lp (0.4) 0 0 where 1/p + 1/q = 1. t)|q dt)1/q ≤ ( 0 |K(s. q}. all of the type in (6. 1) → Lp (0. then for any s ∈ [0. 0 0 0 Thus.5) 0 0 . t)|p dt)1/p . 1). For each n ∈ N.2). Therefore.23. 1).) If we choose K so that the above quantity is finite. 1) for all f ∈ Lp (0. t)|p dt ds f p = Kp f p . 1 1 1/p TK f p ≤ |K(s. Functions of the type in (6.3) are dense in Lr ([0. 1] × [0. t)|q dt ds f p = Kq f p . we saw that 1 1 p/q 1/p TK f p ≤ |K(s. Then 1 1/p 1 1/q TK f p ≤ A(s) ds p f p ≤ A(s)q ds f p 0 0 1 1 1/q = |K(s.4) is bounded by Kr f p .24 Let us explore which kernels we can use to define a Fredholm operator as in (6. 1) to be well-defined. Observe that 1 1/p 1 1 p 1/p TK f p = |TK f (s)|p ds ≤ |K(s.3). t)|2 dt ds f 2 = K2 f 2 . Example 6. ∞). 0 0 Therefore.6. q}. An important case is when p = 2 (and thus q = 2). t)|q dt)1/q . Thus. we have 1 1 p/q 1/p TK f p ≤ |K(s. In order for TK : Lp (0. 0 0 (See (6. 1]).14. then TK will be a well-defined bounded linear operator. 0 0 1 Now suppose p ≤ q and let A(s) = ( 0 |K(s. t) f (t)| dt ds . (6. Suppose that q ≤ p. by Theorem 6. where r = max{p. we 1 1 have that ( 0 |K(s.4). there exists a sequence of kernels (Kn )∞n=1 . 1]). 1]. TK is the limit of a sequence of finite rank operators. where the limit is in the norm on Lr ([0. Consequently. 1] × [0. t)|q dt ds f p . such that K = n→∞ lim Kn . applying Hölder’s Inequality to the inner integral. and so TK is compact. We will show that the right side of (6. 1).1 Compact Operators 139 that K is in Lr ([0. TK f p ≤ Kr f p . we must ensure that TK f ∈ Lp (0. Because TK f p ≤ Kr f p . In this case. the operator n→∞ TKn is a finite rank operator on Lp (0. then trace(A) = 0. t) K(t. i=1 It is known that if A = 0. Proposition 6. For a complete list of problems and prizes. known as the “Approximation Problem.) Perhaps one of the most unexpected equivalences is the following. when the kernel K is continuous. . a live goose was considered very valuable. 1) to L2 (0. Famously.1. 1]).25 Each compact operator between arbitrary Banach spaces is the limit of a sequence of finite rank operators if and only if the following is true: 1 If K ∈ C([0. Recall that the trace of the matrix A is n trace(A) = aii . if K ∈ L2 ([0. 1]. u) dt = 0 for all {s.140 6 Compact Operators and Fredholm Theory Therefore. Per Enflo was awarded a live goose by Stanislaw Mazur for solving the Approximation Problem. 0 1 then K(s. it was one of Grothendieck’s alternate formulations that Enflo disproved. 0 (6. 1]) and K(s. j =1 i=1 Proposition 6. In 1936. (Many problems included in the Scottish Book came with the promise of a reward for a solution. 1] × [0. 1) and TK ≤ K2 .6) comes from the finite-dimensional theory of linear algebra. (In fact. Let n ∈ N and suppose A = (aj k )nj. n} ⇒ aii = 0. In the examples given above. u} ⊆ [0. . Alexander Grothendieck studied compact operators in depth [17]. .” motivated much of the early development of functional analysis. see [24].25 suggests that a similar result might be true for the Fredholm operator TK . We will say more about these operators in Chapter 7. in 1955. s) ds = 0. That is. It was not until 1973 that Per Enflo settled the issue and proved that not every compact operator was the limit of a sequence of finite rank operators [9]. especially those that were considered difficult or important. then TK is a bounded linear operator from L2 (0. We define the trace of the Fredholm operator . k} ⊆ {1. when Mazur included a problem in the Scottish Book [Problem 153] that was equivalent to the Approximation Problem. 1] × [0.k=1 is an n × n matrix. An operator of this type is known as a Hilbert–Schmidt operator. He developed several conditions which were equivalent to the statement that every compact operator is the limit of a sequence of finite rank operators. each operator was shown to be compact by demon- strating that it was the limit (in the operator norm) of a sequence of finite rank operators.6) The motivation behind (6.) Much earlier. . This leads to a natural question: Is it true that every compact operator is a limit (in the operator norm) of a sequence of finite rank operators? This question. 2 n n aij aj k = 0 for all {i. 1] ⇒ K(s. u) dt = 0 for all {s. It can be shown. Proof By the Banach–Alaoglu Theorem (Theorem 5. T is a compact operator.6. Consequently. u} ⊆ [0. where δ > 0. . We need to show that V is a w∗ -neighborhood of y0∗ relative to BY ∗ . 1] × [0. . we know that BY ∗ is w∗ - compact. · ) is a continuous map. Let V denote the preimage of this set in BY ∗ .2 A Rank-Nullity Theorem for Compact Operators We resume the study of compact operators with a theorem of Schauder that guarantees the compactness of the adjoint of a compact operator. that is. Theorem 6. we must find a w∗ -open set W (open in the w∗ -topology of Y ∗ ) containing y0∗ such that W ∩ BY ∗ ⊆ V .12.39). that the implication does hold if K is Hölder continuous with exponent α for all α > 1/2. . 1 1 K(s. this implication does not hold for all K ∈ C([0. V = (T ∗ )−1 (T ∗ y0∗ + δBX∗ ) ∩ BY ∗ . and so T (BX ) is a relatively compact set in Y . s) ds. . that is. and hence there exists a finite set {x1 . then T ∗ : Y ∗ → X∗ is also compact. because then T ∗ (BY ∗ ) will be compact as the continuous image of a compact set. s) ds = 0. (6. (6. xn } ⊆ BX such that n δ T (BX ) ⊆ T x j + BY . 1]). however. then if K is a continuous kernel.26 (Schauder’s Theorem) If X and Y are Banach spaces and the operator T : X → Y is compact. 6.7) 0 Proposition 6. By Theorem 6. 0 0 Since it is not true that each compact operator between arbitrary Banach spaces is the limit of a sequence of finite rank operators. t) K(t. a relatively compact set is totally bounded. Suppose that y0∗ ∈ BY ∗ and consider the closed neighborhood of T ∗ y0∗ given by ∗ ∗ T y0 + δBX∗ . it suffices to show that T ∗ : (BY ∗ . w∗ ) → (X ∗ .8) j =1 3 . By assumption.2 A Rank-Nullity Theorem for Compact Operators 141 TK to be 1 trace(TK ) = K(s.25 states that if each compact operator between arbitrary Banach spaces is the limit of a sequence of finite rank operators. we cannot extend this to the entire space Y ∗ . By the triangle inequality. f ∈ L2 (0. w∗ ) is always continuous. however. (See Exercise 6. 1]. w∗ ) to (X ∗ . 0 This operator. |(T ∗ y ∗ − T ∗ y0∗ )(x)| = |(y ∗ − y0∗ )(T x)| ≤ |(y ∗ −y0∗ )(T xj )| + |(y ∗ −y0∗ )(T x −T xj )|. 1). Define a map V : L2 (0. · ) is continuous.142 6 Compact Operators and Fredholm Theory Define a w∗ -open neighborhood W of y0∗ by ! δ " W = y ∗ ∈ Y ∗ : |y ∗ (T xj ) − y0∗ (T xj )| < .41). is a Hilbert– Schmidt operator with kernel . 3 We claim that W ∩ BY ⊆ V . w∗ ) → (X∗ . j ∈ {1. · ).44. we did not have to take the closure of T ∗ (BY ∗ ) in X ∗ to get a compact set. We conclude that T ∗ : (BY ∗ . x ∈ [0.8). And since y ∗ and y0∗ are in BY ∗ . w∗ ) → (X∗ . and hence T ∗ (BY ∗ ) is compact as the continuous image of the weak∗ -compact set BY ∗ . Because y ∗ ∈ W . In general. Let L2 (0. In the proof of Theorem 6. we conclude that 2δ |(y ∗ − y0∗ )(T x − T xj )| ≤ y ∗ − y0∗ T x − T xj ≤ . To that end.3.27 (Volterra operator) Our purpose for adding structure to our linear operators was to recover some of the properties of linear operators on finite-dimensional vector spaces.26. 1) by x Vf (x) = f (t) dt. by Proposition 5. By (6. . |(T ∗ y ∗ − T ∗ y0∗ )(x)| ≤ δ for all x ∈ BX .26. it follows that |(y ∗ − y0∗ )(T xj )| < δ/3. · ) is continuous. . known as the Volterra operator (see Example 3. as required. we will show ∗ T ∗ (W ∩ BY ∗ ) ⊆ T ∗ y0∗ + δBX∗ . w∗ ) → (X∗ . 1]. We will now provide a compact operator on a complex Banach space with no eigenvalues. which assures us that BY ∗ is always compact in the weak∗ topology. 3 Consequently. That is. there exists some j such that T x − T xj ≤ δ/3. Example 6. we cannot say that T ∗ : (Y ∗ . we do know that T ∗ : (Y ∗ . This is a direct consequence of the Banach– Alaoglu Theorem. Let x ∈ BX . and so T ∗ y ∗ − T ∗ y0∗ ≤ δ. Notice that in the proof of Theorem 6. 1) be the complex Banach space of square-integrable functions on [0. we demonstrated the continuity of T ∗ when viewed as a map from (BY ∗ . n} .) On the other hand. . 1) → L2 (0. . Pick y ∗ ∈ W ∩ BY ∗ . We need to show that T ∗ y ∗ − T ∗ y0∗ ≤ δ. K(x. t) = 0 if t > x. . Since a Hilbert–Schmidt operator is always compact (see Example 6. 1 if t ≤ x. we conclude that V is a compact operator.24). If λ = 0. We refer to the map L = λI as a scaling of the identity. Then xn ≤ 2αn . Therefore. and I : X → X is the identity operator on the Banach space X. where λ is a nonzero scalar.18. (6. This implies that f = 0a. we will be able to prove a version of the Rank-Nullity Theorem. kerT = 1. we can recover some theory from the finite- dimensional case. 1]. and hence T xn xn lim = 0 and d .e. assume that αn → ∞ as n → ∞. we may assume that αn > 0 for all n ∈ N.28 Suppose L is a bounded linear operator. Proof On kerT . f (0) = λ1 Vf (0) = 0.29 Let T = λI − K.6. By the definition of xn . we will need some preparation. In particular. Then we may divide by αn . Furthermore. Then Vf (x) = λ f (x) for almost every x x ∈ [0.. we have d(xn . However. for a constant C ∈ C. The general solution to this differential 1 equation is f (x) = C ex/λ . Let xn = xn − zn . then the operator L − K is called a compact perturbation of L or a perturbation of L by a compact operator. (See Example 6.2 A Rank-Nullity Theorem for Compact Operators 143 Suppose that λ ∈ C is an eigenvalue. kerT ) = d(xn . Proof Recall that the range of T is denoted by ranT .e. x Suppose λ = 0. 1].30 Let T = λI − K.. Let y ∈ ranT . where λ is a nonzero scalar and K is a compact operator on a Banach space X. so that T is a compact perturbation of a scaling of the identity. kerT ) = αn . Lemma 6. f = 0. Without loss of generality. and so 0 f (t) dt = 0 for almost every x. we will be interested in L = λI . we have the equality K = λI . T xn = T xn → y as n → ∞. We claim the sequence (αn )∞n=1 is bounded. For each n ∈ N.9) n→∞ αn αn . let αn = d(xn . then Vf = 0a. Definition 6. Then f is differentiable and f (x) = λ f (x) for all x ∈ [0. Then there exists for each n ∈ N some zn ∈ kerT such that xn − zn ≤ 2αn . Observe that kerT = {x : Kx = λx}. In what follows. and so λ = 0 is not an eigenvalue. To prove this. and so x ∈ kerT if and only if x is an eigenvector of K with eigenvalue λ (when x = 0). and so dim(kerT ) < ∞. Then f (x) = λ1 0 f (t) dt for almost every x. If K is a compact operator. For the moment. and soλ = 0 is not an eigenvalue. Pick (xn )∞ n=1 in X such that T xn → y as n → ∞. Despite the previous example. kerT ) = inf{xn − z : z ∈ kerT }. we let K : X → X be a compact operator and define T = λI − K.) 2 Lemma 6. and so C = 0. The range of T is a closed subspace of X. This implies that I |kerT is a compact operator. where λ is a nonzero scalar and K is a compact operator on a Banach space X. The set kerT is finite-dimensional. In order to articulate the theorem. by Theorem 6.26 (Schauder’s Theorem) and Lemma 6. The quotient X/ranT is finite-dimensional. then limk→∞ d(xn k /αnk .31 Let T = λI − K. It follows that αn → ∞ as n → ∞.9). Consequently. it follows that T ∗ = λI ∗ − K ∗ . there is a subsequence (xn k /αnk )∞ k=1 such that lim k→∞ K(xnk /αnk ) exists. and hence x ∗ ∈ ker(T ∗ ). we see that xn xnk x nk λ· k =T +K . xn /αn ≤ 2. dim(X/ranT )∗ = dim(ker(T ∗ )) < ∞. Then λxn k = T (xn k ) + K(xn k ) −−−→ y + ν. Since T = λI − K.) 2 . y+ν T = lim T (xn k ) = y. (See Exercise 6. But this means T ∗ x ∗ = 0. and using the limit in (6. there is a convergent subsequence (K(xn k ))∞ ∞ k=1 of (K(xn ))n=1 . We know xn ≤ 2C for all n ∈ N. we have that (X/ranT )∗ = (ranT )⊥ = {x ∗ ∈ X∗ : x ∗ (T x) = 0 for all x ∈ X}. and thus we conclude that the range of T is closed. Lemma 6. and so there must be some C > 0 such that |αn | ≤ C for all n ∈ N. Our assumption was that y ∈ ranT . λ k→∞ and hence y ∈ ranT .144 6 Compact Operators and Fredholm Theory Furthermore. Once again. which contradicts the second equality in (6. (X/ranT )∗ = ker(T ∗ ). k→∞ αn k If ν ∈ kerT . Thus. Proof By Proposition 3. Thus. since K is a compact operator.11. The statement that x ∗ (T x) = 0 for all x ∈ X is equivalent to the statement that (T ∗ x ∗ )(x) = 0 for all x ∈ X. Call this limit ν. The result follows. k→∞ Miraculously. kerT ) = 0. Hence. Suppose limk→∞ K(xnk ) = ν.51. Therefore. since dim(X/ranT ) = dim(X/ranT )∗ . The same argument shows that no subsequence of (αn )∞ n=1 can tend to infinity. we invoke the compactness of the operator K. αn k αn k αn k Letting k → ∞. where λ is a nonzero scalar and K is a compact operator on a Banach space X. Recalling that T = λI − K.29.9). the set ker(T ∗ ) is finite-dimensional. we see that λ xn k /αnk → ν as k → ∞. But this implies that ν ∈ kerT : x nk T (ν) = lim λ T = 0. we extract a corollary that is of independent interest. for each n ∈ N. Then there exists an e ∈ E such that ε 1 u − e = d(u. the Heine–Borel property asserts the existence of a convergent subsequence.32 If X is a Banach space and T is a compact perturbation of a scaling of the identity. 1−ε 1−ε Observe that u − e > 0. E). say . E) = 1. E) > 1 − ε.33 (Rank-Nullity Theorem) Let X be a Banach space with identity operator I . E) > 0. If T = λI − K. Pick δ = 1−ε ε d(u. Now assume E is finite-dimensional. For each n ∈ N. Theorem 6. then dim(X/ranT ) = dim(ker(T ∗ )).34 (Riesz’s Lemma) Suppose X is a Banach space and E is a proper closed subspace of X. u − e u − e as required. E) > 1 − ε. pick en ∈ E such that 1 u − en < 1 + d(u. and so there exists some u ∈ X\E with d(u. E) = d(u − e. e) < d(u. asserts that any closed proper subspace of a Banach space is always a prescribed distance away from some portion of the unit sphere ∂BX . where λ is a nonzero scalar and K is a compact operator on X. E) ≤ u. Furthermore.2 A Rank-Nullity Theorem for Compact Operators 145 From the preceding proof. E) < d(u. if E is finite-dimensional. Before we can prove this theorem. If we let x = u−e u−e . It follows that. we have en ≤ u + 2 d(u. then 1 1 d(x. we require a few more technical results. the set {en : n ∈ N} is bounded in norm. n Note that 0 ∈ E (because E is a subspace of X). E) = d(u. The first of these.6. and so d(u. Therefore. We will see that it is really a statement about compact perturbations of scalings of the identity. We are now ready to state (although not prove) our version of the Rank-Nullity Theorem for compact operators. Since E is finite-dimensional. there exists some e ∈ E such that d(u. E) + δ. known as Riesz’s Lemma. Proof Denote the metric on X by d. Corollary 6. E is a proper closed subspace of X. E) ≤ 3 u. E) = d(u. By assumption. Given any ε > 0. We may assume without loss of generality that ε < 1. for any δ > 0. E). E). then x ∈ ∂BX can be chosen so that d(x. e) < 1 + d(u. then dim(kerT ) = dim(X/ranT ). Lemma 6. Thus. there exists some x ∈ ∂BX such that d(x. By definition. we have that (λ1 I − K) (λ2 I − K) · · · (λn I − K)(xn ) = 0. d(Kxn . by Lemma 6.29. the product is a compact perturbation of a scaling of the identity. If x = u−e u−e . even if E is infinite-dimensional. then each En is finite-dimensional and there exists some N ∈ N such that Em = EN for all m ≥ N . Consequently. Let n ∈ A. E). Thus. then we can find a sequence (xn )∞ n=1 in BX such that Kxm −Kxn ≥ δ for all m = n. the set A must be finite. En−1 ) = |λn | d(xn . but all the terms are at least δ apart.35 If E is a closed proper subspace of a reflexive Banach space X.) Lemma 6. where lim |λn | = 0. and so Kxn − Kxm = d(Kxn . where K̂ is some compact operator. we have (λ1 I − K) (λ2 I − K) · · · (λn I − K) = λ1 · · · λn I − K̂. Theorem 6. En−1 ) ≥ δ. the sets (En )∞ n=1 form a nested sequence E1 ⊆ E2 ⊆ · · · . or (iii) Λ = (λn )∞ n=1 . Remark 6.146 6 Compact Operators and Fredholm Theory (enk )∞ k=1 . Kxm ) ≥ δ. E) = 1.36 Let X be a Banach space and suppose K : X → X is a compact operator. Let δ > 0 be given and suppose (λn )∞ n=1 is a sequence of scalars (either real or complex) such that |λn | ≥ δ > 0 for all n ∈ N. Consequently.37 Let K be a compact operator and let Λ be the set of nonzero eigenvalues of K. If m ∈ A is chosen such that m < n. then we can always find an x ∈ X such that d(x. Since xn ∈ En . Fix an n ∈ N. Therefore dimEn < ∞. E) = 1. then x ∈ ∂BX and d(x. By Riesz’s Lemma (Lemma 6. Then Kxn ∈ λxn + En−1 . n→∞ . By construction. (ii) Λ is finite. and so (λn I − K)(xn ) ∈ En−1 . We wish to show that the inclusions eventually become equality. then we can pick an element xm ∈ Em such that Kxm ∈ λm xm + Em−1 ⊆ Em . If En = ker (λ1 I − K) (λ2 I − K) · · · (λn I − K) . If A is infinite. The result follows. En−1 ) = 1. Proof Denote the metric on X by d. This contradicts the assumption that K is a compact operator. because we should be able to find a convergent subsequence. n ∈ N. there exists some xn ∈ En such that xn = d(xn . Then there is some e ∈ E such that lim k→∞ enk = e.34).6. Then one of the following three things must be true: (i) Λ = ∅. Define a subset A of the natural numbers to be A = {n : En = En−1 }. Therefore. Kxm ∈ Em ⊆ En−1 . Expanding. (See Exercise 6. it follows that u − e = d(u. . But |μk | ≥ r for all k ∈ N.30). where λ = 0.) Let N = max{N1 . For each n ∈ N. Let v ∈ V .33 Let us begin by recalling the setup. Since V = ran(T N ). if there are infinitely many eigenvalues. Our assumption was that v = T N u. Assume the set is infinite. Proof of Theorem 6. (That is. where the inclusions are proper. there is some v ∈ V such that v = T (v ). define the nested sets En = ker((μ1 I − K) · · · (μn I − K)). Since V = ran(T N ) = ran(T N+1 ). Then we can find a sequence (μk )∞ k=1 in Λ ∩ {z ∈ C : |z| ≥ r} such that each element in the sequence is distinct. there exists an N1 ∈ N such that ker(T m ) = ker(T N1 ) for all m ≥ N1 . N2 }.) Similarly. Let T = λI − K. Thus. there exists some N ∈ N such that En = EN for all n ≥ N . Finally. and hence T N u = 0.30). It follows that either the set of eigenvalues is finite or. We have arrived at a contradiction. By Lemma 6. Since the μk are distinct eigenvalues. (In the lemma.36. We have demonstrated that T |V : V → V is a bounded linear bijection on the Banach space V . the map T |V is onto. we conclude that E1 ⊂ E2 ⊂ E3 ⊂ · · · . Observe that T N = λN I − K̂. for some compact operator K̂. Since V is a closed subspace of a Banach space. for each r > 0.6. We start with a Banach space X and a compact operator K : X → X. we are ready to prove the Rank-Nullity Theorem for compact operators. Thus. it too is a Banach space. by Lemma 6. and consequently v = 0. there is a bounded linear operator S : V → V such that ST |V = T |V S = I |V .36. Suppose v ∈ V is such that T v = 0. we deduce that ran(T m ) = ran(T N2 ) for all m ≥ N2 . Claim 1. It follows that u ∈ ker(T N+1 ) = ker(T N ). For ease of notation. Thus. Consequently. (That is. From the latter equality. by the Bounded Inverse Theorem (Corollary 4. We recall that T |V is the restriction of the map T to the subspace V . there exists some u ∈ X such that v = T N u. the spaces W and V are invariant under T ). they must accumulate to 0. there is some u ∈ X such that v = T N+1 u = T (T N u) ∈ T (V ). W is finite- dimensional (by Lemma 6. The map T |V : V → V is onto. (See Exercise 6. T has a bounded linear inverse on V ).12. Therefore.2 A Rank-Nullity Theorem for Compact Operators 147 Proof We will show that the set Λ∩{z ∈ C : |z| ≥ r} is finite for any r > 0.29) and V is closed (by Lemma 6. Thus. there exists an N2 ∈ N such that ker((T ∗ )m ) = ker((T ∗ )N2 ) for all m ≥ N2 . the map T |V : V → V is one-to-one. The map T |V : V → V is one-to-one. let W = ker(T N ) and V = ran(T N ). let λn = λ for all n ∈ N. there are only finitely many eigenvalues inside the set {z ∈ C : |z| ≥ r}. and so. Then 0 = T v = T (T N u) = T N +1 u. Notice that T (W ) ⊆ W and T (V ) ⊆ V . Claim 2. . and so P is a projection onto V .6 Let X be a reflexive Banach space and let E be an infinite-dimensional closed proper subspace of X. Hence. and so we may apply the finite-dimensional Rank-Nullity Theorem to T |W . . we deduce that kerT = kerT |W . n→∞ Exercise 6. show that T is a compact operator. What are the eigenvalues if q = ∞? Exercise 6. We conclude that kerP = W . η3 . η2 . · ) is continuous only if T ∗ has finite rank. 2 Exercises Exercise 6. if X and Y are Banach spaces such that X is reflexive. w∗ ) → (X ∗ . If x ∈ V . Consequently. . That is. Observe also that P x = 0 if and only if T N x = 0 (because S is an isomorphism on V ). . Exercise 6. If (xn )∞ n=1 is a sequence in X such that xn → 0 weakly.4 Suppose X and Y are Banach spaces and T : X → Y is a compact operator. then show that lim T (xn ) = 0.42. show that T (BX ) is closed in Y . Exercise 6. . n→∞ Exercise 6. by Theorem 4. Making the necessary substitutions. and lim T (xn ) = 0 whenever xn → 0 weakly. .5 Show that the converse to Exercise 6.1 Let 1 ≤ q < ∞ and let S : q → q be the left shift operator given by S(η1 .148 6 Compact Operators and Fredholm Theory Now. define a map P : X → V by P x = S N T N x. Consequently. we have that P x = 0 if and only if x ∈ ker(T N ) = W . Show that λ is an eigenvalue for S if and only if |λ| < 1. Show there exists an x ∈ X with x = 1 such that . We know that W is finite-dimensional.2 Suppose X and Y are Banach spaces and T : X → Y is a bounded linear operator. we have X/ranT = W/ranT |W . since X = V ⊕ W . Since X = V ⊕ W and T |V is invertible. ) = (η2 . Therefore.4 is true if X is a reflexive Ba- nach space. dim(kerT |W ) + dim(ranT |W ) = dimW. If X is reflexive. (ηk )∞ k=1 ∈ q . Additionally. the result follows. η3 . η4 . ). X = V ⊕ W . dim(kerT |W ) = dim(W/ranT |W ). then P x = x.3 Suppose X and Y are Banach spaces and T : X → Y is a bounded linear operator. x ∈ X. Show that T ∗ : (Y ∗ . Exercise 6. show that dim(E) = dim(E ∗ ) if E is finite-dimensional. (Hint: Use the Hahn–Banach Theorem.15 Let X and Y be Banach spaces and suppose T : X → Y is a bounded linear operator. .) Exercise 6. Compute ker(TK ) and ran(TK ). Show that if T : X → X is a compact operator. then there exists an element x ∈ X with x = 1 such that T x = T . f ∈ C[0. Let λ = 0 and let T = λI − K be a compact perturbation of a scaling of the identity. Exercise 6. 0 Show that TK is a compact operator. Show that T is weakly compact if and only if T ∗∗ (X ∗∗ ) ⊆ Y . 1] → C[0.) Exercise 6.7 Let X be a reflexive Banach space.16 (Gantmacher’s Theorem) Prove the following: If T is weakly compact. Exercise 6. 1]. x ∈ [0. 1] by 1 TK f (x) = sin ((x − y)π)f (y) dy. then S + T is a weakly compact operator.17 Let X and Y be Banach spaces.13 Suppose that T : C[0. (Hint: Use Exercise 5. The operator T is called weakly compact if the set T (BX ) is relatively compact in the weak topology on Y . Exercise 6.34. where N is the natural number from (a). Also. Exercise 6. is T a compact operator? Explain your answer. If T (X) = Y . E) = inf{x − e : e ∈ E} = 1. then T ∗ is weakly compact.39. Show that E is infinite-dimensional if and only if E ∗ is infinite-dimensional. (Hint: Find a weak limit point of the sequence (en )∞ n=1 in the proof of Lemma 6. Y ) are weakly compact operators. Show that if S and T in L(X.8 Suppose X and Y are Banach spaces and T : X → Y is a compact operator.12 Let X be a Banach space and suppose K : X → X is a compact operator.20.) Exercise 6.9 Suppose X and Y are Banach spaces and T : X → Y is a compact operator. 1] is a compact operator. Show that there exists a sequence (Tn )∞ n=1 of finite rank operators such that lim T − Tn = 0. 1]. If T (X) is a closed infinite-dimensional subset of Y .) Exercise 6. Exercise 6. 1] → C[0. then show that Y is separable.11 Let E be a Banach space. then show that Y has finite dimension. If T (X) is dense in Y .Exercises 149 d(x. (b) Use (a) to show that ran(T m ) = ran(T N ) for all m ≥ N . n→∞ Exercise 6. (a) Show there exists an N ∈ N such that ker((T ∗ )m ) = ker((T ∗ )N ) for all m ≥ N .14 Define a map TK : C[0.10 Suppose X and Y are Banach spaces and T : X → Y is a bounded linear operator. (Hint: Use Theorem 5. Y . . Show that if A : W → X and B : Y → Z are bounded linear operators. Exercise 6. and Z are Banach spaces and suppose that the map T : X → Y is a weakly compact operator. (b) If X is reflexive and S : c0 → X is a bounded linear operator. then T is compact.150 6 Compact Operators and Fredholm Theory Exercise 6. X.19 Prove the following theorems of Pettis: (a) If X is reflexive and T : X → 1 is a bounded linear operator. then S is compact.18 Assume that W . then BT A : W → Z is weakly compact. where {x.19. y).1007/978-1-4939-1945-1_7 . We will assume H is a complex inner product space.Chapter 7 Hilbert Space Theory In this chapter. y1 . y1 ) + β(x. When a map satisfies (iii) and (iv). y)| ≤ xy. and (iv) (x. and (x. y.2 (Cauchy–Schwarz Inequality) If x and y are elements of an inner product space. Define a norm on H by x = (x. Kalton.) When the underlying vector space is real. y) + β(x2 .1 Let H be a complex vector space. we will consider the spectral theory for compact hermitian operators on a Hilbert space. a map satisfying (iii) and (iv) is called bilinear. (The term sesqui comes from the Latin for one and a half. y) = (y. αy1 + βy2 ) = α(x. Theorem 7.) A real inner product space is similarly defined. β} ⊆ C. x). J. The positive-definiteness and homogeneity of · are clear (from the definition of the inner product). x) for all x ∈ H . it is called a complex inner product space. y2 } ⊆ H and {α. © Springer Science+Business Media. A complex inner product on H is a map (·.1 Basics of Hilbert Spaces Before we begin our discussion of linear operators on a Hilbert space. y2 ). 7. An Introductory Course in Functional Analysis. we recall the basic definitions and the important theorems we will use. then |(x. Bowers. x1 . x) ≥ 0 for all x ∈ H . x) = 0 if and only if x = 0. while a map satisfying (ii) is said to be conjugate symmetric. (See Definition 3. (iii) (αx1 + βx2 . DOI 10. Definition 7. but first we need an important inequality. LLC 2014 151 A. (ii) (x. except the scalars are real. N. We will verify subadditivity in a moment. y) = α(x1 . it is called sesquilinear. Universitext. ·) : H × H → C that satisfies the following properties: (i) (x. A map that satisfies (i) is said to be positive definite. unless otherwise stated. When H is equipped with a complex inner product. √ Let H be a complex inner product space. x2 . y) + t 2 y2 > 0. where x = (x. y). + . If x + ty = 0 for some t ∈ R.20 and the comments in between. Since (y. y) + (y. y)| = |t| y2 = x y. y)| = |eiθ (x. Suppose now that x + ty2 > 0 for all t ∈ R. x) + t(x. x) + y2 . x) = (x.1) for all t ∈ R. x) + t 2 (y.152 7 Hilbert Space Theory Proof Observe that x + ty2 ≥ 0 for all t ∈ R.19 and 3. y)| ≤ x y. subadditivity of · follows from the next inequality. Therefore. (x. y) + (y. Expanding the norm as an inner product. y) ∈ R. and consequently the discriminant must be negative. By the Cauchy–Schwarz Inequality. it follows that (x. By assumption. y)). and so x + y2 ≤ x2 + 2 x y + y2 = (x + y)2 . It follows that |((x. 2π) so that eiθ (x. y) + t(y. The left side of (7. (7. · ) is Definition 7. y) | ≤ eiθ x y = x y. We recall that the norm on H is said to be induced by the inner product on H . we obtain (x. Choose θ ∈ [0. y)| = | (eiθ x. 2 √ is an inner product space H such that (H . from which we conclude that · is a norm. The result follows by taking square roots. x) = 2 ((x. y) > 0. 2 In the case of an inner product space. (See Definitions 3.1) is a quadratic in t with real coefficients that has no real zeros. y) ∈ C. y) − 4x2 y2 < 0. then x = |t| y. y))| ≤ x y. and so x2 + 2t (x. Then |(x.2 4 (x.) .3 (Minkowski’s Inequality) If x and y are elements of an inner product space. and so |(x. Theorem 7. |(x. y)| = |( − ty.4 A Hilbert space a Banach space. as required. This is true for all x and y in H . Proof Observe that x + y2 = (x + y. then x + y ≤ x + y. x) for all x ∈ H . x + y) = x2 + (x. the inner product is given by . We will use the notation x ⊥ y to indicate x is orthogonal to y. then there exists a unique inner product that gives rise to the norm. then em ⊥ En whenever m < n. ·) is a complex inner product in each of these cases. g} ⊆ L2 (0. and so y = y . then x + y2 + x − y2 = 2 x2 + y2 . and the result follows directly. The inner product on 2 is given by ∞ (x.6 Let H be an inner product space. Two elements x and y in H are said to be orthogonal if (x. then (x. then x + y2 = x2 + y2 . and so x + y2 = x2 + 2 ((x.10. y ) for all x ∈ H . 1) is given by 1 (f . {x. Example 7. As a consequence. y) = xi yi .1. if y and y are elements of H such that (x. g) = f (s) g(s) ds. If y ⊥ H . e) = 0 for all e ∈ E. y) = (x. In particular. en+1 .1 Basics of Hilbert Spaces 153 Example 7. {f . We will see that these two examples are (in some sense) typical. y)) + y2 = x2 + y2 . y) = 0. . Theorem 7. and zero everywhere else. it must be the case that y − y ⊥ H . Then en ⊥ em for all n = m. Definition 7.) 2 In fact. If H is a real normed space. the space of all square-summable sequences. the Parallelogram Law characterizes inner product spaces. it must be the case that (y. then x is said to be orthogonal to E if (x. and x ⊥ E to indicate x is orthogonal to E. Example 7. (x. If E is a closed subspace of H . 1). y) = 0 for all x ∈ H . and if x ⊥ y. y) = 0. i=1 where x = (xi )∞ ∞ i=1 and y = (yi )i=1 . if H is a normed vector space which satisfies the Parallelogram Law. (Parallelogram Law) If x and y are elements of an inner product space.5 The classical examples of Hilbert space are the sequence space 2 and the function space L2 (0. 0 It is not hard to check that (·. The inner product on L2 (0.9 (Pythagorean Theorem) If x and y are elements of an inner product space. as required. and so y = 0. . if En = span{en . That is. (See Exercise 3.8 Let H be a Hilbert space and suppose y ∈ H .7. This is a fact we shall exploit repeatedly. y) = 0. let en denote the sequence with 1 in the nth coordinate.7 Suppose H = 2 . Proof We merely need to expand the expression on the left. y} ⊆ 2 . . For each n ∈ N. 1). Furthermore. Proof By assumption. }. 2 Theorem 7. 154 7 Hilbert Space Theory 1 (x. By the Parallelogram Law. and so 1 x − yn + x − ym ≥ 2 δ + ym − yn . Consequently. 2 is in C. C) = inf x − z. z∈C Proof Let δ = d(x. y) = x + y2 − x − y2 . . Thus. 4 These are known as the polarization formulas. there exists some y ∈ C (because C is closed) such that y = lim yn . C) is an infimum. pick yn ∈ C such that 1 δ ≤ x − yn ≤ δ + . because δ = d(x. by (7. n n m m Combining these inequalities. 2 2 ym +yn ym +yn Because C is convex. For each n ∈ N. (See Exercise 7. x − 2 ≥ δ. y) = x + y2 − x − y2 + ix + iy2 − ix − iy2 . 4 In the case of a complex normed space. (See Exercise 7. determine an inner product that agrees with the given norm. We leave it to the interested reader to verify that these formulas. n→∞ To show uniqueness. It then follows from (7.2) that x − y = δ. y m + yn x − yn 2 + x − ym 2 = 2 x − 2 + ym − yn 2 . n m n2 m2 The right side of this inequality tends to zero as m and n approach ∞. there is a unique point y ∈ C such that x − y = d(x. Again using the Parallelogram Law.) One consequence of the Parallelogram Law is that any Hilbert space is uniformly convex. Given x ∈ H .2).2) n This we can do. 1 (x.25.) Lemma 7. (yn )∞ n=1 is a Cauchy sequence. along with the Parallelogram Law. C). 2 22 2 4 On the other hand. Therefore.11 (Closest Point Lemma) Let H be a Hilbert space and suppose C is a nonempty closed convex subset of H . (7. suppose y and y are in C such that x − y = x − y = δ. 2δ 1 2δ 1 x − yn 2 + x − ym 2 ≤ δ2 + + 2 + δ2 + + 2 . we conclude 1 1 1 1 ym − yn 2 ≤ 4 δ + +2 + .25. e)) ≥ 0. Proof Let x ∈ H be given.7. e) has zero real and imaginary parts for all e ∈ E. it follows that ie ∈ E. Suppose e ∈ E. t ∈ R. e)) ≤ 0. e)) > −t e2 /2. We have established that ((x − y. Consequently. . 1 x − y2 + x − y 2 ≥ 2 δ 2 + y − y 2 . it follows that y−y ⊥ E. if t < 0. we conclude that ((x −y. e)) + t 2 e2 ≥ x − y2 . as required. which exists (and is unique) by the Closest Point Lemma (Lemma 7. since y is the unique element of E that is closest to x. Therefore. we see that ((x − y.1 Basics of Hilbert Spaces 155 y + y x − y2 + x − y 2 = 2 x − 2 + y − y 2 .11). and so y−y 2 = (y−y .12 In a reflexive space. Furthermore. e)) = 0 for all e ∈ E. e)) < |t| e2 /2. Pick y ∈ E to be the closest point to x. we obtain: x − y2 + 2t ((x − y. x − y ⊥ E. Similarly. but it might not be unique. ((x − y. Thus. we can conclude that equality holds only when t = 0. as required. y = y . then there exists a unique y ∈ E such that x − y ⊥ E. 2 We have established that for any x ∈ H . Because y was chosen to be the point in E closest to x. y = y . Since y − y = (x − y ) − (x − y). Therefore. (x − y. However. by convexity (and the definition of δ).13 Let H be a Hilbert space with closed subspace E. e)) = 0 for all e ∈ E. 1 2 δ2 ≥ 2 δ2 + y − y 2 . and so ((x − y. there exists a unique y ∈ E such that x − y ⊥ E. Taking the limit as t → 0− . Proposition 7. t ∈ R. and any closed subspace E of H . Because E is a linear subspace. ((x − y. then ((x − y. We have thus demonstrated that (x − y. Suppose y ∈ E also has the property that x − y ⊥ E. 4 Therefore. It remains to show that y is the unique element of E such that x −y ⊥ E. Squaring both sides of the inequality. Taking the limit as t → 0+ . we have ((x −y. y−y ) = 0. If t > 0. e) = 0 for all e ∈ E. If x ∈ H . 2 2 And so. e)). ie)) = −((x − y. ie)) = 0 for all e ∈ E. Hence. e)) = 0. y−y ∈ E. However. e)) + t 2 e2 ≥ 0. as required. Consequently. it follows that x − y + te ≥ x − y for all t ∈ R. This motivates the next definition. 2 and so y − y = 0. a nonempty closed convex subset has a closest point. 2t ((x − y. We will verify that x −y ⊥ E. ((x − y. Consequently. 2 Remark 7. By the definition of w. Proof We prove (i) first. w). and so (x.16 (Riesz–Fréchet Theorem) Let H be a Hilbert space. . Thence. Lemma 7. then PE is a bounded linear projection with PE = 1. we have (u. then the symbol PE x denotes the unique element of E with the property x − PE x ⊥ E. then φv ∈ H ∗ and φv = v. The element u was chosen so that φ(u) = 1. e) = α(x − PE x. The result follows. then (αx + βy − (αPE x + βPE y). Certainly. Now suppose x ∈ H . (ii) If φ ∈ H ∗ . w) = (w. w) = φ(x)(u. If e ∈ E. (i) If v ∈ H . is one of the key results in Hilbert space theory.156 7 Hilbert Space Theory Definition 7. However. e) + β(y − PE y. then v = 0 satisfies the conclusions of (ii). Then φ(u) = 0. Let x and y be elements of H and let α and β be complex numbers. φ is a continuous linear functional. and consequently E = kerφ is a closed subspace of H . Therefore.14 Let H be a Hilbert space and let E be a closed subspace of H . and so PE = 1. and hence w is orthogonal to x − φ(x)u. It follows that φ(x − φ(x)u) = 0. and so x − φ(x)u ∈ E. x2 = PE x2 + x − PE x2 . known as the Riesz–Fréchet Theorem (and sometimes the Riesz Representation Theorem for Hilbert Spaces). Since φv (v) = (v. w) = (x. By definition. we know that w ⊥ E. w) = w2 . Proof First we show linearity. and so we may assume φ = 0. and so φv ≤ v. 0 = (x − φ(x)u. assume φ(u) = 1. and φv (x) = (x.15 Let H be a Hilbert space with closed subspace E. PE (αx + βy) is the unique element of E with this property. Theorem 7. Let w = u−PE u and suppose x ∈ H . Certainly φv is linear. v) = v2 . then there exists a unique v ∈ H such that φ(x) = (x. we have that PE x ≤ x. By assumption. w). Since u = w + PE u and w ⊥ PE u. PE x ∈ E and so x − PE x ⊥ PE x. and φ = v. w) − φ(x)(u. w) + (PE u. This remains true for all e ∈ E. PE (αx + βy) = α PE x + β PE y. v) for all x ∈ H . Consequently. v) for all x ∈ H . 2 The next theorem. if x ∈ E. we conclude that φv = v. φv (x) ≤ x v. by the Pythagorean Theorem. and thus αPE x + βPE y is an element of E such that αx + βy − (αPE x + βPE y) ⊥ E. Therefore. If x ∈ H . e) = 0. By the Cauchy-Schwarz Inequality. then PE x = x. If φ = 0. Pick u ∈ E. If PE : H → H is defined by PE (x) = PE x for all x ∈ H . and so PE is linear. Now we turn our attention to (ii). Without loss of generality. 2 Operators on Hilbert Space 157 Thus. By the uniqueness of T ∗ y in H . w) = φ(x)w2 . we have found a v ∈ H such that φ = φv . T ∗ y). it remains an isometry. x ∈ H. {x. and such that φy = T ∗ y. 2 The significance of Theorem 7. by the Cauchy-Schwarz Inequality. Let y ∈ H and define φy : H → C by φy (x) = (T x. When H is a real Hilbert space. Specifically. We focus our attention not on general operators. y) = (x. y) = (x. If T : H → H is a bounded linear operator. such that φy (x) = (x. the uniqueness of T ∗ is clear.) We now show the existence of T ∗ .16 is the identification of H ∗ with H . Sy) = (x. (See Example 7. Therefore. T ∗ y) for all x and y in H . Before giving the definition of a hermitian operator. w) = φ(x)(u. if φ and ψ in H ∗ correspond to u and v in H (respectively).8. In the notation of (i). using the conclusion of (i). the map T ∗ is well defined. Proof Provided we can show such a map exists. We assumed that u ∈ E. Regardless of the underlying scalar field (real or complex). called hermitian. Certainly. y} ⊆ H . Let v = w/w2 . φy is a linear functional and φy ≤ T y.17 Let H be a Hilbert space. Then φ(x) = (x. T ∗ y). T ∗ y) for all x ∈ H . Lemma 7. but on a special type of operator. then there exists a unique linear operator T ∗ : H → H such that (T x. we discuss operators on Hilbert space. we have φ = v. It follows that Sy = T ∗ y for all y ∈ H .7. y} ⊆ H. a consequence of the identification of H with H ∗ is that any Hilbert space is reflexive. and so w = 0. If there is another such map S : H → H that satisfies the conclusion of the lemma. then (x. we consider the following existence lemma. Sometimes this is called conjugate linearity or anti-isomorphism. y). and (by construction) we have (T x. Define a map T ∗ : H → H by T ∗ (y) = T ∗ y for all y ∈ H . but the correspondence is given by what is sometimes called a conjugate isomorphism. v) for all x ∈ X. By the Riesz–Fréchet Theorem. When H is complex. and so S = T ∗ . 7. then αφ + βψ corresponds to αu + βv.2 Operators on Hilbert Space In this section. there exists a unique element in H (depending on y). and such that T ∗ = T . {x. the identification with H ∗ is an isometric isomorphism. (x. . which we call T ∗ y. It suffices then to show that T ∗∗ = T . Then (x. T ∗ (αy + βy ) = T x. T ∗∗ y). and retains many features of its finite-dimensional cousin.18.36 and 7. T ∗ (αy + βy ) = α x. x. αT ∗ y + βT ∗ y . and so we need only show the reverse inequality. . (x. A real hermitian operator is also known as a symmetric operator. T ∗ y)| = sup |(T x. Then for all x ∈ H .19 Let H be a Hilbert space and suppose T : H → H is a bounded linear operator. via the Riesz–Fréchet Theorem. and so T ∗ (αy + βy ) = αT ∗ y + βT ∗ y . T ∗ ≤ T . and so T = T ∗∗ . Then. x. It remains to show that T ∗ = T . y) = (x. and so T ∗ is bounded. invoking the Riesz–Fréchet Theorem. let y ∈ H . This is true for all x ∈ H . x∈BH x∈BH where the last inequality comes from the Cauchy-Schwarz Inequality.158 7 Hilbert Space Theory We now show that T ∗ is linear. T ∗ y) for all x and y in H . 2 Definition 7.18 Let H be a Hilbert space and suppose T : H → H is a bounded linear operator. T ∗∗ y) = (T ∗ x. The unique linear operator T ∗ : H → H such that (T x. To see that T ∗ is bounded. y)| ≤ T y. then T is called a hermitian operator. If T = T ∗ . T ∗ y = x. T ∗ is linear. x) = (x. It may seem as though there is a risk of confusion. T ∗∗ y) = (x.11. (See Exercise 7. y) = (y. T ∗ x) = (T y.2.) Definition 7. The Hilbert space adjoint is analogous to the matrix adjoint from linear algebra. and such that T ∗∗ ≤ T ∗ .) We have now defined two adjoints for the operator T : H → H . (See Exercise 7. Therefore. y . but there is a natural correspondence between the two. T ∗ y + β x. This completes the proof. T ∗ y) for all x and y in H is called the Hilbert space adjoint of T . We have already established T ∗ ≤ T . To that end. (See Definitions 3. T y) for all x and y in H . since (T x. y) = (x. Let x and y be in H . y + β T x. y} ⊆ H . {x.) Recall that the (operator) adjoint of T is the bounded linear map T ∗ : H ∗ → H ∗ defined by (T ∗ x ∗ )(x) = x ∗ (T x) for all x ∈ H and x ∗ ∈ H ∗ . Then. we repeat the above construction on T ∗ to define an operator T ∗∗ : H → H with the property that (T ∗ x. T ∗ y = sup |(x. Let y and y be elements in H and let α and β be complex numbers. y) = (x. Therefore. both denoted T ∗ . Therefore. αy + βy = α T x. T y). . y) g(x) dx f (y) dy.21 Let I be a (possibly uncountable) index set.1. 0 0 Since T ∗ is the unique operator with the property that (Tf . x ∈ [0. Definition 7.2 Operators on Hilbert Space 159 Example 7. x) for almost every x and y. 0 Taking complex conjugates. f ∈ L2 (0. 0 Now that we have obtained a formula for the Hilbert space adjoint T ∗ . this equals 1 1 K(x. g) = K(x. n}.k=1 is called hermitian if aj k = akj for each j and k in the set {1. 1]). 0 where K ∈ L2 ([0. We recall that the Kronecker delta is defined to be . This is reminiscent of the n-dimensional case (n < ∞).7. g) = (f . . 0 0 By Fubini’s Theorem. 1]. 1). 1) → L2 (0. we see that T is hermitian if K(x.20 Suppose T : L2 (0. 1) is a bounded linear operator given by the formula 1 Tf (x) = K(x. . where a matrix (aj k )nj. Then 1 1 (Tf . A subset (ej )j ∈I of H is said to be orthonormal if ej = 1 for all j ∈ I . We established that T is well-defined in Section 6. T ∗ g). y) f (y) dy. Let f and g be functions in L2 (0. ek ) = 0 for all j = k. and if (ej . . 1). y) = K(y. 1] × [0. y) f (y) dy g(x) dx. x) g(y) dy. y) g(x) dx. we conclude that 1 ∗ T g(y) = K(x. we obtain 1 ∗ T g(x) = K(y. and swapping the roles of x and y. 1 if j = k. 2 . √ Observe that if (ej )j ∈I is an orthonormal set in H . δj k = 0 if j = k. Lemma 7. Proof This follows from Zorn’s Lemma. Using this notation.22 Every orthonormal set is contained in a maximal orthonormal set. From this we conclude that any orthonormal subset of a separable Hilbert space is necessarily countable. ek ) = δj k for all indices j and k in I . the subset (ej )j ∈I is called orthonormal if (ej . then ej − ek = 2 whenever j = k (by the Pythagorean Theorem). ej )|2 ≤ x2 . . ek ) = |(x. ej )|2 . (ii) If N < ∞. implies that y = x. Theorem 7. we will restrict our attention to countable orthonormal sets. j =1 j =1 Therefore.4). For any positive integers m and n such that m < n. then the assumption. arguing as in (7. ej ) (ej . (7. ej ) (x. ek ) (ej . and hence x = y.3). ej ) ej = |(x. ek ). In order to complete the proof of the first part of the theorem. we have n 2 n (x. and so x − y ⊥ y. ej )| ≤ x 2 2 for all m ≤ N. ek ) = 0 for all k ∈ {1. then (x. Now suppose that N = ∞. ej ) δj k = (x. ek ) = δj k for all indices j and k. where N ∈ N ∪ {∞}. N (i) If x ∈ H . we have x − y = 0.4) j =1 k=1 j =1 recalling that (ej . ej )|2 ≤ lim x2 = x2 . as well. Let y = j =1 (x. . but much of what we say will remain true for uncountable sets. (7. . mj =1 |(x. y) = 2 (x. (y − x. Choose m ∈ N such that m ≤ N . Thus. m}. m→∞ m→∞ j =1 j =1 This proves part (i) of the theorem. j =1 j =1 m Proof (i) Let x ∈ H . ej )|2 .160 7 Hilbert Space Theory In what follows. From this we conclude that y ≤ x. Computing the norm of y: m m m y = (y. It follows that (y − x. . then ∞ m |(x. If N = ∞. ej )|2 = x2 . j =1 N N (ii) If x ∈ H and |(x. m m (y.23 (Bessel’s Inequality) Suppose (ej )N j =1 is a countable orthonormal set. . ej )|2 = lim |(x.4). x2 = x − y2 + y2 . we consider two cases.3) by the Pythagorean Theorem. ej ) ej . m}. For each k ∈ {1. then |(x. y) = 0. ej ) ej = x. because of (7. when m is finite. ek ) = (x. Consequently. . together with (7. then let m = N and the proof is complete. If N ∈ N. Therefore. . ek ) = (x. . j =m+1 j =m+1 . the sequence n ∞ (x. ej ) = (x.) j =1 (iii) (ej )∞ j =1 is a maximal orthonormal sequence.25. en ) = 0 for all n ∈ N. By assumption. Let (ei )i∈I be an orthonormal basis for H . there exists a u ∈ H such is a Cauchy that u = ∞ j =1 (x. by Corollary 7. . Proof We begin by assuming (i). ej ) ej converges to x and that x2 = ∞ j =1 |(x. ∞ (ii) If x ∈ H . Thus. by the Pythagorean Theorem. It follows that x − u = 0. but that (ej )∞ j =1 is not maximal.24 Let H be a Hilbert space. ej ) ej n=1 j =1 sequence in the norm on H . (See (7. Therefore. Let x ∈ H be such that ∞ j =1 (x. a countable maximal orthonormal set is an orthonormal basis. Theorem 7. ej ) for all j ∈ N. as required. Thus. Then H has a maximal orthonormal set.2 Operators on Hilbert Space 161 ∞ By assumption. x = u. Let (ej )∞ j =1 be an orthonormal basis for H and let x ∈ H .4) and the text following it. Then there exists some x ∈ H with x = 1 such that (x. Proof Let H be a nonzero separable Hilbert space. and so (ej )∞ j =1 must be maximal. This set is countable.22. x − u ⊥ ej for all j ∈ N. (Parseval’s Identity.) By the Pythagorean Theorem. ej ) ej converges to some u ∈ H .26.23 (Bessel’s Inequality)). because H is separable. we showed that the series ∞ j =1 (x.22. we see that (u.27 (Identification of separable Hilbert spaces). (ej )∞j =1 is an orthonormal basis for H . then x2 = |(x. we have that x − u ⊥ u.25 Let H be a Hilbert space and suppose (ej )∞ j =1 is an orthonormal sequence in H . (i) implies (ii). Since x = u (by assumption). (See the proof of Theorem 7. ej )|2 converges.23 (Bessel’s Inequality). ej ) ej for every x ∈ H . Consequently. ej ) ej . By completeness. the series j =1 |(x. ej )|2 . 2 ∞ Definition 7. as required.26 Every nonzero separable Hilbert space has an orthonormal basis. ∞An orthonormal sequence (en )n=1 is called an orthonormal basis for H if x = j =1 (x. and so (iii) implies (i). Assume now that (ii) is true. The series ∞ j =1 (x. This is a violation of (ii). This proves that (ii) implies (iii). the element x−u x−u has norm one and is orthogonal to ej for each j ∈ N.) By construction. e j ) ej is not x. By direct computation. by Lemma 7. (Such a basis is known to exist. Let H be an infinite- dimensional separable Hilbert space. ej )| 2 .7. The following are equivalent: (i) The sequence (ej )∞ j =1 is an orthonormal basis for H . (See the comments preceding Lemma 7.) Therefore. In the proof of Bessel’s Inequality (Theorem 7. and so we conclude that x = u. we have that x2 = x − u2 + u2 .) By Theorem 7. but suppose that (ej )∞ j =1 is not a basis for H . 2 Corollary 7. This contradicts the maximally of (ej )∞ j =1 . Now assume (iii). 2 Example 7. this map is well-defined. We claim that (en )n∈Z is an orthonormal basis for L2 (T). and indeed T x = x. every infinite-dimensional separable Hilbert space has a countable orthonormal basis. 2π 0 (Compare to Example 7. x ∈ H. Therefore. there is only one infinite-dimensional separable Hilbert space. θ ∈ T. We define a map T : H → 2 by T x = ((x. 2π dθ . define en : T → C by en (θ ) = einθ . Theorem 7. 2π ). it must be the case that I is a countable set. Therefore. as far as Banach spaces are concerned. T is an isometry. Example 7. Since H is separable. g} ⊆ L2 (T). It follows that all infinite-dimensional separable Hilbert spaces are isometrically isomorphic to 2 .29 (The Fourier transform and separability of L2 ) Let L2 (T) denote the space of (equivalence classes of) complex-valued square-integrable functions on the probability space T. Without loss of generality. Choosing an orthonormal basis for H is essentially representing H as 2 .5.28 Every infinite-dimensional separable Hilbert space is isometrically isomorphic to 2 . 2 The next example can be seen as a special case of the previous. we can uniquely express every x ∈ H as x = ∞ j =1 (x. The space L2 (T) is a Hilbert space with inner product given by 2π 1 (f . Proof See the discussion preceding the statement of the theorem. because 2π . ej ) ej .) For each n ∈ Z. {f . Conse- quently. ej ))∞ j =1 . Since(ei )i∈N is an orthonormal basis for H . To emphasize this result. we may assume I = N. we place it in a theorem. By Parseval’s Identity. g) = f (θ ) g(θ ) dθ. where T = [0. The sequence (en )n∈Z is an orthonormal set.162 7 Hilbert Space Theory √ we have ei − ej = 2 whenever i = j . and Lusin’s Theorem (Theorem A. (en . 1 inθ −imθ 1 if n = m.36) states that the set of continuous functions is dense in L2 (T). . The maximality of (en )n∈Z follows from the density of trigonometric polynomials: The Weierstrass Approximation Theorem states that the set of trigonometric polynomials is dense in C(T). em ) = e e dθ = 2π 0 0 if n = m. ) Before proceeding with the proof of Theorem 7. there exists an x ∈ H with x = 1 such that T x = T . also called the Fourier transform. for any x ∈ H . To see this precisely. we see that f L2 (T) = F(f )2 (Z) . Also. after F. Therefore. that is.31 If T is a compact hermitian operator. If T : H → H is a compact hermitian operator. The Fourier transform represents L2 (T) as 2 (Z). and keeping in mind that x has norm one. f ∈ L2 (T). and so L2 (T) and 2 (Z) are identical as Banach spaces.30].30 Let H be an infinite-dimensional separable Hilbert space. It follows that T 2 ≤ T 2 . each of whom proved it (independently) in 1907 [11. observe that the sequence of eigenvalues (λn )∞n=1 promised in Theorem 7. we have T 2 (x) = T (T x) ≤ T T x ≤ T 2 x. Proof First.2 Operators on Hilbert Space 163 If f ∈ L2 (T).30 must converge to 0. we require some preliminary results.7. By Parseval’s Identity (Theorem 7. λx) = λ. which nicely mirrors the matrix theory of finite-dimensional vector spaces. (We proved this for a general compact operator in Theorem 6.32 Let H be a nonzero Hilbert space. en ) for n ∈ Z. Remark 7.37. T x) = (x. then the Fourier series of f is given by the series f . define a map F : L2 (T) → 2 (Z). Riesz and E. we see that the Fourier series of f ∈ L2 (T) will always converge to f in the L2 (T)-norm. To see this. there exists a sequence (λn )∞ n=1 such that T en = λn en for all n ∈ N. we claim that T 2 = T 2 . We are now ready to state the main result of this section. Lemma 7. n∈Z Since the sequence (en )n∈Z is an orthonormal basis for L2 (T). x) = (x. x) = (T x. by F(f ) = (fˆ(n))n∈Z . Fischer. F determines an isometric isomorphism. The identification of L2 (T) and 2 (Z) is sometimes known as the Riesz–Fischer Theorem. Using the properties of inner products. There exists an orthonormal basis (en )∞ n=1 of H such that en is an eigenvector of T for each n ∈ N. While we state (and prove) the result for separable Hilbert spaces. Suppose T : H → H is a compact hermitian operator.30. we see that λ = (λx. .) We recall that the Fourier transform of f is given by the formula fˆ(n) = (f . en en . suppose x ∈ H is an eigenvector with x = 1 such that T x = λx. Theorem 7.17. It is not hard to show that F is an isomorphism. it remains true for general Hilbert spaces. Certainly. f ∈ L2 (T).S.25). (Compare this to Example 4. then any eigenvalue λ of T must be real. it follows that u2 + (T 2 x. Since T xn ≤ T for all n ∈ N.32 holds for any compact operator T : X → X. we have (u. as claimed. Therefore. x) = (T 2 x. If T : H → H is a compact hermitian operator. x). x) (x. the conclusion of Lemma 7.) 2 Remark 7. Then x ≤ 1. and so T (BH ) is relatively compact in H . (7.7) . (7. x) x2 = 0. T x) = (T 2 x. passing to a subsequence if necessary. by (7. Notice that u = T 2 x − T 2 x. x) − (T 2 x. T n→∞ T Therefore.7. and hence T 2 = T 2 .5) n→∞ We assumed T was compact. (Note this implies x = 1. 1 1 2 Tx = T y = lim T xn . x) x2 . x) x. since T is a hermitian operator. Taking the supremum over all x ∈ BH . and in particular u ⊥ (T 2 x. and such that lim T 2 xn = T 2 = T 2 .33 The hermitian assumption is not required in Lemma 7. we may assume (without loss of generality) that (T xn )∞ n=1 converges to some y ∈ H .32. Consequently. We make the observation that. we conclude that T 2 ≤ T 2 . for any x ∈ H .164 7 Hilbert Space Theory By the hermitian assumption on T . x) x.6) Define u = T 2 x − (T 2 x. Now. By Lemma 7. Thus. by (7. By the definition of u. it must be the case that y ≤ T . pick a sequence (xn )∞ n=1 in H such that xn ≤ 1 for all n ∈ N. In fact. T x2 = (T x. Let x = y/T . Furthermore. T x) = T x2 = T 2 . (T 2 x.) We require one more lemma. (7. By the Pythagorean Theorem. x) = (T 2 x. x) − (T 2 x. then either T or −T is an eigenvalue.34 Let H be a nonzero Hilbert space. whenever X is a reflexive Banach space. we have u ⊥ x. x) x2 = u + (T 2 x. there exists some x ∈ H with x = 1 such that T x = T . it follows that T x2 ≤ T 2 x x. as required. x) = (T x. Lemma 7.5). T x = T .6).32. by the Cauchy-Schwarz Inequality. Proof We may assume T = 0. (See Exercise 6. 34. 2 We are now prepared to prove Theorem 7. x) x2 = T 2 x2 ≤ T 4 . T en ) = (x. otherwise let A = N.7). Computing: T y + T y = (T 2 x − T T x) + (T T x − T 2 x) = 0.7). If N < ∞. Let (en )N n=1 be a maximal set of orthonormal eigenvectors. we observe (T 2 x. Suppose that x ∈ H0 . 2 .2 Operators on Hilbert Space 165 On the left side of (7. This completes the proof. then T |H0 : H0 → H0 has an eigenvector x0 .34. T x ∈ H0 . Let H0 = {x : (x. let A = {1. λn en ) = 0. and so y = 0 is an eigenvector for T with eigenvalue −T . We have established that either T or −T is an eigenvalue for T . Observe that H0 is a Hilbert space. and we are done. by Lemma 7. en ) = 0 for all n ∈ A}. Suppose instead that T x = T x. and consequently T |H0 : H0 → H0 is a well-defined compact hermitian operator. We have established that no nonzero element of H is orthogonal to the orthonormal set (en )n∈A . Therefore. u = 0. Consequently. . Thus. and so there exists a maximal set. T y = −T y.30 By Lemma 7. by Theorem 7. We will show that H0 = {0}. en ) = (x. We begin by claiming that T (H0 ) ⊆ H0 . x) x2 = |(T 2 x.6). This contradicts the maximality of (en )n∈A . Then for all n ∈ A. If T x = T x. there exists a nonempty set of orthonormal eigenvectors. if H0 = {0}. where N ∈ N ∪ {∞}. On the right side of (7. But then (en )n∈A ∪ {x0 /x0 } forms an orthonormal set of eigenvectors for T : H → H . recalling the definition of u. let λn denote the eigenvalue corresponding to the eigenvector en . and so T 2 x = T 2 x. and thus it is an orthonormal basis for H .25. . (T x. then x is an eigenvector for T with eigenvalue T . Proof of Theorem 7. For each n ∈ A.7. . we have u + (T 2 x. Then y = T x − T x is a nonzero element of H . by Zorn’s Lemma. . It follows that (en )n∈A is maximal amongst all orthonormal sets. It follows that u2 + T 4 ≤ T 4 . Therefore.30. N}. by (7. and so H0 = {0}. x)|2 x2 = T 4 . 8) is that a |K(x. Recall that L2 (a. b) denotes the set of (equivalence classes of) Borel measurable functions f : [a. Consequently. The function K is the kernel associated with T . b) → L2 (a.9) a We will show that T is a well-defined bounded linear operator on L2 (a. b). a b This space is a complex Hilbert space with the inner product (f . From (7. We use K to define a map T : L2 (a. b b 1/2 b 1/2 |K(x.b) . b). Tf L2 (a.b) .10) a a a This quantity is finite for almost every x in [a. so that b b K2L2 = |K(x. b). a By squaring. Therefore. (7. Linearity of T is now evident. or simply the kernel of T . b]. x ∈ [a.166 7 Hilbert Space Theory 7. b] → C is in L2 ([a. b] whenever f ∈ L2 (a. b). b]. the integral in (7. y) f (y) dy. a for some K ∈ L2 ([a. b]).9) exists for almost every x in [a. y)|2 dy dx < ∞. It remains to show that T is bounded. x ∈ [a.b) ≤ KL2 f L2 (a. and then integrating with respect to x. b) → L2 (a. g) = a f (x) g(x) dx for f and g in L2 (a. y) f (y) dy. y)|2 dy |f (y)|2 dy . (7. Suppose that K : [a. and so T is a well-defined linear operator. (7. b) is a Hilbert-Schmidt operator if b Tf (x) = K(x.3 Hilbert–Schmidt Operators Let a and b be real numbers such that a < b. provided that the function f is an element of L2 (a. b) by b Tf (x) = K(x. b]). . we discover b b b |Tf (x)|2 dx ≤ |K(x. b] × [a.8) a a b A consequence of (7.b) = |f (x)| dx 2 < ∞. a a a Therefore. we see that b 1/2 |Tf (x)| ≤ |K(x. f ∈ L2 (a. y)|2 dy dx f 2L2 (a. y)|2 dy < ∞ for almost every x ∈ [a. b] → C such that b 1/2 f L2 (a. b). b] × [a. b).35 A map T : L2 (a. y) f (y)| dy ≤ |K(x. b]. y)|2 dy f L2 (a. b]. f ∈ L2 (a.b) . b] × [a.10). the map T is a well- defined bounded linear operator and T ≤ KL2 . Definition 7. By Hölder’s Inequality. let L2 = L2 ([a. Proof For ease of notation. Let g ∈ L2 and suppose for all m and n in N. We wish to show that (fmn )∞ m. This motivates the next definition. then g = 0 in L2 . (en )∞ n=1 will denote an orthonormal basis for L2 (a. x) for almost every x and y in [a. b]. b b 1/2 |g(x. (We know an orthonormal basis exists by Corollary 7. b] × [a.4) with the definition of T in (7. b]). fmn ) = 0 for all natural numbers m and n. Definition 7. The basis is countable by an argument similar to that used in Example 7.n=1 is an orthonormal basis for L2 . To that end.) Proposition 7. g) = K(x. b). we see that T ∗ = T precisely when K(x.12) a a By Hölder’s Inequality. y) em (x)| dx ≤ |g(x.3 Hilbert–Schmidt Operators 167 We wish to identify T ∗ . the Hilbert space adjoint of T . We say that K is a hermitian kernel.) Comparing (7. For each m and n in N. and therefore b T ∗ g(x) = K(y.11) a (See Example 7. y) dx dy. T ∗ g) for all f and g in L2 (a. b].37 Let (en )∞ n=1 be an orthonormal basis for L2 (a. b]). b b (g. Throughout the remainder of this section. define a function fmn : [a. It suffices to show that if g ∈ L2 and (g.n=1 is an orthonormal basis for L2 ([a.36 Suppose T is a Hilbert-Schmidt operator on L2 (a. b] × [a. x) g(y) dy. y) = em (x) en (y).20. y) g(x) dx f (y) dy. y) = K(y. b b b b (Tf . (7. fmn ) = g(x.3. f ) = g(x.26. (7.9). b). y) f (x. or simply that K is hermitian. (7. if K(x. y} ⊆ [a. b). x) for almost every x and y in [a. The inner product on L2 is given by b b (g. y) em (x) en (y) dx dy = 0. suppose f and g are in L2 (a. b] × [a. b] → C by fmn (x.29. g) = (f .13) a a . y) f (y) dy g(x) dx = K(x. b). a a a a ∗ The Hilbert space adjoint T is the unique operator that satisfies the equation (Tf . The set (fmn )∞ m. {x. y)|2 dx < ∞. a a where g and f are functions in L2 . y) = K(y. By Fubini’s Theorem. A real-valued hermitian kernel is also known as a symmetric kernel. b) with kernel K. b].7. b] by b hm (y) = g(x. Therefore. b].13). . and thus the equality in (7. by (7. a The function hm is well-defined by (7. Proof Let (en )∞ n=1 be an orthonormal basis for the Hilbert space L2 (a. y) em (x) dx = 0 a. b] × [a. a for every m ∈ N.14) implies for each m ∈ N that hm = 0a.(x) for almost every y. b] × [a.b) = |hm (y)|2 dy ≤ |g(x. (7. we conclude that g(x. b b b hm 2L2 (a. b) is a compact operator. b) and let (fmn )∞ m. so too is (en )n=1 . n=1 m=1 then K − KN L2 → 0 as N → ∞. Because (fmn )∞ m.14) a Since (en )∞ ∞ n=1 is an orthonormal basis for L2 (a. a a a It follows that hm ∈ L2 (a. y) em (x) dx. and so for almost every y. en ) = hm (y) en (y) dy = 0. For each m ∈ N.38 A Hilbert-Schmidt operator on L2 (a. To be precise. y)|2 dx dy = g2L2 . if for each N ∈ N.n=1 is an orthonormal basis for L2 ([a.(y). (7.(y). y) em (x) dx = 0. as required. we have b (hm .12). b).e. b]).n=1 be the orthonormal basis for L2 ([a. g = 0 in L2 . For every n ∈ N.e. b]). b g(x. Thus. b]) given in Proposition 7. y ∈ [a. for each m ∈ N.15) n=1 m=1 where the series converges in the norm on L2 ([a. b] × [a. define a function hm on [a. b) for each m ∈ N.168 7 Hilbert Space Theory for almost every y. y) = 0a. fmn ) fmn .e. 2 Proposition 7. N N KN = (K. Suppose that TK is a Hilbert-Schmidt operator with kernel K. a A countable collection of measure zero sets is still measure zero. fmn ) fmn . we have that ∞ ∞ K= (K. b g(x. Again invoking the fact that (en )∞ n=1 is an orthonormal basis.37. and furthermore. It follows that TK − TKN → 0 as N → ∞. y) g(y) dy. in addition to being square-integrable. fmn )|2 = |λn δmn |2 = |λn |2 . as the limit of a sequence of finite rank operators. n=1 m=1 a n=1 m=1 Consequently. fmn ) = K(x. m=1 n=1 Therefore. we know that there exists an orthonormal basis for L2 (a. (In particular. and hence b (K. that K is a hermitian kernel. fmn ) fmn (x. n=1 m=1 n=1 m=1 n=1 We summarize in the following theorem. y) = K(y. b]×[a.30. b]) given in Proposition 7. Denote this set of orthonormal eigenvectors by (en )∞ n=1 and let the corresponding sequence of eigenvalues be (λn )∞ n=1 . en ) em (x). we compute: b b b (K. 2 Now assume. so that K(x. N N TKN g(x) = (K. the range of TKN is at most N -dimensional. a n=1 m=1 Recalling the definition of fmn (x. fmn ) (g. T en = λn en for all n ∈ N.25). From Theorem 7. b) composed of eigenvectors for TK . By Proposition 7. y). Therefore.) Observe that TK − TKN = TK−KN . y) en (y) dy em (x) dx = T en (x) · em (x) dx. by Parseval’s Identity (Theorem 7. ∞ ∞ ∞ ∞ ∞ K2L2 = |(K. We know that TK−KN ≤ K − KN L2 (see the comments at the start of this section) and K − KN L2 → 0 as N → ∞ (by construction).3 Hilbert–Schmidt Operators 169 The Hilbert-Schmidt operator with kernel KN is given by b N N TKN g(x) = (K.38. TK is a compact operator. fmn ) em (x) en (y) g(y) dy = (K.37. (7. en ) em (x). fmn ) = λn en (x) · em (x) dx = λn (en . a Thus. this becomes N N b N N (K. y) = em (x) en (y) for all x and y in [a. Let (fmn ) ∞ m. fmn ) (g.7. .n=1 be the orthonormal basis for L2 ([a. the rank of TKN is finite. x) for almost every x and y. and so the rank of TKN is at most N. Using the fact that fmn (x.16) a a a By assumption. the map TK is a compact hermitian operator. em ) = λn δmn . b]. . a a n=1 and TK = sup |λn |. While we start with the assumption f ∈ C[a. {f .e. (I C1 ) . we will see an example of how to apply the theory of compact hermitian operators to solve differential equations. The only way that this can satisfy (BC ) is if α = β = 0. Let us return our attention to the Sturm-Liouville system in (7.17) ⎩y(a) = y(b) = 0. b] (a twice continuously differentiable function) that satisfies the differential equation (DE) subject to the boundary conditions (BC). we will make the following assumption on the homogeneous system (i. Example 7. f ∈ C[0. The system we will consider is ⎧ ⎨y + q(x)y = f (x). Suppose that a = 0 and b = π. then b b ∞ |K(x. b]. (DE) (7.17). n∈N Proof The proof of the integral equation can be found in the discussion preceding the statement of the theorem. q ∈ C[a. If (λn )∞ n=1 is the sequence of eigenvalues of TK . the system with f = 0): Assumption 1 The only solution to the homogeneous system is y = 0. y)|2 dx dy = |λn |2 < ∞. Suppose u is a solution to the initial value problem ⎧ ⎨y + q(x)y = 0. y (a) = 1. b]. (DE0 ) ⎩y(a) = 0. the homogeneous equation y = 0 has general solution y(t) = α + βt. we will later extend to f ∈ L2 (a.4 Sturm–Liouville Systems In this section.40 Our basic model of a Sturm-Liouville system comes from a vibrating string. Consider the following differential system: ⎧ ⎨y = f (x). (BC) We wish to find y ∈ C (2) [a. For simplicity. We will consider a special case of a system of differential equations known as a Sturm-Liouville system. (DE ) ⎩y(0) = y(π) = 0. 2 7.170 7 Hilbert Space Theory Theorem 7. Throughout what follows. q} ⊆ C[a. and so the vibrating string model satisfies our basic assumption. π ]. we will suppose the scalar field is R. b). The equality TK = supn∈N |λn | follows from the fact that T en = λn en for all en in the orthonormal basis (en )∞ n=1 . (BC ) For the differential equation (DE ).39 Let TK be a Hilbert-Schmidt operator with hermitian kernel K. b]. ψ(a) = 0. u and v satisfy (DE0 ).17). b]. Keeping Assumption 2 in mind. From the preceding calculations.18) . (DE0 ) ⎩y(b) = 0. ⎩φ u + ψ v = f . we see that this will be accomplished if we can solve the following differential system: ⎧ ⎨φ u + ψ v = 0. It follows that y + qy = φ u + ψ v + (u + qu) φ + (v + qv) ψ. Our goal is to find a solution to the differential system in (7. then u is a solution to the homogeneous system. q ∈ C[a. A little algebraic manipulation reveals: (u v − u v ) φ = f v and (u v − u v) ψ = f u. where φ and ψ are differentiable functions to be determined at a later time. and so u(b) = 0. x ∈ [a. Furthermore. b].7. It would follow that u = 0. (I C2 ) As before. φ(b) = 0. If u(b) = 0. let v be a solution to the initial value problem ⎧ ⎨y + q(x)y = 0. Continue by computing the second derivative of y: y = φ u + φ u + ψ v + ψ v . This violates the initial conditions in (I C1 ). (7. let us differentiate y: y = φ u + φ u + ψ v + ψ v = φ u + ψ v . v(a) = 0. By assumption. y (b) = 1. We will insist only that φ and ψ satisfy the following assumption: Assumption 2 φ and ψ are differentiable functions such that φ u + ψ v = 0.4 Sturm–Liouville Systems 171 We know that a solution u to this system exists by the general theory of ordinary differential equations. and so we conclude that y + qy = φ u + ψ v . violating the initial conditions in (I C2 ). Next. or else it would be trivial. such a solution is know to exist by general theory. Consider the function y(x) = φ(x) u(x) + ψ(x) v(x). by Assumption 1. (7. α a x Define a map K : [a. Let α = W (a) be the value of this constant.172 7 Hilbert Space Theory Let the Wronskian of u and v be given by the formula ⎛ ⎞ u v W = u v − u v = det ⎝ ⎠. We will show that W is a nonzero constant function. To solve for φ and ψ. b W (t) α x We now have an integral formula for y: 1 x b y(x) = v(x) f (t) u(t) dt + u(x) f (t)v(t) dt . (7. We know that v(a) = 0 and u(b) = 0. b].20). The results are x f (t)u(t) 1 x ψ(x) = dt = f (t)u(t) dt.19) u v Then. we recall that u and v satisfy the initial conditions (I C1 ) and (I C2 ). Therefore. b] × [a.20) W W provided that W (x) = 0 for any x ∈ [a. t) = α (7.18) becomes fv fu φ = − and ψ = . we compute the derivative: W = u v + u v − u v − u v = u v − u v. b] → R by ⎧ ⎨ 1 v(x) u(t) if t ≤ x. we now integrate the equations in (7. K(x. Observe that W (a) = u(a) v (a) − u (a) v(a) = −v(a) and W (b) = u(b) v (b) − u (b) v(b) = u(b). u + qu = 0 and v + qv = 0. By assumption. α . and so W = u(− qv) − (− qu)v = 0. W is a constant function. a W (t) α a and x b f (t)v(t) 1 φ(x) = − dt = f (t)v(t) dt. and so W is a nonzero constant function.21) ⎩ 1 u(x) v(t) if t ≥ x. respectively. To verify that W is nonzero. (7. To that end. b). b]. whenever |x−y| < δ.7. for any f ∈ L2 (a. To see this. y) = K(y. a By Hölder’s Inequality. b]. b] × [a. b]. Then there exists a nonzero scalar λ such that TK e = λe. let > 0 be given and observe that for all x and y in [a. b |TK f (x) − TK f (y)| = K(x. by Hölder’s Inequality.22) a Furthermore.b] ≤ b − a KC([a. For each x ∈ [a.b) .b) . f ∈ L2 (a. a a The function K is continuous on the compact set [a. t)| < √ . K(x. Since K is a symmetric (hermitian) kernel.b]×[a. |TK f (x)−TK f (y)| ≤ whenever |x − y| < δ. Consequently. (7. f L2 (a.4 Sturm–Liouville Systems 173 Then b y(x) = K(x. The argument to show TK is bounded on L2 (a. x ∈ [a. √ TK f C[a. b]. t) − K(y. b]. b]. and hence it is bounded on [a. In this case. b]. t) − K(y. t)f (t) dt. this is bounded by b 1/2 |K(x. b) → C[a. since K is continuous. t) f (t) dt . we can actually improve this to the statement TK : L2 (a. b] and is symmetric. a The map K is continuous on [a. and so TK f is continuous on [a.. b] is bounded and TK ≤ b − a KC([a. b).17). t)|2 dt f L2 (a. b]. b] × [a. x) for all x and y in [a. K is continuous on a compact set. b] × [a. b). (Note that λ = 0 because . t)f (t) dt.b) . b) → C[a.e. we have b b 1/2 |TK f (x)| = K(x.b]) f L2 (a.b]×[a. t)|2 dt f L2 (a. i. Therefore. we know that TK defines a bounded linear map into L2 (a. t)f (t) dt ≤ |K(x. b]. b) is similar. We first show that TK f is continuous for all f ∈ L2 (a. It follows that K defines a compact symmetric (hermitian) operator b TK f (x) = K(x. Suppose now that e is an eigenvector for TK (an eigenfunction in this case). a By assumption. b) we have that y = TK f is a solution to the system in (7. √ Therefore.b]) . t) − K(y.b) b − a for all t ∈ [a. and so there exists a δ > 0 such that |K(x. TK : L2 (a. Since λn → 0 as n → ∞ (by Theorem 6. en ) en . The result follows because λn = 1/αn . for all n ∈ N. the solution to the system of differential equations ⎧ ⎨y + q(x)y = f (x). There is a sequence of scalars (αn )∞ n=1 such that |αn | → ∞ as n → ∞ and such that. b).174 7 Hilbert Space Theory y = TK e is a solution to y + qy = e. Since (en )∞ n=1 is an orthonormal basis for L2 (a. α n=1 n Proof Let TK be the operator defined by (7. we conclude that ∞ y = TK f = (TK f . TK en ) = λn (f . suppose f ∈ L2 (a.22). b]. λ This leads us to the following theorem. en ). n=1 Since TK is symmetric (hermitian). there exists a twice-differentiable function en ∈ C[a. which contradicts the assumption that e is an eigenvector.39). b).30. or 1 e + qe = e. Theorem 7. is given by ∞ 1 y= (f . b). en (a) = 0. If y = TK e = 0. 2 .21) and (7. it follows that |αn | → ∞ as n → ∞. b] that satisfies en + qen = αn en . b). and define αn = 1/λn . let λn be the eigenvalue associated with en . en (b) = 0. en ) = (f .41 Let a and b be real numbers such that a < b and let q ∈ C[a. For each n ∈ N. b). en ) en . Finally. the operator TK has a sequence of eigenvectors (en )∞ n=1 that form an orthonormal basis for L2 (a.37. By Theorem 7. We have demonstrated that y = TK f is a solution to the given differential system.) We know that y = TK e is a solution to y + qy = e (by construction) and that TK e = λe. the differential equation implies that e = 0. and (en )∞ n=1 forms an orthonormal basis for L2 (a. Substituting y = λe into the differential equation. 2 ⎩y(a) = y(b) = 0. or even Theorem 7. where λ = 0. Furthermore. we have (λe ) + q(λe) = e. for each n ∈ N. we have (TK f . We know that the eigenvectors satisfy the differential equation because of the discussion prior to the statement of the theorem. f ∈ L (a. The differential equation becomes y − β 2 y = 0.23) ⎩ 1 x (t − π). 0 The next step is to calculate the eigenvectors and eigenvalues for TK . which is described by the following system of differential equations: ⎧ ⎨y = f (x). we consider the initial value problem at the right endpoint: y = 0. y (0) = 1. where A and B are real numbers. First consider the initial value problem at the left endpoint: y = 0. (DE ) ⎩y(0) = y(π) = 0. we can consider solutions to the equation y − α y = 0. We know that α = 0. and the general solution to this differential equation is y(x) = A eβx + B e−βx . (BC ) We will parallel the argument used in the general case to see how it works in this example.24) The differential equation in (7. π ]. π As usual. The boundary conditions imply that A = 0 and B = 0. K(x.21): ⎧ ⎨ 1 (x − π) t. y(π ) = 0. t) = π (7. y(0) = 0.4 Sturm–Liouville Systems 175 Example 7. we can calculate it explicitly: W = u v − u v = π. The solution to this initial value problem is v(x) = x − π for all x ∈ [0. x ∈ [0.41. π ]. y (π ) = 1.42 We return now to the example of the vibrating string. Suppose α > 0. y(π) = 0. .7. f ∈ L2 (0. because the only solution to the homogeneous equation is y = 0. y(0) = 0. Then α = β 2 for some β > 0. if t ≥ x. and so there are no positive eigenvalues. (7. From the general case. Next. t) f (t) dt. According to Theorem 7. We now define the kernel K as in (7.19) is a constant. In this example. and the solution depends on the sign of α. π ].24) is a homogeneous linear second order ordinary differential equation. π ). f ∈ L2 (0. The solution to this initial value problem is u(x) = x for all x ∈ [0. we know that the value of W in (7. if t ≤ x. π ). we let TK be the Hilbert-Schmidt operator with kernel K: π TK f (x) = K(x. 39: ∞ π π 1 4 = |K(x.) . The boundary condition y(0) = 0 implies that A = 0.26) n=1 n 0 0 π x π π 1 1 2 = 2 (x − π) 2 2 t dt dx + 2 x (t − π )2 dt dx.π ) = 2 sin (nx) dx = .21. the eigenvalues of TK are 1 λn = − . We wish to normalize y. 0 2 Therefore. Computing the L2 (0. 0 0 π 0 x π We leave it to the reader to verify this equality. √ Theorem 7.176 7 Hilbert Space Theory Now let α = −β 2 .43 ( 2/π sin (nx))∞ n=1 is an orthonormal basis for L2 (0. Consequently.42. We have established that the solutions to (7. (7. Proof See Theorem 7. Since we have an explicit formula for the kernel K of TK (see (7. α = −n2 .41 and Example 7. for each n ∈ N. where β > 0. t)|2 dt dx (7. π )-norm of y: π 1/2 . we also get an interesting summation formula via Theorem 7. The differential equation becomes y +β 2 y = 0. π ).23)). the condition y(π ) = 0 implies that sin (βπ ) = 0.25) n2 where λn is the eigenvalue corresponding to en . At the other endpoint (and setting A = 0). n ∈ N. n ∈ N. (See Exercise 7.π yL2 (0. and this has general solution y(x) = A cos (βx) + B sin (βx). This happens whenever β ∈ N (since β > 0). 2 The eigenvalues of TK are the reciprocals of the αn values. π We now have the following interesting result. we let - 2 αn = −n 2 and en = sin (nx).24) are: y = sin (nx). where A and B are real numbers. vφ ) for all x ∈ H . Show that x + y = x + y if and only if y = cx.) Exercise 7.7 Suppose A is a bounded linear operator on the complex inner product space H .9 Let H be an inner product space and assume T : H → H is a bounded linear operator. (This equivalence cannot hold in a real Hilbert space because it is necessarily true that (T x.. Show that (Ax. x) = (Bx. x −y −i A(x +iy).3 Suppose T and S are hermitian operators on a Hilbert space. x) ∈ R for all x ∈ H . (b) Show that if (Ax. Exercise 7. Show that H is an inner product space. 4 Exercise 7. where c > 0. x) for all x ∈ H and H is a complex inner product space. T (y)) = (x.) Exercise 7.10 Let T : H → H be such that T (0) = 0 and T (x) − T (y) = x − y for all x and y in H .).4 Let H be a normed space that satisfies the Parallelogram Law (The- orem 7. Exercise 7. (Hint: Use the polarization formulas given after the statement of Theorem 7. y) for all x and y in H . x +y − A(x −y).10. let vφ ∈ H be the unique element (from the Riesz–Fr échet Theorem) satisfying φ(x) = (x. and (ST )∗ = T ∗ S ∗ .Exercises 177 Exercises Exercise 7. (c) What assumptions need to be added to A and B in order for (b) to hold when H is a real inner product space? Exercise 7. x∈BH Show that the same formula holds in a real inner product space if A is hermitian. (αT )∗ = α T ∗ . y) for all x and y in H ). Exercise 7. x −iy . x +iy +i A(x −iy). Show that T = T ∗ if and only if (T x. (Hint: Show first that (T (x).2 Let H be a Hilbert space. Exercise 7. then A = B.5 Let H be a complex inner product space and assume A : H → H is a bounded linear operator. For each φ ∈ H ∗ . Show that T is a linear isometry from H to itself. y) = (Bx.11 Let H be a Hilbert space. . Show that A = sup |(Ax. y) can be written as 1 . then A = B. / A(x +y).8 Let H be a complex Hilbert space and let T : H → H be a bounded linear operator.6 Let H be an inner product space and assume A and B are bounded linear operators on H . Exercise 7. Exercise 7. (a) Show that if (Ax.1 Let H be a Hilbert space and suppose x and y are nonzero elements of H .10. x)|. Show that T S is hermitian if and only if T S = ST . y) ∈ R for all x and y in H when H is a real Hilbert space. Show that T ∗ T = T 2 . Suppose T and S are operators on H and let α ∈ C. Show that (T + S)∗ = T ∗ + S ∗ . 178 7 Hilbert Space Theory If TO∗ : H ∗ → H ∗ is the operator adjoint of T (see Definition 3. Exercise 7.17 Let H be an infinite-dimensional Hilbert space with inner product (·. w) × (H .14 Let H be a Hilbert space and suppose (xn )∞ ∞ n=1 and (yn )n=1 are sequences ∞ ∞ in H .36) and TA∗ : H → H is the Hilbert space adjoint of T (see Definition 7.15 (Hellinger–Toeplitz Theorem) Let H be a Hilbert space. show that limn→∞ (xn . If (xn )∞ ∞ n=1 and (yn )n=1 are sequences in BH . and lim n→∞ (xn . respectively. Prove the following: If T : H → H is a linear map that satisfies the equation (T x. then T is continuous. yn ) = 1. show that vTO∗ (φ) = TA∗ vφ for all φ ∈ H ∗ . Show that the function (·. yn ) = (x.13 If V is a closed subspace of a Hilbert space H . show that limn→∞ xn − yn = 0. What is (V ⊥ )⊥ if V is not closed? Exercise 7. but that it is not continuous on the product (H . If (xn )n=1 and (yn )n=1 converge (in norm) to x and y. w).18 Solve the system of differential equations . show (V ⊥ )⊥ = V . ·). y) = (x. Exercise 7. Exercise 7. T y) for all x and y in H .16 Let H be a Hilbert space with inner product (·. ·) and norm · . w) × (H . (In this problem. (H . Exercise 7. w) → C is continuous in each argument separately. w) denotes the Hilbert space H endowed with the weak topology. ·) : (H . show H = V ⊕ V ⊥ . Exercise 7.18).12 If V is a closed subspace of a Hilbert space H .) Exercise 7. y). 20 A theorem from linear algebra states that the trace of a square matrix equals the sum of its eigenvalues. b) is a Hilbert–Schmidt b operator defined by the formula TK f (x) = a K(x.7).25).43. y(2π) = 1. Let K be the kernel in (7. ∞ 1 π2 Exercise 7. λ ∈ R. 2π ).1. dθ Use your answer to show that (en )n∈Z is an orthonormal basis for L2 T. y + λ2 y = 0. where en (θ) = einθ and θ ∈ T = [0. 2π . b) → L2 (a.18. 2π) and Exercise 7.) . π ) and Theorem 7. x) dx. (b) Use the function f (θ) = θ for all θ ∈ [0. (Compare to (6. Exercise 7.19 Use Parseval’s Identity (Theorem 7. If TK : L2 (a.23) and show trace(TK ) = ∞ n=1 λn . then the trace of the Hilbert–Schmidt operator TK is defined to be b trace(TK ) = K(x.25) to show that = : n=1 n2 6 (a) Use the function f (x) = x for all x ∈ [0. where (λn )∞ n=1 is the sequence of eigenvalues for TK given in (7. y(0) = 1. a whenever it exists. y)f (y) dy. {f . (a) Show there exists an x ∗∗ ∈ X∗∗ such that x ∗∗ = 1 and d(x ∗∗ . 1] → R is called absolutely continuous on [0.26) to show that = . y ≤ 1. Find the element φa ∈ H such that Λa (f ) = (f . (c) Let x and x ∗ be as found in (b). Show that there exists x ∗ ∈ X ∗ with x ∗ = 1 and x ∗∗ (x ∗ ) > 1 − /2. 1] such that f (0) = 0 and f ∈ L2 (0. Exercise 7. Show that Λa is a bounded linear functional. (a) Let H denote the collection of all (real-valued) absolutely continuous functions on [0. Σ).26 Suppose X is a non-reflexive Banach space and let > 0 be given. Show that T is the limit (in operator norm) of a sequence of finite-rank operators. Define a map V : L2 (μ) → L2 (ν) by V (f ) = φf for all f ∈ L2 (μ). 1). Use Goldstine’s Theorem to show that there exists a y ∈ X with y ≤ 1 such that x ∗ (y) > 1 − /2 and y ∗ (y − x) > 1 − . φa ) for all f ∈ H . d is the metric induced by the norm on X∗∗ . 1] if f is differentiable almost everywhere (with respect x to Lebesgue measure).) (b) Let x ∗∗ be as found in (a). Exercise 7. n=1 n4 90 Exercise 7. (d) Deduce that every uniformly convex space is reflexive. Show that this map is a well-defined isometry. x) : x ∈ X} > 1 − . 0 defines a complete inner product on H . (Here. g} ⊆ H . and if f ∈ L1 (0. (b) Show that the map T : H → L2 (0.22 Suppose μ and ν are probability measures on a measure space (Ω. defined by Tf = f for all f ∈ H . (c) Fix a ∈ (0. Find T −1 .25 A Banach space X is called uniformly convex if given > 0 there exists a δ > 0 such that x − y < whenever x ≤ 1. Exercise 7. Pick x ∈ X with x ≤ 1 such that x ∗ (x) > 1 − /2. 1]. Deduce that x + y > 2 − and x − y > 1 − . 1).24 Let H be a Hilbert space and suppose T : H → H is a compact operator.21 Compute the integral in (7. is an isomorphism. Show that 1 (f . 1) satisfies the equation f (x)−f (0) = 0 f (t) dt for all x ∈ [0. Exercise 7. Show that V is an isomorphism if and only if ν μ. g) = f (t) g (t) dt. X) = inf{d(x ∗∗ . and x + y > 2 − δ. 1) and define a map Λa : H → R by Λa (f ) = f (a) for all f ∈ H .Exercises 179 ∞ 1 π4 Exercise 7. Show that there exists y ∗ ∈ X∗ with y ∗ = 1 and y ∗ (x ∗∗ − x) > 1 − .23 Recall that a function f : [0. Assume that μ ν and φ is the Radon–Nikodým √ derivative of μ with respect to ν. Show that a Hilbert space is uniformly convex. . 2 (Hint: Show that the function |1+t| +|1−t| −2 p p |t|p is bounded below.) (b) Deduce from (a) that if f and g are functions in Lp (0.27 For the following questions. assume 2 < p < ∞.180 7 Hilbert Space Theory Exercise 7. (a) Show that there is a constant c > 0 such that 1 |1 + t|p + |1 − t|p ≥ 1 + cp |t|p . t ∈ R. 1). but it is a little more tricky to show. then 1 f + gp + f − gp ≥ f p + cp gp . 1) is uniformly convex. 2 (c) Conclude that Lp (0.) . (This is also true if 1 < p < 2. Definition 8. Universitext. © Springer Science+Business Media.2 in the appendix.2 A Banach algebra is a Banach space that is also an associative algebra with a submultiplicative norm. see Sect. A norm on an associative algebra is submultiplicative if a · b ≤ a b. Throughout this section. An Introductory Course in Functional Analysis. Example 8. (iii) a · (b + c) = a · b + a · c (left-distribution). (ii) (a + b) · c = a · c + b · c (right-distribution). we will confine ourselves to considering complex Banach spaces. LLC 2014 181 A. B. Bowers. and one in which the norm is what is called submultiplicative.3 Suppose X is a Banach space.Chapter 8 Banach Algebras In this chapter. (iv) λ(a · b) = (λa) · b = a · (λb) (bilinearity of scalar multiplication). (For a brief review of results from complex analysis. where a. for all elements a and b in the Banach space. which will allow us to make use of powerful theorems from complex analysis. Kalton. We are interested in a type of associative algebra that is also a Banach space.1 The set A is called an associative algebra (over the scalar field K) if it is a vector space over K together with an operation called multiplication (often denoted either by · or juxtaposition) that satisfies the following operations: (i) a · (b · c) = (a · b) · c (associativity).1007/978-1-4939-1945-1_8 . Definition 8. b. DOI 10. These spaces are called Banach algebras.) 8. we will study Banach spaces with a multiplication. J. and c are elements of the set A and λ is a scalar in K. Then the space L(X) of linear operators on X forms a Banach algebra with multiplication given by composition. N.1 The Spectral Radius We start with some basic definitions and properties. If there is a risk of confusing the identity element 1 in A with the number 1 in the scalar field. b ∈ A. s ∈ K. we can then think of A as a closed subalgebra of a unital Banach algebra. known as the identity operator. (b) Let D = {z ∈ C : |z| < 1}. then ka ≤ La ≤ a. a ∈ A.13). Observe that La ≤ a. We call an algebra unital if there is an identity 1 such that 1 = 1. {f . and so a = b. the function that is constantly 1 on K. has norm one and is such that I ◦ T = T = T ◦ I for all T ∈ L(X). if we let k = 1 1A . To see this. suppose A is a Banach algebra with identity 1A . Elements of a Banach algebra with this property play a special role. For each a ∈ A. Definition 8. Notice also that if La = Lb for a and b in A. but 1A = 1.6 We now consider several examples of commutative Banach algebras. Therefore.4 An element 1 in a Banach algebra A is called an identity element if 1 · a = a = a · 1. Example 8. then La (1A ) = Lb (1A ). Since L1A = 1.) We remark that A(D) can be viewed as a subalgebra . that is. known as the disk algebra. We denote by A(D) the collection of all analytic functions f : D → C that are extendable to elements of C(D). Additionally. When equipped with the supremum norm and pointwise multiplication. a = La (1A ) ≤ La 1A .5 A Banach algebra is commutative if ab = ba for all elements a and b in the algebra. we will assume 1 = 1. then C(K) forms a commutative Banach algebra under pointwise multiplication: (f · g)(s) = f (s) g(s). Definition 8. let La denote left-multiplication: La (b) = ab. because the norm is submultiplicative.182 8 Banach Algebras In the space L(X) we distinguish a particular operator I that has the property I (x) = x for all x ∈ X. the space A(D) is a Ba- nach algebra. we will denote the identity element in A by 1A . (a) If K is a compact Hausdorff space. g} ⊆ C(K). We will assume that all Banach algebras with an identity are unital. We can make this assumption without any loss of generality. This operator. the map a → La determines an embedding A → L(A). (Completeness follows from Morera’s Theorem (Theorem B. Thus. The identity in C(K) is χK . . a1 b0 + a0 b1 . −∞ −∞ and so we have f ∗ g1 ≤ f 1 g1 . .).11). }. It is worth noting that this Banach algebra lacks an identity element. From (8. ∞ ∞ |f (s − t)| ds = |f (s)| ds = f 1 .1 The Spectral Radius 183 of both C(D) and C(T). we deduce that a ∗ b1 ≤ a1 b1 . −∞ −∞ −∞ −∞ By the translation-invariance of Lebesgue measure. (a ∗ b)k = j =0 ak−j bj for k ∈ Z+ .8. Suppose a = (ak )∞ ∞ k=0 and b = (bk )k=0 are sequences in 1 . 2. In the context of the complex plane. s ∈ R. . 1. we define the convolution of f with g to be the function ∞ (f ∗ g)(s) = f (s − t) g(t) dt. −∞ To compute a bound for the L1 -norm of f ∗ g. The inclusion of A(D) in C(T) follows from the Maximum Modulus Theorem (Theorem B. we use T to denote the unit circle T = ∂D = {z ∈ C : |z| = 1}. ∞ ∞ ∞ ∞ |f (s − t) g(t)| dt ds = |f (s − t)| ds |g(t)| dt. We will turn this space into a convolution algebra. we observe that ∞ ∞ ∞ |(f ∗ g)(s)| ds ≤ |f (s − t) g(t)| dt ds. and so 1 becomes a Banach algebra with this multiplication. Define the product a ∗ b to be the sequence of coefficients of the formal power series ∞ ∞ k k ak t bk t . −∞ −∞ −∞ By Fubini’s Theorem. (c) Consider the sequence space 1 = 1 (Z+ ). . We will define a product ∗ on 1 . (d) Now consider the function space L1 (R). where Z+ = {0. . and so a ∗ b = (a0 b0 . a2 b0 + a1 b1 + a0 b2 . . The space L1 (R) becomes a Banach algebra with multiplication given by convolution. This example is known as a convolution algebra because the product is known as a convolution. If f and g are functions in L1 (R). (8. .1) k=0 k=0 k Then.1). Equivalently. Then A is a unital Banach algebra with identity element (0. A function f ∈ C(K) is invertible (in the algebraic sense) if f (s) = 0 for any s ∈ K. . we may simply add an identity element to A. This is a convention we will adopt throughout the remainder of this chapter. but not every element of a Banach algebra is invertible. consider the following example. ξ1 .9 Consider the Banach algebra L(2 ) of operators on 2 with identity operator I . it is necessarily unique. ξ2 . Neither R nor L has an inverse: R is not onto. only unital Banach algebras.8 Consider the algebra C(K) of continuous functions on the compact Hausdorff space K.6(d)) the Banach algebra A will lack an identity element. b} ⊆ A and {λ1 . without loss of generality. Observe that L and R do satisfy the relationship LR = I . the algebra can always be embedded in an algebra with an identity.6 are commutative. L is the left-inverse of R and R is the right-inverse of L. a −1 is called the inverse of a.) and R(ξ1 . ξ4 . Let A = A ⊕ C. where {a. 1). Recall the left shift operator L and the right shift operator R on 2 : L(ξ1 . . where X is a Banach space of dimension greater than one. . Definition 8. where (again) {a. . but not all Banach algebras are commutative. . λ1 ) · (b. . let 1 be an identity element and let A = A ⊕ 1C. we call b the left-inverse of a whenever b · a = 1. λ1 λ2 ). ξ3 .). . . If it exists.184 8 Banach Algebras All of the Banach algebras listed in Example 8. ξ3 .) = (0. however. . In such a circumstance. and define a multiplication on A by (a. An example of a noncommutative Banach algebra is L(X). This time. we call a the right-inverse of b.7 Let A be a unital Banach algebra. λ2 } ⊆ C. To illustrate this point. . In some cases (such as Example 8. b} ⊆ A and {λ1 . λ2 } ⊆ C. ξ2 . An element a ∈ A is said to be invertible if there exists some a −1 ∈ A such that a a −1 = 1 = a −1 a. ξ3 . the inverse of f is given by the function 1/f . λ2 ) = (ab + λ1 b + λ2 a. . Example 8. when this equation is satisfied. To do this. ξ2 . and L has a zero eigenvalue. define a multiplication by (a + 1λ1 ) · (b + 1λ2 ) = ab + λ1 b + λ2 a + 1λ1 λ2 . Correspondingly. In a Banach algebra A. In the preceding example. . Example 8. . they are not inverses of one another because RL = I . If an inverse exists. where (ξn )∞ n=1 is an element of 2 . The above discussion allows us to consider.) = (ξ2 . In this case. a is invertible. (See Exercise 8. (Sn )n=1 is a Cauchy n n sequence.) The next proposition. Proof Let A be a unital Banach algebra and let G be the group of invertible elements in A. let Sn = 1 + a + · · · + a n . Proposition 8. first due to the mathematician Carl Neumann (in the context of operators). we also have a −1 (a − b) ≤ a −1 a − b < 1.11 Let A be a unital Banach algebra. 2 The collection of invertible elements in a Banach algebra turns out to be very important.12 The group of invertible elements in a unital Banach algebra is an open set. as required. By the choice of b. we have (1 − a) S = S (1 − a) = 1. . it also has interesting topological properties. n=0 Proof For each n ∈ N.) In addition to being a group. Proposition 8. Observe that the series ∞n=0 a n ∞ converges.8. We will find an open ball centered at a that is contained in G.1. Observe that 1 − a and Sn commute for each n ∈ N. then they must coincide (and hence a is invertible).10 (Neumann Series) Let A be a unital Banach algebra and let a ∈ A.1 The Spectral Radius 185 Simple algebra shows that if a has both a left-inverse and a right-inverse. and (1 − a) Sn = Sn (1 − a) = 1 − a n+1 . (See Exercise 8. Definition 8. Taking limits. By assumption. and hence converges to some S ∈ A. then 1 − a is invertible and the inverse is given by the Neumann series: ∞ (1 − a)−1 = an = 1 + a + a2 + a3 + · · · . since a ≤ a and a < 1. Let 1 B = b ∈ A : a − b < −1 . Observe that b = a − (a − b) = a (1 − a −1 (a − b) . will prove central to all that follows. The set G = {a ∈ A : a −1 exists} is called the group of invertible elements in A. a Let b ∈ B. ∞ and so S = n=0 a n is the inverse of 1 − a. If a < 1.5. Consequently. Let a ∈ G. We leave it to the reader to verify that the group of invertible elements is indeed a group. the Banach algebra of n × n complex matrices. . Consequently. B is an open set such that a ∈ B and B ⊆ G. (8. . Since C is algebraically closed. we can even provide a formula for the inverse of b: −1 −1 b−1 = 1 − a −1 (a − b) a = 1 + a (a − b) + a −1 (a − b)a −1 (a − b) + · · · a −1 −1 = a −1 + a −1 (a − b)a −1 + a −1 (a − b)a −1 (a − b)a −1 + · · · .186 8 Banach Algebras Thus.10. A complex scalar λ is in Sp(T ) provided λI −T is a non-invertible matrix. and if b ∈ A is sufficiently close to a. 1]. Example 8. then b−1 = a −1 + (a −1 )2 (a − b) + (a −1 )3 (a − b)2 + · · · . Sp(T ) = {λ1 . by Proposition 8. Consequently. and so Sp(T ) is nonempty for all T ∈ Mn (C). If f ∈ C[0. the operator λ1−T is not invertible. . In the infinite-dimensional case. 2 In the proof of Proposition 8. both a and 1 − a −1 (a − b) are elements of the group G. Since the choice of b ∈ B was arbitrary.13 Suppose A is a complex unital Banach algebra with group of invertible elements G. the spectrum of T is not empty. then b is invertible. for any λ ∈ [0. In fact.15 Define an operator T on C[0. f ∈ C[0. i. when det (λI −T ) = 0. When A is a finite-dimensional space of operators. Example 8. we conclude that 1 − a −1 (a − b) is invertible. The spectrum of a is the subset of C defined by Sp(a) = {λ : λ1 − a ∈ G}. we have λ ∈ Sp(T ) if and only if λ is an eigenvalue for T . Even so. T will always have at least one eigenvalue. the spectrum of an element is well-understood in terms of eigenvalues. x ∈ [0. In fact. λr }. thanks to Proposition 8. 1] by Tf (x) = x f (x). Definition 8.14 Consider Mn (C). Thus. Therefore. we can identify the spectrum of T as the collection of eigenvalues of T .. The resolvent of a is the subset of C defined by Res(a) = {λ : λ1 − a ∈ G}. too. b = a (1 − a −1 (a − b) ∈ G. Therefore.e.10. Let T ∈ Mn (C) be a square matrix.2) If A is a commutative Banach algebra. . for a finite sequence of scalars. we conclude that B ⊆ G. Therefore. however. things are more subtle. 1]. 1].12. where n ∈ N. We saw in Example 6. as the product of two elements in the group. 1]. . Let a ∈ A. in this case. that is.3 that T has no eigenvalues. we saw that if a ∈ A is invertible. where r ≤ n. G is an open set in A. (Note that S is bounded if λ ∈ [0. there exists a nonzero x ∈ X such that (λ1 − T )(x) = 0. but λ1 − T is not bounded below. 1]. 1]. There are three (not necessarily disjoint) possibilities: (1) Suppose λ1 − T is not one-to-one. we can find a sequence (xn )∞ n=1 in X such that xn = 1 for all n ∈ N and such that lim λxn − T xn = 0.3) n→∞ (See Exercise 1.18. It follows that λ is an eigenvalue of T . the spectrum of T is Sp(T ) = [0. (2) Suppose λ1−T is one-to-one. We will reserve the term approximate eigenvalue to describe those λ ∈ Sp(T ) which are not eigenvalues of T . 1]. Recall that an operator is not invertible either because it is not injective or not surjective (or both).20. we call it an approximate eigenvalue of T . there is some x ∗ ∈ X∗ such that x ∗ = 0. but x ∗ ((λ1 − T )(X)) = 0. x ∈ [0. (8. (See Example 8.3) for some sequence (xn )∞ n=1 . and so λ is an eigenvalue of T ∗ . Therefore. λ−x but S is not a bounded operator on C[0.16 Let X be a Banach space and suppose T ∈ L(X).) Example 8. By definition. We conclude that if λ ∈ Sp(T ). The approximate point spectrum of T is the set of all approximate eigenvalues of T . By the Hahn–Banach Theorem. the complex scalar λ is in Sp(T ) whenever λ1 − T is not invertible. (See Exercise 5.8. 1]. It may happen. and so T x = λx. there does not exist a constant c > 0 such that λx − T x ≥ c x for all x ∈ X. Therefore.) It follows that (T ∗ x ∗ )(x) = (λx ∗ )(x) for every x ∈ X. however. In this case.17 Let T ∈ L(X) for a Banach space X.12. but (λ1−T )(X) is a proper closed linear subspace of X. x ∈ [0. The point spectrum of T is the set of all eigenvalues of T . 1].) Definition 8. that is. that λ is both an approximate eigenvalue of T and an eigenvalue of T ∗ . The inverse operator would have to be g(x) Sg(x) = .) When λ satisfies (8.1 The Spectral Radius 187 then (λ1 − T )f (x) = λf (x) − xf (x) = (λ − x)f (x). . then λ is either an eigenvalue of T or T ∗ . (3) Suppose λ1 − T is one-to-one. 1] if λ ∈ [0. or an approximate eigenvalue of T . g ∈ C[0. 1]. then 1 1 2 |(λ − x) fn (x)| ≤ |λ − x| (n|λ − x| + 1) < · n · + 1 = . and so each λ ∈ [0. n n n Therefore. There are many such sequences. Therefore. We already mentioned that T has no eigenvalues. 1]. Then there exists some μ ∈ M[0. the function fn is continuous and fn ∞ = 1.15 that Sp(T ) = [0. It suffices to find a sequence (fn )∞n=1 for which fn is a function that peaks with value 1 at λ and then decreases to zero. then fn (x) = 0. 1] as follows: ⎧ ⎪ ⎪0 if 0 ≤ x < λ − n1 . fn (x) = n ⎪ ⎪ n(λ − x) + 1 if λ ≤ x < λ + n1 . 1]) is defined by the formula: Tf (x) = x f (x). We established in Example 8. 1 For each n ∈ N. x ∈ [0. if |x − λ| < 1/n. On the other hand. 2 λ fn − Tfn ∞ = sup |λ fn (x) − x fn (x)| ≤ . 1] such that T ∗ μ(dx) = λ μ(dx). 1] is an eigenvalue of T ∗ .188 8 Banach Algebras Example 8. ⎪ ⎨n(x − λ) + 1 if λ − 1 ≤ x < λ. then 1 T ∗ μ(f ) = μ(Tf ) = x f (x) μ(dx).e. If a ∈ A. Then for every Borel set A. define fn ∈ C[0. If μ ∈ M[0. 0 Therefore. 1] → M[0. T ∗ δλ = λδλ for each λ ∈ [0.15. then Sp(a) is a nonempty compact set.1] n Naturally. f ∈ C[0. 1]. which means μ = δλ . 1]. For each λ ∈ [0. ⎪ ⎩ 0 if λ + n ≤ x ≤ 1.(μ). 1] be given and for each n ∈ N. . We now show that each point in [0. We start by computing T ∗ : M[0. this tends to zero. 1] is also an approximate eigenvalue of T . 1] and we have shown that each point in [0. Suppose that λ is an eigenvalue of T ∗ .19 Let A be a complex unital Banach algebra. Suppose T ∈ L(C[0. We now identify the eigenvalues of T ∗ . T ∗ μ(dx) = x μ(dx). x∈[0. If |x − λ| > 1/n. let λ ∈ [0. x μ(dx) = λ μ(dx). A A This implies x = λa. and it follows that λ is an approximate eigenvalue for the operator T . we need a sequence (fn )∞ n=1 of continuous functions such that fn ∞ = 1 and λ fn − Tfn ∞ → 0 as n → ∞. Theorem 8. As an example. 1]. the Dirac measure at λ.18 Let us revisit Example 8. 1] is an eigenvalue of T ∗ . 1]. 1]. where fn decreases more rapidly to zero as n increases. Let φ ∈ A∗ and define f : C → C by f (λ) = φ((λ1 − a)−1 ). Define a map ρ : C → A by ρ(λ) = λ1 − a.10) there is a Neumann series for the inverse of 1 + (μ − λ)(λ1 − a)−1 : ∞ −1 1 + (μ − λ)(λ1 − a)−1 = ( − 1)n (μ − λ)n (λ1 − a)−n .8.e. Now we show that Sp(a) is nonempty. then |λ| ≤ a. λ ∈ C. by Proposition 8. f is an entire function).. By assumption. Let G denote the set of invertible elements in A. Then ρ is a continuous map. Consequently. and so −1 (μ1 − a)−1 = 1 + (μ − λ)(λ1 − a)−1 (λ1 − a)−1 Because of (8. Consequently. G is an open set in A. . if λ1 − a is not invertible. Sp(a) ⊆ a BC . n=0 Therefore.1 The Spectral Radius 189 Proof We will first show that Sp(a) is compact. Sp(a) = {λ : λ1 − a ∈ G} = C \ ρ −1 (G) is closed in C. ∞ −1 (μ1 − a) = ( − 1)n (μ − λ)n (λ1 − a)−n−1 . λ ∈ C.4) (We call f the resolvent function. ∞ f (μ) = f (λ) + ( − 1)n (μ − λ)n φ (λ1 − a)−n−1 . we know (μ − λ)(λ1 − a)−1 < 1. μ1 − a is invertible. Suppose μ ∈ C is such that |μ − λ| < (λ1 − a)−1 −1 .5). (8. If |λ| > a. Assume to the contrary that Sp(a) = ∅. Therefore. n=0 Since φ ∈ A∗ .5) Using a bit of algebra: μ1 − a = (λ1 − a) + (μ − λ)1 = (λ1 − a)(1 + (μ − λ)(λ1 − a)−1 ).12. (8. and so is compact by the Heine–BorelTheorem. then 1 − (a/λ) is invertible.10. λ1 − a is invertible. and thus (by Proposition 8. and so ρ −1 (G) is an open set in C. Then λ1 − a is invertible for all λ ∈ C. n=1 It follows that f is analytic on C. as well. Let λ ∈ C. We will show that there is a neighborhood of λ in which f has a power series expansion. Thus. By Proposition 8.) We claim that f is analytic on C (i. |λ|→∞ |λ| − a we conclude that f = 0. if |λ| ≥ 2a.6). This is a contradiction. we view C as a real Banach algebra. because of (8. Theorem 8.10. a = λ1.6) |λ| 1 − a |λ| |λ| − a Because f is analytic on C.190 8 Banach Algebras We claim that f is a bounded function. ∞ ∞ 1 −n φ a n |f (λ)| ≤ |λ| φ a = n . 2 Theorem 8. To that end. there exists some M ≥ 0 such that |f (λ)| ≤ M for all |λ| ≤ 2a. The . n=0 Consequently. by assumption. ∞ f (λ) = φ((λ1 − a)−1 ) = λ−1 λ−n φ(a n ). then A is isometrically isomorphic to C. the geometric series converges. f is constant. λ1 − a is invertible and has Neumann series ∞ −1 −1 (λ1 − a) = λ λ−n a n . We have established that f (λ) = 0 for all λ ∈ C. or H. (8. for |λ| > a. we recall that an algebra is a field if it is commutative and if every nonzero element is invertible. n=0 Thus. Proof Let a ∈ A. by Proposition 8. 2 For the next theorem. C.4) was arbitrary. for |λ| > a. then A is R. the only noninvertible element is 0.20 remains true if we replace field with skew-field (also known as a noncommutative field or division algebra).19. By assumption. By Theorem 8. Therefore. If A is a real Banach algebra which is a skew-field. Thus there exists some λ ∈ C such that λ1 − a is not invertible. The only way this can happen is if (λ1 − a)−1 = 0 (because of the Hahn–Banach Theorem). |λ| n=0 |λ| n=0 |λ| Since a/|λ| < 1. suppose that |λ| > a. In fact. Here. Sp(a) is nonempty. It follows that f is bounded by max{M. Therefore. Hence. If A is a field. and so φ 1 φ |f (λ)| ≤ · = . Then 1 − a/λ is invertible. by Liouville’s Theorem (Theorem B. since φ lim = 0. then |f (λ)| ≤ φ/a. and so we conclude that φ((λ1 − a)−1 ) = 0 for all φ ∈ A∗ . φ/a}. We have established that f is a bounded entire function. and so λ1 − a = 0. On the other hand. and so we conclude that Sp(a) is nonempty.20 (Gelfand–Mazur Theorem) Suppose A is a complex Banach algebra.14). and the result follows. The choice of φ ∈ A∗ in (8. and so we know there is some λ ∈ Sp(a) which achieves the maximum in Definition 8. More precisely. If a ∈ A. however.) Let n ∈ N and λ ∈ Sp(a). or H [25]. then λn ∈ Sp(a n ) for all n ∈ N.21 Let A be a unital Banach algebra. (We use H in honor of the Irish mathematician Sir William Hamilton. (Because then |λ|n ≤ a n . His original proof did not appear until much later. it follows that λ1 − a is invertible.) This contradicts the assumption that λ ∈ Sp(a).1.) We remark that H is not a complex Banach algebra because it does not satisfy property (iv) of Definition 8. C.23 (Spectral Radius Formula) Let A be a unital Banach algebra. then r(a) ≤ a n 1/n for all n ∈ N. one which provides a formula for computing the spectral radius. then the number r(a) = max{|λ| : λ ∈ Sp(a)} is called the spectral radius of a. The Gelfand–Mazur Theorem dates back to 1938. Observe that λn 1 − a n = (λ1 − a)(λn−1 1 + λn−2 a + · · · + λa n−2 + a n−1 ) = (λn−1 1 + λn−2 a + · · · + λa n−2 + a n−1 )(λ1 − a). (See Exercise 8. 2 We now come to a key result. we have (λ1 − a)−1 = (λn 1 − a n )−1 (λn−1 1 + λn−2 a + · · · + λa n−2 + a n−1 ).10 would imply that λ1 − a is invertible. n∈N n→∞ .21. Thus. We also know that r(a) ≤ a. Proof It will suffice to show that if λ ∈ Sp(a). then a/λ < 1.8. Theorem 8.1 The Spectral Radius 191 symbol H denotes the quaternions. then r(a) = lim a n 1/n = inf a n 1/n . Lemma 8. If a ∈ A.1. From this. the spectrum of a is nonempty and compact. By Theorem 8. which is a contradiction. If a ∈ A. Then 1 = (λ1 − a)(λn−1 1 + λn−2 a + · · · + λa n−2 + a n−1 )(λn 1 − a n )−1 = (λn 1 − a n )−1 (λn−1 1 + λn−2 a + · · · + λa n−2 + a n−1 )(λ1 − a). λn 1 − a n is not invertible. in 1973 [37]. when Mazur stated in an article that every normed division algebra over R is either R. and so λn ∈ Sp(a n ). Suppose λn 1 − a n is invertible. The theorem now called the Gelfand–Mazur Theorem was proved by Gelfand in 1941 [14]. we know that r(a) ≤ inf a n 1/n ≤ lim inf a n 1/n .19. because if λ > a.22. who first described the quaternions in 1843.22 Let A be a unital Banach algebra. and so Proposition 8. Definition 8. n→∞ n∈N Proof By Lemma 8. as required. . Let ρ ∈ R be such that ρ > r(a). we observe that lim |ξ n φ(a n )| = 0 n→∞ for each ξ ∈ B. there exists a constant Cρ (that depends on the choice of ρ) such that Cρ > 0 and ρ −n a n ≤ Cρ . Then 1/ρ ∈ B. Then ξ a < 1. Suppose that ξ ∈ C is such that |ξ | < 1/a. In particular. (There will be a singularity at any λ ∈ Sp(a). λ ∈ C. to show that lim sup a n 1/n ≤ r(a). for any |ξ | < 1/a. ∞ ∞ −1 F (ξ ) = φ((1 − ξ a) ) = φ n n ξ a = ξ n φ(a n ). and so we conclude that ∞ F (ξ ) = ξ n φ(a n ). however. (8.7) (This is the resolvent function defined in (8. and so (by Proposition 8.10) there is a Neumann series for (1 − ξ a)−1 : ∞ (1 − ξ a)−1 = 1 + ξ a + ξ 2 a 2 + · · · = ξ nan. n∈N n∈N ∗ This is true for all φ ∈ A .1.6) converges for all ξ ∈ B.8) n=0 Since the series in (8. and so by the Uniform Boundedness Principle. n ∈ N.19 that f is an analytic function in the region |λ| > r(a). Then. n→∞ Suppose φ ∈ A∗ and let f (λ) = φ((λ1 − a)−1 ). ξ ∈ C. we conclude that F is an analytic function on the open disk B = {ξ : |ξ | < 1/r(a)}.) We know from the proof of Theo- rem 8. but otherwise arbitrary. we have a n 1/n ≤ Cρ1/n ρ.4). Since F (ξ ) = ξ −1 f (ξ −1 ). We will demonstrate.) Now define a new function: F (ξ ) = φ((1 − ξ a)−1 ).192 8 Banach Algebras It will suffice. we have that a n ≤ Cρ ρ n . except possibly at the origin ξ = 0. for all n ∈ N. that F has a power series on B centered at ξ = 0. ξ ∈ B. (8. n=0 Then. and so sup |φ(ρ −n a n )| = sup |ρ −n φ(a n )| < ∞. therefore. n=0 n=0 The power series expansion is unique. due to Fredholm. then either λ1−K is invertible or λ is an eigenvalue of K. dim(X/ran(λ1 − K)) = dim(ker(λ1 − K)) = 0. Certainly.25 (FredholmAlternative) Let K be a compact operator on the Banach space X. If a ∈ A is such that r(a) = 0. x ∈ [0. the proof is n→∞ n→∞ complete. n→∞ The choice of ρ > r(a) was arbitrary. Proof Let λ be a nonzero scalar and assume that λ is not an eigenvalue of K. it is compact. by the Rank-Nullity Theorem (Theorem 6.41 and 6. 1]. By the initial assumption. n→∞ n∈N n→∞ Since it is always the case that lim inf a n 1/n ≤ lim sup a n 1/n .37. Theorem 8.33). 1) → L2 (0.27. and so λ1 − K is a surjection. and so 0 ∈ Sp(V ). 2 Observe that if a is such that r(a) = 0.25 (the . We know that all eigenvalues of K are in Sp(K). by assumption. If λ is a nonzero scalar. Then. 2 Example 8.1 The Spectral Radius 193 Therefore. 0 Since V is a Hilbert–Schmidt operator. K cannot be invertible. it was established that V has no eigenvalues.27. This motivates the next definition. λ1−K is an injection.8. and so lim sup a n 1/n ≤ r(a). lim sup a n 1/n ≤ ρ. 1) defined by x Vf (x) = f (t) dt. then a is called quasinilpotent. By Theorem 8. n→∞ It follows that lim sup a n 1/n ≤ r(a) ≤ inf a n 1/n ≤ lim inf a n 1/n . f ∈ L2 (0. In Ex- ample 6. From Theorem 6. Let us now return our attention to compact operators on a Banach space. Thus.24 Let A be a unital Banach algebra. we also know that the only possible limit point of the eigenvalues of K is 0. then (by Theorem 8. and so 0 ∈ Sp(K). and therefore λ1−K is invertible. 1). We are left with the question: What other elements of the spectrum are not eigenvalues? This leads us to a classical theorem. It follows that ran(λ1 − K) = X. The Volterra operator is a map V : L2 (0. Let X be an infinite-dimensional Banach space and suppose K ∈ L(X) is a compact operator.23) it must be the case that limn→∞ a n 1/n = 0.26 Recall the Volterra operator from Examples 3. Definition 8. ker(λ1 − K) = {0}. . ) = ξ1∗ (1. ) = (λξ1∗ . It follows that ξ ∗ = (ξ1∗ . ξ = (ξn )n∈Z ∈ p (Z). .16. For ease of notation. denote the dual action of q on p by ξ ∗ (ξ ) = ξ . ξ ∗ ∈ q . where ξ = (ξk )∞ k=1 is any sequence indexed by the natural numbers. and hence R ∗ = L. Then ∗p = q . .194 8 Banach Algebras Fredholm Alternative). More precisely. the set Sp(R) is closed. Then (ξ2∗ . ξ ∈ p . . R is actually invertible. A moments thought will reveal that Rξ . so R −1 = L (where L is defined in the obvious way). We will compute the eigenvalues of R ∗ : q → q .27 Suppose R and L are the right and left shift operators. we conclude that r(R) = 1. . λ. Consequently.) Example 8. and so it follows that R n = 1 for all n ∈ N.19.). We wish to identify Sp(R). and consequently ξn+1 = λn ξ1∗ . . By the Spectral Radius Formula (Theorem 8. . r(V ) = 0 and V is quasinilpotent. By Theorem 8. ξ2 . then the operator λ1 − R is one-to-one and does not have closed range. λξ3∗ . . it must be the case that every point in T = ∂D is an approximate eigenvalue of R. we have established mutual inclusion. ∞) and consider the Banach space p (Z) of doubly infinite p-summable sequences. λξ2∗ . Suppose that λ ∈ C is an eigenvalue of L : q → q . Rξ p = ξ p for all ξ ∈ p . Thus. ξ4∗ . it follows that Sp(V ) = {0}. ξ3 . we have ξn+1 = λξn∗ . ξ3∗ . λ2 . Let q be the conjugate exponent for p.). ξ1 . . . ∗ ∗ Therefore. . Therefore. . ξ ∗ for all ξ ∈ p and ξ ∗ ∈ q . Suppose ξ ∗ = (ξn∗ )∞ n=1 . it follows that if |λ| = 1. . . We have observed that R has no eigenvalues.). for all n ∈ N. Example 8. ) and Lξ = (ξ2 . The first step is to identify R ∗ . . The right shift operator R : p (Z) → p (Z) is now given by the formula R(ξ ) = (ξn−1 )n∈Z . and so we must look elsewhere to find elements in the spectrum of R. (Recall that D denotes the open unit disk in C. Such a sequence is in q if and only if |λ| < 1. Then Rξ = (0. In this case. (See Example 8. Then there is some ξ ∗ ∈ q such that Lξ ∗ = λξ ∗ .23). and so D ⊆ Sp(R). and hence Sp(R) = D. so that R = 1. Since no λ ∈ T is an eigenvalue of L. and the inverse is the left shift operator. ξ ∗ = ξ . We have that R n ξ p = ξ p for all ξ ∈ p . .) It is easy to see that R has no eigenvalues. and we have shown that every point in D is an eigenvalue of R ∗ = L. . λ2 ξ1∗ . and hence Sp(R) ⊆ D. λ ξ1∗ . we conclude that D ⊆ Sp(R). . We have determined that λ is an eigenvalue for R ∗ = L if |λ| < 1.28 Let p ∈ (1. If 1 < p < ∞. . then R : p → p is a bounded linear operator. respectively. Lξ ∗ . Since the spectrum of R is the closed unit disk. neither R nor L has any eigenvalues.4. In fact. Since ξ ∈ p (Z). and so (ξn−1 )n∈Z = (λξn )n∈Z . . we will use T to mean the unit circle. 2π )}.) Example 8.29. (See Exercise 8. Define the multiplier operator M : L2 → L2 by Mf (θ) = eiθ f (θ ). The conclusion is that Sp(R) ⊆ T.29. given by the Fourier transform f → (fˆ(n))n∈Z . 2π dθ ) and 2 (Z). if λ ∈ Sp(R). (Note that λ = 0. θ ∈ [0. ξn−1 = λ ξn for all n ∈ Z. the two sets are readily identifiable. making use of the identification ∂D = {e2π iθ : θ ∈ [0. there is an isometric isomorphism between the Hilbert spaces L2 = L2 ([0.) Then. by Parseval’s Identity. (This usage indicates the origin of the name torus for the symbol T. 2π ). To see that R has no eigenvalues. we used the symbol T to denote the interval [0. In fact. 2π ). however. ξ = ξ0 (λ−n )n∈Z . then λ ∈ D and 1/λ ∈ D. Similarly. this cannot happen.10. In the context of complex analysis.) Remark 8. λn Therefore. it is still the case that R n = 1 for all n ∈ N. however. r(L) = 1. Despite the previous remarks. 1)}. we adopt the convention that T denotes the unit circle ∂D = {z ∈ C : |z| = 1}. we saw that. 1). f ∈ L2 . we have R ξ = λ ξ .) Although this may seem to be an overuse of notation. Consequently. it must be that both |λ| < 1 and 1 |λ| < 1.) Therefore. the multiplier operator M on L2 corresponds to the shift operator R on 2 (Z). Naturally.30 In Example 7. (Consider sequences ξ with ξ−n = 0 for all n ≥ 0. It is routine to show that λ ∈ Sp(R) if and only if 1/λ ∈ Sp(R −1 ). For the remainder of this text.29 In Example 7. Sp(R) = T. (See Exercise 8. and consequently r(R) = 1. From this we conclude that 1 ξ−n = λn ξ0 and ξn = ξ0 . for some ξ = (ξn )n∈Z . suppose λ ∈ C is an eigenvalue of R. since the unit circle can be written as {z ∈ C : |z| = 1} = {eiθ : θ ∈ [0.8. n ∈ N. With regards to the isometric isomorphism. 2π ). some authors prefer to use the symbol T to denote the unit interval [0. and so we have both Sp(R) ⊆ D and Sp(R −1 ) ⊆ D.1 The Spectral Radius 195 This time. A similar argument shows that L has no eigenvalues. because of the previous example. Calkin. as required. It follows that ab+I ≤ a+I b+I . n n Since this bound holds for all n ∈ N. by making some definitions in the general context of an algebra. it is called a two-sided ideal. Let a and b be fixed elements in A.31 A nonempty linear subspace I of an algebra A is called a left ideal of A if ax ∈ I whenever a ∈ A and x ∈ I .47. In any case. a proper ideal cannot contain an element that is invertible. we may select xn ∈ a + I and yn ∈ b + I so that 1 1 xn < a + I + and yn < b + I + . which contradicts the fact that I is a proper ideal.33 Let H be an infinite-dimensional separable Hilbert space. and hence 1 1 ab + I ≤ xn yn < a + I + b + I + . {a.2 Commutative Algebras In this section (as before). I is called proper if I = A. and {0} is always a proper two-sided ideal in any unital algebra. We now define a multiplication on A/I : (a + I ) · (b + I ) = ab + I . The space L(H ) of operators on H is a Banach algebra. then it must contain every element of A. It remains to show that ab + I ≤ a + I b + I for all a and b in A. If x ∈ I and 1 − x < 1. after the mathematician J.10). n n For each n ∈ N.17. then x is invertible (by Proposition 8. however. I is called a right ideal if xa ∈ I whenever a ∈ A and x ∈ I . 1 + I = 1. it is clear that 1 + I is an identity in A/I . Proposition 8. Similarly. Definition 8. 2 Example 8.196 8 Banach Algebras 8. we may take the limit as n → ∞. and the subspace K(H ) of compact operators on H is a closed two-sided ideal. and so A/I is unital. This added structure leads to some remarkable consequences. If I is both a left ideal and a right ideal. we consider complex Banach algebras that are unital. b} ⊆ A. by Proposition 3. we will impose on our Banach algebras the additional restriction of commutativity. W. Therefore. called the quotient algebra. We begin. With this multiplication. (See Theorem 6.) The quotient algebra L(H )/K(H ) is called the Calkin algebra. and so for each n ∈ N. If an ideal contains the identity element of A. where 1 is the identity in A.32 If I is a proper closed two-sided ideal in a unital Banach algebra A. The Calkin algebra is significant because it is not isomorphic to an algebra of operators on a separable Hilbert space (even though H is itself separable) [5]. then A/I is a unital Banach algebra. Proof We know that A/I is a Banach space. Notice that an ideal necessarily contains 0. we have xn yn ∈ ab + I . a proper ideal does not contain the identity element. Shortly. The norm on A/I is an infimum. Thus. Similarly. This . We insist on the condition that φ(1) = 1 in order to disqualify the trivial linear functional φ = 0.8. Example 8.) Proposition 8. (See Exercise 8. this implies that any ideal is necessarily two-sided. By Proposition 8.49. and so φ(1) = 1.35 If φ is a multiplicative linear functional on a commutative complex unital Banach algebra.) In particular. |α| > 1.38 (Krull’s Theorem) Any proper ideal is contained in a maximal ideal. It is a consequence of the multiplicative property of φ that kerφ is an ideal. and so it must be that either φ(1) = 1 or φ(1) = 0. when A is a commutative algebra. and so α −1 x < 1. This is a contradiction. any unital algebra has a maximal ideal.2 Commutative Algebras 197 is remarkable because the Calkin algebra is also a C ∗ -algebra. in particular. In particular. Consequently. φ(α −1 x) = 1. A linear functional φ : A→ C is called a multiplicative linear functional if it is a ring homomorphism. . if φ(1) = 1 and φ(xy) = φ(x) φ(y) for all x and y in A. the quotient A/kerφ is isomorphic to C. then φ is continuous and φ = 1. φ is a multiplicative linear functional. because φ(1 − α −1 x) = 0.10. The multiplicative property of a multiplicative linear functional φ guarantees that φ(1)2 = φ(1). every chain of proper ideals has an upper bound. This means that the codimension of kerφ is one and. that is. but |φ(x)| > 1. Theorem 8. a two-sided ideal is called an ideal. To see this. A proper ideal I is said to be maximal in A if I = J whenever J is a proper ideal such that I ⊆ J .5. and so φ 1 − α −1 x · φ (1 − α −1 x)−1 = 1. (See Definition 8. Proof If (Ji )i∈I is a chain of proper ideals. we will suppose that all Banach algebras are commutative. and so 1 ∈ J . Therefore. The Gelfand-Naimark Theorem dates back to the work of Gelfand and Naimark in 1943 [16]. On the other hand. We claim that the kernel of φ is a closed maximal ideal in A.36 Let A be a commutative algebra. It follows that φ ≥ 1. then J = i∈I Ji is also a proper ideal. It is a proper ideal because 1 ∈ kerφ. and so φ(1 − α −1 x) = 0. it must be the case that φ = 1. By definition. we have that 1 − α −1 x is invertible. Recall that the kernel of φ is the set kerφ = {x : φ(x) = 0}. and from the Gelfand– Naimark Theorem it is known that every C ∗ -algebra is isomorphic to an algebra of operators on some Hilbert space. this tells us that kerφ is a maximal ideal. simply note that 1 ∈ Ji for any i ∈ I . Definition 8.11. Proof By assumption. Let α = φ(x). Consequently. Suppose there exists some x ∈ A such that x ≤ 1. We remind the reader that an algebra A is commutative if ab = ba for all a and b in A. 2 Definition 8. and so the result follows from Zorn’s Lemma. For the remainder of this section.) By Proposition 3. (This is why we assumed φ(1) = 1 in the definition of a multiplicative linear functional.34 Let A be a commutative complex unital Banach algebra.37 Suppose φ is a multiplicative linear functional on a commutative complex unital Banach algebra A. J ) = 1. The proof is the same. Since J [x] = A. Specifically. observe that {0} is a proper ideal. like Zorn’s Lemma.37 and Theorem 8. this theorem is not true (in general) for real Banach algebras. Denote this isometric isomorphism by i : A/J → C. Then J [x] is an ideal which is strictly larger than J . by Proposition 8. and so 1 ∈ J . It follows that A/J is a unital Banach algebra. By the continuity of multiplication. and consequently 1 + J = ax + J = (a + J )(x + J ). Through Example 8. J ) < 1. It will suffice to show that 1 ∈ J . we conclude that A/J is isometrically isomorphic to C. In fact. Proposition 8. Suppose that d(1. Define a subset of A by J [x] = {ax + y : a ∈ A. Proof Let J be a maximal ideal in a commutative complex unital Banach algebra A. J is also an ideal. π (1) = π (ax + y). Consequently.40 Each maximal ideal in a commutative complex unital Banach algebra is the kernel of some multiplicative linear functional. Therefore. who proved the general version of the result in 1929 [22]. Krull’s Theorem asserts the existence of maximal ideals in any unital ring. and so every nonzero element of A/J is invertible. d(1. It follows that (x + J )−1 exists whenever x ∈ J . Krull’s Theorem is equivalent to the Axiom of Choice [18]. By Proposition 8. J is a proper ideal in A. Krull’s Theorem is named after Wolfgang Krull. J ) = 1. we have established a correspondence between the multiplicative linear functionals on a commutative complex unital Banach algebra A and the maximal ideals of A. it must be that 1 = ax+y for some a ∈ A and y ∈ J . y ∈ J }. and so A = J [x]. 2 Theorem 8.32. and so must be contained in a maximal ideal. and so there exists some x ∈ A such that x ∈ J .39 Every maximal ideal in a commutative unital Banach algebra is closed. 2 We are working within the context of Banach algebras. which contradicts the assumption that J is a proper ideal. By assumption. By the Gelfand–Mazur Theorem (Theorem 8. First. because 0 ∈ J .10) that x is invertible. and therefore J = J . 2 Since the proof of the above theorem uses the Gelfand–Mazur Theorem. Indeed. it only works for complex Banach algebras. Proof Suppose J is a maximal ideal in a commutative unital Banach algebra A. and relies on Zorn’s Lemma. Then there exists an element x ∈ J such that d(1. the set J is a closed subset of A. This implies (by Proposition 8.198 8 Banach Algebras To show every unital algebra has a maximal ideal. Then φ = i ◦ π is a multiplicative linear functional on A and kerφ = J . Let π : A → A/J be the quotient map.20). Therefore. J ) ≤ 1. .39. but Krull’s Theorem remains true in the more general setting of ring theory. We will show this by showing that d(1. J is a proper ideal containing the maximal ideal J . We need only show that J is a proper ideal. observe that d(1. x) = 1−x < 1.40. then Sp(x) = x̂(Σ) and x̂C(Σ) = r(x) = lim x n 1/n . The next proposition explains why we call Σ(A) the spectrum of A. the set Σ(A) is sometimes called the maximal ideal space of A. Proposition 8.8. there is some multiplicative linear functional for which this maximal ideal is the kernel. It remains to show that Σ(A) is w∗ -compact. then φ(a)φ(a −1 ) = 1. and so the result follows. φ(xy) = φ(x)φ(y) for all {x. is the collection of all multiplicative linear functionals on A. is a norm-decreasing algebra homomorphism. If x ∈ A.38. (8.) It follows that λ ∈ Sp(x). Σ(A) is nonempty. and so φ(a) = 0.39 (the Banach-Alaoglu Theorem). there exists a maximal ideal in A. we know that BA∗ is a w∗ -compact set. n→∞ Proof Suppose λ ∈ x̂(Σ). and so λ1 − x is not invertible. The Gelfand transform of x is well-defined. The spectrum of A.2 Commutative Algebras 199 Definition 8. then we define the Gelfand transform of x to be the (complex-valued) continuous function x̂ ∈ C(Σ(A)) given by x̂(φ) = φ(x). Consequently.44 Let A be a commutative complex unital Banach algebra and let Σ = Σ(A). Therefore. By Proposition 8. Due to the correspondence between multiplicative linear functionals and maximal ideals. Theorem 8. By Theorem 5. From Theorem 8. which we denote by Σ(A). we deduce that Σ(A) ⊆ BA∗ . (If a is invertible. . it suffices to show that Σ(A) is w∗ -closed. If x ∈ A. y} ⊆ A} (8.43 Let A be a commutative complex unital Banach algebra. Proof Let A be a commutative complex unital Banach algebra. φ ∈ Σ(A).40. Observe that Σ(A) = {φ ∈ BA∗ : φ(1) = 1. By Theorem 8.35.42 The spectrum of a commutative complex unital Banach algebra is a nonempty w∗ -compact set. which is called the Gelfand transform. One must be careful to not confuse the spectrum of an algebra Σ(A) with the spectrum of an element Sp(x). where x is an element in the algebra A. φ(λ1 − x) = 0. We will discover shortly the reason behind this naming.9) ⎛ ⎞ = BA∗ ∩ {φ : φ(1) = 1} ∩ ⎝ {φ : φ(xy) = φ(x)φ(y)}⎠ .10) x∈A y∈A All of these sets are closed in the weak∗ -topology. 2 Definition 8. Observe also that the map x → x̂.41 Let A be a commutative complex unital Banach algebra. Then there exists some φ ∈ Σ such that φ(x) = x̂(φ) = λ = φ(λ1). because Σ(A) is a w∗ -compact subset of BA∗ . Thus. 1]. consider the Volterra operator: x Vf (x) = f (t) dt. we saw that Sp(V ) = {0}. Example 8.200 8 Banach Algebras Now suppose λ ∈ Sp(x). By Theorem 8.) Let A denote the algebra given by the closure of all polynomials in V . 1]. where x is an element in the Banach algebra A.38. there exists some multiplicative linear functional φ such that J = kerφ.45 To emphasize the relationship between Σ(A) and Sp(x). f ∈ C[0. 0 In Example 8.26.40. By Proposition 8. the spectrum of x is sometimes denoted σ (x).46 Once again. that is. (This was a result of the Fredholm Alternative.) 2 Remark 8. (See Theorem 8. and so λ = φ(x) = x̂(φ). there exists a maximal ideal J containing the ideal (λ1 − x)A. The rest of the proposition is the Spectral Radius Formula. x ∈ [0. We conclude that φ(λ1 − x) = 0. . as required. and hence (λ1 − a)A is a proper ideal. because V is a compact operator with no eigenvalues.23. The fact that x̂C(Σ) = r(x) follows from the identification Sp(x) = x̂(Σ). Then λ1 − x is not invertible. By Proposition 8. an ) ∈ Cn+1 . By induction.47 Let A be a commutative complex unital Banach algebra and let Σ = Σ(A). . the Gelfand transform is an isometry. n ∈ N . x 2 = x2 for all x ∈ A. k k First. by Proposition 8. assume x 2 = x2 for all x ∈ A. . the Gelfand transform is an isometry if and only if x 2 = x2 for all x ∈ A. Σ(A) contains only one element φ and n φ ak V k = a0 . .44. Theorem 8. The Gelfand transform is an algebra homomorphism between A and a subalgebra of C(Σ). . n A= ak V k : (a0 . an ) ∈ Cn+1 . Therefore. Therefore. Furthermore. . It follows that φ(V ) = 0 for all multiplicative linear functionals on A. we have V (Σ(A)) = Sp(V ) = {0}. where n ∈ N. k=0 for all (a0 . k=0 We know that A is a commutative Banach algebra. k→∞ Thus. . . . Proof It is clear the Gelfand transform is an algebra homomorphism onto its image. k k x̂C(Σ) = lim x 2 1/2 = x. We show that it is an isometry if and only if x 2 = x2 for all x ∈ A.44. . 47). . an > 0. Now define g ∈ C(K) by n n g= gj g j = |gj |2 . We would like to identify Σ = Σ(C(K)). . By Theorem 8. . assume x̂C(Σ) = x for all x ∈ A. If s ∈ K. all multiplicative linear functionals on C(K) can be achieved as point evaluation. On the other hand. which is a w∗ -open set in Σ. fn } ⊆ C(K) so that {μ ∈ C(K)∗ : |μ(f1 ) − φ(f1 )| < . 2 We now consider some examples of Banach algebras. We will show that. φ(gj ) = 0 for all j ∈ {1. . and so it is a trivial calculation to show it satisfies the relationship x2 = x 2 for every x in the algebra. . define a function gj ∈ C(K) by gj = fj − φ(fj )1. . and we identify the spectrum in each case.66.48 Suppose K is a compact Hausdorff space. and so it must be that Σ = {δs : s ∈ K}. and so g is invertible (in the algebra C(K)). Here.) By definition. j =1 which implies that g is not invertible. and so there exists an n ∈ N. .2 Commutative Algebras 201 Conversely. |μ(fn ) − φ(fn )| < } ⊆ W. . n} such that |gj (s)| ≥ . for any s ∈ K. . Observe that. . . as required. In each of the following examples. the map δs (f ) = f (s). in each case. j =1 j =1 For each s ∈ K. there exists some j ∈ {1. . . (Notice that 1 = χK . is a multiplicative linear functional. f ∈ C(K). the spectrum of C(K). n φ(g) = φ(gj ) φ(gj ) = 0. C(K) denotes the space of complex- valued continuous functions on K. where 1 is the unit element in the algebra C(K). . . Suppose that φ ∈ Σ\{δs : s ∈ K}.8. then. in fact. . the Gelfand transform is an isometry (by Theorem 8. n}. and so there exists some j ∈ {1. n}. . For each j ∈ {1.23 and Proposition 8. Therefore. we conclude that Σ is homeomorphic . Example 8. There exists. . a w∗ -neighborhood W of φ so that W ∩ {δs : s ∈ K} = ∅. . . We have derived a contradiction. The set W is a neighborhood of φ in the w∗ -topology. Consequently. n} such that |fj (s) − φ(fj )| = |δs (fj ) − φ(fj )| ≥ . we will be looking at a Banach algebra with the supremum (or essential supremum) norm. . . 2 2 Therefore. and functions {f1 . the function that is constantly 1 for all s ∈ K. By Lemma 5.44. x = x̂C(Σ) = r(x) = inf x n 1/n ≤ x 2 1/2 . n∈N and hence x ≤ x . By submultiplicativity of the norm. |g(s)| ≥ 2 for all s ∈ K. . then δs ∈ W . x 2 ≤ x2 . . x2 = x 2 . Certainly.50 Consider the space L∞ (0. but it is very difficult to describe.3 The Wiener Algebra Consider the space 1 (Z) of doubly infinite absolutely summable sequences. making 1 (Z) into a convolution algebra. and so we will omit it. a similar theorem does hold for Banach algebras over R. The space Ω is called the Stone space of the measure algebra. then. We will equip 1 (Z) with a multiplication akin to the one in Example 8. Example 8. It turns out that the Gelfand transform is actually an isomorphism. the spectrum of ∞ is the set Σ = βN. however. because there is no (constructive) way to define the value of f ∈ L∞ (0.6(c). the algebra ∞ is isometrically isomorphic to C(Σ). too. Therefore. (See Theorem 4. we obtain N ⊆ Σ.5 in [2]. Certainly.48. j (8. in this example. and so the Gelfand transform is an isometry (by Theorem 8. We remark. (See Sect. This inclusion cannot be equality.11) k=−∞ j =−∞ j =−∞ . so that ξ · η is the sequence with components (ξ · η)n = ξn ηn . it relies on the Gelfand–Mazur Theorem and requires the underlying scalar field to be C. the spectrum of C(K) is in fact K.47 is a statement about complex Banach algebras. 1)). 1) of (equivalence classes of) essentially bounded measurable functions on the unit interval [0. 1) at a given point in [0. however. f 2 ∞ = f 2∞ for all f ∈ L∞ (0.) 8. n ∈ N. We denote the spectrum by Ω = Σ(L∞ (0. For each n ∈ N. where ξ = (ξn )∞ ∞ n=1 and η = (ηn )n=1 are sequences in ∞ .1 for the definition of the Stone–Čech compactification. B.202 8 Banach Algebras to K.) In this example.49 Consider the sequence space ∞ of bounded complex-valued sequences. Theorem 8. that it relies on the Axiom of Choice.47). 1). The proof of this fact is beyond the scope of our current discussion. Ultimately. We define a multiplication in ∞ component-wise. Define the product of ξ and η to be the sequence ξ ∗ η = ((ξ ∗ η)k )k∈Z of coefficients of the formal Laurent series ⎛ ⎞⎛ ⎞ ∞ ∞ ∞ (ξ ∗ η)k t = ⎝ k ξj t ⎠ ⎝ j ηj t ⎠ . however. the Stone–Čech compactification of N.2. Example 8. the projection onto the nth coordinate ξ → ξn determines a multiplicative linear functional. We again wish to identify Σ = Σ(∞ ). because N is not a w∗ -compact set. 1]. and so L∞ (0. we have that C(K) is isometrically isomorphic to C(Σ). In fact. 1) is identical (as a Banach algebra) to C(Ω). Suppose ξ = (ξj )j ∈Z and η = (ηj )j ∈Z are sequences in 1 (Z). Consequently. we cannot realize the multiplicative linear functionals in this case as point evaluation. Unlike Example 8. 1] (because f represents a class of functions that are equal almost everywhere). The arguments given here cannot be applied to real Banach algebras. Suppose φ is a multiplicative linear functional. Furthermore. (8. (The argument is similar to the proof of Lemma 5. If φ(t) = λ. the spectrum of W .. As a result. the space 1 (Z) becomes a commutative Banach algebra. we say that ξ ∗ η is the convolution of ξ with η. Therefore. let φλ denote the multiplicative linear functional determined by φλ (t) = λ. k ∈ Z. We have also verified that the norm is submultiplicative. and zero elsewhere. ξ ∗ η ∈ 1 (Z). then φ(t k ) = λk for all k ∈ Z. and so equipped with this multiplication. by Fubini’s Theorem. We now identify Σ(W ). Certainly.3 The Wiener Algebra 203 Then ∞ (ξ ∗ η)k = ξk−j ηj . and so the multiplication is well-defined. This Banach algebra is called the Wiener algebra (after Norbert Wiener). i. Then Σ = {φλ : λ ∈ T}. It will be convenient to identify W with the space of formal Laurent series. ∞ ∞ ∞ ξ ∗ η1 ≤ |ηj | |ξk−j | = |ηj | ξ 1 = η1 ξ 1 .8.12).13) k=−∞ The series in (8.13) is doubly infinite. From (8. ∞ φ(ξ ) = λk ξk .e. (8.13).66. where the sequence η = (ηk )k∈Z is defined so that ηk = λk ξk for each k ∈ Z. Observe that φ is completely determined by what it does on t. We may thus identify Σ with T.) Notice that φλ (ξ ) = φ1 (η).12) j =−∞ We must verify that this multiplication is well-defined. For each λ ∈ T. This makes sense because of how we defined the multiplication in W . ξ ∈ W. that ξ ∗ η is in fact an absolutely summable sequence. |λ| = 1. the linear functionals that are multiplicative linear functionals are those given by φ(t) = λ for λ ∈ T. j =−∞ k=−∞ j =−∞ Therefore. the unit circle in C. k=−∞ j =−∞ k=−∞ j =−∞ For a fixed j ∈ Z. and so λ ∈ T. we have ∞ k=−∞ |ξk−j | = ξ 1 . any λ ∈ T determines a multiplicative linear functional according to (8. Consequently. We call W a convolution algebra because the multiplication ∗ is called a convolution. Thus. so that t k corresponds to the sequence with 1 in the k th position (k ∈ Z). . We wish to identify the multiplicative linear functionals φ on W . Consequently. and is denoted by W . it will converge for all ξ ∈ W only if both |λ| ≤ 1 (for the series with k ≥ 0) and |λ| ≥ 1 (for the series with k ≤ 0). we have ⎛ ⎞ ∞ ∞ ∞ ∞ ξ ∗ η1 = ξk−j ηj ≤ ⎝ |ξk−j | |ηj |⎠ . In particular. There exists some a ∈ W such that â = f . (8. and we provide it with the norm ∞ f A(T) = |fˆ(k)|. it follows that 0 ∈ Sp(a). 2π ). Therefore. 2π ). Proof By assumption. f ∈ A(T).204 8 Banach Algebras Since Σ is homeomorphic to T. Since it is isometrically isomorphic to W (as an algebra) we also call A(T) the Wiener Algebra. 2π ). and hence ∞ f (θ) = ξk eikθ . Suppose that ξ ∈ W . 2π). If f (θ ) = 0 for any θ ∈ [0.44.51 In the above discussion. 2π) so that λ = eiθ . we once again made use of the correspondence between T and [0. we let A(T) denote the collection of all functions on T with absolutely convergent Fourier series. The conclusion of the theorem is not at all obvious. and so a is invertible in W .14) k=−∞ Consequently. Remark 8.) Then A(T) is a Banach algebra.15) k=−∞ where fˆ(k) is the k th Fourier coefficient of f . and W is homeomorphic to A(T) via the Gelfand transform. (See (4.2. One should not mistake the brevity of the proof as a mark of triviality—it is brief because of the power behind our machinery. f is invertible in A(T).52 (Wiener Inversion Theorem) Let f ∈ C(T) have an absolutely convergent Fourier series. then 1/f has an absolutely converging Fourier series.1). f ∈ A(T). (8. Therefore. It is for this reason that we can say f ∈ C(T) when it is actually an element of C[0. Then ξ̂ ∈ C(T) via the map ∞ ξ̂ (λ) = ξ̂ (φλ ) = λk ξk . each ξ ∈ W corresponds to an f ∈ C(T) with absolutely convergent Fourier series. 2π). θ ∈ [0. we identified the continuous functions on T with the continuous functions on [0. there is a θ ∈ [0. we can embed W into C(T) via the Gelfand transform. and the result follows. 2π) → C by f (θ ) = ξ̂ (eiθ ). k=−∞ For every λ ∈ T. By Proposition 8. The next theorem gives an indication of the power behind the tools we have developed. but the proof itself is quite simple (and short). 2 . λ ∈ T. Motivated by the above discussion. Since 0 ∈ f (T). Theorem 8. we define a function f : [0. f (T) = â(T) = Sp(a). Suppose a ∈ A is such that ∞n=0 a < ∞. Show that in a commutative unital Banach algebra any maximal ideal is prime.9 Let X be a commutative Banach algebra. Exercise 8.8 Let A be a unital Banach algebra and let G(A) be the group of invertible elements. (Hint: See the proof of Theorem 8. Show λ ∈ Sp(T ) if and only if 1/λ ∈ Sp T −1 . Suppose x and y are elements in A such that both xy and yx are invertible. (Recall that fˆ(n) = 0 f (θ) e−inθ 2π dθ for all n ∈ Z. n Exercise 8.Exercises 205 Exercises Exercise 8. Show that if a ∈ A has an inverse.6 Let A be a commutative unital Banach algebra and let G(A) be the group of invertible elements of A. Show that both x and y are invertible and show that (xy)−1 = y −1 x −1 . Show that x ∈ G(A) if and only if φ(x) = 0 for all multiplicative linear functionals φ. Find an example of a Banach algebra with a n→∞ n→∞ topological divisor of zero that is not a divisor of zero. Exercise 8.) . Exercise 8. Show that the Fourier iθ 0 (n) = fˆ(n − 1) for all transform of the multiplier operator satisfies the equation Mf f ∈ L2 and each n ∈ Z. Show that every x ∈ ∂G(A) is a topological divisor of zero.3 Let Abe a unital Banach algebra with identity element 1. An element x ∈ A is called a topological divisor of zero if there exists a sequence (yn )∞ n=1 in A with yn = 1 for all n ∈ N such that lim xyn = 0 = lim yn x. Show that 1 − a is an invertible element and n −1 ∞ that (1 − a) = n=0 a .40. Exercise 8.7 Let A be a Banach algebra.5 Let A be a unital Banach algebra. Exercise 8. Exercise 8. 2π ). Show that the set of invertible elements of A is a group.) Exercise 8. Show that an ideal of a complex commutative unital Banach algebra X is maximal if and only if it has codimension 1.1 Let A be a unital Banach algebra.11 If N is a subspace of a vector space X. 2π dθ ) given by the formula Mf (θ ) = e f (θ) for all f ∈ L2 and θ ∈ [0.4 Let X be a Banach space and suppose T ∈ L(X) is invertible. Exercise 8. then a is invertible. A proper ideal I of X is called prime if for any elements a and b in X the product ab ∈ I implies that either a ∈ I or b ∈ I . 2π ). Explicitly write out the relationship between M and the 2π shift operator R on 2 (Z).10 Let M be the multiplier operator on L2 = L2 ([0. Exercise 8. then the codimension of N in X is the dimension of X/N. Show that if a has both a left-inverse and a right-inverse. then it must be unique.2 Let A be a noncommutative unital Banach algebra. where ∂G(A) is the boundary of G(A). and show that C(Σ) is the collection of all functions having an absolutely convergent Taylor series in D. where a = (aj )∞ ∞ j =0 and b = (bj )j =0 are sequences in A. and if f (z) = 0 for any z ∈ D.15 If f has an absolutely convergent Taylor series in D.) Exercise 8. Recall that the multiplication in A is given by the formula (a ∗ b)k = kj =0 ak−j bj for all k ∈ Z+ . n→∞ n∈N (Hint: Let θn = log a n .206 8 Banach Algebras Exercise 8. (Hint: Use the previous problem. Show that A = C.13 Let A be a Banach algebra and suppose a ∈ A. the spectrum of A. then show that 1/f has an absolutely convergent Taylor series in D. Use Fekete’s Lemma (Exercise 3.12 Let A be a complex unital Banach algebra.) . Identify Σ = Σ(A).10) to show that lim a n 1/n = inf a n 1/n .) Exercise 8. Suppose there exists some M < ∞ such that xy ≤ Mxy for all x and y in A.14 Let A = 1 (Z+ ) be the convolution algebra from Example 8.6(c). Exercise 8. (Hint: A is a subalgebra of W . Bowers. then ∞ n=1 An is always measurable. An Introductory Course in Functional Analysis. ∅ ∈ A. because c ∞ ∞ An = Acn .1. we give a brief overview of the basics of measure theory. If we require only finite unions of measurable sets to be in A. Definition A. Rudin [33]. The σ in the name σ -algebra refers to the countability assumption in (c) of Defi- nition A. The elements A ∈ A are known as measurable sets.1 Let X be a set. (c) If An ∈ A for all n ∈ N. © Springer Science+Business Media. a good source is Real and Complex Analysis by W. the set A\B = A ∩ B c is measurable. we can easily see that %the empty set is always measurable. Kalton. A).1 Measurability Definition A. A. A collection A of subsets of X is called a σ -algebra of sets in X if the following three statements are true: (a) The set X is in A. then A ∈ A. for measurable sets A and B. A set X together with a σ -algebra A is called a measurable space and is denoted by (X. that is. we can now conclude that. A scalar-valued function f with domain X is called measurable if f −1 (V ) ∈ A whenever V is an open set in the scalar field. or simply by X when there is no risk of confusion. Universitext. DOI 10.2 Let (X. From (a) and (b). A) be a measurable space. then Ac ∈ A. then A is called an algebra of sets in X. n=1 n=1 Among other things. where Ac is the complement of A in X. If An ∈ A for all n ∈ N.1007/978-1-4939-1945-1 .Appendix A Basics of Measure Theory In this appendix. J. LLC 2014 207 A. For a detailed discussion of measure theory. and if A = ∞n=1 An . by (b) and (c). (b) If A ∈ A. N. 3 Let (X. then the characteristic function (or indicator function) of E is the function defined by . If E ⊆ X.208 Appendix A Basics of Measure Theory Example A. A) be a measurable space. For example. The characteristic function of E is measurable if and only if E ∈ A. if (fn )∞n=1 is a sequence of real-valued measurable functions. Measurable functions are very well-behaved. Then f = u + iv is a complex measurable function if and only if u and v are real measurable functions. Such functions are called set functions. . Theorem A. . an are scalars. x ∈ X. Furthermore. A) be a measurable space. j =1 where n ∈ N. The next theorem is key to the further development of the subject. ∞]. Suppose u and v are real-valued functions. Definition A. . then there exists a sequence (sn )∞ n=1 of simple functions such that: 0 ≤ s1 ≤ s2 ≤ · · · ≤ f . 1 if x ∈ E. if f and g are measurable functions. En are measurable sets. and f (x) = lim sn (x). a1 .2 Positive Measures and Integration In this section. . If f is a positive measurable function. The characteristic function χE is also denoted by 1E . For now. and E1 . n→∞ A. χE (x) = 0 if x ∈ E. . . . Certainly. we will consider functions defined on a σ -algebra A. then so is f + g. since it is a finite sum of measurable functions. then the functions sup fn and lim sup fn n∈N n→∞ are measurable whenever the function values are finite. . . then the limit of the sequence is also a scalar-valued measurable function. In particular.5 Let (X. we will restrict our attention to set functions taking values in the interval [0. that is n aj χE j . if (fn )∞ n=1 is a convergent sequence of scalar-valued measurable functions. any simple function is measurable.4 A simple function is any linear combination of characteristic functions. ∞] such that μ(∅) = 0 is called a positive measure on A.) We now define a complete measure on R that agrees with m for all Borel measurable sets in B. The Lebesgue σ -algebra on R is the smallest σ -algebra on R that contains the Borel measurable sets and all subsets of m-measure-zero sets. Equivalently. A set function μ : A → [0. we define n(A) to be the size (or cardinality) of the set A. which is the family of all subsets of N. Example A.9 (Counting measure) Another classic example of a positive measure is the so-called counting measure on N. . we often write (X.2 Positive Measures and Integration 209 Definition A. The Lebesgue σ -algebra on R is the collection of all sets of the form B ∪N . A μ-negligible set may or may not be measurable. Example A.1) j =1 j =1 whenever (Aj )∞ j =1 is a sequence of pairwise disjoint measurable sets. The assumption that μ(∅) = 0 is to avoid the trivial case where the set function is always ∞. If the σ -algebra is understood. If every μ-negligible set is measurable (and hence a μ-null set). A subset of a μ-null set is called μ-negligible. Then n is a positive measure on N and n(N) = ∞. μ is complete if every subset of a μ-null set is μ-measurable and has μ-measure zero. A) be a measurable space. In order to do this. then the measure μ is called complete. we call the triple (X. often denoted P(N) or 2N . (Such a measure can be shown to exist. Let μ be a positive measure on a σ -algebra. Notice that we allow μ to attain infinite values.6 Let (X. b)) = b − a whenever a and b are real numbers such that a < b. we could assume that there exists some A ∈ A such that μ(A) < ∞.A. but there are Lebesgue measurable sets that are not Borel measurable.) Example A. Define a measure λ on this collection of sets by the rule λ(B ∪ N ) = m(B) for any m-measurable set B and any m-negligible set N . we will enlarge our σ -algebra. A measurable set that has zero measure with respect to μ is called a μ-null set or a μ-measure-zero set. A measure defined on the Borel σ -algebra on R is called a Borel measure on R.8 (Lebesgue measure) Let B be the Borel σ -algebra on R and let m be a positive Borel measure on B defined so that m((a. μ) a positive measure space. The measure λ is called Lebesgue measure on R. For any subset A ⊆ N. (That is. μ) for the positive measure space and say μ is a positive measure on X. We let A be the power set. all Borel measurable sets are Lebesgue measurable. A. It can be shown that λ is a complete countably additive measure on the Lebesgue σ -algebra on R.7 The Borel σ -algebra on R is the smallest σ -algebra that contains all of the open intervals in R. ∞] is said to be countably additive if ∞ ∞ μ Aj = μ(Aj ). Consequently. A countably additive set function μ : A → [0. where B ∈ B and N is an m-negligible set. By pairwise disjoint we mean Aj ∩ Ak = ∅ whenever j = k. In such a case. (A. then we write f ∈ L1 (μ). Define the integral of f over A by + fdμ = f dμ − f − dμ. A A where the supremum is taken over all simple functions s such that s(x) ≤ f (x) for all x ∈ X. Definition A. μ) be a positive measure space. 1] is a probability measure. Example A. then we define the integral of f over A with respect to μ to be fdμ = sup sdμ . let (X. A.12 Let (X.11 The restriction of the measure λ from Example A. Now. Definition A. If A and B are measurable sets such that A ⊆ B. A.) If f : X → R is a nonnegative measurable function.13 Let (X. we first define the class of functions for which our integral will exist. Also. Let us make some obvious comments about positive measures. A. then we call μ a probability measure. A scalar-valued measurable function f is called Lebesgue integrable (or just integrable) with respect to μ if |f | dμ < ∞. then μ( · )/μ(X) is a probability measure.10 A positive measure that attains only finite values is called a finite measure. A j =1 (The value of this integral does not depend on the representation of s that is used.8 to the unit interval [0. A. Suppose (X. then μ(A) ≤ μ(B). In order to meaningfully extend our notion of an integral to a wider class of functions. X If f is integrable with respect to μ. μ) is a positive measure space.210 Appendix A Basics of Measure Theory Definition A. For any A ∈ A. we define the integral of s over A with respect to μ to be n sdμ = aj μ(A ∩ Ej ). If μ is a positive measure on X such that μ(X) = 1. μ) be a positive measure space and let s = nj=1 aj χEj be a simple function. observe that if μ is a finite positive measure. μ) be a positive measure space and let f : X → R be a real-valued integrable function. A A A where . + f (x) if f (x) ≥ 0. . f (x) = 0 otherwise. 3 Convergence Theorems and Fatou’s Lemma 211 is the positive part of f .A. and . If f : X → C is a complex-valued integrable function. then fdμ = gdμ. A. A A A. −f (x) if f (x) ≤ 0. if f = ga. and in such a situation we write fdμ = f (x) μ(dx). With this in mind. A function f is defined a. Every measurable function (in the sense of Definition A. we extend our notion of a measurable function. it is useful to explicitly show the variable dependence in an integration. A A This is true because the integral over a set of measure zero is always zero. is the negative part of f . we adopt the convention that 0·∞ = 0. A. The previous definition is significant because. Furthermore. In some cases. Two measurable functions f and g are said to be equal almost everywhere if μ{x ∈ X : f (x) = g(x)} = 0. f − (x) = 0 otherwise. Consequently. Let (X.(μ) on X if the domain D of f is a subset of X and X \ D is a μ-negligible set. If f and g are equal almost everywhere.3 Convergence Theorems and Fatou’s Lemma Definition A.e. μ) be a positive measure space. then A f dμ = 0 for all measurable functions f . We say that a μ-measurable function f is integrable if it is . define the integral of f over A by fdμ = (f ) dμ + i (f ) dμ. A ∈ A. Sometimes we say f (x) = g(x) for almost every x. μ) be a measure space.(μ). To avoid certain arithmetic difficulties which arise from the definitions above. if μ(A) = 0 for a measurable set A ∈ A.2) is μ-measurable and every μ-measurable function is equal almost everywhere (with respect to μ) to a measurable function. we write f = ga. A f dμ = 0 whenever f (x) = 0 for all x ∈ A.e. then f is called μ-measurable.e. A A A where (f ) and (f ) denote the real and imaginary parts of f . Functions that are equal almost everywhere are indistinguishable from the point of view of integration. respectively.e.(μ) on X and there is a μ-null set E ∈ A such that f −1 (V ) \ E is measurable for every open set V .(μ). If f is defined a. The notion of μ-measurability allows us to extend our constructions to a wider collection of functions.14 Let (X. even if μ(A) = ∞. Theorem A. The next result is a direct consequence of Lebesgue’s Dominated Convergence Theorem.(μ) to a measurable function g that is integrable with respect to μ and we define A f dμ to be A g dμ for every measurable set A. n→∞ X In particular. If (fn )∞ n=1 is a sequence of scalar-valued measurable functions.(μ). named after the mathematician Pierre Fatou.(μ). μ) be a positive finite measure space and suppose (fn )∞ n=1 is a sequence of uniformly bounded scalar-valued measurable functions. X n→∞ X .212 Appendix A Basics of Measure Theory equal a.17 (Lebesgue’s Dominated Convergence Theorem) Let (X. then lim inf |fn | dμ ≤ lim inf |fn | dμ. we write fn → f a. A. A. X n→∞ n→∞ X Perhaps the most important use of Fatou’s Lemma is in proving the next theorem.16 (Fatou’s Lemma) Let (X. μ) be a positive measure space. X n→∞ X The last equality follows from the fact that | X f dμ| ≤ X |f | dμ whenever f is an integrable function.e. Theorem A.(μ) for all n ∈ N. If fn → f a. A sequence of measurable functions (fn )∞ n=1 is said to converge almost everywhere to a function f if the set {x ∈ X : fn (x) → f (x)} has μ-measure zero.15 (Monotone Convergence Theorem) Let (X. and one of the better known consequences of it is Fatou’s Lemma. If fn → f a. X n→∞ X The Monotone Convergence Theorem is a very useful tool. Suppose (fn )∞ n=1 is a sequence of measurable functions such that 0 ≤ f1 (x) ≤ f2 (x) ≤ f3 (x) ≤ · · · for almost every x.18 (Bounded Convergence Theorem) Let (X. then f ∈ L1 (μ) and fdμ = lim fn dμ. μ) be a positive measure space and suppose (fn )∞ n=1 is a sequence of scalar-valued measurable functions that converge almost everywhere to f . fdμ = lim fn dμ. μ) be a positive measure space.(μ)-limit of measurable functions is μ-measurable. then f ∈ L1 (μ) and lim |fn − f | dμ = 0. then f is an integrable function and fdμ = lim fn dμ. A function f which is the a. A.e. A. Corollary A.e.e. If there exists a function g ∈ L1 (μ) such that |fn | ≤ ga. which is one of the cornerstones of measure theory. In this case.(μ). Theorem A.e.e. 2) j =1 j =1 whenever (Aj )∞ j =1 is a sequence of pairwise disjoint measurable sets in A.4 Complex Measures and Absolute Continuity Definition A.A. ⎩ ⎭ j =1 where the supremum is taken over all finite sequences of pairwise-disjoint measurable sets (Ej )nj=1 . if |μ|(E) = 0 for some set E ∈ A. (Compare to Definition A.19. If μ is a complex measure on A. This is an instance of a property called absolute continuity. and so we will confine ourselves to complex measures and positive measures. where the series in (A. we mean ⎛ ⎞ ∞ ∞ μ ⎝ Aj ⎠ = μ(Aj ).2) is absolutely convergent. we require that a complex measure μ be finite. such generality is not necessary. μ) is a complex measure space. it must be the case that μ(E) = 0. then the triple (X. Definition A. it is a finite measure—a fact that we state in the next theorem. then |μ| is a positive measure on A and |μ|(X) < ∞. A complex measure which takes values in R is called a real measure or a signed measure. for all n ∈ N. a relationship between a measure and its total variation measure. Theorem A. There is. we can define a measure to be a countably additive set function that takes values in any topological vector space.21 If (X. A) be a measurable space. We say that μ is absolutely continuous with respect to λ if. There is a significant difference between positive measures and complex measures. Additionally. For our purposes. such that E = nj=1 Ej . For example. |μ(A)| < ∞ for all A ∈ A. In DefinitionA. then we may write (X. A countably additive set function μ : A → C is called a complex measure.20 Let μ be a complex measure on the σ -algebra A. More generally. (A. μ). If the σ -algebra is understood. Definition A.4 Complex Measures and Absolute Continuity 213 A. for all . A.19 Let (X. μ) is called a (complex) measure space.2) makes sense. so long as the convergence of the series in (A. A. When we say μ is countably additive. the total variation measure is a measure. As the name suggests.6). naturally.22 Let μ be a complex measure on A and suppose λ is a positive measure on A. We define the total variation measure of μ to be the set function |μ| : A → R given by ⎧ ⎫ ⎨ n ⎬ |μ|(E) = sup |μ(Ej )| . that is. This was not a requirement for a positive measure. ) One can also have a counting measure on R. μ) is a complex measure space. The function g ∈ L1 (λ) is called the Radon–Nikodým derivative of μ with respect to λ and is sometimes denoted dμ/dλ. When μ is absolutely continuous with respect to λ. Important examples of σ -finite measures include Lebesgue measure on R and counting measure on N. one of the key results in measure theory. A) be a measurable space. Proposition A.(μ)). then there exists a unique g ∈ L1 (λ) such that μ(E) = g(x) λ(dx). We seek to define an integral with respect to a complex measure. The σ -finite assumption on μ in Theorem A. we write μ λ.23 Let (X. Definition A. (A. and Lebesgue’s Dominated Convergence Theorem hold for integrals with respect to complex measures. respectively. μ) be a complex measure space. but it is not σ -finite. (A. A positive measure μ on A is said to be σ -finite if there exists a countable sequence (Ej )∞ j =1 of measurable sets such that X = ∞ E j =1 j and μ(E j ) < ∞ for each j ∈ N. we will state the Radon–Nikodým Theorem. E ∈ A. We can. therefore.4). A. we have that μ(A) = 0 whenever λ(A) = 0. then g = ha.(λ) (and consequently a.3) E The equation in (A. If μ is a complex measure on A that is absolutely continuous with respect to λ. E ∈ A. The following proposition is a corollary of Theorem A. As a consequence of Proposition A.24 and is often called the polar representation or polar decomposition of μ. Before that. we introduce a definition. That is.8 and A. versions of the Monotone Convergence Theorem. When we say the Radon–Nikodým derivative is unique. then there exists a measurable function g such that |g(x)| = 1 for all x ∈ X and such that dμ = g d|μ|.3) is sometimes written μ(dx) = g(x) λ(dx) or dμ = g dλ. define the integral of a measurable function f : X → C by fdμ = fgd|μ|.e. To that end. Theorem A. let (X.214 Appendix A Basics of Measure Theory A ∈ A. We apply .24 cannot be relaxed. (See Examples A. we mean up to a set of measure zero. A.25 If (X.e. however.9. Fatou’s Lemma. Since μ |μ|. A) is a measurable space and let λ be a positive σ -finite measure on A.4) E E More can be said about the Radon–Nikodým derivative of μ with respect to |μ|.25 and the definition in (A.24 (Radon–Nikodým Theorem) Suppose (X. Shortly. if g and h are both Radon– Nikodým derivatives of μ with respect to λ. there exists (by the Radon– Nikodým Theorem) a unique (up to sets of measure zero) function g ∈ L1 (|μ|) such that dμ = g d|μ|. For p ∈ [1. In particular. A. Theorem A.A. μ).(μ). A. ∞). X This is the space of p-integrable functions. let ! " Lp (μ) = f a measurable function : |f |p dμ < ∞ . The proof of TheoremA. we consider a measure space (X. also known as Lp -functions. The set L∞ (μ) = {f a measurable function : f ∞ < ∞} is the space of essentially bounded measurable functions on X.5 Lp -spaces 215 the existing theorems to the positive finite measure |μ| and observe that f dμ ≤ |f | d|μ|.26 Let (X. where μ is a positive measure. In fact.e.26 relies heavily on the following fundamental inequality.e. Lp (μ) is a Banach space. For the case p = ∞. If 1 ≤ p ≤ ∞. For each p ∈ [1. μ) be a positive measure space. for all measurable functions f . Two functions f and g in Lp (μ) are considered equivalent if f = ga. which provides the triangle inequality for Lp (μ). X where f is a μ-measurable function on X. who proved the theorem for the general case in 1930 [28]. and is the smallest number having the property that |f | ≤ f ∞ a.(μ). who proved the theorem for Rn in 1913 (n ∈ N). Observe that f p < ∞ if and only if f ∈ Lp (μ). let ! " f ∞ = inf K : μ(|f | > K) = 0 . and for Otton Nikodým. the Lp -spaces are collections of equivalence classes of measurable functions. . then · p is a complete norm on Lp (μ). We will identify certain spaces of measurable functions on X. ∞). we will generally speak of the elements in Lp -spaces as functions. we let 1/p f p = |f |p dμ . rather than equivalence classes of functions. on X. The Radon–Nikodým Theorem is named after Johann Radon. In spite of this.5 Lp -spaces In this section. The quantity f ∞ is called the essential supremum norm of f . X X for all measurable functions f on X. 29 (Hölder’s Inequality) Let (X. which follows from Theorem A. which is ubiquitous in measure theory. in order to prove something about Lp -spaces.27 (Minkowski’s Inequality) Let (X. The conjugate exponent of p = 1 is defined to be q = ∞ (and vice versa). If p ∈ [1. This is a consequence of the next theorem. X defines a bounded linear functional on Lp (μ) whenever 1 ≤ p < ∞. Sup- pose 1 ≤ p < ∞ and let q be conjugate to p. If f and g are in Lp (μ). If p ∈ (1. or even that p and q are conjugate to each other. Before we give another theorem. The set of simple functions is dense in Lp (μ) whenever 1 ≤ p ≤ ∞.5 and Theorem A.31 (Density of Simple Functions) Let (X. which is the content of the next theorem. Hölder’s Inequality ensures that. μ) be a positive σ -finite measure space. it is sufficient to prove it for simple functions. then q is called the conjugate exponent of p. A. μ) be a positive measure space. we must introduce a definition.30 Let (X. It turns out that any bounded linear functional on Lp (μ) can be achieved in this way. Definition A. μ) be a positive measure space. Theorem A. then f + g ∈ Lp (μ) and f + gp ≤ f p + gp . The following theorem. Frequently.216 Appendix A Basics of Measure Theory Theorem A.15 (the Monotone Convergence Theorem). and if q is conjugate to p. it is sometimes convenient to write q = p−1 p . If 1 ≤ p < ∞. then fg ∈ L1 (μ) and f g1 ≤ f p gq . ∞) and q is conjugate to p. A function which is measurable with respect to the Borel σ -algebra is called a Borel . The smallest σ -algebra on X containing the open sets in X is called the Borel σ -algebra. f ∈ Lp (μ).32 Suppose X is a topological space. or the Borel field.6 Borel Measurability and Measures Definition A.28 If 1 < p < ∞ and p1 + q1 = 1. μ) be a positive measure space. can be seen as a generalization of the Cauchy–Schwarz inequality. given any g ∈ Lq (μ). the map f → f (x) g(x) μ(dx). Theorem A. Theorem A. where 1 ≤ p ≤ ∞. then Lp (μ)∗ = Lq (μ). then we can also say that q is conjugate to p. on X. If f ∈ Lp (μ) and g ∈ Lq (μ). ∞] and q is the conjugate exponent of p. or a Borel function. functions in L∞ (μ) need not be continuous.35 (Riesz Representation Theorem) Let X be a locally compact Hausdorff space. Theorem A.35 is named after F. The set C0 (X) is a Banach space under the supremum norm. who originally proved the theorem for the special case X = [0. A complex measure is called regular if |μ| is regular. For example. Convergence in the L∞ -norm is the same as uniform convergence. If X is a locally compact Hausdorff space. Example A. Indeed. f ∈ C0 (X). however. The space C0 (X) is dense in Lp (μ) for all p ∈ [1. who also worked in the context of the real line [23].36 (Lusin’s Theorem) Let X be a locally compact Hausdorff space and suppose μ is a finite positive regular Borel measure on X.34 allows us to describe the dual space of C0 (X). Definition A. A measure on the Borel σ -algebra is called a Borel measure. but not compact.A. K a compact set}.34 Let X be a locally compact Hausdorff space and suppose μ is a positive Borel measure on X. We say f vanishes at infinity if for every ε > 0.33 The Banach space C0 (N) is the classical sequence space c0 of sequences tending to zero. . for every measurable set E. the theorem holds for any positive Borel measures that are finite on compact sets. Naturally. Theorem A. Theorem A. then μ is called inner regular. then we call μ a regular measure. Note that Lusin’s Theorem is not true when p = ∞. Definition A. and the uniform limit of a sequence of continuous functions is continuous. μ(E) = sup{μ(K) : K ⊆ E. the real line R with its standard topology is locally compact. there exists a compact set K such that |f (x)| < ε for all x ∈ K. but the converse need not be true. 1] [31]. μ(E) = inf{μ(V ) : E ⊆ V . for every measurable set E. Riesz.6 Borel Measurability and Measures 217 measurable function. then we denote by C0 (X) the collection of all continuous functions that vanish at infinity. If μ is both inner regular and outer regular. If Λ is a bounded linear functional on C0 (X). then μ is called outer regular. If. The next theorem is named after Nikolai Lusin (or Luzin). all compact spaces are locally compact. ∞). then there exists a unique regular complex Borel measure μ such that Λ(f ) = f dμ. X and Λ = |μ|(X). V an open set}. The hypotheses of Lusin’s Theorem can be relaxed somewhat. If. We recall that a topological space is said to be locally compact if every point has a compact neighborhood. y) is B-measurable for all x ∈ X. Theorem A. B) be two measurable spaces.37.218 Appendix A Basics of Measure Theory Proposition A.37 If X is a locally compact metrizable space. . who proved a version of Theorem A. then any finite Borel measure on X is necessarily regular. Now let (X. A.39 in 1909 [35]. We denote by σ (A × B) the smallest σ -algebra containing all measurable rectangles in X × Y . We define the product measure on σ (A × B) to be the set function μ × ν given by the formula (μ × ν)(Q) = χQ (x. A. In light of Proposition A. in which case the term “regular” is omitted from the conclusion. It is named after Guido Fubini. B. y) ν(dy) μ(dx) X Y = χQ (x. Proposition A. Again. Y X The same conclusion holds if f is a σ (A × B)-measurable function such that |f (x. y) is A-measurable for all y ∈ Y . y) μ(dx) ν(dy). B) be two measurable spaces. for example. μ × ν is a measure on σ (A × B).) A.7 Product Measures Let (X. y) ν(dy) μ(dx) X×Y X Y = f (x. The fundamental result in this section is known as Fubini’s Theorem. As the name implies. where A ∈ A and B ∈ B. If a scalar- valued function f on X × Y is σ (A × B)-measurable. A measurable rectangle in X × Y is any set of the form A × B. and y → f (x. Theorem A. (This was done. A) and (Y . then f d(μ × ν) = f (x. y)| ν(dy) μ(dx) < ∞.35 is often stated for a metrizable space X. y) μ(dx) ν(dy). in Theorem 2. A) and (Y . If f ∈ L1 (μ × ν). B. who proved a version of the theorem in 1907 [13].38 Let (X. after Leonida Tonelli. then the map x → f (x. μ) and (Y .20 of the current text. the Borel measure in question need only be finite on compact sets for the conclusion to hold. μ) and (Y . the measure μ × ν is such that (μ × ν)(E × F ) = μ(E)ν(F ) for all E ∈ A and F ∈ B. Furthermore. X Y Fubini’s Theorem is sometimes called the Fubini-Tonelli theorem. ν) be two σ -finite measure spaces. Y X for all Q ∈ σ (A × B). ν) be two σ -finite measure spaces.39 (Fubini’s Theorem) Let (X. and so appear in Appendix 8. For more. Many of these come from measure theory.3. see (for example) Topology: a first course by James Munkres [27] and Functions of One Complex Variable by John Conway [7]. where K is compact in X.1 Let X be a locally compact Hausdorff space that is not compact. No- tice that N is locally compact but not compact. we have invoked important results (usually by name). B.1 The One-Point and Stone-Čech Compactifications We will give only a brief discussion of the necessary topological concepts.15. J. We include the rest here. The classic example of a one-point compactification is N ∪ {∞}. (Recall that a space is locally compact if every point has a compact neighborhood containing it. we refer the interested reader to Topology: a first course. as well as more discussion on these topics.1007/978-1-4939-1945-1 . called the point at infinity. For proofs of these theorems. N. The space N ∪ {∞} is homeomorphic to the subspace {1/n : n ∈ N} ∪ {0} of R. Kalton. Define a topology on Y by declaring a set U to be open in Y if either U is open in X or U = Y \K. Bowers. for easy reference. The space Y is called the one-point compactification of X. to form a set Y = X ∪ {∞}. LLC 2014 219 A. We start by defining the one-point compactification of a locally compact Hausdorff space that is not compact. Adjoin to X an element ∞. (See Exercise 2. Universitext. DOI 10.) Definition B.Appendix B Results From Other Areas of Mathematics Throughout the course of this text. the one-point compactification of the natural numbers.) © Springer Science+Business Media. where N is given the discrete topology. sometimes without explicitly writing out the statement of the result being used. An Introductory Course in Functional Analysis. by James Munkres [27]. 1] such that f (x) = 1 and f (A) = {0}. Cn } of C.44 in [3]. .220 Appendix B Results From Other Areas of Mathematics Theorem B.3 Let X be a locally compact Hausdorff space that is not compact. If {Kα }α∈J is a collection of compact topological spaces. however.6 A topological space X is compact if and only if every collection C of closed % sets in X satisfying the Finite Intersection Property is such that the intersection C∈C C of all elements in C is nonempty. The inquisitive reader is directed to Theorem 3. The above theorem leads directly to the following useful corollary. . there are some circumstances under which a metric does exist. . we require some background. Then Y is a compact Hausdorff space. any locally compact separable metric space that is not compact (such as N) will have a metrizable one-point compactification. Before introducing this concept. then the intersection ∞ n=1 Cn of all sets in the sequence is nonempty.2 Let X be a locally compact Hausdorff space that is not compact and let Y be its one-point compactification.7 (Nested Interval Property) Let X be a topological space. we can profit from some knowledge of the methods used within. We now state an alternative formulation of compactness. which we will use presently. The next type of compactification we wish to define is the Stone-Čech compactification. the intersection C1 ∩ · · · ∩ Cn is nonempty. The standard proof of Tychonoff’s Theorem uses an alternate characterization of compactness. Definition B. We will not provide a proof of Theorem B. the one-point compactification of a locally compact metric space is not itself metrizable. Theorem B. If (Cn )∞ n=1 is a sequence%of nonempty compact sets such that Cn+1 ⊆ Cn for all n ∈ N. .4 (Tychonoff’s Theorem) Let J $ be an index set. Tychonoff’s Theorem is a statement about compactness. X is a subspace of Y . Theorem B. then α∈J Kα is compact in the product topology.38). In general. The one-point compactification of X is metrizable if and only if X is second countable. Corollary B. Theorem B. We begin by restating Tychonoff’s Theorem (Theorem 5. and Y is the closure of X (in Y ). . however. A space X is called completely regular if all singletons (single-point sets) in X are closed and if for each x ∈ X and each closed subset A not containing x there exists a bounded continuous function f : X → [0.3.5 (Finite Intersection Property) Let X be a topological space. A collection C of subsets of X is said to satisfy the Finite Intersection Property if for every finite subcollection {C1 . Y \X contains exactly one point. one for which we need a definition. In particular. and while we do not provide a proof of this theorem. We remark here that a space X is called regular if any closed subset A and point x ∈ A can be separated by open neighborhoods. it is both necessary and sufficient that X be completely regular. There are generally many compactifications for a completely regular space X. x ∈ X. That is to say. then the collection of bounded real-valued continuous functions on X separates the points of X. That is. We have a bijective mapping Φ : Y → φ(X) given by the rule . See Theorem 4. 1] such that f (x) = 1 but f (y) = 0. $ Define a map φ : X → α∈J Iα by φ(x) = (fα (x))α∈J . A completely regular space is regular. Let {fα }α∈J be the collection of all bounded continuous real-valued functions on X. fα ∞ . For each α ∈ J .) In order for a noncompact space X to have a compactification. x∈X x∈X or + . by Tychonoff’s Theorem. respectively. The one we consider now is the Stone-Čech compactification. there is some continuous function f : X → [0. but there exist regular spaces that are not completely regular. Proposition B. sup fα (x) . It can be shown that φ is an embedding. Let A = φ(X)\φ(X) and define a space Y by Y = X ∪ A. Iα = − fα ∞ . (An example is the one-point compactification given earlier.2 in [27] for the details.4. A compactification of a space X is a compact Hausdorff space Y that contains X as a dense subset.) We conclude that φ(X) is a compact Hausdorff space. say 1 2 Iα = inf fα (x). Each of the intervals Iα is compact in R. Let X be a completely regular space.1 The One-Point and Stone-Čech Compactifications 221 In particular. let Iα be any closed interval containing the range of fα . because the collection {fα }α∈J separates the points of X. and so. whenever x = y. (This follows from the complete regularity of X. there exist disjoint open neighborhoods containing A and x.B. indexed by some (possibly uncountable) set J .8 A topological space X is completely regular if and only if it is homeomorphic to a subspace of a compact Hausdorff space. the product $ α∈J Iα is a compact space. if X is completely regular. It follows that Φ is a homeomorphism. . It may seem that β(X) is not well-defined. The topological space Y constructed above is known as the Stone-Čech compactification of X. Φ(y) = y if y ∈ A. φ(y) if y ∈ X. and X is a subspace of the compact Hausdorff space Y . Define a topology on Y by letting U be open in Y whenever Φ(U ) is open in φ(X). and is generally denoted β(X). Let us begin with some definitions. the notation z → z0 means that |z−z0 | → 0. . Note to Appendix A]: . The Stone-Čech compactification is named after Marshall Harvey Stone and Ed- uard Čech.222 Appendix B Results From Other Areas of Mathematics since one could construct it with a different choice of sets. It is not clear that these theorems are named quite as they should be.11 (Maximum Modulus Theorem) Let D be a bounded open set in the complex plane. . A complex-valued function is called holomorphic (or analytic) if it is differentiable in a neighborhood of every point in its domain. a fact which was remarked upon by Walter Rudin in his Functional Analysis [34. (See Example 8. however. the space of continuous functions on the unit circle. Any bounded continuous real-valued function on X can be uniquely extended to a continuous real-valued function on β(X). A complex-valued function is called differentiable at z0 in the complex plane C if f (z) − f (z0 ) f (z0 ) = lim z→z0 z − z0 exists (and thus is a complex number). The next theorem is one of the key results of complex analysis. . then we say f is differentiable in the set (or on the set).10 A holomorphic function is infinitely differentiable.6(b). Unfortunately. we only encounter a small portion of the subject in our current undertaking. B. If f is differentiable at every point in a set. If f is a holomorphic function in D that is extendable to a continuous function on D.2 Complex Analysis The subject of complex analysis is overflowing with fantastic and improbable theorems. Thus it appears that Čech proved the Tychonoff theorem. The next theorem is a frequently cited result and is used to show. z∈D z∈∂D Holomorphic functions display a variety of interesting and useful properties. One such property is stated in the next theorem. Theorem B. Tychonoff’s Theorem is named after Andrey Nikolayevich Tychonoff. for example. the Stone-Čech compactification is unique up to homeomorphism (and a homeomorphism can be chosen so that it is the identity on X). we will state a very important property that it possesses. In this context. Theorem B. whereas Tychonoff found the Čech compactification—a good illustration of the historical reliability of mathematical nomenclature.9 Let X be a completely regular space with Stone-Čech compactification β(X). then max|f (z)| = max|f (z)|. While a full discussion of the Stone-Čech compactification is beyond the scope of this appendix. that the disk algebra A(D) is a subalgebra of C(T).) Theorem B. A classic example is the function f (z) = sinz z . Theorem B. which has a removable singularity at z = 0. and suppose f = limn→∞ fn . then we have f γ n (z) dz = 0 for all n ∈ N. According to Liouville’s Theorem. if γ is any continuously differentiable closedcurve. something even stronger can be said. On any neighborhood of z0 . then f is holomorphic on D. If f : D → C is a continuous function such that γ f (z) dz = 0 for every closed piecewise continuously differentiable curve γ in D.16 (Picard’s Greater Theorem) Let f be an analytic function with an essential singularity at z0 . it follows from Morera’s Theorem that f is holomorphic.13 (Morera’s Theorem) Let D be a connected open set in the complex plane.15 was proved by Charles Émile Picard in 1879. we mean that any two points in D can be connected by a path.12 is a standard line integral over a path γ . Theorem B. then γ f (z) dz = 0. A singularity z0 is a pole of f if there is some n ∈ N such that lim (z − z0 )n f (z) (B.15 (Picard’s Lesser Theorem) If an entire function is not constant. In fact. with the possible exception of one point. then it must be unbounded. then its image is the entire complex plane. The range of this function is C\{0}. then we call it a removable singularity. Theorem B. There are some truly remarkable theorems related to entire functions. The classic example of such a non-constant function is f (z) = ez .B. with the possible exception of one point. the function f will attain every value of C. The integral appearing in Theorem B. Morera’s Theorem is used to show (among other things) that the uniform limit of holomorphic functions is holomorphic: Suppose (fn )∞ n=1 is a uniformly convergent sequence of holomorphic functions. infinitely often. and that any path connecting those two points can be continuously transformed into any other.5) z→z0 . A function is said to have a singularity at the point z0 if f is analytic in a neighborhood containing z0 . Cauchy’s Integral Formula has a converse. if an entire function is not constant. By Cauchy’s Integral Formula. If γ is a closed curve in D. A function that is holomorphic on all of C is called an entire function.14 (Liouville’s Theorem) A bounded entire function is constant.12 (Cauchy’s Integral Formula) Let f : D → C be a holomorphic function on the simply connected domain D. Theorem B. Another significant theorem due to Picard concerns essential singularities. except f (z0 ) does not exist. which is named after the mathematician and engineer Giacinto Morera. If limz→z0 f (z) exists. When we say D is simply connected. A less precise way of saying that is to say D has no “holes” in it. Since γ f (z) dz = lim n→∞ γ fn (z) dz = 0 (by uniform convergence).2 Complex Analysis 223 Theorem B. Theorem B. not being born until 1856. If the limit in (B. Such a function has a pole of order n ∈ N at z0 if a−n = 0. We call z0 a pole of order n if n is the smallest integer such that (B.224 Appendix B Results From Other Areas of Mathematics exists. were contemporaries. only one year before Cauchy passed away (on 23 May 1857). then f has an essential singularity at z0 . A simple example of a function with a polar singularity of order n ∈ N at z = 0 is f (z) = z1n . . Augustin-Louis Cauchy did fundamental research in complex analysis in the first half of the nineteenth century. An alternate characterization of singularities can be obtained from the Laurent series of the function f .14 is named. Charles Émile Picard came later. Any function with a singularity at z0 will have a Laurent series there. If a−k = 0 for infinitely many k ∈ N. A function f has a Laurent series about z0 if there exists a doubly infinite sequence of scalars (an )n∈Z such that ∞ f (z) = an (z − z0 )n . after whom Theorem B. Examples of functions with an essential singularity at z = 0 are f (z) = e1/z and f (z) = sin (1/z). we call z0 an essential singularity of f .5) exists. n=−∞ There is a formula to compute an for n ∈ Z: 1 f (z) an = dz. He and Joseph Liouville. but a−k = 0 for all k > n.5) does not exist for any n ∈ N. 2πi γ (z − z0 )n+1 where γ is a circle centered at z0 and contained in an annular region in which f is holomorphic. 23. 252–267 (1940) 2. 2004) 21. Sci. Ann. A. LLC 2014 225 A. Topics in Banach Space Theory. J. I.S. Acad. (Mat. New York. (Springer-Verlag. Soc.C. G. Krull Implies Zorn. DOI 10. 365–390 (1903) 13. Math. Am. Enflo. Fischer. M.References 1. Kolmogoroff. 205–216 (1963/1964) 20. 41. (Peter G. Sur la convergence en moyenne. E. (Paris) 154. Produits tensoriels topologiques et espaces nucléaires. 130. J.H. A geometric form of the axiom of choice.1007/978-1-4939-1945-1 . Fekete. Rend.L. Hodges. 27. Bowers. C. 839–873 (1941) 6. On rings of continuous functions on topological spaces. Grothendieck. Sur les propriétés des fonctions mesurables. Acta.J. vol. Normierte Ringe. 2006) 4. Casazza. An Introduction to Harmonic Analysis. Ann. 9. Acad.W. F. C. P. vol. Gelfand. Math. 21 of Graduate Studies in Mathematics (American Mathematical Society. Acta Math. 2000) 9. N. Calkin. Fundam. Two-sided ideals and congruences in the ring of bounded operators in Hilbert space. R. Berlin. Rec. Conway. I. Studia Math. Providence. 233 of Graduate Texts in Mathematics (Springer. M. Casazza. 101. (2) 42. Lincei. A tribute to Nigel J. (2). M. N. Fredholm. James. On extreme points of regular convex sets. I. J. coordinating editor) 7. Conway. (Mat. An Introductory Course in Functional Analysis. Sbornik) 12(54). A Course in Operator Theory.G. 167–170 (1972) 5. Aliprantis. D. Zeitschrift 17. Soc. 608–614 (1907) 14. Math. Functions of One Complex Variable. Milman. Über die Verteilung der Wurzeln bei gewissen algebraischen Gleichungen mit ganzzahligen Koeffizienten. Bell. A counterexample to the approximation problem in Banach spaces. (Springer. Rom. 942–951 (2012). Math. Akad. Neumark. R. 133–138 (1940) 22. Nauk. J. Fubini. 1978) 8. P. 309–317 (1973) 10. 1688–1690 (1912) © Springer Science+Business Media. Fremlin. J. Katznelson. Gelfand. 140 (1955) 18. Rec.B. Albiac. New York. Characterizations of Reflexivity. Universitext. D. Math. K. Kalton (1946–2010). 3rd edn. Cambridge. N. L. 2006) 3. (Paris) 144. Sur une classe d’équations fonctionnelles. (Cambridge University Press. R. (2) 19. 3rd edn. 11 of Graduate Texts in Mathematics. Notices Am. 1955. Ann. J. Math. 1022–1024 (1907) 12. Alaoglu. 197–213 (1943) 17. 228–249 (1923) 11. Lusin. Math. Soc. 59. Kalton. 11–15 (1939) 16. Y. 3–24 (1941) 15. Border.S. Sugli integrali multipli. Studia Math. Math. Math. 285–287 (1979) 19. A. (5) 16(1). I. Kalton. N. C. London Math. SSSR 22. Gelfand. 729–744 (1929) 23. Krull. Infinite Dimensional Analysis: A Hitchhiker’s Guide. N.B. Dokl. Idealtheorie in Ringen ohne Endlichkeitsbedingung. Math. W. Weak topologies of normed linear spaces. Sbornik) 9(51).C. Sci. 77. 2nd edn. Cambridge Mathematical Library. vol. Mem. Accad. W. On the imbedding of normed rings into the ring of operators in Hilbert space.D. Krein. Żelazko.. R. Accad. G. Sci. (Paris) 144. R. (McGraw-Hill Inc. W. Acad. Moore. Sci. The dual space of L∞ is L1 . (5).H. Including selected papers presented at the Scottish Book Conference held at North Texas State University. and Influence. (Translated from the Polish by Marcin E. 1982) 27. 8 of Studies in the History of Mathematics and Physical Sciences (Springer-Verlag..C.. J. (McGraw-Hill Book Co. F. Kuczma. S. New York. J. Munkres. 1973). 3rd edn. Sur les anneaux linéaires. W. 1025–1027 (1938) 26. Math. R. Birkhäuser Boston. M. 974–977 (1909) 32. Amsterdam. Sci. Mazur. L. New York. (Paris) 207. 2nd edn. 619–625 (1998) 37. Rudin.W. Mass. Nikodym. Radon. 18(2). 9. Sur les systèmes orthogonaux de fonctions. The Scottish Book. Zermelo’s Axiom of Choice. Development. W. 2 of Graduate Texts in Mathematics.. Fundam. Oxtoby. N. (Paris) 149. A compact convex set with no extreme points. Acad. Tex. Acad. Sur les opérations fonctionnelles linéaires. vol. Riesz.) . 615–619 (1907) 31. 60. 1981. 1975) 28. Indag. Sur une généralisation des intégrales de M. Real and Complex Analysis. 131–179 (1930) 29. C. Its Origins. Roberts. International Series in Pure and Applied Mathematics. Banach Algebras (Elsevier Publishing Co.). F. C. Sull’ integrazione per parti. Tonelli. 1980) 30. Math. Rend. (Springer-Verlag.S. Measure and Category: A Survey of the Analogies Between Topological and Measure Spaces. J. Englewood Cliffs.R. O. Mathematics from the Scottish Café. Mauldin (ed. Rudin. 1987) 34.. 15. Functional Analysis. vol.D. R. Riesz. Denton. 2nd edn. New York. 1991) 35. (May 1979) 25. Väth. Rom. C. J. Studia Math. 255–266 (1977) 33. 246–253 (1909) 36. Lincei.226 References 24. New York. Topology: A First Course (Prentice-Hall Inc. 211 measure. 59 field. 49 Banach space adjoint. 213 Banach-Tarski Paradox. Universitext. Bowers. 7. Wiener Algebra 204 Spaces. 86 C(K). 212 Approximation Problem. holomorphic function 222 bounded annihilator. 216 equality. 2 Calkin algebra. J. 121 absolutely continuous Banach-Alaoglu Theorem. 31. An Introductory Course in Functional Analysis. 105. 198 bounded sequence. 179 Banach-Steinhaus Theorem. 106 σ -algebra.Index A Banach. 196 © Springer Science+Business Media. 14 B C β(X). 65. 42 C0 (X). 127 approximate point spectrum. 26 commutative. 86 Bessel’s Inequality. 187 set. 64 has no extreme points. 38 Banach limit. 199 function.1007/978-1-4939-1945-1 . 57 function. René-Louis. 106 A(D). Stefan. 209 almost open. 20. 211 measure on R. 84 absorbent. 45 Banach–Stone Theorem. Kalton. 158 bilinear map. 72 σ -algebra. 2 Axiom of Choice. 181 completeness of. 36 balanced set. 113 Baire. 74 abelian group. DOI 10. 136. 8 anti-isomorphism. 15 Baire Category Theorem. 5 Hilbert space adjoint. 7 approximate eigenvalue. N. 93 base for a topology. disk algebra 182 Banach–Mazur Characterization of Separable A(T). LLC 2014 227 A. 61. 157 sequence. 209 almost convergent sequence. 216 Banach space. 74 associative algebra. 160 adjoint bidual. 201 Banach algebra. Leonidas. 216 analytic function. 216 almost everywhere measurable function. 50. 182. 50 Borel Alaoglu. 187 Bounded Convergence Theorem. 31 absolutely convex set. 66 measure. 181 bounded linear map. 216 convergence. 1]. 108 bijection. 140 Bounded Inverse Theorem. 196 separability for K = [0. 61 c. 151 of a matrix. Stone-Čech compactification 221 c0 . 43. 190 set. 87 in a metric space. 151 diffuse measure. compact 16. 76 disk algebra. Pierre. 85 essentially bounded function. 14. 2 defined a. 41. 43. 126. 61 Cauchy’s Integral Formula. 124 division algebra.e. 16. 64 convergence in measure. 119 equivalent norms. 6. 212 exponent. 5 dual space action. 67 hull. 5. 79 characteristic function. Per. 21. 215 Fatou. 189. 220 continuous. 28 D Cauchy sequence. 122 chain. 163 . 223 commutative algebra. 182 map. 6 Enflo. 70. 79 operator. 94 eigenfunction. 163. 81 complete normed space. 196 convolution. 2 F completely regular topological space. 173 Closed Graph Theorem. 32 direct sum. 208 Dirichlet kernel. 212 closed ball dual space. 109 series. 83 finite rank operator. 85 closed topology. 88. Ernst Sigismund. 91 dense set. 203 Cantor function. 83. 71 linearity. 205 entire function. 129. 140 codimension. 111 complemented subspace. matrix adjoint 50 Finite Intersection Property. 23 C ∗ -algebra. see trus45 metric. J. 209 Cantor measure. 182. 76 eigenvalue. 183. 206 symmetric inner product. 221 Dirac mass. 215 set. 215 topological space. 24. 78. see Dirac measure Cesàro means. 162. 23 Čech. 53. 6. 110 coefficient. 67. 23. 31 discrete circle group. 215 group. 59.228 Index Calkin. 6 E in 2p . 187 Closest Point Lemma. 224 differentiable function. 15. 129 closure. 5. 85 graph. 222 Density of Simple Functions. 16.. 82. 54 Cantor group. 157 Fekete’s Lemma. 220 Fσ -set.W. 222 Cauchy-Schwarz Inequality. 28 Cantor’s Middle Thirds Set. 85 convergent sequence. 82 Dirac measure. 6 Dominated Convergence Theorem. 151 field. 221 extreme point. 157 Fejér kernel. 197 Cantor set. 133 essential supremum. 8 first category. 131. 49 in a normed vector space. 85 extremal set. 114 compactification. 130 continuously differentiable. 23 coset. 45 countable additivity. 190 transpose. 2. 216 Cauchy. 69 choice function. 85 Fischer. Eduard.. 93 transform. 61 conjugate Fatou’s Lemma. 90 first countable. 212 isomorphism. 16. Augustin-Louis. 70. 211 Cauchy Summability Criterion. 116. 197 equivalence classes of measurable functions. 163 convex Fourier function. 154 eigenvector. 204 set. 196 Fredholm. 79. 137 maximal. 88. 90. Erik Ivar. 84 real. 210 axioms. 127 Krein. 183 Hilbert. characteristic function 208 Gelfand transform. 32 kernel (nullspace). 159 kernel. 119. 193 ideal. 152 L real. Wolfgang. 61 incomplete normed space. 166 Hausdorff. 197 complex. 190 inner product Gelfand-Naimark Theorem. 114 hermitian Kronecker delta. 210 graph of a function. 137 left. Joseph. 85 of a Hilbert-Schmidt operator. 184 invertible element. 5.. 2 Gantmacher’s Theorem. 112 operations. 108 for real normed spaces. 25 holomorphic function. 82 L1 (R) convolution algebra. 38 James. 16. 197 operator. 6. 222 dual space. Felix. 197 injection. 166 for 0 < p < 1. 151 generated topology. 214 group. 32 Fredholm kernel. 116 Hellinger–Toeplitz Theorem. Alexander. 38 . 196 Fubini. Robert C. 216 homeomorphism. 79 Hahn–Banach Theorem Extension Theorem. 89. 125 separable (identification of). 184 H isometry.Index 229 Fourier. 196 Fundamental Theorem of Algebra. M. 33 J for complex normed spaces. 203. 215 Hilbert-Schmidt operator. 11 homogeneous system. David. 124 Hölder’s Inequality. 158 Krull. 38 homogeneity. 218 proper. 5. 56. 129 two-sided. 170 separability. Guido. 47. 140 Lp . 54 separability. 182 G image. 218 right. 1. 167 Krull’s Theorem. 196 identity. 43 L0 . 76 integral Grothendieck. 24. 197 Fredholm operator. 129 Gδ -set. Israel. 199 indiscrete topology. 149 indicator function. 183. 161 1 convolution algebra. 140 with respect to a complex measure. 94 kernel Hausdorff Maximality Principle. 7 L2 (a. 6 I Fredholm Alternative. 42 Goldstine’s Theorem. 197 Fubini’s Theorem. 83 Gelfand. 40 K Separation Theorem. 202 Hilbert transform. 198 Hilbert space complex. 129 Heine–Borel property. 44 inverse. 5 Gelfand–Mazur Theorem. 185 Intermediate Value Theorem. 106 integrable function. 137 Hausdorff space. 44 with respect to a positive measure. 121 Haar measure. 166 Hilbert–Schmidt operator. 121 isomorphism. 96 p . 44 interior.. 6 of invertible elements. b). 178 Krein-Milman Theorem. 216 completeness of. 36 invariant version. spectrum of a Banach nonatomic measure. 3 Morera. 119 Open Mapping Theorem. 209 integrable function. 1 Mazur. 7 natural embedding. 108 normed space. 22 Neumann series. 207. 23 complete. 205 locally compact. 64 for a compact hermitian operator. James. 25 regular. 83 Lebesgue measure on T. 71 invariant. D. 85 Minkowski functional. 48 meager set. 210 metric. 224 Morera’s Theorem. 84 multiplier operator.230 Index L∞ . 209 open ball complex. Nikolai. 216. 83 M Nested Interval Property. 32 Nikodým. 23 algebra 199 norm. 222 Lipschitz constant. 210 operator. 84 counting measure. 50 Neumann. 222 Lipschitz continuous. 195. 185 maximal element. 207 space. 152. 209 in a normed vector space. 124 Mazur’s Theorem. 163 measurable nowhere dense set. 209 orthogonal . 185 matrix adjoint. 182. 215 Liouville’s Theorem. 5 Minkowski’s Inequality. 212 Liouville. 197 Lwów. 211 null set. 110. 122. 91 absolutely continuous. 207 O measure ω. 117 Lebesgue. 202. 214 maximal ideal space. 140 norming element. 1 Maximum Modulus Theorem. Giacinto. 210 open map. 223 support of a measure. 209 open set. 5. 55. 64 function. 6. metrizable. Henri. 217 Naimark. 183. 220 M(K). 117 Lebesgue measure space. 209 neighborhood. 209 measure-zero set. 23 topology. 3 multiplicative linear functional. 85 212 Milman’s Theorem. 209 group. 210 space. 219 locally convex. 219 Munkres. 116 limit in a topological space. 116. 1 measure on R. 6 finite measure. 217 nullspace. Otton. 213 in a metric space. 213 one-point compactification. 84 Lebesgue’s Devil Staircase. 223 Monotone Convergence Theorem. 190. 213 σ -algebra. 1.. 209 rectangle. 16. kernel 129 sets. 214 Laurent series. 215 probability measure. 7 Milman. 217 ∞ . 73 Lebesgue measure on R. 202 product measure. 218 completeness of. Joseph. 222 normal. 44 measure on T. 14 σ -finite. Stanislaw. 219 Borel measure on R. Mark. 84 Lebesgue’s Dominated Convergence Theorem. 93 Lusin’s Theorem. 209 of N. 197 local base for a topology. 97 linear functional. 85. Carl. 49 negligible set. 3 positive. 217 N Lusin. 210 as a Banach algebra. 220 orthonormal set. 85 pole of order n. 77. Johann. 130 orthogonal operator. 163 regular topological space. 161 Rank-Nullity Theorem. 151 separable space. 1 as a quotient of 1 . 207 Q similar to an orthogonal operator. 208 projection. 205 sequential space. 186 P (K). 215 Riesz Representation Theorem. 129. 57 sequential compactness. 193 simple function. 38 pre-annihilator. 153 shift-invariance. 116 reverse triangle inequality. 159 relatively compact set. 190 space. 45 orthonormal basis. 54 singularity. 45 partial sum operator. 156 Riemann-Lebesgue Lemma. 43. 196 singular measure. 145 p-summable sequence. 22. 54 skew-field. 100 p-integrable function. 199 . 224 Schauder’s Theorem. 54. 79 spectral radius. 153 range. 43 quasinilpotent. 64 polarization formula. 58. 62 residual set. 43. 16. 58. 223 norm. 96 inner product. 189 P resolvent of an element. 219 Schmidt. 217 p-norm. Charles Émile. 154 second countable. 197 partial order. 32 rotations in an abelian group. 191 simply connected topological space. 161. 74 positive homogeneity. 191 Spectral Radius Formula. 163. 84 set function. 191 R spectral theory. 223 positive definite semi-norm. 214 second category. 49 of eigenvectors. 11 Riesz’s Lemma. 8 PE . 67 Rudin. 141 point at infinity. 220 algebra. 8 product topology sequentially open set. 208 quaternions. 153 Radon–Nikodým Theorem. Frigyes. 214 to a vector. 163 Parseval’s Identity. 130 oscillation of a function. 32 classical examples. 40. 218 sequentially closed metric space. 99 product measure. 214 spectrum Radon–Nikodým derivative. 7. 222 quotient singleton. Walter. 7 point spectrum. 137. 21 map. 83 sesquilinear map.Index 231 to a closed subspace. 151 Radon. 223 S Picard. 195 ring homomorphism. 129 orthogonal group. 143 Picard’s Theorems. 209 Riesz–Fréchet Theorem. 64 resolvent function. 58 rank of a matrix. 86. 11 Riesz. 45 rank. 86 finite product. 6 norm. 7. 145 for L2 (0. Erhard. 134 prime ideal. 39 σ -algebra. 214 of a Banach algebra. 151 infinite product. 187 Scottish Book. 194 Pythagorean Theorem. π). 140 polar representation of a complex measure. 222 perturbation by a compact operator. 217 pairwise disjoint sets. 130. 153 Riesz-Fischer Theorem. 79 shift operator. 156 Parallelogram Law. 176 reflexive Banach space. 122. 32. 21. 205 bounded set. 218 topology. 179 Stone-Čech compactification. 65. Marshall Harvey. 1 W w-topology. 200 operator.232 Index of an element. 14 surjection. Leonida. 45. 221 uniformly convex space. 204 totally bounded set. 32 support of a measure. 195 Weierstrass Approximation Theorem. 6 sublinear functional. 172 of a Hilbert–Schmidt operator. 32 Tychonoff. 124 supremum norm. 131 vector space. 65 group. 84 topology. 202 Uniform Boundedness Principle. 186 U Stone space. 83 continuous operator. 149 space. total variation 137. 52. 86 weak∗ topology. 66 topology generated by a metric. 170 compact. 119 compact operator. 33. 104 subadditivity. 71. 162 measure. 193. 102 T weak ternary expansion. Andrey Nikolayevich. 31 Tychonoff’s Theorem. 67. 167 Volterra operator. 99 Tonelli. 178 triangle inequality. 220 Zorn’s Lemma. 98 topological weakly divisor of zero. 7. 98 w∗ -topology. 102 torus. 38. 83 bounded. 221 Zorn. 142. 82 Zermelo–Fraenkel Axioms. 86 inner product. 96 in a normed space. 22 algebra. 96 Z trigonometric polynomial. 64. 32 . 182 submultiplicative norm. 22 convergence. see masure117 Urysohn’s Lemma. 1. 192 Stone. 7. 203 of a Fredholm operator. 43 vibrating string. 132 Inversion Theorem. Norbert. 181 upper bound. 158 property of a metric. 1. 32 unital algebra. 105. Max. 213 Wiener norm. 141 Wronskian. 221 unit ball Sturm-Liouville system. 5 V symmetric vector topology. 154. 175 kernel. 204 trace Wiener. 202.

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