Banach Space Theory: The Basis for Linear and Nonlinear Analysis

Canadian Mathematical Society Société mathématique du Canada

Editors-in-Chief Rédacteurs-en-chef K. Dilcher K. Taylor

Advisory Board Comité consultatif G. Bluman P. Borwein R. Kane

For other titles published in this series, go to http://www.springer.com/series/4318

Marián Fabian · Petr Habala · Petr Hájek · Vicente Montesinos · Václav Zizler

Banach Space Theory The Basis for Linear and Nonlinear Analysis

123

Marián Fabian Mathematical Institute of the Academy of Sciences of the Czech Republic Žitná 25, Praha 1 11567 Prague, Czech Republic [email protected]

Vicente Montesinos Universidad Politécnica de Valencia Departamento de Matematica Aplicada Camino de Vera s/n 46022 Valencia, Spain [email protected]

Petr Habala Czech Technical University in Prague Department of Mathematics Faculty of Electrical Engineering Technická 2 16627 Prague, Czech Republic [email protected]

Václav Zizler University of Alberta Department of Mathematical and Statistical Sciences Central Academic Building Edmonton T6G 2G1 Alberta, Canada [email protected]

Petr Hájek Mathematical Institute of the Academy of Sciences of the Czech Republic Žitná 25, Praha 1 11567 Prague, Czech Republic [email protected] Editors-in-Chief Rédacteurs-en-chef Canada K. Dilcher K. Taylor Department of Mathematics and Statistics Dalhousie University Halifax, Nova Scotia B3H 3J5 [email protected]

ISSN 1613-5237 ISBN 978-1-4419-7514-0 e-ISBN 978-1-4419-7515-7 DOI 10.1007/978-1-4419-7515-7 Springer New York Dordrecht Heidelberg London Library of Congress Control Number: 2010938895 Mathematics Subject Classicication (2010): Primary: 46Bxx Secondary: 46A03, 46A20, 46A22, 46A25, 46A30, 46A32, 46A50, 46A55, 46B03, 46B04, 46B07, 46B10, 46B15, 46B20, 46B22, 46B25, 46B26, 46B28, 46B45, 46B50, 46B80, 46C05, 46C15, 46G05, 46G12, 47A10, 52A07, 52A21, 52A41, 58C20, 58C25 c Springer Science+Business Media, LLC 2011  All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, LLC, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com)

Preface

Many problems in modern linear and nonlinear analysis are of infinite-dimensional nature. The theory of Banach spaces provides a suitable framework for the study of these areas, as it blends classical analysis, geometry, topology, and linearity. This in turn makes Banach space theory a wonderful and active research area in Mathematics. In infinite dimensions, neighborhoods of points are not relatively compact, continuous functions usually do not attain their extrema and linear operators are not automatically continuous. By introducing weak topologies, compactness can be obtained via Tychonoff’s theorem. Similarly, functions often need to be perturbed so that the problem of finding extrema is solvable. To deal with problems in linear and nonlinear analysis, a good working knowledge of Banach space theory techniques is needed. It is the purpose of this introductory text to help the reader grasp the basic principles of Banach space theory and nonlinear geometric analysis. The text presents the basic principles and techniques that form the core of the theory. It is organized to help the reader proceed from the elementary part of the subject to more recent developments. This task is not easy. Experience shows that working through a large number of exercises, provided with hints that direct the reader, is one of the most efficient ways to master the subject. Exercises are of several levels of difficulty, ranging from simple exercises to important results or examples. They illustrate delicate points in the theory and introduce the reader to additional lines of research. In this respect, they should be considered an integral part of the text. A list of remarks and open problems ends each chapter, presenting further developments and suggesting research paths. An effort has been made to ensure that the book can serve experts in related fields such as Optimization, Partial Differential Equations, Fixed Point Theory, Real Analysis, Topology, and Applied Mathematics, among others. As prerequisites, basic undergraduate courses in calculus, linear algebra, and general topology, should suffice. The text is divided into 17 chapters. In Chapter 1 we present basic notions in Banach space theory and introduce the classical Banach spaces, in particular sequence and function spaces.

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In Chapter 2 we discuss two fundamental principles of Banach space theory, namely the Hahn–Banach Theorem on extension of bounded linear functionals and the Banach Open mapping Theorem, together with some of their applications. In Chapter 3 we discuss weak topologies and their properties related to compactness. Then we prove the third fundamental principle, namely the Banach–Steinhaus Uniform Boundedness principle. Special attention is devoted to weak compactness, in particular to the theorems of Eberlein, Šmulyan, Grothendieck and James, and the theory of reflexive Banach spaces. In Chapter 4 we introduce Schauder bases in Banach spaces. The possibility to represent each element of the space as the sequence of its coefficients in a given Schauder basis transfers the purely geometric techniques of the elementary Banach space theory to the analytic computations of the classical analysis. Although not every separable Banach space admits a Schauder basis, the use of basic sequences and Schauder bases with additional properties is one of the main tools in the investigation of the structural properties of Banach spaces. In Chapter 5 we continue the study of the structure of Banach spaces by adding results on extensions of operators, injectivity, and weak injectivity. The core of the chapter is the theory of separable Banach spaces not containing isomorphic copies of 1 . Chapter 6 is an introduction to some basic results in the geometry of finitedimensional Banach spaces and their connection to the structure of infinitedimensional spaces. We do not discuss the deeper parts of the theory, which essentially depend on measure theoretical techniques. We introduce the notion of finite representability, and prove the principle of local reflexivity. We use the John ellipsoid to prove the Kadec–Snobar theorem and give a proof of Tzafriri’s theorem. We indicate the connection of this result with Dvoretzky’s theorem. Last part of the chapter is devoted to the Grothendieck inequality. In Chapter 7 we present an introduction to nonlinear analysis, namely to variational principles and differentiability. In Chapter 8 we study the interplay between differentiability of norms and the structure of separable Asplund spaces. Chapter 9 introduces the subject of superreflexive spaces, whose structure is nicely described by the behavior of its finite-dimensional subspaces. Chapter 10 studies the impact of the existence of higher order smooth norms on the structure of the underlying space. Special effort is devoted to countable compact spaces and  p spaces. Chapter 11 deals with the property of dentability and results on differentiation of vector measures. We prove some basic results on Banach spaces with the Radon– Nikodým property. Chapter 12 introduces the reader to the nonlinear geometric analysis of Banach spaces. Results on uniform and nonuniform homeomorphisms are presented, including Keller’s theorem and basic fixed points theorems (Brouwer, Schauder, etc). We discuss a proof of the homeomorphisms of Banach spaces and results on uniform, in particular Lipschitz, homeomorphisms.

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Chapter 13 contains a basic study of an important class of non-separable Banach spaces, the weakly compactly generated spaces. In particular, we discuss their decompositions and renormings. We also study weakly compact operators, absolutely summing operators, and the Dunford–Pettis property. Chapter 14 deals with results on weak topologies, focusing on special types of compacta (scattered, Eberlein, Corson, etc.). Chapter 15 presents basic results in the spectral theory of operators. We study compact and self-adjoint operators. Chapter 16 deals with the basic theory of tensor products. We follow the Banach space approach, focusing on the Grothendieck duality theory of tensor products, Schauder bases, applications to spaces of compact operators, etc. We include Enflo’s example of a Banach space without the approximation property. A short appendix (Chapter 17) has been included collecting some very basic definitions and results that are used in the text, for the reader’s immediate access. In writing the text we strived to avoid excessive technicalities, keeping each subject as elementary as reasonably possible. Each chapter ends with a brief section of Remarks and Open Questions, containing further known results and some problems in the area that are—to our best knowledge—open. Several more specialized books and survey articles appeared recently in Banach space theory, as [AlKa], [BeLi], [BoVa], [CasGon], [DJT], [HMVZ], [JoLi3], [Kalt4], [KaKuLP], [LPT], [MOTV2], [Wojt], among others. We hope that the present text can help both the student and the professional mathematician to get acquainted with the techniques needed in these directions. We also made an effort to make this text closer to a reference book in order to help researchers in Banach space theory. We are grateful to many of our colleagues for suggestions, advice, and discussions on the subject of the book. We thank our Institutions: the Institute of Mathematics of the Czech Academy of Sciences, the Czech Technical University in Prague, the Department of Mathematical and Statistical Sciences at the University of Alberta, Edmonton, Canada, the Universidad Politécnica de Valencia, Spain, and its Instituto Universitario de Matemática Pura y Aplicada. This work has been supported by several Grant Agencies: The Czech National Grant Agency and the Institutional Research Plan of the Academy of Sciences (Czech Republic), NSERC Canada, the Ministerio de Educación (Spain) and the Generalitat Valenciana (Valenˇ cia, Spain). The grants involved are IAA 100 190 610, IAA 100 190 901, GACR 201/07/0394, No. AVOZ 101 905 03 (Czech Republic), Proyecto MTM2008-03211 (Spain), BEST/2009/096 (Generalitat Valenciana) and PR2009-0267 (Ministerio de Educación), NSERC-7926 (Canada). We would like to thank the Springer Team for their interest in this project. In particular, we are thankful to Keith F. Taylor, Karl Dilcher, Mark Spencer, Vaishali Damle, and Charlene C. Cerdas. We thank also Eulalia Noguera for her help with the tex file, and to Integra Software Services Pvt Ltd, in particular Sankara Narayanan, for their assistance in editing the final version of this book. Above all, we are indebted to our families for their moral support and encouragement.

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We would be glad if this book inspired some young mathematicians to choose Banach Space Theory and/or Nonlinear Geometric Analysis as their field of interest. We wish the reader a pleasant time spent over this book. Prague, Czech Republic Prague, Czech Republic Prague, Czech Republic Valencia, Spain Edmonton, AB, Canada Spring, 2010

Marián Fabian Petr Habala Petr Hájek Vicente Montesinos Václav Zizler

Contents

1 Basic Concepts in Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Hölder and Minkowski Inequalities, Classical Spaces C[0, 1],  p , c0 , L p [0, 1] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Operators, Quotients, Finite-Dimensional Spaces . . . . . . . . . . . . . . 1.4 Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Remarks and Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises for Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Hahn–Banach and Banach Open Mapping Theorems . . . . . . . . . . . . . 2.1 Hahn–Banach Extension and Separation Theorems . . . . . . . . . . . . 2.2 Duals of Classical Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Banach Open Mapping Theorem, Closed Graph Theorem, Dual Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Remarks and Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises for Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 3 13 24 29 31 53 54 60 65 68 68

3 Weak Topologies and Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 3.1 Dual Pairs, Weak Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 3.2 Topological Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 3.3 Locally Convex Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 3.4 Polarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 3.5 Topologies Compatible with a Dual Pair . . . . . . . . . . . . . . . . . . . . . 100 3.6 Topologies of Subspaces and Quotients . . . . . . . . . . . . . . . . . . . . . . 103 3.7 Weak Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 3.8 Extreme Points, Krein–Milman Theorem . . . . . . . . . . . . . . . . . . . . . 109 3.9 Representation and Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 3.10 The Space of Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 3.11 Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 3.11.1 Banach–Steinhaus Theorem . . . . . . . . . . . . . . . . . . . . . . . 119 3.11.2 Banach–Dieudonné Theorem . . . . . . . . . . . . . . . . . . . . . . 122

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3.11.3 The Bidual Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 3.11.4 The Completion of a Normed Space . . . . . . . . . . . . . . . . 126 3.11.5 Separability and Metrizability . . . . . . . . . . . . . . . . . . . . . 127 3.11.6 Weak Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 3.11.7 Reflexivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 3.11.8 Boundaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 3.12 Remarks and Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 Exercises for Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 4 Schauder Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 4.1 Projections and Complementability, Auerbach Bases . . . . . . . . . . . 179 4.2 Basics on Schauder Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 4.3 Shrinking and Boundedly Complete Bases, Perturbation . . . . . . . . 187 4.4 Block Bases, Bessaga–Pełczy´nski Selection Principle . . . . . . . . . . 194 4.5 Unconditional Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 4.6 Bases in Classical Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 4.7 Subspaces of L p Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 4.8 Markushevich Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 4.9 Remarks and Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 Exercises for Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 5 Structure of Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 5.1 Extension of Operators and Lifting . . . . . . . . . . . . . . . . . . . . . . . . . . 237 5.2 Weak Injectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 5.2.1 Schur Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 5.3 Rosenthal’s 1 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 5.4 Remarks and Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264 Exercises for Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 6 Finite-Dimensional Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 6.1 Finite Representability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 6.2 Spreading Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294 6.3 Complemented Subspaces in Spaces with an Unconditional Schauder Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 6.4 The Complemented-Subspace Result . . . . . . . . . . . . . . . . . . . . . . . . 309 6.5 The John Ellipsoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312 6.6 Kadec–Snobar Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 6.7 Grothendieck’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 6.8 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 Exercises for Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326 7 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 7.2 Subdifferentials: Šmulyan’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . 336

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7.3 Ekeland Principle and Bishop–Phelps Theorem . . . . . . . . . . . . . . . 351 7.4 Smooth Variational Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 7.5 Norm-Attaining Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 7.6 Michael’s Selection Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 7.7 Remarks and Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364 Exercises for Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 8 C 1 -Smoothness in Separable Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 8.1 Smoothness and Renormings in Separable Spaces . . . . . . . . . . . . . 383 8.2 Equivalence of Separable Asplund Spaces . . . . . . . . . . . . . . . . . . . . 385 8.3 Applications in Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394 8.4 Smooth Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402 8.5 Ranges of Smooth Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405 8.6 Remarks and Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408 Exercises for Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 9 Superreflexive Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429 9.1 Uniform Convexity and Uniform Smoothness,  p and L p Spaces . 429 9.2 Finite Representability, Superreflexivity . . . . . . . . . . . . . . . . . . . . . . 435 9.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449 9.4 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453 Exercises for Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453 10 Higher Order Smoothness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465 10.2 Smoothness in  p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466 10.3 Countable James Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468 10.4 Remarks and Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474 Exercises for Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475 11 Dentability and Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479 11.1 Dentability in X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479 11.2 Dentability in X ∗ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486 11.3 The Radon–Nikodým Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490 11.4 Extension of Rademacher’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 504 11.5 Remarks and Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510 Exercises for Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511 12 Basics in Nonlinear Geometric Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 521 12.1 Contractions and Nonexpansive Mappings . . . . . . . . . . . . . . . . . . . . 521 12.2 Brouwer and Schauder Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 526

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Contents

12.3

The Homeomorphisms of Convex Compact Sets: Keller’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533 12.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533 12.3.2 Elliptically Convex Sets . . . . . . . . . . . . . . . . . . . . . . . . . . 535 12.3.3 The Space T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537 12.3.4 Compact Elliptically Convex Subsets of 2 . . . . . . . . . . . 538 12.3.5 Keller Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 541 12.3.6 Applications to Fixed Points . . . . . . . . . . . . . . . . . . . . . . . 541 12.4 Homeomorphisms: Kadec’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 542 12.5 Lipschitz Homeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545 12.6 Remarks and Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 559 Exercises for Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561 13 Weakly Compactly Generated Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 575 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575 13.2 Projectional Resolutions of the Identity . . . . . . . . . . . . . . . . . . . . . . 577 13.3 Consequences of the Existence of a Projectional Resolution . . . . . 581 13.4 Renormings of Weakly Compactly Generated Banach Spaces . . . . 586 13.5 Weakly Compact Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 13.6 Absolutely Summing Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 592 13.7 The Dunford–Pettis Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 596 13.8 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 598 13.9 Remarks and Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 602 Exercises for Chapter 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 14 Topics in Weak Topologies on Banach Spaces . . . . . . . . . . . . . . . . . . . . . 617 14.1 Eberlein Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617 14.2 Uniform Eberlein Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 622 14.3 Scattered Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625 14.4 Weakly Lindelöf Spaces, Property C . . . . . . . . . . . . . . . . . . . . . . . . . 629 14.5 Weak∗ Topology of the Dual Unit Ball . . . . . . . . . . . . . . . . . . . . . . . 634 14.6 Remarks and Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 642 Exercises for Chapter 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 643 15 Compact Operators on Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 657 15.1 Compact Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 657 15.2 Spectral Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 661 15.3 Self-Adjoint Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 668 15.4 Remarks and Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 678 Exercises for Chapter 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 678 16 Tensor Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 687 16.1 Tensor Products and Their Topologies . . . . . . . . . . . . . . . . . . . . . . . 687 16.2 Duality of Injective Tensor Products . . . . . . . . . . . . . . . . . . . . . . . . . 696

Contents

xiii

16.3 Approximation Property and Duality of Spaces of Operators . . . . 700 16.4 The Trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 708 16.5 Banach Spaces Without the Approximation Property . . . . . . . . . . . 711 16.6 The Bounded Approximation Property . . . . . . . . . . . . . . . . . . . . . . . 717 16.7 Schauder Bases in Tensor Products . . . . . . . . . . . . . . . . . . . . . . . . . . 721 16.8 Remarks and Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 726 Exercises for Chapter 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 727 17 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 733 17.1 Basics in Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 733 17.2 Nets and Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 735 17.3 Nets and Filters in Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . 736 17.4 Ultraproducts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 737 17.5 The Order Topology on the Ordinals . . . . . . . . . . . . . . . . . . . . . . . . . 737 17.6 Continuity of Set-Valued Mappings . . . . . . . . . . . . . . . . . . . . . . . . . 738 17.7 The Cantor Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 739 17.8 Baire’s Great Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 741 17.9 Polish Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 741 17.10 Uniform Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 741 17.11 Nets and Filters in Uniform Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 742 17.12 Partitions of Unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 743 17.13 Measure and Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744 17.13.1 Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744 17.13.2 Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 745 17.14 Continued Fractions and the Representation of the Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 746 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 751 Symbol Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 777 Subject Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 781 Author Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 807

Chapter 1

Basic Concepts in Banach Spaces

In this chapter we introduce basic notions and concepts in Banach space theory. As a rule we will work with real scalars, only in a few instances, e.g., in spectral theory, we will use complex scalars. K denotes simultaneously the real (R) or complex (C) scalar field. We use N for the set {1, 2, . . . }. All topologies are assumed to be Hausdorff, unless stated otherwise. In particular, by a compact space we mean a compact Hausdorff space. By a neighborhood of a point x in a topological space T we mean any subset of T that contains an open subset O of T such that x ∈ O.  If (T, T ) is a topological space, and S is a nonempty subset, we shall write T  S for the restriction of the topology T to S (and so (S, T ) becomes a topological space). If there is no possibility of misunderstanding, the restricted topology will be called again T . For a brief review on basic topological notions see, e.g., the Appendix.

1.1 Basic Definitions Definition 1.1 A non-negative function  ·  on a vector (i.e., linear) space X is called a norm on X if (i) x ≥ 0 for every x ∈ X , (ii) x = 0 if and only if x = 0, (iii) λx = |λ| x for every x ∈ X and every scalar λ, (iv) x + y ≤ x + y for every x, y ∈ X (the “triangle inequality”). A vector space X with a norm  ·  is denoted by (X,  · ), and is called a normed linear space (or just a normed space). Note that the function ρ(x, y) := x − y, where x, y ∈ X , is indeed a metric on X . To check the triangle inequality we write ρ(x, z) = x − z = x − y + y − z ≤ x − y + y − z = ρ(x, y) + ρ(y, z). n  n By induction,  i=1 xi  ≤ i=1 xi  for a finite number of vectors x1 , . . . , x n in X .

M. Fabian et al., Banach Space Theory, CMS Books in Mathematics, C Springer Science+Business Media, LLC 2011 DOI 10.1007/978-1-4419-7515-7_1, 

1

2

1 Basic Concepts in Banach Spaces

All topological and uniform notions in normed spaces refer to the canonical metric given by the norm, unless stated otherwise. In situations when more than one normed space is considered, we will sometimes use  ·  X to denote the norm of X . Definition 1.2 A Banach space is a normed linear space (X,  · ) that is complete in the canonical metric defined by ρ(x, y) = x − y for x, y ∈ X , i.e., every Cauchy sequence in X for the metric ρ converges to some point in X . Let (X,  · ) be a normed space. The set B X := {x ∈ X : x ≤ 1} is said to be the closed unit ball of X , and S X := {x ∈ X : x = 1} the unit sphere of (X,  · ). Given x0 ∈ X and r > 0, the set B(x0 , r ) := {x ∈ X : x − x0  ≤ r } is said to be the closed ball centered at x0 with radius r . If M ⊂ X , then span(M) stands for the linear hull—or span—of M, that is, the intersection of all linear subspaces of X containing M. Equivalently, span(M) is the smallest (in the sense of inclusion) linear subspace of X containing M, or the set of all finite linear combinations of elements in M. Similarly, span(M) stands for the closed linear hull of M, i.e., the smallest closed linear subspace of X containing M. If no misunderstanding can arise, by a “subspace” of a vector space we will mean a linear subspace and, in case of normed spaces, a closed linear subspace. Definition 1.3 Let E be a vector space. Given x, y ∈ E, the set [x, y] := {λx + (1 − λ)y : 0 ≤ λ ≤ 1} is called the closed segment defined by x and y. If x = y, the set (x, y) := {λx + (1 − λ)y : 0 < λ < 1} is called the open segment defined by x and y. A subset C of a vector space E is called convex if [x, y] ⊂ C whenever x, y ∈ C. If M ⊂ X , the convex hull of M is the smallest convex subset of X containing M, and will be denoted by conv(M); conv(M) denotes the closed convex hull of M, i.e., the smallest closed convex subset of X containing M. Definition 1.4 Let U be a convex subset  of a vector space V . We say that a function f : U → R is convex if f λx+(1−λ)y ≤ λ f(x)+(1−λ) f(y) for all x, y ∈ U and λ ∈ [0, 1]. We say that f is strictly convex if f λx +(1−λ)y < λ f (x)+(1−λ) f (y) for all x, y ∈ U , x = y, and λ ∈ (0, 1). For instance, every norm of a normed space X is a convex function on X . Observe that a function f : U → R is convex if and only if the epigraph of f , i.e., the set epi f := {(x, r ) ∈ U × R : f (x) ≤ r } ⊂ X × R, is convex (the linear structure of X × R is defined coordinatewise). For subsets A, B of a vector space X and a scalar α we also write A + B := {a + b : a ∈ A, b ∈ B} and α A := {αa : a ∈ A}. A set M ⊂ X is called symmetric if (−1)M ⊂ M, and balanced if α M ⊂ M for all α ∈ K, |α| ≤ 1. Let Y be a subspace of a normed space (X,  · ). By (Y,  · ) we denote Y endowed with the restriction of  ·  to Y if there is no risk of misunderstanding. Fact 1.5 Let Y be a subspace of a Banach space X . Then Y is a Banach space if and only if Y is closed in X .

1.2

Hölder and Minkowski Inequalities, Classical Spaces

3

Proof: Assume that Y is closed. Consider a Cauchy sequence {yn }∞ n=1 in Y . Since the norm on Y is the restriction of the norm of X , the sequence is Cauchy in X and therefore converges to some y ∈ X . As Y is closed, y ∈ Y and yn → y in Y . The other direction is proved by a similar argument. Definition 1.6 A subset M of a normed space (X,  · ) is called bounded if there exists r > 0 such that M ⊂ r B X . M is called totally bounded if for every ε > 0 the set M can be covered by a finite number of translates of ε B X . A sequence {xn } in X is called bounded (totally bounded) if the set {xn : n ∈ N} is bounded (respectively, totally bounded). Note that every totally bounded set is already bounded. See also Exercises 1.47 and 1.48 for a description of total boundedness by using ε-nets, Definition 3.11, and Section 17.10.

1.2 Hölder and Minkowski Inequalities, Classical Spaces C[0, 1],  p , c0 , L p [0, 1] We will now turn to some examples of Banach spaces. Definition 1.7 The symbol C[0, 1] denotes the vector space of all scalar valued continuous functions on the interval [0, 1] (the vector addition and the scalar multiplication being defined pointwise), endowed with the norm  f ∞ := sup{| f (t)| : t ∈ [0, 1]} (= max{| f (t)| : t ∈ [0, 1]}). Proposition 1.8 The function  · ∞ introduced in Definition 1.7 is indeed a norm, and (C[0, 1],  · ∞ ) is a Banach space. Proof: We easily check that C[0, 1] is a normed space. Consider a Cauchy sequence ∞ { f n }∞ n=1 in C[0, 1]. As | f k (t) − fl (t)| ≤  f k − f l ∞ , the sequence { f n (t)}n=1 is a Cauchy sequence for every t ∈ [0, 1]. Set f (t) := lim f n (t). This defines a scalar n→∞ valued function f on [0, 1]. It remains to show that f is continuous and f n → f uniformly (i.e., in  · ∞ ). Given ε > 0, there is n 0 such that | f n (t) − f m (t)| ≤ ε for every t ∈ [0, 1] and every n, m ≥ n 0 . By fixing n ≥ n 0 and letting m → ∞ we get | f n (t) − f (t)| ≤ ε for every n ≥ n 0 and every t ∈ [0, 1]. Let t0 ∈ [0, 1] and ε > 0 be fixed. Choose δ > 0 so that | f n 0 (t) − f n 0 (t0 )| < ε whenever |t − t0 | < δ. Then, whenever |t − t0 | < δ, | f (t) − f (t0 )| ≤ | f (t) − f n 0 (t)| + | f n 0 (t) − f n 0 (t0 )| + | f n 0 (t0 ) − f (t0 )| < 3ε. Therefore f ∈ C[0, 1]. It has been shown above that, for every n ≥ n 0 ,  f n − f ∞ ≤ ε. This proves that  f n − f ∞ → 0, so C[0, 1] is complete. Analogously, the space C(K ) of continuous scalar functions on a compact space K , endowed with the supremum norm, is a Banach space.

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1 Basic Concepts in Banach Spaces

We note that C[0, 1] is an infinite-dimensional Banach space. To see this, it is enough to produce, for any n ∈ N, a linearly independent set of n elements in C[0, 1]. The set of functions {1, t, t 2 , . . . , t n−1 } has this property. More generally, the space C(K ), where K is a compact topological space, is infinite-dimensional as soon as K is infinite; indeed, given a finite set of distinct points S := {ki : i = 1, 2, . . . , n} in K , define the function δki on S for i = 1, 2, . . . , n, where δk is the Kronecker delta function at k, i.e., δk (k) = 1 and δk (k ) = 0 for all k = k. Extend each δki to a continuous function on K by using the Tietze–Urysohn theorem (see Corollary 7.55). The resulting set of extended functions {δki : i = 1, 2, . . . , n} is linearly independent in C(K ). Definition 1.9 The symbol n∞ denotes the n-dimensional vector space of all ntuples of scalars (that is, Rn or Cn ), the vector addition and the scalar multiplication being defined coordinatewise, endowed with the supremum norm  · ∞ defined for x = (x1 , . . . , xn ) ∈ n∞ by x∞ = max{|xi | : i = 1, . . . , n}. Note that n∞ is a special case of a C(K ) space, where K := {1, . . . , k}, endowed with the discrete topology. In order to introduce the class of  p spaces for 1 < p < ∞ we need to prove the following classical inequalities. Theorem 1.10 (Hölder inequality) Let p, q > 1 be such that n ∈ N. Then for all ak , bk ∈ K, k = 1, . . . , n, we have n 

|ak bk | ≤

n 

k=1

|ak | p

1 p

+

1 q

= 1 and let

n 1  1 p q · |bk |q .

k=1

(1.1)

k=1

For p = 2, q = 2, the inequality (1.1) is known as the Cauchy–Schwarz inequality. In the proof of Theorem 1.10 we will use the following statement. Lemma 1.11 Let p, q > 1 be such that a, b ≥ 0.

1 p

+

1 q

= 1. Then ab ≤

ap p

+

bq q

for all

Proof: Consider the graph of the function y = x p−1 , x ≥ 0, and the areas A1 of the region bounded by the curves y = x p−1 , y = 0, x = a, and A2 of the region bounded by the curves y = x p−1 , x = 0, y = b (see Fig. 1.1). Clearly,

a

b p q A1 = 0 x p−1 dx = ap . As x = y 1/( p−1) = y q−1 , we get A2 = 0 y q−1 dy = bq . It follows that ab ≤ A1 + A2 =

ap p

+

bq q .

An alternative proof is by checking extrema of the function ϕ(a) := for a fixed b > 0.

ap p

q

+ bq −ab

Proof of Theorem 1.10: We may assume that ai , bi ≥ 0 and neither all ai nor all bi are zero. For k = 1, . . . , n define

1.2

Hölder and Minkowski Inequalities, Classical Spaces

5

Y

b A2 A1 0

a

X

Fig. 1.1 Two areas and a rectangle in the proof of Lemma 1.11

A k = ak

n 

aj

p

− 1

p

Bk = bk

and

n 

j=1

bjq

− 1 q

.

j=1

n  q We note that nk=1 Ak p = k=1 Bk = 1. By Lemma 1.11, we have for k = 1 1 p q 1, . . . , n that Ak Bk ≤ p Ak + q Bk . Summing up this inequality for k = 1, . . . , n we get n 

Ak Bk ≤

k=1

n n 1 p 1 q 1 1 Ak + Bk = + = 1, p q p q k=1

k=1

which implies the desired inequality. Theorem 1.12 (Minkowski inequality) Let p ∈ [1, ∞) and n ∈ N. Then for all ak , bk ∈ K, k = 1, . . . , n, we have n 

|ak + bk | p

1

p

≤

k=1

n  k=1

|ak | p

1

p

+

n 

|bk | p

1

p

.

(1.2)

k=1

Proof: The statement is trivial for p = 1. If p ∈ (1, ∞), let q ∈ (1, ∞) be such that 1 1 p + q = 1. We may assume that ai , bi ≥ 0. Using the Hölder inequality (1.1) and the fact that ( p − 1)q = p we obtain    (ak + bk ) p−1 (ak + bk ) = (ak + bk ) p−1 ak + (ak + bk ) p−1 bk (ak + bk ) p = 1  1  1  1  q p q p ≤ + ak p bk p (ak + bk )( p−1)q (ak + bk )( p−1)q



=

1  1  1  1  q p q p + ak p bk p . (ak + bk ) p (ak + bk ) p

Dividing by

 1 (ak + bk ) p q we get

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1 Basic Concepts in Banach Spaces



(ak + bk ) p

1

p

=



(ak + bk ) p

1− 1 q

≤



ak p

1

p

+



bk p

1

p

.

Definition 1.13 Let p ∈ [1, ∞). The symbol np denotes the n-dimensional vector space Kn , the vector addition and the scalar multiplication being defined coordinatewise, endowed with the norm defined for x = (x1 , . . . , xn ) ∈ np by x p =

n 

|xi | p

1

p

.

i=1

By Minkowski’s inequality (1.2),  ·  p is indeed a norm on X . The closed unit ball of 21 is the square with vertices ±e1 , ±e2 , where ei are the standard unit vectors, e1 = (1, 0) and e2 = (0, 1). The unit ball of 2∞ is the square with vertices (±e1 ± e2 ). The unit ball of 22 is the disk of radius 1 centered at the origin. Balls for other ps are somehow in between (see Fig. 1.2).

B2 1

B2 3/

2

B2 2

B2 4

2 B∞

Fig. 1.2 Several balls in R2

The difference between n1 and n∞ becomes apparent once we increase the dimension. It is already apparent in three dimensions: The unit ball of 3∞ is a cube, whereas the unit ball of 31 is an octahedron. The unit ball of 32 is a Euclidean ball (see Fig. 1.3).

Fig. 1.3 Several balls in R3

B3 1

B3 2

3 B∞

Definition 1.14 Let p ∈ [1, ∞). The symbol  p =  p (N) the vector space denotes ∞ satisfying |xi | p < ∞, the vector of all scalar valued sequences x = {xi }i=1 addition and the scalar multiplication being defined coordinatewise, endowed with the norm

1.2

Hölder and Minkowski Inequalities, Classical Spaces

x p :=

∞ 

|xi | p

7

1

p

.

i=1 ∞ is considered an element of  , we use the When a scalar sequence x = {xi }i=1 p notation x = (xi ). This applies to other sequence spaces defined below. To see that the definition is correct, we need to show that if x = (xi ), y = (yi ) ∈  p , then x + y ∈  p and x + y p ≤ x p + y p . For every n ≤ m ∈ N we have by the Minkowski inequality (1.2): n 

|xi + yi |

p

1

p

≤

i=1

n 

|xi |

p

1

p

+

i=1

n 

|yi |

p

1

p

≤

i=1

m 

|xi |

p

1

p

+

m 

i=1

|yi | p

1

p

.

i=1

By letting m → ∞ we see that for every n: n  i=1

|xi + yi | p

1

p

≤

∞ 

|xi | p

1

p

+

∞ 

i=1

|yi | p

1

p

i=1

Letting n → ∞ we get x + y ∈  p and the triangle inequality follows. ∞ be a sequence of scalars. We define the support of x by Let x = (xi )i=1 supp(x) = {i ∈ N : x i = 0}. Definition 1.15 The symbol ∞ = ∞ (N) denotes the vector space of all bounded scalar valued sequences endowed with the norm defined for x = (x i ) ∈ ∞ by x∞ = sup{|xi | : i ∈ N}. The symbol c00 = c00 (N) denotes the subspace of ∞ consisting of all x = (xi ) such that supp(x) is finite. The symbol c = c(N) denotes the subspace of ∞ consisting of all x = (xi ) such that lim xi exists and is finite. i→∞

The symbol c0 = c0 (N) denotes the subspace of ∞ consisting of all x = (xi ) such that lim xi = 0. i→∞

In all these cases, the vector addition and the scalar multiplication are defined coordinatewise. Note that c0 is the closure of c00 in ∞ . Note also that if x = (xi ) belongs to c00 or c0 , then x∞ = max{|xi | : i ∈ N}. That the spaces  p (N), ∞ (N), c0 (N), and c00 (N) are infinite-dimensional follows from the fact that this is so for a vector space containing a linearly independent n are linearly independent, where set of n vectors for each n ∈ N. Vectors {ei }i=1 ei = (0, . . . , 0, 1, 0, . . .), and 1 is at the ith position. These vectors are called the canonical unit vectors. Proposition 1.16 (i) For p ∈ [1, ∞], the space  p is a Banach space.

8

1 Basic Concepts in Banach Spaces

(ii) The spaces c and c0 are closed subspaces of ∞ and thus they are Banach spaces. (iii) The space c00 is not complete. In the proof we will use the following lemma. Lemma 1.17 Let X be a normed space. If a sequence {x n }∞ n=1 in X is Cauchy, then it is bounded in X . Proof: Using the Cauchy property of {x n } we find n 0 ∈ N such that xn − xn 0  ≤ 1 for every n ≥ n 0 . Then x n  ≤ xn − xn 0  + x n 0  ≤ 1 + xn 0  for every n ≥ n 0 . Therefore for every n ∈ N: x n  ≤ max{x1 , x2 , . . . , xn 0 −1 , x n 0  + 1}.

Proof of Proposition 1.16: (i) It is easily checked, using a method similar to that for C[0, 1], that ∞ is a Banach space. Now consider p ∈ [1, ∞). k k Let {x k }∞ k=1 be a Cauchy sequence in  p , where x = (xi ). Given ε > 0, find k0 such that ∞ 

|xik − xil | p

1

p

≤ε

(1.3)

i=1

for every k, l ≥ k0 . In particular, |xik − xil | ≤ ε for every k, l ≥ k0 and i ∈ N, hence the sequence {xik }∞ k=1 converges to some xi for every i ∈ N. Put x = (xi ). We will show that x ∈  p . By Lemma 1.17, there is a constant 1 1   ∞ k p p ≤ C for every k. Therefore n |x k | p p ≤ C C > 0 such that i=1 |xi | i=1 i  1 n p p ≤ C for every n ∈ N. for all n, k ∈ N. By letting k → ∞ we get i=1 |x i | 1  ∞ p p ≤ C and x ∈  . |x | Therefore p i=1 i We will now show that x k → x in  p . Given ε > 0, we let l → ∞ in (1.3) and  1 n k p p ≤ ε for every n ∈ N and every k ≥ k . We let n → ∞ to get |x − x | i 0 i=1 i obtain ∞ 

|xik − xi | p

1

p

= x k − x p ≤ ε

i=1

for every k ≥ k0 . Therefore x k → x in  p .

1.2

Hölder and Minkowski Inequalities, Classical Spaces

9

(ii) Let x k = (xik ) ∈ c and x k → x in ∞ . For k ∈ N we denote lk = lim xik . We i→∞

will prove that lim lk exists and is finite by showing that {ln }∞ k=1 is Cauchy. Given k→∞

ε > 0 let n 0 be such that x n − x m ∞ < ε for all n, m ≥ n 0 . Thus |xin − xim | < ε for every i ∈ N and every n, m ≥ n 0 . Fixing n, m ≥ n 0 and letting i → ∞ we get |lm − ln | < ε. Therefore l := lim lk exists and is finite. k→∞

Since x k → x = (xi ) in ∞ , we have lim xik = xi for all i ∈ N. We will show k→∞

that lim xi = l, that is, x ∈ c. Given ε > 0, we find n 0 so that x k − x∞ < ε i→∞

and |lk − l| < ε for k ≥ n 0 , in particular |xik − xi |∞ < ε. Then fix i 0 so that |xin 0 − ln 0 | < ε for i ≥ i 0 . We have, for i ≥ i 0 , n

n

|x i − l| ≤ |x i − xi 0 | + |xi 0 − ln 0 | + |ln 0 − l| < 3ε. Thus lim xi = l and x ∈ c. This shows that c is closed in ∞ . Similarly we show that c0 is a closed subspace of ∞ . By Fact 1.5 the spaces c and c0 are Banach spaces. (iii) By Fact 1.5, it is enough to show that c00 is not closed in c0 . To this end, consider x ∈ c0 defined by x = (xi ), where xi = 1i for every i, and x n =   1 1 →0 1, 2 , . . . , n1 , 0, . . . . Then x n ∈ c00 and x n → x in c0 since x n −x∞ = n+1 as n → ∞. However, x ∈ / c00 . Let p ∈ [1, ∞). More generally, for an abstract nonempty set Γ we introduce spaces  p (Γ ) and c0 (Γ ). The space  p (Γ ) consists of all functions f : Γ → K 1   p p , where such that γ ∈Γ | f (γ )| p < ∞, with the norm  f  p := γ ∈Γ | f (γ )| the sum is defined by  γ ∈Γ

 | f (γ )| p = sup | f (γ )| p : F a finite subset of Γ . γ ∈F

The space ∞ (Γ ) consists of all bounded functions f : Γ → K and is endowed with the supremum norm  · ∞ . Its subspace c0 (Γ ) consists of all functions f ∈ ∞ (Γ ) such that the set {γ ∈ Γ : | f (γ )| ≥ ε} is finite for every ε > 0. We also consider the space c00 (Γ ) of all functions f ∈ ∞ (Γ ) whose support supp( f ) := {γ ∈ Γ : f (γ ) = 0} is finite. Note that c0 (Γ ) = c00 (Γ ) (closure in ∞ (Γ )). Similarly as above we show that  p (Γ ) and c0 (Γ ) are Banach spaces. Note that every element f in c0 (Γ ) has countable support. Indeed, supp( f ) =

∞ n=1 {γ : | f (γ )| ≥ 1/n}. Since  p (Γ ) ⊂ c0 (Γ ) for all 1 ≤ p < ∞, the same is true for every element in such an  p (Γ ). Definition 1.18 Let p ∈ [1, ∞). The symbol L p = L p [0, 1] denotes the vector space of all classes of Lebesgue measurable scalar functions f defined almost everywhere on [0, 1] (we identify functions that are equal almost everywhere) such

1 that 0 | f (t)| p dt < ∞, the vector addition and the scalar multiplication being defined pointwise, endowed with the norm

10

1 Basic Concepts in Banach Spaces

 f  p :=



1

| f (t)| p dt

1

p

.

0

Note that L p is a vector space. Indeed,   | f (t) + g(t)| p ≤ 2 p max(| f (t)|, |g(t)|) p = 2 p max | f (t)| p , |g(t)| p   ≤ 2 p | f (t)| p + |g(t)| p ,

1 so 0 | f + g| p dt < ∞ and ( f + g) ∈ L p whenever f, g ∈ L p . Similarly, α f ∈ L p for every α ∈ K and f ∈ L p . These spaces are also infinite-dimensional; indeed, n given n ∈ N, the set {χ[i−1/n,i/n] }i=1 , where χ S denotes the characteristic function of a set S, is linearly independent. The triangle inequality for  ·  p follows from the following versions of the Hölder and Minkowski inequalities (Theorems 1.19 and 1.20). Theorem 1.19 (Hölder inequality) If p > 1, (1/ p) + (1/q) = 1, f ∈ L p , and g ∈ L q , then f g ∈ L 1 , and 

1 0



1

| f (t)g(t)|dt ≤

1/ p  | f (t)| dt p

0

1

1/q |g(t)| dt q

(=  f  p gq ).

0

(1.4)

Proof: If  f  p = 0 or gq = 0, then the left-hand side of Equation (1.4) is also zero, so the inequality holds. Otherwise, put, for t ∈ [0, 1], a=

| f (t)| ,  f p

and

b=

|g(t)| , gq

and use Lemma 1.11 to obtain | f (t)g(t)| 1 | f (t)| p 1 |g(t)|q ≤ . p +  f  p gq p  f p q gqq

(1.5)

The function f g is measurable and Equation (1.5) shows that its absolute value is dominated by an integrable function, so f g is integrable, i.e., it is an element in L 1 . Integration of both members in inequality (1.5) gives (1.4). Theorem 1.20 (Minkowski inequality) If p ≥ 1 and f, g ∈ L p , then  f + g p ≤  f  p + g p .

(1.6)

Proof: For p = 1 the assertion is trivial. For p > 1 it follows from Hölder inequality (Theorem 1.19). Indeed, f + g ∈ L p and | f + g| p−1 ∈ L q , so

1.2



1

Hölder and Minkowski Inequalities, Classical Spaces

 | f (t) + g(t)| p dt =

0

1

11



0



1

+

1

| f (t) + g(t)|.| f (t) + g(t)| p−1 dt ≤

| f (t)|.| f (t) + g(t)| p−1 dt

0



|g(t)|.| f (t) + g(t)| p−1 dt ≤ ( f  p + g p )

0

1

1/q | f (t) + g(t)| p dt

,

0

so (1.6) holds. Theorem 1.21 If p ∈ [1, ∞), then L p is a Banach space.

∞ In the proof we will use the following lemma. i in a x∞ ∞ We say that a series i=1  < ∞. Recall that normed space X is absolutely convergent if i=1 x i i=1 x i n is called convergent if the sequence {sn }, where sn = i=1 xi , n ∈ N, is convergent in X . Lemma 1.22 A normed space X is a Banach space if and only if every absolutely convergent series in X is convergent. Proof: Assume that X is a Banach space and {xn } is a sequence  in X such that  xn  < ∞. Then given ε > 0 there is n 0 ∈ N such that nk+1xi  < εfor n   every n > k ≥ n 0 ; thus for sn := i=1 xi we have sn − sk  =  nk+1 xi  ≤ n ∞ k+1 xi  < ε. Therefore the sequence {sn }n=1 is Cauchy and thus convergent in X . So every absolutely convergent series in a Banach space is convergent. Now assume that the condition on absolute convergence is true and let {x n }∞ n=1 be has a convergent subsequence. a Cauchy sequence in X . First we show that {x n }∞ n=1 To this end, choose a subsequence n 1 < n 2 < . . . such that x n k − x nl  < 2−k for l ≥ k. This is done by induction: first choose n 1 such that x n 1 −xl  < 2−1 for every l ≥ n 1 . Then choose n 2 > n 1 such that x n 2 − xl  < 2−2 for every l ≥ n 2 , etc. Put k 1−k for x n 0 = 0 and set y k = x n k − xn k−1 . We have x n k = i=1 yi , where  yk  ≤ 2 every k ≥ 2. As yk  < ∞, by our assumption we get that yk is convergent and hence its partial sums x n k form a convergent sequence. To show that the whole sequence {x n }∞ n=1 converges to the same element is standard. This concludes the proof of Lemma 1.22. For a slight improvement of Lemma 1.22, see Exercise 1.26.

 Proof of Theorem 1.21: By Lemma 1.22 we need f n is a conto show that whenever f ∈ L satisfy  f  < ∞. Denote M = vergent series in L p n p ∞ nn p  f  . For n ∈ N we define functions g (t) = | f (t)|. Then g n p np ≤ k=1 k

1

1 p n n=1 n p p k=1  | f k |  p ≤ M, so 0 |gn | dt = 0 gn dt ≤ M . We have gn+1 (t) ≥ gn (t) for every n ∈ N and t ∈ [0, 1]. Thus the limit g(t) := lim gn (t) exists (finite or +∞) for every t ∈ [0, 1]. Since gn ≥ 0, by Fatou’s n→∞

1 p

1 lemma, 0 g p dt ≤ lim inf 0 gn dt ≤ M p . n→∞

∈ L 1 [0, 1], so g(t) < ∞ on a set S of measure Therefore g p 1 in [0, 1]. For f k (t) is finite. t ∈ S we have | f k (t)| = g(t) < ∞. Thus the sum s(t) :=  Define s(t) = +∞ for t ∈ / S. We need to show that s = f k in L p . Denote

12

1 Basic Concepts in Banach Spaces

sn (t) =

n

k=1 f k (t).

Then s = lim sn almost everywhere (on S at least), so it n→∞

is a measurable function on [0, 1]. We also have |sn (t)| ≤ gn (t) ≤ g(t) for every

t ∈ S. pTherefore |s(t)| ≤ g(t) almost everywhere on [0, 1], which implies that |s(t)| dt < ∞ and hence s ∈ L p . Finally, we have   |sn (t) − s(t)| p ≤ 2 p max |sn (t)| p , |s(t)| p ≤ 2 p g p (t)

almost everywhere on [0, 1], 2 p g p dt < ∞, and |sn (t) − s(t)| → 0 almost everywhere. By the Lebesgue dominated convergence theorem (see, e.g., [Rudi2, Theorem 1.34]), 

1

|sn (t) − s(t)| p dt → 0.

0

This means that sn − s p → 0, that is,



f k = s in L p .

Let f be a measurable function on [0, 1]. We define    ess sup( f ) = inf sup f (t) : t ∈ N , where the infimum is taken over all measurable subsets N of [0, 1] of Lebesgue measure 1. One can also use other equivalent definitions, for instance   ess sup( f ) = inf α : |{t ∈ [0, 1] : f (t) > α}| = 0 , where |A| denotes the Lebesgue measure of A. Clearly f ≤ ess sup( f ) outside a set of measure zero. Definition 1.23 The space L ∞ = L ∞ [0, 1] denotes the vector space of all classes of measurable functions f that are “essentially bounded,” i.e., such that ess sup(| f |) < ∞, endowed with the norm  f ∞ := ess sup(| f |). Proposition 1.24 The space L ∞ is a Banach space. Proof: It is easy to check that  · ∞ is a norm on the space L ∞ . We will show that it is a Banach space. First observe that { f n }∞ n=1 being a Cauchy sequence in L ∞ implies that lim ( f n − f m ) = 0 uniformly on a set of measure 1. Indeed, for k, l ∈ N put

m,n→∞

Z k, = {t ∈ [0, 1] : | f k (t) − fl (t)| ≥  f k − fl ∞ }.

1.3

Operators, Quotients, Finite-Dimensional Spaces

On [0, 1]\

13

Z k,l the functions { f n } form a uniformly Cauchy and therefore uni-

k,l

formly convergent sequence. From this and from the completeness of the supremum norm, we obtain that L ∞ is a Banach space. The space L ∞ is infinite-dimensional; to see this follow the same argument as in the L p case. Given a measure space (Ω, μ) and p ∈ [1, ∞], the space L p (Ω, μ) (also denoted L p (μ)) can be introduced in a similar fashion (see, e.g., [DiUh] or [Wojt]).

1.3 Operators, Quotients, Finite-Dimensional Spaces We now begin an investigation of linear mappings between normed spaces. Recall that a mapping T from a vector space X over the field K into another vector space Y over K is called linear if T (α1 x 1 + α2 x2 ) = α1 T (x1 ) + α2 T (x 2 ) for every α1 , α2 ∈ K and x1 , x2 ∈ X . The vector space of all linear mapping from X into Y will be denoted L(X, Y ). A linear functional on X is a linear mapping from X into the field K. Observe that a real linear functional (i.e., a linear functional from a vector space over R into R) is a convex function. Recall, too, that a mapping T : (P, ρ) → (Q, ) between metric spaces  is called Lipschitz, more precisely, C-Lipschitz, if there is C > 0 such that T (x), T (y) ≤ Cρ(x, y) for all x, y ∈ P. In case of normed spaces P, Q this inequality becomes T (x) − T (y) Q ≤ Cx − y P . Note that every Lipschitz mapping is uniformly continuous. Proposition 1.25 Let (X,  ·  X ) and (Y,  · Y ) be normed spaces and let T be a linear mapping from X into Y . The following are equivalent: (i) T is continuous on X . (ii) T is continuous at the origin. (iii) There is C > 0 such that T (x)Y ≤ Cx X for every x ∈ X . (iv) T is Lipschitz. (v) T (B X ) is a bounded set in Y . Proof: (i) ⇐⇒ (ii) follows from the linearity of T . (iii) ⇐⇒ (iv) follows similarly as T (x) − T (y)Y = T (x − y)Y ≤ Cx − y X . If (ii) is true, then given ε > 0, there is δ > 0 such that T (x)Y ≤ ε whenever x x  X = δ and thus T δ x Y ≤ ε. x X ≤ δ. If x ∈ X , x = 0, then δ x X X Therefore T (x)Y ≤ ε/δx X for every x ∈ X , which shows (iii). On the other hand, (iii) clearly implies (ii) with δ := ε/C. Assuming (iii) we obtain that T (x)Y ≤ C whenever x ∈ B X , so T (B X ) is bounded. On the other hand, if T (B X ) is bounded and, say, T (x)  Y ≤ C for x Y ≤ C, so every x ∈ B X , then for every x ∈ X , x = 0, we have T x X T (x)Y ≤ Cx X . This shows the equivalence of (iii) and (v).

14

1 Basic Concepts in Banach Spaces

Definition 1.26 Let X, Y be normed spaces. A linear mapping from X into Y is called an operator. An operator T from X into Y is called bounded if T (B X ) is bounded in Y . We define the operator norm of T by T  = sup{T (x)Y : x ∈ B X }. B(X, Y ) denotes the vector space of all bounded operators from X into Y , endowed with the operator norm. In the case of X = Y we put B(X ) = B(X, X ). By Proposition 1.25, elements in B(X, Y ) are precisely the continuous operators from X into Y . It is easy to check that B(X, Y ) is indeed a normed linear space. Note that T  is the smallest number C that satisfies (iii) in Proposition 1.25 and if T : X → Y is a bounded operator, then the kernel Ker(T ) := {x ∈ X : T (x) = 0} is a closed subspace of X . That B(X, Y ) is an infinite-dimensional space in the case that X is infinite-dimensional and Y is not reduced to {0} follows from the Hahn–Banach Theorem 2.2, see Exercise 2.43. Proposition 1.27 Let X, Y be normed linear spaces. If Y is a Banach space then B(X, Y ) is also a Banach space. Proof: The proof of completeness is similar to that for the space C[0, 1]. In particular, if {Tn } is a sequence of linear mappings from X into Y and Tn (x) → T (x) for all x ∈ X , then T must be linear. Operators in B(X, K) are called continuous linear functionals. The operator norm introduced in Definition 1.26 appears now as  f  = sup{| f (x)| : x ∈ B X } for f ∈ B(X, K). We formally introduce the following definition. Definition 1.28 Let (X,  · ) be a normed space. By X ∗ we denote the vector space B(X, K) of all continuous linear functionals on X , endowed with the operator norm  f  = sup{| f (x)| : x ∈ B X }, called the canonical dual norm. X ∗ is called the dual space of X . Remark: If f ∈ X ∗ and x ∈ X , we shall use indistinctly the notation f (x) or  f, x for the action of f on x. That the dual space does not reduce to {0} if X is different from {0} follows from the Hahn–Banach Theorem 2.2. More precisely, if dim (X ) = n for some n ∈ N, then dim (X ∗ ) = n, and if X is infinite-dimensional, so it is X ∗ , see Exercise 2.23. Note that, by the definition, if f ∈ X ∗ and x ∈ X , then | f (x)| ≤  f  · x. The canonical dual norm  ·  is denoted sometimes by  · ∗ or  ·  X ∗ to emphasize that we work in the dual space. Since K (that is, R or C) is complete, using Proposition 1.27 we readily obtain Proposition 1.29 X ∗ is a Banach space for every normed space X . A mapping ϕ from a set D into a set R is called an injection (also called oneto-one) if ϕ(d1 ) = ϕ(d2 ) whenever d1 = d2 in D. It is called a surjection (also

1.3

Operators, Quotients, Finite-Dimensional Spaces

15

called onto) if for every r ∈ R there exists d ∈ D such that ϕ(d) = r . The mapping ϕ is called a bijection if it is both an injection and a surjection. We use the terms injective, surjective, and bijective mapping, respectively. An operator T ∈ B(X, Y ) is called a linear isomorphism from X onto Y , (or just an isomorphism) if T is a bijection mapping from X to Y , and T −1 ∈ B(Y, X ). If T : X → Y is an isomorphism, we also say that T is invertible. An isomorphism from X onto X is called an automorphism. Note that the inverse mapping of a linear bijection is also a linear mapping. An operator T ∈ L(X, Y ) is an isomorphism from X onto Y if and only if it is surjective and there exist constants C 1 and C2 such that, for all x ∈ X , C1 x ≤ T (x) ≤ C2 x.

(1.7)

This follows from (iii) in Proposition 1.25, and the fact that Equation (1.7) implies that T is one-to-one. Note (Exercise 1.74) that (1.7) is equivalent to simultaneously T  ≤ C2 and T −1  ≤ 1/C1 . In geometrical terms, for an operator from X onto Y , Equation (1.7) is equivalent to C1 BY ⊂ T (B X ) ⊂ C2 BY . Normed spaces X, Y are called linearly isomorphic (or just isomorphic) if there is a linear isomorphism T from X onto Y . It is easy to see that an isomorphism T carries Cauchy (convergent) sequences onto Cauchy (convergent) sequences. Therefore, if X, Y are isomorphic normed spaces and X is a Banach space, then Y is a Banach space as well. An operator T ∈ B(X, Y ) is called a linear isomorphism from X into Y (or just an isomorphism from X into Y , alternatively an isomorphism into) if T is an isomorphism from X onto a subset T (X ) of Y . Clearly T (X ) is a subspace of Y . If X is a Banach space then T (X ) is complete too, so in particular it must be closed in Y , by Fact 1.5. An operator T ∈ B(X, Y ) is called a (linear) isometry from X into Y if T (x)Y = x X for every x ∈ X . Spaces X, Y are called (linearly) isometric if there exists a linear isometry from X onto Y . Definition 1.30 Let X, Y be isomorphic normed spaces. The Banach–Mazur distance between X and Y is defined by d(X, Y ) = inf{T T −1  : T an isomorphism of X onto Y }. Note that d(X, Y ) ≥ 1 and we have d(X, Z ) ≤ d(X, Y ) d(Y, Z ). The fact that d(X, Y ) < d means that there is an isomorphism T of X onto Y such that C11 BY ⊂ T (B X ) ⊂ C2 BY for some positive constants C 1 and C 2 satisfying C1 C2 < d. Let a vector space X be endowed with two norms, denote X 1 = (X,  · 1 ) and X 2 = (X,  · 2 ). Norms  · 1 and  · 2 are called equivalent if the formal identity mapping I X : X → X defined by I X (x) = x is an isomorphism between the spaces X 1 and X 2 , i.e., if there exist constants c, C > 0 such that cx2 ≤ x1 ≤ Cx2 for every x ∈ X (see Equation (1.7)). According to Exercise 1.74, this is equivalent to cB1 ⊂ B2 ⊂ C B1 , where Bi denotes the closed unit ball of X i , i = 1, 2.

16

1 Basic Concepts in Banach Spaces

We say that the space (X, ·2 ) is a renorming of the space (X, ·1 ). To renorm a normed space means to endow the space with an equivalent norm. Definition 1.31 Let X and Y be normed spaces. An operator T ∈ B(X, Y ) is called a compact operator if T (B X ) is compact in Y . The space of all compact operators from X into Y with the norm inherited from B(X, Y ) is denoted by K(X, Y ). If X = Y then we write K(X ) instead of K(X, X ). An operator T ∈ B(X,  Y ) is called a finite rank operator or a finite-dimensional operator if dim T (X ) < ∞. By F(X, Y ) we denote the space of all finite rank operators from X into Y with the norm inherited from B(X, Y ). If X = Y then we write F(X ) instead of F(X, X ). If f ∈ S X ∗ and f does not attain its norm on B X , then f (B X ) = (−1, 1) (see Exercise 3.161). Thus we have f ∈ K(X, R), yet f (B X ) is not compact. Therefore, the closure T (B X ) in the definition of a compact operator cannot be dropped. Example: A compact operator that is not a finite rank operator, but it is the limit (in the operator norm) of a sequence of finite rank operators: Define T ∈ B(2 ) by T (x) = (2−i xi ) for x = (xi ). Then T ∈ K(2 )\F(2 ). Indeed, since T (B2 ) is a closed subset of the Hilbert cube (Exercise 1.51), it is compact. The sequence of finite rank operators we are looking for is {Tn }, where Tn (x) := (2−1 x1 , . . . , 2−n xn , 0, 0, . . .), for all n ∈ N. The identity operator on 2 is an example of a bounded non-compact operator, √ since ei − e j  = 2, if i = j. Let T : X → Y be a finite rank operator. Let n = dim T (X ). A simple result in n in T (X ) × Linear Algebra shows that we can find a biorthogonal system {ei ; gi }i=1  ∗ T (X ) (i.e., ei , g j  = δi, j for i, j ∈ {1, 2, . . . , n}). In particular, {ei : i = 1, 2, . . . , n} is a Hamel basis (i.e., an algebraic basis) of T (X ). We define n f i = ∗ , i = 1, 2, . . . , n. Then for every x ∈ X we have T (x) = gi ◦ T ∈ X i=1 (gi ◦ n ∗ and e , . . . , e ∈ Y , T )(x)ei =  f (x)e . Conversely, given f , . . . , f ∈ X i 1 n 1 n i=1 i n then T : x → i=1 f in(x)ei is a finite rank operator from X into Y . Such an operator is denoted by T = i=1 f i ⊗ ei . The subject of compact operators will be treated extensively in Chapter 15. Definition 1.32 Let Z be a vector space. A (linear) projection P on Z is a linear mapping from Z into Z such that P ◦ P = P. In this situation we say, too, that P is a projection from Z onto P(Z ) parallel to Ker P. We shall often consider P as a mapping from the vector space Z onto the vector space P(Z ). Observe that associated to every linear projection on Z there exists an algebraic direct sum decomposition of Z into two subspaces X and Y , namely X = P Z and Y = Ker P, in the sense that each z ∈ Z can be written in a unique way as z = x + y, with x ∈ P Z and y ∈ Ker P. Note that, in this case, x = Pz and y = z − Pz. We shall write in this case Z = X ⊕ Y , making sure that this is understood only in the algebraic sense, with no topologies involved. Conversely, given two subspaces X and Y of Z such that Z = X ⊕ Y (i.e., each z ∈ Z can be written in a unique way as x + y, with x ∈ X and y ∈ Y ), then the mapping P : Z → Z defined as Pz = x is a linear projection with range X and kernel Y .

1.3

Operators, Quotients, Finite-Dimensional Spaces

17

More generally, given two vector spaces X and Y , the algebraic direct sum X ⊕Y is the vector space of all ordered pairs (x, y), x ∈ X , y ∈ Y , with the vector operations defined coordinatewise. The spaces X and Y are algebraically isomorphic to the subspaces {(x, 0) : x ∈ X } and {(0, y) : y ∈ Y } of X ⊕ Y , respectively. Definition 1.33 Let (X,  ·  X ) and (Y,  · Y ) be normed spaces. The algebraic direct sum X ⊕ Y of X and Y becomes a normed space, called the topological direct sum of X and Y and still denoted X ⊕ Y , when it is endowed with the norm (x, y) := x X + yY . The spaces X and Y are isometric to the subspaces {(x, 0) : x ∈ X } and {(0, y) : y ∈ Y } of X ⊕ Y , respectively. If X and Y are Banach spaces, then so is X ⊕ Y . Let Y be a closed subspace of a normed space X . For x ∈ X we consider the coset xˆ relative to Y , xˆ := {z ∈ X : (x − z) ∈ Y } = {x + y : y ∈ Y }. The space X/Y := {xˆ : x ∈ X } of all cosets, together with the addition and scalar   and λxˆ = λ x, is clearly a vector space. It is multiplication defined by xˆ + yˆ = x+y easy to check that x ˆ := inf{y : y ∈ x} ˆ turns X/Y into a normed space. Indeed, for any z ∈ xˆ we have x ˆ = inf{z − y : y ∈ Y } = dist(z, Y ). Therefore xˆ = 0ˆ if and only if x ∈ Y , as Y is closed. If Y is a subspace of X , then dist(αx, Y ) = |α| dist(x, Y ). Therefore λx ˆ = |λ| x. ˆ The triangle inequality follows since if x1 , x2 are in X and y1 , y2 are in Y , then x 1 + x 2 − (y1 + y2 ) ≤ x 1 − y1  + x2 − y2 . Therefore dist(x1 + x 2 , Y ) ≤ dist(x1 , Y ) + dist(x2 , Y ). Definition 1.34 Let Y be a closed subspace of a normed space X . The space X/Y endowed with the canonical norm x ˆ := inf{x : x ∈ x}, ˆ where xˆ ∈ X/Y , is called the quotient space of X with respect to Y . The canonical quotient mapping q : X → X/Y associates to every x ∈ X the coset xˆ to which it belongs. Obviously, the mapping q : X → X/Y is linear and continuous. In fact, q ≤ 1, as it follows from the definition of the norm in X/Y . We will see later, as a consequence of Lemma 1.37, that q = 1 if Y is a closed subspace of X such that Y = X . Proposition 1.35 Let Y be a closed subspace of a Banach space X . Then X/Y is a Banach space.  Proof: Assume that n xˆn is an absolutely convergent  series in X/Y . For n ∈ N, xn is an absolutely converchoose x n ∈ xˆn such that xn  ≤ xˆn  + 2−n . Then gent series in X that, according to Lemma 1.22, converges. Clearly, the canonical

18

1 Basic Concepts in Banach Spaces

 quotient mapping q : X → X/Y is continuous, so the series xˆn converges, too. It is enough to apply again Lemma 1.22 to conclude that X/Y is a Banach space. It is easy to check that (X ⊕ Y )/ X is isomorphic to Y and (X ⊕ Y )/Y is isomorphic to X . However, if Y is a closed subspace of X , then X may not be isomorphic to Y ⊕ (X/Y ), see Exercise 12.50. Proposition 1.36 Let X be a vector space. If X is finite-dimensional, then any two norms on X are equivalent. In particular, all finite-dimensional normed spaces are Banach spaces and every normed space of dimension n is isomorphic to n2 . Consequently, if X is a Banach space and Y is a finite-dimensional subspace of X , then Y is closed in X by Fact 1.5. Proof: Let {e1 , . . . ,  en } be an algebraic basis of X . We introduce a new norm  · 1 on Xby x1 =  |λi | for x = λi ei . To check the triangle inequality, for x = λi ei and y = βi ei we write x + y1 =



|λi + βi | ≤



|λi | +



|βi | = x1 + y1 .

We will show that  an arbitrary norm   ·  on X is a Lipschitz function on (X,  · 1 ). Indeed, if x = λi ei and y = βi ei , then      |λi − βi | ei  x − y =  (λi − βi )ei  ≤  |λi − βi | = max{ei } · x − y1 . ≤ max{ei }   Therefore x − y ≤ x − y ≤ max{ei }x − y1 . We note that S1 := {x ∈ X : x1 = 1} iscompact in (X,  · 1 ). Indeed, let n n λik ei for k ∈ N. We have i=1 |λik | = 1 for every k and thus x k ∈ S1 , x k = i=1 k ∞ ∞ be such that λkl → λ for {λi }k=1 is bounded for every i. Let a subsequence {kl }l=1 i i n every i. Then i=1 |λikl − λi | → 0 as l → ∞, so we have x kl → x as l → ∞,  n  k where x = λi ei . Since i=1 |λi l | = 1 for every l, we have |λi | = 1 and thus x ∈ S1 . Since  ·  is continuous on the compact set S1 , there exist constants c > 0 and x  < d for every non-zero x ∈ X . From the latter d > 0 such that c ≤  x 1 inequality we have cx1 ≤ x ≤ dx1 for every x ∈ X , so  ·  is equivalent to  · 1 . Consequently, all norms are equivalent. If X is an n-dimensional vector space and T is a linear bijection from X onto n2 , we can define a norm  · 2 on X by x2 = T (x)n2 . Then T is an isometry of (X,  · 2 ) onto n2 and  · 2 is an equivalent norm. We shall show later (see the paragraph after the proof of Theorem 4.49) that no space  p is isomorphic to q for p = q. Since (x, y) = (x X , yY )1 for every (x, y) ∈ X ⊕ Y (see Definition p

p

1

1.33), it follows from Proposition 1.36 that |||(x, y)||| := (x X + yY ) p is an

1.3

Operators, Quotients, Finite-Dimensional Spaces

19

equivalent   renorming of X ⊕ Y for every p ≥ 1. Such a renormed space is denoted X ⊕ Y p. To characterize finite-dimensional normed spaces, we need the following statement. Recall that a subspace Y of a vector space X is called proper if Y = X . Lemma 1.37 (Riesz) Let X be a normed space. If Y is a proper closed subspace of X then for every ε > 0 there is x ∈ S X such that dist(x, Y ) ≥ 1 − ε. Proof: Choose an arbitrary element zˆ ∈ X/Y satisfying 1 > ˆz  > 1 − ε. Now pick any z ∈ zˆ , z ≤ 1, and set x = z/z (see Fig. 1.4). We get dist(x, Y ) = dist(z, Y )/z = ˆz /z ≥ ˆz  > 1 − ε.

x

Fig. 1.4 Riesz’s lemma

1−ε Y

BX

It follows from Lemma 1.37 that q = 1, where q : X → X/Y is the canonical quotient mapping. Theorem 1.38 Let X be a normed space. The space X is finite-dimensional if and only if the unit ball B X of X is compact. Proof: If X is finite-dimensional, then B X is compact by the proof of Proposition 1.36. If X is infinite-dimensional, by Lemma 1.37 we find by induction an infi nite sequence {x n } in S X and such that dist xn , span{x1 , . . . , xn−1 }) > 12 for n = 2, 3, . . .. Thus dist(xn , xm ) > 12 for all n = m and therefore {xn } does not have any convergent subsequence. Hence B X is not compact. Proposition 1.39 Every operator T from a finite-dimensional normed space X into a normed space Y is continuous. Proof: If dim X = n, then X is isomorphic to n2 (Proposition 1.36). It is enough, then, to prove that every operator T from n2 into a normed space is continuous. n n is the canonical basis of n2 and x = (xi ) ∈ n2 , then x = If {ei }i=1 i=1 x i ei .

20

1 Basic Concepts in Banach Spaces

n By linearity, T (x) = i=1 xi T (ei ). For each i, |xi | ≤ x2 , so the ith coordinate mapping x → xi from n2 into K is continuous. It follows that T is continuous. A straightforward consequence of the previous result is that, for every n ∈ N, two n-dimensional normed spaces X and Y are linearly isomorphic. Indeed, if T : X → Y is a linear isomorphism, then both T and T −1 are continuous. This result follows from Proposition 1.36, too. The following statement is an application of the preceding result to finite rank operators. A complementary result on the class of compact operators is included. Proposition 1.40 Let X, Y be normed spaces. Then F(X, Y ) is a linear subspace of K(X, Y ). The space K(X, Y ) is a closed subspace of B(X, Y ); hence if Y is a Banach space, then K(X, Y ) is also a Banach space. Proof: Since (T1 + T2 )(X ) ⊂ T1 (X ) + T2 (X ), F(X, Y ) is a subspace of B(X, Y ). If T is a finite rank operator, then T (B X ) is a bounded set in a finite-dimensional (closed) space T (X ), hence T (B X ) is compact (see the proof of Proposition 1.36). For operators T1 , T2 we also have (αT1 + βT2 )(B X ) ⊂ αT1 (B X ) + βT2 (B X ) ⊂ αT1 (B X ) + βT2 (B X ) and if Ti are compact, i = 1, 2, the right hand side is a compact set (Exercise 1.61). Thus K(X, Y ) is a subspace of B(X, Y ). We will show that it is closed there. Consider Tn ∈ K(X, Y ) such that lim Tn = T in B(X, Y ). To show that T is a compact operator, given ε > 0 we shall find a finite ε-net for T (B X ) (see Exercises 1.47 and 1.48). First note that Tn → T in B(X, Y ) means that lim Tn (x) = T (x) uniformly for x ∈ B X . Thus there exists n 0 such that Tn (x) − T (x) < ε/2 for x ∈ B X and n ≥ n 0 . Since Tn 0 (B X ) is totally bounded in Y , there is a finite ε/2-net F for Tn 0 (B X ). We claim that F is a finite ε-net for T (B X ). Indeed, given x ∈ B X , we find y ∈ F such that Tn 0 (x) − y < ε/2. Then T (x) − y ≤ T (x) − Tn 0 (x) + Tn 0 (x) − y < ε. Therefore T is a compact operator. The last statement follows from this and Fact 1.5. Remarks: 1. In general, F(X, Y ) is not a closed subspace of K(X, Y ). See the example after Definition 1.31. 2. Note that if X is infinite-dimensional, then no isomorphism from X into Y can be a compact operator by Theorem 1.38. In particular, the identity operator I X on an infinite-dimensional normed space X is never compact. The spaces K(X, Y )∗ and K(X, Y )∗∗ are discussed in Chapter 16. Example: Let X = L 2 [0, 1] and K ∈ L 2 ([0, 1]×[0, 1]). Define an operator T from L 2 [0, 1] into L 2 [0, 1] by

1.3

Operators, Quotients, Finite-Dimensional Spaces

 T (x) : t →

1

21

K (t, s)x(s) ds.

0

Note that, indeed, T (x) ∈ L 2 [0, 1] whenever x ∈ L 2 [0, 1]: 2 1  1  1 2   K (s, t)x(s) ds  dt  0 0  1  1 2 1 2 ≤ |K (s, t)||x(s)| ds dt

T (x) L 2 =

0

≤

0

 1  0

=



1



0 1

1

|K (s, t)| ds 2

x 2 (s) ds

0

0

1  1

1

2

0

1 2 |x(s)|2 ds dt

|K (s, t)|2 ds dt

1 2

< ∞.

0



1 2 1 1 Thus also T ∈ B(L 2 [0, 1]) and T  ≤ 0 0 |K (s, t)|2 ds dt . We will show that T is a compact operator. First we will show that if K is continuous on [0, 1] × [0, 1], then T maps L 2 [0, 1] into C[0, 1]: By the continuity of K we have M = sup{|K (s, t)| : (s, t) ∈ [0, 1] × [0, 1]} < ∞ and hence for x ∈ B L 2 we get  1   1   |T (x)(t)| =  K (t, s)x(s) ds  ≤ |K (t, s)| |x(s)| ds 0 0   1 1  1 1 2 2 2 ≤ |K (t, s)| ds x L 2 ≤ M 2 ds x L 2 ≤ M. 0

0

Given ε > 0, find δ > 0 (from the uniform continuity of K on [0, 1] × [0, 1]) such that if t1 , t2 ∈ [0, 1], |t2 − t2 | < δ, then for every s ∈ [0, 1] we have |K (t1 , s) − K (t2 , s)| < ε. Consequently, for every x ∈ B L 2 and |t1 − t2 | < δ we have |T (x)(t1 ) − T (x)(t2 )|   1  K (t1 , s)x(s) ds − = 0

≤



0

1

1 0

   K (t2 , s)x(s) ds  ≤

|K (t1 , s) − K (t2 , s)|2 ds

1 2

x L 2 ≤

 0

1

0

1

ε2

|K (t1 , s) − K (t2 , s)| |x(s)| ds

1 2

x L 2 ≤ ε.

Therefore if K is continuous on [0, 1] × [0, 1], then T (x) ∈ C[0, 1]. In fact we also showed that T (B L 2 ) is a uniformly bounded (by M) and equicontinuous subset of C[0, 1], hence by the Arzelà–Ascoli theorem, T (B L 2 ) is relatively compact in C[0, 1]. Since the norm topology of L 2 [0, 1] is weaker than the topology of C[0, 1], we have that T (B L 2 ) is compact in L 2 [0, 1] and T is a compact operator. If K ∈ L 2 ([0, 1] × [0, 1]) then choose a sequence K n of continuous functions on [0, 1] × [0, 1] such that

22

1 Basic Concepts in Banach Spaces

 1 1  0

K (t, s) − K n (t, s)

2

1 ds dt

2

→ 0 as n → ∞.

0

For n ∈ N define a compact operator Tn : L 2 [0, 1] → L 2 [0, 1] by 

1

Tn (x) : t →

K n (t, s)x(s) ds.

0

By the first estimate of this example, T − Tn  L 2 ≤

 1 0

1

|K (t, s) − K n (t, s)|2 ds dt

1 2

→ 0 as n → ∞.

0

Thus T ∈ K(L 2 [0, 1]) = K(L 2 [0, 1]). A subset M of a topological space X is called dense in X if M, the closure of M in X , is equal to X . A subset M of a normed space X is called linearly dense if span(M) = X . Remark: In general, F(X, Y ) is not a dense subspace of K(X, Y ). See Chapter 16, in particular Theorems 16.35 and 16.54. Definition 1.41 Let X be a topological space. We say that X is separable if there exists a countable dense subset of X . Proposition 1.42 (i) If p ∈ [1, ∞), then the space  p is separable. (ii) The spaces c and c0 are separable. (iii) The space ∞ is not separable. Proof: (i) Consider in  p the family F formed by all finitely supported vectors with rational coefficients. Then F is countable. We will show that F is dense in  p . Given  ∞ p ≤ ε p and then find x ∈  p and ε > 0, choose n 0 ∈ N such that i=n 0 |x i | 2 rational numbers r1 , r2 , . . . , rn 0 −1 such that |xi − ri | p ≤  Then s := r1 , r2 , . . . , rn 0 −1 , 0, . . . ) is in F and p

s − x p =

n 0 −1 i=1

|xi − r2 | p +

∞  i=n 0

|xi | p ≤

n 0 −1 i=1

εp 2n 0

for i = 1, . . . , n 0 − 1.

εp εp + < ε p. 2n 0 2

Therefore F is dense in  p . (ii) The proof of the separability of c0 is similar to (i). The case of the space c needs only one adjustment, namely we define F as the family of vectors with rational coefficients such that the vectors are eventually constant. (iii) Assume that ∞ is separable and let D be a dense countable set in ∞ . The cardinal number of the family F of all subsets of N is the cardinality of the continuum c, in particular it is uncountable. For every F ∈ F, let χ F denote the

1.3

Operators, Quotients, Finite-Dimensional Spaces

23

characteristic function of F in N. If F1 , F2 ∈ F and F1 = F2 , then χ F1 −χ F2 ∞ ≥ |χ F1 (n) − χ F2 (n)| = 1 for some n ∈ F1 \F2 or n ∈ F2 \F1 . For each F ∈ F, let d F ∈ D be chosen such that χ F − d F  < 14 . If F1 = F2 , then d F1 − d F2 ∞ > 14 . Indeed, if we had d F1 − d F2 ∞ ≤ 14 , then χ F1 − χ F2 ∞ ≤ χ F1 − d F1 ∞ + d F1 − d F2 ∞ + d F2 − χ F2  ≤

3 4

< 1,

a contradiction. Therefore the mapping F → d F is one-to-one and maps an uncountable set into a countable set, which is a contradiction. Thus ∞ is not separable. Similarly we prove that c0 (Γ ) and  p (Γ ), p ∈ [1, ∞], are nonseparable for Γ uncountable. Proposition 1.43 (i) The space C[0, 1] is separable. (ii) If p ∈ [1, ∞), then L p is separable. (iii) The space L ∞ is not separable. Proof: We only discuss the real case (real-valued functions). In the complex case, we consider real and imaginary parts of functions separately in order to reduce the problem to the real case. (i) The collection P of polynomials on [0, 1] forms an algebra in C[0, 1] (i.e., a vector subspace closed for multiplication) that separates the points of [0, 1] (i.e., given s = t in [0, 1], there exists p ∈ P such that p(s) = p(t)) and contains a constant function. Therefore the closure P in C[0, 1] is C[0, 1] by the Stone– Weierstrass theorem. Since the countable set of polynomials on [0, 1] with rational coefficients is dense in P, we obtain that C[0, 1] is separable. (ii) It is proven in the theory of Lebesgue integral that C[0, 1] is dense in L p (see, e.g., [Royd]). Therefore the real case of (ii) follows from (i). (iii) Consider the functions f t := χ[0,t] , t ∈ [0, 1], where χ[0,t] is the characteristic function of [0, t]. Then  f t − f t  = 1 if t = t and a similar argument as in the proof of Proposition 1.42 gives that L ∞ is not separable. Proposition 1.44 B(2 ) contains an isometric copy of ∞ and thus it is not separable.   Proof: Define a mapping ϕ from ∞ into B(2 ) as follows: If (ai ) ∈ ∞ , let ϕ (ai ) be the bounded operator from 2 into 2 defined for (xi )i ∈ 2 by   ϕ (ai ) : (xi )i → (ai xi )i . We claim that ϕ is a linear isometry from ∞ into B(2 ). It is enough to check that  if (ai )∞ = 1, then the operator ϕ (ai ) has norm 1 in B(2 ). First note that if x = (xi ) ∈ 2 , then (ai xi )2 =



|ai |2 |xi |2

1 2

≤ (ai )∞



|xi |2

1 2

= (ai )∞ x2

24

1 Basic Concepts in Banach Spaces

  and thus the operator ϕ (ai ) has norm at most  1. On the other hand, given ε > 0, choose n 0 such that |an 0 | > 1 − ε. Then ϕ (ai ) (en 0 ) = an 0 en 0 , which has norm   |an 0 |. Letting ε → 0 we obtain ϕ (ai )  = 1.

1.4 Hilbert Spaces An inner product (or a scalar product or a dot product) on a vector space X is a scalar valued function (·, ·) on X × X such that (1) for every y ∈ X , the function x → (x, y) is linear, (2) (x, y) = (y, x), where the bar denotes the complex conjugation, (3) (x, x) ≥ 0 for every x ∈ X , (4) (x, x) = 0 if and only if x = 0. Note that by (1), (0, y) = 0 for any y ∈ X , hence also (y, 0) = 0 by (2). Theorem 1.45 (Cauchy–Schwarz inequality) Let (x, y) be an inner product on a vector space X . √ √ (i) For x, y ∈ X we have√ |(x, y)| ≤ (x, x) (y, y). (ii) The function x := (x, x) is a norm on X . Proof: (i) If (y, y) = 0, we have y = 0 and the inequality is satisfied. Assume that (y, y) > 0. Then  (x, y) |(x, y)|2 (x, y) y, x − y = (x, x) − 0≤ x− (y, y) (y, y) (y, y) and the statement follows. (ii) We will check the triangle inequality. For x, y ∈ X we have x + y2 = (x + y, x + y) = (x, x) + (y, y) + (x, y) + (y, x) = (x, x) + (y, y) + 2 Re(x, y) ≤ (x, x) + (y, y) + 2|(x, y)|     2  2 ≤ (x, x) + (y, y) + 2 (x, x) (y, y) = (x, x) + (y, y) = x + y .

One immediate consequence of Theorem 1.45 is that (·, ·) is a continuous mapping from (X,  · ) × (X,  · ) to the scalar field. In particular, it implies that for a fixed vector y ∈ X , x → (x, y) is a continuous linear functional on X . Definition 1.46 A Banach space (H,  · ) √ is called a Hilbert space if there is an inner product (·, ·) on H such that x = (x, x) for every x ∈ H . It is straightforward to check that the norm  ·  of a Hilbert space H satisfies the parallelogram equality, namely, for every x, y ∈ H we have x + y2 + x − y2 = 2x2 + 2y2 .

(1.8)

1.4

Hilbert Spaces

25

We also have the polarization identity: (x, y) =

1 4

  x + y2 − x − y2

(1.9)

in the real case and (x, y) =

1 4

  x + y2 − x − y2 + ix + i y2 − ix − i y2

(1.10)

in the complex case. On the other hand, assume that a norm  ·  on a Banach space X satisfies the parallelogram equality. If we define (x, y) by the above equations, it turns out to be an inner product (Exercise 1.92) and x2 = (x, x). Thus X is a Hilbert space. Therefore a Banach space X is a Hilbert space if and only if every twodimensional subspace of X is a Hilbert space. The parallelogram equality (1.8) gives that n2 , 2 , and L 2 are Hilbert spaces. In Chapter 4 (see the paragraph after the proof of Theorem 4.53), we shall show that  p and L p are not even isomorphic to a Hilbert space for p = 2. Let H be a Hilbert space. Let x, y ∈ H . We say that x is orthogonal to y, denoted x ⊥ y, if (x, y) = 0. Let M ⊂ H . We say that x is orthogonal to M, denoted x ⊥ M, if x is orthogonal to every vector y from M. Definition 1.47 Let F be a subspace of a Hilbert space H . The set F ⊥ := {h ∈ H : h ⊥ F} is called the orthogonal complement of F in H . Fact 1.48 If F is a subspace of a Hilbert space H , then F ⊥ is a closed subspace of H . Proof: Clearly, F ⊥ is a subspace. If f, h, h n ∈ H , (h n , f ) = 0 and h n → h, then (h, f ) = 0. This follows from the continuity of the linear functional h → (h, f ) discussed immediately before Definition 1.46. Hence F ⊥ is a closed subspace. Obviously F ∩ F ⊥ = {0}. Therefore every element z ∈ F + F ⊥ has a unique expression in the form z = x + y with x ∈ F, y ∈ F ⊥ (i.e., F + F ⊥ is in fact an algebraic direct sum). We can also see that the orthogonality gives z2 = (x + y, x + y) = x2 + y2 .

(1.11)

It follows that T : F ⊕ F ⊥ → H defined by T (x, y) = x + y is an isomorphism of F ⊕ F ⊥ onto F + F ⊥ ⊂ H , so F + F ⊥ is a topological direct sum. Theorem 1.49 (Riesz) Let F be a subspace of a Hilbert space H . If F is closed, then F + F ⊥ = H . Thus T : F ⊕ F ⊥ → H defined by T (x, y) = x + y is an isomorphism of F ⊕ F ⊥ onto H , and so H is the topological direct sum of F and F ⊥. Proof: Let x ∈ H . We claim there is a closest element x1 to x in F and (x −x1 ) ⊥ F. Let yn ∈ F be such that x − yn 2 < d 2 + n1 , where d := dist(x, F) = inf{x − z :

26

1 Basic Concepts in Banach Spaces

z ∈ F}. We now prove that yn is a Cauchy sequence. Consider x − yn and x − ym in the parallelogram equality (1.8): 2x − (yn + ym )2 + ym − ym 2 = 2x − yn 2 + 2x − ym 2 . We can estimate ym − ym 2 = 2x − yn 2 + 2x − ym 2 − 2x − (yn + ym )2  yn + ym  2  = 2x − yn 2 + 2x − ym 2 − 4x −  2  2 1  2 1 ≤ 2 d + n + 2 d + m − 4d 2 = n2 + m2 , yn + ym ∈ F. Therefore {yn }∞ n=1 converges to some point x 1 ∈ F. 2 Then x − x1  = lim x − yn  = d. Put x2 = x − x 1 , note that x2 2 = d 2 .

where we used

n→∞

Assume that x2 is not orthogonal to F, so there is z ∈ F such that (x 2 , z) > 0. Then for ε > 0 we have x − (x1 + εz)2 = x 2 − εz2 = (x 2 − εz, x2 − εz)   = (x2 , x 2 ) − 2ε(x2 , z) + ε2 (z, z) = d 2 − ε 2(x 2 , z) − εz2 . Since that (x 2 , z) > 0, for ε small enough we have 2(x2 , z) − εz2 > 0 and therefore x − (x1 + εz) < d, a contradiction. Thus x2 ⊥ F. For any x ∈ H we found x 1 ∈ F, x2 ∈ F ⊥ such that x = x1 + x 2 , so F + F ⊥ = H. Note that for a closed subspace F of a Hilbert space H we have (F ⊥ )⊥ = F. Indeed, write H = F ⊕ Z , where Z := F ⊥ . If y ∈ F and z ∈ Z , then (y, z) = 0 and thus F ⊂ Z ⊥ . If x ∈ H , x ∈ / F, write x = y + z with y ∈ F, z ∈ Z . Then z = 0 / Z ⊥ . Therefore F = Z ⊥ = (F ⊥ )⊥ . and (z, x) = (z, y + z) = z2 = 0. Thus x ∈ Corollary 1.50 If F is a closed subspace of a Hilbert space H , then F is onecomplemented in H , i.e., there is a linear projection of norm 1 from H onto F. Proof: Any linear and continuous projection has, clearly, norm greater than or equal to 1. The other inequality follows from Equation (1.11). Proposition 1.51 Let H be a Hilbert space and F be a subspace of H . Then F is linearly isometric to H/F ⊥ . Proof: If x ∈ F, then inf{x − y2 : y ∈ F ⊥ } = inf{x2 + y2 : y ∈ F ⊥ } = x2 , so the mapping x → xˆ ∈ H/F ⊥ from F into H/F ⊥ is a linear isometry onto.

1.4

Hilbert Spaces

27

Definition 1.52 Let H be a Hilbert space and S ⊂ H . S is called an orthonormal set if (s1 , s2 ) = 0 whenever s1 = s2 ∈ S and (s, s) = 1 for every s ∈ S. A maximal orthonormal set (in the sense of inclusion) in H is called an orthonormal basis of H . Theorem 1.53 Every Hilbert space has an orthonormal basis. Proof: If {Mα } is a chain

of orthonormal sets, i.e., Mα ⊂ Mβ or Mβ ⊂ M

α for all indices α, β, then Mα is an orthonormal set. Indeed, whenever x, y ∈ Mα , x = y, there is Mβ such that x, y ∈ Mβ and thus (x, y) = 0. By Zorn’s lemma, given an orthonormal set S0 , there is a maximal orthonormal set S containing S0 . Theorem 1.54 Let H be a Hilbert space, and let H0 be a closed subspace of H . Then every orthonormal basis of H0 can be extended to an orthonormal basis of H , i.e., if {eγ }γ ∈Γ0 is an orthonormal basis of H0 , then there exists an orthonormal set {eγ }γ ∈Γ1 in H such that {eγ }γ ∈Γ0 ∪Γ1 is an orthonormal basis of H . Proof: By Theorem 1.53 the space H0 (a Hilbert space when endowed with the restriction of the inner product on H ) has an orthonormal basis, say {eγ }γ ∈Γ0 . Put M0 = {eγ }γ ∈Γ0 and consider the family M of all orthonormal sets in H containing M0 . As in the proof of Theorem 1.53, every chain in M has an upper bound (the union of all elements in the chain), so by Zorn’s lemma there exists a maximal orthonormal set (i.e., an orthonormal basis of H ) in M. This orthonormal basis extends {eγ }γ ∈Γ0 . As an immediate consequence of Theorem 1.49, we obtain that if {eμ } is an  orthonormal basis of H , then span {eμ } = H . Indeed, otherwise we find a vector  ⊥ x from S H ∩ span {eμ } , a contradiction with the maximality of {eμ }. Theorem 1.55 Every separable infinite-dimensional Hilbert space H has an ∞ . orthonormal basis {ei }i=1 ∞ Moreover, if {ei }i=1 is an orthonormal basis of H , then for every x ∈ H , x=

∞ 

(x, ei )ei .

i=1

The numbers (x, ei ) are called Fourier coefficients and expansion of x or the Fourier series for x.



(x, ei )ei is the Fourier

Proof: Let {eμ } be an orthonormal basis of H by Theorem 1.53. Since eμ − eμ  = √ 2 for μ = μ , by the proof of Proposition 1.43 (iii), {eμ } is countable. Note that {eμ } cannot be finite as H is infinite-dimensional. Choose an ordering of {eμ } into a sequence {e1 , e2 , . . . }. We claim that for x ∈ H and n ∈ N we have n        dist x, span{e1 , . . . , en } = x − (x, ei )ei . i=1

28

1 Basic Concepts in Banach Spaces

To this end, note that for all scalars c1 , c2 , . . . , cn we have n n n  2      ci ei  = (x, x) + |ci − (x, ei )|2 − |(x, ei )|2 . x − i=1

i=1

i=1

 2 n n   x − i=1 ci ei  = (x, x) − i=1 |(x, ei )|2 and the infimum is c1 ,...,cn ∈K ∞ show that x = i=1 (x, ei )ei . Given ε > 0, attained when ci = (x,  ei ). We now n0  ∞ ) = H . For n ≥ n find n 0 such that dist x, span({ei }i=1 ) < ε using span({ei }i=1 0 we have Hence

inf

n          (x, ei )ei  = dist x, span{e1 , . . . , en } ≤ dist x, span{e1 , . . . , en 0 } < ε. x − i=1

∞ be an orthonormal set in a Hilbert space H and x ∈ Proposition 1.56 Let {ei }i=1 H . Then (i) ∞ 

|(x, ei )|2 ≤ x2

(The Bessel inequality).

(1.12)

i=1 ∞ is an orthonormal basis of H , then (ii) If {ei }i=1

x2 =

∞ 

|(x, ei )|2

(The Parseval equality).

(1.13)

i=1 ∞ is an orthonormal (iii) If the Parseval equality holds for every x ∈ H , then {ei }i=1 basis of H .   ∞ ∞ is an orthonormal basis of H . = H , then {ei }i=1 (iv) If span {ei }i=1

Proof: (i) For every n ∈ N we have n n  2     (x, ei )ei  = (x, x) − |(x, ei )|2 . 0 ≤ x − i=1

i=1

From this the Bessel inequality follows. (ii) From the proof of Theorem 1.55 we have n  2  |(x, ei )|2 dist x, span{e1 , . . . , en } = (x, x) − i=1

1.5

Remarks and Open Problems

29

for n every n ∈2 N and dist(x, span{e1 , . . . , en }) → 0 as n → ∞. Thus (x, x) − i=1 |(x i ei )| → 0 as n → ∞. (iii) If {ei } is not an orthonormal basis of H , then span{ei } = H . By Theorem1.49 there is e ∈ H \{0} such that (e, ei ) = 0 for every i ∈ N. Then 0 = |(e, ei )|2 = e2 , a contradiction. (iv) Assume that span{ei } = H and {ei } is not an orthonormal basis. Choose e orthogonal to all ei and e = 0. Let yn ∈ span{ei }, yn → e. Then (e, yn ) = 0 for every n and therefore (e, e) = lim(e, yn ) = 0, a contradiction. Theorem 1.57 (Riesz, Fischer) Every separable infinite-dimensional Hilbert space H is linearly isometric to 2 . ∞ be an orthonormal basis of H . For x ∈ H put T (x) = Proof: Let {ei }i=1 ∞ ∞ ∈  and that T ((x, ei ))i=1 . Parseval’s equality (1.13) implies that ((x, ei ))i=1 2 is an isometry. It remains to prove that T maps H onto 2 . Given (ci ) ∈ 2 , define 2       n n ci ei  = i=1 |ci |2 , therefore ci ei x ∈ H by x = ci ei . Then clearly  i=1 ∞ is Cauchy and thus convergent in H . Next we will show that if x = i=1 ci ei , then ci = (x, ei ) for every i, that is, T (x) = (ci ). This follows from the continuity of the inner product: ∞ 

n  ci ei , e j = lim ci ei , e j = lim c j = c j for every j.

i=1

n

i=1

n

Thus T is an onto mapping.

1.5 Remarks and Open Problems Remarks

1 1. The space L 2 [0, 1] endowed with the inner product ( f, g) := 0 f g¯ dt is a separable Hilbert  space. Similarly 2 is a Hilbert space, the inner product being ¯ (an ), (bn ) := ∞ n=1 an bn . Therefore L 2 is linearly isometric to 2 . Note that in Chapter 4 (see the paragraph after the proof of Theorem 4.53) we will show that  p is not isomorphic to L p for p ∈ [1, ∞), p = 2. The space ∞ is isomorphic (Exercise 4.42) but not isometric (Exercise 7.41) to L ∞ . ∞ 2. We have ∞shown that if {ei }i=1 is an orthonormal basis of a Hilbert space H , then x = i=1 (x, ei )ei for every  x ∈ H and this expression is unique, that is, if for some ci ∈ K we have x = ci ei , then ci = (x, ei ) for every i. ∞ is an orthonormal basis of H and {e 3.  If {ei }i=1 π(i) } is a permutation of {ei }, then (x, eπ(i) )eπi = x and {eπ(i) } are a basis of H . Indeed, given ε > 0, find n 0 ∈ N such that dist(x, span{e1 , . . . , en 0 }) < ε. Then find m 0 such that {1, . . . , n 0 } ⊂ {π(1), . . . , π(m 0 )}. For m ≥ m 0 we have

30

1 Basic Concepts in Banach Spaces m        (x, eπ(i) )eπ(i)  = dist x, span{eπ(i) , . . . , eπ(m 0 ) } x − i=1

  ≤ dist x, span{e1 , . . . , en 0 } < ε.

4. If H is a non-separable Hilbert space and {eγ }γ ∈Γ is an orthonormal basis (see Theorem 1.53), every x ∈ H has only a countable number of non-zero Fourier coefficients (x, eγ ). This follows from the Bessel inequality (1.12). Indeed, for every ε > 0 the set {γ ∈ Γ : |(x, eγ )| > ε} is finite. Then we still have x = γ ∈Γ (x, eγ )eγ (where the sum is in fact a countable one, and it is independent of the particular enumeration of the countable number of non-zero summands). Using this observation and proceeding similarly to how Theorem 1.57 was proven, we can show that every Hilbert space is isometric to some 2 (Γ ) (see Exercise 1.95). 5. The trigonometric system {1, e2πit , e−2πit , e4πit , e−4πit , . . . } in the indicated order is an orthonormal basis in L 2 [0, 1], a separable Hilbert space when endowed with the inner product defined in (i) above. Indeed, it is an orthonormal system by inspection. Completeness of the trigonometric system, i.e., the fact that its closed linear span is L 2 [0, 1], follows from standard results: first, from the Lebesgue dominated convergence theorem, the set of all complex, measurable, simple functions s on [0, 1] such that λ({t ∈ [0, 1] : s(t) = 0}), is dense in L 2 [0, 1]. Lusin’s theorem then shows that the space C[0, 1] is dense in L 2 [0, 1]. The arithmetic means of the partial sums of the Fourier series of a continuous function in [0, 1] converge uniformly to the function (Féjer). This finishes the proof of the completeness of the trigonometric system (see, e.g., [Katz, p. 15]). 6. We refer to [LiTz3] for more on the class of Orlicz and Lorentz Banach spaces defined in Exercises 1.100 and 1.101.

Open Problems 1. If the property (iv) in Definition 1.1 is replaced by: There is a constant C ≥ 1 so that x + y ≤ C(x + y) for every x, y ∈ X , then we call  ·  a quasi-norm on X . It is then possible to replace  ·  by an equivalent quasi-norm | · | so that there is 0 < p ≤ 1 such that |x + y| p ≤ |x| p + |y| p for all x, y ∈ X . Then (X, | · |) is called a quasi-Banach space if it is complete in the metric d(x, y) := x − y p . We refer the reader to, e.g., [Kalt1, pp. 1009–1130] and references therein for the theory of quasi-Banach spaces. We just mention that the lack of convexity is behind the fact that the structure of quasi-Banach spaces often differs from that of Banach spaces. For example, there is a quasiBanach space X so that X contains a vector x = 0 such that every closed infinitedimensional subspace of X contains x. It is an open problem if there is a quasiBanach space which has no proper closed infinite-dimensional subspaces. See also [Maur, pp. 1247–1297]. We will return to this topic in Remarks in Chapters 2 and 12.

Exercises for Chapter 1

31

Exercises for Chapter 1 1.1 Show that if A is a balanced subset of a vector space V , then [0, a] ⊂ A for all a ∈ A. Hint. It follows from the definition. 1.2 Prove that a convex set in a real vector space is symmetric if and only if it is balanced. Hint. Balanced always implies symmetric. On the other hand, if S is symmetric and convex, 0 ∈ S. To prove that αS ⊂ S for all |α| ≤ 1 split the argument into two parts: first for 0 ≤ α ≤ 1 and then for −1 ≤ α < 0, using that (−1)S ⊂ S. 1.3 Show that if V is a real vector space, A ⊂ V is a balanced set and f : V → R is a linear mapping, then, if x0 ∈ V satisfies (x0 + A) ∩ Ker f = ∅ then f has constant sign on x0 + A. Hint. Assume that f (x0 ) > 0 and that for  some x ∈ x0 + A we have f (x) < 0. By Exercise 1.1, [x0 , x] ⊂ x0 + A. Since f [x ,x] is continuous, the intermediate value 0 property gives some x ∈ (x0 , x) with f (x) = 0, a contradiction. 1.4 Show that if A is a balanced subset of a vector space V , then conv(A) is balanced. Hint. Use conv(A) = {λa + (1 − λ)b : a, b ∈ A, λ ∈ [0, 1]}. 1.5 Let V be a vector space. Recall that a mapping ϕ : V → V is called affine if there is a linear mapping T : V → V and a vector x 0 such that ϕ(x) = x0 + T (x) for all x ∈ V .  n λ x = Prove that ϕ is an affine mapping of V into V if and only if ϕ i i i=1 n n i=1 λi ϕ(x i ) for all x 1 , . . . , x n ∈ V and scalars λi such that i=1 λi = 1. In particular, if ϕ is an affine mapping and K ⊂ V is convex, then ϕ(K ) is convex.  Hint. If ϕ is affine, we have x0 = λi x0 and thus ϕ

n  i=1

n n n n     λi xi = λi x0 + T λi xi = λi (x0 + T (xi )) = λi ϕ(xi ). i=1

i=1

i=1

i=1

If ϕ has the stated property, it is enough to show that ψ(x) = ϕ(x) − ϕ(0) is a linear mapping. For x, y ∈ V and scalars α, β we have ψ(αx + βy) = ψ(αx + βy + (1 − α − β)0) = ϕ(αx + βy + (1 − α − β)0) − ϕ(0) = αϕ(x) + βϕ(y) + (1 − α − β)ϕ(0) − ϕ(0) = αϕ(x) + βϕ(y) − αϕ(0) − βϕ(0) = αψ(x) + βψ(y).

 1.6 Let X be a normed linear space. Prove that for any x, y ∈ X we have x − y ≤ x − y. Hint. Triangle inequality, x = (x − y) + y.

32

1 Basic Concepts in Banach Spaces

1.7 Let X be a normed linear space. Assume that for x, y ∈ X we have x + y = x + y. Show that then αx + βy = αx + βy for every α, β ≥ 0. Hint. Assume α ≥ β. Write αx + βy = α(x + y) − (α − β)y ≥ αx + y − (α − β)y = α(x + y) − (α − β)y = αx + βy. 1.8 Calculate the distance in L 2 [0, 1], L 4 [0, 1], and in C[0, 1], from the function x(t) := t 2 to the linear hull of the functions y(t) := t and z(t) = sin t. Hint. Direct computation. 1.9 Prove that the closure of a subspace of a normed space is again a subspace. Hint. A simple continuity argument. 1.10 Show that span(L) = span(L) and conv(M) = conv(M) (look at the given definitions of the closed linear hull and the closed convex hull, respectively). Hint. By the definition, span(L) is the intersection of all closed subspaces containing L, and span(L) is one of such subspaces. From this one inclusion follows. The other inclusion follows similarly as span(L) is the intersection of all subspaces containing L.  1.11 Show that C is a convex set in a vector space if and only if λi x i ∈ C whenever x1 , . . . , xn ∈ C and λ1 , . . . , λn ≥ 0 satisfy λi = 1. 1 2 x1 + λ1λ+λ x2 ) + λ3 x3 and Hint. (a) λ1 x1 + λ2 x2 + λ3 x3 = (λ1 + λ2 )( λ1λ+λ 2 2 induction. 1.12 Let A and B be two convex sets in a normed space X . Show that conv(A∪B) = {λx + (1 − λ)y : x ∈ A, y ∈ B, λ ∈ [0, 1]}. Hint. Show first that the set on the right hand side is convex. 1.13 Let A be a set in a real normed space X . Show that the symmetric convex hull of of all symmetric convex sets containing A, is equal to A, i.e., the intersection n n i=1 λi xi : x i ∈ A, i=1 |λi | ≤ 1, n ∈ N . Hint. Show first that the above set is convex and symmetric. 1.14 Show that the linear subspace R of Riemann integrable functions in L 1 [0, 1] is not closed in L 1 [0, 1]. Hint. If C is the Cantor discontinuum of positive Lebesgue measure and χ (C) is its characteristic function, then there are Riemann integrable functions ψn that ψn → χ (C) pointwise (use the step functions given by the definition of χ (C)). Since the measure of C is positive, it can be shown that there is no Riemann integrable function that is equal to χ (C) almost everywhere (see, e.g., [Stro, p. 273]). 1.15 Let 1 ≤ p ≤ q ≤ ∞. Then xq ≤ x p for x ∈  p and  f  L p ≤  f  L q for f ∈ L q [0, 1].

Exercises for Chapter 1

33

In particular,  p ⊂ q and, if 1 ≤ p < ∞, then  p ⊂ c0 . Moreover, L q [0, 1] ⊂ L p [0, 1]. All the corresponding identity operators have norm one. Hint. Take x = (xi ) ∈  p such that x p = 1. Then |xi | ≤ 1, so |xi |q ≤ |xi | p . q p Consequently xq ≤ x p = 1. If 1 ≤ p < ∞ and x ∈  p then obviously xi → 0. The inequality for function spaces follows from the Hölder inequality (1.1) used with r = q/ p. 1.16 Let f ∈ L p0 [0, 1] for some p0 > 1. Show that lim  f  L p =  f  L 1 . If p→1+

f ∈ L ∞ [0, 1], then lim  f  L p =  f  L ∞ . p→∞

Let x ∈ q for some q ≥ 1. Show that lim x p = x∞ . p→∞

Hint. If f is bounded, we have that f p → f pointwise as p → 1. By the Lebesgue

1

1 dominated convergence theorem, 0 | f | p dt → 0 | f | dt. We may assume that

1

1

1 | f | ≤ 1, then 0 | f | p dt ≤ 1 and hence 0 | f | p dt ≤  f  L p ≤ 0 | f | dt, consequently  f  L p →  f  L 1 as p → 1. If f is not bounded, given ε, we first find a bounded function F ∈ L p0 such that  f − F L p0 < ε. Then also  f − F L p < ε for every p ∈ [1, p0 ] and we   can use the result above to find δ > 0 such that  f  L 1 −  f  L p  < 3ε whenever p ∈ (1, 1 + δ). Let f ∈ L ∞ [0, 1]. For ε > 0, set M = {t ∈ [0, 1] : | f (t)| ≥  f  L ∞ −ε}. By the 

1/ p definition, λ(M) > 0. We have  f  L p ≥ M | f | p dλ ≥ λ(M)1/ p ( f  L ∞ −ε).   1/ p  → 1 as p → ∞, there is p0 such that  f  L p −  f  L ∞  < 2ε for Since λ(M) p ≥ p0 . a finite set M of Now assume that x ∈ q and x∞ = 1. Then there is  coordinates i such that |xi | =  1, denote K = |M|. Note that i∈M |xi | p = K . q Fix n / M, we 0 > max(M) such that i>n 0 |x i | < K /2. As |xi | < 1 for i ∈ p have i>n 0 |xi | < K /2 for every p ≥ q, moreover, there is p0 ≥ q such that |xi | p < 2(n 0K−K ) whenever p ≥ p0 and i = n 0 , i ∈ / M. Consequently, K ≤

∞  i=1

|xi | p ≤ K +

K K + = 2K 2 2

for all p ≥ p0 , that is, K 1/ p ≤ x p ≤ (2K )1/ p . Letting p → ∞ we get lim x p = 1 = x∞ . p→∞

1.17 Let 1 ≤ p1 < p2 < +∞ and let X be a subspace of L p2 [0, 1] on which the L p1 [0, 1] and L p2 [0, 1] norms are equivalent. Show that then all L p [0, 1] norms on X , for 1 < p ≤ p2 , are equivalent. Hint. Due to Exercise 1.16 we may assume that p < p1 . Suppose that, for some C > 0, we have x p2 ≤ Cx p1 for all x ∈ X . Let 1 < p < p1 . Write p1 = 1 , λp + (1 − λ) p2 , 0 < λ < 1. By the Hölder inequality used for numbers λ1 and 1−λ we have

34

1 Basic Concepts in Banach Spaces 

p

x p11 = 

1

≤

|x|λp+(1−λ) p2 =

0

pλ x p



1

|x|λp · |x|(1−λ) p2

0

1

(|x|λp ) λ

0

≤

1

λ  1−λ 1 1 pλ p (1−λ) · (|x|(1−λ) p2 ) 1−λ = x p · x p22 0

· (Cx p1 )

p2 (1−λ)

p (1−λ)

= x p · C p2 (1−λ) · x p21 pλ

p −λp

= x p · C p2 (1−λ) · x p11 pλ

.

Thus λp

−λp

1 ≤ x p · C p2 (1−λ) · x p1 . Hence λp

λp

x p1 ≤ C p2 (1−λ) · x p . Thus x p1 ≤ C

p2 (1−λ) λp

· x p ≤ C

p2 (1−λ) λp

x p1 .

1.18 Show that for every p ≥ 1,  p is linearly isometric to a subspace of L p [0, 1].  1/ p Hint. Consider span{ f n }, where f n := n(n + 1) χ[ 1 , 1 ] . n+1 n

1.19 Let 1 ≤ p ≤ ∞ and let μ be a σ -finite measure. Show that L p (μ) is isometric to L p (ν), where ν is a probability measure.

Hint. Take g an a.e. positive function such that f dμ = 1, and define a measure ν by dν = gdμ. Then the mapping f → f · g −1/ p defines an isometry from L p (μ) onto L p (ν). For p = ∞ the isometry is just the formal identity. 1.20 Show that c0 (Γ ) is the closure of c00 (Γ ) in ∞ (Γ ). 1.21 Let Γ be a set and p ∈ [1, ∞]. Show that c0 (Γ ) and  p (Γ ) are Banach spaces. Hint. The proof of Proposition 1.16. 1.22 Show that  p (I ) is linearly isometric to  p (J ) whenever card(I ) = card(J ). Here card(I ) denotes the cardinality of the set I . Hint. If ϕ is a bijection from I onto J , consider the mapping f → f ◦ ϕ. 1.23 Let C n [0, 1] be the space of all real-valued functions on [0, 1] that have n continuous derivatives on [0, 1], with the norm    f  := max max{| f k (t)| : t ∈ [0, 1]} . 0≤k≤n

Show that C n [0, 1] is a Banach space. Hint. If f n → f uniformly and f n → g uniformly, then f = g.

Exercises for Chapter 1

35

1.24 Let L be the normed space of all Lipschitz functions on a Banach space X that are equal to 0 at the origin, under the norm  | f (x)− f (y)|

 f  := sup

x−y

 : x, y ∈ X .

Show that L is a Banach space. Hint. Cauchy sequences are bounded. 1.25 Let D be the normed space of all bounded Lipschitz Fréchet differentiable functions (for definitions see Chapter 8) on a Banach space X under the norm  f  := sup{| f (x)| : x ∈ X } + sup{ f (x) : x ∈ X }, where  f (x) is the norm of f (x) in X ∗ , i.e.,  f (x) := sup{ f (x)(h) : h ∈ B X }. Show that D is a Banach space. Hint. Classical rules of differentiation.  1.26 Show that a normed space Y is a Banach space if and only if yn converges in Y whenever yn  ≤ 2−n for every n. −k Hint. Use Lemma 1.22.  To prove the necessary condition, note that if yk  ≤ 2 for all k ∈ N, then yk is absolutely convergent. For sufficiency, observe that the sequence {yk } constructed in that lemma satisfies yk  ≤ 21−k for all k ≥ 2. 1.27 Let Y be a closed subspace of a normed space X . Show that if Y and X/Y are both Banach spaces, then X is a Banach space. Note: A property P is said to be a three-space property if the following holds: Let Y be a closed subspace of a space X . If Y and X/Y have P then X has P. Thus the property of being complete is a three-space property in the class of normed linear spaces. ˆ There are {yn } in Y Hint. If {xn } is Cauchy in X , there is x ∈ X such that xˆn → x. such that {xn − x − yn } → 0. Thus {yn } is Cauchy, so yn → y and x n → x + y. 1.28 Let Y, Z be subspaces of a Banach space X such that Y is isomorphic to Z . Are X/Y and X/Z isomorphic? Hint. No. Let X = 2 , Y = {(0, x2 , x 3 , . . . )}, and Z = {(0, 0, x3 , x 4 , . . . )}. 1.29 Show that the distance d(x) of a point x = (xi ) ∈ ∞ to c0 is equal to lim sup |xi |. Thus the norm in ∞ /c0 is x ˆ = lim sup |xi |. i→∞

i→∞

Hint. There is only finitely many i such that |xi | > lim sup |xi | + ε. 1.30 Let  · 1 ,  · 2 be two norms on a vector space X . Let B1 and B2 be the closed unit balls of (X,  · 1 ) and (X,  · 2 ), respectively. Prove that  · 1 ≤ C  · 2 (that is, x1 ≤ C x2 for all x ∈ X ) if and only if C1 B2 ⊂ B1 . 1.31 Let  · 1 and  · 2 be two equivalent norms on a vector space X . Let B1 and B2 be the closed unit balls of (X,  · 1 ) and (X,  · 2 ), respectively. Show that B1 and B2 are homeomorphic.

36

1 Basic Concepts in Banach Spaces

Recall that two topological spaces K and L are called homeomorphic if there exists a bijection ϕ from K onto L such that ϕ and ϕ −1 are continuous. Such ϕ is called a homeomorphism. 1 Hint. Define a mapping φ from B1 onto B2 by φ(0) = 0 and φ(x) = x x2 x for x ∈ B1 \{0}. Clearly φ(x)2 = x1 , continuity at 0 follows from the equivalence of the norms. 1.32  −iLet X be the normed space obtained by taking c0 with the norm x0 := 2 |xi |. Show that X is not a Banach space. Note that this shows that  · 0 is not an equivalent norm on c0 . n

Hint. The sequence {(1, 1, . . . , 1, 0, . . . )}∞ n=1 is Cauchy and not convergent as the only candidate for the limit would be (1, 1, . . . ) ∈ / c0 . 1.33 Let M be a dense (not necessarily countable) subset of a  Banach space X . xk and x k  ≤ Show that for every x ∈ X \{0} there are xk ∈ M such that x = 3x . 2k Hint. Find x 1 ∈ M such that x − x1  ≤

x −(x1 +· · ·+ xk−1 )− x k      x − kn=1 x n  ≤ 2x k−1 +

≤ x . Then 2k x = 3x . 2k 2k

x 2 ,

then by induction xk ∈ M such that  k−1   x= xk and xk  =  x − n=1 xn −

1.34 Show that a Banach space X is separable if and only if S X is separable. Hint. If {x n } is dense in S X , consider {rk xn }k,n for some dense sequence {rk } in K. x If (0 ∈) D is countable and dense in X , consider { x : x ∈ D}. 1.35 Let Y be a closed subspace of a Banach space X . Show that if X is separable, then Y and X/Y are separable. Show that if Y and X/Y are separable, then X is then separable. Thus separability is a three-space property. Hint. If {xˆn } is a dense set in X/Y and {xn } is dense in Y , choosing yn ∈ xˆn and considering {yn + xk : n, k ∈ N} we have a dense set in X . 1.36 Show that the space BC(0, 1) of bounded continuous functions on (0, 1) with the sup-norm is nonseparable. 1 , n1 ) and  f n ∞ = 1 for all n Hint. Fix { f n } in BC(0, 1) such that supp f n ⊂ ( n+1  and then define T : ∞ → BC(0, 1) by T ((an )) = n an f n . Thus BC(0, 1) contains an isometric copy of ∞ . Note that this cannot be done in C[0, 1] as n an f n is not necessarily continuous at 0. 1.37 If F is a finite set of positive integers and {x j : elements in a Banach space X , then

j ∈ F} is a finite set of

Exercises for Chapter 1

37

          ∗ ∗ sup  εi xi  : εi = ±1 = sup |x , xi | : x ∈ S X ∗   i∈F i∈F        ai xi  : max |ai | ≤ 1 . = sup   

(1.14)

i∈F

Hint. Assume that the first listed supremum is ≤ 1 and let x ∗ ∈ S X ∗ . For i ∈ N put si = sign x ∗ , xi . Then 

|x ∗ , xi | =

i∈F



 si x ∗ , xi  = x ∗ ,

i∈F

 i∈F

 si xi

      ≤ si xi  ≤ 1.   i∈F

Assume that the second  listed supremum  is ≤ 1. Let |ai | ≤ 1 for i ∈ F. Pick x ∗ ∈ S X ∗ so that x ∗ , i∈F ai xi  =  i∈F ai xi . Then          ai x i  = ai x ∗ , xi  ≤ |ai ||x ∗ , xi | ≤ |x ∗ , xi |.    i∈F

i∈F

i∈F

i∈F

From these considerations and from homogeneity, it follows that the second supremum is less than or equal to the first supremum, and that the third supremum is less than or equal to the second one. Obviously, the first one is less than or equal to the third one. of all finite subsets of N 1.38 Let X be a normed space. Let P f (N) be the family  (including ∅). Let {x n }∞ xn in X is called n=1 be a sequence in X . The series (i) U-Cauchy (for unconditionally Cauchy) if, given ε > 0 there exists F0 ∈  P f (N) such that  n∈F xn  < ε for every F ∈ P f (N) with F ∩ F0 = ∅. (ii) S-Cauchy (for subseries Cauchy) if, for every sequence n 1 < n 2 < . . . in N,  the series k xn k is Cauchy. (iii) BM-Cauchy (for  bounded-multiplier Cauchy) if, for every bounded sequence (an ) in R, the series an xn is Cauchy.  (iv) R-Cauchy (for reordered Cauchy) if, for every permutation π of N, the series x π(n) is Cauchy. Prove that all these concepts coincide. Hint. BM-Cauchy⇒S-Cauchy:  This is obvious. S-Cauchy⇒U-Cauchy: If xn is S-Cauchy but not U-Cauchy,  there exists ε > 0 and a sequence (Fk ) in P f (N) such that sup F < inf F and  k k+1 n∈Fk x n  ≥ ε

F in increasing order to get a non-Cauchy for all k ∈ N. List the elements in k k  subseries of xn , a contradiction. U-Cauchy⇒BM-Cauchy: Given ε > 0, we can find F0 ∈ P f (N) such that  n∈F xn  < ε for all F ∈ P f (N) with F ∩ F0 = ∅. If εn = ±1 for all n then, for F ∈ P f (N) with F ∩ F0 = ∅,

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                               εn xn  =  x − xn  ≤  xn  +  xn    < 2ε,    + n      n∈F n∈F n∈F − n∈F + n∈F − where F + := {n ∈ F : εn = 1} and F − := {n ∈ F : εn = −1}. In view of equalities (1.14) in Exercise 1.37, we get  n∈F an x n  < 2ε whenever |an | ≤ 1  for all n. This proves that x n is BM-Cauchy.  U-Cauchy⇒R-Cauchy: Observe that if xn is U-Cauchy, then every reorder ing x is obviously U-Cauchy, too. Since U-Cauchy implies Cauchy, the series π(n)  xn is R-Cauchy.  R-Cauchy⇒U-Cauchy: If x n is R-Cauchy and not U-Cauchy, we can find (Fk ) as in the second implication. Put F0 = ∅, max F0 = 0, and Dn = {max Fn−1 + 1, . . . , max Fn } \ Fn for all n ∈ N. Now define a reordering of N by listing in increasing order the elements of D1 , then F1 , then D2 , then F2 , and so on. The reordered sum is not Cauchy, a contradiction. ∞ 1.39  Let X be a normed space, {x n }n=1 a sequence in X and x ∈ X . The series xn is said to be U-convergent (for unconditionally  convergent) to x if, for every ε > 0 there exists F0 ∈ P f (N) such that x − n∈F xn  < ε for all F ∈ P f (N) with F0 ⊂ F. We can also say that the series is S-convergent (for subseries convergent), BMconvergent (for bounded multiplier convergent) R-convergent (for reordered convergent), when (ii), respectively (iii), respectively (iv) in Exercise 1.38, holds with the term “Cauchy” replaced by “convergent.” Prove that every U-Cauchy series in X that converges to some x ∈ X is Uconvergent to x. In particular, prove that in a Banach space, each of the Cauchy concepts introduced in Exercise 1.38 (and then any of them) implies the corresponding convergent concept introduced in this Exercise 1.39 (and then any of them). n 0 xn  < Hint. Given ε > 0 find n0 ∈ N such that F0 ⊂ {1, 2, , . . . , n 0 } and x− n=1 ε. Then, for F ∈ P f (N) such that F0 ⊂ F,

           n0 n0                      xn  ≤ x − xn  +  x − x x − x + x − n n n n  < 3ε.         n∈F0  n∈F n=1 n=1 n∈F0 n∈F This proves the assertion. For the second part, use Exercise 1.38 and the fact that every Cauchy series in a Banach space is convergent. 1.40 Let X be a normed space. (i) Prove that, for a given series in X , BM-convergent⇒S-convergent⇒Uconvergent. (ii) Prove that every series in X that U-converges to some x ∈ X is convergent (to x).

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(iii) Prove that U-convergent⇒R-convergent (and every reordered series has the same sum). Prove that R-convergent⇒U-convergent (hence every reordered series has the same sum). (iv) Give examples to prove that none of the reverse implications in (i) holds. Hint. (i) The first implication is obvious. If xn is S-convergent, then it is SCauchy, hence U-Cauchy (see Exercise 1.38). It is also convergent (say, to x). Use  is U-convergent to x. now Exercise 1.39 to conclude that x n  (ii) If x n is U-convergent to x, given ε > 0 there is F0 ∈ P f (N) such that F0 ⊂ F. Find n 0 ∈ N such x − n∈F xn  < ε for every F ∈ P f (N)  suchthat n that F0 ⊂ {1, 2, . . . , n 0 }. Then, obviously, x − i=1 xi  ≤ ε for all n ≥ n 0 . This proves the assertion. (iii) If x n is U-convergent to x, and π is a permutation of N, then obviously  n x π(n) is U-convergent  to x, hence convergent (to x) by (ii). Assume now that xn is R-convergent. Then itis R-Cauchy. Use Exercise 1.38 to obtain that it is U-Cauchy. Since xn = x for some x ∈ X , use  in particular Exercise 1.39 to conclude that x n is U-convergent (to x). (iv) An example of a U-convergent series that is not S-convergent in the space  c00 of the eventually zero sequences in c0 endowed with the supremum norm, is x n , where x n := n −1 en − (n + 1)−1 en+1 , and en is the nth unit vector of the canonical basis of c0 . An example of a S-convergent series that is not BM-convergent in the space of real sequences that take only a finite number of values, endowed with the norm x := supn n −1 |x(n)|, is en , with en as above.  1.41 Let X be a Banach space. Prove that, for a series xn in X , the concepts U-convergent, S-convergent, BM-convergent, and R-convergent coincide. Hint. This is a consequence of Exercises 1.38, 1.39, and 1.40.  1.42 Prove that a series x n is a normed  space X is xn , F ∈ P f (N)} is totally bounded. (i) U-Cauchy if and only if the set { n∈F  (ii) S-convergent if and only if the set { n∈G xn : G ∈ P(N)} is relatively compact (in fact, S-convergent implies that this set is actually compact).  Hint. Puts(G) = n∈G xn for G ∈ P(N). (i) If xn is U-Cauchy, fix ε > 0 and find F0 ∈ P f (N) such that s(F) < ε for all F ∈ P f (N) with F ∩ F0 = ∅. Given F ∈ P f (N), notice that s(F) − s(F ∩ F0 ) = s(F \ F0  < ε. Since P(F0 ) is a finite set, it follows that {s(F) : F ∈ P f (N)} is totally bounded. bounded (in particular, bounded, Assume now that {s(F) : F ∈ P f (N)} is totally say s(F) ≤ M for all F ∈ P f (N)) and that xn is not U-Cauchy. Then we can find ε > 0 and a sequence (Fk ) in P f (N) such that sup Fk < inf Fk+1 and s(Fk ) ≥ ε for all k ∈ N. The set {s(F) : F ∈ P f (N)} can be covered by a finite number of sets with diameter not greater than ε/2, so at least one of those sets contains an infinite subsequence of {s(Fk )}, denoted by {z 0 , z 1 , . . .}. Let k be an integer greater than 2M/ε. Since z i − z 0  ≤ ε/2 for i ∈ N, we have 1  (z 1 + . . . + z k ) − z 0  ≤ ε/2. k

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However, z 1 + . . . + z k ∈ {s(F) : F ∈ P f (N)}, hence z 1 + . . . + z k  ≤ M. It follows that z 0  ≤ ε/2 + M/k < ε, a contradiction. N N If  (ii) Let 2 = {0, 1} the Cantor space, endowed Nwith the product topology.  xn is S-convergent, we can define a mapping ϕ : 2 → X by ϕ(a) = a(n)x n N N for a = (a(n))  ∈ 2 . The mapping ϕ is continuous. Indeed, fix a ∈ 2 and ε > 0. Then, since x n is U-Cauchy (see Exercise 1.38), there exists F0 ∈ P f (N) such that s(F) < ε for every F ∈ P f (N) with F ∩ F0 = ∅. Hence, if b ∈ 2N satisfies b(n) = a(n) for n ∈ F0 , then ϕ(b) − ϕ(a) < 2ε. Since 2N is compact, ϕ(2N ) (= {s(F) : F ∈ P(N)}) is also compact. Finally, assume that  {s(G) : G ∈ P(N)} is relatively compact. This set is totally moreover, bounded, so, by (i), xn is U-Cauchy and, by Exercise 1.38, S-Cauchy; all partial sums of any subseries lie in a relatively compact set, hence x n is Sconvergent. 1.43 Prove that every absolutely  convergent series in a Banach space is Uconvergent. Note that the series i 1i ei in 2 is U-convergent but not absolutely convergent.  Hint. Given ε > 0, find n 0 ∈ N such that ∞ n 0 +1 x n  < ε. Then,  if F0 := {1, 2, . . . , n 0 } and F ∈ P f (N) satisfies F0 ∩ F = ∅, we have  n∈F xn  ≤  1 n∈F x n < ε. To see that i ei converges unconditionally in 2 , note that   1   G i ei  is small if min(G) is large enough.  1.44 Let xi be an unconditionally convergent  in a Banach space X . Show  series   m that for every ε > 0 there is n 0 such that  i=n εi xi  < ε for every εi = ±1 and     ∞ ε x  < ε. m ≥ n ≥ n 0 ; in particular,  i=n 0 i i Hint. See the proof that U-Cauchy implies BM-Cauchy in Exercise 1.38.  n 1.45 Let ∞ n=1 x n be a series in a Banach space such that the set S := { i=1 εi x i : εi = ±1, n ∈ N} is bounded.  Assume that (an ) is a sequence of real numbers such that lim an = 0. Show that ∞ n=1 an x n is unconditionally convergent. Hint. Put M = sup{s : s ∈ S}. Fix n 0 ∈ N. Let F be a finite subset of [n 0 , +∞). Put m = maxi∈F i. If supi≥n 0 |ai | = 0 we get, due to Exercise 1.37,         ai     ai xi  = sup |ai |.  xi      i≥n 0 supi≥n 0 |ai |  i∈F i∈F      m   m         bi xi  = sup |ai |. sup  εi xi  ≤ sup |ai |. sup   ≤ 2M sup |ai | i≥n 0 i≥n 0 |bi |≤1 i=n εi =±1 i=n  i≥n 0  0 0 (if  supi∈F |ai | = 0 we have obviously the same inequality). It follows that the series an xn is unconditionally Cauchy (see the definition in Exercise 1.38) hence, by Exercise 1.39, unconditionally convergent.

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 1.46 Assume that ∞ n=1 is unconditionally convergent in a Banach space X and that (an ) is a bounded sequence of real numbers. Show that ∞ n=1 an x n is unconditionally convergent. Hint. Similar to the hint in Exercise 1.45. 1.47 Let X be a normed space, M ⊂ X and ε > 0. We say that A ⊂ M is an ε-net in M if for every x ∈ M there is y ∈ A such that x − y < ε. We say that A ⊂ M is ε-separated in M if x − y ≥ ε for all x = y ∈ A. Clearly, a maximal ε-separated subset of M in the sense of inclusion is an ε-net in M. Let X be an n-dimensional real normed space. Show that if {x j } Nj=1 is an ε-net in B X , then N ≥ ε−n . On the other hand, there is an ε-net {x j } Nj=1 for B X with  n N ≤ 2ε + 1 . Hint. Fix a linear isomorphism T of X onto Rn . For a compact subset C of X we define the volume of C by vol(C) = λ T (C) , where λ is the Lebesgue measure on Rn . Let B(x, r ) is the closed ball centered at x with radius r . We have B X ⊂ N

B(x j , ε), so j=1

vol(B X ) ≤ vol

N 

N    B(x j , ε) ≤ vol B(x j , ε) = N ·vol(ε B X ) = N ·ε n ·vol(B X ).

j=1

j=1

Thus N ≥ ε−n . On the other hand, if {x j } Nj=1 is a maximal ε-separated set in B X , then {x j } Nj=1 is an ε-net in B X . Since B(x j , ε/2) ∩ B(xi , ε/2) = ∅ if i = j and B(x j , ε/2) ⊂ (1 + ε/2)B X , we have N ·(ε/2)n ·vol(B X ) = vol

N 

  B(x j , ε/2) ≤ vol (1+ε/2)B X = (1+ε/2)n vol(B X ).

j=1

Thus N ≤

 1+ε/2 n ε/2

.

1.48 Prove that a subset M of a normed space X is totally bounded if and only if for every ε > 0 there exists a finite ε-net in M. Hint. It follows from the definition. 1.49 Prove that the closure of a totally bounded set in a normed space is totally bounded. Hint. It follows from Exercise 1.48. 1.50 Show that a bounded set M in c0 is totally bounded if and only if for every ε > 0 there is n 0 such that |x n | ≤ ε for every x ∈ M and n ≥ n 0 . Formulate and prove the analogous result for  p spaces. Hint. Every bounded subset of Rn 0 is totally bounded.

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1.51 The Hilbert cube Q is defined as {x = (xi ) ∈ 2 : ∀i, |xi | ≤ 2−i }. Show that the Hilbert cube is a compact set in 2 .  ∞ |xi | ≤ ε for every x ∈ Q. Then Hint. Given ε > 0 there is n 0 such that i=n 0 +1 n 0 use finite ε-nets in R . 1.52 Prove the following version of Riesz’s Lemma 1.37: If Y is a finitedimensional proper subspace of a normed space X , there is a point x ∈ S X whose distance from Y is 1. Hint. See the proof of Lemma 1.37: for zˆ ∈ X/Y , the coset zˆ := {z + Y } is the translate of a finite-dimensional subspace, so there is y ∈ Y such that z+ y = ˆz . 1.53 Let X 1  X 2  . . . be finite-dimensional subspaces of a normed space X . Then there are unit vectors x 1 , x2 , . . . such that xn ∈ X n for all n ∈ N and dist(x n , X n−1 ) = 1 for all n ≥ 2. Hint. Use Exercise 1.52. 1.54 This exercise slightly improves Exercise 1.53. Let X 1 ⊂ X 2 ⊂ . . . subspaces of a real normed space X with dim X n = n for all n ∈ N. Then there is a sequence x1 , x2 , . . . of unit vectors such that xi − x j  > 1 if i = j and span{xi , . . . , xn } = X n for all n ∈ N. Hint. Given linearly independent vectors {x 1 , . . . , xn−1 } in a n-dimensional real normed space X , let f ∈ S X ∗ such that f (xi ) = 0 for i = 1, 2, . . . , n − 1. Let g ∈ X ∗ be such that g(xi ) = 1 for i = 1, 2, . . . , n − 1. The set K := {x ∈ S X : f (x) = 1} is compact and nonempty. Choose xn ∈ K such that g(x n ) = min{g(x) : x ∈ K } (see Fig. 1.5). Use f to prove that x n − xi  ≥ 1 for all i ≤ n − 1. If xn − xi  = 1, then xn − xi ∈ K but g(xn − x i ) = g(xn ) − 1 < g(xn ), a contradiction. K {x : g(x)=1}

Fig. 1.5 The construction in Exercise 1.54

x3

{x : f(x)=1}

x2

x1

{x : f(x)=0}

BX

Let us remark that a result of Elton and Odell (see [Dies2, Chapter XIV]) says that for any infinite-dimensional Banach space X there is a constant ε > 0 and a sequence {x n } in S X so that xn − xm  ≥ 1 + ε whenever n = m. 1.55 Let C be a convex set in a normed space X and assume that Int(C) = ∅ (recall that Int(C) denotes the interior of C). Prove that for every x ∈ C and x0 ∈ Int(C), then [x0 , x) ⊂ Int(C). Hint. Use the “cone” argument: there is an open ball BδO such that x 0 + BδO ⊂ C; by the convexity of C, the cone with vertex x and base x0 + BδO is a subset of C and all points in this cone but the vertex x are in Int(C) (draw a picture).

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1.56 Let X be a normed space and C a nonempty convex subset of X . Prove that both Int(C) and C are convex sets. Hint. The first part follows from Exercise 1.55. For the second part use a simple continuity argument. 1.57 Let C be a convex set in a normed space X , assume Int(C) = ∅. Show that Int(C) = C and Int(C) = Int(C). Hint. The first part is a consequence of Exercise 1.55. To prove the second part, let x0 be an element in Int(C). Given c ∈ Int(C), there is ε > 0 such that c + ε B X ⊂ C. Let d = c + ε(c − x0 ). Then d ∈ C, so for every ν > 0 there is y ∈ C with d − y < ν. If ν is small (namely ν ≤ √ 2 εδ ), c is in the cone with base x0 + BδO and vertex y, so 2 δ +x0 −c

c ∈ Int(C). Draw a picture. 1.58 Let A be an open set in a normed space X . Show that conv(A) is open.  n n ∈ A and {λi }i=1 such that λi ≥ 0 for every i and λi = 1, Hint. Given {xi }i=1 note that if λ1 > 0 and O1 is an open set containing x1 and contained in A, then conv(A) contains the open set λ1 O1 + λ2 x2 + · · · + λn xn . 1.59 Is the convex hull of a closed set in R2 closed? Hint. No in general, check A := {(x, x1 ) : x > 0} ∪ {(0, 0)}. 1.60 Let K , C be subsets of a normed space X . (i) Show that if K , C are closed, K + C need not be closed. (ii) Show that if K is compact and C is closed, then K +C is closed. Is conv(K ∪ C) closed? (iii) Show that if K is compact and C is bounded and closed, then conv(K ∪ C) is closed. Hint. (i) Consider K := {(x, 0) : x ∈ R} and C := {(x, 1x ) : x > 0}. (ii) If x n = kn + cn → y for kn ∈ K , cn ∈ C, then by compactness assume kn → k, then also cn = x n − kn → (y − k) and use that C is closed. For negative answer to conv(K ∪ C) see the previous exercise. (iii) If xn = λn kn + (1 − λn )cn → x for kn ∈ K , cn ∈ C, λ ∈ [0, 1], find a subsequence n i such that kn i → k and λn i → λ. If λ = 1, then by boundedness, (1 − λn i )cn i → 0 and x = k ∈ K . If λ = 1, cn i → x−λk 1−λ ∈ C by closedness of C. 1.61 Let A, B be convex compact sets in a Banach space X . Show that conv(A ∪ B) and A + B are compact. Generalize this statement to a finite number of sets. Hint. Using Exercise 1.12, show that conv(A ∪ B) is a continuous image of the compact set {(α, β) : α, β ≥ 0, α + β = 1} × A × B, so it is compact. Similarly, A + B is the image of A × B under the continuous mapping (x, y) → x + y. 1.62 Let A be a totally bounded set in a normed space X . Show that conv(A) is totally bounded. Likewise, prove that the closed convex and balanced hull of A is

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totally bounded. In particular, if A is a compact subset of a Banach space, then the closed convex and balanced hull of A is compact. Hint. Let B be a finite δ-net for A. It is straightforward to check that conv(B) is a δ-net for conv(A). Since conv(B) lies in a finite-dimensional space and is bounded, it is totally bounded, and its δ-net would produce a 2δ-net for conv(A). Note also that the closure of a totally bounded set is totally bounded (see Exercise 1.49). The argument for the convex and balanced hull of A is similar. 1.63 Show that the closed convex hull of a compact subset of an incomplete normed space need not to be compact. Hint. Let S = { n1 en } ∪ {0} ⊂ c00 . This set is compact in c00 . If conv(S) in c00 were compact, it would be equal to conv(S) in c0 (compact spaces are closed in overspaces). However, the vector ( 21n n1 ) shows the contrary. 1.64 Let X be a Banach space and C be a compact set in X . Is it true that conv(C) is compact? Hint. Not in general. Consider C := { 1i ei } ∪ {0} in 2 , where ei are the standard   unit vectors. Clearly C is compact. The vector 2−i 1i is in conv(C) and it is not in conv(C), since any point in conv(C) is finitely supported. 1.65 Let C be a compact set in a finite-dimensional Banach space X . Show that conv(C) is compact. some Hint. If a point x lies in conv(E) ⊂ Rn , then x lies in the convex hull of subset of E that has at most n + 1 points. Indeed, assume that r > n and x = ti xi is a convex combination of some r + 1 vectors xi ∈ E. We will show that then x is actually a convex combination of some r of these vectors. Assume that ti > 0 for dependent since 1 ≤ i ≤ r + 1. The r vectors xi − xr +1 for 1 ≤ i ≤ r are linearly  +1 r > n. Thus there are real numbers ai not all zero such that ri=1 ai xi = 0 and r +1 a = 0. Choose m so that |a /t | ≤ |a /t | for 1 ≤ i ≤ r + 1 and define i i i m m i=1   ci = ti − aai mtm for 1 ≤ i ≤ r + 1. Then ci ≥ 0, ci = ti = 1, x = ci xi and cm = 0. Having this, we can use the fact that in Rn , conv(C) is the image of the n+1 compact set S × C n+1 in Rn+1 × C n+1 , where S is formed by  points {λi }1 such that λi ≥ 0 and λi = 1, under the mapping ({λ}, {xi }) → λi xi . 1.66 (See, e.g., [Jmsn, §22]) Let X be a normed ∞space. Given a sequence {x n } in X , a convex series is a series (convergent or not) n=1 λn x n , where λn ≥ 0 for all n ∈ N  and ∞ λ = 1. A subset A of X is called CS-closed (for convex-series-closed) n=1 n if it contains the sum of every convergent convex series of its elements. A is called CS-compact if every convex series of its elements converges to an element in A. Prove the following statements. (i) Every closed convex set A ⊂ X is CS-closed. (ii) Every open convex set O ⊂ X is CS-closed. (iii) If A ⊂ X is bounded, then every convex series of elements in A is Cauchy. (iv) Every CS-compact set is CSclosed and bounded; the converse is true in Banach spaces. (v) Suppose that A is CS-compact and B is CS-closed. Then A + B and conv (A ∪ B) are CS-closed. (vi)

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45

The continuous linear image of a CS-compact set is CS-compact. (vii) The inverse image of a CS-closed set under a continuous linear mapping is CS-closed. (viii) If A is a CS-closed, then A and A have the same interior (compare with Exercise 1.57). Hint. (i) We may assume that 0 ∈ A. If λ ≥ 0, λk = 1, and ak ∈ A for k    all k, (ii) If x = λk xk then nk=1 λk ak = nk=1 λk ak + (1 − nk=1 λk )0 ∈ A for all n.  is a convergent elements in O, putx = Λn nk=1 λk Λ−1 n xk + ∞ convex series of n −1 λ . Taking n such that (1 − Λn ) k=n+1 λk (1 − Λn ) xk , where Λn := k k=1 0 < Λn < 1 we get x ∈ O (use (i) and Exercise 1.55). (iii) should be clear. (iv) That a CS-compact set A is CS-closed is clear. If itis not bounded, choose an ∈ A with an  > 2n and consider the convex series 2−n an . The converse follows from (iii). (v) is simple; use Exercise 1.12. (vi) and (vii) are easy. (viii) is certainly the most interesting feature of CS-closed sets. It is enough to prove that Int(A) ⊂ A and, by translating, that if 0 ∈ Int(A) then 0 ∈ A. Let δ > 0 such that δ B X ⊂ A. Find a1 ∈ A, a1  ≤ δ/2, and put s1 = a1 /2. Since  − 4s1  ≤ δ, find a2 ∈ A with  − 4s1 − a2  ≤ δ/2. Put s2 = a1 /2 + a2 /4. Then   − 8s2  ≤ δ, hence a3 ∈ A exists with  − 8s2 − a3  ≤ δ/2. In this way, 0 = 2−n an (∈ A). Observe that the argument in (viii) is in the core of the proof of the Banach Open Mapping Theorem 2.25. 1.67 Let A be a subset of a Banach space X . Denote by sconv ∞(A)—the superconvex hull of A—the set of all x ∈ X that can be written as x = i=1 λi xi , where xi ∈ A,  λi ≥ 0 and λi = 1. Show that (i) A is CS-closed (see Exercise 1.66) if and only if A = sconv (A). (ii) sconv (A) ⊂ conv(A). (iii) Let A be the set of all standard / sconv (A). unit vectors ei in 2 ; then 0 ∈ conv(A) and 0 ∈ Hint. (i) is easy. (ii) Obviously, A ⊂ C, where C := conv (A). Then sconv (A) ⊂ sconv (C). Since C is closed and convex, it is CS-closed (see (i) in Exercise 1.66),  hence sconv (C) = C by (i) here. Finally we get sconv (A) ⊂ C. (iii) If λi ei = 0, then λi = 0 for every i. 1.68 Let {K i } be a finite family of convex compact sets in Rn such that every subfamily of them consisting of n + 1 members has a nonempty intersection. Then  K i = ∅. Prove this Helly’s theorem for n = 1. Show an example for n = 2 that n + 1 is necessary. Hint. Consider the interval between the maximum of the left endpoints of K i and the minimum of the right endpoints of K i . For general case see, e.g., [DGK]. Example: three lines forming a triangle. 1.69 Let X be a Banach space. Show that if A ⊂ X is totally bounded, then there is a sequence {xn } ∈ X such that xn → 0 in X and A ⊂ sconv {xn } (see Exercise 1.67). In particular, for every compact subset A of X there exists a sequence {x n } such that x n → 0 and A ⊂ conv{x n } (Grothendieck). Hint. We set A1 = A, let B1 be a finite 2−2 -net in A1 . If Ai and Bi were defined for i ≤ n, let An+1 = (An − Bn ) ∩ 2−2n B X , note that every an ∈ An is of the form an = an+1 + bn , where an+1 ∈ An+1 , bn ∈ Bn . Let Bn+1 be a finite 2−2n -net in An+1 . Therefore every element a ∈ A (= A1 ) is of the form a = b1 + a2 = b1 + b2 + a3 =

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1 Basic Concepts in Banach Spaces

   −i i · · · = n1 bn + an+1 . Since an+1 ∈ 2−2n B X , we have a = ∞ 2 (2 bi ). n=1 bi = Then it suffices to take for {xn } the sequence that contains first all vectors from 21 B1 , then from 22 B2 and so on. Since 2i bi ∈ 2i B X ⊂ 2i Ai , we have 2i bi  ≤ 22−i and x n  → 0. Note that we in fact proved that if x n → 0, then the closed convex symmetric  ∞ ∞ hull of {xn } is n=1 λn x n : n=1 |λn | ≤ 1 . 1.70 Let X, Y be non-zero normed spaces and T ∈ B(X, Y ). Show that T  = sup{T (x)Y : x X < 1} = sup{T (x)Y : x X = 1}. Hint. Clearly, both suprema are not greater √ than T . Given ε > 0, find x ∈ B X √ x such that T (x)Y ≥ 1 − ε T . Then  1 − ε x < 1 and x ∈ S X , both X vectors give T (y)Y ≥ (1 − ε)T . 1.71 Assume that T is an operator from a normed space X into a normed space Y such that {T (xn )} is bounded for every sequence {xn } ⊂ X satisfying xn  → 0. Is T necessarily continuous? √ Hint. Yes. Assuming the contrary, consider a sequence {x n / xn } for {xn } such that x n → 0 and T (xn ) → 0. 1.72 Let T be a one-to-one bounded operator from a normed space X into a normed space Y . Show that T is an isometry onto Y if and only if T (B X ) = BY if and only if T (S X ) = SY if and only if T (B XO ) = BYO , where B XO is the open unit ball in X . Hint. By homogeneity, T is an isometry onto Y if and only if T (S X ) = SY . Assume that T (B X ) = BY . If there is x ∈ S X such that T (x) = C < 1, then x/C > 1 and T (x/C) = 1. But there must be y ∈ B X such that T (y) = T (x/C), a contradiction with T being one-to-one. A slight modification of this argument gives the last equivalence. 1.73 Let X, Y be Banach spaces and T ∈ B(X, Y ). If there is δ > 0 such that T (x) ≥ δx for all x ∈ X , then T (X ) is closed in Y . Moreover, T is an isomorphism from X into Y . Hint. The inequality we assume implies that T is one-to-one, so the inverse T −1 : T (X ) → X is well defined and T −1  ≤ 1δ . The mapping T : X → T X is an isomorphism, so T X is a Banach subspace of Y , hence closed by Fact 1.5. 1.74 Let T be an operator from a normed space X onto another normed space Y . Prove that the following are equivalent: (i) There exist two positive constants C1 , C 2 such that, for every x ∈ X , C1 x ≤ T x ≤ C2 x. (ii) C1 BY ⊂ T (B X ) ⊂ C2 BY . (iii) T is an isomorphism and T  ≤ C2 , T −1  ≤ 1/C1 . Hint. Direct computation.

Exercises for Chapter 1

47

1.75 Let X be a normed space isomorphic to another normed space Y and let d(·, ·) be the Banach–Mazur distance. Let C be a positive constant. Prove that the following are equivalent: (i) d(X, Y ) < C. (ii) There exist positive constants C1 and C 2 and an linear isomorphism from X onto Y such that (1/C1 )BY ⊂ T (B X ) ⊂ C2 BY , and C 1 C2 < C. (iii) There exists a linear isomorphism T from X onto Y , and a  < C such that BY ⊂ T (B X ) ⊂ C BY . positive constant C Hint. Use Exercise 1.74 and the definition of the Banach–Mazur distance. 1.76 Let X be a finite-dimensional normed space and let Y be a normed space. Let T be a (bounded) operator from X onto Y . Assume that there is δ ∈ (0, 14 ) and a finite δ-net M := {xi } in S X (see Exercise 1.47) such that (1 + δ)−1 ≤ T (xi ) ≤ (1 + δ)  1 δ(1+δ) −1 . for every i. Then T −1 exists and T T −1  ≤ θ (δ) = ( 1+δ 1−δ ) 1+δ − 1−δ Hint. Given x ∈ S X , there is xi ∈ M with x − xi  X < δ. Then T x − T xi Y ≤ T .x − x i  X < T δ, hence T xY ≤ T xi Y + T x − T xi Y ≤ (1 + δ) + T δ. By Exercise 1.70, T  ≤ (1 + δ)(1 − δ)−1 . On the other hand, T xY = T x − T xi + T xi Y ≥ T xi Y − T x − T xi Y 1+δ 1 − δ (> 0). ≥ (1 + δ)−1 − T δ ≥ 1+δ 1−δ In particular, T is one-to-one (and onto), so an isomorphism. It follows that  1 −1 T −1  ≤ 1+δ − δ(1+δ) . Both estimates give the conclusion. 1−δ 1.77 Let X, Y be Banach spaces and T ∈ K(X, Y ). Prove that if {xn }∞ n=1 is a w

·

sequence in X , x ∈ X , and x n → x in X , then T (xn ) → T (x) in Y . An operator satisfying the conclusion of the statement is called a completely continuous operator. Thus every compact operator is completely continuous. The example of the identity operator on 1 shows that the converse implication is not valid in general. w Hint. If xn → x, then {x n } is weakly (and hence norm-) bounded, and we may w assume that x, xn ∈ B X . We have T (xn ) → T (x) by the w-w-continuity of T . Since T (B X ) is a compact space in the norm topology and the w-topology is weaker than the norm topology and Hausdorff, these two topologies coincide on T (B X ). ·

Consequently, T (x n ) → T (x). 1.78 Let X be a Banach space and G be a subspace of X that is a G δ set in X . Show that G is closed in X . Hint. Let S = G\G, where Gdenotes the closure of G. Since G is a G δ set in G n , where G n are open subsets in G. Therefore X , G is G δ in G. Hence G =

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1 Basic Concepts in Banach Spaces

S = (G\G n ) and each G n is dense in G as it contains G. Thus each G\G n is nowhere dense in G and so S is of first category in G. We will show that S is an empty set. If x 0 ∈ S, consider the set G ∗ := {x0 + x : x ∈ G}. Note that G ∗ ⊂ S. Indeed, any point x0 + x, x ∈ G, is in G since G is a linear set. If for some x ∈ G we have x0 + x ∈ G then by linearity of G we have x0 ∈ G, a contradiction. Therefore G ∗ ⊂ S and G ∗ is of first category in G. Thus the shift G of G ∗ is also of first category in G. Hence the whole G as a union of G and S, which are both of first category in G, is of first category in itself. This is a contradiction, since G is a complete metric space. 1.79 Let X be a normed linear space X . Show that if X is topologically complete in its norm topology (that is, X is homeomorphic to a complete metric space), then X is a Banach space. Hint. If X is topologically complete, then by the Alexandrov–Hausdorff– Lavrentieff–Mazurkiewicz theorem, see, e.g., [Enge, Theorems 4.3.23 and 4.3.24], X is G δ in the completion of X , which is a Banach space. Note that this means that a non-complete normed space cannot be homeomorphic to a Banach space. 1.80 Let X be an infinite-dimensional Banach space. Show that there is no translation invariant Borel measure μ on X such that μ(U ) > 0 for every open set U and such that μ(U1 ) < ∞ for some open U1 . Hint. Every open ball contains an infinite number of disjoint open balls of equal radii (see Lemma 1.37). 1.81 Let X be an infinite-dimensional Banach space. Show that X admits no countable Hamel basis. Therefore c00 cannot be normed to become a Banach space. Hint. If {ei } is a countable infinite Hamel basis of a Banach space X , put Fn = span{e1 , . . . , en }. Fn are closed and thus by the Baire category theorem, at least one Fn 0 has a nonempty interior, that is, there is x ∈ X and a ball B = δ B X such that x + B ⊂ Fn 0 . Using linearity of Fn 0 we have that −x + B ⊂ Fn 0 , so B ⊂ (x + B) + (−x + B) ⊂ Fn 0 . Thus 0 is an interior point of Fn 0 . This would mean that Fn 0 = X , a contradiction. 1.82 Let X be an infinite-dimensional separable Banach γ } be a Hamel  space and {e basis for X . Define a norm  ·  on X by |||x||| = |x γ | for x = x γ eγ . Show that ||| · ||| is indeed a norm on X . Prove that it is not equivalent to the original norm of X . Hint. Consider the question of separability of X in ||| · ||| using the previous exercise. 1.83 Prove that on every infinite-dimensional Banach space (X, ·) there is a norm | · | that is not equivalent to  · . Hint. If X is separable, use Exercise 1.82. In general, choose an infinite-dimensional separable subspace Y of X and define a non-equivalent norm | · | on Y using the

Exercises for Chapter 1

49

first part. Then, if Z is an algebraic complement of Y in X , put |x| = |y| + z, whenever x = y + z, y ∈ Y , z ∈ Z . This defines a norm on X that cannot be equivalent to  ·  (otherwise the restriction | · | to Y will be also an equivalent norm). 1.84 Show that the norm |x| := norm. ∞ . Hint. Check this norm on {ei }i=1

∞

i=1 2

−i |x

i|

in 2 is not equivalent to the  · 2 -

1.85 Show that the linear dimension (the cardinality of a Hamel basis) of the space  p , p ∈ [1, ∞), is the continuum c. Hint. First note that card( p ) = c. Therefore the linear dimension of  p is less than or equal to c. On the other hand, note that if λ < 1, then the vector (λ, λ2 , . . . ) ∈  p , and these vectors form a linearly independent set. 1.86 Find a vector space X with two norms on it such that both of them are complete norms and they are not equivalent. Hint. Take a vector space V of linear dimension c and let T1 and T2 be linear bijections of V onto 2 and 4 , respectively. Define norms on V by x1 = T1 (x)2 and x2 = T2 (x)4 . Then (V,  · 1 ) is isomorphic to 2 and (V,  · 2 ) is isomorphic to 4 . Since 2 is not isomorphic to 4 (see Exercise 1.99),  · 1 and  · 2 are not equivalent. 1.87 Let X and Y be Banach spaces and T be a bounded operator from X into Y . 1 Show that |x| := (x2 + T x2 ) 2 is an equivalent norm on X . Hint. Hölder inequality.  1.88 Show that the following function f defined on 1 by f (x) = ( |xi |)2 + 1  1 2 2 is an equivalent norm on  . 1 i 2i |x i | Hint. Hölder inequality. 1

1.89 If 0 < p < 1, show that the function f (x, y) := (|x| p + |y| p ) p is not a norm on R2 . Hint. Look at Fig. 1.6, where the “unit ball” for p = 1/2 is shown. 1.90 A family {x α }α∈Γ in a Banach space X is said to be ω0 -independent if, for )∞ every sequence (αn n=1 in Γ of distinct indices and every sequence of real numbers ∞ (λn )∞ , the series n=1 λn x αn converges to zero in X if and only if all λn are zero. n=1 Observe that, since ω0 -independence clearly implies linear independence, no finitedimensional space can contain an infinite ω0 -independent family. The following result is due to Kalton [Kalt3] (see also [HMVZ, Theorem 1.58]): Let X be a Banach space, and let G be a subset of X . Let H be the set of accumulation points of G, and suppose that X is the closed linear span of H . Then, given any x ∈ X and any

50

1 Basic Concepts in Banach Spaces y

Fig. 1.6 The ball for p = 1/2 in Exercise 1.89

0

x

 sequence of numbers (an ) with |an | = ∞ and lim an = 0, there is a sequence of signs n and distinct elements gn ∈ G so that x = ∞ n=1 n an gn . Use this theorem to prove the following result of Fremlin and Sersouri [FrSe], that answers a question of Z. Lipecki: Suppose X is a separable Banach space. Then every ω0 -independent family in X is countable. Hint. By contradiction, let G be an uncountable ω0 -independent family in X . Because G is a subset of the separable (metric) space X , it contains an uncountable set H that is dense in itself. Let Y be the closed linear span ofH . According to Kalton’s theorem quoted above, given any sequence (an ) with n|an | = ∞ and lim an = 0, we can find signs n and distinct gn ∈ G so that 0 = ∞ n=1 n an gn , a contradiction. 1.91 Let H be a Hilbert space. Prove the generalized parallelogram equality, i.e., if x 1 , . . . , xn ∈ H , then n n 2     εi xi  = 2n xi 2 .  εi =±1 i=1

i=1

Hint. Induction on n. 1.92 Let X be a Banach space whose norm  ·  satisfies the parallelogram equality (1.8). Define (x, y) by the polarization identity (1.9), (1.10) and prove that (x, y) is an inner product. Hint. Clearly (·, ·) is continuous in both coordinates, (x, y) = (y, x) and (−x, y) = −(x, y). Using the parallelogram equality show that (x + y, z) = (x, z) + (y, z). Then by induction (nx, y) = n(x, y) for all n ∈ N, hence also for all integers n. n Given mn , write ( mn x, y) = n( m1 x, y) = mn m( m1 x, y) = mn ( m m x, y) = m (x, y). By continuity we get (αx, y) = α(x, y) for all α ∈ R. 1.93 Show that n4 is not a Hilbert space. Hint. Parallelogram equality.

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1.94 Find a Hilbert space H and a linear (not closed) subspace F of it such that H = F + F ⊥ . This shows that the assumption of closedness in Theorem 1.49 is crucial. Hint. Consider the subspace F of finitely supported vectors in 2 . Then F ⊥ = {0} as for every x ∈ H \{0}, (x, ei ) = 0 for i ∈ supp(x). 1.95 Let H be a Hilbert space. Show that there exists an abstract set Γ such that H is isometric to 2 (Γ ). The density character of a normed space will be introduced in Definition 13.2. Prove that the cardinality of Γ is the same as the density character of H . Hint. Take an orthonormal basis {eγ }γ ∈Γ and follow the proof of Theorem 1.57. For √ √ √ the second part, observe that eγ − eδ  = 2, so B(eγ , 2/2) ∩ B(eδ , 2/2) = ∅ for all γ = δ. Let D ⊂ H be a dense subset of H such that card(D), the cardinality of D, is the density of√H . The mapping that to each γ ∈ Γ associates a single element in D ∩ B(eγ , 2/2) is one-to-one, so card(Γ ) ≤ card(D). On the other hand, the set of all finite rational combinations of elements in {eγ : γ ∈ Γ } is dense in H . This gives the other inequality. 1.96 (Orlicz) Show  that an unconditionally convergent series space satisfies xi 2 < ∞. Hint. Use Exercises 1.44 and 1.91.



x i in a Hilbert

k k 1.97 Suppose {x k }∞ k=1 is an orthonormal sequence in 2 , where x := (x i ). Show that lim xik = 0 for every i ∈ N. k→∞

Hint. Use the Bessel inequality (1.12) to show that (ei , x k ) → 0 as k → ∞. 1.98 A Banach space X is said to have cotype 2 if there exists a constant C > 0 such that for all vectors x1 , . . . , x n ∈ X we have n n  1/2 1  1    2 ε x x  . ≥   i i i 2n C εi =±1 i=1

i=1

We say that X has type 2 if there exists a constant C > 0 such that for all vectors x1 , . . . , xn ∈ X we have n n  1/2  1    2 ε x x  . ≤ C   i i i 2n εi =±1 i=1

i=1

Similarly we define type/cotype q. Taking for granted the Kahane–Khintchine inequality (see, e.g., [AlKa, p. 134]) that the average on the left hand side can be replaced by the expression

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1 Basic Concepts in Banach Spaces

 2 1/2  n  ε x , show that if X is a Banach space isomorphic to a  εi =±1 i=1 i i  Hilbert space, then X is both of type 2 and cotype 2. A theorem of Kwapie´n says that every space that is of type 2 and cotype 2 is isomorphic to a Hilbert space ([Kwap], see, e.g., [AlKa, p. 187]). Hint. Square both sides of the inequality and use Exercise 1.91. 

1 2n



1.99 Show that 4 is not isomorphic to a subspace of 2 . Hint. Show that 4 is not of cotype 2 by considering the standard unit vectors. 1.100 An Orlicz function M is a continuous nondecreasing convex function defined for t ≥ 0 such that M(0) = 0 and lim M(t) = ∞. We assume that t→∞

M(t) > 0 for all t > 0 and that M satisfies the Δ2 -condition at zero, i.e., that lim supt→0 M(2t)/M(t) < ∞. Let h M be the space of all sequences of real numbers x = (a1 , a2 , . . . ) such that ∞ for all ρ > 0. (Due to the Δ2 condition for M this is the n=1 M(|an |/ρ) < ∞ same as to require that ∞ n=1 M(|an |/ρ) < ∞ for some ρ > 0.) Define a norm  ·  on h M for x = (a1 , a2 , . . . ) by x = inf{ρ > 0 :

∞ 

M(|an |/ρ) ≤ 1}.

n=1

Show that h M equipped with the norm  ·  is a Banach space. It is called an Orlicz sequence space. Hint. Fatou lemma. 1.101 For every 1 ≤ p < ∞ and every nonincreasing sequence of positive numbers w = (wn )∞ n=1 we consider the space d(w, p) of all sequences of real numbers x = (an )∞ n=1 for which x := sup π

∞ 

1/ p |aπ(n) | p wn

< ∞.

n=1

where π ranges over all permutations of N. Show that d(w, p) with the norm  ·  is a Banach space. It is called a Lorentz sequence space. 1. Show that d(w, p) is isomorphic to  p if inf n wn > 0. 2. Show that d(w, p) is isomorphic to ∞ if ∞ n=1 wn < ∞. Therefore it is usually assumed that w1 = 1, that lim wn = 0, and that n→∞ ∞ w = ∞. n n=1 Hint. Fatou lemma.

Chapter 2

Hahn–Banach and Banach Open Mapping Theorems

The Hahn-Banach theorem, in the geometrical form, states that a closed and convex set can be separated from any external point by means of a hyperplane. This intuitively appealing principle underlines the role of convexity in the theory. It is the first, and most important, of the fundamental principles of functional analysis. The rich duality theory of Banach spaces is one of its direct consequences. The second fundamental principle, the Banach open mapping theorem, is studied in the rest of the chapter. A real-valued function p on a vector space X is called a subadditive if p(x + y) ≤ p(x) + p(y) for all x, y ∈ X . It is called positively homogeneous if for all x ∈ X and α ≥ 0 it satisfies p(αx) = αp(x). If p is subadditive and, moreover, p(αx) = |α| p(x) for all x ∈ X and all scalars α, then p is called a seminorm on X . Note that every norm is a seminorm. Note, too, that every positively homogeneous subadditive function is a convex function. By a linear functional on a vector space X , we mean a linear mapping from X into K. Theorem 2.1 (Hahn, Banach) Let Y be subspace of a real linear space X , and let p be a positively homogeneous subadditive functional on X . If f is a linear functional on Y such that f (x) ≤ p(x) for every x ∈ Y , then there is a linear functional F on X such that F = f on Y and F(x) ≤ p(x) for every x ∈ X . Proof: Let P be the collection of all ordered pairs (M , f ), where M is a subspace of X containing Y and f is a linear functional on M that coincides with f on Y and f ). We partially satisfies f ≤ p on M . P is nonempty as it contains the  pair (Y, ) ≺ (M , f ) if M ⊂ M and f  . If {M , f } is a order P by (M , f = f α α M chain, then M := Mα and a linear functional f on M defined by f (x) = f α (x) for x ∈ Mα satisfy (Mα , f α ) ≺ (M , f ) for all α. By Zorn’s lemma, P has a maximal element (M, F). We need to show that M = X . Assume M = X , pick x1 ∈ X \M and put M1 = span{M, x1 }. We will find (M1 , F1 ) ∈ P such that (M, F) ≺ (M1 , F1 ), a contradiction. For a fixed α ∈ R we define F1 (x + t x1 ) = F(x) + tα for x ∈ M, t ∈ R. Then F is linear. It remains to show that we can choose α so that F1 ≤ p.

M. Fabian et al., Banach Space Theory, CMS Books in Mathematics, C Springer Science+Business Media, LLC 2011 DOI 10.1007/978-1-4419-7515-7_2, 

53

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2 Hahn–Banach and Banach Open Mapping Theorems

Due to the positive homogeneity of p and F, it is enough to choose α such that F1 (x + x1 ) ≤ p(x + x 1 ) for every x ∈ M. F1 (x − x1 ) ≤ p(x − x 1 )

(2.1)

Indeed, for t > 0 we then have F1 (x + t x1 ) = t F1

x t

   + x 1 ≤ t p xt + x 1 = p(x + t x1 )

and for t = −η < 0 we have   F1 (x + t x 1 ) = F1 (x − ηx1 ) = ηF1 ηx − x1   ≤ ηp ηx − x1 = p(x − ηx1 ) = p(x + t x1 ). But (2.1) is equivalent to (α :=) F1 (x1 ) ≤ p(x + x1 ) − F(x) and (−α =) −F1 (x1 ) ≤ p(x − x 1 ) − F(x) for every x ∈ M. This in turn is equivalent to F(y) − p(y − x1 ) ≤ α ≤ p(x + x1 ) − F(x) for every x, y ∈ M. Thus to find a suitable α ∈ R we need to show that sup{F(y) − p(y − x 1 ) : y ∈ M} ≤ inf{ p(x + x1 ) − F(x) : x ∈ M}. This is in turn equivalent to the statement that for every x, y ∈ M we have F(y) − p(y − x1 ) ≤ p(x + x 1 ) − F(x). The latter reads F(x + y) ≤ p(x + x1 ) + p(y − x1 ), which is true as F(x + y) ≤ p(x + y) = p(x + x1 + y − x1 ) ≤ p(x + x 1 ) + p(y − x 1 ). This completes the proof of Theorem 2.1.

2.1 Hahn–Banach Extension and Separation Theorems Before we pass to normed space versions of the Hahn–Banach theorem, we need to establish the relationship between the real and the complex normed spaces. Let X be a complex normed space. The space X is also a real normed space. We will denote this real version of X by X R . On the other hand, if X is a real normed space, then X × X becomes a complex normed space X C when its linear structure and norm are defined for x, y, u, v ∈ X and a, b ∈ R by (x, y) + (u, v) := (x + u, y + v) (a + ib)(x, y) := (ax − by, bx + ay) (x, y)C := sup{ cos(θ )x + sin(θ )y : 0 ≤ θ ≤ 2π }.

2.1

Hahn–Banach Extension and Separation Theorems

55

The set X × {0} := {(x, 0) : x ∈ X } is a closed R-linear subspace of X C which is—as a real space—isometric to  X under the mapping (x, 0) → x. Conversely, X C = h + ik : h, k ∈ X × {0} . We will verify that  · C is actually a norm on X C . It is clear that  · C is nonnegative, satisfies the triangle inequality, and factors real constants to their absolute value. If α is real and z := (x, y) ∈ X C , then   e−iα zC =  cos(α)x + sin(α)y, − sin(α)x + cos(α)y C = sup{ cos(θ )[cos(α)x + sin(α)y] + sin(θ )[− sin(α)x + cos(α)y] : 0 ≤ θ ≤ 2π} = sup{ cos(θ + α)x + sin(θ + α)y : 0 ≤ θ ≤ 2π} = sup{ cos(η)x + sin(η)y : 0 ≤ η ≤ 2π} = zC .

Therefore  · C is a norm on X C . Since max{x, y} ≤ (x, y)C ≤ x + y, we have that the topology induced on X C = X × X by  · C is equivalent to the product topology induced on X × X by  · . We will now relate duals of X and X R . Consider the mapping R : X ∗ → X R ∗ defined by R( f )(x) = Re f (x) for x ∈ X , where Re( f (x)) is the real part of f (x). We claim that it is a norm-preserving mapping from X ∗ onto X R ∗ and is linear as a mapping (X ∗ )R → X R ∗ . To see this claim, note that if X is a complex Banach space and f ∈ X ∗ , then supz∈B X | f (z)| = supz∈B X | Re f (z) |. Indeed, for all z we have | f (z)| ≥   | Re f (z) |, so one inequality is clear. On the other hand, for z ∈ B X we write f (z) = eiα | f (z)| and have f (e−iα z) = e−iα f (z) = | f (z)|. Thus | Re f (e−iα z) | = | f (z)| and e−iα z = z. Now we show that R is onto X R ∗ . To g ∈ X R ∗ we assign the functional defined on X by G(x) = g(x) − ig(i x). Then G is linear over R, but also   G(i x) = g(i x) − ig(−x) = g(i x) + ig(x) = i g(x) − ig(i x) = i G(x). Therefore G is linear over C and hence G ∈ X ∗ . Moreover, R(G) = g. ∗ Theorem 2.2 (Hahn, Banach) Let Y be  a subspace of a normed space X . If f ∈ Y ∗ then there exists F ∈ X such that F Y = f and F X ∗ =  f Y ∗ .

Proof: First assume that X is a real normed space. Define a new norm ||| · ||| on X by |||x||| =  f Y ∗ x, where  ·  is the original norm of X . We have | f (y)| ≤ |||y||| for all y ∈ Y , so by Theorem 2.1 there is a linear functional F on X that extends f and |F(x)| ≤ |||x||| (=  f Y ∗ x) for every x ∈ X . Therefore F X ∗ := sup{|F(x)| : x ≤ 1} ≤  f Y ∗ . Since F extends f , we obviously have F X ∗ ≥  f Y ∗ as well. Consequently F X ∗ =  f Y ∗ . Now assume that X is a complex normed space. Consider the linear functional R( f ) on YR , where R is the isometry defined above. By the first part of this proof, we extend R( f ) to a linear functional g ∈ X R ∗ that satisfies g X R ∗ = R( f )YR ∗ =  f Y ∗ . Then the norm of the linear functional F(x) := g(x) − ig(i x) ∈ X ∗ is equal to g X R ∗ (=  f Y ∗ ).

56

2 Hahn–Banach and Banach Open Mapping Theorems

    The real part of F is g and thus Re F Y = Re( f ), that is, R(F Y ) = R( f ). Since R is a bijection of Y ∗ onto YR ∗ , we get F Y = f . Corollary 2.3 (Hahn, Banach) Let X be a normed space. For every x ∈ X there is f ∈ S X ∗ such that f (x) = x. In particular, x = max{| f (x)| : f ∈ B X ∗ } for every x ∈ X . As a consequence, if X = {0} then X ∗ = {0} as well (see Corollary 3.33). Proof: Put Y = span{x} and define f ∈ Y ∗ by f (t x) = tx. Clearly  f Y ∗ = 1 and f (x) = x. Using Theorem 2.2 we extend f to a linear functional from X ∗ with the same norm. From | f (x)| ≤  f  x we have sup f ∈B X ∗ | f (x)| ≤ x. On the other hand, the linear functional constructed above shows that the supremum is attained and equal to x. n Corollary 2.4 Let {xi }i=1 be a linearly independent set of vectors in a normed n space X and {αi }i=1 be a set of real numbers. Then there is f ∈ X ∗ such that f (x i ) = αi for i = 1, . . . , n.

Proof: Define a linear functional f on span{xi } by f (xi ) = αi for i = 1, . . . , n. Proposition 1.39 shows that f is continuous. The result follows from Theorem 2.2. Definition 2.5 Let C be a convex subset of a normed space X and let x ∈ C. A non-zero linear functional f ∈ X ∗ is called a supporting functional of C at x if f (x) = sup{ f (y) : y ∈ C}. The point x is said to be a support point of C (supported by f ). By Corollary 2.3, for every x ∈ S X there is a supporting functional of B X at x, and so x is a support point of B X . There exists a closed convex and bounded set C in a Banach space, having empty interior, and a point in C that is not a support point (see Exercise 2.17). However, every closed convex and bounded subset of a Banach space must have support points. This follows from Theorem 7.41. Consider a Banach space X . If Y is a subset of X , we define its annihilator by Y ⊥ = { f ∈ X ∗ : f (y) = 0 for all y ∈ Y }. Note that Y ⊥ is a closed subspace of X ∗ . Similarly, for a subset Y of X ∗ we define Y⊥ = {x ∈ X : f (x) = 0 for every f ∈ Y }, which is a closed subspace of X . Note that if F is a subset of a Hilbert space H , then the orthogonal complement F ⊥ when considered a subspace of the dual H ∗ under the canonical duality (see Theorem 2.22) coincides with the annihilator F ⊥ . Proposition 2.6 Let Y be a closed subspace of a Banach space X . Then (X/Y )∗ is isometric to Y ⊥ and Y ∗ is isometric to X ∗ /Y ⊥ . Proof: Consider the mapping δ : Y ⊥ → (X/Y )∗ defined by δ(x ∗ ) : xˆ → x ∗ (x), where x ∈ x. ˆ This definition is correct, since x ∗ (x1 ) = x ∗ (x2 ) whenever x 1 , x2 ∈ xˆ

2.1

Hahn–Banach Extension and Separation Theorems

57

as x ∗ ∈ Y ⊥ . To see that δ maps Y ⊥ onto (X/Y )∗ , given f ∈ (X/Y )∗ , define x ∗ ∈ X ∗ by x ∗ (x) = f (x), ˆ where x ∈ x. ˆ Then x ∗ ∈ Y ⊥ and δ(x ∗ )(x) ˆ = x ∗ (x) = f (x). ˆ To check that δ is an isometry, write ˆ = sup |x ∗ (x)| = x ∗ . δ(x ∗ ) = sup |δ(x ∗ )(x)| x<1 ˆ

x<1

The middle equality follows since given x ˆ < 1, there is x ∈ xˆ such that x < 1. On the other hand, given x < 1, we have x ˆ < 1. To prove the second part of this proposition, define a mapping σ from Y ∗ into X ∗ /Y ⊥ by σ (y ∗ ) = {all extensions of y ∗ on X }. It is easy to see that σ (y ∗ ) is a coset in X ∗ /Y ⊥ and the Hahn–Banach theorem gives that σ (y ∗ ) = y ∗ Y ∗ . It follows that σ is a linear and onto mapping. We will now establish several separation results. Proposition 2.7 Let Y be a closed subspace of a normed space X . If x ∈ / Y then there is f ∈ S X ∗ such that f (y) = 0 for all y ∈ Y and f (x) = dist(x, Y ). Proof: Let (0 <) d = dist(x, Y ). Put Z = span{Y, x} and define  a linear functional f on Z by f (y + t x) = td for y ∈ Y and t ∈ K. Clearly f Y = 0 and f (x) = d. For u := y + t x, where y ∈ Y and t is a scalar such that u = 0, we have | f (u)| = |t| d =

|t|.u u |t|.u d= d= d u y + t x (y/t) + x ud ud   ≤ = u. = dist(x, Y ) x − −(y/t) 

Therefore  f  ≤ 1. On the other hand, there is a sequence yn ∈ Y such that yn − x → d. We have d = | f (yn ) − f (x)| ≤  f  · yn − x, so by passing to the limit when n → ∞ we obtain d ≤  f d. Thus  f  = 1, and f (x) = dist(x, Y ). Extending f on X with the same norm we obtain the desired functional. Proposition 2.8 Let X be a normed space. If X ∗ is separable, then X is separable. Proof: Choose a dense subset { f n } of S X ∗ . For every n ∈ N, pick x n ∈ S X such that f n (xn ) > 12 . Let Y = span{xn }. As Y is separable (finite rational combinations of {xn } are dense in Y ), it is enough to show that X = Y . If Y = X , then there is f ∈ X ∗ ,  f  = 1 such that f (x) = 0 for every x ∈ Y . Let n be such that  f n − f  < 14 . Then | f (xn )| = | f n (xn ) − ( f n (xn ) − f (xn ))| ≥ | f n (xn )| − | f n (xn ) − f (xn )| ≥ | f n (xn )| −  f − f n  · xn  > 12 − 14 = 14 , a contradiction. To prove separation results for sets we need a new notion.

58

2 Hahn–Banach and Banach Open Mapping Theorems

Definition 2.9 Let C be a set in a normed space X . We define the Minkowski functional of C, μC : X → [0, +∞], by  μC (x) =

inf{λ > 0 : x ∈ λC}, if {λ > 0 : x ∈ λC} = ∅, +∞, if {λ > 0 : x ∈ λC} = ∅.

Lemma 2.10 Let μ be a subadditive real function on a real normed space X . Then (i) μ is continuous if and only if it is continuous at 0. (ii) If μ is continuous, every linear functional f : X → R such that f ≤ μ is also continuous. Proof: From the subadditivity of μ it follows easily that, for x, y ∈ X , −μ(y − x) ≤ μ(x) − μ(y) ≤ μ(x − y), so μ is continuous if (and only if) it is continuous at 0. This proves (i). In order to prove (ii), use Exercise 2.1. Lemma 2.11 Let C be a convex neighborhood of 0 in a normed space X . Then its Minkowski functional μC is a finite non-negative positively homogeneous subadditive continuous functional. Moreover, {x : μC (x) < 1} = Int(C) ⊂ C ⊂ C = {x : μC (x) ≤ 1}. Proof: Let Bδ = {x : x ≤ δ} ⊂ C for some δ > 0. Since 0 ∈ C, the point 0 is in x ∈ Bδ ⊂ C, λC for every λ > 0 and thus μC (0) = 0. Given x ∈ X \{0}, we get δ x

so x ∈

x δ C.

Thus (0 ≤) μC (x) ≤

x < ∞. δ

(2.2)

Given α, λ > 0, clearly x ∈ λC if and only if αx ∈ λαC. Therefore μC (αx) = αμC (x) and thus μC is positively homogeneous. We claim that μC (x) < λ implies that x ∈ λC. Indeed, there exists λ0 such that μC (x) ≤ λ0 < λ and x ∈ λ0 C. Then we can find c ∈ C such that x = λ0 c. Therefore   λ0 (1 − λ0 ) (x =) λ0 c = λ c+ 0 . (2.3) λ λ Since C is convex and 0 ∈ C we get x ∈ λC as claimed. To prove subadditivity, let x, y ∈ X and s, t such that μC (x) < s, μC (y) < t. By the former claim, x ∈ sC and y ∈ tC. Then x + y ∈ sC + tC, and thus by the convexity x + y ∈ (t + s)

 s t C+ C ⊂ (t + s)C. t +s t +s

Therefore μC (x + y) ≤ t + s, so by the choice of s and t we have μC (x + y) ≤ μC (x) + μC (y).

2.1

Hahn–Banach Extension and Separation Theorems

59

The continuity of μC at 0 follows from (2.2), so μC is continuous by Lemma 2.10. By the continuity of μC , the set {x ∈ X : μC (x) < 1} is open, and, by the claim, a subset of C, hence a subset of Int(C). It follows that, if μC (x) = 1 and 0 < s < 1 < t, then sx ∈ Int(C) and t x ∈ C. Therefore, if μC (x) = 1 then x belongs to the boundary of C and if μC (x) > 1 then, again by the continuity of μC , x ∈ Int(X \C). This proves the statement. Theorem 2.12 (Hahn, Banach) Let C be a closed convex set in a normed space X .    / C then there is f ∈ X ∗ such that Re f (x0 ) > sup{Re f (x) : x ∈ C}. If x 0 ∈ Proof: First, let X be a real space. We may assume without loss of generality that 0 ∈ C, otherwise we consider (C − x) and x0 − x for some x ∈ C. Let δ = dist(x 0 , C). Then δ is positive as C is closed. Set D = {x ∈ X : dist(x, C) ≤ δ/2}. Since 0 ∈ C, we have 4δ B X ⊂ D and so D contains 0 as an interior point. D is also / D. Let μ D be the Minkowski functional of D. Since D is closed, convex, and x0 ∈ closed and x0 ∈ / D, we have μ D (x0 ) > 1 (Exercise 2.21). Define a linear functional on span{x0 } by f (λx 0 ) = λμ D (x0 ). Then on span{x 0 } we have f (λx 0 ) ≤ μ D (λx0 ). For λ ≥ 0 it is clear from the definition of f , for λ < 0 we have f (λx0 ) = λμ D (x0 ) < 0 while μ D (λx 0 ) ≥ 0. Extend f onto X by Theorem 2.1 and denote this extension by f again. Then f (x) ≤ μ D (x) for every x ∈ X . The continuity of f follows from Lemma 2.10. Since μ D (x0 ) > 1 and f (x 0 ) = μ D (x0 ), we get f (x 0 ) > 1, so f (x0 ) > sup{ f (x) : x ∈ C}. ∗ as in the real case and then If X is a complex space, we construct g from X R define f (x) = g(x) − ig(i x). For simplicity we will state the following result only for the real case. Proposition 2.13 Let X be a real normed space. (i) Let C be an open convex set in X . If x0 ∈ / C then there is f ∈ X ∗ such that f (x) < f (x 0 ) for all x ∈ C. (ii) Let A, B be disjoint convex sets in X . If A is open then there is f ∈ X ∗ such that f (a) < inf f (b): b∈B for all a ∈ A. Proof: (i) We pick some y ∈ C and consider D := C − y, y0 := x 0 − y. Then define μ D and f on span{y0 } as in the proof of Theorem 2.12. Let f denote the extended / D. Also f (x) < 1 for functional as well. We have f (y0 ) = μ D (y0 ) ≥ 1 as y0 ∈ x ∈ D as D is open (Exercise 2.21), and the statement follows. (ii) Applying (i) to the open convex set C := A − B and to x0 := 0 we obtain f such that f (x) < f (0) (= 0) for x ∈ A − B. Thus f (a) < f (b) for every a ∈ A, b ∈ B. It follows that f (a) ≤ inf{ f (b) : b ∈ B} for a ∈ A. If a ∈ A is such that f (a) = inf{ f (b) : b ∈ B}, then, from the openness of A we get f (a + h) > inf{ f (b) : b ∈ B} for some a + h ∈ A, a contradiction. Therefore f (a) < inf{ f (b) : b ∈ B} for all α ∈ A.

60

2 Hahn–Banach and Banach Open Mapping Theorems

Proposition 2.14 Let (X,  · ) be a normed space. Let Y be a subspace of X and let | · | be an equivalent norm on Y . Then there is an equivalent norm | · | on X inducing on Y the norm | · |. Proof: Without loss of generality, we may assume that B(Y,·) ⊂ B(Y,|·|) . The set B := conv {B(Y,|·|) ∪ B(X,·) } is convex and balanced. Obviously, B is bounded and contains B(X,·) , so its Minkowski functional is an equivalent norm | · | on X (see Fig. 2.1). Fig. 2.1 Extending a norm

B (X,

)

B (Y,

)

B(X,|·|)

This norm certainly induces on Y the norm |·|, since B ∩Y = B(Y,|·|) . Indeed, B(Y,|·|) ⊂ B. On the other hand, if y ∈ B ∩ B(Y,|·|) , then y = λy1 + (1 − λ)x, where 0 ≤ λ ≤ 1, y1 ∈ B(Y,|·|) , and x ∈ B(X,·) (see Exercise 1.12). We get (1 − λ)x = y − λy1 ∈ Y . If λ = 1 then x ∈ Y , so x ∈ B(X,·) ∩ Y = B(Y,·) ⊂ B(Y,|·|) . By convexity, y ∈ B(Y,|·|) . If, on the contrary, λ = 1, we obtain again y ∈ B(Y,|·|) . We refer to Exercise 5.95 for an alternative proof of the Hahn–Banach theorem.

2.2 Duals of Classical Spaces In Propositions 2.15, 2.16, 2.17, 2.18, 2.19, and 2.20, we assume the scalar field to be R. in the sense that for every f ∈ c0∗ there is a Proposition 2.15 (Riesz) c0∗ = 1  ai xi for all x = (xi ) ∈ c0 , and the mapping unique (ai ) ∈ 1 such that f (x) = f → (ai ) is a linear isometry from c0∗ onto 1 . i

Proof: Given f ∈ c0∗ , define ai = f (ei ), where ei := (0, . . . , 0, 1, 0, . . . ) are the standard unit vectors in c0 . For n ∈ N we set x n = (sign(a1 ), . . . , sign(an ), 0, . . . ) ∈ c0 . n n n n Then i=1 |ai | ≤  f  · x  ∞ =  f . Therefore ∞ x ∞ = 1 and f (x ) = |a | ≤  f  < ∞, that is, the mapping f  → f (e ) i is a continuous mapping i=1 i into 1 . It is obviously linear.

2.2

Duals of Classical Spaces

61

∞  On the other hand,if i=1 |ai | < ∞ then |ai xi | < ∞ for every x = (xi ) ∈  c0 . Indeed, we have |ai xi | ≤ sup |xi | ·  |ai | = (ai )1 (xi )∞ . Consider the ai xi . Then from the above estimate linear functional h defined on c0 by h(x) = we have h ≤ (a ) and also h(e ) = a , so h ∈ c0∗ and the mapping f → i 1 i i   f (ei ) is thus onto. We also obtain that  f  =  f (ai ) 1 , hence the considered mapping is an isometry onto 1 . Proposition 2.16 (Riesz) ∗1 = ∞  in the sense that for every f ∈ ∗1 there is a unique (ai ) ∈ ∞ such that f (x) = ai xi for all x = (x0 ) ∈ 1 , and the mapping f → (ai ) is a linear isometry from ∗1 onto ∞ . Proof: Given f ∈ ∗1 , put ai = f (ei ) for i ∈ N, where ei are the standard unit vectors in 1 . Then |ai | ≤  f , so (ai )∞ ≤  f . Conversely, for (ai ) ∈ ∞ consider the functional h defined on 1 by h(x) = ai xi . Again, |h(x)| ≤ (ai )∞ x1 , hence h ∈ ∗1 and h ≤ (ai )∞ . Similarly as in the proof of Proposition 2.15, we conclude that the mapping is a linear isometry onto ∞ . Proposition 2.17 (Riesz) Let p, q ∈ (1, ∞) be such that 1p + q1 = 1. Then ∗p = q ∗ in the sense  that for every f ∈  p there exists a unique element (ai ) ∈ q such that f (x) = ai xi for all x =: (xi ) ∈  p , and the mapping f → (ai ) is a linear isometry from ∗p onto q . Proof: For f ∈ ∗p , put ai = f (ei ). Considering   x n := |a1 |q−1 sign(a1 ), . . . , |an |q−1 sign(an ), 0, . . . we see that n 

n n 1    q−1  p 1p p |ai |q = f (x n ) ≤  f  · x n  p =  f  = f· |ai |q . |ai |

i=1

i=1

 n

1 q

≤  f . Hence (ai )q ≤  f  < ∞.  If (ai ) ∈ q and (xi ) ∈  p , then the series xi ai is convergent by the Hölder inequality (1.1) as |xi ai | ≤ (xi ) p (ai )q . Therefore the functional h defined on  p by h(x) = xi ai is well defined and h ≤ (ai )q . The rest of the proof is analogous to those above. This reads

q i=1 |ai |

i=1

Similarly we show that for a set Γ and p ∈ [1, ∞) we have c0 (Γ )∗ = 1 (Γ ) and  p (Γ )∗ = q (Γ ), where 1p + q1 = 1. This applies, in particular, for a finite set Γ . Proposition 2.18 (Riesz) Let p, q ∈ (1, ∞) be such that 1p + q1 = 1. Then L p [0, 1]∗ = L q [0, 1] in the sense that for every F ∈ L ∗p there is a unique f ∈ L q

1 such that F(g) = 0 g f dx for all g ∈ L p , and the mapping F → f is a linear isometry of L ∗p onto L q .

62

2 Hahn–Banach and Banach Open Mapping Theorems

Proof: Let F ∈ L ∗p . For t ∈ [0, 1], let u t = χ[0,t) be the characteristic function of [0, t). Define α(t) = F(u t ). We claim that α is absolutely continuous (see the definition right before Proposition 11.13). Indeed, if [τi , ti ], i = 1, . . . , n, is a collection of non-overlapping intervals, that is, their interiors are pairwise disjoint, put εi = sign(α(ti ) − α(τi )) and estimate: n 

|α(ti ) − α(τi )| =

i=1

n 

εi (α(ti ) − α(τi )) = F

i=1

≤ F L ∗p

n     · εi (u ti − u τi ) i=1

n  ti 

= F

i=1

τi

1

p

1 dx

n 

εi (u ti − u τi )



i=1

n p 1  1  p   = F L ∗p εi (u ti − u τi ) dx 

Lp

0

= F ·

n 

(ti − τi )

i=1

1

p

.

i=1

Therefore α is an absolutely continuous function on [0,

t1]. By the Lebesgue fundamental theorem of calculus, we have α(t) − α(0) = 0 α dx for every t ∈ [0, 1]. Setting f = α and using α(0) = F(u 0 ) = 0 we get  F(u t ) = α(t) =

t

 f dx =

0

1

u t f dx. 0

1 Since F is linear, we also have F(gn ) = 0 gn f dx for all step functions gn :=   n . k=1 ck u nk − u k−1 n Let g be a bounded measurable function on [0, 1]. Then there is a sequence of step functions gn such that gn → g a.e. and {gn } is uniformly bounded. By the Lebesgue dominated convergence theorem, we get  lim F(gn ) = lim

n→∞

n→∞ 0

1

 gn f dx =

1

 lim gn f dx =

0 n→∞

1

g f dx. 0

On the other hand, since gn → g a.e. and gn are uniformly bounded, the same theorem implies gn − g L p → 0 as n → ∞. By the continuity of F on L p , we

1

1 thus have F(g) = lim F(gn ) = 0 g f dx. Hence F(g) = 0 g f dx for every n→∞ bounded measurable function g on [0, 1]. We will show that f ∈ L q and  f q ≤ F. Consider a family of functions gn defined by  gn (x) =

  | f (x)|q−1 sign f (x) if | f (x)| ≤ n, 0 if | f (x)| > n.

The functions gn are bounded and measurable. Thus we have F(gn ) = Note also that |F(gn )| ≤ F gn  p . On the other hand,

1 0

gn f dx.

2.2

Duals of Classical Spaces



1



1

|gn | dx = p

0



63

|gn |

q q−1



0

1

|gn (t)| |gn (t)| q−1 dx

0

1

≤

1

dx = 

1

|gn | | f | dx =

0

gn f dx = F(gn ) = |F(gn )|.

0



1 

1

1 p q 1 1 ≤ Hence 0 |gn | p dx ≤ F · gn  p = F 0 |gn | p dx , so 0 |gn | p dx F. Since f is integrable, we have |gn | → | f |q−1 a.e. By Fatou’s lemma, the last inequality implies that 

1

| f |q dx

1 q

0

=



1

| f |(q−1) p dx

0

1 q

=



1

|gn | p dx

1 q

≤ F.

0

This shows that f ∈ L q . Finally, let g ∈ L p . There exists a sequence {gn } of bounded measurable functions that converges to g in L p . Then F(gn ) → F(g) and

1

1 by Hölder’s inequality (1.1) we have 0 gn f dx → 0 g f dx. We have shown that

1

1 F(gn ) = 0 gn f dx for bounded measurable functions, so F(g) = 0 g f dx as claimed. On the other hand, given a function f ∈ L q , we can define a linear functional

1 on L p by F(g) = 0 g f dx. It follows from the Hölder inequality (1.1) that F is continuous and F ≤  f q . Using similar methods, we obtain an analogous result for the space L 1 . Proposition 2.19 (Riesz) L 1 [0, 1]∗ = L ∞ [0, 1] in the sense that for every F ∈ L ∗1

1 there exists a unique f ∈ L ∞ such that F(g) = 0 g f dx for all g ∈ L 1 , and the mapping F → f is a linear isometry of L ∗1 onto L ∞ . Proposition 2.20 (Riesz) For every F ∈ C[0, 1]∗ there exists a function f on [0, 1]

1 with bounded variation such that F(g) = 0 g d f (Stieltjes integral) for all g ∈ C[0, 1] and F =

1

1

f , where 0

f denotes the variation of f on [0, 1].

0

On the other hand, if f is a function of bounded variation on [0, 1], then F(g) :=

1 0 g d f is a continuous linear functional on C[0, 1]. Proof: Consider the space ∞ [0, 1] of bounded functions on [0, 1] with the supremum norm denoted by  · ∞ . If F ∈ C[0, 1]∗ , we have that |F(g)| ≤ F · g∞ for every g ∈ C[0, 1]. Since C[0, 1] is a subspace of ∞ [0, 1], by the Hahn–  on ∞ [0, 1] such that | F(g)|  Banach theorem we can extend F to a functional F ≤  similarly to the L p setting above. For t ∈ [0, 1], F · g∞ . We will represent F  t ) for t ∈ [0, 1] let u t = χ[0,t) , the characteristic function of [0, t). Put f (t) = F(u (note that F is not defined on u t as u t is not continuous). We will prove that f has bounded variation on [0, 1]. To this end, consider t0 = 0 < t1 < · · · < tn−1 < tn = 1 and put εi = sign( f (ti ) − f (ti−1 )). We have

64

2 Hahn–Banach and Banach Open Mapping Theorems n 

| f (ti ) − f (ti−1 )| =

i=1

n 

n   εi ( f (ti ) − f (ti−1 )) = F εi (u ti − u ti−1 )

i=1

n     · εi (u ti − u ti−1 ) ≤  F 

∞

i=1

i=1

= F · 1.

Hence f has bounded variation [0,1]which is bounded by F.  on n g nk u k − u k−1 we have For g ∈ C[0, 1] and gn := i=1 n

 n) = F(g

n

 n        g nk f nk − f k−1 = n i=1

1

gn d f. 0

  1 ∈ g( nk ) f ( nk ) − f ( k−1 ) = 0 g d f . Since F n n→∞ n→∞

1  n ) = F(g),   so F(g) = 0 gdf. ∞ [0, 1]∗ and gn → g in  · ∞ , we have lim F(g

1  However, for g ∈ C[0, 1] we have F(g) = F(g), hence F(g) = 0 g(t) d f (t).  n ) = lim Therefore lim F(g

n

k=1

1

We have already shown that

f ≤ F. On the other hand, from the theory of

0

Riemann–Stieltjes integral we have that given a function f of bounded variation, 1

1

1 f . Therefore F F : g → 0 g d f is a linear mapping and 0 g(t) d f (t) ≤ g∞ 0

is continuous and F ≤

1

f. 0

In general, if K is a compact set, the space C(K )∗ can be identified with the space of all regular Borel measures

on K of bounded variation. Every such measure μ defines a functional Fμ ( f ) := K f dμ, the correspondence μ → Fμ is a linear isometry ([Rudi2, Theorem 2.14]). Let k ∈ K . We define the corresponding Dirac functional (or Dirac measure) by linear δk ( f ) = f (k) for every f ∈ C(K ). Observe that δk is a continuous   functional of norm one. Indeed, on one hand, δk  = sup δk ( f ) = sup f (k) ≤ 1. By  f ≤1

 f ≤1

considering the constant function f = 1, we obtain δk  = 1. Proposition 2.21 The space C[0, 1]∗ is not separable.

Proof: Consider the Dirac measures δt for t ∈ [0, 1]. We claim that if t1 = t2 then δt1 − δt2  = 2. Indeed, δt1 − δt2  ≤ δt1  + δt2  = 2. On the other hand, choose f 0 ∈ C[0, 1] such that f 0 (t1 ) = 1, f 0 (t2 ) = −1, and  f 0 ∞ = 1. Then δt1 − δt2  ≥ | f 0 (t1 ) − f 0 (t2 )| = 2. Similarly to the case of ∞ we find that C[0, 1]∗ is not separable. Recall that the inner product on a complex Hilbert space 2 , respectively L 2 , is

1    x i y¯i , respectively (g, f ) = 0 g f¯ dx. This motivates defined by (xi ), (yi ) = the following identification of the dual space in case of complex scalars. Recall that

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Banach Open Mapping Theorem, Closed Graph Theorem, Dual Operators

65

a mapping Φ is called conjugate linear if Φ(αx + y) = αΦ(x) ¯ + Φ(y) for all vectors x, y and scalars α. Theorem 2.22 (Riesz) Let H be a Hilbert space. For every f ∈ H ∗ there is a unique a ∈ H such that f (x) = (x, a) for all x ∈ H . The mapping f → a is a conjugate-linear isometry of H ∗ onto H . Proof: The uniqueness of such a is clear. Indeed, if f (x) = (x, a1 ) = (x, a2 ) then using x = a1 − a2 we get (a1 − a2 , a1 ) = (a1 − a2 , a2 ). Thus (a1 − a2 , a1 − a2 ) = 0, so a1 = a2 . By the Cauchy–Schwarz inequality,    f  = sup | f (x)| = sup |(x, a)| ≤ sup a · x ≤ a. x≤1

x≤1

x≤1

On the other hand,  f  = sup{| f (x)| : x ≤ 1} ≥ (a/a, a) = a. Hence  f  = a. To obtain the representation of 0 = f ∈ H ∗ , consider N := Ker( f ). It is a proper closed subspace of H . Choose z 0 ∈ N ⊥ and assume without loss of generality that f (z 0 ) = 1. We claim that H = N ⊕span{z 0 }. Indeed, given h ∈ H , it suffices to find a scalar α such that h − αz 0 ∈ N , that is, f (h − αz 0 ) = 0. This is satisfied for α = f (h). We now show that f (x) = x, zz02 for every x ∈ H . Given x := y + αz 0 , 0 where y ∈ N and α is a scalar, we have f (x) = α f (z 0 ) = α = α(z 0 , z 0 )/z 0 2

 = (y, z 0 )/z 0 2 + (αz 0 , z 0 )/z 0 2 = x,

z0 . z 0 2

2.3 Banach Open Mapping Theorem, Closed Graph Theorem, Dual Operators Definition 2.23 Let ϕ be a mapping from a topological space X into a topological space Y . We say that ϕ is an open mapping if it maps open sets in X onto open sets in Y . Let T be an operator from a normed space X into a normed space Y . Observe that if T is an open mapping, then T is necessarily onto. Indeed, by Exercise 2.37, δ BY ⊂ T (B X ) for some δ > 0 and hence by linearity, Y ⊂ T (X ). We will now establish the converse for bounded operators. By B XO (r ) we denote the open ball with radius r centered at the origin of a Banach space X .

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Lemma 2.24 (Banach) Let X be a Banach space, Y a normed space and T ∈ B(X, Y ). If r, s > 0 satisfy BYO (s) ⊂ T (B XO (r )), then BYO (s) ⊂ T (B XO (r )). Proof: By considering rs T if necessary, we may assume that r = s = 1. Denote B XO = B XO (1) and BYO = BYO (1). Let z ∈ BYO be given. Choose δ > 0 such that zY < 1 − δ < 1 and put y = (1 − δ)−1 z. Note that yY < 1. We will show that y ∈ (1 − δ)−1 T (B XO ), which implies that z ∈ T (B XO ). We start with y0 = 0 and inductively find a sequence yn ∈ Y such that y − yn Y < δ n and (yn − yn−1 ) ∈ T (δ n−1 B XO ). Indeed, having chosen y0 , y1 , . . . , yn−1 ∈ Y , we have (y − yn−1 ) ∈ δ n−1 BYO ⊂ T (δ n−1 B XO ), hence there is w ∈ T (δ n−1 B XO ) such that w − (y − yn−1 )Y < δ n . Setting yn = yn−1 + w we complete the construction. X such that xn  X < δ n−1 and T (xn ) = Next we find a sequence {xn }∞ n=1 ⊂ xi is absolutely convergent, we put x = yn − yn−1 for n ∈ N. Since  ∞the series  ∞ ∞ n−1 = 1 and by the continuity n=1 x n . Then x X ≤ n=1 x n  X < n=1 δ 1−δ and linearity of T , T (x) = lim

N →∞

N  n=1

T (xn ) = lim

N →∞

N  n=1

(yn − yn−1 ) = lim y N = y. N →∞

Note that T (B XO (r )) = T (B X (r )), so the conclusion of the lemma is true if we assume for instance δ BY ⊂ T (B X ). Theorem 2.25 (Banach open mapping principle) Let X, Y be Banach spaces and T ∈ B(X, Y ). If T is onto Y then T is an open mapping. Proof: Put G = T (B XO ). Since T is linear, we only need to prove that G contains a neighborhood of the origin. Note that we have T (B XO (r )) = r G and r G = r G for every r > 0. Therefore T (B XO (r )) = r G for every r > 0. This implies that ∞

nG. By the Baire category theorem, there is n ∈ N such that nG Y = T (X ) = n=1   contains an interior point, so there is x0 ∈ G and δ > 0 such that x0 + BYO (δ) ⊂   nG. Since nG is symmetric, we have −x0 + BYO (δ) ⊂ nG. If x ∈ BYO (δ) then from the convexity of nG we have x = 12 (x0 +x)+ 12 (−x0 +x) ∈ nG. Therefore BYO (δ) ⊂   T (B XO (n)) and consequently BYO nδ ⊂ n1 T (B XO (n)) = T (B XO ). By Lemma 2.24,   we have BYO nδ ⊂ T (B XO ) as claimed. It follows from the proof that if T : X → Y is onto, then there is δ > 0 such that δ BY ⊂ T (B X ). Note that even if T ∈ B(X, Y ) is open, it does not imply that T (M) is closed in Y whenever M is closed in X (Exercise 15.11).

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Banach Open Mapping Theorem, Closed Graph Theorem, Dual Operators

67

In Exercise 2.33, a rewording of the proof of the Banach open mapping principle in the language of convex series is presented. Corollary 2.26 Let X, Y be Banach spaces and let T ∈ B(X, Y ) be onto Y . (i) If T is one-to-one, then T −1 is a bounded operator. (ii) There is a constant M > 0 such that for every y ∈ Y there is x ∈ T −1 (y) satisfying x X ≤ MyY . (iii) Y is isomorphic to X/ Ker(T ). Proof: (i) If O is open in X , then (T −1 )−1 (O) = T (O) is open in Y showing that T −1 is continuous. (ii) By the open mapping theorem, there is δ > 0 such that δ BY ⊂ T (B X ). Therefore for every y ∈ Y such that yY = δ, there is x ∈ B X such that T (x) = y. Thus it is enough to put M = 1/δ. ! from X/ Ker(T ) onto Y by T !(x) (iii) Define a linear mapping T ˆ = T (x) for ! is well defined. Moreover T ! is one-to-one and onto Y . Let x ∈ x. ˆ The mapping T xˆn → 0. Then there is xn ∈ xˆn such that x n  X < xˆ n  + 1/n and therefore !(xˆn ) → 0. Hence T ! xn → 0. Since T is continuous, we have T (xn ) → 0 and thus T is continuous and one-to-one, so by (i) it is an isomorphism of X/ Ker(T ) onto Y . Theorem 2.27 (Banach closed graph theorem) Let X, Y be Banach spaces and let T be an operator  from X into Y . T is a bounded operator if and only if its graph G := { x, T (x) : x ∈ X } is closed in X ⊕ Y . Recall that the norm on X ⊕ Y is defined by (x, y) = x X + yY . In particular, (xn , yn ) → (x, y) if and only if x n → x and yn → y (see Definition 1.33). Note that G, the graph of T , is a subspace of X ⊕ Y .   Proof: If T is continuous and x n , T (xn ) → (x0 , y0 ), then y0 = T (x0 ). Indeed, we of T implies that T (xn ) → have xn → x0 and T (xn ) → y0, while  the continuity  T (x 0 ). This means that (x 0 , y0 ) = x 0 , T (x 0 ) is in the graph of T , showing that G is closed. If G is closed in X ⊕ Y , then G is a Banach space in the norm induced from X ⊕ Y . Consider the mapping p : G → X defined by p x, T (x) = x. By the definition of the norm in X ⊕ Y we see that p is continuous, maps G onto X , and is  one-to-one. By Corollary 2.26, p−1 : x → x, T (x) is a continuous mapping from X onto G. Since also q : X ⊕ Y → Y , q(x, y) := y, is continuous and T = q ◦ p−1 , T must be continuous. Definition 2.28 Let X, Y be Banach spaces and T ∈ B(X, Y ). We define the  dual ∗ ∈ B(Y ∗ , X ∗ ) for f ∈ Y ∗ by T ∗ ( f ) (x) = (also called adjoint) operator T   f T (x) , for all x ∈ X .   It is easy to observe that x → f T (x) is a linear mapping. If x ≤ 1 then   |T ∗ ( f )(x)| = | f T (x) | ≤  f  T . Thus T ∗ ( f ) is also bounded, so T ∗ ( f ) ∈ Y ∗ and T ∗ is well defined. Also the mapping f → T ∗ ( f ) is linear and the above estimate shows that T ∗ ( f ) ≤ T   f . Consequently, T ∗ is a bounded operator from X ∗ into Y ∗ .

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2 Hahn–Banach and Banach Open Mapping Theorems

Proposition 2.29 Let X, Y be Banach spaces. If T ∈ B(X, Y ) then T ∗  = T . Proof: We have T ∗  = sup T ∗ ( f ) X ∗ = sup f ∈BY ∗







sup |T ∗ ( f )(x)|



f ∈BY ∗ x∈B X

  = sup { sup | f T (x) |} = sup { sup | f T (x) |} = sup {T (x)Y } = T . f ∈BY ∗ x∈B X

x∈B X f ∈BY ∗

x∈B X

Let X, Y, Z be Banach spaces and let T ∈ B(X, Y ), S ∈ B(Y, Z ). Then (ST )∗ = Indeed, consider f  ∈ Z ∗ . Then for every x ∈ X we get (ST )∗ ( f )(x) =    ∗ f ST (x) = (S f ) T (x) = T ∗ S ∗ ( f ) (x), so (ST )∗ ( f ) = (T ∗ S ∗ )( f ).

T ∗ S∗.

2.4 Remarks and Open Problems Remarks 1. We mentioned in Open Problem 1 in Chapter 1 that quasi-Banach spaces behave differently from Banach spaces. This is mainly due to the fact that the Hahn– Banach theorem fails in that context, see [Kalt1].

Open Problems 1. It is an open problem if every infinite-dimensional Banach space has a separable infinite-dimensional quotient, i.e., if for every Banach space X there is an infinite-dimensional separable Banach space Y and a bounded operator from X onto Y . This problem is equivalent to the problem whether in every Banach space X there is an increasing sequence {E n }∞ n=1 of distinct closed subspaces such that

E = X (see, e.g., [Muji] and [HMVZ, Chapter 4]). n n

Exercises for Chapter 2 2.1 Let X be a real normed space. If f is a linear functional on E that is dominated by a function p : E → R (i.e., f ≤ p), and p is continuous at 0, then f is continuous. Hint. − f (x) = f (−x) ≤ p(−x), hence − p(−x) ≤ f (x) ≤ p(x) for all x ∈ E. Since this implies that f is continuous at 0, the conclusion follows from Proposition 1.25.

Exercises for Chapter 2

69

2.2 Let C be a convex symmetric set in a Banach space X . Assume that a linear functional f on X is continuous at 0 when restricted to C. Show that the restriction of f to C is uniformly continuous. Hint. Given ε > 0, we look for a neighborhood U of the origin in X such that x, y ∈ C and x − y ∈ U imply | f (x − y)| < ε. We have 12 (x − y) ∈ C, so by homogeneity of f we only need to find an open ball U centered at 0such that | f (w)| < ε/2 for point w ∈ C ∩ U . Such U exists by the continuity of f C at 0. 2.3 Show that if X is a finite-dimensional Banach space, then every linear functional f on X is continuous on X . Hint. Use Proposition 1.39. 2.4 Show that if X is an infinite-dimensional normed space, then X admits a discontinuous linear functional. Hint. Let {eγ } be a Hamel basis formed by vectors of norm 1. Define a linear functional f on {eγ } so that the set { f (eγ )} is unbounded, and extend f on X linearly. Then f is not bounded on the unit ball. 2.5 Show that if f = 0 is a linear functional on a normed space X , then the codimension of f −1 (0) in X is 1. Hint. For x ∈ X write x = (x − ( f (x)/ f (x0 ))x0 ) + ( f (x)/ f (x 0 ))x0 , where x0 is some fixed element in X with f (x0 ) = 0. 2.6 Recall that by a hyperplane of a normed space X we mean any subspace Y of codimension 1 (that is, dim (X/Y ) = 1). Let Y be a subspace of a normed space X . Show that Y is a hyperplane if and only if there is a linear functional f , f = 0, such that Y = f −1 (0). Show that Y is a closed hyperplane if and only if there is f ∈ X ∗ , f = 0, such that Y = f −1 (0). Hint. One direction: Exercise 2.5. Given a closed hyperplane Y , take e ∈ / Y , use Proposition 2.7 to find f . Then Y ⊂ f −1 (0) and since codim(Y ) = 1, equality follows. For a general hyperplane, the proof is similar. 2.7 Let H be a hyperplane in a normed space X , and let F be a two-dimensional subspace of X . Show that dim (F ∩ H ) ≥ 1. Hint. Use algebraic complementability of H in X . 2.8 Let H be a closed hyperplane of a Banach space X . Let x0 ∈ X \H . Prove that there is a linear and continuous projection P from X onto H parallel to x 0 , i.e., such that P x0 = 0 (for a more precise statement, see Exercise 5.7). Hint. Exercise 2.6 gives f ∈ X ∗ such that Ker f = H . By scaling we may assume that f (x0 ) = 1. Let P : X → X be defined by P(x) = x − f (x)x0 . This is the sought projection. 2.9 Let X be a Banach space. Show that all closed hyperplanes of X are mutually isomorphic. By induction we get that given k ∈ N, all closed subspaces of X of

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2 Hahn–Banach and Banach Open Mapping Theorems

codimension k are mutually isomorphic. In fact, Zippin proved that the Banach– Mazur distance of two hyperplanes of the same infinite-dimensional Banach space is less than or equal to 25, see [AlKa, p. 238]. Hint. Let F and G to distinct closed hyperplanes of X . According to Exercise 2.6, there exists f, g ∈ X ∗ such that F := f −1 (0) and G := g −1 (0). Find e ∈ X such that f (e) = g(e) = 1. Let P f (resp., Pg ) be the (linear and continuous) projection of X onto F (resp., onto G) parallel to e (see Exercise 2.8). Clearly, P f ◦ Pg (x) = x for every x ∈ F, and Pg ◦ P f (y) = y for every y ∈ G. From this it follows that Pg  F : F → G is an isomorphism. See Fig. 2.2. Fig. 2.2 All hyperplanes are mutually isomorphic

F

x e

Pf (x)

Pg(x)

0

G

2.10 Let f be a linear functional on a Banach space X . Show that if f is not identically 0, the following are equivalent (see also Proposition 3.19): (i) f is continuous. (ii) f is continuous at 0. (iii) f −1 (0) is closed. (iv) f −1 (0) is not dense in X . Hint. (i)⇒(ii) is obvious, and (ii)⇒(i) is clear from the linearity of f . (i)⇒(iii) is clear. (iii)⇒(iv) is obvious. (iv)⇒(iii) follows from the fact that if f −1 (0) is not closed, then f −1 (0)  f −1 (0) ⊂ X and f −1 (0) is a linear subspace. It is −1 enough to use now Exercise 2.5. (iii)⇒(i):  Since f (R\{0}) = ∅ is open, there  issome ball B = x0 + δ B X such that f B = 0. Assume f (x 0 ) > 0, then also f  B > 0 (connect x0 with points of B, f = 0 on the connecting segments and f is  continuous on each of those segments). Then f  B ≥ − 1δ f (x 0 ), so by symmetry of X

B X we get | f (x)| ≤ 1δ f (x 0 ) for x ∈ B X and f is continuous. Another related approach is the following: two subspaces A and B of a normed space X form an (algebraic) direct sum decomposition of X (written X = A ⊕ B) if A ∩ B = {0} and A + B = X (see the paragraph prior to Definition 1.33). Prove first that if f is a non-zero linear functional and x0 ∈ X such that f (x0 ) = 0, and K := f −1 (0), then X = K ⊕ span{x 0 } (see Exercise 2.5). As a consequence, no subspace S of X exists such that K  S  X . Since K ⊂ K ⊂ X , and the closure of a subspace is also a subspace (see Exercise 1.9) it follows that K is dense if it is

Exercises for Chapter 2

71

not closed (the converse being trivially true). To finish the exercise, notice that if K is closed, then X/K is then a one-dimensional space. Let fˆ : X/K → K a linear functional such that fˆ ◦ q = f , where q : X → X/K is the canonical quotient mapping (see Exercise 2.35). Apply now Exercise 2.3. 2.11 Find a discontinuous linear mapping T from some Banach space X into X such that Ker(T ) is closed. Hint. Let X = c0 and T (x) = ( f (x), x1 , x 2 , . . . ) for x = (xi ), where f is a discontinuous linear functional on X . 2.12 Let X be a Banach space, f ∈ S X ∗ . Show that for every x ∈ X we have dist x, f −1 (0) = | f (x)|. Hint. The result is obviously true if f = 0. If not, put K = f −1 (0). There exists, by Proposition 2.7, g ∈ S X ∗ that vanishes on K and g(x) = dist(x, K ). Since g −1 (0) = f −1 (0), we get g = λ f for some scalar λ (see Exercise 2.5), and |λ| = 1 since both f and g belong to S X ∗ . This proves the assertion. 2.13 (The “parallel-hyperplane lemma".) Let X be a real Banach space, f, g ∈ S X ∗ and ε > 0 be such that | f (x)| ≤ ε for every x ∈ g −1 (0) ∩ B X . Prove that either  f − g ≤ 2ε or  f + g ≤ 2ε (see Fig. 2.3). {x : f(x)=ε} {x : f(x)=0} {x : f(x)=−ε}

BX 0

{x : g(x)=0}

Fig. 2.3 The “parallel-hyperplane lemma”

Hint. Consider f on g −1 (0) and extend it with the same norm (at most ε) on X , −1 ˜ calling this extension f˜. Then f˜ − f = 0 on  g (0) and thus f − f = αg for some α by Lemma 3.21. Note that |1 − |α|| =  f  −  f − f˜ ≤  f˜ ≤ ε. Thus if α ≥ 0, then g + f  = (1 − α)g + f˜ ≤ |1 − α| +  f˜ ≤ 2ε. If α < 0, calculate g − f . 2.14 If X is an infinite-dimensional Banach space, show that there are convex sets C1 and C2 such that C1 ∪ C2 = X , C1 ∩ C2 = ∅, and both C1 and C2 are dense in X. Hint. Take a discontinuous functional f on X (Exercise 2.4), define C 1 = {x : f (x) ≥ 0} and C 2 = {x : f (x) < 0}, use Exercise 2.10. 2.15 Let X be a finite-dimensional Banach space. Let C be a convex subset of X that is dense in X . Prove that C = X .

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2 Hahn–Banach and Banach Open Mapping Theorems

Hint. We may assume that 0 ∈ C. Let {e1 , e2 , . . . , en } be an algebraic basis of X consisting of unit vectors. Fix ε > 0. For each i ∈ {1, 2, . . . , n} we can find vi ∈ C such that ei − vi  < ε. If ε > 0 is small enough, {v1 , . . . , vn } is a linearly independent set in C (look at the determinant of the matrix with columns vi , i = 1, 2, . . . .n). The set conv ({vi : i = 1, 2, . . . , n} ∪ {0}) has a nonempty interior and is contained in C, so C has a nonempty interior. If x0 ∈ X \C, then there exists a closed hyperplane H that separates {x0 } and Int(C), a contradiction with the denseness of C. 2.16 Let N be a maximal ε-separated set in the unit sphere of a Banach space X (see Exercise 1.47). Show that (1 − ε)B X ⊂ conv(N ). Hint. Otherwise, by the separation theorem, we find x ∈ X and f ∈ S X ∗ with x ≤ 1 − ε and f (x) > supconv(N ) ( f ) = sup N ( f ). For δ > 0 choose y ∈ S X such that f (y) > 1 − δ. By the maximality of N , there exists z ∈ N with ε > y − z ≥ f (y) − f (z). Thus sup N ( f ) ≥ f (z) > f (y) − ε > 1 − δ − ε. This holds for any δ > 0, so we have 1 − ε ≤ sup N ( f ) < f (x) ≤ x ≤ 1 − ε, a contradiction. 2.17 Let D = {±ei : i ∈ N} ⊂ 2 , where ei is the ith unit vector. The set C := conv(D) has empty interior, so it coincides with its boundary. Show that 0 is not a support point of C. Hint. If 0 is supported by some f , prove that f must be 0. That the interior of C is empty follows from the fact that C is the unit ball of 1 (use Exercise 3.36 or, more generally, Exercise 3.37). 2.18 Let C be a subset of a Banach space X and f be a Lipschitz real-valued function on C. Show that f can be extended to a Lipschitz function on X . Hint. Assume without loss of generality that f is 1-Lipschitz. Put for x ∈ X , F(x) = inf{ f (z) + z − x; z ∈ C}. To see that F is finite for every x ∈ X , pick an arbitrary z 0 ∈ C. Then for any z ∈ C, f (z) + x − z ≥ f (z0 ) − z − z 0  + x − z ≥ f (z 0 ) − x − z 0 . Thus F(x) ≥ f (z 0 ) − x − z 0 . If x ∈ C, then for every z ∈ C, f (x) ≤ f (z) + z − x. Thus F(x) = f (x). To show that F is 1-Lipschitz, pick x, y ∈ X , ε > 0 and choose z 0 ∈ C so that f (z 0 ) + z 0 − x < F(x) + ε.

Exercises for Chapter 2

73

Then F(y) − F(x) ≤ F(y) − f (z 0 ) − z 0 − x + ε ≤ f (z 0 ) + z 0 − y − f (z 0 ) − z 0 − x + ε ≤ x − y + ε. In Exercises 2.19, 2.20, 2.21, and 2.22, μC denotes the Minkowski functional of a set C. 2.19 Let (X,  · ) be a Banach space. Show that μ B X (x) = x. Hint. Use continuity of the norm. 2.20 Let A, B be convex sets in a Banach space X . Show that if A ⊂ B then μ B ≤ μ A . Show that μc A (x) = 1c μ A (x) for c > 0. Hint. Follows from the definition. 2.21 Let C be a convex neighborhood of 0 in a real Banach space X (then μC is a non-negative positive homogeneous subadditive continuous functional on X , see Lemma 2.11). Prove the following: (i) If C is also open, then C = {x : μC (x) < 1}. If C is closed instead, then C = {x : μC (x) ≤ 1}. (ii) There is c > 0 such that μC (x) ≤ cx. (iii) If C is moreover symmetric, then μC is a continuous seminorm, that is, it is a continuous homogeneous subadditive functional. (iv) If C is moreover symmetric and bounded, then μC is a norm that is equivalent to  ·  X . In particular, it is complete, that is, (X, μC ) is a Banach space. Note that the symmetry condition is good only for the real case. In a complex normed space X we have to replace it by C being balanced. Hint. (i) It follows from Lemma 2.11. (ii) See Equation (2.2). (iii) Observing that μC (−x) = μC (x) and positive homogeneity are enough to prove μC (λx) = |λ|μC (x) for all λ ∈ R, x ∈ X . (iv) From (iii) we already have the homogeneity and the triangle inequality. We need to show that μC (x) = 0 implies x = 0 (the other direction is obvious). Indeed, μC (x) = 0 implies that x ∈ λC for all λ > 0, which by the boundedness of C only allows for x = 0. In (ii) we proved μC (x) ≤ cx, an upper estimate follows from C ⊂ d B X . The equivalence then implies completeness of the new norm. 2.22 Let K be a bounded closed convex and symmetric set in a Banach space X . Denote by Y the linear hull of K . Let | · | on Y be defined as the Minkowski functional of K . Show that (Y, | · |) is a Banach space, i.e., K is a Banach disc. For an extension of this result to the setting of locally convex spaces and for some of its consequences see Exercises 3.71, 3.72, 3.73, and 3.74.

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Hint. If (xn ) is a Cauchy sequence in (Y, | · |) it is Cauchy in X and converges, say to x0 , in X . Given a closed ball U in (Y, | · |), note that U is closed in X . As (xn ) is Cauchy in (Y, | · |), there is n 0 ∈ N such that xn − x m ∈ U for all n, m ≥ n 0 . As U is closed in X , x n − x 0 ∈ U for n ≥ n o (in particular, x0 ∈ Y ). It follows that x n → x0 in (Y, | · |). 2.23 Prove that, if n ∈ N, the dual space of a n-dimensional Banach space is again n-dimensional. Prove that the dual space of an infinite-dimensional normed space is again infinite-dimensional. Hint. Use Propositions 1.36 and 2.17—this last one for a finite index set. The infinite-dimensional assertion follows from this. 2.24 Show that if Y is a subspace of a Banach space X and X ∗ is separable then so is Y ∗ . Hint. Y ∗ is isomorphic to the separable space X ∗ /Y ⊥ . 2.25 Show that 1 is not isomorphic to a subspace of c0 . Hint. The dual of 1 is nonseparable. Use now Exercise 2.24. 2.26 Show that c0 is not isomorphic to C[0, 1]. Hint. Check the separability of their duals—Proposition 2.21. 2.27 Let X be a Banach space. (i) Show that in X ∗ we have X ⊥ = {0} and {0}⊥ = X ∗ . Show that in X we have ∗ (X )⊥ = {0} and {0}⊥ = X . (ii) Let A ⊂ B be subsets of X . Show that B ⊥ is a subspace of A⊥ . Hint. Follows from the definition. 2.28 Let X be a Banach space. Show that: (i) span(A) = (A⊥ )⊥ for A ⊂ X . (ii) span(B) ⊂ (B⊥ )⊥ for B ⊂ X ∗ . Note that in general we cannot put equality.   ⊥  (iii) A⊥ = (A⊥ )⊥ for A ⊂ X and B⊥ = (B⊥ )⊥ ⊥ for B ⊂ X ∗ . Hint. (i) Using definition, show that A ⊂ (A⊥ )⊥ . Then use that B⊥ is a closed subspace for any B ⊂ X ∗ , proving that span(A) ⊂ (A⊥ )⊥ . Take any x ∈ / span(A). subspace, by the separation theorem there is f ∈ X ∗ such Since span(A) is a closed   that f (x) > 0 and f span(A) = 0. But then f  A = 0, hence f ∈ A⊥ , also f (x) > 0, so x ∈ / (A⊥ )⊥ . (ii) Similar to (i).  ⊥ (iii) Applying (i) to A⊥ we get A⊥ ⊂ (A⊥ )⊥ . On the other hand, using  ⊥ A ⊂ (A⊥ )⊥ and the previous exercise we get (A⊥ )⊥ ⊂ A⊥ . The dual statement is proved in the same way.

2.29 Let X = R2 with the norm x = (|x1 |4 + |x2 |4 )1/4 . Calculate directly the dual norm on X ∗ using the Lagrange multipliers.

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Hint. The dual norm of (a, b) ∈ X ∗ is sup{ax1 + bx 2 : x14 + x 24 = 1}. Define F(x1 , x2 , λ) = ax1 + bx 2 − λ(x14 + x24 − 1) and multiply by x 1 and x 2 , respectively, the equations you get from ∂∂xF1 = 0 and ∂∂xF2 = 0. 2.30 Let Γ be a set and let p ∈ [1, ∞), q ∈ (1, ∞] be such that that c0 (Γ )∗ = 1 (Γ ) and  p (Γ )∗ = q (Γ ). Hint. See the proofs of Propositions 2.15, and 2.16, 2.17.

1 p

+

1 q

= 1. Show

2.31 Show that c∗ is linearly isometric to 1 . Hint. We observe that c = c0 ⊕ span{e}, where e := (1, 1, . . . ) (express x := (ξi ) ∈ c in the form x = ξ0 e + x0 with ξ0 := lim ξi and x0 ∈ c0 ). If u ∈ c∗ , i→∞

put v0 =u(e) and vi = u(ei ) for i ≥ 1. Then we have u(x) = u(ξ0 e) + u(x0 ) = ∞ ξ0 v0 + i=1 vi (ξi − ξ0 ) and (v1 , v2 , . . . ) ∈ 1 as in Proposition 2.15. Put u˜ = ∞ − v := v (v0 , v1 , . . . ), where 0 i=1 vi , and write x˜ := (ξ0 , ξ1 , . . . ). We have 0 ∞ u(x) = ξ0 v0 + i=1 vi ξi = u( ˜ x). ˜ above rule gives a continuous linear functional u on Conversely, if u˜ ∈ 1 then the  ∞ |v | sup |ξi | = u ˜ sup |ξi | = u ˜ 1 x∞ . c with u ≤ u, ˜ as |u( ˜ x)| ˜ ≤ i=0 i i≥0

i≥0

The inequality u ˜ ≤ u follows like this: Let ξi be such that |vi | = ξi vi if vi = 0 and ξi = 1 otherwise, i = 0, 1, . . . . Set x n = (ξ1 , . . . , ξn , ξ0 , ξ0 , . . . ). n ∞ n n n ˜ x˜  )| ≥ |v0 | + i=1 |vi | − i=n+1 |vi |. Since Then x ∞ = 1 and |u(x )| = |u( n ∞ n |u(x )| ≤ u, we have u ≥ |v0 | + i=1 |vi | − i=n+1 |vi |. By letting n → ∞ we get u ˜ ≤ u.   X n p we 2.32 Let p ∈ (1, ∞) and X n be Banach spaces for n ∈ N. By X := ∞ denote the normed linear space of all sequences x = {xi }i=1 , xi ∈ X i , such that   p p 1 xi  X i < ∞, with the norm x := xi  X i p .  ∗  X i q (where Show that X is a Banach space and that X ∗ is isometric to ∞ such that f ∈ X ∗ + q1 = 1) in the following sense: to f ∈ X ∗ we assign { f i }i=1 i i    ∞ and f {xi }i=1 = f i (xi ).    Xn p. Remark: Sometimes the notation  p X n will be used instead of Hint. Follow the proof for  p , which is the case of X i = R. 1 p

2.33 Prove the open mapping theorem by using the concept of convex series (Exercise 1.66) and the Baire category theorem. Hint. Let T : X → Y be a bounded linear and onto mapping between Banach spaces. According to Exercise 1.66, B X is

a CS-compact set, so T B X is again CScompact, hence CS-closed. Since Y = ∞ n=1 nT B X , the Baire category theorem ensures that ∅ = Int(T B X ). According to Exercise 1.66, Int(T B X ) = Int(T B X ). Again a “cone argument” (see Exercise 1.55) concludes that 0 ∈ Int(T B X ), so T is an open mapping. 2.34 We proved the closed graph theorem using the open mapping theorem. Now prove the open mapping principle using the closed graph theorem.

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  Hint. First prove it for one-to-one mappings using the fact that { y, T −1 (y) } is closed. For the general case, note that the quotient mapping is an open mapping by the definition of the quotient topology. ! : X/ Ker(T ) → Y 2.35 Let X, Y be normed spaces, T ∈ B(X, Y ). Show that T !(x) defined by T ˆ = T (x) is a bounded operator onto T (X ). 2.36 (i) Prove directly that if X is a Banach space and f is a non-zero linear functional on X , then f is an open mapping from X onto the scalars.   (ii) Let the operator T from c0 into c0 be defined by T (xi ) = ( 1i xi ). Is T a bounded operator? Is T an open map? Does T map c0 onto a dense subset in c0 ? Hint. (i) If f (x) = δ > 0 for some x ∈ B XO , then (−δ, δ) ⊂ f (B XO ). (ii) Yes. No. Yes (use finitely supported vectors). 2.37 Let T be an operator (not necessarily bounded) from a normed space X into a normed space Y . Show that the following are equivalent: (i) T is an open mapping. (ii) There is δ > 0 such that δ BY ⊂ T (B X ). (iii) There is M > 0 such that for every y ∈ Y there is x ∈ T −1 (y) satisfying x X ≤ MyY . Hint. (i)⇒(ii): T (B XO ) is open and contains 0; hence it contains a closed ball centered at 0. (ii)⇒(iii): Let 0 = y ∈ Y . We have δy−1 Y y ∈ δ BY (⊂ T (B X )). We can find then u ∈ B X such that δy−1 y = T u, so y = T (x), where x := yY δ −1 u. Y −1 Certainly x X ≤ MyY , where M := δ . (iii)⇒(i): If y ∈ M −1 BY there exists x ∈ X such that T x = y and x X ≤ MyY (≤ 1), so y ∈ T (B X ). This proves that M −1 BY ⊂ T (B X ). By linearity, T is open. 2.38 Let X, Y be normed spaces, T ∈ B(X, Y ). Show that if X is complete and T is an open mapping, then Y is complete. Hint. Use (iii) in the previous exercise and Exercise 1.26. 2.39 Let X, Y be Banach spaces, T ∈ B(X, Y ). Show that if T is one-to-one and BYO ⊂ T (B X ) ⊂ BY , then T is an isometry onto Y . Hint. Since BYO ⊂ T (B X ), T is onto (Exercise 2.37) and hence invertible. From T (B X ) ⊂ BY we get T  ≤ 1. Assume that there is x ∈ S X such that T (x) < x. Pick δ > 1 such that δT (x) < 1. Then T (δx) ∈ BYO ⊂ T (B X ). Thus there / B X , a contradiction must be z ∈ B X such that T (z) = T (δx) but it cannot be δx ∈ with T being one-to-one. 2.40 Let X, Y be Banach spaces and T ∈ B(X, Y ). Show that the following are equivalent: (i) T (X ) is closed.

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(ii) T is an open mapping when considered as a mapping from X onto T (X ). (iii) There is M > 0 such that for every y ∈ T (X ) there is x ∈ T −1 (y) satisfying x X ≤ MyY . Hint. (i)⇒(ii): Theorem 2.25. (ii)⇒(iii): Exercise 2.37. (iii)⇒(i): By Exercise 2.37, T : X → T (X ) is an open mapping. Now use Exercise 2.38 and Fact 1.5. 2.41 Let X, Y be Banach spaces and T ∈ B(X, Y ). Show that if T maps bounded closed sets in X onto closed sets in Y , then T (X ) is closed in Y . Hint. Assume T (x n ) → y ∈ / T (X ). Put M = Ker(T ), set dn = dist(x n , M) and find wn ∈ M such that dn ≤ xn − wn  ≤ 2dn . If {x n − wn } is bounded then T (xn −wn ) → y ∈ T (X ), since the closure of {xn −wn } is mapped onto a closed set containing y, a contradiction. Therefore we may assume that xn −wn  → ∞. Since n T (xn − wn ) → y, we have T ( xxnn −w −wn  ) → 0. By the hypothesis, M must contain x n −wn a point w from the closure of { xn −wn  } as 0 lies in the closure of the image of this   n   sequence. Fix n so that  xxnn −w −wn  − w < 1/3. Then x n − wn − x n − wn w ≤ 1 3 x n

− wn  < (2/3)dn and wn + x n − wn w ∈ M, a contradiction.

2.42 Let X and Y be Banach spaces. Then K(X, Y ) contains isomorphic copies of Y and X ∗ . Hint. T (x) = f ∗ (x)y. 2.43 Let X and Y be normed spaces. Prove that B(X, Y ) is an infinite-dimensional space if X is infinite-dimensional and Y is not reduced to {0}. Hint. The space B(X, Y ) contains an isometric copy of X ∗ . Use now Exercise 2.23. 2.44 Let T ∈ B(X, Y ). Prove the following: (i) Ker(T ) = T ∗ (Y ∗ )⊥ and Ker(T ∗ ) = T (X )⊥ . (ii) T (X ) = Ker(T ∗ )⊥ and T ∗ (Y ∗ ) ⊂ Ker(T )⊥ .   Hint. (i) Assume x ∈ T ∗ (Y ∗ )⊥ . Then for any g ∈ Y ∗ we have g T (x) = T ∗ (g)(x) = 0, hence T (x) = 0. Thus x ∈ Ker(T ). (ii) T (X ) = span T (X ) = (T (X )⊥ )⊥ = Ker(T ∗ )⊥ . !(x) 2.45 Let X, Y be normed spaces, T ∈ B(X, Y ). Consider T ˆ := T (x), where !∗ : T (X )∗ → x ∈ x, ˆ as an operator from X/ Ker(T ) into T (X ). Then we get T ⊥ (X/ Ker(T ))∗ . Using Proposition 2.6 and T (X ) = T (X )⊥ = Ker(T ∗ ) we may ∗ ∗ ! is a bounded operator from Y / Ker(T ∗ ) into Ker(T )⊥ ⊂ X ∗ . On assume that T ∗ : Y ∗ / Ker(T ∗ ) → X ∗ . Show the other hand, for T ∗ : Y ∗ → X ∗ we may consider T ∗ ∗ . ! =T that T Hint. Take any yˆ ∈ Y ∗ / Ker(T ∗ ) and x ∈ X . Then using the above identifications we obtain     ∗ ( y!∗ )(x). !(x) !∗ ( y!∗ )(x) ˆ = y!∗ T ˆ ˆ = y ∗ T (x) = T ∗ (y ∗ )(x) = T T

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2.46 Let X, Y be Banach spaces and T ∈ B(X, Y ). Show that T maps X onto a dense set in Y if and only if T ∗ maps Y ∗ one-to-one into X ∗ . Also, if T ∗ maps onto a dense set, then T is one-to-one. Hint. If T (X ) = Y , let f ∈ Y ∗ \{0} be such that f = 0 on T (X ). Then T ∗ ( f ) = 0. The other implications are straightforward. 2.47 Let X, Y be Banach spaces and T ∈ B(X, Y ). If T is one-to-one, is T ∗ necessarily onto? Hint. No, consider the identity mapping from 1 into 2 . 2.48 Let X, Y be Banach spaces and T ∈ B(X, Y ). If T is an isomorphism into Y , is T ∗ necessarily an isomorphism into X ∗ ? Hint. No, embed R into R2 . 2.49 Let X, Y be Banach spaces and T ∈ B(X, Y ). Show that: (i) T ∗ is onto if and only if T is an isomorphism into Y . (ii) T is onto if and only if T ∗ is an isomorphism into X ∗ . (iii) T (X ) is closed in Y if and only if T ∗ (Y ∗ ) is closed in X ∗ . Hint. (i) If T ∗ is onto, it is an open mapping (Theorem 2.25) and by Exercise 2.37 there is δ > 0 so that δ B X ∗ ⊂ T ∗ (BY ∗ ). Then   T (x)Y = sup y ∗ T (x) = sup T ∗ (y ∗ )(x) = y ∗ ∈BY ∗

≥

sup



x ∗ ∈δ B X ∗



y ∗ ∈BY ∗

sup



x ∗ ∈T ∗ (BY ∗ )

x ∗ (x)



x ∗ (x) = δx X

and use Exercise 1.73. If T is an isomorphism into, then T −1 is a bounded  operator  from T (X ) into X . Given x ∗ ∈ X ∗ , define y ∗ on T (X ) by y ∗ (y) = x ∗ T −1 (y) . Clearly y ∗ ∈ T (X )∗ , extend it to a functional in Y ∗ . Then T ∗ (y ∗ ) = x ∗ . (ii) If T is onto, as in (i) we find δ > 0 such that δ BY ⊂ T (B X ), then T ∗ (y ∗ ) X ∗ ≥ δy ∗ Y ∗ and use Exercise 1.73. Assume T ∗ is an isomorphism into. By Exercise 2.37 and Lemma 2.24, it is enough to find δ > 0 so that δ BY ⊂ T (B X ). Assume by contradiction that no such δ exists. Then find yn → 0 such that yn ∈ / T (B X ). The set is closed, so dn := dist(yn , T (B X )) > 0.  Fix n, set Vn = / y + BYO ( d2n ) . Then Vn is an open convex set and yn ∈ y∈T (B X )

Vn , so by Proposition 2.13 there is y ∗ ∈ Y ∗ such that |y ∗ | < 1 on Vn and y ∗ (yn ) = 1. Since T (B X ) ⊂ Vn , we get   T ∗ (y ∗ ) = sup T ∗ (y ∗ )(x) = sup y ∗ T (x) = x∈B X

x∈B X

sup



y∈T (B X )

 y ∗ (y) ≤ 1,

so y ∗  ≤ (T ∗ )−1  T ∗ (y ∗ ) ≤ (T ∗ )−1 , 1 = y ∗ (yn ) ≤ (T ∗ )−1  yn . This shows that yn  ≥ 1/(T ∗ )−1  for every n, contradicting yn → 0.

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(iii) If T (X ) is closed and q : X → X/ Ker(T ) is the canonical quo! such that T ! ◦ q, is an operator from X/ Ker(T ) onto a tient mapping, then T !∗ is an isomorphism into, in particuBanach space T (X ), hence by (ii) above, T ∗ ∗ ∗ ! ∗ (Y ∗ / Ker(T ∗ )) = T ∗ (Y ∗ ) is lar T (Y / Ker(T )) is closed. By Exercise 2.45, T closed. ! : X → T (X ). Then T !∗ (Y ∗ / Ker(T ∗ )) = If T ∗ (Y ∗ ) is closed, consider T ∗ ∗ ∗ ∗ ∗ ∗ !  T (Y / Ker(T )) = T (Y ) is closed and T is one-to-one, hence it is an iso! must be onto, that is, T (X ) = T (X ). morphism into. By (ii), T 2.50 Show that there is no T ∈ B(2 , 1 ) such that T is an onto mapping. Hint. By Exercise 2.49, T ∗ would be an isomorphism of ∞ into 2 , which is impossible as ∞ is nonseparable and 2 is separable. 2.51 Let X, Y be Banach spaces, T ∈ B(X, Y ). Show that: (i) T is an isomorphism of X onto Y if and only if T ∗ is an isomorphism of Y ∗ onto X ∗ . (ii) T is an isometry of X onto Y if and only if T ∗ is an isometry of Y ∗ onto X ∗ . Hint. (i) Follows from Exercise 2.49. (ii) If T is an isometry, then by (i), T ∗ is an isomorphism. Also T (B X ) = BY , so T ∗ (y ∗ ) = sup T ∗ (y ∗ )(x) = y ∗ . The other direction is similar. x∈B X

2.52 We have T  = T ∗  for a bounded operator on a Banach space. So, if for a sequence of bounded operators Tn we have Tn  → 0, then Tn∗  → 0. Find an example of a sequence of bounded operators Tn on a Banach space X such that Tn (x) → 0 for every x ∈ X but it is not true that Tn∗ (x ∗ ) → 0 for every x ∗ ∈ X ∗. Hint. Let Tn (x) = (xn , xn+1 , . . . ) in 2 . Then Tn∗ (x) = (0, . . . , 0, x1 , x 2 , . . . ), where x1 is on the nth place. 2.53 Let X be a normed space with two norms  · 1 and  · 2 such that X in both of them is a complete space. Assume that  · 1 is not equivalent to  · 2 . Let I1 be the identity mapping from (X,  · 1 ) onto (X,  · 2 ) and I2 be the identity mapping from (X,  · 2 ) onto (X,  · 1 ). Show that neither I1 nor I2 are continuous. Hint. The Banach open mapping theorem. 2.54 Let L be a closed subset of a compact space K . Show that C(L) is isomorphic to a quotient of C(K ).  Hint. Let T : C(K ) → C(L) be defined for f ∈ C(K ) by T ( f ) = f  L . Then T is onto by Tietze’s theorem, use Corollary 2.26. 2.55 Let X, Y be Banach spaces and T ∈ B(X, Y ). Show that if Y is separable and T is onto Y , then there is a separable closed subspace Z of X such that T (Z ) = Y . Hint. Let {yn } be dense in BY and take x n ∈ X such that T (x n ) = yn and xn  < K for some K > 0 (Corollary 2.26). Set Z = span{x n }, clearly T (Z ) ⊂ Y . By density

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of {yn }, BYO ⊂ T (K B ZO ), hence by Lemma 2.24 we have BYO ⊂ T (K B ZO ). Thus Y ⊂ T (Z ). 2.56 Let Y be a closed subspace of a Banach space X . Assume that X/Y is separable. Denote by q the canonical quotient mapping of X onto X/Y . Show that there is a separable closed subspace Z ⊂ X such that q(Z ) = X/Y . Hint. Apply the previous exercise. 2.57 Let X be a Banach space and let Y be a separable closed subspace of X ∗ . Then there is a separable closed subspace Z ⊂ X such that Y is isometric to a subspace of Z ∗ . Hint. Let { f n } be dense in SY ∗ . For every n, let {xnk }k ⊂ S X be such that f n (xnk ) → 1 as k → ∞. Put Z = span{x nk : n, k ∈ N}. 2.58 Let X be the normed space of all real-valued functions on [0, 1] with continuous derivative, endowed with the supremum norm. Define a linear mapping T from X into C[0, 1] by T ( f ) = f . Show that T has closed graph. Prove that T is not bounded. Explain why the closed graph theorem cannot be used here. Hint. The graph of T is closed: let ( f n , f n ) → ( f, g) in X ⊕ C[0, 1]. Then f n → f uniformly on [0, 1] and f n → g uniformly. Hence by a standard result of real analysis, f = g. T is not bounded: use { f n } bounded with { f n } unbounded. The space in question is not complete. 2.59 Let X be a closed subspace of C[0, 1] such that every element of X is a continuously differentiable function on [0, 1]. Show that X is finite-dimensional. Hint. Let T : X → C[0, 1] be defined for f ∈ X by T ( f ) = f . The graph of T is closed (see the previous exercise). Therefore T is continuous by the closed graph theorem. Thus for some n ∈ N we have  f ∞ ≤ n whenever f ∈ X satisfies  f ∞ ≤ 1. i for i = 0, 1, . . . , 4n. Define an operator S : X → R4n+1 by S( f ) = Let xi = 4n { f (xi )}. We claim that S is one-to-one. It is enough to show that if  f ∞ = 1, then for some i, S( f )(xi ) = 0. Assume that this is not true. If f (x) = 1 and i i+1 i , 4n ), then by the Lagrange mean value theorem we have | f (x)− f ( 4n )| = x ∈ ( 4n i 1 | f (ξ )||x − 4n | ≤ n · 4n , a contradiction. Therefore dim (X ) ≤ 4n + 1. 2.60 (Grothendieck) Let X be a closed subspace of L 2 [0, 1] whose every element belongs also to L ∞ [0, 1]. Show that dim (X ) < ∞. Hint. The identity mapping from X to (L ∞ [0, 1],  · ∞ ) has a closed graph, so . . , f n } be an for some α we get  f ∞ ≤ α f 2 for every f ∈ X . Let { f 1 , . orthonormal set in X . For every x := {x1 , . . . xn } ∈ Cn we put f x = xk f k . Then | f x (t)| ≤ α f x 2 ≤ αx2 for almost all t ∈ [0, 1] and so if Λ is a countable dense set in Cn , there exists a set of measure zero N such that | f x (t)| ≤ αx2 for every x ∈ Λ and every t ∈ [0, 1]\N . Each mapping x → f x (t) from Cn to C is linear

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81

and continuous, so | f x (t)| ≤ αx2 for all x ∈ Cn and t ∈ [0, 1]\N . In particular, | f x (t)| ≤ α for x ∈ BCn and t ∈ [0, 1]\N . The choice ( f 1 (t),

. . . , f2n (t)) gives  x := f k 22 = | f k (t)| dt ≤ α 2 . us | f k (t)|2 ≤ α 2 . Integration then gives n =  2.61 Show that the bounded linear one-to-one mapping φ from L 1 [0, 2π ] into c0 defined by T ( f ) = fˆ(n), where fˆ(n) are Fourier coefficients of f , is not onto c0 . Hint. If T were onto c0 , then by the Banach open mapping theorem, T −1 would be bounded, which is not the case as the sequence {χ{1,...,n} } shows (note that we have Dn 1 → ∞, where Dn is the Dirichlet kernel). 2.62 Show that there is a linear functional L on ∞ with the following properties: (1) L = 1, (2) if x := (xi ) ∈ c, then L(x) = lim xi , i→∞

(3) if x := (xi ) ∈ ∞ and xi ≥ 0 for all i, then L(x) ≥ 0, (4) if x := (xi ) ∈ ∞ and x = (x2 , x3 , . . . ), then L(x) = L(x ). This functional is called a Banach limit or a generalized limit. Hint. We propose several approaches. (a) For simplicity we consider only the real scalars setting. Let M be the subspace of ∞ formed by elements x −x for x ∈ ∞ and x as above. Let 1 denote the vector (1, 1, . . . ). We claim that dist(1, M) = 1. Note that 0 ∈ M and thus dist(1, M) ≤ 1. Let x ∈ ∞ . If (x −x )i ≤ 0 for any of i then 1−(x −x )∞ ≥ 1. If (x −x )i ≥ 0 for all i, then xi ≥ xi+1 for all i, meaning that lim xi exists. Therefore lim(xi − xi ) = 0 and thus 1 − (x − x ) ≥ 1. By the Hahn–Banach theorem, there is L ∈ ∗∞ with L = 1, L(1) = 1, and L(m) = 0 for all m ∈ M. This functional satisfies (1) and (4). To prove (2), it is enough to show that c0 ⊂ L −1 (0). To see this, for x ∈ ∞ we inductively define x (1) = x and x (n+1) = (x (n) ) and note that by telescopic argument we have x (n) − x ∈ M. Hence L(x) = L(x (n) ) for every x ∈ ∞ and every n. If x ∈ c0 then x (n)  → 0 and thus L(x) = 0. To show (3), assume that for some x = (xn ) we have xi ≥ 0 for all i and L(x) < 0. By scaling, we may assume that 1 ≥ xi ≥ 0 for all i. Then 1 − x∞ ≤ 1 and L(1 − x) = 1 − L(x) > 1, a contradiction with L = 1. (b) For x:= (xi ) ∈ ∞ , k ∈ N and n 1 < . . . < n k in N, put π(x; n 1 , . . . , n k ) = k lim supn k1 i=1 xn+n i and p(x) = inf{π(x; n 1 , . . . , n k ) : n 1 < . . . < n k , k ∈ N}. Then p is a convex function on ∞ . Deduce the existence of a continuous linear functional L : ∞ → R such that L ≤ p and check the sought properties of L.

Chapter 3

Weak Topologies and Banach Spaces

An indispensable tool in the study of deeper structural properties of a Banach space X is its weak topology, i.e., the topology on X of the pointwise convergence on elements of the dual space X ∗ , or the weak∗ topology on X ∗ , i.e., the topology on X ∗ of the pointwise convergence on elements of X . The topology on X ∗ of the uniform convergence on the family of all convex balanced and weakly compact subsets of X plays also an important role. All those topologies can be efficiently studied in the general framework of topological vector spaces. This allows the use of Tychonoff’s compactness theorem for weak∗ topologies in duals to Banach spaces and the results of Banach–Dieudonné type. We also study extreme points, the Choquet representation theorem, properties of James boundaries, and characterizations of weakly compact sets. We briefly discuss nonlocally convex spaces and the space of distributions. We then study the third fundamental principle of Functional Analysis—the Hahn–Banach theorem and the Banach open mapping principle being the subject of the previous chapter—, namely the Banach–Steinhaus uniform boundedness principle. Finally we introduce and study reflexive Banach spaces.

3.1 Dual Pairs, Weak Topologies Definition 3.1 Let E be a vector space over the field K ( R or C). By E # we denote the vector space of all linear functionals on E, i.e., L(E, K). This space is called the algebraic dual of E. Let E and F be vector spaces over the field K of real or complex numbers. A bilinear form on E × F is a mapping ·, · : E × F → K that is linear in each variable separately, i.e., for every e1 , e2 , e ∈ E, f 1 , f 2 , f ∈ F, α1 , α2 ∈ K, (i) α1 e1 + α2 e2 , f  = α1 e1 , f  + α2 e2 , f , and (ii) e, α1 f 1 + α2 f 2  = α1 e, f 1  + α2 e, f 2 . The choice of the notation is done in order to emphasize the symmetry between the roles played by E and F.

M. Fabian et al., Banach Space Theory, CMS Books in Mathematics, C Springer Science+Business Media, LLC 2011 DOI 10.1007/978-1-4419-7515-7_3, 

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A subset S of F separates points of E under the bilinear form ·, · if, given e ∈ E such that e, s = 0 for all s ∈ S, then e = 0. If the space E and the bilinear form are understood, we also say that S is separating. Definition 3.2 A couple of vector spaces E, F over the field K, together with a bilinear form ·, · : E × F → K, is said to form a dual pair if, under the bilinear form, E separates points of F and F separates points of E. For simplicity, we shall write E, F to denote a dual pair. Remark: Given a dual pair E, F, the mapping φ that to an element e ∈ E associates the element φ(e) ∈ F # given by φ(e)( f ) = e, f  for every f ∈ F is linear and one-to-one thanks to the fact that F separates points of E. In this way, E is identified with a (linear) subspace of F # . In the same way, F is identified with a (linear) subspace of E # . Indeed, the roles of E and F can be reversed. These natural identifications will be made in the sequel and we shall use indistinctly the notation e, f ,  f, e, e( f ), or f (e), where e ∈ E and f ∈ F, when dealing with a dual pair E, F, if there is no risk of misunderstanding. Usually, the notation f (e) or, equivalently,  f, e, will be adopted when it is important to stress that f is understood as a function of the variable e. Recall that, given"a nonempty set Γ and a family {Sγ : γ ∈ Γ } of nonempty sets Sγ , the product γ ∈Γ Sγ is the set of all elements (sγ )γ ∈Γ , where sγ ∈ Sγ for all γ ∈ Γ . A particular case is when all Sγ coincide with a certain set S. In this case the product is called the power set S Γ , and it can be identified with the set of all functions f : Γ → S. Dealing with elements in the power set, we shall adopt the functional notation f (γ ), γ ∈ Γ or, alternatively, the vector notation ( f γ )γ ∈Γ , where f γ := f (γ ) for all γ ∈ Γ . If S is a vector space, the support of an element f ∈ S Γ is the set supp( f ) := {γ ∈ Γ : f (γ ) = 0}. Some examples of dual pairs follow: 1. Let E be a vector space. Then E, E #  is a dual pair under the bilinear form  f, e := f (e) for e ∈ E and f ∈ E # . Obviously E separates points of E # . That E # separates points of E follows from a simple linear algebra argument. 2. Let X be a normed space and let X ∗ be its dual space (see Definition 1.28). Then X, X ∗  is a dual pair under the bilinear form x ∗ , x := x ∗ (x), where (x, x ∗ ) ∈ X × X ∗ . Obviously X separates points of X ∗ ; the fact that X ∗ separates points of X is a consequence of the Hahn–Banach theorem. 3. Let K be a compact space, and let C(K ) be the Banach space of all continuous scalar-valued functions on K , equipped with the supremum norm (see Definition 1.7, Proposition 1.8, and the subsequent note). For k ∈ K define the element δk in C(K )∗ by δk ( f ) = f (k) for all f ∈ C(K ). Then C(K ), span({δk : k ∈ K }), where the bilinear form is the one induced by the bilinear form on C(K ) × C(K )∗ defined in Example 1 above, is a dual pair. The fact that C(K ) separates points of span({δk : k ∈ K }) is a consequence of the Tietze–Urysohn theorem. Obviously, span({δk : k ∈ K }) separates points of C(K ).

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4. Let Γ be a nonempty set and ϕ(Γ ) be the vector space of all the finitely supΓ Γ ported vectors  in K . Then K , ϕ(Γ ) is Γa dual pair under the bilinear form x, y := γ ∈Γ xγ yγ , where x = (xγ ) ∈ K and y = (yγ ) ∈ ϕ(Γ ). We introduce now a natural topology associated to a dual pair on one of the vector spaces involved. We stress that the symmetric character of the dual pair allows us to work either in one or the other of the two vector spaces involved. Since it is a particular case of a product topology (also called topology of the pointwise convergence), let us recall here first this well-known concept. Definition 3.3 " Let {(Sγ , Tγ )}γ ∈Γ be a family of Hausdorff topological spaces. The product set γ ∈Γ Sγ becomes a (Hausdorff) topological space when equipped with the product topology T p (sometimes called the topology of the pointwise convergence). A base for this" topology is given by the family of sets (called elementary open sets) of the form γ ∈Γ Oγ , where Oγ is an open set in Sγ and each Oγ is equal to the corresponding Sγ but for a finite number of γ ∈ Γ . In particular, if all (Sγ , Tγ ) are equal (to some (S, T )) we speak of the power topological space (S Γ , T p ). " That the topology T p on γ ∈Γ Sγ is Hausdorff as soon as the topology Tγ on Sγ is Hausdorff for every γ ∈ Γ can be easily proved by looking at the way the elementary open sets are defined. " It " is standard to check that a net { fi }i∈I in γ ∈Γ Sγ T p -converges to an element f ∈ γ ∈Γ Sγ if and only if { f i (γ )}i∈I Tγ -converges to f (γ ) for every γ ∈ Γ . If S is a topological space and nothing is said on the contrary, the power set S Γ will be always considered as a topological space in the pointwise topology T p . Tychonoff’s theorem (see, e.g., [Enge, Theorem 3.2.4]) states that an arbitrary product of compact spaces is itself compact. (This statement is equivalent to Zorn’s lemma, or to the Axiom of Choice, see, e.g., [Kell].) The following simple example justifies the use of nets in convergence (an introduction to the concept of net is done in Section 17.2), since sequences are not enough in this case for describing the topology. Let Γ be an uncountable set. Let ψ(Γ ) be the subspace of RΓ of all the countably supported elements in RΓ . Then ψ(Γ ) = RΓ , while there is no sequence in ψ(Γ ) that converges to the element f ∈ RΓ defined by f (γ ) = 1 for every γ ∈ Γ . If E, F is a dual pair, we agreed to identify E with a (linear) subspace of F # , in turn a (linear) subspace of K F (see the paragraph after Definition 3.2). In this way, we have the following definition. Definition 3.4 Let E, F be a dual pair. The weak topology on E associated to the dual pair, denoted w(E, F), is the restriction to E of the topology on K F of the pointwise convergence. The following proposition gives two more descriptions of the topology w(E, F), and it is a straightforward consequence of the nature of the product, i.e., pointwise, topology on K F . Proposition 3.5 Let E, F be a dual pair. Then

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(i) A base of neighborhoods of an element e0 ∈ E in the topology w(E, F) is given by {U (e0 ; f 1 , . . . , f n ; ε) : f 1 , . . . , f n ∈ F, ε > 0, n ∈ N},

(3.1)

where U (e0 ; f 1 , . . . , f n ; ε) := {e ∈ E : |e − e0 , f i | < ε, i = 1, 2, . . . , n}. (ii) A net {ei } in E is w(E, F)-convergent to an element e ∈ E if and only if ei , f  → e, f  for every f ∈ F. Since K F is Hausdorff, the topology w(E, F) is Hausdorff, too, whenever E, F is a dual pair. If E is a vector space, we saw in Example 1 above that E, E #  is a dual pair, so we can endow E # with the topology w(E # , E). The following proposition gives a precise description of the sort of space we obtain.   Proposition 3.6 Let E be a vector space not reduced to {0}. Then, E # , w(E # , E) is linearly isomorphic to some (K B , T p ), where card(B) is the algebraic dimension of E. Proof: Let B be an algebraic basis of E. We claim that the mapping φ : (E # , w(E # , E)) → (K B , T p ) given by φ( f ) = f  B for f ∈ E # is a linear isomorphism. Linearity is clear, and the fact that B is an algebraic basis gives that φ is one-to-one and onto. If a net { fα } in E # is w(E # , E)-convergent to some f ∈ E # , then {φ( f α )} is T p -convergent to φ( f ), since f α (b) →α f (b) for all b ∈ B (see Proposition 3.5). On the other hand, if f α (b) →α f (b) for all b ∈ B, then linearity gives f α (e) →α f (e) for all e ∈ E. Again Proposition 3.5 implies that f α → f in the topology w(E # , E).

3.2 Topological Vector Spaces Definition 3.7 Let E be a vector space and let T be a Hausdorff topology on E such that the operations (x, y) ∈ E × E → x + y and (x, α) ∈ E × K → αx are continuous on E × E and E × K, respectively. Then (E, T ) is called a topological vector space. If the topology T is clear from the context, we shall write briefly E for the topological vector space (E, T ). A normed space in its norm topology is a particular instance of a topological vector space. Many of the concepts introduced in the context of normed spaces extend naturally to the more general setting of topological vector spaces. For example, a subset M of a topological vector space (E, T ) is said to be linearly dense if span(M) = E. Two topological vector spaces E and F are called linearly isomorphic (or just isomorphic) if there exists a continuous and one-to-one operator from E onto F such that its inverse mapping is also continuous (if isomorphism is

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meant only in the topological sense, with no linearity involved, we will mention this explicitly). If E is a topological vector space, for every x 1 , x2 ∈ E and a neighborhood W of x1 + x2 there are neighborhoods V1 and V2 of x1 and x 2 respectively such that V1 + V2 ⊂ W , where V1 + V2 := {x + y : x ∈ V1 , y ∈ V2 }. Also, for every x ∈ E, α ∈ K and a neighborhood W of αx there is a neighborhood V of x in E and δ > 0 such that βV ⊂ W for every |β − α| < δ, where βV := {βv : v ∈ V }. We have the following statement. Fact 3.8 Let X be a topological vector space. (i) For every a ∈ E, the translation operator Ta defined for x ∈ E by Ta (x) = x + a is a homeomorphism of E onto E. (ii) For every α ∈ K, α = 0, the multiplication operator Mα defined for x ∈ E by Mα (x) = αx is a homeomorphism of E onto E. By a local base for a topological vector space (E, T ) we mean a base of neighborhoods of the origin 0, that is, a collection B of neighborhoods of 0 such that every neighborhood of 0 contains an element of B. Lemma 3.9 If B is a local base of a topological vector space E, then every set from B contains the closure of some set from B. In particular, E has a local base consisting of closed sets. Proof: Let U ∈ B. Since (x, y) → x − y is continuous, we find an open neighborhood V of zero in E such that V − V ⊂ U , that is, V ∩ (E \ U ) + V = ∅.   Since (E \ U ) + V is an open set, we have V ∩ (E \ U ) + V = ∅, in particular V ∩ (E \ U ) = ∅ as 0 ∈ V . Thus V ⊂ U , and it is enough to choose an element B ∈ B such that B ⊂ V to obtain the conclusion. The fact that for every a ∈ E the translation operator Ta is a homeomorphism implies that, if B is a local base for the topological vector space (E, T ), then {a+ B : B ∈ B} is a base of neighborhoods of a. Symmetric and balanced sets were introduced in Chapter 1 in the context of normed spaces. Those definitions make sense in any vector space over the field K. Note that a convex set in a real vector space is symmetric (see Exercise 1.2) if and only if it is balanced. The two concepts do not agree in complex vector spaces. Proposition 3.10 Every topological vector space E has a local base consisting of balanced sets. Proof: Let U be a neighborhood of 0 in E. By the continuity of scalar multiplication, there is δ >

0 and a neighborhood V of 0 in E such that αV ⊂ U for every |α| < δ. Put W = |α|<δ αV . Then W is a balanced neighborhood of 0 in E and W ⊂ U . A topological vector space (E, T ) is an example of a uniform space (see Section 17.10). Henceforth, all concepts associated to uniformities (like system of

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vicinities, Cauchy nets, completeness, and so on) are applicable to topological vector spaces. For example, a net {xα }α∈I in E is Cauchy if for every neighborhood U of 0 in E there is α0 ∈ I such that (xα −xβ ) ∈ U whenever α0 ≤ α, β. E is complete if every Cauchy net is convergent. In the case of normed spaces, completeness can be checked just by using sequences (see Exercise 3.11). Thus, a normed space X is complete, i.e., it is a Banach space, if every Cauchy sequence in X converges to some point in X . Similarly to the case of normed spaces, every topological vector space (E, T ) can be embedded in a smallest complete topological vector space, its  T). It is unique up to linear isomorphisms. The closures in ( E,  T) completion ( E,   of the elements of a local base of (E, T ) form a local base of ( E, T ). In the same way that in the case of normed spaces (see the discussion after Proposition 1.29), to every topological vector space (E, T ) over the field C we associate a topological vector space (E R , T ) over the field R. There is a one-to-one correspondence between complex and real linear functionals that to f : E → C associates Re f . In this way, to a real linear functional u : E R → R it corresponds the complex linear functional f : E → C defined by f (x) = u(x) − iu(i x) for x ∈ E (so Re f = u). Obviously, f is continuous if and only if Re f is continuous. Given a topological vector space (E, T ) and a vector subspace G of E, clearly G becomes a topological vector space when endowed with the restriction of the topology T . Instead of introducing a new symbol for this restricted topology, we shall denote it again T , and so the topological vector subspace will be denoted (G, T ). Let" {(E γ , Tγ ) : γ ∈ Γ } be a collection of topological vector spaces. The vector space γ ∈Γ E γ becomes a topological vector space when endowed with the product topology introduced in Definition 3.3. Checking this fact is a simple exercise. In particular, if (E, T ) is a topological vector space and Γ is a nonempty set, then (E Γ , T p ) is a topological vector space, where T p denotes the product topology. Let (E, T ) be a topological vector space, and let F be a closed (linear) subspace. The set E/F, called the quotient of E and F, is the vector space of all the cosets x + F of E, x ∈ E. The mapping q : E → E/F that maps x ∈ E to the coset x + F is called the canonical quotient mapping. When the vector space E/F is endowed with the quotient topology Tq , i.e., the (Hausdorff) topology on E/F consisting ! such that q −1 (U !) ∈ T , it becomes a topological vector space, on all the sets U the quotient space (E/F, Tq ). (Observe that the elements of Tq are precisely the images by q of the elements of T .) That the topology Tq is indeed a topology and that, moreover, it makes E/F a topological vector space can be checked easily. The mapping q : E → E/F is onto, continuous and open. We provide now several natural examples of topological vector spaces. 1. A normed space in the topology defined by the norm is an example of a topological vector space. 2. Let X be a normed space, Γ a nonempty set. Then, the space (X Γ , T p ) is a topological vector space.   3. Given a dual pair E, F, the  space E,w(E, F) is a topological vector space. It is enough to recall that E, w(E, F) is just a (linear) subspace of the topo-

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logical vector space (K F , T p ). In particular, every normed space X in its weak topology, or its dual in its weak∗ topology (see Definition 3.20), is topological vector spaces. ∞ |xn | p < ∞}. The space  p , with 4. Given 0 < p < 1, put  p = {(x n ) ∈ KN : i=1 p 1/ p , x = (x ) ∈  , the topology induced by q p , where q p (x) := ( ∞ n p n=1 |x n | ) is a topological vector space. However, it is not a normed space, see Exercise 3.6. Bounded and totally bounded sets were already introduced in the setting of normed spaces. The corresponding notions for topological vector spaces are defined below. See also Section 17.10. Definition 3.11 Let A be a subset of a topological vector space E. A is called bounded if for every neighborhood U of 0 in E there is α > 0 such that αA ⊂ U. A is called totally bounded if for every neighborhood U of zero in E there is a finite set F ⊂ E such that A ⊂ F + U . Note that every totally bounded set A in a topological vector space E is bounded. Indeed, given a balanced neighborhood W of zero in E, we use the continuity of the addition to obtain a balanced neighborhood V of zero in E such that V + V ⊂ W . Since A is totally bounded, there is a finite set F ⊂ E such that A ⊂ F +V . Let F = {x1 , . . . , xn }. From the continuity of the scalar multiplication, we find 1 > δi > 0 such that δx i ∈ V whenever 0 < δ ≤ δi . Set δ = min{δi }, then δxi ∈ V for every i. From this and the balancedness of V we obtain δ A ⊂ δ F + δV ⊂ V + V ⊂ W . This shows that A is bounded in E. Obviously, every compact subset of a topological vector space is totally bounded. From the description of a local base for the topology w (see Proposition 3.5), we get easily the following result. Proposition 3.12 Let E, F be a dual pair. A subset A of E is w(E, F)-bounded if and only if supx∈A |x, f | < +∞ for every f ∈ F. The following result extends Proposition 1.36 to the class of topological vector spaces. Proposition 3.13 Let E be a topological vector space. If dim (E) = n then E is linearly homeomorphic to n2 . } be a basis of E. We define a mapping u from n2 onto E for Proof: Let {e1 , . . . , en n x := (xi ) by u(x) = i=1 xi ei . From the continuity of vector operations in E, it follows that u is a continuous linear bijection of n2 onto E. To complete the proof we need to show that u −1 is continuous. Let B1 denote the unit ball of n2 , we will show that u(B1 ) contains a neighborhood of 0 in E. By the linearity of u and of the topologies of E and n2 , this will prove continuity at all points of E. Let S1 be the unit sphere in n2 . Then S1 is compact in n2 and since u is continu/ u(S1 ). Since E is ous, u(S1 ) is compact in E. As u is one-to-one, we have that 0 ∈

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a Hausdorff space, for every s ∈ u(S1 ) there are neighborhoods Us and Vs of s and 0 respectively such that Us ∩ Vs = ∅.

p By compactness, there are s1 , . . . , s p ∈ u(S1 ) be such that u(S1 ) ⊂ i=1 Usi . p Put V = i=1 Vsi . Then V is a neighborhood of zero in E such that V ∩ u(S1 ) = ∅. From Proposition 3.10 it follows that there is a balanced neighborhood W of zero in E such that W ⊂ V and thus W ∩ u(S1 ) = ∅. We claim that W ⊂ u(B1 ). Indeed, / B1 . Then if for some w ∈ W we have w ∈ / u(B1 ), then w = u(v) for some v ∈ v v w ∈ S and u ∈ W since W is balanced, w ∈ W and v > 1. The = 1 v v v proof is complete. Corollary 3.14 Let F be a subspace of a topological vector space E. If F is finitedimensional then F is closed in E. Proof: Since n2 is a complete normed space, by Exercise 3.11 it is also complete as a topological vector space. By Proposition 3.13 the same is true for F. Let y ∈ F. Then there is a net {x α } ⊂ F such that lim x α = y. Clearly {xα } is a Cauchy net in F, so it converges to some element in F. Hence y ∈ F. Corollary 3.15 Let E be a finite-dimensional vector space. Then, all the topological vector space topologies on E coincide (i.e., they have the same family of open sets). Corollary 3.16 Let E be a finite-dimensional vector space and let F be a topological vector space. Then, every linear mapping from E into F is continuous. Proof: Let T : E → F be a linear mapping. It is enough to prove that T : E → T (E) is continuous. The result follows from Propositions 3.13 and 1.39, since T (E) is a finite-dimensional topological vector space. Proposition 3.17 Let E be a topological vector space. If E has a totally bounded neighborhood of zero, then E is finite-dimensional (and conversely). Proof: U be a totally bounded neighborhood of 0. It is also bounded, hence  −n Let ∞ 2 U n=1 forms a local base of E. By the total boundedness of U , there is a finite set A ⊂ E such that U ⊂ A + 12 U . Let F = span(A). Then U ⊂ F + 12 U . Multiplying this inclusion by 12 we have 12 U ⊂ F + 14 U . Combining these inclusions we have U ⊂ F + F + 14 U = F + 14 U . By induction we obtain U ⊂ F + 2−n U for  every n. Therefore U ⊂ (F + 2−n U ) ⊂ F. The last inclusion follows from the fact that {2−n U } forms a local base in E. Since F is closed in E by Corollary 3.14, we have U ⊂ F. Given x ∈ E, we have 0 · x ∈ U ; thus, by continuity there is δ > 0 such that δx ⊂ U ⊂ F and hence x ∈ F. Therefore F = E and E is finite-dimensional. If E is finite-dimensional, then E is linearly isomorphic to n2 for some n ∈ N (see Proposition 3.13). Use now Theorem 1.38. The following notion was introduced in Definition 1.28 for the case of a normed space.

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Definition 3.18 Let E be a topological vector space. The space E ∗ denotes the set of all continuous linear functionals on E, and it is called the (topological) dual space of E. Proposition 3.19 Let E be a topological vector space and let f be a linear functional on E. The following are equivalent: (i) f is continuous on E, that is, f ∈ E ∗ . (ii) f is continuous at some point of E. (iii) There is a neighborhood U (x 0 ) of some x0 ∈ E such that | f | is bounded on U (x 0 ). (iv) Ker f is a closed subspace of E. (v) Ker f is not dense in E. Proof: The implications (i)⇒(ii)⇒(iii) are trivial. (iii)⇒(iv): By translating we get that | f | is bounded (say by M) on some neighborhood U (0) of 0. Let (xi ) be a net in Ker f that converges to some x ∈ E. Fix n ∈ N. Since (nxi ) → nx, there exists i 0 such that (nxi − nx) ∈ U (0) for all i ≥ i 0 . In particular, | f (nxi ) − f (nx)| ≤ M, hence | f (x)| = | f (xi ) − f (x)| ≤ M/n for all i ≥ i 0 . Since n ∈ N is arbitrary, f (x) = 0, so x ∈ Ker f , hence Ker f is closed. (iv)⇒(v) is obvious. (v)⇒(iv): We have Ker f ⊂ Ker f  E. Since Ker f is a subspace of E, and Ker f is a one-codimensional subspace of E, we necessarily have Ker f = Ker f . (iv)⇒(i): Without loss of generality we may assume that E is a real topological vector space. If Ker f is closed, it follows from Exercise 3.1 that for every α ∈ R, {x ∈ E : f (x) ≤ α} and {x ∈ E : f (x) ≥ α} are both closed. We shall prove that f is continuous at 0 (and this will obviously suffice, by linearity). Assume, on the contrary, that (x i ) is a net in E that converges to 0 and that the net ( f (xi )) does not converge to 0. By passing to a subnet if necessary, we may assume that , for some ε > 0, | f (xi )| ≥ ε for all i. Since {x ∈ E : f (x) ≤ −ε} ∪ {x ∈ E : f (x) ≥ ε} is closed, we get | f (0)| ≥ ε, a contradiction. In particular, a closed hyperplane in E is the kernel of a continuous linear functional on E and conversely. Indeed, if H is a closed hyperplane of E, then (algebraically) E = H ⊕ [x0 ] for some 0 = x0 ∈ E, so the mapping f : E → K defined by f (x) = λ for x := h + λx 0 , h ∈ H , λ ∈ K, is linear and Ker f = H . By the previous proposition, f is continuous. The converse is trivial. Definition 3.20 If E is a topological vector space, the topology w(E, E ∗ ) is called the weak topology of E, and will be denoted briefly by w. The topology w(E ∗ , E) is called the weak∗ topology of E ∗ , and will be denoted briefly by w∗ . If (E, T ) is a topological vector space, the weak topology of E is weaker that the topology T . This follows directly from the description of a local base for the topology w (see Proposition 3.5, all w-neighborhoods of the form (3.1) are clearly T -open sets) or, alternatively, from the fact that if a net {xi } in E T -converges to x ∈ E then f (xi ) → f (x) for every f ∈ E ∗ (see Proposition 3.5 again).

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If E is a vector space, we already mentioned that E, E #  is a dual pair. On the other hand, if E, F is a dual pair, we can always consider F as a linear subspace of E # (see Section 3.1). From the very definition of a dual pair, the subspace F of E # separates points of E, so the topology w(E, F) on E is Hausdorff. By a slight abuse of the notation, we can also consider the case of a nonempty (otherwise arbitrary) subset F of E # (although F would not separate points of E), and the topology w(E, F) on E of the pointwise convergence on all points in F (which may not be Hausdorff). In Proposition 3.2 we shall describe the dual space of a space endowed with a weak topology of this type. We shall need the following lemma. Lemma 3.21 Let X be a vector space and let f, f1 , f 2 , . . . , f n be linear functionals n  on X . If f i−1 (0) ⊂ f −1 (0), then f is a linear combination of f 1 , . . . , f n (and i=1

conversely). We use the following fact from linear algebra. Claim: Let E 1 , E 2 , E 3 be vector spaces, let f : E 1 → E 3 and g : E 1 → E 2 be linear mappings. There is a linear mapping h : E 2 → E 3 such that f = h ◦ g if and only if g −1 (0) ⊂ f −1 (0). Proof of the Claim: Suppose that g −1 (0) ⊂ f −1 (0). Define h : g[E 1 ] → E 3 by h(g(x)) = f (x) for x ∈ E. To check the consistence, assume g(x1 ) = g(x2 ). Then (x 1 − x2 ) ∈ g −1 (0) ⊂ f −1 (0), so f (x 1 ) = f (x2 ). Extend h to a linear mapping on E 2 , clearly h(g) = f . The other implication is clear.  n Proof of Lemma  3.21: Using the Claim with E 1 := X , E 2 := R , E 3 := R, nf := f :R →R and g(x) := f 1 (x), . . . , f n (x) we get that there is a linear mapping h such that f (x) = h(g(x)). The mapping h can be written as  h(y) = αi yi for αi f i (x) for every some α1 , . . . , αn and every y = (yi ) ∈ Rn . Therefore f (x) = x ∈ X.

Let E be a vector space. For every subset F of E # , Proposition 3.22 ∗ E, w(E, F) = span(F).  ∗ Proof: By the definition of the w(E, F)-topology, F ⊂ E, w(E, F) . We need to  ∗  show that given f ∈ E, w(E, F) there are f 1 , . . . , f n such that f = αi f i . Since f is continuous in the w(E, F)-topology, there is an open neighborhood V of 0 such that | f (x)| < 1 for all x ∈ V . We may assume that there are f 1 , . . . , f n ∈ F and ε > 0 such that V := {x ∈ E : | f i (x)| < ε}. Take any z such that f i (z) = 0 for all i. Then also | f i (nz)| < ε for all n ∈ N. By the linearity this means that | f (z)| < n1 for all n ∈ N, that is, f (z) = 0. This shows  that Ker( f i ) ⊂ Ker( f ) and the proposition follows from Lemma 3.21. i

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93

Proposition 3.23 Let X be an infinite-dimensional Banach space. Then the wtopology of X is not first countable, in particular, it is not metrizable. The same is true for the w ∗ -topology of X ∗ . Proof: If {Vn } were a countable base of neighborhoods at 0, assume without loss of generality that Vn = {x ∈ X : max | f in (x)| < εn }. Since X ∗ has no countable i=1,...,Nn

Hamel basis, there is f ∈ X ∗ such that f is not in the span of all f in . We claim contain any of that U = f −1 (−1, 1) (which is a weak neighborhood of 0) does  not f i−1 (0) ⊂ f −1 (0) these Vn . Indeed, if Vn = {x : max | f i (x)| < εn } ⊂ U , then and f would be a linear combination of { f i }. The proof for X ∗ is similar. A simple consequence of the Hahn–Banach theorem is the following. Proposition 3.24 There exists a non-zero continuous functional on a topological vector space E if and only if E has a convex neighborhood of the origin which is different of the whole space E. Proof: Without loss of generality, we may assume that E is a real space. If f is a non-zero continuous functional on E, the set {x ∈ E : | f (x)| ≤ 1} is a convex neighborhood of 0 which differs from E. Conversely, assume that U is a convex neighborhood of 0 that differs from E. Its Minkowski functional pU is a continuous positively homogeneous sublinear function on E (see Exercise 3.2). Let x0 ∈ E such that pU (x0 ) = 0. Put f (λx 0 ) = λpU (x0 ) for all λ ∈ R. If λ ≥ 0, f (λx 0 ) = pU (λx0 ). Otherwise, f (λx0 ) < 0 ≤ pU (λx0 ). It follows that f (λx 0 ) ≤ pU (λx0 ) for all λ ∈ R. Use the Hahn–Banach theorem 2.1 to extend f to a linear functional  f is continuous (see Exercise 3.3). f on E such that  f ≤ pU . Then  The following example shows that there are topological vector spaces where Proposition 3.24 does not apply, i.e., topological vector spaces E for which the space E ∗ may be degenerate. Example: Fix p ∈ (0, 1). Let L p denote the vector space of Lebesgue measurable real-valued functions on [0, 1] for which 

1

q( f ) :=

| f (t)| p dt < ∞.

0

Since for a ≥ 0 and b ≥ 0 we have (a + b) p ≤ a p + b p , it follows that q( f + g) ≤ q( f ) + q(g). Therefore the formula d( f, g) = q( f − g) defines a translation invariant metric on L p in which L p is complete. This follows in the same way as for L p , p ≥ 1 (see Theorem 1.21). We claim that if O is a nonempty open convex set in L p , then O = L p . To prove this claim, assume that V = ∅ is an open convex set in L p such that 0 ∈ V . Choose f ∈ L p . We will show that f ∈ V . Choose r > 0 such that Br ⊂ V , where Br := { f ∈ L p : q( f ) < r }, then choose

x n ∈ N such that n p−1 q( f ) < r . By the continuity of the indefinite integral 0 | f | p dt, there are

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points 0 =: x0 < x1 < · · · < x n := 1 such that i = 1, . . . , n. Define for i = 1, . . . , n functions gi on [0, 1] by  gi (t) = Then gi ∈ V as q(gi ) =

xi x i−1

xi xi−1

| f | p dt = n −1 q( f ) for

n f (t) for xi−1 < t ≤ xi 0 otherwise.

|n f | p dt = n p−1 q( f ) < r and Br ⊂ V .

Since V is convex and f = n1 (g1 + · · · + gn ), it follows that f ∈ V and the claim is proved. Consequently, the only open convex neighborhood of 0 is the whole L p , so the topology is not locally convex (see Definition 3.25).  Also, let F ∈ L ∗p . As F −1 (−α, α) is a convex neighborhood of zero in L p for   every α > 0, we have F −1 (−α, α) = L p . Thus F = 0, and L ∗p = {0}.

3.3 Locally Convex Spaces We saw in the example after Proposition 3.24 that there are topological vector spaces with no continuous linear functional but 0. In order to avoid this “pathological” behavior, and motivated by Proposition 3.24, we introduce now a particular subclass. Definition 3.25 A topological vector space E is called a locally convex space if E has a local base consisting of convex sets. A topology T on E is called a locally convex topology if (E, T ) is a locally convex space. If (E, T ) is a locally convex space and F is a vector subspace of E, then F becomes a locally convex space when endowed with the restriction of the topology T . " If (E γ , Tγ )γ ∈Γ is a family of locally convex space, then ( γ ∈Γ E γ , T p ) is a locally convex space. If (E, T ) is a locally convex space and F is a closed (linear) subspace, then (E/F, Tq ) is a locally convex space. These three assertions are easy to verify. The following are natural examples of locally convex spaces. 1. Every normed space E is a locally convex space. Indeed, every ball is a convex subset of E. 2. If Γ is a nonempty set, the space " (KΓ , T p ) is a locally convex space. Indeed, elementary open sets of the form γ ∈Γ Oγ , where Oγ := K but for a finite number of them, all equal to (−ε, ε), where ε > 0, form a local base of convex subsets of KΓ . 3. Let E, F be a dual pair. Then (E, w(E, F)) is a locally convex space, since it is a linear subspace of the locally convex space (K F , T p ). In other words, every

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95

element U (0; f 1 , . . . , f n ; ε) (see Equation (3.1)) of the local base described in Proposition 3.5 is convex. 4. In particular, every normed space X in its weak topology is a locally convex space, and so it is X ∗ endowed with its weak∗ topology (see Definition 3.20). Some more examples of locally convex spaces will be given later. See in particular Section 3.10. The class of spaces in Example 4, Section 3.2, consists of non-locally convex topological vector spaces, see Exercise 3.6. Another class of non-locally convex topological vector spaces was given in the example after Proposition 3.24. Proposition 3.26 Every locally convex space E has a local base consisting of convex balanced sets. Proof: Let U be a convex neighborhood of 0 in E. First we construct W as in the proof of Proposition 3.10. Since W is balanced, then V := conv(W ) is a balanced (see Exercise 1.4) and convex neighborhood of 0 and V ⊂ U as U is convex. A topological space (S, T ) is completely regular if it is Haussdorf and, given any x ∈ S and any nonempty closed subset F of S such that x ∈ F, there exists a continuous function f : S → [0, 1] such that f (x) = 0 and f (y) = 1 for all y ∈ F. It is known that every uniform space is, as a topological space, completely regular, see, e.g., [Enge, Theorem 8.1.20]. Since every topological vector space (E, T ) has a uniformity on it whose associated topology is T (see Section 17.10), we get that every topological vector space is completely regular. We can prove this result for the class of locally convex spaces without relying on the result about uniformities. Proposition 3.27 Every locally convex space is completely regular. Proof: Let (E, T ) be a locally convex space. Certainly, the topology T is Haussdorf. It is enough to prove that, given a closed and convex neighborhood U of 0, there is a continuous function f : E → [0, 1] such that f (0) = 0 and f (x) = 1 for all x ∈ U . Let pU be the Minkowski functional of U (a continuous function according to Lemma 2.11). The sought function is f (x) := min{ pU (x), 1}, x ∈ E. A metric d(·, ·) on a vector space E is called (translation) invariant if d(x + z, y + z) = d(x, y) for all x, y, z ∈ E. Definition 3.28 A locally convex space E is called a Fréchet space if its topology is induced by a complete invariant metric. A locally convex space E is called normable if its topology is induced by a norm on E. In particular, if X is a Banach space then the metric d(x, y) := x − y makes it a Fréchet space. Proposition 3.29 Let E be a locally convex space. If E has a countable local base, then the topology of E is induced by a translation invariant metric (and conversely).

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Proof: Let {Un } be a countable local base in E. By Proposition 3.26 we can assume that every Un is a convex and balanced set. For every n, let pn be the Minkowski functional of Un . Then pn is a seminorm, i.e., pn (x + y) ≤ pn (x) + pn (y) for x, y ∈ E and pn (αx) = |α| p(x) for x ∈ E and α ∈ R (see Exercise 2.21 for the Banach space case). Define a metric d on E by d(x, y) =

∞  n=1

2−n

pn (x − y) . 1 + pn (x − y)

To verify that d is indeed a metric, note that

t 1+t

(3.2)

is increasing. Thus for x, y, z ∈ E,

pn (x − y) + pn (y − z) pn (x − z) ≤ 1 + pn (x − z) 1 + pn (x − y) + pn (y − z) pn (x − y) pn (y − z) = + 1 + pn (x − y) + pn (y − z) 1 + pn (x − y) + pn (y − z) pn (y − z) pn (x − y) + . ≤ 1 + pn (x − y) 1 + pn (y − z) Since the series in (3.2) converges uniformly and every pn is continuous (see Exercise 3.2), d is also continuous. As the metric d is translation invariant, to prove that d induces the topology of E it is enough to show that every Un contains the ball pn (x) {x ∈ E : d(x, 0) < 2−(n+1) }. If d(x, 0) < 2−(n+1) then < 12 , and since 1 + pn (x) t is increasing, we have pn (x) < 1 and thus x ∈ Un . the function 1+t Proposition 3.30 Let E be a locally convex space. E is normable if and only if E has a bounded neighborhood of zero. Proof: Clearly, if E is normable then it has a bounded neighborhood of zero, namely the open unit ball centered at the origin. Assume that U is a bounded balanced convex neighborhood of zero in E. Then the Minkowski functional p of U is a continuous norm on E which induces the topology of E. (See Exercise 3.2.) Example: The space RN (denoted also by s), i.e., the vector space of all sequences of real numbers with the product topology, is a separable Fréchet space that is not normable, see Exercise 3.12 (for another example, see Fact 3.76). In Section 12.4 we will discuss the result that all infinite-dimensional separable Banach spaces are (non-linearly) homeomorphic to the space RN . Theorem 3.31 Let E be a locally convex space. Let F be a (not necessarily closed) subspace of E. Let f : F → K be a continuous linear functional. Then there exists a continuous linear functional  f : E → K that extends f .

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97

Proof: Assume first that E is real. Put B = {r ∈ R : r ≤ 1}. The set V := f −1 (B) is a neighborhood of 0 in F. There exists a closed convex and balanced neighborhood U of 0 in E such that U ∩ F ⊂ V . Let pV : F → R be the Minkowski functional of V on F. If y ∈ λV for some λ > 0, then y = λv for some v ∈ V , hence f (y) = λ f (v) ≤ λ. It follows that f (y) ≤ pV (y) (≤ pU (y)) for all y ∈ F, where pU : E → R is the Minkowski functional of U on E. It is enough to use Theorem 2.1 to conclude that there exists a linear functional  f : E → R that extends f is continuous by Lemma 2.10, f and such that  f (x) ≤ pU (x) for all x ∈ E (so  that holds in this context, too). If E is complex, the previous argument gives an extension of Re f to a continf uous real functional u on E R , and u is the real part of a (continuous) functional  that extends f (see the paragraph in page 88). Note that this extension theorem does not work in general for topological vector spaces. Indeed, we can define a non-zero functional on a given one-dimensional subspace of any non-zero topological vector space. By Corollary 3.16, this functional is continuous. If it would be always possible to extend such a functional to the whole space, then the dual space will never be reduced to 0. The example in Section 3.2 proves that in general this is not the case. The following separation results are the topological vector space version of those that were formulated in the context of Banach spaces (Theorem 2.12 and Proposition 2.13). Theorem 3.32 Let E be a topological vector space. (i) If A is a nonempty open convex set in E and x0 ∈ E \ A, then there is f ∈ E ∗ such that Re f (a) < Re f (x 0 ) for all a ∈ A. (ii) If E is a locally convex space, C a nonempty closed convex set in E, and x 0 ∈ E \ C, then there exists f ∈ E ∗ such that supc∈C Re f (c) < Re f (x 0 ). Proof: Without loss of generality, we may assume that E is a real space. (i) By considering a shift of A we may also assume that 0 ∈ A. The Minkowski functional μ A of A is a non-negative continuous positively homogeneous and subadditive functional on E, and A = {x ∈ E : μ A (x) < 1} (see Exercise 3.2). Define f on span{x0 } by f (αx0 ) = αμ A (x0 ) for α ∈ R and proceed like in the proof of Proposition 3.24 to obtain a linear functional on E (denoted by f again) that agrees with f on span{x0 } and such that f ≤ μ A . Then f ∈ E ∗ (see Exercise 3.3) and we have f (a) ≤ μ A (a) < 1 ≤ μ A (x0 ) = f (x0 ) for every a ∈ A. (ii) Assume now that E is locally convex. Let U be an open convex and balanced neighborhood of 0 such that (x 0 + U ) ∩ C = ∅. Since x 0 ∈ C + U , and C + U is an open and convex set, by (i) we can find f ∈ E ∗ such that f (c + u) < f (x0 ) for all c ∈ C and u ∈ U . Then supc∈C ≤ f (x 0 ) − f (u) for all u ∈ U . It is enough to choose u ∈ U such that f (u) > 0 to obtain the conclusion. The following corollary is immediate. Corollary 3.33 Let E be a locally convex space and x, y ∈ E. If x = y then there is f ∈ E ∗ such that f (x) = f (y). In other words, E, E ∗  is a dual pair. In particular, E ∗ = ∅.

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We introduced in Definition 3.20 the weak∗ topology (in short, w ∗ ) in the dual of a locally convex space E. The space (E ∗ , w ∗ ) is a topological vector space (even more, a locally convex space) whose dual space is E, and all the separation theorems formulated above in the context of topological (and locally convex) vector spaces hold true. For example, we have the following result. E∗

Corollary 3.34 Let E be a locally convex space. Then, given a nonempty w∗ -closed convex subset C of E ∗ and an element x0∗ ∈ E ∗ \ C, there exists x ∈ E such that supx ∗ ∈C Rex ∗ , x < x0∗ , x. Another useful separation result is the following. Theorem 3.35 Let E be a topological vector space. Let A, B be two convex subsets of E such that A ∩ B = ∅. Then (i) If A is open, then there exist f ∈ E ∗ such that Re f (a) < infb∈B Re f (b) for all a ∈ A, b ∈ B. (ii) If E is a locally convex space, A is closed and B is compact, then there exists f ∈ E ∗ and λ ∈ R such that Re f (a) < λ < infb∈B Re f (b) for all a ∈ A. Proof: Without loss of generality, we may assume that E is a real space. (i) The set A − B is an open convex subset of E and 0 ∈ A − B. Use (i) in Theorem 3.32 to obtain f ∈ E ∗ such that f (a − b) < f (0) (= 0) for all a ∈ A, b ∈ B. Then f (a) < f (b) for all a ∈ A, b ∈ B, and this implies f (a) ≤ λ := infb∈B f (b) for all a ∈ A. If f (a0 ) = λ for some a0 ∈ A, by the fact that A is open we can find a ∈ A such that f (a) > λ, a contradiction. (ii) Assume now that E is a locally convex space. Given b ∈ B there exists an open convex and balanced neighborhood Ub of 0 such that (b + Ub ) ∩ A = ∅. Since {b + Ub : b ∈ B} is an open there exists

cover of B and B is compact,

a finite set B0 ⊂ B such that B ⊂ b∈B0 b + Ub . Set U = b∈B0 Ub . Then U is an open convex and balanced neighborhood of 0 and (A + U ) ∩ B = ∅. By (ii) in Theorem 3.32 applied to (A + U ) and B, there exists f ∈ E ∗ such that f (a + u) < infb∈B f (b) for all a ∈ A, u ∈ U . It is enough to choose some u ∈ U with f (u) > 0 to obtain the conclusion. For further separation result we refer, e.g., to the remark following Theorem 3.122.

3.4 Polarity Definition 3.36 Let E, F be a dual pair. Let A be a subset of E. By the polar of A in F we mean A◦ := { f ∈ F : Rex, f  ≤ 1 for all x ∈ A}. The set { f ∈ F : |x, f | ≤ 1, for all x ∈ A} is called the absolute polar of A. It easily follows that, for every A ⊂ E, A◦ is a convex and w(F, E)-closed subset of F that contains 0. If A is balanced, so is the corresponding polar set; we then have that the polar and the absolute polar sets of A coincide.

3.4

Polarity

99

If E is a locally convex space, and A ⊂ E ∗ , it is customary to write A◦ for the polar set of A in E, i.e., the polar set of A with respect to the dual pair E, E ∗ . If we are dealing with a dual pair, say E, F, we shall not do this distinction, so, for example, the polar of a set A ⊂ G will be denoted A◦◦ . Note, too, that the polar set of a linear subspace G of a vector space E with respect to a dual pair E, F is the annihilator G ⊥ of G in F (introduced for Banach spaces right before Proposition 2.6), i.e., G ◦ = G ⊥ := { f ∈ F : g, f  = 0 for all g ∈ G}. Observe that if X is a normed space, then the polar set (B X )◦ of B X in X ∗ is just B X ∗ , and the polar set (B X ∗ )◦ of B X ∗ in X is simply B X . Theorem 3.37 (Alaoglu, Bourbaki) Let E be a locally convex space. If A ⊂ E is a neighborhood of 0, then A◦ is w∗ -compact. Proof: Assume first that A is an convex balanced neighborhood of 0. Consider the dual pair E, E #  (see Example 1 in Section 3.1). The set A◦ (⊂ E # ) is bounded (and closed) in (E # , w(E # , E)). Indeed, given x ∈ E there exists λ > 0 such that λx ∈ A. Then, sup f ∈A◦ | f (x)| ≤ λ−1 and we can use Proposition 3.12. Since (E # , w(E # , E)) is linearly homeomorphic to (K B , T p ) for some B (see Proposition 3.6), the Tychonoff theorem implies that A◦ is compact in (E # , w(E # , E)). Observe now that A◦ (⊂ E # ) is indeed a subset of E ∗ (hence a (w ∗ =) w(E ∗ , E)compact set). This follows from Proposition 3.19, since every f ∈ A◦ is bounded on A, so f ∈ E ∗ . This finishes the proof for the case of an convex balanced set A. For a general neighborhood U of 0 we find a convex balanced neighborhood A of 0 such that A ⊂ U (Proposition 3.26). Then U ◦ in E ∗ is a w ∗ -closed subset of the w ∗ -compact set A◦ and the theorem is proved. In particular, consider a dual pair E, F and A ⊂ E a w(E, F)-neighborhood of 0; then A◦ is w(F, E)-compact in F. If X is a Banach space, then B X◦ , that is, B X ∗ , is a w ∗ -compact subset of X ∗ . Theorem 3.38 (Bipolar theorem) Let E, F be a dual pair, and denote w := w(E, F). Then, for every subset A of E we have A◦◦ (:= (A◦ )◦ ) = convw (A ∪ {0}). In particular, let E be a locally convex space. If A ⊂ E is w-closed, convex and 0 ∈ A, then A = (A◦ )◦ . Similarly, if B ⊂ E ∗ is w∗ -closed, convex and 0 ∈ B, then B = (B◦ )◦ . Proof: Let C = convw (A ∪ {0}). Since A ⊂ A◦◦ , 0 ∈ A◦◦ and A◦◦ is w-closed and convex, we have C ⊂ A◦◦ . Assume that there is x ∈ A◦◦ \C. By Theorem 3.32 there is f ∈ F such that Re f (x) > supC (Re f ). Since 0 ∈ C, we have supC (Re f ) ≥ 0, so by scaling we may assume that supC (Re f ) ≤ 1 and Re f (x) > 1. Then sup A (Re f ) ≤ 1, hence f ∈ A◦ , also x ∈ A◦◦ , so we have Re f (x) ≤ 1, a contradiction. We now show an application of the bipolar Theorem 3.38. Proposition 3.39 Let E, F be a dual pair. Let G be a linear subspace of F. Then G separates points of E (and so E, G is a dual pair) if and only if G is w(F, E)-dense in F.

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Proof: Recall that, with respect to a dual pair, the polar set of a linear subspace is its annihilator. If G separates points of E then G ◦ = {0}, so G ◦◦ = F. By w(F,E) Theorem 3.38, F = G . If G is w(F, E)-dense in F and for some e ∈ E we have e, g = 0 for all g ∈ G, then e, f  = 0 for all f ∈ F, since e is, by Proposition 3.2, w(F, E)-continuous on F. This implies that e = 0 and so G separates points of E.

3.5 Topologies Compatible with a Dual Pair Let E, F be a dual pair. Let M be a family of w(F, E)-bounded subsets of F

that separates points of E, i.e., if e, f  = 0 for all f ∈ M M, then e = 0. A locally convex topology on E associated to the family M can be defined by giving a local base. It consists of all sets of the form M ◦ , where M ∈ M, then their scalar multiples, and finally taking the finite intersections of these. It is simple to prove that the topology so generated is indeed a (Hausdorff) locally convex topology on E. It is called the topology on E of the uniform convergence on the sets of M or, briefly, the topology on E of the uniform convergence on M. Let E, F be a dual pair. A family G of w(F, E)-bounded sets in F is called saturated if G is closed with respect to taking subsets, scalar multiples and balanced convex w(F, E)-closed hulls of the union of any two of its members. If a saturated family G separates points of E, then a local base for the topology TG is given by the polars of the w(F, E)-closed convex and balanced elements of G. The saturation of a family M of w(F, E)-bounded subsets of F is, by definition, the smallest saturated family of subsets of F that contains M. It is simple to prove that if S(M) is the saturation of a family M of w(F, E)-bounded subsets of F that separates points of E, then the topologies TM and TS (M) on E coincide. Examples: 1. Let E, F be a dual pair. The topology w(E, F) on E coincides with the topology TG , where G is the family of all w(F, E)-closed convex balanced hulls of finite subsets of F and their subsets. This family is certainly saturated and covers F. 2. Let X be a normed space. Then, the norm topology on X is the topology TG , where G is the family {r B X ∗ : r ∈ R, r > 0} and their subsets, i.e., the family of all bounded subsets of X ∗ . This is again a saturated family of subsets of X ∗ that covers X ∗ . 3. Let E, F be a dual pair. The family of all convex balanced σ (F, E)-compact subsets of F and their subsets is saturated and covers F (see Exercise 3.14 and Definition 3.43). Definition 3.40 Let E, F be a dual pair. A topology T on E that makes E a topological vector space is called compatible with the dual pair if (E, T )∗ = F.

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Topologies Compatible with a Dual Pair

101

Note that Proposition 3.2 implies that if (E, T ) is a locally convex space, then the weak topology w on E is compatible with the dual pair E, E ∗ , and so it is the topology w ∗ on E ∗ . The following result characterizes the topologies compatible with a dual pair. Theorem 3.41 (Mackey, Arens) Let E, F be a dual pair. A locally convex topology T on E is compatible with the dual pair E, F if and only if T is the topology of uniform convergence on some saturated family G that covers F and consists of w(F, E)-compact convex balanced sets in F and their subsets. Proof: Let G be a saturated family satisfying the assumptions. If f ∈ F, then f ∈ G for some G ∈ G convex, balanced, and w(F, E)-closed. Then | f | ≤ 1 on G ◦ , hence by Proposition 3.19, f ∈ (E, TG )∗ . Conversely, if f ∈ (E, TG )∗ , then there is G ∈ G, convex, balanced, and w(F, E)-closed such that | f | ≤ 1 on G ◦ , i.e., f ∈ G ◦◦ . By the bipolar theorem, f ∈ G ◦◦ = G ⊂ F. Now assume that a locally convex topology T on E is such that (E, T )∗ = F. Define G = {G : G ⊂ U ◦ , U ∈ U} , where U is the family of all closed convex balanced T -neighborhood of 0. Note that U ◦ is a w(F, E)-compact set in F by Alaoglu’s theorem and it is also balanced and convex. The separation theorem shows that the w(F, E)-closed convex and balanced hull of U1◦ ∩ U2◦ is (U1 ∩ U2 )◦ for any U1 and U2 in U. This proves that G is a saturated family. If f ∈ F, then | f | ≤ 1 on some U ∈ U, so f ∈ U ◦ ∈ G, hence G covers F. We will show that T = TG . Let U be a T -neighborhood of 0. By Lemma 3.9 and Proposition 3.26 we may assume that U is T -closed, convex, and balanced. Thus U ◦ ∈ G. Since U = U ◦◦ by the bipolar theorem, U is a TG -neighborhood of 0. Let U be a TG -neighborhood of 0. We may assume that U = V ◦◦ , where V is some closed convex balanced T -neighborhood of 0. Then U = V (so U is a T -neighborhood of 0). In the proof of Theorem 3.41 is shown the following result. Recall the definition of an equicontinuous set of functions (see Section 17.10). It is obvious that a subset M of the dual E ∗ of a locally convex space (E, T ) is uniformly equicontinuous if and only if it equicontinuous, if and only if it is equicontinuous at 0, if and only if it is a subset of the polar of a closed convex and balanced neighborhood of 0 in (E, T ). Proposition 3.42 The topology of a locally convex space coincides with the topology TG , where G is the (saturated) family of all the equicontinuous subsets of the dual space E ∗ . A consequence of Theorem 3.41 is that every locally convex topology on E compatible with a dual pair E, F is stronger than w(E, F). Definition 3.43 Let E, F be a dual pair. The locally convex topology on E of uniform convergence on all w(F, E)-compact convex balanced sets in F is called the Mackey topology on E associated to the dual pair E, F, and is denoted μ(E, F). Observe that the family of all w(F, E)-compact convex balanced sets in F and their subsets is saturated (see Example 3 after Definition 3.40).

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From Proposition 3.2, Theorem 3.41 and the note after this theorem we obtain the following corollary. Corollary 3.44 Let E, F be a dual pair. Then the strongest locally convex topology T on E compatible with the dual pair E, F is the Mackey topology μ(E, F), and the weakest locally convex topology w on E compatible with this same dual pair is the weak topology w(E, F) on E. If E is a locally convex space, a closed (respectively, open) half-space in E is a set of the form {x ∈ E : Re f (x) ≤ α} (respectively, {x ∈ E : Re f (x) < α}), where f ∈ E ∗ and α ∈ R. As a consequence of Theorem 3.32, the following result holds. Theorem 3.45 (Mazur) The closure C of a convex set C in a locally convex space E T T is the intersection of all closed half-spaces that contain C. In particular, C 1 = C 2 if T1 and T2 are two compatible locally convex topologies topology! compatible with a dual pair on E. If E, F is a dual pair, all compatible locally convex topologies on E define the same family of closed convex sets. Proof: Let x 0 ∈ E\C. By Theorem 3.32 we can find f ∈ X ∗ such that Re f (x0 ) > sup{Re f (x) : x ∈ C}. If Re f (x0 ) > α > sup{Re f (x) : x ∈ C}, the closed half-space {x ∈ E : Re f (x) ≤ α} contains C and excludes x0 . This proves one inclusion. For the other, just recall that an arbitrary intersection of closed sets is itself closed. The particular case and the last statement follow from the fact that all compatible topologies give, by the very definition, the same dual space. Remark: In particular, if X is a normed space, every  · -closed convex subset of X is w-closed. so the family of all w-closed convex subsets of X and the family of all  · -closed convex subsets of X coincide. Another way to formulate this is to say that the w-closure and the  · -closure of any convex subset of a normed space coincide. For non-convex sets in normed spaces, the norm and weak closure may differ, see Exercise 3.46. w In an infinite-dimensional locally convex space (E, T ), xi → x does not in T

w

general imply xi → x (for instance, en → 0 in c0 or  p , p > 1, and en  = 1 for all n ∈ N). However, we have the following. Corollary 3.46 Let (E, T ) be a locally convex space. If a net {x i }i∈I in E is wconvergent to some x ∈ E, there exists a net {y j } j∈J in conv {xα } that T -converges to x. If E is a normed space, we can find a sequence in conv {xα } that ·-converges to x. w

Proof: Define C = convT {xi : i ∈ I }. Since xi → x, we have x ∈ C C is convex and T -closed. The statement follows.

w

= C, since

Lemma 3.47 Let E be a locally convex space with a countable local base. Assume that A is a balanced convex set in E such that for every bounded set B ⊂ E there is α > 0 satisfying α B ⊂ A. Then A is a neighborhood of zero in E.

3.6

Topologies of Subspaces and Quotients

103

Proof: Let {Un } be a countable local base for X . We may assume that Un+1 ⊂ Un for every n. By contradiction, assume that there is no n ∈ N for which Un ⊂ n A. Then / n A. Clearly lim x n = 0. Since every convergent there are xn ∈ Un such that x n ∈ sequence is bounded, there is K > 0 such that x n ∈ K A for every n and thus / n A for all n. x n ∈ n A for n ≥ K . This is a contradiction with x n ∈ ∗ be the dual space of Let (E, T ) be a complex locally convex space. Let E R its associated real locally convex space. If A is a nonempty subset of E, the set ∗ (see Definition 3.36). Re A◦ := {Re f : f ∈ A◦ } is the polar set of A in E R This implies, in particular, that the topology T can be obtained as a TG -topology by using the saturated family G in E ∗ as in the proof of Theorem 3.41 or the family ∗ (a saturated family in E ∗ ). In particular, Re G of all the sets Re G, G ∈ G, in E R R ∗ ∗ the topologies w(E, E ) and μ(E, E ) coincide, respectively, with the topologies ∗ ) and μ(E, E ∗ ). The same happens in the case of a normed complex space w(E, E R R and its norm topology.

3.6 Topologies of Subspaces and Quotients Let E 1 , E 2  be a dual pair. Let G be a linear subspace of E 1 . The couple (G, E 2 /G ⊥ ) forms a dual pair under the bilinear form ·, · defined for all g ∈ G, eˆ2 ∈ E 2 /G ⊥ , by g, eˆ2  = g, e2 , where e2 ∈ eˆ2 . This bilinear form is well defined and, clearly, under ·, · the space G separates points of E 2 /G ⊥ and the space E 2 /G ⊥ separates points of G. Let q : E 2 → E 2 /G ⊥ be the canonical quotient mapping. Let M be a saturated family consisting of w(E 2 , E 1 )-bounded subsets of E 2 and such that M M separates points of E 1 . This family defines a locally convex  := {q(M) : M ∈ M} consists of topology TM on E 1 . Observe that the family M

⊥ ⊥ w(E 2 /G , G)-bounded subsets of E 2 /G and M∈M q(M) separates points of G. On G we may, then, consider two natural locally convex topologies associated to the family M: (i) the restriction TM of TM to G, and (ii) the topology T M  on G. Proposition 3.48 Let (E, T ) be a locally convex spaces and let G be a (not necessarily closed) subspace of E. Then, in the situation described in the two paragraphs above, we have TM = T M . Proof: It is enough to check that M ◦ ∩ G = (q(M))◦ for every M ∈ M. Then,  is a TM -neighborhood of 0 in G, there exists a TM -neighborhood of 0 in E 1 if U . We can find a closed convex and balanced set M ∈ M such that U ∩ G = U ◦ , so U  is such that M ⊂ U . It follows that (q(M))◦ = M ◦ ∩ G ⊂ U ∩ G = U a TM  -neighborhood of 0. On the other hand, given M ∈ M, again by the fact that M -neighborhood of 0. This proves M ◦ ∩ G = (q(M))◦ we get that (q(M))◦ is a T M -neighborhood of 0. that, conversely, every T M  neighborhood of 0 is a T Corollary 3.49 Let E be a locally convex space and let G be a (non necessarily closed) subspace. Then, the topology w(E, E ∗ ) on E induces the topology w(G, G ∗ ) on G.

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Let (E, T ) be a locally convex space and let M be the family of all the equicontinuous subsets of E ∗ . This family is saturated and covers E ∗ (see the paragraph preceding Proposition 3.42). We know from the same Proposition 3.42 that T = TM . Let G be a closed subspace of E. The natural embedding I from (E/G)∗ into E ∗ is an algebraic isomorphism from (E/G)∗ onto G ⊥ (⊂ E ∗ ). That I is onto follows from the fact that, given e∗ ∈ G ⊥ , a mapping (e) ˆ ∗ from E/G into K can be defined by ˆ = e∗ , e, for all e ∈ eˆ ∈ E/G. (e) ˆ ∗ , e The mapping (e) ˆ ∗ is continuous: indeed, given ε > 0 there exists a neighborhood of ˆ ∗ , q(u)| ≤ 1 for all u ∈ U . 0 in E such that |e∗ , u| ≤ 1 for all u ∈ U . Then |(e) Continuity follows from the fact that q(U ) is a neighborhood of 0 in (E/G). From now on we shall identify (E/G)∗ and G ⊥ in the sense described. On E/G we may consider two natural topologies: the quotient topology of T ,  is the family of all elements in denoted by T!, and the topology TM  , where M ⊥  M that are in G . This family M consists of w(G ⊥ , E/G)-bounded sets, is also saturated with respect to the pair E/G, G ⊥ , and covers G ⊥ , so it defines a locally convex topology TM  on E/G. Proposition 3.50 Let (E, T ) be a locally convex space. Let G be a closed subspace of E. Then, in the situation described in the two paragraphs above, the quotient topology T! of T on E/G coincides with the topology TM . ! be a T!-equicontinuous subset of (E/G)∗ . There exists an open convex Proof: Let M   ! ⊂ q(U ) ◦ . Therefore and balanced neighborhood U of 0 in (X, T )) such that M  Conversely, if M  then there is a T ! ∈ M. ! ∈ M, M ⊂ U ◦ ∩ G ⊥ , that is, M     ◦ ⊥ ! ⊂ q(U ) ◦ ; ! neighborhood U of 0 such that M ⊂ (U ∩ G ) = (U + G)⊥ , so M ! is T !-equicontinuous. thus, M Corollary 3.51 Let E be a locally convex space, and let G be a closed subspace. Then, the topology w(E, E ∗ ) induces on G the topology w(G, G ∗ ). Likewise, the topology μ(E, E ∗ ) induces on G the topology μ(G, G ∗ ).

3.7 Weak Compactness A subset A of a topological space T is said to be relatively compact if A is compact. In the case of topological vector spaces, this is equivalent to the fact that every net in A has a subnet that converges to some point in A, see Exercise 3.10. A is said to be (relatively) countably compact if every sequence in A has a subnet that converges to some point in A (in A). A is said to be (relatively) sequentially compact if every sequence in A has a subsequence that converges to some point in A (in A). Obviously, every (relatively) compact set is (relatively) countably compact, and every (relatively) sequentially compact set is (relatively) countably compact.

3.7

Weak Compactness

105

Let K be a compact topological space. Let C p (K ) denote the space of all real continuous functions on K endowed with the restriction of the topology T p on R K of the pointwise convergence on K (see Definition 3.3). The restriction of this topology to C(K ) will be denoted also by T p . Theorem 3.52 (Eberlein, Grothendieck, see, e.g., [Todo1]) Let K be a compact topological space. In C p (K ), every relatively countably compact set is relatively compact (and conversely). Proof: The space C p (K ) is a subspace of (R K , T p ). Let M be a (nonempty) relatively countably compact subset of C p (K ). Note that { f (k) : f ∈ M} is bounded for every k ∈ K ; then, by Tychonoff’s theorem, M RK

RK

is compact in (R K , T p ).

Claim: M ⊂ C(K ). In order to prove the claim we proceed by contradiction. Assume that there exists RK

that is discontinuous (at some point k0 ∈ K ). We can find then a net f ∈ M {ti }i∈I in K and some ε > 0 such that ti → k0 , and | f (ti ) − f (k0 )| ≥ ε for all i ∈ I.

(3.3)

Step 1: ⎧ RK ⎪ ∃ f ∈ M so that | f (k ) − f (k )| < ε/2 (since f ∈ M ). ⎪ 1 1 0 0 ⎨ Put U1 = {k ∈ K : | f 1 (k) − f 1 (k0 )| ≤ ε/2}, ⎪ a closed neighborhood of k0 (since f 1 is continuous). ⎪ ⎩ ∃ k1 ∈ U1 such that | f (k1 ) − f (k0 )| ≥ ε (since (3.3) holds). Step 2: ⎧ RK 2 ⎪ ⎪ ∃ f 2 ∈ M so that | f 2 (ki ) − f (ki )| < ε/2 , i = 0, 1 (since f ∈ M ). ⎨ Put U2 = {k ∈ K : | fi (k) − f i (k0 )| ≤ ε/22 , i = 1, 2}, ⎪ a closed neighborhood of k0 (since fi , i = 1, 2, are continuous). ⎪ ⎩ ∃ k2 ∈ U2 so that | f (k2 ) − f (k0 )| ≥ ε (since (3.3) holds). Step n: ⎧ RK n ⎪ (n1) ⎪ ⎨ ∃ f n ∈ M so that | f n (ki ) − f (ki )| < ε/2 ,n0 ≤ i < n (since f ∈ M ). Put Un = {k ∈ K : | fi (k) − f i (k0 )| ≤ ε/2 , 1 ≤ i ≤ n}, ⎪ a closed neighborhood of k0 (since f i , 1 ≤ i ≤ n, are continuous). (n2) ⎪ ⎩ (n3) ∃ kn ∈ Un so that | f (kn ) − f (k0 )| ≥ ε (since (3.3) holds).

Note that Un+1 ⊂ Un for all n ∈ N. Also, for i = 1, 2, . . . , n and n ∈ N, | f n+1 (ki ) − f (k0 )| ≥ | f (ki ) − f (k0 )| − | f n+1 (ki ) − f (ki )| ≥ ε − ε/2n+1 > ε/2.

(n4)

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3 Weak Topologies and Banach Spaces

The sequence {kn } has a cluster point k∞ ∈ K . Observe that k∞ ∈ f n (k∞ ) = f n (k0 ), for all n ∈ N.



n∈N Un ,

hence (3.4)

The sequence { f n } has a cluster point g in C p (K ). In particular, g(k0 ) is a cluster point of { f n (k0 )}. There exists a subsequence of { f n } (denoted again { fn }) such that f n (k0 ) → g(k0 ). In view of (3.4), f n (k∞ ) → g(k0 ). At the same time, g(k∞ ) is a cluster point of the sequence { f n (k∞ )}, and we get g(k0 ) = g(k∞ ).

(3.5)

Fixing i ∈ {0, 1, 2, . . .} and letting n → ∞ in (n1) we get g(ki ) = f (ki ), for all i = 0, 1, 2, . . .

(3.6)

Fixing i ∈ N and letting n → ∞ in (n4) we get |g(ki ) − f (k0 )| ≥ ε/2. By (3.6) we get |g(ki ) − g(k0 )| ≥ ε/2 for all i ∈ N. By (3.5) we get |g(ki ) − g(k∞ )| ≥ ε/2 for all i ∈ N. This contradicts the continuity of g at x ∞ and proves the Claim.  Since M

RK

is T p -compact, the conclusion follows.

Definition 3.53 A topological space T is said to have countable tightness if for every A ⊂ T and x ∈ A there is a countable set S ⊂ A such that x ∈ S. A topological space T is called a Fréchet–Urysohn space if for every A ⊂ T and x ∈ A there is a sequence {xn } ⊂ A such that x n → x. A topological space T is called angelic if for every relatively countably compact subset A of T , the two following properties hold: (i) A is relatively compact, and (ii) given x ∈ A there is a sequence {xn } ⊂ A such that xn → x. Obviously, if K is a compact topological space, K is a Fréchet–Urysohn space if and only if it is angelic. √ Example: Let xn = nen ∈ 2 , where en is the standard nth unit vector in 2 . w √ Then 0 ∈ { nen } and there is no subsequence of xn that weakly converges to 0 (Exercise 3.55). Thus 2 in its weak topology is not a Fréchet–Urysohn space. It is simple to prove that in angelic spaces the classes of relatively compact, relatively countably compact and relatively sequentially compact sets coincide; the same is true if the word “relatively” is dropped (Exercise 3.9). Example: Let Γ be uncountable. The element a = (1, 1, . . . ) ∈ ∞ (Γ ) is in the closure of Bc0 (Γ ) in the w ∗ -topology by Goldstine’s theorem. However, there is w∗

w∗

no countable S ⊂ Bc0 (Γ ) such that a ∈ S . Indeed, assume that a ∈ S for some countable S ⊂ Bc0 (Γ ) . Every element of c0 (Γ ) has a countable support and therefore there is a countable set M ⊂ Γ such that s(γ ) = 0 for every γ ∈ / M and w∗ / M, a contradiction. every s ∈ S. Then for a ∈ S we have a(γ ) = 0 for γ ∈ Therefore B∞ (Γ ) (= B∗1 (Γ ) ) in its w ∗ -topology does not have countable tightness.

3.7

Weak Compactness

107

Theorem 3.54 (Kaplansky, see, e.g., [Koth, §24.1(6)]) Let (E, T ) be a locally con∗ vex space such that E ∗ = ∞ n=1 Mn , where each Mn is a w(E , E)-compact subset ∗ of E . Then, (E, w) has countable tightness. Proof: Let A be an arbitrary nonempty subset of E. Take a ∈ A. We shall prove that there exists a countable subset A0 of A such that a ∈ A0 . Without loss of generality, we may assume that the sequence {Mn }∞ n=1 is increasing. Fix n, m, k ∈ N. Given x 1∗ , . . . , xk∗ ∈ Mn , define a w-neighborhood of a of the form {x ∈ E : |x ∗ − i, x − a| < 1/m, i = 1, 2, . . . , k}. Associate to this an element a ∈ A in this neighborhood. The set V := {x ∗ ∈ Mn : |x ∗ , a − a| < 1/m} is open in (Mn , w ∗ ) and contains xi∗ , i = 1, 2, . . . , k, so V × .(k) . . ×V is a neighborhood of (x1∗ , . . . , x k∗ ) k ∗ ∗ in Mn . This is done for every (x1 , . . . , xk ) ∈ Mnk ; we obtain an open covering of the set Mnk (a compact set in the product topology of w ∗ , by Tychonoff’s theorem). Therefore we can find a finite subcovering, which leads to a finite subset An,m,k

of A. The sought set is A0 := n,m,k An,m.k . It is clear that this set satisfies the requirement. ! = {δk : k ∈ K } (⊂ C(K )∗ ), where Let K be a compact topological space. Put K δk is the Dirac functional defined by the element k ∈ K . Lemma 3.55 Let K be a compact topological space. Then K is homeomorphic ! of all Dirac functionals endowed with the topology induced by w∗ in to the set K ∗ C(K ) . Proof: The mapping δ : K → C(K )∗ that associates δk to each k ∈ K is continuous for the topology of K and the w∗ -topology of C(K )∗ . Indeed, if (ki ) is a net in K that converges to some k ∈ K , then f (ki ) → f (k) for every f ∈ C(K ). It is one-to-one thanks to Urysohn’s lemma. Since K is compact, δ is a homeomorphism !). onto δ(K ) (= K !) (⊂ C(K )∗ ). The space L separates points of C(K ) (in fact, it is Let L = span( K a 1-norming subspace of C(K )∗ ), hence we may consider the dual pair C(K ), L, see Example 3 after Definition 3.2 and the paragraph preceding Exercise 3.88. The following lemma is almost obvious. Lemma 3.56 The topology w(C(K ), L) on C(K ) coincides with the topology T p of the pointwise convergence on C(K ). Proof: Let { f i } be a net in C(K ) that T p -converges to some f ∈ C(K ). Fix n ∈ N, scalars λ j , and k j ∈ K , for j = 1, 2, . . . , n. Then  n  j=1

 λ j δk j , f i =

n  j=1

i

λ j δk j , f i  →

n  j=1

 λ j δk j , f  =

n 

 λ j δk j , f .

j=1

On the other hand, if { fi } is a net in C(K ) that w(C(K ), L)-converges to some f ∈ C(K ), certainly it T p -converges to f .

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3 Weak Topologies and Banach Spaces

Corollary 3.57 Let K be a compact topological space. Then, the space C p (K ) has countable tightness. Proof: The space L is the topological dual of the locally convex space C(K ), w(C(K ), L) . Observe that, for n, m ∈ N, the sets

K n,m :=

⎧ n ⎨ ⎩

j=1

λ j δk j :

n 

|λ j | ≤ m, k1 , . . . , kn ∈ K

j=1

⎫ ⎬ ⎭

 

are w L , C(K ) -compact, due to Lemma 3.55, and that L = n,m∈N K n,m . It is enough to apply Theorem 3.54 and to note that Lemma 3.56 ensures that   C(K ), w(C(K ), L) = C p (K ). Theorem 3.58 (Eberlein, Šmulyan, Grothendieck, see, e.g., [Todo1]) For a given nonempty compact topological space K , the space C p (K ) is angelic. In particular, the following classes of subsets of C p (K ) coincide: (relatively) countably compact, (relatively) sequentially compact, and (relatively) compact. T

Proof: Let A ⊂ C(K ) be T p -relatively countably compact, and let a ∈ A p . T A0 p .

Let By Corollary 3.57 there exists a countable set A0 ⊂ A such that a ∈ F = spanT p (A0 ), a T p -separable subspace of C(K ). We consider the dual pair !), see the paragraph before Lemma 3.56. Recall C(K ), L, where L := span( K   that T p on C(K ) is just the topology w C(K ), L (Lemma 3.56). The space F carries the topology induced by T p , denoted again T p . The dual space of (F, T p ) is L/F ⊥ . The canonical quotient mapping     q : L , w(L , C(K )) → L/F ⊥ , w(L/F ⊥ , F)   !) is compact in the space L/F ⊥ , w(L/F ⊥ , F) , is continuous. Therefore q( K   ! := {δk : k ∈ K}, see Lemma 3.55. Since the space F, w(F, L/F ⊥ ) is where K !) is w(L/F ⊥ , F)-metrizable, hence w(L/F ⊥ , F)-separable. separable, the set q( K ! Certainly, q( K ) separates points of F, hence it is w(L/F ⊥ , F)-linearly dense in   T L/F ⊥ , so L/F ⊥ , w(L/F ⊥ , F) is separable. The set A0 p (⊂ F) is T p -compact, according to Theorem 3.52. It follows that it is T p -metrizable. Henceforth there exists a sequence {an } in A0 that T p -converges to a. The angelicity of C p (K ) follows from this and from Theorem 3.52. The identity of the classes of T p -compacta mentioned in the statement follows from Exercise 3.9. For the case that the sets involved in Theorem 3.58 would be uniformly bounded, see Theorem 3.139.

3.8

Extreme Points, Krein–Milman Theorem

109

3.8 Extreme Points, Krein–Milman Theorem We will now study extreme points in real locally convex spaces. This subject, in the case of Banach spaces, will be considered again in Subsection 3.11.8. Some stronger results in this direction will be discussed in Chapters 8 and 11. Definition 3.59 Let E be a topological vector space and C be a nonempty convex subset of E. We say that a point x 0 ∈ C is an extreme point of C if x1 = x2 = x0 whenever x1 ∈ C, x 2 ∈ C and x0 = (1/2)(x1 + x 2 ). This is a particular case of the following concept. Definition 3.60 Let E be a locally convex space and S a nonempty subset of E. A real linear manifold H ⊂ E (i.e., the translate of a real linear subspace of E) is called a supporting manifold of S if H ∩ S = ∅ and if every open interval (x, y) ⊂ S which contains a point of H satisfies (x, y) ⊂ H . An extreme point of S is a supporting manifold of dimension 0 of S. A simple example of a supporting manifold is given by the following lemma. Lemma 3.61 Let S be a nonempty subset of a locally convex space E. Let f be a real linear functional f : E → R that attains the supremum on S. Then the set H := {y ∈ E : f (y) = s}, where s := supx∈S f (x), is a supporting manifold of S. In particular, if H ∩ S is a single point, this point is an extreme point of S. Proof: Certainly, H ∩ S = ∅. Let (x, y) ⊂ S be an open interval such that (x, y) ∩ H = ∅. Choose x0 , y0 ∈ (x, y) such that x0 = y0 and [x0 , y0 ] ∩ H = ∅. If [x0 , y0 ] ⊂ H then either f (x0 ) > s or f (y0 ) > s, in both cases a contradiction. It follows that (x, y) ⊂ H . The supporting manifolds of a set S have a certain “transitive” property. Lemma 3.62 Let H be a supporting manifold of a nonempty set S in a locally convex space E. A real linear manifold H1 ⊂ H is a supporting manifold of S if and only in it is a supporting manifold of H ∩ S. Proof: Assume first that H1 is a supporting manifold of S. Then H1 ∩ (H ∩ S) = H1 ∩ S = ∅. Let (x, y) ⊂ H ∩ S (⊂ S) be an open interval such that (x, y)∩ H1 = ∅. Then (x, y) ⊂ H1 and so H1 is a supporting manifold of H ∩ S. Conversely, assume that H1 is a supporting manifold of H ∩ S. Let (x, y) ⊂ S be an open interval such that (x, y) ∩ H1 = ∅. Then (x, y) ∩ H = ∅, hence (x, y) ⊂ H , so (x, y) ⊂ H ∩ S. It follows that (x, y) ⊂ H1 and so H1 is a supporting manifold of S. Lemma 3.63 Let K be a nonempty compact subset of a locally convex space E. Let H = {Hγ }γ ∈Γ be a family of closed supporting manifolds of K directed by inclusion, i.e., given Hγ 1 and Hγ2 in H, there exists Hγ3 ∈ H such that Hγ3 ⊂ Hγ1 ∩ Hγ2 . Then H := γ ∈Γ Hγ is a closed supporting manifold of K .

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3 Weak Topologies and Banach Spaces

Proof: Let (x, y) ⊂ K be an open interval such that (x, y) ∩ H = ∅. Then, for every γ ∈ Γ we have (x, y) ∩ Hγ = ∅. Since Hγ is a supporting manifold of K we get (x, y) ⊂ Hγ . This is true for all γ ∈ Γ , so (x, y) ⊂ H . We claim that H ∩ K = ∅. In order to prove the claim, define a partial order in Γ by γ1 ≤ γ2 whenever Hγ1 ⊃ Hγ2 . With this order, Γ is directed upward. By choosing, for every γ ∈ Γ , some xγ ∈ Hγ ∩ K , we obtain a net {x γ }Γ in K . It has a cluster point in K (that belongs also to H ), and this proves the claim. It follows that H is a supporting manifold of K . Proposition 3.64 Let K be a nonempty compact subset of a locally convex space E. Let H be a closed supporting manifold of K . Then H contains at least one extreme point of K . Proof: By Lemmas 3.62 and 3.63, the (nonempty) family of closed supporting manifolds of K contained in H is directed by inclusion and inductive (i.e., every decreasing chain has a least element). By Zorn lemma, it has a minimal element, say H1 . Assume that H1 does not reduces to a point. By translating, we may always assume 0 ∈ H1 . The set H1 ∩ K is a compact nonempty subset of the locally convex space H1 . Let H0 ⊂ H1 be a closed hyperplane of H1 that supports H1 ∩ K (use the compactness). Then, Lemma 3.62 proves that H0 ( H1 ) is a closed supporting manifold of K . This violates the minimality of H1 . Since H1 reduces to a point and it is a supporting manifold of K , this point is an extreme point of K . Theorem 3.65 (Krein–Milman) Let E be a locally convex space and let K be a nonempty convex and compact subset of E. Then (i) K has at least an extreme point. (ii) K = conv (Ext(K )), where Ext(K ) is the set of all extreme points of K . Proof: (i) Every real continuous linear functional f : E → R attains its supremum on K . By Lemma 3.61, K has a supporting manifold. By Proposition 3.64, K has an extreme point. (ii) Assume that there exists x 0 ∈ K \conv (Ext(K )). By Theorem 3.35 (ii), we can find a real continuous linear functional f such that f (x0 ) > supx∈conv (Ext(K )) f (x). The set {y ∈ E : f (y) = supx∈K f (x)} is, according to Lemma 3.61, a supporting manifold of K ; by Proposition 3.64, it has at least one extreme point of K . This is a contradiction. Remark: If 0 < p < 1, then, in the L p -space introduced in the example after Proposition 3.24, there is a convex compact set with no extreme point ([Robe1] and [Robe2]). This is not in contradiction with Theorem 3.65, since, in this case, L p is not a locally convex space. The following result is a kind of converse of Theorem 3.65. Theorem 3.66 (Milman) Let E be a locally convex space and B be a nonempty subset of E such that C := conv (B) is compact. Then Ext(C) ⊂ B (and so every extreme point of C is also an extreme point of B).

3.8

Extreme Points, Krein–Milman Theorem

111

Proof: It is enough to prove that every extreme point of C lies in B + U , where U is an arbitrary closed convex and balanced neighborhood of 0 in E. Fix such a set U . Since B is totally bounded, it can be covered by a finite number of sets xi + U , where xi ∈ B for i = 1, 2, . . . , n. The sets Bi := conv (B ∩ (xi + U )) are convex n It follows easily and compact, hence conv ( i=1 Bi ) is also convex and

n compact. n that C = conv ( i=1 Bi ). If e ∈ Ext(C), then e = λ b , where bi ∈ Bi , i i i=1 n λi = 1. Necessarily e = bi for some bi . λi ≥ 0 for all i = 1, 2, . . . , n, and i=1 This proves that Ext(C) ⊂ B + U . For a proof of Theorem 3.66 for the weak topology of a locally convex space, based on Lemma 3.69, see Exercise 3.145. Let E and F be two locally convex spaces, and T : E → F be a continuous linear mapping. If S is a nonempty subset of E and x 0 ∈ Ext(S), it is not always true that T x 0 ∈ Ext(T (S)) (see Exercise 3.21). It is easy to prove that if H is a closed supporting manifold of T (S) then T −1 (H ) is a (closed) supporting manifold of S (see Exercise 3.22). The following result is a consequence of this observation and Lemma 3.64. Proposition 3.67 Let E and F be two locally convex spaces, T : E → F be a continuous linear mapping and K be a nonempty compact subset of E. Then, every extreme point of T (K ) is the image of an extreme point of K . Definition 3.68 Let C be a set in a locally convex space X . A slice of C is a nonempty intersection of C with an open half-space of X . We observed that finite intersections of slices form a basis of the weak topology. Due to the remark at the end of Section 3.5, this is true both in the real or complex case. The definition of the concept of supporting manifold (in particular, extreme point) (see Definition 3.60) uses only real linear manifolds and segments (x, y) defined by using only real scalars. In this section we will assume that all spaces considered are real. Lemma 3.69 (Choquet) Let C be a w-compact convex set in a locally convex space X . For every x ∈ Ext(C), the slices of C containing x form a neighborhood base of x in the relative weak topology of C. Proof: Let V be a neighborhood of x in the relative weak topology of C of the V˜i := Vi ∩ C and Vi are form V = V˜1 ∩ · · · ∩ V˜k , where V˜i are slices of C, i.e., k  open half-spaces in X (see Fig. 3.1). Then x ∈ / i=1 (X \Vi ) ∩ C . Consequently 

k  x∈ / conv i=1 (X \Vi ) ∩ C as x is an extreme point of C (Exercise 3.127). As the convex hull of a finite number of weakly convex  sets is weakly compact 

k compact (X \Vk ) ∩ C . By Theorem 2.12, there (Exercise 3.14), we have that x ∈ / conv i=1

k is f ∈ X ∗ and α ∈ R such that f (x) > α > sup f (x) : x ∈ i=1 (X \Vi ) ∩ C . Then the slice C ∩ {x ∈ X : f (x) > α} contains x and is contained in V .

112

3 Weak Topologies and Banach Spaces X \ V1

Fig. 3.1 The proof of Choquet’s Lemma 3.69

x

X \ V2

{x : f (x) = α}

C

X \ V3

3.9 Representation and Compactness In the case of finite-dimensional topological vector space, Krein–Milman Theorem 3.65 can be strengthened. This is the content of the following result. Proposition 3.70 (Carathéodory, see, e.g., [Rudi3]) Let D be a compact convex subset of an n-dimensional topological vector space E. Every x ∈ D is a convex combination of at most n + 1 extreme points of D. Proof: By induction on n. For n = 1 we have D = [a, b] and the statement follows. Assume that the result holds for n − 1. Consider D ⊂ E with dim (E) = n and Int(D) = ∅ (Exercise 3.23). If x is a boundary point of D, let H be a supporting hyperplane of D such that x ∈ H ∩ D. By the induction hypothesis, x is a convex combination of at most n extreme points of H ∩ D which are extreme points of D (see the proof of the Krein–Milman theorem). If x ∈ Int(D), choose an extreme point z 0 ∈ D and let y be another boundary point of D that lies on the line going through x and z 0 . By the same argument as above, y  is a convex combination of n , . . . , z of D. Thus y = at most n extreme points z 1 n i=1 αi z i with αi ≥ 0 and n i=1 αi = 1. n ααi z i + (1 − α)z 0 , Now x = αy + (1 − α)z 0 for some α ∈ (0, 1), so x = i=1 which is a convex combination of n + 1 extreme points z 0 , . . . , z n . A related argument was suggested in the Hint of Exercise 1.65. We will now reformulate Carathéodory’s theorem in the language of integral representation. Let D be a compact convex subset of an n-dimensional topological vector space E. For y ∈ D, let δ y be the Dirac measure at y. Let x ∈ D be given. By Carathéodory’s k theorem, there are extreme k points αi = 1 such that x = i=1 αi xi . x1 , . . . , xk of D, k ≤ n + 1, and αi ≥ 0, i=1 k Put μ = i=1 αi δxi . Then μ is a probability measure on D. If f is a continuous function on D, we have       f dμ = f d( αi δxi ) = αi f dδxi = αi f (xi ). D

D

D



If f is linear, we have f (x) = f (

αi xi ) =



αi f (xi ) =

D

f dμ.

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Representation and Compactness

113

This motivates the following definition. Definition 3.71 Let D be a compact convex subset of a locally convex space E. We say that a probability measure μ on D represents x ∈ D if for all f ∈ E ∗ ,  f (x) =

f dμ. D

If S is a Borel subset of D, we say that a measure μ is supported by S if μ(D\S) = 0. Proposition 3.72 Let D be a convex compact set in a locally convex space E. For every x ∈ D there is a probability measure μ on D supported by Ext(D) that represents x. = D. Hence is a net Proof: By the Krein–Milman theorem, n αconv(Ext(D))  there α α α α {yα } such that yα → x and yα = α x , where α ≥ 0, α = 1, and i i i=1 i i  αiα δxiα . By xiα ∈ Ext(D). We represent each yα by a probability measure μα = the Riesz representation theorem, the set of all probability measures on the compact set Ext(D) is identified with a w ∗ -compact set in BC(Ext(D))∗ . Therefore there is a subnet {yβ } of {yα } convergent in the w∗ -topology to a probability measure μ on Ext(D). If f is a continuous linear functional on E, then its restriction to Ext(D) is in C(Ext(D)) and thus we have   f dμβ = f dμ. f (x) = lim f (yβ ) = lim Ext(D)

Ext(D)

To finish the proof, we extend μ to μ˜ on D by μ(B) ˜ = μ(B ∩ Ext(D)) for every Borel set B in D. A natural question is whether a representing measure is supported by Ext(D) alone, as in Carathéodory’s theorem. One of the difficulties is that Ext(D) may not in general be a Borel set ([Phel1]). However, if D is metrizable, we have the following result. Lemma 3.73 Let D be a compact convex set in a locally convex space E. If D is metrizable then the set of extreme points of D is a G δ set in D. Proof: Let d be a metric on D that induces the topology of D. For n ∈ N, put  Fn = x ∈ D : x =

y+z 2 ,

y, z ∈ D, d(y, z) ≥

1 n



.

k Then Fn is a closed set in D. Indeed, let xk ∈ Fn and lim x k = x, where xk = yk +z 2 1 for yk , z k ∈ D, d(yk , z k ) ≥ n . Since D is metrizable and compact, we may assume

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3 Weak Topologies and Banach Spaces

that lim yk = y and lim z k = z. Thus we have x = y+z 2 . Moreover, d(yk , z k ) → d(y, z), so d(y, z) ≥ n1 . This shows that x ∈ Fn , proving that Fn is closed. To conclude the proof, observe that x ∈ D is not an extreme point of D if and only if x ∈ Fn for some n. Theorem 3.74 (Choquet representation theorem, see, e.g., [Phel1]) Let D be a metrizable compact convex set in a locally convex space E. For every x 0 ∈ D there is a probability measure μ on D supported by Ext(D) that represents x0 . Proof: Note that Ext(D) is a Borel set by Lemma 3.73. Since D is a metrizable compact set, the space C(D) of all continuous functions on D with the supremum norm  ·  is separable (Lemma 3.102). Let A denote the subspace of C(D) formed by all continuous functions f on D such that f (αx + βy) = α f (x) + β f (y) for every x, y ∈ D and α, β ≥ 0 with α + β = 1 and αx + βy ∈ D. Note that f ∈ A need not, in general, be a restriction onto D of some F(x) = k + ϕ(x) for ϕ ∈ E ∗ (Exercise Let S A = { f ∈ A :  f  = 1}, let {h n } be a dense set in S A . Put  3.26). −n 2 h= ∞ n=1 2 h n . We claim that h is a strictly convex continuous function on D. Let x = y ∈ D be given. Choose n such that h n (x) = h n (y). Using the affine property of h n and the fact that the real function t → t 2 is strictly convex, for α ∈ (0, 1) we have h 2n (αx + (1 − α)y) < αh 2n (x) + (1 − α)h 2n (y). Therefore h(αx + (1 − α)y) = 2−n h 2n (αx + (1 − α)y) + ≤

2−n h 2n (αx

+ (1 − α)y) +





2− j h 2j (αx + (1 − α)y)

j =n

2

−j

j =n

< α2−n h 2n (x) + (1 − α)2−n h 2n (y) +

  2 αh j (x) + (1 − α)h 2j (y) 

α2− j h 2j (x) +

j =n



(1 − α)2− j h 2j (y)

j =n

= αh(x) + (1 − α)h(y). Let Y be the subspace of C(D) spanned by A and h, i.e., Y = A ⊕ Rh. For f ∈ C(D) the concave envelope f¯ : D → R is defined by f¯(x) = inf{g(x) : g ∈ A, g ≥ f }. We list a few basic properties of f¯: (1) f¯ is concave, bounded, upper semicontinuous and thus Borel measurable, (2) f ≤ f¯, and f = f¯ if f is concave, ¯ f + g = f¯ + g if g ∈ A; r f = r f¯ if r > 0, (3) f + g ≤ f¯ + g; (4) | f¯ − g| ¯ ≤  f − g. ¯ (to see (4), write f¯ = f − g + g ≤ f − g + g¯ ≤  f − g + g).

3.10

The Space of Distributions

115

Define a positively homogeneous subadditive functional p on C(D) by p(g) = ¯ 0 ) for g ∈ A and r ∈ R. g(x ¯ 0 ) and define ϕ ∈ Y ∗ by ϕ(g + r h) = g(x 0 ) + r h(x ¯ 0) ≤ Note that ϕ is dominated by p on Y . We only need to verify that g(x0 ) + r h(x g + r h(x0 ), which follows from (3) if r > 0. If r < 0, then g + r h = g + r h ≥ g + r h¯ by the concavity of g + r h. By the Hahn–Banach theorem, there is an extension m of ϕ onto the whole C(D) such that m(g) ≤ g(x ¯ 0 ) for every g ∈ C(D). If g ∈ C(D) and g ≤ 0, then 0 ≥ g(x ¯ 0 ) ≥ m(g). Thus if g ≤ 1, then g −1 ≤ 0 and m(g) ≤ m(1). Since m(1) = ϕ(1) = 1, we obtain that m is a norm-one functional on C(D). By Riesz’ measure μ

representation theorem, there is a probability

¯ 0) on D such that m(g) = D g dμ for every g ∈ C(D). Note that D h dμ = h(x and g(x 0 ) = m(x0 ) = D g dμ for every g ∈ A, in particular for every g ∈ E ∗ . To conclude

it is enough to show that:

the proof, (a) D h dμ = D h¯ dμ, ¯ (b) {x ∈ D : h(x) = h(x)} ⊂ Ext(D). ¯ ¯ Indeed, since h ≥ h on D, from (a) we get μ{x ∈ D : h(x) < h(x)} = 0. From   ¯ (b) it follows that D\ Ext(D) ⊂ {x ∈ D : h(x) > h(x)} and thus μ D\ Ext(D) = 0. To verify (a), write 

 D

 g dμ : g ∈ A, g ≥ h¯ = inf g dμ : g ∈ A, g ≥ h D D  ¯ 0 ) = ϕ(h) = m(h) = = inf{g(x 0 ) : g ∈ A, g ≥ h} = h(x h dμ.

h¯ dμ ≤ inf

h dμ ≤ D



D

To prove (b), assume that x ∈ D\ Ext(D). Then x = y = z. Hence using concavity of h¯ ¯ ¯ + 12 h(z) ≤ h¯ h(x) < 12 h(y) + 12 h(z) ≤ 12 h(y)

y+z 2 ,

 y+z  2

where y, z ∈ D,

¯ = h(x).

3.10 The Space of Distributions Let Ω be a nonempty open set in Rn . We will now construct the space of distributions and establish some of its properties. Choose any sequence {K n }n∈N of compact subsets of Ω such that K n ⊂ Int(K n+1 ) and K n = Ω; for example K n = {x : dist(x, Rn \Ω) ≥ n1 , x2 ≤ n}, where  · 2 is the canonical norm in n2 . Definition 3.75 Let C(Ω) denote the vector space of all real-valued continuous functions on Ω with the topology κ determined by the local base {Un }, where Un :=



f ∈ C(Ω) : sup | f (x)| < x∈K n

1 n



.

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3 Weak Topologies and Banach Spaces

  Fact 3.76 C(Ω), κ is a locally convex space whose topology does not depend on the choice of {K n }. The topology κ is the topology of uniform convergence on every compact subset of Ω. That is, f α → f in κ if and only if f α → f uniformly on every compact subset of Ω.  C(Ω), κ is a Fréchet space but the topology κ is not normable. Proof: Local convexity of κ is routine to verify. Let {K n } be another family of compact sets with the required properties. Then for n ∈ N, {Int(K m )}m is an open cover of K n , hence using K n ⊂ Int(K n+1 ), there is K m such that K n ⊂ Int(K m ) ⊂ K m . This shows the independence of κ on the choice of {K n } and also that κ is the topology of uniform convergence on every  compact  set in Ω. By a standard argument we see that C(Ω), κ is a Fréchet space. Since every Un contains functions f for which sup K n+1 | f | can be arbitrary large, we have that no Un is bounded. Therefore (C(Ω), κ) is not normable.  In particular, any set of the form f ∈ C(Ω) : supx∈K | f (x)| < ε with ε > 0 and K a compact set in Ω is an open neighborhood of 0 in C(Ω). Recall that for every multi-index α := (α1 , . . . , αn ), where αi are non-negative integers, the corresponding derivative is defined by Dα =



 ∂ α1 ∂ x1

···



The degree of the derivative is defined as |α| =

 ∂ αn . ∂ xn 

αi .

Definition 3.77 By C ∞ (Ω) we denote the set of all real-valued functions f on Ω such that D α f ∈ C(Ω) for every α. A topology on C ∞ (Ω) is defined by its local base {Un,α } Un,α :=



f ∈ C ∞ (Ω) : sup |D α f (x)| < x∈K n

1 . n

Note that the base {Un,α } is countable. Fact 3.78 C ∞ (Ω) is a Fréchet space whose topology does not depend on the choice of {K n }. f β → f in C ∞ (Ω) if and only if D α f β → D α f uniformly on K for every compact set K in Ω and every multi-index α. C ∞ (Ω) has the Heine–Borel property. Recall that a space has the Heine–Borel property if all of its closed bounded subsets are compact.

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The Space of Distributions

117

Proof: We will show the Heine–Borel property. Let A be a bounded and closed set in C ∞ (Ω). Then for every n there is an Mn > 0 such that |D α f (x)| ≤ Mn for all |α| ≤ n, f ∈ A and x ∈ K n . Therefore {D α f : f ∈ A} is equicontinuous on K n−1 for |α| ≤ n − 1. From the Arzelà–Ascoli theorem and the Cantor diagonal process, it follows that every sequence { f i } ⊂ A contains a subsequence f i j for which all D α f i j converge uniformly on every compact subset of Ω. Hence fi j is convergent in C ∞ (Ω). This proves that A is compact. The differentiation mapping D α : C ∞ (Ω) → C ∞ (Ω) is continuous and linear. In this section we shall depart from the previously defined concept of support and use the symbol supp( f ) and the word support of a real-valued function on Ω to denote the set {x ∈ Ω : f (x) = 0}. Definition 3.79 Let K be a compact set in Ω such that Int(K ) = ∅. We define D K as the closed subspace of C ∞ (Ω) formed by functions f such that supp( f ) ⊂ K . We define D(Ω) as the union of all D K (Ω) for compact sets K as above. Note that D(Ω) is not a complete subspace of C ∞ (Ω). For Ω = R this is seen as follows: Choose ϕ ∈ C ∞ (R) with supp(ϕ) ⊂ [0, 1] and ϕ > 0 on (0, 1). For m ∈ N put ϕm (x) = ϕ(x − 1) + 12 ϕ(x − 2) + · · · +

1 m ϕ(x

− m).

Then {ϕm } is a Cauchy sequence in D(R) ⊂ C ∞ (Ω) and lim ϕm does not have compact support. We now introduce a locally convex topology on D(Ω). For every compact set K ⊂ Ω with nonempty interior, let τ K denote the topology of D K inherited from C ∞ (Ω). Let B denote the collection of all convex balanced sets W ⊂ D(Ω) such that D K ∩ W is τ K -open in D K for every such compact set K ⊂ Ω. Then B is a local base for a locally convex topology on D(Ω), which will be denoted by τ . Proposition 3.80 (i) The topology on D K inherited from D(Ω) coincides with τ K for every compact set K ⊂ Ω with Int(K ) = ∅. (ii) If A is a bounded subset of D(Ω), then there is a compact set K ⊂ Ω such that A ⊂ DK . (iii) D(Ω) has the Heine–Borel property. (iv) Every Cauchy sequence in D(Ω) is convergent. (v) D(Ω) is not metrizable. Proof: (Sketch) To prove (i) it is enough to show that given an open set O in D K , there exists an open set V in D(Ω) such that O = V ∩ D K . Let {Un } be the local base in D K as above and pn be the Minkowski functional of Un . For every ϕ ∈ O there is n and δ > 0 such that {ψ ∈ D K : pn (ψ − ϕ) < δ} ⊂ O.

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3 Weak Topologies and Banach Spaces

Put Wϕ = {ψ ∈ D(Ω) : pn (ψ) < δ}. Then Wϕ ∈ B and D K ∩ (ϕ + Wϕ ) = ϕ + (D K ∩ Wϕ ) ⊂ O. Let V =

ϕ∈O

(ϕ + Wϕ ). Then V has the desired property.

(ii) Assume that A is bounded in D(Ω) and A lies in no D K . Then there are ϕm ∈ A and distinct points xm ∈ Ω without a limit point in Ω such that ϕm (xm ) = 0, m ∈ N. Let W be the set of all ϕ ∈ D(Ω) that satisfy |ϕ(x m )| < m1 |ϕm (xm )| for all m ∈ N. Note that W is convex and balanced. Since every compact set K contains only finitely many xm , it is easy to see that D K ∩ W is open in D K . Therefore W ∈ B. / mW , no multiple of A is contained in W . Therefore A is not bounded Since ϕm ∈ in D(Ω). (iii) follows from (ii) and the fact that D K has the Heine–Borel property. (iv) follows from the fact that every Cauchy sequence is bounded, from (ii) and from the completeness of D K . (v) It is easy to see that D K has no interior point in D(Ω). Assume that D(Ω) is metrizable. Then D(Ω) is a complete metrizable space as any Cauchy sequence in D(Ω) is convergent. Note that Dk are closed subspaces of D(Ω). This leads to a contradiction with the Baire category theorem. We now characterize continuity of linear mappings on D(Ω). Proposition 3.81 A linear functional f on D(Ω) is continuous if and only if f is sequentially continuous if and only if the restriction of f to every D K is continuous. Similarly, a linear mapping T : D(Ω) → E, where E is a locally convex space, is continuous if and only if T D is continuous for every D K . K

Proof: Assume  that the restrictions of f to all D K are continuous. Consider the set V := f −1 (−δ, δ) . Then V is balanced and convex. By Proposition 3.80, V is open in D(Ω) if and only if D K ∩ V is open in D K for every K . But this is the case as the restriction of f to each D K is continuous. If f is sequentially continuous on D(Ω), then the restriction of f to every D K is continuous as D K is metrizable for every K . Definition 3.82 Continuous linear functionals on D(Ω) are called distributions. It is traditional to denote D(Ω)∗ by D (Ω). We will show some important examples of distributions:

Recall that a measurable function f in Ω is locally integrable if K | f | < ∞ for every compact subset K . The space of locally integrable functions L 1loc (Ω) can be embedded into D (Ω).

Indeed, let f ∈ L 1loc (Ω). We define T f (ϕ) = Rn f ϕ dω. Then T f is a distribution. The continuity on every D K follows from      f ϕ  dω ≤ | f | max |ϕ| dω.  K

K

K

3.11

Banach Spaces

119

In particular, we may consider C(Ω) a subset of D (Ω). The space of all Borel measures on Ω can be also embedded into D (Ω). It is easy to see that the functional Tμ (ϕ) := Rn ϕ dμ is a distribution. In particular, Dirac measures are distributions. Let α be a multi-index. For F ∈ D (Ω) we define the derivative (in the distributional sense) by D α F : ϕ → (−1)|α| F(D α (ϕ)) It is easy to see that D α F ∈ D (Ω). The definition is motivated by the following. Let T be the distribution corresponding to some function f , that is, T := T f . Assume that f has continuous derivatives of order |α|. Using integration by parts and the fact that ϕ ∈ D(Ω) vanish on the boundary of Ω we obtain D α T f (ϕ) = TD α f (ϕ) for all ϕ ∈ D(Ω). So for differentiable functions, the distributional derivative D α agrees with the usual partial derivative. Fact 3.83 D α : D (Ω) → D (Ω) is a continuous operator. Proof: Linearity is obvious. Let K be a compact set in Ω and let F ∈ D (Ω) be bounded by C > 0 on a neighborhood Un from the local base in D K . Let pn be the Minkowski functional of Un . Then |F(ϕ)| ≤ C pn (ϕ) for every ϕ ∈ D K . Consequently |(D α F)(ϕ)| ≤ C pn (D α ϕ) ≤ C pn+|α| (ϕ). Therefore D α F is a continuous linear functional when restricted to an arbitrary D K , which by Proposition 3.81 means that D α F ∈ D (Ω). The space D (Ω) is important in the theory of generalized solutions to differential equations. Note that every continuous function on Ω has all partial derivatives in the sense of distributions.

3.11 Banach Spaces 3.11.1 Banach–Steinhaus Theorem Definition 3.84 Let X, Y be normed spaces and A ⊂ B(X, Y ). We say that A is pointwise bounded if sup{T (x)Y : T ∈ A} < ∞ for all x ∈ X . Observe that a set A ⊂ B(X, Y ) is pointwise bounded precisely when it is bounded in B(X, Y ) equipped with the topology T p of the pointwise convergence, see Definitions 3.3 and 3.11.

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3 Weak Topologies and Banach Spaces

If A is bounded in B(X, Y ) equipped with the operator norm, that is, there is C > 0 so that T  ≤ C for all T ∈ A, then A is pointwise bounded. Indeed, for x ∈ X we have T (x)Y ≤ T  x X ≤ Cx X , so sup{T (x)Y : T ∈ A} ≤ Cx X . The opposite direction is often called the Banach–Steinhaus Uniform Boundedness principle. Theorem 3.85 (Banach, Steinhaus) Let X, Y be Banach spaces and A ⊂ B(X, Y ). If A is pointwise bounded then A is bounded in B(X, Y ).   Proof: For n ∈ N set Nn = x ∈ X : supT ∈A T (x)Y ≤ n . We claim that Nn is closed, convex, and balanced in X . The balancedness of Nn is obvious. To check closedness, let xk ∈ Nn and xk → x ∈ X . Given T ∈ A, we have T (xk )Y ≤ n, so T (x)Y ≤ n by continuity. To see that Nn is convex, let x1 , x2 ∈ Nn and λ ∈ [0, 1]. Then for every T ∈ A we have T (λx1 + (1 − λ)x2 )Y ≤ λT (x 1 )Y + (1 − λ)T (x 2 )Y ≤ λn + (1 − λ)n = n. Since for every x ∈ X we have supT ∈A T (x)

Y < ∞, there is some n ∈ N greater than the supremum, hence x ∈ Nn . So ∞ n=1 Nn = X . By the Baire category theorem, there is n 0 such that the set Nn 0 contains an interior point x0 . Thus there is δ > 0 such that x 0 + δ B X ⊂ Nn 0 . Because of the symmetry of Nn 0 we have −x0 + δ B X ⊂ Nn 0 . If b ∈ B X , then by the convexity of Nn 0 we have b = 12 (x0 + b) + 12 (−x 0 + b) ∈ Nn 0 . Hence δ B X ⊂ Nn 0 . Consequently, given T ∈ A, for every x ∈ B X we have T (δx)Y ≤ n 0 , that is, T  ≤ nδ0 . This means that sup T  ≤ nδ0 . T ∈A

For an alternative proof of Theorem 3.85 that makes no use of the Baire category argument see Exercise 3.68. Corollary 3.86 Let X, Y be Banach spaces and Tn ∈ B(X, Y ) for n ∈ N. Assume that for every x ∈ X there exists the limit T (x) := lim Tn (x). Then T ∈ B(X, Y ) and T  ≤ lim inf Tn  (< ∞).

n→∞

n→∞

Proof: The linearity of T is easy:   T (x+y) = lim Tn (x+y) = lim Tn (x)+Tn (y) = lim Tn (x)+lim Tn (y) = T (x)+T (y).

From Tn (x) → T (x) we have Tn (x) → T (x), in particular {Tn (x) : n ∈ N} is bounded for all x ∈ X . By the Banach–Steinhaus theorem, {Tn } is bounded in B(X, Y ). Denote C = lim inf Tn . Then T (x) = lim Tn (x) = lim inf Tn (x) ≤ lim inf(x Tn ). This shows that T (x) ≤ C x, that is, T is bounded and T  ≤ C.

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Corollary 3.87 Let X, Y be Banach spaces and T1 , T2 , · · · ∈ B(X, Y ). If lim Tn (x) = T (x) for every x ∈ X , then for every compact set K in X we have Tn (x) → T (x) uniformly on K . Proof: By contradiction, assume that there is a compact set K in X , ε > 0, a subsequence of {Tn } (denoted by {Tn } again) and x n ∈ K such that Tn (xn )−T (xn ) ≥ ε. Since {x n } ⊂ K , we may assume that xn → x for some x ∈ K . Note that by the Banach–Steinhaus theorem we have M := sup{T , T1 , T2 , . . . } < ∞. Thus (Tn − T )(xn ) ≤ (Tn − T )(x) + (Tn − T )(x n − x) ≤ (Tn − T )(x)+Tn − T  · xn − x ≤ (Tn − T )(x)+2M · x n − x → 0, a contradiction with Tn (xn ) − T (xn ) ≥ ε. Theorem 3.88 (Banach, Steinhaus) Let X be a Banach space. If M ⊂ X ∗ is w∗ bounded then M is bounded. If M ⊂ X is w-bounded then M is bounded. Proof: The space (X ∗ , w ∗ ) is B(X, K) endowed with the topology of the pointwise convergence. The result follows from Theorem 3.85. The mapping π : X → B(X ∗ , K) given by π(x)(x ∗ ) = x ∗ , x for every x ∗ ∈ X ∗ , x ∈ X , is an isometry from (X,  · ) into B(X ∗ , K) endowed with the operator norm. Moreover, the topology T p on B(X ∗ , K) of the pointwise convergence induces on π(X ) a topology, denoted again T p , that makes π : (X, w) → (π(X ), T p ) an isomorphism. The result follows from Theorem 3.85. For an extension of Theorem 3.88 to the setting of locally convex spaces, see Exercises 3.71, 3.72, 3.73, and 3.74. We saw in Corollary 3.15 that all topological vector space topologies on a finitedimensional vector space coincide. In particular, if X is a finite-dimensional normed space, then on X the weak topology coincides with the norm topology. However, for infinite-dimensional spaces X , the weak topology never coincides with the norm topology of X . This is proved in the next result. Proposition 3.89 Let X be an infinite-dimensional normed space and O ⊂ X . If O = ∅ is w-open, then O is not bounded. In particular, the weak and norm topology on X do not coincide. Proof: We can assume that 0 ∈ O. Then there is ε > 0, f 1 , . . . , f n ∈ X ∗ such that {x : | f i (x)| < ε} ⊂ O. Clearly, the set N := {x ∈ X : fi (x) = 0 for i = 1, . . . , n} =

*

f i−1 (0)

is contained in O. We claim that N = {0}. Indeed, suppose that N = {0}. Then for every f ∈ X ∗ we have N ⊂ f −1 (0), so by Lemma 3.21, f is a linear combination of f 1 , . . . , f n . Thus X ∗ = span{ fi }, a contradiction.

122

3 Weak Topologies and Banach Spaces {x : f1(x)=−ε} {x : f1(x)=0} {x : f1(x)=ε}

N

{x : f2(x)=ε}

0

{x : f2(x)=0}

N O

{x : f2(x)=−ε}

Fig. 3.2 Non-empty w-open sets in infinite dimension are unbounded

Therefore we can find 0 = x ∈ N . Then for every scalar λ we have λx ∈ N , so O contains a line through x. Hence O cannot be bounded in X (see Fig. 3.2). Proposition 3.90 Let X be a Banach space. Then the original norm topology on X coincides with the Mackey topology on X . Proof: By Theorem 3.41, μ(X, X ∗ ) is stronger than the norm topology. Conversely, let V be a μ(X, X ∗ )-neighborhood of zero. Then there is a w ∗ -compact, convex, and balanced set S in X ∗ such that S◦ ⊂ V . Since S is w ∗ -bounded, it is also norm-bounded in X ∗ and there is α > 0 such that S ⊂ α B X ∗ . Then V ⊃ S◦ ⊃   1 1 ∗ α BX ◦ = α BX . Proposition 3.91 Let X be a Banach space and T be a locally convex topology on X such that (X, T )∗ = X ∗ . If T has a countable local base, then T coincides with the original norm topology of X . Proof: By the Banach–Steinhaus theorem, the systems of all bounded sets in (X,  · ) and (X, w) coincide. By Corollary 3.44 and Proposition 3.90, this system is the same for all locally convex topologies T on X for which (X, T )∗ = X ∗ . Applying Lemma 3.47 for the set A = B X completes the proof.

3.11.2 Banach–Dieudonné Theorem Theorem 3.92 (Banach, Dieudonné) Let X be a Banach space and let A be a convex set in X ∗ . If A ∩ n B X ∗ is w ∗ -closed for every n ∈ N, then A is w ∗ -closed.

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Proof: First note that under our assumption, A ∩ B is w ∗ -closed for every closed ball B in X ∗ (centered at a general point of X ∗ ). Indeed, if n is large enough so that B ⊂ n B X ∗ , then A ∩ B = (A ∩ n B X ∗ ) ∩ B, where both B and A ∩ n B X ∗ are w∗ -closed sets. Also note that A is norm closed. Indeed, if f n ∈ A, fn → f in X ∗ , then f n is bounded and thus for some j we have that f n ∈ j B X ∗ for every n. Since A∩ j B X ∗ is w∗

w∗ -closed, f n → f , and f n ∈ A ∩ j B X ∗ for every n, we have f ∈ A ∩ j B X ∗ ⊂ A. We may assume without 0 ∈ A. For n ∈ N, put An =

loss of generality that An and each An is w∗ -closed, convex and contains 0, A ∩ 2n B X ∗ . Then A = therefore An = ((An )◦ )◦ by the bipolar theorem. Put G = (An )◦ . We will show that A = G ◦ and this will finish the proof, as every polar set is ∗ ◦ ◦ w -closed. First observe that G ⊂

(An )◦ for◦ every n and thus G ⊃ ((An )◦ ) = An for every n. Consequently, A = An ⊂ G . It remains to show that G ◦ ⊂ A. This will be done in several steps. Claim A (An )◦ ⊂ (An+1 )◦ + 2−n B X for every n ∈ N. Assume x ∈ / (An+1 )◦ + 2−n B X . By the separation theorem, we obtain f ∈ X ∗ such that f (x) ≥ 1 ≥ sup{ f (x) : x ∈ (An+1 )◦ + 2−n B X}. By the definition ofAn we get (An+1 )◦ ⊃ 2−(n+1) B X and f 3 · 2−(n+1) B X = f 2−(n+1) B X + 2−n B X ⊂   f (An+1 )◦ + 2−n B X . Therefore  f  ≤ 23 2n . Also, sup{ f (x) : x ∈ (An+1 )◦ + 2−n B X } = sup ( f ) + sup ( f ) ≤ 1 and thus (An+1 )◦

2−n B X

sup ( f ) ≤ 1 − 2−n  f . (An+1 )◦   1 f . From ε < 2−n  f  and Choose ε ∈ 0, min( 13 , 2−n  f ) and put g = 1−ε −n sup ( f ) ≤ 1 − 2  f  we get sup (g) ≤ 1, that is, g ∈ ((An+1 )◦ )◦ = An+1 . (An+1 )◦

(An+1 )◦ ≤ 2n (as

In particular, g ∈ A. Since g  f  ≤ 23 2n and ε < 13 ) and g ∈ A, we / (An )◦ . This proves Claim A.  have g ∈ An . Moreover, g(x) > 1 and thus x ∈ Claim B (An )◦ ⊂ G + 2−(n−1) B X for every n ∈ N. Given n ∈ N and x n ∈ (An )◦ , define inductively, using Claim A, vectors xn+1 , xn+2 , . . . in such a way that for m = n, n + 1, . . . we have xm+1 ∈ (Am+1 )◦ and xm − x m+1  ≤ 2−m . Then {xm } is norm Cauchy and thus convergent to some  x ∈ X . Since (Am )◦ is norm closed for all m, we have that x ∈ (Am )◦ = G.  1 1  Moreover, x n − x ≤ ∞ j=0 n+ j ≤ n−1 . 2 2 Claim C (1 + ε)A. A= ε>0

  1 1 Since A is convex and 0 ∈ A, write a = 1+ε (1 + ε)a + 1 − 1+ε 0 for a ∈ A (1 + ε)A. and ε > 0 to see that A ⊂ (1 + ε)A. Therefore A ⊂ ε>0

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   If x ∈ (1+ε)A and x = 1+ n1 xn with xn ∈ A for every n, then x − xn = n1 xn and xn  ≤ x. Thus xn → x and since A is closed, we have x ∈ A. Claim C is proved.  Claim D  (1 + ε)A. G◦ ⊂ ε>0

Let ε > 0 and n ∈ N be given. For every x and y in X , 

x ε y x + y = (1 + ε) + 1+ε 1+ε ε

 .

  1 B X . Therefore by From this and Claim B we have (An )◦ ⊂ (1 + ε) conv G ∪ 2n−1 ε  ◦  1 G ∩ 2n−1 ε B X ∗ ⊂ An ⊂ A for every n ∈ N taking polars we have 1+ε 1 G ◦ ⊂ A, which means Keeping ε > 0 and taking the union over nwe have 1+ε ◦ ◦ (1 + ε)A. This proves Claim D,  G ⊂ (1 + ε)A for every ε > 0. Thus G ⊂ and Theorem 3.92 follows.

ε>0

Let X be a Banach space. A set C ⊂ X ∗ is called w ∗ -sequentially closed if w∗

{ f n } ⊂ C, f n → f implies that f ∈ C. Corollary 3.93 Let X be a separable Banach space and let C be a convex set in X ∗ . If C is w ∗ -sequentially closed then it is w ∗ -closed. Proof: The w∗ -topology of B X ∗ is metrizable (see Proposition 3.101). In metrizable spaces the closures and sequential closures coincide, so since C ∩ n B X ∗ is w ∗ sequentially closed (in n B X ∗ ), C ∩ n B X ∗ is w∗ -closed in n B X ∗ and thus also in X ∗ as n B X ∗ itself is w ∗ -closed. It is enough now to apply Theorem 3.92 to finish the proof. √ Remark: If X = 2 and xn := nen , where en is the standard unit vector in w 2 , then 0 ∈ {x n } (Exercise 3.55). Thus the set {xn } is not w-closed in 2 and yet {xn }∩ j B X is finite and thus weakly closed in X . Since on 2 the w- and w∗ -topology coincide as 2 is reflexive (see Definition 3.110 and examples after it), we see that the assumption of convexity of the set A in the Banach–Dieudonné theorem cannot be dropped. Corollary 3.94 Let X be a Banach space and let F be a linear functional on X ∗ . The following are equivalent: (i) F is w ∗ -continuous. (ii) F ∈ X , i.e., there is x ∈ X such that F( f ) = f (x) for every f ∈ X ∗ . (iii) The restriction of F to B X ∗ is w ∗ -continuous. (iv) F −1 (0) ∩ B X ∗ is w ∗ -closed. Proof: (i) is equivalent to (ii) by Proposition 3.2 and the note after Definition 3.40.

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(i)⇒(iii) and (iii)⇒(iv) are trivial. (iv)⇒(i): If (iv) holds, then by Theorem 3.92 and linearity of F, F −1 (0) is w ∗ -closed. It is enough now to use Proposition 3.19.

3.11.3 The Bidual Space Given a normed space (X,  · ), we endow X ∗ with its dual norm. This space becomes a Banach space (see Proposition 1.29). By X ∗∗ we denote the space (X ∗ )∗ with the norm F := sup f ∈B X ∗ |F( f )|. We define higher duals by induction as X ∗∗∗ = (X ∗∗ )∗ , etc. Definition 3.95 Let X be a normed space. The canonical embedding π of X into X ∗∗ is defined for x ∈ X by π(x) : f → f (x), for all f ∈ X ∗ . Note that π is a linear isometry into. Indeed, π(αx + βy)( f ) = f (αx + βy) = α f (x) + β f (y) = [απ(x) + βπ(y)]( f ). Also, for x ∈ X we have π(x) = sup f ∈B X ∗ | f (x)| ≤ sup f ∈B X ∗  f  · x ≤ x. Considering f 0 ∈ S X ∗ such that f 0 (x) = x we have π(x) ≥ | f 0 (x)| = x. Therefore π(x) = x. Using this natural identification, for x ∈ X we often write x ∈ X ∗∗ instead of π(x) ∈ X ∗∗ , and we identify X with π(X ) ⊂ X ∗∗ . In particular, π(x)( f ) = x( f ) for x ∈ X and f ∈ X ∗ . Note too that if F ∈ X ∗∗ is w ∗ -continuous on X ∗ , then there exists x ∈ X such that π(x) = F (see Corollary 3.94). We can consider then two dual pairs, namely X, X ∗  and X ∗ , X ∗∗ . By the very definition, the topology w(X ∗∗ , X ∗ ) (i.e., the w ∗ topology on X ∗∗ ) induces on X the topology w(X, X ∗ ) (i.e., the w topology on X ). A consequence of Corollary 3.49 is that the weak topology of X ∗∗ (i.e., the topology w(X ∗∗ , X ∗∗∗ ), a topology stronger than w(X ∗∗ , X ∗ )) induces again the topology w on X . Proposition 3.39 implies that X is w(X ∗∗ , X ∗ )-dense in X ∗∗ . The following result proves something more precise, namely that the closed unit ball of X is w ∗ dense in the closed unit ball of X ∗∗ . Theorem 3.96 (Goldstine) Let X be a Banach space. The w∗ -closure of B X in X ∗∗ is B X ∗∗ .

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Proof: We consider the dual pair X ∗ , X ∗∗ . According to the bipolar theorem 3.38, w(X ∗∗ ,X ∗ ) . It is elementary that (B X )◦ = B X ∗ and, similarly, (B X ∗ )◦ = (B X )◦◦ = B X ∗∗ BX . The following result will be used later. Lemma 3.97 Let (X,  · ) be a Banach space and let ||| · ||| be an equivalent norm on X ∗ . ||| · ||| is a dual norm to some equivalent norm on X if and only if ||| · ||| is w ∗ -lower semicontinuous. Recall that a norm | · |∗ on X ∗ is a dual norm to some norm | · | on X if for every f ∈ X ∗ we have | f |∗ = sup{ f (x) : |x| ≤ 1}. Proof: As a supremum of w ∗ -continuous functions x : f → f (x) for x ∈ S X , every dual norm is w ∗ -lower semicontinuous (see Exercise 3.31). Assume that ||| · ||| is a w ∗ -lower semicontinuous equivalent norm on X ∗ and denote by B its closed unit ball. Then B is w ∗ -closed and, by the bipolar theorem, B = (B0 )0 . It follows that ||| · ||| is the dual norm to the (equivalent) norm given by the Minkowski functional of B0 . Given normed spaces X, Y and T ∈ B(X, Y ), we introduced in Definition 2.28 the adjoint operator T ∗ ∈ B(Y ∗ , X ∗ ). If the procedure is applied to T ∗ , we obtain T ∗∗ := (T ∗ )∗ , an element in B(X ∗∗ , Y ∗∗ ). Note that for x ∈ X we have T ∗∗ (π(x)) = π(T (x)), in particular T ∗∗ (π(X )) ⊂ π(Y ). By using the natural identification between X and π(X ) (and between Y and π(Y )) described above, this  amounts to say that T ∗∗ (X ) ⊂ Y and T ∗∗  X = T .

3.11.4 The Completion of a Normed Space Definition 3.98 Let X be a normed space. We say that a Banach space Z is a completion of X if X is linearly isometric to a dense subspace of Z . Given a completion  X of a normed space X , we shall always assume that X is a (linear) subspace of  X. Fact 3.99 Every normed space has a completion. Proof: It is enough to recall that X ∗∗ is a Banach space (see Proposition 1.29) and that X is isometric to a subspace of X ∗∗ via the canonical mapping π : X → X ∗∗ X ∗∗ (see Definition 3.95). As  X we can then use π(X ) . The following simple result will be used later. Lemma 3.100 Let X be  a normed space. Then every  x of its completion element ∞ ∞  x , where x ∈ X , x  X can be  written  x = n n < ∞. Moreover, n=1 n n=1 x , where the infimum is taken over all series in X summing up  x  = inf ∞ n n=1 to  x.

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Proof: Obviously, B X = B  x ∈ X . We may assume, without loss of generality, that  B . There exists x ∈ B such that  x − x  ≤ 1/2. Since 2( x − x ) ∈ B , we can  1 X 1 1 X X 2. x− x 1 ) − x2  ≤ 1/2, i.e.,  x − x − (1/2)x  ≤ 1/2 find x 2 ∈ B X such that 2( 1 2  2 Find x 3 ∈ B X such that 2  x − x 1 − (1/2)x 2 − x3  ≤ 1/2. Proceed recursively to obtain a sequence {xn } in B X such that  x − x1 − (1/2)x2 − . . . − (1/2n )xn+1  ≤ 1/2n+1 , for all n ∈ N.  n−1 )x . To prove the second assertion, modify the This proves that  x = ∞ n n=1 (1/2 construction in the following way: Fix m ∈ N, and start choosing x1 ∈ B X such that  x − x 1  ≤ 1/2m . Then choose x 2 ∈ B X such that  x − x 1 − (1/2m )x2  ≤ m+1 1/2  . Proceed recursively to obtain a sequence {x n }. Since x 1  ≤  x  + 1/2m ∞ m m−1 and n=1 1/2 = 1/2 , we get the conclusion.

3.11.5 Separability and Metrizability Proposition 3.101 Let X be a separable Banach space. Then (B X ∗ , w ∗ ) is a metrizable compact space. A metric on B X ∗ compatible with the topology w ∗ is given by ρ( f, g) :=

∞ 

2−i |( f − g)(xi )|,

(3.7)

i=1

where {xn }∞ n=1 ⊂ S X is dense in S X . Proof: Let {x n }∞ n=1 ⊂ S X be a dense subset of S X . Since D := span{x n : n ∈ N} is a dense subspace of X , X ∗ , D is a dual pair (see Proposition 3.39). The topology w(X ∗ , D) is metrizable (see Proposition 3.29), Hausdorff, and coarser than w∗ . Therefore it agrees with w ∗ on B X ∗ , since (B X ∗ , w ∗ ) is compact (Theorem 3.37). That ρ defined by (3.7) is a metric is clear. In order to prove that the associated topology is w(X ∗ , D), let f, f 1 , f2 , . . . elements in B X ∗ . Assume first n ) is ∞that ( f−i w(X ∗ , D)-convergent to f . Fix ε > 0. There exists i 0 ∈ N such that i=i 2 < 0 +1 i0 −i ε/4. Find then n 0 ∈ N such that i=1 2 |( f − f n )(xi )| < ε/2 for every n ≥ n 0 . Since |( f − f n )(xi )| ≤ 2 for all n and i, we get ρ( f, f n ) < ε for every n ≥ n 0 . Conversely, assume that ρ( f − f n ) → 0. Then, for each i ∈ N we have ( f − f n )(xi ) →n 0. This proves that f n → f in the topology w(X ∗ , D). Thus if X is separable, then the topological space (B X ∗ , w ∗ ) is metrizable. In fact, the converse is also true. This will be proved in Proposition 3.103. In its proof we shall need the following general result concerning the space C(K ) of continuous functions on a compact ! := {δk : k ∈ K } topological space K . Recall that the set of all Dirac functionals K is a (w ∗ -compact) subset of SC(K )∗ (see the paragraph right before Proposition 2.21 and Lemma 3.55).

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Lemma 3.102 Let K be a compact metric space. The Banach space C(K ) is separable if and only if K is metrizable. Proof: If C(K ) is separable then BC(K )∗ in the w ∗ -topology is metrizable by Propo! ⊂ (BC(K )∗ , w ∗ ) is metrizable, so is K , see Lemma 3.55. sition 3.101. Since K Assume now that K is a compact metric space. Let {xn }∞ n=1 be a dense sequence in K . Let f n,m (x) = m1 −dist(x, xn ) if dist(x, xn ) ≤ m1 and f n,m = 0 if dist(x, xn ) > 1 m . Then the family { f n,m } together with a constant function generates a separable algebra that separates the points of K and thus its closure is C(K ) by the Stone– Weierstrass theorem. Proposition 3.103 Let X be a Banach space. (B X ∗ , w ∗ ) is metrizable if and only if X is separable. Proof: Assume that (B X ∗ , w ∗ ) is metrizable. Lemma 3.102 implies that C(B X ∗ , w ∗ ) is separable in its supremum norm. The mapping I : X → C(B X ∗ , w ∗ ), I (x) : f → f (x), is an isometry, hence X is separable as a subspace of a separable metric space. The other implication is contained in Proposition 3.101.

Since X ∗ = n∈N n B X ∗ , and compact metrizable space are separable, from Proposition 3.101 we also have Corollary 3.104 Let X be a Banach space. If X is separable then B X ∗ , and so X ∗ , are w ∗ -separable. We note that (X ∗ , w∗ ) may be separable although (B X ∗ , w∗ ) may be not, [JoLi1]. Also, (B X ∗ , w∗ ) may be separable although X may be not, [JoLi1]. For more in this direction, we refer to [HMVZ, Chapter 7]. Proposition 3.105 Let X be a Banach space. If X is w-separable then X is separable. Proof: Let S be a countable set that is weakly dense in X , denote D = span(S). The countable set C of rational linear combinations of S is dense in D, hence D is normw separable. Since D is convex and dense in X , by Theorem 3.45, D = D = X , so C is dense in X (see also Exercise 3.100). We will now investigate (B X , w). Proposition 3.106 Let X be a Banach space. (B X , w) is metrizable if and only if X ∗ is separable. If X ∗ is separable, then (B X ∗∗ , w ∗ ) is metrizable by Proposition 3.101. Since (B X , w) is a topological subspace of (B X ∗∗ , w∗ ) (see Corollary 3.49 and the paragraph after Definition 3.95), (B X , w) is also metrizable. For the other direction, let U = {Un : n ∈ N} be a countable base of neighborhoods of 0 in (B X , w). Without loss of generality we may assume that Un := {x ∈ B X : | f (x)| < 1, f ∈ Fn }, where Fn is a finite subset of X ∗ , n ∈ N.

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To finish the proof it will be enough to ensure that span{ n∈N Fn } = X ∗ . To this end, let x ∗∗ ∈ X ∗∗ that vanishes on n∈N Fn . By Theorem 3.96 there exists a net {xα }α∈I in B X that w ∗ -converges to x ∗∗ . Fix n ∈ N. Then, there exists α0 ∈ I such that | f (xα )| = | f (xα − x ∗∗ )| < 1 for α ≥ α0 and for all f ∈ Fn . Then xα ∈ Un for α ≥ α0 . This holds for all n ∈ N, so w- lim xα = 0, hence x ∗∗ = 0. Proposition 3.107 Let C be a weakly compact set in a Banach space X . If X ∗ is w∗ -separable, then (C, w) is metrizable. In particular, if X is separable then (C, w) is metrizable. Proof: If D is a countable separating set in X ∗ , then the topology w(X, D) is metrizable and coincides with the weak topology on C as C is weakly compact. Such a separating set exists if X ∗ is w ∗ -separable (take any w ∗ -dense countable set).

3.11.6 Weak Compactness Note that a subset A of a Banach space is relatively weakly compact if and only if A is bounded and its w∗ -closure in X ∗∗ is a subset of X . Indeed, if A is relatively w weakly compact, then certainly A is bounded and A is w-compact in X , hence ∗ w w A = A (⊂ X ). On the other hand, if A is bounded, then A◦ is a neighborhood of 0 in X ∗ and so, with respect to the dual pair X ∗ , X ∗∗ , A◦◦ is w ∗ -compact by Alaoglu’s theorem. w∗ w Since A ⊂ A◦◦ we get that A is w ∗ -compact; if it is a subset of X , then A (= w∗ A ) is w-compact. For the concept of angelic space see Definition 3.53. Proposition 3.108 Let X be a Banach space. Then (X, w) is an angelic space. Proof: It is enough to observe that subspaces of angelic spaces are themselves angelic, and to identify (X, w) with a subset of the topological space C p (B X ∗ ) where B X ∗ carries the topology induced by w∗ . Use then Theorem 3.58. From the angelicity of a Banach space in its weak topology, we obtain the following result (see Exercise 3.9). Theorem 3.109 (Eberlein, Šmulyan) Let A be a subset of a Banach space X . The following are equivalent: (i) A is (relatively) w-compact. (ii) A is (relatively) w-sequentially compact. (iii) A is (relatively) w-countably compact.

3.11.7 Reflexivity Definition 3.110 A Banach space X is said to be reflexive if the mapping π in Definition 3.95 maps X onto X ∗∗ .

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In particular, a reflexive space X is isometric to X ∗∗ . On the other hand, there are examples of nonreflexive spaces X such that X is linearly isometric to X ∗∗ —of course, not by means of π (James’ space J , see Definition 4.43). We will see in Proposition 3.113 that X is reflexive if and only if the w-topology and the w ∗ -topology agree on X ∗ . Examples: (i)  p and L p spaces for 1 < p < ∞ are reflexive. Indeed, if 1p + q1 = 1 then ∗ ∗ ∗∗ p = q=  p in the sense that that for every f ∈ q there is g ∈  p such that f (x) = xi g¯i for every x ∈ q . So the action of f on x is the same as the action of g on x and π is thus a mapping from  p onto ∗∗ p . Similarly it is shown for L p . (ii) c0 is not reflexive since c0∗∗ = ∞ , c0 is separable and ∞ is not. (iii) C[0, 1] is not reflexive. Indeed, C[0, 1]∗∗ would then be separable and thus C[0, 1]∗ would be separable by Proposition 2.8, which is not true (Proposition 2.21). Theorem 3.111 A Banach space X is reflexive if and only if B X is weakly compact. Proof: Let X be reflexive. Then X = X ∗∗ and thus B X = B X ∗∗ is w ∗ -compact in X ∗∗ by Alaoglu’s theorem, so it is weakly compact in X . If B X is weakly compact, then it is w ∗ -closed in X ∗∗ . Since the w∗ -closure of B X in X ∗∗ is B X ∗∗ by Theorem 3.96, we get that B X = B X ∗∗ and the space X is reflexive. Proposition 3.112 A Banach space X is reflexive if and only if X ∗ is reflexive. Proof: Let X be reflexive. Then X = X ∗∗ and thus the w∗ -topology and w-topology on X ∗ coincide. Therefore, by the Alaoglu’s theorem, B X ∗ is weakly compact and X ∗ is reflexive by Theorem 3.111. Let X ∗ be reflexive. By the first part of this proof, we have that X ∗∗ is reflexive. Therefore B X ∗∗ is w-compact. B X is a closed and therefore w-closed (Theorem 3.45) subset of B X ∗∗ and thus w-compact in X ∗∗ . Since the weak topology on X ∗∗ induces on X its weak topology (see Corollary 3.49), B X is w-compact, and we are done by Theorem 3.111. Proposition 3.113 A Banach space is reflexive if and only if the w-topology and the w∗ -topology agree on X ∗ . Proof: If X is reflexive, then X ∗ is also reflexive, by Proposition 3.112, so on X ∗ the w- and w∗ -topologies agree. Conversely, if these two topologies agree on X ∗ , then, by Alaoglu’s theorem (Theorem 3.37), (B X ∗ , w ∗ ) (= (B X ∗ , w)) is compact. By Theorem 3.111, the space X ∗ is reflexive. The result follows from Proposition 3.112. Observe that, if X is not reflexive, there is a convex set in X ∗ , namely f −1 (0) for f ∈ X ∗ \X , that is w-closed but not w ∗ -closed—it is w∗ -dense and different from X ∗.

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Proposition 3.114 Let X be a reflexive Banach space. If Y is a closed subspace of X , then Y is a reflexive Banach space. Proof: By Corollary 3.49, the weak topology on Y coincides with the restriction to Y of the weak topology on X . To complete the proof, it suffices to observe that BY is a weakly closed subset of the weakly compact set B X . Therefore BY is weakly compact in Y and thus Y is reflexive by Theorem 3.111. Proposition 3.115 Let X be a Banach space. If X is separable and reflexive, then (B X , w) is a compact metrizable space. Proof: The result follows from Proposition 3.107 and Theorem 3.111.

3.11.8 Boundaries 3.11.8.1 Dirac Deltas and Extreme Points of BC(K )∗

  Lemma 3.116 Let K be a compact topological space. Then Ext BC(K )∗ = {±δk }k∈K , where δk are the Dirac functionals in C(K )∗ . Proof: We will first show that all extreme points of BC(K )∗ are of the form ±δk . ∗ Let A = convw {±δk }k∈K . We claim that A = BC(K )∗ . Indeed, assuming F ∈ BC(K )∗ \A, by Corollary 3.34 there is f ∈ C(K ) such that sup A ( f ) ≤ 1 and F( f ) > 1. Since A is symmetric, sup A | f | ≤ 1, which implies supk∈K |δk ( f )| =  f  ≤ 1. Thus F( f ) ≤ 1, a contradiction. Now by Theorem 3.66 we obtain that   w∗ Ext BC(K )∗ ⊂ {±δk }k∈K . In Lemma 3.55 we showed that ({δk }k∈K , w ∗ ) is homeomorphic to K , so {δk }k∈K is w∗ -compact. Thus also {±δk }k∈K = {δk }k∈K ∪ {−δk }k∈K is w ∗ -compact, hence w ∗ -closed, so Ext(BC(K )∗ ) ⊂ {±δk }k∈K . It remains to show that every Dirac measure δk is an extreme point of BC(K )∗ . This will complete the proof as then −δk are extreme by the symmetry of BC(K )∗ . Given k0 ∈ K , consider the family U of all open neighborhoods of k0 in K . Given U ∈ U, by Urysohn’s lemma there is fU ∈ BC(K ) such that fU (k0 ) = 1 ∗ ∗ and fU = 0 on K \U . Then HU := {F ∈ C(K  ) : F( fU ) = 1} is a w -closed supporting manifold for BC(K )∗ . Define H = U ∈U HU . Since δk0 ∈ HU for all U ∈ U, we have that H = ∅, and it is a w∗ -closed supporting manifold for BC(K )∗ . Since H ∩ BC(K )∗ is w ∗ -compact, it contains an extreme point, which is also an extreme point of BC(K )∗ (see Lemma 3.62). How/ H ever, the only candidates are then ±δk for k ∈ K and we easily check that δk ∈ for k = k0 , as the topology of K is Hausdorff. So δk0 must be an extreme point. Theorem 3.117 (Banach, Stone) Let K , L be compact spaces. C(K ) is linearly isometric to C(L) if and only if K and L are homeomorphic.

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Proof: Let ϕ be a homeomorphism of K onto L. Define a mapping T : C(K ) → C(L) by T ( f ) = f ◦ ϕ −1 . It is standard to check that T is an isometry of C(K ) onto C(L). Assume now that T is an isometry of C(K ) onto C(L). Then T ∗ is an isometry (Exercise 2.51). Thus T (BC(K )∗ ) = BC(L)∗ and by Lemma 3.116 and Exercise 3.126, for l ∈ L we have T ∗ (δl ) = ε(l)δkl , where kl ∈ K and ε(l) ∈ {−1, 1}. The mapping : l → ε(l)kl is continuous as T ∗ is w∗ -w∗ -continuous. We now show that the function ε : L → {−1, 1} is continuous. Indeed, for the constant function 1, denoted x, we can write ε(l) = ε(l)δkl (x) = T ∗ (δl )(x) = T (x)(l), so ε = T (x), hence ε is continuous. Then the mapping ρ : L → K defined by ρ = /ε, i.e., ρ(l) = kl for l ∈ L, is a continuous one-to-one mapping from L onto K . Remark: The space C(K ) is isomorphic to C(L) whenever K and L are metrizable uncountable compact spaces (Milyutin, see, e.g., [Rose10] and [AlKa]). 3.11.8.2 James Boundaries Definition 3.118 Let (X,  · ) be a Banach space. Let C be a bounded subset of X ∗ . A set B ⊂ C is called a James boundary for C if for every x ∈ X there is g ∈ B such that g(x) = sup{ f (x) : f ∈ C}. A set B ⊂ B X ∗ is called a James boundary of X if it is a James boundary for B X ∗ . Note that the condition is homogeneous in x. For instance, B ⊂ B X ∗ is a James boundary of X if for every x ∈ S X there is f ∈ B such that f (x) = 1. Fact 3.119 Let X be a Banach space. Let C be a bounded w ∗ -closed subset of X ∗ . Then Ext(C) is a James boundary for C. In particular, Ext(B X ∗ ) is a James boundary of X . Proof: Given x ∈ S X , consider H := { f ∈ X ∗ : f (x) = supg∈C g(x)}. Then H is a w ∗ -closed supporting manifold for C which contains an extreme point of C by Proposition 3.64. Note that there are James boundaries B of X such that B ∩ Ext(B X ∗ ) = ∅ (see Exercise 3.151). Another simple observation is contained in the following result. Fact 3.120 Let X be a Banach space. Let C be a convex w ∗ -compact subset of X ∗ ∗ and B ⊂ C be a James boundary for C. Then conv w (B) = C. ∗

Proof: Assume that there exists g ∈ C such that g ∈ conv w (B) (⊂ C). By the separation theorem, there exists x ∈ X such that sup f ∈B f (x) < g(x). It follows

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that there is no point in B where x attains its supremum on C, a contradiction with the definition of a James boundary. It is important to be able to check weak compactness of a bounded subset S of a Banach space X with a minimum of requirements. An example of the kind of result we are looking for is the following: A uniformly bounded subset S of C[0, 1] is weakly compact if (and only if) it is pointwise sequentially compact. This is a consequence of the Lebesgue’s dominated convergence theorem. Indeed, let { f n } be a sequence in S. By hypothesis, it has a subsequence { f n k } that pointwise converges to some f ∈ S. Since there is a constant M ≥ 0 such that | f n k | ≤ M for all

1 k ∈ N, it happens that 0 f n k (t) dμ(t) → 0 as k → ∞, for every regular Borel measure μ on [0, 1]. This can be stated, by the Riesz’s representation theorem, as w μ, f n k  → 0 as k → ∞ for every μ ∈ C[0, 1]∗ or, in other words, f n k → 0. Since this is true for every sequence { f n } in S, the set S is w-sequentially compact. By the Eberlein–Šmulyan theorem, S is w-compact. Observe that, in the former example, the set B := {±δt : t ∈ [0, 1]} is a James boundary of X := C[0, 1], due to Lemma 3.116 and Fact 3.119. We just proved, in this case, that a bounded subset S of X is w-compact whenever S is w(X, B)-sequentially compact. This is an instance of a theorem of Pfitzner. Theorem 3.121 (Pfitzner, [Pfi]) Let B be a James boundary of a Banach space X . Then a bounded subset S of a Banach space is w-compact whenever it is w(X, B)compact. We are not going to prove Pfitzner’s theorem in full generality, but we will present some important special cases (obtained earlier by various authors) in this section and in Exercises 3.155, 3.156, and 3.157. A closely related problem is to provide sufficient conditions on a Banach space X , on a convex w∗ -compact subset C of X ∗ , or on a James boundary B for C, so that B would be a strong James boundary for C, i.e., conv · (B) = C. Fact 3.120 says ∗ is strong. that conv w (B) = C always hold. However, not every James boundary  For example, consider X := C[0, 1] and B = ±δt : t ∈ [0, 1] ⊂ B X ∗ . Then conv· (B) = B X ∗ (Exercise 3.130) and yet, B is a James boundary of X . Note that, for a strong James boundary B of a Banach space X , the conclusion of Pfitzner’s theorem holds. This is a consequence of the following trivial fact: if A is a bounded subset of a normed space, B is a subset of X ∗ , and a net in A converges to some point x ∈ X pointwise on elements in B, then it converges to x pointwise · on elements in B . 3.11.8.3 Strong James Boundaries The following result gives a sufficient condition for a James boundary to be strong. Variants of it were obtained independently by Rodé and Godefroy using different methods. Here we will follow Godefroy’s approach using Simons’ inequality (Lemma 3.123), which, in turn, originated in the work of James, [Jame2] and [Jame5].

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Theorem 3.122 (Godefroy [Gode2]) Let X be a Banach space and let C be a bounded closed convex subset of X ∗ . If B is a norm-separable James boundary for C, then C = conv· (B). In particular, if a norm-separable subset B of B X ∗ is such that conv(B) = B X ∗ , then there is x ∈ S X such that x( f ) < 1 for all f ∈ B. Theorem 3.122 does not hold in general if the assumption on the separability of B is dropped, as we noted above (see again Exercise 3.130). Let B be a nonempty set. Consider the space ∞ (B) with its canonical sup-norm. If x = (x b ) ∈ ∞ (B), we denote sup B x = sup{xb : b ∈ B}. If there is b0 ∈ B such that x b0 = sup B x, we say that x attains its supremum over B. In the proof of Theorem 3.122 we will use the following results. Lemma 3.123 (Simons’ inequality [Simo]) Let B be a nonempty set. Let {xn } be abounded sequence in ∞ (B). ∞ Assume that for all Λn ≥ 0, n ∈ N, satisfying ∞ Λ = 1, the vector n=1 n n=1 Λn x n attains its supremum over B (i.e., every element in the superconvex hull of {xn : n ∈ N} attains its supremum over B). Then, if u(b) := lim supn xn (b) for all b ∈ B,   sup u ≥ inf sup x : x ∈ conv{xn } . B

B

Proof: (Oja [Oja1]) Set Ck = It will be enough to prove that

 ∞

n=k

Λn xn : Λn ≥ 0,

∞

n=k

Λn = 1 for k ∈ N.

inf sup x ≤ sup u.

x∈C1 B

(3.8)

B

Let ε > 0. Choose inductively z k ∈ Ck such that for k = 0, 1, . . . , sup(2k vk + z k+1 ) ≤ inf sup(2k vk + z) + z∈Ck+1 B

B

ε 2k+1

,

 zn zn for k ∈ N. Put v = ∞ n=1 n . n 2 2 Since z k+1 = 2k+1 vk+1 − 2k+1 vk , we get 2k+1 vk+1 − 2k vk = 2k vk + z k+1 , so  zn using 2k v − 2k vk = 2k ∞ n=k+1 n ∈ C k+1 we get for all k = 0, 1, . . . , 2 where v0 = 0 and vk :=

k

n=1

  sup(2k+1 vk+1 − 2k vk ) ≤ sup 2k vk + (2k v − 2k vk ) + B

B

= sup 2k v + B

ε 2k+1

ε 2k+1 ε = 2k sup v + k+1 . 2 B

As v ∈ C1 , we can choose t ∈ B such that v(t) = sup v. Since from the last inequality we have for m ∈ N:

B

m−1 k=0

2k = 2m − 1,

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2m vm (t) =

m−1 

(2k+1 vk+1 − 2k vk )(t)

k=0

≤ (2m − 1) sup v + ε = 2m v(t) + ε − sup v. B

B

Thus sup v ≤ 2m v(t) − 2m vm (t) + ε, hence B

inf sup x ≤ sup v ≤ lim sup(2m v − 2m vm )(t) + ε

x∈C1 B

m

B

≤ lim sup xm (t) + ε = u(t) + ε. m

as 2m v − 2m vm ∈ Cm+1 . Inequality (3.8) follows as ε > 0 was arbitrary. If X is a Banach space and B ⊂ X ∗ is bounded, we can consider X a subset of ∞ (B) by identifying x ∈ X with f (x) f ∈B . Note that if B is a James boundary of X , then sup x = x and the supremum is attained. Thus we get B

Theorem 3.124 (Simons [Simo]) Let B be a James boundary for a bounded subset M of the dual of a Banach space X . If {xn } is a bounded sequence in X , and u(x ∗ ) := lim supn x ∗ (xn ) for every x ∗ ∈ X ∗ , then sup u = sup u. B

M

Proof: Assume sup B u < α < sup M u. Let f ∈ M be such that u( f ) > α. Assume without loss of generality that f (xn ) > α for all n ∈ N. If x ∈ conv{xn }, from astandard convexity argument we have f (x) > α. Moreover, every element x := ∞ ∞ λ x , where λ ≥ 0 for all n and n n=1 n n n=1 λn = 1, belongs to X (in other language, sconv(x n ) ⊂ X , see Exercise 1.67), so x attains its supremum on B (and sup B x = sup M x). From Simons’inequality, (α >) sup u ≥ B

=

inf

inf

sup x

x∈conv (xn ) B

sup x ≥

x∈conv (xn ) M

inf

x∈conv (x n )

f (x) ≥ α,

a contradiction. We are now ready to prove Godefroy’s Theorem 3.122. Proof of Theorem 3.122: [Gode2] By contradiction, assume that conv(B) = C (see Fig. 3.3). By the separation theorem, we can find F ∈ S X ∗∗ , α < β and y0∗ ∈ C\conv(B) such that F( f ) ≤ α for all f ∈ B and F(y0∗ ) > β. Let S = {x ∈ B X : w∗

y0∗ (x) > β}. Using Goldstine’s theorem we get that F ∈ S . As B is separable, the topology on bounded sets in X ∗∗ of pointwise convergence on B is metrizable.

136

3 Weak Topologies and Banach Spaces {x∗∗:y0∗(x∗∗)=β}

BX ∗∗ C

F

y0∗ {x∗ : F(x∗) = β}

BX

S

{x∗ : F(x∗) = α}

conv(B) X∗

X

Fig. 3.3 The proof of Theorem 3.122

Thus there is a sequence {x n } in S which converges to F on points of B ∪ {y0∗ }. Put u(x ∗ ) = lim sup x ∗ (xn ) for x ∗ ∈ X ∗ . Then, by Theorem 3.124, we get α ≥ sup F = sup u = sup u ≥ u(y0∗ ) = F(y0∗ ) > β, B

B

C

a contradiction. Corollary 3.125 (Godefroy) Let X be a Banach space X . If X has a separable James boundary, then X ∗ is separable. Proof: If B is a separable James boundary of X , then by Theorem 3.122, conv(B) = B X ∗ . Thus B X ∗ and also X ∗ are separable. By Fact 3.119 we have the following result, which was obtained independently by several authors (for references see, e.g., [LoMo]). Corollary 3.126 Let X be a Banach space. If Ext(B X ∗ ) is separable, then X ∗ is separable. 3.11.8.4 James Boundaries and James Theorem Using Simons’ lemma we also see how James boundaries interact with second duals. Corollary 3.127 Let B be a James boundary for a bounded subset M of the dual of a Banach space X . If x ∗∗ ∈ X ∗∗ is the w∗ -limit of a bounded sequence in X , then supx ∗ ∈B x ∗∗ (x ∗ ) = supx ∗ ∈M x ∗∗ (x ∗ ). Proof: It is enough to observe that x ∗∗ = lim sup xn when we consider x ∗∗ and xn , n ∈ N, functions on X ∗ , and apply then Theorem 3.124. In particular we obtain.

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Corollary 3.128 Let B be a James boundary of a Banach space X . If x ∗∗ ∈ X ∗∗ is the w∗ -limit of a bounded sequence in X , then x ∗∗  = sup x ∗∗ (x ∗ ). x ∗ ∈B

Corollary 3.129 Let X be a Banach space and let C be a closed convex and bounded subset of X ∗ . If every x ∗∗ ∈ S X ∗∗ is the w∗ -limit of a sequence in B X , then C = conv· (B) for every James boundary B for C. In particular, if X is a separable Banach space that does not contain an isomorphic copy of 1 , then C = conv · (B) for every closed convex and bounded subset of X ∗ and every James boundary B for C. Proof: Assume that there exists x ∗ ∈ C\conv · (B). The separation theorem provides x ∗∗ ∈ S X ∗∗ and α ∈ R such that sup B x ∗∗ < α < x ∗∗ (x ∗ ). This contradicts Corollary 3.127. The particular case follows from the first part by using Theorem 5.40. The fundamental James’ theorem below (Theorem 3.130) can be seen as a boundary problem: If every f ∗ ∈ X ∗ attains its supremum over a set A ⊂ X at some point of A, then the set A is certainly bounded, and it is a James boundary for the ∗ w ∗ -compact and convex set conv w (A) ⊂ X ∗∗ . Theorem 3.130 (James [Jame2]) Let A be a w-closed subset of a Banach space X . A is w-compact if and only if every f ∈ X ∗ attains its supremum over A at some point of A. Proof: Assume that A is w-compact. Since every f ∈ X ∗ is w-continuous, the image f (A) of the w-compact set A must be compact in R, hence closed, and the supremum is attained. We present the proof of the other direction only for a separable set A (in particular, if the space X is separable). Let C = conv (A). Observe that, for f ∈ X ∗ , sup A f = supC f = sup w∗ f . By our assumption, A is a separable James boundary w∗

C

w∗

for C in X ∗∗ , hence by Theorem 3.122, (X ⊃) conv(A) = C . Note that if every f ∈ X ∗ attains its supremum over C, then C is bounded by Theorem 3.88. It follows that C (a convex and w-closed subset of X ) is w-compact (see the paragraph before Proposition 3.108). In particular, A is w-compact. The proof in the nonseparable case is much more difficult (see, e.g., [Dies2] or [Flor]). Corollary 3.131 (James [Jame2]) A Banach space X is reflexive if and only if every f ∈ X ∗ attains its norm. We will now show another James’ characterization of reflexivity. Theorem 3.132 (James [Jame1]) A Banach space X is reflexive if and only if there exists θ ∈ (0, 1) such that whenever {xn } is a sequence in S X with inf{u : u ∈ conv{xn }} ≥ θ , then there are n 0 ∈ N, u ∈ conv{x 1 , . . . , x n 0 } and v ∈ conv{x n 0 +1 , xn 0 +2 , . . . } such that u − v ≤ θ .

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3 Weak Topologies and Banach Spaces

Proof: (Oja [Oja2]) Assume that X is reflexive. Fix θ ∈ (0, 1), let {xn } be a sequence in S X . For an integer n ≥ 0 let K n = conv{xn+1 , xn+2 , . . . }.  Since {K n } is a nested sequence of nonempty weakly compact sets, there is x ∈ K n . As x ∈ K 0 , there is n 0 and u ∈ conv{x1 , . . . , xn 0 } such that x − u < θ/2. As x ∈ K n 0 , there is v ∈ conv{xn 0 +1 , x n 0 +2 , . . . , } such that x − v < θ/2. Then v − u < θ . Let X = X ∗∗ and any θ ∈ (0, 1) be given. We will construct a sequence {x n } failing the above condition. Denote Bθ = {F ∈ X ∗∗ : F ≤ θ }. By Lemma 1.37,

(x + Bθ ) (see Fig. 3.4). there is Fθ ∈ S X ∗∗ \ x∈X

BX ∗∗

Fθ

Fig. 3.4 The first step in the proof of Theorem 3.132

Bθ

V1

X

x1

0

As Fθ ∈ Bθ , there exists a convex w∗ -neighborhood V1 of Fθ in X ∗∗ such that V1 ∩ Bθ = ∅. Pick x 1 ∈ V1 ∩ S X . As Fθ ∈ x 1 + Bθ , there is a convex w∗ neighborhood V2 ⊂ V1 of Fθ which is disjoint with x 1 + Bθ . Choose x2 ∈ V2 ∩ S X . As Fθ ∈ / (conv{x 1 , x2 } + Bθ ), there is a convex w ∗ -neighborhood V3 ⊂ V2 of Fθ such that V3 ∩(conv{x 1 , x2 }+ Bθ ) = ∅. Pick x 3 ∈ V3 ∩ S X and continue by induction to get a decreasing sequence {Vn } of convex sets and a sequence {xn } of vectors with the following properties: conv{x n } ⊂ V1 , hence inf{u : u ∈ conv{xn }} ≥ θ . For all n ∈ N, conv{xn+1 , xn+2 , . . .} ⊂ Vn+1 , and Vn+1 ∩ (conv{x1 , . . . , xn } + Bθ ) = ∅, so given u ∈ conv{x 1 , . . . , xn } and v ∈ conv{xn+1 , xn+2 , . . .}, u − v > θ. We will now show several applications of Theorem 3.130. In Exercise 1.62 we mentioned that the closed convex hull of a  · -compact subset of a Banach space is  · -compact. The following theorem extends this result in a significant way. Theorem 3.133 (Krein) Let X be a Banach space. If A is a weakly compact set in X , then C := conv(A) is weakly compact in X . Proof: If f ∈ X ∗ , then sup A ( f ) = supC ( f ), so f attains its supremum over C (a w-closed set) at some point of C. It is enough now to apply Theorem 3.130. It is possible to prove Theorem 3.133 in full generality (see Exercise 3.177) by using just the separable version of James Theorem 3.130, whose proof, presented above, is much easier than the general case.

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Theorem 3.130 can be used to give another proof of (iii)⇒(i) in Theorem 3.109 (however, observe that the proof of Theorem 3.130 was only completed in the separable case). See Exercise 3.178. Note the similarity of the following result with the Lebesgue dominated convergence theorem for continuous functions. 3.11.8.5 The Rainwater–Simons Theorem Theorem 3.134 (Rainwater [Rain1], Simons [Simo]) Let B be a James boundary for a bounded subset M of the dual of a Banach space X . Let {xn } be a bounded sequence in X and x ∈ X . If f (x n ) → f (x) for all f ∈ B, then f (xn ) → f (x) for all f ∈ M. In particular, if B is a James boundary of a Banach space X and w f (x n ) → f (x) for all f ∈ B, then xn → x. Proof: Without loss of generality, we may assume that {xn } converges to 0 on elements in B. Let u = lim sup xn , where each x n is considered as a map from X ∗ into R. According to Theorem 3.124, sup M u = 0. If there exists x ∗ ∈ M such that x ∗ (xn ) → 0, we can find ε > 0 and a subsequence {x n k } such that |x ∗ (xn k )| > ε for all k ∈ N. If x ∗ (xn k ) > ε for infinitely many k s, we get a contradiction. Otherwise, use the same argument for the sequence {−xn } to get again a contradiction. Corollary 3.135 Let B be a James boundary of a Banach space X . Let A ⊂ X be a bounded and (relatively) w(X, B)-sequentially compact subset of X . Then A is (relatively) w-compact. Proof: Every sequence {an } in A has a subsequence {an k } that w(X, B)-converges to some point a ∈ A (a ∈ X ). By Theorem 3.134, {an k } is w-convergent to a. The set A is then (relatively) w-sequentially compact. The Eberlein–Šmulyan theorem gives the conclusion. Corollary 3.136 Let X be a separable Banach space. Let B be a James boundary of X . Let A be a bounded and (relatively) w(X, B)-compact subset of X . Then A is (relatively) w-compact. Proof: The set (B X ∗ , w ∗ ) is compact and metrizable, hence it is separable, and so it is its subset B. Let B0 be a countable and w∗ -dense subset of B. Given a sequence {an } in A, a diagonal argument gives a subsequence {an k } of {an } such that {b0 , an }n converges for every b0 ∈ B0 . The set B0 separates points of X , and (A, w(X, B))  w(X,B)  ( A , w(X, B) ) is compact, so the restrictions of the topologies w(X, B) and w(X,B)

) coincide. The sequence {an } has a w(X, B)-cluster w(X, B0 ) to A (to A w(X,B) point a ∈ A (a ∈ A ); hence the sequence {an k } is w(X, B)-convergent to a. This proves that A is (relatively) w(X, B)-sequentially compact. Corollary 3.135 gives the conclusion.

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Corollary 3.137 (Rainwater [Rain1]) Let X be a Banach space, let {xn } be a bounded sequence in X and x ∈ X . If f (xn ) → f (x) for every f ∈ Ext(B X ∗ ), w then x n → x. Corollary 3.138 Let K be a compact topological space. Let { f n } be a bounded w sequence in C(K ) and f ∈ C(K ). Then, if f n → f pointwise, we have f n → f . Proof: Use Corollary 3.137 and Lemma 3.116. The following result contains the conclusion of Pfitzner’s theorem in the particular setting of C(K ) spaces and the James boundary consisting of the extreme points of the closed dual unit ball. Theorem 3.139 (Eberlein, Šmulyan, Grothendieck) Let K be a compact space. Then, if A is a uniformly bounded subset of C(K ), the following are equivalent: The set A is (i-p) (relatively) T p -countably compact, (ii-p) (relatively) T p -sequentially compact, (iii-p) (relatively) T p -compact, (i-w) (relatively) w-countably compact, (ii-w) (relatively) w-sequentially compact, and (iii-w) (relatively) w-compact. Proof: The equivalence between (i-p), (ii-p), and (iii-p) holds for any subset of C(K ), see Theorem 3.52. Assume now that A is uniformly bounded. If A is (relatively) T p -sequentially compact and { f n } is a sequence in A, there exists a subsequence { f n k } of { f n } that T p -converges to some f ∈ A ( f ∈ C(K )). By Corollary 3.138, the sequence { f n k } is w-convergent to f . This proves that A is (relatively) w-sequentially compact. It is enough now to apply Theorem 3.109. Remark: The version of Theorem 3.139 for an arbitrary James boundary of C(K ) is given in Exercises 3.155, 3.156, and 3.157. Another important special instance of Pfitzner’s theorem is given in the next result, that covers the case of convex sets and arbitrary James boundaries. Theorem 3.140 (See [Gode3] and references therein) Let X be a Banach space. Let B be a James boundary of X . Then, a convex and w(X, B)-compact subset of X is w-compact. Proof: Let A be a convex and w(X, B)-compact subset of X . Consider the   locally convex space X, w(X, B) and the fact that B is bounded in the space   span(B), w(span(B), X ) ; then, according to Exercises 3.73 and 3.74, the set A is ∗ uniformly bounded on the set B. Since conv w (B) = B X ∗ , A is uniformly bounded on B X ∗ , i.e., A is bounded in X .

3.12

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141

In order to prove that A is w-compact, it is enough to show, by the Eberlein– Šmulyan theorem, that every countable subset {an : n ∈ N} of A is w-relatively compact. The boundedness of A allows us to define a linear and continuous mapping S : 1 → X by S((λn )) =

∞ 

λn an , for all (λn ) ∈ 1 .

n=1

We shall show that S is a w-compact mapping; this will finish the proof. By Theorem 13.34, it is enough to show that S ∗ : X ∗ → ∞ is w-compact. Let w∗ Y = span· (B), a closed linear subspace of (X ∗ , ·). By Fact 3.120, BY = B X ∗ . Since S ∗ is w∗ -w∗ -continuous, it suffices to show that S ∗ (BY ) is w-relatively compact. To that end, we shall prove that every μ ∈ B∗∞ attains its supremum on S ∗ (BY ) at some point of S ∗ (BY ) (or, equivalently, that every z := S ∗∗ (μ) attains its supremum on BY at some point of BY ). James’ Theorem 3.130 will give then the conclusion. There is a net {li }i∈I in B1 that w ∗ -converges to μ. For i ∈ I , put xi = S(li ). The net {xi }i∈I is in A, since A is convex and w-closed, so it has a w(X, B)-cluster point a ∈ A. Notice that, moreover, {xi }i∈I is w∗ -convergent to z. Obviously, b, a = b, z for every b ∈ B, so a and z coincide on Y . Since B is a James boundary, a attains its supremum on BY at some point of BY , and so does z. This concludes the proof. We refer to [FLP] for an approach to James boundaries via the Choquet representation theory (see Theorem 3.74).

3.12 Remarks and Open Problems Remarks 1. The dual Mackey topology, i.e., the topology on X ∗ of the uniform convergence on the family of all convex, balanced and w-compact subsets of X , is discussed, e.g., in [HMVZ, Section 3.1]. See also Exercises 3.40 and 3.41. 2. We refer, e.g., to the article [PeWo, pp. 1361–1423] for a discussion on Sobolev spaces. 3. For a proof of the Eberlein–Šmulyan theorem by using Schauder bases’ techniques we refer to [Pelc3d], see [AlKa, p. 23]. 4. Bourgain and Talagrand [BoTa] gave the positive solution to the boundary problem for an arbitrary Banach space and its James boundary consisting of all extreme points of the closed dual unit ball. Godefroy’s boundary problem appears in [Gode2], see also [Gode3]. For a convex set (and an arbitrary James boundary) the result was implicit in Simons [Simo], see also [Gode3, p. 44] and [Flor]. This is Theorem 3.140. The answer was also known to be positive for Banach spaces X such that (B X ∗ , w ∗ ) is angelic [CasVe]. For the case

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of C(K ) spaces, the positive solution was in [CasGod]. This is presented in Exercise 3.157. Finally, a complete positive solution has been given by Pfitzner in [Pfi].

Open Problems 1. Assume that for a Banach space X ,  f − f n  → 0 whenever f, f n ∈ S X ∗ satisfy w∗

f n → f . It is apparently unknown whether it is then true that Y ∗ is separable for all separable subspaces Y of X . 2. A separable Banach space X is reflexive if and only if it admits an equivalent 2Rnorm, that is, a norm · such that a sequence {xn } in S X is convergent whenever lim xn + xm  = 2, [OdSc3]. This is unknown for general nonseparable n,m→∞ spaces.

Exercises for Chapter 3 3.1 Let f : E → R be a linear functional on the real topological vector space E such that Ker f is closed and f = 0. Show, without using Proposition 3.19, that, for every α ∈ R, the set {x ∈ E : f (x) < 0} is open, and that we have {x ∈ E : f (x) < 0} = {x ∈ E : f (x) ≤ 0}. Conclude that the same is true whenever 0 is replaced by any α ∈ R and the < and ≤ symbols are replaced, respectively, by > and ≥. Hint. Let x0 ∈ E such that f (x 0 ) < 0. Since Ker f is closed, we can find a balanced neighborhood U of 0 such that (x 0 +U )∩Ker f = ∅. By Exercise 1.3 we get f (x) < 0 for all x ∈ x0 + U . This proves that {x ∈ E : f (x) < 0} is open. Its complement, i.e., the set {x ∈ E : f (x) ≥ 0}, is closed. It follows that {x ∈ E : f (x) > 0} ⊂ {x ∈ E : f (x) ≥ 0}. Let x 0 ∈ E such that f (x0 ) = 0, and let y ∈ E such that f (y) > 0. Then [y, x0 ) ⊂ {x ∈ E : f (x) > 0}, hence x0 ∈ {x ∈ E : f (x) > 0}; this proves that {x ∈ E : f (x) ≥ 0} ⊂ {x ∈ E : f (x) > 0}. 3.2 Let (E, T ) be a topological vector space. Assume that there exists a convex neighborhood C of 0. Let pC be the Minkowski functional of C (use Definition 2.9 now in the setting of topological vector spaces). Prove the following assertions. (i) pC is a finite non-negative positively homogeneous and subadditive continuous functional on E. (ii) Int(C) = {x ∈ E : pC (x) < 1} ⊂ C ⊂ {x ∈ E : pC (x) ≤ 1} = C. (iii) If C is moreover balanced then pC is a continuous seminorm (a continuous norm if C is also bounded; in this case, this norm induces the topology T ). Hint. (i) for the positive homogeneity and the subadditivity proceed as in the proof of Lemma 2.11; to prove that pC (0) = 0, follow the argument there, too. If {xi } is a net converging to 0, then for every λ > 0 we have λxi → 0, hence there exists i 0 such that λxi ∈ C for i ≥ i 0 . Then pC (λxi ) ≤ 1, i.e., pC (xi ) ≤ 1/λ. This proves the continuity of pC at 0. Lemma 2.10 holds in this context with the same proof, so

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pC is continuous. (ii) Follow again the proof of Lemma 2.11. (iii) Follow the hint in Exercise 2.21. If C is bounded and pC (x) = 0 it follows that x ∈ U for every neighborhood U of 0, hence x = 0. 3.3 Let E be a real topological vector space. Let f : E → R be a linear functional and p : E → R a function that is continuous at 0, and such that f ≤ p. Then f is continuous. Hint. Since − f (x) = f (−x) ≤ p(−x), we get − p(−x) ≤ f (x) ≤ p(x) for all x ∈ E, hence f is continuous at 0. The conclusion follows from Proposition 3.19. 3.4 Let L 0 be the vector space of all Lebesgue measurable functions on [0, 1] with

1 |x(t)−y(t)| the metric ρ(x, y) = 0 1+|x(t)−y(t)| dt. Show that ρ(x n , x) → 0 if and only if xn → x in measure, and that L 0 is a topological vector space. For the definition of convergence in measure, see the proof of Theorem 4.55. Hint. Let λ be the Lebesgue measure in [0, 1]. Denote Aε = {t : |(x n − x)(t)| ≥ ε}, Bε = {t : |(x n − x)(t)| ≤ ε}, then 

1

0

|xn − x| dt = 1 + |xn − x|

 Aε

|xn − x| dt + 1 + |xn − x|

 Bε

|xn − x| dt 1 + |xn − x|

≤ λ{t : |(xn − x)(t)| ≥ ε} + ε. If xn does not converge to 0 in measure, there exists δ > 0 such that λ(Aε ) ≥ δ for all ε > 0. Then 

1 0

|xn − x| dt ≥ 1 + |xn − x|

 Aε

|xn − x| dt ≥ 1 + |xn − x|

 Aε

ε εδ dt = , 1+ε 1+ε

and the latter does not converge to 0. We used here the monotonicity of the function t 1+t . For the reverse implication, use a similar argument. To observe that the convergence in measure generates a topology in which both operations are continuous, write |αn xn − α0 x0 | ≤ |αn | |xn − x0 | + |αn − α0 | |x0 | and note that {t : |αn xn − α0 x0 | ≥ ε} ⊂ {t : |αn | |x m − x 0 | ≥ ε/2} ∪ {t : |αn − α0 | |x 0 | ≥ ε/2}. Measures of the sets on the right hand side go to zero (for the second summand use the Chebyshev inequality). 3.5 Let X be the vector space of all continuous functions on (0, 1). For f ∈ X and r > 0, let V ( f, r ) = {g ∈ X : |g(x) − f (x)| < r for all x ∈ (0, 1)}. Let T be the

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topology on X that these sets generate. Show that the addition is T -continuous but scalar multiplication is not. Hint. { n1 x1 } → 0. 3.6 Let 0  < p < 1. Let  p be the space of all sequences  of realpnumbers x = (xi ) |xi − yi | . Show that d is a such that |x i | p < ∞, with the metric d(x, y) = complete metric and  p is a topological vector space but not a locally convex space.  Show that ∗p = ∞ and the set U = {(xi ) : |xi | ≤ 1} has the property that f (U ) is bounded for every f ∈ ∗p but U is unbounded in  p . Hint. To show that ∗p = ∞ , note that the topology of  p is stronger than that of 1 and thus every element of ∞ is a continuous linear functional on  p . If f ∈ ∞ , we find x ∈  p such that fi xi diverges. To see that  p is not locally convex, consider the set A = {x n } ⊂  p , where x n ∈  p is defined by xin = n p−1 if i = n and xin = 0 otherwise. This is a relatively compact set in  p whose convex hull is not bounded as the sequence {y n } defined by y n = n −1 nk=1 x k is unbounded in  p . The latter is seen by an elementary calculation, as d(0, y 2n ) can be estimated from below by (2n)− p

2n  k=n

n (1− p) → ∞. 2p 2

kp

2− p

≥ (2n)− p n n p

2− p

=

3.7 Show that if {xn } is a sequence in a metrizable topological vector space X and x n → 0, then there are positive numbers γn such that γn → ∞ and γn xn → 0. However, if X is the vector space of all real-valued functions on [0, 1] with the topology of pointwise convergence on [0, 1], then there is a sequence { f n } ⊂ X that pointwise converges to zero in X and if γn is any sequence of numbers such that γn → ∞, then {γn f n } does not converge to zero. Hint. First part is standard, γn = √ρ(x1 ,0) . To see the second part, note that the n cardinality of the family of all sequences converging to zero is c. 3.8 Let X, Y be topological vector spaces. Assume that T is a linear mapping from X onto Y such that T −1 (0) is closed in X and Y is finite-dimensional. Show that T is a continuous and open mapping. Hint. If U is a balanced neighborhood of zero in X and {ei } is a basis for Y , then for some n, T (nU ) ⊃ {ei }. The convex hull of {±ei } contains 0 as an interior point. This shows that T is an open mapping. T is an algebraic isomorphism of X/T −1 (0) and Y . Since all linear vector topologies on finite-dimensional spaces coincide, and since the mapping x → xˆ of X onto X/T −1 (0) is continuous, T continuous. 3.9 Prove that in an angelic topological space, the following classes of sets coincide: (relatively) compact, (relatively) countably compact, (relatively) sequentially compact. Hint. (1) (Relative) compactness implies (relative) countable compactness. (2) Relative countable compactness implies relative compactness by definition of angelicity. (3) Let A be a relatively compact set. Given a sequence {an } in A, if the set

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A0 := {an : n ∈ N} has no accumulation point, then A0 is finite, and so the sequence {an } has a convergent subsequence. Otherwise, let z be an accumulation point of A0 . If z = an infinitely often, then {an } has a convergent subsequence. If not, we may assume z = an for all n ∈ N. The angelicity gives a sequence in {an : n ∈ N} that converges to z, hence a subsequence of {an } that also converges to z. This proves that A is relatively sequentially compact. (4) Obviously, (relative) sequential compactness implies (relative) countable compactness. (5) If A is countably compact then it is relatively compact; an element z in A is the limit of a sequence in A, so z ∈ A. We get that A is closed, hence compact. (5) If A is compact it is, by (3), relatively sequentially compact; it is also closed, hence sequentially compact. 3.10 Let (E, T ) be a topological vector space. Prove that a set A ⊂ E is relatively compact if and only if every net in A has a convergent subnet. Hint. The necessary condition is clear. To prove sufficiency, let {a i : i ∈ I, ≤} be a net in A. We shall prove that it has a subnet that converges to some point of A. Let U be a local base of balanced neighborhoods in E. Given i ∈ I and U ∈ U, find ai,U ∈ (a i + U ) ∩ A. Define in I × U a partial order by (i 1 , U1 ) ≺ (i 2 , U2 ) whenever i 1 ≤ i 2 and U1 ⊃ U2 . Prove that (I × U, ≺) is directed upwards. Since {ai,U : (i, U ) ∈ I × U, ≺} is a net in A, there exists a subnet {aφ( j) : j ∈ J, } that converges (to an element a ∈ A). If p1 : I × U → I denotes the first coordinate mapping, p1 ◦ φ defines a subnet of {a i : i ∈ I, ≤} that converges to a. Indeed, the mapping p1 ◦ φ satisfies the condition for defining a subnet; moreover, given U ∈ U there exists j0 ∈ J such that for all j0  j, aφ( j) ∈ a + U . Recall that aφ( j) ∈ a p1 ◦φ( j) + U , hence a p1 ◦φ( j) ∈ aφ( j) + U ∈ a + 2U for all j0  j. This proves the assertion. 3.11 Let X be a normed space. Prove that X is complete in the canonical metric given by its norm if and only if it is complete in its norm topology as a topological vector space. Hint. If every Cauchy net in X converges, then also every Cauchy sequence (which is also a net) converges. Assume that X is a complete normed space and {x α }α∈I is a Cauchy net in X . Given n ∈ N, find i n ∈ I , i n > i n−1 , such that (x α − xβ ) ∈ n1 B X for all α, β ≥ i n , in particular, x α ∈ B(xin , n1 ) for all α ≥ i n . Then {xin } is a Cauchy sequence which must converge to some x ∈ X . We claim that x is the limit of the net {xα }α∈I : Choose n ∈ N. There exists n 0 ∈ N such that n 0 ≥ n and xim ∈ B(x, n1 ) for all m ≥ n 0 , in particular, xin0 ∈ B(x, n1 ). Then for all α ∈ I such that α ≥ i n 0 we get xα ∈ B(xin0 , n10 ), hence xα ∈ B(x, n2 ). Consequently, xα → x. 3.12 Prove that the space RN (equipped with the product topology) is a separable Fréchet space that is not normable.

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Hint. The separability follows from the fact that it is a countable product of separable spaces. To show the completeness is also standard. The fact that it is not normable follows from the non-boundedness of the elements in the local base. X . Show 3.13 (Klee)  Let K 1 , . . . , K n be open convex sets in a locally convex space n−1 such that that if K = ∅ then there is a bounded operator L from X into R i  L(K i ) = ∅. Hint. Let K = {(x1 − x2 , x1 − x3 , . . . , x 1 − x n ) : xi ∈ K i , i = 1, . . . , n}. Then K ⊂ X n−1 is an open convex set that does not contain 0. Thus by the Hahn–Banach n−1 ∗ ∗ ∈ X ∗ such that i=1 xi (x1 − xi ) > 0 for all theorem, there exist x1∗ , x2∗ , . . . , xn−1  ∗  ∗ xi ∈ K i , i = 1, . . . , n. Put L(x) = x 1 (x), . . . , xn−1 (x) for x ∈ X . 3.14 Let (E, T ) be a locally convex space. Let K 1 , . . . , K n be convex (respectively convex and balanced) compact of E. Then the convex hull (respectively the

subsets n convex and balanced hull) of i=1 K i is again compact. Hint. For the convex case, observe that  conv

n 

i=1

 Ki

=

 n  i=1

αi ki : αi ≥ 0, ki ∈ K , for i = 1, 2, . . . , n,

n 

 αi = 1 .

i=1

n Note then that {(α1 , . . . , αn ) : αi ≥ 0 for i = 1, 2, . . . , n, i=1 αi = 1} is a n . It is enough now to use a convergence argument on nets in compact subset of K   n n αi = 1 and balanced case change αi ≥ 0 and i=1 conv i=1 K i . For the convex  n to (α1 , . . . , αn ) ∈ Kn with i=1 |αi | ≤ 1. + 3.15 Given a family " {E γ }γ ∈Γ of vector spaces, its algebraic direct sum Γ E γ is the subspace of Γ E γ consisting of all elements + x = (xγ ) having finitely many non-zero coordinates. For γ ∈ Γ , let Iγ : E γ → Γ E γ be the natural embedding. γ If each E γ is a locally convex space and {Bβ } is a base of neighborhoods of 0 + of 0 in E γ , endow Γ E γ with the topology having as a base  of neighborhoods γ the collection of sets that are convex and balanced hulls γ ∈Γ Iγ (Bβ ). Prove that + Γ E γ endowed with this topology is a locally convex space (called the locally convex direct sum of the family {E γ }Γ . + + Prove that each E γ is a subspace of Γ E γ . Prove that Γ E γ is complete if each E γ is complete. Prove that the locally convex direct sum and the topological product of a finite family of locally convex spaces agree. Hint. Standard computation. 3.16 Let E be a locally convex space and let {E γ }γ ∈Γ be a family of subspaces of E. We say that E is the locally convex direct + sum of the family {E γ }γ ∈Γ if E is isomorphic to the locally convex direct sum Γ E γ defined in Exercise 3.15. Prove that if E is a locally convex space and F is a finite-dimensional subspace of E, there exists a subspace G of E such that E = F ⊕ G (in this case, E/F is isomorphic

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to G and E/G is isomorphic to F). Observe that the result is no longer true in the class of topological vector spaces. Hint. Since F is isomorphic to n2 if dim (F) = n (see Proposition 3.13), we can find { f i : i = 1, 2, . . . , n} in F and { fi∗ : i = 1, 2, . . . , n} in F ∗ such that  f j∗ , f i  = n Ker( f i∗ ). It is easy to prove that E = F ⊕ G. If the δi j for all i, j. Put G = i=1 result should be true in the class of topological vector spaces, by taking a non-zero (continuous) linear functional on F and extending this to be 0 on G we shall always obtain non-zero continuous linear functionals on every topological vector space, against the example in Section 3.2. 3.17 Let E be a locally convex space and A a nonempty subset of E. Prove directly ∗ -closed set. that A◦ is a w ◦ { f : x( f ) ≤ 1} and the sets are w∗ -closed, since vectors from X Hint. A = x∈A

are w∗ -continuous as elements of (X ∗ , w ∗ )∗ . 3.18 Let E be a locally convex space, α } a nonempty and {A ◦  family of nonempty ◦  Aα , Aα = (Aα )◦ and, if λ = 0 subsets of E. Show that (Aα )◦ ⊂ and A is a nonempty subset of E, (λA)◦ = λ1 A◦ . Show that if A1 and A2 are two nonempty subsets of E such that A1 ⊂ A2 then (A2 )◦ ⊂ (A1 )◦ . Hint. Use the definition. 3.19 Show that if X is a normed space, then (B X )◦ = B X ∗ , and (B X ∗ )◦ = B X . Hint. Use the definition. 3.20 (Corson) Let K be a convex compact set in a locally convex space E. Show that if Ext(K ) is countable then K is metrizable. Hint. Let X be the Banach space of all continuous affine functions on K with the sup-norm. Identify K with its image in X ∗ . By the bipolar theorem, B X ∗ is the w ∗ closed convex hull of K ∪(−K ). Hence Ext(B X ∗ ) is countable by Theorem 3.66 and thus X ∗ is separable by Corollary 3.126. Thus X is separable and hence (B X ∗ , w ∗ ) is metrizable, so K is metrizable. 3.21 Give an example of two locally convex spaces E and F, a continuous linear mapping T : E → F, a compact convex subset K of E and an extreme point e0 of K such that T (x 0 ) is not an extreme point of T (K ). Hint. E := R2 , F := R, T the orthogonal projection. 3.22 Let E, F two locally convex spaces, T : E → F a continuous linear mapping, S a nonempty subset of E, H ⊂ F a closed supporting manifold of T (S). Prove that T −1 (H ) is a supporting manifold of S. Hint. Obviously, T −1 (H ) ∩ S = ∅. Let (x, y) ⊂ S such that (x, y) ∩ T −1 (H ) = ∅. Consider first the case T (x) = T (y) and then the case T (x) = T (y). Use the definitions.

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3.23 Let D be a convex set in a Rn . Show that either Int(D) = 0 or D lies in a proper subspace of Rn . m be a maximal linearly independent set in D. Hint. Assume that 0 ∈ D, let {ei }i=1 m Since D ⊂ span{ei }i=1 , it remains to prove that Int(D) = ∅ if n = m. Consider    C = conv {ei } ∪ {0} ⊂ D, note that { αi ei : 0 ≤ αi ≤ m1 } ⊂ C. Let x0 =   1 1 1 αi ei : |αi − 2m | < 2m } is an open neighborhood of x 0 and 2m ei . Then B = { B ⊂ C ⊂ D. 3.24 Let K be a nonempty convex compact subset of a finite-dimensional topological vector space, and let f : K → R be a continuous convex function. Prove that f attains its supremum at an extreme point of K . Hint. That f is bounded and attains its supremum is clear. Use now Carathéodory’s result 3.70. 3.25 Assume that C is a convex compact set in a locally convex space X and D is a closed subset of C. Show that if μ is a probability measure on D, then there is a unique point x ∈ C represented by μ. Moreover, the mapping μ → x is continuous from the w∗ -topology in C(D)∗ into the topology of X . Prove this result. Hint. For f ∈ X ∗ put H f = {x ∈ D : f (x) = μ( f )}. We will show that  f ∈X ∗ H f ∩ C = ∅. Since  C is compact, by the finite intersection property it is enough to show that ( H f i ) ∩ C = ∅ for every finite set f 1 , . . . , f n ∈ X ∗ . Define a mapping T from X into Rn by T (x) = ( f 1 (x), . . . , f n (x)). Then T (C) is compact

and convex. It suffices to show that p ∈ T (C), where p = / T (C). Then there is a linear functional ( D f 1 dμ, . . . , D f n dμ). Assume that p ∈ on Rn which strictly separates p from the compact convex set T (C), i.e., there is a = (a1 , . . . , an ) such that (a, p) > sup{ a, T (x) : x ∈ C}, where (., .) is the inner product on Rn . Define g ∈ X ∗ by g = ai f i . Then we have 

  ai f i dμ D D   ai f i (x) = sup(g). = ai μ( f i ) = (a, p) > sup

g dμ = D

 



ai f i

dμ =

C

C

This is a contradiction as μ(D) = 1. If the mapping in question is not continuous as stated, then there are measures μα and μ such that μα → μ and their corresponding points xα and x do not converge in X . Since C is compact, assume without loss of generality that x α → y = x. Then for every f ∈ X ∗ we have f (xα ) = μα ( f ) → μ( f ) = f (x). Because of continuity of f we also have f (y) = lim f (xα ). Thus f (x) = f (y). Since X ∗ separates points, we have a contradiction. 3.26 Give an example of an affine function, in the sense as in the proof of Choquet’s theorem, defined on a metrizable compact subset D of a Banach space X , that is not a restriction to D of a function of the form f (x) = k + ϕ(x) for k ∈ R, ϕ ∈ X ∗ .

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Hint. X = 1 , D = {x ∈ 1 : |xi | ≤ 4−i }, f (x) =



2i xi .

3.27 Let X be a separable Banach space. Assume that (B X ∗∗ , w∗ ) is angelic.  Show that then every w∗ -compact convex set C in X ∗ satisfies C = conv Ext(C) . Hint. Use Choquet’s theorem in (X, w), then the representation also holds for f ∈ X ∗∗ by the Lebesgue dominated convergence theorem. Then use the separation theorem. Note that the result follows directly from Corollary 3.129. 3.28 Assume that f j ∈ D (Ω) and f (ϕ) = lim f j (ϕ) exists finite for every ϕ ∈ j

D(Ω). Show that f ∈ D (Ω). Hint. By the Banach–Steinhaus theorem, f is continuous on every D K . 3.29 Let Ω be an open set in Rn and { f j } ⊂ D (Ω) be such that { f j (ϕ)} is bounded for every ϕ ∈ D(Ω). Show that there is a subsequence { f jk } and f ∈ D (Ω) such that lim f jk (ϕ) = f (ϕ) uniformly for ϕ on every bounded subset of D(Ω). Hint. By the Banach–Steinhaus theorem, the restrictions of f j to D K are equicontinuous. Then use the Arzelà–Ascoli theorem. 3.30 Let C ∞ [0, 1] be the vector space of all real-valued functions on [0, 1] whose all derivatives are continuous on [0, 1]. Let k ∈ N and p ∈ [1, ∞). Define the 1  (l)  p p , where f (l) denotes the lth  f norm  · kp on C ∞ by  f kp = p 0≤l≤k derivative of f and  ·  p is the norm from L p . The Sobolev space W pk [0, 1] is the completion of the normed space (C ∞ [0, 1],  · kp ). Show that W pk [0, 1] is isomorphic to L p [0, 1]. w

3.31 Let {xn } be a sequence in a Banach space X . Prove that if xn → x then {x n } is bounded and x ≤ lim inf xn  (the second statement has a more general formulan tion: the norm is a w-lower semicontinuous function, see Definition 17.6). w∗

Let { f n } be a sequence in X ∗ . Prove that if f n → f then { f n } is bounded and  f  ≤ lim inf  f n  (the second statement has a more general formulation: the dual n

norm in X ∗ is a w∗ -lower semicontinuous function). Hint. Proof is similar to that of Corollary 3.86. To prove lower semicontinuity in full generality consider the w-closedness (w ∗ -closedness) of the level sets, i.e., the balls centered at 0 in X (in X ∗ ), see the paragraph after Definition 17.6. For X use Mazur’s Theorem 3.45, for X ∗ the fact that the polar of a subset of X in X ∗ is w∗ -closed. 3.32 Let X be a normed space. (i) Let A be a nonempty bounded subset of X . Prove w that diam A = diam A . (ii) Let B be a nonempty bounded subset of X ∗ . Prove that ∗ w diam B = diam B . Hint. (i) The w-lower semicontinuity of the norm. (ii) The w∗ -lower semicontinuity of the dual norm.

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3.33 Let {xn } be a sequence in a Banach space X . Prove that x n → x if and only if {x n } is bounded and the set { f ∈ X ∗ : f (xn ) → f (x)} is dense in X ∗ . w∗

Similarly, let { f n } be a sequence in X ∗ . Prove that f n → f if and only if { f n } is bounded and the set {x ∈ X : fn (x) → f (x)} is dense in X . In particular, we have the following corollary. Assume that {x n } is bounded and M ⊂ X ∗ is such that span(M) = X ∗ and f (xn ) → f (x) for all f ∈ M. Then w x n → x. Analogous statement is true for w∗ -convergence. Hint. One direction is a direct consequence of the Banach–Steinhaus Theorem 3.85. The other one is standard: Let f ∈ X ∗ . Given ε > 0, approximate f by g such that g − f  < ε and g(x n ) → g(x). In the estimate | f (xn ) − f (x)| ≤ | f (xn ) − g(xn )| + |g(xn ) − g(x)| + |g(x) − f (x)| we pass to the limit and obtain lim sup | f (xn ) − f (x)| ≤ ε sup x n  + 0 + εx. Since ε > 0 was arbitrary, we get f (x n ) → f (x) as needed. 3.34 Let {xn } be a sequence in a Banach space X . Assume that f (xn ) → 0 for every w f ∈ A ⊂ X ∗ , where A is a set of second Baire category in X ∗ . Show that xn → 0. Hint. See the proof of the Banach–Steinhaus theorem. w

3.35 Let X be a normed space. Show that if {x n } is Cauchy and xn → 0, then xn → 0 in norm. Hint. xn ∈ xm + ε B X and xm + ε B X is weakly closed. 3.36 Let X be a Banach space and let {xn } be a sequence in X that w-converges to 0. Prove that the and balanced hull M of {xn : n ∈ N} is equal  closed convex ∞ to the set B := { ∞ λ x : n n n=1 n=1 |λn | ≤ 1}, and that B is w-compact ([Koth, 20.9(6)]). Hint. The w-compactness n of M∞follows from Krein’s theorem. Given λ = (λn ) ∈ B1 , the  sequence { i=1 λi xi }n=1 is  · -Cauchy, so it converges in X (to the ∞ λi xi ) and we get B ⊂ M. Let ϕ : 1 → X be defined as element  i=1 ∞ ϕ(λ) = i=1 λi xi for λ = (λi ) ∈ 1 ; it is well defined,  · 1 - · -continuous and linear; moreover, M = ϕ(B1 ). Let ϕ ∗ : X ∗ → ∞ be its adjoint mapping. Since, for every x ∗ ∈ X ∗ , we have ϕ ∗ (x ∗ ), en  = x ∗ , ϕ(en ) = x ∗ , xn  →n 0, we get ϕ ∗ (x ∗ ) ∈ c0 , hence ϕ is w(1 , c0 )-w-continuous. Thus, by Alaoglu’s theorem, M is w-compact, in particular w-closed. This proves that M = B. 3.37 Assume that {xn } is a weakly convergent sequence in an infinite-dimensional Banach space X . Show that conv{x n } does not have any interior point. This is not the case for the w ∗ -convergence as the standard unit vectors of 1 show.

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Hint. (Veselý, Zanco) Assume that xn → 0 and that K = conv{xn } contains an interior point 0. Let x be an arbitrary point of the topological boundary of K . Using the Minkowski functional of K , we construct a functional f supporting K at x. Let N be such that f (x n ) < f (x)/2 for n > N . Put A = conv{xn : n ≤ N } and B = conv{x n : n > N }. Then there are λk ∈ [0, 1], ak ∈ A a bk ∈ B such that the points yk = (1 − λk )ak + λk bk tend to x. By applying f we can see that λk → 0 as otherwise f could not have a maximum at x. Hence ak → x and A is closed. Thus x ∈ A ⊂ conv{x n }. Hence X would have a countable Hamel basis, a contradiction. 3.38 Assume that B X ∗ is angelic in its w ∗ -topology and assume that M is a linearly dense set in X . Assume that for a w ∗ -linearly dense set D ⊂ X ∗ , D has a countable support on M. Then all elements of X ∗ have countable support on M. Hint. Let S be the collection of all elements of X ∗ that have countable support on M. Then observe that S ∩ B X ∗ is w∗ -closed and use Theorem 3.92 to see that S is w ∗ -closed in X ∗ . The set S is assumed to be w∗ -dense in X ∗ . 3.39 The Josefson–Nissenzweig theorem asserts that for every Banach space X , w∗

there is a sequence { f n } ⊂ S X ∗ such that f n → 0 ([Dies2]).

w∗

Show that given f ∈ B X ∗ , there is a sequence { f n } ⊂ S X ∗ such that f n → f . Hint. Consider the sphere S centered at f with radius 1. Then there are gn ∈ S such w∗

that (gn − f ) → 0. Project all gn onto S X in a radial way; namely, given gn , let f n be the intersection of S X and the ray emanating from f and going through gn . Then f n = f + cn (gn − f ) for some cn ≥ 0 (use the fact that f ∈ B X ∗ , draw a picture). w∗

We also have cn ≤ 2 and hence ( f n − f ) → 0 as needed. 3.40 Let X be a Banach space and let TK be the topology on X ∗ of the uniform convergence on the family K of all  · -compact subsets of X . Prove that (X ∗ , TK ) is complete. Hint. Let {x α∗ } be a TK -Cauchy net in X ∗ ; it is also w ∗ -Cauchy. Since (X # , w(X # , X )) is complete (see Proposition 3.6), the net {xα } is w(X # , X )convergent to some x0# ∈ X # (here X # denotes the space of all linear forms on X ). Fix a  · -compact convex and balanced subset K of X and some ε > 0. There exists α0 such that supx∈K |x α∗ − xβ∗ , x| < ε/2 for all α, β ≥ α0 . Then supx∈K |xα∗ − x0# , x| ≤ ε/2 for all α ≥ α0 . Since x α∗0 is continuous, there exists a neighborhood U of 0 in X such that supx∈K ∩U |x α∗0 , x| ≤ ε/2. It follows that  supx∈K ∩U |x 0# , x| ≤ ε. This proves that x #  K is continuous at 0. Then, by a standard convexity argument, it is (uniformly) continuous on K . Assume now that x0# is not continuous on X . We can then find a sequence {xn } in X such that x n → 0 and x # , xn  → 0. This is a contradiction, since the closed convex and balanced hull of {xn : n ∈ N} is a  · -compact subset of X . 3.41 Prove that, for a Banach space X , the space (X ∗ , μ(X ∗ , X )) is complete, where μ(X ∗ , X ) denotes the Mackey topology on X ∗ associated to the dual pair X ∗ , X .

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Hint. Use Exercise 3.40 and Proposition 17.17. 3.42 [Groth3] A Banach space is said to have the Grothendieck property if every w ∗ -convergent sequence in X ∗ is w-convergent. (i) Show that if X has the Grothendieck property and T is a bounded operator onto a Banach space Y , then Y has the Grothendieck property. (ii) Show that a quotient of a Banach space with the Grothendieck property has the Grothendieck property. (iii) Show that every separable Banach space with the Grothendieck property is reflexive. (iv) Show that every separable quotient space of ∞ is reflexive. Hint. (i) Direct proof, use the fact that T ∗ is w ∗ -w ∗ -continuous and T ∗∗ is onto (Exercise 2.49). (ii) It follows from (i). (iii) The Eberlein–Šmulyan Theorem 3.109 for the dual unit ball. (iv) ∞ has the Grothendieck property ([Dies2, Chapter VII], [HMVZ, Chapter VII]). −1 ∗ ∗ 3.43 Prove that if F ∈ ∗∗ ∞ \∞ , then F (0) (⊂ ∞ ) is w -sequentially closed but not w ∗ -closed. Hint. The w∗ -sequential closedness follows from the fact that in ∗∞ , every w ∗ convergent sequence is w-convergent (the Grothendieck property of ∞ , see Exercise 3.42).

3.44 Does there exist a bounded operator from ∞ onto c0 ? Does there exist a bounded operator from ∞ onto 1 ? Does there exist a bounded operator from ∞ onto C[0, 1]? Hint. No. It is known that ∞ has the Grothendieck property ([Dies2]). Use Exercise 3.42. 3.45 Is c0 , 1 or C[0, 1] isomorphic to a quotient of ∞ ? Hint. No, see Exercise 3.44. w

3.46 Show that if X is infinite-dimensional, then 0 ∈ S X . w Using this, as in the previous exercise we can show that S X = B X for infinitedimensional spaces. Hint. Consider a neighborhood V = {x : | f i (x)| < ε for i = 1, . . . , n}. Since all f i−1 (0) are 1-codimensional (Exercise 2.5), f i−1 (0) is finite-codimensional. As X is infinite-dimensional, there is x ∈ S X ∩ V . 3.47 If p > 1 show that a sequence { f n } in L p [0, 1] converges to zero weakly if x and only if it is bounded in L p [0, 1] and 0 f n →n 0 for every x ∈ [0, 1]. Hint. Linear denseness of step functions.

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3.48 In c0 , let xn = nen , where en is the standard nth unit vector, n ∈ N. Show that xn → 0 pointwise, but not weakly by finding a concrete f ∈ 1 so that { f (xn )} is unbounded.  Hint. Put f = n 2nn e2n . w

3.49 Let {ei } be an orthonormal basis of a Hilbert space H . Show that ei → 0. Hint. (ei , e j ) → 0 as i → ∞ for each j, and span{e j } = H = H ∗ . 3.50 Show that in the space 1 , 0 is not in conv{ei }. Hint. Consider (1, 1, 1, . . . ) ∈ ∞ . w∗

3.51 Let X = 1 ∼ = c0∗ . Show that the standard unit vectors ei satisfy ei → 0 but w not ei → 0. Hint. Show that ei (x) → 0 for x ∈ c0 . For the second part, use the previous exercise. 3.52 Show that in the space ∞ , the set {ei } ∪ {0} is weakly compact but not norm compact. w Hint. The set is weakly compact in c0 as ei → 0 in c0 . 3.53 Show that the set {ei } of the standard unit vectors in 2 is norm closed but not w-closed in 2 . Hint. If x n ∈ {ei } and x n → x in 2 , then {x n } is eventually constant. Recall (Exercise 3.49) that the origin is in the w-closure of {ei }. ∞ be the standard unit vectors in  . Show that A = {e }∞ ∪ {0} is 3.54 Let {ei }i=1 2 i i=1 a weakly compact set in 2 . w Hint. By Exercise 3.49, ei → 0.

3.55 Let {ei } be the sequence of the standard unit vectors in 2 . Show that 0 ∈ w √ √ { nen } and that no subsequence of { nen } converges weakly to 0. Note that this fact also shows that the weak topology of 2 is not metrizable. 1 2 n Hint. Let V be the neighborhood of 0 given by vectors nx , x k, . . . , x in 2 and ε > 0. Consider the element z ∈ 2 defined by z i = k=1 |x i |. Note that for an / 2 . Therefore infinite number of indexes i we have |zi |2 < εi since otherwise z ∈ √ k )| < ε for k = 1, . . . , n, in for an infinite number of indexes i we have | ie (x i √ particular V ∩ { iei } = ∅. The second part follows from Exercise 3.31. w

3.56 Let f n ∈ C[0, 1],  f n  ≤ 1. Show that f n → 0 in C[0, 1] if and only if f n (x) → 0 for every x ∈ [0, 1]. This shows that for bounded sequences in C[0, 1], weak convergence is equivalent to pointwise convergence. An analogous statement is true for c0 . Hint. Use Corollary 3.137 and Lemma 3.116.

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3.57 Let x, xn ∈ 2 be such that xn → x in 2 . Show that there is a subsequence {xn k } such that the Cesàro means xn 1 + xn 2 + · · · + x n k k converge to x in 2 (the Banach–Saks theorem). Show that an analogous statement is true for c0 . Spaces that have this property are called spaces with the weak Banach–Saks property. On the other hand, a Banach space X is said to have the Banach–Saks property if the former statement holds for every bounded sequence in X . Since every bounded sequence in 2 has a w-convergent subsequence, we are proving in fact that 2 has the Banach–Saks property. Hint. 2 case: Assume x = 0. Let xn  ≤ M for all n. Put n 1 = 1 and if n 1 , . . . , n k were chosen, pick n k+1 such that |(x n j , x n k+1 )| ≤ 1k for every j ≤ k. Then (k + 1)M 2 + 2 · 1 + 2 · xn 1 + xn 2 + · · · + xn k+1 2 ≤ 2 (k + 1) (k + 1)2

2 2

+ ··· +

2k k

→ 0.

c0 case: The sliding hump argument.  x i for x = (xi ). Show that f is norm 3.58 Define a functional f on 1 by f (x) = ∗ continuous on 1 and that f is not w -continuous on 1 ∼ = c0∗ (that is, f ∈ ∞ \c0 ). w∗

Hint. Consider the standard basis {en } of 1 . Note that en → 0 in 1 . 3.59 Let X and Y be Banach spaces. Show that if an operator T from X into Y is w–w-continuous then T ∈ B(X, Y ). On the other hand, every bounded operator is w–w-continuous. Hint. Given f ∈ Y ∗ , consider V = f −1 (−1, 1). Then V is a w-neighborhood of of zero in X . Therefore zero in Y . By our assumption, T −1 (V ) is a w-neighborhood  λB X ⊂ T −1 (V ) for some λ, that is, | f T (B X ) | ≤ λ1 . Thus T (B X ) is w-bounded w

and also bounded in Y . On the other hand, if xα → 0 in X and f ∈ Y ∗ , then w f ◦ T ∈ X ∗ and thus T (xα ) → 0 in Y .

3.60 Let T be a bounded operator from Y ∗ into X ∗ . Show that T is a dual operator if and only if T is w∗ -w∗ -continuous. Hint. Assume T is w ∗ -w ∗ -continuous. Given x ∈ X , the function π(x) ◦ T is w∗ -continuous on Y ∗ , so π(x) ◦ T = x(T ) ∈ Y . The assignment x → π(x) ◦ T defines the predual operator. The other implication is straightforward. 3.61 Let X, Y be Banach spaces and T ∈ B(X, Y ). Show that (i) T is an isomorphism into if and only if T ∗∗ is an isomorphism into. (ii) T is an isometry into if and only if T ∗∗ is an isometry into.

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Hint. (i) If T is an isomorphism into, it follows directly from Exercise 2.49 using (i) and then (ii) applied to T ∗ . If T ∗∗ is an isomorphism into, we have T ∗∗ (x ∗∗ )Y ∗∗ ≥ δx ∗∗  X ∗∗ for some δ, hence the same holds for T , then use Exercise 1.73. (ii) Going through the proof of Exercise 2.49 we see that T ∗ (BY ∗ ) = B X ∗ , which implies T ∗∗ (F)Y ∗∗ = F X ∗∗ . 3.62 Let X, Y be Banach spaces, T ∈ B(X, Y ). Show that if T is an isomorphism of X onto Y , then T ∗∗ (X ) = Y and (T ∗∗ )−1 (Y ) = X . Hint. T is onto Y , hence T ∗∗ (X ) = T (X ) = Y . Also T ∗∗ is an isomorphism, in particular T ∗∗ is one-to-one, so (T ∗∗ )−1 (Y ) ⊂ X . 3.63 Let X, Y be Banach spaces and T ∈ B(X, Y ). If T : X → Y is one-to one, is T ∗∗ necessarily one-to-one? Hint. No. Consider the identity mapping from 1 into 2 . Its dual operator cannot be onto a dense set in ∞ as ∞ is not separable. Since T ∗ is not onto a dense set, T ∗∗ is not one-to-one by Exercise 2.46. 3.64 Let X be a separable Banach space. Find T ∈ B(2 , X ) such that T (2 ) is dense in X . Note that then T ∗ : X ∗ → 2 is a bounded w ∗ -w-continuous one-to-one operator. Hint.  −i Let {yi } be dense in B X and T be defined for x = (xi ) ∈ 2 by T (x) = 2 xi yi . Show that yi ∈ T (2 ) for every i. 3.65 Assume that X and Y are Banach spaces. Assume that T is a bounded operator from X ∗ into Y ∗ that is w ∗ -w ∗ -continuous when restricted to B X ∗ . Show that then T is w ∗ -w∗ -continuous, thus T is a dual operator to a bounded operator T0 from Y into X . Hint. We need to show that if y ∈ Y , then y(T ) is a w∗ -continuous linear functional on X ∗ , i.e., that its kernel, i.e., y(T )−1 (0) is w∗ -closed (Corollary 3.94). This fact follows from Banach–Dieudonné Theorem 3.92. The operator T0 is defined as follows: For every f ∈ X ∗ and for every y ∈ Y , f (T0 (y)) = f (y(T ). 3.66 Assume that X and Y are Banach spaces, that M be a linearly dense set in Y , i.e., that its norm-closed linear hull is Y . Assume that T is a bounded operator T from X ∗ into Y ∗ such that y(T ) is a w∗ -continuous linear functional on X ∗ for every y ∈ M. Prove that then T is a w∗ -w∗ -continuous operator from X ∗ into Y∗ . Hint. Observe that y(T ) = T ∗ (y) for all y ∈ M. Then T ∗ (Y ) = T ∗ span(M) ⊂   T ∗ span(M) ⊂ X = X , and the conclusion follows. 3.67 Let X be a separable Banach space, {xi } be a sequence in B X and T be a bounded operator from X ∗ into 2 defined by (T x ∗ )i = 21i x ∗ (xi ). Show that T is a w ∗ -w∗ -continuous compact operator. Hint. Use Exercise 3.66.

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3.68 (Hahn, see [Megg, p. 49]) Prove the Banach–Steinhaus Theorem 3.85 by providing the details of the following “gliding hump" argument (the notation refers to the aforesaid theorem): Assume that A is unbounded. Then there exists a sequence {Tn } in A and a sequence {xn } in X such that 

 T1 x1  ≥ 1, Tn xn  ≥ n + n−1 j=1 Tn x j , for n = 2, 3, . . . , x 1  ≤ 2−1 , xn  ≤ 2−n min j
(3.9)

 It follows that (a) the series ∞ n=1 x n converges (to some element x ∈ X ), that (b)  ∞ T x  ≤ 1 for all n ∈ N, and that (c) Tn x ≥ n − 1 for all n ∈ N (so n j j=n+1 conclude that supT ∈A T x = ∞). Hint. Assume that supT ∈A T  = ∞ and that, at the same time, we have supT ∈A T z < ∞ for every z ∈ X . Choose first T1 ∈ A such that T1  > 2. Let s1 ∈ S X so that T1 s1  > 2 and put x 1 = s1 /2. Since M1 := supT ∈A T x 1  < ∞, we can choose T2 ∈ A so that T2  > (2 + M1 )22 T1 . Find s2 ∈ S X such that T2 s2  > (2 + M1 )22 T1  and put x 2 = 2−2 T1 −1 s2 . Since M2 := supT ∈A T x2  < ∞, we can choose T3 ∈ A so that T3  > (3 + M1 + M2 )23 max{T1 , T2 }. Find s3 ∈ S X such that T3 s3  > (3 + M1 + M2 )23 max{T1 , T2 } and put x 3 = 2−3 min{T1 −1 , T2 −1 }s2 . Continue in this way to define the two sequences {Tn } and {x n } verifying (3.9). Now (a) should be clear. To prove (b) observe that Tn x j  ≤ Tn .x j  ≤  Tn 2− j Tn −1 = 2− j for j > n and n ∈ N. (c) follows from Tn x = Tn ∞ j=1 x j  ≥ Tn x n  − ∞ n−1 j=n+1 Tn x j  and the estimates above. j=1 Tn x j  − 3.69 Show that there exists a sequence { f n }∞ n=1 in the dual of some normed linear space X such that { f n (x)}∞ n=1 is bounded for every x ∈ X and yet { f n } is unbounded. This shows that the assumption of completeness of X cannot be dropped in the Banach–Steinhaus theorem. ∗ put f = ne , where e is the standard unit vector. Then check that for Hint. In c00 n n n every x ∈ c00 we have f n (x) = 0 for n large enough. 3.70 Show that every weakly compact set C in a Banach space is a bounded set. Hint. If f ∈ X ∗ , then f (C) is compact in K since every f ∈ X ∗ is w-continuous. Then use the Banach–Steinhaus theorem. 3.71 (Banach discs, I) Let (E, T ) be a locally convex space. A bounded convex and balanced subset of E is called a disc. The Minkowski functional μ D of a disc D ⊂ E in the linear space span(D) is a norm. A disc D in E such that (span(D), μ D ) is a Banach space is called a Banach disc. This concept was introduced in the context of Banach spaces in Exercise 2.22. Prove that a Banach disc D is uniformly bounded on every w ∗ -bounded subset of E ∗ .  Hint. Given x ∗ ∈ E ∗ , the set {x ∗ (d) : d ∈ D} is bounded, so x ∗ span(D) belongs to (span(D), μ D )∗ . Let B be a w∗ -bounded subset of E ∗ . The family

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 {x ∗ span(M) : x ∗ ∈ B} is pointwise bounded, and so μ D -bounded by the Banach– Steinhaus theorem. This gives the conclusion. 3.72 (Banach discs, II, see, e.g., [Koth, §20.11(2)]) Let D be a closed disc in a locally convex space (E, T ). Prove that if D is sequentially complete (i.e., every Cauchy sequence in D converges to some point in D), then D is a Banach disc; moreover, D is the closed unit ball of (span(D), μ D ). Hint. Obviously, on span(D) the topology induced by T is coarser than the norm topology. It follows that if {xn } is a μ D -Cauchy sequence in D, then it is T -Cauchy, hence T -convergent (to some element d ∈ D). Since D is T -closed, it follows that {xn } is also μ D -convergent (to d). This gives the conclusion. 3.73 (Banach discs, III) Let (E, T ) be a locally convex space. Prove that every w-bounded subset of E is T -bounded. Hint. Let A be a w-bounded subset of E. Let U ⊂ E be a neighborhood of 0. Then U ◦ is a w∗ -compact convex balanced subset of E ∗ . Exercise 3.72 gives that U ◦ is a Banach disc. Now Exercise 3.71 ensures that U ◦ is uniformly bounded on A. This gives the conclusion. 3.74 (Banach discs, IV, see, e.g., [Flor, p. 17]) Let (E, T ) be a locally convex space. Prove that every convex and relatively countably compact subset A of E is contained in a Banach disc.  denotes the completion of E, the mapping Hint. A is clearly T -bounded. If E   T : 1 (A) → E given by T (λ) = a∈A λ(a)a—observe that this sum has only a countable number of non-zero summands—is well defined, continuous, and there that contains A. Actually, T (B1 (A) ) ⊂ E. In fore T (B1 (A) ) is a Banach disc in E N order to show this, take first λn ≥ 0 for all n ∈ N such that s N := n=1 λn → s, and an ∈ A for all n ∈ N. Then ∞  n=1

λn an = lim s N N

N N   λn λn an = s. lim an ∈ E, N sN sN n=1

n=1

since A is convex and relatively countably compact. Now, for arbitrary signs of the coefficients, split the sum collecting the positive or negative ones. This proves that T (B1 (A) ) ⊂ E. 3.75 Let { f n } ⊂ X ∗ . Show that if there exists εn → 0 such that for every x ∈ X there is K x > 0 with | f n (x)| ≤ K x εn , then  f n  → 0. Hint. The Baire category. w∗

3.76 Let X be a Banach space. Assume that x n → x in X and f n → f in X ∗ . Show that f n (xn ) → f (x).

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Hint. Write f n (xn ) − f (x) = f n (xn ) − f n (x) + f n (x) − f (x) and note that { f n } is bounded by the Banach–Steinhaus theorem. 3.77 Assume that  · 1 and  · 2 are norms on a normed space X that are not equivalent. Show that then there is a linear functional on X that is continuous in one of the norms and is not continuous in the other. Hint. One of the unit balls in these norms is not a bounded set in the normed space whose unit ball is the other unit ball. This unbounded ball is not weakly bounded in the topology given by the other norm. Thus there is a functional continuous in the other norm and not in this one. 3.78 As an application of the Banach–Steinhaus uniform boundedness principle show that there is a continuous 2π -periodic function whose Fourier series diverges at 0. Hint. Let C be the Banach space of 2π -periodic continuous functions on R with n c , where ck is the kth the supremum norm. For f ∈ C put σn ( f ) = k k=−n Fourier coefficient of f (σn ( f ) is the value at 0 of the nth symmetric partial π 1 −ikt sum of the Fourier series of f ). As ck = 2π −π f (t)e  dt, we have σn ( f ) =

π sin (n + 12 )t n 1 −ikt = . Thus σn is a f (t)D (t) dt, where D (t) = e n n k=−n 2π −π sin( 12 t)

π 1 continuous linear functional on C and σn  = 2π −π |Dn (t)| dt. We estimate       1 π  sin (n + 12 )t  1 π  sin (n + 12 )t   t  σn  = ·  dt =  dt   π 0 π 0 t sin( 12 t) sin( 12 t)   π   (n+ 1 )  1  2  sin(u)   sin (n + 2 )t  ≥c    dt = c  du t u 0 0

∞ and this expression tends to infinity as n → ∞ since 0 | sinu u | du diverges. Thus supn σn  = ∞. By the Banach–Steinhaus theorem, there is f ∈ C such that {σn ( f )}∞ n=1 is unbounded, so the Fourier series for f diverges at 0. w∗

3.79 Find a Banach space X and f n ∈ X ∗ such that fn → 0 and yet for every convex combination h of { f n } we have h = 1. This shows that Mazur’s theorem does not hold for the w ∗ -topology. n

Hint. Put X = c0 , f n = (0, . . . , 0, 1, 0, . . . ) ∈ 1 . 3.80 Let C be a bounded set in a Banach space X . Show that C is relatively ww∗ compact if and only if C ⊂ X . w∗ Hint. C is w ∗ -compact (Alaoglu).

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3.81 Let X be a Banach space. Assume that C is a w ∗ -compact set in X ∗ . Is ∗ convw (C) also w∗ -compact? ∗ Hint. Yes, as convw (C) is a w ∗ -closed subset of some dual ball, which is w ∗ compact by Alaoglu’s theorem. 3.82 Can we replace B X and B X ∗∗ in the Goldstine’s theorem by S X and S X ∗∗ ? Hint. Yes. Use the w ∗ -lower semicontinuity of the second dual norm (see Exercise 3.31). 3.83 Show that if X is an infinite-dimensional Banach space, then there is a set S in X ∗ such that the set of all w∗ -sequential limits of S is not norm closed. Compare this with the situation with Baire 1 functions and find the reason for the difference. Hint. Let f n ∈ X ∗ with  f n  = 1/n for each n. Put f n = w∗ -lim f n,k so that k

 f n,k  = n for all n and k. By the Banach–Steinhaus principle, no sequence in { f n,k : n, k ∈ N} can w∗ -converge to 0. Concerning Baire 1 functions: a lattice structure. 3.84 Show the following Helly’s theorem: Let X be a Banach space. If x ∗∗ ∈ S X ∗∗ , F is a finite set in X ∗ and ε > 0, then there is x ∈ X with x = (1 + ε) such that f (x) = x ∗∗ ( f ) for every f ∈ F. Note the difference between Helly’s and Goldstine’s theorems. Hint. Put Z = F⊥ . Let q : X → X/Z be the canonical quotient mapping. Consider q ∗∗ x ∗∗ ∈ (X/Z )∗∗ (= (X/Z ), since X/Z is finite-dimensional). Then find in the coset q ∗∗ x ∗∗ an element x ∈ X of norm 1 + ε. Note that ε cannot in general be taken 0. Let f ∈ S X ∗ be such that it does not attain its norm and take x ∗∗ ∈ S X ∗∗ such that x ∗∗ ( f ) = 1. Then there does not exist any x ∈ B X such that f (x) = x ∗∗ ( f ) = 1. 3.85 Let X be a Banach space. In Exercise 2.28 we proved that for a subset Y of X we have (Y ⊥ )⊥ = span(Y ). Since span(Y ) is convex, by Mazur’s theorem it also reads (Y ⊥ )⊥ = spanw (Y ). ∗ Prove that for a subset Y of X ∗ we have (Y⊥ )⊥ = spanw (Y ). w w∗ For T ∈ B(X, Y ) we then have T (X ) = Ker(T ∗ )⊥ and T ∗ (Y ∗ ) = Ker(T )⊥ . ∗

Hint. Show that (Y⊥ )⊥ is w∗ -closed, so spanw (Y ) ⊂ (Y⊥ )⊥ . Assuming that the inclusion is strict, derive a contradiction using Corollary 3.34. ! as an operator 3.86 Let X, Y be Banach spaces and T ∈ B(X, Y ). Consider T !∗∗ can be considered an operator from X/ Ker(T ) into T (X ). Show that then T !∗∗ (x) ˆ = T (x) for all x ∈ X from X ∗∗ / Ker(T ∗∗ ) into T (X )∗∗ ; in particular, T (see Exercise 2.45). Show also that under the isometry of (X/ Ker(T ))∗∗ onto X ∗∗ / Ker(T ∗∗ ), the subspace X/ Ker(T ) of (X/ Ker(T ))∗∗ corresponds to the subspace {x + Ker(T ∗∗ ) : x ∈ X } of X ∗∗ / Ker(T ∗∗ ).

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!∗ : T (X )∗ → (X/ Ker(T ))∗ , so T !∗∗ : (X/ Ker(T ))∗∗ → T (X )∗∗ . Hint. We have T But  ∗  , w ∗ ∗ w∗ ⊥ = X ∗∗ T ∗ (X ∗ ) . (X/ Ker(T ))∗∗ = Ker(T )⊥ = T ∗ (X ∗ ) w∗ ⊥

= Z ⊥ for any subspace Z of X ∗ , so (X/ Ker(T ))∗∗ = But in general, Z ∗∗ ∗ ∗ ⊥ ∗∗ X /T (X ) = X / Ker(T ∗∗ ). 3.87 Let X, Y be Banach spaces, T ∈ B(X, Y ). Assume that T (X ) is closed, let x ∗∗ ∈ X ∗∗ . If T ∗∗ (x ∗∗ ) ∈ T (X ), then for every δ > 0 there exists x ∈ X such that T (x) = T ∗∗ (x ∗∗ ) and x ≤ (1 + δ)x ∗∗ . ! : X/ Ker(T ) → Y is an isomorphism and Hint. Assume that T (X ) = Y , then T !∗∗ : X ∗∗ / Ker(T ∗∗ ) → Y ∗∗ is an isomorphism (Exercises 3.86 and 2.51). thus T ∗∗ ) ∈ Y , we have x ∗∗ = x + Ker(T ∗∗ ) for some x ∈ X . Thus x ∗∗ !∗∗ (x Since T corresponds to some xˆ ∈ X/ Ker(T ) (Exercise 3.86), hence there is x ∈ xˆ such that 1 ∗∗  ≤ x ∗∗ . ˆ = x 1+δ x ≤  x Let X be a Banach space and A ⊂ X ∗ . Let λ be a positive number. We say that A is λ-norming if sup{ f (x) : f ∈ A ∩ B X ∗ } ≥ λ1 x for every x ∈ X (if this is the case, certainly λ ≥ 1). We say that A is norming if it is λ-norming for some λ ≥ 1. By the Hahn–Banach theorem, both X ∗ and B X ∗ are 1-norming. Also, X is 1norming for X ∗ . Observe that a norming subset of X ∗ separates points of X (see Exercise 3.92). 3.88 Let X be a Banach space and let F ∈ X ∗∗ \X . Show that F −1 (0) is a norming subspace of X ∗ . If x ∈ X \{0} ⊂ X ∗∗ , can x −1 (0) ⊂ X ∗ be norming? Hint. Let the distance of F to X be 2δ > 0. Assume that for some x ∈ S X we have sup{ f (x) : f ∈ F −1 (0),  f  ≤ 1} ≤ δ/2. Then by Exercise 2.13 we have x ± F < δ, a contradiction with the distance of F to X . If x ∈ X \{0}, x −1 (0) cannot be norming as it is w∗ -closed. If it were norming, it would be separating, hence w ∗ -dense in X ∗ and thus x −1 (0) = X ∗ , a contradiction. 3.89 Let (X,  · ) be a Banach space, let Y be a closed subspace of X ∗ . For x ∈ X define xY = sup{|y ∗ (x)| : y ∗ ∈ BY }. Clearly  · Y is a seminorm on X and  · Y ≤  · . Show that: (i) xY = dist(x, Y ⊥ ) (distance in X ∗∗ ). (ii) Y is separating if and only if  · Y is a norm on X if and only if X ∩Y ⊥ = {0} in X ∗∗ . (iii) Y is norming if and only if  · Y is an equivalent norm on X if and only if dist(S X , Y ⊥ ) > 0 (distance in X ∗∗ ). (iv) Y is 1-norming if and only if  · Y =  ·  X if and only if dist(S X , Y ⊥ ) = 1 (distance in X ∗∗ ). Hint. (i) xY = xY ∗ = x X ∗∗ /Y ⊥ = dist(x, Y ⊥ ). (ii) through (iv) are routine.

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3.90 Let X be a Banach space, let Y be a separating closed subspace of X ∗ . The topology w(X, Y ) on X is locally convex and Hausdorff. Denote B Y = {x ∈ X : xY ≤ 1} (see Exercise 3.89). Clearly B X ⊂ B Y . Show that: w(X,Y ) (i) B Y = B X . w(X,Y ) is bounded. (ii) Y is norming if and only if B X (iii) Y is 1-norming if and only if B X is w(X, Y )-closed if and only if the norm of X is w(X, Y )-lower  semicontinuous.   ∗ Hint. (i) Consider X, w(X, Y ) , then X, w(X, Y ) = Y . We have BY = (B X )◦ w(X,Y )

(polar in Y ) and B Y = ((B X )◦ )◦ = B X . (ii) and the first two equivalences in (iii) then follow using Exercise 3.89. The latter two conditions are equivalent since α B X =  · −1 ((−∞, α]) for α ≥ 0. 3.91 Assume that B X is w(X, Y )-closed for every norming subspace Y of X ∗ . Show that X is then reflexive. Hint. By contradiction. Let G ∈ S X ∗∗ \X be such that dist(G, X ) < 14 . Put Y = G −1 (0). Then Y is norming (Exercise 3.88), hence B X is w(X, Y )-closed and by the previous exercise, Y is 1-norming. This is a contradiction: Since dist(G, X ) < 14 , we have f (x) − G( f ) < 14 for all f ∈ B X ∗ . Thus f (x) < 14 for all f ∈ BY and Y cannot be 1-norming. 3.92 Show that every norming subspace of X ∗ is a separating set for X . Find an example of a separating subspace that is not norming. Hint. For the secondstatement, take a partition I, I1 , I2 , . . . of N into disjoint infi nite sets. Let S = y ∈ 1 : for each k ∈ I, yk = 1k n∈Ik yn . Then S is separating for c0 but not norming (consider ek for k ∈ I ). Note that if X is separable and X ∗∗ / X is infinite-dimensional, then there is a separating subspace of X ∗ that is not norming (see [DaLi]). 3.93 Let X be a Banach space and Y be a Banach space such that X ∗ ⊂ Y . Let F ⊂ Y ∗ be the closed linear subset formed by all functionals whose restriction to X ∗ are w ∗ -continuous. Show that F is a norming subspace in Y ∗ . Hint. Observe first that F is closed due to the Banach–Dieudonné theorem (more precisely, to Corollary 3.94). Consider y ∈ SY . If dist(y, X ∗ ) < 14 , choose y1 ∈ X ∗ such that y − y1  < 14 and then choose f ∈ X such that  f  = 1 and f (y1 ) > y − 14 . Extend f to Y with norm 1 and call this extension again f . Then f (y) ≥ f (y1 ) − | f (y − y1 )| ≥ y − 14 − y − y1  ≥ 12 ≥ 14 . If, on the other side, dist(y, X ∗ ) ≥ 14 , then choose f ∈ Y ∗ such that f = 0 on X ∗ ,  f  = dist(y, X ∗ )−1 , and f (y) = 1. Then dist(y, X ∗ ) f is a norm-1 functional on Y which belong to F, and dist(y, X ∗ ) f (y) ≥ 14 . 3.94 Let X be a Banach space and A ⊂ X ∗ . Show that A separates the points of X ∗ if and only if A⊥ = {0} if and only if spanw (A) = X ∗ . Hint. Easy from definition and Exercises 3.85 and 2.27.

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3.95 Let X be a separable Banach space. Show that there is a sequence { f n } ⊂ S X ∗ such that { f n } is separating in X . Hint. Let {x n } be dense in S X . For each n choose f n ∈ S X ∗ such that f n (xn ) = 1. Given x ∈ S X , take n such that x − xn  < 12 , then f n (x) ≥ 12 . 3.96 Let X be a reflexive space and let Y ⊂ X ∗ be a closed subspace of X ∗ . Show that if Y separates points of X then Y = X ∗ . w∗ w Hint. We have Y = X ∗ . By reflexivity Y = X ∗ , then use Theorem 3.45. 3.97 Let T be an operator from a Banach space X into a Banach space Y . Suppose that F is a subset of Y ∗ that separates points of Y . Assume that f (T (xn )) → 0 whenever f ∈ F and {x n } ⊂ X is such that x n  → 0. Show that T is a bounded operator. Hint. The closed graph theorem. 3.98 Let C be a weakly separable subset of a Banach space. Show that C is norm separable. Hint. The norm closed linear hull is weakly and thus norm-separable, thus C is norm separable. 3.99 Prove directly that if X is a separable Banach space, then X ∗ is w∗ -separable. Hint. Take the sequence { f n } ⊂ S X ∗ from Exercise 3.95. Then span{ f n } is w ∗ dense. Therefore the rational combinations form a w ∗ -dense set in X ∗ . 3.100 Let X be a Banach space. Show that the following are equivalent: (i) B X is w-separable, (ii) S X is w-separable, (iii) X is separable. Hint. (i)⇒(ii): If a countable C ⊂ B X is w-dense in B X and x ∈ S X , then there w is a net {xα } ⊂ C such that xα → x. We can assume that xα  → 1 as the norm is w x : x ∈ C} weakly lower semicontinuous (see Exercise 3.31), hence xxαα  → x. { x is w-dense in S X . Every weak neighborhood of a point from B X intersect S X if X is infinitedimensional. Thus (ii)⇒(i). Clearly, (i) implies that X is weakly and thus normseparable (Proposition 3.105). (iii) implies that the ball and the sphere are separable and hence w-separable. 3.101 Let X be a Banach space. Show that B X ∗ is w ∗ -separable if and only if S X ∗ is w∗ -separable. Note that the w ∗ -separability of B X ∗ is not preserved when passing to an equivalent norm (Exercise 14.42). Also, there is a space X for which X ∗ is w ∗ -separable and no equivalent norm on X exists so that B X ∗ is w ∗ -separable ([JoLi1]). Hint. Similar to the previous exercise.

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3.102 Show that ∗∞ is w ∗ -separable. ∗ Hint. The Goldstine’s theorem and ∗∗ 1 = ∞ . 3.103 Find a Banach space X such that X ∗ is w ∗ -separable and a subspace Y of X such that Y ∗ is a w ∗ -dense subspace of X ∗ that is not w∗ -separable. Hint. Take X := 1 (Γ ) with card(Γ ) = c, and Y := c0 (Γ ). The w ∗ -separability of X ∗ follows from the fact that 1 (Γ ) ⊂ C[0, 1]∗ ⊂ ∞ and the Goldstine’s theorem (see Exercise 3.102). The restriction of the w∗ -topology of X ∗ to c0 (Γ ) is the w-topology on c0 (Γ ), and c0 (Γ ) is not norm-separable. The same proof gives that [0, 1][0,1] is pointwise separable. 3.104 Show that every weakly compact set in ∞ is norm-separable. Hint. Let C be w-compact in ∞ . Since ∗∞ is w∗ -separable, C is w-metrizable by Proposition 3.107, hence also w-separable. Thus span(C) is weakly and thus norm-separable. 3.105 Use the following hint to show that the weak topology of an infinitedimensional Banach space is never metrizable. Hint. Assume that the weak topology of a Banach space X is metrizable by a metric ρ. We know that 0 is in the weak closure of nS X for every n. Therefore we can find a point x n ∈ nS X for which ρ(xn , 0) < n1 . Then ρ(xn , 0) → 0, which means that x n → 0 in the weak topology. Thus by Banach–Steinhaus, {x n } must be bounded, which is not the case. 3.106 Let Y be a closed subspace of a Banach space X . Show that the weak topology of X/Y is the factor topology of the weak topology of X , that is, U is weakly open in X/Y if and only if q −1 (U ) is weakly open in X , where q is the canonical quotient mapping. Show that if T is a bounded operator from a Banach space X onto a Banach space Y , then T is an open mapping in the respective weak topologies of X and Y . Hint. (X/Y )∗ = Y ⊥ and every bounded linear mapping is w–w-continuous. 3.107 Let X be an infinite-dimensional Banach space. Show that (X ∗ , w ∗ ) is not complete in the uniformity of the topological vector space (see the paragraph preceding Proposition 3.27, and Section 17.10). Similar result holds for the w-topology. Recall that a sequence {x i } is w-Cauchy if { f (xi )} is Cauchy for every f ∈ X ∗ . Hint. Consider a linear discontinuous functional f on X (it always exists, see Exercise 2.4). For every finite-dimensional subspace of X with a basis B and ε > 0 consider a continuous linear functional g on X such that f and g differ by at most ε on vectors in B. By ordering B by inclusion one can get a net of continuous linear functionals converging to f . This net is thus w ∗ -Cauchy and not w ∗ -convergent in X ∗.

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3.108 Let X be an infinite-dimensional Banach space. Show that S X is a dense G δ set in (B X , w). Deduce from this that the norm of an infinite-dimensional space is never a weakly continuous function on X . Hint. Given x0  < 1, let U = {x : | f i (x − x0 )| < ε}. Take x = 0 such that x ∈ f i−1 (0) and note that the continuous function  ·  considered on the set {x0 + t x : t ∈ R} attains the value 1 somewhere, showing that U ∩ S X = ∅. B X . Put G n = {x ∈ B X : x > 1 − n1 }. Then G n are Therefore S X is w-dense in open in (B X , w) and S X = G n . 3.109 Construct an equivalent norm on 1 ∼ = c0∗ that is not w∗ -lower semicontinuous.   |xi | + 2| xi |. Then take ai < 0 such that Hint.  Consider the norm (xi ) = ai = −1 and consider the vectors xn = (1, 0, . . . , 0, a1 , a2 , . . . ), where a1 is at w∗

the nth place. Show that x n → (1, 0, . . . ) and compare the norms. 3.110 Let X be an infinite-dimensional Banach space. Can X be a Baire space in its weak topology? Recall that a topological space T is called a Baire space if the intersection of a countable family of open dense subsets of T is dense in T . Hint. No. n B X are nowhere dense in (X, w). 3.111 Show that 0 is not a G δ -point of B2 (Γ ) in the weak  topology if Γ is uncountable. Recall that p ∈ W is a G δ -point of W if { p} = On , where On are open in W.  weak topology, then Hint. If { p} = On and On are basic neighborhoods in the  each On restricts only a countable number of coordinates, so On restrict only a countable number of coordinates and thus cannot be a singleton if Γ is uncountable. 3.112 Let X be a reflexive Banach space. Show that if Y is isomorphic to X then Y is reflexive. Hint. BY is w-compact as a w-closed (see Theorem 3.45) subset of T (T −1 B X ), which is w-compact by the w-w-continuity of T . 3.113 Show that c0 is not isomorphic to a subspace of  p , p ∈ (1, ∞). Hint. c0 is not reflexive. 3.114 Let Y be a closed subspace of a reflexive Banach space X . Show that X/Y is reflexive. Hint. (X/Y )∗ = Y ⊥ ⊂ X ∗ and X ∗ is reflexive. 3.115 Let X be a Banach space. Show that if every separable subspace of X is reflexive, then X is reflexive. Hint. The Eberlein–Šmulyan theorem for the weak compactness of B X .

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3.116 Let Y be a closed subspace of a Banach space X . Show that if Y and X/Y are reflexive, then X is reflexive. Thus reflexivity is a three-space property. Hint. If Y is reflexive, then Y is w∗ -closed in X ∗∗ . Thus Y = (Y ⊥ )⊥ . Then X ∗∗ /Y ⊥⊥ = (Y ⊥ )∗ and Y ∗ = (X/Y )∗ , so (Y ⊥ )∗ = (X/Y )∗∗ = X/Y . Consequently X = X ∗∗ . Note that this is a delicate proof although it looks easy. We have to prove not X = X ∗∗ but actually π(X ) = X ∗∗ , that is, we have to check that the mappings in the above equalities work as supposed to. 3.117 Let X be a reflexive Banach space and let T be a bounded operator from X onto a Banach space Y . Show that Y is reflexive. Hint. Y is isomorphic to X/ Ker(T ), which is a reflexive Banach space. 3.118 Show that if T is a bounded one-to-one operator from 1 into 2 then T (1 ) is not closed in 2 . Hint. Otherwise T would be an isomorphism of 1 onto a reflexive space T (1 ), hence 1 would be reflexive. 3.119 Prove that a Banach space is reflexive if and only if the weak and weak∗ topologies on B X ∗ coincide. Compare with Proposition 3.113 and with Exercise 3.121. Hint. One direction is obvious. For the other, use Corollary 3.94. 3.120 ([DaJo]) Show that a Banach space X is reflexive if and only if the closed unit ball of every equivalent norm on X ∗ is w∗ -closed. Hint. If the condition on w∗ -closedness holds and F ∈ X ∗∗ is given, consider the set Vn = { f ∈ B X ∗ : |F( f )| ≤ n1 }. The Minkowski functional of Vn is an equivalent norm on X ∗ and Vn is its closed By our assumption, Vn is w∗ -closed.  unit ball. −1 ∗ ∗ Therefore F (0) ∩ B X = Vn is w -closed and thus F ∈ X , meaning that X is reflexive. If X is reflexive, then the closed unit ball of any equivalent norm is w-compact and thus w ∗ -compact. 3.121 Let X be a Banach space. Show that if the w-topology and the w∗ -topology coincide on X ∗∗ , then X is reflexive. (See also Exercise 3.119.) Hint. Apply Proposition 3.113 and then Proposition 3.112. 3.122 Recall that, given a Banach space X , a biorthogonal system {x n ; x n∗ }∞ n=1 in ∗ }∞ in X ∗ such that X × X ∗ consists of two sequences, {x n }∞ in X and {x n n=1 n=1 x m∗ , xn  = δn,m for all n, m ∈ N. It is called total if {x n∗ : n ∈ N} is w∗ -linearly dense in X ∗ . Prove that if X is a Banach space X and {x n ; x n∗ }∞ n=1 is a biorthogonal system in X × X ∗ such that the set {x n : n ∈ N} has the property that for ∗ ∗ ∗ every n x ∗∈ X and every c > 0, the set {n ∈ N : |x , xn | > c} is finite, then  i=1 xi  →n ∞.

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n p ∗ Hint. If not, there exists an increasing sequence {n p } in N such that { i=1 xi  : p∈ N} is bounded in X ∗ . Let x ∗ be a w ∗ -cluster point of this sequence. Note that np  i=1 xi∗ , x j  = 1 for every j ≤ n p . Then x ∗ , x j  = 1 for all j ∈ N. This is a contradiction. 3.123 Using Exercise 3.122 and Theorem 3.132, prove the following result: (Pták [Ptak2]) Let X be a Banach space. Then the following assertions are equivalent: (i) ∗ X is reflexive. (ii) For every biorthogonal system {x n ; x n∗ }∞ n=1 in X × X such that ∞ n ∗ {x n : n ∈ N} is bounded, the sequence { k=1 xk n=1 is unbounded. (iii) For every ∗ biorthogonal system {xn ; xn∗ }∞ n=1 in X × X such that {x n : n ∈ N} is bounded, the ∞ n ∗ sequence { k=1 xk n=1 is unbounded. Hint. [FGM] (i)⇒(ii): Assume that the space X is reflexive, and let {xn ; xn∗ }n∈N be a biorthogonal system in X × X ∗ such that {xn∗ : n ∈ N} is bounded. Let Y = span{xn : n ∈ N}; this is a reflexive space. Let q : X ∗ → X ∗ /Y ⊥ the canonical quotient mapping. Since (X ∗ /Y ⊥ )∗ = Y , the system {q(x n∗ ); xn }n∈N is total and biorthogonal in (X ∗ /Y ⊥ ) × Y . We claim that the set {q(xn∗ ) : |q(x n∗ ), y| > c} is finite for every y ∈ Y and every c > 0. Otherwise, since {q(xn∗ ) : n ∈ N} is a weakly relatively compact subset of X ∗ /Y ⊥ , the set {q(xn∗ ) : |q(x n∗ ), y| > c} would have a w-cluster point xˆ ∗ in X ∗ /Y ; by the orthogonality and the fact that {x n : n ∈ N} is linearly dense in Y , ∗ xˆ ∗ = 0, a contradiction n with|xˆ , y| ≥ c. The claim holds, and it follows from Lemma 3.122 that  k=1 xk  →n ∞. (i)⇒(iii): If X is reflexive, so is X ∗ . Given a biorthogonal system {xn ; xn∗ }n∈N ∗ ∗∗ in X × X ∗ , it can also be seen as a biorthogonal system {xn∗ ; xn }n∈ Nnin X ∗ × X . If the set {xn : n ∈ N} is bounded, it follows from (i)⇒(ii) that  k=1 xk  →n ∞. (ii)⇒(i): Assume that X is not reflexive. Theorem 3.132 says, in particular, that there exist two sequences (xn ) in S X and (xn∗ ) in S X ∗ such that x m∗ , xn  = 12 if n ≥ m, and x m∗ , x n  = 0 if n < m. Let d1 = 2x1 , dn = 2(xn − xn−1 ), n = 2, 3, . . . Then, it is clear that the family dn : x n∗ n∈N is a biorthogonal system in X × X ∗ .  Moreover, {x n∗ : n ∈ N} is bounded. Observe, too, that nk=1 dk = 2xn for all n ∈ N. We obtain thus a contradiction with (ii). (iii)⇒(i): Starting from the assumption that X is not reflexive, we proceed as in the proof of (iii)⇒(i). Once we have the two sequences (xn ) and (xn∗ ), put dn∗ = ∗ ) for n ∈ N. The system {x ; d ∗ } 2(xn∗ − xn+1 n n n∈N is again a biorthogonal system and ∗ ) for all n ∈ N. We the set {x n : n ∈ N} is bounded. Now nk=1 dn∗ = 2(x1∗ − xn+1 obtain again a contradiction, this time with (iii). 3.124 Let X, Y be Banach spaces, T ∈ B(X, Y ). Show that if X is reflexive then T (B X ) is closed in Y . Hint. B X is weakly compact and T is weakly continuous. 3.125 Assume that the unit sphere of a given Banach space X contains an infinite sequence of points such that the mutual distance of two different members of the sequence is 2. Does it imply that X is nonreflexive?

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Hint. No. Consider the equivalent norm on 2 defined for x = {x i } by   x = max x2 , max{|xi | + |x j | : i = j} . Check the standard unit vectors. 3.126 (i) Let X, Y be Banach spaces and let T be a one-to-one   mapping  from  X onto Y . Show that if C is a convex subset of X , then T Ext(C) = Ext T (C) . In    particular, if T is an isometry of X into Y then T Ext(B X ) = Ext T (B X ) . (ii) Let T be a bounded   operator  from X into Y . Show that if C is w-compact in X , then Ext T (C) ⊂ T Ext(C) . Hint. (i) If x ∈ C\ Ext(C), then x = 12 (y + z) for some y, z ∈ C. But then also    T (x) = 12 T (y) + T (z) . Since T C is invertible, the other direction follows.   (ii) Let d ∈ Ext T (C) . Denote K = C ∩ T −1 (d). Since T is w–w-continuous, K is w-compact and hence there is c ∈ Ext(K ) such that  T (c) = d. Check that c ∈ Ext(C): If c = 12 (x + y), x, y ∈ C, then d = 12 T (x) + T (y) . Since d is extreme, T (x) = T (y) = T (d), so x, y ∈ K . Since c is extreme in K , x = y. 3.127 Let C be a convex n set in a Banach  space X and let c be an extreme point in C. Show that if c = i=1 λi xi with λi = 1, λi ≥ 0, and xi ∈ C for i = 1, . . . , n then c = xi for some i. 1 2 Hint. λ1 x1 + λ2 x2 + λ3 x3 = (λ1 + λ2 )( λ1λ+λ x1 + λ1λ+λ x2 ) + λ3 x3 , use induction 2 2 on n. 3.128 Let m 0 be the set of all sequences assuming only finitely many different values. Show that m 0 is dense in ∞ . A more precise statement is given in Exercise 3.129. Hint. Let α = (an )∞ n=1 be a sequence such that a∞ ≤ 1. For any ε > 0, pick N ∈ N such that 1/N < ε. Define the sequence b = (bn )∞ n=1 ∈ m 0 by bn = (sign an ) j/N , if j/N ≤ |an | < ( j + 1)/N , j = 1, 2, . . . , N . Then a − b∞ ≤ 1/N < ε.   3.129 Show directly that B∞ = conv Ext(B∞ ) . Hint. Note that (xn ) is extreme in B∞ if and only if |xn | = 1 for each n. Given x ∈ B∞ and ε > 0, find a simple function y on N such that x − y < ε. If k k are the values of y, write {ai }i=1 as a convex combination of extreme points {ai }i=1 in Bk∞ . Then express y as the same combination of the characteristic functions corresponding to {ai }k1 , which are extreme points in B∞ .   3.130 Show that conv Ext(BC[0,1]∗ ) = BC[0,1]∗ .   Hint. Show that theLebesgue measure λ is not in conv Ext(BC[0,1]∗ ) . Indeed, if λ is norm-close to αi δti , find f ∈ BC[0,1] such that f (ti ) = 0 but λ( f ) is close to 1.

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3.131 Show that Bc0 has no extreme point. Hint. If x = (xi ) ∈ Bc0 , let i 0 be such that |xi0 | < 14 . Then x is the center of a line segment with endpoints (xi ) ± 14 ei0 . 3.132 Show that extreme points of Bc are vectors x = (x n ) ∈ c such that |xn | = 1 for all n ∈ N. Show that c and c0 are not isometric. Hint. Isometry preserves extreme points, the previous exercise. 3.133 Show that the extreme points of BC[0,1] are exactly two functions: constant functions 1 and −1. Show that BC(K ) , for an arbitrary compact set K , has some extreme points. Hint. Use the definition. 3.134 Show that c0 is not isometric to any C(K ) space. Hint. Lack of extreme points in Bc0 unlike in BC(K ) . 3.135 Show that the extreme points of B1 are exactly the vectors ±ei . Hint. Do it first in 21 . 3.136 Show that B L 1 [0,1] has no extreme points. c Hint. For c with 0 | f | dt = 1/2 put f 1 = 2 f · χ[0,c] and f 2 = 2 f · χ[c,1] . 3.137 Find a norm-closed convex and bounded subset of C[0, 1]∗ that has no extreme point. Hint. First note that L 1 [0, 1] ⊂ C[0, 1]∗ by using the mapping f →

1 0 f (x)g(x)dx, g ∈ C[0, 1]. Then use Exercise 3.136. 3.138 Show that the extreme points of B L ∞ [0,1] are exactly functions for which | f |=1 almost everywhere. Hint. Use the definition. 3.139 Show that none of C[0, 1], c0 , L 1 [0, 1] is isometric to a dual space. Hint. Isometry preserves extreme points, the Krein–Milman theorem. 3.140 Show that C([0, 1] ∪ [2, 3]) is not isometric to C[0, 1]. Hint. Use the Banach–Stone theorem and the connectedness. 3.141 Use the Banach–Stone theorem to show that c ⊕c is not isometric to c. Recall that these spaces are isomorphic. Hint. c ⊕ c is isometric to the space of continuous functions on the disjoint union of two copies of one-point-compactifications of N, see Section 17.1. This set has two limit points. c is isometric to the space of continuous functions on the one-point-

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compactification of natural numbers. This set has only one limit point. (For another approach to the same result see Exercise 5.21.) 3.142 Let K be a compact metric space such that the identity I K is the only homeomorphism of K onto K . Show that all linear isometries of C(K ) onto C(K ) are ±IC(K ) . Such K does exist. In fact, there is a connected compact set K in R2 such that the only continuous mappings K → K are constants and the identity. Hint. The Banach–Stone theorem. 3.143 Show that 1 (c) is isometric to a subspace of C[0, 1]∗ , where c denotes a set of cardinality continuum.  Hint. Given (at ) ∈ 1 (c), consider the functional at δ(t). 3.144 Find an example of a compact convex set C in R3 such that extreme points of C do not form a closed set. Can that happen in R2 ? Hint. Let D denote the disk of radius 1 in the (x, y)-plane centered at (1, 0, 0). Let L be the line segment on the z-axis from the point (0, 0, −1) to the point (0, 0, 1). For the set C take conv{D ∪ L}. This cannot happen in R2 , since if x is not an extreme point then points close to x are not extreme as well. 3.145 (Milman) Use Lemma 3.69 to prove that if C is a weakly compact convex set in a locally convex space X and B ⊂ C is such that conv(B) = C, then Ext(C) ⊂ w B . w Hint. Assume that for some x ∈ Ext(C) we have x ∈ / B . By Lemma 3.69, there is a slice S of C that contains x ∈ C and such that S ∩ B = ∅. Thus there are f ∈ X ∗ and α ∈ R such that f (x) > α ≥ sup B ( f ). By the linearity of f , sup B ( f ) = supconv(B) ( f ), so f (x) > α ≥ supconv(B) ( f ). This contradicts C = conv(B). 3.146 (Troyanski–Lin) Let x ∈ Ext(B X ) and assume that at x the relative norm and weak topology of B X coincide. Show that the slices form a neighborhood base of the norm topology at x. Hint. (Rosenthal) Show that x is an extreme point of B X ∗∗ . To this end, assume that x = 12 (x1 + x2 ) with xi ∈ B X ∗∗ . By a geometric argument, show that the w∗ -topology and the norm topology of B X ∗∗ coincide at xi . By Goldstine, xi ∈ B X , a contradiction. Having shown that x is an extreme point of the second dual ball, use Choquet’s lemma. 3.147 Show that Bl∞ does not have arbitrarily small slices. Hint. Given F ∈ S∗∞ and ε > 0, consider the slice Sε = {x ∈ B∞ : F(x) > 1−ε}. We claim that diam Sε ≥ 2. To this end, it is enough to prove that Sε contains at least two extreme points of B∞ . Note that functionals not attaining their norm are dense in B∗∞ (if there was a ball of norm-attaining, all would attain and ∞ would be reflexive). So assume that F does not attain its norm.

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By Exercise 3.129, Sε contains at least one extreme point x. If it was the only one, it would be in all slices Sδ for δ ∈ (0, ε). Then F(x) = 1, so F attains its norm, a contradiction. 3.148 Let X be a Banach space and A (respectively B) be a bounded subset of X (respectively, of X ∗ ). Show that A has arbitrary small slices if and only if conv(A) ∗ does. Show that B has arbitrarily small w ∗ -slices if and only if convw (B) does. . Hint. Assume that A has small slices. Pick a small ε-slice A1 of A that results from a cut of A by a hyperplane H and put D1 = conv(A\A1 ), A2 = conv(A1 ). Then conv A ⊂ conv(A2 ∪ D1 ). Show that if you move H in the parallel way, the cuts of conv(A2 ∪ D1 ) have eventually diameter smaller than 2ε. The argument for B is similar. Details can be found in [Bour], see also [Dies2, pp. 157–158]. 3.149 Find a compact convex set C in a Banach space such that C is not the convex hull of its extreme points. Hint. Put C= conv({2−i ei } ∪ {0}), where ei are the standard unit vectors in 2 . Then c =  4−i ei ∈ C. By Theorem 3.66, Ext(C) ⊂ {2−i ei } ∪ {0}. Therefore conv Ext(C) is formed by points that are all finitely supported and thus it does not contain c. ∗ 3.150 Let Y be a closed subspace of a Banach  space X . We know that Y is isomet∗ ⊥ ric to X /Y . Show that for every xˆ ∈ Ext B X ∗ /Y ⊥ there exists x ∈ xˆ such that x ∈ Ext(B X ∗ ). Hint. Exercise 3.126.

3.151 Find a James boundary B for a Banach space X with the property that B ∩ Ext(B X ∗ ) = ∅. Hint. Take X := 1 (Γ ), where Γ is uncountable. Let B = {x ∗ ∈ X ∗ : x ∗ (γ ) ∈ {−1, 0, 1} for all γ ∈ Γ, supp(x ∗ ) countable}. 3.152 Consider X = R2 with the dual norm on X ∗ being the norm whose unit ball is   D = B((1, 0), 10) ∩ B((−1, 0), 10)) ∩ (x, y) ∈ R2 : −1 ≤ y ≤ 1 , where B(x, r ) is a closed ball centered at x with radius r . Prove that B = Ext(D\{upper left corner}) is a James boundary of X which is not closed. w∗

3.153 Let X be a Banach space, let B be a separable subset of X ∗ such that B = B X ∗ . Can we conclude as in Theorem 3.122 that B is not a James boundary of X ? Hint. No. Let X be R2 with the maximum norm and let B = {±e1 , ±e2 }, where e1 are the standard unit vectors in R2 . Then for each x ∈ S X we have max(x) = 1. B

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3.154 Show that if B is a James boundary, then Ext(B X ∗ ) ⊂ B . ∗ Hint. By separation, convw (B) = B X ∗ . Then use Theorem 3.66. 3.155 (Cascales, Godefroy [CasGod]) Prove the following result: Let K be a compact space and B ⊂ SC(K )∗ a James boundary of C(K ). If { f n } is an arbitrary sequence in C(K ) and x ∈ K , then there exists μ ∈ B such that f n (x) = f n (μ) for every n ∈ N. Hint. The set F := {y ∈ K : f n (y) = f n (x) for all n ∈ N} is a closed G δ -subset of K . It follows from Urysohn’s lemma (see [Dugu2, p. 148]) that there exists a continuous function h from K to [0, 1] such that F = {x ∈ K : h(x) = 1}. Since B is a James boundary, there exists μ ∈ B such that μ(h) = 1. But this clearly implies that μ is a probability measure supported by F, and thus that μ works. 3.156 (Cascales, Godefroy [CasGod]) Prove the following result: Let K be a compact space and B ⊂ SC(K )∗ a James boundary of C(K ). (i) If { f n } is an arbitrary sequence of C(K ), then any w(C(K ), B)-cluster point of { f n } in C(K ) is also a T p -cluster point. (ii) If L is a w(C(K ), B)-relatively countably compact subset of C(K ), then it is T p - relatively compact in C(K ). (iii) If L is a w(C(K ), B)-compact subset of C(K ), then it is T p -compact. Hint. (i) is a straightforward consequence of Lemma 3.155. (ii) follows from (i) together with the fact that the T p -relatively countably compact subsets of C(K ) are actually T p -relatively compact, see Theorem 3.139. To prove (iii) let us fix a T p -compact subset L of C(K ). From (ii) we get that L is T p -relatively compact in C(K ). To finish the proof it will be enough to show that L is T p -closed. Pick g in the T p -closure of L. Using that (C(K ), T p ) is angelic (see Theorem 3.58), there is a sequence {gn } in L with g = T p - lim gn . n→∞

(3.10)

By compactness, there is h ∈ L which is a w(C(K ), B)-cluster point of {gn }. Statement (i) implies that h is a T p -cluster point of {gn }, hence, by (3.10), h = g and thus g ∈ L. This shows that L is T p -compact and the proof is complete. 3.157 (Cascales, Godefroy [CasGod]) Prove the following result, the case of Pfitzner’s theorem for the case of C(K ) spaces: Let K be a compact space and B ⊂ SC(K )∗ a James boundary of C(K ). Then a subset A of C(K ) is weakly compact if, and only if, it is norm bounded and w(C(K ), B)-compact. Hint. Obviously any weakly compact subset of C(K ) is norm bounded and w(C(K ), B)-compact. Conversely, let A be a bounded and w(C(K ), B)-compact subset of C(K ). Statement (iii) in Proposition 3.156 ensures that A is T p -compact. Since A is bounded, Theorem 3.139 implies that A is weakly compact.

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3.158 Let C be a bounded convex set in a Banach space X such that every continuous linear functional on X attains its supremum over C. Is C necessarily closed? Hint. No. Consider in the plane 22 the convex hull of two circles of radius one centered at (2, 0) and (−2, 0), remove the points (±2, ±1). The resulting set is bounded and not closed, one can show that every bounded linear functional attains its supremum over this set (level sets of linear functionals are lines in R2 ). 3.159 Find a Banach space X and a convex continuous bounded below function f on X such that lim f (x) = ∞ and f does not attain its infimum on X . Can X x→∞

be reflexive? Can X be finite-dimensional if we drop the requirement on the limit of f? Hint. Let X = c0 and let f be in S X ∗ such that f does not attain its norm. Define φ for x ∈ X by φ(x) = x2 − f (x). Then φ is bounded below. Indeed, if x ≥ 2, then φ(x) ≥ x2 − x ≥ 2. If x ≤ 2, then |φ(x)| ≤ x2 + | f (x)| ≤ x2 + 2 f  ≤ 6. Let x < 12 be such that 12 ≥ f (x) > 14 . Then φ(x) = x2 − f (x) < 1 4 − f (x) < 0. Hence φ attains negative values at some points of X . Assume that φ attains its minimum on X at x0 . As φ(0) = 0, we get x 0 = 0. φ(x 0 ) ≤ φ(x) reads x0 2 − f (x 0 ) ≤ x2 − f (x) for all x ∈ X , hence f (x) ≤ f (x 0 ) + x2 − x 0 2 for all x ∈ X . If x ∈ X is such that x ≤ x 0 , then f (x) ≤ f (x0 ). Thus f attains its maximum on the set {x ∈ X : x ≤ x 0 }. Hence f attains its maximum on B X , a contradiction. The function e x on R can be used to finish the solution of this exercise. The space X cannot be reflexive, as on such space the weakly lower semicontinuity of φ can be used to produce the minimum of φ, by using the limit condition on φ. 3.160 Find an example of an incomplete infinite-dimensional normed space X such that every f ∈ X∗ attains its norm ([Jame4]). ∞ n Hint. Let X = n=1 ∞ . Note that if x = (x n ) ∈ X is an extreme point of 2 B X , then xn is an extreme  point of Bn∞ for every n, so xn = (±1, ±1, . . . , ±1). Let Z = span Ext(B X ) . X is reflexive, so by the Krein–Milman theorem we have   B X = conv Ext(B X ) and thus Z = X . Note that Z = X . To see this, consider the following x ∈ X : x1 = 1, x2 = ( 12 , 13 ), m x3 = ( 14 , 15 , 16 ), . . . Assume that x ∈ Z , that is, x = l=1 αl y l , where y 1 , y 2 ,. . . ,  m m m l y ∈ Ext(B X ). Let n > 2 . Then xn = l=1 αl yn . Since every coordinate of each vector ynl is ±1, we have at most 2m different coordinates of the vector xn . As n > 2m , at least two coordinates of xn should be equal, which is not the case. Hence x∈ / Z. Now pick any f ∈ Z ∗ . Extend it on X and call it f again. As B X is weakly compact, f attains its norm on B X . From the proof of Krein–Milman theorem, we have that f attains its norm at an extreme point of B X . Using the fact that the norm of f on X and Z is the same we get that f attains its norm on B Z .

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3.161 Show using Theorem 3.130 that Bc0 is not weakly compact. Hint. (2−i ) ∈ 1 does not  attain its norm on Bc0 . Indeed, if x ∈ Bc0 then eventually |xi | < 1, so | 2−i xi | < 2−i = 1. n Direct proof of the exercise: i=1 ei has no w-cluster point. Note that exactly finitely supported elements in 1 attain their norms. 3.162 Show that the functionals in c0∗ that attain their norm do not form a residual set in c0∗ . Hint. Recall that it is exactly finitely supported functionals that attain their suprema over Bc0 (see the previous exercise). Then find the Baire category of the set of finitely supported vectors. Use the fact that finite-dimensional subspaces of 1 are closed and have empty interiors in 1 . 3.163 Let X be a Banach space and f ∈ S X ∗ . Does there exist f n ∈ X ∗ that attain their norm such that  f n − f  → 0 and f n lie on one line through f ? Hint. Not in general. Consider in c0∗ = 1 an element with infinite support and use the fact that precisely finitely supported vectors in 1 attain their norm. 3.164 Assume that C is a separable closed convex subset of X ∗ such that each x ∈ X attains its supremum over C. Is C w ∗ -compact? Hint. Yes, by the proof of Theorem 3.122, C is equal to its w∗ -closure. 3.165 Show that a Banach space X is reflexive if and only if the distance of any point to a given closed convex subset of X is attained. Hint. Assume there always is a closest point. Let f ∈ S X ∗ and consider the closed convex set F := {x ∈ X : f (x) = 1}. The distance of F to 0 is 1. A closest point in F to 0 is then an element in B X ∗ where f attains its norm. It is enough to apply Corollary 3.131. Assume now that X is reflexive. Let C be a nonempty closed convex subset of X , x 0 ∈ X \C and d := dist(x0 , C) (> 0). Fix ε > 0. The set B(x0 , d + ε) ∩ C is nonempty, closed, convex and bounded (so w-compact). The function x → x − x 0  is w-lower semicontinuous, hence it attains its minimum on B(x0 , d + ε) ∩ C at some x, also where the distance from x0 to C is attained. 3.166 Show the following Šmulyan’s theorem ([Smul1]): A Banach space X is reflexive if each nested sequence C n ⊃ Cn+1 of closed convex subsets of B X has a nonempty intersection. Hint. Show that then every f ∈ X ∗ attains its norm by considering the sets Cn := {x ∈ B X : f (x) ≥ sup B X f − n1 }. 3.167 Let X be a Banach space, let C be a separable set in X ∗ . If C is w∗ -compact, is D = conv(C) also w∗ -compact? What if the separability assumption is dropped? Hint. Yes for the first part. Each x ∈ X attains its supremum over C which is the supremum of x over D. Then use Exercise 3.164.

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If the separability is dropped the situation is different. Consider the Dirac measures in BC[0,1]∗ , see Exercise 3.130. 3.168 Let X be a separable Banach space and let B be a James boundary for X . Show that if A ⊂ X is a bounded set in X compact in the topology of pointwise convergence on B, then A is w-compact. w∗ Hint. We may assume that B is convex. Then B = B X ∗ . As B X ∗ is metrizable in the w ∗ -topology, B is w∗ -separable and there is a countable w ∗ -dense subset D w∗ of B, in particular D = B X ∗ . The topology of the pointwise convergence on D is a metrizable topology on A which, by the compactness assumption, coincides with the pointwise topology on B. Thus every sequence in A has a subsequence that is convergent in the topology of the pointwise convergence on B, which by Theorem 3.134 gives that A is weakly compact. 3.169 (i) Show that Theorem 3.134 does not in general hold for nets. (ii) Show that Theorem 3.134 does not in general hold for unbounded sequences. (iii) Show that Theorem 3.134 does not in general hold for x ∈ X ∗∗ . Hint. (i) For finite F ⊂ [0, 1] consider the following function f F ∈ SC[0,1] : Assume F = {xi }, 0 ≤ x 1 < · · · < x n ≤ 1. Define f F as the piece-wise linear function with nodes in xi , where f F (xi ) = 1, and nodes in 0,1, and midpoints of [xi , xi+1 ], where f F = 0. Then for the Lebesgue measure λ we have λ( f F ) = 12 , but considering the net { f F }, F partially ordered by the inclusion, we see that δt ( f F ) → 1 for all Dirac measures. (ii) Put x i = iei , where ei denotes the standard unit vector in c0 . Then x i → 0 pointwise in c0 , but x i does not converge weakly to 0 in c0 as it is unbounded. (iii) Let F ∈ C[0, 1]∗∗ be zero on all Dirac measures but F = 0. Take x n = 0 ∈ C[0, 1]. 3.170 Show that Simons’s Lemma 3.123 does not hold if the requirement that elements in the superconvex hull of a sequence attains their supremum is dropped. Hint. Take B := [0, +∞) and xn := χ[n,+∞) , n ∈ N. 3.171 Show that Lemma 3.123 does not hold in general if the assumption on the completeness of X is dropped. Hint. Consider the subspace X of ∞ formed by all sequences that are eventually constant. Then the standard unit vectors {ei } form a James boundary B of X . Con sider x i = χ{i,i+1,... } . Then sup B lim supi x i = 0. If x ∈ conv{x i }, then for large n, the nth coordinate of x equals 1. Note that this is caused by the fact that infinite convex combinations are used in the proof of the Simons’ inequality. 3.172 Show the following Lyapunoff theorem: Suppose that μ1 , . . . , μn are realvalued nonatomic measures on a σ -algebra M. Define μ(E) = (μ1 (E), . . . , μn (E))

(E ∈ M).

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Then μ is a function with domain M whose range is a compact convex set in IR n . Note that μ is called nonatomic if every set E ∈ M with |λ|(E) > 0 contains a set A ∈ M with 0 < |λ|(A) < |λ|(E). Here |λ| denotes the variation measure of λ. Hint. (Lindenstrauss, see [Rudi3, p. 113]) Put |σ | = |μ1 | + · · · + |μn | and K = {g ∈ L ∞ (σ ) : 0 ≤ g ≤ 1}. Define Λ : L ∞ (σ ) → IR n by Λ(g) =



 gdμ1 , . . . ,

gdμn .

The mapping Λ is a weak∗ -continuous linear mapping from L ∞ (σ ) into IR n . Moreover, K is a weak∗ -compact convex set in L ∞ (σ ) (= L 1 (σ )∗ ). Therefore Λ(K ) is a compact convex set in IR n . So, it remains to show that μ(M) = Λ(K ). It is easy to show that μ(M) ⊂ Λ(K ). To obtain the opposite inclusion, pick a point p ∈ Λ(K ) and define K p = {g ∈ K : Λg = p}. We have to show that K p contains some χ E , a characteristic function of E ∈ M, for then p = μ(E). The set K p is a weak∗ -compact convex set. By the Krein–Milman theorem, K p has an extreme point g0 . Assume that g0 is not a characteristic function in L ∞ (σ ). Then there is a set E ∈ M and a constant 0 < r < 1/2 such that σ (E) > 0 and r ≤ g0 ≤ 1 − r on E. Put Y = χ E · L ∞ (σ ). Since σ (E) > 0 and σ is nonatomic, dim Y > n. Hence there exists g ∈ Y , not the zero element in L ∞ (σ ), such that Λg = 0 and such that −r < g < r . Therefore g0 + g and g0 − g are both in K p . Thus g0 is not an extreme point of K p . This shows that all extreme points of K p are characteristic functions. This finishes the proof. 3.173 Prove that an operator T from a Banach space X into a Banach space Y is completely continuous if and only if T (K ) is  · -compact for every w-compact subset K of X . Hint. Assume that T is completely continuous. Note first that T is continuous, since it maps ·-null sequences into ·-null sequences. Let K be a w-compact subset of X . Given a sequence {T xn } in T K , the sequence {x n } has a subsequence {xn k } that w-converges to some point x ∈ K , due to the Eberlein–Šmulyan Theorem 3.109. Then {T x n k } is  · -convergent to T x, and this proves that T (K ) is  · -compact. Conversely, assume that T maps w-compact subsets of X onto  · -compact subsets of Y . Note that T is continuous: otherwise there would exist a sequence {xn } in B X such that T xn  ≥ n 2 for all n ∈ N. The set {x n /n : n ∈ N} ∪ {0} is  · -compact, and {T x n : n ∈ N} is not even bounded, a contradiction. Let {x n } be a w-null sequence. The set {xn } ∪ {0} is w-compact, hence {T xn } ∪ {0} is  · -compact. Let {T xn k } be an arbitrary subsequence of {T x n }. There exists a subsequence {T xn km } w

of {T x n k } that  · -converges to some T x, where x ∈ K . Since x n km → 0, we get w

T xn km → 0, and so T x = 0. Since this happens for every subsequence {T xn k } of ·

{T xn }, we get T xn → 0. 3.174 Show that B∗∞ is not w∗ -sequentially compact.

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3 Weak Topologies and Banach Spaces

  Hint. Define a sequence of elements of B∗∞ by f n (xi ) = xn . This sequence has no w ∗ -convergent subsequence. To see it, assume that { f n k } is a w∗ -convergent subsequence. Define a vector x = (xi ) ∈ ∞ by x n k = (−1)k and xi = 0 if i∈ / {n k }. Then f n k (x) is not convergent. [0,2π ] has no pointwise con3.175 Show that the sequence {sin nx}∞ n=1 in [−1, 1] vergent subsequence (this proves, in particular, that [−1, 1][0,2π ] , endowed with the topology of the pointwise convergence, is not sequentially compact).

2π Hint. Otherwise, {sinn k+1 − sinn k }k → 0 pointwise, hence 0 (sinn k+1 (x) − sinn k (x))2 dx → 0 because of the Lebesgue dominated convergence theorem. How 2π ever, 0 (sinn k+1 (x) − sinn k (x))2 dx = 2π for all k.

3.176 (i) Let { f n } be a sequence in the dual X ∗ of a Banach space X . Show directly that if X is separable and { f n } is bounded, then there exists a w ∗ -convergent subsequence of { f n }. (ii) Let {xn } be a sequence in a Banach space X . Show directly that if X is reflexive and {xn } is bounded, then there exists a w-convergent subsequence of {xn }. Hint. (i) Let {xi } be a dense countable subset of X . Since { f n } is bounded, so is { f n (x1 )} and there exists a convergent subsequence { f n1 (x1 )}. Applying this trick to { f n1 } and x2 we get a subsequence { f n2 } such that { f n2 (xi )} converges for i = 1, 2. Continue in this manner and set f n k = f kk . Then for every xi the limit lim f n k (xi ) k

exists. Using the boundedness of { f n } and density of {xi }, show that lim f n k (x) exists k

for all x ∈ X , call it f (x). f is linear and assuming  f n  ≤ C we show that | f (xi )| ≤ Cxi , so f is continuous. (ii) Define Y = span{xn }. Then Y is reflexive and separable, hence also Y ∗∗ = Y and consequently Y ∗ are separable. Since {xn } is bounded in Y ∗∗ , by the first part it has a subsequence {x n k } that w ∗ -converges in Y ∗∗ to some x ∈ Y ∗∗ = Y . So w f (x n k ) → f (x) for all f ∈ Y ∗ , hence also for all f ∈ X ∗ and x n k → x. 3.177 Prove Krein’s Theorem 3.133 by using only the separable version of James’ Theorem 3.130. Hint. Let A be a weakly compact subset of a Banach space X and let C = conv (A). Take a sequence {cn } in C. There exists a countable subset S of A such that {cn : n ∈ N} ⊂ conv (S). The set conv (S) is  · -separable and every f ∈ X ∗ attains its w supremum over conv (S) at some point in the weakly compact set S (⊂ conv (S)). By the separable version of Theorem 3.130 (a complete proof of this was provided), conv (S) is w-compact. Therefore, C is weakly countably compact. By the Eberlein– Šmulyan Theorem 3.109, C is weakly compact. 3.178 Prove (iii)⇒(i) in Theorem 3.109 by using Theorem 3.130.

Exercises for Chapter 3

177

Hint. Let A be a relatively weakly countably compact subset of a Banach space X . Certainly A is bounded. Given f ∈ X ∗ , there is a sequence {an } in A such that f (an ) → sup A f (= sup Aw f ). The sequence {an } has a w-cluster point in w w w A . This proves that f attains its supremum over A at some point in A . Use w Theorem 3.130 to conclude that A is w-compact. 3.179 Show that the convexity assumption in Theorem 3.130 cannot be dropped. This shows that any characterization of general weakly compact sets via attainment of maxima (for instance Theorem 14.49) must consider more functions than just elements of X ∗ . Hint. Consider B2 \ 12 B2 .

Chapter 4

Schauder Bases

In this chapter we shall introduce Schauder bases, an important concept in Banach space theory. Elements of a Banach space with a Schauder basis may be represented as infinite sequences of “coordinates,” which is very natural and useful for analytical work. Although not every separable Banach space admits a Schauder basis (this is a deep and difficult result of Enflo), basic sequences exist in every infinite-dimensional Banach space (Mazur) and are ideal for the study of linear subspaces and quotients. In many respects, the assumption of having a Schauder basis is not very restrictive, and in fact, for naturally defined separable Banach spaces, it is usually easy to find their Schauder basis. This notion has proved to be an extremely useful tool in the study of the structure of classical as well as abstract Banach spaces. Among the results in this chapter let us mention the Pełczy´nski decomposition principle, leading to a complete characterization of complemented subspaces of the classical sequence spaces, James’ structural theory of spaces with an unconditional Schauder basis, Bessaga–Pełczy´nski characterization of spaces containing c0 , etc.

4.1 Projections and Complementability, Auerbach Bases Let V be a vector space. In Definition 1.32 we introduced the notion of linear projection in V , and we mentioned, right after it, that associated to a projection P in V there is an algebraic direct sum (or algebraic direct decomposition) V = P X ⊕Ker P. Conversely, if V = M1 ⊕ M2 is an algebraic direct decomposition of V , the mappings Pi : V → V given by Pi (v) = vi , where v = v1 + v2 , vi ∈ Mi , i = 1, 2, are linear projection onto Mi (parallel to the other subspace) for i = 1, 2, and P1 + P2 = I V (the identity mapping on V ). If T is a mapping from a vector space into itself, we denote T 2 = T ◦ T and we define by induction T n for n ∈ N. Given a subspace M1 of a vector space V , there is another subspace M2 such that V = M1 ⊕ M2 . Such a subspace M2 is called an algebraic complement of M1 in V . Definition 4.1 A closed subspace Y of a Banach space X is said to be complemented in X if there is a bounded linear projection P of X onto Y . If λ is a real number (necessarily λ ≥ 1) such that P ≤ λ we say that Y is λ-complemented in X .

M. Fabian et al., Banach Space Theory, CMS Books in Mathematics, C Springer Science+Business Media, LLC 2011 DOI 10.1007/978-1-4419-7515-7_4, 

179

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4 Schauder Bases

Let Y1 be a closed subspace of X . We say that Y2 is a (topological) complement of Y1 in X if X = Y1 ⊕ Y2 and Y2 is a closed subspace of X . Note that if Y is a complemented subspace of a Banach space X , then Y is closed. Indeed, if Y = P(X ) for some projection P on X , then Y = (I − P)−1 (0). On the other hand, not every closed subspace of X is complemented. However, finitedimensional and finite-codimensional closed subspaces are complemented (Theorem 4.5 and Exercise 5.8, respectively). Proposition 4.2 Let Y be a closed subspace of a Banach space X . Y is complemented in X if and only if there exists a topological complement of Y in X . Proof: Assume that there is a closed subspace Z of X such that X = Y ⊕ Z . Let P be thelinear projection onto Y such that Z = Ker(P). We will show that the graph  G = { x, P(x) : x ∈ X } is closed.    Let xn ∈ X be such that xn , P(x n ) → x, y . Since Y is closed, y ∈ Y . Note     that x n − P(x n ) ∈ Z = Ker(P). Indeed, P xn − P(xn ) = P(xn ) − P 2 (xn ) = 0. Since Z is closed, we get x − y ∈ Z . By the uniqueness of the decomposition we get y = P(x). Thus (x, y) ∈ G and by the closed graph theorem, P is a bounded projection onto Y . To prove the opposite implication, assume P 2 = P, P(X ) = Y . We will show that Ker(P) is a topological complement of Y . If x ∈ Ker(P) ∩ P(X ), then x = P(x) = 0, so Y ∩ Ker(P) = {0}. Consider any x ∈ X . Since P is a projection, x − P(x) ∈ Ker(P) and we get a decomposition  x = P(x) + x − P(x) . This shows that X = Z ⊕ Ker(P). Since P is continuous, Ker(P) is a closed subspace of X . Observe that if H is a Hilbert space and F its closed subspace, the orthogonal complement F ⊥ of F is a complement of F in the sense of Definition 4.1, and the orthogonal projection of H to F is a norm-one projection onto F (see Corollary 1.50). Fact 4.3 Let X, Y be Banach spaces, let T be an isomorphism of X onto Y . If X 1 is a complemented subspace of X with topological complement X 2 , then T (X 1 ) is a complemented subspace of Y with topological complement T (X 2 ). Proof: Clearly Y = T (X 1 ) ⊕ T (X 2 ) and both subspaces are closed. Note that if P is a projection of X onto X 1 with Ker(P) = X 2 , then Q = T P T −1 is a projection of Y onto T (X 1 ) with Ker(Q) = T (X 2 ). Let X be a normed space such that X = Y ⊕ Z for complemented subspaces Y, Z . Then X is isomorphic to the direct sum (Y ⊕ Z )∞ of spaces Y, Z with the  norm (y, z) = max y, z . Proposition 4.4 Let Y be a closed subspace of a Banach space X . If Y is complemented and Z is a complement of Y in X , then X/Y is isomorphic to Z . Moreover,

4.1

Projections and Complementability, Auerbach Bases

181

if Q : X → X is the projection onto Z parallel to Y and Q = 1, then X/Y is isometric to Z The dual X ∗ is isomorphic to Y ∗ ⊕ Z ∗ ; in short, (Y ⊕ Z )∗ = Y ∗ ⊕ Z ∗ . If P : X → X is the projection onto Y parallel to Z , then Y ∗ is isomorphic to P ∗ (X ∗ ). Moreover, if P = 1, then Y ∗ is isometric to P ∗ (X ∗ ). If Y is a closed subspace of a Hilbert space H and Z is the orthogonal complement of Y in H , then H/Y is isometric to Z . Proof: Consider the operator I : Z → X/Y defined by I (z) = zˆ (∈ X/Y ), where zˆ denotes the equivalence class containing z. From the definition of the norm of X/Y , it follows that I (z) ≤ z for every z ∈ Z , so I is a continuous operator. We claim that I is one-to-one. To see this, assume that for some z ∈ Z we have zˆ = 0ˆ (= Y ). Since z ∈ zˆ , we get z ∈ Z ∩ Y and thus z = 0. I is also an onto mapping. Indeed, let xˆ ∈ X/Y . Pick an arbitrary x ∈ x. ˆ Write x = y + z, where y ∈ Y and z ∈ Z . Then xˆ = zˆ (= I (z)). Therefore I is an isomorphism of Z onto X/Y by the open mapping theorem. Assume now that Q = 1. Take z ∈ Z . If x ∈ X is an arbitrary element in zˆ , then x = y + z for some y ∈ Y . Thus z = Q(x) ≤ x. By the definition of the norm in X/Y we get z ≤ ˆz . This proves the reverse inequality and so I is an isometry. Let P, respectively Q, be projections of X onto Y , respectively Z . By Exercise 4.10, Y ∗ is isomorphic to P ∗ (X ∗ ) and Z ∗ is isomorphic to Q ∗ (X ∗ ). Moreover, from I X = P + Q we get I X ∗ = P ∗ + Q ∗ , that is, P ∗ (X ∗ ) + Q ∗ (X ∗ ) = X ∗ . It remains to show that P ∗ (X ∗ ) ∩ Q ∗ (X ∗ ) = {0}. Take f ∈ P ∗ (X ∗ ) ∩ Q ∗ (X ∗ ). For y ∈ Y we have f (y) = Q ∗ ( f )(y) = f Q(y) = 0, and for z ∈ Z ,  ∗ f (z) = P ( f )(z) = f P(z) = 0, hence f = 0. Since P ∗ (X ∗ ) and Q ∗ (X ∗ ) are closed subspaces, it follows from Proposition 4.2 that X ∗ = P ∗ (X ∗ ) ⊕ Q ∗ (X ∗ ) is a topological direct sum. If P = 1, use again Exercise 4.10. If H is a Hilbert space, then Q = 1 by Corollary 1.50, and the conclusion follows from the first part of the statement. Recall the definition of the Kronecker delta: δi j = 0 if i = j and δi j = 1 if i = j. Let X be a Banach space, consider vectors {e1 , . . . , en } ⊂ X . Functionals n { f 1 , . . . , f n } ⊂ X ∗ are called biorthogonal to {ei }i=1 if f i (e j ) = δi j for i, j = n 1, . . . , n. In other words, the set {ei ; f i }i=1 is a biorthogonal system in X × X ∗ . n n A biorthogonal system {ei ; f i }i=1 is called an Auerbach basis of X if {ei }i=1 is a basis of X and ei  =  f i  = 1 for every i. Theorem 4.5 (Auerbach, see, e.g., [LiTz3]) (i) Let X be a Banach space. If n dim (X ) = n then there exists an Auerbach basis {ei ; f i }i=1 of X . (ii) Let Y be a subspace of a Banach space X . If dim (Y ) = n then there exists a projection P of X onto Y such that P ≤ n. Part (ii) in Theorem 4.5 gives that the so called projection constant λ(Y ) of an n-dimensional Banach space Y (see Definition 5.15) is less or equal than n. This will be substantially improved in Theorem 6.28.

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Proof of Theorem 4.5: (i) Let {x1 , . . . , xn } be an algebraic basis of X . For u 1 , . . . , u n ∈ B X let v(u 1 , . . . , u n ) be the determinant of the matrix whose jth row is formed by the coordinates of u j in the basis {x1 , . . . , x n }. The function |v| is continuous on the compact set B X × · · · × B X , therefore there is (e1 , . . . , en ) ∈ B X × · · · × B X such that v(e1 , . . . , en ) = max{|v(x 1 , . . . , xn )| : (x1 , . . . , xn ) ∈ B X × · · · × B X }. Since determinants are homogeneous in each coordinate, we have ei ∈ S X . As v(e1 , . . . , en ) = 0, the vectors {ei } are linearly independent, so they form a basis of X . For i = 1, . . . , n, define fi ∈ X ∗ by f i (x) =

v(e1 , . . . , ei−1 , x, ei+1 , . . . , en ) . v(e1 , e2 , . . . , en )

  Then f i (e j ) = δi j , so {e1 , . . . , en } : { f 1 , . . . , f n } is a biorthogonal system. Moren over, sup{| f j (x)| : x ∈ B X } ≤ 1 for each j, therefore  f j  = 1. Thus {ei ; f i }i=1 is an Auerbach basis of X . n be an Auerbach basis of Y . We extend f i tonorm-one func(ii) Let {ei ; f i }i=1 n tionals on X . Then we define an operator P : X → Y by P(x) = i=1 f i (x)ei for x ∈ X.  We claim that P is a projection onto Y . Indeed, for every y = αi ei ∈ Y we n P(y) = f (y)e = y. Finally, if x ∈ X and have αi = f i (y). Therefore i i i=1  n n x ≤ 1, then P(x) ≤ i=1 | f i (x)| ei  ≤ i=1 1 = n.

4.2 Basics on Schauder Bases Definition 4.6 Let X be an infinite-dimensional normed linear space. A sequence ∞ in X is called a Schauder basis of X if for every x ∈ X there is a unique {ei }i=1 ∞ , called the coordinates of x, such that x = ∞ a e . sequence of scalars {ai }i=1 i=1 i i ∞ is a linearly independent set in X . Clearly {ei }i=1 If a Banach space X has a Schauder basis, then it is separable as finite rational combinations of elements of the basis form a countable dense set. If X is n-dimensional, the notion of Schauder basis coincides with the notion of algebraic basis. We will write {ei }, span{ei }, etc. to cover both the finite- and infinite-dimensional case. The condition in the definition can be weakened. If every vector x ∈ X has a n  w ai ei → x as n → ∞, it already unique decomposition ai ei such that i=1 implies that {ei } is a Schauder basis (see, e.g., [Karl]). projections If {ei } is a Schauder basis of a normed space  X , then the canonical n ∞ Pn : X → X are defined for n ∈ N by Pn = i=1 ai ei i=1 ai ei . Each Pn

4.2

Basics on Schauder Bases

183

is a linear projections from X onto the linear subspace spanned by {ei : i = 1, 2, . . . , n}. Lemma 4.7 Let {ei } be a Schauder basis of a normed space X . The canonical projections  Pn satisfy: (i) dim Pn (X ) = n, (ii) Pn Pm = Pm Pn = Pmin(m,n) , (iii) Pn (x) → x in X for every x ∈ X . Conversely, if bounded linear projections {Pn }∞ n=1 in a normed space X satisfy (i)– (iii), then Pn are canonical projections associated with some Schauder basis of X .

Proof: The set {ei : n ∈ N} is a linearly independent set in X . Thus (i) follows, (ii) is obvious. The property (iii) follows directly from the definition of the Schauder basis. Conversely, if projections Pn satisfy (i)–(iii) put P0 = 0 and choose 0 = ei ∈ Pi (X ) ∩ Ker(Pi−1 ) for i ∈ N. Then n      Pi (x) − Pi−1 (x) x = lim Pn (x) = lim Pn (x) − P0 (x) = lim n→∞

=

∞   i=1

n→∞

 Pi (x) − Pi−1 (x) =

n→∞

∞ 

i=1

αi ei

i=1

  for some scalars αi as dim Pn (X )/Pn−1 (X ) = 1, i ∈ N. ∞ The uniqueness of {αi : i ∈ N} follows nfrom the fact that if x = i=1 βi ei , then by the continuity of Pn we get Pn (x) = i=1 βi ei , hence βi ei = Pi (x)− Pi−1 (x) = αi ei , for all n ∈ N. Thus {ei } is a Schauder basis of X and Pn are the projections associated with {ei }. Fact 4.8 Let {ei } be a Schauder basis of a normed linear space X with canonical projections Pn . If supn Pn  < ∞ (we say that Pn are uniformly bounded), then {ei } is also a Schauder basis of the completion  X of X . n of Pn on  X satisfy (i)–(iii) Proof: We will show that the extensions P   of Lemma 4.7. n  Since Pn (X ) is finite-dimensional, it is closed in  X and thus P X = Pn (X ), so (i) n by the continuity of Pn . Since Pn (x) → x follows. (ii) is extended from Pn to P n (x)  for all x in a dense set X and Pn are uniformly bounded, we have P ˜ → x˜ in X,    so (iii) is also true. Since en ∈ Pn (X ) ∩ Ker(Pn−1 ), we get en ∈ Pn X ∩ Ker Pn−1 n are canonical projections associated with the Schauder for every n. Therefore P X. basis {ei } of 

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4 Schauder Bases

Lemma 4.9 Let {ei} be a Schauder basis of a Banach space (X,  · ). Define ||| · |||  ∞ n  on X by |||x||| = sup i=1 ai ei  for x = i=1 ai ei . Then: n

(i) ||| · ||| is a norm on X , {ei } is a Schauder basis of (X, ||| · |||) and the canonical projections Pn are uniformly bounded by 1 in ||| · |||, (ii) ||| · ||| is an equivalent norm on X . Proof: (i) The triangle inequality and homogeneity of   |||·||| are simple to check. Since  n  ai ei  by Lemma 4.7, we obtain that for every x ∈ X we have x = lim  i=1 n→∞ |||x||| ≥ x for every x ∈ X . This in particular means that ||| · ||| is a norm on X . To show that {ei } is a Schauder basis of (X, ||| · |||), we use Lemma 4.7. The properties (i) and (ii) are straightforward. To check (iii), we note that for x ∈ X we have |||x − Pm (x)||| = sup Pn (x) − Pn Pm (x) = sup Pn (x) − Pm (x) → 0 n

n≥m

as m → ∞. Finally, for m ∈ N we estimate |||Pm ||| = sup |||Pm (x)||| = sup sup Pn Pm (x) = sup sup Pn Pm (x) |||x|||≤1

|||x|||≤1 n

n |||x|||≤1

   = sup sup Pn Pm (x) : x with sup Pi (x) ≤ 1 ≤ 1. n

i

(ii) We will show that X is complete in the norm ||| · |||, i.e., that  X ⊂ X , where  X basis is the completion of X in ||| · |||. By (i) we already know that {ei } is a Schauder  of  X . Given x ∈  X , there is a unique sequence of scalars αi such that x = αi ei , where the convergence is in the norm ||| · |||. Since ||| · ||| ≥  ·  on X , we get that  αi ei is Cauchy in  ·  andthus convergent to some element x ∈ X in the norm  · . As shown in part (i), αi ei then converges to x in the norm ||| · |||. Thus x = x ∈ X . This means that X is complete in ||| · |||.    The formal identity mapping I : X, ||| · ||| →  X,  ·  is a linear bijection X   of a Banach space X, ||| · ||| onto a Banach space X,  ·  which is continuous as ||| · ||| ≥  · . From the Banach open mapping principle it follows that I X−1 is continuous, which means that ||| · ||| is an equivalent norm on X . Theorem 4.10 (Banach) Let {ei } be a Schauder basis of a Banach space X . The canonical projections Pn associated with {ei } are uniformly bounded. The value bc{ei } = supn Pn  is called the basis constant of {ei }. Proof: Define ||| · ||| as in Lemma 4.9. Then |||Pn ||| ≤ 1 for every n and since ||| · ||| is an equivalent norm, the result follows. Considering the vectors en we see that Pn  ≥ 1, in particular bc{ei } ≥ 1. A Schauder basis {ei } is called normalized if ei  = 1 for every n. A Schauder basis {xi } of a Banach space X is called seminormalized if 0 < inf xi  ≤ sup xi  <

4.2

Basics on Schauder Bases

185

∞. It is called monotone if bc{ei } = 1, that is, its associated projections satisfy Pn  = 1 for every n. We proved in Lemma 4.9 that every Banach space with a Schauder basis can be renormed in such a way that, in the new norm the basis becomes monotone. It follows from Theorem 4.10 that, if {ei } is a Schauder basis of a Banach space X and n ∈ N, then X = span{ei : i = 1, 2, . . . , n} ⊕ span{ei : i = n + 1, n + 2, . . .} is a topological direct decomposition. Indeed, Pn is continuous, and certainly (I − Pn )X = Ker Pn . ∞ ai ei Let {ei } be a Schauder basis of a Banach space X . For j ∈ N and x = i=1 denote f j (x) = a j . Then P j (x) − P j−1 (x) =  f j (x)e j  = | f j (x)| · e j  and thus  f j  = sup | f j (x)| = e j −1 sup  f j (x)e j  ≤ 2e j −1 sup Pn . x∈B X

n

x∈B X

Therefore f j ∈ X ∗ for every j. The functionals { f i } are called the associated biorthogonal functionals (or coordinate functionals) to {ei } and we have x = ∞ i=1 f i (x)ei for every x ∈ X . We will denote the biorthogonal functionals f i by ei∗ and say that {ei ; ei∗ } is a Schauder basis of the Banach space X. It is a biorthogonal system and we have just proved that ei  ei∗  ≤ 2 bc{ei }. Note that {ei∗ } is separating for X . Examples: (i) Any linear basis of a finite-dimensional Banach space X is a Schauder basis of X . In particular, any Auerbach basis (see the definition right before Theorem 4.5) is a Schauder basis. (ii) Any orthogonal basis of a Hilbert space H is a Schauder basis of H . (iii) If X = c0 or X =  p for p ∈ [1, ∞), then the sequence {ei } of the standard i

unit vectors ei = (0, . . . , 0, 1, 0, . . . ) is a Schauder basis of X . All the above statements are easy to verify. (iv) Let {t j }∞ j=1 be a sequence of distinct points in [0, 1] such that t1 = 0, t2 = 1, and {t j } = [0, 1]. Define projections Pn from C[0, 1] into C[0, 1] by P1 ( f ) = f (0) and Pn ( f ) to be the piecewise linear function with nodes at t j , j = 1, . . . , n and such that Pn ( f )(t j ) = f (t j ) for j = 1, . . . , n. By Lemma 4.7 and the uniform continuity of continuous functions on [0, 1], it follows that the projections Pn determine a monotone Schauder basis of C[0, 1]. This basis is called the Faber–Schauder basis of C[0, 1]. Figure 4.1 shows the first elements of this basis for the particular election of the dyadic points. 1

0

e1

e2

1

e3

e4

1/2

1/4

e5

e6

3/4 1/8

Fig. 4.1 The first elements of the Faber–Schauder basis for dyadic nodes

186

4 Schauder Bases

(v) Let the functions h i be defined on [0, 1] as follows - (see  Fig. 4.2): h 0 (x) = 1 for x ∈ [0, 1]; h 1 (x) = 1 for x ∈ 0, 12 and h 1 (x) = −1 for  . . x ∈ 12 , 1 ; h 2 (x) = 1 for x ∈ 0, 14 , h 2 (x) = −1 for x ∈ 14 , 12 , h 2 (x) = .  .   0 for x ∈ 12 , 1 ; h 3 (x) = 1 for x ∈ 12 , 34 , h 3 (x) = −1 for x ∈ 34 , 1 and h 3 (x) = 0 elsewhere, etc. Fix p ∈ [1, ∞). The set {h i } is linearly independent. Since H = span{h i } contains the characteristic functions of dyadic intervals, we get L H p = L p [0, m1]. n For x = i=0 αi h i ∈ H define Pn (x) = i=0 αi h i (we can always assume that m ≥ n by adding 0h j to the sum). These projections satisfy (i)–(iii) of Lemma 4.7, so by Fact 4.8, to show that {h i } is a Schauder basis (called the Haar basis) of L p [0, 1] we only need to prove that Pn are uniformly bounded on (H,  ·  p ). n n+1 Assume that f = i=1 ai h i and g = i=1 ai h i for some real numbers ai . Then f and g differ only on some dyadic interval I where f has a constant value, say b, and g has value b + an+1 on the first half of I and the value b − an+1 on the second half of I . Since for every p ∈ [1, ∞) we have |b| p = | 12 (b+an+1 )+ 12 (b−an−1 )| p ≤ 1 1 p p p 2 |b + an+1 | + 2 |b − an+1 | by convexity of |x| , we get  f  ≤ g in L p and it follows that the projections Pn have norm at most 1. Thus {h i } is a monotone Schauder basis of L p [0, 1]. By inspection, we see that {h i } is an orthonormal basis of L 2 [0, 1]. Note that, for the Faber–Schauder basis, f 0 = 1 and f n , for n ≥ 1, is the indefinite integral of h n−1 , where {h n } is the Haar basis.

h1

1

0

1

h2

h3

h4

h5

h6

1/2

1/4

3/4

1/8

3/8

−1 Fig. 4.2 The first elements of the Haar basis

(vi) The trigonometric system defined in Note (v) after Theorem 1.57 is a Schauder basis in L p [0, 1], p > 1 (see, e.g., [Katz, p. 50]) but it is not a Schauder basis in L 1 [0, 1] (see, e.g., [Katz, p. 47]). It is not a Schauder basis in the space of continuous functions (see, e.g., [AlKa, p. 9]). (vii) On the space c, define projections Pn for x = (xi ) ∈ c by P1 (x) = (x1 , x1 , . . . ), P2 (x) = (x1 , x2 , x2 , . . . ), . . . , Pn (x) = (x1 , x2 , . . . , xn , xn , . . . ). Then for x ∈ c and n ∈ N, x − Pn (x) = (0, . . . , 0, x n+1 − x n , xn+2 − xn , . . . ) = max |x j − x n | → 0 j>n

4.3

Shrinking and Boundedly Complete Bases, Perturbation

187

as n → ∞. By Lemma 4.7, Pn generate a Schauder basis x1 = (1, 1, . . . ), n

x2 = (0, 1, 1, . . . ), . . . , xn = (0, . . . , 0, 1, 1, . . . ). Given z = (z 1 , z 2 , . . . ) ∈ c and (α1 , α2 , . . . , αn ) such that (z 1 , z 2 , . . . , z n , z n , . . . ) = α1 x1 + · · · + αn xn , we calculate that z 1 = α1 , z 2 = α1 + α2 , . . . , z n = α1 + · · · + αn . Therefore Pn (z) = α1 x1 + · · · + αn xn and Pn (z) = max j≤n |z j | = max |α1 | : |α1 + α2 |, . . . , |α1 + · · · + αn | . The basis {xi } is called the summing basis of c.

4.3 Shrinking and Boundedly Complete Bases, Perturbation Let {ei } be a Schauder basis of X . Considering ei ∈ X ∗∗ we see that {ei∗ ; ei } is a biorthogonal system in X ∗ . Fact 4.11 Let {ei ; ei∗ } be a Schauder basis of a Banach space X with the canonical projections Pn . n n f (ei )ei∗ = i=1 ei ( f )ei∗ for every f ∈ X ∗ . (i) For n ∈ N, Pn∗ ( f ) = i=1 w∗

(ii) Pn∗ ( f ) → f in X ∗ for every f ∈ X ∗ . (iii) {ei∗ ; ei } is a Schauder basis of span{ei∗ } with the canonical projections Pn∗ . In particular, Pn∗ ( f ) → f for every f ∈ span{ei∗ }. Proof: (i) For n ∈ N, f ∈ X ∗ and x =

∞

∗ i=1 ei (x)ei

we have

n n     ei∗ (x)ei = f (ei )ei∗ (x). Pn∗ ( f )(x) = f Pn (x) = f i=1

i=1

(ii) By the continuity of f ∈ X ∗ , lim Pn∗ ( f )(x) = lim

n→∞

n→∞

n  i=1

ei∗ (x) f (ei ) = f

 lim

n→∞

n 

ei∗ (x)ei = f (x).

i=1

∗ (iii) We easily check that Pn∗ Pm∗ = Pmin(n,m) . If f ∈ span{ei∗ }, then Pn∗ ( f ) = f ∗ for n large enough and thus lim Pn ( f ) − f  = 0. Since Pn  = Pn∗ , {Pn∗ } are uniformly bounded and we can apply Lemma 4.7 and Fact 4.8.

Definition 4.12 Let {ei ; ei∗ } be a Schauder basis of a Banach space X . It is called shrinking if span{ei∗ } =X ∗ . ∞ ai ei converges whenever the scalars ai are It is called boundedly complete if i=1    n  such that sup i=1 ai ei  < ∞. n

188

4 Schauder Bases w

Every normalized shrinking basis {ei } has the property that ei → 0 because lim ek∗ (ei ) = 0 for every k ∈ N. However, there exists a Banach space with a

i→∞

w

normalized non-shrinking Schauder basis {ei } such that ei → 0 in X ([PeSz]). Fact 4.13 Let {ei ; ei∗ } be a Schauder basis of a Banach space X with the canonical projections Pn . The following are equivalent: (i) {ei ; ei∗ } is shrinking. (ii) {ei∗ } isa Schauder basis of X ∗ . (iii) lim  f span{e }  = 0 for every f ∈ X ∗ . n→∞

i i>n

Proof: (i)⇒(ii): Fact 4.11. (ii)⇒(i): If projections {Pn∗ } generate a Schauder basis of X ∗ , then Pn∗ ( f ) → f for every f and thus X ∗ = span{ei∗ }. (i) ⇐⇒ (iii): Note that if P isa bounded  linear projection of a Banach space X onto P(X ), then sup ( f ) = sup P ∗ ( f ) = P ∗ ( f ) and P(B X )

BX

B P(X ) ⊂ P(B X ) ⊂ PB X ∩ P(X ) ⊂ PB P(X ) . Thus we have for every f ∈ X ∗ and n,   f (I

X −Pn )(X )

 = sup{ f (x) : x ∈ B(I X −Pn )(X ) }

≤ sup{ f (x) : x ∈ (I X − Pn )(B X )} ≤ sup{ f (x) : x ∈ (Pn  + 1)B(I X −Pn )(X ) }.

  Hence  f (I −Pn )(X )  ≤  f − Pn∗ ( f ) ≤ (Pn  + 1) f (I −Pn )(X ) . Thus {ei } is X X  shrinking if and only if  f (I −P )(X )  → 0 for every f ∈ X ∗ . X

n

For a sufficient condition for shrinkingness in terms of the smoothness of the norm, see Proposition 8.22. The standard unit vector basis of c0 and  p , p ∈ (1, ∞), is shrinking, while the standard unit vector basis of 1 is not. The standard unit vector basis of  p is boundedly complete for p ∈ [1, ∞) while the standard unit vector basis of c0 is not, n

as the vectors (1, . . . , 1, 0, . . . ) show. basis Proposition 4.14 Let {ei ; ei∗ } be a Schauder   of a Banach space X .∗∗If {ei } is shrinking, then the mapping T (x ∗∗ ) = x ∗∗ (ei∗ ) is an isomorphism   of X onto to  n  ai ei  < ∞. the space of all sequences (ai ) such that |||(ai )||| = supn  i=1 Moreover, if {ei } is monotone, then T is an isometry. Proof: It is routine to check that ||| · ||| defines a norm on the vector space of (ai ) such that |||(ai )||| < ∞. Denote K = bc{ei }, let Pn be the canonical projections associated with {ei }. For x ∈ X , x ∗ ∈X ∗ and x ∗∗ ∈ X ∗∗ we have n n ∗ (e )e∗ and P ∗∗ (x ∗∗ )(x ∗ ) = ∗∗ ∗ ∗ x Pn∗ (x ∗ ) = n i=1  i i i=1 x (ei )x (ei ), so we can n ∗ ∗∗ ∗∗ ∗∗ write Pn (x ) = i=1 x (ei )ei . Thus

4.3

Shrinking and Boundedly Complete Bases, Perturbation

189

n     |||T (x ∗∗ )||| = sup x ∗∗ (ei∗ )ei  = sup Pn∗∗ (x ∗∗ ) ≤ K x ∗∗  n

n

i=1

and T is a bounded operator with T  ≤ K . ∞ such that |||(a )||| < ∞. Since X ∗ is separable (as Now consider (ai )i=1 i  n ∗∗ ∗ {ei } is shrinking) and i=1 ai ei is bounded in X , there exists a w -cluster  n which satisfies x ∗∗ (ei∗ ) = ai . Moreover, x ∗∗  ≤ point x ∗∗ of i=1 ai ei    n  ai ei  ≤ |||(ai )|||. Thus T (x ∗∗ ) = (ai ) and T (x ∗∗ ) ≥ x ∗∗ , which lim sup i=1 completes the proof. Theorem 4.15 Let {ei ; ei∗ } be a Schauder basis of a Banach space X . If {ei } is ∗  boundedly complete, then X is isomorphic to span{ei∗ } . Proof: Let Pn be the canonical projections associated with {ei }. Denote Z = span{ei∗ } and define J : X → Z ∗ by J (x) : z → z(x), then J is a bounded operator. We will show that J is an isomorphism of X onto Z ∗ . Let x ∈ X . Then for every z ∈ Z we have |J (x)(z)| = |z(x)| ≤ z x, so J for n ∈ N find x ∗ ∈ S X ∗  (x)  ≤ x. On the other hand, ∗ ∗ ∗ ∗ n , we have such that x Pn (x) = Pn (x). Since Pn (X ) = span{e  i }i=1   ∗ ∗ ∗ ∗ Pn (x ) ∈ Z , and Pn (x ) ≤ K = bc{ei }. By definition, J Pn (x) Pn∗ (x ∗ ) =       ∗ ∗  Pn (x ) Pn (x) = x ∗ Pn2 (x) = x ∗ Pn (x) = Pn (x) and therefore    Pn∗ (x ∗ )   1 1 J Pn (x)  Z ∗ ≥ J Pn (x) P ∗ (x ∗ ) = P ∗ (x ∗ ) Pn (x) ≥ K Pn (x). By the n

n

continuity of J , we have K1 x ≤ J (x) Z ∗ ≤ x for every x ∈ X . We will now show that J maps X onto Z ∗ . To this end, observe first that ∗ n denote its canonical projections. Then {ei , J (ei )} is a Schauder basis of Z , let P ∗

w n∗  = supn  P n  ≤ K < ∞, hence for every n∗ (z ∗ ) → z ∗ in Z ∗ and supn  P P ∗ ∗ z ∈ Z and n ∈ N we have n      z ∗ (ei∗ )ei  J i=1

Z∗

n     = z ∗ (ei∗ )J (ei ) i=1

Z∗

n∗ (z ∗ ) ≤ K · z ∗ . = P

       n   n ∗ ∗ 2 ∗ Thus we have  i=1 z ∗ (ei∗ )ei  ≤ K · J i=1 z (ei )ei  ∗ ≤ K · z . Since ∞ ∗ ∗ Z the basis {ei } is boundedly complete, the series i=1 z (ei )ei is convergent in X to some x ∈ X . J is a continuous mapping, therefore J (x) = lim J n→∞

n  i=1

n  n∗ (z ∗ ) z ∗ (ei∗ )ei = lim z ∗ (ei∗ )J (ei ) = lim P n→∞

i=1

n→∞

190

4 Schauder Bases ∗

w n∗ (z ∗ ) → in the norm topology of Z ∗ . But P z ∗ in Z ∗ , so z ∗ = J (x) and J is an onto mapping. This concludes the proof.

Theorem 4.16 (James, see, e.g., [LiTz3]) Let X be a Banach space with a Schauder basis {ei }. The space X is reflexive if and only if {ei } is both shrinking and boundedly complete. w∗

Proof: Let X be reflexive. By Fact 4.11, for every f ∈ X ∗ we have Pn∗ ( f ) → f and w thus Pn∗ ( f ) → f in X ∗ as X is reflexive. Therefore X ∗ = spanw {ei∗ } = span{ei∗ } by Mazur’s theorem and {ei } is a shrinking basis of X . to the (Banach) space Y := space X∗∗ is isomorphic  By Proposition 4.14,the n  (ai ) : |||(ai )||| := maxn  i=1 ai ei  < ∞ . Under this correspondence, X ⊂ X ∗∗  ∞ corresponds to Y1 = (ai ) : i=1 ai ei converges . Since X is reflexive, we get Y = Y1 and {ei } is boundedly complete. Conversely, if {ei } is shrinking and boundedly complete, then we have the above identification and Y1 = Y , thus X = X ∗∗ and X is reflexive. Definition 4.17 A sequence {ei } in a Banach space X is called a basic sequence if {ei } is a Schauder basis of span{ei }. A basic sequence {ei } is called shrinking (respectively, boundedly complete) if it is a shrinking (respectively boundedly complete) basis of span{ei }. Proposition 4.18 (Banach) Let {ei } be a sequence of non-zero vectors in a Banach space X . The sequence {ei } is a basic sequence if and only if there is K > 0 such that for all n < m and scalars a1 , . . . , am we have n m         ≤ K a e ai ei .   i i i=1

(4.1)

i=1

Moreover, the smallest such K is equal to bc{ei }. Proof: One implication is clear from n m m m                  P = ≤ bc{e ≤ P a e a e  a e } ai ei .   n i i i i  n  i i i  i=1

i=1

i=1

i=1

     n  m   ai ei  ≤ K  i=1 ai ei  for On the other hand, suppose that K satisfies  i=1  m := all ai and n ≤ m. We define projections Pn on span{ei } by Pn i=1 ai ei n a e for m > n and scalars a and observe that P have norm at most K . We i n i=1 i i check that Pn satisfy (i)–(iii) of Lemma 4.7 on span{ei }, so by Fact 4.8, {ei } is a Schauder basis of span{ei } and bc{ei } ≤ K . Not every separable Banach space admits a Schauder basis (Enflo, see Section 16.5 and Theorem 16.54). But we have the following.

4.3

Shrinking and Boundedly Complete Bases, Perturbation

191

Theorem 4.19 (Mazur) Every infinite-dimensional Banach space contains a basic sequence. If X ∗ is separable, then X contains a shrinking basic sequence. In the proof, we use the following lemma. Lemma 4.20 Let Y be a finite-dimensional subspace of an infinite-dimensional Banach space X . For every ε > 0 there is x ∈ S X such that y ≤ (1 + ε)y + λx for every y ∈ Y and every scalar λ. m Proof: (see Fig. 4.3) Let ε ∈ (0, 1). Let {yi }i=1 be an 2ε -net in SY . For i ∈ {1, . . . , m} ∗ ∗ choose yi ∈ S X ∗ with yi (yi ) = 1. Since X is infinite-dimensional, there is x ∈ S X such that yi∗ (x) = 0 for every i = 1, . . . , m. We claim that x has the desired property. Indeed, let y ∈ SY . Choose i ∈ {1, . . . , m} such that yi − y < ε2 . Let λ be a scalar. Then

y + λx ≥ yi + λx −

ε 2

≥ yi∗ (yi + λx) −

y Thus, given y ∈ Y \{0} and a scalar λ, we have  y +

ε 2

=1−

λ y x

≥

ε 2

≥

1 1+ε .

1 1+ε .

∗ {x : y1(x) = 0}

BX {x : y1∗(x) = 1}

x

0

Y ym

y1

y2

Fig. 4.3 Proof of Lemma 4.20

Proof of Theorem 4.19: Given ε > 0, find εn > 0 with

∞ "

(1 + εn ) ≤ 1 + ε.

n=1

Let x 1 ∈ S X be an arbitrary element. Using Lemma 4.20, construct inductively a sequence {xn }∞ n=2 in S X such that for every n ≥ 1, y ≤ (1 + εn )y + λxn+1  for all y ∈ span{x1 , . . . , xn }. By induction, for n < m and scalars a1 , . . . , am we have

192

4 Schauder Bases n m m             ai xi  ≤ (1 + εn ) · · · (1 + εm−1 ) ai xi  ≤ (1 + ε) ai xi .  i=1

i=1

i=1

By Proposition 4.18, {x i } is a basic sequence and bc{xi } ≤ 1 + ε. ∞ " Note that Pn  ≤ (1 + εi ) → 1 as n → ∞. i=n

An easy modification of Mazur’s construction in Lemma 4.20 and then the proof of Theorem 4.19 gives the last part: let D = {dn∗ : n ∈ N} be a  · -dense countable subset of X ∗ . At the nth step in the construction of the basic sequence {xn } and for εn , consider in X ∗ not only the (finite) set {yi∗ : i = 1, 2, . . . , m} (see the proof 4.20) the set {di∗ : i = 1, 2, . . . , n} in order to mof Lemma n but also ∗ ∗ find xn+1 ∈ i=1 Ker yi ∩ i=1 Kerdi ∩ S X . Now, an appeal to Fact 4.13 gives the conclusion: indeed, given x ∗ ∈ X ∗ and ε > 0, find n 0 ∈ N such that dn∗0 −x ∗  < ε. Then, dn∗0 span{x : n≥n }  = 0, so x ∗ span{x : n≥n }  < ε for every n 1 > n 0 . This n n 1 1 proves that the basic sequence is shrinking. For another proof of the last part in the statement of Theorem 4.19, see the paragraph after Proposition 8.22. It is not known whether every separable Banach space X contains a closed subspace Y such that both Y and X/Y have a Schauder basis. We note that if X is separable and nonreflexive, there is a nonreflexive closed subspace Y of X such that Y has a Schauder basis (Pełczy´nski, see, e.g., [Dies2]). Definition 4.21 Let {ei } be a basic sequence in a Banach space X and let { f i } be a to { fi } if for all basic sequence in a Banach space Y . We say that {ei } is equivalent  sequences of scalars (ai ), ai ei converges if and only if ai f i converges. Fact 4.22 Let {ei } be a basic sequence in a Banach space X and let { fi } be a sequence in a Banach space Y . The following are equivalent: (i) { f i } is a basic sequence equivalent to {ei }. (ii) There is an isomorphism T of span{ei } onto span{ f i } such that T (ei ) = f i for every i. (iii) There are C1 , C 2 > 0 such that for all scalars a1 , . . . , an we have n n   1     a e ≤ ai   i i X C1 i=1

i=1

n       fi  ≤ C2  ai ei  . Y

i=1

X

Proof: Define a mapping T from span{ei } into span{ f i } by  (i)⇒(ii): ∞ ∞ = T i=1 ai ei i=1 ai f i . From the equivalence of {ei } and {xi } we have that T is well defined, one-to-one and onto span{ f i }. ∞ k ai ei converge We will now show that T has a closed graph. Indeed, if x k = i=1 ∞ k ∞ ∞ k) = a e and T (x a f converge to c f to x = i=1 i i i=1 i i i=1 i i , then by the

4.3

Shrinking and Boundedly Complete Bases, Perturbation

193

continuity of the coordinate functionals we have aik → ai and in the same way  ai f i = T (x). aik → ci for every i. Hence ai = ci for every i and thus T (x k ) → By the closed graph theorem, T is continuous, and by the open mapping theorem, T −1 is continuous as well. (ii)⇒(iii): This follows easily with C2 = T , C1 = T −1 .     n ai f i  ≤ (iii)⇒(i): For n < m and scalars a1 , . . . , am we have  i=1     m C1 C2 bc{ei } i=1 ai f i , so by Proposition 4.18, { f i } is a basic sequence. From   (iii) we also have that ai f i is Cauchy if and only if ai ei is Cauchy. The following result is sometimes called the small perturbation lemma. Theorem 4.23 (Krein, Milman, Rutman [LiTz3], Valdivia [Vald5]) Let {ei } be a basic sequence in a Banach space X and let {ei∗ } be the coefficient functionals of ∞ the basis {ei } of span{ei }. Assume that { f i } is a sequence in X such that i=1 ei − fi  ei∗  = C < 1. Then: (i) { f i } is a basic sequence in X equivalent to {ei }. (ii) If span{ei } is complemented in X , then so is span{ f i }. (iii) If {ei } is a Schauder basis of X , then so is { f i }. Moreover, let { f i∗ } be the coefficient functionals of the basis { f i } of X . Then span{ei∗ } = span{ f i∗ }. Proof:(i) Extend ei∗ to functionals on ∞X of∗ the same norm. For x ∈ X we ∞ ∗ (x)(e − f ) ≤ x e have i i i=1 ei  ei − f i  = Cx, so S(x) = ∞ ∗i=1 i e (x)(e − f ) defines a bounded operator from X into X with S ≤ C < 1. i i i=1 i Let T = I X − S. We have x − T (x) = S(x) ≤ Cx, hence T (x) ≥ (1 − C)x. Since 1 − C > 0, by Exercise 1.73, T is an isomorphism into, in particular T (X ) is closed. We claim that T (X ) = X . Assume T (X ) = X . Since C < 1, by Lemma 1.37 there is x ∈ S X such that dist(x, T (X )) > C, which contradicts x − T (x) ≤ C. So T is an isomorphism of X onto X . Then using T (ei ) = f i we obtain that T maps span{ei } onto span{ f i } and (i) is proved. (ii) follows from the proof of (i) using T , Fact 4.3 and the note following this fact. (iii) Let T be the isomorphism of X onto X from (i). Since {ei } is a Schauder basis of X , we get X = T (X ) = T (span{ei }) = span{ f i }, so { f i } is a Schauder basis of X .  Fix some i ∈ N and denote xk∗ = kj=1 f i∗ (e j )e∗j for k ∈ N. Then x k∗ ∈ span{e∗j }  w∗ ∗ ∗ and by Fact 4.11, x k∗ → f i∗ . For x ∈ B X we then have fi∗ (x) = ∞ j=1 f i (e j )e j (x), so for k ≥ i we estimate |( f i∗

−

x k∗ )(x)|

∞ ∞           ∗ ∗ = f i (e j )e j (x) =  fi∗ (e j − f j )e∗j (x) j=k+1

≤  f i∗ 

j=k+1 ∞ 

j=k+1

e j − f j  e∗j  → 0 as k → ∞.

194

4 Schauder Bases

Since the estimate is independent of x ∈ B X , we get  f i∗ − xk∗  → 0 as k → ∞ and thus f i∗ ∈ span{ei∗ }. Therefore span{ f i∗ } ⊂ span{ei∗ }. ∗ ∗ ∗ −1 ∗ ∗ −1 ∗ Note that∗(T ) (ei ) = fi , so  fi  ≤ (T )  ei  for i ∈ N. Thus the series  f i  ei − f i  converges and we can reverse the roles of ei and f i in the last paragraph, obtaining span{ei∗ } ⊂ span{ fi∗ }. The following corollary will be used in Chapter 5. Corollary 4.24 Assume that a finite-dimensional subspace E of a Banach space X is complemented in X by a projection P. If {ei : i = 1, 2, . . . , n} is a basis in E, then there is ε > 0 so that if fi ∈ X , i = 1, 2, . . . , n, satisfy ei − f i  < ε for all i = 1, 2, . . . , n, then span{ fi : i = 1, 2, . . . , n} is complemented in X with a projection of norm less or equal than 2P.

´ 4.4 Block Bases, Bessaga–Pełczynski Selection Principle Definition 4.25 Let {ei } be a basic sequence in a Banach space X . A sequence of p j+1 non-zero vectors {u j } in X of the form u j = i= p j +1 ai ei with scalars ai and p1 < p2 < . . . is called a block basic sequence of {ei }. Note that a block basic sequence of {ei } is a basic sequence with basis constant not greater than bc{ei }. Indeed, for k ≤ l we have p j+1 p j+1 k k k               αju j =  αj ai ei  =  α j ai ei   j=1

i= p j +1

j=1

l 

 ≤ bc{ei }

p j+1



j=1 i= p j +1 l       α j ai ei  = bc{ei } α j u j .

j=1 i= p j +1

j=1

The following result is often used when investigating subspaces of a given space. Theorem 4.26 (Pełczy´nski [Pelc3]) Let X be a Banach space with a Schauder basis {ei }. If Y is an infinite-dimensional closed subspace of X , then Y contains an infinite-dimensional closed subspace Z with a Schauder basis that is equivalent to a block basic sequence of {ei }. Proof: Let K = bc{ei }. Given p ∈ N, let W p be the finite-codimensional subspace of X defined by ∞   Wp = x ∈ X : x = ai ei = span{ei }i> p . i= p+1

Then W p ∩ Y is infinite-dimensional, so there is y ∈ SY ∩ W p . We will inductively construct the two equivalent basic sequences.

4.4

Block Bases, Bessaga–Pełczy´nski Selection Principle

195

∞ 1 Choose an arbitrary y1 = i=1 ai ei ∈ Y with y1  = 1. Find p1 ∈ N such that  p1 1 ∞ 1 2 . Choose y2 = i= for u 1 = i=1 ai ei ∈ X we have y1 − u 1  < 4K p1 +1 ai ei ∈  p2 SY ∩ W p1 and fix p2 ∈ N such that for u 2 = i= p1 +1 ai2 ei we have y2 − u 2  < 1 . Continue in this manner. Then {u j } is a block basic sequence of {ei }. Since 2 · 22 K  ∞ 1 ∗ j=1 y j − u j  < 2K and u j  ≤ 2 bc{u j } ≤ 2K , by Theorem 4.23, {y j } is a basic sequence in Y equivalent to {u j }, so the subspace Z = span{y j } has the desired property. The above procedure of finding vectors with almost successive supports is called the “sliding hump argument.” An easy modification provides the following. Corollary 4.27 (Bessaga–Pełczy´nski selection principle, see, e.g., [LiTz3]) Let X be a Banach space with a Schauder basis {ei }. If a sequence {xn } satisfies inf x n  > w 0 and x n → 0, then some subsequence {xn k } of {xn } is a basic sequence equivalent to a block basic sequence of {ei }. Theorem 4.28 (Johnson–Rosenthal, [JoRo], see also [LiTz3, p. 10]) Let X be a separable Banach space. Then X has a quotient with a Schauder basis.

Proof: Here [L] will denote the norm-closed linear hull of a set L. Let {yk } ⊂ S X ∗ ,  w∗ of positive numbers less than 1 such that ∞ yk → 0. Let {εn } be a sequence n=1 εn < "∞ ∞ and, consequently, n=1 (1 − εn )−1 < ∞. Using Helly’s theorem, the compactness of the unit ball of a finite-dimensional space and the separability of X , we may choose an increasing sequence k1 < k2 < . . . of positive

∞ integers and finite subsets F1 ⊂ F2 ⊂ . . . of S X such that the linear Fi is dense in X and that for each n ∈ N, span of i=1 n (i) for each f ∈ [{yki }i=1 ]∗ with  f  = 1, there is x ∈ Fn such that n |y(x) − f (y)| ≤ (εn /3)y, for all y ∈ [(yki )i=1 ],

(ii) |ykn+1 (x)| < εn /3, for all x ∈ Fn . We will show that (yki ) is a basic let numbers nsequence. For it, fix n ∈ N and n a2 , . . . , an be given such that  i=1 ai yki  = 1. Pick f ∈ [{yki }i=1 ]∗ such that a1 , n f ( i=1 ai yki ) = 1 =  f  and choose x ∈ Fn that satisfies (i) for this f . It follows that  n      εn   ai yki (x) ≥ 1 − .    3 i=1

Then, for any number λ,

196

4 Schauder Bases

 n   n          ai yki + λykn+1  ≥  ai yki (x) + λykn+1 (x)      i=1 i=1  εn εn εn εn ≥ 1 − 3 − λ 3 ≥ 1 − 3 − 2 3 = 1 − εn , if λ ≤ 2, ≥ 1, otherwise. Thus  n    1   ai yki  ≤    1 − εn i=1

n+1      ai yki     i=1

holds for any numbers a1 , a2 , . . . , an+1 . Then, by induction, we get that for any k and for any numbers a1 , . . . , an+k , ⎞  ⎛n+k−1   n n+k    1 ⎠  ⎝ 1    ai yki  ≤ ai yki       1 − εj  i=1

j=1

i=1

∞ denote the functionals in [(y )∞ ] and (yki ) is thus a basic sequence. Let ( fi )i=1 ki i=1 ∞ biorthogonal to (yki )i=1 . ∞ ] → [( f )∞ ] by Define the projections [( f i )i=1 i i=1

Pm f =

m 

∞ f (yki ) f i , for f ∈ [( f i )i=1 ].

i=1

" −1 We have Pm  ≤ ∞ n=m (1 − εn ) , and thus Pm  → 1. ∞ ∞ ]. We will show that [( f )∞ ] Therefore ( f i )i=1 is a Schauder basis for [( f i )i=1 i i=1 is a quotient of X . This will finish the proof. ∞ ] is a quotient of X , we need to find a bounded In order to show that [( f i )i=1 ∞ ∞ ]∗ be defined for operator from X onto [( f i )i=1 ]. For it, let T : X → [(yki )i=1 x ∈ X by ∞ ]. T x(y) = y(x), for y ∈ [(yki )i=1 ∞ ]. We will first show that T (X ) ⊂ [( f )∞ ]. We need to show that T (X ) = [( f i )i=1 i i=1 ∞ |yki (x)| < ∞ and thus For this, let x ∈ Fn for some n ∈ N. Then, by (ii), i=1

T x(y) =

∞ 

∞ yki (x) f i ∈ [( fi )i=1 ].

i=1

∞ Fi is dense in X and T is a bounded operator, we get Since the linear hull of i=1 ∞ ]. We will now show the following. T (X ) ⊂ [( f i )i=1

4.4

Block Bases, Bessaga–Pełczy´nski Selection Principle

197

Claim: ∞ with g = 1 and for every ε > 0, there is For every g in the linear hull of ( f i )i=1 x ∈ X with x = 1 such that T x − g < 4ε. Having this claim proved, the open mapping principle gives that T (X ) = ∞ ] and finishes the proof. [( f i )i=1  To show the Claim, let 0 < ε < 1 and choose N such that ∞ ε < ε and  j=n j Pn  ≤ 1 + ε for all n > N . Fix n > N . Define  f 1 =  f [(y )n ]  for f ∈ k j j=1

[( f j )nj=1 ]. Observe that  f 1 ≤  f  ≤ Pn . f 1 ≤ 2 f 1 for all f ∈ [( f j )∞ j=1 ]. n Now fix g ∈ [( f j ) j=1 ] with g = 1 and put f = g/g1 . Choose x ∈ Fn satisfying (i) for f . We have, by (i),       n  ≤ εn .  y (x) f − f kj j   3   j=1 1

 Hence  nj=1 yki (x) f j − f  ≤ 2εn /3 < (2/3)ε. Moreover,  f j  = P j − P j−1  ≤ 4 for all j > n and hence, by (ii),     ∞ ∞     εj 4   y (x) f ≤ 4 < ε. kj j  3 3   j=n+1 j=n Thus T x − f  ≤

2 4 ε + ε = 2ε. 3 3

Moreover, since as above, 1 = g ≤ Pn .g1 ≤ (1 + ε)g1 . Thus g1 ≥ (1 + ε)−1 and      g  1  = g  f − g =  − g − 1 ≤ ε.  g  g1 1 Therefore T x − g ≤ T x − f  +  f − g ≤ 2ε + ε = 3ε. This completes the proof of the theorem. Corollary 4.29 (Johnson–Rosenthal, [JoRo], see also [LiTz3, p. 14]) Let X be a Banach space whose dual is separable. Assume that Y is an infinite-dimensional subspace of X ∗ with separable dual Y ∗ . Then Y has an infinite-dimensional reflexive subspace.

198

4 Schauder Bases

Proof: By Theorem 4.19, Y contains a shrinking basic sequence {yk }∞ k=1 ⊂ SY . w

w

w∗

Then yk → 0 in Y and thus yk → 0 in X ∗ ; hence yk → 0 in X ∗ . By the proof ∞ of Theorem 4.28, there is a subsequence {ykn }∞ n=1 of {yk }k=1 which is a boundedly ∞ complete basic sequence. Since {yk }k=1 is shrinking, {ykn }∞ n=1 is shrinking as well. Therefore, the norm-closed linear hull of {ykn }∞ is reflexive by Theorem 4.16. n=1 Corollary 4.30 (Johnson–Rosenthal, [JoRo], see also [LiTz3, p. 14]) Assume that X is an infinite-dimensional Banach space such that X ∗∗ is separable. Then every infinite-dimensional subspace of X or of X ∗ contains an infinite-dimensional reflexive subspace. Proof: The second statement follows directly from Corollary 4.29. To prove the first statement, assume that Y is an infinite-dimensional subspace of X . Then Y is also an infinite-dimensional subspace of the separable space X ∗∗ . Moreover Y ∗ is separable as X ∗ is separable. Therefore, by Corollary 4.29, Y contains an infinite-dimensional reflexive subspace. Definition 4.31 Let X be a Banach space. A sequence {X n }∞ n=1 of finitedimensional subspaces of X is called a finite-dimensional decomposition of X ( FDD, in short) if every x ∈ X has a unique representation of the form x = ∞ n=1 x n with xn ∈ X n for every n ∈ N. If dim X n = 1 for each n ∈ N, i.e., if X n = span{xn } for some xn ∈ X n , then ∞ {X n }∞ n=1 is an FDD for X if and only if {x n }n=1 is a Schauder basis of X . ∞ If{X n }n=1 isan FDD for a Banach space X , define projections Pn on X by ∞ n xi ) = i=1 xi . Pn ( i=1 The following result is the finite-dimensional-decomposition version of Lemma 4.7. Proposition 4.32 If {X n }∞ n=1 is an FDD for a Banach space X , then supn Pn  < ∞. Conversely, If {Pn } is a sequence of finite rank projections on X such that Pn Pm = Pmin(m,n) and lim Pn x = x for every x ∈ X , then {X n }∞ n=1 determines n

a unique FDD on X by putting X 1 = P1 X and X n = (Pn − Pn−1 )X for n > 1. Definition 4.33 An FDD {X n }∞ n=1 of a Banach space X is called shrinking if the associated projections satisfy lim Pn∗ x ∗ − x ∗  = 0 for every x ∗ ∈ X ∗ . n

Theorem 4.34 (Johnson–Rosenthal, [JoRo], see also [LiTz3, p. 480]) If X is a separable Banach space, then there is a subspace Y of X such that both Y and X/Y have FDD. If X ∗ is separable, Y may be chosen so that both Y and X/Y have a shrinking FDD. Proof: Let {x n ; x n∗ } be a 1-norming Markushevich basis for X , i.e., a biorthogonal system with span{xn : n ∈ N} = X and span{xn∗ : n ∈ N} a 1-norming subspace of X ∗ (see Definition 4.58). Choose finite sets σ1 ⊂ σ2 ⊂ . . . and Δ1 ⊂ Δ2 ⊂ . . . so

4.4

Block Bases, Bessaga–Pełczy´nski Selection Principle

199

that σ := σn and Δ := Δn are complementary infinite subsets of N and that, for n ∈ N, (i) If x ∗ ∈ span{xi∗ : i ∈ Δn } there is x ∈ span{xi : i ∈ Δn ∪ σn+1 } so that x = 1 and |x ∗ (x)| > (1 − 1/(n + 1))x ∗ ; (ii) If x ∈ span{xi : i ∈ σn } there is x ∗ ∈ span{xi∗ : i ∈ σn ∪ Δn } so that ∗ x  = 1 and |x ∗ (x)| > (1 − 1/(n + 1))x.  For n∈ N define Sn : X → X and Tn : X → X by Sn x = i∈σn xi∗ (x)xi and Tn x = i=Δn xi∗ (x)xi . Then, for n ∈ N,  Tn∗ {x : i∈σ i

⊥ n+1 }

≤1+

1 n

and  1 Sn∗ {x ∗ : i∈Δ }  ≤ 1 + . n i n Indeed, to see the first inequality, suppose y ∈ {xi : i ∈ σn+1 }⊥ . Pick x ∈ span{xi : i ∈ Δn ∪ σn+1 } so that x = 1 and Tn∗ y(x) ≥ (1 − 1/(n + 1))Tn∗ y. Since y ∈ {xi : i ∈ σn+1 }⊥ , y(x) = Tn∗ y(x) and hence  |y(x)| ≥ 1 −

 1 Tn∗ y, n+1

which gives  1 Tn∗ y, y ≥ 1 − n+1 

so   1 y. Tn∗ y ≤ 1 − n The other inequality follows similarly. Put Y = span{xi : i ∈ σ }. We will show ∗ ∗ that Y ⊥ = spanw {xi∗ : i ∈ Δ}. The only nontrivial inclusion is Y ⊥ ⊂ spanw {xi∗ : w∗

i ∈ Δ}. For it , let y ∈ {xi : i ∈ σ }⊥ . It is enough to show that Tn∗ y −→ y. Since w∗

{Tn∗ y : n ∈ N} is bounded, assume that Tn∗ y −→ x ∗ for some x ∗ ∈ X ∗ . Then Tn∗ x ∗ = Tn∗ y for every n and hence x ∗ − y ∈ {xi : i ∈ Δn }⊥ for each n. Thus x ∗ − y ∈ {xi : i ∈ Δ}⊥ . However, obviously, x ∗ − y ∈ {xi : i ∈ σ }⊥ . Hence, w∗

∗

y = x ∗ and Tn∗ y −→ y. Hence Y ⊂ spanw {xi∗ : i ∈ Δ}. Now, ∗

span{xi : i ∈ σ } = {xi : i ∈ σ }⊥# = (spanw {xi∗ : i ∈ Δ})# = {xi∗ : i ∈ Δ}# . We have

200

4 Schauder Bases ∗

(X/Y )∗ = Y ⊥ = spanw {xi∗ : i ∈ Δ}  and (Tn∗ ) produce by duality an FDD for X/Y . The sequence of projections (Sn Y ) then in turn produce an FDD for Y . If X ∗ is separable and the dual norm is locally uniformly rotund (see Definition 7.9 and Theorem 8.7), which is shared by subspaces and quotients of X , we get that there is a shrinking FDD both in Y and X/Y (see the situation with Schauder bases).

4.5 Unconditional Bases  Recall that a series xi in a Banach space X is unconditionally convergent if  εi xi converges for all choices of signs εi = ±1 (see Exercise 1.39). Definition 4.35 A Schauder basis {ei } of a  Banach space X is said to be unconditional if for every x ∈ X , its expansion x = ai ei converges unconditionally. A sequence {ei } in a Banach space X is called an unconditional basic sequence if it is an unconditional basis of span{ei }. It is straightforward to check that the canonical basis {ei } of c0 or  p , p ∈ [1, ∞), is an unconditional basis. Also, as we saw in the notes following Theorem 1.57, every orthonormal basis of a separable Hilbert space is unconditional. The Haar basis is an unconditional basis for L p [0, 1] for p > 1 (Paley and Marcinkiewicz, see, e.g., [AlKa, p. 130]), but not for L 1 [0, 1], as the latter space does not admit any unconditional basis, see Corollary 4.39. The trigonometric system is not an unconditional basis for L p [0, 1] if p = 2 (see, e.g., [LiTz2, p. 143]). Note that every basis equivalent to an unconditional basis is also unconditional. unit vector basis of  c0 . For n ∈ N set x n = Example: Let {ei } be the standard ∞ n ∞ e . We check that if x = α e , then x = , where βn = i i i i=1 i=1 n=1 βn x n ∞ ∞ αn − αn+1 . It follows that {β } forms a convergent series with n n=1 βn = α1   n=1   ∞  and x∞ = supk  n=k βn . On the other hand, for every convergent series βk we have x=

∞  k=1

βk xk =

∞  ∞ 

βk ei ∈ c0 .

i=1 k=i

We just showed that (c0 ,  · ∞ ) is isometric  to the space of convergent series  ∞  {βn } with the norm (βn ) = supk  n=k βk . The uniqueness of expansion x =  ∞ βn xn follows by induction, using the fact that i=i βk = αi , where αi is the standard ith coordinate of x (βn is then necessarily αn − αn+1 ). It follows that {xn }  (−1)n is a Schauder basis of c0 which is not unconditional. Indeed, n x n ∈ c0 while 1 x ∈ / c . The basis {x } is called the summing basis of c . 0 n 0 n n

4.5

Unconditional Bases

201

Proposition 4.36 Let {ei } be a sequence in a Banach space X . The following are equivalent: (i) {ei } is an unconditional basic sequence. (ii) There is a constant K such that for all scalars a1 , . . . , am and signs εi = ±1 we have m m         ε a e ai ei . ≤ K   i i i i=1

i=1

(iii) There is a constant L such that for all scalars a1 , . . . , am and every subset σ of {1, . . . , m} we have m         ai ei  ≤ L  ai ei .  i∈σ

i=1

    We claim that the condition (iii) can be equivalently stated as  i∈σ ai ei  ≤    ∞  L  i=1 ai ei  whenever σ ⊂ N.  Indeed, let σ be any (even infinite) subset of N. Assuming (iii) we show that i∈σ ai ei is   Cauchy: Given ε > 0, there is n 0 such that for m > n > n 0 we have  m   i=n ai ei  < Lε . Considering σ = σ ∩ {n, . . . , m} and bi = ai for n ≤ i ≤ m,     bi = 0 otherwise, we use the condition (iii) to see that  i∈σ,n≤i≤m ai ei  < ε.  Thus i∈σ ai ei is convergent. Passing to the limit we get the claim. However, in most applications the finite-sum statement is easier to handle as one does not need to discuss the convergence. Similar observation can be made about the condition (ii). Proof: (i)⇒(iii): Let Y = span{ei }. Given  σ ⊂ N, define an operator Pσ from Y into Y by P (x) = a e for x = ai ei . The operator Pσ is well defined as σ i∈σ i i   a e converges whenever a e converges (use (i) and Exercise 1.41). i i i∈σ i i  We now check that Pσ has a closed graph. Indeed, let x k → x in Y for x k = i aik ei ,    bi ei . From the continuity of the x = ai ei , and Pσ (x k ) = i∈σ aik ei → y = biorthogonal functionals in Schauder bases, we have aik → ai for every i and for the same reason, aik → bi for every i. Thus bi = ai for every i and hence Pσ (x) = y, meaning that Pσ has a closed graph and is thus continuous. Consider now the family of operators Pσ , σ running through all subsets of N. We  claim that for every fixed x = ai ei ∈ X , the family {Pσ (x)} is bounded. Indeed, fromthe unconditionality we get that given ε > 0, there is a finite set F ⊂ N such    that  i∈A ai ei  < ε whenever A ∩ F = ∅ (see exercises in Chapter 1). From this the boundedness of {Pσ (x)} for every x ∈ X follows. Now the Banach–Steinhaus uniform boundedness principle gives that the operators Pσ are uniformly bounded by some L.

202

4 Schauder Bases

(iii)⇒(ii): Given scalars a1 , . . . , am and signs εi = ±1, we define σ = {i : εi = 1} and σ = {1, . . . , m}\σ . Then m          εi ai ei  =  ai ei − ai ei   i∈σ

i=1

i∈σ

m             ≤ ai ei  +  ai ei  ≤ 2L  ai ei . i∈σ

i∈σ

i=1

(ii)⇒(iii): Given a1 , . . . , am and σ ⊂ {1, . . . , m}, we define εi = 1 if i ∈ σ and εi = −1 for i ∈ {1, . . . , m}\σ . Then m    1      ai ei  =  (εi ai ei + ai ei )  2 i∈σ

i=1

m  1   εi ai ei  + ≤  2 i=1

m m    1     ai ei  ≤ K  ai ei .  2 i=1

i=1

(ii) and (iii)⇒(i): Given n < m and scalars a1 , . . . , am , we use (iii) with σ = {1, . . . , n} see that by Proposition 4.18, {ei } is a basic sequence with bc{ei } ≤ L. to ∞ convergent series. Given εi = ±1, using (ii) we show that Now let i=1ai ei be a     m  m   εi ai ei is Cauchy, hence convergent.  i=n εi ai ei  ≤ 2K  i=n ai ei . Thus  This shows that ai ei converges unconditionally. The best possible constant K from the condition (ii) in Proposition 4.36 is called the unconditional basis constant of {ei } and is denoted by ubc{ei }. In the the proof above, we have shown that L ≤ K and bc{ei } ≤ ubc{ei }. A natural question is whether every Banach space contains an unconditional basic sequence (see Theorem 4.19). This long-standing problem was answered in the negative by Gowers and Maurey ([GoMa]). Theorem 4.37 (James, see, e.g., [LiTz3]) Let X be a separable Banach space. If X has an unconditional Schauder basis that is not boundedly complete, then X contains an isomorphic copy of c0 . In the proof, we will use the following statement. basic sequence in a Banach space X . Lemma 4.38 Let {ei } be an unconditional  Then for all scalars (ai ) such that ai ei converges and all bounded sequences of scalars {λi } we have ∞ ∞         λi ai ei  ≤ ubc{ei }(sup |λi |) ai ei .  i=1

i

i=1

4.5

Unconditional Bases

203

     m m Proof: Given m ∈ N, pick x ∗ ∈ S X ∗ so that x ∗ i=1 λi ai ei =  i=1 λi ai ei  and define εi by εi = 1 if ai x ∗ (ei ) ≥ 0 and εi = −1 if ai x ∗ (ei ) < 0. Then m m m        ∗ λ a e |λ | |a x (e )| ≤ sup |λ | εi ai x ∗ (ei ) ≤  i i i i i i i i=1

1≤i≤m

i=1

i=1

m m             εi ai ei  ≤ sup |λi | ubc{ei } ·  ai ei . ≤ sup |λi | x ∗  1≤i≤m

1≤i≤m

i=1

i=1

Proof of Theorem 4.37: Let {ei } be an unconditional basis of X that is not boundedly  n  ai ei  ≤ 1 for every n and complete. Then there are scalars (ai ) such that  i=1 ∞ i=1 ai ei does not converge. By the Cauchy criterion, qithere are ε > 0 and natural numbers p1 < q1 < p2 < q2 · · · such that for u j = j= p j a j e j we have u j  ≥ ε      m  ∞   u i  ≤ K  i=1 ai ei  ≤ K , where K = ubc{ei }. for every j, yet  i=1 By the previous lemma, for every sequence {λ j }mj=1 of scalars we have m m            λ u |λ | u j  ≤ K 2 sup |λ j | = K 2 (λ j )∞ . ≤ K sup  j j j  j=1

j

j=1

j

On the other hand, from the unconditionality of {ei } we have for each ε   i ∈ {1, . . . , m}:  mj=1 λ j u j  ≥ K1 λi u i  ≥ Kε |λi |, that is, (λ j )∞ ≤ K    m   j=1 λ j u j . Thus {u j } is equivalent to the canonical basis of c0 . Corollary 4.39 The space L 1 [0, 1] does not have an unconditional Schauder basis.

Proof: The space L 1 [0, 1] is weakly sequentially complete (a classical Steinhaus theorem, see Definition 5.35 and Exercises 13.48 and 13.49 for a proof). Let {ei } n be the standard unit vector basis of c0 . Then the sequence x n = i=1 ei is weakly Cauchy but not weakly convergent, therefore c0 is not isomorphic to a subspace of L 1 [0, 1]. If L 1 [0, 1] had an unconditional basis {x i }, {xi } would be boundedly complete by Theorem 4.37 and thus L 1 [0, 1] would be isomorphic to a dual space by Theorem 4.15. But this is not possible: In a separable dual space every closed convex bounded set is the closed convex hull of its extreme points (see Theorem 3.122), while the unit ball of L 1 [0, 1] has no extreme points (Exercise 3.136). For another approach, see Exercise 8.23. Theorem 4.40 The space C[0, 1] does not have an unconditional basis.

204

4 Schauder Bases

Proof: (Kadec) Assume C[0, 1] has an unconditional basis. Let α = sup PA , where PA are the canonical projections onto {ei }i∈A for finite A ⊂ N. Pick A such that PA  > α − 12 . Then by the Daugavet result in Exercise 9.27, I − PA  = I  + PA  = 1 + PA  > 1 + α −

1 2

= α + 12 .

On the other hand, I − PA  = PN\A0  ≤ α. Hence α ≥ α + 12 , a contradiction. Theorem 4.41 (James, see, e.g., [LiTz3]) Let X be a separable Banach space. If X has an unconditional Schauder basis that is not shrinking, then X contains an isomorphic copy of 1 . In particular, a separable Banach space with an unconditional Schauder basis contains an isomorphic copy of 1 if X ∗ is nonseparable. Proof: Let {ei } be the assumedbasis, let K = bc{ei }. Since {ei } is not shrinking, there is f ∈ S X ∗ such that sup f (z) : z ∈ B X ∩ span{ei }i≥n → 0. This means that there is ε > 0 and a sequence n k < n k+1 such that sup{ f (z) : z ∈ B X ∩ span{en k , en k +1 , . . . }} ≥ ε for every k. Using the sliding hump technique, we can construct a normalized block ε/2 for every j. Let m ∈ N and basic sequence u j of {ei } suchthat f (u j ) ≥  a1 , . . . , am be scalars. Assume i≤m,ai ≥0 ai ≥ i≤m,ai <0 −ai . Then m   4     ai u i  ≥  4 K

i≤m,ai ≥0

i=1

≥2 

ε K



ai ≥

i≤m,ai ≥0

ai u i

i≤m,ai ≥0

 i≤m,ai ≥0

ai +

ε K



−ai ≥

i≤m,ai <0

m ε  |ai |. K i=1



we consider −ai to obtain the same inequal    m  m ai u i  ≤ i=1 |ai |. ity. On the other hand, since {u j } is normalized, we have  i=1     m ε Therefore 4K (ai )1 ≤  i=1 ai u i  ≤ (ai )1 , that is, {u j } is equivalent to the canonical basis of 1 . To see the second statement, note that if {ei ; ei∗ } is shrinking, then X ∗ = span{ei } and thus X ∗ is separable. If

i≤m,ai ≥0 ai

<

ε K

   4  ai u i  ≥ · f K

i≤m,ai <0 −ai ,

Corollary 4.42 (James, see, e.g., [LiTz3]) Let X be a Banach space with an unconditional basis. Then X is reflexive if and only if X contains no isomorphic copy of 1 or c0 . Proof: It follows directly from Theorems 4.16, 4.41, and 4.37. The following definition is due to James ([Jame1] and [Jame1b]).

4.6

Bases in Classical Spaces

205

Definition 4.43 The James space J consists of all sequences (αi ) of real numbers such that lim αi = 0 and (ai ) < ∞, where the norm (ai ) is defined by (ai ) =

 sup

n 1 <···
(αn 1 − αn 2 )2 + (αn 2 − αn 3 )2 + · · · + (αn k−1 − αn k )2

1 2

.

It is standard to verify that (J,  · ) is a Banach space. Moreover, the sequence n

{ei } of the standard unit vectors en = (0, . . . , 0, 1, 0, . . . ) is a monotone Schauder basis of J . Claim: The sequence {ei } is a shrinking Schauder basis of J and J ∗∗ span{(1, 1, 1, . . . )}, in particular, J is not reflexive.

=

J ⊕

Proof: Assume that the basis is not shrinking. As in the proof of Proposition 4.41 we find x ∗ ∈ J ∗ , ε > 0, and a normalized block basic sequence {u j } of {ei } such that  1 x ∗ (u j ) > ε for every j. Consider the vector u = ∞ j=1 j u j . By Hölder’s inequality (1.1) applied to the norm of J and by considering the division points of u j s we find  1  1  ∗ u j 2 ≤ 2 < ∞. On the other hand, x ∗ (u) = x (u j )/j that u2 ≤ 2 j2 j2 is not finite, a contradiction.     k  ∗∗ By Proposition 4.14, J is identified with (βi ) such that supk  i=1 βi ei  < ∞   via the mapping T (x ∗∗ ) = x ∗∗ (ei∗ ) . From the definition of the norm  ·  of J k  and from the fact that supk  i=1 βi ei  < ∞ it follows that lim βi = β exists. i→∞

Thus (βi − β)i corresponds to an element of J as lim (βi − β) = 0. Therefore J ∗∗ = J ⊕ span{(1, 1, . . . )}.

i→∞

Since J ∗∗ is separable, J does not contain an isomorphic copy of c0 or 1 . Indeed, if J contained an isomorphic copy of c0 , then J ∗ would have a quotient isomorphic to c0∗ and J ∗∗ would have a subspace isomorphic to c0∗∗ = ∞ . Similar argument shows that J does not contain an isomorphic copy of 1 . By Corollary 4.42, J does not have an unconditional basis. However, J contains an isomorphic copy of 2 , in particular it contains an unconditional basic sequence. Bessaga and Pełczy´nski proved (see, e.g., [LiTz3]) that if a Banach space X has an unconditional basis and Y is a nonreflexive closed subspace of X , then Y contains an isomorphic copy of c0 or 1 . Using this result and the fact that J (as every separable Banach space) is isometric to a subspace of C[0, 1] (Theorem 5.8), we see again that C[0, 1] does not admit any unconditional basis.

4.6 Bases in Classical Spaces It was a long-standing problem whether every Banach space contains either a reflexive subspace or an isomorphic copy of c0 or 1 . This problem was recently

206

4 Schauder Bases

answered in the negative by Gowers. In fact, Gowers showed ([Gowe4]) that there is a separable Banach space X such that every infinite-dimensional closed subspace of X has nonseparable dual and yet X contains no isomorphic copy of 1 . Theorem 4.44 (Bessaga, Pełczy´nski, see, e.g., [LiTz3]) Let X be a Banach space. If X ∗ has a subspace isomorphic to c0 , then X has a complemented subspace isomorphic to l 1 . In particular, X ∗ has a subspace isomorphic to ∞ . Proof: Let T be an isomorphism from c0 into X ∗ . Then T ∗ maps X ∗∗ onto 1 (Exercise 2.49). Since B X is w ∗ -dense in B X ∗∗ by Goldstine’s theorem, using Corollary 2.26 we find K > 0 and xn ∈ X for n ∈ N such that xn  ≤ K , T ∗ (xn )(en ) = 1 n−1 and i=1 |T ∗ (xn )(ei )| < n1 , where ei are the standard unit vectors in c0 . From Corollary 4.52 and Proposition 4.45, it follows that {T ∗ (xn )} has a subsequence {T ∗ (xn k )} that is equivalent to the canonical basis of 1 and whose span is complemented in 1 by a projection P. Therefore for some constant M > 0, and every  |ak | < ∞ we have choice of scalars {ak }∞ k=1 satisfying ∞ ∞ ∞ ∞              ak x n k  ≤ K |ak | ≤ K M  ak T ∗ (xn k ) ≤ K MT ∗   ak xn k .  k=1

k=1

k=1

k=1

Hence T ∗ is an isomorphism of Y = span{xn k } onto span{T ∗ (xn k )}. Thus Y is  = (T ∗ )−1 P T ∗ is a projection of X onto Y . Finally, P ∗ (X ∗ ) isomorphic to 1 and P ∗ ∗  ) =  = ∞ (Exercise 4.10). is isomorphic to P(X 1

Given a bounded sequence in a Banach space with a separable dual, we can extract a weakly Cauchy subsequence by a standard Cantor diagonal procedure. Rosenthal characterized spaces share this property as spaces not containing an isomorphic copy of 1 . Precisely, he proved that if {x n } is a bounded sequence in a Banach space X , then either {xn } has a weak Cauchy subsequence or contains a subsequence that is equivalent to the standard unit vector basis of 1 (see Theorem 5.37). There are separable spaces whose dual is nonseparable and yet they do not contain an isomorphic copy of 1 (e.g., the space J T discussed in exercises). We will now investigate the structure of subspaces of  p spaces. Proposition 4.45 Let X be c0 or  p with p ∈ [1, ∞). If {u j } is a normalized block basic sequence of the standard unit vector basis {ei }, then {u i } is equivalent to {ei }, span{u i } is isometric to X , and there is a projection of norm one of X onto span{u i }.

Proof: We present the proof for  p . Let u j = for j ∈ N. Then

 p j+1

i= p j +1 λi ei

with

 p j+1

i= p j +1 |λi |

p

=1

4.6

Bases in Classical Spaces

207

p j+1 p j+1 m m m    1 1    p p   aju j = |a j | p |λi | p = |a j | p |λi | p  j=1 i= p j +1

j=1

=

m 

|a j | p

i= p j +1

j=1

1

p

m 

 =

j=1

  a j e j .

j=1

  So {ei } and {u i } are equivalent and T ( a j u j ) = a j e j is the isometry. p ∗ To find a projection, for every j ∈ N choose u ∗j ∈ span{ei }i=j+1 p j +1 ⊂  p for ∗ ∗ ∗ which u j  = u j (u j ) = 1. Then u j (u k ) = 0 for k = j and the operator P from  ∗ X into span{u j } defined by P(x) = ∞ j=1 u j (x)u j is a linear projection of X onto  p p span{u j }. For x = ai ei ∈ X we have |u ∗j (x)| p ≤ i=j+1 p j +1 |ai | for every j as ∗ u j  = 1 and thus P(x) p =

j+1 ∞ p 

|u ∗j (x)| p · |λi | p =

j=1 i= p j +1 ∞  

∞ 

|u ∗j (x)| p

j=1

p j+1

≤

|ai | p = x p .

j=1 i= p j +1

This shows that P = 1. Combining Theorem 4.26 and Proposition 4.45 we have Theorem 4.46 Let X be c0 or  p with p ∈ [1, ∞). If Y is an infinite-dimensional closed subspace of X , then Y contains a subspace Z which is isomorphic to X and complemented in X . The following result is called the Pełczy´nski decomposition method. Theorem 4.47 (Pełczy´nski, see, e.g., [LiTz3]) Let X and Y be Banach spaces so that X is isomorphic to a complemented subspace of Y and Y is isomorphic to a complemented subspace of X . Assume that either: (i) X is isomorphic to X ⊕ X and Y is isomorphic to Y  ⊕ Y , or (ii)  X is isomorphic to ( X )c0 or X is isomorphic to ( X )∞ or X is isomorphic to ( X ) p for some 1 ≤ p < ∞. Then X is isomorphic to Y . Proof: Put Y ≈ X ⊕ E and X ≈ Y ⊕ F, where ≈ denotes isomorphism. If (i) holds, then we have Y ≈ X ⊕ E ≈ X ⊕ X ⊕ E ≈ X ⊕ Y . Similarly, X ≈ Y ⊕ F ≈ Y ⊕ Y ⊕ F ≈ Y ⊕ X . Thus X ≈ Y . If (ii) holds, we have in particular X ≈ X ⊕X and Y ≈ X ⊕E ⊕X ≈ X ⊕X ⊕E ≈ X ⊕ Y . Furthermore,     Y⊕F Y F ≈ ≈ ⊕ . X p

p

p

p

208

4 Schauder Bases

Hence if X ≈

  X  , we have p

     ≈ Y ⊕ F ≈Y ⊕ Y ⊕ F X p p p p p   Y⊕F X ≈Y⊕ ≈Y⊕ ≈ Y ⊕ X ≈ Y.

X≈

p

p

A similar argument is used for the c0 or for the ∞ sums. Corollary 4.48 (Pełczy´nski) Let X be c0 or  p , p ∈ [1, ∞). If Y is an infinitedimensional complemented subspace of X , then Y is isomorphic to X . Proof: By Theorem 4.46, Y has a subspace Z that is isomorphic to X and complemented in X (hence complemented in Y ). So X is isomorphic to a complemented subspace of Y (and  Y is a complemented subspace of X ). If X := c0 then X is isomorphic to ( X )c0 . Similarly, if X :=  p (1 ≤ p < ∞) then X is isomorphic  to ( X ) p (Exercise 4.39). Using (ii) in Theorem 4.47, we get the conclusion. We note that for p ∈ (1, ∞), p = 2, there are closed subspaces of  p that are isomorphic to  p and not complemented in  p [BDGJN], [Rose4], and [Rose1]. Also, if p ∈ [1, ∞)\{2} then there are infinite-dimensional closed subspaces of  p that are not isomorphic to  p [Pelc3]. It seems to be unknown whether every isomorphic copy of 1 in 1 has to be complemented in 1 , [BDGJN]. When comparing structure of Banach spaces, an important role is played by certain classes of operators. Let X, Y be Banach spaces and T ∈ B(X, Y ). We say that T is strictly singular if there is no infinite-dimensional subspace Z of X such that T  Z is an isomorphism into Y . We recall that T is a compact operator if T (B X ) is compact in Y . Compact operators will be treated extensively in Chapter 15. Note that every compact operator is strictly singular (Exercise 4.45). Proposition 4.49 (Pitt, see, e.g., [LiTz3, p. 76]) Let 1 ≤ p < r < ∞. Every bounded operator T from r into  p or from c0 into  p is compact. Proof: Assume that T : r →  p is not compact. Since r is reflexive, there is a w sequence {x n } in r such that xn → 0 in r and T (xn ) ≥ ε for some ε > 0 (Exercise 15.8). Then inf xn  > 0, so using Corollary 4.27 and Proposition 4.45 first in r and then in  p we find a subsequence {xn k } of {xn } such that {x n k } is equivalent to the canonical basis of r and {T (xn k )} is equivalent to the canonical basis of  p .  If (αk ) ∈ r \ p , then αk xn k is convergent, hence T( Thus





ak x n k ) =



ak T (xnk ) ∈  p .

|αk | p < ∞, which is a contradiction.

4.6

Bases in Classical Spaces

209

The c0 case follows from the first part by using Theorem 15.3. We thus obtain that  p , q are not isomorphic for p = q. Using Theorems 4.46 and 4.49 we deduce that if p, q ∈ [1, ∞) and p = q, then the spaces  p and q do not have isomorphic infinite-dimensional closed subspaces. Such spaces are called totally incomparable. Proposition 4.50 Assume that T is a bounded operator from c0 into a Banach space X . Then the following are equivalent. (i) Tis compact. ∞ ∞ is the canonical (ii) i=1 T ei converges unconditionally in X , where {ei }i=1 basis of c0 .  Proof: Assume that i T ei is unconditionally convergent. Then given ε > 0, there is n 0 ∈ N such that  i∈F ai T ei  ≤ ε whenever F is a finite set in {n 0 , n 0 +1, . . .} and |ai | ≤ 1 for i ∈ F. Let Pn 0 denote the canonical projection onto the span of the first n 0 coordinates in c0 . Then it follows that the compact set T Pn 0 Bc0 provides a 2ε-net for T Bc0 . Therefore T is a compact operator. Assume, conversely, that T is compact. Consider T ∗∗ : c0∗∗ (= ∞ ) → X (⊂ X ∗∗ ). The restriction of T ∗∗ to B∞ is w∗ - · -continuous because  on the  · -compact set T Bc0the w ∗ - and the norm-topologies coincide. Since eπ(n) is w ∗ -convergent  in T e is  · -convergent for any permutation π , which gives that T ei ∞ , ∞ π(n) n=1 is unconditionally convergent in X . Theorem 4.51 (Pełcy´nski, [Pelc3b]) Let T be a bounded noncompact operator from c0 into a Banach space  X . Then there is a subspace Z of c0 isomorphic to c0 such that the restriction T  Z of T to Z is an isomorphism of Z into X .  Proof: By Proposition 4.50, T ei is not unconditionally convergent in X , where {ei } denotes the canonical basis of c0 . By Exercises 1.38 and 1.39, there  is ε > 0 and a sequence (Fn )∞ of mutually disjoint finite subsets of N so that  n=1  k∈Fn T ek  ≥ ε for every n. Put xn = k∈Fn T ek for n ∈ N. Since the sequence { k∈Fn ek }∞ n=1 in c0 is a weakly null sequence, and T is a bounded operator, the sequence {xn } is weakly null in X . By passing possibly to a subsequence we can assume, by Corollary 4.27, that the sequence {xn } is basic in X , with a basis constant K , say. Then, for ξ = (ξ(n))∞ n=1 ∈ c00 , we have ⎛ ⎞ ∞    ∞           ⎝ ⎠ ξ(n)xn  = T ξ(n) ek    ≤ T . max{|ξ(n)| : n ∈ N}.     k∈Fn n=1 n=1 On the other hand,

210

4 Schauder Bases

 ∞     max{|ξ(n)| : n ∈ N} ≤ 2K  ξ(n)xn  ,   n=1

as {xn } is basic with basis constant K , and Pn+1 − Pn  ≤ 2K , where Pn denotes the nth canonical projection for the basis {x n } of span{xn : n ∈ N}. Thus {x n∞} is equivalent to the canonical basis of c0 and therefore also to the basis { k∈Fn ek }n=1 .  Thus to finish the proof, we put Z = span{ k∈Fn ek : n ∈ N}. Corollary 4.52 (Bessaga, Pełczy´nski [BePe0]) Let X be a Banach space. Then the following are equivalent. (i) X does not contain an isomorphic copy of c0 . n (ii) If {x n } is a sequence in X such that the set S := { i=1 εi xi : εi = ±1, n ∈ N} is bounded, then S is  · -relatively compact. n (iii) If {xn } is a sequence ∞ in X such that the set S := { i=1 εi xi : εi = ±1, n ∈ N} is bounded, then i=1 xi is unconditionally convergent. Proof: (i)⇒(iii): By Exercise 1.37, the operator T : c00 → X defined by T ei = xi , where ei is the ith unit vector, is bounded and thus has a bounded extension on c0 . If  xi is not unconditionally convergent, then T is not compact (see Exercise 1.42), and thus, by Theorem 4.51, X contains an isomorphic copy of c0 . (iii)⇒(ii): Use Exercise 1.42. (ii)⇒(i): Assume that X contains an isomorphic copy of c0 , and let T be an ∞ denotes the canonical basis of c , the set isomorphism of c0 into X . If {ei }i=1 0  S :=

n 

 εi ei : εi = ±1, n ∈ N

(⊂ c0 )

i=1

n ei , is bounded and not  · -relatively compact (it contains for instance x n := i=1 where dist(x n , x m ) = 1 if n = m). So the set T S is not  · -relatively compact but bounded in X . Note that the canonical basis of c0 satisfies the condition (iii) in Corollary 4.52 and is not unconditionally convergent. We will now show that the structure of subspaces of L p is different from that of p. Theorem 4.53 (Khintchine, see, e.g., [LiTz3]) If p ∈ (1, ∞) then L p [0, 1] contains a complemented subspace isomorphic to 2 . L 1 [0, 1] contains a subspace isomorphic to 2 . Note that L 1 [0, 1] has no complemented subspace isomorphic to 2 (Exercise 13.43). In the proof 4.53, we will use Rademacher’s functions defined by  of Theorem  rn (t) = sign sin(2n π t) for t ∈ [0, 1] and n ∈ N. Clearly rn ∈ SL p [0,1] for every p ≥ 1 (see Fig. 4.4).

4.6

Bases in Classical Spaces

211 1

Fig. 4.4 The first Rademacher functions

0

0

1

1

0

1

−1

1

rl (t)r k (t) dt = δ kl . Consequently, {rn } is an orthonor  m  mal set in L 2 [0, 1]. In particular,  n=1 an rn  = (an )2 , that is, {rn } considIt is easy to observe that

0

L2

ered in L 2 is a basic sequence that is equivalent (with constant 1) to the canonical basis of 2 . The behavior of {rn } in L p spaces is described in the following result. Lemma 4.54 (Khintchine inequality) Let rn be the Rademacher functions on [0, 1]. For every p ∈ [1, ∞) there exist positive constants A p and B p such that for every a 1 , . . . , am , Ap

m 

|an |2

1 2

≤

m m p 1  1  1  p 2   an rn (t) dt ≤ Bp |an |2 .  0

n=1

n=1

(4.2)

n=1

By A p and B p , we denote the best possible constants in this inequality. They are called Khintchine’s constants and their values are known. We observed that A2 = B2 = 1. From the Hölder inequality (1.1) it follows that if p > r , then 1 

1 

1 1 p dt p ≥ r dt r . Consequently A ≤ A and B ≤ B . | f | | f | r p r p 0 0 Proof: By the last remark, it is enough to show that there exist A1 > 0 and B2k < ∞ for all k ∈ N. We start with B2k .  1   1  m m 2k 2k   an rn (t) dt = an rn (t) dt  0

0

n=1

=



n=1 α

Aα1 ,...,α j anα11 · · · an jj



1 0

α

rnα11 (t) · · · rn jj (t) dt,

here the summation runs through all multi-indices (α1 , . . . , α j ) with

j

= 2k i=1 α i ( α j )! Aα1 ,...,α j = (α1 )!···(α j )! .

and 1 ≤ n 1 ≤ · · · ≤ n j ≤ m. By elementary combinatorics,

1 " Observe that 0 rnαki (t) dt = 1 if all αi are even and it is equal to 0 otherwise. n 1 <···
Using βi = αi /2 we therefore write  1  n 2k  2β   an rn (t) dt = A2β1 ,...,2β j an2β1 1 · · · an j j ,  0

n=1

212

4 Schauder Bases

where the summation runs through subsets (β1 , . . . , β j ) of N such that and 1 ≤ n 1 ≤ · · · ≤ n j ≤ m. Thus we have m 

|an |2

k

=



2β

Aβ1 ,...,β j an2β1 1 · · · an j j =

 Aβ1 ,...,β j A2β1 ,...,2β j

n=1

j

i=1 βi

=k

2β

A2β1 ,...,2β j an2β1 1 · · · an j j

m 2k  Aβ ,...,β   Aβ ,...,β  1  2β   1 j 1 j A2β1 ,...,2β j an2β1 · · · an j j = min ≥ min an rn (t) dt.  A2β1 ,...,2β j A2β1 ,...,2β j 0 n=1

j Since i=1 βi = k, the minimum above is a minimum of a finite set of positive numbers and so it is positive. Hence B2k exists finite. To prove the existence of A1 , put f (t) = m n=1 an rn (t). By Hölder’s inequality (1.1) used for p = 32 and q = 3, by the first part of the proof and our observation before this lemma, we have 

1

0

≤



1

| f (t)| dt = 2

2 3

| f (t)| | f (t)| dt ≤

0



1

| f (t)| dt

0

2 3

4 3

4 3

B4

m 

|an |2

2 3

=



1 0



1 0

| f (t)| dt

2 3

− 43

≥ B4

| f (t)| dt ≥ B4−2



1

0 1

| f (t)| dt

| f (t)| dt

 0



1 0

1

| f (t)|2 dt

| f (t)|2 dt

1 2

2 

1

3

2

0

n=1

Therefore



3

| f (t)|4 dt

1 3

0

4 3

B4



1

| f (t)|2 dt

2 3

.

0

1 3

, that is,

= B4−2

m 

|an |2

1 2

.

n=1

Hence A1 ≥ B4−2 .   Proof of Theorem 4.53: Define a mapping T from 2 into L p [0, 1] by T (an ) =  ∞ n=1 an rn , where {rn } are the Rademacher functions. From Lemma 4.54, it follows that T is an isomorphism from 2 into L p [0, 1]. If p ≥ 2, then L p [0, 1] is a subspace of L 2 [0, 1] (Exercise 1.15). Let P be the restriction to L p [0, 1] of the orthogonal projection of L 2 [0, 1] onto span{rn }. We have ∞ 1 P( f )(t) = n=1 ( 0 f (s)r n (s) ds)rn (t). This projection is bounded in L p [0, 1], since P( f ) L p ≤ B p P( f ) L 2 ≤ B p  f  L 2 ≤ B p  f  L p by the Khintchine and Hölder inequalities. If p ∈ (1, 2) and q > 2 satisfies 1p + q1 = 1, there is a subspace Y of L q [0, 1] isomorphic to 2 complemented by a projection P. Then P ∗ (Y ∗ ) is a complemented subspace of L p [0, 1] isomorphic to 2 (Exercise 4.10). One of the consequences is that  p is not isomorphic to L p [0, 1] for p = 2, ∞. Indeed, by duality it is enough to show that  p is not isomorphic to L p [0, 1] for p < 2. Suppose that  p is isomorphic to L p [0, 1] for some p < 2. By Theorem 4.53, 2

4.7

Subspaces of L p Spaces

213

must be isomorphic to a subspace of  p , which is a contradiction with Pitt’s theorem. The situation is different if p = ∞ as ∞ is isomorphic to L ∞ (Exercise 4.42). Also, L 2 [0, 1] is isometric to 2 by Theorem 1.57. The spaces L p and L q are not isomorphic if p = q, p, q ∈ [1, ∞). This follows for instance using the notion of type and cotype. If 1 ≤ p < q ≤ 2 then  p is not isomorphic to a subspace of L q (this can be proved for instance using the notion of type). However, L p contains a subspace isometric to L q (see, e.g., [AlKa, p. 140]).

4.7 Subspaces of L p Spaces Theorem 4.55 (Kadec and Pełczy´nski, [KaPe], see, e.g., [AlKa, p. 148]) Let 2 < p < ∞ and X be an infinite-dimensional subspace of L p . Then either X is isomorphic to 2 (in which case is complemented in L p ) or X contains a subspace that is isomorphic to  p and complemented in L p . Before starting on the proof, we remark that if 1 < p < 2, then L p contains an uncomplemented isomorphic copy of 2 , [BDGJN]. Proof of Theorem 4.55: Assume that X is isomorphic to 2 . Since p > 2, L p ⊂ L 2 . Denote the continuous inclusion mapping from L p into L 2 by T . The restriction of T to X is, by our assumption, an isomorphism of X onto T X ⊂ L 2 ∩ X , which is of course complemented in L 2 by a projection P. Then the mapping T −1 P T is a bounded linear projection of L p onto X . Assume that X is not isomorphic to 2 . Then there is a sequence ( f n ) of vectors in X such that  f n  p = 1 and  f n 2 → 0. Then, obviously, f n → 0 in measure (a sequence { f n } of measurable functions is said to converge in measure to a function f if, for every ε > 0, μ{x : | f n (x) − f (x)| ≥ ε} → 0; for the result, see, e.g., [Rudi2, Ex. 3.18], or [WhZy, Theorem 4.21]). By passing to a subsequence if needed, we can assume that f n → 0 almost everywhere, see, e.g., [Stro, p. 315]. We will prove the following. Claim 1: There is a sequence ( f n k ) of ( f n ) and a sequence of disjoint measurable sets {Ak } such that  f n .χ Ak  p → 1, where χ Ak is the characteristic function of Ak . Assuming Claim 1 proved, by normalizing we can see that it is enough to prove the following Claim 2: Let ( f n ) be a sequence of norm-one disjointly supported functions in L p . Then the closed linear hull of ( f n ) is a 1-complemented subspace of L p that is isomorphic to  p . Proof of Claim 2: By the disjointness of the supports of f i s, for any sequence ∞ ⊂ c we have (ai )i=1 00

214

4 Schauder Bases

∞   ai   i=1

=

p    fi  = 

  ∞

p

p ∞     ai f i (t) dt    i=1

|ai f i (t)| p dt =

i=1

∞ 

 |ai | p

| f i (t)| p dt =

i=1

∞ 

|ai | p .

i=1

Therefore the closed linear hull of { f n : n ∈ N} is even isometric to  p . We will now prove that the closed linear hull of { f n : n ∈ N} is complemented in L p . Let p −1 + q −1 = 1. By the Hahn–Banach theorem, for i ∈ N, there is gi ∈ L q

with gi q = 1 and 1 =  f i  p = f i (t)gi (t)dt. We can assume that the support of gi is the same as that of f i for all i. Define an operator P from L p into the closed linear hull of { f n : n ∈ N} for f ∈ L p by P( f ) =

∞  

 f (t)gi (t)dt

fi .

i=1

It is clear that P maps L p into the closed linear hull of { f n : n ∈ N} and that P is an operator. Moreover, P f i = f i for all i. Thus P is a linear projection. Its norm can be estimated as follows: P f  p =

 ∞   p 1/ p    f (t)gi (t)dt   

 ∞    =  i=1

i=1

supp fi

 p 1/ p  ∞    f (t)gi (t)dt  ≤ i=1

1/ p | f (t)| dt p

 ≤

1/ p | f (t)| p dt

.

supp fi

 Therefore, it remains to prove Claim 1. In doing so, let f n ∈ L p ,  f n  = 1 for all n and f n → 0 almost everywhere. Note that then f n → 0 in measure μ. We use now a variant of a classical “gliding hump” argument: Put f n 1 = f 1 and take F1 := {w : | f n 1 (w)| p > 1/2}. Since f n 1 ∈ L p , there is δ1 > 0 such that E | f n 1 | p < 1/2 whenever μ(E) < δ1 . Then pick n 2 > n 1 such that μ{w : | f n 2 (w)| > (1/22 )} < δ1 and put F2 = {w : | f n 2 (w)| p > 1/22 }. In the same way, there is δ2 > 0 such that E | f n i | p < 1/22 for i = 1, 2 whenever μ(E) < δ2 . Then pick n 3 > n 2 such that μ{w : | f n 3 (w)| p > 1/23 } < δ2 and put F3 = {w : | f n 3 (w)| p > 1/23 }. By induction, we find a subsequence { f n k } of { f n } and a sequence of sets {Fk } such that  f n k − f n k χ Fk  <

1 , for all k. 2k



Put A1 = F1 \ k>1 Fk , A2 = F2 \ k>2 Fk , . . . , A j = F j \ k> j Fk . Then the sets A j , j ∈ N, are mutually disjoint. We have

4.7

Subspaces of L p Spaces



215

 Fk

| fnk | p −

Ak

| fnk | p ≤

 j>k

Fj

| fnk | p ≤

 j>k

1 2 j−1

=

1 2k−1

.

This means p

 f n k χ Fk − f n k χ Ak  p <

1 2k−1

for all k. Hence   f n k − f n k χ Ak  p ≤  f n k − f n k χ Fk  p + f n k χ Fnk − f n k χ Ak  p ≤

1 2k

1/ p   1 1/ p + k−1 . 2

Thus  f n k χ Ak  p ≥  f n k  −  f n k − f n k χ Ak  → 1.

Theorem 4.56 If 1 < p, q < ∞, then  p is isomorphic to a complemented subspace of L q if and only if either p = 2 or p = q. Proof: First of all, 2 is isomorphic to a complemented subspace of L q for all q > 1 (Theorem 4.53). If  p is isomorphic to a subspace of L q for q > 2, then p = 2 or p = q by Theorem 4.55. If  p , p = 2, is isomorphic to a complemented subspace of L q , q < 2, then ∗p (isomorphic to  p∗ ) is isomorphic to a subspace of L q ∗ , where p ∗ is the dual index of p, i.e., p −1 + ( p∗ )−1 = 1. By Theorem 4.55, p∗ = q ∗ , i.e., p = q. Theorem 4.57 L 1 does not contain any complemented subspace isomorphic to  p , p > 1. For any 1 < q ≤ 2, L 1 contains a subspace isomorphic to q . For no q > 2, L 1 contains a subspace isomorphic to q . Of course, L 1 contains a complemented subspace isomorphic to 1 . Proof: The first statement is a consequence of the fact that L 1 has the Dunford– Pettis property (since its dual space does) and thus L 1 does not contain any infinitedimensional reflexive complemented subspace (Proposition 13.44 and the remark after Proposition 13.42). That L 1 contains subspaces isomorphic to q for all 1 < q < 2 is a result of Kadec [Kade2]. For q = 2 use Theorem 4.53. That L 1 cannot contain any subspace isomorphic to q for q > 2 follows from the fact that L 1 has cotype 2 and q , q > 2, not (see [AlKa, p. 140]). That L 1 contains a complemented subspace isomorphic to 1 is shown in [AlKa, p. 121]. It is an open problem whether every infinite-dimensional complemented subspace of L 1 is isomorphic either to L 1 or to 1 , see, e.g., [AlKa, p. 122]. If 1 ≤ p < q ≤ 2, then L p contains an isomorphic copy of L q , [LiPe].

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4.8 Markushevich Bases Definition 4.58 Let X be a Banach space. A biorthogonal system {xα ; f α }α∈Γ in X is called a Markushevich basis of X if span{xα }α∈Γ = X and { f α }α∈Γ separates the points of X . A Markushevich basis {xα ; f α }α∈Γ is called shrinking if span{ f α } = X ∗ . Sometimes, a Markushevich basis {xα ; f α }α∈Γ will be denoted by {xα }α∈Γ if the set of functional coefficients is understood. Clearly, every Schauder basis of a Banach space X is a Markushevich basis of X . An example of a Markushevich basis that is not a Schauder basis is the sequence  1] of of trigonometric polynomials {ei2π nt : n = 0, ±1, ±2, . . . } in the space C[0, complex continuous functions on [0, 1] whose values at 0 and 1 are equal, with the sup-norm. Theorem 4.59 (Markushevich, see, e.g., [LiTz3, Section 1.f] or [HMVZ, p. 8]) Let X be a separable Banach space. If {z i }i ⊂ X satisfies span{zi }i = X and {gi }i ⊂ X ∗ separates points of X , then there is a Markushevich basis {xi ; f i } of X such that span{xi } = span{z i } and span{ fi } = span{gi }. Proof: Define x1 = z 1 and f 1 = gk1 /gk1 (z 1 ), where k1 ∈ N is such that gk1 (z 1 ) = 0. Then find the smallest integer h 2 such that gh 2 ∈ / span{ f 1 }. Define f 2 = gh 2 − gh 2 (x1 ) f 1 . Find an index k2 such that f 2 (z k2 ) = 0, and set x2 = (z k2 − / span{x 1 , x2 }. Put f 1 (z k2 )x1 )/ f 2 (z k2 ). Let h 3 be the smallest integer such that z h 3 ∈ x3 = z h 3 − f 1 (z h 3 )x1 − f 2 (z h 3 )x2 and f 3 = (gk3 − gk3 (x1 ) f 1 − gk3 (x2 ) f 2 )/gk3 (x3 ), where k3 is an index such that gk3 (x3 ) = 0. Continue by induction. At the step 2n we construct f 2n first, at the step 2n + 1 we start by constructing x2n+1 . It follows  2n  n  2n  n that span z i 1 ⊂ span xi 1 and span gi 1 ⊂ span f i 1 . Clearly f i (x j ) = δi j , span{xi } ⊂ span{z i } and span{ f i } ⊂ span{gi }. It is an open problem whether every separable Banach space X admits a Marku∞ with x  =  f  = 1 for all i. It is known that shevich basis {xi ; f i }i=1 i i given a separable space X and ε > 0, a Markushevich basis of X exists so that sup x i   f i  < 1 + ε (Ovsepian–Pełczy´nski, see, e.g., [HMVZ, p. 14]). Theorem 4.60 (Gurarii, Kadec [GuKa], see, e.g., [HMVZ, p. 30]) Let Z be a closed subspace of a separable Banach space X . Any Markushevich basis {xi ; f i } of Z can be extended to a Markushevich basis of X . Precisely, there are z j ∈ X , gj ∈ X ∗ , and extensions of f i to functionals on X such  that {xi } ∪ {z j }; { f i } ∪ {g j } is a Markushevich basis of X . Proof: Extend all f i onto X and denote these extensions by f˜i . Let { yˆ j , φ j } be a Markushevich basis of X/Z (it is separable). For all j choose y j ∈ yˆ j and define ϕ j (x) = φ j (x) ˆ for x ∈ X , note that ϕ j (xi ) = 0 for all i. We have span {xi } ∪  {y j } = X and { f˜i } ∪ {ϕ j } is a family separating points of X . j  Put z j = y j − i=1 λi j xi and ψi = f˜i − ij=1 λi j ϕ j , where λi j = f˜i (y j ) for   i = j and λii = 12 ( f i (yi ) − 1). Then {xi } ∪ {z j }; {ψi } ∪ {ϕ j } is a Markushevich

4.8

Markushevich Bases

217

basis of X that extends {xi , f i }. Indeed, from the definition of z j and ψi it is clear that span{xi , z j } = X , {ψi } ∪ {ϕ j } is separating points of X , and ψi extend fi onto X . It is routine to check that the system is biorthogonal. Theorem 4.61 (Johnson, [Johns1]) The space ∞ does not admit a Markushevich basis. We will need the following. Lemma 4.62 (Rosenthal) Let X be a Banach space. Every reflexive subspace of X ∗ is w∗ -closed in X ∗ . Proof: Let Y be a reflexive subspace of X ∗ . By the Banach–Dieudonné theorem, we only need to show that BY is w∗ -closed in X ∗ . Since BY is w-compact in Y , it is w-compact as a subset of X ∗ . Therefore BY is w ∗ -compact in X ∗ and thus w∗ -closed in X ∗ .   Proof of Theorem 4.61: Assume that x α ; f α α∈Γ is a Markushevich basis of ∞ .   Put Y = span f α α∈Γ . We claim that Y is reflexive. It is enough to prove that BY is weakly sequentially compact. Let {yn } be a sequence in BY . Since every element of span{ f α } has a countable support over {x α } (as a limit in norm of a sequence of points in span{ f α } which have finite support), there is a countable subset N of Γ such that for α ∈ Γ \N and n ∈ N we have yn (xα ) = 0. By the Cantor diagonal argument, let {yn k } be a subsequence of {yn } such that lim yn k (eα ) exists for every α ∈ N . Since yn (eα ) = 0 for every α ∈ Γ \N and {yn } k→∞

is bounded, we have that lim yn k (x) exists and is finite for every x ∈ X . Denote k→∞

w∗

this limit by y(x). Then y is a bounded linear functional on ∞ and yn k → y in ∗∞ . By the Grothendieck property of ∞ (see Exercises 3.42 and 3.44), this means that w yn k → y in X ∗ and y ∈ BY as BY is w-closed in X ∗ . Therefore BY is w-compact in X ∗ and thus also w-compact in Y . Consequently, Y is reflexive. By Lemma 4.62, Y is w ∗ -closed in X ∗ , and since Y is separating for X , it is ∗ w -dense in X ∗ . Thus Y = ∗∞ . Consequently ∗∞ and thus ∞ is reflexive, a contradiction. Theorem 4.63 (Plichko) Let X be a Banach space. If X is separable then X ∗ is a complemented subspace of a Banach space with Markushevich basis. In particular, ∞ is a complemented subspace of a Banach space with Markushevich basis.  and Proof: Let I be a set with card(I ) = dens(X ∗ ). Define U = I c0 1 (I )  ∗  Z = X ⊕ U 1 . Let {en ; f n } be a Markushevich basis of X scaled so that en  ≤ 1 for every n. Pick {y i }i∈I dense in B X ∗ . For i ∈ I and n ∈ N, let

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yni = y i −

n 

y i (e j ) f j .

j=1

  For a fixed n ∈ N define h n = {yki (en )}k i . We have yki (en ) = 0 for k ≥ n and |yki (en )| ≤ 1 for k < n, hence hn ∈

 I

1

∞ (Γ )

= U ∗.

Let u inbe the standard unit vector in U and gni be the standard unit vector in the  space = U ∗ . For n ∈ N and i ∈ I put xni = (yni , u in ) ∈ Z and I 1 ∞ (I )  } en = (en , −h n ) ∈ Z ∗ . We claim that {xni }n∈N,i∈I ∪ {( f m , 0)}n,m∈N ; {em m∈N ∪  i {(0, gn )}n∈N,i∈I is a Markushevich basis of Z . It is easy to see that this is a biorthogonal system and that the functionals separate points of Z (en = en on X ∗ ). It remains to show that the span of vectors {xni , f j }n, j∈N,i∈I is dense in Z . To see this, we note that for n ∈ N and i ∈ I we have u in + y i = u in + yni +

n 

y i (e j ) f j = x ni +

j=1

n 

 ∞ y i (e j ) f j ∈ span x ni , f j j=1 .

j=1

Also, for every i ∈ I we have k  1  u in + y i   − y i  = → 0 as k → ∞.  k k n=1

  Therefore y i ∈ span xni , f j . Furthermore, for every n ∈ N and i ∈ I we have    u in = xni − yni = x ni − y i + nj=1 y i (e j ) f j ∈ span xni , f j .   Thus span x ni , x j n, j∈N,i∈I = Z and the proof is complete.

4.9 Remarks and Open Problems Remarks 1. In [Zipp3], M. Zippin proved that every Banach space with separable dual is isomorphic to a subspace of a Banach space with a shrinking Schauder basis. It follows that every separable reflexive Banach space is isomorphic to a subspace of a separable reflexive space with a Schauder basis. By a dual argument, it follows that every separable reflexive space is a quotient of a reflexive space with a Schauder basis.

4.9

Remarks and Open Problems

219

2. In [DFJP] it is shown that every Banach space with a separable dual is a quotient space of a space with a shrinking basis. 3. The “convexified” Tsirelson space T2 (see, e.g., [Casa, p. 276]) is an example of a space that, in particular, does not contain an isomorphic copy of 2 and every subspace of it has a Schauder basis. 4. If K is a compact metric space, then C(K ) has a Schauder basis (Vakher [Vakh]). This at present time follows since, if K is uncountable, then C(K ) is isomorphic to C[0, 1] by Miljutin’s theorem (see, e.g., [AlKa, p. 94]) and, if K is countable, then K is homeomorphic to an ordinal segment [0, α] for some countable α by Mazurkiewicz–Sierpi´nski theorem (see, e.g., [HMVZ, p. 73]). 5. Pełczy´nski and Singer proved in [PeSi] that if an infinite-dimensional Banach space X has a basis, then it has a basis that is not unconditional, see, e.g., [AlKa, Theorem 9.5.6]. 6. Let us remark that we did not discuss symmetric bases in this text. We refer to [LiTz3] for this topic. 7. Let us remark that there is a Banach space X with a Schauder basis whose dual is separable and X does not have any shrinking Schauder basis (see [LiTz3, p. 10] and Theorem 16.56). 8. Let us remark that if a Banach space X has the property that X ∗ has a Schauder basis, then X has a shrinking Schauder basis, [JRZ], see, e.g., [LiTz3, p. 10]. 9. Zippin showed in [Zipp0] that if a nonreflexive space X has a basis, then it has a non-shrinking basis, see, e.g., [AlKa, p. 59]. 10. There is a separable Banach space with a normalized non-shrinking uncondiw tional basis {en }∞ n=1 such that en → 0, [PeSz]. 11. Pełczy´nski showed in [Pelc3c] that a Banach space X is reflexive if every subspace with a Schauder basis is reflexive. 12. We showed in Theorem 4.40 that C[0, 1] does not admit any unconditional basis. More generally, let us refer to [AlKa, p. 96], where it is shown that if for a compact metric space K , C(K ) is a subspace of a Banach space with unconditional basis, then C(K ) is isomorphic to c0 . Let us also refer to [Pisi3, p. 112] for the proof that the space K(2 ) of compact operators on 2 does not admit any unconditional basis. One way to check whether a Banach space fails to have an unconditional basis is to use the so-called property (u), introduced by Pełczy´nski in [Pelc1b]. A Banach space X is said to have property (u) if whenever {x n } is a weakly there is a weakly uncondi Cauchy sequence in X ,  n u in X so that x − tionally Cauchy series ∞ k n k=1 k=1 u k → 0 in the weak topology. In [Pelc1b], it was proved that every Banach space with unconditional basis has property (u). We refer to [AlKa, Section 3.5] for more on Pełczy´nski’s property (u). 13. Let L p (μ) for 1 ≤ p < ∞ be a separable space and let μ be a purely nonatomic probability measure (i.e., a probability measure such that for every measurable set of positive measure there is a measurable subset with strictly smaller positive measure). Then L p (μ) is isometric to L p (0, 1) (see, e.g., [JoLi3, p. 15]). If the measure μ is purely atomic with atoms {Aγ }γ ∈Γ , then the function which maps for each γ the unit vector eγ to μ(Aγ )−1/ p times the indicator function of Aγ , extends to an isometry from  p (Γ ) onto L p (μ). From these facts, it follows (see

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4 Schauder Bases

[JoLi3, p. 15]) that for 1 ≤ p < ∞, np ,  p , L p (0, 1),  p ⊕ p L p (0, 1), np ⊕ p L p (0, 1), n = 1, 2, ..., is a complete listing, up to isometry, of the separable L p (μ)-spaces when p = 2 and these are all mutually nonisometric. Of course, in the Hilbertian case p = 2, np , n = 1, 2, . . ., 2 , is the appropriate listing. By the Pełczy´nski decomposition method, we thus get that np ,  p , L p (0, 1) is a complete listing, up to isomorphism, of the separable L p -spaces. Moreover, all these spaces are mutually nonisomorphic. In nonseparable cases, the situation is much more complicated. For example, we will show in Exercise 13.52 that 1 (ω1 ) is not a subspace of any L 1 (μ), where μ is a σ -finite measure. 14. We refer to, e.g., [HMVZ] for more on Markushevich bases and their applications in the geometry of Banach spaces.

Open Problems 1. It is an open problem whether every infinite-dimensional complemented subspace of L 1 is isomorphic either to L 1 or to 1 , see, e.g., [AlKa, p. 122]. 2. It is an open problem whether every complemented subspace of a space with unconditional basis has an unconditional basis, see [Casa, p. 279]. 3. It is an open problem whether every separable Banach space contains a Marku∞ with x  =  f  = 1 for all i ∈ N (i.e., an Auerbach shevich basis {xi ; f i }i=1 i i basis). It is known that given a separable space X and ε > 0, a Markushevich basis of X exists so that sup xi   f i  < 1 + ε (Ovsepian–Pełczy´nski, see, e.g., [HMVZ, p. 14]). See also, e.g., [HMVZ, Section 1.2]. 4. It is apparently an open problem whether every Banach space contains a monotone subbasis. 5. It is not known whether every separable Banach space X contains a closed subspace Y such that both Y and X/Y have a Schauder basis, see, e.g., [LiTz3, p. 12]. See Remark 4.11.

Exercises for Chapter 4 4.1 (i) Let M be a subspace  of a vector space V . Show that there is a linear projection of V onto M, i.e., P  M = I M and P(V ) = M. (ii) Show that if a linear mapping P : V → V satisfies P 2 = P, then V = P(V ) ⊕ Ker(P). Moreover, Q = I V − P is a projection such that Q(V ) = Ker(P) and Ker(Q) = P(V ). Hint. (i) Extend an algebraic basis {x α } of M to an algebraic basis {x α , yβ } of V and define P by P(x α ) = xα , P(yβ ) = 0.   (ii) Any x ∈ V can be written as x = P(x) + x − P(x) . 4.2 Let P be a bounded linear projection in a Banach space X . Show that P ∗ is a (bounded) projection in X ∗ .

Exercises for Chapter 4

221

Hint. (P ∗ )2 = (P 2 )∗ = P ∗ . 4.3 Let P, Q be projections in a Banach space X . Show that the following are equivalent: (i) P(X ) ⊂ Q(X ) and P ∗ (X ∗ ) ⊂ Q ∗ (X ∗ ). (ii) P Q = Q P = P. (iii) P(X ) ⊂ Q(X ) and Ker(Q) ⊂ Ker(P). Hint. P(X ) ⊂ Q(X ) if and only if Q P = P. Using dual projections, (i) and (ii) are therefore equivalent. Assume (ii) holds, let Q(x) = 0. Then P(x) = P Q(x) = P(0)  P. Moreover, given x ∈ X , P Q(x) − P(x) =  = 0. If (iii) holds, then Q P = P Q(x) − x = 0 as Q Q(x) − x = 0. Thus (ii) holds. 4.4 Let P be a bounded linear projection of a Banach space X onto P(X ). Show that for every x ∈ X ,     dist x, P(X ) ≤ x − P(x) ≤ (P + 1) dist x, P(X ) .   Hint. Obviously, x − P(x) ≥ dist x, P(X ) . On the other hand, if y ∈ P(X ), write x − P(x) = x − y + y − P(x) = x − y + P(y) − P(x). Thus x − P(x) ≤ x − y + P(x) − P(y) ≤ (1 + P)x − y. 4.5 Let Y be a subspace of a Banach space X . Show that if there is a bounded projection P onto Y then Y is closed. Hint. Take yn ∈ Y such that yn → y. Then P(yn ) → P(y), so also yn → P(y). By the uniqueness of limit, y = P(y) and y ∈ Y . 4.6 Assume that Yi are subspaces of a Banach space X such that X = Y1 ⊕ Y2 (algebraic sum). Let Pi be the associated linear projections onto Yi (so P1 + P2 = I X ). Show that: (i) P1 (X ) = Ker(P2 ). (ii) P1 is bounded if and only if P2 is bounded. (iii) Both Y1 and Y2 are closed if and only if both Pi are bounded. Find an example when Y1 is closed but Y2 is not. The complement of a subspace Y is sometimes defined as a subspace Z such that X = Y ⊕ Z and the corresponding projections are bounded. (iii) shows that it is an equivalent definition, the previous exercise shows that closedness of Y is necessary for the existence of a complement. Hint. (i) x ∈ Ker(P2 ) satisfies x = (P1 + P2 )(x) = P1 (x), similarly the other inclusion. (ii) P2 = I X − P1 . (iii) Proposition 4.2. For the example, Y1 = R, Y2 = Ker( f ) for some discontinuous linear functional f. 4.7 Let X be a Banach space. Show that X ∗ is complemented in X ∗∗∗ .

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4 Schauder Bases

 Hint. Define P : X ∗∗∗ → X ∗ by P( f ) = f  X ∗ . This is called Dixmier’s projection. Note that P = 1. 4.8 Let Y be a complemented subspace of a Banach space X . Let P be a projection of X onto Y . Show that the dual operator P ∗ is a mapping that extends elements of Y ∗ to elements in X ∗ . If P = 1, we get a linear Hahn–Banach extension. Let Y be a closed subspace of a reflexive Banach space X . Assume that there is a bounded operator E : Y ∗ → X ∗ such that that E( f ) is an extension of f on X with the same norm. Show that Y is then complemented in X . Hint. Show that E ∗ is a projection of X onto Y . 4.9 Let X be a Banach space. Show that if Y is a complemented subspace of X (by a projection P), then Ker(P)⊥ is a complemented subspace of X ∗ (by the projection P ∗ ) and Y ⊥ = Ker(P ∗ ) is a complemented subspace of X ∗ . Hint. Clearly P ∗ (X ∗ ) ⊂ Ker(P)⊥ . On the other hand, fix y ∗ ∈ Ker(P)⊥ . For x ∈ X write x = y + z, y ∈ Y and z ∈ Ker(P), then P ∗ (y ∗ )(x) = y ∗ (x), so P ∗ (y ∗ ) = y ∗ . Ker(P ∗ ) = Y ⊥ is straightforward. 4.10 Let X be a Banach space and P be a bounded linear projection of X onto P(X ). Show that P(X )∗ is isomorphic to P ∗ (X ∗ ). If, moreover, P = 1 then P(X )∗ is isometric to P ∗ (X ∗ ). Hint. Note that P can be considered as a mapping from X onto P(X ), so P ∗ maps P(X )∗ into X ∗ . Check this operator for the isomorphism. If P = 1, then P ∗  = 1, so P ∗ (g) ≤ g for every g ∈ P(X )∗ . Moreover, P ∗ (g) (= g ◦ P) extends g to X (see Exercise 4.8). 4.11 Show that if Y is isomorphic to a complemented subspace of X , then Y ∗ is isomorphic to a complemented subspace of X ∗ . Hint. Let Y be isomorphic to Z and P be a projection of X onto Z . By Exercise 4.10, P ∗ (Z ∗ ) is closed in X ∗ and isomorphic to Z ∗ , hence to Y ∗ . By Exercise 4.2, P ∗ (Z ∗ ) is complemented in X ∗ . 4.12 Let P : X → X be a norm-one projection onto a subspace E of a Banach space X . Is P B X closed in E? Hint. Yes: P B X = B X ∩ E. 4.13 Let P : X → X be a bounded projection onto a subspace E of a Banach space X . Is P B X necessarily closed in E? Hint. No. Let f ∈ 2S X ∗ , e ∈ S X , f not attaining its norm. Put P x = f (x)e. 4.14 Calculate the norm of the projection on the line y = 12 x that has as kernel the y-axis, in spaces 2p for p ∈ [1, ∞), and in 2∞ . Hint. Draw a picture. The results: (1 +

1

1 p 2p ) ,

and 1 in the case 2∞ .

Exercises for Chapter 4

223

4.15 Let X be a normed space, Y, Z subspaces of X such that X = Y ⊕ Z . Assume that there is a bounded linear projection P of X onto Y with Ker(P) = Z . Show that: (i) X is isomorphic to (Y ⊕ Z )∞ (hence to (Y ⊕ Z ) p for p ≥ 1). (ii) If Y, Z are complete spaces, then so is X . Hint. (i) Define T (y, z) = y + z. Since X = Y + Z and Y ∩ Z = {0}, this is a bijection. We have T (y, z) = y + z ≤ y + z ≤ 2 max(y, z). On the other hand, y ≤ P y + z and z ≤ I X − P y + z, so max(y, z) ≤ (1 + P)T (y, z). (ii) If Y, Z are complete then so is (Y ⊕ Z )∞ 4.16 Let Y, Z be closed subspaces of a Banach space X . Show that X = Y ⊕ Z (topological sum) if and only if X ∗ = Y ⊥ ⊕ Z ⊥ (topological sum). Hint. If X = Y ⊕ Z , the statement follows from Exercise 4.9. Assume X ∗ = Y ⊥ ⊕ Z ⊥ . Then f (x) = 0 for all f ∈ X ∗ and x ∈ Y ∩ Z , hence Y ∩ Z = {0}. Using the separation theorem and Y ⊥ ∩ Z ⊥ = {0} we find that Y + Z is dense in X . We claim Y + Z is closed. Let Q be the linear projection of Y ⊕ Z (algebraic sum) onto Y . We claim that it is bounded. Let P be the bounded projection of Y ⊥ ⊕ Z ⊥ onto Y ⊥ . Fix y ∈ Y , z ∈ Z . Then for x ∗ ∈ X ∗ write x ∗ = y ∗ + z ∗ and x ∗ (y) = y ∗ (y) = y ∗ (y + z) ≤ y ∗  y + z ≤ P x ∗  y + z, so y ≤ P y + z. Thus Q ≤ P. By Exercise 4.15, Y ⊕ Z is closed. 4.17 Find two closed subspaces Y, Z of a Banach space X so that Y + Z is not closed in X . Hint. Consider Y = span{e2i−1 }, Z = span{e2i−1 +2−i e2i } inX = 1 . Clearly both N are closed subspaces, but Y + Z is not closed. Indeed, z N = i=1 2−i e2i ∈ Y + Z N   N ∞ −i as z N = i=1 (e2i −1 + 2 e2i ) − i=1 e2i −1 . But z N → z = i=1 2−i e2i , which cannot be in Y + Z as (1, 1, . . . ) ∈ / 1 . 4.18 Let Y, Z be closed subspaces of a Banach space X such that Y ∩ Z = {0}. Denote d = dist(SY , S Z ). Show that the subspace Y + Z is closed in X if and only if d > 0. Hint. Assume that Y + Z is closed in X . Then Y is complemented in the Banach space Y + Z , so there is a bounded linear projection of Y + Z onto Z . For every y ∈ SY , z ∈ S Z we have Py − z ≥ P(y − z) = y = 1, hence dist(SY , S Z ) ≥ P−1 . Assume that Y + Z is not closed in X . By the first part of this proof, there is no C > 0 such that Cy − z ≥ y for every y ∈ Y , z ∈ Z . By scaling we have the existence of yn ∈ Y and z n ∈ Z , n ∈ N such that yn  = 1 and yn + z n  ≤ n1 for      n  −  − z n  ≤ yn + z n  ≤ 1 . Therefore by the  all n. Hence 1 − z n   n  = y−1 triangle inequality,  yn + z n  z n  → 0, so d = 0.

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4 Schauder Bases

4.19 Let X be a Banach space and Y be a subspace of X . Show that Y is 1complemented in X if and only if there is a w∗ -closed subspace Q of X ∗ such that Q 1-norms Y and whenever for some q ∈ Q we have q(y) = 0 for all y ∈ Y , then q = 0. Hint. For the necessary condition, let X = Y ⊕ Z , with P : X → Y the associated projection onto Y , and P = 1. Put Q = P ∗ X ∗ . Check that Q = Z ⊥ . This proves that Q is w ∗ -closed. The other properties of Q are also easy to check. For the sufficient condition, let Z = Q ⊥ . Prove first that Y ∩ Z = {0}. This shows that Y + Z is an algebraic direct sum. Let P : Y + Z → Y be the associated projection. Given y ∈ Y , z ∈ Z , note that P(y + z) = y = sup{|q, y| : q ∈ Q, q ≤ 1} = sup{|q, y + z| : q ∈ Q, q ≤ 1} ≤ y + z, hence P is continuous and P = 1, since P is a projection. This proves, in particular, that Y ⊕ Z is closed. The other condition shows that Y ⊕ Z is dense in X , hence X = Y ⊕ Z .

4.20 Let Y and Z be closed subspaces of a Banach space X such that Y ∩ Z = {0}. Define a norm ||| · ||| on Y ⊕ Z by |||y + z||| = y + z. (i) Show that ||| · ||| is a complete norm on Y ⊕ Z . (ii) Show that the following are equivalent: (1) ||| · ||| is equivalent to the original norm on Y ⊕ Z . (2) Y + Z is closed. (3) Y is complemented in Y + Z , that is, Y ⊕ Z is a topological sum. (4) There is k > 0 such that y ≤ ky + z for every y ∈ Y . Hint. (i) {yn + z n } is Cauchy (respectively convergent) if and only if both {yn } and {z n } are Cauchy (respectively convergent). (ii) If ||| · ||| is an equivalent norm, then (Y ⊕ Z ,  · ) is complete, hence Y + Z is closed. But then Y ⊕ Z is an algebraic sum of two closed subspaces of a Banach space, hence it is a topological sum. In particular, Y is complemented by projection P such that y ≤ P y +z. Finally, the inequality implies z ≤ (1+k)y +z and so |||y + z||| ≤ (2k + 1)y + z, clearly y + z ≤ |||y + z|||.

4.21 Show that there are two closed subspaces M1 and M2 of a Hilbert space such that M1 ∩ M2 = {0} and M1 + M2 is not closed, and thus there is no projection on M1 that has kernel M2 . Hint. Define T ∈ B(2 ) by T (x) = (2−i xi ) for x = (xi ). Show that T maps 2 onto a dense subset of 2 but does not map it onto 2 . Denote by G the graph of T in 2 ⊕ 2 and let M = 2 ⊕ {0} ⊂ 2 ⊕ 2 . Show that G ∩ M = {0}, G + M is dense in 2 ⊕ 2 but that G + M = 2 ⊕ 2 .

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4.22 Let X be a Banach space. Assume that Y, Z are w ∗ -closed subspaces of X ∗ such that Z ⊕ Y = X ∗ (algebraic sum). Show that the corresponding (algebraic) projection Q of X ∗ onto Y is w ∗ –w ∗ -continuous. There is a Banach space X such that an isomorphic copy Y of a Hilbert space is a complemented subspace of X ∗ ; however, there is no w ∗ -w∗ -continuous projection of X ∗ onto Y (see [JoLi1]). Hint. Note that Y⊥ ⊕ Z ⊥ = X (algebraic sum). Let P be the corresponding (algebraic) projection of X onto Z ⊥ . Then P ∗ (y + z)(y⊥ + z ⊥ ) = Q(y + z)(y⊥ + z ⊥ ) = y(z ⊥ ). 4.23 Let {eγ } be a Hamel basis of an infinite-dimensional Banach space X . Show that some of the coordinate functionals associated with this basis are not continuous. Hint. Pick an infinite sequence {en i } in {eγ }. Consider the vector x = ∞ −i en i  . Since {eγ } is a Hamel basis of X , we have x = F x j e j , where F i=1 2 en i  m −i en i is a finite set. Let n p be such that n p ∈ / F, i.e., x n p = 0. For every m, i=1 2 en i  has the n p -coordinate equal to 2−n p /en p . If the n p -coordinate functional were continuous, we would have xn p = 2−n p /en p  = 0, a contradiction. 4.24 Why do we not use (iii) in Lemma 4.7 and the Banach–Steinhaus theorem to conclude that Pn are uniformly bounded? Hint. Do we know that they are bounded operators? 4.25 Show that the canonical projections of a Schauder basis of a normed space X need not be uniformly bounded if X is not a Banach space. Hint. Consider the trigonometric polynomials in the space of continuous functions on [0, 2π ]. 4.26 Use the notion of basic sequence to prove that a Hamel basis of an infinitedimensional Banach space has cardinality at least the continuum. Hint. Take any basic sequence {xn } in X . Let {Nα }α ∈ Γ be a collection of infinite subsets of N such Nα ∩ Nβ is finite if α = β and card(Γ ) ≥ c (Lemma 5.7).  that 2−i xi . Then {yα } is a linearly independent set of cardinality at Define yα = least c.

i∈Nα

4.27 Let {ei } be a Schauder basis of a Banach space X . Prove that there is an equivalent norm on X in which {ei } is monotone. Hint. Put |||x||| = supn Pn (x). Then |||Pm (x)||| = sup Pn Pm (x) = sup Pn (x) ≤ sup Pn (x). n

n≤m

n

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4 Schauder Bases

 ∞ 4.28 Let {ei } be a Schauder basis. For n ≤ m ∈ N define Tn,m = i=1 ai ei m a e . We say that {e } is bimonotone if T  = 1 for all n, m. i n,m i=n i i (i) Show that Tn,m  ≤ 2 bc{ei }. (ii) Show that there is an equivalent norm |||·||| on X such that {ei } is a bimonotone basis of (X, ||| · |||). Hint. (i) Tn,m = Pm − Pn−1 . (ii) |||x||| = sup Tn,m (x). n≤m

4.29 Let {ei } be a Schauder basis of a Banach space X . Show that {ei } is monotone if and only if Pn+1 (x) ≥ Pn (x) for every x ∈ X and n ∈ N. Hint. If Pn  = 1 for all n, use Pn (x) = Pn Pn+1 (x). If the condition holds, prove first Pk (x) ≥ Pn (x) for all k ≥ n and then take the limit for k → ∞. 4.30 Show that C[0, 1] has a Schauder basis consisting of polynomials. Hint. Consider any Schauder basis {ei } of C[0, 1] and approximate {ei } by polynomials using the Stone–Weierstrass theorem. Use the stability theorem for Schauder bases. 4.31 Find two dense linear subspaces of a Banach space having intersection {0}. Hint. Let X = { f ∈ C[0, 1] : f (0) = f (1) = 0}. This space, endowed with the supremum norm, is a Banach space. The set {en : n = 3, 4, . . .} is a Schauder basis of X , where {en }n∈N is the Faber–Schauder basis of C[0, 1] defined in Example (iv) after Theorem 4.10 by using the dyadic points ∞in [0, 1]. Indeed, {en : n = 3, 4, . . .} ⊂ X , and, if f ∈ X is written as f = n=1 an en , then 0 = f (0) = a1 ,  ∞ and 0 = f (1) = a1 + a2 , hence f = ∞ a e n=3 n n . Choose a C -function gn in n [0, 1] such that gn (0) = gn (1) = 0 and gn − en  < 1/2 , n = 3, 4, . . .. Then {gn : n = 3, 4, . . .} is a Schauder basis (see Theorem 4.23). Every element in span{gn : n = 3, 4, . . .} is in the class C ∞ , but no element in span{en : n = 3, 4, . . .} \ {0} is. 4.32 Let X, Y be closed subspaces of a separable Banach space Z such that X ∩Y = {0}. Assume that the algebraic sum X + Y is not closed in Z . Show that there exist a basic sequence {xi } ⊂ X and a basic sequence {yi } ⊂ Y such that xi  = yi  = 1 and xi − yi  < 4−i for n ∈ N. In particular, X and Y have infinite-dimensional closed subspaces that are isomorphic. Thus if X and Y are totally incomparable spaces (for example  p , q for p = q), then X + Y is closed in every overspace. Hint. If X + Y is not closed, then (Exercise 4.18) there exists x ∈ X , y ∈ Y , x = y = 1 and x − y arbitrarily small. Then use the proof of Propositions 4.19 and 4.23. 4.33 Let {ei ; ei∗ } be a Schauder basis of a Banach space X . Show that if {ei } is shrinking, then {ei∗ } is a boundedly complete basis of X ∗ .

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    n Hint. {ei∗ } is a Schauder basis of X ∗ . Consider ai such that sup i=1 ai ei∗  < ∞. X is separable, so there is a sequence {n k } ⊂ N and x ∗ ∈ X ∗ such that xn∗k = n k w∗ ∗ → a e x ∗ (Exercise 3.176). Since {ei∗ } is a basis of X ∗ , we have x ∗ = i i i=1 ∞ ∗ ∗ ∗ ∗ i=1 βi ei . Fix j ∈ N. Then lim x n k (e j ) = x (e j ) = β j and x n k (e j ) = a j for k∞ n k ≥ j, hence a j = β j . Thus i=1 ai ei∗ = x ∗ . 4.34 Let {ei ; ei∗ } be a Schauder basis of a Banach space X . Show that span{ei∗ } is a norming subspace of X ∗ . Hint. Let |||x||| = sup Pn (x). Then |||Pn ||| = 1 for every n, so |||Pn∗ ||| = 1. Given w∗

f ∈ S X ∗ , we have |||Pn∗ ( f )||| ≤ 1, Pn∗ ( f ) ∈ span{e1∗ , . . . , en∗ } and Pn∗ ( f ) → f . w∗

Hence span{ei∗ } ∩ B X ∗ = B X ∗ , where B X ∗ is the dual unit ball in the dual norm to ||| · |||. Thus span{ei∗ } is 1-norming for (X, ||| · |||). Since ||| · ||| is an equivalent norm on X , span{ei∗ } is a norming set.  n  4.35 Let X = ∞ 2 . Show that every infinite-dimensional closed subspace of X contains an isomorphic copy of 2 . Hint. Given a closed subspace Z of X , there is a subsequence in Z which is equivalent to a block basic sequence of the canonical basis of X such that the “nods” are at the ends of the canonical blocks. This block basis is equivalent to the standard unit vector basis of 2 . 4.36 Let X be a Banach space. Assume that there is λ ≥ 1 so that for every ε > 0 and every finite set F ⊂ X , there is a finite rank operator T : X → X so that T x − x < ε for every x ∈ F and T  ≤ λ. Must X have the bounded approximation property? Hint. Yes, given a compact set K and ε > 0 consider a finite set F ⊂ X such that K ⊂ x∈F B(x, ε/(3λ)). If T is a finite rank operator on X such that T x − x < ε/3 for x ∈ F, then T x − x < ε for every x ∈ K . 4.37 Show that every Banach space X has the following property: Given ε > 0 and given a finite set F ⊂ X , there is a finite rank operator T on X such that x − T x < ε for every x ∈ F. Hint. Any projection from X onto span(F). Note that ε can be taken 0. 4.38 Show that



L p [0, 1]



is isometric to L p [0, 1].

p 1 Hint. Put An = [ n+1 , n1 ] and identify the isometry of L p [0, 1] and L p (An ).

 f ∈ L p [0, 1] with the sequence { f  A }. Use

    c0 c is isometric to c0 . 4.39 Show that  p  is isometric to  p and p 0 Hint. Direct computation. 4.40 Show that



C[0, 1]

 c0

is isomorphic to C[0, 1].

n

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4 Schauder Bases

  Hint. Use the Pełczy´nski’s decomposition method. Represent C[0, 1] c as a 0 complemented subspace of C[0, 1] using an infinite number of nods in [0, 1] and the closed subspace of C[0, 1] of functions that vanish at these nodes. 4.41 Let p ∈ (1, ∞). Show that L p [0, 1] ⊕ 2 is isomorphic to L p [0, 1]. Hint. Use that 2 is isomorphic to a complemented subspace of L p [0, 1]. 4.42 Use that L ∞ [0, 1] is complemented in every overspace (see Exercise 5.91) to show that ∞ is isomorphic to L ∞ [0, 1] ([Pelc2]). As 1 is not isomorphic to L 1 [0, 1], this provides an example of two non-isomorphic spaces whose duals are isomorphic. This also shows that no isomorphism of ∞ and L ∞ [0, 1] can be w∗ w ∗ -continuous (Exercises 2.51 and 3.60). Hint. As the dual of a separable space, L ∞ is isomorphic to a subspace of ∞ . ∞ is complemented in every overspace, so we use the Pełczy´nski decomposition method. 4.43 Show that 2 is isomorphic to a quotient of ∞ . Compare this with Exercise 13.40. Hint. The space 2 is isomorphic to a subspace of L 1 [0, 1] by Theorem 4.53. Therefore ∗2 ∼ = 2 is isomorphic to a quotient of L ∗1 = L ∞ and then use the previous exercise. 4.44 Assume that T is a bounded operator from a Banach space X into X such that T (X ) is not closed. Show that on no finite-codimensional subspace Y of X , T is an isomorphism from Y into X . Hint. Assume that T is an isomorphism from a finite-codimensional Y into X . We have X = Y ⊕ Z as Y is complemented. Then T (X ) = T (Y ) + T (Z ). If T is an isomorphism on Y then T (Y ) is closed. Since Z is finite-dimensional, we have that T (Y ) + T (Z ) is closed. 4.45 Let X, Y be Banach spaces. Show that every compact operator from X into Y is strictly singular. Give an example of a strictly singular operator that is not compact. Hint. Let Z be a subspace of X such that T is an isomorphism of Z onto T (Z ). Then T (B Z ) is a closed subset of T (B X ), hence it is compact. As T  Z is an isomorphism, B Z must be compact, hence Z is finite-dimensional. For the second question, consider the formal identity mapping from 1 into 2 . 4.46 (KaTo) Let X be an infinite-dimensional Banach space and let T be a bounded operator from X into X such that the restriction to every infinite-dimensional closed subspace of X is not compact. Show that there is a finite-codimensional subspace Z of X such that the restriction of T on Z is an isomorphism. Hint. Assume that the restriction of T to every finite-codimensional subspace is not m ⊂ X ∗ there is x ∈ S X with an isomorphism. Then for every δ > 0 and {xi∗ }i=1 ∗ T (x) ≤ δ and xi (x) = 0 for i = 1, . . . , m. By the proof of Mazur’s theorem,

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229

there is a basic sequence {xi } ⊂ X with bc{xi } ≤2 so that T (xi ) < 8−i . Let Z = span{xi }. Then T (B Z ) is contained in the set { ai T (xi ) : |ai | ≤ 2}, which is totally bounded due to T (xi ) ≤ 8−i (see Exercise 1.50). 4.47 Show that the sum of two strictly singular operators T, S from a Banach space X into a Banach space Y is strictly singular. Hint. By repeated use of the previous exercise, first for T and then for S get an infinite-dimensional closed subspace Z of X on which both T and S are compact. Then T + S is compact on Z and thus not an isomorphism on Z . 4.48 Let T be a strictly singular operator from a Banach space X into a Banach space Y . Is T ∗ necessarily strictly singular? Hint. No. Consider a bounded operator T from 1 onto 2 (Theorem 5.1). T is strictly singular as 2 is not isomorphic to any subspace of 1 (Pitt’s theorem). However, T ∗ is an isomorphism into (Exercise 2.49) and thus T ∗ is not strictly singular. 4.49 Let Γ be uncountable and 1 ≤ p < q < ∞. Show that there is no bounded one-to-one operator from q (Γ ) into  p (Γ ). Similarly, there is no bounded one-toone operator from c0 (Γ ) into  p (Γ ), p ∈ [1, ∞). Hint. Assume that such an operator T : q (Γ ) →  p (Γ ) exists. Let ε > 0 and an uncountable set Γ1 ⊂ Γ be such that T (eγ ) ≥ ε for γ ∈ Γ1 . An infinite sequence in {eγ }γ ∈Γ1 tends weakly to zero and we use Pitt’s theorem for its closed span. 4.50 Show that every strictly singular operator T from  p into  p is necessarily compact for p ∈ [1, ∞). Hint. Examine the proof of Pitt’s theorem. 4.51 Does there exist a bounded operator from C[0, 1] onto 2 ? Note that there is no isomorphic copy of 2 complemented in C[0, 1] (this follows using the Dunford–Pettis property, see Chapter 13). Hint. Yes. ∗2 is isomorphic to a subspace of L 1 [0, 1], which is isomorphic to a subspace of C[0, 1]∗ by the Riesz representation theorem. By Lemma 4.62, ∗2 is w∗ -closed in C[0, 1]∗ . 4.52 Does there exist a bounded operator from  p onto q , p = q ∈ [1, ∞)? Hint. No. Pitt’s theorem, duality. 4.53 Does there exist a bounded operator from  p onto c0 ? Hint. If and only if p = 1. For p = 1 see Theorem 5.1. For p > 1 use the reflexivity of  p . 4.54 Show that a block basic sequence of an unconditional basis is unconditional.

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4.55 Show that every infinite-dimensional subspace of a Banach space with unconditional basis has an infinite-dimensional unconditional basic sequence. Hint. A block basis of an unconditional basis is unconditional. Then use the Bessaga–Pełczy´nski selection principle. 4.56 Show that the vectors e1 , e2 − e1 , e3 − e2 , . . . form a basis of 1 that is not unconditional (ei denotes the standard ith unit vector). Hint. Standard. 4.57 (Köthe, Lorch) Show that every normalized unconditional basis {xi } of a Hilbert space is equivalent to the canonical basis of 2 .  Hint. We need to prove scalars, the series λi xi con that2for a sequence {λi } of verges if and only if |λi | < ∞. Assume that λ x converges. It converges i i   λi xi 2 converges. On the unconditionally so, by Exercise 1.96, ( |λi |2 =) other hand     1/2           1   2 λi xi  ≤ K  εi λi xi  |λi |   2n =K     F

εi =±1 F

F

for every nonempty finite set F ⊂ N, where K is the unconditional basis constant (the inequality follows from the unconditionality, the equality from the parallelogram identity). Note that an analogous result holds for c0 and 1 . However, in these cases the proof is more difficult (see, e.g., [LiTz3, paraghaph preceeding Proposition 2.b.9]). 4.58 Assume that an X admits an unconditional basis and that X ∗ is separable. Is it true that then X ∗ has an unconditional basis? Hint. Yes, James’ characterization of spaces with unconditional bases not containing 1 . 4.59 Assume that X ∗ has an unconditional basis. Does X necessarily admit an unconditional basis? ω Hint. No. C(ω0 0 ), see [AlKa, p. 96]. 4.60 Let X be a Banach space (not necessarily separable). A family {eγ }γ ∈Γ in X is called an unconditional Schauder basis of X if for every x ∈ X there is a unique aγ eγ , where the summation is family of real numbers {aγ }γ ∈Γ such that x = meant in the sense that for every ε > 0 there is a finite set F ⊂ Γ such that      x − γ ∈F aγ eγ  ≤ ε for every F ⊃ F. Note that for every x ∈ X only countably many coordinates aγ are nonzero. Indeed, given n ∈ N, there is a finite set F ⊂ Γ    such that  γ ∈F aγ eγ  ≤ n1 for every finite F disjoint from F. Applying this to F = {γ } and assuming eγ  = 1 we get {γ : |aγ | ≥ n1 } ⊂ F.

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231

Similarly as for the case of countable Schauder bases it can be shown that the coordinate functionals aγ∗ are in fact in X ∗ . Show that the closed linear span of the coordinate functionals of an unconditional basis is a norming subspace in X ∗ . Hint. Use (PF )∗ . 4.61 Let X be a Banach space. Show that if X ∗ is separable, then X admits a shrinking Markushevich basis. Hint. The proof of Theorem 4.59. 4.62 Let X, Y be separable Banach spaces. Show that there is a bounded operator mapping X onto a linearly dense subset in Y . Hint. Let {xi ; f i } and {z i ; gi } be Markushevich bases of X and Y respectively, with  { f i } and {z i } bounded. Put T (x) = 2−i f i (x)z i . 4.63 Let X be an infinite-dimensional separable Banach space. Show that there is a biorthogonal system {x i ; f i } such that span{xi } = X and { f i } is not separating. Note that this cannot be done in finite-dimensional spaces. Hint. Let {yi } be a linearly independent sequence such that span{yi } = X . Pick x ∈ X \ span{yi } (infinite-dimensional spaces cannot have a countable Hamel basis, Exercise 1.81). Since {x, y1 , . . . } is a linearly independent set, using Hahn–Banach we find gi ∈ S X ∗ such that gi (x) = 0, gi (yi ) = 1, and gi (y j ) = 0 for j = 1, . . . , i − 1. Then {gi } separates points of span{yi } and as in Theorem 4.59 we find a biorthogonal system {xi ; f i } so that span{xi } = span{yi } and span{ fi } = span{gi }. Then span{xi } = X and { f i } does not separate points. Indeed, gi (x) = 0 for all i, hence also f (x) = 0 for all f ∈ span{gi }. 4.64 Let X be a Banach space and {xγ ; x γ∗ }γ ∈Γ be a biorthogonal system in X × X ∗ such that {xγ : γ ∈ Γ } is linearly dense in X . Prove that card(Γ ) = dens X . Hint. Let D ⊂ X be a dense subset of X with card(D) = dens X . Given d ∈ D : γ ∈ Γd }. Then there exists a countable

set Γd ⊂ Γ such that d ∈ span{xγ span{x : γ ∈ Γ }, hence X = span{x : γ ∈ D ⊂ γ d γ d∈D d∈D Γd }. If γ0 ∈

Γ \ d∈D Γd , note that x γ∗0 , xγ  = 0 for all γ ∈ d∈D Γd , hence xγ0 ∈ span{x γ : γ ∈ d∈D Γd } (= X ), a contradiction. Thus d∈D Γd = Γ and we get card(Γ ) = card(D) = dens X . 4.65 Show that James’ space J (Definition 4.43) contains an isomorphic copy of 2 . Hint. Let H denote the subspace of J consisting of vectors with even coordinates equal to zero. Show that on H , James’ norm is equivalent to the 2 norm. 4.66 Let u i = ei − ei+1 , where {ei } is the shrinking basis of J . Show that {u i } is a boundedly complete basis of J which is not shrinking, and

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4 Schauder Bases

  m−1  n j+1 −1 2 12   ξi u i  = sup ξi : 1 ≤ n1 < · · · < nm .  j=1

i=n j

4.67 Let {u i } be the boundedly complete basis of J from the previous exercise. Let {vi } be the biorthogonal functionals to {u i }, which span the predual space∗ J∗ to J by Theorem 4.15. Define g ∈ J ∗ by g( ξi u i ) = ξi . Show that J = span({vi } ∪ {g}) = span(J∗ ∪ {g}). Hint. J∗ = span{vi }. Clearly, g ∈ J ∗ \J∗ , then use dim (J ∗ /J∗ ) = 1. 4.68 Assume that we replace the exponent 2 in the definition of James’ space J by 1. Is the resulting space isomorphic to 1 ? Can c0 be obtained in a similar way? Hint. Yes. Yes. 4.69 Is the following an equivalent norm on James’ space J ? z20 =

sup

n 1 <...
m 

|z n 2i−1 − z n 2i |2



i=1

Hint. Yes. 4.70 Define a norm ||| · ||| on James’ space J by 1 1  |||x||| = sup √ (xn 1 − x n 2 )2 + ... + (xn m−1 − xn m )2 + (xn m − x n 1 )2 2 . 2 Show that ||| · ||| is an equivalent norm on J and in this norm, J ∗∗ is isometric to J. Hint. Consider the mapping U : J ∗∗ → J defined by U (x ∗∗ ) = (−λ, x ∗∗ (e1 ) − λ, x ∗∗ (e2 ) − λ, . . . ), where λ = lim x ∗∗ (ei∗ ). 4.71 Find a Schauder basis {ei } of a Banach space X with separable dual such that {ei } is not shrinking. Hint. Consider the biorthogonal functionals { f i } to the standard unit vector basis {ei } in James’ space J . Since the standard basis is shrinking, { fi } is a basis of J ∗ . However, { f i } cannot be shrinking as its biorthogonal functionals are {ei } in J and span{ei } = J = J ∗∗ . 4.72 Does there exist a Banach space X not isomorphic to a Hilbert space and such that X ∗ is isomorphic to X ?

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233

  Hint. Consider  p ⊕ q 2 , where p ∈ (1, ∞) and 1p + q1 = 1. To obtain a nonreflexive example, consider J ⊕ J ∗ , where J is James’ space. 4.73 Does there exist a Banach space X such that X ⊕ X is not isomorphic to X ? Hint. Consider James’ space. What is the codimension of J ⊕ J in (J ⊕ J )∗∗ ? Note that the first reflexive space X that is not isomorphic to X ⊕ X was constructed by T. Figiel [Figi2]. In Exercises 4.74, 4.75, 4.76, 4.77, 4.78, and 4.79 we define and investigate the James tree space J T , [Jame2] and [LiSt]. All notation and definitions made in one exercise carry to the following ones. 4.74 Let T = {(n, i) : 0 ≤ i ≤ 2n − 1, n ∈ N0 } be equipped with a partial ordering < determined by the relation (n, i) < (n + 1, j) if, and only if, j ∈ {2i, 2i + 1}. Then (T, <) is a dyadic tree. By a segment we mean a subset S = {t ∈ T : (n, i) ≤ t ≤ (m, j)} for some (n, i), (m, j) ∈ T . A maximal linearly ordered subset of T is called a branch. Let Γ denote the set of all branches in T . Show that card(Γ ) = c, the cardinality of the continuum. 4.75 The James tree space J T consists of all real functions defined on T with the norm k   2 1  2 x(n, i) , x = sup j=1 (n,i)∈S j

where the supremum is taken over all finite sets of pairwise disjoint segments in T . Verify that J T is a Banach space with a boundedly complete basis {e(n,i) } ordered lexicographically, that is, (0, 0) ≤ (1, 0) ≤ (1, 1) ≤ · · · ≤ (n, i) ≤ (n, i + 1) ≤ · · · ≤ (n, 2n − 1) ≤ (n + 1, 0) ≤ . . .

and the functions e(n,i) are defined by e(n,i) (m, j) = 1 if (n, i) = (m, j) and zero otherwise (i.e., e(n,i) (m, j) = δ(m, j)(n,i) ). In particular, there exists a predual J T∗ to J T with a shrinking basis { f (n,i) } with e(n,i) ( f (m, j) ) = δ(m, j),(n,i) . Thus J T ∗ can be represented as a space of real  2n −1 functions F on T satisfying F = w∗ - lim ∞ n=0 i=0 F(n, i) f (n,i) . Hint. By contradiction. Consider x ∈ J T and ε > 0 such that for every n ∈ N there exists a finite set {S nj }kj=1 of pairwise disjoint segments satisfying   2 1 2 k n x(n, i) ≥ ε. Then there exists a fast growing sequence {nl } (n,i)∈S j j=1 2   nl nl n of integers such that Sl1 1 ∩ Sl2 2 = ∅ for l1 = l2 . Thus l = (n,i)∈Sl l x(n, i) ∞, a contradiction.

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4 Schauder Bases

4.76 For every branch γ ∈ Γ , let Yγ = {x ∈ J T : supp(x) ⊂ γ }. Show that Yγ is a subspace of J T isomorphic to James’ space J . Show that the operator Pγ : J T → J T defined by Pγ

n −1 ∞ 2 

 α(n,i) e(n,i) = αt e t t∈γ

n=0 i=0

is a norm-one projection of J T onto Y . Hint. See Exercise 4.66 4.77 For F ∈ J T ∗ define S(F) ∈ 2 (Γ ) by S(F) = JT∗



lim

n→∞,(n,i)∈γ

F(n, i)

 γ

for

γ ∈ Γ . Show that S : → 2 (Γ ) is a bounded linear mapping. Hint. F = sup{F(x) : x ∈ B J T }. Using this one can get S = 1. To show n that S is onto, given {γi }i=1 ⊂ Γ and scalars a1 , . . . , an , find F ∈ J T ∗ such that 4 n ai γi and F = S(F) = i=1 ai2 . 4.78 Show that Ker(S) = J T∗ . Hint. For m = 0, 1, 2, . . . , define for j ∈ {0, . . . , 2m − 1} Pm, j

n −1 ∞ 2 

n=0 i=0

∞  t(n,i) e(n,i) =



t(n,i) e(n,i)

n=m (m, j)≤(n,i))

m and Pm = 2j=0−1 Pm, j . Then Pm are projections on J T . First show that for x ∗ ∈ ∗ (x ∗ ) = 0. If it were not the case, there would be a Ker(S), lim maxn Pn,i n→∞ 0≤i≤2 −1

∗ (x ∗ ) > ε. One can show that sequence {(n k , i k )}k∈N and ε > 0 such that P(n k ,i k ) there exists only a limited number of incomparable elements in {(n k , i k )}k∈N , thus there exists a branch γ ∈ Γ containing infinitely many elements from {(n k , i k )}k∈N . ∗ ∗ We may assume that {(n k , i k )}k∈N ⊂ γ and P(n (x ∗ ) − P(n (x ∗ ) > ε k ,i k ) k+1 ,i k+1 ) for all k. By the previous exercise, we have Pγ (x ∗ ) ∈ span{ f (n,i) : (n, i) ∈ γ } and for k large enough, (Pn∗k − Pn∗k+1 )Pγ (x ∗ ) < 12 ε. Define for k ∈ N mappings ∗ ∗ − P(n − (Pn∗k − Pn∗k+1 )Pγ∗ . Uk∗ = P(n k ,i k ) k+1 ,i k+1 )

Then Uk∗ is a dual projection to some Uk and Uk∗ (x ∗ ) > 12 ε. The supports of the subspaces Uk (J T ) are mutually disjoint and no branch γ ∈ Γ intersects 2     j j more than one of them. Thus  k=1 Uk∗ (x ∗ ) = k=1 Uk∗ (x ∗ )2 and for j large enough we obtain a contradiction. Now assume that there is v ∗ ∈ Ker(S)\J T∗ , let dist(v ∗ , J T∗ ) > (1 − δ)v ∗  for some δ. To get a contradiction, we find x, y ∈ S J T such that v(x) > 1 − δ,

Exercises for Chapter 4

235

v(y) > 1 − δ, and Pm 1 (x) = 0, Pm 2 (y) = y for some m 2 much larger than m 1 . For δ small enough we use our limit formula and obtain by calculating the norm of x + y from definition that x + y < 3.5. Details may be found in [LiSt]. 4.79 Show that J T is a separable space with non-separable dual containing no isomorphic copy of 1 . Hint. Using Dixmier’s projection (see Exercise 4.7), from the above exercise we obtain that J T ∗∗ ∼ J T ⊕ 2 (Γ ). Thus card(J T ∗∗ ) = c. If there was an isomorphic copy of 1 in J T , we would find an isomorphic copy of 1 (c) in J T ∗ by Exercise 5.35 and thus ∞ (c) would be isomorphic to some quotient of J T ∗∗ . This would imply that card(J T ∗∗ ) > 2c , which is a contradiction. Remark: The space J T has the property of being saturated by copies of 2 , see Remark 3 in Chapter 5. Hagler constructed a modification of J T : a separable Banach space X with non-separable dual such that X is saturated by copies of c0 and X ∗ is saturated by copies of 1 , [Hag]. We mentioned in the introduction to Section 4.6 a stronger result of Gowers in [Gowe4]. The space X constructed there contains no c0 or 1 or reflexive subspace. This was a solution to a long-standing problem. 4.80 (Lindenstrauss, [Lind12b]) Let X be a separable space. Then there exists a separable Banach space Z , such that X = Z ∗∗ /Z , and Z ∗∗ has a boundedly complete basis. ∞ be a norm dense sequence from S . We let Hint. Let {xi }i=1 X ∞   ∞ Z = (ai )i=1 : ai xi = 0 i=1

to be a space of scalar sequences with the norm ⎛

 2 ⎞ 12  pj m      ⎟ ⎜  ai x i  z = sup ⎝   ⎠ < ∞. 0= p0 < p1 < p2 <···< pm  j=1 i= p j−1 +1

(4.3)

Show that Z ∗∗ consists of all scalar sequences satisfying (4.3), and such that  ∞ ai xi converges. Next show that T : Z ∗∗ → X , T (a1 , a2 , . . . ) = i=1 ∞ i=1 ai xi is a quotient mapping. Finally, show that the canonical sequence en = (0, . . . , 1n , 0, . . . ) is a boundedly complete basis of Z ∗∗ .

Chapter 5

Structure of Banach Spaces

The focus of the study in the present chapter is on Banach spaces containing subspaces isomorphic to c0 or 1 . We prove Sobczyk’s theorem on complementability of c0 in separable overspaces, lifting property of 1 and Pełczy´nski’s characterization of separable Banach spaces containing 1 . We present Rosenthal’s 1 theorem, Odell–Rosenthal theorem and the Rosenthal–Bourgain–Fremlin–Talagrand theory of Baire-1 functions on Polish spaces.

5.1 Extension of Operators and Lifting Theorem 5.1 Every separable Banach space is linearly isometric to a quotient space of 1 . Proof: Let X be a separable Banach space. Let {xi : i ∈ N} be a dense subset of B X . Define a mapping T from 1 into X by T (ξ ) =

∞ 

ξi xi , for ξ = (ξ )i ∈ 1 .

i=1

 Note that, for  every ξ ∈ 1 , the series i ξi xi is absolutely convergent since  i ξi x i  ≤ i |ξi | = ξ 1 . We will show that T B1 = B X . To that end, given x ∈ B X choose a sequence {xn k } as follows: First choose n 1 such that x − xn 1  < 2−1 . Since 2(x − xn 1 ) ∈ B X , by the density of {xn : n ∈ N} in B X we can choose n 2 > n 1 such that 2(x −xn 1 )−xn 2  < 2−1 , i.e., x −xn 1 −(1/2)xn 2  < 2−1 xn 2 ) − xn 3  < 2−1 , i.e., 2−2 . Then pick n 3 > n 2 such that 22 (x − x n 1 −  −1 −2 −3 x − x n 1 − 2 xn 2 − 2 xn 3  < 2 , etc. We have k 2−k xn k = x. The element ∞ such that ξ = 0 for i ∈ {n : k ∈ N} and ξ −k for k ∈ N ξ = (ξi )i=1 i k nk = 2 is in B1 , and T ξ = x. Therefore TB1 = B X and hence X is linearly isometric to 1 /T −1 (0). Similarly we can prove that given a Banach space X , there is a set Γ such that X is isometric to a quotient of 1 (Γ ).

M. Fabian et al., Banach Space Theory, CMS Books in Mathematics, C Springer Science+Business Media, LLC 2011 DOI 10.1007/978-1-4419-7515-7_5, 

237

238

5 Structure of Banach Spaces

Note that if X is not isomorphic to 1 and T1 , T2 are bounded operators from 1 onto X , then Ker(T1 ) is isomorphic to Ker(T2 ) ([LiRo1], see also [LiTz2]). We say that a Banach space X has the lifting property if for all Banach spaces Y, Z such that there is an onto operator S ∈ B(Y, Z ) and for all T ∈ B(X, Z ) there  ∈ B(X, Y ) such that T = S ◦ T . is T Proposition 5.2 1 has the lifting property. Proof: Let Y and Z be two Banach spaces and S : Y → Z an onto continuous linear mapping. If T : 1 → Z is a continuous linear mapping, put xi = T (ei ) for i ∈ N, where ei denote the standard unit vectors of 1 . By the open mapping theorem, there is a constant C such that for every i there  is yi ∈ Y with S(yi ) = xi and yi  ≤ C. (ei ) = S(yi ) = T (ei ) for all i ∈ N, and (ei ) = yi for i ∈ N. Then S T We define T   ≤ C. by linearity, S ◦ T = T . From B1 = conv{ei : i ∈ N} we then get T Corollary 5.3 If there is a bounded operator from a Banach space X onto 1 then 1 is isomorphic to a subspace of X . Proof: It follows from Proposition 5.2, for Z = X , T = I X , and S being the canonical quotient mapping. Proposition 5.4 Every separable Banach space is linearly isometric to a subspace of ∞ . ∗ Proof: By Proposition 3.101  there  is a sequence { f i } that is w -dense in B X ∗ . Define T : X → ∞ by T (x) = f i (x) i . Then x = sup | f i (x)| = T (x)∞ and T is a linear isometry into ∞ .

Similarly we prove that every Banach space is linearly isometric to a subspace of ∞ (Γ ) for some Γ . Note (Exercise 5.33) that a dual space to an arbitrary separable Banach space is isometric to a subspace of ∞ . Proposition 5.5 Every separable Banach space is linearly isometric to a subspace of ∞ /c0 . Proof: Let {xi∗ } be a w∗ -dense sequence in B X ∗ . Define T : X → ∞ by T (x) = (x 1∗ (x), x2∗ (x), x1∗ (x), x2∗ (x), x3∗ (x), x 1∗ (x), . . . ). As in Proposition 5.4, we prove that T is a linear isometry into ∞ . Let q : ∞ → ∞ /c0 be the canonical quotient operator. We claim that q ◦ T is an isometry from X into ∞ /c0 . This follows from the fact that q(xi ) = (xi )∞ /c0 = lim sup |xi | (Exercise 1.29). Theorem 5.6 (Phillips [Phil], Sobczyk [Sobc]) The space c0 is not complemented in ∞ .

5.1

Extension of Operators and Lifting

239

In the proof of Theorem 5.6, we will use the following lemma. Lemma 5.7 (Sierpi´nski) There exists a family F of cardinality continuum formed by infinite subsets of N such that if F1 , F2 ∈ F, F1 = F2 , then card(F1 ∩ F2 ) < ∞. Proof: Let i : N → Q be a one-to-one and onto mapping. For every real number r choose a sequence of distinct rationals {gnr } convergent to r . Define a mapping ∞ N −1 r ϕ : R → N by ϕ(r ) = i (gn ) n=1 .   For r1 , r2 ∈ R, r1 = r2 , we have card {i −1 (gnr1 )}n∈N ∩ {i −1 (gnr2 )}n∈N < ∞. Putting F = {ϕ(r ) : r ∈ R} concludes the proof of the lemma. Proof of Theorem 5.6: (Whitley [Whitl], see, e.g., [Jmsn]) Assume, by contradiction, that P is a bounded linear projection of ∞ onto c0 , and put Q = I − P, where I is the identity mapping of ∞ . For γ ∈ R put xγ = χ Fγ (the characteristic function of Fγ in N), where {Fγ }γ ∈R is the family constructed above. Claim: For every ε > 0 and n ∈ N we have card{γ ∈ R : |Q(xγ )n | > ε} < ∞. Indeed, assume that |Q(x γi )n | ≥ ε for i = 1, . . . , k. Define xγ i = χ F

γi \

Then x γi − x γ i = χ F

.

{Fγ j : j =i, j=1,...,k}

 ∈ c0 as the intersection

{Fγ j : j =i, j=1,...,k} = Q(xγ i ) for i = 1, . . . , k. We have

γi ∩

Thus we obtain that Q(x γi )

Q

k 

k  . . sign Q(x γ i )n xγ i = sign Q(x γ i )n Q(x γ i )n

i=1

n

i=1

=

k 

|Q(xγ i )n | =

i=1

of

∞

i = 1, . . . , k. We thus have

kε ≤ Q

k  i=1

k 

|Q(xγi )n | ≥ kε.

i=1

 .   k  sign Q(x γ i )n xγ i  On the other hand,  i=1

x γ i ,

is finite.

≤ 1 due to the disjoint supports

k  - .  .    sign Q(x γ i )n xγ i ≤ Q  sign Qx γ i n xγ i  n

i=1

∞

≤ Q.

 Therefore k ≤ Q ε .

{γ ∈ R : Q(x ) = 0} is countable. Using the Claim, we see that A = ∞ γ n n=1 Therefore there is γ ∈ R\A for which we have Q(xγ )n = 0 for every n. Hence

240

5 Structure of Banach Spaces

Q(xγ ) = 0 and thus xγ = P(xγ ). This implies that xγ ∈ c0 , a contradiction with the fact that xγ is a characteristic function of an infinite set. In fact, we saw in Exercise 3.44 that c0 is not even isomorphic to a quotient of ∞ . However, there exists a Lipschitz mapping from ∞ onto c0 whose restriction to c0 is the identity ([Lind1], Exercise 5.29). Theorem 5.8 (Banach, Mazur) Every separable Banach space is linearly isometric to a subspace of C[0, 1]. Proof: Let X be a separable Banach space. Then B X ∗ is a compact metric space in the w∗ -topology, therefore there exists a continuous mapping ϕ from the Cantor  from X into set Δ ⊂ [0, 1] onto (B X ∗ , w ∗ ) (Theorem 17.11). Define a mapping T (x) = x ◦ ϕ, where x ∈ X C(Δ), the space of continuous functions on Δ, by T (x) is a conis considered as a (continuous) mapping acting on (X ∗ , w ∗ ). Then T tinuous function on Δ. Now we define a mapping T from X into C[0, 1]. Given (x) defined on Δ onto [0, 1] as follows: for r ∈ x ∈ X , extend the function T / Δ (x)(r1 ) + β T (x)(r2 ), where r1 = max{ p ∈ Δ : p < r }, put T (x)(r ) = α T r2 = min{ p ∈ Δ : r < p} and r = αr1 + βr2 . T is a linear mapping from X into C[0, 1] and (x)(d)| = sup |(x ◦φ)(d)| = sup |x( f )| = x. T (x) := sup |T (x)(r )| = sup |T r ∈[0,1]

d∈Δ

d∈Δ

f ∈B X ∗

Therefore T is a linear isometry from X into C[0, 1]. We say that C[0, 1] is isometrically universal for all separable Banach spaces. Note that there is no finite-dimensional Banach space isometrically universal for all two-dimensional Banach spaces ([Bess1]). It is conjectured that every complemented subspace of C[0, 1] is isomorphic to C[0, 1] or to C(K ) for some countable metric space K . Corollary 5.9 (Banach, Fréchet, Mazur) Every separable metric space (P, ρ) is isometric to a subset of C[0, 1]. Proof: Let {xn }∞ n=1 be a dense sequence  in (P, ρ). Choose x0 ∈ P and define a mapping ϕ from P into ∞ by ϕ(x) = ρ(x, x n ) − ρ(x 0 , x n ) n . Note that given x ∈ P, by the triangle inequality |ρ(x, xn ) − ρ(x0 , xn )| ≤ ρ(x, x 0 ) and therefore indeed ϕ(x) ∈ ∞ . We now show that ϕ is an isometry from P into ∞ . First, if x and x are in P then     ϕ(x) − ϕ(x )∞ = sup | ρ(x, xn ) − ρ(x0 , x n ) − ρ(x , xn ) − ρ(x0 , xn ) | n

= sup |ρ(x, xn ) − ρ(x , xn )| ≤ ρ(x, x ). n

5.1

Extension of Operators and Lifting

241

We claim that ϕ(x) − ϕ(x )∞ ≥ ρ(x, x ) for every x, x in P. Given ε > 0, by density choose n ∈ N such that ρ(x, x n ) < ε/2. Then ρ(x , x n ) ≥ ρ(x , x) − ρ(x, xn ) ≥ ρ(x , x) − ε/2. Therefore ϕ(x) − ϕ(x )∞ ≥ |ρ(x , xn ) − ρ(x, xn )| ≥ ρ(x , xn ) − ρ(x, x n ) ≥ ρ(x , x) − ε/2 − ε/2 = ρ(x , x) − ε. Thus ϕ(x) − ϕ(x )∞ = ρ(x , x) as ε > 0 was arbitrary. We see that ϕ is an  isometric bijection of P onto a separable subset ϕ(P) of ∞ . Let X = span ϕ(P) in ∞ . Then X is separable and there is a linear isometry T of X onto a separable subspace of C[0, 1]. Then the composition T ◦ ϕ is an isometric bijection of P onto a subset of C[0, 1]. Proposition 5.10 Let T be a bounded operator from a Banach space X into a Banach space ∞ (Γ ) for some Γ . Let Y be a Banach space such that Y ⊃ X . Then T can be extended to a bounded operator from Y into ∞ (Γ ) with the same norm as T . Proof: Assume without loss of generality that T  = 1. For γ ∈ Γ , let f γ ∈ B X ∗ be defined by f γ (x) = (T x)γ , where (T x)γ is the γ -coordinate of T x in ∞ (Γ ). Let  :Y → f γ ∈ BY ∗ be a Hahn–Banach extension of f γ on Y . Define an operator T ∞ (Γ ) by   y = f˜γ (y) , for y ∈ Y. T γ  ≤ 1 and T  extends T (so T  = T ). Then T Theorem 5.11 (Sobczyk [Sobc], see also [LiTz3, p. 106]) Let c0 ⊂ X where X is a separable Banach space. Then there is a projection P from X onto c0 with P ≤ 2. Proof: (Veech [Veec]) Let ρ be a translation-invariant metric that induces the w∗ topology on B X ∗ and let F = B X ∗ ∩ c0⊥ . For n ∈ N, let xn∗ be a Hahn–Banach extension of the n-th standard unit vector of c0∗ = 1 . Note that every w ∗ -limit point of (xn∗ ) belongs to F. Since (B X ∗ , ρ) is a compact space, it follows that ρ(x n∗ , F) → 0 as n → ∞. Thus we can find wn∗ ∈ F such that ρ(x n∗ , wn∗ ) → 0 as n → ∞. This means that xn∗ − wn∗ → 0 in the w∗ -topology of X ∗ . Define an operator P : X → c0 by   P x = (xn∗ − wn∗ )(x)

242

5 Structure of Banach Spaces

∗ for x ∈ X . Then  P ≤ 2 and, if x = (x n ) ∈ c0 , then wn (x) = 0 for all n, and thus ∗ P x = xn (x) = (xn ). Hence P is a projection from X onto c0 .

For the complementability of c0 (Γ ) in overspaces, see, e.g., [GKL1] and [ACGJM]. Definition 5.12 A Banach space X is said to be injective (respectively separably injective) if for every Banach space Y containing X (respectively for every separable Banach space Y containing X ) there is a bounded projection from Y onto X . If there is λ ≥ 1 so that such projection exists with norm ≤ λ, then X is called λ-injective (respectively λ-separably injective). Proposition 5.13 Let X be a Banach space and λ ≥ 1. Then the following are equivalent. (i) X is λ-injective. (ii) For every pair of Banach spaces Z ⊃ Y , and every bounded operator T : Y →  : Z → X that extends T and T  ≤ λT . X there is a bounded operator T (iii) For every Banach space Y ⊃ X , for every Banach space Z and for every  : Y → Z that bounded operator T : X → Z there is a bounded operator T  extends T and T  ≤ λT . Proof: (ii)⇒(i) and (iii)⇒(i): If Y is a Banach space such that X ⊂ Y , extend the I : Y → X with  I  ≤ λ. identity operator I : X → X to Y to obtain a projection  (i)⇒(iii): Let Y, Z and T be as in (iii). Let P be a bounded linear projection  := T P has the desired from Y onto X such that P ≤ λ. Then the operator T property. (i)⇒(ii): Let Y, Z and T as in (ii). Let {xγ∗ }γ ∈Γ be a subset of S X ∗ such that x = supγ |xγ∗ (x)| for x ∈ X . Define an isometry S from X into ∞ (Γ ) by Sx(γ ) = x γ∗ (x), for γ ∈ Γ, x ∈ X. 7  = ST  (= 7 : Z → ∞ (Γ ) with  ST Use Proposition 5.10 to extend ST to ST 7 is the T ). Let P : ∞ (Γ ) → X be a projection such that P ≤ λ. Then P ST desired extension. Note that, for every nonempty set Γ , the space ∞ (Γ ) is 1-injective. This follows from Proposition 5.10 and (ii) in Proposition 5.13. Proposition 5.14 Let X be an injective space. Then there is λ ≥ 1 such that X is λ-injective. Proof: Suppose that no such λ exists. Then, by Proposition 5.13, for every n ∈ N there are spaces Z n ⊃ Yn and operators Tn : Yn → X with Tn  = 1 such that 3   any  bounded linear extension Tn of Tn from Z n to X satisfies Tn  ≥ n . Let Y = ( Yn )c0 and T be a bounded operator from Y into X defined by

5.1

Extension of Operators and Lifting

T (y1 , y2 , . . . , yn , . . .) =

243 ∞  Tn yn . n2 n=1

  be a bounded linear extension of T from ( Z n )c0 to X . The restriction of Let T  to Z n (i.e., to elements (0, 0, . . . , z n , 0, . . .)) is, by the definition of T , an extenn2 T (0, 0, . . . , yn , 0, . . .) = T (0, 0, . . . , yn , 0, . . .) = sion of Tn , since for yn ∈ Yn , T 2  to Z n is ≤ n 2 T  < n 3 for n > T . Tn (yn )/n . The norm of this restriction of n 2 T This contradiction finishes the proof. Definition 5.15 Let X be a Banach space. Let Z be a Banach space such that X ⊂ Z . If X is not complemented in Z put λ(X, Z ) = ∞. Otherwise, λ(X, Z ) := inf{P : P : Z → X a linear and bounded projection}.

(5.1)

Define the projection constant of X as λ(X ) = sup{λ(X, Z ) : Z an overspace of X }.

(5.2)

The following result proves that in order to compute λ(X ) for any Banach space X , it is enough to restrict the class of overspaces of X to any (in fact, to a single) overspace of type ∞ (Γ ). Recall that every Banach space is isometric to a subspace of ∞ (B X ∗ ). Proposition 5.16 Let X be a subspace of ∞ (Γ ) for some nonempty index set Γ . Then either (a) X is not complemented in ∞ (Γ ) (and then λ(X ) = ∞), or (b) X is complemented in ∞ (Γ ) (and, in such a case, λ(X ) = λ(X, ∞ (Γ ))). Proof: If (a) holds, certainly λ(X ) = ∞. Assume that, on the contrary, X is complemented in ∞ (Γ ) (and let P : ∞ (Γ ) → X be a projection). Then X is injective. Indeed, let Y and Z be two Banach spaces such that Y ⊂ Z , and let T : Y → X be a bounded operator. Let J : X → ∞ (Γ ) be the inclusion mapping. By Theorem 5.10, we obtain a bounded linear extension  S to Z of the mapping S := J ◦ T .  := P ◦  Then T S (: Z → X ) is an extension of T to Z . Moreover,  ≤ P. T S = P.S = P.T , hence, according to Definition 5.12, X is P-injective. In particular, λ(X, Z ) ≤ P for every overspace Z of X , so λ(X ) ≤ P. This holds for every projection P : ∞ (Γ ) → X , so λ(X ) ≤ λ(X, ∞ (Γ )). On the other side, λ(X, ∞ (Γ )) ≤ λ(X ). This proves the assertion. Remarks: 1. From Proposition 5.16 we get that a Banach space is injective if and only if it is complemented in any (and then in all) overspace of the form ∞ (Γ ) (and that

244

5 Structure of Banach Spaces

λ(X ) can be computed by using just one of those overspaces). A space of the form ∞ (Γ ) is a particular case of a space of type C(K ) for a compact space K ˇ (just endow Γ with the discrete topology and consider its Stone–Cech compactification βΓ ; then ∞ (Γ ) = C(βΓ ), see Section 17.1). For finite-dimensional spaces, the computation of λ(X ) will be reduced to checking λ(X, C(K )) for any overspace C(K ) (see Proposition 5.26). 2. Assume that dim (X ) = n. Let Z be an overspace of X . We can use the fact that an Auerbach basis in X exists to prove that we can always find a projection P0 : Z → X such that P0  ≤ n (see Theorem 4.5). It follows that λ(X, Z ) ≤ n. Since this is true for every overspace Z , we get λ(X ) ≤ n. This result will be substantially improved in Theorem 6.28. The three following lemmas are simple. We collect them here for future references. Lemma 5.17 Let X be an injective Banach space. (i) If, for some λ ≥ 1, X is λ-injective, then λ(X ) ≤ λ. (ii) The space X is, for every ε > 0, (λ(X ) + ε)-injective. (iii) If X is finite-dimensional, then X is λ(X )-injective. Proof: (i) Assume that, for some λ ≥ 1, X is λ-injective. Let Z be a Banach space with X ⊂ Z . There exists a projection P from Z onto X such that P ≤ λ. Then, λ(X, Z ) ≤ λ. Since this is true for all such Z , we get λ(X ) ≤ λ. (ii) Given a Banach space Z with X ⊂ Z we have λ(X, Z ) ≤ λ(X ). It follows that for each ε > 0 we can find a projection P : Z → X such that P < λ(X ) + ε. This shows that X is (λ(X ) + ε)-injective. (iii) Assume now that X is finite-dimensional. Let Z be a Banach space such that X ⊂ Z . According to (ii), for each n ∈ N there exists a projection Pn from Z onto X such that Pn  ≤ λ(X ) + (1/n). Choose an algebraic basis {ei : i = 1, 2, . . . , n} of nX and let πi : X → K be the mapping that associates αi to an element x = i=1 αi ei , for i = 1, 2, . . . , n. Fix i ∈ {1, 2, . . . , n}. The mapping πi ◦ Pn is an element of Z ∗ for each n ∈ N. Moreover, {πi ◦ Pn : n ∈ N} is a bounded subset of Z ∗ , hence we can extract a w ∗ -convergent subnet. If this is done consecutively for i = 1, 2, . . . , n, it follows that there exists a single subnet {Pn j } j of {Pn } such that {πi ◦ Pn j } j is w ∗ -convergent for all i ∈ {1, 2, . . . , n}. This means that {Pn j } j is pointwise-to-norm convergent to an element P that is, certainly, a projection from Z onto X , and P ≤ λ(X ). This proves that X is λ(X )-injective. Lemma 5.18 Assume that, for some ε > 0, the Banach–Mazur distance d(X, X ) between two Banach spaces X and X satisfy d(X, X ) < 1 + ε. Then, if X is λ-injective, then X is (1 + ε)λ-injective. Proof: Let Y and Z be two Banach spaces such that Y ⊂ Z . Let T : Y → X be a bounded operator, and let S : X → X be an isomorphism such that S.S −1  ≤ (1 + ε). The mapping S ◦ T : Y → X has an extension S ◦ T : Z → X such that  := S −1 ◦ S ◦ T , a bounded linear  S ◦ T  ≤ λS ◦ T  (≤ λS.T ). Then, T

5.1

Extension of Operators and Lifting

245

 ≤ (1 + ε)λT . This proves mapping from Z into X , is an extension of T , and T that X is (1 + ε)λ-injective. Lemma 5.19 For some ε > 0, let X and X be two finite-dimensional Banach spaces such that d(X, X ) < 1 + ε. Then, (1 + ε)−1 λ(X ) ≤ λ(X ) ≤ (1 + ε)λ(X ).

(5.3)

Proof: According to (iii) in Lemma 5.17, X is λ(X )-injective. By Lemma 5.18, X is (1 + ε)λ(X )-injective. Use now (i) in Lemma 5.17 to conclude that λ(X ) ≤ (1+ε)λ(X ). Since the Banach–Mazur distance is symmetric we obtain, by reversing the roles of X and X , the second inequality in (5.3). Theorem 5.20 Let Y and Z be Banach spaces, Y ⊂ Z and let T be a bounded operator from Y into c0 . Assume that Z /Y is separable. Then T extends to a bounded operator Tˆ from Z into c0 with Tˆ  ≤ 2T . : Proof: Since ∞ is 1-injective and c0 ⊂ ∞ , T extends to a bounded operator T  = T . Since moreover Z /Y is separable, T Z is separable. Z → ∞ such that T (Z ) ∪ c0 } (⊂ ∞ ). This is a separable Banach space that contains Put E = span{T c0 . By Sobczyk’s Theorem 5.11, there is a projection P from E onto c0 such that  ≤ 2T   maps Z into c0 , Tˆ  ≤ P · T P ≤ 2. Then the operator Tˆ := P T and if y ∈ Y , then Tˆ y = P T y = T y since T y ∈ c0 . Corollary 5.21 If X is isomorphic to c0 , then X is separably injective. Proof: Let T : X → c0 be an isomorphism from X onto c0 . Let Z be a separable Banach space containing X . By Theorem 5.20, T extends to a bounded operator  : Z → X . If x ∈ X , then T x = T x and thus  : Z → c0 . Put P = T −1 T T P x = T −1 T x = x. Hence P is a bounded linear projection from Z onto X . Definition 5.22 A bounded operator T from a Banach space X into a Banach space Y is called weakly compact if T (B X ) is a weakly compact subset of Y . We shall study weakly compact operators in Chapter 13. Note: The space 2 is, as every separable Banach space, linearly isometric to a subspace of C[0, 1] by the Banach–Mazur Theorem 5.8, and certainly C[0, 1]/2 is separable. The formal identity operator T : 2 → c0 , which is weakly compact as 2 is reflexive, cannot be extended to a weakly compact operator from C[0, 1] into  existed, then T  would carry weakly compact sets to c0 . Indeed, if such extension T norm compact sets, as C[0, 1] has the Dunford–Pettis property (see Theorem 13.43). w This is not true as ei → 0 in C[0, 1], where ei are the standard unit vectors in 2 , and T ei  = 1 → 0.

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5 Structure of Banach Spaces

For compact operators, the situation is different, as the next theorem shows. However, note that if X is a Banach space that is not isomorphic to a Hilbert space, then there is a subspace Y of X and a compact operator T from Y into Y such that  from X into Y that extends T . To see this, modify there is no bounded operator T accordingly the proof of Theorem 6.16 (see [LiTz1]). Theorem 5.23 Let X ⊂ Y be Banach spaces, K be a compact space and let T be a  from compact operator from X into C(K ). Then T extends to a compact operator T Y into C(K ). In the proof, the following proposition will be used. The technique used in its proof is sometimes called the “small perturbation principle.” Proposition 5.24 Let K be a compact space. Let F be a finite-dimensional subspace of C(K ) and η > 0. Then there is a finite-dimensional subspace G ⊃ F of C(K ) such that d(G, n∞ ) < 1 + η, where d(·, ·) denotes the Banach–Mazur distance. Moreover, there is a projection P from C(K ) onto G with P < 1 + η. Therefore, for all compact spaces K , the space C(K ) has the bounded approximation property. Proof: For simplicity, we will prove the statement only for metrizable spaces K . Let f 1 , f 2 , . . . , f n be a basis of F with  f i  = 1 for all i. Given ε > 0, choose for every y ∈ K a neighborhood U y of y such that the oscillation of all f 1 , f 2 , . . . , f n on U y is smaller than ε. Let U1 , U2 , . . . , Um be a finite subcollection of {U y : y ∈ K } that covers K . Let xi ∈ Ui with xi = x j if i = j. We claim that there is a peaked partition of unity  Φ1 , Φ2 , . . . , Φm on K (i.e., nonnegative continuous functions on K such that i Φi (x) = 1 for all x ∈ K , Φi vanishes outside Ui , Φi  = 1, and Φi (xi ) = 1 for all i = 1, 2, . . . , m). Indeed, put . . . , m. Then V1 , V2 , . . . , Vm is an open cover Vi = Ui \{x j : j = i}, i = 1, 2, of K . Put Φi (x) = ρ(x, X \Vi )/ mj=1 ρ(x, X \V j ), where ρ is a metric on K . Then m Φ j (xi ) = 0 for j = i, so Φi (xi ) = 1 for all i = 1, 2, . . . , m. We note that  i=1 αi Φi  = sup{|αi | : i = 1, 2, . . . , m} for all real numbers α1 , . . . , αm . Indeed, if x ∈ K then, as Φi ≥ 0,   m m      αi Φi (x) ≤ |αi |Φi (x)    i=1

i=1

≤ sup{|αi | : i = 1, 2, . . . , m}

m 

Φi (x) = sup{|αi | : i = 1, 2, . . . , m}.

i=1

On the other hand,    m  n         αi Φi  ≥  αi Φi (x j ) = |α j |      i=1

for all j.

i=1

5.1

Extension of Operators and Lifting

247

  Let G = span{Φi : i = 1, 2, . . . , m} ⊂ C(K ) . Moreover, it easily follows that the mapping P : C(K ) → G defined by Pf =

n 

f (xi )Φi , f ∈ C(K )

i=1

ia a norm-one projection from C(K ) onto G. Given f i , an element of the basis { f 1 , f 2 , . . . , f m } for F, let gi ∈ G be defined by gi (x) =

m 

Φ j (x) f i (x j ).

j=1

Then, for all x ∈ K ,    m  m     Φ j (x) f i (x) − f i (x j )  ≤ | f i (x) − f i (x j )|Φ j (x) ≤ ε, | f i (x) − gi (x)| =   j=1  j=1 since a term in the sum is not zero only if x and x j belong to the same U y where | f i (x) − f i (x j )| ≤ ε. We will now show how to “shift” G to an overspace of F. This method is called “the principle of small perturbations.” Since all norms on a finite-dimensional space F are equivalent, there is K ≥ 1 such that  m      −1 λi f i  ≤ K max |λi | K max |λi | ≤    i=1

for all real numbers λ1 , λ2 , . . . , λm . Let 1 > δ > 0 such that (1+δ)/(1−δ) < 1+η, and put ε = δ/(2n K ). For i = 1, 2, . . . , n, let gi ∈ G be such that gi  = 1 and  f i − gi  < ε. Then  m      (2K )−1 max |λi | ≤  λi gi  ≤ 2K max |λi |.   i=1

By the Hahn–Banach theorem, there exists xi∗ ∈ C(K )∗ with xi∗  ≤ 2K and such that xi∗ (g j ) = δi, j , the Kronecker delta, i = 1, 2, . . . , n. Let an operator T : C(K ) → C(K ) be defined by Th = h +

n 

xi∗ (h)( f i − gi )

i=1

for h ∈ C(K ). Then T gi = f i for i = 1, 2, . . . , n and

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5 Structure of Banach Spaces

(1 − δ)h ≤ T h ≤ (1 + δ)h, h ∈ C(K ). Indeed, to see for example the right-hand-side inequality, estimate T h − h ≤

n 

xi∗ .h. fi − gi  ≤ h2K nε ≤ δh, for h ∈ C(K ).

i=1

Thus T G ⊃ F and d(T G, n∞ ) ≤ (1 + δ)/(1 − δ) < 1 + η. If P : C(K ) → G is the norm-one projection defined above in this proof, then P1 := T P T −1 is a projection from C(K ) onto T (G) with norm less or equal than 1 + η. This implies that C(K ) has the bounded approximation property. Indeed, if C is a compact set in C(K ) and η > 0 is given, find a finite set { fi : i = 1, 2, . . . , m} so that C⊂

   B f i , η/(1 + ε) . i

By the above, let P be a projection from C(K ) onto span{ f i : i = 1, 2, . . . m} with P ≤ 1 + ε. If f ∈ C then, for some i, P f − f  = (I − P) f | ≤ (I − P) f i  + I − P. f − f i  ≤ (2 + ε) f − f i  ≤

2+ε η = η. 2+ε

Proof of Theorem 5.23: Given a compact operator T from X into C(K ), by using the bounded approximation property of C(K ) (see Proposition 5.24) we can write  T T = ∞ n=0 n in the operator-norm topology, where Tn are finite rank operators and Tn  ≤ 2−n for n ≥ 1. n from Y to C(K ) By Proposition 5.24, each Tn extends to a finite rank operator T   := ∞ T n  ≤ (1 + ε)/2n for n = 1, 2, . . . Thus the operator T  with T n=0 n is a compact operator from Y into C(K ) that extends T . Corollary 5.25 If Y ⊃ X are Banach spaces and T is a compact operator from X into c0 , then T can be extended to a compact operator from Y into c0 .  from Y into C[0, 1] (⊃ Proof: By Theorem 5.23, T extends to a compact operator T c0 ). If P is a projection from C[0, 1] onto c0 (see Sobczyk Theorem 5.11), then the ! := P T  extends the operator T . compact operator T Another consequence of Proposition 5.24 is the following result. Although it is of interest on its own, we present it here since it will be used in the proof of Theorem 6.28. Proposition 5.26 Let X be a finite-dimensional Banach space. Let K be a compact space such that X ⊂ C(K ). Then λ(X ) = λ(X, C(K )).

5.1

Extension of Operators and Lifting

249

Proof: Fix ε > 0. By Proposition 5.24, there exists a finite-dimensional Banach space G such that X ⊂ G ⊂ C(K ) and d(G, n∞ ) < 1 + ε. Let S : G → n∞ be an isomorphism such that S.S −1  ≤ 1 + ε. Given an arbitrary projection P from G onto X , the mapping P := S ◦ P ◦ S −1 is a projection from n∞ onto X := S(X ), and P  ≤ S.P.S −1  ≤ (1 + ε)P. Then λ(X , n∞ ) ≤ P  ≤ (1 + ε)P. This holds for all projections P from G onto X , so (1 + ε)−1 λ(X , n∞ ) ≤ λ(X, G). By Proposition 5.16 we know that λ(X ) = λ(X , n∞ ). Moreover, if P is a projection from C(K ) onto X , then P G is a projection from G onto X , and   P G  ≤ P; this proves that λ(X, G) ≤ λ(X, C(K )). Finally, we get λ(X, C(K )) ≥ λ(X, G) ≥ (1+ε)−1 λ(X , n∞ ) = (1+ε)−1 λ(X ) ≥ (1+ε)−2 λ(X ), where the last inequality follows from Lemma 5.19. Since ε > 0 was arbitrary and, certainly, λ(X, C(K )) ≤ λ(X ), we obtain the conclusion. Theorem 5.27 (Michael [Mich]) Assume that C is a closed convex subset of a Banach space X . Then there is a continuous mapping from X into C that is the identity on C. Proof: Define a set-valued mapping Φ from X into the subsets of C by  Φ(x) =

x, if x ∈ C, C, if x ∈ C.

Since C is closed, the mapping Φ is lower semicontinuous (see Definition 17.6). Indeed, if x 0 ∈ X and G is an open subset of X that intersects Φ(x 0 ), the set O := {x ∈ X : Φ(x) ∩ G = ∅} is open, since O = (X \C) ∪ G. By Michael’s selection theorem (Theorem 7.53), Φ admits a continuous selection, and this is the desired mapping. Corollary 5.28 For every closed subspace Y of a Banach space X there is a (nonlinear) continuous projection from X onto Y , i.e., there is a continuous mapping ϕ : X → Y such that ϕ in the identity mapping on Y . Lindenstrauss’ result on uniformly continuous (non-linear) projections will be discussed in Chapter 12. Let S1 ⊂ S2 be two topological spaces. A function (in general, non-linear) T : C(S1 ) → C(S2 ) is called an extension operator if, for every f ∈ C(S1 ), T ( f ) is an extension of f to S2 , i.e., T ( f )(s1 ) = f (s1 ) for all s1 ∈ S1 .

250

5 Structure of Banach Spaces

Theorem 5.29 (Arens, Borsuk [Bors], Dugundji [Dugu1], Kakutani [Kaku2]) Let K be a compact metric space and H be a closed subset of K . Then there is a linear isometry extension operator T from C(H ) onto a norm-one complemented subspace of C(K ) such that T 1 = 1. In particular, C(H ) is linearly isometric to a 1-complemented subspace of C(K ). Proof: The space C(H ) is separable and thus BC(H )∗ in its w ∗ -topology is affinely homeomorphic to a compact convex subset of 2 in its norm topology (see the remark after Definition 12.24 and Exercise 12.20), and so is the set P ⊂ BC(H )∗ of probability measures on H . Define a set-valued mapping Φ from K into P for every k ∈ K by  Φ(k) =

δk , if k ∈ H, P, if k ∈ K \H,

where δk ∈ C(H )∗ is the Dirac measure at k in the compact set H . Then Φ is lower semicontinuous and Φ(k) is closed and convex for every k ∈ K . Thus, by Michael’s selection Theorem 7.53, the mapping Φ has a continuous selection ϕ from K into P. Given f ∈ C(H ), define T f (k) = f (ϕ(k)) for k ∈ K . If kn ∈ K , and kn → k in K , then kn → k in the w ∗ -topology of C(K )∗ , hence ϕ(kn ) → ϕ(k) in the w∗ -topology of C(H )∗ , and thus f (ϕ(kn )) → f (ϕ(k)) as f ∈ C(H ). This proves that T is a (linear) mapping from C(H ) into C(K ). Moreover, T f (h) = f (h) for all h ∈ H , T 1 = 1 and T is a linear isometry. If R denotes the operator of restriction from K to H , then the operator Q = T (R) is a norm-one linear operator from C(K ) into C(K ). Moreover, if f ∈ C(K ) and g = R( f ) then Q(Q( f )) = T R(T R( f )) = T R(T g) = T (g) = T (R f ) = Q( f ). Therefore Q is a linear projection. Remark: We refer to [Pelc5] for more in this area.

5.2 Weak Injectivity We say that a separable topological space (X, T ) is a Polish space, if its topology T is generated by a complete metric. So, by a Polish space we will simply mean a separable complete metric space (see Section 17.9). We denote by Δ the Cantor set. Lemma 5.30 Let (P, ρ) be a Polish space and φ : P → Δ a continuous and onto mapping. Then there exists a subset S of P such that S is homeomorphic to Δ and  φ  S is a homeomorphism onto φ(S).

5.2

Weak Injectivity

251

Proof: We will construct by induction a sequence {Δk }k∈N of relatively open subsets of P, with the following properties.

k 2k is a family of pairwise (i) For k ∈ N we have Δk = 2i=1 Δik , where {Δik }i=1 disjoint relatively open and nonempty subsets of P. 1 2i i i k (ii) Δ2i−1 k+1 ∪ Δk+1 ⊂ Δk and ρ-diam Δk < k , for 1 ≤ i ≤ 2 and k ∈ N. 2i i k (iii) φ(Δ2i−1 k+1 ) ∩ φ(Δk+1 ) = ∅ and all φ(Δk+1 ) are uncountable, for 1 ≤ i ≤ 2 and k ∈ N. Let us describe the inductive step (the initial step is similar). Having constructed {Δl }l≤k , and given i, choose a countable base {B j } j∈N of the open set Δik consisting

1 . Clearly, φ(Δik ) = j∈N φ(B j ) so there exists of sets of ρ-diameter at most k+1 some j ∈ N such that φ(B j ) is uncountable. There exist points t = r , t, r ∈ φ(B j ), such that every open neighborhood of either t, r in φ(B j ) is uncountable. Fix two ∩ φ −1 (U ) disjoint open neighborhoods U of t and V of r , and set Δ2i−1 k+1 = B j −1 and Δ2i that S := ∞ k=1 Δk is k+1 = B j ∩ φ (V ). The conditions (i)–(iii) guarantee   homeomorphic to the Cantor set Δ , and moreover φ S is injective.

The following result is sometimes called the weak injectivity theorem. Theorem 5.31 (Pełczy´nski [Pelc6]) Let X be a separable Banach space containing a subspace Y that is isomorphic to C[0, 1]. Then Y has a subspace Z that is isomorphic to C[0, 1] and complemented in X . Proof: Let Δ be the Cantor set. By Theorem 5.29, the subspace Y contains a further subspace Y0 that is isomorphic to C(Δ). There is an equivalent norm | · |Y0 on Y0 such that (Y0 , | · |Y0 ) is isometric to C(Δ). Use Proposition 2.14 to define an equivalent norm | · | X on X that induces | · |Y0 on Y0 . The (separable) space (X, | · | X ) is, by Theorem 5.8, isometric to a subspace of C[0, 1], so there is no loss of generality assuming that X is actually isometric to C[0, 1] (and that the subspace Y0 is isometric to C(Δ)). So T ∗ (B X ∗ ) → BY0∗ is onto, and by Proposition 3.67, extBY0∗ ⊂ T ∗ (extB X ∗ ). We have extB X ∗ = {±δt : t ∈ [0, 1]}, extBY0∗ = {±δt : t ∈ Δ}, where δt is the Dirac functional at a point t. Use a simple argument and Lemma 5.30 to obtain some S ⊂ [0, 1], homeomorphic to S := T ∗ (S) ⊂ Δ. Δ, such that T ∗  S → Δ is a homeomorphism onto the image  Let Φ : C(S) → C( S) be an isometry onto resulting from the action of T ∗ . By the Borsuk–Dugundji Theorem 5.29 there exists an isometric extension operator Z ). It is E : C( S) → C(Δ), with the image  Z := E(C( S)) → Y0 . Let Z = T(  standard to check that P : C[0, 1] → Z defined as P( f ) = T ◦ E ◦ Φ( f  S ), is the sought projection onto Z . Corollary 5.32 Let X be a separable Banach space. If C[0, 1] is isomorphic to a quotient of a subspace of X , then it is isomorphic to a quotient of X . Proof: Let Y be a subspace of X such that C[0, 1] is isomorphic to a quotient Z of Y . Let T : Y → Z be a quotient mapping. The space Z is isometric to a subspace of

252

5 Structure of Banach Spaces

: ∞ . Since ∞ is an injective space (see Proposition 5.10), there is an extension T (X ) is a separable space with Z ⊂ E. By Theorem 5.31 X → ∞ of T , and E := T C[0, 1], that is complemented in E, say by a projection P. there is Z 1 → Z , Z 1 ∼ =  : X → Z 1 is the sought quotient mapping. Then P ◦ T Corollary 5.33 (Pełczy´nski [Pelc6]) Let X be a separable Banach space. The following are equivalent. (i) C[0, 1] is isomorphic to a quotient of X . (ii) 1 → X . Proof: (i)⇒(ii): By Theorem 5.8, 1 is isometric to a subspace of C[0, 1]. By the lifting property of 1 (see Proposition 5.2) we obtain (ii). (ii)⇒(i): By Theorem 5.1, the space C[0, 1] is isometric to a quotient of 1 . By Corollary 5.32, C[0, 1] is isomorphic to a quotient of X . The following result is a particular case of Milyutin theorem (see, e.g., [Rose10]). Corollary 5.34 If Δ denotes the Cantor ternary set, the spaces C[0, 1] and C(Δ) are linearly isomorphic. Proof: Since Δ is a compact subset of [0, 1], Theorem 5.29 shows that C(Δ) is isomorphic to a complemented subspace of C[0, 1]. On the other hand, there is a continuous onto function φ : Δ → [0, 1]. This implies that C[0, 1] is linearly isometric to a subspace Y of the separable Banach space C(Δ). Theorem 5.31 implies that Y has a further subspace Z that is isomorphic to C[0, 1] and complemented in C(Δ). It is enough now to apply Pełczy´nski’s decomposition method (Theorem 4.47) to get the conclusion.

5.2.1 Schur Property ∞ is Recall that a sequence {xn } ⊂ X is weakly Cauchy if and only if { f (x n )}i=1 Cauchy (that is, convergent) for every f ∈ X ∗ .

Definition 5.35 A Banach space X is called weakly sequentially complete if every weakly Cauchy sequence in X is weakly convergent to some point in X . Note that there are Banach spaces which nare not∞weakly sequentially complete. An example is c0 ; indeed, the sequence { i=1 ei }n=1 is weakly Cauchy, although is not weakly convergent in c0 . The space 1 has a stronger property than being weakly sequentially complete. This is the content of the following result. Theorem 5.36 (Schur) Let {x n } be a sequence in 1 . If {x n } is weakly Cauchy then {x n } is norm convergent in 1 .

Rosenthal’s 1 Theorem

5.3

253 w

Proof: First we show that if x n → 0 in 1 , then x n → 0. By contradiction, assume that for some 0 there is an increasing  sequence n 1 < n 2 < . . . such that ∞ ε > n ∞ |xi j | > ε. Choose N1 so that i=N |xin 1 | < 5ε . x n j  = i=1 1 +1  N1 n 1  N1 n 1 n 1 n Then i=1 |xi | ≥ 45 ε, which is equivalent to i=1 εi xi ≥ 45 ε, where εi 1 = n1 sign(xi ) for i = 1, . . . , N1 . Note that if we choose an arbitrary sequence of signs {εi = ±1} such that εi = εin 1 for i ≤ N1 , we have N1 N1 ∞ ∞ ∞         1 3 4       εi xin 1  =  εin 1 x in 1 + εi xi  ≥  εin 1 xin 1 − |xi | ≥ ε− ε = ε.  5 5 5 i=1

i=N1 +1

i=1

i=N1 +1

i=1

 N1 n j2 Next, find n j2 such that i=1 |x | < ε5 . This is possible since xin → 0. Then we ∞i  N2 n j2 nj choose N2 > N1 such that i=N2 +1 |xi 2 | ≤ ε5 and consequently i=1 |xi | ≥ n1 4 5 ε. Then for arbitrary choice of signs {εi = ±1} satisfying εi = εi for i ≤ N1 and n εi = εi 2 for N1 < i ≤ N2 we have N2 N1 ∞ ∞        n j2  n j2  nj nj   εi xi  ≥  εi xi  − |xi 2 | − |xi 2 |  i=N1 +1

i=1

i=N2 +1

i=1

N2 N2 N1    2   2 4 ε 2 ε nj  nj nj  εi xi 2  − ε = |xi 2 | − |xi 2 | − ε ≥ ε − − ε = . ≥ 5 5 5 5 5 5 i=N1 +1

i=1

i=1

nk Repeating this process, we obtain a vector u = (u  i )n ∈ ∞ εdefined by u i = εi jk for Nk−1 < i ≤ Nk , which has the property that u x ≥ 5 for all k. This is a w

contradiction with x n → 0. For a weakly Cauchy sequence we proceed in an analogous way: If {x n } is weakly Cauchy and not norm Cauchy, there exists ε > 0 and indices n j , m j → ∞ such that x n j − x m j  ≥ ε. Then consider the sequence {x n j − x m j } j , which converges weakly to zero. We say that a Banach space X has the Schur property if every weakly convergent sequence is norm convergent. We have just shown that 1 has this property. The same is true for 1 (Γ ) for any nonempty set Γ . Observe, too, that the technique of proof of Theorem 5.36 is another instance of what has been called the “sliding hump argument.”

5.3 Rosenthal’s 1 Theorem In the proof of Theorem 5.37, we shall use the following notation: The family of all infinite subsets of a given infinite set S is denoted P∞ (S), while the family of all finite subsets (including the empty set) of a set M is denoted P f (M). Given Z ∈ P∞ (N), we will always consider Z as an increasing sequence Z := {s1 , s2 , . . .}.

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5 Structure of Banach Spaces

Put then Z 1 = {s1 , s3 , s5 , . . .} and Z 2 = {s2 , s4 , s6 , . . .}. Given a sequence {x n } in a Banach space, and ε > 0, let Sε denote the set of all infinite subsets A of N such that there exists f ∈ B X ∗ with the property inf f (xn ) ≥ sup f (x n ) + 2ε.

n∈A2

n∈A1

Theorem 5.37 (Rosenthal [Rose6]) Let {xn }∞ n=1 be a bounded sequence in an infinite-dimensional Banach space X . Then there is a subsequence {xn k }∞ k=1 that is either weakly Cauchy or it is equivalent to the unit basis of 1 . Proof: For simplicity, we give the proof for the case of real scalars. Let us start with a simple but important observation. The 1 -criterion: Let X be a Banach space, Γ be a set, {xi }i∈Γ ⊂ B X . Then {xi }i∈Γ is equivalent to the unit basis of 1 (Γ ) if and only if there is ε > 0 such that, for every disjoint sets A, B ⊂ Γ , there exists f ∈ B X ∗ such that infi∈A f (xi ) ≥ ε and supi∈B f (xi ) ≤ −ε.   Necessity is clear. To prove sufficiency, let x = i∈Γ ai xi , i∈Γ |ai | = 1, and let A = {i : ai ≥ 0}, B = Γ \A. Choose f ∈ B X ∗ such that infi∈A f (xi ) ≥ ε, supi∈B f (xi ) ≤ −ε. Then x ≥ f (x) =

 i∈Γ

ai f (xi ) =

 i∈A

ai f (xi ) +



ai f (xi ) ≥ ε.

i∈B

This proves the criterion. Suppose that supn xn  < K , for some K ≥ 1. Claim 1: A bounded sequence {xn }∞ n=1 contains a subsequence equivalent to the 1 -basis if and only if there exists some infinite set M ⊂ N and ε > 0 such that B ∈ Sε for every infinite set B ⊂ M. Formally     ∃M ∈ P∞ (N) (∃ε > 0) ∀B ∈ P∞ (M) (B ∈ Sε )

(5.4)

The proof of necessity is immediate. To prove sufficiency, apply (5.4) first to B := M. We get f 0 ∈ B X ∗ such that infm∈M2 f 0 (xm ) ≥ supm∈M1 f 0 (xm ) + 2ε. By changing f 0 to − f 0 if necessary, we may assume that for some M ∈ P∞ (M) we have infm∈M f 0 (xm ) ≥ ε. Let l ∈ [ε, K ] be a cluster point of { f 0 (xm ) : m ∈ M }. Put g = (ε/2l) f 0 . It follows that {g(x m ) : m ∈ M } clusters to ε/2, and we can take M ∈ P∞ (M ) such that |g(x m )−ε/2| < ε2 /(4K ) for all m ∈ M . Put M = M1 . We shall prove, by using the criterion above, that {x n : n ∈ M } is equivalent to the 1 -basis. To this end, let A and B be two mutually disjoint elements in P(M ). We can always construct an infinite subset C of M such that A ⊂ C1 and B ⊂ C2 . By using (5.4) again, we can find f ∈ B X ∗ such that infm∈C2 f (x m ) ≥ supm∈C1 f (xm ) + 2ε, so, in particular, β := infm∈B f (x m ) ≥ supm∈A f (x m ) + 2ε =: α + 2ε. Take λ := −(α + β)/ε. It is easy to verify that the function h 0 := f + λg satisfies the following properties.

5.3

Rosenthal’s 1 Theorem

(i) h 0  ≤ 1 + K /(2ε). (ii) infm∈B h 0 (xm ) ≥ ε/2, supm∈A h 0 (xm ) < −ε/2.  −1 Hence, the function h := 1 + K /(2ε) h 0 satisfies (i) h ∈ B X ∗ , (ii) infm∈B h(xm ) ≥ ε 2 /(2ε + K ), supm∈A h(xm ) < −ε 2 /(2ε + K ). The conclusion follows by using the 1 -criterion above.

255



Claim 2: A bounded sequence {x n }∞ n=1 contains a weakly Cauchy subsequence if and only if there exists some infinite set M ⊂ N such that for every infinite set L ⊂ M / Sε for every and for every ε > 0 there exists an infinite set Aε,L ⊂ L, such that B ∈ infinite set B ⊂ A ε,L . Using the logical symbols, this means that (∃M ∈ P∞ (N))(∀ε > 0)(∀L ∈ P∞ (M)) (∃Aε,L ∈ P∞ (L))(∀B ∈ P∞ (Aε,L ))(B ∈ / Sε ).

(5.5)

Necessity is immediate as the “natural” formulation of the existence of a weakly Cauchy subsequence is / Sε ). (∃M ∈ P∞ (N))(∀ε > 0)(∀L ∈ P∞ (M))(L ∈

(5.6)

To prove sufficiency, construct a nested sequence of infinite sets {An }∞ n=1 , A0 := M, An+1 := A 1 ,An for n ∈ N by using (5.5) repeatedly. Then, for n ∈ N and every n+1 ∞ be a diagonal increasing sequence, m ∈ A B ∈ P∞ (An ), B ∈ S1/n . Let {m i }i=1 i i ∞ is weakly Cauchy, which establishes for i ∈ N. It is clear (use (5.6) that {xm i }i=1 the claim.  We proceed with the proof of the theorem. If {xn }∞ n=1 contains no weakly Cauchy subsequence, then by negating (5.5) we obtain (∀M ∈ P∞ (N ))(∃ε > 0)(∃L ∈ P∞ (M)) (∀A ∈ P∞ (L))(∃B ∈ P∞ (A))(B ∈ Sε ).

(5.7)

For notational convenience assume that L = N in (5.7). So there exists ε > 0, to be fixed from now on, such that for every infinite A ⊂ N there exists an infinite B ⊂ A, B ∈ S := Sε , i.e., (note that we have achieved a favorable change of the order of quantifiers compared to the negation of (5.6)) (∃ε > 0)(∀A ∈ P∞ (N ))(∃B ∈ P∞ (A))(B ∈ Sε ).

(5.8)

The strategy of the proof is to build inductively the “initial finite segments” of a sequence M that is ultimately going to satisfy (5.4). Heuristically, we are building an almost basis of 1 (although our condition is weaker and to get an 1 basis we need to pass to a subsequence, at the end). The main point here is to make sure that the initial finite sequence can be extended arbitrarily many times. This is achieved by controlling simultaneously a candidate set of tails for the future extensions. More

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5 Structure of Banach Spaces

precisely, we shall prove that as soon as we have a finite set F ⊂ N and an infinite set I ⊂ N with max F < min I and satisfying (∀J ∈ P∞ (I ))(∀G ∈ P(F))(∃L ∈ P∞ (J ))(G ∪ L ∈ S),

(5.9)

then we can enlarge F to a finite superset and reduce I to an infinite subset in such a way that (5.9) still holds for those two new sets. Note that the couple (F := ∅, I := N) satisfies (5.9), due to (5.8). Assume now that we have a couple (F, I ) that satisfies (5.9). To produce the new couple, we shall prove the following Claim: (∃n ∈ I )(∃J ∈ P∞ (I ), n < minJ ) (∀L ∈ P∞ (J ))(∀G ∈ P(F ∪ {n}))(∃M ∈ P∞ (L))(G ∪ M ∈ S) (5.10) Proceeding by contradiction, suppose that the Claim does not hold. This means (∀n ∈ I )(∀J ∈ P∞ (I ), n < minJ ) (∃L n,J ∈ P∞ (J )) (∃G n,J ∈ P(F ∪ {n})) (∀M ∈ P∞ (L n,J ))(G n,J ∪ M ∈ /(5.11) S) By applying (5.9), note that n ∈ G n,J for all n ∈ I . Using a simple inductive ∞ , a sequence {G }∞ of argument, there exists an increasing sequence J := {n i }i=1 n i i=1 ∞ finite sets, and a sequence of infinite sets {L i }i=1 such that n 1 = minI, G n 1 = G n 1 ,I \{n1 } , L 1 = L n 1 ,I \{n 1 } ,

(5.12)

and, for i = 2, 3, . . ., n i = minL i−1 , G n i = G n i , L i−1 \{ni } , L i = L n i ,L i−1 \{ni } ,

(5.13)

such that for i ∈ N and for all M ∈ P∞ (L i ), we have G ni ∪ M ∈ S. By passing to an infinite subsequence of J , we may, without loss of generality, assume that G n \{n} = G is independent of the choice of n ∈ J . According to (5.9), there exists an infinite subset L ⊂ J such that G ∪ L ∈ S. Put nl = min L. On the other hand, G ∪ L = (G ∪ {nl }) ∪ (L\{nl }) = G n l ∪ (L\{nl }), and L\{nl } ∈ P∞ (L l ), so G ∪ L ∈ S. This is a contradiction and proves (5.10).  ∞ , and an The inductive procedure leads to an increasing sequence M := {n i }i=1 ∞ auxiliary decreasing sequence of infinite sets {In i }i=1 , such that, for all k ∈ N, the sets F := {n 1 , . . . , n k } and I := In k satisfy (5.9), and n k+1 ∈ In k . Given any finite set F ⊂ M and an infinite set I ⊂ M such that maxF < minI , we have that there exists some infinite set L ⊂ I such that F ∪ L ∈ S, i.e.,

(∃M ∈ P∞ (N))(∀F ∈ P f (M)) (∀I ∈ P∞ (M), maxF < minI )(∃L ∈ P∞ (I ))(F ∪ L ∈ S)

(5.14)

5.3

Rosenthal’s 1 Theorem

257

∞ ⊂ M be any infinite subset written as an increasing sequence. Let A = {ai }i=1 For each n ∈ N there exists some L n ∈ P∞ (N) such that max{a1 , . . . , a2n } < min L n and {a1 , . . . , a2n } ∪ L n ∈ S; in particular, there exists gn ∈ B X ∗ such that infi∈{1,...,n} gn (xa2i ) ≥ supi∈{1,...,n} gn (xa2i−1 ) + 2ε. Let g be a w ∗ -cluster point of {gn }∞ n=1 . Clearly, infi∈N g(xa2i ) ≥ supi∈N g(xa2i−1 ) + 2ε, so A ∈ S. The proof is finished by an appeal to Claim 1.

We are going now to investigate further properties of Banach spaces related to 1 . As we shall see later, it will be natural to work in the setting of Baire 1 functions on Polish spaces. If X is a topological space, we denote by T p the topology of pointwise convergence in the space R X of all real-valued functions on X . Definition 5.38 Let X be a metric space, f be a real function on X . We say that f is a Baire 1 class function if f is a T p -limit of a sequence of elements from C(X ). By B1 (X ) we denote the space of all Baire 1 real functions, equipped with the topology Tp. Remark: The usual definition of a Baire 1 mapping f : X → Y is that the inverse image of an open set is an Fσ set. This turns out to be equivalent to our definition whenever X is a metric space and Y = R (see, e.g., [DGZ3, p. 18]). If X is a separable Banach space, it is natural to consider the Polish space P := (B X ∗ , w∗ ), and the embedding X ∗∗ ⊂ R P , where the T p -topology corresponds to the w ∗ -topology on X ∗∗ . Tp

Lemma 5.39 Let (P, ρ) be a Polish space, F be a subset of R P , and f ∈ F . If f is not a Baire 1 function, then there is a sequence { f k }∞ k=1 in F that has no T p -convergent subsequence. Proof: By Baire’s Great Theorem 17.12, there exists a closed and nonempty set  K ⊂ P such that f  K has no point of continuity. For each n ∈ N and m ∈ Z, we define K n,m ⊂ K by putting r ∈ K n,m if and only if for every neighborhood U of r in K , there exist t, s ∈ U such that f (t) ≥ m+3 n , m f (s) ≤ n . Clearly, K n,m are closed sets and K = n∈N,m∈Z K n,m . By the Baire category theorem some K n,m has a nonempty interior. For simplicity of notation let K = K n,m . We are going to construct inductively a pair of sequences { f k }∞ k=1 from F and {Dk }∞ k=1 of relatively open subsets of K , with the following properties.

2k Dki , where {Dki }i is a collection of pairwise disjoint relatively (i) Dk = i=1

2k−1 2i 0 = i=1 open and nonempty subsets of K . For convenience we also denote D Dk , k

2k−1 2i−1 1 = i=1 D . D k k 2i−1 2i (ii) Dk+1 ∪ Dk+1 ⊂ Dki , ρ-diam Dki < 1k . m+1 m+2 (iii) sup D 1 f k ≥ n . 0 f k ≤ n and inf D k k Let us describe the inductive step (the initial step is similar). Having constructed { f k }k≤l and {Dk }k≤l , we find points {tli : 1 ≤ i ≤ 2l+1 }, such that tl2i ∈ Dli , tl2i−1 ∈

258

5 Structure of Banach Spaces

Dli , and f (tli ) ≤ mn for i even and f (tli ) ≥ m+3 n for i odd. There exists fl+1 ∈ F m+2 i for i even and f such that fl+1 (tli ) < m+1 l+1 (tl ) > n n for i odd. It remains i to fix some small enough pairwise disjoint neighborhoods Dl+1 of the points tli ,

l+1 2 i i = 1, . . . 2l+1 , so that (i)–(iii) will be satisfied for Dl+1 = i=1 Dl+1 . Note that ∞ i(k) condition (ii) implies that k=1 Dk = ∅ if and only if i(k +1) ∈ {2i(k), 2i(k)−1} ∞ for all k ∈ N. To finish the proof, let { f k( j) }∞ j=1 be any subsequence of { f k }k=1 . ∞ Choose a sequence {i(k)}k=1 , satisfying i(k + 1) ∈ {2i(k), 2i(k) − 1} for all k ∈ N, and moreover such that i(k( j)) is even if and only if j is even. There exists some  i(k) m+1 m+2 t∈ ∞ k=1 Dk . We have now f k( j) (t) ≤ n for all j even, while f k( j) (t) ≥ n ∞ for all j odd. This implies that { f k( j) } j=1 is not T p -convergent and the proof is finished. Theorem 5.40 (Odell and Rosenthal [OdRo], Rosenthal [Rose7]) Let X be a separable infinite-dimensional Banach space. The following are equivalent. (i) X does not contain an isomorphic copy of 1 . (ii) Every element of B X ∗∗ is the w∗ -limit of a sequence in B X . (iii) card(X ∗∗ ) = card(X ) = c. Proof: (ii)⇒(iii): Choose a norm-dense sequence {x n }∞ n=1 in B X . Since each element of B X ∗∗ is the w∗ -limit of a subsequence of {x n }∞ n=1 , the conclusion follows. (iii)⇒(i): By contradiction, assume that 1 → X . Then ∗∞ ⊂ X ∗∗ . It is known that 1 (c) → ∞ (Exercise 5.34), and thus ∗∞ has ∞ (c) as a quotient. This gives the conclusion. (i)⇒(ii): We proceed by contradiction, assuming that x ∗∗ ∈ B X ∗∗ is not a w ∗ limit of any sequence from B X . By [DGZ3, Lemma 3.4, p. 112], this implies that x ∗∗ is not a Baire 1 function. We apply Lemma 5.39 to P = (B X ∗ , w∗ ), F = B X ⊂ B X ∗ , ∞ contains no T -convergent subsequence, which f = x ∗∗ . Then the sequence { f i }i=1 p is equivalent to saying that it has no weakly Cauchy subsequence. By Rosenthal’s Theorem 5.37, we obtain the conclusion. Theorem 5.41 (Rosenthal [Rose7]) Let (P, ρ) be a Polish space, and F be a subset of B1 (P). The following are equivalent in the T p -topology. (i) F is relatively compact. (ii) F is relatively countably compact. (iii) F is relatively sequentially compact. Proof: (i)⇒(ii) is immediate. We proceed with the proof that (ii)⇒(iii). Lemma 5.42 Let (P, ρ) be a Polish space and { f k }∞ k=1 a pointwise bounded sequence from R P that has no T p -convergent subsequence. Then there is an infinite subset M of N, r ∈ R, δ > 0 such that for every infinite set A ⊂ M, there exists t A ∈ P such that lim inf f ai (t A ) < r, lim sup f ai (t A ) > r + δ. i∈N

i∈N

(5.15)

5.3

Rosenthal’s 1 Theorem

259

∞ be an enumeration of all rational pairs with δ > 0. Proceed Proof: Let {(ri , δi )}i=1 i ∞ by contradiction. Using induction, we are going to construct a sequence {Mi }i=1 of infinite subsets of the integers. M0 = N, and Mi+1 ⊂ Mi is chosen so that letting M = Mi , A = Mi+1 , r = ri and δ = δi , (5.15) fails for every t ∈ P. Let ∞ be the diagonal set for the sequence {M }∞ . We obtain that for every M = {m i }i=1 i i=1 ∞ is convergent, a contradiction. t ∈ P, { f m i (t)}i=1

Lemma 5.43 Let (P, ρ) be a Polish space and { f k }∞ k=1 be a pointwise bounded P sequence from R , that has no T p -convergent subsequence. Then there is an infinite M ⊂ N such that { f k }k∈M has no cluster point in B1 (P). Proof: By Lemma 5.42 there exists M ⊂ N, r, δ such that for every A ⊂ M there exists t ∈ P such that f m (t) > r + δ and f m (t) < r happens infinitely often for m ∈ A. For every A ⊂ M, denote K (A) = {t A : satisfies (5.15)}. Lemma 5.44 There exists some infinite set N such that K (N ) = K (L) (=: K ) for all infinite L ⊂ N , where K (·) is defined in the proof of Lemma 5.43. Proof: Clearly, L ⊂ M implies ∅ = K (L) ⊂ K (M). If for every L there exists N ⊂ L with K (N )  K (L), then we can construct an ordinal sequence {L α }α<ω1 , such that β > α implies that L β \L α is finite and K (L β )  K (L α ). Indeed, having ∞ , and put obtained L α for all α < β < ω1 , reindex the set {L α }α<β as { L i }i=1  ∞  L β = {ai }i=1 , where ai ∈  j≤i L j . This contradicts the Lindelöf property of the separable metric space P \ α<ω1 K (L α ). ∞ from K and infinite sets {M }∞ , We proceed by constructing sequences {ti }i=1 i i=1 with the properties: 1. N = M1 ⊃ M2 . . . 2. f i (t j ) > r + ρ for all j even and i ∈ M j , f i (t j ) < r for all j odd and i ∈ M j , ∞ and {t ∞ 3. {t2i }i=1 2i+1 }i=1 are both dense in K . ∞ . It remains to see that Denote by M the diagonal set of the system {Mi }i=1

T

p { f i }i∈M has no Baire 1 T p -cluster point. Assuming the contrary, let f ∈ { f i }i∈M . By Baire’s Great Theorem 17.12, f has a point of continuity t ∈ K . We have f (t j ) ≥ r + ρ for every j even, and f (t j ) ≤ r for all j odd. We have obtained a contradiction. This finishes the proof of Lemma 5.43 and the proof of (ii)⇒(iii).

(iii)⇒(i) follows from Lemma 5.39. This finishes the proof of Theorem 5.41. The next abstract lemma will be useful in what follows. Lemma 5.45 Let (V, ⊆) be a system of nonempty closed subsets of a Polish space (P, ρ), partially ordered by inclusion, with the following properties: 1. For every pair A, B ∈ V there exists some C ∈ V, such that C ⊂ A ∩ B. ∞ 2.  For every sequence {Vn }n=1 ⊂ V, Vn+1 ⊂ Vn there exists a V ∈ V, V ⊂ V . n n Then V has a minimal element.

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5 Structure of Banach Spaces

Proof: P has a countable base {Un }n∈N . If for all V ∈ V we have V = P there is nothing to prove. We may assume then that V = P for all V ∈ V. Let M = {n ∈ N : Un ∩ V = ∅ for some V ∈ V}. Given n ∈ M, let V n ∈ V such that Un ∩ Vn = ∅. By  ∈ V such that V  ⊂ n∈M Vn . We claim that V  is the the assumption, there exists V sought minimal element. Indeed, assume that, on the contrary, there is V ∈ V such  \ V ) = ∅. This implies . Certainly, there exists n ∈ N such that Un ∩ (V that V  V  \V ) ⊂ V  (⊂ Vn ), so that n ∈ M. Then we have Un ∩ (V \V ) ⊂ Un , and Un ∩ (V Un ∩ Vn = ∅, a contradiction. Theorem 5.46 (Rosenthal [Rose7]) Let (P, ρ) be a Polish space, F be a T p relatively compact subset of B1 (P). Then F has the property that every f ∈ F lies in the T p -closure of a countable subset of F.

Tp

Proof: We start with an auxiliary lemma. T

Lemma 5.47 Let S be a T p -relatively compact subset of B1 (P), so that 0 ∈ S p . Then for every δ > 0 there is a countable subset H ⊂ S so that infh∈H h(t) < δ for all t ∈ P. Proof: By contradiction, there exists δ > 0 such that for every countable subset H ⊂ S it holds infh∈H h(t) ≥ δ for some t ∈ P. We define V (H ) = {t ∈ P : inf f ∈H f (t) ≥ δ}, where H is a countable subset of S. Define V = {V (H ) : H ⊂ S, H countable} a system of closed subsets of P. By Lemma 5.45, there exists in V a nonempty minimal element V P = V (H P ). Choose a dense set ∞ ⊂ V , and find a sequence { f }∞ ∈ S such that lim f (t ) = 0 for all {ti }i=1 P k k=1 k i k→∞

i ∈ N. By Theorem 5.41, S is T p -sequentially compact, so we may, without loss of generality, assume that { f k }∞ k=1 is T p -convergent to a Baire 1 function f . Thus f restricted to V P has a point of continuity t ∈ V P by Baire’s Great Theorem 17.12, and necessarily f (t) = 0. This however implies that V (H P ∪ { f k }∞ k=1 )  V P , a contradiction. To finish the proof of the theorem, suppose that F is T p -relatively compact, and for each m ∈ N define the mapping φm : B1 (P) → B1 (P m ) by m T φm ( f )(t1 , . . . , tm ) = i=1 | f (ti )|. Let g ∈ F p . We may, without loss of generality, assume that g = 0. Hence Sm = φm (F) are T p -relatively compact in B1 (P m ), T

with 0 ∈ Sm p . Using Lemma 5.47, there is a countable subset Hm of F so that T 1 inf{φm (h)(y) : h ∈ Hm } for all y ∈ P m . It follows that 0 ∈ H p , where m > ∞ H = m=1 Hm is a countable subset of F. We need the following topological lemma. Lemma 5.48 Let (X , T ) be a regular topological space, that is sequentially com∞ be a sequence in X and {I }∞ be a pact and countably tight. Let x ∈ X , {xi }i=1 n n=1 T

decreasing sequence of infinite subsets of N, such that x ∈ {xi }i∈In for all n ∈ N.

5.3

Rosenthal’s 1 Theorem

261

Then there is an infinite set I ⊂ N such that I \In is finite for all n ∈ N, and T x ∈ {xi }i∈I . Proof: Let F = {lim xi : I infinite, (∀n ∈ N) card(I \In ) < ∞, lim xi exists}. i∈I

i∈I

(5.16)

Let U be an open neighborhood of x. Then J := {i : xi ∈ U } intersects every In in an infinite set. As In is decreasing, there is some K ⊂ J such that K \In is finite for all n ∈ N. As X is sequentially compact, there is some I ⊂ K such that z I = lim xi ∈ U exists. Since X is regular and U was arbitrary, we have i∈I

x ∈ F. Since X is countably tight, there exists a sequence {z m }∞ m=1 ∈ F such that T

x ∈ {z m }m∈N . Using that z m = z Jm = lim xi , where Jm \In is finite for all n ∈ N, let i∈Jm

I = ∞ n=1 (In ∩ Jn ). Because In is decreasing, we have that I \In and Jn \I are finite T

T

for every n ∈ N. It follows that z m ∈ {xi }i∈I for all m ∈ N. Therefore x ∈ {xi }i∈I , as requested. Theorem 5.49 (Bourgain, Fremlin, Talagrand [BFT]) Let (P, ρ) be a Polish space. Then B1 (P) is T p -angelic. Proof: In view of Theorems 5.41 and 5.46, it remains to show that the closure of a countable and relatively compact set in B1 (P) coincides with its sequential closure. We begin with a variant of Lemma 5.39. Lemma 5.50 Let (P, ρ) be a Polish space, { f n }∞ n=1 be a T p -relatively compact T

sequence in B1 (P) of continuous functions, 0 ∈ { f n }n∈p N . Let W ⊂ P be a nonempty closed subset, ε > 0. Then there is a nonempty relatively open U ⊂ W and an Tp infinite I ⊂ N such that 0 ∈ { fi }i∈I , and moreover lim supi∈I | f i (t)| ≤ ε for all t ∈ U. Proof: Proceeding by contradiction, there exists a closed W ⊂ P and ε > 0 so that for every relatively open U ⊂ W and every infinite I ⊂ N, lim supi∈I | f i (t)| > ε for all t ∈ U . We are going to construct I ⊂ N so that { f i }i∈I has no T p -convergent subsequence. By Theorem 5.41 this will be a contradiction. We start by making a simple observation. Suppose that E i , i = 1, . . . , n be nonempty relatively open subsets of W . Denote Ii = { j : | f j (t)| < ε, for all t ∈ E i }. We have that T

p for all i = 1, . . . , n. Consequently, we have 0∈ / { f j } j∈I i

T

0 ∈ { f j } j∈p N\ n

i=1 Ii

.

(5.17)

We construct a pair of sequences consisting of {gk }∞ k=1 , a subsequence of the origi∞ , consisting of relatively open subsets of W , with the , and {D } nal one { f n }∞ k k=1 n=1 following properties.

262

(i) Dk =

5 Structure of Banach Spaces

2k

Dki , Dki pairwise disjoint relatively open and nonempty subsets

2k−1 2i 1 2k−1 2i−1  = i=1 D 0 = i=1 Dk , D . of K . For convenience, we also denote D k k k i=1

2i−1 2i (ii) Dk+1 ∪ Dk+1 ⊂ Dki , ρ-diam Dki < 1k . ε (iii) sup D 0 gk ≤ 2 and inf D 1 gk ≥ ε. k k Let us describe the inductive step (the initial step is similar). Assume that we have constructed {gk }k≤l and {Dk }k≤l . We use (5.17) to find points {tli : 1 ≤ i ≤ 2l+1 }, ε i such that tl2i ∈ Dli , tl2i−1 ∈ Dli , and a function gl+1 ∈ { f n }∞ n=1 , so that gl+1 (tl ) ≤ 2 for i even and gl+1 (tli ) > ε for i odd. Using the continuity of gl+1 at this point of the proof, it remains to fix some small enough pairwise disjoint neighborhoods i of the points tli , i = 1, . . . 2l+1 , so that (i)–(iii) will be satisfied for Dl+1 = Dl+1

2l+1 i ∞ i(k) = ∅ if and only if i=1 Dl+1 . Note that condition (ii) implies that k=1 Dk i(k + 1) ∈ {2i(k), 2i(k) − 1} for all k ∈ N. To finish the proof, it suffices to show ∞ that {gk }∞ k=1 has no T p -convergent subsequence. Let {gk( j) } j=1 be any subsequence ∞ ∞ of {gk }k=1 . Choose a sequence {i(k)}k=1 , satisfying i(k + 1) ∈ {2i(k), 2i(k) − 1} for all k ∈ N, and moreover such that i(k( j)) is even if and only if j is even.  i(k) ε There exists some t ∈ ∞ k=1 Dk . We have now gk( j) (t) ≤ 2 for all j even, while ∞ gk( j) (t) ≥ ε for all j odd. This implies that {gk( j) } j=1 is not T p -convergent and the proof is finished.

Lemma 5.51 Under the assumptions of Lemma 5.50, for every ε > 0 there is an Tp , and moreover lim supi∈I | f i (t)| ≤ ε for all infinite set I ⊂ N such that 0 ∈ { f i }i∈I t ∈ P. Proof: For each infinite set I ⊂ N put U (I ) = int{t : lim supi∈I | f i (t)| ≤ ε}, and T

p let A(I ) = { f i }i∈I . Note that I \J finite implies that U (J ) ⊂ U (I ). Let {Vk }∞ k=1 be a base of the topology of P, and choose {Ik }∞ , 0 ∈ A(I ) as follows. I k 0 := N. k=1 Given Ik , we choose Ik+1 ⊂ Ik such that either 0 ∈ A(Ik+1 ) and Vk ⊂ U (Ik+1 ), if such a set Ik+1 exists, or else we put Ik+1 = Ik . We obtain a decreasing sequence {Ik }∞ k=1 to which Lemma 5.48 applies. Thus for some I ⊂ N, such that I \Ik is

T

T

p p finite, we have 0 ∈ { f i }i∈I . Let J ⊂ I be infinite and such that 0 ∈ { f i }i∈J . We claim that U (J ) = U (I ). Indeed, assuming the contrary there is some Vk ⊂ U (J ) but Vk  U (I ). Since J \Ik is finite, J ∩ Ik is an infinite subset of Ik such that 0 ∈ A(J ∩ Ik ) and Vk ⊂ U (J ∩ Ik ). Accordingly, we must have chosen Ik+1 such that Vk ⊂ U (Ik+1 ) = U (I ). As I \Ik+1 is finite, we reach a contradiction. It remains to show that U (I ) = P. Assuming the contrary again, let W = X \U (I ) be a nonempty closed set. By Theorems 5.41 and 5.46, the assumptions of Lemma 5.48 are satisfied. Thus there is a nonempty relatively open U ⊂ W and an infinite J ⊂ I , Tp such that 0 ∈ { f i }i∈J and moreover lim supi∈J |xi (t)| ≤ ε for all t ∈ U . Thus U ⊂ U (J )\U (I ) = ∅, a contradiction which finishes the proof.

Repeating the lemma above for a decreasing sequence εn → 0, and choosing a diagonal sequence I ⊂ N, we obtain that lim f i (t) = 0 holds for every t ∈ P. So far i∈I

5.3

Rosenthal’s 1 Theorem

263

we have proved that if the countable set { f n }∞ n=1 consists of continuous functions, and a cluster point f is also a continuous function (passing from a zero function to a general continuous function is trivial), then f can be reached by a T p -convergent subsequence of { f n }∞ n=1 . In order to dispose of the continuity assumption, we use Theorem 17.16. As a consequence, if F ⊂ B1 (P) is a countable T p -relatively compact set with a cluster point f (without loss of generality, f = 0), then there exists a metric ρ  on P, stronger than the original ρ, which turns P into a Polish space (P, ρ ), and moreover all g ∈ F are continuous in the topology coming from ρ . It is immediate )) is T p -relatively compact. This finishes the proof. that F ⊂ B1 ((P, ρ We finish this section by collecting, in Theorem 5.52, some previous results that characterized the absence of copies of 1 in the space in terms of the extreme points of weak∗ -compact convex sets in the dual space. Another characterization, this time in terms of convergence of sequences in the dual space, is given in Theorem 5.53. Theorem 5.52 (Odell and Rosenthal [OdRo], Rosenthal [Rose7]) Let X be a separable Banach space. Then, the following statements are equivalent: (i) X contains no isomorphic copy of 1 . (ii) Every x ∗∗ ∈ B X ∗∗ is the w∗ -limit of a sequence in B X . (iii) Every weak∗ -compact convex set in X ∗ is the norm-closed convex hull of its extreme points. Proof: (i)⇒(ii) is in Theorem 5.40. (ii)⇒(iii): This is Corollary 3.129. (iii)⇒(i): If 1 → X , by Theorem 5.44 C[0, 1] is isomorphic to a quotient of X , hence (BC[0,1]∗ , w ∗ ) is a subspace of (B X ∗ , w ∗ ). Therefore it suffices to show that λ ∈ conv· (Ext(BC[0,1]∗ )), where λ denotes the Lebesgue measure on [0, 1].  To see this, note that if λ is norm-close to a finite convex sum ai δi of Dirac’s measures, we can find f ∈ BC[0,1] such that f (ti ) = 0 and λ( f ) is close to 1. Recall that the Mackey topology μ(X ∗ , X ) on X ∗ associated to the dual pair X, X ∗  is the topology on X ∗ of the uniform convergence on the family of all convex balanced and w-compact subsets of X (see Definition 3.43). The following theorem was motivated by a result of Borwein [Bor]. Theorem 5.53 (Orno [Orn91], Valdivia [Vald6], see, e.g., [HMVZ, p. 94]) A Banach space X does not contain an isomorphic copy of 1 if and only if the two topologies μ(X ∗ , X ) and  ·  agree sequentially on X ∗ . Proof: Assume that X does not contain an isomorphic copy of 1 and { f n } is a sequence in X ∗ that converges to 0 uniformly on every weakly compact set but not in norm of X ∗ . Then there is a subsequence, which we will denote again by { f n }, vectors xn ∈ B X and ε > 0 such that f n (xn ) ≥ ε for every n. By Rosenthal Theorem 5.37, there is a subsequence of {xn }, which we will again denote by {x n }, w∗

such that {xn } is weakly Cauchy. Since f n → 0, it is easy to see that there is a subsequence of { f n }, which we will again denote by { f n }, such that f n (x2n ) − f n (x2n−1 )

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stays bounded away from zero. The sequence {x 2n − x 2n−1 } is weakly null. This is a contradiction with the fact that f n converge to 0 uniformly on weak compact sets. To prove that the condition suffices, suppose that Y is a subspace of X which is isomorphic to 1 and let {en }∞ n=1 be the image of the unit vector basis under some isomorphism from 1 onto Y . Define a bounded operator from Y into L ∞ [0, 1] by mapping en to the nth Rademacher function rn . By the injective property of L ∞ [0, 1] (see Exercise 5.91), this operator extends to a bounded linear operator T from X into L ∞ [0, 1]. Let rn∗ be the nth Rademacher function in L 1 [0, 1] considered as a subspace of (L ∞ [0, 1])∗ . Thus the sequence {rn∗ }, being equivalent to an orthonormal sequence in a Hilbert space, converges weakly to zero. Since L ∞ [0, 1] has the Dunford–Pettis property (see Exercise 4.42 and Theorem 13.43, and observe that ∞ is a C(K )-space), {rn∗ } converges in the topology μ(X ∗ , X ) to zero, a fortiori {T ∗rn∗ } is μ(X ∗ , X )-convergent to zero. But T ∗rn∗ , en  = rn∗ , rn  = 1, so {T ∗rn∗ } does not converge to zero in norm.

5.4 Remarks and Open Problems Remarks 1. If 0 < p < 1, then there is in L p a subspace such that all bounded operators on it are multiples of the identity [KaRo]. Shelah and Steprans proved that there is a nonseparable Banach space X for which every bounded operator from X into X has the form S + ρ I , where S is an operator with separable range, I is the identity operator and ρ is a real number [ShSt1]. 2. An infinite-dimensional Banach space X is called hereditarily indecomposable (HI in short) if no infinite-dimensional subspace Y of X can be written as a topological direct sum of two infinite-dimensional closed subspaces Y1 and Y2 of Y . Gowers and Maurey [GoMa] proved that there exists a reflexive HI space. In this direction see the subsequent paper [Fere]. Note that an HI space cannot contain an infinite-dimensional subspace with unconditional basis (the existence of such space was a longstanding famous open problem). Also, if X is an HI space, then X is not isomorphic to any proper subspace of X , in particular it is not isomorphic to any of its hyperplanes (see [Maur, p. 1265]). This solved another longstanding open problem. Gowers showed in [Gowe6] that if X is an arbitrary infinite-dimensional Banach space, then either X contains an infinite-dimensional subspace with unconditional basis or X contains an HI subspace. We refer also to [ArTo], where it is shown that there is a separable Banach space Y with separable dual Y ∗ and non-separable Y ∗∗ such that all Y , Y ∗ and Y ∗∗ are hereditarily indecomposable and any bounded operator on Y ∗∗ is of the form λI + S, where S has a separable range. These results are beyond the scope of this text and we refer to the article [Maur, pp. 1247–1297] for information on these topics.

5.4

Remarks and Open Problems

265

3. A Banach space (X, ·) is called distortable if there is λ > 1 and an equivalent norm | · | on X so that for all infinite-dimensional subspaces Y of X ,  sup

8 |y1 | : y1 , y2 ∈ S(X,·) ≥ λ. |y2 |

We will see, in Exercise 5.42, James’ classical result that if X is isomorphic to c0 or to 1 , then X does not contain a distortable subspace. Milman showed in [Milm2] that if X does not contain isomorphic copies of c0 or some  p , p ∈ [1, ∞), then X contains a distortable subspace (for the proof see [OdSc4]). So, the existence of a distortable space was clear after Tsirelson space appeared in 1974 [Tsir], see Exercises 9.29, 9.30, and 9.31. Now it is known that any  p , 1 < p < ∞ is distortable, and moreover, if X does not contain a distortable subspace, then X is saturated by isomorphic copies of c0 or 1 (we say that a Banach space is saturated by isomorphic copies of a Banach space Y if every infinite-dimensional subspace of X contains an isomorphic copy of Y ), see [OdSc4]. If X is a Banach space, we will say that a function f : S X → R stabilizes if for every infinite-dimensional subspace Y → X and every ε > 0, there exists an infinite-dimensional subspace Z → Y so that osc ( f, S Z ) := sup{ f (z 1 ) − f (z 2 ) : z 1 , z 2 ∈ S Z } < ε. Note that X does not contain a distortable subspace if and only if every equivalent norm on X stabilizes. Gowers showed in [Gowe1] that every Lipschitz function f : Sc0 → R stabilizes. These topics are beyond the scope of this text and we refer to [OdSc4] for information in this direction. 4. Talagrand proved in [Tala4] the following result: if | · | is an equivalent norm on ∞ , then there is δ > 0 such that for every n ∈ N, there is a subspace X n of ∞ isomorphic to ∞ and such that on X n , (δ − 2−n ) · ∞ ≤ | · | ≤ (δ + 2−n ) · ∞ . Partington showed in [Part] that if Γ is uncountable, then ∞ (Γ ) in any equivalent norm contains an isometric copy of ∞ in the supremum norm. Thus, for Γ uncountable, ∞ (Γ ) admits no equivalent strictly convex norm. 5. Gowers showed in [Gowe5] that there is a Banach space Z that is isomorphic to Z ⊕Z ⊕Z but not isomorphic to Z ⊕Z . Then Z is isomorphic to a complemented subspace of Z ⊕ Z and Z ⊕ Z is isomorphic to a complemented subspace of Z , since the latter is isomorphic to Z ⊕ Z ⊕ Z . This is a negative solution to the Schroeder–Bernstein problem for Banach spaces and should be compared with the Pełczy´nski decomposition method. Note that if we do not insist on the complementability, then for a negative solution of this problem, we can just consider an infinite-dimensional subspace of c0 that is not isomorphic to c0

266

6.

7.

8.

9. 10.

11. 12.

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(see Exercise 5.67), as this subspace has to contain an isomorphic copy of c0 by Theorem 4.46. Let us remark in passing that Lindenstrauss solved Köthe’s problem in [Lind2] by showing that there are two nonisomorphic spaces that have isomorphically the same family of all subspaces and isomorphically the same family of all quotients, see Exercise 5.70. Concerning a problem on isometrically universal spaces, let us remark that Szankowski showed in [Szan1] that there is a reflexive separable space X such that every finite-dimensional space W is isometric to a one-complemented subspace of X . Lindenstrauss proved in [Lind6] that any two-dimensional space is isometric to a subspace of L 1 , and Bessaga showed in [Bess1] that there is no finitedimensional space that would be isometrically universal for all 2-dimensional spaces. Godefroy and Kalton showed in [GoKa2] that if a separable Banach space Z contains isometric copies of all strictly convex separable Banach spaces, then Z contains an isometric copy of all separable Banach spaces. Maurey showed in [Maur0] that if X is a Banach space of type 2 and Y is a subspace of X which is isomorphic to a Hilbert space, then Y is complemented in X . Lindenstrauss and Pełczy´nski, in [LiPe2], proved that if E is a subspace of c0 and K is a compact space, then any bounded operator T : E → C(K ) extends to a bounded operator from c0 into C(K ). For generalizations, see [JoZi]. For a characterization of spaces that do not contain a copy of 1 in terms of renorming, we refer to [DGZ3, p. 106]. Stegall proved in [Steg1] the following result: If X is a separable Banach space such that X ∗ is not separable and ε > 0 is given, then there is a subset Δ ⊂ S X ∗ w∗ -homeomorphic to the Cantorset and a sequence {x n } in X with xn  < 1 + ε for all n ∈ N such that ∞ n=1 T x n − χ An  < ε, where T : X → C(Δ) is the evaluation operator (i.e., T x(x ∗ ) = x ∗ (x)) and the sets An are the (homeomorphic images of) the dyadic intervals of the Cantor set. (For more in this direction see, e.g., [LoMo].) We refer to, e.g., [HMVZ, Ch. 7] for more on fixing c0 by operators. There is a Banach space X not isomorphic to c0 such that X ∗ is isometric to 1 [BeLi0].

Open Problems 1. Is it true that every complemented subspace in a C(K ) space, for K a compact topological space, is isomorphic to a C(L) space, L a compact topological space? See, e.g., [Rose10, p. 1593]. 2. Assume that a Banach space X is complemented in all its overspaces. Is X isomorphic to C(K ) where K is a extremely disconnected compact space (i.e., the closure of any open subset is open)? See, e.g., [Zipp2, p. 1714].

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3. Assume that a separable Banach space X has the property that all of its subspaces with Schauder basis are complemented in X . Is X isomorphic to 2 ? [Pel8, Problem 2.3]. 4. Assume that a separable Banach space X has the property that all of its infinitedimensional subspaces with basis are isomorphic. Is X isomorphic to 2 ? [Pel8, Problem 2.4]. It is known that a separable infinite-dimensional Banach space all of whose infinite-dimensional subspaces are isomorphic is necessarily isomorphic to a Hilbert space, [Gowe6] and [KoTo]. 5. Assume that every subspace of a separable Banach space X has an unconditional basis. Is X isomorphic to a Hilbert space? See, e.g., [Casa, p. 279]. 6. Does the space H∞ of all bounded analytical functions on {z ∈ C : |z| < 1} with the supremum norm have the approximation property? (See Definition 16.34.) See, e.g., [Casa, p. 285].

Exercises for Chapter 5 5.1 Consider the space L p (Ω, Σ, μ) for 1 ≤ p < ∞ and fix ε > 0. Let F be a finite-dimensional subspace of L p (Ω, Σ, μ). Prove that there is a finite and d( F,  mp ) < 1 + ε for  ⊂ L p (Ω, Σ, μ) such that F ⊂ F dimensional space F  Therefore some m and that there is a norm-2 projection from L p (Ω, Σ, μ) onto F. L p (Ω, Σ, μ) has the bounded approximation property. n Hint. Let { fi }i=1 be a Hamel basis for F. Since every function in L p is the limit of simple functions, we can find a partition of Ω into sets {A j }mj=1 so that 0 < n μ(A j ) < ∞ for each j and so that there are {gi }i=1 in G := span{χ A j : j = 1, 2, . . . , m} such that  fi − gi  < ε for all j. Thus F is “almost” contained in G and G is isometric to mp .  of L p which is closed We will now show that G can be replaced by a subspace F  of norm less or to G, contains F and, moreover, there is a projection of L p onto F equal than 2. (See also Proposition 5.24.) Since all norms on a finite-dimensional space F are equivalent, there is K ≥ 1 such that  m      K −1 max |λi | ≤  λi f i  ≤ K max |λi |   i=1

for all real numbers λ1 , λ2 , . . . , λm . Let 1 > δ > 0 such that (1+δ)/(1−δ) < 1+η, and put ε = δ/(2n K ). For i = 1, 2, . . . , n, let gi ∈ G be such that gi  = 1 and  f i − gi  < ε. Then   m     (2K )−1 max |λi | ≤  λi gi  ≤ 2K max |λi |.   i=1

By the Hahn–Banach theorem, there exists x i∗ ∈ L ∗p with xi∗  ≤ 2K and such that xi∗ (g j ) = δi, j , the Kronecker delta, i = 1, 2, . . . , n. Let an operator T : L p → L p

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be defined by Th = h +

n 

xi∗ (h)( f i − gi )

i=1

for h ∈ L p . Then T gi = f i for i = 1, 2, . . . , n and (1 − δ)h ≤ T h ≤ (1 + δ)h. Indeed, to see for example the right-hand-side inequality, estimate T h − h ≤

n 

xi∗ .h. f i − gi  ≤ h2K nε ≤ δh.

i=1

Thus T G ⊃ F and ρ(T G, m ∞ ) ≤ (1 + δ)/(1 − δ) < 1 + ε.  = T G. Let P be a norm-one projection from L p onto G. Then P1 := Put F  T P T −1 is the sought projection onto F. 5.2 Show that a Banach space X is isomorphic to a Hilbert space if every separable subspace of X is isomorphic to a Hilbert space. Hint. For separable Y ⊂ X , put CY = sup{C > 0 : there exists a bilinear form bY on Y × Y such that Cy2 ≤ bY (y, y) ≤ y2 , for all y ∈ Y }. Show that C := inf{CY : Y separable subspace of X } > 0 (otherwise there are separable subspaces Yn , n ∈ N, such that C Yn → 0 and consider the closed linear span of n Yn ). Given a separable subspace Y of X define on X × X a bilinear form that is bY on Y × Y and such that (C/2)y2 ≤ bY (y, y) ≤ y2 for y ∈ Y . Consider [−1, 1] B X ×X and Tychonoff’s theorem to get a limit point, obtaining the desired bilinear form on X to give a dot product on X . The assertion follows also from Kwapie´n’s theorem (see Exercise 1.98). 5.3 Let X be a Banach space, let Y be a closed separating subspace of X ∗ . For x ∈ X define xY = sup{|y ∗ (x)| : y ∗ ∈ BY } (see Exercise 3.89). Show that: (i) Y is norming if and only if X + Y ⊥ is closed in X ∗∗ (that is, it is a topological sum). (ii) Y is 1-norming if and only if X + Y ⊥ is closed in X ∗∗ and the projection P of X + Y ⊥ onto X has P = 1. Hint. (i) Assume Y is c-norming. Then for x ∈ X , y ∈ Y ⊥ we have x ≤ cxY = cx + yY ≤ cx + y

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269

and the claim follows by Exercise 4.20. If the sum X ⊕ Y ⊥ is topological, then X is isomorphic to (X + Y ⊥ )/Y ⊥ . Thus for some c > 0, we have 1c x ≤ x ˆ = dist(x, Y ⊥ ) = xY . (ii) If Y is 1-norming, by (i) we have x ≤ x + y. Now assume there is a !: Z /Ker(P) → X . We norm-one projection P of Z = X + Y ⊥ onto X . Consider P ! Z / Ker(P) ) = P(B Z ) = B X and P ! is one-to-one, hence it is an isometry of have P(B ˆ = dist(x, Y ⊥ ) = xY . Z /Ker(P) = Z /Y ⊥ onto X . We get x = x 5.4 Let x ∗∗ ∈ X ∗∗ \X . Show that Ker(x ∗∗ ) is 1-norming if and only if x ∗∗ + x ≥ x for all x ∈ X (see Exercise 3.88). Hint. If Ker(x ∗∗ ) is 1-norming, then the inequality follows by the previous exercise. On the other hand, having the inequality we obtain a norm-one projection of X + Ker(x ∗∗ )⊥ = X + span{x ∗∗ } onto X (note that span{x ∗∗ } is w∗ -closed, so Ker(x ∗∗ )⊥ = (span{x ∗∗ }⊥ )⊥ = span{x ∗∗ }). 5.5 Assume that the codimension of X in X ∗∗ is finite. Show that then every w ∗ dense closed subspace in X ∗ is norming. Hint. Since X is finite-codimensional in X + Y ⊥ , X + Y ⊥ is closed in X ∗∗ . Use the preceding exercises. 5.6 Let Y be a closed subspace of a Banach space, recall that Y ∗ is isometric to X ∗ /Y ⊥ . Let q be the canonical quotient mapping X ∗ → X ∗ /Y ⊥ . Show that if Z a closed norming subspace of Y ∗ = X ∗ /Y ⊥ , then W = q −1 (Z ) is a norming subspace of X ∗ . Hint. Assume without loss of generality that Z is a 1-norming subspace of Y ∗ (consider  ·  Z on Y , see Exercise 5.3, and extend it to an equivalent norm in X , see Proposition 2.14). Take x ∈ S X . If dist(x, Y ) < 14 , take y ∈ Y such that x − y < 14 , so y > 34 . Find y ∗ ∈ B Z such that y ∗ (y) > 34 . Thus there is w ∗ ∈ 32 BW such that q(w∗ ) = y ∗ , that is, w ∗ (y) > 34 . Then x ∗ := 23 w∗ ∈ BW and x ∗ (x) = 23 [w ∗ (y) + w ∗ (x − y)] > 14 . Now assume that dist(x, Y ) ≥ 14 . Then x ˆ ≥ 14 (xˆ is the coset containing x and the norm is calculated in X/Y ), so there is x ∗ ∈ Y ⊥ such that x ∗ (x) ≥ 14 and x ∗  ≤ 1. Since q(x ∗ ) = 0 ∈ Z , we get that x ∗ ∈ BW . So in any case, for every x ∈ S X we get sup{x ∗ (x) : x ∗ ∈ BW } ≥ 14 . 5.7 Let Y be a closed hyperplane of a Banach space X . Show that for any ε > 0 there is a projection P of X onto Y with P ≤ 2 + ε (see Exercise 5.8 for an extension). In particular, X is isomorphic to Y ⊕ R. Hint. Write X = Y ⊕ Z (algebraic sum), where Z = span(e). Using fˆ ∈ (X/Y )∗ ∼ = Y ⊥ such that fˆ(e) ˆ = 1, find f ∈ (1 + ε)B X ∗ such that f (e) = 1 and f (y) = 0 for y ∈ Y . Let P(x) = x − f (x)e. The last part follows as Y is complemented.

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5.8 Let Y be a closed subspace of codimension n in a Banach space X . Show that for every ε > 0 there is a projection P of X onto Y such that P < n + 1 + ε. In particular, Y is complemented. Hint. Let { f i ; Fi }n1 ⊂ (Y ⊥ × (Y ⊥ )∗ ) be an Auerbach basis. Identify (Y ⊥ )∗ with  ∗∗ f i (x)x i . (X/Y ) . Take xi ∈ Fi with xi  < 1 + ε/n. Let P(x) = x − 5.9 Let Z be a hyperplane of a Banach space X . Show that if there exists a bounded operator T from some Banach space Y onto Z then Z is closed in X . Hint. By taking quotients assume that T : Y → Z is one-to-one. Consider the mapping T1 from Y ⊕ R onto X defined by T1 (y, r ) = T (y) + r z 0 , where z 0 is in the algebraic complement of Z in X . Then T1 ∈ B(X ⊕ R, X ) and is onto, hence it is an isomorphism and so T (Y ) must be closed in X . 5.10 Let Y be a finite-codimensional subspace of a Banach space X . Show that if there is a bounded operator T from some Banach space Z onto Y , then Y is closed. Hint. Induction on the previous exercise. / Y . Show that if Y is closed 5.11 Let Y of a Banach space X , let x0 ∈   be a subspace then span Y ∪ {x 0 } is closed in X . This gives by induction that if Y is a closed subspace of X and F is a finitedimensional subspace of X , then span(Y ∪ F) is closed in X . Hint. Let f ∈ X ∗ be such that f (z) = 0 for all y ∈ Y and f (x 0 ) = 1. Let yn + λn x0 → x ∈ X , yn ∈ Y . Then λn = f (yn + λn x0 ) → f (x). Therefore yn → x − f (x)x  0 and x − f (x)x 0 ∈ Y since Y is closed. Then x = x − f (x)x 0 + f (x)x0 ∈ span Y ∪ {x 0 } . 5.12 Let X, Y, Z be Banach spaces, let T be a bounded operator from X into Y such that T (X ) is closed in Y , and let S be a finite rank operator from X into  Z (that is, dim S(X ) is finite). Define U : X → Y ⊕ Z by U (x) = T (x), S(x) . Show that U (X ) is closed in Y ⊕ Z . Hint. If y ∗ ∈ Y ∗ , z ∗ ∈ Z ∗ and x ∈ X , we have     U ∗ (y ∗ , z ∗ )(x) = (y ∗ , z ∗ ) U (x) = (y ∗ , z ∗ ) T (x), S(x)       = y ∗ T (x) + z ∗ S(x) − T ∗ (y ∗ ) + S ∗ (z ∗ ) (x), so U ∗ (y ∗ , z ∗ ) = T ∗ (y ∗ ) + S ∗ (z ∗ ). Since S ∗ is a finite rank operator and T ∗ has a closed range by Exercise 2.49, we get that U ∗ has a closed range T ∗ (Y ∗ ) + S ∗ (Z ∗ ) by Exercise 5.11. By Exercise 2.49, we obtain that U has a closed range as well. 5.13 Show that the intersection of a finite-codimensional subspace and an infinitedimensional subspace in a Banach space is an infinite-dimensional subspace. Hint. The intersection of a subspace with a hyperplane has codimension at most one in the subspace. Then use induction.

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5.14 Let X be a Banach space. Show that X ⊕ R is isomorphic to X if and only if X is isomorphic to all its closed hyperplanes. Hint. Assume that X is isomorphic to a closed hyperplane H . Then H ⊕ R is isomorphic to X ⊕ R and to X as well (Exercise 5.7). If X is isomorphic to X ⊕ R by an isomorphism T : X ⊕ R → X , put H = T (X ⊕ {0}) and observe that X is isomorphic to H , which is a closed hyperplane in X . Then use Exercise 2.9. The question whether every Banach space is isomorphic to its closed hyperplanes was answered in the negative by Gowers ([Gowe3]). 5.15 Let f ∈ X ∗ \{0}, where X is c0 , c, or the space  p for 1 < p < ∞. Show that f −1 (0) is isomorphic to the original space. Equivalently, all closed hyperplanes in c0 or  p are isomorphic to the whole space. Hint. The closed hyperplane {(0, x 1 , x2 , . . . ) : (xi ) ∈ X } in X is isomorphic to X by the mapping (x 1 , x 2 , . . . ) → (0, x1 , x 2 , . . . ), then use Exercise 2.9. 5.16 Show that the spaces c0 and c are isomorphic. Recall that they are not isometric (Exercise 3.132) Hint. c0 is a hyperplane in c given by the functional f (x) = lim xn . Alternatively, consider T (x) = (lim x n , x1 − lim xn , x2 − lim xn , . . . ). 5.17 Show that all closed hyperplanes of C[0, 1] are isomorphic to C[0, 1] and C[0, 1] ⊕ R is isomorphic to C[0, 1]. Consider the subspace (C[0, 1])0 formed by all functions in C[0, 1] that vanish at 0. Use it to show directly that C[0, 1] ⊕ R is isomorphic to C[0, 1]. Hint. By the Sobczyk and Banach–Mazur theorems, there is an isometric copy of c0 complemented in C[0, 1], therefore C[0, 1] ∼ c0 ⊕ Z for some Z . Show that the hyperplane {(0, x1 , x2 , . . . ) : (xi ) ∈ c0 } ⊕ Z is isomorphic to c0 ⊕ Z . Then use Exercise 5.14. For the second part: (C[0, 1])0 ⊕ R is isomorphic to C[0, 1] by the mapping ( f, r ) → f + r for f ∈ (C[0, 1])0 , r ∈ R. 5.18 Show that C[0, 1] ⊕ C[0, 1] is isomorphic to C[0, 1]. Hint. Let C0 be a subspace of C[0, 1] formed by functions equal to 0 at 0 and let C1 be a subspace of C[0, 1] formed by functions equal to 0 at 1. By the previous exercise, C0 , C1 are isomorphic to C[0, 1]. Putting together graphs of functions, we can see that C1 ⊕ C0 is isomorphic to a subspace of C[0, 2] consisting of functions equal to 0 at 1. By scaling, this is in turn isometric to a subspace C2 of C[0, 1] consisting of functions that are 0 at 12 . This is a hyperplane in C[0, 1], hence isomorphic to C[0, 1] itself. 5.19 Show that C[0, 1] ⊕ c0 is isomorphic to C[0, 1]. Hint. Use previous exercises and the Pełczy´nski decomposition method. 5.20 Show that c0 is isometric to c0 ⊕ c0 with max-norm.

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  Hint. Map (x1 , x2 , . . . ), (y1 , y2 , . . . ) to (x 1 , y1 , x2 , y2 , . . . ). 5.21 Show that c ⊕ c is not isometric to c. Hint. Assume that T is an isometry of c ⊕ c onto c. Consider x, y ∈(c ⊕ c) defined  by x = (1, 1, . . . ), (1, 1, . . . )) and y = ((1, 1, . . . ), (−1, −1, . . . ) . Then x, y are x−y extreme points of Bc⊕c . Both h 1 = x+y 2 , h 2 = 2 have norm one and for each of them there exists an infinite-dimensional closed subspace Yi of c ⊕ c such that h i +yi  = 1 for every yi ∈ Yi ∩Bc⊕c . Isometry would have to carry these properties to c (in particular, it carries extreme points to extreme points), but such behavior is impossible in c. (For another approach to the same result see Exercise 3.141.) 5.22 Show that C([0, 1] ∪ [2, 3]) is isomorphic to C[0, 1]. Note that [0, 1] ∪ [2, 3] is not homeomorphic to [0, 1] (connectedness). Hint. Similar to the previous exercises. 5.23 Show that L p [0, 1] ⊕  p is isomorphic to L p . Hint. Similar to the previous exercises. 5.24 Show that C 1 [0, 1] is isomorphic to C[0, 1]. Hint. Consider the mapping f → ( f , f (0)) and use that C[0, 1]⊕R is isomorphic to C[0, 1]. 5.25 (Klee, [Klee1]) Let Γ be an infinite set. Prove, by a suitable modification of the proof of Theorem 5.1, that 1 (Γ ) has an equivalent rotund norm | · | such that every Banach space of density character less or equal than card(Γ ) is isometric to a quotient of (1 , | · |). Hint. For each n ∈ N, let gn : [0, 1] → [0, 1] be a continuous strictly

convex function such that nt/(n + 1) ≤ gn (t) ≤ t, for all t ∈ [0, 1]. Put Γ = ∞ n=1 Γn , where the sets Γn are pairwise disjoint and card(Γn ) = card(Γ ) for all n ∈ N. For n ∈ N and γ ∈ Γn , put gγ = gn . Let B = {x ∈ 1 (Γ ) : |xγ | ≤ 1,



gγ (|x γ |) < 1}.

γ ∈Γ

The set B is absolutely convex, and B1 (Γ ) ⊂ B ⊂ 2B1 (Γ ) , hence  ·  B , the Minkowski functional of B, is an equivalent norm on 1 (Γ ). It is easy to show that  ·  B is rotund. Now consider a Banach space Y of density character less than or equal to card(Γ ). It is easy to define a mapping ϕ from Γ onto a dense subset of the open unit ball BYO of Y and  such that ((n + 1)/n)ϕ(Γn ) ⊂ BYO for each n ∈ N. For x ∈ 1 (Γ ) put T (x) = γ ∈Γ xγ ϕ(γ ). Then T is a linear mapping from 1 (Γ ) into Y . It is easy to see that BYO ⊂ T (B). Conversely, if x ∈ B,

Exercises for Chapter 5

T x ≤

 n∈N

⎛ ⎝

 γ ∈Γn

273

⎞ |xγ |.ϕ(γ )⎠ ≤

  n∈N γ ∈Γn

 n |xγ | ≤ gγ (|xγ |) < 1, n+1 γ ∈Γ

hence T (B) ⊂ BYO , and the conclusion follows now as in the proof of Theorem 5.1. 5.26 Show that for any infinite compact space K , C(K ) contains a subspace isometric to c0 . If K is metrizable, then this subspace is complemented in C(K ). Hint. Let {Un } be a sequence of nonempty disjoint open subsets of K . Such a sequence can be found by induction: Pick U1 such that K 1 := K \U 1 is infinite, and take U2 ⊂ K 1 such that K 2 := K 1 \U 2 is infinite, and so on. Then pick a sequence {ϕn } of continuous functions form K into [0, 1] such that maxs∈K ϕn (s) = 1 and {s∈ K : ϕn (s) > 0} ⊂ Un for all n. Then, for any (an ) ∈ c00 , we have  ∞ n=1 an ϕn ∞ = maxn |an |. Then (ϕn ) is a basic sequence in C(K ) isometrically equivalent to the unit vector basis in c0 . For the second statement, note that C(K ) is then separable; use now Sobczyk’s Theorem 5.11. 5.27 Prove that a Banach space is injective if and only if it is isomorphic to a complemented subspace of some ∞ (Γ ). Hint. Every Banach space is isometric to a subspace of some ∞ (Γ ) (see the comments after Proposition 5.4 and Exercise 5.30). If it is injective, then it must be complemented there. Conversely, use (iii) in Proposition 5.13. 5.28 Let c be the subspace of ∞ of all convergent sequences. Show that every bounded projection P : c → c0 satisfies P ≥ 2. This proves that 2 is the best possible constant in Sobczyk’s theorem. Compare with Exercise 5.7. Hint. The space c0 is the kernel of the bounded functional f : C → K given by f (x) = lim xn , for x = (xn ) ∈ c. Every bounded projection P : c → c0 is written P(x) = x − f (x)a, for some a ∈ c such that f (a) = 1. It is easy to find x ∈ c such that f (x) = 1, x = 1 and x − a (= P(x)) is almost 2. 5.29 Recall that the distance d(x) of a point x = (xi ) ∈ ∞ to c0 is equal to lim sup |xi |. Define a mapping φ from ∞ onto c0 by φ(x)i = 0 if |xi | ≤ d(x) and φ(x)i = sign(xi )(|xi | − d(x)) if |xi | > d(x). Show that φ is a Lipschitz retraction of ∞ onto c0 , that is, a Lipschitz mapping from ∞ onto c0 that is the identity on c0 . Note that φ cannot be linear as c0 is not complemented in ∞ . Hint. Direct calculation. 5.30 Let X be a Banach space. Show that there is a set Γ such that X is isometric to a quotient of 1 (Γ ). Show that there is a set Γ such that X is isometric to a subspace of ∞ (Γ ). Hint. The proofs of Theorem 5.1 and Proposition 5.4.

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5 Structure of Banach Spaces

  ∗ 5.31 We have conv Ext(B∞ ) = B∞ (see Exercise 3.129). Show  that∞ = 1 ∗ contains a w -compact convex subset C that is not equal to conv Ext(C) . Hint. As any separable space, C[0, 1] is a quotient of 1 and we thus can carry the unit ball of C[0, 1]∗ into ∗1 . Then use Exercise 3.130. 5.32 Show that n2 is isometric to a subspace of ∞ but not to any subspace of c0 . Hint. Any separable Banach space is isometric to a subspace of ∞ (Proposition 5.4). n2 has uncountably many extreme points, while finite-dimensional subspaces of c0 have only finitely many of them. To see the latter, use the uniform convergence to zero of elements of compact sets in c0 . 5.33 Let X be a Banach space. Show that if X is separable then X ∗ is isometric to a subspace of ∞ . Hint. The proof of Theorem 5.1. 5.34 Recall that 1 (c) is isometric to a subspace of C[0, 1]∗ (Exercise 3.143). Use it to show that 1 (c) is isometric to a subspace of ∞ . Show that this is not the case for c0 (c). Note that ∗∞ contains an isometric copy of 2 (c) ([Rose3]). Hint. Use Exercise 5.33. For the second part, there is no countable separating family F in the dual of c0 (c). Indeed, if such an F exists, every

functional in F has a countable support, hence there is r ∈ c which is not in supp( f ). Clearly f (er ) = 0 for every f ∈ F.

f ∈F

5.35 Suppose a Banach space X contains an isomorphic copy of 1 . Show that then X ∗ contains an isomorphic copy of 1 (2N ). Hint. We have ∞ as a quotient of X ∗ . The space ∞ has a subspace isometric to 1 (c) = 1 (2N ). So X ∗ has a subspace Y such that 1 (2N ) is a quotient of it. By the lifting property, 1 (2N ) is a subspace of Y , hence of X ∗ . 5.36 Show that ∞ /c0 contains a subspace isometric to c0 (c). Hint. Consider a system of characteristic functions of subsets Aλ of N with the property card(Aλ ) = ∞ and card(Aλ1 ∩ Aλ2 ) < ∞ whenever λ1 = λ2 (Lemma 5.7). Then the cosets containing χ Aλ form the standard unit vector basis of c0 (c) in the space ∞ /c0 . Note that this implies that ∞ /c0 is not isomorphic to a subspace of ∞ (since ∗∞ is w ∗ -separable and c0 (c) is not). 5.37 Show that 1 is of first Baire category in c0 . Hint. B1 is closed in c0 : If {a k } ⊂ B1 , a k = (aik ), and a k → a in c0 , then a k → a n n |aik | ≤ 1. Passing k → ∞ we get i=1 |ai | ≤ 1, pointwise. Given n, we have i=1 so a ∈ B1 . Then observe that B1 does not contain any open ball of c0 . 5.38 Show that L 2 [0, 1] is a set of the first Baire category in L 1 [0, 1].

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275

Hint. B L 2 is a weakly compact set in L 1 , so it is closed. Show by constructing an appropriate function that B L 2 contains no interior point as a set in L 1 .   5.39 (Lindenstrauss [Lind4]) Let X be a Banach space and assume that X ⊕ R ∞ is isometric to a subspace of X . Show that X has a subspace isometric to c0 . Hint. X contains a vector x 1 ∈ S X and a subspace Y1 that is isometric to X such that x1 + y = max(1, y). Continuing by induction we get for every n ∈ N a vector xn ∈ SYn and a subspace Yn−1 of Yn isometric to X such that x n +y = max(1,  y) for every y ∈ Yn . It follows that for all real λ1 , . . . , λn we  n λi xi  = max |λi |. Hence the closed subspace of X spanned by {xi } is have  i=1 isometric to c0 . 5.40 (Lindenstrauss [Lind8]) Show that there is no separable reflexive Banach space X 0 such that every separable reflexive Banach space is isometric to a subspace of X. Szlenk proved that there is no separable reflexive Banach space X so that every separable reflexive Banach space is isomorphic to a subspace of X ([Szle]). Hint. Use the previous exercise for X 0 and the fact that c0 is not reflexive. Note that this method works for many other classes of Banach spaces. 5.41 Numbers C1 , C2 in the definition of equivalence of basic sequences may serve as a measure of “closedness” of basic sequences. Show that if basic sequences {ei }, { f i } are equivalent with constants C1 , C2 , then the spaces span{ei }, span{ f i } are isomorphic and we can estimate d(span{ei }, span{ f i }) ≤ C 1 C2 , where d(·, ·) denotes the Banach–Mazur distance. Hint. Standard. 5.42 (James [Jame3], see also [LiTz3, p. 97]) Let (X,  · ) be the space 1 or c0 with its usual norm. Let | · | be an equivalent norm on X . Show that then, for every ε > 0 there is a subspace Y of X with d((Y, | · |), (X,  · )) < 1 + ε, where d(·, ·) is the Banach–Mazur distance. Hint. Assume X = 1 and that α|x| ≤ x ≤ |x| for some α and all x ∈ X . Let ε > 0 and let (Pn ) be the sequence of natural projections in X . For every n put λn = sup{x : |x| = 1, Pn x = 0}. Clearly, λn ↓ λ for some 1 ≥ λ ≥ α. Let n 0 be such that λn 0 < λ(1+ε). By the definition of λn , there is a block basis (yk ) of the unit vector basis of X so that for all k, |yk | = 1,Pn 0 yk = 0 and yk  > λ/(1 + ε). For every choice of numbers (ak ), we have Pn 0 ( ∞ k=1 ak yk ) = 0 and thus ∞ ∞          ak yk  ≥ λ−1 a y   k k n0 k=1

= λ−1 n0

k=1 ∞  k=1

|ak |.yk  ≥ λ−1 n0

∞ ∞  1 λ  |ak | ≥ |ak |. 1+ε (1 + ε)2 k=1

k=1

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5 Structure of Banach Spaces

On the other hand, by the triangle inequality, ∞ ∞ ∞        ak yk  ≤ |ak |.|yk | ≤ |ak |.  k=1

k=1

k=1

So,  d (span{yk : k ∈ N}, | · |), (1 ,  · ) < (1 + ε)2 ,    (consider the mapping T i ai ei := i ai yi from 1 onto span{yk : k ∈ N}). The proof for c0 is similar, we only replace the sup by the inf in the definition of λn . 5.43 Assume X and Y are two Banach spaces so that there is an isomorphism T of X onto Y such that T  · T −1  = 1. Show that X is isometric to Y .  := T /T . Note that T itself need not be an isometry: X = Y Hint. Consider T and T x = 2x for x ∈ X . 5.44 Find an example of two Banach spaces whose Banach–Mazur distance is 1 but these spaces are not isometric (compare with Exercise 5.43). Hint. (see [PeBe]) For X take c0 with the norm ∞ 1   2 x1 = max |xi |2 + 2−i |xi |2 i=1

and for Y take c0 with the norm ∞ 1   2 2 2−i+1 |xi |2 . x2 = max |xi | + i=2

Define Tn ∈ B(X, Y ) by Tn (x) = (xn , x1 , . . . , xn−1 , xn+1 , . . . ). It is clearly onto, Tn  → 1 and Tn−1  → 1. We find that X is strictly convex (see the proof of Theorem 8.2). However, Y is not strictly convex (consider (−1, 1, 0, 0, . . . ) and (1, 1, 0, 0, . . . )). Therefore X and Y cannot be isometric. 5.45 Let Γ be an abstract set. Show that 1 (Γ ) has the Schur property.  w Hint. Let xn → x in 1 (Γ ). Since x n ∈ 1 (Γ ), we have that γ ∈Γ |x n (γ )| < ∞.

Therefore supp(xn ) and supp(x) is countable. Let Γ = supp(xn ) ∪ supp(x). Then Γ is countable, hence 1 (Γ ) is isometric to 1 . Since xn , x ∈ 1 (Γ ), Theorem 5.36 implies that xn → x. 5.46 Show that on S1 the w ∗ - and the norm topology coincide.

Exercises for Chapter 5

277 w∗

Hint. By metrizability, it is enough to check sequences. Let x, xn ∈ S1 and xn → x. ∞ w∗ Given ε > 0, fix k0 such that i=k |xi | < ε. Using x n → x, find n 0 such that 0 +1 k 0 ∞ k0 n i| + k0 =1 |x i | = 1, we i=1 |x i − x i | < ε for every n ≥ n 0 . Since i=1 |x  k0 k0  k 0  n also have i=1 |xi | > 1 − ε. Moreover, for n ≥ n 0 ,  i=1 |xi | − i=1 |xi | ≤ k0 k0 k 0 |xin −xi | < ε and hence i=1 |xin | ≥ i=1 |xi |−ε > 1−2ε and also because i=1  k 0 ∞ ∞ n n that i=k0 +1 |xi | = 1− i=1 |xin | ≤ 1−(1−2ε) = 2ε. i=1 |x i | = 1 for every n,   ∞  ∞ Therefore, for n ≥ n 0 , i=k0 +1 |xin − xi | ≤ i=k0 +1 |xin | + |xi | ≤ 3ε and thus k 0 ∞ ∞ n n n i=1 |x i − xi | = i=k0 +1 |xi − xi | ≤ 4ε. i=1 |x i − xi | + 5.47 Let C be a weakly compact set in 1 (Γ ). Show that C is compact. Hint. By the Eberlein–Šmulyan theorem, C is weakly sequentially compact. If {x n } is a sequence in C, there is a subsequence {xn k } that is weakly convergent, hence— by the Schur property—norm convergent. 5.48 Let X be an infinite-dimensional closed subspace of 1 . Show that X ∗ is nonseparable. w Hint. 0 ∈ S X . If X ∗ were separable, then (B X , w) would be metrizable. Thus there w w would be {x n } ⊂ S X such that x n → 0 in X , therefore x n → 0 in 1 . By Schur’s theorem, x n → 0, a contradiction. 5.49 Let f, f 1 , f 2 , · · · ∈ L 1 [0, 1]. Show that if f n → f almost everywhere and  f n 1 →  f 1 , then f n → f in L 1 [0, 1] (Vitali). Hint. | f n | + | f | − | f n − f | → 2| f | almost everywhere. Thus by Fatou’s lemma, 

1

2

 | f (t)| dt ≤ lim inf

0

1

| f n (t)| + | f (t)| − |( f n − f )(t)| dt

0



1

=2

| f (t)| dt − lim sup

0

Hence lim sup

1 0



1

|( f n − f )(t)| dt.

0

|( f n − f )(t)| dt = 0.

5.50 Assume that X is an infinite-dimensional Schur space. Then it contains an isomorphic copy of 1 . Hint. If not then from any bounded sequence can be extracted a subsequence that is weak- and thus norm-Cauchy, so X is finite-dimensional. 5.51 Let X be c0 or  p for p ∈ [1, ∞]\{2}. Let T be an isometry of X onto X . Show that there exists of signs εi = ±1 and a permutation π of natural   a sequence numbers such that T (xi ) = (εi x π(i) ). Hint.  p case: Note that if vectors x, y satisfy x + y p = x − y p = x p + y p , then they have disjoint supports. This is easy to see for p = 1 and requires

278

5 Structure of Banach Spaces

elementary calculations for p > 1. Note also that this fails for p = 2, since then any two orthogonal vectors satisfy the condition. Using this, we have that T (ei ) and T (e j ) have disjoint supports for i = j. Since T is onto, we readily obtain that supports of T (ei ) must be singletons and span{T (ei )} =  p . If X = c0 , we consider T ∗∗  , which is an isometry of ∞ onto ∞ , so we use the previous result and T = T ∗∗  X . 5.52 Let E and F be vector spaces and B(E × F) be the vector space of all bilinear forms on E × F. Let the mapping χ : E × F → B(E × F)# be defined by χ (e, f )(b) = b(e, f ) for all b ∈ B(E × F) and all e ∈ E, f ∈ F. Then χ is a bilinear mapping from E × F into B(E × F)# . The linear hull of χ (E × F) in B(E × F)# is called the tensor product of E and F and is denoted by E ⊗ F. The mapping χ is called the canonical bilinear mapping from E × F into E ⊗ F and χ (e, f ) is denoted by e ⊗ f . The tensor product E ⊗ F is thus the vector space of finite sums λi (ei ⊗ f i ) for ei ∈ E, f i ∈ F, and λi real numbers. Show that, for e ∈ E and f ∈ F, λ(e ⊗ f ) = (λe) ⊗ f = e ⊗ (λ f ) (e1 + e2 ) ⊗ f = e1 ⊗ f + e2 ⊗ f e ⊗ ( f 1 + f2 ) = e ⊗ f1 + e ⊗ f2 .  Note that thus every element u ∈ E ⊗ F is of the form u = ei ⊗ f i . If we assume that both sets {ei } and { f i } are linearly independent, then show that the number of terms in the sum for u is uniquely determined. This number is called the rank of u. Show, too, that the ⊗-product of two linearly independent sets is linearly independent. This means that, in the finite-dimensional case, dim(E ⊗ F) = dimE · dimF. Show that the mapping U → U ◦ χ is a linear isomorphism from (E ⊗ F)# onto B(E × F). Hint. Easy computation. We will study tensor products in Chapter 16. 5.53 Let X, Y be Banach spaces and let b be a bilinear form b on X × Y . Show that the following are equivalent: (i) b is continuous at the origin (0, 0) of X × Y . (ii) b is uniformly continuous on bounded sets of X × Y . (iii) There is K > 0 such that |b(x, y)| ≤ K x y for all x ∈ X and y ∈ Y . (iv) b is separately continuous on X ×Y , i.e., b(x, y0 ) is continuous in x for every y0 ∈ Y and b(x0 , y) is continuous in y for every x0 ∈ X .

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279

Hint. (i)⇒(iii): Assume that there is δ > 0 such that |b(x, y)| ≤ 1 whenever x ∈ δ B X and y ∈ δ BY . By a homogeneity argument, we get that |b(x, y)| ≤ δ −2 x y for every x ∈ X and y ∈ Y . (iii)⇒(ii): If x, x ∈ B X and y, y ∈ BY are such that x − x  < δ and y − y  < δ, then |b(x, y) − b(x , y )| ≤ |b(x − x , y − y ) + b(x , y − y ) + b(x − x , y )| ≤ K x − x  y − y  + K x  y − y  + K y  x − x  ≤ K δ 2 + 2K δ. (ii)⇒(i) and (ii)⇒(iv) are trivial. (iv)⇒(iii): Consider the family F = {b(x0 , ·) : x 0 ∈ B X } of continuous linear functionals on Y . For every y0 ∈ Y we have supx∈B X |b(x, y0 )| < ∞ as this functional is continuous. So {b(x 0 , y) : x 0 ∈ B X } is bounded for every y ∈ Y and the uniform boundedness principle, sup{|b(x, y)| : x ∈ B X , y ∈ BY } < ∞, which gives (iii). 5.54 For Banach spaces X and Y , we say that a bilinear form b on X ×Y is bounded if it satisfies (iii) (and so all the other equivalent conditions) in Exercise 5.53. In this case, its norm b is defined by b = sup{|b(x, y)| : x ∈ B X , y ∈ BY }. Show that the normed space of all bounded bilinear forms on X × Y with this norm is a Banach space. We will denote it by (Bil(X × Y ),  · ). Show that (Bil(X × Y ),  · ) is isometric to the Banach space B(X, Y ∗ ) via the mapping b → T defined by 

 T (x), · = b(x, ·)

Note that from these facts it follows that Bil(2 × 2 ) is nonseparable. Hint. Note that the space of bounded operators on a Hilbert space is nonseparable (Proposition 1.44). The rest is standard. 5.55 Find a basis for the linear space Bil(n2 × m 2 ). n is a basis for n2 and { f j }mj=1 is a basis for m Hint. If {ei }i=1 2 , let bi 0 , j0 be a bilinear   form defined by bi0 , j0 ( αi ei , β j f ≤ i 0 ≤ n, 1 ≤ j0 ≤ m. j ) = αi 0 β j0 for 1  Then if b ∈ Bil(n2 × m ), and x = i αi ei and y = j β j f j , then b(x, y) = 2      α β b(e , f ) = b (x, y)b(ei , f j ). Moreover, b αi ei , β j f j = i j i j i, j i, j i, j {bi, j : i = 1, 2, . . . , n, j = 1, 2, . . . , m} is a linearly independent set, as for given (i 0 , j0 ), bi0 , j0 (i 0 , j0 ) = 1 and bi, j (i 0 , j0 ) = 0 if (i, j) = (i 0 , j0 ). 5.56 Let p ∈ [1, ∞), x ∈  p . Let a > 0 and h j ∈  p , j ∈ N, be such that h j  = a w p p and h j → 0 in  p . Show that then x + h j  p → x p + a p . Hint. For z = (z i ) ∈  p and n ∈ N, denote z˜ n = (z 1 , . . . , z n , 0, . . . ) and z˜ n = (0, . . . , 0, z n+1 , z n+2 , . . . ). p Given ε > 0, from the uniform continuity of  ·  p on bounded sets there is n   w p p such that x + h j  p − x˜ n + h j  p  < ε for every j. Since h j → 0, there is

280

5 Structure of Banach Spaces

 p p  9 for every j ≥ j0 we have x˜ n + h j  p − x˜ n + (h j )n  p < ε and  j0 such that p p p p n n p (h 9 9 9 j )n  p − h j  p < ε. Finally note that  x˜ + (h j )n  p =  x˜  p + (h j )n  p for every j. 5.57 A function Q on a Banach space X is called a continuous quadratic form if there is a continuous bilinear form b on X × X such that Q(x) = b(x, x) for x ∈ X and b is symmetric, that is, b(x, y) = b(y, x) for every x, y ∈ X . Let p ∈ (2, ∞), let Q be a continuous quadratic form on  p . Show that Q is w w-sequentially continuous, i.e., Q(x j ) → Q(x) whenever x j → x in  p . Hint. Fixing x ∈  p , write Q(x + h) = Q(x) + Q(h) + A(x, h) for h ∈ X . Since A(x, h) is a continuous linear functional and thus weakly continuous, we have that w A(x, h j ) → 0 for every x. We need to prove that Q(h j ) → 0 whenever h j → 0. Assume that this is not true and for some ε > 0 and h j  = 1, j ∈ N, we have w h j → 0 and Q(h j ) ≥ ε for all j. Put n 1 = 1 and xn 1 = x 1 = h 1 . Find n 2 such p p that |A(xn 1 , h n 2 )| < ε2 and x n 1 + h n 2  p ≤ xn 1  p + 2 (use Exercise 5.56). Put p p ε x 2 = xn 1 + h n 2 , find n 3 such that |A(x2 , h n 3 )| < 2 and x2 + h n 3  p ≤ x 2  p + 2, etc. p Then we have x j  p ≤ 2 j for every j and Q(x j ) = Q(x j−1 ) + A(x j−1 , h n j ) + Q(h n j ) ≥ Q(x j−1 ) + ε −

ε 2

≥ Q(x j−1 ) + 2ε .

Since Q(x1 ) = Q(h 1 ) > ε, we get by induction that Q(x j ) > j 2ε for every  ε  1−2/ p Q(x ) j. Then x j2 ≥ j ε2 (2 j)−2/ p = 21+2/ → ∞, which contradicts the p j j p

continuity of Q using Exercise 5.53. Similarly, all polynomials of degree smaller than p on  p and all polynomials on c0 are weakly sequentially continuous ([BoFr] and Exercise 10.5). 5.58 Show that the property to be isomorphic to a subspace of c0 is a three-space property, i.e., if a subspace Y of X and X/Y both have it, then X has it. Hint. Let Y ⊂ X and let both Y and X/Y be isomorphic to subspaces of c0 . By Theorem 5.20, the isomorphism of Y into c0 extends to a bounded operator T from X into c0 . If j : X/Y → c0 denotes the isomorphism into c0 and q : X → X/Y is the quotient mapping, then T ⊕ ( j ◦ q) : X → c0 ⊕ c0 is an isomorphism into c0 . 5.59 Let T be a bounded operator from 1 onto c0 . Show that T −1 (0) is not complemented in 1 . Hint. If 1 = T −1 (0) ⊕ Z , then, by the open mapping theorem, Z would be isomorphic to c0 , a contradiction with Theorem 4.46. 5.60 Show that the Dirac measures δ1/n do not converge to δ0 in the weak topology of C[0, 1]∗ . Note that this gives another proof of the fact that C[0, 1] does not have the Grothendieck property. Hint. They are in 1 [0, 1] ⊂ C[0, 1]∗ ; use then the Schur property of 1 spaces.

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281

5.61 Show that the sequence (χ[n,∞) ) of characteristic functions in N converge to 0 in the w∗ -topology of ∞ (= ∗1 ), but not in the w-topology of ∞ . Hint. 0 ∈ conv · {χ[n,∞) : n ∈ N}. 5.62 Show that if X is an infinite-dimensional Banach space and Y is a finitedimensional Banach space, then there is a bounded operator from X onto Y . Hint. If dim Y = n, choose an n-dimensional subspace E of X . Then E is isomorphic to Y by an isomorphism T . Moreover, E is complemented in X by a projection P. Then T ◦ P maps X onto Y . 5.63 If X is an infinite-dimensional Banach space, show that there is a closed infinite-dimensional subspace Y of X such that X/Y is infinite-dimensional. Hint. If X is separable, take a Markushevich basis {x n ; x n∗ } for X and put Y = span{x2n : n ∈ N}. The following few exercises are taken from [Lind2]. 5.64 Assume that (X n ) is a strictly increasing sequence of finite-dimensional subspaces of ∞ such that there is λ > 0 so that, for every n ∈ N,

X n is complemented in ∞ by a projection of norm less or equal than λ. Show that n X n is not reflexive.. Hint. Assume that it is reflexive. If (Pn ) is the sequence of the associated uniformly bounded projections, consider its limit in the weak operator topology, i.e., Tn x → T x in the weak topology of the target space, for every x in the

domain space (see the definition right before Exercise 16.17). Then, for x ∈ n X n , Pn x = x for

sufficiently large n, and thus P x = x for x ∈ n X n . Thus P is a projection onto

n X n . This is impossible by Proposition 13.45.  n  5.65 Show that the space 2 0 is not isomorphic to c0 .  n  Hint. If it were, put X n = n2 ⊂ 2 ⊂ ∞ and note that then 2 ∞ is isomorphic to ∞ and so 2 would be isomorphic to a complemented subspace of ∞ , which is impossible as ∞ has the Dunford–Pettis property. Banach space for each n ∈ N, show that the 5.66 If X n is a finite-dimensional  canonical norm of ( n X n )0 is a Lipschitz Kadec–Klee smooth norm. Hint. For the notion of Lipschitz Kadec–Klee smooth norm, see Definition 12.60. Follow the proof of Theorem 12.64.  5.67 Show that c0 and ( n n2 )0 have the same linear dimension, i.e., both these spaces  have, up to a isomorphism, the same family of closed subspaces. Hint. ( n n2 )0 is isomorphic  to a subspace of c0 by Exercise 5.66 and c0 is clearly isomorphic to a subspace of ( n n2 )0 (take from each n2 a one-dimensional subspace). 5.68 Show that (



n n 2 )1

is isomorphic to a subspace of 1 .

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Hint. The space n2 is 2-isomorphic Mn of 1 by to an n-dimensional subspace  Dvoretzky’s Theorem 6.15. Then ( n n2 )1 is isomorphic to ( n Mn )1 , which is isomorphic to a subspace of 1 .  to 1 . 5.69 Show that ( n n2 )1 is not isomorphic  Hint. Otherwise, its dual space ( n n2 )∞ would be isomorphic to ∞ , which is impossible by Exercise 5.64.  5.70 Show that ( n2 )1 and 1 are two non-isomorphic spaces that have, up to isomorphisms, the same family of subspaces and quotient spaces. quotient of 1 . If we take in each Hint. As any separable space, ( n n2 )1 is a get ( n2 aone-dimensional subspace X n , we n X n ), a complemented subspace  of ( n n2 )1 , that is thus a quotient of ( n n2 )1 and is isomorphic to 1 . 5.71 (Lindenstrauss [Lind5b]) Let T be an operator from 1 onto L 1 [0, 1] and X := T −1 (0) → 1 . Show that X is not isomorphic to a dual space—and thus X does not have an unconditional basis, although 1 does have an unconditional basis. Compare with Exercise 4.55. Hint. The rough idea behind the proof is the following. Under the negation of the conclusion, we shall define an isomorphism from L 1 [0, 1] into 1 . Since L 1 [0, 1] contains a subspace isomorphic to 2 (see Theorem 4.53), the space 2 will be isomorphic to a subspace of 1 , hence it will contain a subspace isomorphic to 1 (see Theorem 4.46), something impossible due to the reflexivity of 2 . In order to construct the isomorphism into under the negation of the conclusion, define a family of finite-dimensional subspaces of L 1 [0, 1] and, for each one of them, a “linear local inverse” of T . Define, too, a “global bounded inverse” (not even linear or continuous) of T . Then, the difference of this two mappings has range in X . Were X isomorphic to a dual space, it would carry a locally convex topology T making the (original) ball of X T -relatively compact. A pointwise T -limit of those differences will exist, and the sought linear mapping will be obtained by adding the “non-linear part” that was previously subtracted. To be precise, let A be the set of all partitions of [0, 1] into a finite number of disjoint measurable sets, and for every α ∈ A, let Yα be the subspace of L 1 [0, 1] spanned by the characteristic functions of the sets in the decomposition. Then each Yα is isometric to an n1 α for some positive integer n α —and we may nα identify n a this two spaces. n a For each α ∈ A, define an operator Sα from 1 into 1 by Sa ( i=1 βi ei ) = i=1 βi xi , for β1 , . . . , βn α ∈ R, where e1 , . . . , en α are the standard unit vectors in n1 a and xi ∈ 1 satisfy xi  ≤ 2 and T xi = ei , i = 1, 2, . . . , n α . Then Sα  ≤ 2 and T Sα = IYα , the identity on Yα , for all α ∈ A. Let ϕ be a mapping  (not necessarily linear or continuous) from L 1 [0, 1] to 1 such that T ϕ = I  L [0,1] and, for all y ∈ L 1 [0, 1], ϕ(y) ≤ ηy for some η < ∞ (use the 1 Banach Open Mapping theorem). Then for each α ∈ A and each y ∈ Yα , we have Sα (y) − ϕ(y) ∈ X .

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Put, for a ∈ A and y ∈

α∈A Yα ,

 πa (y) =

Sα (y) − ϕ(y) if y ∈ Yα , 0 otherwise.

" This defines an element πα ∈ y∈ Yα (2 + η)yB X . If A is partially ordered by (, where α ( β means that every element in α is a union of elements in β, we obtain a net {πα ; α ∈ A, (}. Assume that X is isomorphic to a dual Banach space. Then a locally convex topology T exists on X such that B X is relatively " " compact in (X, T ). The set y∈ Yα (2 + η)yB X is then relatively compact in α∈A (X, T ) " by Tychonov’s theorem, hence the net {πα } has a cluster point π in α∈A (X, T ).

Fix y ∈ α∈A Yα . Then y ∈ Yα0 for some α0 ∈ A and, for every α ) α0 we have Sα = πα + ϕ (a linear mapping). It follows that S := π + ϕ is a linear mapping and, obviously, S(y) ≤ (2 + η)K y for some K > 0 independent of y. The mapping S can be extended to a continuous operator  S from (L 1 [0, 1] =) α∈

A Yα into 1 . Moreover, T S(y) = T π(y) + T ϕ(y) = T ϕ(y) = y for all y ∈ α∈A Yα , since π(y) ∈ X . It follows that the mapping  S is an isomorphism from L 1 [0, 1] into 1 . This finishes the proof. Had the space X an unconditional basis, it will certainly be not boundedly complete, since X is not isomorphic to a dual space, see Theorem 4.15. Then, the space X would have an isomorphic copy of c0 , see Theorem 4.37, and this is impossible. 5.72 Let X be a subspace of a Banach space Y , Z a finite-dimensional Banach space and T a bounded operator from X into Z . Show that T can be extended to a bounded operator from Y into Z . n be a Hamel basis for Z and define elements fi ∈ X ∗ by Hint. Let (ei , ei∗ )i=1 ∗ ∗ f i be a Hahn–Banach extension of f i to Y, f i = T ei , i = 1, 2, . . . , n. Let  n  : Y → Z by T y =  i = 1, 2, . . . , n. Define an operator T i=1 f i (y)ei , for y ∈ Y . Then for x ∈ X we have x = T

n  i=1

f i (x)ei =

n  i=1

(T ∗ ei∗ )(x)ei

=

n 

ei∗ (T x)ei = T x.

i=1

 can be taken in Exer5.73 Does there exist an absolute constant K > 0 so that T  cise 5.72 so that T  ≤ K T ? Hint. No. There is an infinite-dimensional Banach space X and a number δ > 0 such that any finite rank projection P from X into X satisfies P ≥ δ(rank P)1/2 (Pisier, see Remarks in Chapter 6). So if X n are n-dimensional subspaces of this X , then the identity mappings on them cannot be extended to mappings from X into X n with norms less than or equal to K . 5.74 Let X be isomorphic to a Hilbert space, Y ⊂ X be a subspace of X and T a  bounded operator from Y into a Banach space Z . Show that there is an operator T

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from X into Z that extends T . Note that the Hahn–Banach theorem follows from this in the case of Hilbert spaces. Hint. Let P be a projection from X into Y . Consider the operator T ◦ P. 5.75 Let X and Y be infinite-dimensional separable Banach spaces. Show that there is a one-to-one compact operator T from X into Y such that T (X ) is linearly dense in Y . ∞ and {y ; g }∞ be Markushevich bases for X and Y , respecHint. Let {xi ; f i }i=1 i i i=1 tively.  Assume that { f i : i ∈ N} and {yi : i ∈ N} are bounded. Put for x ∈ X , T x = i (1/2i ) f i (x)yi . Then T is compact and one-to-one (if x = 0 there is i such that f i (x) = 0 and then gi (T x) =

1 1 f i (x)gi (yi ) = i f i (x) = 0 2i 2

by the biorthogonality of {yi } and {gi }). Moreover, T xi =

 1 1 f j (xi )y j = i yi 2j 2

by the biorthogonality of {xi } and { f i }. 5.76 Let X and Y be infinite-dimensional separable Banach spaces. Then there is a compact one-to-one operator T from X ∗ into Y such that T (X ∗ ) is linearly dense in Y. Hint. Follow the hint in Exercise 5.75. 5.77 Assume that Y is a subspace of X and W is a subspace of Z . Assume that Y is isomorphic to W and X/Y is isomorphic to Z /W . Is X necessarily isomorphic to Z? Hint. No. There is a space Z that is not isomorphic to a Hilbert space and a subspace W such that both W and Z /W are isomorphic to a Hilbert space (Enflo, Lindenstrauss, and Pisier [ELP]). Alternatively one can use the Ciesielski–Pol space in [DGZ3, p. 260]. 5.78 Find a non-separable space X with a non-complemented subspace isomorphic to c0 . Hint. Use Theorem 14.54. 5.79 Find an example of an element of ∞ that does not attain its norm. Hint. x n = (1 − 1/n).  5.80 Show that ( n n2 )0 is isomorphic to a quotient of c0 . Hint. Use the following dual version of the Dvoretzky Theorem 6.15: Let k be an integer. Then every infinite-dimensional Banach space has a k-dimensional quotient

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space Mk with the quotient norm  ·  and an Euclidean norm · | in it such that  | n for  every x ∈ Mk we have x ≤ |x| ≤ 2x. Then (  ) is 2-isomorphic 0 n 2  c (= ( c ) ). Note that c to ( n Mn )0 which is a quotient of 0 0 0 0 is isomorphic n  n to a complemented subspace of (  ) (consider the first coordinates in n2 ). 0 n 2  n Therefore the spaces ( n 2 )0 and c0 have, up to isomorphisms, the same quotient spaces. 5.81 Let T be a completely regular topological space and BC(T ) be the space of bounded continuous real-valued functions on T with the supremum norm. Then BC(T ) is a Banach space. Show that the mapping ϕ : t → δt , where δt is the Dirac measure at t, is a homeomorphic mapping from T into (B BC(T )∗ , w ∗ ) and that the w∗ ˇ Cech–Stone compactification of T is just ϕ(T ) ⊂ B BC(T )∗ . Hint. Definition of complete regularity and Alaoglu’s theorem.  5.82 Show that ( n n2 ) p is isomorphic to  p if 1 < p < ∞. Note that this gives that the image of the standard basis in ( n2 ) p by this isomorphism provides an unconditional basis in  p that is not equivalent to the unit vector basis in  p if p = 2. Hint. Let p > 2. Let Fn be the subspace of L p [0, 1] spanned by the characteristic functions of the intervals [k/2n , (k + 1)/2n ], k = 0, 1, . . . , 2n−1 . Then Fn is n isometric to 2p and P Fn = span{ri : i = 1, . . . , n}, where P is the projection ∞ 1 P f := n=1 ( 0 f (s)rn (s) ds)r n , and r n are the Rademacher functions (see the proof of Khintchine’s inequality, Lemma 4.54). There is a constant K p so that for n every n, there is a subspace Cn of 2p with dist(Cn , n2 ) ≤ K and the projection  n from 2p onto C n is of norm ≤ K p . Thus the space ( n n2 ) p is isomorphic to a  n  complemented subspace of ( n 2p ) p ≈  p . Therefore ( n n2 ) p is isomorphic to  p by Corollary 4.48. By duality the result holds for 1 < p < 2. 5.83 Assume that T is a noncompact operator from X into 1 . Show that X contains an isomorphic copy of 1 . If T is a nonweakly compact operator from X into L 1 then X contains an isomorphic copy of 1 . Hint. Since T (B X ) is not  · -relatively compact, there exists a sequence {x n } in B X and some ε > 0 such that T (xn ) − T (x m ) ≥ ε for all n = m. Schur’s Theorem 5.36 and Rosenthal’s Theorem 5.37 give the conclusion. For the second part, if T (B X ) is not w-relatively compact, there exists, by the Eberlein–Šmulyan theorem, a sequence {x n } in B X such that {T (xn )} has no w-convergent subsequence. The sequence {xn } cannot be w-Cauchy due to the weak sequential completeness of L 1 . Use again Rosenthal’s Theorem 5.37. 5.84 Let X be a Banach space. Let {x n } be a bounded sequence in X such that {xn } has no weakly Cauchy subsequence. It follows from Rosenthal’s Theorem 5.37 that X contains an isomorphic copy of 1 . Prove this result for a Banach space having an unconditional basis without using Rosenthal’s theorem.

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Hint. X ∗ must be non-separable, otherwise we use the Cantor diagonal procedure to produce a weak Cauchy subsequence. If X ∗ is non-separable, use Theorem 4.42. 5.85 An example of a weakly sequentially complete space is any reflexive space (show this) and any general L 1 (μ) space (this is a classical Steinhaus theorem, see the proof in Exercises 13.48 and 13.49). Show that if X is weakly sequentially complete and nonreflexive, then X must contain an isomorphic copy of 1 . Hint. The first problem: use the Eberlein–Šmulyan theorem. The second problem: If X is not reflexive, by the Eberlein–Šmulyan theorem there exists a bounded sequence {xn } without w-convergent subsequence. Since X is weakly sequentially complete, {xn } has no w-Cauchy subsequence. Use now Rosenthal’s Theorem 5.37. 5.86 A set D in a Banach space X is called limited in X if f n (x) → 0 uniformly on w∗

x ∈ D whenever f n ∈ X ∗ and f n → 0. The space X is called a Gelfand–Phillips space if every limited set in X is ·-relatively compact. (Note that any ·-relatively compact subset of X is limited in X by the Banach–Steinhaus uniform boundedness principle). Show that (i) Every limited set in X is bounded. (ii) Every separable Banach space is a Gelfand–Phillips space. (iii) The space ∞ is not a Gelfand–Phillips space. Hint. (i) If not, there are di ∈ D with di  > i for all i ∈ N. Pick f i ∈ S X ∗ with w∗

f i (di ) = di . Then (1/i) f i → 0 and (1/i) f i (di ) > 1 for all i ∈ N. (ii) If D is not  · -relatively compact, then there are x n ∈ D, n ∈ N, so that lim infn dist(x n , E n−1 ) > 0, where E n−1 := span{x1 , x2 , . . . , x n−1 }. This follows from the compactness of balls in finite-dimensional spaces. Choose ε > 0 and f n ∈ S X ∗ so that f n (xn ) > ε and f n (xk ) = 0 for k < n. Since B X ∗ is w∗ -sequentially compact, let lim f n k = f in the w∗ -topology of X ∗ , where n k is a subsequence of k

the integers. Then f n k (xn k ) > ε and f (xn k ) = lim f n j (xn k ) = 0 for all k. j

∗∞

w∗

are such that f n → 0, then, by the (iii) Put D = Bc0 ⊂ ∞ . If f n ∈ w f n of f n to c0 converge Grothendieck property of ∞ , f n → 0, and the restrictions  weakly to 0 in 1 . By the Schur property of 1 (= c0∗ ),  f n → 0 in the norm topology of c0∗ and thus f n (x) → 0 uniformly on x ∈ D (= Bc0 ). 5.87 Let X be an infinite-dimensional Banach space. Show that B X is not limited in X. Hint. This is a reformulation of the Josefson–Nissenzweig theorem (see Exercise 3.39). 5.88 Assume that K is a compact space and X is a separable subspace of C(K ). Show that there is a separable subspace X˜ of C(K ) such that X˜ ⊃ X and X˜ is isometric to a C(S) space for some compact metric space S.

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Hint. Let  X be a closed separable subalgebra in C(K ) that contains X and the constant functions on K . Let S be formed by all nonzero continuous linear funcX ∗ for which x ∗ ( f g) = (x ∗ ( f ))(x ∗ (g)) for all tionals x ∗ in the closed unit ball of   f, g ∈ X . Then it is shoed in Dunford–Schwartz, p. 275 that it follows from the Stone–Weierstrass theorem that  X is isometric to C(S). 5.89 Assume that Y → X is such that both Y and X/Y are isomorphic to subspaces of ∞ . Show that X is isomorphic to a subspace of ∞ . Hint. Let q : X → X/Y be the quotient mapping, T be an isomorphism of Y into ∞ , and S be an isomorphism of X/Y into ∞ .  is x) is an isomorphism from X into ∞ ⊕ ∞ , where T Then X * x → (Sq x, T an extension of T to X . ∞ ⊂ X, 5.90 There exists a Banach space X , ε > 0 and a bounded sequence {xi }i=1 such that for every disjoint A, B ⊂ N, there exists f ∈ B X ∗ with infi∈A f (xi ) ≥ supi∈B f (xi ) + 2ε, yet the sequence {xi } is not equivalent to the unit basis of 1 . Hint. The idea is to define the norm on the space c00 by setting the norming set consisting of aχ A + bχ B choosing a and b conveniently in order that ( n1 , n1 , . . . , n1 , 0 . . . ) will evaluate almost zero.

5.91 Show that L ∞ (Ω, μ) = L i∗ (Ω, μ) is 1-injective, if μ is σ -finite. Hint. We use a well-known compactness argument: Let ω = (Ω1 , Ω2 , · · · , Ωm ) denote a finite partition of Ω into mutually disjoint measurable sets of positive measure. Let A denote the collection of all such partitions, partially ordered as follows: For ω, γ ∈ A, ω = (Ω1 , Ω2 , · · · , Ωm ) < γ = (Γ1 , Γ2 , · · · , Γn ) if n > m and each Ωi is a union of members of γ . Then m of L (Ω, μ) is isometric A is directed by < and each subspace E ω = [χΩi ]i=1 ∞ to ∞ (1, 2, · · · , m). Hence E ω is 1-injective and thus, whenever X ⊃ L ∞ (Ω,

μ), there is a projection Pω from X onto E ω with Pω  = 1. It is clear that ω E ω is dense in L ∞ (Ω, μ). The unit ball U := BB(X,L ∞ (Ω,μ)) of the space of bounded operators from X into L ∞ (Ω, μ), under the pointwise w∗ -topology, is compact. Since {Pω } is a net in U directed by <, it contains a convergent subnet with limit P ∈ U . Then P is a projection from X onto L ∞ (Ω, μ) with P = 1, and hence L ∞ (Ω, μ) is a 1-injective space. Note that this gives, by Pełczy´nski Decomposition Method, that L ∞ is isomorphic to ∞ . 5.92 Show the following result [Zipp2]: Let X be a Banach space, let E be a subspace of X and let λ ≥ 1. Let K be a compact space and let T : E → C(K ) be a bounded operator. Then T can be extended to a bounded operator Tˆ : X → C(K ) so that Tˆ  ≤ λT  if there is a w∗ -w∗ -continuous function ϕ : B E ∗ → λB X ∗ which extends functionals (i.e., ϕ(e∗ )(e) = e∗ (e) for all e ∈ E and e∗ ∈ B E ∗ ). Hint. Let T : E → C(K ) be an operator of T  = 1. Let the function ψT : K → B E ∗ be defined by ψT (k)(e) = (T e)(k), k ∈ K . Put ψ = ϕ(ψT ) : K → λB X ∗ . Then ψ is w ∗ -continuous. Define Tˆ : X → C(K ) by Tˆ x(k) = ψ(k)(x). Then Tˆ is

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5 Structure of Banach Spaces

linear as for each k ∈ K , ψ is a linear functional; Tˆ  ≤ λ because ψ(k) ∈ λB X ∗ and so ψ(k) ≤ λ and Tˆ extends T because ϕ extends functionals from E ∗ : if e ∈ E then (Tˆ e)(k) = ψ(k)(e) = (ϕ(ψT (k))(e) = ψT (k)(e) = (T e)(k) for all k ∈ K . 5.93 Show that if T is a bounded operator from a subspace of  p into C(K ) space for p > 1, then T can be extended to  p with the same norm. Hint. Exercise 5.92. 5.94 A Banach space X is said to have the binary intersection property if every family of mutually intersecting closed balls has a common point. Assume that K is an extremely disconnected compact space (i.e., the closure of every open subset of K is open). This is known to be equivalent to saying that every nonempty subset of C(K ) that has an upper bound (in the natural order of C(K )) has a least upper bound (see, e.g., [GiJe]). Show that then C(K ) has the binary intersection property. Hint. Let u denote the constant function 1 on K . A closed ball B(x, r ) in C(K ) with center x and radius r is exactly the order segment [x − r u, x + r u] = {y ∈ C(K ) : x − r u ≤ y ≤ x + r u}. If A = [x α , yα ]α∈A is a collection of mutually intersecting segments, then, for every α, β ∈ A, there is a z α,β such that x α , xβ ≤ z α,β ≤ yα , yβ . Hence {x α }α∈A is order bounded from above and thus x = sup{x α : α ∈ A} exists and is a common point of all the segments in A.

5.95 Assume that X has the binary intersection property. Show that X is a 1injective space. In particular, by taking X = R, this gives an alternative proof of the Hahn–Banach theorem. Hint. By Zorn’s lemma, it suffices to show that if Z ⊃ Y , dim(Z /Y ) = 1 and T : Y → X is an operator with norm T  = 1, then T admits an extension Tˆ : Z → X with Tˆ  = 1. Let z ∈ Z /Y and consider the family {B(T y, z − y) : y ∈ Y } of balls in X . Any two of these balls intersect because for y1 , y2 ∈ Y , T y1 − T y2  ≤ y1 − y2  ≤ z − y1  + z − y2 . Therefore, there is a point e common to all the balls of this family. Define Tˆ : Z → X by Tˆ (az + y) = ae + T y for all az + y ∈ Z . It is routine to check that Tˆ extends T and Tˆ  = T  = 1. 5.96 Let Γ be a set with card(Γ ) = 2ℵ0 . Show that the duals of the spaces c0 (Γ ) ⊕∞ C([0, 1]) and C([0, 1]) are isometric, although one of these spaces is separable and the other is not. Hint. Duality of classical spaces.

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5.97 Let X be a hereditarily indecomposable Banach space. Show that X does not contain any infinite-dimensional subspace with an unconditional basis; a fortiori it contains no subspace isomorphic to either c0 or  p . Hint. Projections on spans of infinite subsets of the unconditional basis. 5.98 Show that a Banach space X is an HI space if and only if for all subspaces Y and Z of X , we have inf{y − z : y ∈ SY , z ∈ S Z } = 0. Hint. Exercise 4.18. 5.99 Show that X is an HI space if and only if for every subspace Y → X , the quotient mapping πY : X → X/Y is strictly singular. Hint. See Exercise 4.32.

Chapter 6

Finite-Dimensional Spaces

The interplay between the structure of an infinite-dimensional Banach space and properties of its finite-dimensional subspaces belongs to the subject of the local theory of Banach spaces. It is a vast and deep part of Banach space theory intimately related to probability and combinatorics. Our goal is to familiarize the reader with some of its basic notions and results that are accessible without the use of deep probabilistic tools. We begin with the notion of finite representability, the principle of local reflexivity and more finite representability results such as the Brunel–Sucheston spreading models technique. We prove Tzafriri’s theorem and combine it with previous results to get the crude version of Dvoretzky’s Theorem 6.15, which is sufficient for the proof of the Lindenstrauss–Tzafriri theorem on complemented subspaces. We introduce the John ellipsoid of maximal volume, and apply it to obtain the Kadec– Snobar estimate on norms of projections onto finite-dimensional subspaces. Finally we prove the Grothendieck inequality.

6.1 Finite Representability Definition 6.1 Let X, Y be Banach spaces. We say that Y is crudely finitely representable in X if there is K > 0 such that for every finite-dimensional subspace F of Y there is a linear isomorphism T of F onto T (F) ⊂ X so that T  · T −1  < K . We say that Y is finitely representable in X if for every ε > 0, Y is crudely finitely representable in X with constant K = 1 + ε. We observe that if Y is finitely representable in X and W is finitely representable in Y , then W is finitely representable in X . The concept of finite representability was first studied by James ([Jame7], [Jame8]). We begin with some simple results. Theorem 6.2 (i) Every Banach space is finitely representable  n  in c0 . (ii) Every Banach space is finitely representable in ∞ 2 . (iii) Let p ≥ 1. Then L p [0, 1] is finitely representable in  p . (iv) If X is crudely finitely representable in a Hilbert space, then X is isomorphic to a Hilbert space.

M. Fabian et al., Banach Space Theory, CMS Books in Mathematics, C Springer Science+Business Media, LLC 2011 DOI 10.1007/978-1-4419-7515-7_6, 

291

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6 Finite-Dimensional Spaces

Proof: (i) and (ii) Let Y be a Banach space, ε > 0, and F be a finite-dimensional n subspace of Y . Let {yi }i=1 be an ε-net in S F . Find f i ∈ SY ∗ such that f i (yi ) = 1, i = 1,. . . , n. Consider the mapping T : F → n∞ defined for y ∈ F by T (y) = n f i (y) i=1 . Then   T (y) = max{| f i (y)|} ≤ max  f i  · y ≤ y. i

i

Given y ∈ S F , we find i ∈ {1, . . . , n} such that y − yi  < ε. Then T (y) ≥ | f i (y)| = | f i (yi ) + f i (y − y0 )| ≥ | f i (yi )| −  f i  · y − yi  ≥ 1 − ε. Therefore T (y) ≥ (1 − ε)y for every y ∈ F, which means that T −1  ≤ (1 − ε)−1 . To prove (i) it is enough to consider the mapping   ST of F into c0 , where S is a linear isometry from n∞ into c0 defined by S (xi ) = (x1 , . . . , x n , 0, 0, . . . ).  ∞ n  by For (ii) we consider an isometry  S from n∞ into ∞ 2 defined n=1      ∞ n  S (xi ) = 0, . . . , 0, (xi ), 0, . . . , where (xi ) is in the nth block in n=1 ∞ . 2

For (iii) assume that E is a finite-dimensional subspace of L p [0, 1], with a linear basis {b1 , . . . , bn }. By perturbing (as closely as we wish) the basis functions into a system of simple functions { f 1 , . . . , f n }, we see that those form a linear basis of a subspace F of L p [0, 1] which is (as closely as we wish) almost isometric to E (i.e., d(E, F) is as small as we wish). There exists a system of disjoint measurable subsets E i , 1 ≤ i ≤ N , of [0, 1], such that, for k = 1, 2, . . . , n, f k ∈ span{χ Ei ; 1 ≤ i ≤ N } := Y . It is easy to see that Y is isometric to  Np . For (iv) assume for simplicity that X is separable, X 1 ⊂ . . . ⊂ X n ⊂ X n+1 ⊂ . . .

are subspaces of X such that X = ∞ n=1 X n and, for n ∈ N, the space X n is ndimensional. By assumption, there exists K > 0 and operators Tn : X n → n2 such that Tn−1  > K , Tn  ≤ 1. For n ∈ N, let G n : X n × X n → R be defined by G n (x, y) = Tn x, Tn y. The mapping G n defines an inner product on the respective space X n , such that G n (x, x) ≤ x2 ≤ K 2 G n (x, x). To finish the proof, it suffices lim G n k (x, y) exists for to pass to a subsequence {n k }∞ k=1 , so that G(x, y) := k→∞

∞ every x, y ∈ n=1 X n . Indeed, it is clear that G is an inner product extendible to the whole X , whose corresponding norm is equivalent to the original norm on X . The following fundamental principle due to Lindenstrauss and Rosenthal plays an important role in the theory. It implies, of course, that for every Banach space X , X ∗∗ is finitely representable in X , but the statement contains much more information. Theorem 6.3 (principle of local reflexivity, [LiRo2], [JRZ]) Let X be a Banach space. For every finite-dimensional subspace E of X ∗∗ , finite-dimensional subspace F of X ∗ and ε > 0, there exists  isomorphism T of E onto T (E) ⊂ X such  a linear that T  · T −1  ≤ 1 + ε, x ∗ T (x ∗∗ ) = x ∗∗ (x ∗ ) for x ∗ ∈ F and x ∗∗ ∈ E, and T is the identity on E ∩ X .

6.1

Finite Representability

293

Proof: (Stegall [Steg3]) Choose δ > 0 so that θ (δ) < (1 + ε), where θ (δ) is the ∗ ∈ S ∗ containing a basis of F and function in Exercise 1.76. Choose a1∗ , . . . , am X ∗∗ ∗∗ ∗ such that x  ≤ (1 + δ) sup j |x (a j )| for all x ∗∗ ∈ E. Finally, choose a δ-net {b1∗∗ , . . . , bn∗∗ } in S E such that {b1∗∗ , . . . , bk∗∗ } is a basis of E ∩ X and {b1∗∗ , . . . , br∗∗ }, r ≥ k, is a basis of E. Let q = n − r . Then for every p ∈ {1, . . . , q} we have unique scalars t p,i , 1 ≤ i ≤ r such that br∗∗+ p = ri=1 t p,i bi∗∗ . Define

s p,i

Denote X m =

+ m i=1

X

∞

⎧ ⎨ t p,i i ≤ r, = −1 i = r + p, ⎩ 0, r < i ≤ n and i = r + p. for m ∈ N and define A0 : X n → X k+q by

n n    A0 (x1 , . . . , xn ) = x1 , . . . , xk , s1,i xi , . . . , sq,i xi . i=1

i=1

n s j,i bi∗∗ = The operator A0 is onto, since the matrix (s p,i ) has rank q. Note that i=1 ∗∗ ∗∗ ∗∗ k+q n . Define U : X → Rnm by 0 for j = 1, . . . , q, so A0 (b1 , . . . , bn ) ∈ X n,m ∗ k+q nm U (x 1 , . . . , xn ) = {a j (xi )}i, j=1 . Let Z = X × R , define A : X n → Z by  A(x1 , . . . , xn ) = A0 (x1 , . . . , xn ), U (x 1 , . . . , xn ) . By Exercise 5.12, A is an operator with a closed range. Since A is onto and ∗∗ ∗∗ k+q , there exist d , . . . , d such that A (d , . . . , d ) = A∗∗ 1 n 0 1 n 0 (b1 , . . . , bn ) ∈ X ∗∗ ∗∗ ∗∗ A0 (b1 , . . . , bn ). We claim that in fact there exist e1 , . . . , en such that A(e1 , . . . , en ) = A∗∗ (b1∗∗ , . . . , bn∗∗ ).   Indeed, since U is a finite rank operator, U (Ker(A0 )) = U Ker(A0 )∗∗ = ∗∗ ∗∗ ∗∗ U (Ker(A∗∗ 0 )). Because (b1 , . . . , bn ) ∈ (d1 , . . . , dn ) + Ker(A0 ), there exists (e1 , . . . , en ) ∈ (d1 , . . . , dn ) + Ker(A0 ) so that U (e1 , . . . , en ) = U (b1∗∗ , . . . , bn∗∗ ) and the claim follows. In particular, A∗∗ (b1∗∗ , . . . , bn∗∗ ) ∈ A(X n ). By Exercise 3.87, there is (b1 , . . . , bn ) ∈ X such that A(b1 , . . . , bn ) = A∗∗ (b1∗∗ , . . . , bn∗∗ ) and moreover sup bi  ≤ (1 + δ) sup bi∗∗  = 1 + δ. ∗∗ Define an operator i ) = bi for 1 ≤ i ≤ r . Note that for rT : E →∗∗ X by T (b 1 ≤ p ≤ q we have i=1 s p,i bi = 0 and ri=1 s p,i bi = 0. Thus we have also that T (bi∗∗ ) = bi for r < i ≤ n. Note that for each i we have   T (bi∗∗ ) ≥ sup |a ∗j T (bi∗∗ ) | = sup |bi∗∗ (a ∗j )| ≥ (1 + δ)−1 j

Thus T  · T −1  ≤ 1 + ε by the choice of δ.

j

294

6 Finite-Dimensional Spaces

6.2 Spreading Models Given a set X , we let X (n) to be the set of all subsets of X of cardinality n. We k say that a system of k disjoint sets {Si }i=1 forms a partitioning of X (n) whenever

k (n) X = i=1 Si . k Proposition 6.4 (Ramsey [Rams]) Let k, n ∈ N. Then for every partitioning {Si }i=1 of N(n) there exists i ∈ {1, . . . , k} and an infinite set M ⊂ N, such that M (n) ⊂ Si . An equivalent formulation is the following. Let n be a natural number. Let ψ be a mapping from N(n) to some finite set C. Then there is an infinite subset M of N such that ψ is constant on M (n) .

Still a paraphrasis of the previous statement is the following: If a coloring (with finite number of colors) of sets of natural numbers of a given length n is defined, then there is an infinite subset M of N such that all subsets of M of length n have the same color. Proof of Proposition 6.4: By induction on n. For n = 1 the result is obvious. Assume that for some n > 1 the statement has been proved for 1, 2, . . . , n−1. We shall prove it for n. The argument is based on the following observation: if we fix some j ∈ N, we may consider all elements in N(n) that contain j. Define a mapping ψ from (N\{ j})(n−1) into C as ψ (F) = ψ(F ∪ { j}) for all F ∈ (N\{ j})(n−1) . This is a coloring of all finite subsets of length n − 1 in (N\{ j})(n−1) , hence, by the induction hypothesis, there exists an infinite subset M1 of N\{ j} such that all subsets of M1 of length n − 1 get the same color (i.e., ψ is constant on them). This means that ψ is constant (the same constant) on all sets F ∪ { j}, where F ∈ (N\{ j})(n−1) . To prove the assertion for n, we iterate the construction above: let us start with n 0 := 1, and find an infinite subset M1 of N\{n 0 } such that ψ is constant on all sets of the form {n 0 } ∪ F, for F a subset of length n − 1 of M1 . Let n 1 = min M1 (> n 0 ). Find an infinite subset M2 of M1 \{n 1 } such that ψ is constant (maybe a different constant) on all sets of the form {n 1 }∪F, for F a subset of length n−1 of M2 , and put n 2 = min M2 (> n 1 ). Continue in this way to obtain a sequence M1 ⊃ M2 ⊃ . . . of infinite sets (and the sequence n 0 < n 1 < n 2 < . . .). By passing to a subsequence if necessary (denoted again {Mi }), we may assume that the same constant is associated ∞ . to all Mi s. The sought set is then {n i }i=1 Lemma 6.5 Let {xn }∞ n=1 be a seminormalized basic sequence in a Banach space X , ∞ k ∈ N, ε > 0. Then there exists a subsequence {yn }∞ n=1 of {x n }n=1 such that for all scalars ai , 1 ≤ i ≤ k,  k   k    k             ai yi  ≤  ai yn i  ≤ (1+ε)  ai yi  , whenever n 1 < · · · < n k . (1−ε)        i=1 i=1 i=1 (6.1) Proof: Since the sequence {xn }∞ n=1 is basic and normalized, we can find C > 0 that satisfies (4.1) in Proposition 4.18, and R > 0 such that R −1 ≤ xn  ≤ R for all

6.2

Spreading Models

295

n ∈ N. In particular, given scalars a1 , . . . , ak , we have, for every i ∈ {1, 2, . . . , k},   k   ≤ R ai x n i  |ai | ≤ Rk max |ai |.  1≤i≤k  i=1 i=1   k 

C −1 R −1 |ai | ≤ C −1 |ai |.x n i  = C −1 ai x n i  ≤  

This implies the existence of K > 0 such that, for all scalars a1 , . . . , ak , K

−1

 k      max |ai | ≤  ai xn i  ≤ K max |ai |, whenever n 1 < · · · < n k . (6.2)   1≤i≤k 1≤i≤k i=1

j

j

k Fix δ > 0. Consider a δ-net {(a1 , . . . , ak )} M j=1 in the unit ball of ∞ . Let N0 = N and proceed by finite induction in j = 1, . . . , M as follows. Assume that for some j ∈ {1, 2, . . . , M}, N0 , N1 , . . . , N j−1 have been already (k) defined. Partition N j−1 into {Sm }1≤m≤ p , for p := [δ −1 (K − K −1 ) max1≤i≤k |ai |], where [·] denotes here the integer part of a number, and

Sm := {(n 1 , . . . , n k ) : n 1 < n 2 < . . . < n k ,  k       j −1 −1 max |ai | + (m − 1)δ ≤  ai xn i  < K max |ai | + mδ . K   1≤i≤k 1≤i≤k i=1

Then, Proposition 6.4 gives m j ∈ {1, 2, . . . , p} and an infinite subset N j of N j−1 (k) such that N j ⊂ Sm j . This defines N j and finishes the finite induction process. ∞ and let y = x Denote N M = {m i }i=1 i m i for i ∈ N. It follows that  k  k         j j  ai yni  −  ai yi  < δ, for every n 1 < . . . < n k .    i=1

(6.3)

i=1

Observe that  k  k          j ai yn i  −  ai yn i     i=1

i=1

k     j j (ai − ai )yn i  ≤ Rkδ, whenever (ai )k1 − (ai )k1  < δ. ≤

(6.4)

i=1 k k A similar estimate holds replacing {yn i }i=1 by {yi }i=1 . It follows from (6.3) and (6.4) that

  k k          ai yn i  −  ai yi  < δ(1 + 2Rk).    i=1

i=1

(6.5)

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6 Finite-Dimensional Spaces

Take now an arbitrary (a1 , . . . , ak ) ∈ k∞ . From (6.5) we get          k k k          i=1 ai yi  i=1 ai yn i  i=1 ai yi   −δ(1+2Rk) ≤  ≤ +δ(1+2Rk),  max1≤i≤k |ai |   max1≤i≤k |ai |   max1≤i≤k |ai |  so  k  k    k              ai yi  − δ(1 + 2Rk) max |ai | ≤  ai yn i  ≤  ai yi  + δ(1 + 2Rk) max |ai |.        1≤i≤k 1≤i≤k i=1

i=1

i=1

Taking in account (6.2), we get  k   k    k               ai yi  ≤  ai yn i  ≤ 1 + K δ(1 + 2Rk)  ai yi  . 1 − K δ(1 + 2Rk)        i=1

i=1

i=1

Choosing δ small enough leads to the estimate (6.1). ∞ that a basis ∞ {en }n=1 of a Banach space is called subsymmetric if Recall ∞  i=1 ai ei  =  i=1 ai eki  for every infinite increasing sequence of integers ∞ . (ki )i=1

Theorem 6.6 (Brunel, Sucheston [BrSu2]) Let εn - 0, {x n }∞ n=1 be a normalized basic sequence in a Banach space X . Then there exists a subsequence {yn }∞ n=1 of ∞ , such and a Banach space (Y, | · |) with a subsymmetric basis {e } {xn }∞ n n=1 n=1 that for all k ∈ N and all scalars a1 , a2 , . . . , ak ,                  k    k  k         (1−εk )  ai ei  ≤  ai yn i  ≤ (1+εk )  ai ei   , whenever k ≤ n 1 < · · · < n k .   i=1    i=1  i=1 (6.6)

Proof: A repeated application of Lemma 6.5 leads to a decreasing sequence of infi∞ , such that nite subsets M1 ⊃ M2 ⊃ . . . of the integers, Mk := (m ik )i=1  k  k   k               ai xm k  ≤  ai xni  ≤ (1 + εk )  ai xm k  (1 − εk )  i i     i=1

i=1

(6.7)

i=1

∞ whenever n 1 < · · · < n k belong to Mk . Pass to the diagonal sequence M = (m ii )i=1 ∞ of the system {Mn }n=1 , and let yi = x m i for i ∈ N. Finally, let Y be the completion i of the normed space of all finitely supported elements of c0 equipped with the norm

  k   k          ai ei  := ai yn i  . lim   n 1 <···
i=1

It is clear from (6.7) that the limit exists and (6.6) is satisfied.

(6.8)

6.2

Spreading Models

297

The Banach space Y constructed above is called the spreading model built from the sequence {x n }∞ n=1 . Typically, a given basic sequence leads to many mutually non-isomorphic spreading models. We also say that Y is a spreading model of X provided that Y results as a spreading model built on some basic sequence in X . Theorem 6.7 Every spreading model Y of a Banach space X is finitely repre∞ sentable in X . Moreover, if {x n }∞ n=1 is weakly null then {ei }n=1 is an unconditional ∞ basis of Y , where {ei }n=1 is the basis constructed in Theorem 6.6. w

Proof: The finite representability is obvious from (6.6). Suppose that ∞ xn → 0. k Given a finite set S = (s ) ⊂ N, we introduce the projection P ( i S i=1 ai ei ) := i=1  Proposition 4.36, it suffices to prove that given any finitely supported i∈S ai ei . By N ∞ vector v = i=1 ai ei , PS ( i=1 ai ei ) is bounded by |v|. Let {yn }∞ n=1 be the sub∞ sequence of {xn }n=1 constructed in Theorem 6.6. Given a finite increasing sequence N N of integers n 1 < n 2 < . . . < n N , put M = (n i )i=1 and consider v M = i=1 ai yn i  S and v M = i∈S ai yn i . By (6.6), S lim v M  = |PS (v)|, lim v M  = |v|.

n 1 →∞

(6.9)

n 1 →∞

S S ) = v M . Now fix δ > 0 and observe that Choose f MS ∈ B X ∗ such that f MS (v M for a fixed n 1 , we have that for n 2 large enough it holds that card{i : n 1 < i < n 2 , | f MS (yi )| < δ} > N . Indeed, assuming the contrary there exist for arbitrarily large n 2 ∈ N some f n 2 ∈ B X ∗ such that card{i : n 1 < i < n 2 , | f n 2 (yi )| < δ} ≤ N . w∗

w

Taking f ∈ { f n 2 } , we reach a contradiction with the fact that yi → 0. A similar N argument applies to all n i ∈ M. Consequently, there exist an M = (n i )i=1 and N S L = (m i )i=1 , such that m i = n i if, and only if, i ∈ S, and such that | f M (ym i )| < δ for all m i , i ∈ / S. So, f MS (v L ) ≥ f MS (v LS ) − N δ.

(6.10)

Since δ can be chosen arbitrarily small, and N is fixed, using (6.9) leads to the desired estimate |PS (v)| ≤ |v|. Corollary 6.8 Let X be an infinite-dimensional Banach space. Then there exists an infinite-dimensional Banach space with an unconditional Schauder basis which is crudely finitely representable in X . Proof: Mazur’s technique (see Theorem 4.19 and Lemma 4.20) gives a subspace Y of X with a Schauder basis. Applying Rosenthal’s 1 Theorem 5.37 to a separate sequence on the unit sphere of Y (see the proof of Theorem 1.38), Y contains either an 1 basic sequence, or a weakly Cauchy sequence. By taking consecutive differences and applying Corollary 4.27, we get a normalized basic sequence. Theorem 6.6 gives then the assertion.

298

6 Finite-Dimensional Spaces

6.3 Complemented Subspaces in Spaces with an Unconditional Schauder Basis In this section, we prove a theorem of Tzafriri on sequences of complemented finitely-dimensional subspaces—behaving like  p spaces—of Banach spaces with an unconditional basis. In subsequent sections, this theorem will be referred to as the “crude Dvoretzky theorem," in view of the theorem of Dvoretzky—Theorem 6.15 below. Theorem 6.9 (Tzafriri [Tzaf]) Let X be an infinite-dimensional Banach space with an unconditional basis. Then there exist p ∈ {1, 2, ∞}, a constant M > 0, and a n sequence {Pn }∞ n=1 of projections in X such that Pn  ≤ M and d(Pn X,  p ) ≤ M for all n ∈ N. The proof will be carried out for real Banach spaces. The complex version follows with some minor adjustments. Proposition 6.10 Let X be a Banach space with a normalized monotone Schauder basis {en }. Then, for every ε > 0 there exists a sequence {λ( j)} of positive real numbers and a subsequence {z k := en k } of {en } such that 0 < λ( j) − z k1 + z k2 + . . . + z k j  < ε

(6.11)

for every set of indices j < k1 < k2 < . . . < k j , j ∈ N. In addition, 1 ≤ λ( j) < λ(m) + ε

(6.12)

for every j, m ∈ N such that j < m. Proof: Fix ε > 0. For every integer k > 1, we consider a fixed partition of the interval [1, k + 1] (k)

(k)

(k)

(k)

λ0 := 1 < λ1 < λ2 < . . . < λs(k) := k + 1 (k)

(k)

with the property that λ j − λ j−1 < ε for j = 1, 2, . . . , s(k). Define now a function ϕk from all the unordered k-tuples of different positive integers into the set (k) {1, 2, . . . , s(k)} by setting ϕk {n 1 , n 2 , . . . , n k } = j if λ j−1 ≤ en 1 + en 2 + . . . + (k)

en k  < λ j . Now proceed inductively: put N (1) = N and λ(1) = 1. Apply Proposition 6.4 to the function ϕ2 to obtain an infinite set N (2) ⊂ N (1) such that ϕ2 restricted to all unordered couples from N (2) is constant, say j, for some j ∈ {1, 2, . . . , s(2)}. (2) and the In this case, put λ(2) = λ(2) j . Apply again Proposition 6.4 to the set N (2) function ϕ3 restricted to unordered 3-tuples from elements in N : we obtain an infinite set N (3) ⊂ N (2) such that ϕ3 restricted to all unordered 3-tuples from N (3) is constant, say k, for some k ∈ {1, 2, . . . , s(3)}. In this case, put λ(3) = λ(3) k .

6.3

Complemented Subspaces in Spaces with an Unconditional Schauder Basis

299

Continue in this way to obtain a sequence N (1) := N ⊃ N (2) ⊃ N (3) . . . of infinite sets and a sequence {λ( j)}. A standard diagonal argument gives the sequence n 1 < n 2 < n 3 , . . . and the result. The last estimate follows clearly from the monotonicity of the basis. Two estimates follow trivially from inequality (6.11), the monotonicity of the basis and the fact that all of its vectors are normalized. We single them out for future references. 1 = zk1  ≤ z k1 + z k2 + . . . + z k j  ≤ λ( j),

(6.13)

and, if ε ≤ 1, λ( j) ≤ z k1 + z k2 + . . . + z k j  + ε ≤ 2z k1 + z k2 + . . . + z k j ,

(6.14)

for all j < k1 < k2 < . . . < k j , j ∈ N. Proposition 6.11 Fix 0 < ε < 1/2 in Proposition 6.10 and let {λ( j)} and {z k } be the sequences of numbers and vectors, respectively, obtained there. Assume that there exists an integer h > 1 such that λ(hn) ≥ 1 + ε, n ∈ N. λ(n)

(6.15)

Then there exists a constant A and a number q > 2 such that   ⎞1/q ⎛   n n     1  −1/q ⎝ a j z m+k j  A |a j |q ⎠ ≤n λ(n)    j=1 j=1

(6.16)

for every n ∈ N, m ∈ N such that m > n, integers 0 < k1 < k2 < . . . < kn , and every sequence of scalars {a1 , a2 , . . . , an }. Proof: Find r > 2 such that 1 < h 1/r < 1 + ε. Then λ(hn)/λ(n) > h 1/r for all n ∈ N. It follows that, for n, s ∈ N, λ(h s n) > h 1/r λ(h s−1 n) > (h 2 )1/r λ(h s−2 n) . . . > (h s−1 )1/r λ(hn) > (h s )1/r λ(n), i.e., λ(h s n) > (h s )1/r . λ(n)

(6.17)

Let α and β be two integers such that h j−1 < β ≤ h j and h i−1 < α ≤ h i for some i ≤ j in N. We consider two cases.

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6 Finite-Dimensional Spaces

(1) If i < j we get h j−1 < β ≤ h j ≤ h i−1 < α ≤ h i . Then we have λ(β) < λ(h j ) + ε and λ(h i−1 ) < λ(α) + ε. Hence, putting ρ = λ(h i−1 )/λ(h j ), λ(h i−1 ) − ε ρ − ε/λ(h j ) ρ−ε λ(α) > = ≥ j j λ(β) λ(h ) + ε 1 + ε/λ(h ) 1+ε ρ − 1/2 2ρ − 1 2ρ − ρ ρ ρ > = > = > 1 + 1/2 3 3 3 4  1/r  i− j−1 j h ) α 1 1 i− j−1 1/r 1 λ(h ≥ , ≥ h = j 2/r 4 λ(h ) 4 4h β where the last but one inequality comes from (6.17) and the last one from the fact that α/β ≤ h i / h j−1 . (2) Assume now that i = j. Then h j−1 < β ≤ α ≤ h j and we get

λ(β) − ε ε 1 1 λ(α) ≥ = 1− ≥ 1−ε ≥ ≥ 1/r λ(β) λ(β) λ(β) 2 2h



hj h j−1

1/r ≥

1 2h 1/r

 1/r α , β

where the first inequality follows from (6.12) and the last one from the fact that α/β ≤ h j / h j−1 . Thus, in both cases we have

λ(α) 1 ≥ 2/r λ(β) 4h

 1/r α . β

(6.18)

Fix positive integers n < m and k1 < k2 < . . . < kn , and a sequence of scalars (1) (2) (s) {a1 , a2 , . . . , an }. Put a j = b j − b j , where 0 ≤ b j ≤ |a j | for s = 1, 2, and j = 1, 2, . . . , n. Then        n 2     n (s)      a j z m+k j  ≤ b j z m+k j   .   s=1  j=1   j=1

(6.19)

 Let nj=1 b j z m+k j be one of the two sums in the right hand side of (6.19). Let π be a permutation of the integers {1, 2, . . . , n} such that bπ(1) ≥ bπ(2) ≥ . . . ≥ bπ(n) ≥ 0. Then, by using (6.13),

6.3

Complemented Subspaces in Spaces with an Unconditional Schauder Basis

301

     n   n         b j z m+k j  =  bπ( j) z m+kπ( j)     j=1   j=1   1 2     = (bπ(1) − bπ(2) ) z m+kπ(s) + (bπ(2) − bπ(3) ) z m+kπ(s) + . . .  s=1 s=1  n n−1     +(bπ(n−1) − bπ(n) ) z m+kπ(s) + bπ(n) z m+kπ(s)   s=1

s=1

≤ (bπ(1) − bπ(2) )λ(1) + (bπ(2) − bπ(3) )λ(2) + . . . +(bπ(n−1) − bπ(n) )λ(n − 1) + bπ(n) λ(n). Thus, using (6.18),     n   1   b j z m+k j    λ(n)   j=1 

 1/r  1/r 1 2 + (bπ(2) − bπ(3) ) + ... n n    n − 1 1/r +(bπ(n−1) − bπ(n) ) + bπ(n) n :   ;  n  j 1/r j − 1 1/r 2/r = 4h . (6.20) bπ( j) − n n

≤ 4h 2/r (bπ(1) − bπ(2) )

j=1

Set q > r , and let q and r be the conjugate indices to q and r , respectively, i.e., 1/q + 1/q = 1 and 1/r + 1/r = 1. Observe that, for j ∈ N, we have  1/r  1/r ( j − 1)/j ≤ ( j − 1)/j . This implies 1 − ( j − 1)/j ≤ 1/j, hence

j 1/r − ( j − 1)1/r ≤ j −1/r .

(6.21)

Carry (6.21) to (6.20) and use Hölder inequality for q and q to get    n    1   b j z m+k j    λ(n)   j=1 ⎛ ⎞1/q ⎛ ⎞1/q n n n 2/r 2/r    bπ( j) 1 ⎠ 4h 4h ⎝ ≤ 1/r |bπ( j) |q ⎠ ⎝ ≤ n n 1/r j 1/r j q /r j=1

≤

4h 2/r n 1/r

⎛ ⎞1/q  n  ⎝ |a j |q ⎠ j=1

j=1

1 1 − q /r

j=1



1/q





n (1/q −1/r ) ,

(6.22)

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6 Finite-Dimensional Spaces



where the last inequality is obtained by integrating the function 1/x q /r between 0 and n. This amounts to   ⎛ ⎞1/q  < n 2/r   n  4h  ⎝ b j z m+k j  |a j |q ⎠ n −1/q .   λ(n) ≤ (1 − q /r )1/q  j=1  j=1

(6.23)

Since this , is true for any of the sums at the right hand side of (6.19), by putting A = 8h 2/r (1 − q /r )1/q we obtain inequality (6.16). Proposition 6.12 Let V be a 2n -dimensional Banach space generated by a system of vectors {v1 , v2 , . . . , v2n }. Suppose that there exist constants K > 1 and p > 2 such that ⎛ K −1 ⎝

n

2 

 n −1  n ⎞1/ p ⎛ n  2  2 2      p  ⎝ ajvj vj |a j | ⎠ (2−n )1/ p   ≤K    j=1   j=1 j=1 (6.24)

⎞1/ p |a j | p ⎠

(2−n )1/ p ≤ 

j=1

for every set of scalars {a1 , a2 , . . . , a2n }, where p is the conjugate index to p, i.e., 1/ p + 1/ p = 1. Then there is a constant M := M(K , p) (and so independent of n and V while  ·  in V satisfies (6.24)), and a projection P : V → V such that P ≤ M and d(P V, n2 ) ≤ M. Proof: For j = 1, 2, . . . , 2n , k = 1, 2, . . . , n and h = 1, 2, . . . , 2k−1 , put  εk j =

1 if (2h − 2)2n−k + 1 ≤ j ≤ (2h − 1)2n−k , −1 if (2h − 1)2n−k + 1 ≤ j ≤ 2h2n−k .

(6.25)

Let χ S denote the characteristic function of a set S ⊂ [0, 1]. Observe that the functions n

rk :=

2 

εk j χ[( j−1)/2n , j/2n ] ,

k = 1, 2, . . . , n

(6.26)

j=1

are the first n Rademacher functions (see the proof of Lemma 4.54). Put n

wk =

2 

εk j v j ,

k = 1, 2, . . . , n.

j=1

Vectors w1 , w2 , . . . , wn are the n first elements of the so-called Rademacher system in V associated to the basis {v1 , v2 , . . . , vn }. We shall prove that the space W := span{w1 , w2 , . . . , wn } is the range of a projection in V that satisfies the requirements.

6.3

Complemented Subspaces in Spaces with an Unconditional Schauder Basis

303

Khintchine inequalities (Lemma 4.54) for this index p say that there exists K p > 0 such that K −1 p

 n 

1/2 |ak |2

k=1

 n   n 1/2      ≤ ak r k  ≤ K p |ak |2 .   k=1

(6.27)

k=1

p

Observe that n 

ak wk =

k=1

n 

n

ak

k=1

2  j=1

εk j v j =

 n 2n   j=1

 εk j ak v j .

(6.28)

k=1

Hence, putting together (6.24), (6.27), and (6.28), we get −1   −1   n  n      2n   2n 2n                  ak w k   vj =  εk j ak v j   vj       j=1   j=1 k=1 k=1 j=1  ⎛ n   n  p ⎞1/ p  2  n         −n 1/ p ≤K⎝ εk j ak  ⎠ (2 ) =K ak r k       j=1 k=1

≤ K Kp

 n 

k=1

1/2

|ak |2

p

,

(6.29)

k=1

  where the computation of the norm  nk=1 ak rk  p is done in Exercise 6.15. Analogously, −1   n  n 1/2  2n         −1 −1 2 ak wk   vj |ak | .   ≥ K K p    k=1

j=1

(6.30)

k=1

Inequalities (6.29) and (6.30) together show that d(W, n2 ) ≤ K 2 K p K p . Let Q be the orthogonal projection in L 2 [0, 1] whose range is span{r1 , r2 , . . . , rn }. In the proof of Theorem 4.53, we showed that Q acts as a bounded linear projection in every L r [0, 1], r > 1, and that the norm Qr of Q in L r [0, 1] is independent of n. If ⎛ Q⎝

n

2  j=1

then we set

⎞ a j χ[( j−1)/2n , j/2n ] ⎠ =

n  k=1

bk r k ,

(6.31)

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6 Finite-Dimensional Spaces

⎛ P⎝

n

2  j=1

⎞ ajvj⎠ =

n 

bk wk .

k=1

The mapping P so defined is a linear projection in V whose range is W . Moreover, by using consecutively (6.29), (6.27), (6.31) and, finally, (6.24),    ⎛ ⎞   1/2  2n  n