Integral representation theory: Applications to convexity, Banach spaces and potential theory

de Gruyter Studies in Mathematics 35 Editors: Carsten Carstensen · Nicola Fusco Niels Jacob · Karl-Hermann Neeb de Gr...

Author: Jaroslav Lukes

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de Gruyter Studies in Mathematics 35 Editors: Carsten Carstensen · Nicola Fusco Niels Jacob · Karl-Hermann Neeb

de Gruyter Studies in Mathematics 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

Riemannian Geometry, 2nd rev. ed., Wilhelm P. A. Klingenberg Semimartingales, Michel Me´tivier Holomorphic Functions of Several Variables, Ludger Kaup and Burchard Kaup Spaces of Measures, Corneliu Constantinescu Knots, 2nd rev. and ext. ed., Gerhard Burde and Heiner Zieschang Ergodic Theorems, Ulrich Krengel Mathematical Theory of Statistics, Helmut Strasser Transformation Groups, Tammo tom Dieck Gibbs Measures and Phase Transitions, Hans-Otto Georgii Analyticity in Infinite Dimensional Spaces, Michel Herve´ Elementary Geometry in Hyperbolic Space, Werner Fenchel Transcendental Numbers, Andrei B. Shidlovskii Ordinary Differential Equations, Herbert Amann Dirichlet Forms and Analysis on Wiener Space, Nicolas Bouleau and Francis Hirsch Nevanlinna Theory and Complex Differential Equations, Ilpo Laine Rational Iteration, Norbert Steinmetz Korovkin-type Approximation Theory and its Applications, Francesco Altomare and Michele Campiti Quantum Invariants of Knots and 3-Manifolds, Vladimir G. Turaev Dirichlet Forms and Symmetric Markov Processes, Masatoshi Fukushima, Yoichi Oshima and Masayoshi Takeda Harmonic Analysis of Probability Measures on Hypergroups, Walter R. Bloom and Herbert Heyer Potential Theory on Infinite-Dimensional Abelian Groups, Alexander Bendikov Methods of Noncommutative Analysis, Vladimir E. Nazaikinskii, Victor E. Shatalov and Boris Yu. Sternin Probability Theory, Heinz Bauer Variational Methods for Potential Operator Equations, Jan Chabrowski The Structure of Compact Groups, 2nd rev. and aug. ed., Karl H. Hofmann and Sidney A. Morris Measure and Integration Theory, Heinz Bauer Stochastic Finance, 2nd rev. and ext. ed., Hans Föllmer and Alexander Schied Painleve´ Differential Equations in the Complex Plane, Valerii I. Gromak, Ilpo Laine and Shun Shimomura Discontinuous Groups of Isometries in the Hyperbolic Plane, Werner Fenchel and Jakob Nielsen The Reidemeister Torsion of 3-Manifolds, Liviu I. Nicolaescu Elliptic Curves, Susanne Schmitt and Horst G. Zimmer Circle-valued Morse Theory, Andrei V. Pajitnov Computer Arithmetic and Validity, Ulrich Kulisch Feynman-Kac-Type Theorems and Gibbs Measures on Path Space, Jo´zsef Lörinczi, Fumio Hiroshima and Volker Betz Integral Representation Theory, Jaroslas Lukesˇ, Jan Maly´, Ivan Netuka and Jirˇ´ı Spurny´ Introduction to Harmonic Analysis and Generalized Gelfand Pairs, Gerrit van Dijk Bernstein Functions, Rene´ Schilling, Renming Song and Zoran Vondracˇek

Jaroslav Lukesˇ Jan Maly´ Ivan Netuka Jirˇ´ı Spurny´

Integral Representation Theory Applications to Convexity, Banach Spaces and Potential Theory

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Walter de Gruyter Berlin · New York

Authors Jaroslav Lukesˇ Department of Mathematical Analysis Faculty of Mathematics and Physics Charles University Sokolovska´ 83 18675 Prague 8, Czech Republic E-Mail: [email protected]

Jan Maly´ Department of Mathematical Analysis Faculty of Mathematics and Physics Charles University Sokolovska´ 83 18675 Prague 8, Czech Republic and Department of Mathematics Faculty of Science J. E. Purkyneˇ University ˇ eske´ mla´dezˇe 8 C ´ stı´ nad Labem, Czech Republic 40096 U E-Mail: [email protected]

Ivan Netuka Mathematical Institute Faculty of Mathematics and Physics Charles University Sokolovska´ 83 18675 Prague 8, Czech Republic E-Mail: [email protected]

Jirˇ´ı Spurny´ Department of Mathematical Analysis Faculty of Mathematics and Physics Charles University Sokolovska´ 83 18675 Prague 8, Czech Republic E-Mail: [email protected]

Series Editors Carsten Carstensen Department of Mathematics Humboldt University of Berlin Unter den Linden 6 10099 Berlin, Germany E-Mail: [email protected]

Niels Jacob Department of Mathematics Swansea University Singleton Park Swansea SA2 8PP, Wales, United Kingdom E-Mail: [email protected]

Nicola Fusco Dipartimento di Matematica Universita` di Napoli Frederico II Via Cintia 80126 Napoli, Italy E-Mail: [email protected]

Karl-Hermann Neeb Department of Mathematics Technische Universität Darmstadt Schloßgartenstraße 7 64289 Darmstadt, Germany E-Mail: [email protected]

Mathematics Subject Classification 2000: 46-02, 31-02, 52-02, 46A55, 52A07, 46B99, 31B05, 31A05, 31A25, 31B20, 31C05, 31D05, 35K05, 35K20, 28A05, 54H05 Keywords: Convex sets, Choquet theory, Banach spaces, descriptive set theory, measure theory, potential theory, Dirichlet problem

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Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available in the Internet at http://dnb.d-nb.de.

ISBN 978-3-11-020320-2 쑔 Copyright 2010 by Walter de Gruyter GmbH & Co. KG, 10785 Berlin, Germany. All rights reserved, including those of translation into foreign languages. No part of this book may be reproduced in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the publisher. Printed in Germany. Cover design: Martin Zech, Bremen. Printing and binding: Hubert & Co. GmbH & Co. KG, Göttingen.

Introduction

In many branches of mathematics, one encounters the question of how to reconstruct a convex set from information on its vertices. This idea successfully emerged as the Krein–Milman theorem for compact convex subsets of locally convex spaces since any such set has plenty of extreme points. For any point of a compact convex set, a reformulation of the Krein–Milman theorem provides a representing measure that is concentrated in some sense on the set of extreme points. The goal of our book is to present a more general approach to integral representation theory based upon a notion of a function space and apply the obtained results to the theory of convex sets, Banach spaces and potential theory. We point out that this approach is far from being new, but we hope that our exposition may be profitable both for students interested in the basics of integral representation theory as well as for more advanced readers. The former group could be attracted by a self-contained presentation of the Choquet theory, the latter by a substantial amount of results of fairly recent origin or appearing in a book form for the first time. We also try to incorporate more techniques from descriptive set theory into subject, which further supports our belief that the book will be worth reading even for those well acquainted with the monographs by E. M. Alfsen [5], R. R. Phelps [374], Z. Semadeni [414], L. Asimow and A. J. Ellis [24] or V. P. Fonf, J. Lindenstrauss and R. R. Phelps [179]. Let us continue by looking briefly at the contents of the book. After a prologue on the Korovkin theorem, we present basic facts on the extremal structure of finitedimensional compact convex sets. Then we move on to infinite-dimensional spaces and prove the Krein–Milman theorem and several of its consequences. The second part of Chapter 2 studies the concept of measure convex and measure extremal sets. Chapter 3 is devoted to cornerstones of the Choquet theory of functions spaces such as the Choquet order and its properties and integral representation theorems due to G. Choquet and E. Bishop and K. de Leeuw. Even though the results are standard, the key limiting process is established by means of the Simons lemma, which allows us to present later on several of its applications. The chapter is finished by a discussion on deeper properties of the Choquet ordering. The next chapter studies basic properties of affine functions on compact convex sets and characterizations of functions satisfying the barycentric formula. A link between the theory of function spaces and compact convex sets starts to emerge at the end of the chapter. Chapter 5 is crucial for the subsequent application of descriptive set theory; it describes a hierarchy of Borel sets and functions in topological spaces and proves their

vi

Introduction

basic properties. The most important fact is that many descriptive properties are stable with respect to perfect mappings, which allows us to transfer abstract Borel affine functions to the setting of compact convex sets. Simplicial function spaces are studied in Chapter 6. We discuss several classes of simplicial function spaces, namely the Bauer and Markov simplicial function spaces and spaces with boundary of type Fσ . Among other results, the abstract Dirichlet problem for continuous and non-continuous functions is considered. Choquet simplices are presented at the end of the chapter. Next we generalize the basic concepts for function cones since they are indispensable in potential theory. We focus in particular on ordered compact convex sets. Analogues of faces in a non-convex setting, so-called Choquet sets, are investigated in Chapter 8. The main result is a characterization of simplicial spaces by means of Choquet sets. Suitably chosen families of closed extremal sets generate interesting boundary topologies on the set of extreme points. Chapter 9 studies these topologies and functions continuous with respect to them. It turns out that maximal measures induce measures on sets of extreme points that are regular with respect to boundary topologies. The last section is devoted to a study of a facial topology and facially continuous functions. Chapter 10 collects several deeper results on function spaces and compact convex sets. Among others, study of Shilov and James boundaries, Lazar’s improvement of the Banach–Stone theorem, results on automatic boundedness of affine and convex functions, embedding of `1 in Banach spaces, metrizability of compact convex sets and their open images and some topological properties of the set of extreme points. The Lazar selection theorem and its consequences occupy the first part of Chapter 11. The second part is devoted to a presentation of Debs’ proof of Talagrand’s theorem on measurable selectors. Chapter 12 is concerned with two methods of constructing new function spaces: products and inverse limits. We show that both operations preserve simpliciality and describe resulting boundaries. The inverse limits lead to an interesting description of metrizable simplices as inverse limits of finite-dimensional simplices. The general results are illustrated by a construction of the Poulsen simplex and a couple of compact convex sets due to Talagrand. In Chapter 13, general results from the Choquet theory are applied to potential theory and several of its basic notions are investigated from this perspective. Important function cones and spaces appearing in potential theory are studied in detail, in particular, in connection to various solution methods for the Dirichlet problem. The functional analysis approach makes it possible to provide an interesting interpretation, for instance, of balayage and regular points in terms of representing measures and the Choquet boundary of suitable spaces and cones. The exposition covers potential the-

Introduction

vii

ory for the Laplace equation and the heat equation as well as a more general setting (harmonic spaces, fine potential theory etc.). The final Chapter 14 presents several applications of the integral representation theorems, such as for doubly stochastic matrices, the Riesz–Herglotz theorem, the Lyapunov theorem on the range of a vector measure, the Stone–Weierstrass theorem, positive-definite functions and invariant and ergodic measures. Each chapter concludes with a series of exercises with sketches of proofs and with concluding notes and comments where we try to give precise references and due credits for the results presented in the main body of the text, and discuss additional material which is related to the topics of the chapter in question, but was not included with complete proofs. Open problems are also mentioned. Since the presented material originates in an amalgamation of functional analysis, measure theory, topology, descriptive set theory and potential theory, we collect the needed notions and facts in the Appendix, sometimes even with proofs. We selected the following books for each subject as the key references: W. Rudin [403] and M. Fabian, P. Habala, P. Hájek, V. Montesinos Santaluc´ıa, J. Pelant and V. Zizler [173] for functional analysis, D. H. Fremlin [182], [181] and [183] for measure theory, R. Engelking [169] and K. Kuratowski [285] for topology, A. S. Kechris [262] and C. A. Rogers and J. E. Jayne [394] for descriptive set theory, D. H. Armitage and S. J. Gardiner [21] for classical potential theory and J. Bliedtner and W. Hansen [66] for abstract potential theory. Next we point out what is omitted from the book. First, we focus on integral representation theorems for compact sets, and thus the readers interested in theory of sets with the Radon–Nikodym property are referred to R. D. Bourgin [82], and those interested in Choquet theory in sets of measures are referred to G. Winkler [473]. Second, although we consider several geometric aspects of simplicial spaces, they are not at the center of our attention. They are thoroughly investigated in H. E. Lacey [290] and P. Harmand, D. Werner and W. Werner [216]. Further, we do not pursue applications of integral representation theory in C ∗ -algebras and thus we refer the interested reader to E. M. Alfsen and F. W. Schultz [10] and [9], M. Rørdam [395] and M. Rørdam and E. Størmer [396], B. Blackadar [59] and H. Lin [303] and the references therein. And last but not least, our applications to potential theory do not require the full strength of abstract potential theory and thus we restrict ourselves to a less general framework than the one presented in J. Bliedtner and W. Hansen [66]. Except on a few explicitly stated occasions, we consider only real vector spaces and apart from Chapter 9 we deal only with Hausdorff topologies and Radon measures. We use the standard notation and terminology: •

N, Q, Z, R, C denote the usual sets of numbers,

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Re z and Im z denote the real and imaginary part of a complex number z, respectively,

viii

Introduction

•

cA is the characteristic function of a set A (sometimes we write 1 for the characteristic function of a space),

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A4B is the symmetric difference of sets A and B,

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Ac is the complement of a set A,

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•

• •

f ∧ g, f ∨ g denote the infimum and supremum of functions f, g, respectively, (usually they are considered pointwise), f + , f − , |f | denote the positive and negative parts, and absolute value of a function f , respectively, f |A is the restriction of a function f to a set A, if F is a system of functions, F b and F + are the families of all bounded and positive elements from F, respectively,

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ω0 and ω1 are the first infinite and first uncountable ordinals, respectively,

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A, Int A, ∂A are the closure, interior and boundary of a set A in a topological space, respectively,

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dist(A, B) denotes the distance of sets in a metric space,

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diam A is the diameter of a set A in a metric space,

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U (x, r), B(x, r) and S(x, r) are the open ball, closed ball and sphere centered at x with radius r > 0, respectively,

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co A and span A are the convex and linear hull of a set A in a vector space, respectively, co A is the closed convex hull of a set A in a topological vector space,

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ker T denotes the kernel of an operator between linear spaces,

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BE , UE and SE are the closed unit ball, open unit ball and sphere of a normed linear space E, respectively,

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E/F is the quotient space of a locally convex space with respect to a closed subspace F ,

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E ⊕ F is the sum of locally convex spaces E and F ,

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E ∗ is the dual space of a topological linear space E,

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(x, y) stands for the scalar product of vectors x, y in a Hilbert space,

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c0 is the space of sequences converging to 0,

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C(X) is the space of real-valued continuous functions on a topological space X,

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C b (X) is the space of bounded continuous functions on a topological space X,

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`p and Lp (µ), p ∈ [1, ∞], are the usual Lebesgue spaces (see Section A.3),

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C n (U ), C n , C ∞ (U ), C ∞ stand for the space of n-times continuously differentiable functions on U or infinitely differentiable functions on U , respectively,

Introduction • • •

ix

R −A f (y) dy is the integral mean value of f over a set A, R d S(x,r) f (y) dS(y) is the surface integral of f over the sphere S(x, r) ⊂ R , ∇f is the gradient of f .

A function f is positive if f ≥ 0, it is strictly positive if f > 0. Similarly we use increasing, strictly increasing and so on. If µ is a measure, we often write µ(f ) for R the integral f dµ. In preparation of the present book, we have received many valuable suggestions from many colleagues. In particular, we would like to express our thanks to P. Hájek, P. Holický, M. Johanis, O. Kalenda, P. Kaplický, M. Kraus, E. Murtinová, P. Simon, J. Tiˇser, L. Zaj´ıcˇ ek and M. Zelený for stimulating and fruitful discussions, and to E. Crooks for linguistic assistance. We are also indebted to the publishers for their care and cooperation. The preparation of the manuscript was supported by the grant 201/07/0388 of the Grant Agency of the Czech Republic and partly by the grant MSM21620839 of the Czech Ministry of Education. Finally, our thanks go to Jana, Jarka, Hana and Hanka for encouragement and patience during the preparation of this book. Prague, Summer 2009

Jaroslav Lukeˇs Jan Malý Ivan Netuka Jiˇr´ı Spurný

Contents

Introduction

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1

Prologue 1.1 The Korovkin theorem . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Notes and comments . . . . . . . . . . . . . . . . . . . . . . . . . .

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Compact convex sets 2.1 Geometry of convex sets . . . . . . 2.1.A Finite-dimensional case . . 2.1.B The Krein–Milman theorem 2.1.C Exposed points . . . . . . . 2.2 Interlude: On the space M(K) . . . 2.3 Structures in convex sets . . . . . . 2.3.A Extremal sets and faces . . . 2.3.B Measure convex sets . . . . 2.3.C Measure extremal sets . . . 2.4 Exercises . . . . . . . . . . . . . . 2.5 Notes and comments . . . . . . . .

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Choquet theory of function spaces 3.1 Function spaces . . . . . . . . . . . 3.2 More about Korovkin theorems . . . 3.3 On the H-barycenter mapping . . . 3.4 The Choquet representation theorem 3.5 In-between theorems . . . . . . . . 3.6 Maximal measures . . . . . . . . . 3.7 Boundaries and the Simons lemma . 3.8 The Bishop–de Leeuw theorem . . . 3.9 Minimum principles . . . . . . . . 3.10 Orderings and dilations . . . . . . . 3.11 Exercises . . . . . . . . . . . . . . 3.12 Notes and comments . . . . . . . .

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Affine functions on compact convex sets 107 4.1 Affine functions and the barycentric formula . . . . . . . . . . . . . . 107 4.2 Barycentric theorem and strongly affine functions . . . . . . . . . . . 113

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4.3 4.4 4.5 4.6 5

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State space and representation of affine functions Affine Baire-one functions on dual unit balls . . Exercises . . . . . . . . . . . . . . . . . . . . . Notes and comments . . . . . . . . . . . . . . .

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Perfect classes of functions and representation of affine functions 5.1 Generation of sets and functions . . . . . . . . . . . . . . . . 5.2 Baire and Borel sets . . . . . . . . . . . . . . . . . . . . . . . 5.3 Baire and Borel mappings . . . . . . . . . . . . . . . . . . . 5.4 Perfect classes of functions . . . . . . . . . . . . . . . . . . . 5.5 Affinely perfect classes of functions . . . . . . . . . . . . . . 5.6 Representation of H-affine functions . . . . . . . . . . . . . . 5.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Notes and comments . . . . . . . . . . . . . . . . . . . . . .

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135 136 142 146 149 150 154 159 166

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Simplicial function spaces 6.1 Basic properties of simplicial spaces . . . . . . . . . . . . . . . 6.2 Characterizations of simplicial spaces . . . . . . . . . . . . . . 6.3 Simplicial spaces as L1 -preduals . . . . . . . . . . . . . . . . . 6.4 The weak Dirichlet problem and Ac (H)-exposed points . . . . . 6.5 The Dirichlet problem for a single function . . . . . . . . . . . 6.6 Special classes of simplicial spaces . . . . . . . . . . . . . . . 6.6.A Bauer simplicial spaces . . . . . . . . . . . . . . . . . . 6.6.B Markov simplicial spaces . . . . . . . . . . . . . . . . . 6.6.C Simplicial spaces with Lindelöf boundaries . . . . . . . 6.6.D Simplicial spaces with boundaries of type Fσ . . . . . . 6.7 The Daugavet property of simplicial spaces . . . . . . . . . . . 6.8 Choquet simplices . . . . . . . . . . . . . . . . . . . . . . . . 6.8.A Simplicial function spaces and the classical definition of Choquet simplices . . . . . . . . . . . . . . . . . . . . 6.8.B Prime function spaces and prime compact convex sets . 6.8.C Characterization of Bauer simplices by faces . . . . . . 6.8.D Fakhoury’s theorem . . . . . . . . . . . . . . . . . . . 6.9 Restriction of function spaces . . . . . . . . . . . . . . . . . . 6.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.11 Notes and comments . . . . . . . . . . . . . . . . . . . . . . . Choquet theory of function cones 7.1 Function cones . . . . . . . 7.2 Maximal measures . . . . . 7.3 Representation theorem . . . 7.4 Simplicial cones . . . . . .

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7.5 7.6 7.7 8

Ordered compact convex sets and simplicial measures . . . . . . . . . 232 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 Notes and comments . . . . . . . . . . . . . . . . . . . . . . . . . . 243

Choquet-like sets 8.1 Split and parallel faces . . . . . . . . . . . . . . 8.2 H-extremal and H-convex sets . . . . . . . . . . 8.3 Choquet sets, M -sets and P -sets . . . . . . . . . 8.4 H-exposed sets . . . . . . . . . . . . . . . . . . 8.5 Weak topology on boundary measures . . . . . . 8.6 Characterizations of simpliciality by Choquet sets 8.7 Exercises . . . . . . . . . . . . . . . . . . . . . 8.8 Notes and comments . . . . . . . . . . . . . . .

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244 244 246 250 257 259 262 268 273

Topologies on boundaries 9.1 Topologies generated by extremal sets . . . . . . 9.2 Induced measures on Choquet boundaries . . . . 9.3 Functions continuous in σext and σmax topologies 9.4 Strongly universally measurable functions . . . . 9.5 Facial topology generated by M -sets . . . . . . . 9.6 Exercises . . . . . . . . . . . . . . . . . . . . . 9.7 Notes and comments . . . . . . . . . . . . . . .

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274 274 278 284 288 296 303 308

10 Deeper results on function spaces and compact convex sets 10.1 Boundaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.A Shilov boundary . . . . . . . . . . . . . . . . . . . . 10.1.B Boundaries in Banach spaces . . . . . . . . . . . . . . 10.2 Isometries of spaces of affine continuous functions . . . . . . 10.3 Baire measurability and boundedness of affine functions . . . 10.3.A The Cantor set and its properties . . . . . . . . . . . . 10.3.B Automatic boundedness of affine and convex functions 10.4 Embedding of `1 . . . . . . . . . . . . . . . . . . . . . . . . 10.5 Metrizability of compact convex sets . . . . . . . . . . . . . . 10.6 Continuous affine images . . . . . . . . . . . . . . . . . . . . 10.7 Several topological results on Choquet boundaries . . . . . . . 10.7.A The Choquet boundary as a Baire space . . . . . . . . 10.7.B Polish spaces as Choquet boundaries . . . . . . . . . . 10.7.C K-countably determined boundaries . . . . . . . . . . 10.8 Convex Baire-one functions . . . . . . . . . . . . . . . . . . 10.9 Function spaces with continuous envelopes . . . . . . . . . . 10.9.A Stable compact convex sets . . . . . . . . . . . . . . . 10.9.B CE-function spaces . . . . . . . . . . . . . . . . . . .

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10.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378 10.11 Notes and comments . . . . . . . . . . . . . . . . . . . . . . . . . . 384 11 Continuous and measurable selectors 11.1 The Lazar selection theorem . . . . . . . . . . 11.2 Applications of the Lazar selection theorem . . 11.3 The weak Dirichlet problem for Baire functions 11.4 Pointwise approximation of maximal measures 11.5 Measurable selectors . . . . . . . . . . . . . . 11.5.A Multivalued mappings . . . . . . . . . 11.5.B Selection theorem . . . . . . . . . . . . 11.5.C Applications of the selection theorem . 11.6 Exercises . . . . . . . . . . . . . . . . . . . . 11.7 Notes and comments . . . . . . . . . . . . . .

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389 389 394 398 400 402 402 406 409 412 416

12 Constructions of function spaces 12.1 Products of function spaces . . . . . . . . . . . . . . . . . . . . . . . 12.1.A Definitions and basic properties . . . . . . . . . . . . . . . . 12.1.B Maximal measures and extremal sets . . . . . . . . . . . . . 12.1.C Partitions of unity and approximation in products of function spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1.D Products of simplicial spaces . . . . . . . . . . . . . . . . . . 12.2 Inverse limits of function spaces . . . . . . . . . . . . . . . . . . . . 12.2.A Admissible mappings . . . . . . . . . . . . . . . . . . . . . . 12.2.B Construction of inverse limits . . . . . . . . . . . . . . . . . 12.2.C Inverse limits of simplicial function spaces . . . . . . . . . . 12.2.D Structure of simplices . . . . . . . . . . . . . . . . . . . . . 12.3 Several examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3.A The Poulsen simplex . . . . . . . . . . . . . . . . . . . . . . 12.3.B A big simplicial space . . . . . . . . . . . . . . . . . . . . . 12.3.C Functions of affine classes and Talagrand’s example . . . . . 12.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5 Notes and comments . . . . . . . . . . . . . . . . . . . . . . . . . .

428 436 440 440 442 445 447 455 455 465 470 477 486

13 Function spaces in potential theory and the Dirichlet problem 13.1 Balayage and the Dirichlet problem . . . . . . . . . . . . . . 13.1.A Essential solution of the generalized Dirichlet problem 13.2 Boundary behavior of solutions . . . . . . . . . . . . . . . . 13.2.A Regular points for the Laplace equation . . . . . . . . 13.2.B Regular points for the heat equation . . . . . . . . . . 13.3 Function spaces and cones in potential theory . . . . . . . . . 13.3.A Function spaces and cones: Laplace equation . . . . .

489 491 494 496 497 503 504 505

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419 420 420 424

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13.4

13.5

13.6 13.7

13.3.B Function spaces and cones in parabolic potential theory and harmonic spaces . . . . . . . . . . . . . . . . . . . . . . . . 510 13.3.C Continuity properties of H(U )-concave functions . . . . . . . 513 13.3.D Separation by functions from H(U ) . . . . . . . . . . . . . . 515 Dirichlet problem: solution methods . . . . . . . . . . . . . . . . . . 517 13.4.A PWB solution of the Dirichlet problem . . . . . . . . . . . . 518 13.4.B Cornea’s approach to the Dirichlet problem . . . . . . . . . . 523 13.4.C The Wiener solution . . . . . . . . . . . . . . . . . . . . . . 530 13.4.D Fine Wiener solution . . . . . . . . . . . . . . . . . . . . . . 532 13.4.E PDE solutions in Sobolev spaces . . . . . . . . . . . . . . . . 534 Generalized Dirichlet problem and uniqueness questions . . . . . . . 537 13.5.A Lattice approach . . . . . . . . . . . . . . . . . . . . . . . . 538 13.5.B Uniqueness for the Laplace equation . . . . . . . . . . . . . . 540 13.5.C Keldysh theorems in parabolic and axiomatic potential theories 542 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546 Notes and comments . . . . . . . . . . . . . . . . . . . . . . . . . . 555

14 Applications 14.1 Representation of convex functions . . . . . . 14.2 Representation of concave functions . . . . . . 14.3 Doubly stochastic matrices . . . . . . . . . . . 14.4 The Riesz–Herglotz theorem . . . . . . . . . . 14.5 Typically real holomorphic functions . . . . . . 14.6 Holomorphic functions with positive real part . 14.7 Completely monotonic functions . . . . . . . . 14.8 Positive definite functions on discrete groups . 14.9 Range of vector measures . . . . . . . . . . . . 14.10 The Stone–Weierstrass approximation theorem 14.11 Invariant and ergodic measures . . . . . . . . . 14.12 Exercises . . . . . . . . . . . . . . . . . . . . 14.13 Notes and comments . . . . . . . . . . . . . .

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563 564 567 572 573 575 580 586 589 593 595 597 603 605

A Appendix A.1 Functional analysis . . . . . . . . . . . . . . . . . . . . . A.1.A Locally convex spaces . . . . . . . . . . . . . . . A.1.B Banach spaces . . . . . . . . . . . . . . . . . . . A.1.C Ordered Banach spaces and lattices . . . . . . . . A.2 Topology . . . . . . . . . . . . . . . . . . . . . . . . . . ˇ A.2.A Compact spaces and Cech–Stone compactification A.2.B Baire and Borel sets . . . . . . . . . . . . . . . . A.2.C Semicontinuous functions . . . . . . . . . . . . . A.2.D Baire spaces and sets with the Baire property . . .

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608 608 608 609 610 615 616 619 621 623

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Contents

A.3 Measure theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.3.A Measure spaces . . . . . . . . . . . . . . . . . . . . . . . . A.3.B Radon measures on locally compact σ-compact spaces . . . A.3.C Images, products and inverse limits of Radon measures . . . A.3.D Kernels and disintegration of measures . . . . . . . . . . . A.4 Descriptive set theory . . . . . . . . . . . . . . . . . . . . . . . . . A.5 Resolvable sets and Baire-one functions . . . . . . . . . . . . . . . A.6 The Laplace equation . . . . . . . . . . . . . . . . . . . . . . . . . A.6.A Weak solutions of the Laplace equation . . . . . . . . . . . A.7 The heat equation . . . . . . . . . . . . . . . . . . . . . . . . . . . A.8 Axiomatic potential theory . . . . . . . . . . . . . . . . . . . . . . A.8.A Bauer’s axiomatic theory . . . . . . . . . . . . . . . . . . . A.8.B Hyperharmonic and superharmonic functions . . . . . . . . A.8.C Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . A.8.D Superharmonic functions and Green potentials . . . . . . . A.8.E Superharmonic functions and potentials for the heat equation A.8.F Balayage . . . . . . . . . . . . . . . . . . . . . . . . . . . A.8.G Thinness, base and fine topology . . . . . . . . . . . . . . . A.8.H Polar and semipolar sets . . . . . . . . . . . . . . . . . . .

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624 624 626 632 636 637 640 645 647 649 652 653 654 656 657 660 661 663 665

Bibliography

669

List of symbols

695

Index

703

Chapter 1

Prologue

1.1

The Korovkin theorem

We start with the famous Weierstrass approximation theorem. Theorem 1.1 (The Weierstrass approximation theorem). The space of all polynomial functions on the interval [0, 1] is uniformly dense in the space C([0, 1]). There are several different proofs of this result and several methods for how to associate to a given continuous function f ∈ C([0, 1]) a sequence of polynomials {Pn } that converges uniformly to f on [0, 1]. For example, given f ∈ C([0, 1]) and n ∈ N, we define the corresponding Bernstein polynomial Bn f by n X n j j Bn f : x 7→ f x (1 − x)n−j , x ∈ [0, 1]. j n j=0

The task is to show that the sequence {Bn f } converges uniformly to f on [0, 1]. This can be easily verified in the case when f (x) = 1, x or x2 , since

n−1 2 1 x + x. n n Surprisingly, this is all that we need to compute, since these three tests are enough to guarantee the uniform convergence of Bn f to f for all f in C([0, 1]). Indeed, one of the current proofs of the classical Weierstrass approximation theorem is based on the Korovkin theorem about linear operators. The Weierstrass theorem is an easy consequence since the mappings Bn 1 = 1,

Bn x = x

and

Bn : f 7→ Bn f,

Bn x2 =

f ∈ C([0, 1]),

are positive linear operators on C([0, 1]). Theorem 1.2 (Korovkin). Let pj , j = 0, 1, 2, denote the monomial function pj : x 7→ xj and let {Tn } be a sequence of positive linear operators on the space C([0, 1]). Assume that Tn pj → pj uniformly on [0, 1] as n → ∞ for j = 0, 1, 2. Then Tn f → f uniformly on [0, 1] as n → ∞ for all f ∈ C([0, 1]).

2

1 Prologue

Proof. Pick f ∈ C([0, 1]) and ε > 0. By the uniform continuity of f , there exists δ ∈ (0, 1) such that |f (s) − f (t)| ≤ ε for any s, t ∈ [0, 1], |s − t| ≤ δ. Now fix t ∈ [0, 1] and set p∗ (x) := f (t) − ε − (x − t)2

2kf k , δ2

x ∈ [0, 1],

and

2kf k , x ∈ [0, 1]. δ2 Dealing separately with the cases |x − t| ≤ δ and |x − t| > δ, we get p∗ (x) := f (t) + ε + (x − t)2

|f (x) − f (t)| ≤ ε + 2(x − t)2

kf k δ2

for each x ∈ [0, 1]. Hence, p∗ (x) ≤ f (x) ≤ p∗ (x),

x ∈ [0, 1],

and, therefore, for any n ∈ N, Tn p∗ ≤ Tn f ≤ Tn p∗ . We find N ∈ N such that for any n ≥ N

Tn pj − pj < εδ 2 ,

j = 0, 1, 2.

Since 2kf k 4kf kt 2kf k p∗ (x) = f (t) + ε + t2 2 − x + 2 x2 , 2 δ δ δ

x ∈ [0, 1],

for n ≥ N we have the estimate kTn p∗ − p∗ k ≤ Cε

with C := 9kf k + ε.

In particular, Tn f (t) ≤ Tn p∗ (t) ≤ p∗ (t) + Cε = f (t) + Cε + ε,

n ≥ N,

and similarly Tn f (t) ≥ Tn p∗ (t) ≥ p∗ (t) − Cε = f (t) − Cε − ε, Hence kTn f − f k ≤ ε(C + 1), and the proof is complete.

n ≥ N,

n ≥ N.

1.2 Notes and comments

3

In stating the Korovkin theorem, which sometimes bears the name the first Korovkin theorem, it is possible to go further, replacing the interval [0, 1] by a suitable space, and the set of three functions {p0 , p1 , p2 } by a more general family of functions. In the sequel, we will take a deeper look at this issue. Definition 1.3 (Korovkin closure). Let K be a (metrizable) compact space and P a family of continuous functions on K (sometimes called test functions). We say that a sequence {Tn } of positive operators on C(K) is P-admissible if kTn ϕ − ϕk → 0 for any ϕ ∈ P, and define the Korovkin closure of P as Kor(P) := {f ∈ C(K) : kTn f − f k → 0 for any P-admissible sequence {Tn }} . Let H be the linear span of P. It is simple to check that H ⊂ Kor(H) = Kor(P). Two questions immediately arise: (a) How can Kor(P) be characterized ? (b) Under what conditions does the equality Kor(P) = C(K) hold ? In what follows, we will give answers to these questions and will also study analogous problems. To these ends, the framework of abstract linearity and convexity will turn out to be useful and efficient.

1.2

Notes and comments

The Korovkin theorem was proved independently by H. Bohman in [74] for a kind of special positive operators, and by P. P. Korovkin in [277] for integral-type operators. Korovkin extended his theory in [278] and we followed his proof from this monograph. The Korovkin theorem 1.2 sometimes bears the name of the Bohman– Korovkin theorem. Excellent sources for the Korovkin material are the monograph of F. Altomare and M. Campiti [13], and Chauvenet’s prize paper of H. Bauer [42].

Chapter 2

Compact convex sets

We begin our exposition with classical results on convex sets in finite-dimensional spaces. After showing Carathéodory’s theorem 2.6, we define extreme points and prove Minkowski’s theorem 2.11 stating that any compact convex set in Rd is the convex hull of its extreme points. An amalgamation of these two results contained in Theorem 2.12 is a starting point leading to generalizations in infinite-dimensional spaces. So the next section is devoted to the study of the Krein–Milman theorem and related results. In particular we are interested in its reformulation known as the Integral representation theorem. The basic idea of representing points of a compact convex set as barycenters of probability measures is a central topic of the whole book. Thus after the proof of the Krein–Milman theorem 2.22 and Bauer’s minimum principle 2.24 we define the barycenter of a probability measure on a compact convex set and show its existence and uniqueness (see Theorem 2.29). Then the Integral representation theorem 2.31 and properties of the barycentric mapping are proved. We finish this part by some classical facts: Bauer’s characterization 2.40 of extreme points of a compact convex set, Choquet’s observation on extreme points of a compact convex set contained in Proposition 2.41 and the Milman theorem 2.43. The aim of Subsection 2.1.C is to show that a metrizable compact convex set has abundance of exposed points, namely, that a metrizable compact convex set is the closed convex hull of its exposed points and that exposed points are dense in the set of extreme points. Section 2.2 prepares the ground for examples concerning extremal sets and faces of compact convex sets presented in Section 2.3. We prove several facts on probability measures on compact spaces and show how they lead to a construction of affine functions on compact convex sets that do not satisfy the barycentric formula (see Proposition 2.63). Subsection 2.3.A investigates more closely extremal sets and faces of compact convex sets. The main result contained in Proposition 2.69 shows that a closed extremal set is a union of closed faces. We generalize the concept of convexity and extremality in Subsections 2.3.B and 2.3.C by introducing measure convex and measure extremal sets. The main tool is Theorem 2.75 due to D. H. Fremlin and J. D. Pryce that characterizes measure convex sets. Then we show that convex sets of low Borel complexity are also measure convex, but that there are examples of Fσ or Gδ faces that are not measure convex. Analogous results are proved in the next section for extremal and measure extremal sets.

2.1 Geometry of convex sets

2.1 2.1.A

5

Geometry of convex sets Finite-dimensional case

Throughout this subsection, let W be a real vector space. Definition 2.1 (Convex sets in vector spaces). A set C ⊂ W is convex if λx + (1 − λ)y ∈ C whenever x, y ∈ C and λ ∈ (0, 1). Let A be an arbitrary subset of W . The convex hull of A, denoted by co A, is the intersection of all convex sets of W that contain A. Since W is a convex set and the intersection of any family of convex sets is convex, the set co A is the smallest convex set containing A. It is easy to check that co A =

n nX

λj xj : n ∈ N, x1 , . . . , xn ∈ A and λ1 , . . . , λn ≥ 0,

j=1

n X

o λj = 1 .

j=1

Definition 2.2 (Affine independence and n-simplices). Recall that vectors e0 , . . . , en of W are said to be affinely independent if e1 −e0 , . . . , en −e0 are linearly independent. In other words, if whenever λ0 e0 + · · · + λn en = 0

and λ0 + · · · + λn = 0,

then λ0 = · · · = λn = 0. In this case, the convex hull co {e0 , . . . , en } is termed an n-simplex with vertices e0 , . . . , en . In Rd , there exist at most d + 1 affinely independent points. Definition 2.3 (Affine hulls and subspaces, hyperplanes). For a set A ⊂ W , the affine hull of A, denoted by aff A, is the set of all affine combinations of points of A. (A linear combination α1 x1 + · · · + αn xn , where α1 + · · · + αn = 1, is called an affine combination of points x1 , . . . , xn .) A set A ⊂ W is said to be an affine subspace of W if aff A = A. Affine subspaces are just the translations (of type) x + F , where F is a linear subspace of W and x ∈ W . By definition, the dimension (or, the codimension) of x + F is the dimension (or, the codimension, respectively) of F . A set H is a hyperplane if there exists a nonzero linear functional f on W and α ∈ R such that H = {w ∈ W : f (w) = α}. Since a subspace F of W is a maximal proper subspace of W if and only if there exists a nonzero linear functional f on W such that F = ker f , we see that H is a hyperplane if and only if there exist a maximal proper subspace F of W and w ∈ W such that H = w + F . In other words, hyperplanes are exactly affine subspaces of codimension 1.

6

2 Compact convex sets

Let C be a subset of W and H := {w ∈ W : f (w) = α} be a hyperplane. We say that H is a support hyperplane of C if C ∩ H 6= ∅ and either C ⊂ {w ∈ W : f (w) ≤ α}

or C ⊂ {w ∈ W : f (w) ≥ α}.

Any point c ∈ C ∩ H is called an H-support point of C. We also say that H supports C at c. In the sequel, we need the following assertion. Proposition 2.4. Let C be a closed convex subset of Rd with a nonempty interior and c ∈ ∂C. Then there exists a hyperplane H such that H supports C at c. Proof. See, for example, A. Barvinok [31], Corollary 2.8. Remark 2.5. In what follows, we direct our attention to geometry of compact convex sets in the Euclidean d-dimensional space Rd . All results of this subsection remain valid in any finite-dimensional topological vector space, since any such space is isomorphic to a suitable space Rd . Theorem 2.6 (Carathéodory). Let A be an arbitrary subset of Rd . Then each point of co A is a convex combination of at most d + 1 points of A which are affinely independent. Proof. Assume that x ∈ co A, x = λ1 x1 + · · · + λn xn where x1 , . . . , xn ∈ A, λj > 0 for all j = 1, . . . , n (which we may suppose) and λ1 + · · · + λn = 1. If the vectors x1 , . . . , xn are affinely independent, then n ≤ d + 1 and we are done. Otherwise, n > d + 1 and there is (α1 , . . . , αn ) 6= (0, . . . , 0) such that α1 x1 + · · · + αn xn = 0 and α1 + · · · + αn = 0. Let k ∈ {1, . . . , n} be such that α α j k ≤ λj λk

for all j = 1, . . . , n.

Setting ηj := λj −

λk αj , αk

j = 1, . . . , n,

we have x=

X j6=k

η j xj ,

X

ηj = 1

and ηj ≥ 0 for j = 1, . . . , n.

j6=k

Thus, x is a convex combination of n − 1 points. If these points are affinely independent, the proof is finished. If not, the above argument can be repeated, and after finitely many steps x can be represented as a convex combination of affinely independent points of A.


7

Corollary 2.7. The convex hull co A of any set A ⊂ Rd is the union of all n-simplices (n ≤ d) with vertices in A. Proof. Obviously, any n-simplex with vertices in A, where n ≤ d, is a subset of co A. The reverse inclusion immediately follows from Carathéodory’s theorem 2.6. Corollary 2.8. The convex hull of any compact subset of Rd is compact. Proof. Let K be a compact subset of Rd and d n o X d+1 D := λ ∈ R : λ = (λ0 , . . . , λd ), λj = 1 and λj ≥ 0 for j = 0, . . . , d . j=0

The mapping F : D × K × · · · × K → K defined as F : (λ, x0 , . . . , xd ) 7→

d X

λ j xj

j=0

is continuous. By Carathéodory’s theorem 2.6, co K = F (D × K × · · · × K). Hence, co K, as a continuous image of the compact set D × K × · · · × K, is compact. Definition 2.9 (Extreme points). A point z of a set C ⊂ W is called an extreme point of C if z is not an internal point of any segment having its endpoints in C. In other words, z is an extreme point of C if x, y ∈ C, λ ∈ (0, 1) and z = λx + (1 − λ)y, implies x = y. It is easy to check that z is an extreme point of a convex set C if and only if the set C \ {z} is convex, and this is the case if and only if z is not a midpoint of any nondegenerate segment having its endpoints in C. We denote by ext C the set of all extreme points of C. Lemma 2.10. Let S be an n-simplex in Rd with vertices e0 , . . . , en . Then ext S = {e0 , . . . , en } . Proof. Let x ∈ ext S. Since S = co {e0 , . . . , en } and the set S \ {x} is convex, x = ek for some k ∈ {0, 1, . . . , n}. Conversely, select ek and assume that 1 1 ek = s + t 2 2 where s, t ∈ S = co {e0 , . . . , en }. Write s=

n X j=0

αj ej ,

t=

n X j=0

βj ej ,

8


with αj , βj ≥ 0,

n X j=0

Then ek =

n X 1 j=0

or

X1 j6=k

2

2

αj =

n X

βj = 1.

j=0

(αj + βj )ej

(αj + βj )(ej − ek ) = 0.

Consequently, αj + β j = 0

for j ∈ {0, . . . , n} \ {k} ,

thus s = t = ek . The following assertion shows the prominent role of extreme points in finite-dimensional compact convex sets. Infinite-dimensional situation is more complicated, see Example 2.15 and the Krein–Milman theorem 2.22. Theorem 2.11 (Minkowski). Each point of a compact convex set C ⊂ Rd is a convex combination of extreme points of C. Proof. We proceed by induction on the dimension d. For the dimension d = 0, the set C reduces to a one-point set and the assertion holds. So assume that d > 0 and that the assertion is valid for compact convex sets in spaces of dimension smaller than d. We may also assume that the interior of C is nonempty, for otherwise C is a subset of an affine subspace of a smaller dimension (cf. Exercise 2.107(c)) and the assertion follows by the induction assumption. We distinguish two cases. If x is a boundary point of C, then by Proposition 2.4 there exists a support hyperplane L of C at x. Then the compact convex set F := C ∩ L lies in the affine subspace L of dimension smaller than d. By the induction assumption, x ∈ co ext F . Since obviously ext F ⊂ ext C, the induction step is finished. Now suppose that x ∈ Int C. There exists a segment [a, b] ⊂ C such that x ∈ (a, b) and a, b ∈ ∂C. Since a, b ∈ co ext C by the previous argument, we see that x can be expressed as a convex combination of extreme points of C. Theorem 2.12 (Minkowski–Carathéodory). Each point of a compact convex set C ⊂ Rd is a convex combination of (at most d + 1) affinely independent extreme points of C. Proof. A consequence of Theorems 2.6 and 2.11. Indeed, given a point c ∈ C, by the Minkowski theorem 2.11, there exists a set A ⊂ ext C such that c ∈ co A. Now, it suffices to apply Carathéodory’s theorem 2.6.


9

Corollary 2.13. Let x be a point of a compact convex set C ⊂ Rd . Then there exists an n-simplex S, n ≤ d, such that x ∈ S ⊂ C and ext S ⊂ ext C. Proof. The assertion is a rewording of the previous Minkowski–Carathéodory theorem. Remark 2.14. In [1] E. M. Alfsen constructed a non-simplicial compact polyhedron in `1 , showing that the conclusion of the previous Corollary 2.13 in infinite-dimensional spaces fails. At the same time, he posed a question that “it would be of some interest to find sufficient conditions for a compact convex set X to admit a decomposition” as in Corollary 2.13: Given x ∈ X, there would exists a set S ⊂ X such that x ∈ S, ext S ⊂ ext C, and a unique representing measure for x carried by ext S. We present in Exercise 6.93 an example illustrating this phenomenon.

2.1.B

The Krein–Milman theorem

Recall that a point z of a subset C of a vector space W is an extreme point of C if z is not an internal point of any nondegenerate segment having endpoints in C and that ext C denotes the set of all extreme points of C. In the Euclidean space Rd , the set ext C is of fundamental importance. The Minkowski theorem 2.11 says that each point of a compact convex set C ⊂ Rd is a convex combination of extreme points of C. Thus, C = co(ext C). The aim of this subsection is to examine an analogous result and its relatives in the framework of infinite-dimensional spaces. Note, that in infinite-dimensional spaces, a compact convex set need not be a convex hull of its extreme points, as Example 2.15 shows. 2 Example 2.15. Let {en }∞ n=1 be the orthonormal basis in ` formed by the standard unit vectors en , and let

1 1 B := {0, e1 , e2 , e3 , . . . } and C := co B. 2 3 Since B is clearly compact, it is easy to see that C is a compact convex set (see [173], Exercise 1.56). By the Milman theorem 2.43, ext C ⊂ B. (In fact, it is easy to verify that ext C = B.) Defining xn := (1 − 2−n )−1

n X k=1

1 2−k ek , k

n ∈ N,

we have xn ∈ co ext C and xn → x :=

∞ X k=1

1 2−k ek ∈ C. k

Since every element of co ext C has only a finite number of nonzero coordinates, x∈ / co ext C.

10


Definition 2.16 (Extremal sets and faces). A generalization of the notion of extreme points leads to an important concept: A nonempty subset F of a set C ⊂ W is an extremal subset of C if x, y ∈ F provided that x, y ∈ C and λx + (1 − λ)y ∈ F for some λ ∈ (0, 1). It is needless to say that one-point extremal sets are exactly extreme points of C, and that C itself is an extremal set. Convex extremal sets are called faces. Definition 2.17 (Affine, concave and convex functions). Let C be a convex subset of a vector space W . A real-valued function s on C is said to be concave if s λx + (1 − λ)y ≥ λs(x) + (1 − λ)s(y) for each x, y ∈ C and λ ∈ [0, 1]. A real-valued function f on C is convex if −f is concave and f is called affine if both f and −f are concave. Obviously, the restriction of any linear functional on W to C is an affine function. Lemma 2.18. If H is an extremal subset of F and F is an extremal subset of D, then H is an extremal subset of D. Proof. Obvious. Lemma 2.19. If X is a nonempty compact convex subset of a locally convex space E, s is a concave lower semicontinuous function on X and L := {x ∈ X : s(x) = min s(X)}, then L is a compact extremal subset of X. If K is a nonempty compact subset of E, f ∈ E ∗ and H := {x ∈ K : f (x) = min f (K)}, then H is a compact extremal subset of K. Proof. It is clear that L is compact and nonempty. Choose x, y ∈ X, λ ∈ (0, 1). If a := λx + (1 − λ)y and a ∈ L, then s(a) ≥ λs(x) + (1 − λ)s(y) ≥ λs(a) + (1 − λ)s(a) = s(a). From this it easily follows that x, y ∈ L. The proof of the second assertion is similar. Proposition 2.20. Let K be a nonempty compact subset of a locally convex space E and F a compact extremal subset of K. Then F ∩ ext K 6= ∅. In particular, ext K 6= ∅.


11

Proof. Consider the family F of all closed extremalTsubsets of F ordered by the reverse inclusion. If R is a chain in F, then Y := {R : R ∈ R} is nonempty in view of the compactness of K. Since it is easy to check that Y is an extremal subset of K, Y is an upper bound for R. Zorn’s lemma now provides a maximal element of this family, call it D. An appeal to the Hahn–Banach theorem reveals that D is a one-point set. Indeed, assuming that D were to contain distinct points x and y, the Hahn–Banach theorem would provide f ∈ E ∗ such that f (x) < f (y). By Lemma 2.19, the set {z ∈ D : f (z) = min f (D)} would be a proper closed extremal subset of D and, in view of Lemma 2.18, also a closed extremal subset of F . This contradicts the maximality of D. Since, as mentioned above, one-point extremal sets are exactly extreme points of F , we can find a point x ∈ ext F . Again, by Lemma 2.18, x ∈ ext K and we are done. Theorem 2.21. Let K be a compact subset of a locally convex space E. Then K ⊂ co ext K. Proof. Assume that there exists a point x ∈ K \co ext K. Using the geometric version of the Hahn–Banach theorem, there exists f ∈ E ∗ such that f (t) > f (x) for any t ∈ co ext K. If H := {z ∈ K : f (z) = min f (K)}, then H is a (nonempty) extremal subset of K by Lemma 2.19. Hence, by Proposition 2.20, H ∩ ext K 6= ∅. Since H ∩ co ext K = ∅, this is impossible. Therefore, K ⊂ co ext K. Theorem 2.22 (Krein–Milman). Let X be a nonempty compact convex subset of a locally convex space E. Then X = co ext X. Proof. It is pretty clear that co ext X ⊂ X. The reverse inclusion follows from Theorem 2.21. Remarks 2.23. (a) The Krein–Milman theorem also holds in locally convex spaces over complex numbers. For a proof see, for example, W. Rudin [403], Theorem 3.21. (b) If K is a compact subset of a locally convex space, then co K = co ext K. This is an immediate consequence of Theorem 2.21. Corollary 2.24 (Bauer’s concave minimum principle). Let s be a lower semicontinuous concave function on a nonempty compact convex subset X of a locally convex space. Then there exists z ∈ ext X such that s(z) = min s(X).

12


Proof. Denote D := {x ∈ X : s(x) = min s(X)} . Then D is a nonempty compact subset of X and, by Lemma 2.19, an extremal subset of X. By Proposition 2.20, D ∩ ext X 6= ∅, and thus the proof is complete. Remarks 2.25. (a) The Krein–Milman theorem is an easy consequence of Bauer’s concave minimum principle. Indeed, if X is a nonempty compact convex subset of a locally convex space E, the constant function 1 on X attains its minimum on ext X. Thus the set ext X is nonempty. If x ∈ X \ co ext X, then, by the geometric version of the Hahn–Banach theorem, there exists f ∈ E ∗ such that f (x) < 1

and f ≥ 1

on

co ext X.

This contradicts Bauer’s minimum principle since f |X is a continuous affine function. (b) In Section 3.9 we prove a generalization of Bauer’s concave minimum principle by a different method. Integral representation. Now we would like to show how to use the Krein–Milman theorem for establishing integral-type representation theorems. Let x be a point of a compact convex set X in a locally convex space. Our aim is to find a Radon measure µ on X so that Z f (x) = f dµ X

for any continuous affine function f on X. Of course, the Dirac measure εx at x is one such measure. However, we try to find other “representing” measures, preferably concentrated on a very small part of X. More precisely, we are looking for a measure µ such that (a) the support spt µ of µ is contained in the closure ext X of the set of all extreme points of X, or even (b) µ is carried by the set ext X. Definition 2.26 (Barycenter of a measure). Let X be a compact convex set in a locally convex space E. Denote by Ac (X) the set of all continuous affine functions on X. A point x ∈ X is said to be the barycenter of a probability Radon measure µ ∈ M1 (X) if the following barycentric formula Z f (x) = f dµ X

holds for any f ∈ Since the functionals from E ∗ separate the points of E, and since the restrictions to X of such functionals are elements of Ac (X), we see that the Ac (X).


13

barycenter r(µ) of µ (which exists by Theorem 2.29), is uniquely determined. Note that Z r(µ) = t dµ(t), X

where the integral is to be understood as the Pettis integral. In the case when r(µ) = x, we also say that the measure µ represents the point x. In other words, the equality x = r(µ) means that the Integral representation theorem holds for the point x. We denote by Mx (Ac (X)), or for short Mx (X), the set of all measures representing the point x. Obviously, as noted above, the Dirac measure εx always represents the point x. In what follows, we answer the following questions: (a) Does any Radon measure have a barycenter ? (b) Is any point of X a barycenter of a Radon measure carried by ext X ? Proposition 2.27. The space M1 (K) consisting of all probability measures on a compact space K is a compact convex subset of M(K) and ext M1 (K) = {εx : x ∈ K} . The mapping ε : x 7→ εx , x ∈ K, is a homeomorphism of K onto ext M1 (K). Proof. It is easy to verify that M1 (K) is a convex subset of M(K). By Theorem A.85(a), it is compact. If x ∈ K and εx = αµ + (1 − α)ν

where µ, ν ∈ M1 (K) and α ∈ (0, 1),

then 1 = αµ({x}) + (1 − α)ν({x}). This implies that µ({x}) = ν({x}) = 1, and therefore µ = ν = εx . If µ ∈ M1 (K) is not a Dirac measure, then there exists a compact set F ⊂ K such that the measures ν := µ|F and λ := µ|K\F are nontrivial and distinct from µ. Since µ = ν(F )

λ ν + λ(K \ F ) , ν(F ) λ(K \ F )

we see that µ is not an extreme point of M1 (K). The mapping x 7→ εx , x ∈ K, is an injective continuous mapping from K onto {εx : x ∈ K} and hence K and ext M1 (K) are homeomorphic. Corollary 2.28. The set of all convex combinations of Dirac measures is dense in M1 (K).

14


Proof. As an easy consequence of the Krein–Milman theorem 2.22 and the characterization of extreme points given by Proposition 2.27, we have M1 (K) = co ext M1 (K) = co {εx : x ∈ K} , which finishes the proof. Theorem 2.29. Let X 6= ∅ be a compact convex subset of a locally convex space E. Then each Radon measure from M1 (X) has a (unique) barycenter in X. Proof. With the uniqueness part already out of the way, we nowPconcentrate on an λj εxj , where existence proof. If a measure µPin question is molecular, µ = nj=1 P xj ∈ X, λj ≥ 0, j = 1, . . . , n, nj=1 λj = 1, then obviously r(µ) := nj=1 λj xj ∈ X is the barycenter of µ. Given a measure µ ∈ M1 (X), there exists a net {µγ } of molecular measures on X such that µγ → µ. Since X is a compact set, there exists a subnet {r(µα )} of {r(µγ )} converging to an element z ∈ X. Pick f ∈ Ac (X). Then Z Z f dµ, f dµα = f (z) = lim f (r(µα )) = lim α

α

X

X

and therefore z is a barycenter of µ. Definition 2.30 (Barycenter mapping). Let X be a compact convex subset of a locally convex space. The mapping r : µ 7→ r(µ), assigning to each measure µ ∈ M1 (X) its barycenter, is called the barycenter mapping. In Proposition 2.38 we show that the barycenter mapping is a continuous and affine mapping from M1 (X) into X. This mapping is surjective since r(εx ) = x for any x ∈ X. Theorem 2.31 (Integral representation theorem). Let X be a compact convex subset of a locally convex space E and let x ∈ X. Then there exists a measure µ ∈ M1 (X) such that r(µ) = x and spt µ ⊂ ext K. Proof. From the Krein–Milman theorem 2.22, we can see the following fact: if f ∈ Ac (X) and f = 0 on ext X, then f = 0 on X. We denote by B the subspace of C(ext X) consisting of all restrictions of functions in Ac (X) to ext X. Then, for every h ∈ B, there exists, by the above mentioned fact, a unique function b h ∈ Ac (X) which coincides with h on ext X. We fix x ∈ X and set ϕ : h 7→ b h(x),

h ∈ B.

Evidently ϕ ∈ B ∗ and kϕkB = 1. The functional ϕ can be extended by the Hahn– Banach theorem from B to a functional Φ ∈ (C(ext X))∗ with the same norm. Since,


15

in addition, Φ(1) = ϕ(1) = 1, Φ is a positive functional. Indeed, if f ∈ C(ext X), f ≥ 0, a = 12 sup f (ext X), then ka − f k ≤ a and so a − Φ(f ) = Φ(a) − Φ(f ) = Φ(a − f ) ≤ ka − f k ≤ a. This yields Φ(f ) ≥ 0. By the Riesz representation theorem, there exists a probability R measure µ on ext X such that Φ(f ) = X f dµ for every function f ∈ C(ext X). The measure µ can be regarded as a measure on X carried by the set ext X. Since obviously Z X

g dµ = Φ(g) = ϕ(g|ext X ) = gb(x) = g(x)

for every g ∈ Ac (X), we see that the barycenter of the measure µ is exactly the point x. Remarks 2.32. (a) Theorem 2.31 can be proved by an alternative manner. In fact, we can follow the proof of Proposition 2.39 step by step. (b) In concrete applications, we are often able to characterize the set ext X. However, the character of the elements of the set ext X \ ext X is generally rather obscure. Consequently, the information concerning the support of the measure from the theorem on integral representation is problematic, unless the set ext X is closed. Moreover, there is another problem. Let us imagine that the set of extreme points of a compact convex set X is dense in this set, that is, ext X = X. Then, naturally, the Krein–Milman theorem says nothing, and equally useless is the theorem on integral representation. Indeed, it suffices to take the Dirac measure εx at the point x for the measure representing the point x . This situation can actually occur. As an example, we can take the closed unit ball B in an arbitrary infinite-dimensional Hilbert space, which we of course consider to be equipped with the weak topology. The extreme points of B are then the points of the unit sphere, and its (weak) closure is equal to the whole ball B. A more sophisticated example with ext X = X is the Poulsen simplex in the Hilbert space `2 (see Subsection 12.3.A). However, much more is known. Namely, if we consider the so-called Hausdorff metric on the set F of all nonempty compact convex subsets ofa given Banach space X of infinite dimension, the space F is complete and the set C ∈ F : ext C 6= C is merely meager in F. This assertion was proved by V. L. Klee in [271]. Thus, in a certain sense, for the majority of compact convex sets we have ext C = C. Hence the problem of whether it is possible to find a measure µ which is carried just on the set of extreme points in the theorem on integral representation is crucial. This problem was solved successfully by G. Choquet in the fifties of the 20th century and laid the foundations of the Choquet theory. We will devote the next chapters to it, in the more general setting of function spaces. The Choquet theory has provided many insights for abstract analysis, infinite-dimensional geometry, descriptive set theory,

16


potential theory and other fields of mathematics. It has remained fruitful ever since and has found new applications again and again in deriving new results. (c) In Sections 14.5 and 14.6, we exceptionally consider locally convex spaces over the field of complex numbers. Note that the Integral representation theorem 2.31 extends trivially to the complex case. Indeed, consider a compact convex subset X of a complex locally convex space E. Of course, E can be regarded as a locally convex space over the field of real numbers. Then, given x ∈ X, there exists, by Theorem 2.31, a measure µ ∈ M1 (X) carried by ext X such that Z f dµ (2.1) f (x) = X

for every real continuous functional f on E. Given a complex continuous functional F on E, we can apply (2.1) to Re F and Im F to conclude that Z F dµ. F (x) = X

The Integral representation theorem will be applied several times in the sequel. For this purpose, the following easy consequence of the previous theorem will be useful. Proposition 2.33 (Krein–Milman theorem with transfer). Let E be a locally convex space of real-valued functions on a set M such that, for every x ∈ M , the evaluation functional Fx : f 7→ f (x) is continuous on E. Let K ⊂ E be a compact convex set. Let Q be a compact space and Φ : y 7→ ϕy , y ∈ Q, be an injective continuous mapping of Q onto ext K. Then there exists a probability measure µ on Q such that, for every f ∈ K, Z f (x) =

ϕy (x) dµ(y),

x ∈ M.

Q

Proof. Since ext K is a continuous image of a compact set, it is closed. Let f ∈ K. By Theorem 2.31, there exists a probability measure µ˜ carried by ext K such that Z F (f ) = F dµ˜ (2.2) ext K

for each continuous linear functional F on E. Let us define µ = Φ−1 ] µ. ˜ By Proposition A.92, Z Z Z F dµ˜ = (F ◦ Φ) ◦ Φ−1 dµ˜ = F ◦ Φ dΦ−1 ] µ˜ ext K

ext K

Z F ◦ ϕy dµ(y).

= Q

Q

(2.3)


17

Applying (2.2) and (2.3) to the evaluation functional Fx , we obtain Z Z ϕy (x) dµ(y), x ∈ M. Fx dµ˜ = f (x) = Q

ext K

Lemma 2.34. Let X be a compact convex subset of a locally convex space E. Then the space (E ∗ + R)|X is dense in Ac (X). Proof. Let X be a nonempty compact convex set. Fix a function h ∈ Ac (X) and ε > 0. Denote J1 := {(x, t) ∈ X × R : t = h(x)} and J2 := {(x, t) ∈ X × R : t = h(x) + ε} . Then J1 , J2 is a pair of disjoint nonempty compact convex subsets of E × R. By the Hahn–Banach theorem, there exist a functional F ∈ (E × R)∗ and λ ∈ R such that sup F (J1 ) < λ < inf F (J2 ). There are ϕ ∈ E ∗ and β ∈ R such that F (t, r) = ϕ(t) + βr for any t ∈ E and any r ∈ R. From the separation of J1 and J2 it easily follows that β 6= 0. Put ψ(t) := β1 (λ − ϕ(t)) for t ∈ E. Since ϕ(t) + βh(t) < λ < ϕ(t) + βh(t) + βε,

t ∈ X,

we easily get kh − ψk < ε. Remark 2.35. In general, there might exist a continuous affine function on a compact convex set X that is not of the form (E ∗ + R)|X ; see Exercise 2.111. Proposition 2.36. A Radon measure µ ∈ M1 (X) represents x ∈ X if and only if Z ϕ(x) = ϕ dµ for any ϕ ∈ E ∗ . X

R Proof. Recall that, by definition, µ represents x ∈ X if h(x) = X h dµ for any h ∈ Ac (X). Hence the assertion is an easy consequence of Lemma 2.34 and the Lebesgue dominated convergence theorem. Definition 2.37 (The barycenter revisited). Proposition 2.36 enables us to extend slightly the definition of a barycenter to the case of nonconvex sets and points not belonging to this set. Let K be a compact subset of a locally convex space E. We say that x ∈ E is the barycenter of a Radon measure µ ∈ M1 (K), or that µ represents x, in a symbol x = r(µ), if Z ϕ(x) =

ϕ dµ for any K

ϕ ∈ E∗.

18


Proposition 2.38. If X is a compact convex subset of a locally convex space, the barycenter mapping r : M1 (X) → X is affine and continuous. Hint. Obviously, r is affine. If {µγ } is a net in M1 (X), µγ → µ and f ∈ E ∗ , then f (r(µγ )) = µγ (f ) → µ(f ) = f (r(µ)). Consequently, the net {r(µγ )} is weakly converging in X to r(µ). Since X is a compact set, r(µγ ) → r(µ) by Proposition A.28. Proposition 2.39. If K is a compact subset of a locally convex space E and x ∈ E, then the following statements are equivalent: (i) x ∈ co K, (ii) there exists a Radon measure µ ∈ M1 (K) such that r(µ) = x. Proof. Let µ ∈ M1 (K) satisfy r(µ) = x. Assuming that x ∈ / co K, by the Hahn– Banach theorem there exist ϕ ∈ E ∗ and λ ∈ R such that ϕ(x) < λ ≤ ϕ(t)Rfor any t ∈ co K. Then, obviously, no Radon measure µ ∈ M1 (K) with ϕ(x) = K ϕ dµ exists. Conversely, assume that x ∈ co K. There exists a net {xα } of points in co K such that xα → x. We can write xα =

nα X

λαj xαj ,

where nα ∈ N, xαj ∈ K, λαj ≥ 0,

nα X

λαj = 1.

j=1

j=1

We define for each α µα :=

nα X

λαj εxαj .

j=1

Then µα ∈ M1 (K) and r(µα ) = xα . Since the set M1 (K) is compact, we may assume that {µα } converges to µ ∈ M1 (K). For any ϕ ∈ E ∗ , ϕ(r(µα )) = µα (ϕ) → µ(ϕ) and

ϕ(r(µα )) = ϕ(xα ) → ϕ(x).

Hence ϕ(x) = µ(ϕ) and r(µ) = x. Theorem 2.40 (Bauer’s characterization of ext X). Let X be a compact convex subset of a locally convex space and x ∈ X. Then x ∈ ext X if and only if the Dirac measure εx is the only measure from M1 (X) with a barycenter x. Proof. If x = 12 (a+b) where a, b ∈ X, a 6= b, then the measure 12 (εa +εb ) ∈ M1 (X), which differs from εx , has the barycenter x. Assume now that x ∈ ext X and µ ∈ M1 (X) with r(µ) = x are given. It must be shown that µ = εx . To see this, it suffices to show that spt µ = {x}. Assume that


19

z ∈ spt µ \ {x}. Let U ⊂ X be a closed convex neighborhood of z in X such that x∈ / U . The set U is compact and convex, µ(U ) < 1 in view of Proposition 2.39, and obviously µ(U ) > 0. Set µ1 :=

1 µ|U µ(U )

and

µ2 :=

1 µ| . µ(X \ U ) X\U

Then µ1 , µ2 are in M1 (X). Since µ1 (U ) = 1, we get r(µ1 ) ∈ U (again from Proposition 2.39). Hence r(µ1 ) 6= x. We have µ = µ(U )µ1 + (1 − µ(U ))µ2 and we see that x = µ(U )r(µ1 ) + (1 − µ(U ))r(µ2 ) is a convex combination of r(µ1 ) and r(µ2 ). This contradicts the assumption that x is an extreme point of X and yields the required conclusion. Proposition 2.41 (Choquet). Let X be a compact convex subset of a locally convex space E and x ∈ ext X. Then the family {y ∈ X : f (y) < λ},

f ∈ E ∗ and λ ∈ R with f (x) < λ,

form a base of neighborhoods of x in X. Proof. Let U be an open neighborhood of x. Then X \ U is a compact set so that x∈ / co(X \ U ). Indeed, if this were not true, then Proposition 2.39 would provide a probability measure µ carried by X \ U with r(µ) = x. But this would contradict our assumption that x is an extreme point because the only representing measure for x is the Dirac measure εx (see Theorem 2.40). Now the Hahn–Banach separation theorem yields the existence of a continuous linear functional f and λ ∈ R such that f (x) < λ and f > λ on co(X \ U ). Thus x ∈ {y ∈ X : f (y) < λ} ⊂ U and the proof is finished. Proposition 2.42. Let K1 , . . . , Kn be compact convex subsets of a topological vector space. Then co(K1 ∪ · · · ∪ Kn ) is compact. Proof. We follow the same lines as in the proof of Corollary 2.8. Since co(K1 ∪ · · · ∪ Kn ) = F (D × K1 × · · · × Kn ) where n n o X D := λ ∈ Rn : λ = (λ1 , . . . , λn ), λj = 1 and λj ≥ 0 for j = 1, . . . , n j=1

20


and F : (λ, x1 , . . . , xn ) 7→

n X

(λ, x1 , . . . , xn ) ∈ D × K1 × · · · × Kn

λ j xj ,

j=1

is continuous, it follows that co(K1 ∪ · · · ∪ Kn ) is compact. Theorem 2.43 (Milman). Let B be a subset of a locally convex space E such that the set X := co B is compact. Then ext X ⊂ B. Proof. Assume that x ∈ ext X \ B. By Proposition 2.41, there exist f ∈ E ∗ and λ ∈ R such that f (x) > λ ≥ inf f (B). Since f ∈ E ∗ , inf f (B) = inf f (co B). Therefore we get x ∈ / co B = X, which is a contradiction. Proposition 2.44. Let X be a compact convex subset of a locally convex space E and ∅ 6= B ⊂ X. Then the following assertions are equivalent: (i) co B = X, (ii) inf f (B) = min f (X) for each f ∈ E ∗ , (iii) ext X ⊂ B. Proof. Assertions (i) and (iii) are equivalent by the Krein–Milman and Milman theorems 2.22 and 2.43. The implication (i) =⇒ (ii) is obvious. Conversely, the proof of the implication (ii) =⇒ (i) follows the same lines as the end of the proof of the Krein–Milman theorem: if co B 6= X, the Hahn–Banach separation theorem yields the assertion. Proposition 2.45. Any metrizable compact convex subset X of a locally convex is affinely homeomorphic to a compact convex subset of the Hilbert space `2 . Proof. Since X is a metrizable compact set, the space C(X) is separable, so it is the space Ac (X) of all continuous affine functions on X. Let {fn : n ∈ N} be a dense subset of the closed unit ball of Ac (X). If T : x 7→

n1 n

o fn (x) ,

x ∈ X,

then T is a continuous affine injective mapping of X into `2 . Thus T is an affine homeomorphism of X onto a compact convex subset T (X) of `2 .


2.1.C

21

Exposed points

Definition 2.46 (Exposed points). If X is a compact convex set, a point x ∈ X is exposed if there exists a function f ∈ Ac (X) such that f (x) > f (y) for each y ∈ X \ {x}. We call such a function f an exposing function for x and the set of all exposed points of X is denoted as exp X. We point out the following simple, but important fact. Proposition 2.47. For any compact convex set, exp X ⊂ ext X. Proof. The proof follows by a straightforward verification. Definition 2.48 (Farthest points). Let D be a subset of a normed linear space E. A point z ∈ D is called a farthest point of D if there exists x ∈ E such that kx − zk = sup{kx − tk : t ∈ D}. The set of all farthest points of D is denoted by far D. Lemma 2.49. If X is a nonempty compact convex subset of a Hilbert space H, then far X ⊂ exp X and it is a nonempty set. Proof. The inclusion far X ⊂ exp X follows from the fact that any point x in the closed unit ball BH of norm 1 is an exposed point of BH (we recall that BH is a compact convex set in the weak topology of H). Indeed, f (y) := (y, x), y ∈ H, is an exposing function for x. If y ∈ H is chosen arbitrarily, a simple compactness argument yields the existence of a point z ∈ X such that ky − zk = sup{ky − xk : x ∈ X}. Hence far X is nonempty. Proposition 2.50. Let X be a nonempty compact convex subset of a Hilbert space H. Then X = co far X = co exp X. Proof. Since far X ⊂ exp X, it suffices to show that X = co far X. Assume that x ∈ X \co far X. By the Hahn–Banach theorem and the Fréchet–Riesz representation theorem, there exist h ∈ H and λ ∈ R such that (x, h) < λ ≤ (y, h) for each y ∈ co far X. Denote s := sup{ktk : t ∈ X}. Let α > 0 be such that 2α λ − (h, x) > s2 − kxk2 . By compactness, there exists z ∈ X such that kαh − zk = sup{kαh − tk : t ∈ X}. Of course, z ∈ far X. Since kαh − zk2 = α2 khk2 − 2α(h, z) + kzk2 ≤ α2 khk2 − 2αλ + s2 < α2 khk2 − 2α(h, x) + kxk2 = kαh − xk2 , we get kαh − zk < kαh − xk, which is a contradiction.

22


Theorem 2.51. Let X be a metrizable compact convex subset of a locally convex space. Then X = co exp X. Proof. By Proposition 2.45, X is affinely homeomorphic to a compact convex subset of `2 . Obviously, exposed points are preserved by an affine homeomorphism and thus the assertion follows from Proposition 2.50. Corollary 2.52. Let X be a metrizable compact convex subset of a locally convex space. Then ext X ⊂ exp X. Proof. The assertion is an immediate consequence of Theorem 2.51 and Proposition 2.44.

2.2

Interlude: On the space M(K)

Let K be a compact space and M(K) the space of all signed Radon measures on K. We emphasize that on the space M(K) we always consider the weak∗ -topology given by the duality of C(K) and M(K). In this chapter we collect some properties of this space and its subspaces. Example 2.53 (Image of a measure from M1 (K)). If K is a compact space, it follows from Proposition 2.27 that the mapping ε : x 7→ εx , x ∈ K, is a homeomorphism of K onto ext M1 (K). Let Λ := ε] λ be the image of a probability measure λ with spt λ ⊂ K under ε. The following proposition will be useful in many examples. Proposition 2.54. The measure Λ from Example 2.53 is carried by the (closed) set ε(spt λ) and its barycenter equals λ. Proof. Since Λ(ε(spt λ)) = λ(ε−1 (ε(spt λ))) = λ(spt λ) = 1, we see that Λ is carried ∗ by ε(spt λ). Pick ϕ ∈ M(K) . By duality theory, there exists f ∈ C(K) such that ϕ(µ) = µ(f ) for any µ ∈ M(K). Then (cf. Proposition A.92), Z Λ(ϕ) = (ε] λ)(ϕ) = λ(ϕ ◦ ε) = K

= λ(f ) = ϕ(λ), and r(Λ) = λ.

Z ϕ(εx ) dλ(x) =

f (x) dλ(x) K

2.2 Interlude: On the space M(K)

23

Proposition 2.55. Let F be a closed subset of a compact space K and b > 0. Then the function ϕb : µ 7→ µ({x ∈ F : µ({x}) ≥ b}),

µ ∈ M1 (K),

is upper semicontinuous on M1 (K). Proof. Let c > 0 and µ ∈ G := {ν ∈ M1 (K) : ϕb (ν) < c} be given. We will show that G contains a neighborhood W of µ. The set L := {x ∈ F : µ({x}) ≥ b} is finite. Let U be an open subset of K such that L ⊂ U and µ(U ) < c. For every x ∈ F \ U we find an open neighborhood Vx of x so that µ(V x ) < b. Using compactness, we select finitely many points x1 , . . . , xn such that F \ U ⊂ V x1 ∪ · · · ∪ V xn . Since the function ν 7→ ν(H) is upper semicontinuous on M1 (K) for every closed set H ⊂ K (see Theorem A.85(b)), the set W := {ν ∈ M1 (K) : ν(U ) < c, ν(V xi ) < b, i = 1, . . . , n} is open and contains µ. It remains to show that ϕb (ν) < c for every ν ∈ W . Given a measure ν ∈ W , let Lν := {x ∈ F : ν({x}) ≥ b}. It follows from the choice of W that Lν ⊂ U . Thus ϕb (ν) = ν(Lν ) ≤ ν(U ) < c, and ϕb is upper semicontinuous. Proposition 2.56. Let F be a closed subset of a compact space K. Then the function X ψ : µ 7→ µd (F ) := µ({x}), µ ∈ M1 (K), x∈F

is a limit of an increasing sequence of positive upper semicontinuous functions on M1 (K). Proof. For n ∈ N, let ψn be the function ϕb from Proposition 2.55 for b = 1/n. Then it is easy to check that ψn % ψ on M1 (K). As ψn are upper semicontinuous and positive functions, the proof is finished. Definition 2.57 (Fσ and Gδ faces). A face which is simultaneously an Fσ set is called an Fσ face. Analogously, a Gδ face is a face which is a Gδ set.

24


Proposition 2.58. Let F be a closed subset of a compact space K. Then the set G := µ ∈ M1 (K) : µ|F is continuous is a Gδ face of M1 (K) such that G ∩ ext M1 (K) = ∅. Proof. Let ψ : µ 7→ µd (F ), µ ∈ M1 (K), where µd is the discrete part of µ. According to Proposition 2.56, there is a sequence {ψn } of positive upper semicontinuous functions such that ψn % ψ on M1 (K). Then ∞ \ ∞ \ 1 G = µ ∈ M1 (K) : ψ(µ) = 0 = µ ∈ M1 (K) : ψn (µ) < . k n=1 k=1

It follows that G is a Gδ set which is obviously convex and extremal. Since ext M1 (K) = {εx : x ∈ K} (cf. Proposition 2.27) and Dirac measures are discrete, we have G ∩ ext M1 (K) = ∅. Proposition 2.59. Let H :=

∞ [

µ ∈ M1 ([0, 1]) : spt µ ⊂

1

n, 1

.

n=2

Then H is an Fσ face of M1 ([0, 1]) which is not of type Gδ . Proof. Obviously, H is a convex set and extremal. Moreover, H is of type Fσ . Since both sets H and M1 ([0, 1]) \ H are dense in M1 ([0, 1]), H is not a Gδ set (cf. Theorem A.58). Proposition 2.60. Let K be a compact space and ω ∈ M1 (K). Define ψ : µ 7→ µs (K),

µ ∈ M1 (K),

where µs is the singular part of µ with respect to the measure ω. Then ψ is a limit of a decreasing sequence of lower semicontinuous functions on M1 (K). Proof. For n ∈ N, set 1 ψn (µ) := sup µ(G) : G ⊂ K open and ω(G) < , n

µ ∈ M1 (K).

Obviously, {ψn } is a decreasing sequence of lower semicontinuous functions. Recall that, by Theorem A.85(b), the function µ 7→ µ(G),

µ ∈ M1 (K),

is lower semicontinuous on M1 (K) for any open set G ⊂ K.

2.2 Interlude: On the space M(K)

25

Pick n ∈ N and µ ∈ M1 (K). There exists a Borel set B ⊂ K such that µs (B) = µs (K) = ψ(µ) and ω(B) = 0. Let G ⊂ K be an open set containing B for which ω(B) < n1 . Then ψ(µ) = µs (B) ≤ µs (G) ≤ µ(G) ≤ ψn (µ). Hence, ψ ≤ ψn for any n ∈ N. It remains to show that limn→∞ ψn = ψ. To this end, pick µ ∈ M1 (K) and c > ψ(µ). Since µac ω (recall that µac denotes the absolutely continuous part of µ with respect to ω, see Proposition A.65), there exists n ∈ N so that µac (B) < c − ψ(µ) whenever B is a Borel set with ω(B) < ω(G) < n1 , then

1 n.

Now, if G ⊂ K is an open set satisfying

µ(G) = µs (G) + µac (G) ≤ µs (K) + c − µs (K) = c. Thus, ψn (µ) ≤ c, and therefore ψn → ψ. Proposition 2.61. If λ denotes Lebesgue measure on [0, 1] and L := µ ∈ M1 ([0, 1]) : µ ⊥ λ , then L is a Gδ face of M1 ([0, 1]) which is not of type Fσ . Proof. It is easy to check that L is convex, extremal and dense in M1 ([0, 1]). Therefore all that needs to be proved is that L is a Gδ set. Let {ψn } be a sequence of functions as in Proposition 2.60 for K := [0, 1] and ω := λ. The assertion then follows from the following equalities ∞ \ L = µ ∈ M1 ([0, 1]) : µs ([0, 1]) = 1 = µ ∈ M1 ([0, 1]) : ψn (µ) = 1 n=1

=

∞ \ ∞ \ n=1 k=1

1 µ ∈ M ([0, 1]) : ψn (µ) > 1 − k 1

,

because the functions ψn are lower semicontinuous. Since L is dense, it is not an Fσ set. Remark 2.62. In Exercise 2.118 we indicate another reasoning of the fact that L is a Gδ set.

26


Proposition 2.63 (Choquet’s examples). Let ϕ : µ 7→ µd ([0, 1]),

µ ∈ M1 ([0, 1]),

ψ : µ 7→ µs ([0, 1]),

µ ∈ M1 ([0, 1]).

and Then ϕ and ψ are bounded affine functions on M1 ([0, 1]) of the second Baire class and there exists a probability measure Λ on M1 ([0, 1]) such that Λ(ϕ) 6= ϕ(r(Λ)) and

Λ(ψ) 6= ψ(r(Λ)).

Proof. Obviously, ϕ and ψ are bounded and affine. By Propositions 2.60, 2.56 and A.53, both functions ϕ and ψ are of the second Baire class on M1 ([0, 1]) because M1 ([0, 1]) is metrizable by Theorem A.85. Let Λ be the image of Lebesgue measure λ on [0, 1] under ε. According to Proposition 2.54, Λ is carried by the (closed) set ε([0, 1]) and its barycenter is equal to λ. It remains to show that the “barycentric formula” Z ϕ dΛ Λ(ϕ) = M1 (K)

does not hold. Indeed, Z

Z

Λ(ϕ) =

ϕ dΛ = M1 (K)

ϕ dΛ ε([0,1])

Z 1 dΛ = 1 6= 0 = ϕ(λ).

= ε([0,1])

The same argument can be used in the case of the function ψ.

2.3

Structures in convex sets

Throughout this section, X will be a compact convex subset of a locally convex space E.

2.3.A

Extremal sets and faces

Recall that a nonempty set F ⊂ X is called extremal if x, y ∈ F whenever x, y ∈ X, λ ∈ (0, 1) and

λx + (1 − λ)y ∈ F.

One-point extremal sets are just extreme points of X. It is simply checked that a set F is extremal if and only if [ {(λF − (λ − 1)X) ∩ X : λ ≥ 1} ⊂ F.

2.3 Structures in convex sets

27

Proposition 2.64. Let F be a subset of X. Then (a) F ∩ ext X ⊂ ext F , (b) if F is extremal, then ext F = F ∩ ext X. Proof. The assertion is a straightforward consequence of definitions. Recall that convex extremal sets are called faces. Closed extremal sets occasionally bear the name absorbent sets. Definition 2.65 (Generated faces). It is easy to see that any intersection of faces of X is again a face. Hence, given a set A ⊂ X, there exists the smallest face of X containing A. It equals the intersection of all faces containing A and is denoted by face A. Given x ∈ X, we will write simply face x instead of face {x} where no confusion can arise. Proposition 2.66. If F is a convex subset of X, then [ face F = {(λF − (λ − 1)X) ∩ X : λ ≥ 1} . Moreover, if F is closed, face F is an Fσ set. Proof. Let Fλ := (λF + (λ − 1)X) ∩ X,

λ ≥ 1.

First we notice that, given λ > 1, y ∈ Fλ if and only if there exists x ∈ F such that y + (λ − 1)−1 (x − y) ∈ X. S Hence Fλ ⊂ Fλ0 if 1S≤ λ ≤ λ0 . It follows that λ≥1 Fλ is a convex set. Since it is easy to observe that λ≥1 Fλ is extremal, we have [ face F ⊂ Fλ . λ≥1

On the other hand, Sit is immediate to verify from the definition the converse inclusion. Hence face F = λ≥1 Fλ . If F is closed, a routine verification yields that each Fλ is closed as well. Hence face F =

∞ [

Fn

n=1

is an Fσ set. Corollary 2.67. If x ∈ X, then face x = {y ∈ X : there exists z ∈ X and λ ∈ [0, 1) such that x = λz + (1 − λ)y} and face x is an Fσ set.

28


Proof. Use Proposition 2.66. Proposition 2.68. Let F be a subset of X. Then the following assertions are equivalent: (i) F is extremal, (ii) the characteristic function cF of F is convex, (iii) F is a union of faces. Proof. The equivalence of (i) and (ii) is clear from the definition. Since it is easy to check that any union of extremal sets is extremal, we have (iii) =⇒ (i). If (i) holds, then [ F = face x, x∈F

since face x ⊂ F for any x ∈ F . Proposition 2.69. Let F be a subset of X. The following assertions are equivalent: (i) F is a closed extremal set, (ii) the characteristic function cF of F is upper semicontinuous and convex, (iii) there exists a positive, lower semicontinuous and concave function f on X such that F = {x ∈ X : f (x) = 0}, (iv) F is closed and spt µ ⊂ F whenever µ ∈ M1 (X) and r(µ) ∈ F , (v) F is closed and a union of faces, (vi) F is closed and a union of closed faces. Proof. The equivalence of (i), (ii) and (v) follows immediately; see Proposition 2.68. Assertions (ii) and (iii) are obviously equivalent. To see that (i) =⇒ (iv), suppose that F is a closed extremal set and µ ∈ M1 (X) such that r(µ) ∈ F . If spt µ is not a subset of F , then there exists a closed convex set C ⊂ X \ F such that µ(C) > 0. We have µ(C) < 1, since otherwise r(µ) = r(µ|C ) ∈ / F . Set µ1 :=

1 µ|C µ(C)

and

µ2 :=

1 µ| . 1 − µ(C) X\C

Now, µ = µ(C)µ1 + (1 − µ(C))µ2 , hence r(µ) = µ(C)r(µ1 ) + (1 − µ(C))r(µ2 ) ∈ F. Since F is extremal, r(µ1 ), r(µ2 ) ∈ F , which is a contradiction.


29

Now let F be a closed set and let x ∈ F , x = λy + (1 − λ)z where y, z ∈ X, λ ∈ (0, 1). If µ := λεy + (1 − λ)εz , then r(µ) = x and spt µ = {y, z}. Hence y, z ∈ F , which shows that F is extremal and proves that (iv) =⇒ (i). Since (vi) obviously implies (v), all that remains to be proved is that (i) =⇒ (vi). So let F be a closed extremal subset of X and x ∈ X. The set F := {C ⊂ F : x ∈ C, C convex} ordered by inclusion satisfies the assumptions of Zorn’s lemma. Indeed, if R ⊂ F is a chain, then the set [ {R : R ∈ R} belongs to F and it is an upper bound of R. Hence there exists a maximal element C ∈ F. Since x ∈ C ⊂ C ⊂ F and C ∈ F, the maximality of C implies that C = C. It remains to show that C is a face. Since C is convex, we must verify only that C is extremal. So, we are given a, b ∈ X and λ ∈ (0, 1) such that z := λa + (1 − λ)b ∈ C, and we wish to show that e be the convex hull of C and {a, b}. The proof a, b ∈ C. In order to prove this, let C e ⊂ F . Then, due to the maximality that a, b ∈ C will be achieved by showing that C e of C, C = C. e Then Choose e c ∈ C. e c = λ1 (λ2 a + (1 − λ2 )b) + (1 − λ1 )c, where c ∈ C and λ1 , λ2 ∈ [0, 1]. It is clearly sufficient to assume that λ1 ∈ (0, 1). (If λ1 = 0, e c = c. If λ1 = 1, e c belongs to the segment joining a and b, and thus e c ∈ F by the extremality of F .) Moreover, we may assume that λ2 ≥ λ (otherwise, 1 − λ2 ≥ λ). With λ1 and λ2 chosen in this manner, we set α :=

λ1 λ2 λ1 λ2 + λ(1 − λ1 )

and

β :=

λ . λ1 λ2 + λ(1 − λ1 )

Since β > 0, βe c + (1 − β)b = αz + (1 − α)c ∈ C ⊂ F and since F is extremal, we get e c ∈ F. Remarks 2.70. (a) Let F be an extremal set and x ∈ F . If µ ∈ M1 (X) and x = r(µ), then spt µ ⊂ F . The proof of this assertion can be obtained as a slight modification of the proof of the implication (i) =⇒ (iv) in Proposition 2.69. (b) In general, the closure of a face need not be a face. An example can be found in E. M. Alfsen [1], Theorem 1. In Exercise 4.52 we present an example of a compact convex set X and a point x ∈ X such that face x is not a face.

30


Corollary 2.71. Let F be a closed convex subset of X. Then F is a face if and only if for every measure µ ∈ M1 (X) with barycenter r(µ) in F , we have spt µ ⊂ F . Proof. A closed convex set is a face if and only if it is extremal. Hence the assertion follows immediately from Proposition 2.69, (i) ⇐⇒ (iv). Proposition 2.72. Let ϕ : X → Y be a continuous affine surjection of a compact convex set X onto a compact convex set Y . (a) If H ⊂ Y is extremal, then ϕ−1 (H) is an extremal set of X. (b) If H ⊂ Y is a face, then ϕ−1 (H) is a face of X. (c) ϕ(ext X) ⊃ ext Y . Proof. Let H ⊂ Y be an extremal set and αx1 + (1 − α)x2 ∈ ϕ−1 (H), where x1 , x2 ∈ X, α ∈ [0, 1]. Then αϕ(x1 ) + (1 − α)ϕ(x2 ) ∈ H, which yields ϕ(x1 ), ϕ(x2 ) ∈ H. Hence x1 , x2 ∈ ϕ−1 (H), concluding the proof of (a). Since (b) is a straightforward consequence of (a), we proceed to the proof of (c). Given a point y ∈ ext Y , the set ϕ−1 (y) is a closed extremal set by (a). By Proposition 2.20, it intersects ext X. Hence y ∈ ϕ(ext X), and we are done.

2.3.B

Measure convex sets

In the sequel, stronger versions of convexity and extremality are investigated. Constructions of counterexamples use properties of sets of probability measures studied in Section 2.2. As above, throughout this subsection, X will be a compact convex subset of a locally convex space E. Definition 2.73 (Measure convex sets). A universally measurable set F ⊂ X is measure convex if the barycenter r(µ) belongs to F for any measure µ ∈ M1 (X) carried by F . We will show that, for a universally measurable set F ⊂ X, the relations between “F measure convex” (labelled as MC) and “F convex” (labelled as C) are as follows: MC ⇒ C MC : C MC ⇐ C MC ⇐ C MC ⇐ C MC ⇐ C MC : C MC : C

F is closed or open F is resolvable dim E < ∞ F is a resolvable face F is Fσ face F is Gδ face

(2.74) (2.81) (2.74, 2.76) (2.80) (2.77) (2.91) (2.82, 2.83) (2.84)


31

Proposition 2.74. Every measure convex universally measurable subset of X is convex, and every closed convex subset of X is measure convex. Proof. Let F ⊂ X be measure convex, x, y ∈ F , λ ∈ [0, 1] and z = λx + (1 − λ)y. Since the measure µ := λεx + (1 − λ)εy is carried by F and r(µ) = z, we get z ∈ F . Therefore, F is convex. If F is a closed convex set and if a measure µ ∈ M1 (X) has its support contained in F , then the implication (ii) =⇒ (i) of Proposition 2.39 tells us that r(µ) ∈ F . Accordingly, F is measure convex. Theorem 2.75. Let A be a universally measurable subset of X. Then A is measure convex if and only if co K ⊂ A for any compact set K ⊂ A. Proof. Let A be a measure convex subset of X and K ⊂ A a compact set. If x ∈ co K, then there exists a measure µ ∈ M1 (K) such that x = r(µ) (see Proposition 2.39). By the assumption, x ∈ A. For the proof of the converse implication, let µ ∈ M1 (X) be a probability measure with µ(A) = 1. If µ(K) = 1 for some compact set K ⊂ A, then by Proposition 2.39 and by the assumption, r(µ) ∈ co K ⊂ A. So assume that µ(K) < 1 for each compact set K ⊂ A. In this case, there exists an increasing sequence {Kn }∞ n=0 of compact sets in A satisfying K0 = ∅,

αn := µ(Kn+1 \ Kn ) > 0, n ≥ 0,

and

µ(Kn ) → 1.

P ∞ Since ∞ n=1 αn = 1 − α0 < 1, there is a sequence {βn }n=1 of real numbers in (0, 1] such that ∞ X αn = 1. βn → 0 and βn n=1

Put L := K1 ∪

∞ [

(βn Kn+1 + (1 − βn )K1 ) .

n=1

Then L ⊂ A. Since each Kn is compact and βn → 0, L is compact as well. Pick f ∈ E ∗ and set sn := sup f (Kn+1 ), n ≥ 0. Then s := sup f (L) = max s0 , sup βn sn + (1 − βn )s0 n∈N

= s0 + sup (sn − s0 )βn . n∈N

32


Hence sn ≤ s0 + βn−1 (s − s0 ) for each n ∈ N. Since f ≤ sn on Kn+1 \ Kn , we get Z f dµ =

f (r(µ)) = X

≤

∞ X

∞ Z X

f dµ

Kn+1 \Kn

n=0

αn s n ≤ α0 s 0 +

n=0

∞ X

αn (s0 + βn−1 (s − s0 ))

n=1

= α0 s0 + (s − s0 )

∞ X n=1

∞

X αn + s0 αn βn n=1

= s0 + s − s0 = s. Thus f (r(µ)) ≤ sup f (L). As f is arbitrary, Proposition 2.44 gives r(µ) ∈ co L. Since our assumption ensures that co L ⊂ A, the proof is finished. Proposition 2.76. Any open convex subset of X is measure convex. Proof. Let G ⊂ X be an open convex set. By Theorem 2.75, it is enough to show that co K ⊂ G whenever K ⊂ G is a compact set. So fix such K. For every x ∈ K there exists a compact convex neighborhood Vx such that x ∈ Vx ⊂ G. By compactness, the set K can be covered by finitely many compact convex sets Vx1 , . . . , Vxn . Then, by Proposition 2.42, co K ⊂ co (Vx1 ∪ · · · ∪ Vxn ) = co (Vx1 ∪ · · · ∪ Vxn ) ⊂ G, which is the required inclusion. Proposition 2.77. Let X be a subset of a finite-dimensional space. Then any universally measurable convex set A ⊂ X is measure convex. Proof. We again use Theorem 2.75. If K ⊂ A is a compact set, then co K ⊂ A is compact by Theorem 2.8 (see also Remark 2.5). Lemma 2.78. Let λ be a probability measure on X. If T := {µ ∈ M+ (X) : µ ≤ λ, µ 6= 0} and S := {r( then the closure of S equals co spt λ.

µ ) : µ ∈ T }, kµk


33

Proof. It is easy to see that S = {r(µ) : µ ∈ M1 (X), there exists c ∈ R so that µ ≤ cλ}, from which it follows that S is convex. Set L := co spt λ. To show that S ⊂ L, let µ be a nontrivial measure on X with µ ) ∈ L because L is a closed convex set. µ ≤ λ. Then µ is carried by L and thus r( kµk Thus S ⊂ L and consequently S ⊂ L. Conversely, assuming that λ(S) < 1, we can find a compact set K ⊂ X \ S such that λ(K) > 0. For every x ∈ K we choose its closed convex neighborhood Vx not intersecting S. Using a compactness argument we select finitely many points x1 , . . . , xn of K so that Vx1 ∪· · ·∪Vxn covers K. As λ(K) > 0, there is i ∈ {1, . . . , n} so that λ(Vxi ) > 0. We set V := Vxi and µ := λ|V . Then µ is nontrivial and µ ≤ λ. µ µ belongs to S. On the other hand, r( kµk ) ∈ V because Hence the barycenter of kµk V is a closed convex set. This contradiction shows that λ(S) = 1. Thus spt λ ⊂ S which gives L ⊂ S. We recall that resolvable sets are defined and their basic properties presented in Section A.5. (In particular we note that any resolvable set in a compact space is universally measurable by Proposition A.118.) Lemma 2.79. Let F ⊂ X be a resolvable convex set and let λ ∈ M1 (X) be carried by F . Then there exists a nonempty set G ⊂ F ∩ co spt λ which is open in co spt λ. Proof. Let L := co spt λ. In order to find the required set G we note that L = F ∩ L because the latter set is a closed convex set containing the support of λ. In particular, F ∩ L is a dense resolvable set in L. Due to Proposition A.117(c), F ∩ L has a nonempty interior (relative to L). Hence, the interior of F ∩ L is the sought set G. Proposition 2.80. Any resolvable convex subset of X is measure convex. Proof. Let F be a resolvable convex subset of X and let λ be a probability measure on X carried by F . We set λ0 := λ and let L0 := co spt λ0 . Let S0 , T 0 and G0 be sets obtained from Lemma 2.78 and Lemma 2.79 when applied to the measure λ0 . Since S0 is dense in L0 and G0 is nonempty and open in L0 , there is a measure µ0 ∈ T 0 with µ0 ) ∈ G0 ⊂ F. r( kµ0 k We set λ1 := λ0 − µ0 and construct by transfinite induction a sequence {λα } of positive measures on X such that, for every ordinal number α ≥ 1, (i) λα+1 ≤ λα , (ii) either λα = 0 or kλα+1 k < kλα k,

34


(iii) if λα − λα+1 6= 0, then r(

λα − λα+1 ) ∈ F. kλα − λα+1 k

Suppose that the construction has been completed up to an ordinal α. If λα = 0, we set λα+1 := 0. If λα is nontrivial, we apply Lemma 2.78 and Lemma 2.79 to the measure kλλαα k (which is carried by F ) and get relevant sets Lα , T α , Sα and Gα with the properties described there. In particular, we have Gα ⊂ F ∩ L0 . As in the first step of the proof we choose a nontrivial measure ν ∈ T α such that ν ∈ Gα . r kνk By setting λα+1 := λα − ν we finish the inductive step for an isolated ordinal number. Let α be a limit ordinal number. Assume that λβ has been defined for every β < α. Since {λβ }β 0 for each j = 1, 2, . . . , m. Exercise 2.99. Let X 6= ∅ be a compact convex subset of Rd . Prove directly (do not use the Minkowski theorem 2.11) that ext X 6= ∅. Hint. Find x, y ∈ X such that |x − y| = diam X (here |x − y| denotes the Euclidean distance between points x and y). It easily follows that x, y ∈ ext X. Indeed, assume that 1 x = (x1 + x2 ) where x1 , x2 ∈ X, x1 6= x2 . 2 Since the vectors x1 − y and x2 − y are linearly independent, we get 1 1 diam X = |x − y| = (x1 − y) + (x2 − y) 2 2 1 < |x1 − y| + |x2 − y| ≤ diam X. 2

Another hint. Prove that any point z ∈ X having the property that |z| ≥ |x| for any x ∈ X is an extreme point of X. Exercise 2.100 (Radon). Assume that a set M ⊂ Rd contains at least d + 2 points. Then M = M1 ∪ M2 where M1 ∩ M2 = ∅ and co M1 ∩ co M2 6= ∅.

2.4 Exercises

41

Hint. Suppose that x1 , . . . , xn ∈ M , n ≥ d+2. Then there exists a nontrivial solution (α1 , . . . , αn ) to the system of d + 1 equations n X

αj = 0

n X

and

j=1

αj xj = 0.

j=1

Set I + := {j : αj ≥ 0}

and

I − := {j : αj < 0}

M1 := xj : j ∈ I +

and

M2 := xj : j ∈ I − .

and Then M1 ∩ M2 = ∅. Since λ :=

X

αj > 0,

j∈I +

we get co M1 3

X αj X αj xj = − xj ∈ co M2 . λ λ + −

j∈I

j∈I

Exercise 2.101 (Helly). Assume that K is a family of at least d + 1 convex sets in Rd such that either K is finite or the sets of K are in addition closed and one of them is compact. If each d + 1 sets of K have nonempty intersection, then \ {K : K ∈ K} 6= ∅. Hint. For a finite family K = {K1 , . . . , Kn }, suppose first that n = d + 2. By our assumption, there exist xi ∈

\

Kj ,

1 ≤ i ≤ n.

j∈{1,...,n}\{i}

T If there exist indices i 6= k with xi = xk , then xi ∈ nj=1 Ki . Otherwise we use Exercise 2.100 to find disjoint sets M1 , M2 such that M1 ∪ M2 = {x1 , . . . , xn } and co M1 ∩ co M2 contains a point y. Let i ∈ {1, . . . , n} be arbitrary. If xi ∈ M1 , then Ki ⊃ M2 . Hence y ∈Tco M2 ⊂ Ki . Analogously we get that y ∈ Ki in the case when xi ∈ M2 . Hence y ∈ ni=1 Ki . Assume now that the assertion has been proved for each family K in Rd consisting of n − 1 sets, where n ≥ d + 2. Let K = {K1 , . . . , Kn }. By the first part, each family K0 ⊂ K of at most d + 2 elements has nonempty intersection. Thus the family {K1 , . . . , Kn−2 , Kn−1 ∩ Kn }

42


satisfies that any subfamily with d + 1 elements has nonempty intersection. Hence the inductive assumption yields that the latter, and, consequently, the former family has nonempty intersection. Assume now that the family K consisting of closed convex sets is infinite and a set Z ∈ K is compact. By the first part, every finite subfamily of K has nonempty intersection. Hence K, and consequently {Z ∩K : K ∈ K}, has the finite intersection property. By compactness, \ \ {Z ∩ K : K ∈ K} = {K : K ∈ K} is nonempty. Exercise 2.102. (a) Let G be an open subset of Rd . Prove that co G is an open set. (b) If F is a closed subset of Rd , the convex hull co F need not be closed. Hint. Consider, for example, the following sets {(x, y) ∈ R2 : x > 0, xy = 1 or xy = −1} or {(x, y) ∈ R2 : x = 0} ∪ (1, 0). Exercise 2.103. (a) Let C be a compact convex subset of R2 . Prove that the set ext C is closed. (b) The set ext C need not be closed if C is a compact convex subset of Rd for d ≥ 3. Hint. Let {xn } be a sequence of points in ext C converging to x. Assuming that x is not an extreme point of C, let x = 21 (a + b) for some points a, b ∈ C, a 6= b. By passing to a subsequence, we may assume that all points xn are contained in the same open halfplane determined by the line passing through a and b. Then the interior of the triangle co{x1 , a, b} contains xn for a suitable n ∈ N, a contradiction with xn ∈ ext C. For (b), consider the convex hull in R3 of the set (x, y, z) : (x − 1)2 + y 2 = 1, z = 0 ∪ {(0, 0, 1) ∪ (0, 0, −1)} .

Exercise 2.104. Let C be a closed convex subset of Rd containing a line. Prove that ext C = ∅. If C contains no line, then ext C 6= ∅. Hint. If C contains a line L and x ∈ C, then C contains also a line passing through x parallel to L. If C contains no line then use induction on the dimension d. Find a boundary point of C and follow the reasoning of the proof of the Minkowski theorem 2.11.

2.4 Exercises

43

Exercise 2.105 (Closed convex hulls). Let C be a subset of a locally convex space E. Prove that the closed convex hull co C defined as \ co C := {F : F is a convex closed subset of E, F ⊃ C} is the closure of the convex hull co C, that is, co C = co C. Exercise 2.106 (Exposed points in Rd ). (a) Construct a compact convex set K ⊂ R2 for which ext K \ exp K 6= ∅. (b) Construct an example of a nonempty closed convex subset C of Rd such that the set exp C is not closed and co exp C 6= C. Hint. For the proof of (a) consider K to be the convex hull of the union of two circles {(x, y) ∈ R2 : (x + 1)2 + y 2 = 1} ∪ {(x, y) ∈ R2 : (x − 1)2 + y 2 = 1} then (−1, 1) ∈ ext K \ exp K. To show (b) consider again the example from (a). Exercise 2.107. (a) Let S be a d-simplex in Rd determined by affinely independent points e0 , . . . , ed . Prove that S has nonempty interior. (b) Using (a) prove that there are no d + 1 affinely independent points in any convex subset of Rd with empty interior. (c) Let C be a convex subset of Rd with empty interior. Prove that there is an affine subspace A of Rd containing C such that dim A < d. Hint. For (a) show that

e0 + · · · + ed ∈ Int S. d+1

To verify (c), let e0 , . . . , en be affinely independent points in C where n < d + 1 is the maximum number of affinely independent points in C. Consider now the affine hull of e0 , . . . , en . Exercise 2.108. Let X be a nonempty compact convex subset of a locally convex space E and x ∈ X. Prove that Mx (X) is a nonempty convex compact subset of M1 (X). Hint. A straightforward verification. Exercise 2.109. Let F be a closed subset of a compact convex set X such that ext X ⊂ F . Assume that for any x ∈ X there exists a unique measure µ ∈ M1 (F ) such that r(µ) = x. Prove that ext X = F .

44


Hint. Let x ∈ F \ ext X. Then x = y+z 2 , y 6= z, for some y, z ∈ X. The Integral representation theorem 2.31 yields measures µy , µz ∈ M1 (ext X) such that r(µy ) = y and r(µz ) = z. If µ = 12 (µy + µz ), then µ ∈ M1 (ext X),

r(µ) = x

and

µ 6= εx .

This contradicts the assumption, hence ext X = F . Exercise 2.110 (Proof of the Milman theorem 2.43). Verify the following indication of an alternative proof of Theorem 2.43. Hint. Let x ∈ ext X. We note that co B = co B. By Proposition 2.39 applied to K = B, there exists a measure µ ∈ M1 (B) representing the point x. Since x ∈ X, Bauer’s characterization in 2.40 asserts that µ = εx . The measure µ is carried by the (closed) set B, which yields that x ∈ B. Exercise 2.111. Find an example of a compact convex set X in a locally convex space E such that (E ∗ + R)|X 6= Ac (X). P n Hint. Let E := `2 , X := {x ∈ E : 0 ≤ xn ≤ 4−n , n ∈ N} and f (x) := ∞ n=1 2 xn , x ∈ X. Exercise 2.112. Prove that in any infinite-dimensional Banach space E, there exists a compact convex set X such that (E ∗ + R)|X 6= Ac (X). Hint. Find inductively points xn ∈ SE and functionals ϕn ∈ E ∗ , n ∈ N, such that ( 1, n = m, ϕn (xm ) = n, m ∈ N. 0, n 6= m, P −n on X, n Set X := co{4−n xn : n ∈ N} and f := ∞ n=1 2 ϕn . Since 0 ≤ ϕn ≤ 4 n ∈ N, f is well defined. If f = ϕ + c on X for some ϕ ∈ E ∗ and c ∈ R, then c = 0, because f (0) = 0. Further, since f (4−n xn ) = 2−n , we get ϕ(xn ) = 2n for each n ∈ N. But this is impossible as ϕ is bounded on BE . Exercise 2.113. Find a compact convex set X ⊂ R2 such that far X 6= exp X. Hint. Let X = {(x, y) ∈ R2 : |x|3 ≤ y ≤ 1, −1 ≤ x ≤ 1}. Then the point (0, 0) is obviously exposed. However, (0, 0) ∈ / far X as an easy geometrical argument shows. Exercise 2.114. Find an example of a compact convex subset X of a locally convex space E and a point z ∈ X that is exposed but there is no f ∈ E ∗ exposing z.

2.4 Exercises

45

First hint. Let H be a Hilbert space, M be a dense proper subspace of H ∗ and let Φ : H ∗ → H be the mapping assigning to each ϕ ∈ H ∗ a point y ∈ H such that ϕ(h) = (h, y) for any h ∈ H. Let X := BH be equipped with the σ(H, M )-topology. Since BH is w-compact and σ(H, M ) is weaker and Hausdorff, w = σ(H, M ) on X. Choose x ∈ SH \ Φ(M ) and denote ψ := Φ−1 (x). Show that ψ is not σ(H, M )continuous. Further show that ψ|X ∈ Ac ((X, σ(H, M )) and that ψ exposes x. On the other hand, no functional from (H, σ(H, M ))∗ = M exposes x. (This easily follows from the fact that, given x, y ∈ SH , (x, y) = 1 if and only if y = x.) Second hint. Let {en } be a sequence of standard unit vector in c0 . For 1 ≤ i ≤ j denote ui,j := −ei + 2ej . Let further X := cok·k C

where

C := 2−i−j ui,j : 1 ≤ i ≤ j .

Define the function h on X as h : x 7→

∞ X

xn ,

x = {xn } ∈ X.

n=1

It is easy to show that h is an affine continuous function on X. Obviously, C ∪ {0} is a compact subset of c0 and C ⊂ {{xn } ∈ c0 : |xn | ≤ 2−n }. It easily follows that X is compact. Let x ∈P X \ {0}. Since ext X ⊂ C ∪ {0}, by the Krein–Milman theorem 2.22 we have x = i≤j αi,j 2−i−j ui,j , where αi,j ≥ 0 and their sum is smaller or equal to 1. There exists a pair i ≤ j such that αi,j > 0. Hence h(x) =

X

αi,j 2−i−j h(ui,j ) =

i≤j

X

αi,j 2−i−j > 0.

i≤j

It follows that 0 is an exposed point of X. On the other hand, 0 is not exposed by any functional from (c0 )∗ = `1 . Indeed, assume that 0 6= f = {fn } ∈ (c0 )∗ = `1 . There exists i such that fi 6= 0. If fi < 0, then f (ui,i ) = f (ei ) = fi < 0. If fi > 0, then f (ui,j ) = f (−ei + 2ej ) = −fi + 2fj . Since limj→∞ fj = 0, we have limj→∞ f (ui,j ) = −fi . Hence, there exists j ≥ i such that f (ui,j ) < 0. If x := 2−i−j ui,j , then x ∈ X, x 6= 0, and f (x) < 0 = f (0). Exercise 2.115. Keeping the notation of the first hint in Exercise 2.114, consider locally convex spaces E1 := (H, w) and E2 := (H, σ(H, M )). Prove that the compact convex sets (BH , σ(H, M )) and (BH , w) are affinely homeomorphic (by the identity mapping), and that the point x is exposed by a functional from E1∗ whereas it is not exposed by a functional from E2∗ . (Compare with the proof of Theorem 2.51.) Hint. Follow the reasoning of Exercise 2.114.

46


Exercise 2.116. Let X be a nonempty compact convex set and K ⊂ X \ ext X be compact. Then co K ∩ ext X = ∅. Hint. For any x ∈ co K \ K there exists µ ∈ Mx (K) (see Proposition 2.39). Obviously, µ 6= εx , and thus x ∈ / ext X by Theorem 2.40. Exercise 2.117. Let F be a closed face of a compact convex set X and let U ⊂ X be an open set containing F . Then F ∩ co(X \ U ) = ∅. Hint. Proceed as in Exercise 2.116. Exercise 2.118. Let ν be a Radon probability measure on a compact space K. Let L := µ ∈ M1 (K) : µ ⊥ ν . Prove that L is a Gδ set. Hint. For each n ∈ N and each open subset G of K, let L(n, G) := µ ∈ M1 (K) : µ(G) > 1 − 2−n and Ln :=

[

L(n, G) : G ⊂ K open, ν(G) < 2−n .

Since each set L(n, G) is open in M1 (K) by Theorem A.85(b), the set Gδ set. The proof will be complete once we show that L=

∞ \

T∞

n=1 Ln

is a

Ln .

n=1

Pick an arbitrary index n ∈ N and µ ∈ L. As µ ⊥ ν, there is a Borel set A ⊂ K such that ν(A) = 0 and µ(A) = 1. Due to the regularity of ν it follows that there exists an open set G ⊃ A such that ν(G) < 2−n . Since µ(G) ≥ µ(A) = 1 > 1 − 2−n , we get µ ∈ Ln . T Conversely, assume that µ ∈ ∞ n=1 Ln . There is a sequence {Gn } of open subsets of K such that ν(Gn ) < 2−n and µ(Gn ) > 1 − 2−n T S∞ for each n ∈ N. Set G := ∞ m=1 n=m Gn . Then ν(G) = 0. Since 1 ≥ µ(G) = lim µ( m→∞

∞ [

n=m

Gn ) ≥ lim sup µ(Gm ) ≥ lim sup(1 − 2−m ) = 1, m→∞

m→∞

it follows that µ ⊥ ν. Hence, µ ∈ L, as required. Exercise 2.119. If F is a face of X and G is a face of F , then G is a face of X.

2.4 Exercises

47

Hint. A straightforward verification. Exercise 2.120. Prove that a set A ⊂ X is convex if X \ A is extremal. Hint. A straightforward verification. Exercise 2.121. Find an Fσ face that is not a countable union of closed faces. Hint. Use the set F from Proposition 2.94. If F were a countable union of closed faces, F would be measure extremal by Proposition 2.86. But this is not the case. Exercise 2.122. If F is a closed face of X, then there exists a set A ⊂ ext X such that F = co A. In particular, if F is a nonempty closed face of X, then F ∩ ext X 6= ∅ (cf. Theorem 2.20). Hint. Set A := F ∩ ext X. Then A = ext F by Proposition 2.64(b) and it suffices to apply the Krein–Milman theorem 2.22. Exercise 2.123. Let X be a convex subset of a locally convex space E. Prove that the boundary ∂X is an extremal subset of X. Hint. If Int X = ∅, the assertion is obvious. Hence, assume that Int X 6= ∅ and that ∂X is not extremal. In this case, there exist z ∈ ∂X, x ∈ X and y ∈ Int X so that z = λx + (1 − λ)y for some λ ∈ (0, 1). Find a neighborhood V of 0 such that y + V ⊂ Int X. For every ε > 0, there exists zε such that zε ∈ (z + εV ) \ X. Let wε :=

zε − λx . 1−λ

Show that we can choose ε small enough such that wε ∈ y + V . Since zε = λx + (1 − λ)wε , this implies that zε ∈ X, which is a contradiction. Exercise 2.124. Let X be a compact convex set and x ∈ X. Prove that \ D := {A ⊂ X : A is a closed extremal set containing x} is convex. Hint. The set D is obviously closed and extremal. Then use characterization (vi) of Proposition 2.69. Exercise 2.125. Find a continuous affine surjection of a compact convex set X onto a compact convex set Y such that ext Y 6= ϕ(ext X). Hint. Consider a triangle given as the convex hull of points (0, 0), (1, 0), ( 12 , 1) and its projection onto the unit segment [0, 1].

48


Exercise 2.126. Prove that the set G := µ ∈ M1 ([0, 1]) : µ = µc of all continuous Radon measures on [0, 1] is measure convex (cf. also Proposition 2.93). Hint. Let Ω ∈ M1 M1 ([0, 1]) , Ω(G) = 1 with r(Ω) = ω be given. For a bounded Borel function f on [0, 1], define an affine function If on M1 ([0, 1]) by the formula If (µ) := µ(f ),

µ ∈ M1 ([0, 1]).

Obviously, If is continuous whenever f is. Thus Ω(If ) = If (r(Ω)) = ω(f ) for any f ∈ C[0, 1].

(2.4)

Now, use the Lebesgue dominated convergence theorem to show that equality (2.4) holds for any bounded Borel function f on [0, 1]. Pick x ∈ [0, 1]. Then Z Ic{x} (µ) dΩ(µ) = 0. ω({x}) = Ic{x} (ω) = G

Thus ω({x}) = 0 for any x ∈ [0, 1], and therefore ω ∈ G. Exercise 2.127. Let K be a compact space, X := M1 (K), ω ∈ M1 (K), n µ o A1 := : µ ∈ M+ (K), µ ≤ ω and µ(K) > 0 µ(K) and A2 :=

n fω o : 0 ≤ f ≤ 1 Borel, ω(f ) > 0 . ω(f )

Prove that (a) face ω = A1 = A2 , (b) face ω = M1 ([0, 1]) if K = [0, 1] and ω is Lebesgue measure on [0, 1]. Hint. (a) Let µ ∈ face ω be given. Then there exist α ∈ (0, 1] and ν ∈ M1 (K) such that ω = αµ + (1 − α)ν. Then 0 < αµ ≤ ω. If µ ≤ ω, by the Radon–Nikodym theorem there exists a Borel measurable function f such that µ = f ω. Then 0 ≤ f ≤ 1 ω-almost everywhere and ω(f ) = µ(K) > 0. Hence µ fω = µ(K) ω(f ) and µ ∈ A2 .


49

fω Finally, let µ = ω(f ) for some Borel function f with 0 ≤ f ≤ 1 and ω(f ) > 0. We may assume that ω(f ) < 1. Then

ω = ω(f )

(1 − f )ω fω + ω(1 − f ) , ω(f ) ω(1 − f )

and µ ∈ face ω. (b) Assume that λ is Lebesgue measure on [0, 1]. Given a point x ∈ [0, 1], it is easy to find a sequence of measures in face λ that converges to εx . Hence face λ is a convex set containing all extreme points of M1 ([0, 1]), and thus face λ = M1 ([0, 1]).

2.5

Notes and comments

The material of the Subsection 2.1.A on convexity in finite-dimensional spaces is standard and can be found in many textbooks; see, for example, Barvinok’s monograph [31]. Besides Minkowski’s result, at the beginning of 20th century, three important theorems on convex sets in Euclidean spaces associated with the names of Carathéodory, Helly and Radon appeared. H. Minkowski proved Theorem 2.11 in the period 1901-03. The result appeared for the first time in a chapter on convex bodies (where the origin of the notion of extreme points can be traced) included in his collected works published in 1911; see J. J. Saccoman [405]. An exhaustive survey on Helly’s theorem 2.101 and its proofs, variants and applications is presented in the paper by L. Dantzer, B. Grünbaum and V. Klee [130]. The Krein–Milman theorem 2.22 was proved by M. Krein and D. Milman in [283] for the case of w∗ -compact convex sets in the dual to a Banach space using the transfinite induction. The proof of a more general version contained in Theorem 2.22, based on an application of Zorn’s lemma, goes back to J. L. Kelley [266]. A similar proof was given by E. Artin (a letter from Artin to his former student M. Zorn published in the Picayune Sentinel of Indiana University in 1950; cf. [23]), A. Hotta [240] and by K. Yosida and M. Fukamiya [477]. The Krein–Milman theorem is one of the fundamental theorems of functional analysis and has rich applications. For instance, recall its use in de Branges’ proof of the Stone–Weierstrass theorem, Lindenstrauss’ proof of the Lyapunov theorem on the range of a vector measure, and in the proof of the Banach–Stone theorem on isometrically isomorphic spaces of continuous functions. See Chapter 14 for more applications. Bauer’s concave minimum principle 2.24 (even in a more general form) appeared in H. Bauer [36]. The Integral representation theorem 2.31 is a reformulation of the Krein–Milman theorem. It has wide applications in several areas of analysis. We gave two different proofs of this theorem (besides Theorem 2.31 it is its more general form in Proposition 2.39). Still another proof of Theorem 2.29, avoiding the notion of a net, is presented in Phelp’s book [374], Proposition 1.1.

50


Bauer’s characterization of ext X in Theorem 2.40 presented in [37] is used later on as the definition of the Choquet boundary of function spaces in 3.4. Proposition 2.41 can be found as Proposition 25.13 in G. Choquet [108] and it is used several times in our text. For example, Milman’s converse 2.43 of the Krein–Milman theorem, which goes to V. P. Milman [346], is an easy consequence of it. Proposition 2.45 enables to solve certain problems concerning metrizable compact convex sets reducing them to subsets of Hilbert spaces. It was shown by O. -H. Keller [265] that any infinite-dimensional compact convex subset of a Hilbert space is N homeomorphic to the Hilbert cube [0, 1] (which is affinely homeomorphic to the set 1 2 {xn } ∈ ` : |xn | ≤ n , n ∈ N ). Moreover, V. L. Klee observed that any compact (convex) subset of a Banach space is affinely homeomorphic to a subset of a Hilbert space. An infinite-dimensional compact convex subset C of a topological vector space is said to be a Keller set if C is affinely homeomorphic to a compact convex subset of a Hilbert space `2 . Hence, Proposition 2.45 shows that any infinite-dimensional metrizable compact convex set is a Keller set. Moreover, it has been shown by C. Bessaga and T. Dobrowolski in [56] that any locally compact convex subset C of a topological vector space with a countable family of continuous affine functions on C separating points of C can be affinely embedded into `2 . The concept of exposed points in the case of Euclidean spaces was introduced by S. Straszewicz [441] in 1935. In this case, Corollary 2.52 and example in Exercise 2.103(b) are due to him. The proof of Proposition 2.50 is taken from V. P. Fonf, J. Lindenstrauss and R. R. Phelps [179]. In the paper [53] by S. K. Berberian, the Krein–Milman theorem in Hilbert spaces is derived. This result can be deduced from more general statements concerning the exposed points in normed linear spaces; see, for example, the paper of V. L. Klee [272]. In fact, from the proof of Klee’s result it follows that in any smooth and strictly convex normed linear space any compact convex set is the closed convex hull of its set of so-called bare points (cf. a review of the paper [53] in Mathematical Reviews). See also the paper by M. V. Balashov [30]. We also refer the reader to the paper [151] by M. Edelstein and J. E. Lewis on exposed and farthest points. In [18], R. F. Arens and J. L. Kelley described extreme points of the unit ball of the ∗ space C(K) as Dirac measures εx and their antipodes −εx (cf. Proposition 2.27). For an alternative proof of Corollary 2.28 see, for example, Theorem 30.4 and Corollary 30.5 in Bauer’s monograph [45]. In Proposition 2.56 we present a simplified version of the proof of [5], Proposition I.2.8. Examples described in Proposition 2.63 are due to G. Choquet [106]; see also Alfsen’s monograph [5], Example I.2.10. Any union of faces was labelled in Goullet de Rugy in [200] as a σ-face. Closed extremal sets were studied by E. M. Alfsen in [1] under the name “stable sets”. The equivalence of (i) and (v) in Proposition 2.69 was proved by E. M. Alfsen in [1]; the idea of the proof (i) =⇒ (vi) is from D. P. Milman [346] (§ 4, Theorem 7).


51

In part, the material Subsections 2.3.B and 2.3.C concerning a more detailed study of measure convex and measure extremal sets is taken from the paper [146] by P. Dostál, J. Lukeˇs and J. Spurný. Theorem 2.75 and its proof are taken from D. H. Fremlin and J. D. Pryce [185]. Alfsen’s example 2.81 is taken from [5], p. 130. The proof of Proposition 2.92 can also be proved by means of a result of J. Saint Raymond 10.75 without recourse to Proposition 2.80. Counterexamples contained in Propositions 2.93 and 2.94 are just suitable modifications of the example by G. Choquet in [106]. The examples of Propositions 2.95 and 2.96 partially use a construction of H. v. Weizsäcker (see [463]). We also refer the reader to Lecture Notes [473] by G. Winkler where a thorough investigation of measure convex sets is given. The result of Exercise 2.103(a) is due to G. B. Price [377]. A characterization of continuous affine functions on X (cf. Lemma 2.34 and Exercise 2.112) belonging to E ∗ + R|X even for noncompact convex sets X is given in a paper [281] by M. Kraus. Examples of Exercises 2.114 and 2.115 are due to M. Kraus and O. Kurka.

Chapter 3

Choquet theory of function spaces

This chapter lays the groundwork for the rest of the book by presenting the foundations of the Choquet theory of function spaces. The central concept of a function space is defined and its basic properties investigated in Section 3.1. We generalize the framework of spaces of affine continuous functions on compact convex sets by taking a subspace H of the space C(K) of continuous functions on a compact space K such that H contains constants and separates points of K. Then we introduce H-representing measures, H-affine and H-convex functions, and so on. A suitable substitute for the set of extreme points turns out to be the Choquet boundary and we show in Proposition 3.15 its nonemptiness and prove a minimum principle in Theorem 3.16. A crucial notion for obtaining integral representation theorems is the Choquet ordering introduced in Definition 3.19. This ordering somehow indicates how close to the Choquet boundary a measure is situated. A key tool for handling function spaces is Lemma 3.21 which serves as a substitute for the Hahn–Banach theorem. Several of its applications are shown afterwards, along with Bauer’s characterization 3.24 of the Choquet boundary. These abstract results are then applied to a reexamination of Korovkin’s theorems; Theorems 3.32, 3.34 and 3.36 are proved by means of the Choquet theory. After generalizing the concept of the barycenter mapping in Section 3.3, we turn our attention to the Choquet representation theorem 3.45 for function spaces on metrizable compact spaces. Our approach is to use the existence of a “strictly convex” function. Next, Section 3.5 indicates how the Key lemma 3.21 enables us to prove analogues of classical results on approximation of semicontinuous convex functions on compact convex sets. More precisely, we show that semicontinuous H-convex functions can be approximated by continuous H-convex functions (see Propositions 3.48 and 3.54). An important corollary is the fact that the Choquet ordering of measures can be extended to semicontinuous H-convex functions (see Proposition 3.56). Measures maximal with respect to the Choquet ordering are investigated in Section 3.6. First we prove Mokobodzki’s characterization in Theorem 3.58. Then we show that the set of maximal measures is rich enough (see Theorem 3.65) and we finish the section with Theorem 3.70, which describes the space of boundary measures. In order to prove the most important properties of maximal measures, namely, that they are carried by any Baire set containing the Choquet boundary, we need some kind of Fatou’s lemma (see Lemma 3.77). This task is accomplished in Section 3.7 by presenting the important Simons inequality 3.74 and a couple of its applications.

3.1 Function spaces

53

Then we can prove the integral representation theorem for nonmetrizable spaces in Theorem 3.81. The existence of representing measures “carried” by the Choquet boundary opens the way for the proof of several variants of the minimum principle, as shown in Section 3.9. The last section is devoted to a characterization of the fact that a pair of measures µ, ν is related in the Choquet ordering. Theorem 3.92 shows that the important notion of a dilation plays the key role here.

3.1

Function spaces

Definition 3.1 (Function spaces). By a function space H on a compact topological space K we mean a (not necessarily closed) linear subspace of C(K) containing the constant functions and separating points of K. We introduce some main examples of function spaces. Examples 3.2. (a) Continuous functions. The space C(K) of all continuous functions on a compact space K represents a simple example of a function space. Clearly, the space C(K) separates points of K. (b) Quadratic polynomials. The space P2 ([0, 1]) of all quadratic polynomials on the interval [0, 1] is a further example of a function space. We considered this example in Chapter 1. (c) Convex case – affine functions. Let X be a compact convex subset of a locally convex space E. The linear space Ac (X) consisting of all continuous affine functions on X is a function space. (d) Harmonic case – harmonic functions. Let U be a bounded open subset of the Euclidean space Rd . The function space H(U ) consisting of all continuous functions on U which are harmonic on U is another example of a function space. More generally, we can consider a relatively compact open subset U of a Bauer harmonic space (cf. Section A.8) and the function space H(U ), the linear subspace of C(U ) of functions which are harmonic on U . We tacitly assume that constant functions are harmonic and H(U ) separates points of U . (e) If K is a compact subset of Rd , we define [ H0 (K) := {H(U )|K : U is relatively compact open set, K ⊂ U }. Generally, the function space H0 (K) is not a closed subspace of C(K) (see Exercise 13.155). (f) Further examples can be found in 3.47, 3.82, 3.83, 3.103, 3.106, 3.111, 3.119, 6.67, 6.76, 6.77, 6.94, 7.65, 8.11(a), 8.29, 8.80, 9.11, 9.56, 10.97, 10.98, 10.99 and 12.3.B.

54

3 Choquet theory of function spaces

(g) Let K be a compact space and T be a Markov operator on K (cf. Subsection 6.6.B). Let HT := {h ∈ C(K) : T (h) = h} . If HT separates points of K, then HT is a function space. Convention.

In what follows, H denotes a function space on a compact space K.

Definition 3.3 (H-representing measures). Recall that M1 (K) denotes the set of all probability Radon measures on K. We denote by Mx (H) the set of all H-representing measures for x ∈ K, that is, Z 1 f dµ for any f ∈ H}. Mx (H) := {µ ∈ M (K) : f (x) = K

Of course, the Dirac measure εx at the point x always belongs to Mx (H). Definition 3.4 (Choquet boundary). Define the Choquet boundary ChH (K) of a function space H as the set of those points x ∈ K for which the Dirac measure εx is the only H-representing measure for x; that is, ChH (K) := {x ∈ K : Mx (H) = {εx }} . Examples 3.5. We describe the Choquet boundary of our main examples from 3.2. We postpone the proofs to later sections. (a) Continuous functions. In the case when H = C(K) (K is a compact space), the equality ChH (K) = K follows immediately from the definition. (b) Quadratic polynomials. Let H = P2 ([0, 1]) be the function space of all quadratic polynomials on the interval [0, 1]. Then ChH ([0, 1]) = [0, 1], as easily follows by Proposition 3.7. (c) Convex case – affine functions. If H is the linear space Ac (X) of all continuous affine functions on a compact convex set X, then ChAc (X) (X) = ext X, by Bauer’s characterization of ext X in Theorem 2.40. (d) Harmonic case – harmonic functions. If H(U ) consists of all continuous functions on U ⊂ Rd which are harmonic on U , then ChH(U ) U = ∂ reg U (cf. Theorem 13.35). In the general situation of an abstract Bauer harmonic space, the situation is more delicate, cf. Theorem 13.41. (e) The Choquet boundary of H0 (K) (see Example 3.2(e)) consists of stable points of K (see Definition 13.4 and Theorem 13.35).

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(f) Let H := {f ∈ C([−1, 1]) : 2f (0) = f (−1) + f (1)}. Then ChH ([−1, 1]) = [−1, 1] \ {0}. Further examples of Choquet boundaries can be found in the examples of function spaces mentioned in Examples 3.2(f). Definition 3.6 (H-exposing functions and H-exposed points). Let x ∈ K. A function h ∈ H such that 0 = h(x) < h(t) for any t ∈ K, t 6= x, is said to be an Hexposing function for x. A point x ∈ K is called an H-exposed point if there exists an H-exposing function for x. The set of all H-exposed points of K will be denoted by expH (K). Proposition 3.7. Any H-exposed point belongs to the Choquet boundary of H. Proof. Let x ∈ K and let h ∈ H for which 0 = h(x) < h(t) for any t ∈ K \ {x}. If µ ∈ Mx (H), then 0 = h(x) = µ(h). Hence spt µ ⊂ {x}, and therefore µ = εx . We see that x ∈ ChH (K). Definition 3.8 (H-affine, H-convex and H-concave functions). We define the family A(H) of all H-affine functions as the family of all universally measurable functions f : K → [−∞, ∞] such that µ(f ) exists for every µ ∈ Mx (H), x ∈ K, and the following barycentric formula holds: Z f (x) = f dµ for each x ∈ K, µ ∈ Mx (H). K

Further, let Ac (H) be the family of all continuous H-affine functions on K. Similarly, we say that a universally measurable function f : K → [−∞, ∞] is H-convex if µ(f ) exists for every µ ∈ Mx (H), x ∈ K, and f (x) ≤ µ(f ). A function f is H-concave if −f is H-convex. We denote by Kc (H), Kusc (H), and Klsc (H) the family of all continuous, upper semicontinuous, and lower semicontinuous H-convex functions, respectively. We write S c (H), S usc (H) and S lsc (H) for the analogous families of H-concave functions. Remark 3.9. As we will see later in Chapter 4, in the convex case of Example 3.2(c), a continuous function is Ac (X)-concave if and only if it is concave in the usual sense. Definition 3.10 (Cone W(H)). We denote by W(H) the smallest min-stable cone generated by H, that is, W(H) consists of all functions of the form h1 ∧ · · · ∧ hn , h1 , . . . , hn ∈ H, n ∈ N. The following proposition collects several easy facts. Proposition 3.11. (a) Ac (H) is a closed function space on K containing H.

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(b) The families W(H), S c (H), S usc (H) and S lsc (H) form min-stable convex cones. (c) The space W(H) − W(H) is dense in C(K). Proof. Since (a) is obvious, we proceed to the proof of (b). Let F be any of the considered families. Obviously, F is a convex cone and the required topological property is stable with respect to taking finite minima (for F = W(H) we use identity f1 ∧ · · · ∧ fn + g1 ∧ · · · ∧ gk =

n ^ k ^

(fi + gj ).)

i=1 j=1

If k1 , k2 ∈ F and µ ∈ Mx (H), then µ(k1 ∧ k2 ) exists and µ(k1 ∧ k2 ) ≤ µ(k1 ) ∧ µ(k2 ) ≤ (k1 ∧ k2 )(x). For the proof of (c), we just use the lattice version of the Stone–Weierstrass theorem from Proposition A.31. Definition 3.12 (H-extremal sets). Let H be a function space on a compact space K. A universally measurable set F ⊂ K is called H-extremal if any measure representing a point in F is carried by F . If X is a compact convex set in a locally convex space and H = Ac (X), then a closed set F ⊂ X is Ac (X)-extremal if and only if F is extremal (see Proposition 2.69). Lemma 3.13. Let F be a closed H-extremal set and f ∈ S lsc (H). Then the set H := {x ∈ F : f (x) = min f (F )} is H-extremal. Proof. Obviously, H is closed. To show that H is H-extremal, pick x ∈ H and µ ∈ Mx (H). Since F is H-extremal, spt µ ⊂ F . Then from the inequalities Z min f (F ) = f (x) ≥

f (t) dµ(t) ≥ min f (F ), F

it follows that spt µ ⊂ H. Proposition 3.14. The family F of all closed H-extremal sets is stable under finite unions and arbitrary intersection.

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Proof. Obviously, F is stable under finite unions. T If {Fa }a∈A are closed H-extremal sets, then their intersection F := a∈A Fa is a closed set as well. Pick x ∈ F and µ ∈ Mx (H) and choose a compact set H ⊂K \F =

[

(K \ Fa )

a∈A

arbitrarily. There are sets Fa1 , . . . , Fan so that H ⊂ (K \ Fa1 ) ∪ · · · ∪ (K \ Fan ). Since µ(K \ Faj ) = 0 for each j = 1, . . . , n, we get µ(H) = 0. From the inner regularity of µ we obtain µ(K \ F ) = 0. The proof of Proposition 3.15 follows along the lines to that of the Krein–Milman theorem 2.22. Proposition 3.15. The Choquet boundary ChH (K) intersects any nonempty closed H-extremal set. In particular, the Choquet boundary ChH (K) is nonempty if K 6= ∅. Proof. Let F be a nonempty closed H-extremal set. We partially order the family S of all nonempty closed H-extremal subsets of F by the reverse inclusion. If Z is a T chain in S, then {C : C ∈ Z} is nonempty since it is the intersection of a downdirected family of nonempty compact sets. Moreover, it is H-extremal according to the previous Proposition 3.14. Zorn’s lemma now provides a maximal element H ∈ S. Assume that H contains two distinct points x and y. Since H is a function space, there exists a function h ∈ H such that h(x) 6= h(y). According to Lemma 3.13, the set {z ∈ H : h(z) = min h(H)} is a closed H-extremal set strictly contained in H, which contradicts the fact that H is maximal. Hence, H = {x} for some x ∈ F . Since one-point H-extremal sets are in ChH (K), it follows that x ∈ ChH (K). Theorem 3.16 (Minimum principle for S lsc (H)). Let f ∈ S lsc (H) be a lower semicontinuous H-concave function, f ≥ 0 on ChH (K). Then f ≥ 0 on K. Proof. Let K 6= ∅ and f ∈ S lsc (H) satisfy f ≥ 0 on ChH (K). Then F := {x ∈ K : f (x) = min f (K)} is a closed H-extremal set by Lemma 3.13. By Proposition 3.15, F ∩ ChH (K) 6= ∅, which implies that f ≥ 0 on K.

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Definition 3.17 (Envelopes and sublinear functionals). Let µ be a Radon measure on K and F be a subset of K containing ChH (K). Then for every extended real-valued function f on F , we define Qµ,F (f ) := inf{µ(k) : k ∈ S c (H), k ≥ f on F }. If x ∈ K and µ = εx , we define F ∗

f (x) := Qεx ,F (f ) and

f∗ (x) := −(F(−f )∗ (x)).

F

If F = K, we write for short Qµ (f ), f ∗ and f∗ . Lemma 3.18. Let µ ∈ M+ (K), F ⊃ ChH (K), and f be an upper bounded function on F . Then (a) Ff ∗ is an upper semicontinuous H-concave function, (b) Qµ,F (f ) = µ(Ff ∗ ), (c) if f is bounded, then −kf k ≤ Ff∗ ≤ Ff ∗ ≤ kf k, (d) if f is upper semicontinuous on F and  f (x), g := lim sup f (y), 

x ∈ F, x ∈ F \ F,

y→x,y∈F

then g is upper semicontinuous on F and Qµ,F (f ) = Qµ,F (g). (e) g 7→ Qµ,F (g) is a sublinear functional on `∞ (F ). Proof. The proof of (a) is obvious. To prove (b), let µ ∈ M+ (K) and an upper bounded function f on F be given. By Lemma 3.11(b), the family {k ∈ S c (H) : k ≥ f on F } is down-directed. Hence by Theorem A.84, Qµ,F (f ) = µ(Ff ∗ ). We proceed to (c). Let −k1 , k2 ∈ S c (H) be such that k1 ≤ f ≤ k2 on F . By Theorem 3.16, k1 ≤ k2 . Hence Ff∗ ≤ Ff ∗ on K. Since H contains constant functions, −kf k ≤ Ff∗ ≤ Ff ∗ ≤ kf k. Let g be defined as in (d). By a routine verification, g is upper semicontinuous on F . If k ∈ S c (H), then k ≥ f on F if and only if k ≥ g on F . This proves (d). Since the proof of (e) is straightforward, the proof is finished.

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Definition 3.19 (Choquet ordering). Let H be a function space on a compact space K. The convex cone Kc (H) of all continuous H-convex functions on K determines the Choquet ordering on M+ (K): µ ≺H ν

if

µ(f ) ≤ ν(f ) for each f ∈ Kc (H).

(Where no confusion can occur, µ ≺H ν will be abbreviated by µ ≺ ν.) Since, by Proposition A.31, the set Kc (H) − Kc (H) is uniformly dense in C(K), µ = ν whenever µ ≺ ν and ν ≺ µ. Proposition 3.20. Let µ ∈ M+ (K). (a) If ν satisfies µ ≺ ν, then µ − ν ∈ H⊥ and kµk = kνk. (b) If x is a point of K and εx ≺ µ, then µ ∈ Mx (H). Proof. Let h ∈ H. If µ ≺ ν, then µ(h) ≤ ν(h) and µ(−h) ≤ ν(−h). Hence µ(h) = ν(h) and kµk = µ(1) = ν(1) = kνk. This proves (a). Since (b) follows from (a), the proof is finished. Lemma 3.21 (Key lemma). Let µ ∈ M+ (K), F be a closed set in K containing ChH (K) and f be an upper semicontinuous function on K. Then there exists a measure ν ∈ M+ (F ) such that µ ≺ ν and Qµ,F (f ) = ν(f ). In particular, for any x ∈ K there exists a measure ν ∈ Mx (H) ∩ M+ (F ) such that Ff ∗ (x) = ν(f ). Proof. Given µ ∈ M+ (K), we assume first that f ∈ C(K). From Lemma 3.18(e) we know that the mapping p : g 7→ Qµ,F (g) is a sublinear functional on C(F ). We define a functional ϕ : span{f } → R as ϕ(tf ) = tQµ,F (f ),

t ∈ R.

Since 0 = Qµ,F (0) = Qµ,F (f − f ) ≤ Qµ,F (f ) + Qµ,F (−f ), we get ϕ(−f ) = −Qµ,F (f ) ≤ Qµ,F (−f ). Hence ϕ ≤ p on span{f }. The Hahn–Banach theorem provides a linear functional ν on C(F ) such that ν(f ) = Qµ,F (f ) and ν ≤ p on C(F ). Since ν(g) ≤ p(g) ≤ 0 whenever g ∈ C(F ) and g ≤ 0, we see that ν is, according to the Riesz representation theorem A.75, a Radon measure on F . Let k ∈ S c (H). Then ν(k) ≤ p(k) = Qµ,F (k) ≤ µ(k). Hence µ ≺ ν and the proof is finished in the case when f is continuous on F .

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Now let f be an upper semicontinuous function on F . Denote by G the downdirected set {g ∈ C(F ) : g ≥ f on F }. By the first part of the proof, for any g ∈ G there exists a measure νg ∈ M+ (F ) such that µ ≺ νg and νg (g) = Qµ,F (g). Given ψ ∈ G, let Mψ = {νg : g ∈ G, g ≤ ψ}. Since Mψ ⊂ {λ ∈ M+ (F ) : kλk = kµk} and the family {Mψ : ψ ∈ G} has the T finite intersection property, a compactness argument yields the existence of ν ∈ {M ψ : ψ ∈ G}. A moment’s reflection shows that µ ≺ ν. We observe that inf λ(ψ) : λ ∈ Mψ = inf λ(ψ) : λ ∈ M ψ ≤ ν(ψ) for each ψ ∈ G. Hence, using Theorem A.84 for verification of the equalities, Qµ,F (f ) ≤ inf Qµ,F (g) : g ∈ G = inf {νg (g) : g ∈ G} ≤ inf {inf {νg (ψ) : g ∈ G, g ≤ ψ} : ψ ∈ G} ≤ inf {ν(ψ) : ψ ∈ G} = ν(f ) ≤ inf {ν(k) : k ∈ S c (H), k ≥ f on F } ≤ inf {µ(k) : k ∈ S c (H), k ≥ f on F } = Qµ,F (f ), which yields the required equality. The second part of the assertion follows by taking µ = εx and using Proposition 3.20. Lemma 3.22. If f ∈ C(F ), F ⊃ ChH (K) is a closed set and x ∈ K, then F f∗ (x), Ff ∗ (x) = {µ(f ) : µ ∈ Mx (H) ∩ M+ (F )}. Proof. If µ ∈ Mx (H) ∩ M+ (F ) and k ∈ S c (H) satisfies k ≥ f on F , then µ(f ) ≤ µ(k) ≤ k(x). Thus µ(f ) ≤ Ff ∗ (x). Analogously, we get the reverse inequality. Conversely, if α ∈ [Ff∗ (x), Ff ∗ (x)] is given, we proceed as in the proof of Lemma 3.21. We set p(g) = Fg ∗ (x), g ∈ C(F ), and notice that the functional ϕ : span{f } → R defined as ϕ(tf ) = tα, t ∈ R, satisfies ϕ ≤ p on span{f }. Now we can finish the proof as in Lemma 3.21. Corollary 3.23. The following assertions hold:

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61

(a) Ac (H) = {f ∈ C(K) : f ∗ = f∗ }. (b) For an upper bounded upper semicontinuous function f on ChH (K), we have ChH (K) f ∗ = f on Ch (K). H Proof. For the proof of (a), let f ∈ C(K) satisfy f∗ = f ∗ . Then f ∈ S usc (H) ∩ Klsc (H) = Ac (H). Conversely, if f ∈ Ac (H), then f∗ = f ∗ by definition. For the proof of (b), let f be an upper bounded upper semicontinuous function on ChH (K). Let g be the upper semicontinuous extension of f to ChH (K) defined as in Lemma 3.18(d). By Lemma 3.22, for each x ∈ ChH (K) there exists µ ∈ Mx (H) carried by ChH (K) such that µ(g) = ChH (K)g ∗ (x). Since x ∈ ChH (K), µ = εx and f (x) = g(x) = ChH (K)g ∗ (x) ≥ ChH (K)f ∗ (x). This concludes the proof. Theorem 3.24 (Bauer’s characterization of the Choquet boundary). Let x be a point of K. Then the following assertions are equivalent: (i) x ∈ ChH (K), (ii) f ∗ (x) = f (x) for each f ∈ −W(H), (iii) f ∗ (x) = f (x) for each f ∈ Kc (H), (iv) f ∗ (x) = f (x) for each f ∈ C(K), (v) f ∗ (x) = f (x) for each upper semicontinuous function f on K. Proof. For the proof of (i) =⇒ (v), let x ∈ ChH (K) be given and f be an upper semicontinuous function on K. By Lemma 3.21, there exists a measure µ ∈ Mx (H) such that f ∗ (x) = µ(f ). Then µ = εx , and thus f ∗ (x) = µ(f ) = f (x). Since (v) =⇒ (iv) =⇒ (iii) =⇒ (ii) are obvious, we close the chain of implications by proving (ii) =⇒ (i). To this end, let a point x ∈ K satisfy f ∗ (x) = f (x) for each f ∈ −W(H). Then for each µ ∈ Mx (H) and f ∈ −W(H), f ∗ (x) = f (x) ≤ µ(f ) ≤ f ∗ (x). By Lemma 3.11(c), µ = εx and we are done. Proposition 3.25. Let f be an upper bounded function on K and F ⊃ ChH (K) be a closed set.

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(a) Then F ∗

f = inf {g ∈ H : g ≥ f on F } = inf {g ∈ Ac (H) : g ≥ f on F } = inf {g ∈ W(H) : g ≥ f on F } = inf {g ∈ S usc (H) : g ≥ f on F } .

(b) If f is upper semicontinuous, then F ∗ f = inf k ∈ S lsc (H) : k ≥ f on F . Proof. For the proof of (a), we first use Lemma 3.21 to conclude that inf {g ∈ S usc (H) : g ≥ f on F } = inf {g ∈ S c (H) : g ≥ f on F } .

(3.1)

Indeed, given g ∈ S usc (H) with g ≥ f on F , x ∈ K and ε > 0, by Lemma 3.21 there exists a measure µ ∈ Mx (H) ∩ M+ (F ) such that Fg ∗ (x) = µ(g) ≤ g(x). Thus there exists a function g 0 ∈ S c (H) with g 0 ≥ g on F such that g 0 (x) ≤ g(x) + ε. Hence (3.1) follows. Further, we fix a point x ∈ K and define p : C(F ) → R as p(g) = inf {g(x) : g ∈ H, g ≥ f on F } . Then p is a sublinear functional on C(F ) and we may follow the beginning of the proof of Lemma 3.21 to conclude that for any function g ∈ C(F ), there exists a measure µ in Mx (H) ∩ M+ (F ) such that µ(g) = p(g). Thus for every k ∈ S c (H), p(k) = µ(k) ≤ k(x), so that k(x) = inf{h(x) : h ∈ H, h ≥ f on F }. Hence inf{h ∈ H : h ≥ f on F } = inf{k ∈ S c (H) : k ≥ f on F }. Obviously, inf{k ∈ S c (H) : k ≥ f on F } ≤ inf{k ∈ W(H) : k ≥ f on F } ≤ inf{k ∈ H : k ≥ f on F }. Thus F ∗

f ≤ inf{k ∈ S usc (H) : k ≥ f on F } ≤ inf{h ∈ Ac (H) : h ≥ f on F } ≤ inf{h ∈ H : h ≥ f on F } = inf{k ∈ W(H) : k ≥ f on F } = inf{k ∈ S c (H) : k ≥ f on F } = Ff ∗ .

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63

This concludes the proof of (a). Let f be upper semicontinuous and x ∈ K. By Lemma 3.21, there exists a measure µ ∈ Mx (H) ∩ M+ (F ) such that Ff ∗ (x) = µ(f ). Then, for k ∈ S lsc (H) with k ≥ f on F , we have F ∗ f (x) = µ(f ) ≤ µ(k) ≤ k(x). Hence F ∗

f = inf k ∈ S lsc (H) : k ≥ f on F ,

which concludes the proof. Corollary 3.26. If x ∈ K and µ ∈ Mx (H), then εx ≺ µ. Proof. Given a function k ∈ S c (H) and ε > 0, let h ∈ H be such that k ≤ h and h(x) ≤ k(x) + ε (use Proposition 3.25(a)). Then εx (k) = k(x) ≥ h(x) − ε = µ(h) − ε ≥ µ(k) − ε. Hence εx ≺ µ, as needed. Theorem 3.27 (Bauer’s theorem). Let H be a function space on K. Then H = Ac (H) if and only if there exists a min-stable closed set W ⊂ C(K) such that H = W ∩ (−W). Proof. Since the family S c (H) is min-stable and closed in C(K) and Ac (H) = S c (H) ∩ (−S c (H)), necessity immediately follows. As for the sufficiency, we show that S c (H) ⊂ W. Then H = W ∩ (−W) ⊃ S c (H) ∩ (−S c (H)) = Ac (H) ⊃ H. Take any s ∈ S c (H). Then, with the aid of Lemma 3.25(a), s = s∗ = inf {ϕ ∈ H : ϕ ≥ s} ≥ inf {ϕ ∈ W : ϕ ≥ s} ≥ s. Hence, given x ∈ K and ε > 0, there exists wx ∈ W such that s(x) ≤ wx (x) < s(x) + ε. There exists an open neighborhood Ux of x such that wx < s + ε on Ux . By the compactness of K, there are x1 , . . . , xn ∈ K such that K = Ux1 ∪ · · · ∪ Uxn . Set w := wx1 ∧ · · · ∧ wxn . Then w ∈ W and w ≥ s on K. Given y ∈ K, there exists j ∈ {1, . . . , n} so that y ∈ Uxj . Then s(y) ≤ w(y) ≤ wxj (y) < s(y) + ε. Therefore ks − wk ≤ ε. We see that s ∈ W = W.

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3.2

More about Korovkin theorems

Definition 3.28 (Korovkin closures). Let P be a family of continuous functions on a compact space K. A sequence {Ln } of positive linear operators on C(K) is said to be P-admissible if kLn ϕ − ϕk → 0 for any ϕ ∈ P. Recall that the Korovkin closure Kor(P) of P is the set of those functions f ∈ C(K) such that kLn f −f k → 0 for any P-admissible sequence {Ln } of operators on C(K). Proposition 3.29. If H is a function space on a metrizable compact space K, then Kor(H) = Ac (H). Proof. Pick f ∈ Ac (H) and ε > 0. By Proposition 3.25(a) and Corollary 3.23, for any x ∈ K, there exists a function hx ∈ H and a neighborhood Ux of x such that f ≤ hx

and hx (t) − f (t) < ε for any t ∈ Ux .

Select x1 , . . . , xk ∈ K so that Ux1 ∪ · · · ∪ Uxk = K and denote hj := hxj for j = 1, . . . , k. If h := h1 ∧ · · · ∧ hk , then f ≤ h and h − f < ε on K. To show that f ∈ Kor(H), choose an H-admissible sequence {Ln } of positive linear operators on C(K). There exists n0 ∈ N such that kLn hj − hj k < ε for all n ≥ n0

and j = 1, . . . , k.

Now fix n ≥ n0 . Using the monotonicity of Ln , we have Ln f ≤ Ln hj ≤ hj + ε,

j = 1, . . . , k,

and therefore Ln f ≤ h1 ∧ · · · ∧ hk + ε = h + ε. Given x ∈ K, we see that Ln f (x) − f (x) ≤ (Ln f (x) − h(x)) + (h(x) − f (x)) ≤ 2ε. Analogously, using lower envelopes we can prove the existence of n1 ∈ N such that f (x) − Ln f (x) ≤ 2ε for all n ≥ n1 . It follows that kLn f − f k → 0, so f ∈ Kor(H). Now let f be in Kor(H). Given x ∈ K and µ ∈ Mx (H), our aim is to prove that µ(f ) = f (x). Let ρ be a metric on K compatible with the topology of K. For n ∈ N let Bn := {t ∈ K : ρ(x, t) < 1/n}

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65

and define gn (t) := 1 − nρ(x, t) ∨ 0,

t ∈ K.

Then gn ∈ C(K) and 0 ≤ gn ≤ 1,

gn (x) = 1

and gn = 0 on K \ Bn .

If Ln : ϕ 7→ gn µ(ϕ) + (1 − gn )ϕ,

ϕ ∈ C(K),

then Ln : C(K) → C(K) is a positive linear operator. Let h ∈ H and t ∈ K. Since |Ln h(t) − h(t)| = |gn (t)µ(h) + (1 − gn (t))h(t) − h(t)| = |gn (t)µ(h) − gn (t)h(t)| = gn (t)|h(x) − h(t)|, it follows that kLn h − hk ≤ sup {|h(x) − h(t)| : t ∈ Bn } . Since the function h is continuous at x, we get kLn h − hk → 0. We see that the sequence {Ln } is H-admissible, and therefore kLn f − f k → 0. In particular, Ln f (x) → f (x). But Ln f (x) = µ(f ), and the proof is complete. Remark 3.30. Notice that in the proof of the inclusion Ac (H) ⊂ Kor(H) we did not use linearity of Ln , while the metrizability condition was used only in the course of the proof that Kor(H) ⊂ Ac (H). Now, we are in a position to answer our query from Chapter 1, where we asked under which conditions the equality Kor(H) = C(K) holds. Proposition 3.31. Let H be a function space on a metrizable compact space K. Then Kor(H) = C(K) if and only if ChH (K) = K. Proof. The preceding Proposition 3.29 shows that Kor(H) = Ac (H), and therefore we only need to examine when Ac (H) = C(K). According to Bauer’s characterization 3.24 of the Choquet boundary, we know that x ∈ ChH (K) if and only if f∗ (x) = f ∗ (x) for any f ∈ C(K). Hence, ChH (K) = K if and only if f∗ = f ∗ on K for any f ∈ C(K); that is, if and only if Ac (H) = C(K), by Corollary 3.23. (First Korovkin theorem revisited). Let H be the linear span of funcTheorem 3.322 tions 1, id, id on [a, b]. Then Kor(H) = C([a, b]). Proof. Let t ∈ [a, b]. Since x 7→ (t − x)2 , x ∈ [a, b], is an H-exposing function for t, we may apply Proposition 3.7 to see that ChH ([a, b]) = [a, b]. Proposition 3.31 then shows that Kor(H) = C([a, b]).

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Theorem 3.33. Let S := {x ∈ Rd : |x| = 1} and

ϕi (x) := xi , x ∈ S,

i = 1, . . . , d.

If P := {1, ϕ1 , . . . , ϕd }, then Kor(P) = C(S). Proof. For the function space H = Lin{1, ϕ1 , . . . , ϕd } we have ChH (S) = S. Indeed, given w ∈ S, it is straightforward to check that the function x 7→ 1 − x · w,

x ∈ S,

is a P-exposing function for w. Corollary 3.34 (Second Korovkin theorem). Let H be the linear span of {1, cos, sin} on [0, 2π]. Then the Korovkin closure Kor(H) equals the Banach space of all continuous functions f on [0, 2π] with f (0) = f (2π) (equipped with the sup-norm). Proof. The circle S := {x ∈ R2 : |x| = 1} can be topologically identified with the interval [0, 2π] after having identified the points 0 and 2π via the mapping ω : t 7→ (cos t, sin t), t ∈ [0, 2π]. Then the functions cos, sin correspond to the functions ϕ1 , ϕ2 from Theorem 3.33. Remark 3.35. There is also a direct proof of this corollary following the lines of the proof of Theorem 1.2, considering the function x 7→ sin2

t−x 1 = (1 − cos x cos t − sin x sin t), 2 2

x ∈ [0, 2π],

instead of the function x 7→ (x − t)2 . Theorem 3.36 (Third Korovkin theorem). Let H be a linear span of {1, ϕ} on [a, b], where ϕ is a continuous function on [a, b]. Then Kor(H) 6= C([a, b]). Proof. The assertion is obvious if the function ϕ does not separate points of [a, b]. Otherwise, there exists x ∈ (a, b) such that ϕ(x) = 21 ϕ(a) + ϕ(b) . Then the measure 12 (εa + εb ) represents x, and therefore x does not belong to ChH ([a, b]). In view of Proposition 3.31, Kor(H) 6= C([a, b]).

3.3

On the H-barycenter mapping

Proposition 3.37. Let H be a function space on a compact space K. Denote [ M(H) := Mx (H). x∈K

Then M(H) is a compact subset of M1 (K).

3.4 The Choquet representation theorem

67

Proof. It is easy to show that M(H) is a closed subset of M1 (K). Remark 3.38. Note that in the “convex case”, M(Ac (X)) = M1 (X) (see Theorem 2.29). If H = C([0, 1]), then M(H) = {εx : x ∈ [0, 1]} 6= M1 ([0, 1]). See also Exercise 3.106. Let us observe that if µ ∈ Mx (H) ∩ My (H) for some x, y ∈ K, then h(x) = µ(h) = h(y) for any h ∈ H, and hence x = y since H separates points of K. Definition 3.39 (H-barycenter mapping). We define the H-barycenter mapping r from M(H) onto K by r : µ ∈ Mx (H) 7→ x,

µ ∈ M(H).

(By the previous observation, the mapping r is defined correctly.) Proposition 3.40. The H-barycenter mapping r : M(H) → K is continuous. Proof. The mapping µ 7→ h(r(µ)),

µ ∈ M(H),

is continuous for any h ∈ H. Hence r is continuous as a mapping from M(H) into K, where on K we consider the topology generated by the family H. But this one coincides with the topology of K.

3.4

The Choquet representation theorem

In this section we will give a proof of the Choquet representation theorem in the metrizable case. First we establish a construction of strictly H-convex functions. Definition 3.41 (Strictly H-convex functions). A continuous H-convex function f on a compact space K is called strictly H-convex if Z f (x)
h(t)} = 0. In view of Proposition 3.43, ChH (K) = {t ∈ K : h(t) = h∗ (t)} , and the proof is complete. Remark 3.46. By Theorem 3.42, any metrizable compact set admits a continuous strictly H-convex function. Therefore, Proposition 3.43 and Theorem 3.45 hold under the assumption that the compact space K is metrizable. On the other hand, there are function spaces on nonmetrizable compact spaces which admit a continuous strictly H-convex function. As a trivial example consider a nonmetrizable compact space K and H = C(K). In this example, any continuous function on K is strictly H-convex. We present the following slightly less trivial example.

70


Example 3.47. There exists a function space on a nonmetrizable compact space K that admits a strictly H-convex continuous function. Proof. Let K := D ∪ {d∞ } be the one-point compactification of an uncountable discrete topological space D and x and y be distinct points of D. Then K is not metrizable (since K is not separable). If h(x) + h(y) H := h ∈ C(K) : h(d∞ ) = , 2 then c{x,y} is a continuous strictly H-convex function on K.

3.5

In-between theorems

Throughout this section, H will be a function space on a compact space K. There are plenty of insertion problems occurring in various fields of analysis. They can be formulated as follows: Given families of functions S, T and F, find f ∈ F such that s ≤ f ≤ t (or, s < f < t) whenever s ∈ S, t ∈ T and s ≤ t (or, s < t). Perhaps the most famous result in this direction goes back to H. Hahn, who proved that for every pair of real-valued functions s, t on a metric space, where s is upper semicontinuous, t is lower semicontinuous and s ≤ t, there exists a continuous function f such that s ≤ f ≤ t. In this section, we will examine in-between theorems and approximation theorems for the case of H-convex and H-concave functions. Proposition 3.48. Let f be an upper semicontinuous H-concave function on K. If K := {g ∈ S c (H) : g ≥ f on K} and L := {g ∈ S c (H) : g > f on K} , then K and L are down-directed families and f = inf K = inf L. Proof. Since the cone S c (H) is min-stable (cf. Proposition 3.11), the sets K and L are down-directed. It suffices to show that f = inf L. To this end, choose x ∈ K and numbers f (x) < c < d. By Proposition 3.25(a), f ∗ = f and there exists h ∈ H such that h≥f

and

h(x) < c.

Now, if g := h+d−c, then g ∈ H, g(x) < d and g > f . Since d is chosen arbitrarily, f = inf L. Corollary 3.49. If g is a lower semicontinuous function on K, f ∈ S usc (H), f < g on K, then there exists a function k ∈ S c (H) such that f < k < g on K.

3.5 In-between theorems

71

Proof. By Proposition 3.25(a), f ∗ = f . With the aid of Proposition 3.48, for any x ∈ K we select a function gx ∈ S c (H) such that gx ≥ f on K

and gx (x) < g(x).

Adding a small constant to gx , we may suppose that gx > f on K

and gx (x) < g(x).

Now appeal to the upper semicontinuity of f and the lower semicontinuity of g to find an open neighborhood Ux of x such that f (y) < gx (y) < g(y) for any y ∈ Ux . By the compactness of K, there exist x1 , . . . , xn ∈ K such that K = Ux1 ∪ · · · ∪ Uxn . Then the function k := gx1 ∧ · · · ∧ gxn belongs to S c (H) and it is a routine argument to show that f < k < g on K. Indeed, given y ∈ K, there exists j ∈ {1, . . . , n} such that y ∈ Uxj . Then f (y) < k(y) ≤ gxj (y) < g(y).

Remark 3.50. There exist a function space H on a metrizable compact space K, a lower semicontinuous function g on K and a function f ∈ S usc (H) such that f ≤ g on K and there is no function from S c (H) between f and g. See Exercise 3.111. Proposition 3.51. If g is an upper semicontinuous function on K, f ∈ S lsc (H), g < f on K, then there exists a function k ∈ S c (H) such that g < k < f on K. Proof. Fix x ∈ K. By Lemma 3.21, there exists a measure µ ∈ Mx (H) such that g ∗ (x) = µ(g). Then g ∗ (x) = µ(g) < µ(f ) ≤ f (x). Therefore, by Proposition 3.25(a) there exists hx ∈ H such that hx ≥ g on K

and

hx (x) < f (x).

Adding a small constant to hx , we may assume that hx > g everywhere on K and we still have hx (x) < f (x). We infer from the lower semicontinuity of f − hx and a compactness argument that there exist points x1 , . . . , xn ∈ K such that k := hx1 ∧ · · · ∧ hxn < f . The function k has all the required properties. Remark 3.52. A hint to an alternative (slightly different) proof of Proposition 3.51 can be found in Exercise 3.110.

72


Proposition 3.53. If g is an upper semicontinuous function on K, f ∈ S lsc (H), g ≤ f on K, then there exists a function k ∈ S c (H) such that g ≤ k ≤ f on K. Proof. The first order of business is to construct, for each ε > 0, a pair of functions uε and vε such that uε is upper semicontinuous, vε ∈ S lsc (H), g ≤ uε ≤ vε ≤ f

on K

and vε − uε < ε on K. This can be done quite easily. Namely, by Proposition 3.51, there exists a function k ∈ S c (H) such that g < k < f + ε. It is easy to check that the functions uε := (k − ε) ∨ g

and

vε := f ∧ k

do the job. Now, using the above construction we can construct inductively two sequences {un } and {vn } so that, for each n ∈ N, un is upper semicontinuous, vn ∈ S lsc (H), g ≤ un ≤ un+1 ≤ vn+1 ≤ vn ≤ f and

1 . 2n Then the sequence {vn } converges uniformly to an H-concave continuous function k. The inequalities 1 1 g ≤ vn ≤ un + n ≤ f + n 2 2 ensure that g ≤ k ≤ f . vn − un
g1 ∨ g2 . Since (g1 ∨ g2 )∗ < f (cf. the proof of Proposition 3.51), a new application of

3.6 Maximal measures

73

Proposition 3.51 asserts the existence of g ∈ S c (H) such that g1 ∨ g2 ≤ (g1 ∨ g2 )∗ < g < f. All that is left to be shown is that T is up-directed. To finish the proof, consider g1 , g2 ∈ T. By Proposition 3.53, there exists g ∈ S c (H) such that g1 ∨g2 ≤ g ≤ f . Proposition 3.55. Let f ∈ S c (H) and ε > 0. Then there exists h ∈ W(H) such that f ≤ h ≤ f + ε. Proof. By Proposition 3.25, f = inf {h ∈ W(H) : h ≥ f } , where the family W(H) is down-directed. Since f is supposed to be continuous, we may use Dini’s theorem to obtain the desired function. Proposition 3.56. Let H be a function space on K. Then the following assertions are equivalent for µ, ν ∈ M+ (K): (i) µ ≺ ν, (ii) µ(f ) ≤ ν(f ) for any f ∈ −W(H), (iii) µ(f ) ≤ ν(f ) for any f ∈ Kusc (H), (iv) µ(f ) ≤ ν(f ) for any f ∈ Klsc (H). Proof. The implications (i) =⇒ (iii) and (i) =⇒ (iv) follow from Proposition 3.54, Corollary 3.49 and Theorem A.84. Obviously, (i) =⇒ (ii), (iii) =⇒ (i) and (iv) =⇒ (i). By Proposition 3.55, (ii) =⇒ (i).

3.6

Maximal measures

Throughout this section, H will be a function space on a compact space K. Definition 3.57 (H-maximal measures). If H is a function space on a compact space K, we recall that the convex cone Kc (H) of all continuous H-convex functions on K determines the Choquet ordering ≺. Maximal elements of M+ (K) with respect to the Choquet ordering are called maximal measures (or, more precisely, H-maximal measures). The set of all H-maximal measures on K is denoted by Mmax (H). We denote by Mbnd (H) the set of all signed Radon measures on K such that the variation |µ| of µ is a maximal measure. Elements of Mbnd (H) are called Hboundary measures or, simply, boundary measures. Theorem 3.58 (Mokobodzki’s maximality test). The following assertions on a measure µ ∈ M+ (K) are equivalent: (i) µ is maximal,

74


(ii) µ(f ) = µ(f ∗ ) for any f ∈ −W(H), (iii) µ(f ) = µ(f ∗ ) for any f ∈ Kc (H), (iv) µ(f ) = µ(f ∗ ) for any f ∈ C(K), (v) µ(f ) = µ(Ff ∗ ) for any upper semicontinuous function f on K and any set F ⊃ ChH (K). Proof. Let µ ∈ M+ (K) be maximal, F ⊃ ChH (K) and f be an upper semicontinuous function on K. By Lemma 3.18(d), Qµ,F (f ) = Qµ,F (f ). By Lemma 3.21, there exists a measure ν ∈ M+ (F ) such that µ ≺ ν and ν(f ) = Qµ,F (f ). Since µ is maximal, we have µ = ν, and thus by Lemma 3.18(b), µ(f ) = ν(f ) = Qµ,F (f ) = Qµ,F (f ) = µ(Ff ∗ ). Hence (i) =⇒ (v). It is obvious that (v) =⇒ (iv) =⇒ (iii) =⇒ (ii). To see that (ii) =⇒ (i), assume that a measure µ ∈ M+ (K) satisfies µ(f ) = µ(f ∗ ) for each f ∈ −W(H). Let ν ∈ M+ (K), µ ≺ ν and fix f ∈ −W(H). Then µ(f ) ≤ ν(f ) and, using Lemma 3.18(b) and Theorem A.84, we get µ(f ) = µ(f ∗ ) = µ(inf {k ∈ S c (H) : k ≥ f }) = inf {µ(k) : k ∈ S c (H), k ≥ f } ≥ inf {ν(k) : k ∈ S c (H), k ≥ f } = ν(f ∗ ) ≥ ν(f ). Hence µ(f ) = ν(f ). Since the space W(H) − W(H) is uniformly dense in C(K), we conclude that µ = ν. We see that µ is maximal. Corollary 3.59. A measure µ ∈ M+ (K) is maximal if and only if µ({x ∈ K : f (x) < f ∗ (x)}) = 0

for any f ∈ −W(H).

(3.3)

Proof. The assertion follows immediately from Mokobodzki’s maximality test 3.58. Corollary 3.60. If µ ∈ M+ (K) and µ(F ) = 0 for any closed F ⊂ K \ ChH (K), then µ is a maximal. Proof. Let µ be as in the hypothesis. For any f ∈ −W(H), B := {x ∈ K : f ∗ (x) = f (x)} is a Borel set containing ChH (K). By the assumption and regularity of µ, µ(K \ B) = 0 and µ is maximal. Corollary 3.61. Let h be a continuous strictly H-convex function on K and µ ∈ M+ (K). Then the following assertions are equivalent: (i) µ is maximal,


75

(ii) µ(h) = µ(h∗ ) , (iii) µ(K \ ChH (K)) = 0. Proof. Recall that, by Proposition 3.43, ChH (K) is a Gδ set. The implication (i) =⇒ (ii) follows from Mokobodzki’s maximality test 3.58. Assume that µ(h) = µ(h∗ ). Then µ({x ∈ K : h∗ (x) − h(x) > 0}) = 0. Since by Proposition 3.43 K \ ChH (K) = {x ∈ K : h∗ (x) − h(x) > 0} , we get (ii) =⇒ (iii). Finally, assume (iii) and choose v ∈ Kc (H). Since A := {x ∈ K : v ∗ (x) > v(x)} ⊂ K \ ChH (K), we have

Z

Z v dµ +

µ(v) = A

v dµ = µ(v ∗ ),

B

where B := {x ∈ K : v ∗ (x) = v(x)}. Thus (iii) =⇒ (i) by Mokobodzki’s maximality test 3.58. Corollary 3.62. Let K be a metrizable compact space and µ ∈ M+ (K). Then µ is maximal if and only if µ is carried by the Choquet boundary ChH (K). Proof. The assertion is a direct consequence of Theorem 3.42 and Corollary 3.61. Remark 3.63. We know from Corollary 3.62 that in the metrizable case any maximal measure is carried by the Choquet boundary, which is a Borel set. This assertion is no longer valid for nonmetrizable situations. Section 3.8 focuses on the Choquet theory in the nonmetrizable setting. Nevertheless, in this case any maximal measure is carried at least by the closure of the Choquet boundary; see the next Proposition 3.64. On the other hand, a measure carried by ChH (K) need not be maximal. Indeed, consider a function space H on a metrizable compact space K with a nonclosed Choquet boundary (cf. Exercise 2.103(b)). For a point x ∈ ChH (K) \ ChH (K), the Dirac measure at x is clearly carried by ChH (K) and it fails to be maximal in view of Corollary 3.62. A more striking example of this phenomenon is the Poulsen simplex (see Subsection 12.3.A), where the set of extreme points is dense in the whole set. Proposition 3.64. Any maximal measure is carried by ChH (K). Proof. Let µ be a maximal measure on K and C a compact subset of K \ ChH (K). Urysohn’s lemma ensures the existence of a function f ∈ C(K), 0 ≤ f ≤ 1 on K, such that f = 0 on ChH (K) and f = 1 on C.

76


Since f∗ = sup {h ∈ H : h ≤ f }, we have h ≤ 0 on ChH (K) for any h ∈ H with h ≤ f . By Theorem 3.16, h ≤ 0 on K, and therefore f∗ ≤ 0 on K. Because obviously f∗ ≥ 0, we have f∗ = 0 on K. Mokobodzki’s maximality test 3.58 then yields µ(C) ≤ µ(f ) = µ(f∗ ) = 0. By regularity of µ we get µ(K \ ChH (K)) = 0. Theorem 3.65. For every measure µ ∈ M+ (K) there exists a maximal measure λ such that µ ≺ λ. Proof. Set M := {ν ∈ M+ (K) : µ ≺ ν}. Since for any ν ∈ M we have kνk = kµk, it follows that M is contained in the compact set {η ∈ M+ (K) : kηk = kµk}. The assertion will follow from Zorn’s lemma once it is shown that every chain in M has an upper bound. Let R be such a chain. Then R is a net in a compact set (directed by ≺) and hence there exists a subnet J of R which converges to some element ν0 ∈ M+ (K). Since M is closed, ν0 ∈ M . It remains to show that ν ≺ ν0 for any ν ∈ R. To this end, fix ν ∈ R, f ∈ Kc (H) and ε > 0. There is η ∈ J such that ν ≺ η and |ν0 (f ) − η(f )| < ε. Then ν0 (f ) ≥ η(f )−ε ≥ ν(f )−ε. Therefore ν ≺ ν0 since ε is an arbitrary positive number. We conclude by Zorn’s lemma: the ordered set M has a maximal element. Proposition 3.66. If µ ∈ M+ (K) is maximal, then µ({x}) = 0 for each x ∈ K \ ChH (K). Proof. Let µ ∈ M+ (K) be maximal and x ∈ K \ ChH (K). By Theorem 3.24, there exists a function f ∈ −W(H) such that f ∗ (x) > f (x). Since µ({y ∈ K : f ∗ (y) > f (y)}) = 0 (see Corollary 3.59), µ({x}) = 0. Proposition 3.67. If µ, ν ∈ M+ (K), then µ ≺H ν if and only if µ ≺Ac (H) ν. Hence, a measure µ ∈ M+ (K) is H-maximal if and only if it is Ac (H)-maximal. Proof. Since Mx (H) = Mx (Ac (H)) (see Exercise 3.95), a function f on K is H-convex if and only if it is Ac (H)-convex. Now the first assertion is a direct consequence of this fact, and the second assertion follows immediately from the first one. Theorem 3.68. The following assertions on a measure µ ∈ M(K) are equivalent: (i) µ is a boundary measure, (ii) µ(f ) = µ(f ∗ ) for any f ∈ C(K), (iii) µ(f ) = µ(f ∗ ) for any upper semicontinuous function f on K.


77

Proof. By Theorem 3.58, (i) =⇒ (ii). Assume (ii), choose an upper semicontinuous function f on K and ε > 0. Using Theorem 3.48, we can find a function g ∈ S c (H) such that g > f ∗ and µ+ (g) < µ+ (f ∗ ) + ε

and

µ− (g) < µ− (f ∗ ) + ε.

Since f is upper semicontinuous, by Proposition A.50 and Theorem A.84 there exists h ∈ C(K) such that h > f and µ+ (h) < µ+ (f ) + ε

and

µ− (h) < µ− (f ) + ε.

Set ϕ := g ∧ h. Then ϕ ∈ C(K), ϕ > f , f ∗ ≤ ϕ∗ ≤ g and µ+ (ϕ) < µ+ (f ) + ε

and

µ− (ϕ) < µ− (f ) + ε.

µ+ (ϕ∗ ) ≤ µ+ (g) < µ+ (f ∗ ) + ε

and

µ− (ϕ∗ ) ≤ µ− (g) < µ− (f ∗ ) + ε,

It follows,

and therefore,

−2ε = µ(ϕ∗ ) − µ(ϕ) − 2ε ≤ µ(f ∗ ) − µ(f ) ≤ µ(ϕ∗ ) − µ(ϕ) + 2ε = 2ε.

We see that µ(f ) = µ(f ∗ ); hence (ii) =⇒ (iii). It remains to show that (iii) =⇒ (i). We have to verify that |µ| = µ+ + µ− is carried by the set Af := {x ∈ K : f ∗ (x) = f (x)} for any f ∈ C(K). To this end, let f ∈ C(K) be given. Let again µ = µ+ − µ− be the decomposition of µ into its positive and negative part, µ+ be carried by a set P and µ− be carried by a set N , where P ∩ N = ∅. It suffices to prove that µ+ (L) = 0 whenever L ⊂ P \ Af is a compact set. So, fix such a set L and set ( f on K \ L, g := ∗ f on L. ∗ ∗ Then g is upper semicontinuous on µ(g) = R K∗ and g =+ f . By the assumption, ∗ ∗ µ(g ) = µ(f ) = µ(f ), hence L (f − f ) dµ = 0. Since f ∗ − f > 0 on L, µ+ (L) = 0. The proof is complete.

We recall that the space M(K), as a dual space to the ordered Banach space C(K), is an ordered Banach space. Moreover, it is a lattice. Theorem 3.70 below shows that Mbnd (H) is a lattice with the structure inherited from M(K). Proposition 3.69. If µ ∈ Mmax (H) and ν ∈ M(K) is absolutely continuous with respect to µ, then ν ∈ Mbnd (H).

78


Proof. It is enough to consider the case when ν is positive. If g is a continuous function, let A := {x ∈ K : g ∗ (x) = g(x)}. Since µ is a boundary R measure, |µ|(K \ A) = 0 (see Theorem 3.58). Let h ∈ L1 (µ) be such that ν(B) = B h dµ for any Borel set B ⊂ K. Then Z ν(K \ A) = h dµ = 0. K\A

Hence ν(g) = ν(g ∗ ) for any g ∈ C(K), and thus ν is maximal by Theorem 3.58. Theorem 3.70. The set Mbnd (H) is a norm closed subspace of M(K). Moreover, it is a lattice considered with the order inherited from M(K). Proof. We first show that Mbnd (H) is a norm closed subspace of M(K). To this end, let µ1 , µ2 ∈ Mbnd (H) and g ∈ C(K) be given. Since |µ1 | + |µ2 | is carried by the set {x ∈ K : g ∗ (x) = g(x)}, it is easy to see that |µ1 + µ2 | is carried by the set {x ∈ K : g ∗ (x) = g(x)}. Thus µ1 + µ2 is a boundary measure by Theorem 3.58. Analogously we show that cµ ∈ Mbnd (H) for any c ∈ R and µ ∈ Mbnd (H). If {µn } is a sequence of boundary measures norm converging to µ ∈ M(K), we use again Theorem 3.58 to show that µ is a boundary measure. To verify the lattice property we use Proposition A.15. Obviously, any measure µ ∈ Mbnd (H) can be written as a difference of positive maximal measures, and hence Mbnd (H) = (Mbnd (H))+ − (Mbnd (H))+ = Mmax (H) − Mmax (H). Let µ1 , µ2 ∈ Mmax (H) be given. Since M(K) is a lattice, there exists a measure ν ∈ M+ (K) such that ν = µ1 ∧ µ2 (with respect to the ordering of M(K)). Since ν ≤ µ1 + µ2 , ν is absolutely continuous with respect to the maximal measure µ1 + µ2 . By Proposition 3.69, ν ∈ Mmax (H) as well. Hence ν is a supremum of {µ1 , µ2 } in Mbnd (H) and Mbnd (H) is a lattice.

3.7

Boundaries and the Simons lemma

Definition 3.71 (σ-convex hull). Let A be a bounded subset of a Banach space E. We denote by ∞ ∞ nX o X coσ (A) := cn xn : cn ≥ 0, cn = 1, xn ∈ A n=1

n=1

the σ-convex hull of the set A. We notice the following elementary fact. Lemma 3.72. Let A be a bounded subset of a Banach space E. Then coσ (coσ (A)) = coσ (A).

79

3.7 Boundaries and the Simons lemma

Proof. It follows by a straightforward verification. Definition 3.73 (Boundary). Let K be a set and let F be a subset of `∞ (K). We say that a nonempty set B in K is a boundary for F if every function from F attains its maximum at a point of B. Lemma 3.74 (Simons inequality). Let {fn } be a bounded sequence of functions on a set K. Let B ⊂ K be a boundary for coσ ({fn : n ∈ N}). Then sup lim sup fn (b) ≥ inf sup f (K) : f ∈ coσ ({fn : n ∈ N}) .

b∈B n→∞

Proof. Let {fn } be a bounded sequence on K. We start theP proof by fixing ε > 0 and a sequence {ρi } of strictly positive numbers such that ∞ i = 1. We set i=1 ρP P∞ εk σk := i=k ρi and find strictly positive numbers εk , k ∈ N, such that ∞ k=1 σk+1 < ε. Let Ck := coσ ({fn : n ≥ k}), k ∈ N. For a bounded function f on K, we write p(f ) := sup f (B) and q(f ) := sup f (K). We inductively find functions gi ∈ Ci , i ∈ N, such that p(ρ1 g1 ) ≤ infh∈C1 p(ρ1 h) + ε1 , Pk−1 Pk • p( i=1 ρi gi ) ≤ infh∈Ck p( i=1 ρi gi + ρk h) + εk , k ≥ 2. P We set g := ∞ i=1 ρi gi . P Claim. For every n ∈ N, p(g) ≥ p( nk=1 ρk gk ) + σn+1 (p(g) − ε). •

Proof of the claim. Let n ∈ N be given. Throughout the proof we interpret 0. We pick a number k ∈ {1, . . . , n}. Since ∞

k−1

i=1

i=1

P0

i=1

as

X 1 X ρi gi − ρi gi ∈ Ck , σk we get p

k X

ρi g i ≤ p

i=1

k−1 X

ρi gi + ρk

i=1

∞ k−1 X 1 X ( ρi gi − ρi gi ) σk i=1

! + εk

i=1

k−1 ρk X ρk p(g) + 1 − p ρi gi + εk . ≤ σk σk i=1

Hence p(g) 1 −

k k−1 X σk+1 X σk+1 ≥p ρi gi − p ρi g i − ε k . σk σk i=1

i=1

(3.4)

80


We divide (3.4) by σk+1 and sum these inequalities for k = 1, . . . , n. Using p(0) = 0, we have n n X X 1 1 εk p(g) −1 ≥ p ρk gk − . σn+1 σn+1 σk+1 k=1

k=1

By the choice of the numbers εk , we get p(g) ≥ p

n X

ρk gk + σn+1 p(g) − ε .

k=1

This finishes the proof of the claim. We apply the assumption for the function g, and find b ∈ B such that g(b) = p(g) = q(g). Then, for n ∈ N, the claim yields ∞ X

ρi gi (b) ≥

i=1

Since

ρi gi (b) + σn+1 (q(g) − ε).

i=1

P∞ 1 σn+1 ( i=n+1 ρi gi ) q(g) − ε ≤

n X

∈ Cn+1 , we get 1

σn+1

∞ X

ρi gi (b)

i=n+1

≤ sup{gi (b) : i ≥ n + 1} ≤ sup{fi (b) : i ≥ n + 1}, since, for every i ≥ n + 1, gi ∈ Ci , and hence gi (b) ≤ sup{fi (b) : i ≥ n + 1}. By letting n tend to infinity, we get q(g) − ε ≤ lim sup fn (b) ≤ p(lim sup fn ). n→∞

n→∞

This concludes the proof. Corollary 3.75 (Simons lemma). Let {fn } and B ⊂ K be as in Lemma 3.74. Then sup lim sup fn (b) = sup lim sup fn (x). b∈B n→∞

x∈K n→∞

Proof. Let {fn } be as in Lemma 3.74. We again adopt the notation p(f ) = sup f (B) and q(f ) = sup f (K) for each f ∈ `∞ (K). Assume that p(lim supn→∞ fn ) < c < d < q(lim supn→∞ fn ). Let x ∈ K satisfy lim supn→∞ fn (x) > d. Let n1 < n2 < . . . be natural numbers such that fnk (x) > d, k ∈ N. Clearly, B is a boundary for coσ {fnk : k ∈ N}. By Lemma 3.74, there exists f ∈ coσ {fnk : k ∈ N} such that q(f ) ≤ p(lim sup fnk ) + d − c ≤ p(lim sup fn ) + d − c < d. k→∞

n→∞

Obviously, q(f ) ≥ f (x) > d. This contradiction finishes the proof.

3.8 The Bishop–de Leeuw theorem

81

Corollary 3.76. Let {fn } be a bounded sequence of functions in `∞ (K), g ∈ `∞ (K), A := {fn : n ∈ N} ∪ {g} and let B ⊂ K be a boundary for span A. (a) If {fn (b)} is convergent for each b ∈ B, then {fn (x)} is convergent for each x ∈ K. (b) If fn → g on B, then fn → g on K. Proof. (a) Let {fn } and B ⊂ K be as in the statement of the corollary. Note that B is a boundary for coσ (span A). We again write p(f ) := sup f (B) for f ∈ `∞ (K). Let {fn (b)} be convergent for each b ∈ B, and assume that there exists x ∈ K such that {fn (x)} is not Cauchy. By choosing a subsequence if necessary, we may assume that there exists a strictly increasing sequence {nk } of natural numbers and c > 0, such that c < fn2k (x) − fn2k−1 (x), k ∈ N. Then the functions gk := fn2k − fn2k−1 , k ∈ N, satisfy p(lim supk→∞ gk ) = 0, yet 0 < c ≤ lim supk→∞ gk (x). This contradicts Corollary 3.75 and proves (a). (b) Let fn → g on B. Then, for each x ∈ K, Corollary 3.75 gives lim sup(fn (x) − g(x)) ≤ p(lim sup(fn − g)) = 0. n→∞

n→∞

If we apply Corollary 3.75 to g − fn , we get lim inf(fn (x) − g(x)) ≥ 0. n→∞

This concludes the proof.

3.8

The Bishop–de Leeuw theorem

Throughout this section, H will be a function space on a compact space K. Lemma 3.77. Let {fn } be an upper bounded sequence of lower semicontinuous Hconvex functions on K such that lim supn→∞ fn ≤ 0 on ChH (K). Then we have lim supn→∞ fn ≤ 0 on K. Proof. Let {fn } be as in the statement of the lemma. Without loss of generality we may assume that {fn } is a bounded sequence, sup{kfn k : n ∈ N} ≤ C (otherwise we would consider the functions −1∨fn ). Fix x ∈ K and ε > 0. Using Proposition 3.48, we find functions kn ∈ Kc (H) such that kn ≤ fn and kn (x) ≥ fn (x) − ε. Without loss of generality we may assume that sup{kkn k : n ∈ N} ≤ C. By Theorem 3.16, each function in coσ ({kn : n ∈ N}), as an H-convex continuous function, attains its maximum on ChH (K). Since lim sup kn ≤ lim sup fn ≤ 0 n→∞

n→∞

on ChH (K),

82


the Simons lemma 3.75 yields lim sup fn (x) ≤ ε + lim sup kn (x) ≤ ε. n→∞

n→∞

Since ε > 0 is arbitrary, the proof is finished. Lemma 3.78. Let µ be a maximal measure on K and F ⊂ K \ ChH (K) be a compact Gδ set. Then µ(F ) = 0. Proof. Given a maximal measure µ ∈ M+ (K) and a compact Gδ set F ⊂ K \ ChH (K), we find a sequence {fn } of continuous functions such that 0 ≤ fn ≤ 1, n ∈ N, and fn → cF . Then {(fn )∗ } is a bounded sequence of H-convex lower semicontinuous functions such that lim sup(fn )∗ ≤ 0 on ChH (K). By Lemma 3.77, lim supn→∞ (fn )∗ ≤ 0 on K. Thus Theorem 3.58(v) together with Fatou’s lemma yields 0 ≤ µ(F ) = lim µ(fn ) = lim µ((fn )∗ ) n→∞

n→∞

≤ lim sup µ((fn )∗ ) ≤ µ(lim sup(fn )∗ ) ≤ 0. n→∞

n→∞

This concludes the proof. Theorem 3.79. Let µ ∈ M+ (K) be a maximal measure on K. Then (a) µ(B) = 0 for every Baire set B disjoint from ChH (K), (b) µ∗ (K \ L) = 0 for every Lindelöf set L ⊃ ChH (K) (here µ∗ denotes the inner measure induced by µ, see Definition A.62), (c) µ(K \ A) = 0 for every K-analytic set A ⊃ ChH (K). Proof. Let µ ∈ M+ (K) be a maximal measure and B ⊂ K \ ChH (K) be a Baire set. Then µ(B) = 0 by Lemma A.89 and Lemma 3.78. This shows (a). Assume now that the set L ⊃ ChH (K) is Lindelöf and F ⊂ K \ L is a compact set. For each x ∈ L we find an open Fσ set U (x) such that x ∈ Ux ⊂ K \ F . Using the Lindel¨ select countably many points xn ∈ L, n ∈ N, such that S of property we S∞ U L⊂ ∞ . Since n=1 xn n=1 Uxn is a Baire set containing ChH (K), µ(F ) = 0, by (a). Hence µ∗ (K \ L) = 0. For the proof of (c), let A ⊃ ChH (K) be a K-analytic set. By Theorem A.111(d), (g) we get that A is Lindelöf and µ-measurable. Hence µ(K \ A) = 0, by (b). Proposition 3.80. Assume that ChH (K) is a Lindelöf set and µ ∈ M+ (K). Then the following assertions are equivalent: (i) µ∗ K \ ChH (K) = 0, (ii) µ is maximal.

3.8 The Bishop–de Leeuw theorem

83

Proof. Since (ii) =⇒ (i) is proved in Theorem 3.79, it remains to show the reverse implication. Let µ ∈ M+ (K) be a measure satisfying µ∗ (K \ ChH (K)) = 0 and let f ∈ Kc (H) be fixed. By Theorem 3.24, {x ∈ K : f (x) < f ∗ (x)} is disjoint from ChH (K). Since µ({x ∈ K : f (x) < f ∗ (x)}) = 0, by assumption, Corollary 3.59 concludes the proof. Theorem 3.81 (Bishop–de Leeuw). For any point x ∈ K, there exists a measure µ ∈ Mx (H) such that µ(A) = 1 for every K-analytic set A ⊃ ChH (K). Proof. Given x ∈ K, Theorem 3.65 provides a maximal measure µ ∈ Mx (H). By Theorem 3.79(c), the proof is complete. Example 3.82. There exist a function space H on a compact space K and a maximal measure on K carried by a compact set disjoint from ChH (K). Proof. Let K := [0, 1]×{−1, 0, 1} be equipped with the topology defined as follows: points of [0, 1] × {−1, 1} are open (hence [0, 1] × {−1, 1} is discrete), and the base of neighborhoods of (x, 0) ∈ [0, 1] × {0} are of type (U × {−1, 0, 1}) \ F , where U ⊂ [0, 1] is a Euclidean neighborhood of x, and F is finite. If f is a continuous function on this compact space K, we notice that, for each δ > 0, the set {x ∈ [0, 1] : |f (x, 0) − f (x, 1)| + |f (x, 0) − f (x, −1)| > δ}

(3.5)

is finite. We define a function space H as 1 H := {f ∈ C(K) : f (x, 0) = (f (x, −1) + f (x, 1)), x ∈ [0, 1]}. 2 Fix x ∈ [0, 1]. Then the function c(x,1) − c(x,−1) shows that {(x, −1), (x, 1)} is a subset of ChH (K). Further, the function (t, j) 7→ |t − x|, (t, j) ∈ K, yields that the only µ ∈ M(x,0) (H) with µ({(x, 0)}) = 0 equals 21 (ε(x,−1) + ε(x,1) ). Thus ChH (K) = K \ ([0, 1] × {0}) and for each (x, 0) ∈ K, there exists a unique measure µ ∈ M(x,0) (H) with µ({(x, 0)}) = 0. Claim. If f ∈ Kc (H) and δ > 0, then {x ∈ [0, 1] : f ∗ (x, 0) − f (x, 0) > δ} is finite. Proof of the claim. Let f ∈ Kc (H) and δ > 0 be given. If x ∈ [0, 1], by Lemma 3.21, there exists a measure µ ∈ M(x,0) (H) such that f ∗ (x, 0) = µ(f ). Since f is Hconvex, µ(f ) ≤ ν(f ) ≤ f ∗ (x, 0) for any ν ∈ Mx (H) with µ ≺ ν. Hence we may assume that µ is maximal. By Proposition 3.66, µ({(x, 0)}) = 0. As shown above, µ = 21 (ε(x,−1) + ε(x,1) ). Thus 1 f ∗ (x, 0) = (f (x, −1) + f (x, 1)), 2 By (3.5), the proof of the claim is finished.

x ∈ [0, 1].

84


Now let µ ∈ M+ ([0, 1] × {0}) be a continuous measure. By the claim, µ({x ∈ [0, 1] : f ∗ (x, 0) − f (x, 0) > 0}) = 0. Thus by Corollary 3.59, µ is maximal. On the other hand, µ is carried by the compact set [0, 1] × {0}, which is disjoint from ChH (K). Example 3.83. There exists a function space H on a compact space K and a measure on K carried by any Baire set containing ChH (K) which is not maximal. Proof. Let H be the function space from Example 3.47. Then ChH (K) = K \ {d∞ } and εd∞ is not a maximal measure. On the other hand, if A ⊃ ChH (K) is a Baire set, then A = K. Indeed, any Baire set in K is Lindelöf (see Theorem A.111(i) and (d)) and hence K \ {d∞ } is not a Baire set. Otherwise, as an open set it would be an Fσ set, which it obviously is not. Hence εd∞ (A) = 1 for each Baire set A containing ChH (K).

3.9

Minimum principles

In the preceding sections, we already met several minimum principles. Recall Bauer’s convex minimum principle in 2.24 or the minimum principle for lower semicontinuous H-concave functions in Theorem 3.16. Definition 3.84 (Minimum principles). Let H be a function space on a compact space K. We say that a function f on K satisfies the minimum principle if f ≥ 0 on K provided f ≥ 0 on ChH (K), and f satisfies the strict minimum principle if f > 0 on K whenever f > 0 on ChH (K). Theorem 3.85 (Minimum principle for S usc (H)). Let f ∈ S usc (H) be an upper semicontinuous H-concave function. Then f satisfies the minimum principle. Proof. Since f is upper semicontinuous, f ≥ 0 on ChH (K). Pick x ∈ K. By the Choquet–Bishop–de Leeuw theorem 3.81, there exists a maximal measure µ in Mx (H) carried by ChH (K). Hence Z Z f dµ ≥ 0. f (x) ≥ f dµ = K

ChH (K)

Theorem 3.86 (Minimum principle for Baire H-concave functions). Let f be a Baire H-concave function. Then f satisfies the minimum principle.

3.9 Minimum principles

85

Proof. Assume that f ≥ 0 on ChH (K) and f (x) < 0 for some x ∈ K. The set F := {y ∈ K : f (y) ≤ f (x)} is a nonempty Baire subset of K, disjoint from ChH (K). Let µ ∈ Mx (H) be a maximal measure. By the Choquet–Bishop–de Leeuw theorem 3.81, µ(F ) = 0. Hence Z Z f (x) dµ(y) = f (x), f dµ > f (x) ≥ µ(f ) = K\F

K\F

which is a contradiction. Proposition 3.87 (Strict minimum principles). Let f be a lower bounded H-concave function on K. If f is lower or upper semicontinuous, or if f is a Baire function on K, then f satisfies the strict minimum principle. Proof. Let f ∈ S lsc (H), f > 0 on ChH (K). By Theorem 3.16, f ≥ 0 on K. If Z := {x ∈ K : f (x) = 0}, then Z is a closed H-extremal set. Since Z ∩ ChH (K) = ∅, it follows from Proposition 3.15 that Z = ∅. Assume now that f ∈ S(H) is a Baire function. The set Z := {x ∈ K : f (x) = 0} is a Baire set disjoint from ChH (K). If Z contains a point x, choose a maximal measure µ ∈ Mx (H). Then Z 0 = f (x) ≥ µ(f ) = f dµ > 0, K\Z

a contradiction. Finally, let f ∈ S usc (H) be upper semicontinuous, and let f (x) = 0 for some x ∈ K. By Proposition 3.48, there exists a sequence {fn } in S c (H) such that fn ≥ fn+1 ≥ f for each n ∈ N and fn (x) → 0. If h := inf {fn : n ∈ N}, then h is a Baire H-concave function such that h > 0 on ChH (K). By the strict minimum principle for H-concave Baire functions which we have just proved, h > 0 on K. However, h(x) = 0, a contradiction that finishes the proof. Proposition 3.88 (Minimum principle for semicontinuous functions). Assume that H is a function space on a compact space K and f , g are bounded functions on K such that f , −g are H-convex, f ≤ g on ChH (K) and f, g are semicontinuous. Then f ≤ g on K. Proof. Assume, for example, that both functions are upper semicontinuous. We fix x ∈ K and ε > 0 and find a continuous H-concave function g 0 on K such that g ≤ g 0 and g 0 (x) ≤ g(x) + ε (see Proposition 3.48). Then g 0 − f ≥ 0 on ChH (K) and thus g 0 − f ≥ 0 on K by Theorem 3.16. Hence f (x) ≤ g 0 (x) ≤ g(x) + ε. Since ε is arbitrary, the proof is complete. The remaining cases can be dealt with by a similar argument.

86

3.10


Orderings and dilations

Proposition 3.89. Let F1 , F2 be closed sets in K such that F2 ⊃ ChH (K) and µ ∈ M1 (F1 ), ν ∈ M1 (F2 ). Let M := {(εx , λ) ∈ M1 (F1 ) × M1 (F2 ) : εx ≺ λ}. Then M is closed and the following assertions are equivalent: (i) µ ≺ ν, (ii) there exists Λ ∈ M1 (M ) such that (µ, ν) is the barycenter of Λ. Proof. For the proof (i) =⇒ (ii), we show that (µ, ν) ∈ co M. Assume that this is not the case. Then we use the Hahn–Banach theorem to find continuous functions f1 ∈ C(F1 ), f2 ∈ C(F2 ) and c ∈ R such that µ(f1 ) + ν(f2 ) > c > εz (f1 ) + λ(f2 ),

(εz , λ) ∈ M.

The assumption µ ≺ ν yields ν(F2f2∗ ) ≤ µ(F2f2∗ ). We fix x ∈ F1 . By Lemma 3.21, sup{εx (f1 ) + λ(f2 ) : λ ∈ M1 (F2 ), εx ≺ λ} = f1 (x) + sup{λ(f2 ) : λ ∈ Mx (H) ∩ M1 (F2 )} = f1 (x) + F2f2∗ (x). Hence

µ(f1 + F2f2∗ ) ≥ µ(f1 ) + ν(F2f2∗ ) ≥ µ(f1 ) + ν(f2 ) > c ≥ sup{εz (f1 ) + λ(f2 ) : (εz , λ) ∈ M } ≥ f1 (x) + F2f2∗ (x).

Thus µ(f1 + F2f2∗ ) > c ≥ sup (f1 (x) + Ff2∗ (x)). x∈F1

Since µ is a probability measure, this is impossible. Hence (µ, ν) ∈ co M , which yields the desired measure Λ ∈ M1 (M ) in view of Proposition 2.39. For the proof of the converse implication, let Λ ∈ M1 (M ) represent (µ, ν). Then for each f ∈ Kc (H), Z Z µ(f ) = εx (f ) dΛ(εx , λ) ≤ λ(f ) dΛ(εx , λ) = ν(f ). M


M

3.10 Orderings and dilations

87

Proposition 3.90. Let P := M1 (K) × M1 (K) and Λ ∈ M1 (P ) satisfy r(Λ) = (µ1 , µ2 ). Then, for any bounded universally measurable functions f1 , f2 on K, the function (λ1 , λ2 ) 7→ λ1 (f1 ) + λ2 (f2 ), (λ1 , λ2 ) ∈ P, is universally measurable on P , and Z µ1 (f1 ) + µ2 (f2 ) = (λ1 (f1 ) + λ2 (f2 )) dΛ(λ1 , λ2 ).

(3.6)

P

Proof. Let us first notice that (3.6) is clear for f1 , f2 ∈ C(K). To extend this formula to universally measurable functions, we will first consider characteristic functions. So, let A be the family of all universally measurable sets A ⊂ K such that (3.6) holds for pairs of functions (cA , 0) and (0, cA ). Step 1. We show that the family A contains all Borel subsets of K. Clearly, K ∈ A. If G ⊂ K is open, Proposition A.50 yields cG = sup{ϕ ∈ C(K) : 0 ≤ ϕ ≤ cG }. By Theorem A.84, ω(G) = sup{ω(ϕ) : ϕ ∈ C(K), 0 ≤ ϕ ≤ cG },

ω ∈ M1 (K).

Thus the functions (λ1 , λ2 ) 7→ λ1 (G),

(λ1 , λ2 ) 7→ λ2 (G),

(λ1 , λ2 ) ∈ P,

are suprema of up-directed families of continuous functions. Hence we may use Theorem A.84 again for the measure Λ to obtain validity of (3.6) for the pairs (cG , 0) and (0, cG ). Hence A contains all open subsets of K. Obviously K ∈ A and K \A ∈ A whenever A ∈ A. Since A is closed with respect to the formation of unions of two pairwise disjoint elements of A, and the Lebesgue monotone convergence theorem implies that the countable union of an increasing sequence of sets from A also belongs to A, we see that A is a Dynkin system. Since A contains all open subsets of K, the Dynkin lemma A.68 yields that A contains any Borel subset of K. Step 2. To show that any universally measurable subset of K belongs to A, let A be such a set. We write A = H ∪ N,

H is an Fσ set,

µ1 (N ) = µ2 (N ) = 0,

H ∩ N = ∅.

Since we know that H ∈ A, to finish the argument we need to prove that the functions (λ1 , λ2 ) 7→ λ1 (N ), equal zero Λ-almost everywhere.

(λ1 , λ2 ) 7→ λ2 (N ),

(λ1 , λ2 ) ∈ P,

88


We find a decreasing sequence {Gn } of open subsets of K such that N ⊂ Gn and (µ1 + µ2 )(Gn ) → 0. For a rational number q ∈ (0, 1) and n ∈ N, let Nq := {(λ1 , λ2 ) ∈ P : λ1 (N ) ≥ q} ∪ {(λ1 , λ2 ) ∈ P : λ2 (N ) ≥ q} and

Nqn :={(λ1 , λ2 ) ∈ P : λ1 (Gn ) ≥ q} ∪ {(λ1 , λ2 ) ∈ P : λ2 (Gn ) ≥ q}.

Since the sets Gn belong to A, for any n ∈ N and q ∈ (0, 1) ∩ Q, we get Z Z 1 n (λ1 (Gn ) + λ2 (Gn )) dΛ(λ1 , λ2 ) 1 dΛ(λ1 , λ2 ) ≤ Λ(Nq ) = q P Nqn 1 = (µ1 + µ2 )(Gn ). q Since Nq ⊂ Nqn , n ∈ N, the choice of the sets Gn yields that Nq is Λ-measurable and Λ(Nq ) = 0. Thus Λ ({(λ1 , λ2 ) ∈ P : λ1 (N ) + λ2 (N ) > 0}) [ = Λ(Nq ) = 0, q∈Q∩(0,1)

which is the desired conclusion. Step 3. Since A contains all universally measurable subsets of K, (3.6) holds for any pair of bounded universally measurable functions on K, by a standard approximation argument. This finishes the proof. Definition 3.91 (µ-dilations and dilations). If µ ∈ M+ (K) is given, a mapping T = {Tx }x∈K : K → M1 (K) is called a µ-dilation, if • x 7→ T (f ), x ∈ K, is µ-measurable for any bounded universally measurable x function f on K, for each h ∈ H, µ({x ∈ K : h(x) 6= Tx (h)}) = 0. If f is universally measurable, we define a function T f by •

T f (x) = Tx (f ). By the conditions above, the formula Z (T µ)(f ) = T f dµ, K

defines a measure T µ ∈ M+ (K).

f ∈ C(K),


89

A mapping T = {Tx }x∈K : K → M1 (K) is termed a dilation, if • x 7→ T (f ), x ∈ K, is Borel for any f ∈ C(K), x Tx ∈ Mx (H), x ∈ K. Hence T is a particular example of a kernel as defined in Subsection A.3.D. As above, we use the notation T f : x 7→ Tx (f ) if f is universally measurable. We can extend the mapping T to a mapping from M(K) to M(K) by the formula Z T f dν, f ∈ C(K), ν ∈ M(K). (T ν)(f ) = •

K

Before stating our next S theorem, we will recall and introduce some notation. Remember that M(H) := x∈K Mx (H) (cf. Proposition 3.37) and r : M(H) → K is the H-barycenter mapping (cf. Definition 3.39). Let P := M1 (K) × M1 (K). We will also consider the barycenter mapping rP : M(P ) → P , here the subscript P serves only for easier distinguishing which barycenter mapping we are just operating with. We denote by πi : K × K → K the i-the projection, i = 1, 2. Then they induce mappings (πi )] : M1 (K × K) → M1 (K), i = 1, 2. Analogously, let pi : P → M1 (K) be the i-th projection, i = 1, 2. As above, we obtain mappings (pi )] : M1 (P ) → M1 (M1 (K)),

i = 1, 2.

If L is a compact set and F ⊂ L is closed, we write µ ∈ M1 (F ) ⊂ M1 (L) if µ ∈ M1 (L) is carried by F and we want to emphasize that we consider µ both as an element of M1 (F ) and M1 (L). Theorem 3.92. Let µ, ν be probability measures on K and M := {(εx , λ) ∈ P : εx ≺ λ, x ∈ K}. Then the following assertions are equivalent: (i) µ ≺ ν, (ii) there exists Ω ∈ M1 (M ) ⊂ M1 (P ) such that (µ, ν) = rP (Ω), (iii) there exists λ ∈ M1 (K × K) such that • •

(π1 )] λ = µ, (π2 )] λ = ν, R R A×K h ◦ π1 dλ = A×K h ◦ π2 dλ, whenever A ⊂ K is universally measurable and h ∈ H,

(iv) there exists a µ-dilation such that T µ = ν, (v) there exists Λ ∈ M1 (M(H)) such that

90

3 Choquet theory of function spaces •

•

R ν(f ) = M(H) λ(f ) dΛ(λ), for any bounded universally measurable function f on K, r] Λ = µ,

P µi ∈ M+ (K), there exist measures νi ∈ M+ (K), (vi) whenever µ = ni=1 µi withP i = 1, . . . , n, such that ν = ni=1 νi and µi − νi ∈ H⊥ . If, moreover, K is metrizable, then (i) is equivalent with the following assertion: (iv’) there exists a dilation T such that T µ = ν. Proof. We already know from Proposition 3.89 that (i) ⇐⇒ (ii). We will prove (i) =⇒ (iii) =⇒ (iv) =⇒ (i), (ii) =⇒ (v) =⇒ (i), (ii) =⇒ (vi) =⇒ (i) and (ii) =⇒ (iv’) =⇒ (i) in the case when K is metrizable. For the proof of (i) =⇒ (iii), assume that µ ≺ ν. For any functions f, g : K → R, we denote by f ⊗ g : K × K → R, f ⊗ g : (x, y) 7→ f (x)g(y),

(x, y) ∈ K × K.

If f is a function on K × K and x ∈ K, we write fx for the function y 7→ f (x, y), y ∈ K. Analogously define f y , y ∈ K. We define a subspace V ⊂ C(K × K) by V := {f ⊗ 1 + 1 ⊗ g : f, g ∈ C(K)} and a linear functional ρ : V → R by ρ : f ⊗ 1 + 1 ⊗ g 7→ µ(f ) + ν(g), We define σ : C(K × K) → R by Z ∗ σ(f ) := (fx )∗ (x) dµ(x),

f, g ∈ C(K).

f ∈ C(K × K).

K

R∗ ( is the upper integral, see Definition A.69). Then σ is a sublinear functional on C(K × K) satisfying σ(f ) ≤ 0 for f ≤ 0. (3.7) Further, ρ ≤ σ on V . Indeed, if f, g ∈ C(K) are given, we have (f ⊗ 1 + 1 ⊗ g)∗x (x) = f (x) + g ∗ (x),

x ∈ K,

and µ(g ∗ ) ≥ ν(g ∗ ) (see Proposition 3.56 and Lemma 3.18). Thus Z ∗ σ(f ⊗1 + 1 ⊗ g) = ((f ⊗ 1 + 1 ⊗ g)x )∗ (x) dµ(x) K Z = (f (x) + g ∗ (x)) dµ(x) ≥ µ(f ) + ν(g ∗ ) K

≥ µ(f ) + ν(g) = ρ(f ⊗ 1 + 1 ⊗ g).


91

By the Hahn–Banach theorem, there exists a functional λ ∈ (C(K × K))∗ such that λ = ρ on V and λ ≤ σ on C(K × K). By (3.7), λ is a Radon measure on K × K. If k ∈ S c (H) and g ∈ C(K) positive are given, then Z ∗ ((g ⊗ k)x )∗ (x) dµ(x) λ(g ⊗ k) ≤ σ(g ⊗ k) = K Z Z g(x)k(x) dµ(x) = λ(gk ⊗ 1). g(x)k ∗ (x) dµ(x) = = K

K

This gives λ(gh ⊗ 1) = λ(g ⊗ h),

g ∈ C(K), h ∈ H.

(3.8)

(If g is not positive, let c ≥ 0 be such that g + c is positive, and apply (3.8) to (g + c) − c.) We claim that λ ∈ M+ (K ×K) possesses all the required properties. First, λ(1) = µ(1) = 1, and hence λ ∈ M1 (K × K). Further, for f ∈ C(K), we have (π1 )] λ(f ) = λ(f ◦ π1 ) = λ(f ⊗ 1) = ρ(f ⊗ 1) = µ(f ), and (π2 )] λ(f ) = λ(f ◦ π2 ) = λ(1 ⊗ f ) = ρ(1 ⊗ f ) = ν(f ). Finally, (3.8) and standard measure-theoretic techniques yield Z Z g(x)h(x) dλ(x, y) = g(x)h(y) dλ(x, y) K×K

K×K

for h ∈ H and a bounded universally measurable function g on K. By setting g = cA , where A ⊂ K is universally measurable, we get Z Z h(x) dλ(x, y) = h(y) dλ(x, y), h ∈ H. A×K

A×K

For the proof of (iii) =⇒ (iv), let λ ∈ M1 (K × K) be as in (iii). By Theorem A.106, there exists a family {Tx }x∈K of Radon probability measures on K such that Z Z Z f dλ = f (x, y) dTx (y) dµ(x) (3.9) K×K

K

K

for every λ-integrable function f on K × K. We claim that the mapping T : x 7→ Tx , x ∈ K, is a µ-dilation such that T µ = ν. To prove this, let f be a bounded universally measurable function on K. Then the function fe := 1 ⊗ f , is λ-measurable, and so (3.9) together with the properties of λ give Z Z Z µ(T f ) = f (y) dTx (y) dµ(x) = fe(x, y) dλ(x, y) K

K

= ((π2 )] λ)(f ) = ν(f ). Hence T f is µ-measurable and T µ = ν.

K×K

92


To finish the proof of the implication, let h ∈ H be given. For any A ⊂ K Borel, we define e h := cA ⊗ h. Then e h is λ-measurable and using (iii) we obtain Z Z Z Z e e T h dµ = h(x, y) dTx (y) dµ(x) = h(x, y) dλ(x, y) A

K

K

K×K

Z

Z

h(x) dλ(x, y)

h(y) dλ(x, y) =

=

A×K

A×K

Z

Z h(x) d((π1 )] λ) =

=

h(x) dµ(x). A

A

This shows that T h = h for µ-almost all x ∈ K. We proceed with the proof of (iv) =⇒ (i). Let T = {Tx }x∈K : K → M1 (K) be a µ-dilation such that T µ = ν. Let f ∈ −W(H) be given; that is, f = h1 ∨ · · · ∨ hn , h1 , . . . , hn ∈ H. We set A0i := {x ∈ K : f (x) = hi (x)}, A1 := A01 , Ai := A0i \ (A1 ∪ · · · ∪ Ai−1 ),

i = 1, . . . , n, i = 2, . . . , n.

(3.10)

Let N ⊂ K be a set of µ-measure zero such that T hi (x) = hi (x) for each x ∈ K \ N and i = 1, . . . , n. Then Z n Z X µ(f ) = f dµ = hi dµ K\N

=

n Z X Ai \N

i=1

i=1

Ai \N

T hi dµ ≤

n Z X i=1

T f dµ

Ai \N

Z ≤

T f dµ = T µ(f ) = ν(f ). K

Hence µ ≺ ν by Proposition 3.56. For the proof of (ii) =⇒ (v), let Ω ∈ M1 (M ) ⊂ M1 (P ) satisfy rP (Ω) = (µ, ν). We claim that the measure Λ := (p2 )] Ω satisfies our requirements. First, we observe that p2 (M ) ⊂ M(H), so that Λ ∈ M1 (M(H)) ⊂ M1 (M1 (K)). Let f be a bounded universally measurable function on K. Then Proposition 3.90 yields Z Z ν(f ) = λ(f ) dΩ(εx , λ) = λ(f ) dΛ(λ) M(H)

M

and

Z

Z

(r] Λ)(f ) =

f (r(λ)) dΛ(λ) = M(H)

f (r(λ)) dΩ(εx , λ) M

Z =

εx (f ) dΩ(εx , λ) = µ(f ). M

Hence r] Λ = µ.


93

To prove (v) =⇒ (i), let Λ ∈ M1 (M(H)) be as in (v). For each f ∈ Kc (H), we have Z Z µ(f ) = r] Λ(f ) = f (r(λ)) dΛ(λ) ≤ λ(f ) dΛ(λ) = ν(f ). M(H)

M(H)

Hence µ ≺ ν. 1 1 For the proof of (ii) Pn=⇒ (vi), let Ω ∈ M (M ) ⊂ M (P ) satisfy rP (Ω) = (µ, ν). Suppose that µ = i=1 µi , where Radon measures µi are nontrivial. According to the Radon–Nikodym theorem, there exist positive Borel measurable functions fi , i = 1,P. . . , n, such that µi (ϕ) = µ(fi ϕ) for each bounded Borel function ϕ on K. ThenP ni=1 fi = 1 µ-almost everywhere. Without loss of generality we may assume that ni=1 fi = 1 on K. For each i ∈ {1, . . . , n}, we define the function gi : M → R by gi : (εx , λ) 7→ εx (fi ),

(εx , λ) ∈ M,

and the measure Ωi ∈ M1 (P ) by Ωi :=

gi Ω . Ω(gi )

Note that, by Proposition 3.90, Z Ω(gi ) = gi dΩ(εx , λ) = µ(fi ) = µi (K) > 0. M

Let (λ1i , λ2i ) ∈ P be the barycenter of Ωi , i = 1, . . . , n, and let ϕ be a bounded Borel function on K. Then Z Z 1 1 λi (ϕ) = εx (ϕ) dΩi (εx , λ) = εx (ϕ)gi (εx , λ) dΩ(εx , λ) Ω(gi ) M M Z 1 1 1 = εx (fi ϕ) dΩ(εx , λ) = µ(fi ϕ) = µi (ϕ). Ω(gi ) M Ω(gi ) Ω(gi ) Thus µi = Ω(gi )λ1i ,

i = 1, . . . , n.

νi := Ω(gi )λ2i ,

i = 1, . . . , n.

We set By Proposition 3.89, λ1i ≺ λ2i , and hence µi − νi ∈ H⊥ ,

i = 1, . . . , n.

94


Finally, for any ϕ ∈ C(K) we get (using (3.6) from Proposition 3.90) n X

n X

νi (ϕ) =

i=1

Z

i=1 n Z X

=

λ(ϕ) dΩi (εx , λ) =

Ω(gi ) M

i=1

gi (εx , λ) λ(ϕ) dΩ(εx , λ)

M

Z εx (fi )λ(ϕ) dΩ(εx , λ) =

M

i=1

n Z X

εx

n X fi λ(ϕ) dΩ(εx , λ)

M

i=1

Z λ(ϕ) dΩ(εx , λ) = ν(ϕ).

= M

P Hence ν = ni=1 νi , as required. To show that (vi) =⇒ (i), let µ, ν satisfy (vi). If f = h1 ∨ · · · ∨ hn with h1 , . . . , hn ∈ H, let the sets Ai be defined as in (3.10). We set µi := µ|Ai ,

i = 1, . . . , n,

and use the assumptionP to find Radon measures νi , i = 1, . . . , n, such that µi − νi ∈ ⊥ H , i = 1, . . . , n, and ni=1 νi = ν. Then µ(f ) =

n Z X i=1

Ai

hi dµ =

n X

µi (hi ) =

i=1

n X

νi (hi ) ≤

i=1

n X

νi (f ) = ν(f ).

i=1

Thus µ ≺ ν, by Proposition 3.56. We proceed by showing (ii) =⇒ (iv’). In this case, we use the Disintegration theorem A.107; this is the point where we use metrizability. Assume that Ω ∈ M1 (M ) and rP (Ω) = (µ, ν). We define Ψ : M → K by Ψ(εx , λ) := x,

(εx , λ) ∈ M.

Then Ψ is a continuous surjection of M onto K. We pick f ∈ C(K) and set ˜ λ) := λ(f ˜ ), ϕ(λ,

˜ λ) ∈ P. (λ,

Then ϕ ∈ Ac (P ) and we have Z Z Z ˜ ) dΩ(λ, ˜ λ) λ(f f (Ψ(εx , λ)) dΩ(εx , λ) = εx (f ) dΩ(εx , λ) = Ψ] Ω(f ) = M M M Z ˜ λ) dΩ(λ, ˜ λ) = ϕ(µ, ν) = µ(f ). = ϕ(λ, M

This shows that µ = Ψ] Ω. According to the Disintegration theorem A.107, there exists a mapping S = {Sx }x∈K : K → M1 (M ) ⊂ M1 (P ) such that •

the mapping x 7→ Sx (ψ), x ∈ K, is Borel for each ψ ∈ C(M ),

3.11 Exercises • •

95

spt Sx ⊂ Ψ−1 (x), x ∈ K, R Ω(ψ) = K Sx (ψ) dµ(x), ψ ∈ C(M ).

We define the required mapping T = {Tx }x∈K : K → M1 (K) as T : x 7→ p2 (rP (Sx )),

x ∈ K.

Since T is a composition of S with continuous mappings, T f is Borel for each f ∈ C(K). Further, given f ∈ C(K), we set ˜ λ) := λ(f ), ψ(λ,

˜ λ) ∈ P. (λ,

(3.11)

Then ψ ∈ Ac (P ) and Z Sx (ψ) =

˜ λ) dSx (λ, ˜ λ) = ψ(rP (Sx )) ψ(λ,

M

x ∈ K.

= p2 (rP (Sx ))(f ) = Tx (f ), Hence Z T µ(f ) =

Z Tx (f ) dµ(x) =

Sx (ψ) dµ(x) = Ω(ψ) = ψ(rP (Ω))

K

K

= ψ(µ, ν) = ν(f ). This means T µ = ν. For each x ∈ K we have spt Sx ⊂ Ψ−1 (x) = {εx } × Mx (H). Since {εx } × Mx (H) is closed and convex, it follows rP (Sx ) ∈ {εx } × Mx (H) and

Tx = p2 (rP (Sx )) ∈ Mx (H).

To show that (iv’) =⇒ (i), let T = {Tx }x∈K be a dilation such that T µ = ν. Then, for f ∈ Kc (H), Z Z ν(f ) = (T µ)(f ) = Tx f dµ(x) ≥ f (x) dµ(x) K

K

yields µ ≺ ν. This concludes the proof.

3.11

Exercises

Exercise 3.93 (Bauer functionals). For a measure µ ∈ M+ (K) and an upper bounded real-valued function f on K denote B µ (f ) := inf {µ(h) : h ∈ H, h ≥ f } .

96


The mapping f 7→ B µ (f ) is called a Bauer functional on `∞ (K). Of course, µ(f ) ≤ B µ (f ) and B εx (f ) = f ∗ (x) for x ∈ K. Given µ ∈ M+ (K), we denote n o Mµ (H) := ν ∈ M+ (K) : ν − µ ∈ H⊥ . Prove that Mµ (H) is a convex compact subset of M+ (K) and B µ (f ) = sup {ν(f ) : ν ∈ Mµ (H)} for any upper semicontinuous function f on K. The supremum is attained. Hint. The proof is almost the same as the proof of Lemma 3.21. Exercise 3.94. Given a measure µ ∈ M+ (K), we defined in Definition 3.17 the functional Qµ := inf{µ(k) : k ∈ S c (H), k ≥ f on K}. By Lemma 3.21, Qµ (f ) = sup ν(f ) : ν ∈ M+ (K), µ ≺ ν , where the supremum is attained. Prove that, in general, B µ 6= Qµ . Hint. Let H be the space of all affine continuous functions on the closed interval [−1, 1]. Then S c (H) equals the set of all continuous concave functions on [−1, 1]. Show that for 1 f (x) := −|x| + 1 and µ := (ε−1 + ε1 ) 2 we get Qµ (f ) = 0 and B µ (f ) = 1.

Exercise 3.95. Let H be a function space on K and x ∈ K. Prove that Mx (H) = Mx (Ac (H)) and ChH (K) = ChAc (H) (K). Exercise 3.96. Prove that, for any upper bounded function f on K, the inequalities “≥” in the families of functions in Proposition 3.25(a) may be replaced by “>”. Hint. If F is a family of functions and f = inf{h ∈ F : h ≥ f }, then f = inf{h + ε : h ∈ F, ε > 0}. Now use Proposition 3.25.

3.11 Exercises

97

Exercise 3.97. If f = g on ChH (K), then ChH (K)f ∗ = ChH (K)g ∗ . Hint. Obviously, h ≥ f on ChH (K) if and only h ≥ g on ChH (K). Exercise 3.98. Prove that kFf ∗ − Fg ∗ k ≤ kf − gk`∞ (F ) for any set F ⊂ K containing ChH (K) and f, g ∈ `∞ (F ). Hint. Pick x ∈ F . Then Ff ∗ ≤ kf k since kf k ∈ H and f ≤ kf k. Hence F ∗

f (x) = F((f − g) + g)∗ (x) ≤ F(f − g)∗ (x) + Fg ∗ (x).

Thus F ∗

f (x) − Fg ∗ (x) ≤ F(f − g)∗ (x) ≤ kf − gk,

and the assertion easily follows. Exercise 3.99. Let {fα }α∈A be a down-directed net of upper semicontinuous functions on K, fα & f . Prove that fα∗ & f ∗ . Hint. It is clear that {fα∗ }α∈A is a down-directed net of functions satisfying f ∗ ≤ fα∗ for every α ∈ A. Since by Exercise 3.96 f ∗ = inf {h ∈ H : h > f } , it is enough to verify that for any h > f , h ∈ H, there exists α ∈ A such that fα ≤ h. Since fα & f , we can use the upper semicontinuity of the functions {fα } and the compactness of K to find α1 , . . . , αn ∈ A such that fα1 ∧ · · · ∧ fαn < h. Then fα∗ < h for every α ∈ A satisfying α ≥ αi , i = 1, . . . , n. Exercise 3.100. Let {xα } be a net of points converging to x ∈ ChH (K). Let µα ∈ Mxα (H). Then µα → εx . Hint. If we suppose the contrary, then there exist a measure µ 6= εx and a subnet {µβ } so that µβ → µ. It is straightforward to verify that µ is an H-representing measure for x. Since µ is not the Dirac measure at x, we have arrived at a contradiction with the assumption that x ∈ ChH (K). Exercise 3.101. Prove that a closed set F ⊂ K is H-extremal if and only if cK\F ∈ S lsc (H). Exercise 3.102. Let X be a compact convex subset of a locally convex space and B b (X) be the family of all bounded Borel functions on X.

98


(a) If f is upper semicontinuous on X, then there exists a bounded concave upper semicontinuous function g on X such that f = g on ext X. (b) Let f be a bounded concave upper semicontinuous functions on X. Prove that the set {x ∈ ext X : f (x) > f∗ (x)} is meager (in the relative topology of ext X). (c) Let g be a Borel function on X with 0 ≤ g ≤ 1 and f an upper semicontinuous function on X such that {x ∈ ext X : f (x) − g(x) 6= 0} is meager in ext X. Find an upper semicontinuous concave function h with 0 ≤ h ≤ 1 such that {x ∈ ext X : g(x) − h(x) 6= 0} is meager in ext X. (d) Let g ∈ B b (X). Prove that there exists a bounded concave upper semicontinuous function f on X such that the set {x ∈ ext X : g(x) 6= f (x)} is meager. Hint. For the proof of (a) use Theorem 3.24. To verify (b), for any n ∈ N set Mn := {x ∈ ext X : f (x) − f∗ (x) ≥ 1/n}. Each set Mn is closed, so it suffices to show that Int Mn = ∅. Suppose a contrary. By Proposition 2.41, there exists a ∈ Ac (X), a ≤ 1/n, such that ∅ 6= {x ∈ ext X : a(x) > 0} ⊂ Mn . Show that the compact convex set F := {x ∈ X : f (x) − f∗ (x) − a(x) ≥ 0} contains ext X. Hence, by the Krein–Milman theorem 2.22, F = X and f ≥ f∗ + a on X. Then f∗ ≥ f∗ + a and it follows that a ≤ 0 on X, which is an obvious contradiction. To verify (c), let f and g be as in (c). Then h = (0 ∨ (f ∧ 1))∗ is an upper semicontinuous concave function with 0 ≤ h ≤ 1 that satisfies h = 0 ∨ (f ∧ 1) on ext X. Hence {x ∈ ext X : g(x) − h(x) 6= 0} is meager in ext X. Finally show (d). Let V be the family of all g ∈ B b (X) for which there exists a bounded concave upper semicontinuous function f on X such that the set {x ∈ ext X : g(x) 6= f (x)} is meager in ext X. Obviously, V is closed under addition and multiplication by positive scalars. It follows from (b) that −g ∈ V whenever g ∈ V. If {gn } is a bounded decreasing sequence in V and fn , n ∈ N, are bounded upper semicontinuous functions witnessing it, by (c) we can assume that all fn are bounded by the same constants. Then the function infn∈N fn shows that limn→∞ gn is contained in V. Hence V is closed with respect to taking monotone limits of bounded sequences. By (a), V contains bounded upper semicontinuous functions and thus V contains all bounded Borel functions by Proposition A.46. Exercise 3.103. Let H be the linear span of 1, id, id3 on [−1, 1]. Prove that Kor(H) 6= C([−1, 1]). Find a continuous function on [−1, 1] which does not belong to Kor(H). Hint. Show that ChH ([−1, 1]) = [−1, − 21 ] ∪ [ 12 , 1]. Exercise 3.104. Let H be the linear span of {1, id, ϕ} on an interval [a, b] , where ϕ is a continuous function on [a, b]. Prove that Kor(H) = C([a, b]) if and only if ϕ is either strictly convex or strictly concave on [a, b].

3.11 Exercises

99

Hint. If ϕ is strictly convex and a ∈ [0, 1], then the function ϕ − h exposes a for each affine function h supporting ϕ at a. For the converse, set G = {[a, b, c] ∈ [0, 1]3 : a < c < b}, Φ(a, b, c) = (b − c)ϕ(a) + (c − a)ϕ(b) − (b − a)ϕ(c). The set G is convex, thus connected, and ϕ is continuous on G. If ϕ is neither strictly convex, nor strictly concave, then Φ is neither strictly positive, nor strictly negative. It follows that there exist a, b, c ∈ [0, 1] such that a < c < b and Φ(a, b, c) = 0. If we b−c , then λ ∈ (0, 1), c = λa + (1 − λ)b and ϕ(c) = λϕ(a) + (1 − λ)ϕ(b). denote λ = b−a Then h(c) = λh(a)+(1−λ)h(b) for each h ∈ H and thus the measure λεa +(1−λ)εb represents c. Exercise 3.105 (Fejér’s theorem). Let f be a continuous 2π-periodic function on R and let sk (f ) be the k-th partial sum of the Fourier series of f on R. Prove the following Fejér’s theorem: If Tn f :=

s0 (f ) + s1 (f ) + · · · + sn−1 (f ) , n

then the sequence Tn f converges uniformly to f on R. Hint. Since the operators Tn are obviously linear and positive in view of 1 Tn f (x) = 2πn

Z

2π

f (x + t)

0

sin( nt 2 ) sin( 2t )

2 dt,

x ∈ [0, 2π],

it suffices to use the second Korovkin theorem 3.34 to show that the sequence {Tn } is H-admissible if H equals the linear span of {1, cos, sin}. Exercise 3.106. If Z H := f ∈ C([0, 1]) : f (0) =

1

f

0

and λ is Lebesgue measure on [0, 1], prove that M(H) = co {ε0 , λ} ∪ {εx : x ∈ (0, 1]} 6= M1 ([0, 1]). Exercise 3.107. Let X be a compact convex metrizable subset of a locally convex space. Then there exists a continuous strictly convex function on X. Hint. According to Proposition 2.45, we may assume that X is a norm compact convex subset of `2 . Then x 7→ kxk2 , x ∈ X, is the required strictly convex function.

100


Exercise 3.108 (Rosenthal’s proof). In the following, supply the details of Rosenthal’s proof of the Choquet representation theorem in the setting of metrizable compact convex sets. More precisely, prove that given a metrizable compact convex subset X of a locally convex space E and x ∈ X, then there exists a probability measure µ ∈ Mx (X) such that µ is carried by extreme points of X. Hint. Exercise 3.107 provides a strictly convex function Q on X. For n ∈ N, define 1 1 . Fn := x ∈ X : there are y, z ∈ X, 2x = y + z, (Q(y) + Q(z)) ≥ Q(x) + 2 n Obviously, each Fn is closed. First, prove the following assertion. Claim. Let ρ ∈ M1 (Fn ). Then there exists a measure σ ∈ M1 (X) such that Z Z Z Z 1 Q dρ + . Q dσ ≥ f dρ, f ∈ E ∗ , and f dσ = n Fn X X Fn Proof of the claim. First, the claim holds in the case when ρ = εx for some x ∈ Fn . It suffices to put σ = 21 (εy + εz ), where y, z are as in the definition of the set Fn . The assertion then easily follows for any molecular measure. In general, use approximation by molecular measures (cf. Proposition 2.27). Proof of the Choquet theorem. Using the compactness of M1 (X) (cf. again Proposition 2.27) show that there exists a representing measure µ ∈ Mx (X) such that Z nZ o Q dµ = sup Q dν : ν ∈ Mx (X) =: m. X

K

Then assume that µ(X \ ext X) > 0. Since X \ ext X =

∞ [

Fn ,

n=1

find n ∈ N so that r := µ(Fn ) > 0. By the claim, there exists a measure σ for ρ := r1 µ|Fn . Denote κ := rσ + µ|X\Fn and show that κ ∈ Mx (X). Then deduce that Z Z r r m≥ Q dκ ≥ Q dµ + = m + , n n X X which is an obvious contradiction. Exercise 3.109. Let H be a function space on a compact space K, g an upper semicontinuous function on K, f ∈ S lsc (H), g ≤ f on K and ε > 0. Find a function k ∈ S c (H) such that g ≤ k ≤ f + ε on K.

3.11 Exercises

101

Hint. Look at the set F := {f − k + ε : k ∈ S c (H), k ≥ g} . Then F is a nonempty convex set of lower semicontinuous functions on K. The proof will be completed by showing that F contains a positive element. Let us see why this is so. Take any nonzero positive Radon measure µ on K. By Corollary 3.25, g ∗ ≤ f on K. Appeal to Theorem A.84 to find a function k ∈ S c (H) such that k ≥ g and µ(k) < µ(g ∗ ) + εµ(K). Then µ(k) < µ(g ∗ ) + εµ(K) ≤ µ(f ) + εµ(K) = µ(f + ε). Hence µ(f − k + ε) > 0, and Lemma A.88 concludes the proof. Exercise 3.110. If g is an upper semicontinuous function on K, f ∈ S lsc (H), and g < f on K, then there is a function k ∈ S c (H) such that g < k < f on K. Hint. By compactness, f − g > ε for some ε > 0. Now use Exercise 3.109. Exercise 3.111. Prove that if f < g is replaced by f ≤ g in the hypothesis of Corollary 3.49, then a function k ∈ S c (H) satisfying f ≤ k ≤ g on K need not exist. 1 . Let K be a (metrizable) compact space on R Hint. For n ≥ 2, denote rn = 2n consisting of points 0, 1, rn , 1 − rn , 1 + rn , n ≥ 2,

and let H be the space of all continuous functions f on K satisfying f (1 − rn ) =

1 1 f (rn ) + (1 − )f (1 + rn ), n n

n ≥ 2.

Since the function f (x) = x belongs to H, the vector space H separates points of K. Thus H is a function space on K. Since, for any n ≥ 2, the function   x = rn , 1, 1 f (x) := − n−1 , x = 1 + rn ,   0 elsewhere, exposes the points rn and 1 + rn , these points belong to the Choquet boundary ChH (K). Also the points 0 and 1 are exposed, as the functions f (x) := x and   x ∈ [0, 21 ] ∩ K, 1, f (x) := 3n−1 , x = 1 − rn , n ≥ 2, 2n2   x − 1, x ∈ [1, 2] ∩ K, are exposing 0 and 1, respectively.

102


Any measure from M1−rn (H) is carried by the set {rn , 1 − rn , 1 + rn }. Hence the function ( 1, x = 0, f := 0 elsewhere is H-concave. It is clear that f is also upper semicontinuous. Let g be a lower semicontinuous function defined as ( 1, on K ∩ [0, 12 ], g := 0 elsewhere. Then it is impossible to insert a continuous H-concave function between f and g. Indeed, if h were such a function, then h(rn ) > 0 for some rn , by the continuity of h at the point 0. But then 0 = h(1 − rn ) ≥

1 1 1 h(rn ) + (1 − )h(1 + rn ) ≥ h(rn ) > 0. n n n

This obvious contradiction proves the result. Exercise 3.112. Prove that Mmax (H) is a norm closed cone in M(K). Hint. Use Mokobodzki’s maximality test 3.58. Exercise 3.113. Prove that the set of all probability maximal measures is a norm closed face of M1 (K). Hint. Use Mokobodzki’s maximality test 3.58. Exercise 3.114. Let {fn } be a bounded sequence of continuous functions on [0, 1] R1 such that 0 ≤ fn ≤ 1, n ∈ N, and fn → 0. Prove that 0 fn → 0 without recourse to Lebesgue integration theory. R1 Hint. Assume that there exists δ > 0 such that 0 fn ≥ 2δ for any n ∈ N. By Lemma 3.74, there exist λn ≥ 0, n ∈ N, such that ∞ X

λn = 1

∞ X

and

n=1

Let k ∈ N be such that Z δ≥

λn fn ≤ δ on [0, 1].

n=1

Pk

n=1 λn

∞ 1X

> 12 . Then Z

λn fn ≥

0 n=1

This contradiction finishes the proof.

k 1X

0 n=1

λn fn ≥ 2δ

k X n=1

λn > δ.

3.11 Exercises

103

Exercise 3.115. Let {fn } be a bounded sequence of continuous functions on a compact space K that P pointwise converges toP0. Then, for every ε > 0, there exist λi > 0, i = 1, . . . , n, ni=1 λi = 1, such that k ni=1 λi fi k < ε. Prove this without recourse to Mazur’s theorem A.2. Hint. Use the Simons lemma 3.74. Exercise 3.116. Let {xn } be a sequence in a Banach space E that converges weakly P to x ∈ E. P Then for every ε > 0 there exist λi > 0, i = 1, . . . , n, ni=1 λi = 1, such that k ni=1 λi xi − xk < ε. Prove this without the recourse to Mazur’s lemma (cf. W. Rudin [403], Theorem 3.13). Hint. By the uniform boundedness principle (see Theorem 2.5 in W. Rudin [403]), the sequence {xn } is bounded. If we view elements of E as continuous functions on the dual unit ball BE ∗ endowed with the w∗ -topology, then the assertion follows from Exercise 3.115. Exercise 3.117 (Bishop–de Leeuw preordering). Let H be a function space on a compact space K. Given µ, ν ∈ M+ (K), we define the Bishop–de Leeuw preordering l by declaring that µ l ν if µ(h2 ) ≤ ν(h2 ) and

µ(h) = ν(h)

for any h ∈ H. A measure µ ∈ M+ (K) is called l-maximal if given ν ∈ M+ (K), µ l ν, then µ(h2 ) = ν(h2 ) for any h ∈ H. Prove the following assertions hold. (a) The relation l is reflexive and transitive. (b) If µ ≺ ν, then µ l ν. (c) For any measure µ ∈ M+ (K) there exists a l-maximal measure λ ∈ M+ (K) such that µ l λ. (d) Any l-maximal measure is ≺-maximal. Hint. The verification of reflexivity and transitivity in assertion (a) is straightforward. For the proof of (b), we just note that h2 ∈ Kc (H) for any h ∈ H, and assertion (c) can be proved by an analogous argument to that in Theorem 3.65. For the proof of (d), let µ be a l-maximal measure. According to Theorem 3.58, it is enough to show that µ(f ) = µ(f ∗ ) for any f ∈ −W(H). Let ε > 0 and f ∈ −W(H) be given; that is, f = h1 ∨ · · · ∨ hn ,

h1 , . . . , hn ∈ H.

Let H := {x ∈ K : f ∗ (x) − f (x) ≥ ε}

104


and, for each m ∈ N and i = 1, . . . , n, 1 + h2i (x) . Lm,i := x ∈ K : there exists ν ∈ Mx (H) with ν(h2i ) ≥ m Then each Lm,i is a closed set and H⊂

∞ [ n [

Lm,i .

m=1 i=1

Sn S Indeed, let x ∈ H be given. Suppose that x is not contained in ∞ m=1 i=1 Lm,i . By the Key lemma 3.21, there exists a measure λx ∈ Mx (H) such that λx (f ) = f ∗ (x). Since x ∈ K \ Lm,i , we get, for each i = 1, . . . , n, Z 2 Z 2 2 2 λx (hi ) ≤ hi (x) = hi (t) dλx (t) ≤ |hi (t)| dλx (t) ≤ λx (h2i ). K

K

Hence hi = hi (x) λx -almost everywhere. Thus the set A :=

n \

{y ∈ K : hi (y) = hi (x)}

i=1

has λx -measure 1. We define the sets Ai , i = 1, . . . , n, as in (3.10). Then Z n Z X ∗ f (x) = λx (f ) = f (t) dλx (t) ≤ hi (t) dλx (t) A

i=1

A∩Ai

Z ≤

f (x) dλx (t) = f (x). A

S Sn This contradicts our assumption that f ∗ (x) − f (x) ≥ ε. Thus x ∈ ∞ m=1 i=1 Lm,i . To conclude the proof, it is enough to show that µ(Lm,i ) = 0 for every m ∈ N and i = 1, . . . , n. Suppose that this is not the case for some m ∈ N and some i = 1, . . . , n. For the measure σ := µ|Lm,i find a net of molecular measures {σα }α∈A carried by Lm,i , converging to σ and satisfying σα (1) = σ(1). According to the definition of Lm,i , there exist Radon measures {κα }α∈A such that σα l κα

and

κα (h2i ) ≥

σ(K) + σα (h2i ). m

Without loss of generality we may assume that κα → κ. Then σ l κ and κ(h2i ) ≥ σ(K) 2 m + σ(hi ). Hence we get µ l (µ − σ) + κ and µ 6= (µ − σ) + κ, which is a contradiction with the l-maximality of µ. Thus µ(H) = 0. Since ε > 0 is arbitrary, we get ! ∞ [ 1 ∗ ∗ = 0. µ ({x ∈ K : f (x) − f (x) > 0}) = µ x ∈ K : f (x) − f (x) ≥ n n=1

Hence µ(f ) =

µ(f ∗ )

and µ is ≺-maximal. This concludes the proof.


105

Exercise 3.118. Prove that the preordering l need not be antisymmetric. Hint. Let H be the space of all affine functions on K := [0, 1]. Choose a nonzero measure µ ∈ M([0, 1]) satisfying µ(h) = 0 for any h ∈ {1, id, id2 } and decompose µ into its positive and negative part µ+ and µ− . Then µ+ l µ− and µ− l µ+ but µ+ 6= µ− . Thus l need not be antisymmetric. Exercise 3.119. A ≺-maximal measure µ need not be l-maximal. Hint. Let X ⊂ R2 be defined by 1 1 1 1 X := (0, 0), (0, 1), (1, 0), (1, 1), ( , −δ), ( , 1 + δ), (1 + δ, ), (−δ, ) , 2 2 2 2 where 0 < δ < 81 . Set K := co X = co X and let H consist of all affine functions on K. Then ext K = ChH (K) = X. Set µ :=

1 ε(0,0) + ε(1,0) + ε(1,1) + ε(0,1) 4

and

1 ε( 1 ,−δ) + ε(1+δ, 1 ) + ε( 1 ,1+δ) + ε(−δ, 1 ) . 2 2 2 2 4 Then r(µ) = r(ν), ν is ≺-maximal, νlµ, but there exists h ∈ H with ν(h2 ) < µ(h2 ). Obviously, ν is ≺-maximal because ν is carried by extreme points of K. The verification of ν l µ can be done by a straightforward computation. Finally, if we define h(x, y) := x + y, (x, y) ∈ K, ν :=

then ν(h2 )
0 and f1 , . . . , fn ∈ C(X), that is, it consists of all Radon measures ν for which |µ(fi ) − ν(fi )| ≤ ε,

i = 1, . . . , n.

By the continuity of the functions fi , there exists a closed convex neighborhood W of 0 in E such that |fi (y) − fi (y 0 )| ≤ ε, (4.1) whenever i = 1, . . . , n, y, y 0 ∈ X and y − y 0 ∈ W . By compactness, there exist points x01 , . . . , x0m ∈ X such that X⊂

m [

(x0j + W ).

j=1

Denote W0 := ∅, Wj :=

x0j

+ W for j = 1, . . . , m and set

Aj := X ∩ (Wj \ (W1 ∪ · · · ∪ Wj−1 )),

j = 1, . . . , m.

Let J be the set of indices for which µ(Aj ) 6= 0. Let λj := µ(Aj ) for j = 1, . . . , m and j ∈ J. µj := λ−1 j µ|Aj , Clearly, P µj are probability Radon measures, spt µj ⊂ Wj ∩ X for j ∈ J and µ = j∈J λj µj . Let xj be the barycenter of the measure µj , and observe that xj ∈ X ∩ Wj , since µj is carried by the compact convex set X ∩ Wj for every j ∈ J. If j ∈ J and z ∈ Wj ∩ X, then z − xj = (z − x0j ) + (x0j − xj ), and so by (4.1) |fi (z) − fi (xj )| ≤ 2ε,

i = 1, . . . , n.

Since µj is a probability measure carried by X ∩ Wj , we obtain |µj (fi ) − fi (xj )| ≤ ε,

i = 1, . . . , n, j ∈ J.

(4.2)

P If ν := j∈J λj εxj , then ν is a molecular probability measure with the same barycenter as µ. By (4.2), X Z X fi dµ − λj fi (xj ) ≤ λj |µj (fi ) − fi (xj )| |µ(fi ) − ν(fi )| = j∈J

≤ 2ε, This completes the proof.

Aj

i = 1, . . . , n.

j∈J

110

4 Affine functions on compact convex sets

Definition 4.4 (Barycentric formula). A function f : X → R satisfies the barycentric formula if f is universally measurable and, given x ∈ X and µ ∈ Mx (X), µ(f ) exists and f (x) = µ(f ). Sometimes we label functions satisfying the barycentric formula as strongly affine functions. We denote by Abar (X) the space of all functions satisfying the barycentric formula. Lemma 4.5 yields that any strongly affine function is bounded. Thus, a universally measurable function f satisfies the barycentric formula if and only if f is Ac (X)affine. Lemma 4.5. Any strongly affine function on X is bounded. Proof. Let f be a strongly affine function on X. Assume that f is not bounded from above. Then we can find points xn ∈ X such that f (xn ) ≥ 22n forP any n ∈ N. Then P∞ 1 the measure µ := n=1 21n εxn belongs to Mr(µ) (X), and µ(f ) = ∞ n=1 2n f (xn ) = ∞. This contradicts the requirement that µ(f ) = f (r(µ)). Definition 4.6 (Superbarycentric formula). We say that a universally measurable function f on X satisfies the superbarycentric formula if µ(f ) exists for every µ ∈ Mx (X), x ∈ X, and f (x) ≥ µ(f ). As above, a function satisfies the superbarycentric formula if and only if it is Ac (X)-concave. A universally measurable function f on X satisfies the subbarycentric formula if −f satisfies the superbarycentric formula. Proposition 4.7. Let f be a semicontinuous concave function on X. Then f satisfies the superbarycentric formula. Proof. Let x ∈ X and µ ∈ Mx (X). Assume first that f is lower semicontinuous and pick c < µ(f ). Let g ∈ C(X) be such that g ≤ f and c < µ(g). By Proposition 4.3, there exists a molecular measure µ0 ∈ Mx (X) such that µ0 (g) > c. Then c < µ0 (g) ≤ µ0 (f ) ≤ f (x). Since c is arbitrary, µ(f ) ≤ f (x). Assume now that f is a real-valued upper semicontinuous function. Again choose ε > 0. There exists a function h ∈ Ac (X) such that h ≥ f and h(x) < f (x) + ε. Indeed, the closed convex set F := {(t, y) ∈ X × R : y ≤ f (t)} can be separated by the Hahn–Banach theorem from the point (x, f (x) + ε). Then µ(f ) ≤ µ(h) = h(x) < f (x) + ε, and the conclusion follows. An upper semicontinuous function f may attain the infinite value −∞, and then the reasoning above is not correct; namely, the set F need not be well defined. Thus we must use a more subtle argument. We will prove the following claim.

4.1 Affine functions and the barycentric formula

111

Claim. Let x ∈ X and c > f (x) be given. Then there exists n ∈ N such that c > sup {t ∈ R : (x, t) ∈ co (Fn ∪ Hn )} , where Fn := {(y, t) ∈ X × R : −n ≤ t ≤ f (y)}

and

Hn := X × {−n}.

Proof of the claim. Note that both sets Fn and Hn are compact convex sets in E × R, and hence co (Fn ∪ Hn ) = λu + (1 − λ)v : u ∈ Fn , v ∈ Hn , λ ∈ [0, 1] . Suppose that the claim does not hold. Then, for any n ∈ N, there exists tn ∈ R such that c ≤ tn and (x, tn ) ∈ co (Fn ∪ Hn ) . Hence there exist yn , zn ∈ X, λn ∈ [0, 1], sn ∈ R, satisfying (x, tn ) = (λn yn + (1 − λn )zn , λn sn + (1 − λn )(−n))

and

(yn , sn ) ∈ Fn .

Using compactness we find subnets {yni }, {zni } and {λni } of {yn }, {zn } and {λn }, respectively, and points y, z ∈ X and λ ∈ [0, 1] such that yni → y, zni → z and λni → λ. Then x = λy + (1 − λ)z and, for each index i, f (x) < c ≤ tni = λni sni + (1 − λni )(−ni ) ≤ λni f (yni ) + (1 − λni )(−ni ). (4.3) If λ = 1, we get from (4.3) f (x) < c ≤ f (x). It does not matter whether f (x) is finite or infinite; in both cases we obtain a contradiction. If λ < 1, we get c ≤ −∞, which again is a contradiction with the choice of c. The proof of the claim is thus completed. Now, the proof of the proposition can be finished as in the case of a real-valued upper semicontinuous functions. Let x ∈ X, µ ∈ Mx (X) and c > f (x) be given. Find an integer n ∈ N with the property guaranteed by the claim. Then the point (x, c) can be separated from the set co (Fn ∪ Hn ); in other words, there exists a continuous affine function h on X such that f ≤ h and h(x) < c. Then µ(f ) ≤ µ(h) = h(x) < c, and the proof is finished. Proposition 4.7 asserts that a continuous (or semicontinuous) concave function f on X is Ac (X)-concave, since f satisfies the superbarycentric formula if and only if f ∈ S(Ac (X)). Similarly, as in Proposition 4.2, we can show that any Ac (X)concave function is concave. Moreover we get that any semicontinuous affine function f on X is Ac (X)-affine, because both f and −f are Ac (X)-concave. We bring together these observations in the following corollary.

112


Corollary 4.8. Let X be a compact convex set. Then Slsc (X) = S lsc (Ac (X)),

Susc (X) = S usc (Ac (X)),

Ausc (X) = Ausc (Ac (X)),

Alsc (X) = Alsc (Ac (X)),

and

Sc (X) = S c (Ac (X)).

Corollary 4.8 allows us to apply the general approximation results of Propositions 3.48 and 3.54 to get the following assertions. Proposition 4.9. Let f be an upper semicontinuous concave function on X. If K := {g ∈ Sc (X) : g ≥ f on X}

and

L := {g ∈ Sc (X) : g > f on X} ,

then K and L are down-directed families and f = inf K = inf L. Proposition 4.10. Let f be a lower semicontinuous concave function on K. If S := {g ∈ Sc (X) : g < f on X}

and

T := {g ∈ Sc (X) : g ≤ f on X} ,

then S and T are up-directed families and f = sup S = sup T. Lemma 4.11. Let f , −g be upper semicontinuous functions on X with f < g. Then there exists a function h ∈ (E ∗ +R)|X such that f < h < g if and only if µ(f ) < ν(g) for any µ, ν ∈ M1 (X) satisfying r(µ) = r(ν). Proof. The necessity of the condition is obvious. Concerning the sufficiency, let f, −g be upper semicontinuous functions on X such that µ(f ) < ν(g) for any µ, ν ∈ M1 (X) satisfying r(µ) = r(ν). We claim that f ∗ < g∗ . Indeed, pick x ∈ X. By Lemma 3.21, there exist measures µ, ν ∈ Mx (X) such that f ∗ (x) = µ(f )

and

g∗ (x) = ν(f ).

By the assumption, f ∗ (x) < g∗ (x). By Lemma 3.18(a), f ∗ , −g∗ are upper semicontinuous Ac (X)-concave functions. Using Corollary 3.49, there exist continuous Ac (X)-concave functions f1 , −g1 such that f ∗ < f1 < g1 < g∗ . Let m, M ∈ R be numbers satisfying m ≤ f1 < g1 ≤ M on X. Consider the following compact convex sets in E × R: J1 := {(x, t) ∈ E × R : m ≤ t ≤ f1 (x)} and J2 := {(x, t) ∈ E × R : g1 (x) ≤ t ≤ M }.

4.2 Barycentric theorem and strongly affine functions

113

Since J1 , J2 is a pair of disjoint sets, there exist a functional F ∈ (E × R)∗ and λ ∈ R such that sup F (J1 ) < λ < inf F (J2 ). As in the proof of Lemma 2.34, there exist ϕ ∈ E ∗ and β ∈ R so that F (x, t) = ϕ(x) + βt for any x ∈ E and t ∈ R. Then h : x 7→

1 (λ − ϕ(x)), β

x ∈ X,

fulfils f < h < g, as needed. Proposition 4.12. Let f be a lower semicontinuous affine function on X and A := {g ∈ Ac (X) : g < f on X} . Then A is up-directed and f = sup A. Proof. Let h be a continuous function on X, h < f , and let µ, ν ∈ M1 (X) with r(µ) = r(ν) be given. Then µ(h) < µ(f ) = f (r(µ)) = f (r(ν)) = ν(f ), because f is Ac (X)-affine, by Corollary 4.8. An appeal to Lemma 4.11 provides a continuous affine function g with h < g < f . Since f = sup{h ∈ C(X) : h < f }, we get that f = sup A. It remains to prove that A is up-directed. If g1 , g2 ∈ A are given, according to the previous considerations, there exists g ∈ A such that g1 ∨ g2 < g < f . This observation concludes the proof.

4.2

Barycentric theorem and strongly affine functions

If f : X → R is a function and V ⊂ X, we recall that oscV f (x) := inf{diam f (U ∩ V ) : U is a neighborhood of x},

x ∈ V,

is the oscillation of f on V at x (see Definition A.119). Definition 4.13 (The Haydon condition). Let f be a bounded affine function on a compact convex set X. We say that f satisfies the Haydon condition if for any Radon measure µ on X and for any pair of real numbers a < b, there exists a compact convex set L such that µ(L) > 0 and L is contained either in {x ∈ X : f (x) < b} or in {x ∈ X : f (x) > a}. Later on, we will see that the Haydon condition is strongly related to the following notion.

114


Definition 4.14 (Convex inner measure). For a Radon measure µ on X, let the convex inner measure µ for A ⊂ X be defined as µ(A) := sup {µ(L) : L ⊂ A, L is compact convex} . We note that the convex inner measure µ need not be a measure at all. Lemma 4.15. Let µ ∈ M1 (X) and {An } be a decreasing sequence of convex subsets of X. Then ∞ \ µ An = lim µ(An ). n=1

n→∞

Proof. Since µ(A) ≤ µ(B) whenever A ⊂ B, we obtain that limn→∞ µ(An ) exists and ∞ \ µ( An ) ≤ lim µ(An ). n=1

n→∞

For the converse, fix ε > 0. By induction, we will construct a decreasing sequence of compact convex sets Kn ⊂ An such that µ(Kn ) > µ(An ) − ε,

n ∈ N.

First find a compact convex set K1 ⊂ A1 such that µ(K1 ) > µ(A1 ) − 2ε . If the sets K1 , . . . , Kn−1 satisfying the required condition have been constructed, select a real number δ such that 0 < δ < ε and µ(Kn−1 ) > µ(An−1 ) − ε + δ. There exists a compact convex set L ⊂ An such that µ(L) > µ(An ) − δ. If we set Kn := L ∩ Kn−1 , we obtain the desired compact convex set. Indeed, since µ(L \ Kn−1 ) = µ(L ∪ Kn−1 ) − µ(Kn−1 ) ≤ µ(co(L ∪ Kn−1 )) − µ(Kn−1 ) ≤ µ(An−1 ) − µ(Kn−1 ) < ε − δ, we get µ(Kn ) = µ(Kn−1 ∩ L) = µ(L) − µ(L \ Kn−1 ) > µ(An ) − δ − (ε − δ) = µ(An ) − ε. Since

T∞

n=1 Kn

µ(

is a compact convex set contained in

∞ \

n=1

An ) ≥ µ(

∞ \

n=1

T∞

n=1 An ,

we obtain

Kn ) = lim µ(Kn ) ≥ lim µ(An ) − ε. n→∞

Since ε > 0 is arbitrary, the proof is complete.

n→∞


115

Lemma 4.16. Assume that a bounded affine function f on X satisfies the Haydon condition. Then for any µ ∈ M1 (X) and any b ∈ R, µ ({x ∈ X : f (x) ≥ b}) + µ ({x ∈ X : f (x) < b}) = 1.

(4.4)

Proof. Let µ be a Radon probability measure on X. We will prove that for any pair of real numbers a < b, the inequality µ(A) + µ(B) ≥ 1 holds, where A := {x ∈ X : f (x) > a}, B := {x ∈ X : f (x) < b}. If this is not true, we find compact convex sets Ln , Kn , n ∈ N, such that Ln ⊂ A, Kn ⊂ B, µ(Ln ) → µ(A) and µ(Kn ) → µ(B). Due to the assumption, λ := µ|X\S∞ n=1 (Ln ∪Kn ) is a nonzero Radon measure. Hence we may apply the Haydon condition in order to obtain a compact convex set L such that λ(L) > 0 and such that L is contained either in A or in B. Suppose that L ⊂ A. Since A is convex, co(L ∪ Ln ) is contained in A for any n ∈ N. Thus we obtain µ(A) ≥ µ (co(L ∪ Ln )) ≥ µ(L ∪ Ln ) = µ(Ln ) + µ(L \ Ln ) ∞ [ ≥ µ(Ln ) + µ L \ (Ln ∪ Kn ) = µ(Ln ) + λ(L). n=1

Letting n → ∞, we get a contradiction. In order to obtain the desired equality, it is enough to realize that {x ∈ X : f (x) ≥ b} =

∞ \ n=1

1 x ∈ X : f (x) > b − n

.

Thus using the monotonicity of a convex inner measure (Lemma 4.15), we obtain µ ({x ∈ X : f (x) ≥ b}) + µ ({x ∈ X : f (x) < b}) ≥ 1. Since the converse inequality is obvious, the proof is finished. Definition 4.17 (Saturated sets). Let f be a real-valued function on a compact convex set X. A subset F of X is saturated with respect to f if for any Radon measure µ on X and any ε > 0, there exists a sequence {Kn } of compact convex sets in X such that ∞ [ diam f (Kn ) ≤ ε for each n ∈ N and µ F \ Kn = 0. n=1

116


Lemma 4.18. Let f be a function on a compact convex set X such that X is saturated with respect to f . Then f is universally measurable. Proof. Let µ ∈ M1 (X) be given. We fix a sequence {εk } of strictly positive numbers converging to 0. For each k ∈ N, we find a sequence {Knk }∞ of compact convex n=1 S k k sets in X such that diam f (Kn ) ≤ εk for each n ∈ N and µ(X \ ∞ n=1 µ(Kn )) = 0. Sn−1 k k k k k We set An := Kn \ i=1 Ki , n ∈ N, and pick numbers an ∈ f (An ) if the set Akn is nonempty. We define ( akn , if x ∈ Akn , n ∈ N, fk (x) := S k 0, if x ∈ X \ ∞ n=1 An . Then the functions fk are µ-measurable and fk → f uniformly µ-almost everywhere on X. Hence f is universally measurable. Theorem 4.19. Let f be a bounded affine function on X. Then the following conditions are equivalent: (i) f satisfies the barycentric formula, (ii) for any µ ∈ M+ (X) and any ε > 0, there exists a compact convex set L ⊂ X such that µ(L) > µ(X) − ε and f |L is continuous, (iii) f satisfies the Haydon condition, (iv) X is saturated with respect to f. Proof. (i) =⇒ (ii): Suppose that f satisfies the barycentric formula and µ is a Radon measure on X. Clearly we may assume that µ is a probability measure. For given ε > 0, we use Lusin’s theorem A.76 to find a compact set F ⊂ X such that f |F is continuous and µ(F ) ≥ 1 − ε. If we define L := coF, we obtain the required compact convex set. Indeed, we want to prove that f |L is continuous. Let x be a point in L and {xα }α∈A be a net of points of L converging to x. By the Milman theorem 2.43, ext L is a subset of F. The Integral representation theorem 2.31 ensures the existence of a net of probability measures {µα }α∈A carried by ext L with barycenters xα . Without loss of generality, we may assume that {µα }α∈A converges to a probability measure ν. Since the net {xα }α∈A converges to x, the measure ν represents x. Since f is continuous on ext L and µα and ν are carried by ext L, we get µα (f ) → ν(f ). Now we apply the assumption to obtain f (xα ) = µα (f ) → ν(f ) = f (x), which completes the proof. For the proof of the implication (ii) =⇒ (iii), let a nonzero measure µ ∈ M+ (X) and a < b be given. Using (ii), there exists a compact convex set L such that f |L is


117

continuous and µ(L) > 0. We select a point x ∈ spt µ|L . Without loss of generality we may assume that f (x) < b. By the continuity of f |L we can find a closed convex neighborhood V of x such that f < b on V ∩L. Then V ∩L satisfies our requirements. For the proof of the implication (iii) =⇒ (iv) we will use Lemma 4.16. Let a Radon probability measure µ on X and ε > 0 be given. Fix δ > 0. Since f is bounded, we may suppose that 0 ≤ f < 1. Choose N ∈ N such that N1 < ε and define P n Fn := {x ∈ X : n−1 4.16, N n=1 µ(Fn ) = 1. N ≤ f < N }, n = 1, . . . , N . By Lemma P N Hence there exist compact convex sets Kn ⊂ Fn such that n=1 µ(Kn ) > 1 − δ. Then 1 < ε. diam f (Kn ) ≤ diam f (Fn ) ≤ N If we put together all compact convex sets obtained by choosing δl = 1l , l ∈ N, we get the required sequence. It remains to prove (iv) =⇒ (i). Let a Radon probability measure µ on X with the barycenter x and ε > 0 be given. By condition (iv), there exists a sequence {Kn } of closed convex subsets of X such that diam f (Kn ) < ε

and

∞ [ Kn = 0. µ X\ n=1

Find N ∈ N such that µ X \ (K1 ∪ · · · ∪ KN ) < ε

(4.5)

and define En := Kn \

n−1 [

Ki ,

n = 1, . . . , N,

E0 := X \

N [

Ki ,

i=1

i=1

λn := µ(En ),

n = 0, . . . , N.

We define Radon probability measures µn , n = 0, . . . , N , by ( 1 µ|En if λn > 0, µn := λn εx if λn = 0. Let xn be the barycenter of µn , n = 0, . . . , N . Then xn ∈ co En ⊂ Kn if n = 1, . . . , N satisfies λn > 0. The oscillation properties of f give |µ0 (f ) − f (x0 )| ≤ 2kf k

and

|µn (f ) − f (xn )| < ε,

n = 1, . . . , N. (4.6)

Obviously N X n=0

λn = 1,

N X n=0

λn xn = x

and

N X n=0

λn µn = µ.

(4.7)

118


Since f is affine, by (4.5), (4.6) and (4.7) we have N N N X X X |f (x) − µ(f )| = f λ n xn − λn µn (f ) = λn (f (xn ) − µn (f )) n=0

n=0

≤ λ0 |f (x0 ) − µ0 (f )| +

n=0 N X

λn |f (xn ) − µn (f )| ≤ ε(2kf k + 1).

n=1

Letting ε → 0, we conclude the proof. Lemma 4.20. Let f be a convex real-valued function on a compact convex set X that is lower bounded on some nonempty open subset of X. Then f is lower bounded on X. In particular, if f has a point of continuity, then f is lower bounded on X. Proof. Given f as in the lemma, let U be a nonempty open subset of X such that f ≥ C on U for some C ∈ R. Assume that f is not lower bounded on X; that is, that there exist points xn ∈ X, n ∈ N, such that f (xn ) → −∞. Let x be a cluster point of the sequence {xn } and let y ∈ U be chosen arbitrarily. Then there exists α ∈ (0, 1) such that αx + (1 − α)y ∈ U . Since x is a cluster point of {xn }, for infinitely many indices n ∈ N we have αxn + (1 − α)y ∈ U . Then for these indices we get C ≤ f (αxn + (1 − α)y) ≤ αf (xn ) + (1 − α)f (y), which yields a contradiction. Theorem 4.21. Let f be an affine function on a compact convex set X such that f is fragmented. Then f satisfies the barycentric formula. Proof. First, we note that f is bounded by Lemma 4.20 and Theorem A.121. We will show that X is saturated with respect to f . To this end, let µ ∈ M+ (X) and ε > 0 be given. Let U := {U ⊂ X : U is open and there are compact convex sets Kn ⊂ X such that µ U \

∞ [

Kn = 0 and diam f (Kn ) < ε}.

(4.8)

n=1

S It is not difficult to verify that V := {U : U ∈ U} ∈ U. Indeed, V is obviously open. Due to the inner regularity of µ, there exists a sequence {Hk } of compact sets such that µ(Hk ) % µ(V ). By a compactness argument, we can cover each Hk by a finite family U1 , . . . , Unk of sets contained in U. For every Ui , k ∈ N and 1 ≤ i ≤ nk , we find a countable family of compact convex sets guaranteed by (4.8). Putting together all these families, we obtain a countable family L of compact convex sets which covers µ-almost all of V and diam f (K) < ε for each K ∈ L.


119

Let K be the family of all closed convex subsets of X whose complement in X is contained in U. Let Z be the intersection of K. By the argument above, Z is the smallest element of K. Set Y := {x ∈ Z : oscZ f (x) ≥ ε}. Then Y is a closed convex subset of Z. If x ∈ Z \ Y , then there exists an open convex neighborhood U of x such that U ∩ Y = ∅ and diam f (U ∩ Z) < ε. Since U \ Z ∈ U and U ∩ Z contains U ∩ Z, we observe that U ∈ U. Appealing again to the properties of U, it follows that Y is a closed convex subset of Z whose complement in X is contained in U. By the minimality of Z, we have Y = Z. There is no point of continuity of f |Z , and, by our assumption and Theorem A.121, it follows that Z = ∅. Hence X ∈ U and X is saturated with respect to f , which concludes the proof by Theorem 4.19(iv). Corollary 4.22 (Choquet’s barycentric theorem). Let f be a Baire-one affine function on a compact convex set X. Then f is strongly affine. Proof. By Theorem A.124 and Theorem A.121, any Baire-one function on X is fragmented. Thus the assertion follows from Theorem 4.21. Corollary 4.23 (Affine B1 -minimum principle). Let f be a Baire-one affine function on a compact convex set X. If f ≤ 0 on ext X, then f ≤ 0 on X. Moreover, sup {|f (x)| : x ∈ X} = sup {|f (x)| : x ∈ ext X} . Proof. Choquet’s barycentric theorem 4.22 asserts that any affine Baire-one function is Ac (X)-affine. Thus the affine B1 -minimum principle is a particular case of Theorem 3.86. The second assertion is then an immediate consequence. Another simple but important consequence of Choquet’s barycentric theorem is the following approximation theorem due to G. Mokobodzki (cf. M. Rogalski [392], Appendix). Notice that taking Mokobodzki’s theorem for granted, Choquet’s barycentric theorem is its immediate consequence. Theorem 4.24 (Mokobodzki’s approximation theorem). Let X be a compact convex set and f be a Baire-one affine function on X. Then there exists a bounded sequence of continuous affine functions on X pointwise converging to f on X. Proof. Let f be a Baire-one affine function on X. By Theorem A.121, Lemma 4.20 and Theorem A.124, there are bounded sequences {fn } and {gn } such that both fn and −gn are upper semicontinuous functions, fn % f

and

gn & f.

120


Fix n ∈ N. By Choquet’s barycentric theorem 4.22, µ fn −

1 1 < ν gn + n n

for every µ, ν ∈ M1 (X) with r(µ) = r(ν). Lemma 4.11 yields the existence of a function hn ∈ Ac (X) such that fn −

1 1 < hn < gn + . n n

Then {hn } is a bounded sequence of continuous affine functions such that hn → f on X.

4.3

State space and representation of affine functions

In this section we introduce a notion of the state space and present its basic properties. It represents a natural and efficient link between function spaces and convex analysis. Definition 4.25 (State space). Let H be a function space on a compact space K and S(H), the state space of H, be defined as S(H) := {ϕ ∈ H∗ : ϕ ≥ 0, ϕ(1) = 1} . Clearly, S(H) is a w∗ -compact convex subset of the dual space H∗ . We denote by π the quotient mapping from M(K) onto H∗ ; that is, π(µ) = µ|H , µ ∈ M(K). As a simple consequence of the Hahn–Banach theorem, we obtain that S(H) = π(M1 (K)).

(4.9)

Further, we define the evaluation mapping φ : K → S(H) as φ : x 7→ φx ,

x ∈ K,

where φx : h 7→ h(x),

h ∈ H.

Let Φ : H → Ac (S(H)) be the mapping defined for h ∈ H as Φ(h) : s 7→ s(h),

s ∈ S(H).

Proposition 4.26. The evaluation mapping φ is a homeomorphism of K into S(H) and the following assertions hold. (a) S(H) = co(φ(K)), (b) Φ(H) is dense in Ac (S(H)),

4.3 State space and representation of affine functions

121

(c) r(φ] µ) = π(µ) for any µ ∈ M1 (K); in particular, r(φ] µ) = φ(x) for µ ∈ Mx (H), (d) φ(ChH (K)) = ext S(H), (e) Φ is a positive linear mapping and kΦ(h)k = khk for each h ∈ H, (f) Φ is surjective if and only if H is closed, (g) if Φ is surjective, then the inverse mapping is realized by Φ−1 (F ) = F ◦ φ,

F ∈ Ac (S(H)).

(4.10)

Proof. Since φ is a continuous injective mapping of a compact Hausdorff space K into S(H), it is a homeomorphism between K and φ(K). We prove that S(H) = co(φ(K)). To this end, suppose that there exists a point s ∈ S(H) \ co(φ(K)). By the Hahn–Banach theorem, there exists H ∈ (H ∗ , w∗ )∗ such that H(s) < H(t) for any t ∈ co(φ(K)). Further, there exists h ∈ H such that H(ψ) = ψ(h) for every ψ ∈ H∗ . Thus s(h) < φ(x)(h) = h(x),

x ∈ K.

(4.11)

The Hahn–Banach theorem provides a Radon measure µ on K such that kµk = ksk = 1 and s(g) = µ(g) for each g ∈ H. Since µ(1) = s(1) = 1, the measure µ belongs to M1 (K). Thus s(h) = µ(h) ∈ co h(K). But this contradicts (4.11). Thus S(H) = co(φ(K)). Lemma 2.34 yields (b). Indeed, Lemma 2.34 ensures that ((H ∗ , w∗ )∗ + R)|S(H) is dense in Ac (S(H)). Hence the space Φ(H) + R is dense in Ac (S(H)). Since H contains constant functions, Φ(H) + R = Φ(H). For the proof of (c), let µ ∈ M1 (K) be given. By (b), it is enough to verify that Φ(h)(r(φ] µ)) = Φ(h)(π(µ)) for any h ∈ H. To this end, fix h ∈ H. Then Φ(h)(r(φ] µ)) = φ] µ(Φ(h)) = µ(Φ(h) ◦ φ) = µ(h) = π(µ)(h). We proceed with the proof of (d). Pick s ∈ ext S(H). Since S(H) = co(φ(K)), the Milman theorem 2.43 implies that ext S(H) ⊂ φ(K) = φ(K). Hence s = φ(x) for some point x ∈ K. If µ ∈ M1 (K) is an H-representing measure for the point x, then φ] µ represents φ(x). By Bauer’s characterization of extreme points 2.40, φ] µ = εs . Hence µ = εx and x is in the Choquet boundary ChH (K).

122


Conversely, let x ∈ ChH (K) be given and φ(x) = 12 (s1 + s2 ), s1 , s2 ∈ S(H). Find Radon probability measures µi , i = 1, 2, on K with π(µi ) = si , i = 1, 2. Then µ = 12 (µ1 + µ2 ) represents x. Hence µ1 = µ2 = εx , which implies s1 = s2 = s. This completes the proof of (d). Obviously, Φ is positive and linear. Further, the inequalities kΦ(h)k = sup |Φ(h)(s)| = sup |s(h)| ≥ sup |φx (h)| s∈S

s∈S

x∈K

= sup |h(x)| = khk ≥ sup |s(h)| = kΦ(h)k x∈K

s∈S

give (e). If Φ is onto, the function space H is closed because Ac (S(H)) is a complete space and Φ is an isometric mapping. Conversely, suppose that H is a closed space. Since Φ(H) is a dense subspace of Ac (S(H)), H is complete and Φ is an isometry, the space Φ(H) is closed in Ac (S(H)). Hence Ac (S(H)) ⊂ (Φ(H)) = Φ(H). It is straightforward to check the validity of the inversion formula (4.10), so the proof of the proposition is finished. Lemma 4.27. Let F be a continuous convex function on S(H) and φ : K → S(H) be the evaluation mapping. Then F ◦ φ is an H-convex continuous function on K. Proof. Let F ∈ Sc (S(H)), x ∈ K and µ ∈ Mx (H) be given. Then µ(F ◦ φ) = φ] µ(F ) ≥ F (r(φ] µ)) = F (φ(x)), because r(φ] µ) = φ(x) by Proposition 4.26(c). Proposition 4.28. Let H be a function space on a compact space K and S(H) its state space. Then the following assertions hold. (a) If µ ∈ Mφ(x) (Ac (S(H))) with spt µ ⊂ φ(K), then (φ−1 )] µ ∈ Mx (H). (b) π(M1 (F )) = co φ(F ) for any closed set F ⊂ K. (c) If µ, ν ∈ M+ (K), then µ ≺ ν if and only if φ] µ ≺ φ] ν. (d) A measure Λ ∈ M+ (S(H)) is Ac (S(H))-maximal if and only if Λ = φ] λ for some H-maximal measure λ ∈ M+ (K). (e) A measure Λ ∈ M(S(H)) is a boundary measure if and only if Λ = φ] λ for some H-boundary measure λ ∈ M(K).


123

Proof. Assertion (a) is a direct consequence of the definitions and the density of Φ(H) in Ac (S(H)). For the proof of (b), if s ∈ co φ(F ), then there exists a measure µ ∈ M1 (φ(F )) which represents s (see Proposition 2.39). By assertion (a), π((φ−1 )] µ) = r(φ] ((φ−1 )] µ)) = r(µ) = s, and s ∈ π(M1 (F )). Conversely, let µ ∈ M1 (F ) be such that the point s := π(µ) does not belong to co φ(F ). Then we may separate s and co φ(F ) by a continuous affine function on S(H). Due to the density of Φ(H) in Ac (S(H)), we may suppose that there exists a function h ∈ H such that s(h) > sup{t(h) : t ∈ co φ(F )}. Then we obtain µ(h) = s(h) >

sup t∈co φ(F )

t(h) ≥ sup t(h) = sup h(F ) ≥ µ(h), t∈φ(F )

which is a contradiction. Hence π(M1 (F )) = co φ(F ). To prove (c), suppose that µ, ν ∈ M+ (K) with µ ≺ ν and F ∈ Kc (S(H)) are given. By Lemma 4.27, F ◦ φ ∈ Kc (H). Thus φ] µ(F ) = µ(F ◦ φ) ≤ ν(F ◦ φ) = φ] µ(F ), and hence φ] µ ≺ φ] ν. Conversely, assume that φ] µ ≺ φ] ν. By Proposition 3.56, it is enough to show that µ(f ) ≤ ν(f ) for any f = h1 ∨ · · · ∨ hn , where h1 , . . . , hn ∈ H. By the assumption and the definition of the image of a measure, we get Z Z h1 ∨ · · · ∨ hn dµ = Φ(h1 ) ∨ · · · ∨ Φ(hn ) d(φ] µ) K

φ(K)

Z ≤

Z Φ(h1 ) ∨ · · · ∨ Φ(hn ) d(φ] ν) =

φ(K)

h1 ∨ · · · ∨ hn dν, K

which is the desired conclusion. In order to check assertion (d), pick an H-maximal measure µ in M+ (K). Let Λ ∈ M+ (S(H)) be an Ac (S(H))-maximal measure with φ] µ ≺ Λ. Then Λ is carried by φ(K), and thus Λ = φ] λ for some λ ∈ M+ (K). By assertion (c), µ ≺ λ. Since µ is H-maximal, µ = λ and φ] µ is Ac (S(H))-maximal. Conversely, let φ] µ be an Ac (S(H))-maximal measure and λ ∈ M+ (K) satisfying µ ≺ λ. Then φ] µ ≺ φ] λ, which gives that φ] µ = φ] λ, and consequently µ = λ. Since assertion (e) is a direct consequence of (d), the proof is finished.

124


Proposition 4.29. Let H be a closed function space on a compact space K and X := S(H). Then the following assertions hold: (a) X is contained in a closed hyperplane not containing 0, S (b) if (H∗ )+ denotes the set of all positive elements of H∗ , then (H∗ )+ = c≥0 cX, (H∗ )+ ∩ −(H∗ )+ = {0} and H∗ = (H∗ )+ − (H∗ )+ , (c) for any F ∈ Ac (X) there exists a unique Fb ∈ (H∗ , w∗ )∗ such that F = Fb on X. Proof. The first assertion (a) is obvious since X ⊂ {ϕ ∈ H∗ : ϕ(1) = 1}. S For the proof of (b) we first notice that clearly c≥0 cX ⊂ (H∗ )+ . On the other hand, if ϕSis a nonzero positive functional on H and c = kϕk, then ϕ = c(c−1 ϕ), and thus ϕ ∈ c≥0 cX. Obviously, (H∗ )+ ∩ −(H∗ )+ = {0}. If ϕ ∈ H∗ is given, extend ϕ to a signed measure µ ∈ M(K) with µ = ϕ on H. If we decompose µ into its positive and negative part µ+ and µ− and set ϕ+ := µ+ |H and ϕ− := µ− |H , we obtain the desired decomposition of ϕ. Thus H∗ = (H∗ )+ − (H∗ )+ . For the third part of the proof, pick F ∈ Ac (X). Since H is closed, F = Φ(f ) for some f ∈ H by Proposition 4.26(f). Setting Fb(ϕ) := ϕ(f ),

ϕ ∈ H∗ ,

we obviously obtain an extension of F . We have to show that this extension is unique. But this easily follows from assertion (b). This concludes the proof. Notation 4.30. We recall that τX is the topology of pointwise convergence on a set X; see Notation A.27. Proposition 4.31. Let X be a compact convex set in a locally convex space, E := Ac (X) and φ : X → S(Ac (X)) be the evaluation mapping. Then the following assertions hold: (a) The mapping φ is an affine homeomorphism of X onto the set S(Ac (X)) and S(Ac (X)) ⊂ SE ∗ . (b) For every η ∈ E ∗ there exist numbers a1 , a2 ≥ 0 and points x1 , x2 ∈ X such that η = a1 φ(x1 ) − a2 φ(x2 ) and kηk = a1 + a2 . Moreover, a1 , a2 are uniquely determined. (c) φ(X) ∪ −φ(X) ⊂ SE ∗ and BE ∗ = co(φ(X) ∪ −φ(X)). (d) For the set (E ∗ )+ of all positive elements of E ∗ we have (E ∗ )+ =

S

c≥0 cφ(X).

(e) For every affine continuous function f on X there exists a unique function fb ∈ (E ∗ , w∗ )∗ such that f = fb ◦ φ on X. (f) The mapping id : (Ac (X), τX ) → (E, w) is a homeomorphism.

125


Proof. We already know from Proposition 4.26 that φ is a homeomorphism. A routine argument shows that φ is affine. To check its surjectiveness, note that for any s ∈ S(Ac (X)) there exists µ ∈ M1 (X) such that π(µ) = s. Then x := r(µ) satisfies φ(x) = s. Obviously, S(Ac (X)) ⊂ SE ∗ , which concludes the proof of (a). For the proof of (b), let η ∈ E ∗ be given. Let µ ∈ M(X) be an extension of η with kηk = kµk. We decompose µ into its positive and negative part as µ = a1 µ1 − a2 µ2 , where a1 , a2 ≥ 0 and µ1 , µ2 ∈ M1 (X). Let x1 := r(µ1 ) and x2 := r(µ2 ). Then η(h) = a1 h(x1 ) − a2 h(x2 ) for any h ∈ Ac (X), and thus η = a1 φ(x1 ) − a2 φ(x2 ) and

kηk = kµk = a1 + a2 .

(4.12)

If η = b1 φ(y1 ) − b2 φ(y2 ) is another representation of η with kηk = b1 + b2 , we apply η to the constant function 1 to get a1 − a2 = b1 − b2 . This, along with (4.12), yields a1 = b1 and a2 = b2 . For the proof of (c), we note that φ(X) ∪ −φ(X) ⊂ SE ∗ . Further, SE ∗ ⊂ co(φ(X) ∪ −φ(X)) by (b). Since 0 ∈ co(φ(X) ∪ −φ(X)), BE ∗ ⊂ co(φ(X) ∪ −φ(X)). Since the reverse inclusion is obvious, the proof of (c) is complete. Since (d) and (e) follow from Proposition 4.29(b) and (c), respectively, we proceed to the proof of (f). The topology τX on Ac (X) is generated by functionals h 7→ h(x), h ∈ Ac (X), where x ∈ X. On the other hand, by (b), the weak topology on Ac (X) is generated by functionals of the form h 7→ a1 h(x1 ) − a2 h(x2 ), h ∈ Ac (X), where a1 , a2 ≥ 0 and x1 , x2 ∈ X. It follows that these topologies coincide on Ac (X). This finishes the proof of (f). Proposition 4.32. Let X be a compact convex set in a locally convex space and E := Ac (X). (a) For every η ∈ E ∗∗ there exists a unique fη ∈ Ab (X) such that kηk = kfη k and η(a1 φ(x1 ) − a2 φ(x2 )) = a1 fη (x1 ) − a2 fη (x2 ),

a1 , a2 ≥ 0, x1 , x2 ∈ X.

(b) The mapping η 7→ fη is an isometric isomorphism of E ∗∗ onto Ab (X) which is a (w∗ –τX )-homeomorphism. Moreover, it preserves order (that is, η ≥ 0 if and only if fη ≥ 0). Proof. We start with (a). Given a functional η ∈ E ∗∗ , we define a function fη : X → R as fη : x 7→ η(φ(x)), x ∈ X. Obviously, fη is affine and kfη k ≤ kηk. To prove the converse inequality, Proposition 4.31(c) gives kηk = sup{η(a1 φ(x1 ) − a2 φ(x2 )) : a1 , a2 ≥ 0, a1 + a2 ≤ 1, x1 , x2 ∈ X} = sup{a1 fη (x1 ) − a2 fη (x2 ) : a1 , a2 ≥ 0, a1 + a2 ≤ 1, x1 , x2 ∈ X} ≤ sup{a1 kfη k + a2 kfη k : a1 , a2 ≥ 0, a1 + a2 ≤ 1} ≤ kfη k.

126


Uniqueness of fη immediately follows from Proposition 4.31(d). For the proof of (b), we start by noticing that (a) yields that the mapping T : η 7→ fη is an isometric mapping of (E ∗∗ , k · k) into (Ab (X), k · k). If f ∈ Ab (X) is given, the function η(a1 φ(x1 ) − a2 φ(x2 )) := a1 f (x1 ) − a2 f (x2 ),

a1 , a2 ≥ 0, x1 , x2 ∈ X,

is in E ∗∗ (see Exercise 4.48) and f = fη . Hence T is surjective. As in Proposition 4.31(f), for the proof that T is a (w∗ –τX )-homeomorphism, we look at the defining functionals. By Proposition 4.31(b), the w∗ -topology on E ∗∗ is generated by functionals η 7→ η(a1 φ(x1 ) − a2 φ(x2 )), η ∈ E ∗∗ , where a1 , a2 ≥ 0 and x1 , x2 ∈ X. The topology τX on Ab (X) is generated by functionals of the form f 7→ f (x), f ∈ Ab (X), where x ∈ X. Hence it follows from (a) that T is a (w∗ –τX )homeomorphism. Finally, the fact that the function fη is positive if and only if η is positive, follows from the definition of the mapping T . This concludes the proof. Theorem 4.33 (Goldstine’s lemma). If X is a compact convex set, then BAc (X) is τX -dense in BAb (X) . Proof. The assertion is just a reformulation of Goldstine’s lemma (see Theorem A.4). Indeed, if E stands for the Banach space Ac (X), then the unit ball BE is w∗ -dense in BE ∗∗ . Let ε : E → E ∗∗ be the canonical embedding and T : E ∗∗ → Ab (X) be the mapping from Proposition 4.32(b). If I : Ac (X) → Ab (X) denotes the identity mapping, then I = T ◦ ε. From this the assertion follows. Theorem 4.34. For a real-valued function f on a compact convex set X, the following assertions are equivalent: (i) f is strongly affine, (ii) f is bounded and for any measure µ ∈ M1 (X) there exists a sequence {fn } of continuous affine functions on X such that kfn k ≤ kf k, n ∈ N, and fn → f µ-almost everywhere. Proof. To prove (i) =⇒ (ii), let f be a strongly affine function. Since f is bounded by Lemma 4.5, we may assume without loss of generality that kf k = 1. Given a measure µ ∈ M1 (X), Theorem 4.19 implies the existence of compact convex sets Kn ⊂ X, n ∈ N, such that µ(Kn ) → 1 and f |Kn is continuous. For a fixed natural number n ∈ N, consider the restriction operator R : Ab (X) → Ab (Kn ) defined as Rf := f |Kn ,

f ∈ Ab (X).

By Goldstine’s lemma 4.33, BAc (X) is τX -dense in BAb (X) . Since the operator R is (τX –τKn )-continuous, R(BAc (X) ) is τKn -dense in R(BAb (X) ). Hence we have Rf ∈

4.4 Affine Baire-one functions on dual unit balls

127

τK

R(BAc (X) ) n . Since Rf is continuous on Kn and τKn coincides with the weak topology on Ac (Kn ) (see Proposition 4.31(f)), Mazur’s theorem (see Theorem A.2 or Exercise 3.116) yields k·kAc (Kn )

Rf ∈ co R(BE )

k·kAc (Kn )

= R(BE )

.

It follows that there exists a function g ∈ BAc (X) that is uniformly close to f on Kn as much as we wish. Applying this argument to each Kn , we construct the required sequence {fn } in BAc (X) . To show (ii) =⇒ (i), assume that f satisfies (ii). First we notice that the assumption ensures that given finitely many measures µ1 , . . . , µk in M+ (X), we can find a sequence {fn } of continuous affine functions on X such that kfn k ≤ kf k, n ∈ N, and fn → f µi -almost everywhere for each i = 1, . . . , k (we just apply the assumption to a multiple of µ1 + · · · + µk ). Let µ ∈ M1 (X) with x = r(µ) be given. We find a sequence {fn } of continuous affine functions on X such that kfn k ≤ kf k, n ∈ N, fn → f µ-almost everywhere and fn (x) → f (x) (use the preceding argument for the measures µ and εx ). Then µ(f ) = lim µ(fn ) = lim fn (x) = f (x), n→∞

n→∞

and f is strongly affine. This concludes the proof. Theorem 4.35. Let H be a closed function space on a metrizable compact space K. The the set of H-exposed points is dense in ChH (K). Proof. Let X := S(H). Since K is metrizable, H is separable, and thus X is metrizable as well (see Theorem A.5). Let φ : K → X and Φ : H → Ac (X) be as in Definition 4.25. Since Φ is a positive and surjective isometry (see Proposition 4.26(e) and (f)), it is easy to verify that φ(expH (K)) = exp X. Since φ(ChH (K)) = ext X by Proposition 4.26(d), the conclusion follows from Corollary 2.52.

4.4

Affine Baire-one functions on dual unit balls

In what follows, we present an application of Mokobodzki’s approximation result 4.24. For this reason we need a characterization of affine functions on unit balls in Banach spaces. If E is a Banach space, we recall that E is canonically embedded in E ∗∗ . The first proposition is a particular version of the Banach–Dieudonné theorem (see Theorem A.7). Proposition 4.36. Let E be a Banach space, BE ∗ be the closed unit ball of E ∗ equipped with the w∗ -topology, and let f be a function on BE ∗ . Then f is a continuous affine function on BE ∗ if and only if there exist x ∈ E and λ ∈ R such that f = x + λ on BE ∗ .

128


Proof. Assume that f ∈ Ac (BE ∗ ). Since (E ∗ , w∗ )∗ = E, by Lemma 2.34 there exist xn ∈ E and λn ∈ R, n ∈ N, such that xn + λ n → f

uniformly on BE ∗ .

Since f (0) = lim (xn (0) + λn ) = lim λn , n→∞

n→∞

we see that there exists limn→∞ λn and it equals f (0). Given ε > 0, there are n, k ∈ N such that |xn (ϕ) − xk (ϕ)| < ε for all ϕ ∈ BE ∗ . Hence ||xn − xk || ≤ ε for those n, k, and therefore there exists x := limn→∞ xn . It follows that f = x + f (0) on BE ∗ . Obviously, the function (x + λ)|BE∗ , where x ∈ E and λ ∈ R, is a continuous affine function on BE ∗ . Theorem 4.37. Let f be a function on BE ∗ , f (0) = 0. The following assertions are equivalent: (i) f is an affine Baire-one function on BE ∗ , (ii) there exists a sequence {xn } in E such that xn → f on BE ∗ , (iii) there exists s∗∗ in the w∗ -sequential closure of E such that f = s∗∗ on BE ∗ . Proof. To show (i) =⇒ (ii), let f be an affine Baire-one function on BE ∗ with f (0) = 0. By Mokobodzki’s approximation theorem 4.24, there exists a sequence {fn } of continuous affine functions on BE ∗ such that fn → f . By Proposition 4.36, each fn is of the form xn + λn on BE ∗ for suitable xn ∈ E and λn ∈ R. Since f (0) = 0, we may assume that λn = 0 for each n ∈ N and (ii) follows. We proceed with (ii) =⇒ (iii). If xn ∈ E and xn → f on BE ∗ , then f is an affine Baire-one function on BE ∗ . It is easy to see that f is bounded (cf. Theorem A.121 and Lemma 4.20). By Exercise 4.48, there exists s∗∗ ∈ E ∗∗ such that f = s∗∗ on BE ∗ . Hence s∗∗ (ϕ) = f (ϕ) = lim xn (ϕ) n→∞

s∗∗ (ϕ)

for every ϕ ∈ BE ∗ . It follows that = limn→∞ xn (ϕ) even for each ϕ ∈ E ∗ , ∗∗ ∗ and we see that s belongs to the w -sequential closure of E. The implication (iii) =⇒ (i) is obvious. Corollary 4.38. Let E be a Banach space. Then the w∗ -sequential closure of E is norm closed in E ∗∗ . Proof. By Theorem 4.37, the w∗ -sequential closure of E equals the set M := s∗∗ ∈ E ∗∗ : s∗∗ |BE∗ is a Baire-one function , and the assertion follows since M is norm closed (see Proposition A.126).

4.5 Exercises

4.5

129

Exercises

Exercise 4.39. Let f be a continuous function on a compact convex set X and x ∈ X. Then f ∗ (x) = sup{µ(f ) : µ ∈ Mx (X), µ is molecular}. Hint. Use Lemma 3.21 and Proposition 4.3. Exercise 4.40. If a compact convex set X in a locally convex space E admits a strictly convex continuous function, then X is metrizable. Hint. In order to show that X is metrizable, it suffices to find a countable family B ⊂ C(K) such that B separates points on X (see Lemma 10.45). So, suppose that f is a continuous strictly convex function on X. By Corollary 4.8 and Proposition 3.55, there exist kn ∈ N, n ∈ N, and functions fn = hn1 ∨ · · · ∨ hnkn , hni ∈ Ac (X), i = 1, . . . , kn , such that fn < f < fn + n1 , n ∈ N. We want to show that the family B := {hni : i = 1, . . . , kn , n ∈ N} separates points on X. Assume that there are distinct x, y ∈ X such that a(x) = a(y) for each a ∈ B. Let z := 12 (x + y). For ε > 0, find a1 , a2 ∈ B such that f (x) < a1 (x) + ε and f (y) < a2 (y) + ε. Then 1 1 f (z) ≤ (f (x) + f (y)) ≤ (a1 (x) + a2 (y)) + ε 2 2 1 ≤ (f (z) + f (z)) + ε. 2 1 Hence f (z) = 2 (f (x) + f (y)) and f is not strictly convex. Exercise 4.41. Prove that there exists a strictly convex function f on a compact convex set X that is not Ac (X)-strictly convex. Hint. Consider the compact convex set M1 ([0, 1]) and a strictly convex continuous function f on M1 ([0, 1]), 0 ≤ f ≤ 21 . If g : µ 7→ µd ([0, 1]),

µ ∈ M1 ([0, 1]),

where µd denotes the discrete part of µ, then g is a Borel (in fact, by Proposition 2.63, of the second Baire class) affine function on M1 ([0, 1]). Show that f − g is a strictly convex function on M1 ([0, 1]) which is not strictly Ac (M1 ([0, 1]))-convex. By Proposition 2.27, the mapping ε : x 7→ εx is a homeomorphism of [0, 1] onto ext M1 ([0, 1]). Let Λ := ε] λ be the image of Lebesgue measure λ by ε. By Proposition 2.54, λ is a barycenter of Λ. Moreover, 1 Λ(f − g) = Λ(f ) − Λ(g) ≤ − 1 < 0 ≤ f (λ) = f (λ) − g(λ) 2 = (f − g)(λ).

130


Exercise 4.42. Let f, −g be upper semicontinuous concave functions on a compact convex set X, with f < g. Then there exists a function h ∈ Ac (X) such that f < h < g. Hint. By Corollary 4.8, µ(f ) ≤ f (r(µ)) < g(r(µ)) ≤ ν(g) for any µ, ν ∈ M1 (X) with r(µ) = r(ν). Thus we may apply Lemma 4.11. Exercise 4.43. Prove that f ∈ Susc (X) if and only if f = inf {w ∈ W(Ac (X)) : w ≥ f } . Hint. Combine Corollary 4.8, Proposition 3.25 and Corollary 3.23. Exercise 4.44. If f is a real-valued convex function on compact convex set X, then its lower semicontinuous regularization defined as fb(x) = f (x) ∧ lim infy→x f (y), x ∈ X, is convex as well (see Subsection A.2.C). Hint. Obvious from the definition. Exercise 4.45. Let X be a compact convex set. Prove Lemma 3.77 for this particular case without the Simons lemma 3.75. Hint. Let {fn } be an upper bounded sequence of lower semicontinuous convex functions on X satisfying lim supn→∞ fn ≤ 0 on ext X. Without loss of generality we may assume that {fn } is a bounded sequence. Given x ∈ X and ε > 0, we find continuous affine functions hn , n ∈ N, such that hn ≤ fn and hn (x) > fn (x) − ε (use Corollary 3.23(b) and Proposition 3.25(a)). We define a continuous affine mapping ϕ : X → RN by ϕ(x) = {hn (x)}n∈N , x ∈ X, and let Y := ϕ(X). Then Y is a metrizable compact convex set and ϕ(ext X) ⊃ ext Y (see Proposition 2.72(c)). Find a maximal measure µ ∈ Mϕ(x) (Y ) and let h0n ∈ Ac (Y ) be such that h0n ◦ ϕ = hn , n ∈ N. Then lim supn→∞ h0n ≤ 0 on ext Y . By Corollary 3.61 and Fatou’s lemma, lim sup hn (x) = lim sup h0n (ϕ(x)) = lim sup µ(h0n ) n→∞

n→∞

n→∞

≤ µ(lim sup fn ) ≤ 0. n→∞

Hence lim sup fn (x) ≤ ε + lim sup hn (x) ≤ ε. n→∞


n→∞

4.5 Exercises

131

Exercise 4.46 (Mokobodzki). Let f be a real-valued lower bounded concave function on a compact convex set X. Then f is upper bounded on X. Hint. Without loss of generality we can suppose that f ≥ 0 on X. We suppose that f is not upper bounded and we will deduce a contradiction. There are xn ∈ K, n ∈ N, so that f (xn ) > 2n . Let ∞ X 1 x := r εxj j 2 j=1

and sn :=

∞ X

2−j ,

yn := r

∞ X j=n+1

j=n+1

Then x=

1 εxj , j sn 2

n ∈ N.

n X 1 x j + s n yn . 2j

(4.13)

j=1

Indeed, for any h ∈ Ac (X) we have h(x) =

n ∞ X X 1 1 ε (h) h(xj ) + sn h(yn ). = x 2j j 2j j=1

j=1

As functions from Ac (X) separate points of X, we have equality (4.13). Since x is a convex combination of points x1 , . . . , xn , yn , we get f (x) ≥

n X 1 f (xj ) + sn f (yn ) ≥ n. 2j j=1

Hence f (x) = ∞, which is obviously a contradiction. Exercise 4.47. Prove that there exists a function f on a compact convex set X satisfying the barycentric formula such that f |F has no point of continuity for some nonempty closed F ⊂ X. Hint. Take X := M1 ([0, 1]) and f (µ) = µ(g), µ ∈ X, where g is a bounded Borel function on [0, 1] with no point of continuity on [0, 1]. To prove the barycentric formula, start with a continuous function and then use transfinite induction along with the Lebesgue dominated convergence theorem. Exercise 4.48. Let X be a convex subset of a vector space E and f : X → R be an affine function on X. Assume that •

either 0 ∈ X and f (0) = 0,

•

or, there exists ϕ : E → R linear such that X ⊂ {x ∈ E : ϕ(x) = 1}.

132


Then f has a unique linear extension on span X. Hint. If 0 ∈ X and f (0) = 0, then span X = {a1 x1 − a2 x2 : a1 , a2 ≥ 0, x1 , x2 ∈ X} and the required extension is defined by fb(a1 x1 − a2 x2 ) := a1 f (x1 ) − a2 f (x2 ) a1 , a2 ≥ 0, x1 , x2 ∈ X. Since f is affine, fb is a well-defined unique linear extension of f . If the second assumption is satisfied, then X ∩ −X = ∅ and f has a unique affine extension on co(X ∪ −X) given by fb(ax − (1 − a)y) := af (x) − (1 − a)f (y),

a ∈ [0, 1], x, y ∈ X.

Now the extension to span X can be defined as above. Exercise 4.49. Let X be a compact convex set and A ⊂ Ac (X) be a vector space containing the constant functions and separating points of X. Then A is dense in Ac (X). Hint. By the Hahn–Banach theorem it is enough to show that ϕ = 0 whenever ϕ ∈ (Ac (X))∗ satisfies ϕ = 0 on A. By Proposition 4.31(b) there exist numbers a1 , a2 ≥ 0 and x1 , x2 ∈ X so that ϕ(f ) = a1 f (x1 ) − a2 f (x2 ), f ∈ Ac (X). By the assumption, 0 = ϕ(1) = a1 − a2 . Hence f (x1 ) = f (x2 ) for each f ∈ A. Since A separates points of X, x1 = x2 and ϕ = 0. Exercise 4.50. Let X be a compact convex set and A ⊂ Ac (X) be convex. Prove that τ f ∈ Ac (X) is in the norm closure of A if and only if f ∈ A X . Hint. This follows from Theorem A.2 and Proposition 4.31(f). Exercise 4.51. Let H be a function space on a compact space K and let {1, f1 , . . . , fd } be a basis of H. Let ψ : K → Rd be defined as ψ(x) = (f1 (x), . . . , fd (x)),

x ∈ K.

Prove that S(H) is affinely homeomorphic to co ψ(K). Hint. Let ϕ : S(H) → Rd be defined by ϕ(s) := (s(f1 ), . . . , s(fd )),

s ∈ S(H).

Then ϕ is a homeomorphism of S(H) into Rd such that ϕ ◦ φ = ψ on K. By Proposition 4.26(a), ϕ(S(H)) = ϕ(co φ(K)) = co(ϕ ◦ φ(K)) = co ψ(K). Since co ψ(K) = co ψ(K) by Corollary 2.8, S(H) is affinely homeomorphic to co ψ(K).


133

Exercise 4.52. Find a compact convex set X and a point x ∈ X such that face x is not a face. Hint. Let K := {−1} ∪ [0, 1] ∪ {2} and 1 1 H := {f ∈ C(K) : f ( ) = (f (−1) + f (2))}. 2 2 Let X := S(H) and π : M1 (K) → X be the quotient mapping from Proposition 4.26. Let s := π(λ), where λ is Lebesgue measure on [0, 1]. Consider the face λ in M1 ([0, 1]) ⊂ M1 (K). Then π(face λ) = face s.

(4.14)

Indeed, the inclusion face s ⊂ π(face λ) follows from the fact that π −1 (face s) is a face containing λ (see Proposition 2.72). For the converse inclusion, let t ∈ face s be given. Then there exist α ∈ (0, 1] and t0 ∈ X such that s = αt + (1 − α)t0 . Let µ, µ0 ∈ M1 (K) satisfy π(µ) = t and π(µ0 ) = t0 . We set A := {−1, 12 , 2} and find a sequence {fn } of functions from H such that 0 ≤ fn ≤ 1, n ∈ N, and fn → cA . Then the equality λ(fn ) = αµ(fn ) + (1 − α)µ0 (fn ) yields 0 = αµ(A). Hence µ ∈ M1 ([0, 1]). We claim that αµ ≤ λ. To verify this, let a positive function f ∈ C([0, 1]) be given. We find a positive function g ∈ H such that f = g on [0, 1]. Then λ(f ) = s(g) ≥ αt(g) = αµ(f ). Hence µ ∈ face λ by Exercise 2.127(a) and t ∈ π(face λ). This proves (4.14). By (4.14), Proposition A.40 and Exercise 2.127(b), face s = π(face λ) = π(face λ) = π(M1 ([0, 1])). Hence φ 1 ∈ face s, φ 1 = 21 (φ−1 + φ2 ) and φ−1 , φ2 are not contained in face s. Thus 2

2

face s is not a face.

4.6

Notes and comments

Basic properties of affine functions on compact convex sets are well known and can be found for example in E. M. Alfsen [5] and L. Asimow and A. J. Ellis [24]. The proof of Lemma 4.3 is taken from E. M. Alfsen [5], Proposition I.2.3, Lemma 4.5 is a variant of U. Krause [282], Satz 2.1.

134


Section 4.2 follows mainly the paper R. Haydon [220]. Choquet’s barycentric theorem 4.22 is proved by G. Choquet in [106] (see also E. M. Alfsen [5], Theorem I.2.6, or R. R. Phelps [374]). A slightly simplified proof can be found in J. Lukeˇs, J. Malý, I. Netuka, M. Smrˇcka and J. Spurný [319]. G. Choquet himself also showed that the barycentric formula is no longer valid for affine functions of the second Baire class (see Proposition 2.63). Theorem 4.24 due to G. Mokobodzki can be found in M. Rogalski [392]. Results on the state space contained in Section 4.3 are standard; they are variants of results in E. M. Alfsen [5], Chapter 2, §1 and L. Asimow and A. J. Ellis [24], Chapter 2. Theorem 4.34 is taken from S. Teleman [449]. Section 4.4 is a well-known application of G. Mokobodzki’s theorem 4.24; it can be found for example in S. A. Argyros, G. Godefroy and H. P. Rosenthal [19] and E. Odell and H. P. Rosenthal [366]. Exercise 4.40 is proved in M. Hervé [223]. Exercise 4.45 is the classical way of proving Lemma 3.77, see E. M. Alfsen [5], Proposition I.4.10. In A. Goullet de Rugy [200], Exercise 4.46 is ascribed to G. Mokobodzki.

Chapter 5

Perfect classes of functions and representation of affine functions

The aim of this chapter is to introduce a hierarchy of Borel sets in topological spaces, show its relation to inductive generation of Baire and Borel functions, and apply these facts to get a link between topological properties of function spaces and compact convex sets. The first part is devoted to an abstract generation of classes of sets and functions from a given family of sets. It turns out that a suitable starting point for this procedure is an algebra of sets, because then the standard properties of Borel sets hold (see Proposition 5.4) as well as the characterization of Baire classes of functions via level sets (see Theorem 5.9). The next section employs an abstract procedure for creating classes of Baire and Borel sets in topological spaces. The main result is Theorem 5.16, showing that Borel classes are “quotient” with respect to perfect mappings. The key ingredient of the proof is elementary Lemma 5.15. Abstract results obtained so far are applied in Section 5.3 to get a characterization of Baire and Borel mappings via their measurability (see Corollary 5.22). As another consequence we get Theorem 5.23 characterizing Baire-one functions on compact spaces. The nice behavior of our Borel classes with respect to perfect mappings is pointed out by Theorem 5.26, which is a result widely used throughout the book. The next section begins to describe the interplay between topological results and properties of functions on compact convex sets. To emphasize crucial properties of affine functions, we introduce in Definition 5.28 a notion of “affinely perfect functions”. Theorem 5.31 and Corollary 5.32 show to what extent the behavior of an affine function is determined by its properties on the closure of extreme points and Theorem 5.33 provides examples of affinely perfect classes. The last section is crucial for the transfer of properties of function spaces to the context of compact convex sets. A precise description of this transfer is contained in Theorem 5.40 and Corollary 5.41. First we recall several definitions. Definition 5.1 (Sublattices and algebras of sets). If X is a set and F is a family of its subsets, then F is a sublattice if ∅, X ∈ F and F is closed with respect to finite unions and finite intersections. The family F is an algebra if F is a sublattice closed with respect to complements. If A is a family of sets in a set X, we write Aσ (or, Aδ ) for all countable unions (or, intersections, respectively) of sets from A. If f : X → Y

136

5 Perfect classes of functions and representation of affine functions

is a mapping from X to a topological space Y , we say that f is F-measurable, if f −1 (U ) ∈ F for each open U ⊂ Y .

5.1

Generation of sets and functions

Definition 5.2 (Abstract Borel classes of sets). If F is a family of sets in a set X, we define Borel classes generated by F as follows: Let Σ1 (F) := F, Π1 (F) := {X \ F : F ∈ F}, and for each countable ordinal α ∈ (1, ω1 ), let [ Σα (F) := Πβ (F) σ

β s be arbitrary. For any m ∈ N, x ∈ / Bmn if m = 1, . . . , q, or x ∈ / m j=1 Cjn if m > q. In both cases, x∈ /

n [

(Bpn ∩

p=1

p \

Cjn ) = An .

j=1

This concludes the proof. Definition 5.5 (Abstract Baire classes of mappings). Let Φ be a family of mappings from a set X to a topological space Y . We define abstract Baire classes generated by Φ inductively as follows: Let Φ0 := Φ and for each countable S ordinal α ∈ (0, ω1 ), let Φα be the family of all pointwise limits of sequences from β S 0. Since Y is precompact, we may find a finite set {y1 , . . . , yn } ⊂ Y , so that Y ⊂ ni=1 U (yi , ε). Then {f −1 (U (yi , ε)) : i = 1, . . . , n} is a cover of X consisting of sets in Σα+1 (F).

5.1 Generation of sets and functions

139

Using Lemma 5.4(f), we can find a partition {A1 , . . . , An } of X consisting of sets in ∆α+1 (F) so that Ai ⊂ f −1 (U (yi , ε)), i = 1, . . . , n. Then the function x ∈ Ai , i = 1, . . . , n,

g(x) = yi ,

is Σα+1 (F)-measurable and satisfies ρ(f (x), g(x)) < ε, x ∈ X. The proof of Theorem 5.9 below closely follows the proof of Theorem 24.3 in [262]; however, we include the following lemma. Lemma 5.8. Let F be an algebra of sets in a set X and let 0 < ε < ε0 . Let f, g : X → Y be mappings from a set X to a separable metric space (Y, ρ) such that ρ(f (x), g(x)) < ε, x ∈ X. Let fn , gn : X → Y , n ∈ N, be Σγn (F)-measurable mappings such that fn → f , gn → g and γn ∈ (1, ω1 ). Then there exist Σγn (F)-measurable mappings hn : X → Y , n ∈ N, so that hn → g and ρ(fn (x), hn (x)) < ε0 for each x ∈ X and n ∈ N. Proof. Given the objects as in the lemma, for n ∈ N we set A1n := {x ∈ X : ρ(fn (x), gn (x)) < ε0 },

A2n := {x ∈ X : ρ(fn (x), gn (x)) > ε}.

By the separability of the space Y we get that the sets A1n , A2n are in Σγn (F). Indeed, we select countably many open sets Uk , Vk , k ∈ N, in Y such that ∞ [

(Uk × Vk ) = {(y1 , y2 ) ∈ Y × Y : ρ(y1 , y2 ) < ε0 }.

k=1

Hence A1n =

∞ [

(fn−1 (Uk ) ∩ gn−1 (Vk ))

k=1

is contained in Σγn (F). We similarly argue for the set A2n . Thus we may use Proposition 5.4(f) to find disjoint sets Bn1 , Bn2 ∈ Σγn (F) such that n ∈ N. Bn1 ⊂ A1n , Bn2 ⊂ A2n , and Bn1 ∪ Bn2 = X, By setting ( gn hn := fn

on Bn1 , on Bn2 ,

n ∈ N,

we get Σγn (F)-measurable mappings satisfying our requirements. Theorem 5.9. Let F be an algebra of sets in a set X and let Y be a separable metrizable space. Let Φ1 stand for the family of all Σ2 (F)-measurable mappings from X to Y and, for α ∈ (1, ω1 ), let Φα be defined from Φ1 as in Definition 5.5. Then, for each α ∈ (0, ω1 ) and f : X → Y , the following assertions are equivalent:

140


(i) f ∈ Φα , (ii) f is Σα+1 (F)-measurable. Proof. We prove (i) =⇒ (ii) by transfinite induction. If α = 1, the assertion holds by the definition of Φ1 . We assume its validity for all β < α, where α ∈ (1, ω1 ). If f ∈ Φα , then f = limn→∞ fn , where fn ∈ ΦS αn for some αn < α, n ∈ N. Given an S ∞ U = open set U ⊂ Y , we write U = ∞ m m=1 Um , where the sets Um are open. m=1 Then ∞ [ ∞ \ ∞ [ f −1 (U ) = fk−1 (Um ) m=1 n=1 k=n

and the sets fk−1 (Um ) ∈ Παk +1 (F) by the inductive assumption. Thus f −1 (U ) ∈ Σα+1 (F). For the proof of (ii) =⇒ (i) we use transfinite induction again. The case α = 1 follows from the definition. Assume that α ∈ (1, ω1 ) and the assertion holds for all β < α. Let f : X → Y be a Σα+1 (F)-measurable function. We assume first that f : X → Y has only finitely many values; that is, there exist pairwise disjoint sets Ai ⊂ X and points yi ∈ Y , i = 1, . . . , n, such that Ai ∈ ∆α+1 (F) and f = yi on Ai , i = 1, . . . , n. Assume first that α is isolated, that is, α = α0 + 1 for some α0 ∈ [1, ω1 ). According to Proposition 5.4(g), there exist sequences {Aki }∞ k=1 of sets in ∆α (F), i = 1, . . . , n, such that cAk → cAi . We may assume that {Ak1 , . . . , Akn } is a disjoint family for each i k ∈ N. (Otherwise we would take B1k = Ak1 and Bjk = Akj \ (Ak1 ∪ · · · ∪ Akj−1 ), j = 2, . . . , n.) Then fk (x) = yi , x ∈ Aki , i = 1, . . . , n, k ∈ N, are Σα (F)-measurable mappings that converge pointwise to f . By the inductive assumption, the mappings fk are contained in Φα0 , and so f ∈ Φα . Assume that α is limit. In this case we use Proposition 5.4(h) and find sequences {Aki }∞ k=1 of sets in ∆αk (F), i = 1, . . . , n, where αk < α for every k ∈ N, such that cAk → cAi , i = 1, . . . , n. The rest of the argument is analogous to that above. i Assume now that f : X → Y is an arbitrary Σα+1 (F)-measurable mapping. Since Y is separable and metrizable, we can fix on Y a compatible metric ρ such that (Y, ρ) is precompact. Hence we may find finite sets Y k := {y1k , . . . , ynk k } ⊂ Y , k ∈ N, so S k U (yik , 2−k ). that Y k ⊂ Y k+1 and Y ⊂ ni=1 −1 We fix k ∈ N. Then {f (U (yik , 2−k )) : i = 1, . . . , nk } is a cover of X consisting of sets in Σα+1 (F). Using Proposition 5.4(f) we find a partition {Aki : i = 1, . . . , nk } of X consisting of sets in ∆α+1 (F) so that Aki ⊂ f −1 (U (yik , 2−k )), i = 1, . . . , nk . Then the function f k (x) = yi , x ∈ Aki , i = 1, . . . , nk ,

5.1 Generation of sets and functions

141

is Σα+1 (F)-measurable. According to the reasoning above, we may find mappings fnk ∈ Φαnk so that fnk → f k as n → ∞ and αnk < α. By implication (i) =⇒ (ii), fnk ∈ Σαnk +1 (F),

n, k ∈ N.

(5.1)

If x ∈ X, from ρ(f k (x), f (x)) < 2−k it follows that ρ(f k (x), f k+1 (x)) < 2−k+1 . Now we employ Lemma 5.8 to replace {fnk } by {hkn } in such a way that • h1 = f 1 , n ∈ N, n n •

−k+2 , x ∈ X, n, k ∈ N, ρ(hkn (x), hk+1 n (x)) < 2

•

hkn is Σγnk +1 (F)-measurable for some γnk < α, n, k ∈ N,

hkn → f k , k ∈ N. To do this, we proceed inductively. First we set •

h1n := fn1 ,

n ∈ N,

and having hjn : X → Y defined for all n ∈ N and j = 1, . . . , k − 1, we use Lemma 5.8 to get mappings hkn : X → Y , n ∈ N, so that k −(k−1)+2 ρ(hk−1 , n (x), hn (x)) < 2

x ∈ X,

n ∈ N,

and hkn is Σγnk +1 (F)-measurable, where γnk = max{γn(k−1) , αnk }, n ∈ N. Having done this, we set hk (x) := hkk (x),

x ∈ X,

k ∈ N.

Then the mappings hk are Σγkk +1 (F)-measurable, where γkk < α, and hk → f . By the inductive assumption, hk ∈ Φγkk , k ∈ N. Hence f ∈ Φα , as required. Theorem 5.10. Let F be an algebra of sets in a set X and let Φ1 stand for the family of all Σ2 (F)-measurable functions from X to R and, for α ∈ (1, ω1 ), let Φα be the abstract Baire class defined from Φ1 . Let α ∈ (0, ω1 ). Then Φα is a vector space containing constant functions that is closed with respect to uniform convergence, finite minima, multiplications and inversions (if defined). Proof. Let α ∈ (0, ω1 ) be given. By Theorem 5.9, it is enough to show that the family M of all Σα+1 (F)-measurable functions on X possesses all the required properties. Obviously, M contains constant functions. Further, if f, g are Σα+1 (F)-measurable functions on X, the mapping ψ : X → R2 defined as ψ(x) = (f (x), g(x)),

x ∈ X,

is Σα+1 (F)-measurable. (This follows from the fact that any open set in R2 is a countable union of open rectangles.) If ϕ : R2 → R is any of the following mappings (a, b) 7→ ab,

(a, b) 7→ a + b,

(a, b) 7→ a ∨ b,

142


then ϕ is continuous. Hence ϕ ◦ ψ is Σα+1 (F)-measurable, and thus M is stable with respect to addition, multiplication and finite minima. Using the mappings t 7→ ct,

1 t 7→ , t

t ∈ R,

t ∈ R \ {0},

we get that cf ∈ M provided f ∈ M and c ∈ R, and f1 ∈ M provided f ∈ M and f does not attain 0 on X. Let {fn } be a sequence of functions in M that converges uniformly to f . Without loss of generality we may assume that supx∈X |f (x) − fn (x)| < n1 , n ∈ N. Then, for any c ∈ R, we get {x ∈ X : f (x) > c} =

∞ [ ∞ [ k=1 n=k

1 {x ∈ X : fn (x) > c + }. k

Thus f is Σα+1 (F)-measurable and the proof is complete.

5.2

Baire and Borel sets

Notation 5.11. If X is a topological space, we write G(X) for the sublattice of all open subsets of X. We recall the following notions from Subsection A.2.B. A subset A of X is called a zero set if A = f −1 ({0}) for a continuous real-valued function f on X. It is clear that such a function f can be chosen with values in [0, 1]. We recall that a subset A of a topological space X is Fσ if A can be written as a countable union of closed sets. The complement of an Fσ set is called a Gδ set. Lemma 5.12. Let F be a sublattice of sets in a set X. Then the algebra generated by F consists of all finite disjoint unions of differences of sets from F. Proof. Let B denote the family of all finite disjoint unions of differences of sets from F and let A S be the algebra generated by F. Obviously, B ⊂ A. Let B := ni=1 (Fi \Hi ), where the sets Fi , Hi are contained in F and {Fi \Hi : i = 1, . . . , n} is a disjoint family. Since F is a sublattice, we may assume that Hi ⊂ Fi for each i = 1, . . . , n. Then   n \ [ \ [  Hi \ X \ B = (Hi ∪ (X \ Fi )) = Fi  , i=1

I⊂{1,...,n}

i∈I

i∈{1,...,n}\I

where the union on the right-hand side is disjoint. Indeed, let I, J ⊂ {1, . . . , n} be distinct sets. Assume that j ∈ I \ J. If     \ [ \ [ x ∈  Hi \ Fi  ∩  Hi \ Fi  , i∈I

i∈{1,...,n}\I

i∈J

i∈{1,...,n}\J

5.2 Baire and Borel sets

143

then x ∈ Hj ⊂ Fj

and

[

x∈ /

Fi .

i∈{1,...,n}\J

Hence x ∈ / Fj , a contradiction. Thus X \ B ∈ B and B is stable with respect to complements. Obviously, B is stable with respect to finite unions of disjoint sets. Let {Fi \ Hi : i ∈ I} and {Gj \ Vj : j ∈ J} be disjoint finite families and Fi , Hi , Gj , Vj ∈ F. Then [

(Fi \ Hi ) ∩

i∈I

[

(Gj \ Vj ) =

j∈J

[

(Fi \ Hi ) ∩ (Gj \ Vj )

(i,j)∈I×J

=

[

(Fi ∩ Gj ) \ (Hi ∪ Vj ),

(i,j)∈I×J

where the last set is a finite disjoint union of differences of sets from F. Thus intersection of any pair of sets from B is again in B, and, consequently, B is stable with respect to finite intersections. Thus B is an algebra and A ⊂ B. Definition 5.13 (Baire and Borel sets). We consider the following families of subsets of X. (a) The algebra Bas(X) generated by zero sets. By Lemma 5.12, Bas(X) =

n [

(Fi \ Hi ) : Fi , Hi is a zero set in X, n ∈ N .

i=1

(b) The algebra Bos(X) generated by closed subsets of X. As above, Bos(X) =

n [

(Fi \ Hi ) : Fi , Hi closed in X, n ∈ N .

i=1

(c) If we start the Borel hierarchy as defined in Section 5.1 from the sublattice G(X) of all open subsets of X, for metrizable spaces we get the standard Borel classes Σ0α (X) and Π0α (X) as defined in Definition A.113. We denote as Σ0α (G(X)) and Π0α (G(X)) the families obtained by this procedure. We show below its relation to the families defined in (a), (b). We just mention that a set A belongs to Σ02 (G(X)) if and only if A = F ∪ G, where F is Fσ and G is open. For both families (a) and (b) above we consider the abstract Borel classes defined in Section 5.1 and call them the sets of additive, or multiplicative, Borel class α and the sets of additive, or multiplicative, Baire class α for α < ω1 , respectively.

144


Lemma 5.14. Let X be a topological space. Then the following assertions hold. (a) Σα (Bas(X)) ⊂ Σα (Bos(X)), α ∈ (0, ω1 ). S (b) The family α 0 such that f1 ∨f2 +ε ≤ g1 ∧g2 −ε. Then assumption (vi) yields the existence of a function h ∈ Ac (H) satisfying f1 ∨ f2 < f1 ∨ f2 + ε ≤ h ≤ g1 ∧ g2 − ε < g1 ∧ g2 . To prove (vii) =⇒ (v), pick f1 , . . . , fm ∈ Ac (H) and g1 , . . . , gn ∈ Ac (H) such that f1 ∨ · · · ∨ fm < g1 ∧ · · · ∧ gn . We use assumption (vii) and find a function h1 ∈ Ac (H) with f1 ∨ f2 < h1 < g1 ∧ g2 . Inductively choose functions h2 , . . . , hm−1 ∈ Ac (H) satisfying fl+1 ∨ hl−1 < hl < g1 ∧ g2 ,

2 ≤ l ≤ m − 1.

Then the function b h1 := hm−1 fulfills f1 ∨ · · · ∨ fm < b h1 < g1 ∧ g2 . Thus f1 ∨ · · · ∨ b fm < h1 ∧ g3 . Use the previous construction to find a function b h2 ∈ Ac (H) with b b f1 ∨ · · · ∨ fm < h2 < h1 ∧ g3 . After (n − 1)-th step of this construction we arrive at the desired function h := b hn−1 ∈ Ac (H) satisfying f1 ∨ · · · ∨ fm < h < g1 ∧ · · · ∧ gn .

6.2 Characterizations of simplicial spaces

177

For the proof of (v) =⇒ (iii), let f, −g ∈ Kc (H) with f < g be given. Find ε > 0 so that f + ε < g − ε. By Proposition 3.55, there exist functions fb, −b g ∈ W(Ac (H)) with f − ε ≤ fb ≤ f and g ≤ gb ≤ g + ε. An appeal to (v) provides a function h ∈ Ac (H) satisfying f ≤ fb + ε < h < gb − ε ≤ g, which is the desired function. To finish the proof we verify condition (iii) of Theorem 6.5 provided that (iii) holds. Let f ∈ Kc (H) be given. By Exercise 3.96, f ∗ = inf{h ∈ Ac (H) : h > f }. Since the latter family is down-directed by our assumption, we can employ Theorem A.84 and deduce that f ∗ is H-affine. Hence (iii) =⇒ (i), which concludes the proof. Remark 6.17. The conditions from Theorem 6.16(vi) and (vii) express that the space Ac (H) has the Riesz interpolation property and the weak Riesz interpolation property, respectively. Theorem 6.18 (Riesz decomposition property). For a function space H on a compact space K, the following assertions are equivalent: (i) H is a simplicial function space, (ii) for any positive functions f , g1 , g2 ∈ Ac (H) with f ≤ g1 +g2 there exist positive functions h1 , h2 ∈ Ac (H) such that h1 + h2 = f and h1 ≤ g1 , h2 ≤ g2 . Proof. By Proposition A.11 we know that condition (ii) is equivalent to condition (vi) of Theorem 6.16. Remark 6.19. If the function space Ac (H) satisfies the requirements of Theorem 6.18(ii), we say that Ac (H) has the Riesz decomposition property. Theorem 6.20 (Finite binary intersection property). Let H be a function space on a compact space K. Then the following assertions are equivalent: (i) H is simplicial, (ii) if n ∈ N and {Bi }ni=1 is a family ofTclosed balls in Ac (H) satisfying Bi ∩Bj 6= ∅ whenever i, j ∈ {1, . . . , n}, then ni=1 Bi 6= ∅.

178

6 Simplicial function spaces

Proof. Assume that H is a simplicial function space and {B(hi , ri )}ni=1 is a finite family of closed balls in Ac (H) with centers hi and radii ri , i = 1, . . . , n, such that B(hi , ri ) ∩ B(hj , rj ) 6= ∅, i, j = 1, . . . , n. Then the functions fi = hi − ri and gi = hi + ri , i = 1, . . . , n, satisfy f1 ∨ · · · ∨ fn ≤ g1 ∧ · · · ∧ gn . By Theorem 6.16, there exists a function h ∈ Ac (H) so that f1 ∨ · · · ∨ fn ≤ h ≤ g1 ∧ · · · ∧ gn . T Then h ∈ ni=1 B(hi , ri ), concluding the proof of (i) =⇒ (ii). For the proof of (ii) =⇒ (i), let f, g1 , g2 ∈ Ac (H) be positive functions such that f ≤ g1 + g2 . Without loss of generality we may assume that g1 + g2 ≤ 1 on K. Then the closed balls B1 := B(1, 1), B2 := B(g1 − 1, 1), B3 := B(f − 1, 1), B4 := B(1 + f − g2 , 1), satisfy Bi ∩ Bj 6= ∅, i, j ∈ {1, . . . , 4}. Indeed, it is straightforward to verify that 0 ∈ B1 ∩ B2 ∩ B3 ,

f ∈ B1 ∩ B3 ∩ B4

and

g1 ∈ B2 ∩ B4 .

T By the assumption, there exists f1 ∈ 4i=1 Bi . We set f2 := f − f1 . Then f1 ≥ 0 (because f1 ∈ B1 ), f1 ≤ g1 (because f1 ∈ B2 ), f2 ≥ 0 (because f1 ∈ B3 , and so f1 ≤ f ), and f2 ≤ g2 (because f1 ∈ B4 , and so f1 ≥ f − g2 ). By Theorem 6.18, H is simplicial. Remark 6.21. If the function space Ac (H) satisfies the requirements of Theorem 6.20(ii), we say that Ac (H) has the finite binary intersection property. Corollary 6.22. Let Hi be a function space on a compact space Ki , i = 1, 2, and let the spaces Ac (H1 ) and Ac (H2 ) be isometrically isomorphic. Then H1 is simplicial if and only if H2 is simplicial. Proof. The proof follows from Theorem 6.20, because the finite binary intersection property of a Banach space is preserved by an isometric isomorphism.

6.3

Simplicial spaces as L1 -preduals

Lemma 6.23. Let L ∈ H∗ be a nonzero functional on H. Then there exists a boundary measure µ so that kµk = kLk and µ(h) = L(h) for any h ∈ H. Moreover, if L is positive, the measure µ can be taken to be positive.

6.3 Simplicial spaces as L1 -preduals

179

Proof. Let ν ∈ M(K) be a measure corresponding to a Hahn–Banach extension of L to C(K), kνk = kLk 6= 0. If ν = ν + − ν − is the decomposition of ν into its positive and negative part, let µ1 , µ2 be maximal measures such that ν + ≺ µ1 and ν − ≺ µ2 . If we set µ := µ1 − µ2 , then µ is a boundary measure and ν − µ ∈ H⊥ . If µ = µ+ − µ− is the decomposition of µ into the positive and negative part, we get kµk = µ+ (K) + µ− (K) ≤ µ1 (K) + µ2 (K) = ν + (K) + ν − (K) = kνk. On the other hand, kνk = kLk = sup |L(h)| h∈BH

= sup |ν(h)| = sup |µ(h)| ≤ kµk. h∈BH

h∈BH

Hence kµk = kνk. If L is positive, then kLk = L(1). If µ ∈ Mbnd (H) is the measure obtained by the argument above, we have kµk ≥ µ(1) = L(1) = kLk = kµk. It follows that µ is a positive measure. Proposition 6.24. Let H be a simplicial function space on a compact space K. Then there exists an order preserving isometry R : Mbnd (H) → (Ac (H))∗ . In particular, (Ac (H))∗ is a lattice. Proof. Let R : Mbnd (H) → (Ac (H))∗ be the restriction mapping. By Proposition 6.9, R is injective. Using Lemma 6.23 applied to Ac (H) and Proposition 3.67, we infer that R is a surjective isometry. It is easy to check that Rµ is a positive functional on Ac (H) if and only if µ ∈ Mbnd (H) is a positive measure (use Lemma 6.23). Hence R preserves order, which concludes the proof. Theorem 6.25. Let H be a simplicial function space on a compact space K. Then (Ac (H))∗ is order isometric to a space L1 (X, Σ, σ) for a suitable measure space (X, Σ, σ). Proof. By Proposition 6.24 it is enough to show that Mbnd (H) is order isometric to L1 (X, Σ, σ) for a suitable measure space (X, Σ, σ). We use throughout the proof the identification of Radon measures on K with finite measures defined on Borel sets that are inner regular with respect to compact sets (see Proposition A.73). Using Zorn’s lemma, we find a set {µi : i ∈ I} of pairwise singular probability measures in Mmax (H) that is maximal with respect to inclusion. For each i ∈ I we consider the measure space (K, Σi , σi ), where Σi denotes the σ-algebra of Borel sets

180


in K and σi is the restriction of µi to Σi . Let (X, Σ, σ) be the disjoint union of the spaces (K, Σi , σi ), i ∈ I. (More precisely, let X be the disjoint union of spaces Ki , i ∈ I, where each Ki equals K. Further, P Σ consists of all sets A ⊂ X satisfying A ∩ Ki ∈ Σi for each i ∈ I and let σ(A) = i∈I σi (A ∩ Ki ) for any A ∈ Σ.) To construct an order isometry T : Mbnd (H) → L1 (X, Σ, σ), let ν ∈ Mbnd (H) be arbitrary. For any i ∈ I, let νi denote the absolutely continuous part of the restriction of ν to the σ-algebra of Borel sets with respect to σi and let hi ∈ L1 (K, Σi , σi ) be a Borel function that is the Radon–Nikodym derivative of νi with respect to σi . We define T ν ∈ L1 (X, Σ, σ) as (T ν)|Ki := hi ,

i ∈ I.

Since the measures νi , i ∈ I, are pairwise singular, Z X XZ |hi | dµi = |νi |(K) ≤ |ν|(K). |T ν| dσ = X

i∈I

K

(6.3)

i∈I

Thus the function T ν is indeed in L1 (X, Σ, σ) and kT νk ≤ kνk. Conversely, we need to show P that kνk ≤ kT νk. By (6.3) it is enough to show that P ν = i∈I νi . Let λ := ν − i∈I νi and λ+ , λ− be the positive and negative part of λ, respectively. Assume that λ+ 6= 0. Since λ ⊥ µi for any i ∈ I, (λ+ (K))−1 λ+ is a probability measure contradicting Pthe maximality of {µi : i ∈ I}. Analogously we can show that λ− = 0. Thus ν = i∈I νi and T is an isometry. P To check that T is surjective, let h ∈ L1 (X, Σ, σ) be given. Then ν := i∈I hσi is in M(K) and T ν = h. If ν ∈ Mbnd (H) is positive, each νi is positive as well, and thus T ν is a positive function on X. This finishes the proof.

6.4

The weak Dirichlet problem and Ac (H)-exposed points

Definition 6.26 (The weak Dirichlet problem). We say that the weak Dirichlet problem for a function space H on a compact space K is solvable if for every compact set D ⊂ ChH (K) and every f ∈ C(D) there is a “solution” h ∈ Ac (H) such that h=f

on

D

and khk = kf k.

The solution h of the weak Dirichlet problem for f , if it exists, is by no means uniquely determined. Theorem 6.27. If a function space H on a compact space K is simplicial, then the weak Dirichlet problem for H is solvable.

6.4 The weak Dirichlet problem and Ac (H)-exposed points

181

If, moreover, K is metrizable, then H is simplicial if and only if the weak Dirichlet problem for H is solvable. Proof. The proof of the first part of the theorem is an immediate consequence of the Edwards in-between theorem 6.6. Indeed, assume that H is simplicial, D ⊂ ChH (K) compact and f ∈ C(D). Set ( ( min f (D), x ∈ K \ D, max f (D), x ∈ K \ D, s(x) := t(x) := f (x), x ∈ D, f (x), x ∈ D. Then s ∈ Kusc (H), t ∈ S lsc (H), s ≤ t on K, so by the Edwards in-between theorem 6.6 there exists h ∈ Ac (H) such that s ≤ h ≤ t on K and the proof is complete. Assume now that H is a function space on a metrizable compact space K and x ∈ K. If µ and ν are maximal measures from Mx (H), our aim is to show that µ = ν. Select f ∈ C(K). Now if ε > 0, then Corollary 3.61 and the regularity of measures µ and ν yield the existence of a compact set D ⊂ ChH (K) with (µ + ν)(K \ D) < ε. Since the weak Dirichlet problem for H is solvable, there exists h ∈ Ac (H) such that h=f

on D

and khk = kf k.

Then |µ(f ) − ν(f )| ≤ |µ(f ) − h(x)| + |h(x) − ν(g)| = |µ(f ) − µ(h)| + |ν(h) − ν(f )| Z Z Z ≤ |f − h| dµ + |h − f | dν = |f − h| d(µ + ν) K K K Z = |f − h| d(µ + ν) ≤ 2kf kε. K\D

As ε > 0 is arbitrary, we have µ(f ) = ν(f ), as needed. Theorem 6.28. Let H be a simplicial function space on a metrizable compact set K. Then any point of the Choquet boundary ChH (K) is Ac (H)-exposed. Proof. Let x ∈ ChH (K). For y ∈ ChH (K)\{x}, let hy ∈ Ac (H) denote the solution of the weak Dirichlet problem for the set {x, y} such that 0 ≤ hy ≤ 1,

hy (x) = 0,

hy (y) = 1.

In view of Theorem 6.27, such a solution exists. Then [ ChH (K) \ {x} ⊂ {t ∈ K : hy (t) > 0} . y∈ChH (K)\{x}

182


Since ChH (K) \ {x} is Lindelöf, there are yn ∈ ChH (K), n ∈ N, such that ChH (K) \ {x} ⊂

∞ [

{t ∈ K : hyn (t) > 0} .

n=1

Set h :=

∞ X 1 hy . 2n n n=1

Ac (H),

h ≥ 0 on K and h(x) = 0. It remains to show that h > 0 on Then h ∈ K \ {x}. Let z ∈ K satisfy h(z) = 0. Since h > 0 on ChH (K) \ {x} and the maximal measure δz at z is carried by ChH (K), it follows that spt δz ⊂ {x}. Thus z = x. Remark 6.29. There is a more general theorem even in a nonmetrizable case: If H is a simplicial function space on a compact space K, then any Gδ point of the Choquet boundary ChH (K) is Ac (H)-exposed, cf. Exercise 8.75. In general, there is no hope to have an “exposing function” from H. Consult examples of Exercises 6.76 and 6.77.

6.5

The Dirichlet problem for a single function

If K is a topological space, we recall that B bα (K) is the space of all bounded functions of Baire class α on K (see Definition 5.20). Definition 6.30 (Envelopes defined by non-continuous functions). Let H be a function space on a compact space K and α ∈ [0, ω1 ). For a bounded function f defined at least on the set ChH (K), we set ChH (K) ∗,α

f

:= inf{h ∈ A(H) ∩ B bα (K) : h ≥ f on ChH (K)} and

f∗,α := −ChH (K)(−f )∗,α .

ChH (K)

We remark that ChH (K)f∗,α ≤ ChH (K)f ∗,α due to the Minimum principle 3.86. Theorem 6.31. Let α ∈ [0, ω1 ), f be a bounded function on ChH (K) and let g := ChH (K)f ∗,α . Then the following assertions are equivalent: (i) The function f satisfies (i1)

ChH (K)f ∗,α

= ChH (K)f∗,α on ChH (K),

(i2) g ∈ Bbα (ChH (K)), (i3) if µ1 , µ2 ∈ Mx (H) are maximal measures for x ∈ K \ ChH (K), then µ1 (g) = µ2 (g).

6.5 The Dirichlet problem for a single function

183

(ii) There exists a unique function h ∈ A(H)∩B bα (K) such that h = f on ChH (K). (iii) There exists a function h ∈ A(H) ∩ B bα (K) such that h = f on ChH (K). Proof. Let a bounded function f on ChH (K) be given. To start the proof we notice that (ii) =⇒ (iii) is obvious. For the proof of (iii) =⇒ (i), let h ∈ B bα (K) ∩ A(H) extend f . Let b h ∈ B bα (K) ∩ A(H) be any function such that f ≤ b h on ChH (K). By Ch (K) ∗,α b H the Minimum principle 3.86, h ≤ h on K. Hence h ≤ f . Analogously we obtain ChH (K)f∗,α ≤ h. On the other hand, h ≤ ChH (K)f∗,α ≤ ChH (K)f ∗,α ≤ h by the definition. Hence h = ChH (K)f ∗,α = ChH (K)f∗,α on ChH (K) and property (i1) follows. Since (i2) and (i3) are obvious, the proof of (iii) =⇒ (i) is finished. To close the chain of implications, we have to verify that (i) =⇒ (ii). It follows from the Minimum principle 3.86 that the extension is unique, provided it exists. Let f and g be as in (i). We denote Z := ChH (K). Step 1: For any z ∈ Z and µ ∈ Mz (H) ∩ M1 (Z), we have µ(g) = g(z). Indeed, let z ∈ Z and µ ∈ Mz (H) ∩ M1 (Z) be given. Then µ(g) = µ inf{h ∈ A(H) ∩ B bα (K) : h ≥ f on ChH (K)} ≤ inf{µ(h) : h ∈ A(H) ∩ B bα (K)), h ≥ f on ChH (K)} = inf{h(z) : h ∈ A(H) ∩ B bα (K), h ≥ f on ChH (K)} = ChH (K)f ∗,α (z) = g(z). Similarly we deduce that µ(g) ≥ g(z). This finishes the proof of Step 1. Step 2: If µ, ν ∈ M1 (Z) satisfy µ ≺ ν, then µ(g) = ν(g). Let µ ≺ ν be probability measures on K carried by Z. We denote M := {(εx , λ) ∈ M1 (Z) × M1 (Z) : εx ≺ λ}. By Proposition 3.89, there exists Λ ∈ M1 (M ) such that Z µ(f1 ) + ν(f2 ) = (εx (f1 ) + λ(f2 )) dΛ(εx , λ),

f1 , f2 ∈ C(Z).

(6.4)

M

By Proposition 3.90, formula (6.4) holds for any bounded universally measurable functions on Z. Hence we can apply (6.4) along with Step 1 to pairs (g, 0) and (0, g) consecutively to get Z Z µ(g) = εx (g) dΛ(εx , λ) = λ(g) dΛ(εx , λ) = ν(g). M

This concludes the proof of Step 2.

M

184


Step 3: If µ1 , µ2 ∈ Mx (H) ∩ M1 (Z) for some x ∈ K, then µ1 (g) = µ2 (g). For i = 1, 2, let νi be a maximal measure such that µi ≺ νi . Combining Step 2 and Step 1 with assumption (i3), we get µ1 (g) = ν1 (g) = ν2 (g) = µ2 (g). This finishes the proof of Step 3. Now we can define a desired extension h : K → R as h(x) := µ(g),

µ ∈ M1 (Z) ∩ Mx (H), x ∈ K.

It follows from Step 3 that h is well defined. Obviously, h = f on ChH (K). Finally, Theorem 5.31 yields that h ∈ B bα (K) ∩ A(H). This finishes the proof. Theorem 6.32. Let f be a bounded function on ChH (K) and let g := ChH (K)f ∗ . Then the following assertions are equivalent: (i) The function f satisfies (i1)

ChH (K)f ∗

= ChH (K)f∗ on ChH (K),

(i3) µ(g) = 0 for any boundary measure µ ∈ H⊥ . (ii) There exists a unique function h ∈ H such that h = f on ChH (K). (iii) There exists a function h ∈ H such that h = f on ChH (K). Proof. The proof follows from Theorem 6.31 once we realize that (i1) implies the continuity of g on ChH (K). Hence we can find a function h ∈ Ac (H) extending g. To finish the proof we have to verify that h ∈ H. Since µ(f ) = 0 for any boundary µ ∈ H⊥ , Theorem 5.31 gives h ∈ H⊥⊥ . Hence h ∈ H by the Hahn–Banach theorem. This concludes the proof. We recall that H1,b stands for the space of all functions of H-affine class 1 (see Definition 5.35). Theorem 6.33. Let f be a bounded function on ChH (K) and let g := Then the following assertions are equivalent:

ChH (K)f ∗,1 .

(i) The function f satisfies (i1)

ChH (K)f ∗,1

= ChH (K)f∗,1 on ChH (K),

(i2) g ∈ Bb1 (ChH (K)), (i3) µ(g) = 0 for any boundary measure µ ∈ H⊥ . (ii) There exists a unique function h ∈ H1,b such that h = f on ChH (K). (iii) There exists a function h ∈ H1,b such that h = f on ChH (K).

6.6 Special classes of simplicial spaces

185

Proof. The proof again follows from Theorem 6.31. The only fact we need to verify is that the obtained extension h is contained in H1,b . We already know that h ∈ B b1 (K) ∩ A(H). Using assumption (i3) we get from Theorem 5.31(b) that h ∈ H⊥⊥ . It follows from Proposition 5.42(a) that h ∈ H1,b . This concludes the proof. Remark 6.34. Inspection of the previous proofs shows that the whole procedure can be done for any affinely perfect class C of functions on compact convex sets (see Section 5.5). We define envelopes similarly to above, namely, for a bounded function f defined at least on the set ext X, we set ext X ∗,C

f

:= inf{h ∈ C(X) : h ≥ f on ext X}

and

f∗,C := −ext X(−f )∗,C .

ext X

Thus we obtain the following result. Theorem 6.35. Let X be a compact convex set, C be an affinely perfect class of functions, f be a bounded function on ext X and let g := ext Xf ∗,C . Then the following assertions are equivalent: (i) The function f satisfies (i1)

ext Xf ∗,C

= ext Xf∗,C on ext X,

(i2) the function µ 7→ µ(g), µ ∈ M1 (ext X), is contained in C(M1 (ext X)), (i3) µ(g) = 0 for any boundary measure µ ∈ A(X)⊥ . (ii) There exists a unique function h ∈ C(X) such that h = f on ext X. (iii) There exists a function h ∈ C(X) such that h = f on ext X.

6.6 6.6.A

Special classes of simplicial spaces Bauer simplicial spaces

Definition 6.36 (Bauer simplicial spaces and simplices). A simplicial function space H on K with a closed Choquet boundary ChH (K) is called a Bauer simplicial space. A Bauer simplex X is a Choquet simplex whose set of extreme points is closed. There are plenty of conditions saying that a compact convex set X is a Bauer simplex. In the following proposition we collect some of them for the case of general function spaces. Theorem 6.37. Let H be a function space on a compact space K. Then the following conditions are equivalent: (i) H is a Bauer simplicial space, (ii) for any x ∈ K there exists a unique representing measure µ ∈ Mx (H) such that µ is carried by ChH (K),

186


(iii) H is a simplicial space and the mapping x 7→ δx , x ∈ K, from K to M1 (K) is continuous, (iv) given a function f ∈ Kc (H), the function f ∗ is continuous and H-affine on K, (v) for any function f ∈ C(ChH (K)) there exists a function h ∈ Ac (H) such that f = h on ChH (K), (vi) for any f ∈ C(ChH (K)) there exists a function h ∈ Ac (H) such that f = h on ChH (K), (vii) for any bounded continuous function f on ChH (K) there exists a function h ∈ Ac (H) such that f = h on ChH (K), (viii) the space Ac (H) is a lattice in its natural ordering. Proof. For the proof that (i) =⇒ (ii), suppose that x ∈ K and µ ∈ Mx (H) with spt µ ⊂ ChH (K) are given. If f ∈ C(K), then by Bauer’s characterization of the Choquet boundary 3.24, {x ∈ K : f (x) < f ∗ (x)} ⊂ K \ ChH (K) = K \ ChH (K). Hence µ {x ∈ K : f (x) < f ∗ (x)} ≤ µ K \ ChH (K)) = 0. By Corollary 3.59, µ is maximal, and therefore µ = δx . Suppose that (ii) holds. Since any maximal measure is carried by ChH (K) by Proposition 3.64, (ii) implies that H is a simplicial space. Let {xα }α∈A be a net in K, xα → x. We show that δx is the only cluster point of {δxα }α∈A . We lose no generality by supposing that δxα → ν. As spt δxα ⊂ ChH (K), we have spt ν ⊂ ChH (K). Since ν ∈ Mx (H) by Proposition 3.40, it follows from (ii) that ν = δx , which proves that (ii) =⇒ (iii). Let f ∈ Kc (H). Theorem 6.5 along with Lemma 6.4 assures that f ∗ is an H-affine function and that f ∗ (x) = δx (f ) for any x ∈ K. Since the mapping x 7→ δx is continuous, f ∗ is continuous. This shows that (iii) =⇒ (iv). For the proof (iv) =⇒ (v), suppose that f is a continuous function on ChH (K). By Tietze’s theorem extend f continuously to the whole of K. We know from Theorem 6.5 that H is simplicial. Hence for each ε > 0, Theorem 6.8(a) provides functions f1 , f2 ∈ Kc (H) so that kT f − (T f1 − T f2 )k ≤ ε (for the definition of T see Definition 6.7). Since T (Kc (H)) ⊂ C(K) by our assumption, T f , as a uniform limit of continuous functions, is continuous as well. By Theorem 6.8(c), T f is H-affine. Since f = T f on ChH K and both functions are continuous, f = T f on ChH K. Suppose that (v) holds. It is enough to show that ChH (K) = ChH (K). Assume, in order to obtain a contradiction, that there is x ∈ ChH (K) \ ChH (K). Using Bauer’s


187

characterization of the Choquet boundary 3.24, we find a function f ∈ Kc (H) for which f (x) < f ∗ (x). By the assumption, there is h ∈ Ac (H) such that h = f on ChH (K). Since h − f ≥ 0 on ChH (K), the minimum principle for S c (H) (see Theorem 3.85) yields h−f ≥ 0 on K. Since h ≥ f on K, it follows from Proposition 3.25 that h = h∗ ≥ f ∗ . Hence h(x) = f (x) < f ∗ (x) ≤ h(x). This obvious contradiction yields the required conclusion, and therefore (v) =⇒ (vi). Obviously, (vi) =⇒ (vii). Suppose that (vii) holds and select h1 , h2 ∈ Ac (H). By (vii), there is h ∈ Ac (H) such that h = h1 ∨ h2 on ChH (K). If k ∈ Ac (H), k ≥ h1 ∨ h2 on K, then k ≥ h on ChH (K). Using the Minimum principle 3.85 again, we get k ≥ h on K. We see that every pair of elements of Ac (H) has a least upper bound, in other words, Ac (H) is a lattice in its natural ordering, hence (viii) holds. The proof will be completed once it is shown that (viii) =⇒ (i). To this end, let h1 , . . . , hn ∈ Ac (H) and let h ∈ Ac (H) be their least upper bound in Ac (H). By Proposition 3.25(a), h = (h1 ∨ · · · ∨ hn )∗ . We see that the function (h1 ∨ · · · ∨ hn )∗ is H-affine, and thus Theorem 6.5 yields that H is simplicial. It remains to prove that the Choquet boundary ChH (K) is closed. Assume that x ∈ ChH (K) \ ChH (K). With the aid of Bauer’s characterization of ChH (K) (see Theorem 3.24) we can find functions h1 , . . . , hn ∈ Ac (H) such that f := h1 ∨· · ·∨hn satisfies f (x) < f ∗ (x). As above, f ∗ ∈ Ac (H) is the least upper bound of h1 , . . . , hn in Ac (H) and f ∗ = f on ChH (K). The continuity of f and f ∗ yields the equality f (x) = f ∗ (x), which is an obvious contradiction. Proposition 6.38. The set M1 (K) of all probability Radon measures on a compact space K is a Bauer simplex. Proof. We use condition (vi) of Theorem 6.37 as a characterization of Bauer simplices. So assume that F ∈ C(ext M1 (K)) = C({εx : x ∈ K}) is given. We set H := µ(F ◦ ϕ),

µ ∈ M1 (K),

where ϕ : x 7→ εx is the homeomorphic mapping of K onto ext M1 (K) (cf. Proposition 2.27). Then H ∈ Ac (M1 (K)) by Proposition 5.30, and also H = F on ext M1 (K). This concludes the proof. Proposition 6.39. Let X be a Bauer simplex. Then there exists a compact space K such that X is affinely homeomorphic to M1 (K). Proof. Choose K as ext X and consider the mapping x 7→ δx , x ∈ K, where δx is the unique maximal measure representing x.

188


6.6.B

Markov simplicial spaces

Recall that a function space H on a compact space K is simplicial if and only if (Ac (H))⊥ ∩ Mbnd (H) = {0} (see Proposition 6.9), and that simplicial function spaces H having a closed Choquet boundary ChH (K) were labeled as Bauer simplicial spaces. The equivalence (i) ⇐⇒ (v) of Theorem 6.37 says that H is a Bauer simplicial space if and only if Ac (H)|ChH (K) = C(ChH (K)). In this subsection we will examine function spaces for which H|ChH (K) = C(ChH (K)). Before this we recall the notion of a Markov operator. Definition 6.40 (Markov operators and projections). Let K be a compact space. A Markov operator T on K is a bounded linear operator on C(K) with T (1) = 1 = kT k. It is easy to check that any Markov operator on K is positive (see the proof of Theorem 2.31). If, moreover, T 2 = T , the operator T is called a Markov projection on K. If T is a Markov operator on K and HT := {f ∈ C(K) : T f = f } , then HT is a function space on K provided it separates points of K. It is obvious that T f ∈ HT whenever T is a Markov projection and f ∈ C(K). Let T be a Markov operator on K. Given x ∈ K, the functional Φ : f 7→ T f (x),

f ∈ C(K),

is a positive linear functional on C(K). By the Riesz representation theorem A.72 there exists a unique Radon measure µTx on K such that Z T f (x) = f dµTx for any f ∈ C(K). K

Obviously, µTx ∈ Mx (HT ) for every x ∈ K. Lemma 6.41. Let T be a Markov projection on K such that the space HT separates points of K. Then ChHT (K) = x ∈ K : µTx = εx . In particular, the Choquet boundary ChHT (K) is closed.


189

Proof. Pick x ∈ K. If x ∈ ChHT (K), then Mx (HT ) = {εx }, hence µTx = εx . Assume now that x ∈ K \ ChHT (K). Then there exists s ∈ Kc (HT ) such that s(x) < s∗ (x) (see Theorem 3.24). Then T s ∈ HT and T s ≥ s. It follows that s(x) < s∗ (x) = inf {h(x) : h ∈ HT , h ≥ s} ≤ T s(x) = µTx (s), and therefore εx 6= µTx . Since ChHT (K) =

\

{x ∈ K : T f (x) = f (x)} ,

f ∈C(K)

the set ChHT (K) is closed. Theorem 6.42. Let H be a function space on K. The following assertions are equivalent: (i) there exists a Markov projection on K such that H = HT , (ii) H is a Bauer simplicial space and H = Ac (H), (iii) H|ChH (K) = C(ChH (K)). Proof. If (i) holds and f ∈ C(ChH (K)) = C(ChHT (K)) = C(ChHT (K)) is given, there exists g ∈ C(K) such that g = f on ChH (K). Then T g ∈ HT = H and T g = g = f on ChH (K), so (i) =⇒ (iii). Suppose now that (iii) holds. According to (i) ⇐⇒ (v) of Theorem 6.37, H is a Bauer simplicial space. It remains to show that H = Ac (H). To this end, let f ∈ Ac (H). There exists a function h ∈ H so that h = f on ChH (K). Since h − f ∈ Ac (H), the Minimum principle 3.85 yields the equality h = f on K. Hence, f ∈ H. This shows that (iii) =⇒ (ii). Finally, suppose that H is as in (ii). Given x ∈ K, let δx be the unique maximal measure in Mx (H). By Theorem 6.37, the mapping x 7→ δx is continuous, hence the function T f defined as T f (x) = δx (f ), x ∈ K, is continuous for each f ∈ C(K). Since T f is an H-affine function (cf. Theorem 6.8(c)), T is a Markov projection on K. To show that H = HT , let h ∈ H. Then T h(x) = δx (h) = h(x) for each x ∈ K. Thus, H ⊂ HT . Conversely, pick g ∈ HT . Since the function T g is in Ac (H), we get as above g = T g ∈ Ac (H) = H which shows that HT ⊂ H. This proves that (ii) =⇒ (i) and finishes the proof. Definition 6.43 (Markov simplicial spaces). Function spaces for which one of the equivalent conditions of Theorem 6.42 holds are labeled as Markov simplicial spaces. Corollary 6.44. If H is a Markov simplicial space on K, then the function space H is closed and H⊥ ∩ Mbnd (H) = {0}. Proof. The assertion easily follows from the facts that H = Ac (H) and that Ac (H) is closed (see Proposition 3.11(a)).

190


6.6.C

Simplicial spaces with Lindelöf boundaries

We obtained that the solution T f of the Dirichlet problem for a continuous function f on K is a pointwise limit of a sequence of H-affine continuous functions on K provided K is metrizable (see Definition 6.7 and Theorem 6.8(a)). We are able to extend this result to a general simplicial function space H, if we suppose that ChH (K) is a Lindelöf topological space. The precise statement is provided by Theorem 6.45. Theorem 6.45. Let H be a simplicial function space on a compact space K such that ChH (K) is Lindelöf. (a) Let a bounded function f ∈ Kusc (H) be a pointwise limit of a decreasing sequence of continuous functions. Then f ∗ is a pointwise limit of a decreasing sequence of continuous H-affine functions. (b) Any function f ∈ C b (ChH (K)) can be extended to a Baire-one H-affine function on K. Moreover, this extension is unique. Proof. Let f ∈ Kusc (H) be as in the statement of the theorem. Due to the Edwards inbetween theorem (Theorem 6.6), the set {h ∈ Ac (H) : h ≥ f } is down-directed and its pointwise infimum is f ∗ (see Corollary 3.25(a)). For x ∈ ChH (K), f ∗ (x) = f (x) and so, by Lemma A.54, there is a decreasing bounded sequence {fn } of continuous H-affine functions converging pointwise to f on ChH (K). The set B := {x ∈ K : fn (x) → f (x)} is a Baire set containing ChH (K), and therefore carries any maximal measure on K by Theorem 3.79(a). Hence, for all y ∈ K, Z Z ∗ f (y) = f (x) dδy (x) = lim fn (x) dδy (x) = lim fn (y). n→∞ B

B

n→∞

Thus the sequence {fn } decreases to f ∗ on K and the first part of the proof is finished. Let f be a bounded continuous function on ChH (K). We extend f to an upper semicontinuous function g on X := ChH (K) by the formula

g(x) :=

 

lim sup

f (y),

x ∈ X \ ChH (K),

y→x,y∈ChH (K)

 f (x),

x ∈ ChH (K).

By Lemma 3.18(d), X ∗

g (x) = g(x) = f (x),

x ∈ ChH (K).

By Lemma 3.21 and Theorem 3.24, f (x) = inf{h(x) : h ∈ S c (H), h ≥ g on X},

x ∈ ChH (K).


191

Since the latter set is down-directed, it follows from Lemma A.54 that there exist a decreasing sequence of continuous H-concave functions {gn } and an increasing sequence of continuous H-convex functions {hn } such that hn % f

and

gn & f

on ChH (K).

We have hn ≤ gn and so, by Theorem 6.6, for each n ∈ N, there is a continuous H-affine function fn with hn ≤ fn ≤ gn . Then {fn } is a bounded sequence from Ac (H) pointwise converging on ChH (K) to f , and thus, by Corollary 3.76(a), it converges pointwise on K. Hence the function h := lim fn n→∞

is the required extension of f to a Baire-one H-affine function on K. The uniqueness of the extension follows from the Minimum principle 3.86. Theorem 6.46. The following assertions are equivalent: (i) H is simplicial and T (C(K)) ⊂ (Ac (H))1,b , (ii) H is simplicial and T (C(K)) ⊂ B 1 (K), (iii) H is simplicial and T f is a Baire function for each f ∈ C(K), (iv) f ∗ is an H-affine Baire function for any continuous H-convex function f on K, S (v) F := α 0, the set {x ∈ L : f ∗ (x) − f (x) > δ} is finite. Hence it follows that the sets {x ∈ L : f ∗ (x) − f (x) > c}

and

{x ∈ L : f ∗ (x) − f (x) ≥ c}

are finite for each c ∈ R. Since the points of K are Gδ , the function f ∗ − f is Baire-one by Theorem A.124. Hence f ∗ is Baire-one as well. By Theorem 6.8(a), T (C(K)) ⊂ B 1 (K), and thus H is Lion by Theorem 6.46.

6.6.D

Simplicial spaces with boundaries of type Fσ

We know that (in a simplicial case) the Dirichlet problem on the Choquet boundary is solvable for every continuous function on it exactly when this boundary is closed. More precisely, Theorem 6.37 yields that H is a simplicial space with closed Choquet boundary ChH (K) (Bauer simplicial space) if and only if given a bounded continuous function f on ChH (K) there is a continuous H-affine function h such that h = f on ChH (K). We now consider the question, under what conditions every bounded Baire-one function on ChH (K) can be extended to a bounded Baire-one H-affine function on K. To this end, we need an analogue of envelopes defined similarly as in Section 6.5. For a function f on K we define f ∗,1 := inf{h ∈ A(H) ∩ B b1 (K) : h ≥ f } and

f∗,1 := −(−f )∗,1 .


193

Theorem 6.49. Let H be a function space on a metrizable compact space K. Consider the following assertions: (i) H is simplicial and ChH (K) is an Fσ set, (ii) for any bounded Baire-one function f on ChH (K) there exists a bounded Haffine Baire-one function h such that f = h on ChH (K), (iii) for any bounded Baire-one function f on K there exists a bounded H-affine Baire-one function h such that f = h on ChH (K), (iv) H is simplicial and T f ∈ B 1 (K) for any f ∈ B b1 (K), (v) f ∗,1 is an H-affine Baire-one function for every bounded H-convex function f ∈ B 1 (K), (vi) A(H) ∩ B b1 (K) is a lattice in the natural ordering. Then assertions (i)–(v) are equivalent and (v) =⇒ (vi). Proof. For the proof of the implication (i) =⇒ (ii), suppose S∞ that H is simplicial and ChH (K) is an Fσ set. Thus we can write ChH (K) = n=1 Fn where {Fn } is an increasing sequence of compact sets. Let f be a bounded Baire-one function on ChH (K) and {fn } be a sequence of continuous functions on ChH (K) converging pointwise to f . We may assume that kf k, kfn k are all bounded by a positive number M . By Theorem 6.27, there exists a sequence {hn } of H-affine continuous functions on K such that hn = fn on Fn and khn k = kfn k. The proof will be completed by showing that the sequence {hn } converges pointwise to the function h := T f . Then T f is Baire-one and H-affine by the Lebesgue dominated convergence theorem. Notice that the definition is meaningful since maximal measures are carried by ChH (K) due to Theorem 3.81. R For a fixed point x ∈ K and ε > 0 we find n0 ∈ N such that ChH (K) |f −fn | dδx < ε and δx (Fn ) > 1 − ε for all n ≥ n0 . Then, for n ≥ n0 , we have Z |T f (x) − hn (x)| = (f − hn ) dδx K Z Z ≤ |f − fn | dδx + Fn0

2M dδx

K\Fn0

≤ ε + ε2M, which proves the required statement and concludes the proof of (i) =⇒ (ii). Since the implication (ii) =⇒ (iii) is obvious, we proceed to the proof of (iii) =⇒ (iv). Let H be a function space on a compact space K satisfying condition (iii). First we verify that H is simplicial. Indeed, for a given continuous H-convex function f on K we find an H-affine Baire-one function h with h = f on ChH (K).

194


By Theorem 3.86, h ≥ f on K. For a given x ∈ K, Lemma 3.21 yields the existence of a measure µ ∈ Mx (H) such that µ(f ) = f ∗ (x). Then f ∗ (x) = µ(f ) ≤ µ(h) = h(x). On the other hand, let g ∈ H satisfy g ≥ f . Then g ≥ h on ChH (K), and thus g ≥ h on K, which again follows from Theorem 3.86. Hence h(x) ≤ inf{g(x) : g ∈ H, g ≥ f } = f ∗ (x). Thus f ∗ = h is H-affine for every H-convex continuous function f and H is a simplicial function space by Theorem 6.5. It remains to check that T f is a Baire-one function on K for every f ∈ B b1 (K). If f is a bounded Baire-one function on K, let h be an H-affine Baire one function on K with f = h on ChH (K). Then h = T f on ChH (K) and the application of Theorem 3.86 yields that h = T f on K. Thus T f is a Baire-one function as required. The next step will be a proof of the implication (iv) =⇒ (v). Let f be an H-convex Baire-one function on K. By condition (iv), T f is an H-affine Baire-one function. Moreover, T f = f on ChH (K) and T f ≥ f on K by the Minimum principle 3.86. Thus T f ≥ f ∗,1 . On the other hand, given an H-affine Baire-one function h with h ≥ f , the minimum principle laid down in Theorem 3.86 gives h ≥ T f . Thus T f ≤ f ∗,1 and f ∗,1 = T f is an H-affine Baire-one function. In order to prove (v) =⇒ (vi), let f and g be H-affine Baire-one functions. Since h = (f ∨ g)∗,1 is an H-affine Baire-one function, h is obviously the least upper bound for the pair f and g in A(H) ∩ B b1 (K). Thus A(H) ∩ B b1 (K) is a lattice in its natural ordering. It remains to prove the implication (v) =⇒ (i). First we check that H is simplicial. Let f ∈ Kc (H) be given. Then f ∗,1 ≤ f ∗ by the definition. On the other hand, given x ∈ K, let µ ∈ Mx (H) satisfy µ(f ) = f ∗ (x). Then f ∗ (x) = µ(f ) ≤ inf{µ(h) : h ∈ A(H) ∩ B b1 (K), h ≥ f } = f ∗,1 (x). Hence f ∗ = f ∗,1 is H-affine and H is simplicial by Theorem 6.5. Assume that ChH (K) is not an Fσ set in K. We fix on K a compatible metric ρ. Since ChH (K) is a Gδ subset of a compact metrizable space, Theorem A.115 yields the existence of a homeomorphic copy C ⊂ K of the Cantor set {0, 1}N such that C ∩ ChH (K) = C \ ChH (K) = C

and

C \ ChH (K) is countable.

We enumerate C \ ChH (K) as {qn : n ∈ N} and, using regularity of Radon measures, we find compact sets Kn ⊂ ChH (K), n ∈ N, such that δqn (Kn ) > 1 − n1 , n ∈ N (see Proposition 3.80). Then there is no Fσ set F ⊂ K satisfying C\

∞ [ n=1

Kn ⊂ F ⊂ K \ {qn : n ∈ N}.


195

(Indeed, otherwise the set C ∩ ChH (K) would be an Fσ set, which is absurd.) S Another use of Theorem A.115 provides a homeomorphic copy D ⊂ C \ ∞ n=1 Kn of the Cantor set {0, 1}N such that D ∩ {qn : n ∈ N} = D ∩ ChH (K) = D. Then cD is a Baire-one function on K and T cD (x) = 1, T cD (x) ≤

1 , n

x ∈ D \ ChH (K), x = qn , n ∈ N.

Hence T cD has no point of continuity on D, and thus it is not a Baire-one function (see Theorem A.127). This concludes the proof. Remark 6.50. We remark that only the proof of (v) =⇒ (i) in the preceding theorem used the assumption of metrizability of K. Example 6.51. There exists a simplicial space H on a metrizable space K such that B b1 (K) ∩ A(H) is a lattice in the natural ordering but ChH (K) is not an Fσ set. Proof. We take a suitable Stacey simplicial function space from Definition 6.13. Namely, we take L := [0, 1] and B := Q ∩ [0, 1] and K as in Definition 6.13. We enumerate B as {qn : n ∈ N} and take Lqn := {0, 1} with µqn := n1 ε0 + (1 − n1 )ε1 , n ∈ N. By Lemma 6.14, H is simplicial and ChH (K) = K \ B. Hence ChH (K) is not an Fσ set. Further we know from Lemma 6.14 that δqn = µqn . For brevity we write qn+ for the point 1 in Lqn and qn− for the point 0 in Lqn . In order to check that A(H)∩B b1 (K) is a lattice in the natural ordering, it is enough to prove that T (f ∨g) is a Baire-one function for every pair f and g of H-affine Baireone functions. Let f and g be such functions with values in [0, 1] and set h := f ∨ g. We claim that 2 for every n ∈ N. |h(qn+ ) − h(qn )| ≤ (6.5) n Indeed, for a fixed n ∈ N we have 1 1 f (qn− ) − (1 − )f (qn+ )| n n 2 1 ≤ |f (qn+ ) − f (qn− )| ≤ . n n

|f (qn+ ) − f (qn )| = |f (qn+ ) −

By the same argument, |g(qn+ ) − g(qn )| ≤ n2 . We need to check this inequality for the function h. The only nontrivial case is when h(qn+ ) = f (qn+ ) and h(qn ) = g(qn ) (or

196


vice versa). Then h(qn+ ) = f (qn+ ) = f (qn+ ) − f (qn ) + f (qn ) ≤ f (qn+ ) − f (qn ) + g(qn ) ≤ and

2 2 + g(qn ) = + h(qn ), n n

h(qn ) = g(qn ) = g(qn ) − g(qn+ ) + g(qn+ ) ≤ g(qn ) − g(qn+ ) + f (qn+ )

2 2 + f (qn+ ) = + h(qn+ ). n n Combining these inequalities together we get (6.5). Applying inequality (6.5) we obtain ≤

1 1 |T h(qn ) − h(qn )| = | h(qn− ) + (1 − )h(qn+ ) − h(qn )| n n 1 = | (h(qn− ) − h(qn+ )) + h(qn+ ) − h(qn )| n 2 4 2 ≤ + = . n n n Hence the set {x ∈ K : |T h(x) − h(x)| ≥ ε} = {qn ∈ [0, 1] : |T h(qn ) − h(qn )| ≥ ε, n ∈ N} is finite for every ε > 0. By Lemma A.128, T h is a Baire-one function and the space A(H) ∩ B b1 (K) is a lattice in the natural ordering.

6.7

The Daugavet property of simplicial spaces

In 1963, I. K. Daugavet in [131] showed that any compact operator S on the Banach space C([0, 1]) satisfies the identity kI + Sk = 1 + kSk.

(6.6)

(Here, I stands for the identity operator.) Afterwards, Banach spaces and operators satisfying the equality (6.6) were studied. A Banach space E satisfying (6.6) for any compact operator S on E is said to satisfy the Daugavet property, or, for short, the Daugavet space. Among other properties, it was shown that a Banach space satisfies the equality (6.6) for any compact operator S if and only if the equality (6.6) is satisfied for the class of all rank-one operators, and that the space C(K) has the Daugavet property if and only if the compact space K has no isolated points (see, for example, V. M. Kadets, R. V. Shvidkoy, G. G. Sirotkin and D. Werner [255]).

6.7 The Daugavet property of simplicial spaces

197

Definition 6.52 (The Daugavet property). A Banach space E has the Daugavet property if kI + F k = 1 + kF k for each rank-one operator F on E. Theorem 6.53. Let H be a simplicial function space on a compact space K such that ChH (K) has no isolated points. Then Ac (H) has the Daugavet property. Proof. Let a rank-one operator F : Ac (H) → Ac (H) be given. There exists a functional ϕ ∈ Ac (H)∗ and a function h0 ∈ Ac (H) such that F (h) = ϕ(h)h0 ,

h ∈ Ac (H).

Using Lemma 6.23 we find a boundary measure µ ∈ M(K) such that kµk = kϕk and

µ = ϕ on Ac (H).

Fix ε > 0. By Theorem 3.85, there exists a point y0 ∈ ChH (K) (= ChAc (H) (K) by Exercise 3.95) such that h0 (y0 ) = kh0 k. Let W be an open neighborhood of y0 such that h0 > kh0 k − ε on W . Since ChH (K) has no isolated points, there exists a point x0 ∈ W ∩ ChH (K) and an open neighborhood U of x0 such that |µ|(U ) < ε. Let f be a continuous function on K such that kf k = 1 and µ(f ) > kµk − ε. According to Tietze’s theorem, there exists a continuous function g on K such that kgk = 1,

g=f

on K \ U

and g(x0 ) = 1.

Then µ(g) > kµk − 2ε. Since µ is a boundary measure, µ(g) = µ(g ∗ ). For the positive part µ+ of µ we find by Theorem A.84 a continuous H-concave function k + such that g ≤ k+ ≤ 1

and µ+ (k + ) − µ+ (g ∗ ) < ε.

Similarly we choose a continuous H-concave function k − such that g ≤ k− ≤ 1

and µ− (k − ) − µ− (g ∗ ) < ε.

Then k := k + ∧ k − is a continuous H-concave function satisfying |µ(k) − µ(g ∗ )| < 2ε, kkk ≤ 1 and k(x0 ) = g ∗ (x0 ) = g(x0 ) = 1. Another use of Theorem A.84 along with Theorem 6.6 provides a function h ∈ Ac (H) such that −1 ≤ h ≤ k,

|µ(h) − µ(k)| < ε and

|h(x0 ) − k(x0 )| < ε

198


(recall that k∗ (x0 ) = k(x0 ) by Theorem 3.24). Combining all the estimates together we get (I + F )(h)(x0 ) = h(x0 ) + µ(h)h0 (x0 ) ≥ k(x0 ) − ε + (µ(k) − ε)(kh0 k − ε) ≥ 1 − ε + (µ(g ∗ ) − 3ε)(kh0 k − ε) = 1 − ε + (µ(g) − 3ε)(kh0 k − ε) ≥ 1 − ε + (kµk − 5ε)(kh0 k − ε) = 1 + kµkkh0 k − ε(kµk − 5kh0 k − 1 − 5ε). Since ε > 0 is arbitrary, this shows that kI + F k = 1 + kµkkh0 k = 1 + kF k, finishing the proof.

6.8 6.8.A

Choquet simplices Simplicial function spaces and the classical definition of Choquet simplices

We recall that a compact convex set X is a simplex if the function space Ac (X) is a simplicial function space on X. Theorem 6.54 shows how the simpliciality of a function space is related to properties of the state space of Ac (H). Theorem 6.54. For a function space H, the following assertions are equivalent. (i) H is simplicial. (ii) The ordered Banach space Ac (H)∗ is a vector lattice. (iii) The compact convex set S(Ac (H)) is a simplex. Proof. We start the proof by noticing that (i) =⇒ (ii) due to Theorem 6.18 and Theorem A.24. The next step is showing that (ii) =⇒ (iii). Let X := S(Ac (H)). We want to prove that Ac (X) is simplicial, which means, by virtue of Theorem 6.5 and Corollary 4.8, to check that f ∗ is affine for any continuous convex function f on X. Since f ∗ is always concave, we need to check its convexity. To this end, let s := α1 s1 + α2 s2 be a nontrivial convex combination of points s1 , s2 ∈ X. We fix ε > 0 P and use Lemma 3.21 along with Proposition 4.3 to find a convex combination s = nj=1 βj tj , t1 , . . . , tn ∈ X, β1 , . . . , βn strictly positive, such that n X βj f (tj ) > f ∗ (s) − ε. j=1

6.8 Choquet simplices

199

Using Proposition A.11, we find positive functionals u0ij ∈ (Ac (H)∗ , i = 1, 2, j = 1, . . . , n, so that αi si =

n X

u0ij ,

i = 1, 2,

and

βj tj = u01j + u02j ,

j = 1, . . . , n.

j=1

Let γij ≥ 0 and uij ∈ X satisfy u0ij = γij uij , i = 1, 2, j = 1, . . . , n. Then we get the following convex combinations si =

n X γij

αi

j=1

tj =

uij ,

i = 1, 2,

γ1j γ2j u1j + u2j , βj βj

and

j = 1, . . . , n.

We use the convexity of f , concavity of f ∗ and preceding equalities to get ∗

f (s) − ε
0 (cf. Definition 8.47). A set F ⊂ X is said to be exposed if there is a positive function f ∈ Ac (X) such that f (x) = 0 if and only if x ∈ F . We remark that any exposed set is relatively exposed and any relatively exposed set is a closed face (cf. Proposition 8.48). Definition 6.59 (Prime compact convex sets). A compact convex set X is said to be prime if whenever F and G are closed relatively exposed faces of X such that co(F ∪ G) = X, then either F = X or G = X. Theorem 6.60. A compact convex set X is prime if and only if the function space Ac (X) is an antilattice. Proof. Assume that Ac (X) is an antilattice and X = co(F ∪ G) where F and G are proper closed relatively exposed faces. We find nonzero positive continuous affine functions f and g so that f = 0 on F and g = 0 on G. Due to Minimum principle 3.85, the infimum f f g exists and equals 0. Since Ac (X) is an antilattice, either f ≤ g or g ≤ f . We have arrived at the conclusion that either f = 0 or g = 0 which is a contradiction with our choice of f and g. Conversely, suppose that X is a prime compact convex set and f, g ∈ Ac (X) are such that f f g exists in Ac (X). It follows from the definition of the lower envelope that f f g = (f f g)∗ . By Theorem 3.24, f f g = f ∧ g on ext X. If we set F := {x ∈ X : (f f g)(x) = f (x)}

and

G := {x ∈ X : (f f g)(x) = g(x)},

we obtain two exposed faces satisfying ext X ⊂ F ∪ G. Thus X = co(F ∪ G). Since X is supposed to be prime, either X = F or X = G. The former equality yields f ≤ g, the latter one implies g ≤ f . Thus we have verified that Ac (X) is an antilattice and the proof is finished.


201

Lemma 6.61. Let X be a simplex and F ⊂ X be a closed Gδ face of X. Then F is exposed. Proof. Let F be a closed Gδ face of X. For any x ∈ X \F we find a continuous affine function fx so that sup fx (F ) ≤ 0 < fx (x). Since X \ FSis a Lindelöf space, we can ∞ select countably many points {xn }∞ n=1 so that X \ F ⊂ n=1 {y ∈ X : fxn (y) > 0}. Set ∞ X 1 hn . hn := fxn ∨ 0, n ∈ N, and h := 2n khn k n=1

Then h is a continuous convex function on X such that h = 0 on F and h > 0 on X \ F . If we define g := cX\F , we obtain a lower semicontinuous concave function with h ≤ g. Using Theorem 6.6 we obtain an affine continuous function f ∈ Ac (X) so that h ≤ f ≤ g. Obviously f exposes F , which concludes the proof. The previous Lemma 6.61 shows that the following definition of prime simplices is a particular case of the general definition of prime compact convex sets. Definition 6.62 (Prime simplices). A simplex X is said to be prime if given closed Gδ faces F and G in X with X = co(F ∪ G), then either F = X or G = X. Definition 6.63 (Prime simplicial spaces). A simplicial function space H is prime if Ac (H) is an antilattice. Proposition 6.64. Let H be a simplicial space. The following assertions are equivalent: (i) H is prime, (ii) the simplex X := S(Ac (H)) is prime. Proof. Let Φ : Ac (H) → Ac (X) be the surjective isometric isomorphism from Definition 4.25. By Corollary 5.41, Φ preserves order of functions and hence for any f, g ∈ Ac (H), f f g exists if and only if Φ(f ) f Φ(g) exists. Thus the conclusion follows from Theorem 6.60. Lemma 6.65. Let H be a simplicial space. If there exists a point x ∈ ChH (K) such that ChH (K) ⊂ spt δx , then H is prime. Proof. Suppose that for given f, g ∈ Ac (H) there exists f f g ∈ Ac (H). It follows from the definition of the lower envelope that f fg = (f ∧g)∗ . Thus f fg = f ∧g on ChH (K), by Theorem 3.24. By continuity, this equality also holds for every point in ChH (K). In particular, (f fg)(x) = (f ∧g)(x). Let us assume that (f ∧g)(x) = f (x). From the equalities Z Z (f ∧ g) dδx = (f f g) dδx = (f f g)(x) ChH (K)

ChH (K)

Z = (f ∧ g)(x) = f (x) =

f dδx ChH (K)

202


we get that f = f ∧ g almost everywhere with respect to δx . Since ChH (K) ⊂ spt δx , the equality f = f ∧ g holds on ChH (K). In other words, f ≤ g on ChH (K). The Minimum principle 3.85 yields the desired conclusion that f ≤ g on K. Lemma 6.66. Let X be a compact convex set such that ext X = X. Then X is prime. Proof. Suppose that for functions f, g ∈ Ac (X) there exists f f g in Ac (X). As in the previous proof we have f f g = (f ∧ g)∗ . Since (f ∧ g)∗ = f ∧ g on X and ext X is dense in X, we obtain that f ∧ g = f f g on X. We claim that this is possible only if either f ≤ g or g ≤ f . Indeed, suppose that this is not the case, that is, we can find points x, y ∈ X so that f (x) < g(x) and f (y) > g(y). Then it is obvious that f ∧ g = f f g cannot be affine on the segment joining x and y. This contradiction finishes the proof. Example 6.67. There exists a simplicial function space H on a compact space K such that ChH (K) = K and H is not prime. Proof. Let K := [0, 1] ∪ {2} ∪ {3} and H := {f ∈ C(K) : f (0) = 21 (f (2) + f (3))}. Then H = Ac (H), H is simplicial and ChH (K) = K \ {0}. Nevertheless, H is not an antilattice. Indeed, if ( x(1 − x) x ∈ [0, 1], f (x) := 0 otherwise, and ( g(x) :=

1 4x

0

x ∈ [0, 1], otherwise,

then f, g ∈ Ac (H) and f ∧ g ∈ Ac (H) as well and neither f ≤ g nor g ≤ f . Thus f f g exists although ChH (K) is dense in K.

6.8.C

Characterization of Bauer simplices by faces

In this subsection we present another characterization of Bauer simplices by means of faces. We recall that a compact convex set X is a Bauer simplex if X is a simplex whose set of extreme points is closed (see Definition 6.36), and that δx denotes the unique maximal measure in Mx (X). Lemma 6.68. Let X be a simplex and H ⊂ ext X. Then co H is a face of X. Proof. Since Pnco H is convex, it is enough to show that co H is extremal. So assume that x := j=1 λj hj = αy + (1 − α)z, where hj ∈ H, λj ≥ 0, j = 1, . . . , n,


203

P = 1, α ∈ (0, 1) and y, z ∈ X. Then nj=1 λj εhj ∈ Mx (X) is maximal by Corollary 3.59 and αδy + (1 − α)δz ∈ Mx (X). Hence Pn

j=1 λj

n X

λj εhj = αδy + (1 − α)δz ,

j=1

which yields y, z ∈ co H. Theorem 6.69. Let X be a simplex. The following assertions are equivalent: (i) X is a Bauer simplex, (ii) F is a face for any face F ⊂ X. Proof. For the proof of (i) =⇒ (ii), let X be a Bauer simplex and F ⊂ X be a face. Since F is convex, we only have to show that F is extremal, that is, our aim is to prove that spt δx ⊂ F for every x ∈ F (see Exercise 6.82). First, suppose that x ∈ F and that δx is not carried by F . Let V be a closed convex neighborhood not intersecting F such that δx (V ) > 0. If δx (V ) = 1, then x ∈ V and thus V ∩ F 6= ∅, which is impossible. Setting c := δx (V ) ∈ (0, 1), we have δx = cδx |V + (1 − c)δx |X\V , hence x = cr(δx |V ) + (1 − c)r(δx |X\V ). Since F is a face, r(δx |V ) ∈ F and thus r(δx |V ) ∈ V ∩ F , which is a contradiction. Assuming that x ∈ F , there exists a net {xα } in F such that xα → x. By (iii) of Theorem 6.37, δxα → δx . Since spt δxα ⊂ F , we get spt δx ⊂ F . To show that (ii) =⇒ (i), assume that there exists x ∈ ext X \ ext X. There exist y, z ∈ X, y 6= x, such that 2x = y + z. The Hahn–Banach theorem asserts the existence of f ∈ E ∗ and c ∈ R such that f (y) < c < f (x). Denote H := ext X ∩ {t ∈ X : f (t) ≥ c}. According to Lemma 6.68, co H is a face. On the other hand, co H is not a face. Indeed, co H is not extremal since x ∈ H ⊂ co H and y∈ / co H.

6.8.D

Fakhoury’s theorem

Theorem 6.70. Let X be a compact convex set. Then the following conditions are equivalent: (i) X is a simplex, (ii) there exists an affine mapping T : X → M1 (X) such that T (x) ∈ Mx (X).

204


Proof. If X is a simplex, it is enough to check that the mapping x 7→ δx is affine. Let x and y be points of X and α ∈ [0, 1]. It suffices to verify that δαx+(1−α)y (f ) = αδx (f ) + (1 − α)δy (f ) for any f ∈ Kc (X). By Corollary 4.8 and Lemma 6.4, the previous equality can be rewritten as f ∗ (αx + (1 − α)y) = αf ∗ (x) + (1 − α)f ∗ (y), which holds because the function f ∗ is affine by Theorem 6.5. For the proof of the reverse implication, let T be an affine mapping provided by condition (ii). We write Tx for the measure T (x). We need to prove that f ∗ is affine for any f ∈ Kc (X). Since T is assumed to be affine, it is enough to verify that Tx (f ) = f ∗ (x) for any f ∈ Kc (X) and x ∈ X. Let f and x be as above and ε > 0. By Lemma 3.22 and Proposition 4.3, there ∗ exists − ε. Then µ = Pnthat µ(f ) ≥ fP(x) Pn a molecular measure µ ∈ Mx (X) such n i=1 αi xi = x. Since i=1 αi = 1 and i=1 αi εxi where xi ∈ X, αi > 0 such that f is convex and T is affine, Proposition 3.20(b) gives ∗

f (x) − ε ≤ µ(f ) =

n X

αi εxi (f ) ≤

i=1

n X

αi Txi (f )

i=1

= TPni=1 αi xi (f ) = Tx (f ) ≤ f ∗ (x). As ε is arbitrary, Tx (f ) = f ∗ (x). This concludes the proof.

6.9

Restriction of function spaces

Definition 6.71 (Restricted function space). Let H be a function space on a compact space K and L ⊂ K be a closed subset. Then G := {h|L : h ∈ H} is the restricted function space. Proposition 6.72. The following assertions hold: (a) If µ, ν ∈ M+ (L), then µ ≺H ν if and only if µ ≺G ν. (b) Mx (G) = Mx (H) ∩ M1 (L) for any x ∈ L. (c) ChH (K) ∩ L ⊂ ChG (L). (d) If L ⊃ ChH (K), then G-maximal measures coincide with H-maximal measures and ChG (L) = ChH (K). Proof. For the proof of (a), we notice that for µ, ν ∈ M+ (L), µ ≺H ν if and only µ(f ) ≤ ν(f ) for each f ∈ −W(H) (see Proposition 3.56), and this is the case if and only if µ(f ) ≤ ν(f ) for any f ∈ −W(G). Further, (b) and (c) are obvious.

6.10 Exercises

205

Assume that L ⊃ ChH (K). Let µ ∈ M+ (K) be an H-maximal measure. Then µ is carried by L (see Proposition 3.64). Let ν ∈ M+ (L) be a G-maximal measure with µ ≺G ν. Then µ ≺H ν, and thus µ = ν is G-maximal. Conversely, let µ ∈ M+ (L) be a G-maximal measure. If ν ∈ M+ (K) is Hmaximal measure with µ ≺H ν, then ν ∈ M+ (L). For any function f ∈ −W(H), we have µ(f |L ) = µ(f ) ≤ ν(f ) = ν(f |L ). Thus µ(g) ≤ ν(g) for any function g ∈ −W(G). By Proposition 3.56, µ ≺G ν. Hence µ = ν is H-maximal. Since x ∈ ChH (K) if and only if εx is H-maximal, the second part of (d) follows from the previous argument.

6.10

Exercises

Exercise 6.73. Let H be a simplicial space and x ∈ K. Prove that δx ∈ ext(Mx (H)). Hint. By Exercise 3.113, H-maximal probability measures form a face F in M1 (K). Since Mx (H) ∩ F = {δx }, the assertion follows. Exercise 6.74. Let X be a simplex in Rd . Prove that X has at most d + 1 extreme points and they are affinely independent. Hint. If {x0 , . . . , xd+1 } ⊂ ext X are distinct points, Pd+1dependent. P they are affinely a = 0 and Hence there exist numbers a0 , . . . , ad+1 such that d+1 i i=0 ai xi = 0 i=0 (see Definition 2.2). We assume that a0 , . . . , an ≥ 0 and an+1 , . . . , ad+1 < 0. Then 0 < c :=

n X

ai =

i=0

d+1 X

−ai ,

i=n+1

and x :=

n X i=0

c−1 ai xi =

d+1 X

−c−1 ai xi .

i=n+1

Hence x has two different maximal representing measures, a contradiction. Analogously we get that the extreme points of X are affinely independent. Exercise 6.75. Let X be a compact convex set in Rd . Prove that X is a simplex if and only if X is an n-simplex for some n ≤ d (see Definition 2.2). Hint. If X is an n-simplex with vertices {x0 , . . . , xn }, then ext X = {x0 , . . . , xn } by Lemma Pn 2.10 and any point x ∈ X is expressed by a unique convex combination x = i=0 λi xi (the points {x0 , . . . , xn } are affinely independent).

206


Conversely, let X be a simplex. By Exercise 6.74, ext X = {x0 , . . . , xn } for some n ≤ d and the set of extreme points consists of affinely independent points. Hence X is an n-simplex by Definition 2.2. Exercise 6.76. Let H := {f ∈ C([0, 1]) : there is af > 0 such that f is constant on [0, af ]} . Since the function space H is dense in C([0, 1]), we have ChH ([0, 1]) = ChH ([0, 1]) = ChC([0,1]) ([0, 1]) = [0, 1]. It follows that H is simplicial and {0} ∈ ChH ([0, 1]). Nevertheless, {0} is not Hexposed. Exercise 6.77. Let X be the convex hull of two sets (x, y) ∈ R2 : (x + 1)2 + y 2 = 1 and (x, y) ∈ R2 : (x − 1)2 + y 2 = 1 and let K be the boundary of X. If H consists of restrictions of all affine functions on R2 to K, then H is a closed function space on K. Prove that ChH (K) = {(x, y) ∈ K : x ≤ −1 or x ≥ 1} and that a function f ∈ C(K) belongs to Ac (H) if and only if f is affine on each segment contained in K. Further show that H is simplicial and that the point (1, 1) is Ac (H)-exposed but it is not H-exposed. Hint. In order to show that the point (1, 1) is Ac (H)-exposed, consider, for example, the function (x, y) 7→ | − x − y + 2| + |1 − y|, (x, y) ∈ K.

Exercise 6.78. Let H be a closed function space on a compact space K possessing the weak Riesz interpolation property. Then Ac (H) = H. Hint. If f ∈ Ac (H) and ε > 0, for any x ∈ K we use Corollary 3.23(a) and Proposition 3.25(a) to find a function hx ∈ H such that f < hx and hx (x) < f (x) + ε. A simple compactness argument yields the existence of finitely many functions h1 , . . . , hn ∈ H such that f < h1 ∧ · · · ∧ hn < f + ε. Analogously, we find functions g1 , . . . , gm ∈ H with f > g1 ∨ · · · ∨ gm > f − ε.

6.10 Exercises

207

By the weak Riesz interpolation property, there exists a function h ∈ H such that g1 ∨ · · · ∨ gm < h < f1 ∧ · · · ∧ fn . Hence kf − hk < 2ε, which gives f ∈ H = H. Exercise 6.79. Find an example of a simplicial function space whose state space is not a simplex. Hint. Consider the function space H := P 2 ([0, 1]) from Example 3.2(b). Then ChH ([0, 1]) = [0, 1] (see Example 3.5(b)), and thus H is simplicial. On the other hand, if ψ : [0, 1] → R2 is defined by ψ(t) = (t, t2 ),

t ∈ [0, 1],

S(H) is affinely homeomorphic to co ψ([0, 1]) (use Exercise 4.51). Hence S(H) is not a simplex. Exercise 6.80. Let X be a compact convex set, µ, ν ∈ M+ (X), µ ≺ ν. Prove that spt µ ⊂ co spt ν. Hint. If z ∈ spt µ \ co spt ν, find a continuous affine function f on X such that f (z) = −1 and inf f co spt ν ≥ 0. If k := f ∧ 0, then k is continuous concave on X, k = 0 on spt ν and ν(k) ≤ µ(k) < 0. Exercise 6.81. Let F be a closed extremal subset of a compact convex set X, µ, ν ∈ M+ (X), µ ≺ ν. If spt µ ⊂ F , then spt ν ⊂ F . Hint. The characteristic function cX\F is concave (cf. Proposition 2.68) and lower semicontinuous. By Proposition 3.56, ν(cX\F ) ≤ µ(cX\F ). Exercise 6.82. Let F be a closed convex subset of a compact convex set X. Prove that F is extremal if and only if spt ν ⊂ F for any x ∈ F and any maximal measure ν ∈ Mx (X). Hint. If F is a closed convex extremal set, it carries any maximal measure with barycenter in F by Proposition 2.69. Conversely, let x ∈ F and µ ∈ Mx (X). Let ν ∈ M1 (X) be maximal with µ ≺ ν. Then ν ∈ Mx (X). By Exercise 6.80, spt µ ⊂ co spt ν ⊂ F . Hence spt µ ⊂ F and F is extremal. Exercise 6.83. Let F be a closed face in a compact convex set X. Prove that Ac (F )maximal measures coincide with Ac (X)-maximal measures carried by F .

208


Hint. Let µ ∈ M1 (F ) be Ac (X)-maximal and ν ∈ M1 (F ) satisfy µ ≺Ac (F ) ν. Then µ ≺Ac (X) ν and thus µ = ν. If µ ∈ M1 (F ) is Ac (F )-maximal and ν ∈ M1 (X) with µ ≺Ac (X) ν, then ν is carried by F due to Exercise 6.81. If f is a positive convex function on F , the function g := f on F and 0 on X \ F is upper semicontinuous and convex on X. By Proposition 3.56, µ(f ) = µ(g) ≤ ν(g) = ν(f ). Hence µ ≺Ac (F ) ν, and thus µ = ν. It follows that µ is Ac (X)-maximal. Exercise 6.84. Let X be a simplex and H a closed extremal subset of X. Then co H is a closed face of X (cf. Proposition 8.31 and Lemma 6.68). Hint. By Exercise 6.82, we only have to show that spt δx ⊂ co H for any x ∈ co H. Fix x ∈ co H. Using Proposition 2.39 find a measure µ ∈ Mx (X) carried by H. Then µ ≺ δx . In view of Exercise 6.81, we get spt δx ⊂ H. Exercise 6.85. Let X be a simplex and F ⊂ X be a closed face. Prove that F is a simplex. Hint. This follows from Exercise 6.83. Exercise S 6.86. Let X be a simplex and {Fα } a family of closed faces of X. Then co Fα is a face of X. Hint. Similar to the proof of Lemma 6.68. Exercise 6.87. Let X be a compact convex subset of a locally convex space. (a) If X is metrizable and F is a face of X, then F = co(F ∩ ext X). In particular, F ∩ ext X 6= ∅. (b) Prove that there exists a simplex X and x ∈ X such that face x ∩ ext X = ∅. Hint. For the proof of (a), let x ∈ F . By Theorem 3.45, there exists µ ∈ Mx (X) carried by ext X. Since, by Remark 2.70(a), µ is carried by F , we see that µ is carried by F ∩ ext X. Hence, x ∈ co(F ∩ ext X). It follows that F ⊂ co(F ∩ ext X), and therefore F ⊂ co(F ∩ ext X) ⊂ F . To verify (b), we use the existence of a simplex X and a maximal measure µ ∈ M1 (X) such that µ is carried by a compact set K disjoint from ext X. (The construction of X is based on the state space of the function space H of Example 3.82. In fact, the state space X of S(H) is a simplex by Lemma 6.14 and Theorem 6.54.) By Exercise 2.116, co K ∩ ext X = ∅. Let x be the barycenter of µ. Then x ∈ co K (Proposition 2.39). We have face x ∩ ext X = ∅. Indeed, choose y ∈ face x. By Proposition 2.66, there exist z ∈ X and λ ∈ [0, 1) such that x = λz + (1 − λ)y. Let νy and νz be maximal measures representing points y and z, respectively. Then λνz + (1 − λ)νy is a maximal measure representing x. Since X is a simplex, µ = λνz + (1 − λ)νy , and therefore spt νy ⊂ co K. Hence y ∈ co K.

6.10 Exercises

209

Exercise 6.88. Verify the details of the following example to provide an example of a face in a simplex whose closure is not a face. Hint. Let P := {xn } ∈ c0 : xn ≥ 0 for all n ∈ N and Q := {n−1 en : n ∈ N}

and X := co {−e1 } ∪ Q ,

where en are the standard unit vectors in c0 . Show that X is a simplex and ext X = {−e1 } ∪ Q. Let F := co Q. Then F is a face of X by Lemma 6.68. Since P is a closed convex subset of c0 and F ⊂ P , we have F ⊂ P . On the other hand, F is not a face since 0 ∈ F and 0 = 12 (−e1 ) + 12 e1 . Exercise 6.89. Prove that there exist metrizable Bauer simplices X, Y and affine continuous surjections ϕ : X → Y and ψ : Y → X such that X, Y are not affinely homeomorphic. Hint. Let K := [0, 1]. There exist continuous surjections f : K → K 2 and g : K 2 → K. Let X := M1 (K) and Y := M1 (K 2 ). Then f] is an affine surjection of X onto Y and g] is an affine surjection of Y onto X. The sets X and Y are metrizable Bauer simplices (Proposition 6.38). If we assume that X and Y are affinely homeomorphic, the sets ext X and ext Y are homeomorphic. Then, by Proposition 2.27, K and K 2 are homeomorphic, a contradiction. Exercise 6.90. Let X be a simplex and Y := {s ∈ (Ab (X))∗ : s ≥ 0, s(1) = 1} be endowed with the w∗ -topology Let φ : X → Y be the evaluation embedding, that is, φ(x)(f ) = f (x), f ∈ Ab (X), x ∈ X. Prove the following assertions. (a) The ordered Banach space Ab (X) is a lattice. (b) The set φ(X) is dense in Y . (c) The set Y is a Bauer simplex. Hint. For the proof of (a) use Theorem A.24. More precisely, since X is a simplex, Ac (X) has the Riesz decomposition property and (Ac (X))∗ = ((Ac (X))∗ )+ − ((Ac (X))∗ )+ , (Ac (X))∗ is a lattice by Theorem A.24. By Proposition A.17 and Theorem A.24, (Ac (X))∗∗ is a lattice (we remark that (Ac (X))∗∗ can be identified with Ab (X) by Proposition 4.32, and thus satisfies the condition (Ac (X))∗∗ = ((Ac (X))∗∗ )+ − ((Ac (X))∗∗ )+ .) Hence Ab (X) is a lattice.

210


To show (b), notice that Ab (X) can be identified with Ac (Y ) via the canonical embedding of Ab (X) into (Ab (X))∗∗ . Thus assuming the existence of s ∈ Y \ φ(X), we can find a function f ∈ Ab (X) such that s(f ) > sup f (X). Without loss of generality we may suppose that f is positive. Then s(f ) ≤ kf k = sup f (X), a contradiction. The last assertion (c) follows now from (a) and Theorem 6.37. Indeed, Ac (Y ), as well as Ab (X), is a lattice, and thus (viii) of Theorem 6.37 applies. Exercise 6.91. Let X be a compact convex set, A ⊂ Ac (X) be a subspace containing constant functions and F ⊂ A be a τX -dense in A. Let S := {s ∈ A∗ : s ≥ 0, ksk = 1} and ϕ : X → RF be defined as ϕ(x) = {f (x)}f ∈F ,

x ∈ X.

Let Y := ϕ(X). Prove the following assertions: (a) The set Y is a compact convex set. (b) If g ∈ Ac (Y ), then g ◦ ϕ ∈ A. (c) For any f ∈ A there exists g ∈ Ac (Y ) such that f = g ◦ ϕ. (d) If T : Ac (Y ) → A is defined as g 7→ g ◦ ϕ, then T is a positive isometric isomorphism of Ac (Y ) onto A and T 1 = 1. (e) The sets Y and S are affinely homeomorphic. (f) If A has the weak Riesz interpolation property, then so does Ac (Y ) and Y is a simplex. (g) If A is a lattice, then so is Ac (Y ) and Y is a Bauer simplex. Hint. Obviously, Y is a compact convex set. Let g ∈ Ac (Y ) be given and assume that g ◦ ϕ ∈ / A. By the Hahn–Banach theorem and Proposition 4.31(b) there exist x1 , x2 ∈ X and a1 , a2 ∈ R positive such that a1 f (x1 ) − a2 f (x2 ) = 0 for all f ∈ A and a1 g(ϕ(x1 )) 6= a2 g(ϕ(x2 )). Since 1 ∈ A, a1 = a2 . By the definition of ϕ, ϕ(x1 ) = ϕ(x2 ), which is a contradiction. This proves (b). Let f ∈ A be given. Since F is τX -dense in A, f is constant on every ϕ−1 (y), y ∈ Y . Hence there exists a function g : Y → R such that f = g ◦ ϕ. By Theorem 5.33, g ∈ Ac (Y ) and (c) follows. If T is defined as in (d), then T is obviously a positive isometric isomorphism of c A (Y ) into A. By (c), T is surjective.

6.10 Exercises

211

For the proof of (e), let T ∗ : (A)∗ → (Ac (Y ))∗ be the dual operator of T from (d). Then T ∗ is an affine homeomorphism of S onto S(Ac (Y )), and thus S is affinely homeomorphic to Y by Proposition 4.31(a). If A has the weak Riesz interpolation property, then so does A. Since T from (d) preserves order, Ac (Y ) possesses the weak Riesz interpolation property as well. Consequently, Y is a simplex by Theorem 6.16. Analogously we get that Ac (Y ) is a lattice provided A is. In this case, Y is a Bauer simplex by Theorem 6.37. Exercise 6.92 (The ESP property). A function space H on K has the equal support property, for short ESP, if for any x ∈ K and any two maximal measures µ1 , µ2 ∈ Mx (H) representing the point x we have spt µ1 = spt µ2 . It is clear that any simplicial function space has the ESP. There are examples of function spaces having the ESP without being simplicial. Construct a compact convex set X with ext X closed such that X is not a simplex and X has the ESP. Hint. Let µ be any nonzero signed Radon measure on [0, 1] such that µ([0, 1]) = 0, kµ+ k = kµ− k = 1

and

spt µ+ = spt µ− = [0, 1].

Consider the set H := {f ∈ C([0, 1]) : µ(f ) = 0} which is, obviously, a linear subspace of C([0, 1]). Then H⊥ = {µ ∈ M([0, 1]) : ν = αµ for some α ∈ R}. The space H separates points of [0, 1]: Given x, y ∈ [0, 1], assume that h(x) = h(y) for any h ∈ H. Then εx −εy ∈ H⊥ and, consequently, εx −εy = αµ for some α ∈ R. Since spt µ+ = spt µ− = [0, 1], we get x = y. We see that H is a (closed) function space. Moreover, ChH ([0, 1]) = [0, 1]. To show this, select any x ∈ [0, 1] and ν ∈ Mx (H). Then ν − εx ∈ H⊥ , so that ν − εx = αµ for some α ∈ R. Hence ν = εx , and therefore x ∈ ChH ([0, 1]). Let X be the state space of H. Since s := π(µ+ ) = π(µ− ) ∈ X and r(φ] µ+ ) = π(µ+ ) = π(µ− ) = r(φ] µ− ) by Proposition 4.28(a), we see that the point s has two different representing measures φ] µ+ and φ] µ− carried by ChH ([0, 1]) (hence maximal by Corollary 3.61). Therefore, X is not a simplex. On the other hand, X has the ESP property: Suppose that Λ1 and Λ2 are maximal probability measures on X representing a point x ∈ X. By Proposition 4.28(d), there exist H-maximal measures λ1 , λ2 ∈ M1 ([0, 1]) such that Λ1 = φ] λ1 and Λ2 = φ] λ2 . Then λ1 −λ2 ∈ H⊥ . Hence, there exists α ∈ R such that λ1 −λ2 = αµ = αµ+ −αµ− .

212


We may assume that α > 0. Since λ1 ≥ αµ+ , we have α ≤ 1. Then either λ1 = µ+ and λ2 = µ− (in the case α = 1), or there exists η ∈ M1 ([0, 1]) such that λ1 = αµ+ + (1 − α)η. In this case, [0, 1] = spt µ+ ⊂ spt λ1 ⊂ [0, 1]. Analogously, we get spt λ2 = [0, 1]. In either case, spt λ1 = spt λ2 , and therefore spt Λ1 = spt Λ2 . Notice that Λ1 is maximal and spt Λ1 = spt φ] µ+ = φ([0, 1]) = φ ChH ([0, 1]) = ext X. Exercise 6.93. Let X be a compact convex set as given by Exercise 6.92. Recall that X is a metrizable set with ext X closed such that X is not a simplex and such that X has the ESP property. Prove that there exists x ∈ X such that, if F is a compact convex set, x ∈ F ⊂ X and ext F ⊂ ext X, then F is not a simplex. (Compare with Remark 2.14.) Hint. Keep the notation of Exercise 6.92. Let x := π(µ+ ) and let F ⊂ X be a compact convex set containing x such that ext F ⊂ ext X. There exists Λ ∈ Mx (F ) carried by ext F . Since Λ is carried by (a closed set) ext X, it is maximal. Hence, spt Λ = ext X (see the hint of Exercise 6.92). It follows that ext F ⊃ ext X, and therefore F = X. Example 6.94. Let K = [0, 1] ∪ {2} and Z 1 Z H := f ∈ C(K) : f (2) = 2tf (t) dt = 0

1

2(1 − t)f (t) dt .

0

Prove that (a) ChH (K) = [0, 1], Ac (H) = H and H is not simplicial, (b) H has the ESP, (c) the state space S(H) has not the ESP. Hint. Assertion (a) is easy to verify and (b) follows by a similar argument as to that in Exercise 6.92. The only difference is that H⊥ = span{ε2 − 2tλ, ε2 − 2(1 − t)λ} (here λ denotes Lebesgue measure on [0, 1]). To verify (c), let s := π(c[ 1 ,1] 2(2t − 1)λ) 2

and µ1 := φ] (c[ 1 ,1] 2(2t − 1)λ), 2

µ2 := φ] (c[0, 1 ] (1 − 2t)λ + c[ 1 ,1] (2t − 1)λ). 2

2

Then µi is a maximal measure on S(H) and represents s, i = 1, 2. Hence S(H) has not ESP.


213

Exercise 6.95. If T is a rank-one operator on a Banach space E, then kI + T k = 1 + kT k if and only if kI + αT k = 1 + αkT k for each α ≥ 0. Hint. If kI + T k = 1 + kT k, then 1 + αkT k = (1 + kT k) − (1 − α)kT k = kI + T k − (1 − α)kT k ≤ kI + T − (1 − α)T k = kI + αT k ≤ 1 + αkT k. The reverse implication is obvious.

6.11

Notes and comments

The basic facts on simplicial spaces can be found in many sources; we mention for example E. M. Alfsen [5], Chapter 1, §5, G. Choquet [108], Chapter 6, H. Bauer [37] and [38], N. Boboc and A. Cornea [73], J. Bliedtner and W. Hansen [66] or M. Rogalski [393], [391]. Theorem 6.5 can be found in papers by G. Choquet and P. -A. Meyer [106], [105], [113], [339]. Theorem 6.6 is due to D. A. Edwards [155] (see also [157]) and Theorem 6.8 follows the same argument as E. M. Alfsen [5], Proposition II.3.14. Example 6.15 follows the classical example from E. Bishop and K. de Leeuw [58]. This example was generalized in P. J. Stacey [436]. The question of constructing function spaces with given Choquet boundaries was studied by E. Briem [93] and A. Clausing and G. Mägerl [120]. We also mention that simplicial spaces are called weakly simplicial in H. Bauer [44], whereas spaces H with S(Ac (H)) being a simplex are called simplicial there (see Remark 6.2). Characterizations of simplicial spaces in Section 6.2 can be found in J. Lindenstrauss [304] and Z. Semadeni [412], see also E. M. Alfsen [5], Corollary II.3.11. Section 6.3 follows J. Lindenstrauss [304] and Z. Semadeni [412]. Theorem 6.27 can be found in G. Choquet [108], Theorem 28.6; see also E. M. Alfsen [5], Theorem II.3.12. Proposition II.3.17 of [5] provides an example of a nonmetrizable compact convex set X, that is not a simplex, and yet every continuous function defined on a compact subset of ext X can be extended to an affine continuous function with preservation of norm. Theorem 6.28 can be found, for example, in E. M. Alfsen [5] as Corollary II.5.20; we follow the reasoning of J. Bliedtner and W. Hansen [62]. The results of Section 6.5 for continuous functions (see Theorem 6.32) were proved by E. M. Alfsen [4]. The proof of this theorem can be also found in [5], Theorem II.4.5 and its several predecessors in E. M. Alfsen [2], E. G. Effros [163], Theorem 2.4 and A. J. Lazar [292], Theorem 2.2. We also refer the reader to J. B. Bednar [48], N. Boboc and Gh. Bucur [67] and J. N. McDonald [337] for results on continuous affine extensions in the setting of function spaces. The generalization for Baire functions is due to J. Spurný [426] and [423].

214


Subsection 6.6.A follows the paper by H. Bauer [38], see also E. M. Alfsen [5], Theorem II.4.1. The definition of a Markov simplex appeared in a paper [419] by R. Sine. Theorem 6.45 is taken from F. Jellett [250] (see also H. Fakhoury [175]), Theorem 6.46 from M. Rogalski [392] and A. Goullet de Rugy, C. Schol-Cancelier and B. Taylor-MacGibbon [201]. We mention the following problem on the converse implication in Theorem 6.45. Problem 6.96. Let X be a compact convex set such that any f ∈ C b (ext X) has an affine Baire-one extension. Is then ext X a Lindelöf space? Subsection 6.6.D follows the papers by J. Spurný [424] and [427]. The easy implication of Theorem 6.49 can be found in M. Rogalski [392] and F. Jellett [250]. The case of nonmetrizable spaces was solved in the paper J. Spurný and O. Kalenda [432] by proving the following theorem. For a compact convex set X, the following assertions are equivalent: • X is a simplex and ext X is a Lindel¨ of resolvable set, any bounded Baire-one function on ext X has an affine Baire-one extension. The proof uses a result on extension of Baire-one functions from the set of extreme points proved in O. Kalenda and J. Spurný [259]. We mention here a problem posed in [259]. •

Problem 6.97. Let X be a compact convex set with ext X Lindelöf. Is then ext X hereditarily Baire? It follows from these results and Theorem 6.37 that the question of extending continuous and Baire-one functions from the set of extreme points is determined by a topological quality of the set of extreme points. The following result from J. Spurný [431] shows that this is no longer true for functions of higher Baire classes. There exist metrizable simplices X1 , X2 and a homeomorphism ϕ : ext X1 → ext X2 such that • ϕ(ext X ) = ext X , 1 2 •

there exists a bounded Baire-two function on ext X1 that cannot be extended to a Baire-two affine function on X1 ,

if α ∈ [2, ω1 ), any bounded Baire-α function on ext X2 can be extended to a function of affine class α on X2 . For the Daugavet property and its relatives, it is possible to consult a survey paper by D. Werner [466]; in particular, see Example (e) on p. 79. Prime simplices were introduced in E. G. Effros and J. L. Kazdan [165] and we follow in Subsection 6.8.B their paper. In [1], E. M. Alfsen proved that in Bauer simplices, closures of faces are faces. Prime compact convex sets were also investigated in C. H. Chu [117]. The characterization of Bauer simplices in Theorem 6.69 •


215

was given by E. Størmer in [440] (see also A. K. Roy [399] and [400]). Note that E. StørmerSin [440] proved that a compact convex set X is a Bauer simplex if and only if co( α Fα ) is a closed split face of X whenever {Fα } is any family of closed split faces of X (Størmer’s axiom). Theorem 6.70 is from H. Fakhoury [174] The example of Exercise 6.77 is taken from Bauer’s paper [44]. It represents a slight modification of an earlier Bauer’s example from [38] and is attributed to S. Papadopoulou. Exercise 6.88 is due to M. Kraus. Exercise 6.90 is taken from C. H. Chu ˚ Lima [215], Example 3.3(b). and B. Cohen [119] and P. Harmond and A. The ESP property in Exercise 6.92 was introduced by I. Feinberg in his thesis (cf. J. N. McDonald [336]). In the same paper, J. N. McDonald investigated the ESP property and constructed an example of the simplex which fails the ESP property (Exercise 6.92 follows this example; see also J. N. McDonald [338]). Exercises 6.80 and 6.81 can also be found in [336]. S. Alpay proved in [12] that if X is a compact convex set and x ∈ X, then all maximal measures on X with barycenter x have the same support if and only if x ∈ ext(co spt µ) whenever µ is a maximal measure on X with barycenter x. The ESP property is also studied in M. W. Grossman [204]. Note that J. Köhn [274] localized the notion of a simplex in the convex setting and gave several statements that a point has a unique maximal representing measure. We also refer the reader to C. Cho-Ho [118] for similar results. Examples and a study of points of simpliciality in the context of function spaces is presented in the paper [27] by M. Baˇca´ k.

Chapter 7

Choquet theory of function cones

The notion of a function cone generalizes the concept of a function space. Even though we focus mainly on function spaces, the Choquet theory of function cones is an indispensable tool for us, since later on we investigate typical function cones arising in potential theory. Another motivation is the selection theorem from Section 11.5 and a description of boundary measures contained in Section 8.5. Since the basic results of the Choquet theory of function cones very often use the same techniques as their analogues from the theory of function spaces, we present them in Sections 7.1– 7.6 in a brief but, we hope, comprehensible way. We point out that Theorem 7.27 provides a measure on the Choquet boundary induced by a maximal measure. This concept will be strengthened considerably in Chapter 9. Other results not encountered before are Theorem 7.38 and Theorem 7.40, which clear up the relation between simplicial function cones and function spaces generated by them. An important concept of ordered compact convex sets is studied in Section 7.5. Main facts are summarized in Theorem 7.54 and an application of Theorem 7.27 is presented in Theorem 7.55. We show in Theorem 7.58 how an ordered compact convex set can be recovered from the set of maximal elements by a method imitating the Krein–Milman theorem. The Douglas characterization of simplicial measures is given in Theorem 7.60.

7.1

Function cones

Definition 7.1 (Function cones). Let K be a compact space and S a convex cone of continuous functions on K. We call S a function cone if S contains the constant functions and separates points of K. A function cone S is called min-stable if W(S) = {s1 ∧ · · · ∧ sn : s1 , . . . , sn ∈ S, n ∈ N} = S. In this section, S is a function cone on a compact space K 6= ∅. Examples 7.2. (a) Any function space can serve as an example of a function cone. (b) Let H be a function space on a compact space K. Then W(H) is a function cone on K. Also the family S c (H) of all continuous H-concave functions is an example of a function cone.

7.1 Function cones

217

(c) If S is a function cone on a compact space K, then W(S) is a function cone. If F is a closed subset of K, then S|F is a function cone as well. (d) Let U be a bounded open subset of Rd . Then the family S(U ) := s ∈ C(U ) : s is superharmonic on U is a function cone. Definition 7.3 (S-representing measures, ChS (K) and expS (K)). For a point x ∈ K we denote by Mx (S) the set of all S-representing measures for x; more precisely, Mx (S) := {µ ∈ M+ (K) : s(x) ≥ µ(s) for all s ∈ S}. We remark that Mx (S) ⊂ M1 (K) because S contains constant functions. We define the Choquet boundary ChS (K) of S as ChS (K) := {x ∈ K : Mx (S) = {εx }} and the set of S-exposed points as expS (K) := {x ∈ K : there exists s ∈ S with s(x) < s(y) for all y ∈ K \ {x}}. Definition 7.4 (S-convex, concave and affine functions). We define the system A(S) of all S-affine functions to be the family of all universally measurable functions f : K → [−∞, ∞] such that µ(f ) exists for every µ ∈ Mx (S), x ∈ K, and Z f (x) = f dµ for each x ∈ K and µ ∈ Mx (S). K

Ac (S)

Further, let be the family of all continuous S-affine functions on K. Similarly, we say that a universally measurable function f : K → [−∞, ∞] is Sconvex if µ(f ) exists for every µ ∈ Mx (S), x ∈ K, and f (x) ≤ µ(f ). A function f is S-concave if −f is S-convex. We denote by Kc (S), Kusc (S) and Klsc (S) the family of all continuous, upper semicontinuous, and lower semicontinuous S-convex functions, respectively. We write S c (S), S usc (S) and S lsc (S) for the analogous families of S-concave functions. Definition 7.5 (Choquet ordering, S-maximal and S-boundary measures). Let S be a function cone on a compact space K. We define an ordering ≺S on M+ (K) by setting µ ≺S ν if µ(s) ≤ ν(s) for each s ∈ Kc (S). Sometimes we write ≺ instead of ≺S . A measure µ ∈ M+ (K) is S-maximal (briefly maximal), if it is maximal with respect to the partial order ≺S . A measure µ ∈ M(K) is S-boundary (briefly boundary), if |µ| is S-maximal. The set of all S-maximal measures on K is denoted by Mmax (S), the set of all S-boundary measures is denoted by Mbnd (S).

218

7 Choquet theory of function cones

Proposition 7.6. The following assertions hold: (a) expS (K) ⊂ ChS (K), (b) S ⊂ W(S) ⊂ S c (S), (c) W(S), S c (S), S usc (S), S lsc (S) are min-stable convex cones and, consequently, W(S) and S c (S) are function cones, (d) Mx (S) = Mx (W(S)) = Mx (S c (S)) for any x ∈ K, (e) S c (S) = S c (W(S)) = S c (S c (S)), (f) ChS (K) = ChW(S) (K) = ChS c (S) (K), (g) for µ, ν ∈ M+ (K), µ ≺S ν if and only if µ ≺W(S) ν and this is the case if and only if µ ≺S c (S) ν, (h) the space W(S) − W(S) is dense in C(K). Proof. Assertions (a) and (b) are obvious, the first part of (c) follows by the same argument as in Proposition 3.11(b) and the second part from (b). Assertion (d) follows from the definition, (e), (f) are consequences of (d) and (g) follows from (e). Finally, (h) follows from the lattice version of the Stone–Weierstrass theorem contained in Proposition A.31. Definition 7.7 (Envelopes and sublinear functionals). Given f : K → [−∞, ∞] and a measure µ ∈ M+ (K), we define Qµ (f ) := inf{µ(s) : s ∈ S c (S), s ≥ f }. If µ = εx for x ∈ K, we write f ∗ (x) instead of Qεx (f ) and define f∗ (x) := −(−f )∗ (x). Lemma 7.8. Let µ ∈ M+ (K) and f be an upper bounded function on K. Then (a) f ∗ is an upper semicontinuous S-concave function, (b) Qµ (f ) = µ(f ∗ ), (c) if f is bounded, then −kf k ≤ f∗ ≤ f ∗ ≤ kf k, (d) g 7→ Qµ (g) is a sublinear functional on `∞ (K). Proof. Assertion (a) is obvious, (b) follows from Theorem A.84, (c) holds because S contains constant functions and (d) follows by a straightforward verification. The following observation shows that if we consider a function space H as a function cone, our new objects defined for H so far in Section 7.1 coincide with those defined for H in Section 3.1. Proposition 7.9. Let H be a function space on a compact space K. If we consider H as a function cone S, then

7.1 Function cones •

219

Mx (H) = Mx (S), S c (S) = S c (H), ≺H coincides with ≺S , expS (K) = expH (K),

•

S-maximal measures coincide with H-maximal measures,

•

ChS (K) = ChH (K).

Hint. The proof follows easily from the definitions. Lemma 7.10 (Key lemma). Let µ ∈ M+ (K) and f be an upper semicontinuous function on K. Then there exists a measure ν ∈ M+ (K) such that µ ≺ ν and Qµ (f ) = ν(f ). In particular, for any x ∈ K there exists a measure ν ∈ Mx (S) such that f ∗ (x) = ν(f ). Proof. Given µ ∈ M+ (K), we assume first that f ∈ C(K). We define a functional ϕ : span{f } → R as ϕ(tf ) = tQµ (f ), t ∈ R, and a sublinear functional p : C(K) → R as p : g 7→ Qµ (g),

g ∈ C(K).

Since ϕ ≤ p on span{f }, the Hahn–Banach theorem provides a linear functional ν on C(K) such that ν(f ) = Qµ (f ) and ν ≤ p on C(K). Since ν(g) ≤ p(g) ≤ 0 whenever g ∈ C(K) is negative, ν ∈ M+ (K). Let s ∈ S c (S). Then ν(s) ≤ p(s) = Qµ (s) ≤ µ(s). Hence µ ≺ ν and the proof is finished in the case of a continuous function. Now let f be an upper semicontinuous function. We define a down-directed family G := {g ∈ C(K) : g ≥ f } and use the first part of the proof to find a measure νg ∈ M+ (K) such that µ ≺ νg and νg (g) = Qµ (g), g ∈ G. Given h ∈ G, let Mh := {νg : g ∈ G, g ≤ h}. Since Mh ⊂ {λ ∈ M+ (K) : kλk = kµk} and the family {Mh : h ∈ G} has T the finite intersection property, a compactness argument yields the existence of ν ∈ {M h : h ∈ G}. Then µ ≺ ν and inf λ(h) : λ ∈ Mh = inf λ(h) : λ ∈ M h ≤ ν(h) for each h ∈ G.

220


Hence

Qµ (f ) ≤ inf {Qµ (g) : g ∈ G} = inf {νg (g) : g ∈ G} ≤ inf {inf {νg (h) : g ∈ G, g ≤ h} : h ∈ G} ≤ inf {ν(h) : h ∈ G} = ν(f ) ≤ inf {ν(k) : k ∈ S c (S), k ≥ f } ≤ inf {µ(k) : k ∈ S c (S), k ≥ f } = Qµ (f ),

which yields the required equality. The second part of the assertion follows by taking µ = εx and observing that ν ∈ Mx (S) provided εx ≺ ν. Proposition 7.11. Let f be an upper bounded function on K. Then (a) f ∗ = f if and only if f ∈ S usc (S), (b) we have

f ∗ = inf {g ∈ S : g ≥ f } = inf {g ∈ W(S) : g ≥ f } = inf {g ∈ S c (S) : g ≥ f } = inf {g ∈ S usc (S) : g ≥ f } ,

(c) if f is upper semicontinuous, then f ∗ = inf{k ∈ S lsc (S) : k ≥ g}. Proof. Let f be an upper bounded function on K. If f ∈ S usc (S) and x ∈ K, let µ ∈ Mx (S) be chosen such that f ∗ (x) = µ(f ). Then f ∗ (x) = µ(f ) ≤ f (x) ≤ f ∗ (x). Since the converse implication follows from Lemma 7.8(a), assertion (a) is proved. For the proof of (b) we first show that inf{s(x) : s ∈ S, s ≥ g} = max{µ(g) : µ ∈ Mx (S)},

x ∈ K, g ∈ C(K). (7.1)

The inequality “≥” being obvious, we proceed with the proof of “≤”. Let x ∈ K and g ∈ C(K) be fixed. The formula p : h 7→ inf{s(x) : s ∈ S, s ≥ h},

h ∈ C(K),

defines a sublinear functional on C(K). Analogously as in the proof of Lemma 7.10 we find a measure µ ∈ M+ (K) such that p(g) = µ(g) and µ ≤ p on C(K). Then µ(s) ≤ p(s) ≤ s(x), and thus µ ∈ Mx (S). Hence (7.1) follows.

s ∈ S,

7.1 Function cones

221

Let s ∈ S c (S) be arbitrary. By (7.1), p(s) = s(x),

x ∈ K.

Hence inf{g ∈ S : g ≥ f } = inf{g ∈ S c (S) : g ≥ f }.

(7.2)

inf{g ∈ S c (S) : g ≥ f } = inf{g ∈ S usc (S) : g ≥ f }

(7.3)

Further, by (a). Combining (7.2) and (7.3) we get f ∗ = inf{g ∈ S c (S) : g ≥ f } = inf{g ∈ S usc (S) : g ≥ f } ≤ inf{g ∈ W(S) : g ≥ f } ≤ inf{g ∈ S : g ≥ f } = inf{g ∈ S c (S) : g ≥ f }. To verify (c), let f be upper semicontinuous and s ∈ S lsc (S). Let x ∈ K and µ ∈ Mx (S) be such that f ∗ (x) = µ(f ). Then f ∗ (x) = µ(f ) ≤ µ(s) ≤ s(x), which yields “≤” in (c). The reverse inequality being obvious, the proof is finished. Proposition 7.12. Let µ ∈ M+ (K). (a) If ν ∈ M+ (K) satisfies µ ≺ ν, then kµk = kνk. (b) If x is a point of K, then εx ≺ µ if and only if µ ∈ Mx (S). Proof. Assertion (a) follows from inequalities µ(1) ≤ ν(1) and µ(−1) ≤ ν(−1). To show (b), εx ≺ µ clearly implies µ ∈ Mx (S). Conversely, if µ ∈ Mx (S), s ∈ S c (S) and ε > 0, let t ∈ S be such that s ≤ t and t(x) < s(x) + ε (use Proposition 7.11(a), (b)). Then µ(s) ≤ µ(t) ≤ t(x) ≤ s(x) + ε. Hence εx ≺ µ, and the proof is complete. Proposition 7.13. (a) If f ∈ S usc (S) and g is lower semicontinuous with f < g, then there exists s ∈ W(S) such that f < s < g. (b) If f ∈ S usc (S), then f = inf{g ∈ S c (S) : g > f } and this family is downdirected.

222


(c) If f ∈ S lsc (S) and g is upper semicontinuous with g < f , then there exists s ∈ W(S) such that g < s < f . (d) If f ∈ S lsc (S), then f = sup{g ∈ S c (S) : g < f } and this family is up-directed. (e) W(S) is dense in S c (S). Proof. To show (a), let f < g as in (a) be given. For any x ∈ K we use Proposition 7.11(a),(b) to find a function sx ∈ S such that f < s and f (x) < sx (x) < g(x). By the semicontinuity of f and g there exists an open set Ux 3 x such that f < sx < g on Ux . By the compactness of K, we can select finitely many points x1 , . . . , xn such that f < sx1 ∧ · · · ∧ sxn < g. Hence s := sx1 ∧ · · · ∧ sxn is the required function. Since (b) follows immediately from Proposition 7.11(a),(b) and Proposition 7.6(c), we proceed to the proof of (c). Let f ∈ S lsc (S) and upper semicontinuous g with g < f be given. For a fixed x ∈ K, let µ ∈ Mx (S) be such that g ∗ (x) = µ(g) (use Lemma 7.10). Then g ∗ (x) = µ(g) < µ(f ) ≤ f (x). Hence g ∗ < f , and we may use (a) to find a function s ∈ W(S) with g ∗ < s < f . Since f = sup{g ∈ C(K) : g < f } (see Proposition A.50(ii)), assertion (d) follows from (c). Finally, let f ∈ S c (S) be given. Combining (a) with (b) we see that f = inf{s ∈ W(S) : s > f } and the latter family is down-directed. By Dini’s theorem, for any ε > 0 there exists s ∈ W(S) such that f < s < f + ε. This concludes the proof.

7.2

Maximal measures

In this section, S is a function cone on a compact space K 6= ∅. Definition 7.14 (S-extremal sets). A universally measurable set F ⊂ K is called Sextremal if any measure representing a point in F is carried by F . Theorem 7.15. (a) The family of all closed S-extremal sets is stable with respect to finite unions and arbitrary intersections. (b) A closed set F ⊂ K is S-extremal if and only if cK\F ∈ S lsc (S). (c) If f ∈ S lsc (S), then F := {x ∈ K : f (x) = min f (K)} is S-extremal. (d) A point x ∈ K is in ChS (K) if and only if the set {x} is S-extremal. (e) Any nonempty closed S-extremal set intersects ChS (K); in particular, ChS (K) is nonempty. (f) If f ∈ S lsc (S) is positive on ChS (K), then f ≥ 0 on K.


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Proof. The proof of (a) is analogous to the proof of Proposition 3.14 and (b), (c) and (d) follows by a straightforward verification (cf. Lemma 3.13). To show (e) we use Zorn’s lemma again (cf. Proposition 3.15 and Theorem 2.22). Given a closed Sextremal set F , we consider the family Z of all nonempty closed S-extremal sets contained in F endowed with the partial ordering given by the reverse inclusion. Zorn’s lemma provides a maximal element H ∈ Z. If x1 , x2 ∈ H are distinct points, let s ∈ S be such that s(x1 ) 6= s(x2 ). Then it is easy to see that {z ∈ H : s(z) = min s(H)} is a closed S-extremal set strictly smaller than H, a contradiction with the maximality of H. Hence H is a singleton, say H = {x}, and thus x ∈ ChS (K) by (d). This proves (e). Finally, let f ∈ S lsc (S) be positive on ChS (K). Then F := {x ∈ K : f (x) = min f (K)} is a closed S-extremal set, and thus F ∩ ChS (K) is nonempty by (e). Hence f ≥ 0 on K, which concludes the proof. Proposition 7.16. For measures µ, ν ∈ M+ (K), the following assertions are equivalent: (i) µ ≺ ν, (ii) ν(s) ≤ µ(s) for all s ∈ W(S), (iii) ν(s) ≤ µ(s) for all s ∈ S lsc (S), (iv) ν(s) ≤ µ(s) for all s ∈ S usc (S). Proof. Obviously, (i) =⇒ (ii), (iii) =⇒ (i), (iv) =⇒ (i). Proposition 7.13(b),(d) yields (i) =⇒ (iii) and (i) =⇒ (iv). Finally, Proposition 7.13(e) yields (ii) =⇒ (i). This concludes the proof. Corollary 7.17. Let S be a min-stable function cone on a compact space K and µ, ν ∈ M+ (K). Then µ ≺ ν if and only if ν(s) ≤ µ(s) for each s ∈ S. Proof. Since S = W(S) by the assumption, the proof follows from Proposition 7.16. Theorem 7.18. For every measure µ ∈ M+ (K) there exists a maximal measure λ such that µ ≺ λ. Proof. It suffices to follow step by step the proof of Theorem 3.65. Theorem 7.19. For a measure µ ∈ M+ (K), the following assertions are equivalent: (i) µ is maximal, (ii) µ(f ) = µ(f ∗ ) for any f ∈ −W(S), (iii) µ(f ) = µ(f ∗ ) for any f ∈ Kc (S),

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(iv) µ(f ) = µ(f ∗ ) for any f ∈ Kusc (S), (v) µ(f ) = µ(f ∗ ) for any upper semicontinuous function f on K. Proof. Obviously, (v) =⇒ (iv) =⇒ (iii) =⇒ (ii). To verify (i) =⇒ (v), let f be an upper semicontinuous function. Using Lemma 7.10 we find ν ∈ M+ (K) such that µ ≺ ν and Qµ (f ) = ν(f ). Since µ is maximal, µ = ν. Hence µ(f ) = ν(f ) = Qµ (f ) = µ(f ∗ ) by Lemma 7.8(b). For the proof of (ii) =⇒ (i), let ν ∈ M+ (K) satisfy µ ≺ ν. For any f ∈ −W(S) we get, using Theorem A.84, that µ(f ) = µ(f ∗ ) = inf{µ(s) : s ∈ S c (S), s ≥ f } ≥ inf{ν(s) : s ∈ S c (S), s ≥ f } = ν(f ∗ ) ≥ ν(f ). Since µ(f ) ≤ ν(f ), µ(f ) = ν(f ). Thus µ = ν on W(S) − W(S), which yields µ = ν by Proposition 7.6(h). Lemma 7.20. If x ∈ K, then εx is maximal if and only if x ∈ ChS (K). Proof. The assertion follows from Proposition 7.12(b). Theorem 7.21. For a point x ∈ K, the following assertions are equivalent: (i) x ∈ ChS (K), (ii) f (x) = f ∗ (x) for any f ∈ −W(S), (iii) f (x) = f ∗ (x) for any f ∈ Kc (S), (iv) f (x) = f ∗ (x) for any f ∈ Kusc (S), (v) f (x) = f ∗ (x) for any upper semicontinuous function f on K. Proof. It suffices to combine Lemma 7.20 and Theorem 7.19.

7.3

Representation theorem

In this section, S is a function cone on a compact space K 6= ∅. Lemma 7.22. Let {fn } be an upper bounded sequence of lower semicontinuous Sconvex functions on K such that lim supn→∞ fn ≤ 0 on ChS (K). Then we have lim supn→∞ fn ≤ 0 on K.

7.3 Representation theorem

225

Proof. Let {fn } be as in the lemma. By taking functions −1 ∨ fn if necessary we may assume that {fn } is a sequence bounded by a constant C. For a fixed x ∈ K and ε > 0 we use Proposition 7.11(b) to find functions kn ∈ Kc (S) such that kn ≤ fn and kn (x) ≥ fn (x) − ε. Without loss of generality we may assume that sup{kkn k : n ∈ N} ≤ C. By Theorem 7.15(c)(e), each function in coσ ({kn : n ∈ N}), as an S-convex continuous function, attains its maximum on ChS (K). Since lim sup kn ≤ lim sup fn ≤ 0 n→∞

on ChS (K),

n→∞

the Simons lemma 3.75 yields lim sup fn (x) ≤ ε + lim sup kn (x) ≤ ε. n→∞

n→∞

Since ε > 0 is arbitrary, the proof is finished. Lemma 7.23. Let µ be a maximal measure on K and F ⊂ K \ ChS (K) be a compact Gδ set. Then µ(F ) = 0. Proof. Given a maximal measure µ ∈ M+ (K) and a compact Gδ set F ⊂ K \ ChS (K), we find a sequence {fn } of continuous functions such that 0 ≤ fn ≤ 1, n ∈ N, and fn → cF . Then {(fn )∗ } is a bounded sequence of S-convex lower semicontinuous functions such that lim supn→∞ (fn )∗ ≤ 0 on ChS (K). By Lemma 7.22, lim supn→∞ (fn )∗ ≤ 0 on K. Thus Theorem 7.19 and Fatou’s lemma yield 0 ≤ µ(F ) = lim µ(fn ) = lim µ((fn )∗ ) n→∞

n→∞

≤ lim sup µ((fn )∗ ) ≤ µ(lim sup(fn )∗ ) ≤ 0. n→∞

n→∞

This concludes the proof. Theorem 7.24. Let µ ∈ M+ (K) be a maximal measure on K. Then (a) µ(B) = 0 for every Baire set B disjoint from ChS (K), (b) µ∗ (K \ L) = 0 for every Lindelöf set L ⊃ ChS (K) (here µ∗ denotes the inner measure induced by µ, see Definition A.62), (c) µ(K \ A) = 0 for every K-analytic set A ⊃ ChS (K). Proof. The proof follows from Lemma 7.23 as Theorem 3.79 follows from Lemma 3.78. Corollary 7.25. Let S be a function cone on a compact space K. Then for every x ∈ K there exists µ ∈ Mx (S) such that µ(B) = 1 for any K-analytic set containing ChS (K). Proof. Combine Theorem 7.18 with Theorem 7.24(c).

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Theorem 7.26. Let S be a function cone on a metrizable compact space K. Then the following assertions hold: (a) ChS (K) is of type Gδ . (b) A measure µ ∈ M+ (K) is maximal if and only if µ(K \ ChS (K)) = 0. (c) For any x ∈ K there exists µ ∈ Mx (S) such that µ(K \ ChS (K)) = 0. Proof. We select a dense set {kn : n ∈ N} in Kc (S). To show (a), we verify that ChS (K) =

∞ \

{x ∈ K : kn∗ (x) = kn (x)}.

(7.4)

n=1

Indeed, the inclusion “⊂” follows from Theorem 7.21. Conversely, if x is contained in the right-hand side of (7.4), f ∗ (x) = f (x) for any f ∈ Kc (S) due to the density of {kn : n ∈ N} and Lemma 7.8(c). Thus Theorem 7.21 again implies x ∈ ChS (K). Now (7.4) yields assertion (a). To verify (b), we first notice that any measure µ ∈ M+ (K) satisfying µ(K \ ChS (K)) = 0 is maximal due to Theorem 7.19 and Theorem 7.21. Conversely, any maximal measure µ ∈ M+ (K) satisfies µ(f ) = µ(f ∗ ) for all f ∈ Kc (S), and thus ChS (K), being the intersection of countably many sets {x ∈ K : kn∗ (x) = kn (x)}, n ∈ N, is of full measure µ. The last assertion (c) follows from Theorem 7.18 and (b). Theorem 7.27. Let S be a function cone on a compact space K. Let Σ := {A ∩ ChS (K) : A ⊂ K is a Baire set} and µ ∈ M+ (K) be maximal. Then there exists a measure µ0 on the σ-algebra Σ such that (a) f |ChS (K) is µ0 -measurable for any Baire function f on K, (b) Rf |ChS (K) is µ0 -integrable for any µ-integrable Baire function f on K and µ(f ) = 0 ChS (K) f (t) dµ (t). Proof. Given a maximal measure µ ∈ M+ (K) and A0 ∈ Σ, we define µ0 (A0 ) := µ(A),

A ⊂ K is a Baire set with A ∩ ChS (K) = A0 .

It follows from Theorem 7.24 that the value µ0 (A0 ) does not depend on the choice of A. It is easy to verify that µ0 is a measure on the σ-algebra Σ. If f is a Baire function on K, let f 0 := f |ChS (K) . Given an open set U ⊂ R, {x ∈ ChS (K) : f (x) ∈ U } = ChS (K) ∩ {x ∈ K : f (x) ∈ U } shows that (f 0 )−1 (U ) ∈ Σ, and thus f 0 is µ0 -measurable. If, moreover, f is µ-integrable, we define Ft := {x ∈ K : f (x) ≥ t} for any t ∈ R. Then µ0 (Ft ∩ ChS (K)) = µ(Ft ),

7.4 Simplicial cones

and we get from Fubini’s theorem that Z Z µ(f ) = f (x) dµ(x) = µ(Ft ) dt K R Z Z 0 µ (Ft ∩ ChS (K)) dt = = R

227

f (x) dµ0 (x).

ChS (K)


7.4

Simplicial cones

In this section, S is a function cone on a compact space K 6= ∅. Definition 7.28 (Simplicial function cones). A function cone S on a compact space K is called simplicial if for each x ∈ K there exists a unique maximal measure δx ∈ Mx (S). Lemma 7.29. Let S be a function cone on a compact space K. (a) If S is simplicial, x ∈ K and µ ∈ Mx (S), then µ ≺ δx . In particular, εx ≺ δx . (b) The function cone S is simplicial if and only if W(S) is simplicial and this is the case if and only if S c (S) is simplicial. (c) If S is simplicial, then f ∗ (x) = δx (f ) for any x ∈ K and f ∈ Kusc (S). Proof. To show (a), let µ ∈ Mx (S). By Theorem 7.18, there exists a maximal measure λ such that µ ≺ λ. Using Proposition 7.12(b) we get λ ∈ Mx (S). Since S is simplicial, λ = δx . Hence µ ≺ δx . For the proof of (b), we note that the notions of S-maximal, W(S)-maximal and S c (S)-maximal measures coincide (see Proposition 7.6(g)) and that representing measures are the same for all these cones (see Proposition 7.6(d)). If S is simplicial, f ∈ Kusc (S) and x ∈ K are given, we use Lemma 7.10 to find a measure µ ∈ Mx (S) so that µ(f ) = f ∗ (x). By (a), µ ≺ δx . Using Proposition 7.16 we get f ∗ (x) = µ(f ) ≤ δx (f ) ≤ inf{δx (s) : s ∈ S c (S), s ≥ f } ≤ inf{s(x) : s ∈ S c (S), s ≥ f } = f ∗ (x), which concludes the proof. Theorem 7.30. The following assertions are equivalent: (i) S is simplicial, (ii) f ∗ is S-affine for any f ∈ −W(S), (iii) f ∗ is S-affine for any f ∈ Kc (S), (iv) f ∗ is S-affine for any f ∈ Kusc (S).

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Proof. Obviously, (iv) =⇒ (iii) =⇒ (ii). For the proof of (i) =⇒ (iv), let f ∈ Kusc (S) be given. Let x ∈ K and µ ∈ Mx (S) be given. By Lemma 7.29(a), µ ≺ δx . Using Lemma 7.29(c) we get µ(f ∗ ) = inf{µ(k) : k ∈ S c (S), k ≥ f } ≥ inf{δx (k) : k ∈ S c (S), k ≥ f } = δx (f ∗ ) ≥ δx (f ) = f ∗ (x). By Proposition 7.8(a), µ(f ∗ ) ≤ f ∗ (x). Hence f ∗ is S-affine. Let us assume that (ii) holds, let x ∈ K and µ, ν ∈ Mx (S) be maximal measures. Then, for each f ∈ −W(S), Theorem 7.19 yields µ(f ) = µ(f ∗ ) = f ∗ (x) = ν(f ∗ ) = ν(f ). By Proposition 7.6(h), µ = ν as required. This concludes the proof. Theorem 7.31 (In-between theorem). The following assertions are equivalent: (i) S is simplicial, (ii) for any s, −t ∈ Kc (S) with s < t there exists h ∈ Ac (S) such that s ≤ h ≤ t, (iii) for any s, −t ∈ Kusc (S) with s ≤ t there exists h ∈ Ac (S) such that s ≤ h ≤ t, (iv) for any s, −t ∈ Kusc (S) with s < t there exists h ∈ Ac (S) such that s < h < t. Proof. Clearly, (iii) =⇒ (ii). To show (i) =⇒ (iv), assume that S is simplicial and s, t are as in (iv). Given ε > 0, denote F := {f − g + ε : s < f < g < t, f, −g ∈ Kc (S)} . We want to use Lemma A.88 in order to find a strictly positive function in F. To this end, choose a nonzero measure µ ∈ M+ (K). By Theorem A.84 and Lemma 7.8(b) there exists a function k ∈ S c (S) such that s∗ < k

and µ(k − s∗ ) < εµ(K).

Using Proposition 7.13(a) we find a function g ∈ S c (S) such that s∗ < g < k ∧ t. By Theorem 7.30, s∗ is an S-affine function. Proposition 7.13(c) yields the existence of a function f ∈ Kc (S) such that s∗ < f < g. Then µ(g − f ) ≤ µ(g − s∗ ) < µ(k − s∗ ) < εµ(K),


229

and thus µ(f − g + ε) = εµ(K) + µ(f − g) > 0. This proves the existence of a strictly positive element in F and thus we are able to find a pair of functions f, −g ∈ Kc (S) such that s < f < g < t and g − f < ε. Now we construct inductively functions fn , −gn ∈ Kc (S), n ∈ N, such that • s < f < f n n+1 < gn+1 < gn < t, gn − fn < 2−n . Both sequences {fn } and {gn } converge uniformly to an S-affine continuous function h satisfying s < h < t. To verify (iv) =⇒ (iii), let s, −t ∈ Kc (S) satisfy s ≤ t. We inductively construct a sequence {hn }∞ n=0 of S-affine functions such that • s − 1 < h < t + 1, 0 •

(hn−1 − 2−n ) ∨ (s − 2−n ) < hn < (hn−1 + 2−n ) ∧ (t + 2−n ), n ∈ N. To start the construction, we use (iv) to find h0 ∈ Ac (S) with s − 1 < h0 < t + 1. Assume that we have constructed functions h0 , . . . , hn−1 . From •

s − 2−(n−1) < hn−1 < t + 2−(n−1) we get (hn−1 − 2−n ) ∨ (s − 2−n ) < (hn−1 + 2−n ) ∧ (t + 2−n ). By Lemma 7.29(c), ((hn−1 − 2−n ) ∨ (s − 2−n ))∗ < ((hn−1 + 2−n ) ∧ (t + 2−n ))∗ . Moreover, both functions are S-affine by Theorem 7.30. Hence we can use (iv) again to find a function hn ∈ Ac (S) satisfying (hn−1 − 2−n ) ∨ (s − 2−n ) < hn < (hn−1 + 2−n ) ∧ (t + 2−n ). This completes the construction. It is obvious that {hn } is a Cauchy sequence converging to h ∈ Ac (S) that satisfies s ≤ h ≤ t. To finish the proof we have to verify (ii) =⇒ (i). Pick x ∈ K and consider a maximal measure µ ∈ Mx (S). If f ∈ Kc (S), it follows from (ii) that f ∗ = inf{h ∈ Ac (S) : h ≥ f } and the latter family is down-directed. Hence µ(f ∗ ) = inf{µ(h) : h ∈ Ac (S), h ≥ f } = inf{h(x) : h ∈ Ac (S), h ≥ f } = f ∗ (x). Thus f ∗ is S-affine, which concludes the proof by Theorem 7.30.

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Corollary 7.32. If S is a simplicial function cone and f ∈ Kc (S), then f ∗ = inf{h ∈ Ac (S) : h ≥ f } and this family is down-directed. Hence f ∗ ∈ (Ac (S))⊥⊥ . Proof. The first part follows from Theorem 7.31(iii) and the second part is an immediate consequence of the first one. Definition 7.33 (Kernel T ). Let S be a simplicial function cone on a compact space K. For any bounded universally measurable function f on K, we define a function T f as Z T f : x 7→

f dδx ,

x ∈ K.

K

If f is defined only on a universally measurable subset A of K, we consider f to be defined by 0 on K \ A, and define T f as above. We show in Theorem 7.34 below that the mapping T : x 7→ δx is a kernel as defined in Subsection A.3.D. Theorem 7.34. Let S be simplicial. Then the following assertions hold: (a) If f ∈ C(K) and ε > 0, then there exist functions u, v ∈ Kc (S) so that kT (u − v) − T f k < ε. Hence, there exist sequences {fn }, {gn } of bounded upper semicontinuous functions on K such that the sequence {fn − gn } converges uniformly to T f . In particular, if K is metrizable, T f is a Baire-one function. (b) T f is a Borel function for any bounded Baire function f on K. (c) T f ∈ (Ac (S))⊥⊥ , and hence T f is Ac (S)-affine, for any bounded Baire function f on K. Proof. The proof is identical with the proof of Theorem 6.8; we only have to use Lemma 7.29(c) and Corollary 7.32. Proposition 7.35. The following conditions are equivalent: (i) S is simplicial, (ii) (Ac (S))⊥ ∩ Mbnd (S) = {0}. Proof. The implication (i) =⇒ (ii) follows from Corollary 7.32 and (ii) =⇒ (i) is obvious. Theorem 7.36. Let S be a simplicial cone and T : K → M1 (K) be the kernel from Definition 7.33. Then the following assertions hold: (a) For any µ ∈ M+ (K),


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(a1) T µ ∈ M+ (K), (a2) kT µk = kµk, (a3) T µ is S-maximal, (a4) µ ≺ T µ. (b) T µ is a boundary measure for any µ ∈ M(K). (c) T εx = δx , x ∈ K. Proof. The proof is identical to the proof of Theorem 6.11. Corollary 7.37. Let S be simplicial and f be a bounded universally measurable function on K such that δx (f ) = f (x) for each x ∈ K. Then f ∈ (Ac (S))⊥⊥ . In particular, any bounded function in A(S) belongs to (Ac (S))⊥⊥ . Proof. Follow the proof of Corollary 6.12. Theorem 7.38. Suppose that H := Ac (S) separates points of K. (a) Then H is a closed function space, Mx (S) ⊂ Mx (H) for x ∈ K, S c (H) ⊂ S c (S), Mbnd (H) ⊂ Mbnd (S), ChH (K) ⊂ ChS (K) and Ac (H) = H. (b) If S is moreover simplicial, then H is simplicial, Mbnd (S) = Mbnd (H) and ChS (K) = ChH (K). Proof. We start the proof of (a) by noticing that H is indeed a closed function space. Inclusion Mx (S) ⊂ Mx (H) follows by the definition, inclusions S c (H) ⊂ S c (S), Mbnd (H) ⊂ Mbnd (S) and ChH (K) ⊂ ChS (K) are its consequences and Ac (H) = H follows from Theorem 3.27. Now assume that S is simplicial. Let s, −t ∈ Kc (H) with s ≤ t be given. Since c K (H) ⊂ Kc (S), Theorem 7.31 yields the existence of a function h ∈ H such that s ≤ h ≤ t. By Theorem 6.6, H is simplicial. Let µ ∈ M+ (K) be an S-maximal measure and let ν ∈ M+ (K) satisfy µ ≺H ν. Then µ − ν ∈ H⊥ . For any f ∈ Kc (H) ⊂ Kc (S) we get from Theorem 7.19 and Corollary 7.32 that µ(f ) = inf{µ(h) : h ∈ H, h ≥ f } = inf{ν(h) : h ∈ H, h ≥ f } = ν(f ). Hence µ = ν and µ is H-maximal. Thus Mbnd (S) ⊂ Mbnd (H), which gives Mbnd (S) = Mbnd (H) by (a). Since the equality ChS (K) = ChH (K) is then an immediate consequence, the proof is complete. Lemma 7.39. If f ∈ Kusc (S) and g ∈ S usc (S) with f ≤ g on ChS (K), then f ≤ g on K.

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Proof. Given x ∈ K and ε > 0, Proposition 7.13(b) provides a function g 0 ∈ S c (S) such that g < g 0 and g 0 (x) ≤ g(x) + ε. Then g 0 − f ∈ S lsc (S) and g 0 − f ≥ 0 on ChS (K). By Theorem 7.15(f), g 0 − f ≥ 0 on K. In particular, 0 ≤ g 0 (x) − f (x) ≤ g(x) − f (x) + ε. Since ε is arbitrary, the assertion follows. Theorem 7.40. Suppose that Ac (S) separates points of K. If Ac (S) is simplicial and ChAc (S) (K) = ChS (K), then S is simplicial. Proof. Let H stand for the space Ac (S). The aim of the proof is to show that any Smaximal measure is H-maximal. To this end, let f ∈ Kc (H) ⊂ Kc (S) be given. We temporarily write f ∗,H and f ∗,S for the respective envelopes. We claim that f ∗,H = f ∗,S on K. Indeed, f ∗,H = f on ChH (K) by Theorem 3.24 and f ∗,S = f on ChS (K) by Theorem 7.21. Since H is simplicial, f ∗,H is H-affine (see Theorem 6.5) and thus also S-affine. Since f ∗,S ∈ S usc (S) and f ∗,H = f ∗,S on ChS (K), Lemma 7.39 yields f ∗,H ≤ f ∗,S on K. Since f ∗,H ≥ f ∗,S by the definitions and Theorem 7.38(a), the claim is proved. Let µ ∈ Mmax (S) and f ∈ Kc (H) be given. Then µ(f ) = µ(f ∗,S ) by Theorem 7.19, and thus µ(f ) = µ(f ∗,H ) by the previous step. Theorem 3.58 gives that µ is H-maximal. Since Mx (S) ⊂ Mx (H) for all x ∈ K, the simpliciality of S follows.

7.5

Ordered compact convex sets and simplicial measures

Definition 7.41 (Ordered compact convex set). Let E be a locally convex space provided with a partial ordering ≤ given by a closed convex cone E + ⊂ E that is proper (that is, E + ∩ −E + = {0}). A compact convex set X ⊂ E with the ordering inherited from E is called an ordered compact convex set. A real-valued function l on X is called isotone if l(x) ≤ l(y) whenever x ≤ y, x, y ∈ X. We denote by Lc (X) the convex cone of all continuous isotone affine functions on X. Proposition 7.42. Let X be an ordered compact convex set and x, y ∈ X. Then x ≤ y if and only if l(x) ≤ l(y) for each l ∈ Lc (X). In particular, Lc (X) separates points of X. Proof. The first part is nothing else than Lemma A.20. If x, y ∈ X are distinct points, we may assume that x y. By the first part, f (y) < f (x) for a suitable f ∈ Lc (X). This concludes the proof.

7.5 Ordered compact convex sets and simplicial measures

233

Example 7.43. Let H be a function space on a compact space K. Consider the locally convex space M(K) ordered by the cone {µ ∈ M(K) : µ(f ) ≥ 0 for all f ∈ Kc (H)}. Then the ordering induced by this cone on M+ (K) is nothing else than the Choquet ordering ≺H . The sets M1 (K), Mx (H) where x ∈ K, Mµ (H) := {ν ∈ M+ (K) : µ − ν ∈ H⊥ } and {ν ∈ M+ (K) : µ ≺ ν ∈ H⊥ } where µ ∈ M+ (K) are ordered compact convex sets. In these examples, the cone of affine isotone continuous functions coincides with the set of all H-convex continuous functions on K, where we view C(K) as a subspace of (M(K))∗ . Notation 7.44 (Maximal elements). If X is an ordered compact convex set, the set of all points of X, which are maximal in the given ordering, is denoted by Xmax . Proposition 7.45. Let X be an ordered compact convex set. Then the following assertions hold: (a) For every point x ∈ X, there exists a maximal element z ∈ X so that x ≤ z. (b) The set Xmax is an extremal subset of X and ext Xmax = Xmax ∩ ext X. (c) If Xmax is convex, it is a face of X. Proof. If x ∈ X is given, let F := {y ∈ X : x ≤ y}. This is a partially ordered set and we need to show that any chain in F has an upper bound. But this easily follows from the compactness of X. Indeed, if R ⊂ F is a chain, find a subnet J ⊂ R and a point y0 ∈ X such that J converges to y0 . Fix y ∈ R. Let l ∈ Lc (X) and ε > 0. Find y1 ∈ J so that y ≤ y1 and |l(y1 ) − l(y0 )| ≤ ε. Then l(y0 ) ≥ l(y1 ) − ε ≥ l(y) − ε. Since ε is arbitrary, l(y) ≤ l(y0 ) and thus y ≤ y0 due to Proposition 7.42. Since we have verified that any chain in F has an upper bound, Zorn’s lemma concludes the proof of (a). Concerning (b), let z ∈ Xmax and x, y ∈ X with z = 21 (x + y) be given. Our aim is to show that x, y ∈ Xmax . If not, let y0 be a point of X distinct from y and satisfying y ≤ y0 . Then z ≤ 21 (x + y0 ) and z 6= 21 (x + y0 ), which contradicts maximality of z. Thus Xmax is an extremal subset of X. The second part of (b) follows from Proposition 2.64. Finally, (c) follows from (b). Definition 7.46 (Hereditary upwards sets). A subset F of an ordered compact convex set X is said to be hereditary upwards if x ∈ F , y ∈ X and x ≤ y implies that y ∈ F . We remark that X itself is hereditary upwards.

234


Proposition 7.47. If F is a nonempty closed face which is hereditary upwards, then F ∩ ext Xmax 6= ∅. In particular, ext Xmax is nonempty if X 6= ∅. Proof. Let the collection F := {H ⊂ F : H is a nonempty hereditary upwards closed face } be partially ordered by the reverse T inclusion. Then F is nonempty since F ∈ F. If R ⊂ F is a chain, then M := R is also a nonempty hereditary upwards closed face contained in F . Thus F has a maximal element H by Zorn’s lemma. If we suppose that H contains two distinct points x and y, we can find a continuous isotone affine function l in X such that l(x) 6= l(y), say l(x) < l(y). Then N := {z ∈ H : l(z) = max l(H)} is a hereditary upwards closed face properly contained in H, which contradicts maximality of H. Thus H is a singleton, say H = {z}. Since H is a face, z is an extreme point of X. Moreover, z is maximal because H is hereditary upwards. Thus z ∈ ext Xmax ∩ F as required. Proposition 7.48. Let ϕ : X → Y be a continuous affine surjection of an ordered compact convex set X onto an ordered compact convex set Y such that ϕ(x1 ) ≤ ϕ(x2 ) whenever x1 ≤ x2 are elements of X. Then ϕ(Xmax ) ⊃ Ymax . Proof. Given y ∈ Ymax , we select x ∈ ϕ−1 (y). Using Proposition 7.45(a) we find z ∈ Xmax with x ≤ z. Since ϕ−1 (y) is hereditary upwards due to the assumption on ϕ, z ∈ ϕ−1 (y), and the proof is complete. Proposition 7.49. Let ϕ : X → Y be a continuous affine surjection of a compact convex set X onto a compact convex set Y . Then ϕ] (Mmax (X)) ⊃ Mmax (Y ). Proof. If ϕ is as in the hypothesis, ϕ] : M1 (X) → M1 (Y ) satisfies the assumptions of Proposition 7.48 (here M1 (X) and M1 (Y ) are ordered compact convex sets when considered with the Choquet ordering). Hence any Ac (Y )-maximal measure ν ∈ M1 (Y ) is the image of some measure µ ∈ Mmax (X) ∩ M1 (X), and the assertion follows. In the sequel we apply results of previous sections on function cones. It will be more convenient to work with the function cone −Lc (X) instead of Lc (X). Proposition 7.50. Let X be an ordered compact convex set. Then −Lc (X) is a closed function cone. Proof. A straightforward verification using Proposition 7.42.


235

Definition 7.51 (Monotone envelopes). If X is an ordered compact convex set, we define a monotone envelope of a function f on X by fe(x) := inf{l(x) : l ≥ f, −l ∈ Lc (X)},

x ∈ X.

Example 7.52. If X = M1 (K) as in Example 7.43 with ordering determined by some function space H on K, then the family of continuous isotone affine functions on X consists of functions of the form µ 7→ µ(f ), where f is an H-convex continuous function on K. If F ∈ Ac (X), then F (µ) = µ(f ), µ ∈ M1 (K), for some continuous function f on K. Then Fe(µ) = inf{L(µ) : L ≥ F, −L ∈ Lc (X)} = inf{µ(l) : l ≥ f, l ∈ S c (H)} = µ(f ∗ ). Lemma 7.53. Let X be an ordered compact convex set and f be a function on X. Then fe(x) = Qεx (f ), x ∈ X, where Qεx (f ) is the functional from Definition 7.7 if the role of S is played by the cone −Lc (X). Proof. The proof follows from Proposition 7.11(b). Theorem 7.54. Let X be an ordered compact convex set and −Lc (X) be considered as a function cone on X. Then the following assertions hold: (a) If µ, ν ∈ M1 (X) satisfy µ ≺−Lc (X) ν, then r(µ) ≤ r(ν). (b) The function −fe is isotone for any f ∈ C(X). (c) A point x is in Xmax if and only if fe(x) = f (x) for any f ∈ Ac (X). (d) Ch−Lc (X) (X) = ext Xmax . (e) For every x ∈ X there exists a −Lc (X)-maximal measure µ ∈ Mx (−Lc (X)). Proof. Given µ, ν as in (a), let l ∈ Lc (X) be arbitrary. Then l(r(µ)) = µ(l) ≤ ν(l) = l(r(ν)). By Proposition 7.42, r(µ) ≤ r(ν). Assertion (b) is easy to verify because the supremum of a family of isotone functions is again an isotone function. To verify (c), let x ∈ Xmax be given. For any f ∈ Ac (X), there exists a measure µ ∈ Mx (−Lc (X)) such that fe(x) = µ(f ) (see Lemma 7.10). Then εx ≺−Lc (X) µ (see Proposition 7.12(b)) and thus x ≤ r(µ) by (a). Hence x = r(µ) due to the maximality of x. This yields fe(x) = µ(f ) = f (r(µ)) = f (x).

236


Conversely, let x ∈ X \ Xmax be given. It means that there exists y ∈ X such that x < y. Using Proposition 7.42, we find f ∈ Lc (X) such that f (x) < f (y). Then f (x) < f (y) ≤ fe(y) ≤ fe(x) (the last inequality holds due to (b)). This proves (c). To show (d), assume first that x ∈ ext Xmax . Given an arbitrary function f ∈ C(X) and ε > 0, let h ∈ Ac (X) be such that f ≤ h and h(x) < f (x)+ε (use Theorem 3.24 and Proposition 3.25 for the function space Ac (X)). By (c), e h(x) = h(x), and thus c there exists a function l ∈ −L (X) such that h ≤ l and l(x) < h(x) + ε. Thus f ≤ l and l(x) ≤ f (x) + 2ε, which gives fe(x) = f (x). By Theorem 7.21, x ∈ Ch−Lc (X) (X). Conversely, assume that x ∈ Ch−Lc (X) (X). If y ∈ X satisfies x ≤ y, then εx ≺−Lc (X) εy . Since x ∈ Ch−Lc (X) (X), x = y and thus x ∈ Xmax . Let µ ∈ Mx (X) be arbitrary. Then εx ≺Ac (X) µ by Corollary 3.26. Since W(−Lc (X)) ⊂ W(Ac (X)), εx ≺−Lc (X) µ (use Proposition 3.56 and Proposition 7.16). Hence µ = εx and x ∈ ext X, by Theorem 2.40. Thus x ∈ Xmax ∩ ext X = ext Xmax by Proposition 7.45(b). Since (e) is a consequence of Theorem 7.18, the proof is complete. Theorem 7.55. Let X be an ordered compact convex set and −Lc (X) be considered as a function cone on X. If Σ := {A ∩ ext Xmax : A ⊂ X is a Baire set}, then for every x ∈ Xmax there exists a measure µ0 on the σ-algebra Σ such that Z f (x) = f (t) dµ0 (t), f ∈ Ac (X). ext Xmax

Proof. Given the objects as in the hypothesis, let µ ∈ Mx (−Lc (X)) be a −Lc (X)maximal measure (use Theorem 7.18) and let µ0 be the induced measure on Σ given by Theorem 7.27. Since εx ≺−Lc (X) µ, x ≤ r(µ) by Theorem 7.54(a). Since x is a maximal element of X, x = r(µ). Hence Theorem 7.27 gives Z Z f (t) dµ0 (t) = f (t) dµ0 (t) ext Xmax

Ch−Lc (X)

= µ(f ) = f (r(µ)) = f (x) for any f ∈ Ac (X).


237

Definition 7.56 (Weak topology generated by monotone envelopes). Let X be an ordered compact convex set in a locally convex space E. Assume that, for every h ∈ E ∗ , g the monotone envelope h| X is the restriction of some linear functional ϕh : E → R. ∗ Let D := span(E ∪ {ϕh : h ∈ E ∗ }). We denote by τ the locally convex topology on E generated by D. Example 7.57. If X is any of the ordered compact convex sets from Example 7.43, then D is the linear span of functions of the form µ 7→ µ(f ), µ ∈ M(K),

f ∈ C(K) ∪ {g ∗ : g ∈ C(K)}.

Theorem 7.58. Let X 6= ∅ be an ordered compact convex set such that the topology τ from Definition 7.56 is well defined. Then Xmax is convex and Xmax = coτ ext Xmax . Proof. It follows from Theorem 7.54(c) that \ \ g Xmax = {x ∈ X : h| {x ∈ X : ϕh (x) = h(x)}. X (x) = h(x)} = h∈E ∗

h∈E ∗

Thus Xmax is convex and it is a closed set in the topology τ . Hence we have Xmax ⊃ coτ ext Xmax . We know from Proposition 7.47 that ext Xmax is a nonempty set. Assume that z ∈ Xmax \ coτ ext Xmax . By the Hahn–Banach separation argument, there exists a function f ∈ D such that f (z) > sup f (ext Xmax ). P Then f = h0 + ni=1 ci ϕhi , where h0 , . . . , hn ∈ E ∗ , c1 , . . . , cn ∈ R and ϕhi are linear functionals on E such that ϕhi = hg i |X on X. Thus f = h0 +

n X

ci hg i |X

i=1

on X. By Theorem 7.54(c), f = h0 +

Pn

i=1 ci hi

h := h0 +

n X

on Xmax . Thus

ci hi

i=1

is a function from E ∗ such that h(z) > sup h(ext Xmax ). ^ Since z ∈ Xmax , −h| X (z) = −h(z) by Theorem 7.54(c). Hence there exists an isotone function l ∈ Lc (X) so that l≤h

and

l(z) > sup h(ext Xmax ).

238


Then F := {x ∈ X : l(x) = max l(X)} is a nonempty closed face which is hereditary upwards. By Proposition 7.47, F intersects ext Xmax , say at a point y. Then l(z) > sup h(ext Xmax ) ≥ sup l(ext Xmax ) ≥ l(y) = max l(X) ≥ l(z). This contradiction finishes the proof. Definition 7.59 (Simplicial measures). Let H be a function space on a compact space K. For µ ∈ M+ (K), let Mµ (H) := {ν ∈ M+ (K) : ν − µ ∈ H⊥ }. A measure ν ∈ Mµ (H) is called simplicial if ν is an extreme point of Mµ (H). We remark that Mµ (H) is an ordered compact convex set by Example 7.43. If x ∈ K, then Mεx (H) is nothing else than Mx (H). Theorem 7.60 (Douglas). Let H be a function space on a compact space K and µ ∈ M+ (K). A measure ν ∈ Mµ (H) is simplicial if and only if the space H is dense in L1 (ν). Proof. Notice that H is a subspace of L1 (ν) and that H is dense in L1 (ν) if and only if F ∈ (L1 (ν))∗ : F (h) = 0 for all h ∈ H = {0} . If H is not dense in L1 (ν), then there exists f ∈ L∞ (ν) such that kf kL∞ (ν) = 1 and Z for all h ∈ H.

hf dν = 0 K

Define ν1 and ν2 by setting ν1 := (1 − f )ν

and

ν2 := (1 + f )ν.

Then ν1 , ν2 ∈ M+ (K) and Z Z Z ν1 (h) = (1 − f )h dν = h dν = (1 + f )h dν = ν2 (h) K

K

K

for any h ∈ H. Hence ν1 , ν2 ∈ Mµ (H). Since ν=

ν1 + ν2 2

the measure ν is not simplicial.

and

ν1 6= ν 6= ν2 ,


239

For the converse, assume that H is dense in L1 (ν) and that ν ∈ Mµ (H) satisfies 2 where ν1 , ν2 ∈ Mµ (H). Since 0 ≤ ν1 ≤ 2ν and 0 ≤ ν2 ≤ 2ν, the Radon– ν = ν1 +ν 2 Nikodym theorem yields positive functions f1 , f2 ∈ L∞ (ν) such that f1 + f2 = 2 ν-almost everywhere and νj = fj ν for j = 1, 2. For any h ∈ H we have Z Z f1 h dν = ν1 (h) = µ(h) = ν2 (h) = f2 h dν. K

K

Since H is dense in L1 (ν), the same equalities hold for any h ∈ L1 (ν). We see that f1 = f2 = 1 ν-almost everywhere. Therefore ν = ν1 = ν2 . Theorem 7.61. Let H be a function space on a compact space K, µ ∈ M+ (K) and X := Mµ (H) be the ordered compact convex set from Definition 7.59. Let τ be the locally convex topology on M(K) induced by the family D from Example 7.57. Then (a) ν ∈ Xmax if and only if ν ∈ X is H-maximal, (b) Xmax is a face of X, (c) ext Xmax = Xmax ∩ ext X, (d) Xmax = coτ ext Xmax , (e) for any H-maximal measure ν ∈ X there exists a measure Ω0 defined on the σ-algebra Σ := {A ∩ ext Xmax : A ⊂ X is a Baire set} such that

Z

λ(f ) dΩ0 (λ) = ν(f ),

f ∈ C(K).

ext Xmax

Proof. If ν ∈ Xmax and λ ∈ M+ (K) with ν ≺H λ, then λ ∈ X and thus ν = λ. Hence ν is H-maximal. Obviously, any H-maximal measure ν ∈ X is in Xmax . This proves (a). Assertions (b), (c) and (d) follow from Theorem 7.58 and Proposition 7.45; (e) is a consequence of Theorem 7.55. Theorem 7.62. Let H be a function space on a compact space K. Then for any x ∈ K and f ∈ Kc (H) there exists a simplicial measure µ ∈ Mx (H) such that µ(f ) = f ∗ (x). Proof. Let X denote the ordered compact convex set Mx (H). Then F := {µ ∈ X : µ(f ) = f ∗ (x)} is a closed hereditary upwards face of X. Moreover, it is nonempty by Lemma 3.21. Hence F ∩ ext Xmax 6= ∅ by Proposition 7.47, which finishes the proof.

240

7.6


Exercises

Exercise 7.63. Let X be a compact convex set and S := Sc (X) (see Definition 4.1). Prove that + • ≺ c A (X) coincides with ≺S on M (X), • S-maximal measures coincide with Ac (X)-maximal measures, • Ch (X) = ext X, M (S) = M (X) for x ∈ X. x x S Hint. Since W(S) = S and S is closed, S c (S) = S by Proposition 7.13(e). Hence S c (S) = S = S c (Ac (X)). Thus ≺S coincides with ≺Ac (X) . This implies Mx (X) = Mx (S) for x ∈ X and coincidence of S-maximal and Ac (X)-maximal measures. Exercise 7.64. Consider the function cone S := {f ∈ C([0, 1]) : f is increasing}. Prove that • S is closed and min-stable, M (S) = {µ ∈ M1 ([0, 1]) : spt µ ⊂ [0, x]} for x x ∈ [0, 1], ChS ([0, 1]) = {0}, • Ac (S) is not a function space. Hint. A straightforward verification. Exercise 7.65. Find an example of a function cone S on a compact space K such that Ac (S) is a function space and ChAc (S) (K) 6= ChS (K). Hint. For n ∈ N, consider the following objects in R2 : In := [−2, −1] × {n−1 }, xn := (0, n−1 ), I := [−2, −1] × {0},

Jn := [1, 1 + n−1 ] × {n−1 }, yn := ((2n)−1 , n−1 ), x := (0, 0),

z := (1, 0).

Let µn be one-dimensional Lebesgue measure on In and νn be one-dimensional Lebesgue measure on Jn normalized in such a way that it is a probability measure. Let ∞ [ K := I ∪ {x} ∪ {z} ∪ (In ∪ Jn ∪ {xn } ∪ {yn }) n=1

and

S := {f ∈ C(K) : f (xn ) ≥ (1 − n−1 )f (yn ) + n−1 µn (f ), f (xn ) ≥ (1 − n−1 )f (yn ) + n−1 νn (f ), n ∈ N}.

Then S is a closed min-stable function cone on a compact space K and S c (S) = S. Further, the function  0 at z,  f = n−1 on Jn , n ∈ N,   1 otherwise, shows that z ∈ ChS (K).

7.6 Exercises

241

Further, Ac (S) = {f ∈ C(K) : f (xn ) = (1 − n−1 )f (yn ) + n−1 µn (f ), f (xn ) = (1 − n−1 )f (yn ) + n−1 νn (f ), n ∈ N}. is a function space and z ∈ / ChAc (S) (K). Indeed, for any function f ∈ Ac (S) and n ∈ N we have (1 − n−1 )f (yn ) + n−1 µn (f ) = (1 − n−1 )f (yn ) + n−1 νn (f ). Hence µn (f ) = νn (f ),

n ∈ N,

which in turn yields Z f (t) dt = f (z). I

This concludes the proof. Exercise 7.66. Find an example of an ordered compact convex set X such that Xmax is not convex. Hint. Consider E := R2 ordered by the cone E + = {(x, y) ∈ E : x ≥ 0, y ≥ 0}, and let X := {(x, y) ∈ E : x2 + y 2 ≤ 1}. Then Xmax = {(x, y) ∈ X : x ≥ 0, y ≥ 0, x2 + y 2 = 1} is not convex. Exercise 7.67. If X is an ordered compact convex set, x ∈ Xmax and µ ∈ Mx (X), then µ(fe) = µ(f ) for any f ∈ Ac (X). Hint. Given f ∈ Ac (X), −fe is upper semicontinuous and concave. Using Corollary 4.8 and Theorem 7.54(c), we get f (x) = µ(f ) ≤ µ(fe) ≤ fe(x) = f (x).

Exercise 7.68. If X is a metrizable ordered compact convex set, then Xmax is a measure extremal subset of X. Hint. Let {fn : n ∈ N} be a dense countable subset in BAc (X) . From Lemma 7.8(c) and Theorem 7.54(c) we deduce that Xmax =

∞ \

{x ∈ X : fe(x) = f (x)}.

n=1

Thus Xmax is a Gδ set in X. If x ∈ Xmax and µ ∈ Mx (X), Exercise 7.67 yields µ(fe) = µ(f ). Thus we get µ(Xmax ) = 1 and Xmax is measure extremal.

242


Exercise 7.69. Let X be a metrizable compact convex set and f be a bounded universally measurable function on X such that, for any maximal µ ∈ M1 (X), µ(f ) = f (r(µ)). Prove that f is strongly affine. Hint. Given a measure µ ∈ Mx (X), let ν ∈ M1 (X) be maximal with µ ≺ ν. Then ν ∈ Mx (X) as well. We consider Y := M1 (X) as an ordered compact convex set and let πi : Y × Y → Y be the projections, i = 1, 2. Let M := {(εx , λ) ∈ Y × Y : εx ≺ λ}. By Theorem 3.92(ii), there exists Ω ∈ M1 (M ) such that Ω represents the point (µ, ν) ∈ Y × Y . Then (π2 )] Ω represents ν ∈ Ymax , and thus Exercise 7.68 gives that ((π2 )] Ω)(Ymax ) = 1. By assumption, ν(f ) = f (x). We apply Proposition 3.90 and the assumption to the pairs (f, 0) and (0, f ) to get Z Z µ(f ) = λ1 (f ) dΩ(λ1 , λ2 ) = λ1 (f ) dΩ(λ1 , λ2 ) M ∩(Y ×Ymax )

M

Z =

λ1 (f ) dΩ(λ1 , λ2 ) {(εx ,λ)∈Y ×Y :εx ≺λ,λ∈Ymax }

Z =

λ2 (f ) dΩ(λ1 , λ2 ) {(εx ,λ)∈Y ×Y :εx ≺λ,λ∈Ymax }

Z =

λ2 (f ) dΩ(λ1 , λ2 ) M

= ν(f ) = f (x). This concludes the proof. Exercise 7.70. Find an example of a function space H on a compact space K and an ordered compact convex set X ⊂ M(K) such that Xmax is a convex set that is strictly smaller then co ext Xmax . Hint. Let H be a function space on a compact space K such that there exists x ∈ ChH (K) \ ChH (K). If X := M1 (K), then Xmax = M1 (K) ∩ Mmax (H),

ext Xmax = {εy : y ∈ ChH (K)}.

Hence εx ∈ co ext Xmax \ Xmax . Exercise 7.71. Find an example of a function space H on a compact space K and an ordered compact convex set X ⊂ M(K) such that cok·k ext Xmax is strictly smaller then Xmax .


243

Hint. Let K := [0, 1], H := C(K) and X := M1 (K). Then Xmax = M1 (K),

ext Xmax = {εx : x ∈ K}

and cok·k ext Xmax consists of discrete probability measures on K. Hence Xmax 6= cok·k ext Xmax .

7.7

Notes and comments

In the literature, there are various treatises of Choquet theory of function cones in a diverse level of generality. Let us mention, for example, cones of lower semicontinuous functions, cones on locally compact spaces, weakly complete cones, well-capped cones, or cones of potentials. The reader can find material concerning Choquet’s theory of convex cones in several books such as E. M. Alfsen [5], R. Becker [47], J. Bliedtner and W. Hansen [66], N. Boboc and Gh. Bucur [70], G. Choquet [108], D. Sibony [415], and in papers written by many authors. We mention here J. Bliedtner and W. Hansen [62], N. Boboc and Gh. Bucur [68], N. Boboc and A. Cornea [73], D. A. Edwards [157], G. Mokobodzki [347], G. Mokobodzki and D. Sibony [348]. We incorporated many ideas from these sources in Sections 7.1 – 7.4. Section 7.5 follows E. M. Alfsen [5] (see also E. M. Alfsen and C. F. Skau [11]). The characterization of simplicial measures in Theorem 7.60 is in R. G. Douglas [147] (see also N. Boboc and Gh. Bucur [69]). Recall that the Minkowski–Carathéodory theorem 2.12 states that every point of a compact convex subset of Rd is a convex combination of its at most d + 1 affinely independent extreme points. For a proof based on properties of simplicial measures, see [5], Corollary I.6.13. Problem 7.72. We do not know whether the assertion of Exercise 7.69 holds without the metrizability assumption.

Chapter 8

Choquet-like sets

The aim of this chapter is to transfer the concept of faces of compact convex sets to the framework of general function spaces. The most important examples are P -sets and M -sets, which are analogues of parallel and split faces. We start by recalling in Theorems 8.5 and 8.7 measure theoretic characterizations of split and parallel faces that guide us to the general definition. In order to be able to handle “faces” in function spaces as in the framework of compact convex sets, we need several analogues of classical tools from convex analysis. This task is accomplished in Section 8.2, where we define H-convex and H-extremal sets and prove results imitating the Krein–Milman theorem, the Milman theorem and the Hahn–Banach separation theorem (see Corollary 8.19, Corollary 8.20 and Proposition 8.23). In Section 8.3, we present analogues of faces in function spaces termed Choquet sets. Particular examples of P -sets and M -sets are defined there and their basic characterizations are proved (see Theorem 8.39 and Theorem 8.44). An important result due to C. J. K. Batty on characterization of H-maximal measures is given in Theorem 8.32. H-exposed sets are investigated in Section 8.4. The next section provides preparatory results on boundary measures that are necessary in Section 8.6. The central result of this chapter is Theorem 8.60 and its consequences contained in Theorem 8.62 and Corollary 8.63. They provide characterization of simplicial function spaces by means of Choquet sets. In the context of compact convex sets we get a characterization of simplices by means of split and parallel faces. Throughout this chapter, we frequently use the notation and results from Section 4.3. Hence H is a function space on a compact space K, φ is the evaluation mapping from K to the state space S(H) and X denotes a compact convex subset of a locally convex space.

8.1

Split and parallel faces

In the developing of Choquet’s theory we would like to find a counterpart of the notions of split and parallel faces studied in the convex setting. In Theorems 8.5 and 8.7 we present without proofs known characterizations of these faces fitted to our purposes. Definition 8.1 (Complementary sets). Let F ⊂ X. The union of all faces of X disjoint from F is called the complementary set of F and it is denoted by F 0 .

8.1 Split and parallel faces

245

A complementary set F 0 is always an extremal set, and it is a face if and only if it is convex. Moreover, x ∈ F 0 if and only if F ∩ face x = ∅. Proposition 8.2. If F is a closed face of a compact convex set X, then F = {x ∈ X : (cF )∗ (x) = 1}

and F 0 = {x ∈ X : (cF )∗ (x) = 0} .

Every x ∈ X can be written as a convex combination x = λy + (1 − λ)z,

(8.1)

where y ∈ F , z ∈ F 0 and the “barycentric coefficient” λ = (cF )∗ (x) ∈ [0, 1]. Proof. See [5], Proposition II.6.5. Remarks 8.3. (a) Since (cF )∗ is an upper semicontinuous function, the complementary set F 0 of a closed face F is a Gδ set. (b) Notice that {x ∈ X : cF (x) = c∗F (x)} ⊂ F ∪ F 0 . In general, the decomposition x = λy + (1 − λ)z in (8.1) is not unique. The question of uniqueness of the decomposition leads to notions of parallel and split faces and is investigated next. Definition 8.4 (Split faces). A face F of X is said to be a split face if its complementary set F 0 is convex and if any point in x ∈ X \ (F ∪ F 0 ) can be uniquely represented as a convex combination x = λy + (1 − λ)z,

y ∈ F, z ∈ F 0

and

λ ∈ (0, 1).

The following theorem provides a measure theoretic characterization of split faces. Theorem 8.5. The following statements about a closed face F of X are equivalent: (i) F is a split face, (ii) µ|F ∈ (Ac (X))⊥ for any boundary measure µ ∈ (Ac (X))⊥ . Proof. See E. M. Alfsen [5], Theorem II.6.12. Definition 8.6 (Parallel faces). A face F is called parallel if its complementary set F 0 is convex, X = co(F ∪ F 0 ), and if for every point x ∈ X \ (F ∪ F 0 ) the barycentric coefficient λ in the convex combination x = λy + (1 − λ)z, is uniquely determined.

where

y ∈ F, z ∈ F 0 , λ ∈ (0, 1),

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8 Choquet-like sets

As in the case of split faces we have the following measure theoretic characterization of parallel faces. Theorem 8.7. The following statements about a closed face F of X are equivalent: (i) F is a parallel face, (ii) µ(F ) = 0 for any boundary measure µ ∈ (Ac (X))⊥ . Proof. See B. Hirsberg [226], Theorem 2.12 or L. Asimow and A. J. Ellis [24], Theorem 2.10.10.

8.2

H-extremal and H-convex sets

Definition 8.8 (H-extremal sets). We recall that a universally measurable subset F of K is H-extremal if, given x ∈ F and µ ∈ Mx (H), then µ is carried by F (see Definition 3.12). Example 8.9. Any universally measurable subset of the Choquet boundary ChH (K) is H-extremal. This follows from the fact that Mx (H) = {εx } for any x ∈ ChH (K). Lemma 8.10. A universally measurable set B ⊂ K is H-extremal if and only if its characteristic function cB is H-convex, and B is an H-extremal closed set if and only if cB is upper semicontinuous and H-convex. Moreover, if H is a measure extremal subset of the state space S(H), then φ−1 (H ∩ φ(K)) is H-extremal in K. Proof. The former assertion follows immediately from the definitions. If H ⊂ S(H) is measure extremal and µ is a representing measure for x ∈ F := φ−1 (H ∩ φ(K)), then φ] µ represents φ(x) ∈ H. Since H is measure extremal, φ] µ is carried by H. Therefore, µ is carried by F . Examples 8.11. (a) If F is an H-extremal subset of K, then φ(F ) need not be an extremal subset of S(H). Consider the following example: Let λ1 and λ2 denote the restrictions of Lebesgue measure on [0, 1] and [3, 4], respectively. If K := [0, 1] ∪ {2} ∪ [3, 4] and

H :=

1 f ∈ C(K) : f (2) = (λ1 (f ) + λ2 (f )) , 2

then K is an H-extremal set. On the other hand, sj = π(λj ), j = 1, 2, are elements of S(H), 1 φ(2) = (s1 + s2 ) ∈ φ(K), 2 while s1 and s2 do not belong to φ(K) since φ(K) = {π(εx ) : x ∈ K}.

8.2 H-extremal and H-convex sets

247

(b) Example 8.29 illustrates the fact that there is an H-extremal set H in K such that co φ(H) is not an extremal subset of S(H). If F is a closed face of a compact convex set X, then the complementary set F 0 of F equals to {x ∈ X : (cF )∗ (x) = 0} (see Proposition 8.2). Definition 8.12 (Complementary sets). If H is a function space on a compact space K, we associate to each subset F of K its complementary set F 0 by F 0 := {x ∈ K : (cF )∗ (x) = 0} . Since (cF )∗ is an upper semicontinuous function, the complementary set F 0 is a Gδ set. Notice that {x ∈ K : cF (x) = (cF )∗ (x)} ⊂ F ∪ F 0 . Lemma 8.13. Let F be a closed subset of K. Then any maximal measure on K is carried by F ∪ F 0 . Proof. According to Corollary 3.58, µ(cF ) = µ (cF )∗ . Since K \ (F ∪ F 0 ) ⊂ {x ∈ K : cF (x) < (cF )∗ (x)} , it follows that µ is carried by F ∪ F 0 . Definition 8.14 (H-convex sets). We say that a universally measurable set B ⊂ K is H-convex if x ∈ B whenever x ∈ K and µ ∈ Mx (H) satisfies µ(K \ B) = 0. Definition 8.15 (closed H-convex hull). Let F be a subset of K. The closed H-convex hull of F is the set \ coH F := {C : C ⊃ F, C is closed and H-convex} . Obviously, coH F is a closed H-convex set. Conversely, if F is a closed H-convex set, then coH F = F . Also, it is easy to verify that coH F = coH F . Lemma 8.16. Let F ⊂ K. Then we have: (a) The set F is H-convex or H-extremal if and only if it is Ac (H)-convex or Ac (H)extremal, respectively. (b) coH F = coA

c (H)

F.

Proof. The assertions follow directly from the definitions. Proposition 8.17. Let {Fi : i ∈ I} be a family of H-convex sets. If sally measurable, it is H-convex as well. Proof. The proof follows by a straightforward verification.

T

i∈I

Fi is univer-

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8 Choquet-like sets

Proposition 8.18. Let F be a subset of K. Then coH F = {x ∈ X : |h(x)| ≤ sup |h|(F ) for all h ∈ H} = x ∈ X : there is µ ∈ Mx (H) such that spt µ ⊂ F . Proof. Denote by Fe := {x ∈ X : |h(x)| ≤ sup |h|(F ) for all h ∈ H} and Fb := x ∈ X : there is µ ∈ Mx (H) such that spt µ ⊂ F . The set Fe is closed and contains F . Moreover, Fe is H-convex. Indeed, given x ∈ K and µ ∈ Mx (H) such that spt µ ⊂ Fe, we have Z Z |h(x)| = |µ(h)| = h dµ ≤ |h| dµ ≤ sup {|h(t)| : t ∈ F } Fe

Fe

for every h ∈ H. Hence x ∈ Fe. It follows that coH F ⊂ Fe. Assume that x ∈ Fe. Define Tx : f 7→ hf (x) for

f ∈ H|F ,

where hf is any extension of f to a function from H. The functional Tx is well defined, because if h1 and h2 are functions from H such that h1 |F = h2 |F = f , then h1 − h2 = 0 on F . Since x ∈ Fe, we get h1 (x) = h2 (x). Moreover, Tx is positive on H|F and kTx k = 1. By the Hahn–Banach theorem, there exists a Radon measure µ ∈ M1 (K) such that spt µ ⊂ F and µ(f ) = Tx (f ) for any f ∈ H. Since µ(h) = Tx (h) = h(x) for any h ∈ H, we have µ ∈ Mx (H), and therefore x ∈ Fb. We have shown that Fe ⊂ Fb. The next step is to show that Fb ⊂ coH F . To this end, assume that C ⊃ F is a closed H-convex set. Pick x ∈ Fb. There exists a measure µ ∈ Mx (H) such that spt µ ⊂ F ⊂ C. Hence from the H-convexity of C, it follows that x ∈ C. We can conclude that Fb ⊂ coH F . Since coH F = coH F (see Definition 8.15), from foregoing we get coH F ⊂ Fe ⊂ Fb ⊂ coH F = coH F, which established the assertion and finishes the proof. Corollary 8.19 (Krein–Milman type theorem). For any function space H on K, K = coH (ChH (K)). Proof. Given x ∈ K, let µx be a maximal measure in Mx (H). Since maximal measures are carried by ChH (K) (see Proposition 3.64), x ∈ coH (ChH (K)) by Proposition 8.18. Hence K = coH (ChH (K)) = coH (ChH (K)).

8.2 H-extremal and H-convex sets

249

Corollary 8.20 (Milman type theorem). Let F be a subset of K such that K = coH F . Then ChH (K) ⊂ F . Proof. Take x ∈ ChH (K). Since x ∈ coH F , by Proposition 8.18 there exists µ ∈ Mx (H) such that spt µ ⊂ F . Since x ∈ ChH (K), µ = εx , which implies that x ∈ F. Remarks 8.21. (a) Let F be a subset of a compact convex set X and H = Ac (X). Then coH F is nothing else than the closed convex hull of F . (b) It is not true that, for a Borel subset F of a compact convex set X and H = Ac (X), coH F = {x ∈ X : there is µ ∈ Mx (H) such that µ(X \ F ) = 0} . Consider the compact convex set X := M1 ([0, 1]) and F := {εx : x ∈ S} where S is a countable dense subset of [0, 1]. Then co F = X while for any measure Λ ∈ M1 (X) representing Lebesgue measure λ ∈ X we have Λ(F ) = 0. Proposition 8.22. A closed set F ⊂ K is H-convex if and only if F = φ−1 (co φ(F ) ∩ φ(K)). Proof. Let F be a closed H-convex set and let x be in K such that φ(x) ∈ co φ(F ). By Proposition 4.28(b), there exists a measure µ ∈ M1 (φ(F )) representing the point φ(x). Then (φ−1 )] µ ∈ Mx (H) and spt µ ⊂ F . Thus x ∈ coH F = F . Conversely, let F = φ−1 (co F ∩ φ(K)) and let µ be a measure representing a point x ∈ K such that spt µ ⊂ F . Then spt φ] µ ⊂ φ(F ), and hence φ(x) ∈ co φ(F ). Due to the assumption, φ(x) ∈ φ(F ). Proposition 8.23 (Hahn–Banach type theorem). If F is a nonempty subset of K and x∈ / coH F , then there exists h ∈ H such that h(x) > max h(F ). Proof. This readily follows from Proposition 8.18. Proposition 8.24. If µ and ν are positive measures on K, µ ≺ ν, then spt µ ⊂ coH spt ν. Proof. Let µ ≺ ν and x ∈ spt µ \ coH spt ν. By Proposition 8.23, there exists a function h ∈ H such that h(x) > max h(coH spt ν). By adding a suitable constant, we may assume that max h(coH spt ν) = 0. Then h ∨ 0 is a continuous H-convex function and µ(h ∨ 0) > 0 ≥ ν(h ∨ 0) which contradicts the assumption µ ≺ ν. Lemma 8.25. Let F be a closed set in K. Then F ∩ ChH (K) = (coH F ) ∩ ChH (K).

250

8 Choquet-like sets

Proof. The inclusion “⊂” being obvious, we need to verify the reverse one. Let x ∈ (coH F ) ∩ ChH (K) be given. By Proposition 8.18 there exists a measure µ ∈ Mx (H) ∩ M1 (F ). Since x ∈ ChH (K), µ = εx and x ∈ F , which completes the proof. Lemma 8.26. Let µ be a boundary measure and F a closed subset of K. Then |µ|(coH F \ F ) = 0. In particular, if F is a closed subset of ChH (K), then F = (coH F ) ∩ ChH (K). Proof. Let L be a compact subset of coH F \ F . Given ε > 0, there exists a compact set M ⊂ K \ coH F such that |µ|(M ) > |µ|(K \ coH F ) − ε. Let g ∈ C(K) be such that 0 ≤ g ≤ 1 on K, g=1

on L and

g=0

on F ∪ M.

Then g∗ = 0 on coH F (see Proposition 8.18) and (obviously) on M . Since µ is a boundary measure, using Corollary 3.58 we get Z Z Z g∗ d|µ| + g∗ d|µ| + g∗ d|µ| |µ|(L) ≤ |µ|(g) = coH F

K\(M ∪coH F )

M

H

≤ |µ|(K \ (M ∪ co F )) < ε. As ε > 0 is arbitrary, |µ|(L) = 0. Hence, |µ|(coH F \ F ) = 0. The second assertion follows immediately from Lemma 8.25.

8.3

Choquet sets, M -sets and P -sets

Definition 8.27 (Choquet sets). A universally measurable subset F of K is a Choquet set if F is both H-extremal and H-convex. Hence, a closed set F is a Choquet set if and only if the following two conditions are satisfied: (a) given x ∈ F and µ ∈ Mx (H), then spt µ ⊂ F , (b) if x ∈ K \ F and µ ∈ Mx (H), then spt µ is not contained in F . Example 8.28 (Convex case). Let X be a compact convex subset of a locally convex space. A closed set F ⊂ X is a Choquet set if and only if F is a closed face of X. This assertion follows by the fact that a closed set F in X is convex or extremal if and only if F is Ac (X)-convex or Ac (X)-extremal, respectively.

8.3 Choquet sets, M -sets and P -sets

251

Example 8.29 (State space). Let H be a closed face of S(H). Then the set F := φ−1 (H ∩ φ(K)) is a closed Choquet set. Indeed, F is H-convex due to Proposition 8.22 and H-extremal by Lemma 8.10. The converse implication is false in general. Let K := [0, 1] ∪ [3, 4]. Denote by λ1 and λ2 restrictions of Lebesgue measure on [0, 1] and [3, 4], respectively. Let H := {f ∈ C(K) : λ1 (f ) = λ2 (f )}. Then F := [0, 1] is a Choquet set but the set H := co φ(F ) is not Ac (S(H))-extremal in S(H). Indeed, the state s := π(λ1 ) belongs to H, the measure φ] λ2 represents s and (φ] λ2 )(H) = 0. Proposition 8.30. (a) If F is a closed H-extremal set, then F ⊂ coH (F ∩ ChH (K)). (b) For any closed Choquet set F ⊂ K we have F = coH (F ∩ ChH (K)). (c) The family of closed Choquet sets is stable with respect to arbitrary intersections. Proof. For the proof of (a), let F be a closed H-extremal set and let x ∈ F be given. First, we show that h(x) ≤ sup h(F ∩ ChH (K)) (8.2) holds for any strictly positive h ∈ H. Let h ∈ H be a strictly positive function. Then f := hcF is an upper semicontinuous H-convex function, and hence it attains its maximum at some point y ∈ ChH (K) (see Lemma 3.13 and Proposition 3.15). Since h is strictly positive, y ∈ F ∩ ChH (K), and (8.2) follows. If h ∈ H is arbitrary, we apply the previous argument to h + c for a suitable c ∈ R to obtain (8.2) for any h ∈ H. Given h ∈ H, (8.2) applied to h and −h yields |h(x)| ≤ sup |h|(F ∩ ChH (K)). By Proposition 8.18, x ∈ coH (F ∩ ChH (K)). To verify (b), we first notice that obviously F ⊃ coH (F ∩ ChH (K)). Since the reverse inclusion follows by (a), the proof of (b) is complete. By combining Propositions 3.14 and 8.17 we get (c). Proposition 8.31. If H is a simplicial function space, then coH F is a Choquet set for any compact F ⊂ ChH (K). Proof. If F is a compact subset of ChH (K), coH F is obviously an H-convex set. To show its H-extremality, by Proposition 8.24 it is enough to show δx is carried by coH F for any x ∈ coH F . For such a point x we find ν ∈ Mx (H) ∩ M1 (F ) (see Proposition 8.18). Since ν is carried by ChH (K), it is H-maximal (use Corollary 3.59), and thus ν = δx . Hence δx ∈ M1 (coH F ) and the proof is finished. S Theorem 8.32. For any x ∈ K, let Fx (H) := ν∈Mx (H) spt ν. For a measure µ ∈ M+ (K) the following assertions are equivalent: (i) µ is maximal,

252

8 Choquet-like sets

(ii) there exists a set S ⊂ C(K) separating points of K such that each function in S is constant on Fx (H) for µ-almost all x ∈ K, (iii) any continuous function on K is constant on Fx (H) for µ-almost all x ∈ K. Proof. We start the proof by showing (i) =⇒ (ii). Let µ be a maximal measure and let h ∈ H be arbitrary. Then µ((h2 )∗ ) = µ(h2 ) by Theorem 3.58, and thus (h2 )∗ = h2 on a set A ⊂ K with µ(K \ A) = 0. Let x ∈ A be arbitrary and ν ∈ Mx (H). Then, by the Hölder inequality, ν(h2 ) ≤ (h2 )∗ (x) = h2 (x) = (ν(h))2 ≤ ν(h2 ). Thus

Z

2

Z

(h − h(x)) dν =

2

h dν − 2h(x) K

K

Z

h dν + h2 (x) = 0.

K

Hence h = h(x) on spt ν, and h is constant on Fx (H). Thus we can take H to be the required family S of continuous functions. For the proof of (ii) =⇒ (iii), let S ⊂ C(K) be as in (ii). Then the family F := {f ∈ C(K) : f is constant on Fx (H) for µ-almost all x} is a closed linear space stable with respect to pointwise multiplication and maxima. Since S ⊂ F, F = C(K) by the Stone–Weierstrass theorem. Assume now that (iii) holds. Given a function f ∈ C(K), let x ∈ K be a point such that f is constant on Fx (H). We find a measure ν ∈ Mx (H) such that ν(f ) = f ∗ (x) (see Lemma 3.21). Since x ∈ Fx (H) and spt ν ⊂ Fx (H), f = f (x) on Fx (H). Thus f ∗ (x) = ν(f ) = f (x). Thus f (x) = f ∗ (x) for µ-almost all x ∈ K, and thus µ(f ) = µ(f ∗ ). By Theorem 3.58, µ is maximal. Definition 8.33 (Conditions (M) and (P)). We say that a closed set F ⊂ K satisfies condition (M) if (M)

µ|F ∈ H⊥

for any µ ∈ H⊥ ∩ Mbnd (H),

µ(F ) = 0

for any

and condition (P) if (P)

µ ∈ H⊥ ∩ Mbnd (H).

Proposition 8.34. Let F ⊂ K be a closed set. Then F satisfies condition (M) or (P) if and only if the set coH F satisfies (M) or (P), respectively. Proof. If µ ∈ H⊥ ∩ Mbnd (H) is given, Lemma 8.26 asserts that µ|F = µ|coH F , from which the assertion readily follows.


253

Example 8.35 (State space). If a closed set F satisfies (M), then φ(F ) satisfies (M) in the state space as well. Indeed, given µ ∈ Ac (S(H))⊥ ∩ Mbnd (S(H)), we have (φ−1 )] µ ∈ H⊥ ∩ Mbnd (H) which yields (φ−1 )] µ|F ∈ H⊥ . Then µ|φ(F ) ∈ Ac (S(H))⊥ . Similarly, we see that φ(F ) satisfies (P) provided F satisfies (P). Definition 8.36 (P-sets). We say that a set F is a P -set if F is a closed Choquet set satisfying condition (P). Example 8.37 (Convex case). If F is a closed face in a compact convex set X, then F is a P -set if and only if F is a parallel face. This follows from a measure theoretic characterization of parallel faces stated in Theorem 8.7. Proposition 8.38 (State space). The following assertions hold: (a) If H is a closed parallel face of S(H), then the set F := φ−1 (H ∩ φ(K)) is a P -set. (b) If F is a P -set, then co φ(F ) is a parallel face of S(H). Proof. (a) We already know from Example 8.29 that F is a Choquet set. Let µ ∈ H⊥ ∩ Mbnd (H) be given. Then φ] µ ∈ Ac (S(H))⊥ ∩ Mbnd (S(H)). Thus we get µ(F ) = (φ] µ)(H) = 0, which verifies condition (P) for F . (b) By Example 8.35, the set φ(F ) satisfies condition (P). By Proposition 8.34, co φ(F ) satisfies (P) as well. We have to check that co φ(F ) is extremal. Let a maximal measure Λ ∈ M1 (S(H)) and s ∈ co φ(F ) with r(Λ) = s be given. By Proposition 4.28(d), Λ = φ(λ) for some H-maximal measure λ on K. By Proposition 4.28(b), there exists ν ∈ M1 (F ) such that π(ν) = s. Let µ be an H-maximal measure with ν ≺ µ. Since F is a Choquet set, cF is an upper semicontinuous Hconvex function and thus (see Proposition 3.56) 1 = ν(F ) = ν(cF ) ≤ µ(cF ) = µ(F ). Thus µ is carried by F likewise. Then π(µ) = π(ν) = s = π(λ), in other words, µ − λ ∈ H⊥ . Hence condition (P) implies 1 = µ(F ) = λ(F ), and thus Λ is carried by φ(F ), and therefore also by co φ(F ). Let now Λ ∈ M1 (S(H)) with the barycenter in co φ(F ) be given arbitrarily. We find an Ac (S(H))-maximal measure Ω with Λ ≺ Ω. Then Ω is carried by co φ(F ) by the considerations above. By Proposition 8.24, spt Λ ⊂ co spt Ω ⊂ co φ(F ), which gives the desired conclusion that co φ(F ) is extremal and consequently a face. Hence, it is a parallel face by Example 8.37.

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8 Choquet-like sets

Theorem 8.39. Let F be a closed H-convex subset of K. Then the following assertions are equivalent: (i) F is a P -set, (ii) µ(c∗F ) = ν(c∗F ) for any probability measures µ and ν with µ − ν ∈ H⊥ , (iii) the set {h ∈ H : h > cF } is down-directed, (iv) the function c∗F belongs to H⊥⊥ , (v) there exists a parallel face H ∈ S(H) such that F = φ−1 (H ∩ φ(K)). Proof. The equivalence (i) ⇐⇒ (v) follows from Proposition 8.38 and Proposition 8.22. Assume now that F is a P -set and choose µ, ν ∈ M1 (K) such that µ − ν ∈ H⊥ . Lemma 3.21 provides a measure µ0 ∈ M1 (K) so that µ ≺ µ0 and Qµ (cF ) = µ0 (cF ). Let µm be a maximal measure satisfying µ0 ≺ µm . Since F is a closed Choquet set, cF is upper semicontinuous and H-convex, and thus µ0 (cF ) ≤ µm (cF ). Let ν m be a maximal measure with ν ≺ ν m . Then µm − ν m ∈ H⊥ . Then, using Lemma 3.18, Proposition 3.56, Theorem 3.58, and condition (P) we get µ(c∗F ) = Qµ (cF ) = µ0 (cF ) ≤ µm (cF ) = µm (F ) = ν m (F ) = ν m (cF ) = ν m (c∗F ) ≤ ν(c∗F ). By interchanging µ with ν we get that µ(c∗F ) = ν(c∗F ). Hence, (i) =⇒ (ii). Suppose next that (ii) holds and that h1 , h2 ∈ H, h1 > cF , h2 > cF . Then ∗ cF < h1 ∧ h2 . Let µ, ν ∈ M1 (K) with µ − ν ∈ H⊥ be given. Since µ(c∗F ) = ν(c∗F ) < ν(h1 ∧ h2 ), the assumptions of Lemma 5.43 are satisfied. Hence there exists h ∈ H such that cF ≤ c∗F < h < h1 ∧ h2 . Statement (ii) therefore implies (iii). Further, suppose that (iii) holds. Pick µ ∈ H⊥ , µ = µ1 − µ2 where µ1 and µ2 are positive. Using Exercise 3.96 and Theorem A.84 we get µ1 (c∗F ) = µ1 (inf {h ∈ H : h ≥ cF }) = µ1 (inf {h ∈ H : h > cF }) = inf {µ1 (h) : h ∈ H, h > cF } = inf {µ2 (h) : h ∈ H, h > cF } = · · · = µ2 (c∗F ). Thus µ(c∗F ) = 0 as needed. Finally, let (iv) hold. For a boundary measure µ ∈ H⊥ we get from Theorem 3.58 µ(cF ) = µ(c∗F ) = 0. Thus, µ(F ) = 0, which proves (iv) =⇒ (i).


255

Definition 8.40 (M-sets). A set F ⊂ K is said to be an M -set if F is a closed Choquet set satisfying condition (M). Example 8.41 (Convex case). Let X be a compact convex subset of a locally convex space. Then it follows from Theorem 8.5 that a closed face F ⊂ X is an M -set if and only if F is a split face of X. Proposition 8.42 (State space). The following assertions hold: (a) If H is a closed split face of S(H), then the set F := φ−1 (H ∩φ(K)) is an M -set. (b) If F is an M -set, then co φ(F ) is a closed split face of S(H). Proof. To show (a) we first notice that F is a Choquet set by Example 8.29. If µ ∈ Mbnd (H) ∩ H⊥ , then φ] µ ∈ Mbnd (S(H)) ∩ (Ac (S(H)))⊥ . Thus (φ] µ)|H belongs to (Ac (S(H)))⊥ , which yields µ|F ∈ H⊥ . To verify (b), we use Proposition 8.38(b) to get that co φ(F ) is a (parallel) face. By Example 8.35, φ(F ) possesses condition (M), and thus co φ(F ) has condition (M) by Proposition 8.34. Hence co φ(F ) is a split face by Theorem 8.5. Lemma 8.43. Let F ⊂ K be an M -set, f an upper semicontinuous H-convex function on K, g ∈ H and f0 , g0 ∈ H|F . If f 0

on

K \ F.

A set F is relatively H-exposed if for each x ∈ K \ F there is a function hx ∈ H+ such that hx = 0 on F and hx (x) > 0. Proposition 8.48. A set F is relatively H-exposed if and only if F is the intersection of H-exposed sets. Furthermore, any relatively H-exposed set F is a closed Choquet set.

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Proof. Assume that F is relatively H-exposed. Then \ F = {t ∈ K : hx (t) = 0} , x∈K\F

where, for any x ∈ K \ F , hx is a function from H+ such that hx (x) > 0 and hx = 0 on F . Conversely, any intersection of H-exposed sets is relatively H-exposed. Since any H-exposed set is clearly closed, a relatively H-exposed set F (being an intersection of H-exposed sets) is itself closed. Let µ ∈ M1 (F ) be a measure representing x ∈ K. If x ∈ / F , then there exists a function hx ∈ H+ with hx (x) > 0 and hx = 0 on F . Then 0 = µ(hx ) = hx (x) > 0, which is a contradiction. Thus F is H-convex. Let x ∈ F and µ ∈ Mx (H) be given. If µ(K \F ) > 0, we find a point y ∈ spt µ\F and hy ∈ H+ such that hy = 0 on F and hy (y) > 0. Then the inequalities Z Z hy (t) dµ(t) > 0 0 = hy (x) = hy (t) dµ(t) ≥ spt µ

spt µ∩{z∈K:hy (z)>0}

lead to a contradiction. Thus F is H-extremal. Proposition 8.49. Any H-convex Archimedean set is relatively H-exposed. Proof. If F ⊂ K is an H-convex Archimedean set, pick x ∈ K \ F . The function f := cF − 1 is upper semicontinuous. By Lemma 3.21, there exists a Radon measure ν ∈ Mx (H) so that f ∗ (x) = ν(f ). Since F is H-convex and x ∈ / F , we see that ν is not carried by F . Hence ν(f ) < 0, and therefore f ∗ (x) < 0. The definition of f ∗ yields a function g ∈ H such that g ≥ 0 on F and g(x) < 0. Since F is supposed to be Archimedean, there is h ∈ H+ such that h = g on F and h ≥ g on K. If u := h − g, then u ∈ H is positive, u = 0 on F and u(x) > 0. Corollary 8.50. Let H be a closed function space on a compact space K. Then any M -set is relatively H-exposed. Proof. This follows immediately from Propositions 8.46 and 8.49. Summary. Let H be a closed function space. If we summarize results on various Choquet like sets, we get the following sequence of implications for a closed Hconvex set F : F is an M -set =⇒ F is Archimedean =⇒ F is relatively H-exposed =⇒ F is a Choquet set .

8.5 Weak topology on boundary measures

8.5

259

Weak topology on boundary measures

Definition 8.51 (Ordered set of measures). Given a function space H on a compact space K, let M := {(µ, ν) ∈ M+ (K) × M+ (K) : kµk ≤ 1, kνk ≤ 1, µ − ν ∈ H⊥ }. We define a partial ordering ≺ on M: (µ1 , ν1 ) ≺ (µ2 , ν2 ) if µ1 ≺H µ2 and ν1 ≺H ν2 . Then M is an ordered compact convex set (see Definition 7.41). (The cone E + ⊂ M(K) × M(K) required by Definition 7.41 consists of all pairs (µ, ν) ∈ M(K) × M(K) satisfying µ(f ) ≥ 0 and ν(f ) ≥ 0 whenever f ∈ Kc (H).) It can be easily observed that (µ, ν) is maximal in M if and only if both µ and ν are H-maximal measures. If Mmax denotes the set of all ≺-maximal elements of M, then Mmax is a convex and extremal subset of M, that is, Mmax is a face of M. By Proposition 7.47, the set ext Mmax is nonempty. Lemma 8.52. If (µ, ν) ∈ ext Mmax , then either µ = ν or kµ − νk = 2. Proof. Suppose that (µ, ν) ∈ ext Mmax is a pair of nontrivial measures. First we realize that µ, ν ∈ M1 (K). Indeed, since (µ, ν) ∈ M, then µ(K) = ν(K) ≤ 1. Suppose that 0 < µ(K) = ν(K) < 1. Then µ ν (µ, ν) = µ(K) , + (1 − µ(K))(0, 0), µ(K) µ(K) which contradicts the extremality of (µ, ν). Assume that µ 6= ν and kµ − νk < 2. Let µ − ν = η + − η − be the decomposition of µ − ν into the positive and negative part. Then kη + k = kη − k and 2 > kµ − νk = kη + k + kη − k = 2kη + k. Set λ := ν − η − = µ − η + . Since 1 = kη + k + kλk and kη + k < 1, λ is a nontrivial measure. It follows from the equality + η η− λ λ + (µ, ν) = kη k , + kλk , , kη + k kη + k kλk kλk that the pair (µ, ν) cannot be an extreme point of Mmax . Notation 8.53. We denote ⊥ H⊥ 2 := H ∩ {µ ∈ M(K) : kµk ≤ 2} ⊥ and notice that H⊥ 2 ∩ Mbnd (H) 6= {0} if and only if H ∩ Mbnd (H) 6= {0}.

260

8 Choquet-like sets

Lemma 8.54. If H⊥ 2 ∩ Mbnd (H) 6= {0}, then ext H⊥ ∩ M (H) = {µ − ν : (µ, ν) ∈ ext Mmax , kµ − νk = 2}. bnd 2 + − Proof. Let η = η + − η − ∈ ext(H⊥ 2 ∩ Mbnd (H)) be given (η and η is the positive and negative part of η, respectively). Then kη + k = kη − k and

2 = kηk = kη + k + kη − k. It easily follows that (η + , η − ) ∈ ext Mmax . Indeed, if (η + , η − ) = α(µ1 , µ2 ) + (1 − α)(ν1 , ν2 ) for some α ∈ (0, 1) and (µ1 , µ2 ), (ν1 , ν2 ) ∈ Mmax , then η = α(µ1 − µ2 ) + (1 − α)(ν1 − ν2 ). Thus µ1 − µ2 = ν1 − ν2 = ν. Since µ1 , ν1 ≤ ν + , µ2 , ν2 ≤ ν − and all these measures are probability, µ1 = ν + = ν1 and µ2 = ν − = ν2 . Thus (η + , η − ) ∈ ext Mmax . Conversely, let (µ, ν) ∈ ext Mmax with kµ − νk = 2 be given. Then η := µ − ν ∈ H⊥ . Since η + ≤ µ, η − ≤ ν and all these measures are probability, we have η + = µ, η − = ν. If η = αη1 + (1 − α)η2 for some α ∈ (0, 1) and η1 , η2 ∈ H⊥ 2 ∩ Mbnd (H), then 2 = kη1 k = kη2 k = kηk and kη1+ k = kη1− k = kη2+ k = kη2− k = 1. Since

µ − ν = η = α(η1+ − η1− ) + (1 − α)(η2+ − η2− ) = αη1+ + (1 − α)η2+ − (αη1− + (1 − α)η2− ),

we get µ ≤ αη1+ + (1 − α)η2+

and η ≤ αη1− + (1 − α)η2− .

Since all these measures are probability, (µ, ν) = α(η1+ , η1− ) + (1 − α)(η2+ , η2− ). Since (µ, ν) ∈ ext Mmax , we get η1+ = η2+ and η1− = η2− . Thus η1 = η2 , and hence µ − ν ∈ ext(H⊥ 2 ∩ Mbnd (H)). This finishes the proof. ⊥ Lemma 8.55. If H⊥ 2 ∩ Mbnd (H) 6= {0}, then ext(H2 ∩ Mbnd (H)) 6= ∅.

8.5 Weak topology on boundary measures

261

Proof. By Lemma 8.52 and Lemma 8.54 it suffices to show that the set A := {(µ, ν) ∈ ext Mmax : µ 6= ν} is nonempty. Let D be the linear span of functionals on M(K) × M(K) of the form (µ, ν) 7→ µ(f ) + ν(g),

f, g ∈ C(K),

and (µ, ν) 7→ µ(f ∗ ) + ν(g ∗ ),

f, g ∈ C(K).

Consider the locally convex topology σ on M(K) × M(K) generated by functionals from D. It follows from Theorem 7.58 that Mmax = coσ ext Mmax . Now the proof of the lemma follows easily. If the set A were empty, then every pair (µ, ν) ∈ Mmax would satisfy µ = ν. But our hypothesis assures that there is + a nonzero measure µ ∈ H⊥ 2 ∩ Mbnd (H). Then its positive and negative parts µ , − + − + − µ satisfy (µ , µ ) ∈ Mmax and µ 6= µ , a contradiction. This concludes the proof. Definition 8.56 (Weak topology generated by upper envelopes). Let D be the linear span of the family C(K)∪{f ∗ : f ∈ C(K)} in the space of all bounded Borel functions on K. We denote by τ the (locally convex) topology on M(K) generated by the functionals µ 7→ µ(f ), f ∈ D. Theorem 8.57. Let H be a function space on K. Then τ ⊥ co ext H⊥ ∩ M (H) = H ∩ M (H) . bnd bnd 2 2 Proof. By Theorem 3.68 ⊥ H⊥ 2 ∩ Mbnd (H) = H ∩

\

{µ ∈ M(K) : kµk ≤ 2, µ(f ∗ − f ) = 0} ,

f ∈C(K)

and thus the set H⊥ 2 ∩ Mbnd (H) is τ -closed. ⊥ Suppose that H⊥ 2 ∩ Mbnd (H) 6= {0}. Then ext(H2 ∩ Mbnd (H)) 6= ∅ by Lemma 8.55. Assume that there exists τ ⊥ η ∈ H⊥ 2 ∩ Mbnd (H) \ co (ext(H2 ∩ Mbnd (H)).

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8 Choquet-like sets

Then there exists f ∈ D so that n o η(f ) > s := sup µ(f ) : µ ∈ ext(H⊥ 2 ∩ Mbnd (H)) = sup {(µ − ν)(f ) : (µ, ν) ∈ ext Mmax , µ 6= ν} , where the latter equality follows by Lemma 8.54 and Lemma 8.52. Since (µ, ν) ∈ ext Mmax if and only if (ν, µ) ∈ ext Mmax , it follows that s ≥ 0. By the definition of τ there are continuous functions f0 , . . . , fn on K such that f = f0 +

n X

fi∗ .

i=1

Pn

Set g := f0 + i=1 fi and let η + and η − be the positive and the negative part of η. Since η is a boundary measure, we obtain s < η(f ) = η(g) = η + (g) + η − (−g) = η + (g∗ ) + η − ((−g)∗ ). An appeal to Lemma 3.18(b) provides a continuous H-convex functions k1 and k2 such that k1 ≤ g, k2 ≤ −g and η + (k1 ) + η − (k2 ) > s. (8.3) If M denotes the set from Definition 8.51, let F := {(µ, ν) ∈ M : µ(k1 ) + ν(k2 ) = sup{λ1 (k1 ) + λ2 (k2 ) : (λ1 , λ2 ) ∈ M}} . Then F is a nonempty closed face in M which is hereditary upwards, that is, if (µ, ν) ∈ F , (b µ, νb) ∈ M and (µ, ν) ≺ (b µ, νb), then (b µ, νb) ∈ F . By Proposition 7.47, there exists (µ, ν) ∈ ext Mmax ∩ F . If µ = ν, then 0 ≤ s < µ(k1 ) + µ(k2 ) ≤ µ(g) + µ(−g) = 0, which is impossible. Thus µ 6= ν and µ − ν ∈ ext(H⊥ 2 ∩ Mbnd (H)) by Lemma 8.54. By (8.3) we get µ(k1 ) ≤ µ(g) and ν(k2 ) ≤ ν(−g). Then s ≥ µ(f ) − ν(f ) = µ(g) − ν(g) ≥ µ(k1 ) + ν(k2 ) ≥ η + (k1 ) + η − (k2 ) > s yields a contradiction, and the proof is complete.

8.6

Characterizations of simpliciality by Choquet sets

Lemma 8.58. Let µ be a boundary measure on K and F be a P -set. Then Z |µ(F )| ≤ sup h dµ : h ∈ H and khk ≤ 2 . K

8.6 Characterizations of simpliciality by Choquet sets

263

Proof. Let F be a P -set. Then c∗F = inf{h ∈ H : cF < h < 2}. Indeed, c∗F = inf{h ∈ H : cF ≤ h} = inf{h ∈ H : cF < h}. The latter set is down-directed by Theorem 8.39. Thus for any h ∈ H with cF < h we can find g ∈ H with cF < g < h ∧ 2. Pick ε > 0. An appeal to Theorem A.84 provides a function h ∈ H so that cF < h < 2 and |µ|(c∗F ) − |µ|(h) < ε. Then |µ(F )| = |µ(cF )| = |µ(c∗F )| ≤ |µ(c∗F − h)| + |µ(h)| ≤ |µ|(c∗F − h) + |µ(h)| ≤ |µ(h)| + ε Z ≤ sup h dµ : h ∈ H and khk ≤ 2 + ε. K

As ε is arbitrary, we get the assertion. Lemma 8.59. Let H be a closed function space on a compact space K. Assume that any H-exposed set satisfies condition (P). If µ ∈ Mbnd (H) satisfies µ(F ) = 0 for any H-exposed set F , then µ ∈ H⊥ . Proof. For a contradiction, we suppose µ(h) = 6 0 for some h ∈ H and define a continuous functional on H as Z L(h) := h dµ, h ∈ H. K

Since L 6= 0, we may assume that kLk = 1. By the Bishop–Phelps theorem A.8, there are sequences {Ln } in H∗ and {hn } in H so that kLn − Lk → 0, kLn k = Ln (hn ) and khn k = 1 for each n ∈ N. Set Fn+ := {x ∈ K : hn (x) = 1}

and

Fn− := {x ∈ K : hn (x) = −1}.

Let {µn } be a sequence given by Lemma 6.23, that is, for any n ∈ N we have kµn k = kLn k and µn (h) = Ln (h) for any h ∈ H. Fix now n ∈ N. Since khn k = 1 and kµn k = kLn k = Ln (hn ) = µn (hn ), we obtain + − spt µ+ and spt µ− n ⊂ Fn n ⊂ Fn . As each set Fn+ is H-exposed, it satisfies condition (P), and so by Lemma 8.58 we have Z + + |µ(Fn ) − µn (Fn )| ≤ sup h d(µ − µn ) {h∈H:khk≤2}

=

sup {h∈H:khk≤2}

K

|(L − Ln )(h)| = 2 kL − Ln k.

264

8 Choquet-like sets

Thus |µ(Fn+ ) − µn (Fn+ )| → 0. Since Fn+ is an H-exposed set, µ(Fn+ ) = 0 due to the + − + assumption. Recall that spt µ+ n ⊂ Fn and spt µn ∩ Fn = ∅. Hence, + + + + − + + kµ+ n k = µn (Fn ) = µn (Fn ) − µn (Fn ) = µn (Fn )

= µ(Fn+ ) + µn (Fn+ ) − µ(Fn+ ) ≤ |µn (Fn+ ) − µ(Fn+ )|, from which it follows that kµ+ n k → 0. Similarly we obtain kµ− n k → 0. Therefore − 1 = kLk = lim kLn k = lim kµn k = lim kµ+ n k + kµn k = 0, n→∞

n→∞

n→∞

which is a contradiction. Thus µ ∈ H⊥ as required. Theorem 8.60. Let H be a closed function space on a compact space K. Then the following assertions are equivalent: (i) H⊥ ∩ Mbnd (H) = {0}, (ii) any closed Choquet subset of K is an M -set, (iii) any closed Choquet subset of K is a P -set, (iv) any H-exposed subset of K is a P -set. Proof. The implications (i) =⇒ (ii) =⇒ (iii) =⇒ (iv) are obvious. It remains to prove that (iv) =⇒ (i). By Theorem 8.57 it is enough to show that there are no nonzero measures in ⊥ ext(H⊥ 2 ∩ Mbnd (H)). Suppose that µ is a nonzero element of ext(H2 ∩ Mbnd (H)). Let µ = µ+ − µ− be the decomposition of µ into the positive and negative part. Since spt µ+ cannot be a singleton, we can find a Borel subset B of K so that µ+ (K) 6= µ+ (B) 6= 0. Set λ1 := µ+ |B −

µ+ (B) − µ µ− (K)

and

λ2 := µ − λ1 .

Then kµk = kλ1 k + kλ2 k and both λ1 , λ2 are nontrivial boundary measures. If λ1 ∈ H⊥ , then both λ1 and λ2 are contained in H⊥ 2 ∩ Mbnd (H) and µ is a nontrivial convex combination of suitable multiples of λ1 and λ2 . Since this contradicts the extremality of µ, we have λ1 ∈ / H⊥ . Set Z L(h) := h dλ1 , h ∈ H. K

Since L is a nonzero functional on H, an appeal to the Bishop–Phelps theorem provides a sequence {Ln } of nonzero functionals in H∗ and functions hn ∈ H such that kLn − Lk → 0, kLk = kLn k = Ln (hn ) and khn k = 1.


265

Put Fn+ := {x ∈ K : hn (x) = 1}

and

Fn− := {x ∈ K : hn (x) = −1}.

Let {µn } be a sequence of boundary measures corresponding to Ln as in Lemma 6.23, − in particular kµn k = kLn k for each n ∈ N. Let µ+ n and µn be the positive and negative part of µn . As in the proof of Lemma 8.59 we get + spt µ+ n ⊂ Fn

and

− spt µ− n ⊂ Fn .

Claim 1. There exists n0 ∈ N so that spt µ * Fn+ for any n ≥ n0 . Proof of Claim 1. Suppose the contrary. Hence there exists an increasing sequence {nk } in N so that spt µ ⊂ Fn+k . As in the proof of Lemma 8.59 we obtain that − − − − |µ− nk (Fnk ) − λ1 (Fnk )| = |µnk (Fnk ) − λ1 (Fnk )| → 0.

Since we assume that spt µ ⊂ Fn+k , we have spt λ1 ⊂ Fn+k as well. Thus we get lim kµ− nk k = 0.

k→∞

With this fact in hand, from the equalities − 0 = L(1) = lim Lnk (1) = lim µ+ nk (1) − µnk (1) k→∞

= lim

k→∞

µ+ nk (1)

k→∞

= lim kµnk k = lim kLnk k = kLk k→∞

k→∞

we obtain an obvious contradiction. Thus the proof of Claim 1 is finished. Similarly we deduce that spt µ * Fn− for all but finitely many n ∈ N. Claim 2. For all but finitely many n ∈ N, we have µ|Fn+ ∪Fn− = 0. Proof of Claim 2. Let n0 be an integer provided by Claim 1, n ≥ n0 and let F be an H-exposed set. Since F ∩ Fn+ is also H-exposed, by using the assumption we get µ|Fn+ (F ) = µ(F ∩ Fn+ ) = 0. By Lemma 8.59, we have µ|Fn+ ∈ H⊥ . Suppose that µ|Fn+ 6= 0. By Claim 1, µ is a nontrivial convex combination of suitable multiples of measures µ|Fn+ and µ|K\Fn+ , which contradicts the extremality of µ. Similarly we obtain that µ|Fn− = 0. Hence µ|Fn+ ∪Fn− = 0 and the Claim 2 is proved.

266

8 Choquet-like sets

Thus it follows that λ1 |Fn+ ∪Fn− = 0. According to Lemma 8.58, we get |µn (Fn+ )

−

λ1 (Fn+ )|

n Z o ≤ sup h d(µn − λ1 ) : h ∈ H, khk ≤ 2 K

≤ 2kL − Ln k. As in the proof of Lemma 8.59 we obtain + + + − + + µ+ n (Fn ) = µn (Fn ) − µn (Fn ) = µn (Fn )

≤ λ1 (Fn+ ) + |µn (Fn+ ) − λ1 (Fn+ )|. Since λ1 (Fn+ ) = 0, we get + + lim kµ+ n k = lim µn (Fn ) = 0.

n→∞

n→∞

Similarly it can be deduced that lim kµ− n k = 0. Putting this together we have − kLk = lim kLn k = lim kµn k = lim kµ+ n k + kµn k = 0, n→∞

n→∞

n→∞

which contradicts the hypothesis L 6= 0 and finishes the proof. Remark 8.61. Before proceeding, we have a need of a short note. Given a function space H on K, the space Ac (H) of all continuous H-affine functions on K can be considerable larger than H. Although a lot of objects determined by H and by Ac (H) coincide (for example, representing measures, boundary measures, Choquet sets), this coincidence is no longer valid for M -sets, P -sets and exposed sets. In order to obtain next Theorem 8.62 in a full generality, we must specify whether we think over M sets and P -sets with respect to H or with respect to Ac (H). In order to avoid an inaccuracy, we point out that conditions (iii)–(v) of the next theorem are laid down for the function space Ac (H). More precisely, for example, condition (iv) means that any closed Choquet set F ⊂ K satisfies µ(F ) = 0 for any measure µ ∈ (Ac (H))⊥ ∩ Mbnd (H). With these preliminaries out of the way, we may state the main result of this chapter. Theorem 8.62. Let H be a function space on a compact space K. Then the following assertions are equivalent: (i) H is simplicial, (ii) (Ac (H))⊥ ∩ Mbnd (H) = {0} , (iii) any closed Choquet subset of K is an M -set with respect to Ac (H), (iv) any closed Choquet subset of K is a P -set with respect to Ac (H), (v) any Ac (H)-exposed subset of K is a P -set with respect to Ac (H).


267

Proof. We already know from Theorem 8.60 that assertions (ii)–(v) are equivalent (note that the space Ac (H) is closed). Since (i) ⇐⇒ (ii) by Proposition 6.9, the proof is finished. Corollary 8.63. For a compact convex set X the following assertions are equivalent: (i) X is a simplex, (ii) any closed face of X is a split face, (iii) any closed face of X is a parallel face, (iv) any exposed subset of X is a parallel face. Proof. This follows from Theorem 8.62, if we use the characterization of split and parallel faces from Example 8.41 and Example 8.37. Corollary 8.64. Let H be a simplicial function space. Then closed Choquet sets coincide with relatively Ac (H)-exposed sets. Proof. If F is a closed Choquet set, then by Theorem 8.62, F is an M -set with respect to Ac (H). Corollary 8.50 asserts that F is relatively Ac (H)-exposed. The converse implication follows by Proposition 8.48 and the observation that a set F is a Choquet set with respect to H if and only if F is a Choquet set with respect to Ac (H) (in virtue of the equality Mx (H) = Mx (Ac (H)) for any x ∈ K). Corollary 8.65. Let H be a Markov simplicial function space on a compact space K. Then the class of relatively H-exposed sets coincides with the class of closed Choquet sets. Proof. Since we have H = Ac (H) by Theorem 6.42, the assertion follows from Corollary 8.64. There are simplicial function spaces where supports of maximal measures are big enough. In this case we are able to give a better characterization of both closed Choquet sets and M -sets. Proposition 8.66. Let H be a simplicial function space so that spt δx = ChH (K) for any x ∈ K \ ChH (K). Then a closed set F ⊂ K is Choquet if and only if either F = K or F is a proper subset of ChH (K). Proof. Assume first that F is a Choquet set which is not a proper subset of ChH (K). We show that ChH (K) ⊂ F . If z ∈ F \ ChH (K), then spt δz = ChH (K) by the assumption. Since F is an H-extremal set, we have spt δz ⊂ F . Hence ChH (K) ⊂ F . By Corollary 8.19, K = coH (ChH (K)) ⊂ coH F = F. Conversely, assume that F is a closed proper subset of ChH (K). By Example 8.9, F is H-extremal. We now wish to show that F is H-convex. To this end, fix z ∈ K

268

8 Choquet-like sets

and µ ∈ Mz (H) satisfying spt µ ⊂ F . Our aim is to show that z ∈ F . Since spt µ ⊂ F ⊂ ChH (K), we see that µ is a maximal measure. Because H is a simplicial space, µ = δz . Since F is a proper subset of ChH (K), z ∈ ChH (K) according to the assumption. Hence µ = εz , and therefore z ∈ F . Proposition 8.67. Let H be a simplicial space such that H = Ac (H) and spt δx = ChH (K) for any x ∈ K \ ChH (K). Then a closed set F ⊂ K is an M -set if and only if either F is a proper subset of ChH (K) or equals K. Proof. If F is an M -set, then either F equals K or F is a proper subset of ChH (K) by Proposition 8.66. Conversely, if a closed set F is a proper subset of ChH (K), then F is a Choquet set by Proposition 8.66. By Theorem 8.62, F is an M -set with respect to Ac (H) = H. This finishes the proof.

8.7

Exercises

Exercise 8.68. Let F ⊂ K be a closed H-extremal set and µ a Radon measure carried by F . If ν is a Radon measure, µ ≺ ν, then ν is carried by F as well. Hint. We may assume that µ is a probability measure on K. Since cF ∈ Kusc (H), an appeal to Proposition 3.56 reveals that 1 ≥ ν(F ) = ν(cF ) ≥ µ(cF ) = 1. Hence ν is carried by F . Exercise 8.69. Let F ⊂ K and let µ be a Radon measure carried by the complementary set F 0 . If ν is a Radon measure such that µ ≺ ν, then ν is carried by F 0 . Hint. Since the function c∗F is upper semicontinuous and H-concave, we have by Proposition 3.56 0 ≤ ν(c∗F ) ≤ µ(c∗F ) = 0. It follows that ν(c∗F ) = 0, which yields the conclusion that ν is carried by F 0 . ∗ Exercise 8.70. If F ⊂ K, then c∗F = cco HF .

Hint. Obviously, c∗F ≤ c∗coH F . If h ∈ H, h ≥ cF , then h ≥ ccoH F according to ∗ Proposition 8.18. Hence, c∗F ≥ cco HF . Exercise 8.71. Let F be a P -set. Then c∗F ∈ H if and only if the complementary set F 0 is closed.

8.7 Exercises

269

Hint. Clearly, if c∗F ∈ H, then the complementary set F 0 is closed since c∗F is continuous and F 0 = {x ∈ K : c∗F (x) = 0}. Conversely, suppose that F 0 is closed. Choose ε > 0. By (i) =⇒ (iii) of Theorem 8.39 and Dini’s theorem, there exists a function h ∈ H so that h > cF and h 0 so that αu ≥ − min g(K) on F 0 . Set h := g + αu. Then h ∈ H, h ≥ g on K and h = g on F . Further, u ≥ 0 on F ∪ F 0 . Given x ∈ K, let µ ∈ Mx (H) be maximal. By Lemma 8.13, Z Z u(x) = u dµ = u dµ ≥ 0. K

F ∪F 0

Hence u ∈ H+ , which proves that the set F is Archimedean. Exercise 8.74. Clearly, any H-exposed set F is a Gδ subset of K. Prove the following converse assertions. (a) Let H be a closed function space. Then any relatively H-exposed Gδ set F is H-exposed. (b) Relatively H-exposed sets are H-exposed provided the compact space K is metrizable.

270

8 Choquet-like sets

Hint. Since F is relatively H-exposed, for any x ∈ K \ F we can find a function hx ∈ H+ so that hx = 0 on F and hx (x) > 0. Since the complement K \ F is a Lindelöf space, we may find a countable set {xn : n ∈ N} ⊂ K \ F such that K \F ⊂

∞ [

{z ∈ K : hxn (z) > 0}.

n=1

Then the function h :=

∞ X n=1

1 hx 2n khxn k n

exposes the set F . The second part follows from the fact that closed subsets of metrizable spaces are Gδ . Exercise 8.75. Let H be a simplicial function space on a compact space K. Prove that any Gδ point x ∈ ChH (K) is Ac (H)-exposed (cf. also Exercise 8.74). Hint. You can imitate the proof of Lemma 6.61 for the case of function spaces. A different reasoning could be as follows: By Proposition 8.31, {x} is a (closed) Choquet set. By Theorem 8.62, {x} is an M -set with respect to Ac (H). By Corollary 8.50, {x} is relatively Ac (H)-relatively exposed. Finally, use Exercise 8.74. Exercise 8.76. Let X be a compact convex set, E := Ac (X) and φ : X → S(Ac (X)) be the evaluation mapping. Prove that φ(X) is a parallel face of BE ∗ . Hint. It is easy to show that φ(X) and −φ(X) are closed faces. By Proposition 4.31(c), BE ∗ = co(φ(X) ∪ −φ(X)). Hence, φ(X) and −φ(X) are complementary faces. They are parallel by Proposition 4.31(b). Exercise 8.77 (Weak peak points). If X is a compact convex set, x ∈ X is called a weak peak point, if for any ε > 0 and any neighborhood U of x there exists h ∈ Ac (X) such that khk ≤ 1, |h(x)| > 1 − ε and |h(y)| < ε for each y ∈ ext X \ U . (a) Prove that any weak peak point of X is an extreme point. (b) Prove that any weak peak point of X is a split face. (c) If ext X is a closed set and {x} is a parallel face, x is a weak peak point. Hint. To show (a), let x be a weak peak point and µ ∈ Mx (X) be a maximal measure. If U 3 x is an arbitrary closed neighborhood of x and ε > 0, let h be a function guaranteed by the assumption. Since µ is carried by ext X and |h| ≤ ε on ext X \ U , we get Z Z 1 − ε < |µ(h)| ≤

|h| dµ + U

|h| dµ ≤ µ(U ) + ε. ext X\U

Thus µ is carried by {x}, in other words, x is an extreme point.

8.7 Exercises

271

To verify (b), we use Theorem 8.5. Let µ ∈ (Ac (X))⊥ be a boundary measure of norm 1 and assume that µ = cεx + ν, where c 6= 0 and |ν|({x}) = 0. We select ε ∈ (0, 1) such that |c|(1 − ε) − 2ε > 0 and a closed neighborhood U of x with |ν|(U ) < ε. Let h ∈ Ac (X) be chosen for U and ε. Then |h| ≤ ε on ext X \ U , and hence Z Z Z |h| d|ν| + |h| d|ν| = |h| d|ν| ≤ ε + ε. X

U

ext X\U

Thus 0 = |µ(h)| ≥ |cεx |(h) − |ν|(|h|) ≥ |c|(1 − ε) − 2ε > 0, a contradiction. For the proof of (c), let {x} be a parallel face, ext X closed and U 3 x along with ε > 0 be given. We choose η > 0 with (1 + η)−1 > 1 − ε. By Theorem 8.39, (c{x} )∗ is an affine function, the family {h ∈ Ac (X) : h > c{x} } is down-directed and its infimum equals (c{x} )∗ . Thus also the family {h ∈ Ac (X) : 1 + η > h > c{x} } is down-directed and its infimum equals (c{x} )∗ . Since (c{x} )∗ = 0 on ext X \ {x} and ext X is a closed set, Dini’s theorem yields the existence of a function h ∈ Ac (X) such that c{x} < h < 1 + η and h < η on ext X \ U . Then (1 + η)−1 h is the sought function witnessing that x is a weak peak point. Exercise 8.78. Find an example of an extreme point of a simplex that is not a weak peak point. Hint. Consider an extreme point x of the Poulsen simplex S (see Definition 12.56). Let x0 ∈ S be arbitrary and y = 13 x + 23 x0 . If U is a closed convex neighborhood of x not containing y, then there is no h ∈ Ac (S) such that khk ≤ 1, h(x) > 1 − 71 and |h| < 17 on ext S \ U . Indeed, assume that h is such a function. Since x0 ∈ / U and ext S is dense in S, h(x0 ) ≤ 71 . Then 4 1 2 h(y) = h(x) + h(x0 ) ≥ . 3 3 21 Let {xn } be a sequence in ext S converging to y. Since all but finitely many points xn are in X \ U , h(y) ≤ 71 . Thus we get a contradiction. Exercise 8.79. Let X ⊂ Rd be a compact convex set such that {x} is a split face for each x ∈ ext X. Prove that X is a simplex.

272

8 Choquet-like sets

Hint. First we show To Punique. Pnthat convex combinations of extreme points of X are n this end, let x = i=1 ai xi , where ai arePstrictly positive numbers with i=1 ai = 1 and xi are extreme points of X. If a := ni=2 ai , x = a1 x1 + a

n X

a−1 ai xi .

i=2

P Since {x1 } is a split face and ni=2 a−1 ai xi is contained in the complementary face {x1 }0 , there is only one possibility how to decompose x as a convex combination of x1 and an element from {x1 }0 . Proceeding inductively we get that for any x ∈ X there is only one possibility how to write it as a convex combination of extreme points. Further we show that X has at most d + 1 extreme points. Assume that this is not the case and {x0 , . . . , xd+1 } are distinct extreme points of X. Then they are not affinely thus there exist nonzero real numbers a0 , . . . , ad+1 with Pd+1 independent, and Pd+1 0. We assume that a0 , . . . , aj are positive and i=0 ai = 0 such that i=0 ai xi = P aj+1 , . . . , ad+1 are negative. If a := ji=0 ai , then j X

a−1 ai xi =

i=0

d+1 X

a−1 ai xi

i=j+1

are different convex combinations of extreme points expressing the same point. Hence we have arrived to a contradiction with the first part of the proof. Combining both parts together we get that ext X is finite and any x ∈ X can be written uniquely as a convex combination of ext X. Thus X is a simplex. Exercise 8.80. Find an example of a compact convex set X such that each extreme point of X is a weak peak point and X is not a simplex. Hint. Consider the compact space K := [0, 1] ∪ [2, 3] and the function space Z n H := f ∈ C(K) : 0

1

Z f (t) dt =

3

o f (t) dt ,

2

and let X := S(H). We denote by λ1 and λ2 Lebesgue measure on [0, 1] and [2, 3], respectively. The set X is not a simplex since the point π(λ1 ) has two different measures carried by ext X = φ(ChH (K)), namely φ] λ1 and φ] λ2 (here we use the notation and results from Section 4.3). On the other hand, if x ∈ K, ε > 0 and an open set U 3 x are given, it is not difficult to construct a function f ∈ H such that f (x) = 0, 0 ≤ f ≤ 1 and f > 1 − ε on K \ U . Hence any extreme point of X is a weak peak point.


8.8

273

Notes and comments

The presentation in this chapter closely follows the paper [322] by J. Lukeˇs, T. Mocek, M. Smrˇcka and J. Spurný; however, the method of the proof of Theorem 8.62 is analogous to the techniques used in the convex setting. The notion of a split face was introduced by E. M. Alfsen and T. B. Andersen in [6] and independently by F. Perdrizet in [370] and [371]. The measure theoretic characterization of split faces in Theorem 8.5 can be found in E. M. Alfsen and T. B. Andersen [7] (see also Theorem II.6.12 in E. M. Alfsen [5] and Theorem 2.10.9 in L. Asimow and A. J. Ellis [24]). The idea of parallel faces appears in B. Hirsberg’s paper [226]. Conditions (M) and (P) appeared under the labels (A.1) and (A.2); see, for example, a paper by E. M. Alfsen and B. Hirsberg [8]. Theorem 8.32 is contained in C. J. K. Batty [32], the characterization of P -sets is in [226], Theorem 2.12 or in Theorem 2.10.10 of [24]. Lemma 8.43 is an easier variant of Theorem II.6.15 in [5] and Theorem 2.10.5 in [24] (see also E. M. Alfsen and T. B. Andersen [6] and E.G. Effros [162]). Theorems 8.44 and Proposition 8.46 are variants of Theorem II.6.18 and Corollary II.6.16 in [5] for general function spaces. E. Størmer introduced the concept of an Archimedean face in [440] in the convex setting. In [3], E. M. Alfsen gave a characterization of Archimedean faces which suggested our definition in the framework of function spaces. Proposition 8.49 can be found as Proposition II.5.17 in [5]. We do not know the answer to the following question. Problem 8.81. Let H be a closed function space on a compact space K and F ⊂ K be a P -set. Is F Archimedean? Section 8.5 is based upon E. Briem’s paper [92]. The central Section 8.6 combines some techniques of A. J. Ellis and A. K. Roy [167] (see also Theorem 3.2.5 in [24]), E. Briem [94] and E. G. Effros [163], [164]. We also mention another characterization of simplices given by C. J. K Batty [33] which reads as follows: A compact convex set X is a simplex if and only if there exists C > 0 such that, whenever F ⊂ X is a closed extremal set, each positive f ∈ Ac (coF ) has a positive extension g ∈ Ac (X) with kgk ≤ Ckf k. Corollary 8.64 is a slightly more general version of Theorem 2.5 in R. E. Atalla [25], where it is proved that for Markov simplicial function spaces (see Definition 6.43) the class of relatively H-exposed sets coincides with the class of closed Choquet sets. Exercises 8.77 and 8.78 are taken from C. H. Chu and B. Cohen [119].

Chapter 9

Topologies on boundaries

In this chapter we further explore families of H-extremal and Choquet sets and show that they generate compact topologies on Choquet boundaries. The fact that these topologies are typically non Hausdorff turns out to be a minor inconvenience. The first section introduces the topologies σext and σmax generated by closed H-extremal and H-maximally extremal sets. Theorem 9.10 shows that these topologies are Hausdorff if and only if the Choquet boundary is closed. Theorem 9.12 provides a key tool for introducing measures on Choquet boundaries induced by maximal measures. This procedure is clarified by Theorem 9.19 which shows that the induced measures are regular in some sense. As a consequence we get that a bounded H-affine Baire function satisfies the barycentric calculus with respect to the induced measures (see Corollary 9.20). Next we characterize functions continuous with respect to the topologies σext (see Theorem 9.25) and σmax (see Theorem 9.29). Section 9.4 is devoted to a study of H-strongly universally measurable functions. The main result, Theorem 9.36, proves that upper semicontinuous affine functions on compact convex sets are measurable with respect to the completion of the induced measure from Theorem 9.19 and that they satisfy the barycentric calculus with respect to these measures. As a consequence we obtain that H-strongly universally measurable functions satisfy the barycentric calculus with respect to these measures (see Theorem 9.38). The last section describes properties of the so-called facial topology generated by M -sets. It turns out in Theorem 9.48 that facially continuous functions on Choquet boundaries can be identified with the center of H (see Definition 9.46). In the case of prime simplicial function spaces, there are no nonconstant facially continuous functions (see Corollary 9.50).

9.1

Topologies generated by extremal sets

In the sequel, H is a function S space on a compact space K 6= ∅. We recall from Section 3.3 that M(H) = x∈K Mx (H) and r : M(H) → K is the barycentric mapping. Definition 9.1. Let M be a subset of M(H). We say that a universally measurable set F ⊂ K is M-extremal, if µ(K \ F ) = 0 whenever µ ∈ M and r(µ) ∈ F .

9.1 Topologies generated by extremal sets

275

Examples 9.2. (a) If M = {εx : x ∈ K}, then any universally measurable subset is M-extremal. (b) If M = M(H), then M-extremal sets are exactly H-extremal sets (see Definition 3.12). (c) If M = M(H) ∩ Mmax (H), we get the family of H-maximally extremal sets. Proposition 9.3. If M ⊂ M(H), then the family of all closed M-extremal sets is stable with respect to arbitrary intersections and finite unions. Proof. The proof follows by an easy verification from the definitions. Proposition 9.4. The following assertions hold: (a) If F is a closed H-convex set, then F is H-extremal if and only if F is Hmaximally extremal, and this is the case if and only if F is a closed Choquet set. (b) Any closed H-extremal or H-maximally extremal set intersects ChH (K). Proof. We start the proof by noticing that a closed H-convex set F is H-extremal if and only if F is a closed Choquet set, by definition. To finish the proof we have to check that F is H-extremal provided it is H-maximally extremal. Let x ∈ F and µ ∈ Mx (H) be given. We find an H-maximal measure ν with µ ≺ ν and use the assumption to infer that ν ∈ M1 (F ). Since F is H-convex, Proposition 8.24 yields spt µ ⊂ coH spt ν ⊂ F. This concludes the proof of (a). For the proof of (b), let F be a closed set in K. If F is an H-extremal set, the assertion is proved in Proposition 3.15. If F is an H-maximally extremal set, we consider the family F := {H ⊂ F : H is nonempty closed H-maximally extremal}. By compactness, there exists a minimal element (with respect to inclusion) H ∈ F. If H is not a singleton, we find a function h ∈ H that is nonconstant on H. Then H 0 := {x ∈ H : h(x) = max h(H)} is easily seen to be an H-maximally extremal set. But this contradicts the minimality of H. Hence H = {x} is a singleton, and this is the case if and only if x ∈ ChH (K). This concludes the proof. Definition 9.5 (Topologies generated by M-extremal sets). If M is a subset of the set M(H), we define the following family of sets in ChH (K) as {F ∩ ChH (K) : F is closed and M-extremal}.

(9.1)

It follows from Proposition 9.3 that the family (9.1) is the collection of closed sets for a topology on ChH (K). We denote this topology as τM .

276

9 Topologies on boundaries

Examples 9.6. (a) If M = {εx : x ∈ K}, then the topology τM is the original topology on ChH (K). (b) If M = M(H), we denote the resulting topology as σext . (c) If M = M(H) ∩ Mmax (H), we write σmax for the resulting topology. Proposition 9.7. Both σext and σmax are compact topologies on ChH (K) such that one-point subsets of ChH (K) are closed. Proof. Let M stand for M(H) or M(H) ∩ Mmax (H) and let σ stand for σext or σmax . Since any point of ChH (K) is M-extremal, it is a σ-closed set. To check the compactness of (ChH (K), σ), let {Hi : i ∈ I} be a family of σ-closed sets in ChH (K) having the finite intersection property. Let Fi , i ∈ I, be closed Mextremal sets with Hi = Fi ∩ ChHT(K), i ∈ I. Then {Fi } has the finite intersection property as well, and hence F := i∈I Fi is a nonempty closed M-extremal set (see Proposition 9.3). Thus \ \ Hi = Fi ∩ ChH (K) = F ∩ ChH (K) i∈I

i∈I

is nonempty by Proposition 9.4. Hence ChH (K) is compact in σ as required. Proposition 9.8. Let L be a closed H-maximally extremal set and G := {h|L : h ∈ H} be the restricted function space. Then the following assertions hold: (a) if µ ∈ M+ (L), then µ is G-maximal if and only if it is H-maximal, (b) ChG (L) = ChH (K) ∩ L, (c) if H is simplicial, then G is simplicial. Proof. For the proof of (a), let µ ∈ M+ (L) be an H-maximal measure. If ν ∈ M+ (L) with µ ≺G ν, then µ ≺H ν by Proposition 6.72. Hence µ = ν and µ is G-maximal. Conversely, let µ be a G-maximal measure on L. We pick an arbitrary f ∈ Kc (H) and denote by (f |L )∗,G the upper envelope of f |L with respect to G. For any x ∈ L we find a measure ν1 ∈ Mx (H) such that f ∗ (x) = ν1 (f ) (see Lemma 3.21). If ν2 is an H-maximal measure with ν1 ≺H ν2 , then f ∗ (x) = ν1 (f ) ≤ ν2 (f ) ≤ f ∗ (x). Since L is H-maximally extremal, ν2 is carried by L. Hence f ∗ (x) = ν2 (f ) ≤ (f |L )∗,G (x). Obviously, (f |L )∗,G (x) ≤ f ∗ (x) for x ∈ L, and thus µ(f ) = µ((f |L )∗,G ) = µ(f ∗ ). Hence µ is H-maximal by Theorem 3.58.

9.1 Topologies generated by extremal sets

277

To verify (b), let x ∈ ChG (L). Then εx is G-maximal, and thus H-maximal by (a). It follows that x ∈ ChH (K). Since the reverse inclusion follows from Proposition 6.72(c), assertion (b) is proved. Since (c) follows immediately from (a), the proof is finished. Proposition 9.9. A set H ⊂ ChH (K) is σmax -closed if and only if H is relatively closed in ChH (K) and H is H-maximally extremal. Proof. Let H be a subset of ChH (K). If H is relatively closed and H is H-maximally extremal, then H ∩ ChH (K) = H and H is σmax -closed by definition. Assume that H is σmax -closed and let F ⊂ K be an H-maximally extremal set such that H = F ∩ ChH (K). Let x ∈ H and an H-maximal measure µ ∈ Mx (H) be arbitrary. We consider the restricted function space G := {h|F : h ∈ H}. Since µ ∈ M1 (F ), µ is G-maximal by Proposition 9.8(a). By Proposition 3.64 and Proposition 9.8(b), µ is carried by ChG F = ChH (K) ∩ F = H. Hence H is H-maximally extremal, which finishes the proof. Theorem 9.10. The following assertions are equivalent: (i) ChH (K) is closed, (ii) σext is Hausdorff, (iii) σmax is Hausdorff, (iv) σext coincides with the original topology, (v) σmax coincides with the original topology. Proof. Let τ stand for the original topology on ChH (K). Obviously we have σext ⊂ σmax ⊂ τ.

(9.2)

We prove (i) =⇒ (iv) =⇒ (ii) =⇒ (iii) =⇒ (i) and (iv) =⇒ (v) =⇒ (iii). If ChH (K) is closed, any τ -closed subset of ChH (K) is H-extremal and hence σext -closed. Hence τ = σext , and thus (i) =⇒ (iv). Obviously, (iv) =⇒ (ii) and (ii) =⇒ (iii) because of (9.2). To verify (iii) =⇒ (i), assume that there exists x ∈ ChH (K) \ ChH (K). Let F be the smallest closed H-maximally extremal set containing x (such a set exists by virtue of Proposition 9.3). Then H := F ∩ ChH (K) is σmax -closed and nonempty by Proposition 9.4(b). Moreover, H contains two different points, say x1 , x2 . We claim that they cannot be separated by disjoint σext -open sets. Indeed, assume that U1 , U2 are such sets. Let Fi , i = 1, 2, be closed H-maximally extremal sets such that ChH (K) \ Ui = Fi , i = 1, 2. Then ChH (K) ⊂ F1 ∪ F2 . / F. We suppose that x ∈ F1 . By the minimality of F , F ⊂ F1 and thus x1 ∈

278


Analogously, x ∈ F2 implies x2 ∈ / F . In both cases we get a contradiction. Hence σmax is not Hausdorff and (iii) =⇒ (i). If σext = τ , then σext = σmax by (9.2). Hence (iv) =⇒ (v). Obviously, (v) =⇒ (iii), which finishes the proof. Example 9.11. There exists a compact convex set X such that σext is strictly weaker than σmax . Proof. Let x0 , x2 , x2 be distinct points in [0, 1]. We identify (x0 , 0) and (x2 , 0) in [0, 1]×[0, 21 ] and let K := p([0, 1]×[0, 12 ]), where p is the quotient mapping. We write ω0 for the point p(x0 , 0) and identify K \ {ω0 } with [0, 1] × [0, 21 ] \ {(x0 , 0), (x2 , 0)}. Let H := {f ∈ C(K) : f (ω0 ) = tf (x1 , t) + (1 − t)f (x2 , t), t ∈ (0, 12 ], Z f (x1 , 0) + f (ω0 ) = 4 f (x, t) dx dt}. K

Then H is a closed function space on K whose Choquet boundary equals K \ {ω0 }. Let X stand for the state space of H and let φ : K → X be the homeomorphic embedding from Definition 4.25. Let H := {t(x1 , t) + (1 − t)(x2 , t) : t ∈ (0, 12 ]}

and F := φ({ω0 } ∪ H).

Since any H-maximal measure representing ω0 is carried by H, using Proposition 9.9 we get that F is an Ac (X)-maximally extremal set and hence φ(H) is a σmax -closed subset of X. We show that φ(H) is not a σext -closed set in ext X. To this end, let C be a closed extremal subset of X containing φ(ω0 ). Since C is a union of closed faces by Proposition 2.69, there exists a closed face D ⊂ C containing φ(ω0 ). Since φ(ω0 ) = tφ(x1 , t) + (1 − t)φ(x2 , t),

t ∈ (0, 21 ],

C contains both φ(x1 , t) and φ(x2 , t) for each t ∈ (0, 12 ]. Since D is a face, we get 1 (φ(x1 , t) + φ(x2 , t)) ∈ D, 2

t ∈ (0, 21 ],

as well. Thus 12 (φ(x1 , 0) + φ(ω0 )) ∈ D because D is closed. If µ = p] (2λ), where λ is Lebesgue measure on [0, 1] × [0, 21 ], then 21 (φ(x1 , 0) + φ(ω0 )) is the barycenter of φ] µ. Thus D, and consequently C, contains the support of φ] µ. Hence C ⊃ φ(K) and F is not σext -closed.

9.2

Induced measures on Choquet boundaries

Theorem 9.12. Let H be a function space on a compact space K such that H = Ac (H). Let {µn } be a sequence of H-maximal measures, {An } be a sequence of

9.2 Induced measures on Choquet boundaries

279

Baire subsets of K and F ⊂ C(K) be a norm separable set. Then there exist a metrizable compact space K 0 , a function space H0 on K 0 , a family F 0 ⊂ C(K 0 ), and a continuous surjection ϕ : K → K 0 such that (a) h ◦ ϕ ∈ H for every h ∈ H0 , (b) ϕ] µn is H0 -maximal, n ∈ N, (c) ϕ(An ) is a Baire subset of K 0 and ϕ−1 (ϕ(An )) = An , n ∈ N, (d) for every f ∈ F there exists f 0 ∈ F 0 so that f = f 0 ◦ ϕ. Moreover, if H is simplicial, H0 can be chosen to be simplicial as well and then Ac (H0 ) = H0 . Proof. Given the objects as in the hypotheses, we find a dense countable subset D of F. Next we select a countable family F 1 ⊂ C(K) such that each An is contained in the σ-algebra generated by {f −1 (U ) : U ⊂ R open, f ∈ F 1 } (see Proposition A.48). Without loss of generality we may assume that 1 ∈ F 1 . For each function f ∈ D ∪F 1 we find a countable family Hf ⊂ H such that f ∈ W(Hf ) − W(Hf ). Let [

F 2 :=

Hf .

f ∈D∪F 1

Now we inductively construct countable families F k ⊂ H, k ≥ 3, as follows. Assume that F k has been constructed for some k ≥ 2. Using Theorem 3.58, Proposition 3.25 and Theorem A.84 we construct a family F k+1 in such a way that (e) F k ⊂ F k+1 , Pm (f) i=1 qi fi ∈ F k+1 whenever m ∈ N, q1 , . . . , qm ∈ Q and f1 , . . . , fm ∈ F k , (g) for each function f ∈ −W(F k ), any measure µn and ε > 0, there exists a function g ∈ W(F k+1 ) such that f ≤ g and µn (g − f ) < ε. Finally, we define F ∞ :=

∞ [

Fk

k=2

and ϕ : K → RN as x 7→ {f (x)}f ∈F ∞ . Then K 0 := ϕ(K) is a metrizable compact space and for any function f ∈ F ∞ there exists a function gf ∈ C(K 0 ) such that f = gf ◦ ϕ. Let H0 := {gf : f ∈ F ∞ }. It follows from (f), that F ∞ is a linear space, hence H0 is a linear space as well. Obviously, it contains constant functions and separates points of K 0 , so it is a function space. Since F ∞ ⊂ H, h ◦ ϕ ∈ H for every h ∈ H0 , and (a) therefore holds.

280


In the next step we show that ϕ] µn is H0 -maximal for every n ∈ N. Let n ∈ N be fixed and g ∈ −W(H0 ) be arbitrary. For a fixed ε > 0, we can find a function f ∈ −W(F ∞ ) such that kf − g ◦ ϕk < ε. By (g), there exists f 0 ∈ W(F ∞ ) such that f ≤ f 0 and µn (f 0 − f ) < ε. Let g 0 ∈ W(H0 ) satisfy g 0 ◦ ϕ = f 0 . Then g ≤ g 0 and (ϕ] µn )(g) = µn (g ◦ ϕ) ≥ µn (f ) − ε ≥ µn (f 0 ) − 2ε = µn (g 0 ◦ ϕ) − 2ε = (ϕ] µn )(g 0 ) − 2ε ≥ (ϕ] µn )(g ∗ ) − 2ε. Hence ϕ] µn is H0 -maximal. To verify (c) and (d), we notice that, by the construction, any function from D ∪ F 1 is contained in the closure of the family {g ◦ ϕ : g ∈ W(H0 ) − W(H0 )}. Hence any function from D ∪ F 1 can be expressed as g ◦ ϕ for a suitable function g ∈ C(K 0 ). From this observation the existence of the required family F 0 ⊂ C(K 0 ) as well as (c) and (d) follow. Moreover, if H is simplicial, we adjust the construction of the families F k , k ≥ 3, in the following way. We moreover assure in the inductive step that the family F k+1 possesses the following property: (h) for each f1 , −f2 ∈ −W(F k ) with f1 ≤ f2 there exists f ∈ F k+1 with f1 ≤ f ≤ f2 . Then the resulting family F ∞ , and consequently H0 , possesses the weak Riesz interpolation property, which yields that H0 is simplicial and that Ac (H0 ) = H0 (see Theorem 6.16 and Exercise 6.78). This concludes the proof. Lemma 9.13. Let (K, H), (K 0 , H0 ) be function spaces and let ϕ : K → K 0 be a continuous surjection satisfying h0 ◦ ϕ ∈ H for any h0 ∈ H0 . Then ϕ−1 (F 0 ) is Hextremal for any H0 -extremal set F 0 ⊂ K 0 . In particular, ChH0 (K 0 ) ⊂ ϕ(ChH (K)). Proof. Set F := ϕ−1 (F 0 ), where F 0 ⊂ K 0 is a universally measurable H0 -extremal set. Then F is universally measurable as well. Let x ∈ F and µ ∈ Mx (H) be given. Then ϕ] µ ∈ Mϕ(x) (H0 ), and hence (ϕ] µ)(K 0 \ F 0 ) = 0. Thus µ(K \ F ) = 0. Proposition 9.14. Let µ ∈ M+ (K) be a maximal measure. Then (a) for any Baire set B ⊂ K, µ(B) = sup{µ(F ) : F ⊂ B is Gδ , closed and H-extremal}, (b) for any G ⊂ K of type Gδ , µ(G) = sup{µ(F ) : F ⊂ B is Gδ , closed and H-extremal}.


281

Proof. Given a maximal measure µ ∈ M+ (K), a Baire set B ⊂ K and ε > 0, let (K 0 , H0 ) and ϕ : K → K 0 be constructed as in Theorem 9.12. Then ϕ] µ, as an H0 -maximal measure on a metrizable compact space K 0 , satisfies (ϕ] µ)(ϕ(B)) = (ϕ] µ)(ϕ(B) ∩ ChH0 (K 0 )). Using the regularity of ϕ] µ, we find a compact set F 0 ⊂ ϕ(B) ∩ ChH0 (K 0 ) such that (ϕ] µ)(ϕ(B)) ≤ (ϕ] µ)(F 0 ) + ε. By Example 8.9, F 0 is H0 -extremal. Then F := ϕ−1 (F 0 ) is H-extremal (see Lemma 9.13), of type Gδ and µ(F ) = (ϕ] µ)(F 0 ) ≥ (ϕ] µ)(ϕ(B)) − ε = µ(B) − ε. This finishes the proof of (a). T Let G ⊂ K be a Gδ set, say G = ∞ n=1 Gn , where the sets Gn are open in K. If F is a closed subset of an open set U ⊂ K, we can find a closed Gδ set F 0 with F ⊂ F 0 ⊂ U . Hence it follows from (a) that µ(Gn ) = sup{µ(F ) : F ⊂ B is Gδ , closed and H-extremal},

n ∈ N.

For ε > 0 we find closed H-extremal sets Fn ⊂ Gn of type Gδ such that µ(Gn \Fn ) ≤ T F ε2−n . Then F := ∞ n=1 n is a closed H-extremal set of type Gδ with F ⊂ G, and µ(G \ F ) ≤ µ(

∞ [

(Gn \ Fn )) ≤ ε.

n=1

This concludes the proof. Notation 9.15. Let Σ stand for the σ-algebra generated by the family consisting of all Baire sets and closed H-extremal sets, and let Σ0 := {A ∩ ChH (K) : A ∈ Σ}. Proposition 9.16. For any A ∈ Σ and any maximal measure µ ∈ M+ (K), µ(A) = sup{µ(F ) : F ⊂ A closed H-extremal} = sup{µ(F ) : F closed H-extremal, F ∩ ChH (K) ⊂ A}. Proof. We define µ1 , µ2 : Σ → [0, ∞) as µ1 (A) := sup{µ(F ) : F ⊂ A closed H-extremal}, µ2 (A) := sup{µ(F ) : F closed H-extremal, F ∩ ChH (K) ⊂ A}, for any A ∈ Σ. Obviously, µ1 ≤ µ and µ1 ≤ µ2 on Σ.

(9.3)

282


Let T := {A ∈ Σ : µ1 (A) = µ(A), µ1 (K \ A) = µ(K \ A)}. It follows from Proposition 9.14(a) that T contains all Baire sets. Further, any closed H-extremal set is in T , by Proposition 9.14(b). We show that T is an σ-algebra. Clearly, T is closed with respect to complements. S Let {An } be a sequence of sets in T and let A := ∞ n=1 An . For ε > 0 we find k ∈ N S such that µ(A \ kn=1 An ) < ε. For each n ∈ N we select closed H-extremal sets Fn , Fn0 ⊂ K such that Fn ⊂ An ,

Fn0 ⊂ K \ An

Then F :=

and µ(An \ Fn ) + µ(K \ (An ∪ Fn0 )) < ε2−n . k [

Fn

and

F 0 :=

n=1

∞ \

Fn0

n=1

are closed H-extremal sets. Further, F ⊂ A,

F0

⊂ K \ A and

µ(A \ F ) + µ(K \ (A ∪ F 0 )) < 2ε + ε. Hence A ∈ T , and T is a σ-algebra. It follows that Σ = T , which proves the first equality in (9.3). To prove the second equality, we have µ2 (A) ≥ µ1 (A) = µ(A), A ∈ Σ. (9.4) Let A ∈ Σ and F, F 0 be closed H-extremal set with F ∩ ChH (K) ⊂ A

and

F 0 ∩ ChH (K) ⊂ K \ A.

We claim that F ∩ F 0 = ∅. Indeed, assume that x ∈ F ∩ F 0 . Consider the function space G := {h|F ∩F 0 : h ∈ H}. Let λ be a G-maximal measure in Mx (G). By Propositions 3.64 and 9.8(b), λ is carried by ChG (F ∩ F 0 ) = ChH (K) ∩ F ∩ F 0 = ∅, which is impossible. Hence F ∩ F 0 = ∅. Given A ∈ Σ, let F be a closed H-extremal set with F ∩ ChH (K) ⊂ A. We fix ε > 0 and find a closed H-extremal set F 0 ⊂ K \ A such that µ(K \ A) − µ(F 0 ) < ε. By the argument above, F ∩ F 0 = ∅. Thus µ(F ) ≤ µ(K \ F 0 ) = µ(K) − µ(F 0 ) ≤ µ(K) − µ(K \ A) + ε = µ(A) + ε. Hence µ2 (A) ≤ µ(A) for each A ∈ Σ, which, in conjunction with (9.4), completes the proof.

283


Corollary 9.17. If µ is maximal and A ∈ Σ disjoint from ChH (K), then µ(A) = 0. Proof. This follows from (9.3) and Proposition 9.4(b). Proposition 9.18. Let a measure µ ∈ M+ (K) satisfy µ(A) = sup{µ(F ) : F ⊂ A is Gδ closed and H-extremal},

A ⊂ K Baire.

Then µ is H-maximal. Proof. We verify condition (b) of Theorem 8.32; more precisely, we show that any function h ∈ H is constant on Fx (H) for µ-almost all x ∈ K (we recall that Fx (H) = S ν∈Mx (H) spt ν). Let h ∈ H be arbitrary and let [a, b] be a closed interval in R. BySthe assumption, there exist closed H-extremal sets Fn ⊂ h−1 ([a, b]) such that S ∞ ∞ µ( n=1 Fn ) = µ(h−1 ([a, b])). For a point x ∈ n=1 Fn and ν ∈ Mx (H), spt ν ⊂

∞ [

Fn ⊂ h−1 ([a, b]).

n=1

Hence, Fx (H) ⊂ h−1 ([a, b]) for µ-almost all x ∈ h−1 ([a, b]). Thus there exists a set K 0 ⊂ K satisfying • •

µ(K \ K 0 ) = 0, if x ∈ K 0 and [a, b] is a closed interval with rational endpoints such that x ∈ h−1 ([a, b]), then Fx (H) ⊂ h−1 ([a, b]).

It follows that, for any x ∈ K 0 , the function h is constant on Fx (H). Thus µ is maximal. Theorem 9.19. If µ ∈ M+ (K) is a maximal measure, then there exists a measure µ0 on (ChH (K), Σ0 ) such that µ0 (A ∩ ChH (K)) = µ(A),

A ∈ Σ.

(9.5)

Moreover, µ0 (A) = sup{µ0 (F ) : F ⊂ A is σext -closed},

A ∈ Σ0 ,

and µ0 (F ) = inf{µ0 (B ∩ ChH (K)) : B ∩ ChH (K) ⊃ F, B is Baire in K}, for any σext -closed set F ⊂ ChH (K).

284


Proof. If µ is a maximal measure on K, it follows from Corollary 9.17 that µ0 defined by (9.5) is a well-defined measure on Σ0 . Moreover, for A ∈ Σ we have, by Proposition 9.16, µ0 (A ∩ ChH (K)) = µ(A) = sup{µ(F ) : F closed H-extremal, F ∩ ChH (K) ⊂ A ∩ ChH (K)} = sup{µ0 (F 0 ) : F 0 σext -closed, F 0 ⊂ A ∩ ChH (K)}. For a closed H-extremal set F ⊂ K, µ0 (F ∩ ChH (K)) = µ(F ) = inf{µ(B) : B Baire, F ⊂ B} ≥ inf{µ0 (B ∩ ChH (K)) : B Baire, F ∩ ChH (K) ⊂ B ∩ ChH (K)} ≥ µ0 (F ∩ ChH (K)). This proves the assertion. Corollary 9.20. Let µ ∈ Mx (H) be a maximal measure and f be a bounded Baire H-affine function. If µ0Ris the measure from Theorem 9.19, then f |ChH (K) is µ0 measurable and f (x) = ChH (K) f (t) dµ0 (t). Proof. Given µ ∈ Mx (H) and f as in the corollary, we assume that 0 ≤ f ≤ 1. For any t ∈ R, the set Ft := {y ∈ K : f (y) ≥ t} is a Baire set and µ0 (Ft ∩ ChH (K)) = µ(Ft ) by Theorem 9.19. Hence we get from Fubini’s theorem that Z f (x) =

Z f (y) dµ(y) =

K Z 1

1

µ(Ft ) dt 0

0

Z

µ (Ft ∩ ChH (K))) dt =

= 0

f (y) dµ0 (y).

ChH (K)


9.3

Functions continuous in σext and σmax topologies

Notation 9.21. We write τext for the topology on K whose closed sets are closed Hextremal subsets of K and τmax for the topology on K whose closed sets are precisely closed H-maximally extremal sets. Then σext = τext |ChH (K) and σmax = τmax |ChH (K) .

9.3 Functions continuous in σext and σmax topologies

285

Lemma 9.22. Let f : ChH (K) → [0, 1] be a σext -upper semicontinuous function. Then there exist a closed H-extremal set K 0 ⊃ ChH (K) and a τext -upper semicontinuous function g : K 0 → [0, 1] that extends f . Proof. Let f : ChH (K) → [0, 1] be σext -upper semicontinuous. We fix n ∈ N and find closed H-extremal sets Fin , i = 0, . . . , 2n − 1, such that Fin ∩ ChH (K) = {x ∈ ChH (K) : f (x) ≥ i2−n }, Let Kn :=

S2n −1 i=0

i = 0, . . . , 2n − 1.

Fin and

gn (x) :=

sup i=0,...,2n −1

(i + 1)2−n cFin (x),

x ∈ Kn .

Then Kn ⊃ ChH (K) is a closed H-extremal set and gn , as a finite supremum of τext -upper semicontinuous functions, is τext -upper semicontinuous. By this procedure we get τext -upper semicontinuous functions gn defined on closed −n H-extremal T∞ sets Kn ⊃ ChH (K) such that f ≤ gn ≤ f + 2 on ChH (K). By setting 0 K := n=1 Kn we get a closed H-extremal set containing ChH (K). Then g(x) := inf gn (x), x ∈ K 0 , n∈N

is a τext -upper semicontinuous function extending f . Lemma 9.23. Let K 0 be a closed H-extremal set containing ChH (K) and g : K 0 → R a τext -upper semicontinuous function such that g = 0 on ChH (K). Then g ≤ 0 on K 0 and there exists a closed H-extremal set K 00 ⊂ K 0 containing ChH (K) such that g = 0 on K 00 . Proof. Assume that sup g(K 0 ) > 0. Since g is upper semicontinuous, g attains its maximum on K 0 . Then there exists a closed H-extremal set F ⊂ K such that F ∩ K 0 = {y ∈ K 0 : g(y) = max g(K 0 )}. Since K 0 is H-extremal, F ∩ K 0 is a nonempty closed H-extremal set disjoint from ChH (K), a contradiction with Proposition 3.15. For the proof of the second part, let Fn ⊂ K be closed H-extremal sets such that g −1 ([−n−1 , ∞)) = Fn ∩ K 0 ,

n ∈ N.

T 00 Then F := ∞ n=1 Fn is a closed H-extremal set containing ChH (K), and thus K := 0 K ∩ F is the required set. Proposition 9.24. Let K 0 ⊃ ChH (K) be a closed H-extremal set and g : K 0 → R be a τext -continuous function. Then

286


(a) for every x ∈ K 0 and µ ∈ Mx (H), g = g(x) µ-almost everywhere on K 0 , (b) for every continuous function h : R → R, every x ∈ K 0 and µ ∈ Mx (H), we have Z Z h(g(y)) dµ0 (y), h(g(y)) dµ(y) = h(g(x)) = ChH (K)

K

where

µ0

is the induced measure from Theorem 9.19.

Proof. Let g : K 0 → R be a τext -continuous function on a closed H-extremal set K 0 ⊃ ChH (K). For any closed set H ⊂ R, g −1 (H) is a closed H-extremal set. Thus if x ∈ K 0 and µ ∈ Mx (H), then µ is carried by {y ∈ K 0 : g(y) = g(x)}. In other words, g = g(x) µ-almost everywhere. This shows (a). If h : R → R is a continuous function, x ∈ K 0 and µ ∈ Mx (H), then F := {y ∈ K 0 : g(y) = g(x)} is a closed H-extremal set of µ-measure 1. Thus µ0 (F ∩ ChH (K)) = µ(F ) and Z Z Z h(g(y)) dµ(y) = h(g(y)) dµ(y) = h(g(y)) dµ0 (y) K

F

Z =

F ∩ChH (K)

h(g(x)) dµ0 (y) = h(g(x)).

F ∩ChH (K)

This completes the proof. Theorem 9.25. Let f : ChH (K) → R. Then the following assertions are equivalent: (i) f is σext -continuous. (ii) there exist a closed H-extremal set K 0 ⊃ ChH (K) and a τext -continuous function g : K 0 → R extending f . Proof. Let f : ChH (K) → R be σext -continuous. By the compactness of ChH (K) in σext , f is bounded. Using Lemma 9.22, we find closed H-extremal sets K1 , K2 containing ChH (K), a τext -lower semicontinuous function g1 : K1 → R and a τext upper semicontinuous function g2 : K2 → R that extend f . Then K1 ∩ K2 is a closed H-extremal set containing ChH (K). By Lemma 9.23, there exists a closed H-extremal set K 0 ⊂ K1 ∩ K2 containing ChH (K) such that g1 = g2 on K 0 . Then g := g1 |K 0 is the sought extension. This proves (i) =⇒ (ii). Conversely, if g : K 0 → R is a τext -continuous function on a closed H-extremal set 0 K containing ChH (K), then f := g|ChH (K) is σext -continuous. Indeed, f −1 (H) = g −1 (H) ∩ ChH (K) is σext -closed for any closed H ⊂ R. Hence (ii) =⇒ (i). Lemma 9.26. Let f : ChH (K) → [0, 1] be a σmax -upper continuous function. Then there exists a τmax -upper semicontinuous function g : ChH (K) → [0, 1] extending f .

9.3 Functions continuous in σext and σmax topologies

287

Proof. The proof is similar to that of Lemma 9.22. Let f : ChH (K) → [0, 1] be σmax -upper semicontinuous function. We fix n ∈ N and define Fin := {x ∈ ChH (K) : f (x) ≥ i2−n },

i = 0, . . . , 2n − 1.

By Proposition 9.9, each Fin is an H-maximally extremal set. We define gn : ChH (K) → [0, 1] as gn := sup (i + 1)2−n cFin . i=0,...,2n −1

Then gn , as a supremum of finitely many τmax -upper semicontinuous functions, is τmax -upper semicontinuous as well. By this procedure we get a sequence {gn } of τmax -upper semicontinuous functions on ChH (K) such that f ≤ gn ≤ f + 2−n on ChH (K). Hence g(x) := inf gn (x), x ∈ ChH (K), n∈N

is a τmax -upper semicontinuous function extending f . Lemma 9.27. Let g : ChH (K) → [0, ∞) be a τmax -upper semicontinuous function such that g = 0 on ChH (K). Then g = 0 on ChH (K). Proof. Let g : ChH (K) → [0, ∞) be τmax -upper semicontinuous and assume that g(x) > δ > 0 at some point x ∈ ChH (K). Then {y ∈ ChH (K) : g(y) ≥ δ} is a nonempty closed H-maximally extremal set that does not intersect ChH (K), a contradiction with Proposition 9.4(b). Since g ≥ 0 by upper semicontinuity, the proof is finished. Proposition 9.28. Let g : ChH (K) → R be a τmax -continuous function. Then (a) for every x ∈ ChH (K) and µ ∈ Mx (H) ∩ Mmax (H), g = g(x) µ-almost everywhere on ChH (K), (b) for every continuous function h : R → R, every x ∈ ChH (K) and µ ∈ Mx (H) ∩ Mmax (H), we have Z h(g(x)) =

h(g(y)) dµ(y). K

Proof. Let g : ChH (K) → R be a τmax -continuous function and H be a closed set in R. Then g −1 (H) is a closed H-maximally extremal set. If x ∈ ChH (K) and µ ∈ Mx (H) ∩ Mmax (H), then µ is carried by {y ∈ ChH (K) : g(y) = g(x)}. Hence g = g(x) µ-almost everywhere. This shows (a).

288


If h : R → R is a continuous function, x ∈ ChH (K) and µ ∈ Mx (H)∩Mmax (H), then F := {y ∈ ChH (K) : g(y) = g(x)} is a closed H-maximally extremal set of µ-measure 1. Thus Z Z h(g(y)) dµ(y) = h(g(y)) dµ(y) = h(g(x)). K

F

The proof is complete. Theorem 9.29. Let f : ChH (K) → R. Then the following assertions are equivalent: (i) f is σmax -continuous, (ii) there exists a continuous function g : ChH (K) → R extending f such that for every x ∈ ChH (K) and µ ∈ Mx (H) ∩ Mmax (H), g = g(x) µ-almost everywhere. R Moreover, if g is the extension of f , then h(g(x)) = K h(g(y)) dµ(y) for each x ∈ ChH (K), µ ∈ Mx (H) ∩ Mmax (H) and h : R → R continuous. Proof. To show (i) =⇒ (ii), let f : ChH (K) → R be a σmax -continuous function. Since ChH (K) is σmax -compact, f is bounded. By Lemma 9.26, there exist bounded functions g1 , g2 : ChH (K) → R such that •

g1 = f = g2 on ChH (K),

•

g1 , −g2 are τmax -lower semicontinuous.

Since τmax is weaker than the original topology, g1 , −g2 are lower semicontinuous. Hence g1 ≤ g2 on ChH (K). By Lemma 9.27, g := g1 = g2 on ChH (K) is a τmax -continuous extension of f . By Proposition 9.28, g possesses the property required in (ii). Conversely, let g : ChH (K) → R be as in (ii). We show that F := g −1 ([a, b]) is H-maximally extremal for each closed interval [a, b] ⊂ R. For x ∈ F and µ ∈ Mx (H) ∩ Mmax (H), µ is carried by the set {y ∈ ChH (K) : g(y) = g(x)}. Hence µ is carried by F and F is H-maximally extremal. Thus g|ChH (K) is σmax -continuous. The last part of theorem follows from the fact that a function g satisfying condition (ii) is τmax -continuous, and hence Proposition 9.28 applies.

9.4

Strongly universally measurable functions

Definition 9.30. If H is a function space on a compact space K, a function f on K is called H-strongly universally measurable, if for each µ ∈ M1 (K) and ε > 0 there exist bounded lower semicontinuous functions g, −h in H⊥⊥ such that h ≤ f ≤ g and µ(g) − µ(h) < ε. If X is a compact convex set, we call Ac (X)-strongly universally measurable functions just strongly universally measurable.

9.4 Strongly universally measurable functions

289

Proposition 9.31. Any H-strongly universally measurable function is in H⊥⊥ and any bounded semicontinuous function from H⊥⊥ is H-strongly universally measurable. Proof. If f is H-strongly universally measurable, then f is obviously universally measurable and bounded. Given µ, ν ∈ M1 (K) with µ − ν ∈ H⊥ and ε > 0, let g, h be as in Definition 9.30. Then the inequalities µ(f ) ≤ µ(g) ≤ µ(h) + ε = ν(h) + ε ≤ ν(f ) + ε

and

µ(f ) ≥ µ(h) ≥ µ(g) − ε = ν(g) − ε ≥ ν(f ) − ε show that f ∈ H⊥⊥ . Let f be a bounded lower semicontinuous function contained in H⊥⊥ . By Proposition 5.42(b), f is the supremum of an up-directed family of functions from H that are smaller than f . Hence for any measure µ ∈ M1 (K) and ε > 0, we can find h ≤ f in H such that µ(f − h) < ε. Thus f is H-strongly universally measurable. Proposition 9.32. Let X be a compact convex set. Then a function f on X is strongly universally measurable if and only if for each x ∈ X and ε > 0 there exist lower semicontinuous affine functions g, −h such that h ≤ f ≤ g and g(x) − h(x) < ε. In particular, f is strongly affine. Proof. Necessity of the condition is obvious. To show sufficiency, let f satisfy the condition and µ ∈ M1 (X) and ε > 0 be given. Let g, −h be lower semicontinuous affine functions such that h ≤ f ≤ g and g(r(µ)) − h(r(µ)) < ε. By Proposition 4.7 and Lemma 4.5, both g and h are strongly affine and hence bounded. Thus both functions are in Ac (X)⊥⊥ . Finally, µ(g) − µ(h) = g(r(µ)) − h(r(µ)) < ε, as required. From the considerations above it also follows that µ(f ) = f (r(µ)), and f is strongly affine. Proposition 9.33. Let H be a function space on a compact space K, X := S(H) and I be the mapping from Corollary 5.41. Then a function f on K is H-strongly universally measurable if and only if If is strongly universally measurable on X. Proof. If f is H-strongly universally measurable, s ∈ X and ε > 0, let µ ∈ M1 (K) be such that π(µ) = s (see Definition 4.25). Let g, h be functions given by Definition 9.30. Then Ig, −Ih are lower semicontinuous affine functions by Corollary 5.41, Ig ≤ If ≤ Ih and by Definition 5.38 Ih(s) − Ig(s) = µ(g) − µ(h) < ε. Hence If is strongly universally measurable by Proposition 9.32.

290


Conversely, let If be strongly universally measurable. Given µ ∈ M1 (K) and ε > 0, let g, −h be lower semicontinuous affine functions on X such that f ≤ If ≤ g and g(π(µ)) − h(π(µ)) < ε. Then I −1 g, I −1 h are lower semicontinuous functions in H⊥⊥ by Corollary 5.41, I −1 g ≤ f ≤ I −1 h and µ(I −1 g) − µ(I −1 h) = g(π(µ)) − h(π(µ)) < ε. This concludes the proof. Notation 9.34 (Completion of induced measures). If µ is an H-maximal measure, let µ0 be the induced measure on (ChH (K), Σ0 ) from Theorem 9.19. We denote by (ChH (K), Σ00 , µ00 )) the completion of the measure space (ChH (K), Σ0 , µ0 ). Lemma 9.35. Let h1 , h2 be bounded affine functions on a compact convex set X with h1 ≤ h2 . Let Γ := {(x, t) ∈ X × R : h1 (x) ≤ t ≤ h2 (x)} and let π : X × R → X be the projection. (a) We have ext Γ = {(x, h1 (x)) : x ∈ ext X} ∪ {(x, h2 (x)) : x ∈ ext X}. (b) If h1 , −h2 are lower semicontinuous, then for each x ∈ X, µ ∈ Mx (X) and t ∈ [h1 (x), h2 (x)] there exists ν ∈ M1 (Γ) such that π] ν = µ and ν ∈ M(x,t) (Γ). (c) If h1 , −h2 are lower semicontinuous, then for each x ∈ X, a maximal measure µ ∈ Mx (X) and t ∈ [h1 (x), h2 (x)] there exists a maximal measure ν ∈ M1 (Γ) such that π] ν = µ and ν ∈ M(x,t) (Γ). Proof. Assertion (a) follows by a straightforward verification. To show (b), let µ ∈ Mx (X) be given. We assume first that h1 , h2 are continuous. Then the graph gr hi := {(x, hi (x)) : x ∈ X},

i = 1, 2,

is a compact convex set such that π : gr hi → X is an affine homeomorphism. Hence there exist measures νi ∈ M1 (gr hi ), i = 1, 2, such that π] νi = µ and r(νi ) = (x, hi (x)), i = 1, 2. Let α ∈ [0, 1] be such that t = αh1 (x) + (1 − α)h2 (x). Then ν := αν1 + (1 − α)ν2 ∈ M1 (Γ) satisfies π] ν = µ and (x, t) is the barycenter of ν. Assume now that h1 , −h2 are lower semicontinuous. By Proposition 4.12, h1 is a supremum of a bounded up-directed family F 1 of affine continuous functions. Similarly, h2 = inf F 2 for a bounded down-directed family F 2 ⊂ Ac (X). For each f1 ∈ F 1 , f2 ∈ F 2 , let Γf1 ,f2 := {(x, t) ∈ X × R : f1 (x) ≤ t ≤ f2 (x)}.


291

Then Γ ⊂ Γf1 ,f2 and thus we may use the first part to deduce that µ ∈ π] (M(x,t) (Γf1 ,f2 )). Since Γ=

\

{Γf1 ,f2 : f1 ∈ F 1 , f2 ∈ F 2 }

and the latter family is down-directed, \ M(x,t) (Γ) = {M(x,t) (Γf1 ,f2 ) : f1 ∈ F 1 , f2 ∈ F 2 } and the latter family is down-directed as well. By compactness and the continuity of π] , Proposition A.40 yields \ π] (M(x,t) (Γ)) = {π] (M(x,t) (Γf1 ,f2 )) : f1 ∈ F 1 , f2 ∈ F 2 }, and thus µ ∈ π] (M(x,t) (Γ)). To verify (c), let µ ∈ Mx (X) be a maximal measure. Using (b) we find λ ∈ M(x,t) (Γ) with π] λ = µ. Let ν ∈ M1 (Γ) be a maximal measure with λ ≺ ν. Then ν ∈ M(x,t) (Γ) and µ ≺ π] ν. Indeed, if k is a continuous convex function on X, k ◦ π is a continuous convex function on Γ. Hence µ(k) = π] λ(k) = λ(k ◦ π) ≤ ν(k ◦ π) = π] ν(k). Since µ is maximal, π] ν = µ. This finishes the proof. Theorem 9.36. Let f be a lower semicontinuous affine function on a compact convex set X. If µ ∈ M1 (X) is a maximal measure, let µ00 denote the measure from Notation 9.34. R Then f |ext X is µ00 -measurable and f (r(µ)) = ext X f (t) dµ00 (t). Proof. Since f is bounded (see Proposition 4.7 and Lemma 4.5), we may assume that f is a lower semicontinuous function on X with 0 ≤ f < 1. Let Γ := {(x, t) ∈ X × R : f (x) ≤ t ≤ 1}. Using Lemma 9.35(c), we find a maximal measure ν ∈ M1 (Γ) such that π] ν = µ

and r(ν) = (x, f (x)).

We claim that the set gr f is a Gδ face of Γ and gr 1 = {(x, 1) : x ∈ X} is a closed face of Γ. To verify this, we first notice that both sets are obviously faces of Γ and gr 1 is closed. Since gr f =

∞ n \ 1o (x, t) ∈ Γ : f (x) ≤ t < f (x) + n

n=1

292


and the sets on the right-hand side are open in Γ by the lower semicontinuity of f , gr f is of type Gδ in Γ. Further, ν(gr f ) = 1. To see this, we consider a positive continuous affine function F on Γ defined as F (x, t) := t − f (x), (x, t) ∈ Γ. Then

Z 0 = F (x, f (x)) = F (r(ν)) =

F (y, s) dν(y, s). Γ

Hence F = 0 ν-almost everywhere, and thus ν(gr f ) = 1.

(9.6)

Let a ∈ R be arbitrary. We denote E := {y ∈ X : f (y) ≤ a},

E 0 := E ∩ ext X,

F := {y ∈ X : f (y) > a},

F 0 := F ∩ ext X.

Our aim is to prove that both E 0 and F 0 are µ00 -measurable. Let (µ0 )∗ and (µ0 )∗ denote the outer and inner measure on ext X induced by µ0 , respectively. We first notice that Proposition 9.14(b) and Theorem 9.19 applied to the open set F give µ(F ) = sup{µ(D) : D ⊂ F is closed, Baire and H-extremal} = sup{µ0 (D ∩ ext X) : D ⊂ F is closed, Baire and H-extremal}

(9.7)

≤ (µ0 )∗ (F 0 ). The most important step of the proof is the following claim. Claim. We have µ(E) ≤ (µ0 )∗ (E 0 ). Proof of the claim. Obviously we may assume that 0 ≤ a < 1. Since gr f is a Gδ set in Γ, Proposition 9.14 yields the existence of a closed extremal set D1 ⊂ Γ such that D1 ⊂ gr f

and

ν(D1 ) > 1 − ε.

Let G := {(x, t) ∈ Γ : t ≤ a},

G0 := G ∩ ext Γ.

Then G0 = G ∩ ext gr f and E = π(G),

E 0 = π(G0 ).


293

Since G is a closed Gδ set, Theorem 9.19 provides a closed extremal set D2 ⊂ Γ such that D2 ∩ ext Γ ⊂ G ∩ ext Γ = G0

and

ν(D2 ) = ν 0 (D2 ∩ ext Γ) > ν 0 (G0 ) − ε.

Then D := D1 ∩ D2 is a closed extremal set in Γ contained in gr f such that ν(D) > ν(G) − 2ε.

(9.8)

Since π : gr f → X is an affine bijection, π(D) is an extremal set in X. Moreover, it is closed by the continuity of π. From (9.8) we get µ(π(D)) = ν(π −1 (π(D))) = ν(π −1 (π(D)) ∩ gr f ) = ν(D) > ν 0 (G0 ) − 2ε. Since π(D) ∩ ext X ⊂ E 0 , we get (µ0 )∗ (E 0 ) ≥ µ0 (π(D) ∩ ext X) = µ(π(D)) > ν 0 (G0 ) − 2ε.

(9.9)

Since G is a Baire set, we get ν 0 (G0 ) = ν(G) ≤ ν(π −1 (π(G))) = ν(π −1 (π(G)) ∩ gr f ) = ν(G ∩ gr f ) ≤ ν(G). This implies µ(E) = ν(π −1 (E)) = ν(π −1 (π(G))) = ν 0 (G0 ).

(9.10)

From (9.9) and (9.10) we get (µ0 )∗ (E 0 ) ≥ ν 0 (G0 ) − 2ε = µ(E) − 2ε. Hence (µ0 )∗ (E 0 ) ≥ µ(E) as required. It follows from the claim that the set F 0 is µ00 -measurable. Indeed, the claim, along with (9.7), yields (µ0 )∗ (F 0 ) = 1 − (µ0 )∗ (E 0 ) ≤ 1 − µ(E) = µ(F ) ≤ (µ0 )∗ (F 0 ), which implies µ00 -measurability of F 0 . Moreover, µ00 (F 0 ) = µ(F ).

294


Thus f is µ00 -measurable and to finish the proof we need to verify the barycentric formula. To this end, let Fa := {y ∈ X : f (y) > a},

Fa0 := Fa ∩ ext X,

a ∈ R.

By the first part of the proof, µ00 (Fa0 ) = µ(Fa ),

a ∈ R.

Since µ(f ) = f (x) by Proposition 4.7, Fubini’s theorem gives Z 1 Z µ(Fa ) da f (y) dµ(y) = f (x) = X 1

Z =

0

µ00 (Fa0 ) da =

Z

0

f (y) dµ00 (y).

ext X

This concludes the proof. Lemma 9.37. Let H be a function space on a compact space K, X := S(H) be its state space and φ : K → X be the evaluation mapping as in Definition 4.25. For a measure µ ∈ Mmax (H), let µ00 be the induced measure on ChH (K) as in Notation 9.34. Similarly, let (φ] µ)00 be the induced measure on ext X. We write Σ00 (ChH (K)) or Σ00 (ext X) for the respective σ-algebras. If φ0 : ChH (K) → ext X denotes the restriction of φ, then • φ0 : (Ch (K), Σ00 (Ch (K))) → (ext X, Σ00 (ext X)) is measurable; that is, H H (φ0 )−1 (A) ∈ Σ00 (ChH (K)) for any A ∈ Σ00 (ext X), • (φ µ)00 = (φ0 ) µ00 . ] ] Proof. Given the objects as in the statement of the lemma, let Σ(K) and Σ0 (ChH (K)) stand for the σ-algebras from Notation 9.15, where K indicates their relevance to the space K. Analogously, we write Σ(X) and Σ0 (ext X) for the respective σ-algebras on X. Now (ChH (K), Σ00 (ChH (K)), µ00 ) denotes the completion of the measure space (ChH (K), Σ0 (ChH (K), µ0 )), and (ext X, Σ00 (ext X), (φ] µ)00 ) denotes the completion of (ext X, Σ0 (ext X), (φ] µ)0 ). First we notice that (φ0 )−1 (A ∩ ext X) = φ−1 (A) ∩ ChH (K),

A ⊂ X.

(9.11)

Clearly, φ−1 (A) is a Baire set in K for any Baire set A ⊂ X and φ−1 (A) is a closed H-extremal set for any closed extremal set A ⊂ X (see Lemma 8.10). By (9.11), (φ0 )−1 (A0 ) ∈ Σ0 (ChH (K)),

A ∈ Σ0 (ext X).

From this it follows that φ0 : (ChH (K), Σ0 (ChH (K))) → (ext X, Σ0 (ext X)) is measurable.


295

Further, for any A ∈ Σ(X), we get from Theorem 9.19 and (9.11) that (φ] µ)00 (A ∩ ext X) = (φ] µ)(A) = µ(φ−1 (A)) = µ00 (φ−1 (A) ∩ ChH (K)) = µ00 ((φ0 )−1 (A ∩ ext X)) = ((φ0 )] µ00 )(A ∩ ext X). Hence ((φ0 )] µ00 )(A) = (φ] µ)00 (A),

A ∈ Σ0 (ext X).

(9.12)

It follows from (9.12) that, for any N ∈ Σ00 (ext X) with (φ] µ)00 (N ) = 0, (φ0 )−1 (N ) ∈ Σ00 (ChH (K)) and

µ00 ((φ0 )−1 (N )) = 0.

Hence ((φ0 )] µ00 )(A) = (φ] µ)00 (A),

A ∈ Σ00 (ext X),

which is nothing else than the desired result (φ0 )] µ00 = (φ] µ)00 . This concludes the proof. Theorem 9.38. Let H be a function space on a compact space K and f be an Hstrongly universally measurable function. If µ ∈ Mmax (H) and µ00 is the measure from Notation 9.34, the function f |ChH (K) is µ00 -measurable and Z µ(f ) = f (y) dµ00 (y). ChH (K)

In particular, if µ H-represents a point x ∈ K, then Z f (x) = f (y) dµ00 (y). ChH (K)

Proof. Let f be as in the hypotheses and µ ∈ M1 (K) be maximal. Let X := S(H) be the state space of H and let I be the mapping from Corollary 5.41. First we show that If |ext X is (φ] µ)00 -measurable (here (φ] µ)00 is the measure on ext X given by Notation 9.34). By Proposition 9.33, If is strongly universally measurable. Hence, there exist lower semicontinuous affine functions gn , −hn on X, n ∈ N, such that hn ≤ If ≤ gn

and (φ] µ)(gn − hn ) < n−1 .

We define an := h1 ∨ · · · ∨ hn ,

bn := g1 ∧ · · · ∧ gn ,

n ∈ N.

296


Then sup an ≤ f ≤ inf bn , n∈N

n∈N

and by Theorem 9.36 we can write Z ( inf bn (y) − sup an (y)) d(φ] µ)00 (y) 0≤ ext X n∈N

Z = lim

n→∞ ext X

n∈N

(bn (y) − an (y)) d(φ] µ)00 (y)

≤ lim sup(φ] µ)(gn − hn ) n→∞

= 0. Hence If |ext X is (φ] µ)00 -measurable and hn (r(φ] µ)) = (φ] µ)00 (hn |ext X ) ≤ (φ] µ)00 (If |ext X ) ≤ (φ] µ)00 (gn |ext X ) = gn (r(φ] µ)),

n ∈ N,

gives (φ] µ)00 (If |ext X ) = If (r(φ] µ)) = µ(f ). Let φ0 : ChH (K) → ext X be the restriction of φ. By the first part of the proof, If is (φ] µ)00 -measurable. Lemma 9.37 yields (φ] µ)00 = (φ0 )] µ00 , and thus the function If |ext X is (φ0 )] µ00 -measurable. Hence Z µ(f ) = If (r(φ] µ)) = If (s) d(φ] µ)00 (s) ext X Z Z 0 00 If (s) d((φ )] µ )(s) = If (φ0 (y)) dµ00 (y) = ChH (K)

ext X

Z =

f (y) dµ00 (y),

ChH (K)

which is the desired conclusion. Since the particular assertion obviously follows from the general one, the proof is finished.

9.5

Facial topology generated by M -sets

This section studies properties of the topology generated by M -sets (see Definition 8.40).

9.5 Facial topology generated by M -sets

297

Proposition 9.39. The family of all M -sets is closed with respect to taking arbitrary intersections and H-convex hulls of finite unions. T Proof. Let F be a family of M -sets and F := F ∈F F . Given µ ∈ H⊥ ∩ Mbnd (H) and h ∈ H positive, let F 0 stand for the family of all intersections of finite families in T F. Then F = F ∈F 0 F and (µ|F )(h) = µ( inf 0 cF h) = inf 0 (µ|F )(h) F ∈F

F ∈F

by Theorem A.84. Hence it is enough to show that F possesses condition (M) when F is finite. By a simple induction, it follows that we need to verify that F1 ∩ F2 satisfies condition (M) whenever both F1 and F2 do. But this is obvious, because µ|F1 ∈ H⊥ ∩ Mbnd (H) and hence µ|F1 ∩F2 (h) = (µ|F1 )|F2 (h) = 0. Let F1 , F2 be M -sets and let F := coH (F1 ∪ F2 ). We first verify that F satisfies condition (M). To this end, let µ ∈ H⊥ ∩ Mbnd (H) and h ∈ H be given. Then µ|F1 ∪F2 (h) = µ|F1 (h) + µ|F2 (h) − µ|F1 ∩F2 (h) = 0, because F1 ∩ F2 is an M -set. Hence F = coH (F1 ∪ F2 ) satisfies condition (M) by Proposition 8.34. To finish the proof, we have to verify that F is H-extremal. To this end, let x ∈ F and µ ∈ Mx (H) be given. By Proposition 8.18, there exists a measure µ1 ∈ Mx (H) ∩ M1 (F1 ∪ F2 ). Let µ2 be an H-maximal measure with µ1 ≺ µ2 . Since F1 ∪ F2 is H-extremal (see Proposition 9.3), cF1 ∪F2 is H-convex and hence, by Proposition 3.56, 1 = µ1 (F1 ∪ F2 ) ≤ µ2 (F1 ∪ F2 ). Hence µ2 is carried by F1 ∪ F2 as well. Let ν be an H-maximal measure with µ ≺ ν. Then µ2 − ν ∈ H⊥ ∩ Mbnd (H), and thus (µ2 − ν)|F1 ∪F2 (1) = 0. In other words, 1 = µ2 (F1 ∪ F2 ) = ν(F1 ∪ F2 ). Hence ν is carried by F1 ∪ F2 . Finally, from Proposition 8.24 we get spt µ ⊂ coH spt ν ⊂ F. This finishes the proof.

298


Definition 9.40 (Facial topology). It follows from Propositions 9.39 and 8.30(b) that the family {F ∩ ChH (K) : F is an M -set} defines the family of closed sets for a topology on ChH (K). We call it the facial topology and denote it σfac . In general, the topology σfac is not Hausdorff. Proposition 9.41. The topology σfac is compact. Proof. Let {Fi }i∈I be a family of M -sets such that {Fi ∩ ChH (K)}i∈I has the finite intersectionTproperty. Then {Fi }i∈I has the finite intersection property as well, and thus F := i∈I Fi is a nonempty M -set (see Proposition 9.39). Since F ∩ChH (K) 6= ∅ (see Proposition 9.4(b)), we get \ (Fi ∩ ChH (K)) = F ∩ ChH (K) 6= ∅. i∈I

This concludes the proof. Theorem 9.42. Let f : ChH (K) → R be a bounded σfac -upper semicontinuous function and a ∈ H be a positive function. Then there exists a unique upper semicontinuous function h ∈ H⊥⊥ such that h(x) = a(x)f (x),

x ∈ ChH (K).

Moreover, if we denote g := f · a|ChH (K) , then h = ChH (K)g ∗ . Proof. Assume first that f is as in the hypothesis and f (ChH (K)) ⊂ [0, 1]. We fix n ∈ N and find for i = 0, . . . , n an M -set Fi ⊂ K such that {x ∈ ChH (K) : f (x) ≥

i } = Fi ∩ ChH (K). n

By Theorem 8.44, fi := ( na · cFi )∗ ∈ H⊥⊥ , i = 0, . . . , n. Moreover, ( fi =

1 na

0

on Fi ∩ ChH (K), on ChH (K) \ Fi .

Thus the function hn :=

n X

fi

i=0

is upper semicontinuous, belongs to H⊥⊥ and satisfies g ≤ hn ≤ g +

1 n

on ChH (K).


299

Thus we can construct a sequence {hn } of upper semicontinuous functions in H⊥⊥ that converges to g uniformly on ChH (K). By the minimum principle of Proposition 3.88, the sequence {hn } converges uniformly on K. Then its limit h is the required function. If f (ChH (K)) is contained in an interval [−m, m] for some m > 0, we can apply 1 (f + m) to get a function h0 satisfying h0 = the argument above to the function 2m 1 0 2m (f +m)·a|ChH (K) on ChH (K). Then the function h := 2mh −ma is the required one. Since the question of uniqueness is settled again by Proposition 3.88, we proceed to the last assertion. If u ∈ H satisfies u ≥ g on ChH (K), then u ≥ h on K by Proposition 3.88. Hence h ≤ ChH (K)g ∗ . On the other hand, ChH (K)g ∗ ≤ h by Proposition 3.25(a). This concludes the proof. Theorem 9.43. Let f : ChH (K) → R be a σfac -continuous function and a ∈ H. Then there exists a unique h ∈ H such that h(x) = a(x)f (x),

x ∈ ChH (K).

Proof. Given f and a as in the theorem, we first notice that f is bounded by Proposition 9.41. Let m > 0 satisfy a(K) ⊂ [−m, m]. By Theorem 9.42, there exist upper semicontinuous functions h1 , h2 ∈ H⊥⊥ such that h1 (x) = f (x)(a(x) + m),

and

h2 (x) = −f (x),

x ∈ ChH (K).

Then h := h1 +mh2 is upper semicontinuous, h ∈ H⊥⊥ and h = af on ChH (K). As in the proof of Theorem 9.42 we obtain that h = ChH (K)g ∗ , where g := f · a|ChH (K) . Analogously we get that ChH (K)g∗ is a lower semicontinuous function in H⊥⊥ . By the minimum principle of Proposition 3.88, we get that ChH (K) ∗

g = ChH (K)g∗

on K.

Thus h is the sought function. Since the uniqueness follows trivially from the minimum principle, the proof is finished. Corollary 9.44. Let f : ChH (K) → R be a σfac -continuous function. Then there exists a unique h ∈ H such that h = f on ChH (K). Proof. The assertion follows from Theorem 9.43 by setting a := 1. Theorem 9.45. The following assertions are equivalent: (i) the topology σfac coincides with the original topology on ChH (K), (ii) the topology σfac is Hausdorff, (iii) H is a Markov simplicial space.

300


Proof. We start by proving (ii) =⇒ (iii). First we show that ChH (K) is a closed set. Assume that there exists x ∈ ChH (K) \ ChH (K). Let F be the smallest M -set containing x (see Proposition 9.39). Since x ∈ / ChH (K), F intersects ChH (K) in at least two distinct points, say x1 and x2 (see Proposition 3.15 and Proposition 8.30(b)). For i = 1, 2, let Ui be disjoint σfac -open sets containing xi , respectively. By the definition of σfac , there exist M -sets Fi such that ChH (K) \ Ui = Fi ∩ ChH (K),

i = 1, 2.

Since U1 ∩ U2 = ∅, then x ∈ ChH (K) ⊂ F1 ∪ F2 = F1 ∪ F2 . Assume, for example, that x ∈ F1 . Then F ⊂ F1 , and thus x1 ∈ / F , a contradiction. Analogously we get a contradiction in the case x ∈ F2 . Hence the points x1 and x2 cannot be separated, and σfac is not Hausdorff. Hence ChH (K) is closed. Now we know that σfac is a Hausdorff weaker topology then the original compact topology, and thus it coincides with it. Hence any continuous function on ChH (K) is σfac -continuous as well. Thus Corollary 9.44 yields that any continuous function on ChH (K) can be extended to a function from H. By Theorem 6.42, H is a Markov simplicial space. Assume now that H is a Markov simplicial space and H ⊂ ChH (K) is relatively closed in the original topology. Since ChH (K) is closed, it is a closed set in K. By Proposition 8.31 and Theorem 8.62, F := coH H is an M -set with respect to H = Ac (H). Further, H = F ∩ ChH (K) by Lemma 8.25, and thus H is σfac -closed as needed. Since the proof of the remaining implication (i) =⇒ (ii) is obvious, the proof is finished. Definition 9.46 (Center of H). We say that a function f ∈ H belongs to the center of H, if for every a ∈ H there exists h ∈ H such that h(x) = f (x)a(x),

x ∈ ChH (K).

We write Z(H) for the center of H. Proposition 9.47. If H is a closed function space, then the center Z(H) is a commutative Banach algebra and a vector lattice, where the multiplication and lattice operations are considered pointwise on ChH (K). Proof. First we notice that Z(H) is a linear subspace of H containing constant functions. Step 1. Next we show that Z(H) is closed in H. Let {fn } be a sequence of functions from Z(H) uniformly convergent to f ∈ H. For any a ∈ H and n ∈ N,


301

there exists a function hn ∈ H such that hn = fn a on ChH (K). By the minimum principle of Theorem 3.16, {hn } is uniformly convergent on K, say to a function h ∈ H. Then h = f a on ChH (K). Step 2. Further we show that Z(H) is an algebra. Let f1 , f2 be functions in Z(H). By the definition of the center, there exists a function h ∈ H such that h = f1 f2

on ChH (K).

We claim that h ∈ Z(H) as well. Indeed, for a given a ∈ H we can find h1 ∈ H such that h1 = f2 a on ChH (K). Further we find h2 ∈ H with h2 = f1 h1 on ChH (K). Then h2 (x) = f1 (x)h1 (x) = f1 (x)f2 (x)a(x) = h(x)a(x),

x ∈ ChH (K).

Hence h ∈ Z(H). Step 3. If f ∈ Z(H), then f = (kf k + f ) − kf k and thus Z(H) = (Z(H))+ − (Z(H))+ . We show that for any f ∈ Z(H) there exists h ∈ Z(H) such that h = f ∨ 0 on ChH (K). To this end, let {pn } be a sequence of polynomials on R that converges to the function t 7→ t∨0, t ∈ R, uniformly on f (K). By the second step of the proof, we find functions hn ∈ Z(H), n ∈ N, satisfying hn = pn ◦ f on ChH (K). Again by the minimum principle, we deduce that {hn } converges uniformly on K to an element h ∈ Z(H). It easily follows that h = f ∨ 0 on ChH (K). Proposition A.15 yields that Z(H) is a vector lattice, which completes the proof. Theorem 9.48. The mapping R : Z(H) → C(ChH (K), σfac ) defined as the restriction to ChH (K) is an isometric isomorphism preserving multiplication and lattice operations. Proof. We first show that h|ChH (K) is σfac -continuous for each h ∈ Z(H). Since Z(H) is a linear space, to this end it is enough to show that the set H := {x ∈ ChH (K) : h(x) ≥ α} is σfac -closed for any h ∈ Z(H) with h(K) ⊂ [0, 1] and α ∈ [0, 1]. For any n ∈ N we use the algebraic structure of Z(H) to find a function hn ∈ Z(H) such that hn (x) = (((1 − α) + h(x)) ∧ 1)n ,

x ∈ ChH (K).

By the minimum principle, the sequence {hn } decreases to an upper semicontinuous function h ∈ H⊥⊥ . Moreover, ( 1, x ∈ H, h(x) = (9.13) 0, x ∈ ChH (K) \ H.

302


We set F := {x ∈ K : h(x) = 1}. Obviously, F is a closed Choquet set and F ∩ ChH (K) = H. We need to verify that F is an M -set. To this end, let a positive function a ∈ H and µ ∈ H⊥ ∩ Mbnd (H) be given. We notice that (9.13) yields, using Proposition 3.25(a), that h = c∗F . (9.14) By the definition of the center, for any n ∈ N we can find a function fn ∈ H such that fn = hn a on ChH (K). As in the proof of Theorem 9.42 we infer that fn = ChH (K)(hn a)∗ . By Lemma 8.13, µ is carried by F ∪ F 0 , where F 0 = {x ∈ K : c∗F = 0}. By (9.14) and Theorem 3.58, 0 = lim µ(fn ) = lim µ(ChH (K)(hn a)∗ ) n→∞

n→∞

= lim µ(hn a) = µ(ha) n→∞ Z Z ∗ = µ(acF ) = a · 1 dµ + F

a · 0 dµ

F0

= µ(acF ). Hence µ|F ∈ H⊥ as needed, and F is an M -set. This shows that h|ChH (K) is σfac continuous. It is clear from the minimum principle that R is injective and by Theorem 9.43 we know that it is surjective. Since multiplication and lattice operation are pointwise on ChH (K), we get that R respects both multiplication and lattice operations. Finally, it is an isometry from the minimum principle again. This concludes the proof. Proposition 9.49. Let H be a simplicial function space and let σfac denote the topology on ChH (K) generated by M -sets with respect to Ac (H). Then the following assertions are equivalent: (i) H is prime, (ii) for any pair of nonempty σfac -open sets U1 , U2 ⊂ ChH (K), U1 ∩ U2 6= ∅. Proof. Assume that H is prime and U1 , U2 are nonempty disjoint σfac -open sets in ChH (K). By the definition of σfac we find M -sets F1 , F2 with respect to Ac (H) such that Ui = ChH (K) \ Fi , i = 1, 2. We pick xi ∈ Ui and use Corollary 8.50 to find positive functions hi ∈ Ac (H) such that hi (Fi ) = 0 and hi (xi ) = 1, i = 1, 2. Since h1 ∧ h2 = 0 on ChH (K), the function 0 is the greatest lower bound of h1 ∧ h2 and differs from both h1 and h2 . Hence Ac (H) is not an antilattice. For the proof of (ii) =⇒ (i), let h1 , h2 ∈ Ac (H) be such that h1 ∧ h2 admits the greatest lower bound h ∈ Ac (H) which differs from both h1 and h2 . It follows from Proposition 3.25 that h = (h1 ∧ h2 )∗ . Then Fi := {x ∈ K : h(x) = hi (x)},

i = 1, 2,

9.6 Exercises

303

are closed Choquet sets with ChH (K) ⊂ F1 ∪ F2 . Since H is simplicial, they are M -sets with respect to Ac (H) (see Theorem 8.62). Then Ui := ChH (K) \ Fi ,

i = 1, 2,

are σfac -open disjoint sets in ChH (K). Since neither F1 nor F2 covers ChH (K) (otherwise h = h1 or h = h2 ), the sets U1 , U2 are nonempty. This concludes the proof. Corollary 9.50. Let H be a prime simplicial space. Then the center of Ac (H) consists of constant functions. Proof. The assertion follows from Proposition 9.49 and Theorem 9.48.

9.6

Exercises

Exercise 9.51. Let X be a compact convex set, X 0 a minimal closed extremal set containing ext X and g : X 0 → R τext -continuous. Prove that g is constant on any segment and on any face contained in X 0 . Hint. Let x, y ∈ X 0 be distinct points such that the segment [x, y] ⊂ X 0 and g(x) < g(y). Then 1 F1 := {z ∈ X 0 : g(z) ≤ (g(x) + g(y))} 2 and 1 F2 := {z ∈ X 0 : g(z) ≥ (g(x) + g(y))} 2 are closed extremal subsets of X with X 0 ⊂ F1 ∪ F2 . On the other hand, if z = 1 / F1 ∪ F2 . This 2 (x + y), then it follows from the extremality of F1 and F2 that z ∈ contradiction shows that g is constant on the segment joining points x and y. If F ⊂ X 0 is a face, then g is constant on any segment contained in F . Hence g is constant on F . Exercise 9.52. Let H be a function space on a compact space K, K 0 a minimal closed H-extremal set containing ChH (K) and g : K 0 → R τext -continuous. Prove that g need not be constant on every closed Choquet set contained in K 0 . Hint. Consider H := C(K) on a compact space K (see Example 3.2(a)). Then any continuous function on K is τext -continuous and any closed set F ⊂ K is a closed Choquet set. Exercise 9.53. Let Σ0 be the σ-algebra on ChH (K) generated by the family {F ∩ ChH (K) : F is Baire or a closed H-maximally extremal set}.

304


Prove that any maximal measure µ induces a measure µ0 on Σ0 such that µ0 (A) = sup{µ0 (F ) : F ⊂ A is σmax -closed},

A ∈ Σ0 ,

µ0 (F ) = inf{µ0 (B ∩ ChH (K)) : B ∩ ChH (K) ⊃ F, B is Baire in K}, F is a σmax -closed subset of ChH (K). Hint. Imitate the proof of Proposition 9.16 and Theorem 9.19. Exercise 9.54. Let Σ0 and µ0 be as in the previous Exercise 9.53. Let f : ChH (K) → R be σmax -continuous and g : ChH (K) → R be the extension provided by Theorem 9.29. Let h : R → R be continuous, x ∈ ChH (K) and µ ∈ Mx (H) ∩ Mmax (H). Prove that Z h(g(x)) = h(f (y)) dµ0 (y). ChH (K)

Hint. By Proposition 9.28, g = g(x) µ-almost everywhere and by the proof of Theorem 9.29, g is τmax -continuous. Hence f is µ0 -measurable and F := {y ∈ ChH (K) : g(y) = g(x)} is a τmax -closed set which carries µ. Thus µ0 (F ∩ ChH (K)) = µ(F ) = 1 and Z Z h(f (y)) dµ0 (y) = h(f (y)) dµ0 (y) F ∩ChH (K)

ChH (K)

Z =

h(g(x)) dµ0 (y)

F ∩ChH (K)

= h(g(x)).

Exercise 9.55. Prove that any H-strongly affine function satisfies the minimum principle. Hint. Use either Theorem 9.38 or the definition along with the Minimum principle 3.16. Exercise 9.56. Let φ0 : ChH (K) → ext S(H) be the mapping from Lemma 9.37. Prove that (a) φ0 : (ChH (K), σext ) → (ext S(H), σext ) is continuous, (b) φ0 : (ChH (K), σmax ) → (ext S(H), σmax ) is a homeomorphism, (c) φ0 : (ChH (K), σext ) → (ext S(H), σext ) need not be a homeomorphism.

9.6 Exercises

305

Hint. For the σext -continuity of φ0 use Lemma 8.10. To show that φ0 is a homeomorphism with respect to σmax -topologies, it is enough to show that F ⊂ K is H-maximally extremal if and only if φ(F ) is S(H)-maximally extremal. But this easily follows from the facts that µ ∈ M+ (K) is H-maximal if and only if φ] µ is Ac (S(H))-maximal (see Proposition 4.28(d)) and that Ac (S(H))-maximal measures on S(H) are carried by φ(K) (see Proposition 3.64). To find an example required by (c), consider the following compact space K ⊂ R2 . Let In := [−2, −1] × {n−1 }, Jn := [1, 2] × {n−1 }, n ∈ N, I0 := [−2, −1] × {0},

J0 := [1, 2] × {0},

I−1 := [−2, −1] × {−1}, and let µn , n ∈ N ∪ {0, −1} and νn , n ∈ N ∪ {0}, be copies of one-dimensional Lebesgue measure on the sets In and Jn , respectively. Let z := (1, 1) ∈ R2 . We define [ [ Jn In ∪ K := n∈N∪{0,−1}

and

n∈N∪{0}

n 1 1 H := f ∈ C(K) : f (z) = µn (f ) + (1 − )νn (f ), n ∈ N, n n o µ0 (f ) = µ−1 (f ) .

Then H is a function space on K and it can be easily seen that ChH (K) = K \ {z}. We claim that K \ I−1 is H-extremal. To this end, we consider functions hn defined as  S  on I−1 ∪ I0 ∪ ∞ 1 k=n Ik , S∞ −1 n ≥ 2. hn := −(k − 1) on k=n Jk ,   0 otherwise, Then {hn } is a bounded sequence of functions from H converging pointwise to the function cI0 ∪I−1 . If µ ∈ Mz (H), then 0 = hn (z) = µ(hn ) → µ(I0 ∪ I−1 ). Hence µ is carried by K \ I−1 as needed. Thus F := ChH (K) ∩ (K \ I−1 ) is a σext -closed set. We show that φ0 (F ) is not σext -closed set in ext S(H). To this end, let φ : K → S(H) be the evaluation mapping and let H ⊂ S(H) be a closed Ac (S(H))-extremal set in S(H) such that φ0 (F ) ⊂ H ∩ ext S(H).

306


Then φ(z) ∪ φ(I0 ) ⊂ H. Let π : M1 (K) → S(H) be the restriction mapping and denote sn := π(µn ), n ∈ N ∪ {0, −1}, tn := π(νn ), n ∈ N. Then

1 1 sn + (1 − )tn , n ∈ N, n n and thus sn , tn ∈ H, n ∈ N. Since H is closed, s0 ∈ H as well. We get that the measure φ] µ−1 ∈ Ms0 (S(H)), and hence its support is contained in H (see Proposition 2.86). Thus φ(I−1 ) ⊂ H ∩ ext(S(H)). φ(z) =

Hence φ0 (F ) is strictly smaller than ext(S(H)) ∩ H for any closed extremal set H in S(H) satisfying φ0 (F ) ⊂ ext(S(H)) ∩ H. Exercise 9.57. Let H be a function space on a compact space K, µ ∈ M+ (K) be a maximal measure and (ChH (K), Σ00 , µ00 ) be the measure space from Notation 9.34. Let f : ChH (K) → R be a µ00 -measurable function. Prove that for any ε > 0 there exists a σext -closed set F ⊂ ChH (K) such that µ00 (ChH (K) \ F ) < ε and f |F is σext -continuous. Hint. Let {Un : n ∈ N} be a countable base of open sets in R. Given ε > 0, using Theorem 9.19, we find σext -closed sets Fn ⊂ f −1 (Un ) and Hn ⊂ f −1 (R \ Un ),

n ∈ N,

such that µ00 (f −1 (Un ) \ Fn ) + µ00 (f −1 (R \ Un ) \ Hn ) < 2−n ε, Then F :=

∞ \

n ∈ N.

(Fn ∪ Hn )

n=1

is a σext -closed set such that µ00 (ChH (K) \ F ) < ε and f |F is σext -continuous. Exercise 9.58. Find a compact convex set X, a maximal measure µ ∈ M1 (X) and a strongly affine function f on X such that f |ext X is not µ00 -measurable. Hint. Let H be the function space from Example 3.82, X := S(H) be the state space of H and let f := c[0,1]×{1} − c[0,1]×{−1} . Let λ be Lebesgue measure on [0, 1] × {0} and λ00 be the induced measure on ChH (K) from Notation 9.34. Since H is simplicial, H = Ac (H) (see Lemma 6.14 and its proof) and f is Borel, f ∈ H⊥⊥ (see Corollary 6.12). Thus If is a strongly affine function on X (see Corollary 5.41). By Lemma 9.37, (φ] λ)00 = (φ0 )] λ00 . Further, If (φ(x)) = f (x),

x ∈ ChH (K),

9.6 Exercises

307

and hence (φ] λ)00 -measurability of If |ext X implies λ00 -measurability of f |ChH (K) . Thus it suffices to show that f |ChH (K) is not λ00 -measurable. Since Σ00 is the completion of Σ0 with respect to the measure λ0 , we know from Theorem 9.19 that λ00 (A) = sup{λ00 (F ) : F ⊂ A is σext -closed},

A ∈ Σ00 .

Thus to disprove that f |ChH (K) is µ00 -measurable it is enough to show that λ00 (F ) = 0 for any σext -closed set F ⊂ [0, 1] × {1} (and analogously for [0, 1] × {−1}). If F is such a set, then there exists a closed H-extremal set H ⊂ K such that H ∩ ChH (K) = F. If F is infinite, it is easy to realize that H intersects [0, 1] × {−1} as well. Hence F is finite and H = F . But then λ00 (F ) = λ(H) = 0. Exercise 9.59. Prove that the class of strongly universally measurable functions is not affinely perfect. More precisely, find compact convex sets X, Y , an affine continuous surjection π : X → Y and a function f : Y → R such that f ◦π is strongly universally measurable and f is not strongly universally measurable. Hint. Consider the function space H from Example 3.82 and set Y := S(H), X := M1 (K). Let π : M1 (K) → S(H) be the restriction mapping and let f and λ be as in Exercise 9.58. If I is the mapping from Corollary 5.41, the function If |ext Y is not (φ] λ)00 -measurable. The function g : X → R defined as g(µ) = µ(f ), µ ∈ X, is strongly universally measurable on X. Indeed, given ε > 0 and ν ∈ X, let h1 , −h2 be upper semicontinuous bounded functions on K such that h1 ≤ f ≤ h2 and ν(h2 ) − ν(h1 ) < ε. Then the functions b hi (µ) := µ(hi ), µ ∈ X, i = 1, 2, witness that g is strongly universally measurable. Finally, the formula If (π(µ)) = g(µ), µ ∈ X, shows that If is the desired function. Exercise 9.60 (Semiextremal sets). If X is a compact convex set X, a set F ⊂ X is semiextremal if X \ F is convex. Prove the following assertions. (a) Any extremal set is semiextremal. (b) Semiextremal sets are stable with respect to unions and intersections of downdirected families. (c) An intersection of an extremal and a semiextremal set is semiextremal. (d) A closed nonempty semiextremal set intersects ext X.

308


Hint. Assertions (a), (b) and (c) are easy from the definitions. To prove (d), let F be a closed semiextremal set. We consider the family F := {F ∩ H : H closed extremal, F ∩ H 6= ∅} ordered by reverse inclusion. An application of Zorn’s lemma yields the existence of a minimal set in F. Hence we may assume that F has the property that any closed extremal set intersecting F contains F .

(9.15)

Take any point x ∈ F and a maximal measure µ ∈ Mx (X). Then µ(F ) > 0 (otherwise x ∈ X \ F by Proposition 2.76). Thus there exists a maximal measure ν carried by F . If ν = εy for some y ∈ X, y is an extreme point contained in F and the proof is finished. Now consider the case when ν is not a Dirac measure. Then we can find a pair of disjoint Baire sets A1 , A2 ⊂ K such that ν(Ai ) > 0, i = 1, 2. Using Proposition 9.14 we find closed extremal sets Fi ⊂ Ai such that ν(Fi ) > 0, i = 1, 2. In particular, the sets Fi intersect F . It follows from Proposition 2.69 that there exists a pair of closed faces Hi ⊂ X such that ∅ 6= Hi ∩ F ⊂ Fi ∩ F,

i = 1, 2.

But this contradicts (9.15).

9.7

Notes and comments

Our presentation of topologies on Choquet boundaries follows mainly papers of S. Teleman and C. J. K. Batty (see [449], [450], [453], [452], [451], [34] and [35]). We also refer the reader to the paper of M. Rogalski [393]. The general concept of M-extremal sets is explicitly defined in [35] in order to cover simultaneously notions of the Choquet and maximal topology. Proposition 9.9 and Example 9.11 can be found in [35], Theorem 9.10 in N. Boboc and Gh. Bucur [68]. The factorization theorem 9.12 appears as Theorem 1 in [451], properties of induced measures described in Section 9.2 are presented in [450], [35] and [34]. Results of Section 9.3 on functions continuous with respect to the topologies σext and σmax are taken from papers S. Teleman [455] and C. J. K. Batty [35]. The exposition in Section 9.4 follows the paper of S. Teleman [454]. The results of Section 9.5 are classical and can be found in Chapter II, §6 and 7 in E. M. Alfsen [5] and in L. Asimow and A. J. Ellis [24], Chapter 3, Section 1. We only transferred the techniques into the framework of function spaces. The motivation for the introduction of facial topology came from the theory of C ∗ -algebras. Since we do not pursue this topic in our treatise, we refer the reader to [5] and [24] for the references related to this area.


309

Theorem 9.42 is attributed in [5] to T. Bai Andersen, an independently found proof is contained in a paper W. Wils [472] (see also T. B. Andersen and H. R. Atkinson [16] for a related result on σfac -continuous mappings with values in Banach algebras). Theorems 9.43, 9.45 and Corollary 9.44 are in E. M. Alfsen and T. Bai Andersen [6], Theorem 9.48 and Corollary 9.50 can be found as Corollary II.7.11 and Observation II.7.16 in [5] (see also Theorem 3 in E. Briem’s paper [91]). We refer the reader to papers [197], [196] and [195] of A. Gleit for topological properties of the topology σfac . Results on extending Borel affine functions on analytic split faces can be found in P. J. Stacey [435]. Exercise 9.51 is Théorème 2 in S. Teleman [455], Exercise 9.53 is a particular case of Theorem 3.2 in C. J. K. Batty [35] and Exercise 9.54 is Théorème 3 in [455]. Exercise 9.57 can be found as Theorem 7 in S. Teleman [450]. Assertion (d) of Exercise 9.60 is proved in J. D. Pryce [378] by geometric methods; we present it as a consequence of the results on induced measures on extreme points.

Chapter 10

Deeper results on function spaces and compact convex sets

The goal of this chapter is to present several important results on function spaces and compact convex sets. The first section is devoted to boundaries and to applications of this notion in Banach spaces. The first result is Theorem 10.2, proving the existence of the smallest closed boundary (so-called Shilov boundary) under a rather mild condition. Theorem 10.3 recalls the classical fact that the closure of the Choquet boundary is the Shilov boundary for a function space. Next we prove several results on James’ boundaries which is a concept particularly useful in the theory of Banach spaces. We present an approach based upon the notion of I-envelopes of sets in duals of Banach spaces. This leads us to Theorem 10.7 by V. Fonf and J. Lindenstrauss, which opens a way for the proof of Rode’s result on norm separability of dual spaces (see Theorem 10.8). As a consequence of the Simons lemma, we prove James’ theorem characterizing weakly compact sets in separable Banach spaces (see Theorem 10.9). Then we present Grothendieck’s result on angelicity of spaces of continuous functions on compact spaces when they are endowed with the pointwise topology. As its application we show Khurana’s proof of a theorem by J. Bourgain and M. Talagrand on angelicity of bounded sets in a Banach space with respect to the weak topology generated by extreme points of the dual unit ball (see Theorem 10.12). The next section contains the proof of Lazar’s generalization of the Banach–Stone theorem (see Theorems 10.17 and 10.18). Then we turn our attention to results on automatic boundedness and continuity of affine functions. First we need several important auxiliary facts on the Cantor set, such as the 0–1 laws of Theorem 10.24. The main tool needed later is Theorem 10.26 on countable additivity of finitely additive measures on the Cantor set. Then we present Christensen’s results on strongly linearly independent sequences in locally convex spaces and functionals on L∞ (µ) (see Theorems 10.30 and 10.32). Results on automatic boundedness of affine and convex functions on compact convex sets are proved in Theorems 10.31 and 10.37. Haydon’s characterization of Banach spaces not containing `1 by means of strongly affine functions is contained in Theorem 10.42. Metrizability of compact convex sets is investigated in Section 10.5. The first result in Theorem 10.51 shows that a compact convex set X is metrizable provided it has

10.1 Boundaries

311

a boundary with a countable network. Theorem 10.56 collects several topological conditions on extreme points that are equivalent to metrizability of X. Next we study continuous affine images of compact convex set. First we investigate under what conditions a continuous affine surjection induces a mapping preserving maximal measures (see Theorem 10.57 and 10.59), and then we apply the results to get a variant of the Stone–Weierstrass theorem for function spaces (see Theorem 10.60). The second part of this section presents results on openness of affine continuous surjections (see Theorem 10.66). As a corollary, we get that a continuous mapping ϕ of a compact space K onto a compact space L is open if and only if the induced mapping ϕ] : M1 (K) → M1 (L) is open (see Corollary 10.67). The aim of Section 10.7 is the proof of three topological results on Choquet boundaries. The first one asserts that any Choquet boundary is a Baire space (see Theorem 10.68), the second one constructs a metrizable simplex whose set of extreme points is a prescribed Polish space (see Theorem 10.70) and the last one shows that the set of extreme points is a Borel set of low complexity provided it is K-countably determined. Analogues of Theorems 4.21 and 4.24 for fragmented convex functions are proved in Theorems 10.75 and 10.77. Finally, we characterize stable compact convex sets (see Theorem 10.86) and function spaces for which the barycentric mapping is open (see Theorem 10.88).

10.1 10.1.A

Boundaries Shilov boundary

Definition 10.1 (Shilov sets). Let K be a nonempty compact space and F be a nonempty family of lower semicontinuous functions on K. A closed set S ⊂ K is called a Shilov set for F if for every u ∈ F there exists x ∈ S such that u(x) = inf u(K). If there exists a Shilov set, which is contained in any other Shilov set, it is called the Shilov boundary. Obviously, K is a Shilov set for F. Notice that if F contains the constants and u + c ∈ F whenever u ∈ F and c ∈ R, S is a Shilov set if and only if for every u ∈ F with u ≥ 0 on S we have u ≥ 0 on K. Consider the following simple example of the space K := {x1 , x2 , x3 } with the discrete topology. We identify functions on K with vectors in R3 and denote F := {(1, 0, 0), (0, 1, 0), (0, 0, 1)} . Then every two-point subset of K is a Shilov set for F. Consequently, there exists no smallest Shilov set for F.

312

10 Deeper results on function spaces and compact convex sets

The existence of a smallest Shilov set is guaranteed under additional assumptions imposed on F. It is worth mentioning that this is the case when a minimum principle like in Theorem 3.16 holds. Then the Shilov boundary is simply the closure of the Choquet boundary. In this section, however, we provide a direct proof of the existence of a Shilov boundary. Theorem 10.2. Let K be a nonempty compact space, F be a family of lower semicontinuous functions and S ⊂ F be a convex cone of continuous functions containing the constant functions and separating points of K such that u + s ∈ F whenever u ∈ F and s ∈ S. Then there exists the Shilov boundary for F. Proof. A moment’s reflection shows that we may suppose that both F and S are minstable (otherwise we would consider a new family W(F) and a new convex cone W(S) formed by finite minima of elements of F and S, respectively). Claim. The system of Shilov sets is closed with respect to finite intersections. Proof of the claim. Let S and T be Shilov sets. Suppose that u ∈ F and u ≥ 0 on S ∩ T . Fix ε > 0 and define U := {x ∈ K : u(x) + ε > 0} . Then U is an open set containing S ∩ T . In order to show that S ∩ T is a Shilov set, it is sufficient to prove that v := u + ε ≥ 0 on K. Since L := S \ U and T are disjoint compact sets, there are open sets V and W such that L ⊂ V, T ⊂ W and V ∩ W = ∅. Obviously, D := {s − t : s, t ∈ S} is a vector lattice of continuous functions on K containing constant functions and separating points of K. By the Stone–Weierstrass theorem A.30, there exists g ∈ D such that g ≥ 1 on V and g ≤ −1 on W . Since g + = g ∨0 ∈ D, there exist s, t ∈ D such that s−t = g + . We may suppose that s ≥ 0 and v + s ≥ 0 on K, because S contains positive constants. Choose a ∈ (0, 1) such that (1 − a)s ≤ 1 on K. Then s − t ≥ 1 ≥ (1 − a)s

on V,

hence t ≤ as on V , and obviously s = t on W . Let B := {b ∈ [0, ∞) : v + bs ≥ 0 on K} .

10.1 Boundaries

313

Fix b ∈ B. Then 0 ≤ v + bs = v + bt on W and since T ⊂ W is a Shilov set, we have 0 ≤ v + bt on K. In particular, 0 ≤ v + bt ≤ v + bas

on V.

Since v ≥ 0 on U and s ≥ 0 on K, we have 0 ≤ v + bas

on

U ∪ V ⊃ S.

By the assumption, S is a Shilov set, and thus we have 0 ≤ v + bas on K. We conclude that b ∈ B implies ba ∈ B. Since 1 ∈ B, we get a ∈ B and by iteration an ∈ B, hence 0 ≤ v + an s on K for every n ∈ N. Since a ∈ (0, 1), we get v ≥ 0 on K. This establishes the claim. Define now Σ := {S ⊂ K : S is a Shilov set for F} and denote S0 :=

\

{S : S ∈ Σ}.

We claim that S0 is a Shilov set. Choose u ∈ F, u ≥ 0 on S0 , fix ε > 0 and define, as above, the open set U := {x ∈ K : u(x) + ε > 0} . Since U ⊃ S0 and Σ is closed with respect to finite intersections, there exists T0 ∈ Σ with T0 ⊂ U . It follows that u + ε ≥ 0 on T0 . We conclude that u + ε ≥ 0 on K whenever ε > 0, thus u ≥ 0 on K. Consequently S0 is the smallest Shilov set for F, finishing the proof. Theorem 10.3. If H is a function space on a compact space K, then ChH K is the Shilov boundary for H. Proof. By Theorem 10.2, there exists the Shilov boundary S for H. By the Minimum principle 3.16, ChH (K) is a Shilov set for H. To show its minimality, assume that there exists x ∈ ChH (K) \ S. Then x ∈ / coH S (see Proposition 8.18), and thus there exists a function h ∈ H such that h(x) > max h(S) (see Proposition 8.23). But this contradicts the fact that S is a Shilov set.

314


10.1.B

Boundaries in Banach spaces

Throughout this subsection, E denotes a Banach space. Definition 10.4 (I-envelope). If E is a Banach space and B ⊂ E ∗ is a set, we define its I-envelope as I-env(B) :=

∞ ∞ o \n [ [ ∗ Bn . cok·k cow (Bn ) : Bn ⊂ B, B = n=1

n=1

Lemma 10.5. If B ⊂ I-env(B) =

E∗

is a set, then

∞ ∞ o [ \n [ ∗ cok·k cow (Bn ) : B1 ⊂ B2 ⊂ · · · ⊂ B, B = Bn . n=1

n=1

Proof. Let x∗ ∈ / I-env(B), that is, there exist d > 0 and sets Bn ⊂ B, n ∈ N, such that ! ∞ [ w∗ ∗ co Bn = ∅. B(x , d) ∩ co n=1 ∗ By replacing the sets Bn by Bn ∩ kB SE , n, k ∗∈ N, if necessary, we may assume that the sets Bn are bounded. Since co( nk=1 cow (Bk )) is a w∗ -compact convex set for each n ∈ N, the set ! ! ∞ n [ [ w∗ w∗ w∗ co (B1 ∪ · · · ∪ Bn ) ⊂ co co Bk ⊂ co co Bn

n=1

k=1

does not intersect B(x∗ , d). Hence the sequence {B1 ∪ · · · ∪ Bn }∞ n=1 witnesses that x∗ ∈ /

\

cok·k

∞ [ ∗ Bn . cow (Bn ) : B1 ⊂ B2 ⊂ · · · ⊂ B, B =

∞ [

n=1

n=1

This concludes the proof. Proposition 10.6. For x∗ ∈ E ∗ and B ⊂ E ∗ , the following assertions are equivalent: (i) x∗ is not contained in I-env(B), (ii) there exists a sequence {xn } in BE such that inf x∗ (xn ) > sup lim sup b∗ (xn ).

n∈N

b∗ ∈B n→∞

Proof. For the proof of (ii) =⇒ (i), let x∗ and {xn } be as in (ii). We find c1 , c2 ∈ R satisfying inf x∗ (xn ) > c1 > c2 > sup lim sup b∗ (xn ). n∈N

b∗ ∈B n→∞

10.1 Boundaries

By setting Bn :=

∞ \

{b∗ ∈ E ∗ : b∗ (xk ) ≤ c2 },

315

n ∈ N,

k=n

S we get an increasing sequence of w∗ -closed convex sets satisfying B = ∞ n=1 Bn ∩B. S∞ S ∗ moreover, the norm distance of x∗ to this Hence n=1 cow Bn = ∞ B and, n n=1 union is greater than c1 − c2 . Indeed, assume the existence of b∗ in some Bn with kx∗ − b∗ k ≤ c1 − c2 . Then, for k ≥ n, we have x∗ (xk ) ≤ kx∗ − b∗ k + b∗ (xk ) ≤ c1 − c2 + c2 = c1 , a contradiction. S Hence x∗ is not contained in the norm closure of the convex hull of ∞ n=1 Bn , and ∗ therefore x ∈ / I-env(B). For the proof of the converse implication (i) =⇒ (ii), assume that x∗ ∈ / I-env(B). Without loss of generality we may assume that x∗ = 0. By the assumption and Lemma 10.5, there exist d > 0 and an increasing sequence {Bn } such that ! ∞ ∞ [ [ ∗ cow Bn = ∅. B= Bn and B(0, d) ∩ co n=1

n=1

By the Hahn–Banach separation theorem there exist elements xn ∈ E so that sup

b∗ (xn )
c2 > sup lim sup b∗ (xn ).

n∈N

b∗ ∈B n→∞

By the P Simons inequality 3.74, there exists a sequence {λn } of positive numbers such that ∞ n=1 λn = 1 and ∞ X sup y ∗ λn xn < c2 . y ∗ ∈X

n=1

316


On the other hand, ∗

c1 < x

∞ X

λn xn ≤ sup y

∗

∞ X

y ∗ ∈X

n=1

λ n xn ,

n=1

a contradiction. Hence the assertion follows. Theorem 10.8. Let B be a norm separable subset of a w∗ -compact convex set X in E ∗ such that each element of E attains its maximum on X at some point of B. Then X is norm separable as well. ∗ : n ∈ N} ⊂ B be a norm Proof. Let B ⊂ X be as in the hypothesis. n S∞ Let {b dense subset and let ε > 0. Since B ⊂ n=1 B(b∗n , ε) and closed balls in E ∗ are w∗ -compact and convex, Theorem 10.7 yields

X ⊂ cok·k

∞ [

B(b∗n , ε) .

n=1

Hence the closed convex hull of {b∗n : n ∈ N} is norm dense in X, from which the theorem follows. Theorem 10.9 (James’ theorem for separable spaces). Let B ⊂ E be a closed convex subset of a separable Banach space E such that each element of E ∗ attains its supremum on B. Then B is weakly compact. Proof. If B is as in the hypothesis, we first notice that B is bounded since it is weakly bounded (see the uniform boundedness principle, Theorem 3.15 in [173]) and weakly w∗ closed by the Mazur theorem A.2. Thus we need to show that B ⊂ E ∗∗ is a subset w∗ of E. Indeed, if this is the case, B is a w∗ -compact subset of E considered as a w w∗ subset of E ∗∗ , and thus B = B = B is weakly compact. w∗ To this end, let ϕ ∈ B be given. By Theorem A.7, it is enough to show that ϕ is w∗ -continuous on BE ∗ . Since E is separable, it suffices to check the w∗ -sequential continuity of ϕ|BE∗ . Let {x∗n } be a sequence in BE ∗ w∗ -converging to 0. Assume that {ϕ(x∗n )} does not converge to 0. Without loss of generality we may assume that there exists c > 0 such that ϕ(x∗n ) > c for all n ∈ N. If we consider w∗ w∗ points of E ∗ as functions on B , the assumption ensures that B ⊂ B is a boundary for E ∗ . Hence B is a boundary for any x∗ ∈ coσ {x∗n : n ∈ N}. Since x∗n (b) → 0 for each b ∈ B, Corollary 3.74 gives 0 = sup lim sup x∗n (b) = sup lim sup ψ(x∗n ) ≥ lim sup ϕ(x∗n ) ≥ c, b∈B n→∞

ψ∈B

w∗

n→∞

a contradiction. Thus ϕ(x∗n ) → 0, which is the desired conclusion.

n→∞

10.1 Boundaries

317

Definition 10.10 (Relatively countably compact and angelic spaces). We recall that a subset A of a topological space X is relatively countably compact if any sequence {xn } of elements of A has a cluster point in X. A topological space X is angelic if each relatively countably compact set A ⊂ X is relatively compact and any point x ∈ A can be obtained as a limit of a sequence of points from A. Theorem 10.11. If X is a compact space, then (C(X), τX ) is angelic. Proof. Let A ⊂ C(X) be a relatively τX -countably compact set. Step 1: The set A is relatively τX -compact. It follows from the assumption that {f (x) : f ∈ A} is a bounded subset of R for every x ∈ X. By the Tychonoff theorem, we need to show that the τX -closure of A in RX is contained in C(X). To τ achieve this, assume that g ∈ A X is not continuous, say at a point x0 ∈ X. Then there exists ε > 0 such that for each open neighborhood U of x0 there exists y ∈ U so that |g(x0 )−g(y)| ≥ ε. We construct inductively points xn ∈ X, open neighborhoods Un of x0 and functions fn ∈ A, n ∈ N, such that, for each n ∈ N, (a) |g(xn ) − g(x0 )| ≥ ε, (b) |fn (xi ) − g(xi )| ≤ 2−n , i = 0, . . . , n, (c) xi ∈ Un , i ≥ n + 1, (d) Un+1 ⊂ Un and diam fn (Un ) ≤ 2−n . To start the construction, let x1 ∈ X be a point such that |g(x1 ) − g(x0 )| ≥ ε and let f1 ∈ A satisfy |f1 (xi ) − g(xi )| < 2−1 , i = 0, 1. We find a neighborhood U1 of x0 such that diam f1 (U1 ) ≤ 2−1 . This finishes the first step of the induction. Assume that the construction has been completed up to the n-th stage. We find xn+1 ∈ Un such that |g(xn+1 ) − g(x0 )| ≥ ε. Let fn+1 ∈ A be chosen in such a way that |fn+1 (xi ) − g(xi )| ≤ 2−(n+1) for i = 0, . . . , n + 1. Let Un+1 be a neighborhood of x0 such that Un+1 ⊂ Un and diam fn+1 (Un+1 ) ≤ 2−(n+1) . This concludes the construction. Let f be a τX -cluster point of {fn } and x be a cluster point of {xn }. For any i ∈ N, property (b) gives f (xi ) = g(xi ), and hence |f (xi ) − g(x0 )| ≥ ε by (a). By the continuity of f , |f (x) − g(x0 )| ≥ ε. It follows from property (c) that, for a fixed n ∈ N, x ∈ Un . By (d), |fn (x) − fn (x0 )| ≤ 2−n , and thus f (x) = f (x0 ). But by (b) again, f (x0 ) = g(x0 ). Hence we have arrived at a contradiction, because g(x0 ) = f (x0 ) = f (x) 6= g(x0 ). τ

This concludes the first step of the proof. Let now g ∈ A X be given. Step 2: There exists a sequence {fn } in A pointwise converging to g. To find a desired sequence, we construct countable sets Y ⊂ X and B ⊂ A such that

318


(e) for any I ⊂ B ∪ {g} finite, ε > 0 and x ∈ X there exists y ∈ Y such that |f (y) − f (x)| ≤ ε for each f ∈ I, (f) for any I ⊂ Y finite and ε > 0 there exists f ∈ B such that |f (x) − g(x)| ≤ ε for every x ∈ I. The construction will be done inductively. First we notice that, for any finite set I ⊂ C(X), the set {(f (x))f ∈I : x ∈ X} is a separable subset of RI considered with the supremum norm. Hence there exists a countable set YI ⊂ X such that {(f (y))f ∈I : y ∈ Y } is dense in {(f (x))f ∈I : x ∈ X}. If I ⊂ X is finite, we can choose a sequence {fj(I) } from A pointwise converging to g on I. Let BI consists of the elements of the sequence {fj(I) }. We set inductively Yn :=

n−1 [n [ o YI : I ⊂ {g} ∪ Bi and

Bn :=

n−1 [n [ o BI : I ⊂ Yi ,

i=1

n ∈ N.

i=1

S∞

S∞

At the end of the construction we set Y := n=1 Yn and B := n=1 Bn . Then the requirements (e) and (f) are satisfied. By (f), there exists a sequence {fj } of functions from A pointwise converging to g on Y . We claim that fj (y) → g(y) for all y ∈ X. Assume that this is not the case. Then there exist y ∈ X and ε > 0 such that the set {j ∈ N : |fj (y) − g(y)| ≥ ε} is infinite. For I = {f1 , . . . , fm , g} we use (e) to find ym ∈ Y satisfying |u(ym ) − u(y)| < 2−m ,

u ∈ I.

(10.1)

Let x be a cluster point of {ym : m ∈ N} and h be a τX -cluster point of {fj : j ∈ N}. For each ym , fj (ym ) → g(ym ), and hence h(ym ) = g(ym ). By continuity, h(x) = g(x). Further, for each fi we have from (10.1) that fi (x) = fi (y), and thus h(x) = h(y). Also by (10.1), g(x) = g(y). Finally we notice that |g(y) − h(y)| ≥ ε. Putting all these information together, we get g(y) = g(x) = h(x) = h(y) 6= g(y), a contradiction. Hence fj → g as required, and the proof is finished. Theorem 10.12. Let X be a compact convex set. Then (BAc (X) , τext X ) is an angelic space. In particular, any norm bounded relatively τext X -countably compact set A is relatively τX -compact. Proof. Let A ⊂ BAc (X) be relatively τext X -countably compact. Since (C(X), τX ) is an angelic space, to conclude that A is relatively τX -compact it suffices to show that any sequence {fn } in A has a τX -cluster point. We select a τext X -cluster point

10.1 Boundaries

319

f0 of {fn } and show that f0 is also a τX -cluster point of {fn }. First we notice that kf0 k ≤ 1 by the minimum principle of Corollary 2.24. If we assume a contrary, after omitting finitely many elements from {fn } we may assume that there exist x1 , . . . , xk ∈ X and ε > 0 such that {fn : n ∈ N} ∩ {g ∈ C(X) : |f0 (xi ) − g(xi )| < ε, i = 1, . . . , k} = ∅.

(10.2)

We define ϕ : X → RN as ϕ(x) := {fn (x)}∞ n=0 , x ∈ X, and set Y := ϕ(X). Then Y is a metrizable compact convex set and there exist functions gn ∈ Ac (Y ) such that fn = gn ◦ ϕ, n ≥ 0 (namely, the coordinate functions). For i = 1, . . . , k, let yi := ϕ(xi ) and let νi ∈ Myi (Y ) be a maximal measure. By Theorem 3.79 there exists a compact set K ⊂ ext Y such that νi (K) > 1 − 4ε , i = 1, . . . , k. Pick y ∈ K. Then the set F := ϕ−1 (y) is a face of X (see Proposition 2.72), and thus ext F = F ∩ ext X by Proposition 2.64(b). Further, all functions fn are constant on F . It follows that if g is a continuous affine function on X contained in τext X τF , then g|F ∈ {fn |F : n ∈ N} . {fn : n ∈ N} Thus the set {fn |ϕ−1 (K) : n ∈ N} is relatively τϕ−1 (K) -countably compact. By Theorem 10.11, there exists a subsequence {fnk } of {fn } so that fnk → f0 on ϕ−1 (K). By the Lebesgue dominated convergence theorem there exists k0 ∈ N such that Z Z ε gn (t) dνi (t) − g0 (t) dνi (t) < , i = 1, . . . , k. k0 2 K K Then for i = 1, . . . , k, Z Z gnk0 (t) dνi (t) − g0 (t) dνi (t) |fnk0 (xi ) − f0 (xi )| = ext Y ext Y Z Z ≤ gnk0 (t) dνi (t) − g0 (t) dνi (t) K K Z + |gnk0 (t) − g0 (t)| dνi (t) ext Y \K

ε ε < + 2 = ε. 2 4 But this contradicts (10.2) and shows that A is relatively τX -compact. τ Thus A X is τX -compact, and hence also τext X -compact; in particular τext X -closed. Thus τ τ τ A ext X ⊂ A X ⊂ A ext X . τ

τ

It follows that the identity mapping id : (A ext X , τX ) → (A ext X , τext X ), as a continuous injective mapping, is a homeomorphism. Since the topology τX is angelic, τext X τ is angelic on A ext X , which concludes the proof.

320


Corollary 10.13. Let E be a Banach space and A be a norm bounded relatively σ(E, ext BE ∗ )-countably compact set. Then A is relatively weakly compact. Proof. Let A be a norm bounded relatively σ(E, ext BE ∗ )-countably compact set in a Banach space E. Then A can be viewed as a relatively τext BE∗ -countably compact subset of the space Ac (BE ∗ ) (here BE ∗ is endowed with the w∗ -topology). Moreover, the weak topology coincide with the topology τBE∗ (see Proposition 4.31(f)). Hence the assertion follows from Theorem 10.12.

10.2

Isometries of spaces of affine continuous functions

Theorem 10.14. Let X, Y be compact convex sets and T : Ac (X) → Ac (Y ) be a positive isometry. Then X is affinely homeomorphic to Y . Proof. If T is as in the hypothesis, the dual operator T ∗ : (Ac (Y ))∗ → (Ac (X))∗ is a positive isometry as well and, moreover, it is a (w∗ –w∗ )-homeomorphism. Hence T ∗ is an affine homeomorphism of S(Ac (Y )) to S(Ac (X)). To see its surjectivity, let s ∈ S(Ac (X)) be arbitrary. Then t : g 7→ s(T −1 g), g ∈ Ac (Y ), is in S(Ac (Y )) and T ∗ t = s. Using the identification of X with S(Ac (X)) and Y with S(Ac (Y ) (see Proposition 4.31(a)) we finish the proof. Lemma 10.15. Let X1 , X2 be compact convex sets and, for i = 1, 2, let Fi be closed faces of Xi such that •

Fi is a split face of Xi ,

•

the complementary face Fi0 is closed.

Let ϕ : F1 ∪ F10 → F2 ∪ F20 be a continuous mapping that is affine on F1 and F10 and maps F1 to F2 and F10 to F20 . Then there exists a unique continuous affine mapping ψ : X1 → X2 extending ϕ. Moreover, if ϕ is injective or surjective, then ψ is injective or surjective, respectively. Proof. Given x ∈ X1 , let ψ(x) be defined as follows. Let x = αx1 + (1 − α)x2 be the unique decomposition of x with α ∈ [0, 1] and x1 ∈ F1 , x01 ∈ F10 (see Definition 8.4). We set ψ(x) := αϕ(x1 ) + (1 − α)ϕ(x2 ). It is a matter of routine verification that ψ : X1 → X2 is a well-defined affine mapping. To show its continuity, pick g ∈ Ac (X2 ). Then g ◦ ψ is an affine function that is continuous on F1 ∪ F10 . By Lemma 5.39, g ◦ ψ is continuous on X1 . Hence ψ is continuous with respect to the weak topology on X2 . Since X2 is compact, ψ is continuous.

321

10.2 Isometries of spaces of affine continuous functions

Assume that ϕ is injective on F1 ∪ F10 and let ψ(x) = ψ(y) for some x, y ∈ X1 . Choose α, β ∈ [0, 1], x1 , y1 ∈ F1 and x2 , y2 ∈ F10 such that x = αx1 + (1 − α)x2

and

y = βy1 + (1 − β)y2 .

Then αϕ(x1 ) + (1 − α)ϕ(x2 ) = βϕ(y1 ) + (1 − β)ϕ(y2 ). Since F2 is a split face, this decomposition yields α = β and ϕ(xi ) = ϕ(yi ), i = 1, 2. Hence xi = yi , i = 1, 2, and x = y. If ϕ is surjective, then X2 = co(F2 ∪ F20 ) gives ψ(X1 ) = X2 . Lemma 10.16. Let X, Y be compact convex set and T : Ac (X) → Ac (Y ) be an isometric isomorphism. Then the sets F1 := {y ∈ Y : T 1(y) = 1} and

F2 := {y ∈ Y : T 1(y) = −1}

are closed faces such that F1 is parallel and F2 is its complementary face. Proof. Since kT 1k = 1, both F1 , F2 are closed faces. We use the identifications from Proposition 4.31, namely, that span X = Ac (X)∗ and span Y = Ac (Y )∗ . Then T ∗ is an isometry, and thus T ∗ maps extreme points of BAc (Y )∗ onto ext BAc (X)∗ . Hence for any y ∈ ext Y there exists x ∈ ext X such that T ∗ y = x or T ∗ y = −x. Then either T 1(y) = T ∗ y(1) = x(1) = 1

or

T 1(y) = T ∗ y(1) = (−x)(1) = −1.

Hence Y = co ext Y = co(F1 ∪ F2 ). From this it follows that F10 = F2 (see Exercise 10.93). Let y = αy1 + (1 − α)y2 for yi ∈ Fi , i = 1, 2, and α ∈ [0, 1]. Then T 1(y) = T ∗ y(1) = αT ∗ (y1 )1 + (1 − α)T ∗ (y2 )1 = α − (1 − α) = 2α − 1. Hence the coefficient α is unique and F1 is a parallel face. Theorem 10.17. Let X, Y be compact convex sets with the following property: •

if F1 , F2 are closed faces such that F1 is parallel and F2 = F10 , then F1 is a split face.

Let T : Ac (X) → Ac (Y ) be an isometric isomorphism. Then there exists an affine homeomorphism ϕ : Y → X and a function h ∈ Ac (Y ) such that T f (y) = h(y)f (ϕ(y)),

f ∈ Ac (X), y ∈ ext Y.

(10.3)

322


Proof. If T : Ac (X) → Ac (Y ) is an isometry, we define G1 := {y ∈ Y : T 1(y) = 1},

G2 := {y ∈ Y : T 1(y) = −1}

and F1 := {x ∈ X : T −1 1(x) = 1},

F2 := {x ∈ X : T −1 1(x) = −1}.

By Lemma 10.16, G1 is a closed parallel face of Y such that the closed face G2 is its complementary face. Similarly, F1 is a closed parallel face of X such that F2 = F10 . By the assumption, both F1 and G1 are split faces. We use the identification of span X with Ac (X)∗ and span Y with Ac (Y )∗ (see Proposition 4.31). We claim that T ∗ (G1 ) = F1 and T ∗ (G2 ) = −F2 . Indeed, for any y ∈ G1 we see from T ∗ y(1) = T 1(y) = 1, that T ∗ y is a positive element of norm 1 in Ac (X)∗ and thus it belongs to X. Moreover, (T −1 1)(T ∗ y) = ((T −1 )∗ T ∗ y)(1) = 1. Thus T ∗ y ∈ F1 . If x ∈ F1 is given, the equalities ((T −1 )∗ x)(1) = T −1 1(x) = 1 yield that (T −1 )∗ x ∈ Y . Further, T 1((T −1 )∗ x) = ((T −1 )∗ x)(T 1) = (T ∗ (T −1 )∗ x)(1) = 1. Hence (T −1 )∗ x ∈ G1 and T ∗ (G1 ) = F1 . Similarly we obtain T ∗ (G2 ) = −F2 . Let φ : G1 ∪ G2 → F1 ∪ F2 be defined as ( T ∗ y, y ∈ G1 , φ(y) := ∗ −T y, y ∈ G2 . Then φ is continuous, injective, surjective and affine on both G1 and G2 . Using Lemma 10.15 we find an affine homeomorphism ϕ : Y → X extending φ. We set h := T 1 and show the validity of (10.3). Pick f ∈ Ac (X) and y ∈ ext Y . If y ∈ G1 , ϕ(y) = φ(y) ∈ F1 , and thus T f (y) = T ∗ y(f ) = f (φ(y)) = T 1(y)f (φ(y)). Similarly, for y ∈ G2 we get T f (y) = T ∗ y(f ) = −f (φ(y)) = T 1(y)f (φ(y)). Thus T f (y) = h(y)f (ϕ(y)) for all y ∈ ext Y , and the proof is complete.

10.3 Baire measurability and boundedness of affine functions

323

Theorem 10.18. Let X, Y be simplices and T : Ac (X) → Ac (Y ) be an isometric isomorphism. Then X is affinely homeomorphic to Y . Proof. The proof follows from Theorem 10.17 and Corollary 8.63. Corollary 10.19 (Banach–Stone). Let K, L be compact spaces and T : C(K) → C(L) be an isometric isomorphism. Then K is homeomorphic to L. Proof. If T is as in the hypothesis, it induces an isometry between Ac (M1 (K)) and Ac (M1 (L)). Since both sets M1 (K) and M1 (L) are Bauer simplices (see Proposition 6.38), Theorem 10.18 yields that they are affinely homeomorphic. Hence their sets of extreme points are homeomorphic and Proposition 2.27 finishes the proof.

10.3 10.3.A

Baire measurability and boundedness of affine functions The Cantor set and its properties

We recall the following notation from Section A.4. We write N m0 ≥ pσ (m), and hence |un (k) − x∗ (xpτ (k) )| = |x∗ (xpτ (k) )| ≤

1 . n

Thus τ ∈ Mn , as needed. Having observed that the sets Mn are dense and open, we can set M :=

∞ \

Mn

n=1

to get a comeager set in {0, 1}N \ Q. Since σ 7→ N \ σ, σ ∈ {0, 1}N , is a homeomorphism of {0, 1}N onto itself and {0, 1}N \ Q is comeager in {0, 1}N , the set N := M ∩ {N \ σ : σ ∈ M } is comeager in {0, 1}N , in particular it is nonempty. Let σ ∈ N . Then both σ and N \ σ are contained in M , and these two sets yield the required partition of N. Indeed, let us show that, for example, {xpσ (j) } is strongly linearly independent. We define a subspace of c0 as ∗ ∗ U := {{x∗ (xpσ (k) )}∞ k=1 : x ∈ E }.

Since σ ∈ M , the space U is dense in c0 . Let now λ ∈ `1 satisfy w- lim

j→∞

This means lim

j→∞

or equivalently,

j X

j X

λi xpσ (i) = 0.

i=1

λi x∗ (xpσ (i) ) = 0,

x∗ ∈ E ∗ ,

i=1 ∞ X

λi ui = 0,

u ∈ U.

i=1

Since U is norm dense in c0 and (c0 )∗ = `1 , λ = 0. This concludes the proof.


331

Theorem 10.31. Let f be an affine function on a compact convex subset X of a locally convex space that has the Baire property in the restricted sense. Then f is bounded on X. Proof. Assume that f is not bounded. We translate the set X if necessary to achieve that 0 ∈ X and assume that f (0) = 0. Inductively we find a sequence {xn } in X such that xn+1 ∈ / span{x1 , . . . , xn } and 2n |f (xn )| ≥ 2 . (This is possible because f is bounded on F ∩ X for every finitedimensional subspace F of E.) Set xn x0n := , n ∈ N. n Since X is a bounded subset of E, x∗ (x0n ) → 0 for each x∗ ∈ E ∗ (see W. Rudin [403, Theorem 1.15 and Theorem 1.30]). By Theorem 10.30, we can find a subsequence {ynk } of {x0n } that is strongly independent. We relabel the sequence {ynk } as {yk }. Claim. The following assertions hold: (a) the mapping ϕ : {0, 1}N → X defined as X ϕ(σ) := σk 2−nk yk ,

σ ∈ {0, 1}N ,

k∈N

is a well-defined homeomorphism, (b) the function h : {0, 1}N → R defined as h(σ) := f (ϕ(σ)),

σ ∈ {0, 1}N ,

has the Baire property in the restricted sense on {0, 1}N . Proof of the claim. For the proof of (a), we notice that ϕ isP a well-defined mapping and ϕ({0, 1}N ) ⊂ X. Indeed, if σ ∈ {0, 1}N \ {0} and λ = k∈N σk 2−nk , then X σk 2−nk yk λ k∈N

is a well-defined point contained in X (see Theorem 2.29). Hence, ! X σk −nk 2 yk + (1 − λ)0 ϕ(σ) = λ λ k∈N

is well defined and contained in X as well. Further, for σ, τ ∈ {0, 1}N and x∗ ∈ E ∗ , we get |x∗ (ϕ(σ)) − x∗ (ϕ(τ ))| ≤ sup |x∗ (x)| x∈X

∞ X k=m

2−nk ,

332


where m ∈ N is the least index such that σm 6= τm . Thus x∗ ◦ϕ is continuous for each x∗ ∈ E ∗ which gives that ϕ : {0, 1}N → X is continuous with respect to the weak topology of E. Since the weak topology of E coincides with the original topology on the compact set X, ϕ is continuous. Further, we check the injectivity of ϕ. Let σ, τ ∈ {0, 1}N satisfy ϕ(σ) = ϕ(τ ), that is, X X σk 2−nk yk = τk 2−nk yk . (10.5) k∈N

k∈N

Since the sequence {yk } is strongly independent, (10.5) yields that σ = τ . Hence, ϕ is injective and thus a homeomorphism. For the proof of (b), we have to verify that h has the Baire property on each compact set K ⊂ {0, 1}N . But this is obvious, since ϕ : K → X is a homeomorphic mapping. This concludes the proof of the claim. Now we are ready to finish the proof of Theorem 10.31. If x ∈ X and λ < 1, then f (λx) = f (λx + (1 − λ)0) = λf (x) + (1 − λ)f (0) = λf (x). Thus, for x, y ∈ X with x + y ∈ X, f (x + y) = 2f ( x+y 2 ) and consequently f (x + y) = 2f

x + y 2

1 1 f (x) + f (y) 2 2 = f (x) + f (y).

=2

Hence, if σ, τ ∈ {0, 1}N have disjoint supports, we get h(σ + τ ) = f (ϕ(σ + τ )) = f (ϕ(σ) + ϕ(τ )) = f (ϕ(σ)) + f (ϕ(τ )). Thus, the function h is finitely additive on {0, 1}N . By the claim, h has the Baire property in the restricted sense on {0, 1}N . By Theorem 10.26, the function h is countably additive. If ek = (0, . . . , 0, 1, 0, . . . ) denotes the element in {0, 1}N with digit 1 at k-th coordinate, k ∈ N, from the countable additivity of h we get ∞ X |h(ek )| < ∞. k=1

On the other hand, |h(ek )| = 2−nk |f (yk )| = 2−nk |f (n−1 k xnk )| ≥

2nk , nk

k ∈ N.

P Thus, the series ∞ k=1 h(ek ) cannot be absolutely convergent. This contradiction finishes the proof.


333

Theorem 10.32. Let E = L1 (µ) for a σ-finite measure µ on a measurable space (X, Σ, µ) and let f be a linear functional on E ∗ (identified with L∞ (µ)) that has the Baire property in the restricted sense on RBE ∗ with respect to the w∗ -topology. Then there exists h ∈ L1 (µ) such that f (g) = X g(t)h(t) dµ(t), g ∈ L∞ (µ). Proof. Let f : E ∗ → R be as in the hypothesis. For a set A ∈ Σ, we set ν(A) := f (cA ). Then ν is a finitely additive signed measure on Σ. We want to prove that ν is countably additive. To this end, let {An : nS ∈ N} be a countable family of pairwise P∞ ∞ disjoint sets in Σ. Our aim is to show that ν( n=1 An ) = n=1 ν(An ). Let M := {n ∈ N : µ(An ) > 0}. Then ν(An ) = 0 for every n ∈ N \ M and cSn∈N An = cSn∈M An

µ-almost everywhere.

Thus ν

∞ [ n=1

X

[

An = f (cSn∈N An ) = f (cSn∈M An ) = ν

An

and

n∈M

ν(An ) =

X

ν(An ).

n∈M

n∈N

Hence we may assume that all the sets An are of strictly positive measure. Then the mapping ϕ : {0, 1}N → BE ∗ defined as ϕ : σ 7→

∞ X

σn cAn ,

σ ∈ {0, 1}N ,

n=1

is injective and continuous with respect to the w∗ -topology of BE ∗ . Thus h(σ) = f (ϕ(σ)), σ ∈ {0, 1}N , is finitely additive and has the Baire property in the restricted sense on {0, 1}N . By Theorem 10.26, h is countably additive. Thus ν

[ n∈N

X X X An = h en = f (cAn ) = ν(An ). n∈N

n∈N

n∈N

Since ν is absolutely continuous with respect to µ, the R Radon–Nikodym theorem provides a function h ∈ L1 (µ) such that ν(A) = A h dµ, A ∈ Σ. By Theorem 10.31, f is bounded on BE ∗ and thus continuous on E ∗ with respect to the norm topology. R Hence we can use the standard approximation techniques to deduce that f (g) = X gh dµ, g ∈ L∞ (µ). This concludes the proof. The Banach space `1 is also endowed with a natural order, precisely x ≤ y, x, y ∈ if x(n) ≤ y(n) for each n ∈ N. We recall that en is the n-th standard unit vector in `1 . `1 ,

334


Notation 10.33 (Positive cone of `1 ). Let P stand for the positive cone in `1 , that is, P := {x ∈ `1 : x ≥ 0}. Apart from its usual norm topology, we consider on P a finer topology τ whose basis of neighborhoods for x ∈ P consists of the sets of the form V (x, ε) := {y ∈ P : y ≥ x, ky − xk ≤ ε}. Lemma 10.34. The cone P is a Baire space in the topology τ . Proof. Let {Vn } be a sequence of dense open sets in (P, τ ), and let V0 be a given nonempty τ -open set. Inductively we find points xn ∈ P and εn ∈ (0, 2−n ), n ≥ 0, such that V (x0 , ε0 ) ⊂ V0 and V (xn+1 , εn+1 ) ⊂ Vn+1 ∩ V (xn , εn ),

n ≥ 0.

Then the sequence {xn } satisfies xn ≤ xn+1 , n ≥ 0, and kxn − xn+1 k ≤ 2−n . It follows that the sequence {xn } converges to some x ∈ P . Since {xn } is increasing, x ∈ V0 ∩

∞ \

Vn ,

n=1

which concludes the proof. Lemma 10.35. Let f : (P, τ ) → R be a sublinear function with the Baire property. Then infn∈N f (en ) > −∞. Proof. Since the function f has the Baire property with respect to the Baire topology τ , there exists a τ -meager set M ⊂ P such that f |P \M is τ -continuous (see Lemma 10.34 and Proposition A.61). We pick x ∈ P \ M and find ε > 0 such that |f (y) − f (x)| < 1 whenever y ∈ V (x, ε) \ M . Let z ∈ P with kzk ≤ 1 be arbitrary. Since the mapping ϕ : P → P defined as ε ϕ(y) := y + z, y ∈ P, 2 is a homeomorphism of (P, τ ) onto (ϕ(P ), τ ), the set ϕ−1 (M ) is τ -meager. Since a nonempty open set in a Baire space is a Baire space again, we may select y ∈ V (x, 12 ε) \ (M ∪ ϕ−1 (M )). Then ε y + z ∈ V (x, ε) \ M, 2 and hence ε |f (y) − f (x)| < 1 and |f (y + z) − f (x)| ≤ 1. 2 By sublinearity, ε ε −2 ≤ f (y + z) − f (y) ≤ f (z). 2 2 4 Hence f (z) ≥ − ε . Since z with kzk ≤ 1 is arbitrary, we get f (en ) ≥ − 4ε , n ∈ N.

10.4 Embedding of `1

335

Lemma 10.36. Let f : X → R be a convex Borel function, where X := {λ ∈ P : kλk = 1} considered with the norm topology. Then f is lower bounded. Proof. Given f as in the hypothesis, we extend it on the whole P by setting ( 0, g(x) := x kxkf ( kxk ),

x = 0, x 6= 0,

x ∈ P.

x ) is a homeomorphism of P \{0} onto (0, ∞)×{λ ∈ Since the mapping x 7→ (kxk, kxk P : kλk = 1}, the function g is Borel on P . Moreover, g is sublinear. Hence the assertion follows from Lemma 10.35.

Theorem 10.37. Let f be a Borel convex function on a compact convex set X. Then f is lower bounded. Proof. Let f : X → R be as in the hypothesis. If we assume that f is not lower bounded, then there exist points xn ∈ X, n ∈ N, such that f (xn ) ≤ −n. Let Y := {λ ∈ P : kλk = 1} be considered with the w∗ -topology and ϕ : Y → X be defined as ∞ X ϕ(λ) := λn xn , λ ∈ Y. n=1

Then f ◦ ϕ is a convex Borel function on Y that is not lower bounded, a contradiction with Lemma 10.36. This concludes the proof.

10.4

Embedding of `1

Definition 10.38. If X is a set and {fn } a bounded sequence in `∞ (X), we say that {fn } is equivalent to `1 -basis (with constant C) if there exists C > 0 such that n

X

ci fi

`∞ (X)

≥C

i=1

n X

|ci |

i=1

for every finite sequence {ci }ni=1 of real numbers. Lemma 10.39. If {fn } is equivalent to `1 -basis, then span{fn : n ∈ N} is isomorphic to `1 via the mapping T : span{fn : n ∈ N} → `1 satisfying T fn = en (here {en } is the canonical basis of `1 ). Proof. The proof follows by a straightforward verification.

336


Lemma 10.40. Let X be a set, α < β and {fn } be a bounded sequence of functions on X that satisfies the following condition: for every pair M, N of finite disjoint sets in N there exists x ∈ X satisfying x∈

\

{y ∈ X : fi (y) < α} ∩

i∈M

\

{y ∈ X : fi (y) > β}.

i∈N

Then {fn } is equivalent to `1 -basis with constant 21 (β − α). P Proof. Let c1 , . . . , cn be real numbers. Assume first that (α + β) ni=1 ci ≥ 0. We set M := {i : ci ≥ 0} and N := {i : ci < 0} and find x ∈ X such that

Then

n X

fi (x) > β,

i ∈ M,

fi (x) < α,

i ∈ N.

ci fi (x) =

i=1

X

ci fi (x) +

i∈M

≥β

X

≥ Hence

n

X

ci fi

`∞ (X)

ci + α

If (α + β) cases we get

i=1 ci

X

ci

i∈N n

n

i=1 n X

i=1

β−αX β+αX ci + |ci | 2 2 β−α 2

≥

i=1

Pn

ci fi (x)

i∈N

i∈M

=

X

n X

|ci |.

i=1 n

ci fi (x) ≥

β−αX |ci |. 2 i=1

i=1

< 0, we use the previous estimate for the numbers −ci . In both n

X

ci fi

`∞ (X)

i=1

n

≥

β−αX |ci |, 2 i=1

which concludes the proof. Lemma 10.41. Let E be a Banach space, K ⊂ E ∗ bounded, f ∈ BE ∗∗ and α < β. Assume that for every w∗ -open set U ⊂ E ∗ intersecting K there exist x∗ , y ∗ ∈ ∗ cow (U ∩ K) so that f (x∗ ) > β and f (y ∗ ) < α. Then E contains an isomorphic copy of `1 .

10.4 Embedding of `1

337

Proof. Let K be as in the hypothesis. We construct inductively points xn ∈ BE , n ∈ N, so that \ \ K∩ {z ∗ ∈ K : z ∗ (xn ) > β} ∩ {z ∗ ∈ K : z ∗ (xn ) < α} 6= ∅ (10.6) n∈M

n∈N

for any pair M, N of disjoint nonempty finite subsets of N. ∗ In the first step, we find x∗ , y ∗ ∈ cow K such that f (x∗ ) > β and f (y ∗ ) < α. Let x1 ∈ BE approximates f pointwise on x∗ , y ∗ ; precisely we need that x∗ (x1 ) > β and y ∗ (x1 ) < α. ∗ Since x∗ , y ∗ ∈ cow K, both sets {z ∗ ∈ K : z ∗ (x1 ) > β}

and {z ∗ ∈ K : z ∗ (x1 ) < α}

are nonempty. Assume now that the construction has been completed up to the n-th stage, that is, we have x1 , . . . , xn ∈ BE such that the set \ \ {z ∗ ∈ K : z ∗ (xi ) < α} {z ∗ ∈ K : z ∗ (xi ) > β} ∩ UM,N := i∈N

i∈M

intersects K for each pair of disjoint nonempty finite sets M, N in {1, . . . , n}. ∗ ∈ K ∩ UM,N and use the assumption For every such pair we select a point zM,N w∗ ∗ ∗ ∗ to find xM,N , yM,N ∈ co (K ∩ UM,N ) such that f (x∗M,N ) > β and f (yM,N ) < α. We find xn+1 ∈ BE satisfying x∗M,N (xn+1 ) > β Then

∗ and yM,N (xn+1 ) < α.

K ∩ UM,N ∩ {z ∗ ∈ K : z ∗ (xn+1 ) > β} 6= ∅ and K ∩ UM,N ∩ {z ∗ ∈ K : z ∗ (xn+1 ) < α} 6= ∅.

It follows that K∩

\

{z ∗ ∈ K : z ∗ (xi ) > β} ∩

i∈M

\

{z ∗ ∈ K : z ∗ (xi ) < α} 6= ∅

i∈N

for each pair of disjoint nonempty subsets of {1, . . . , n + 1}. This finishes the construction. By (10.6), {xn |K } satisfies the assumptions of Lemma 10.40. If C > 0 satisfies K ⊂ CBE ∗ , for any numbers c1 , . . . , cn we get n n

X

X

C ci xi E ≥ ci xi |K i=1

i=1

`∞ (K)

n

≥

β−αX |ci |. 2 i=1

338


Thus

n

n

i=1

i=1

X

β−αX |ci | ≤ ci xi E , 2C which shows that the sequence {xn } is equivalent to `1 -basis and the proof is finished. Theorem 10.42. For a Banach space E the following assertions are equivalent: (i) E does not contain an isomorphic copy of `1 , (ii) f |BE∗ is universally measurable for any f ∈ E ∗∗ , (iii) for any f ∈ BE ∗∗ and µ ∈ M1 (BE ∗ ) there exists a sequence {xn } in BE such that xn → f on BE ∗ µ-almost everywhere, (iv) f |BE∗ satisfies the barycentric formula for any f ∈ E ∗∗ . Proof. For the proof of (i) =⇒ (iv), let f be an element of E ∗∗ . For a measure µ ∈ M1 (BE ∗ ) given, pick any numbers β > α. We set K := spt µ and find β 0 > α0 such that [α0 , β 0 ] ⊂ (α, β). By Lemma 10.41, there exists a w∗ -open set U intersecting K such that ∗

cow (U ∩ K) ⊂ {z ∗ ∈ K : f (z ∗ ) ≥ α0 } ⊂ {z ∗ ∈ K : f (z ∗ ) > α} w∗

co

∗

∗

0

∗

or

∗

(U ∩ K) ⊂ {z ∈ K : f (z ) ≤ β } ⊂ {z ∈ K : f (z ) < β}.

In both cases we get a w∗ -compact convex set L with strictly positive measure such that either f > α on L or f < β on L. Hence f satisfies the Haydon condition (Definition 4.13) and, by Theorem 4.19, f is strongly affine on BE ∗ . For the proof of (iv) =⇒ (iii) use Theorem 4.34. Since the implication (iii) =⇒ (ii) is obvious, we proceed to the last implication (ii) =⇒ (i). Assume that T : `1 → E is an isomorphic embedding satisfying kT −1 k ≤ 1. Then the dual operator T ∗ : E ∗ → `∞ is surjective and T ∗ (BE ∗ ) ⊃ B`∞ . Let g ∈ (`∞ )∗ be an element whose restriction to B`∞ is not universally measurable (see Lemma 10.28). Then f := g ◦ T ∗ is in E ∗∗ and it is not universally measurable by Corollary 5.27. This concludes the proof.

10.5

Metrizability of compact convex sets

In this section we will need notions and facts contained in Section A.4 and Subsection A.2.A. Definition 10.43 (Networks of topological spaces and distinguishable sets). For a topological space X, a family N of subsets of X is a network if for any open set U ⊂ X we have [ U = {N ∈ N : N ⊂ U }.

10.5 Metrizability of compact convex sets

339

A subset A of a completely regular space X is called distinguishable in X if there exists a separable metric space M and a continuous mapping f : X → M such that A = f −1 (f (A)). A completely regular space X is called distinguishable if X is distinguishable in ˇ its Cech–Stone compactification. Lemma 10.44. Let X be a completely regular space and ∆ := {(x, x) : x ∈ X} ⊂ X × X be the diagonal of X × X. Consider the following assertions: (i) (X × X) \ ∆ is Lindelöf, (ii) there exists a countable family of continuous functions on X separating points of X, (iii) ∆ is a Gδ set in X × X, Then (i) =⇒ (ii) =⇒ (iii). Proof. Let X and ∆ be as in the statement and assume (i). For each x, y ∈ X, x 6= y, let fx,y be a continuous function on X such that fx,y (x) 6= fx,y (y). Then −1 −1 {fx,y (U ) × fx,y (V ) : (x, y) ∈ (X × X) \ ∆, U, V ⊂ R disjoint and open}

is a family of open subsets of (X ×X)\∆ that covers (X ×X)\∆. By the assumption, there exist countably many pairs (xn , yn ) ∈ (X × X) \ ∆ and open sets Un , Vn ⊂ R, n ∈ N, such that Un ∩ Vn = ∅ and (X × X) \ ∆ ⊂

∞ [

fx−1 (Un ) × fx−1 (Vn ) . n ,yn n ,yn

n=1

Then the family {fxn ,yn : n ∈ N} separates points of X. This proves (i) =⇒ (ii). For the proof of (ii) =⇒ (iii), let {fn : n ∈ N} be a countable family of continuous functions on X separating points of X. Then [ (X × X) \ ∆ = {fn−1 (U ) × fn−1 (V ) : U, V ⊂ R are disjoint closed intervals with rational endpoints} is an Fσ set. Thus ∆ is Gδ and the proof is complete. Lemma 10.45. Let X be a compact space and ∆ = {(x, x) : x ∈ X} ⊂ X × X be the diagonal of X × X. Then the following assertions are equivalent. (i) X is metrizable, (ii) (X × X) \ ∆ is Lindelöf, (iii) there exists a countable family of continuous functions on X separating points of X, (iv) ∆ is a Gδ set in X × X.

340


Proof. Obviously, (i) =⇒ (ii). By Lemma 10.44, (ii) =⇒ (iii) =⇒ (iv). Assume that ∆ is a Gδ subset of X × X. Then (X × X) \ ∆, as an Fσ subset of the Lindelöf space X × X, is Lindelöf. By Lemma 10.44, there exists a countable family {fn : n ∈ N} in C(X) separating points of X. We assume that kfn k ≤ 1, n ∈ N, and define ∞ X 1 |fn (x) − fn (y)|, x, y ∈ X. ρ(x, y) := 2n n=1

Then ρ is a continuous metric on X × X and thus ρ generates a weaker topology on X than the original one. By Proposition A.28, these topologies coincide. Lemma 10.46. Let X be a completely regular space. (a) A set A ⊂ X is distinguishable in X if and only if there exists a countable family {fn : n ∈ N} of continuous functions on X such that for every x ∈ A, y ∈ X \ A there exists n ∈ N such that fn (x) 6= fn (y). (b) The family of all distinguishable subsets of X is a σ-algebra. (c) The space X is distinguishable if and only if X is distinguishable in some compact space. (d) If X is distinguishable, then X is K-countably determined. (e) If X is compact and A ⊂ X Baire, then A is distinguishable in X. Proof. For the proof of (a), let f : X → M be a continuous mapping to a metric separable space M such that A = f −1 (f (A)). Let {gn : n ∈ N} be a countable family of continuous functions separating points of M . Then the family {gn ◦ f : n ∈ N} satisfies our requirement. Conversely, let {fn : n ∈ N} be as in (a). Then f : X → RN defined as f (x) = {fn (x)}, x ∈ X, is the required mapping with M = f (X). As the verification of (b) is routine, we proceed to the proof of (c). Y Assume that X is distinguishable in a compact space Y . By replacing Y with X if necessary we may assume that Y is a compactification of X. Let M be a separable metric space and f : Y → M be a continuous mapping ˇ such that X = f −1 (f (X)). Let βX be the Cech–Stone compactification of X and g : βX → Y be the continuous mapping which equals identity on X (see Theorem A.36). It is easy to verify that g ◦ f : βX → M distinguishes X in βX. To check (d), assume that X is distinguishable in βX. Let M be a separable metric space and f : βX → M be continuous such that X = f −1 (f (X)). Since f (βX) is compact, X = ϕ(f (X)), where ϕ : f (βX) → 2βX is defined as ϕ(y) := f −1 (y),

y ∈ βX.

Obviously, ϕ is a usco mapping. Hence X is K-countably determined.


341

Finally, let A be a Baire subset of a compact space X. Let fn ∈ C(X), n ∈ N, be such that A is contained in the σ-algebra generated by {fn−1 (U ) : U ⊂ R open} (see Proposition A.48). Then {fn : n ∈ N} separates A from X \ A and A is distinguishable in X. Lemma 10.47. Let X, Y be completely regular spaces and ϕ : X → 2Y be a usco mapping. Then (a) graph of ϕ (that is, gr ϕ := {(x, y) ∈ X × Y : y ∈ ϕ(x)}) is a closed subset of X ×Y, (b) the projection πX : gr ϕ → X is a perfect mapping, (c) the mapping ψ : X → 2X×Y , x 7→ {x} × ϕ(x), is a usco mapping of X onto gr ϕ. Proof. For the proof of (a), let (x0 , y0 ) be a point of (X × Y ) \ gr ϕ. As ϕ(x0 ) is compact, we can find a neighborhood V of y0 such that V ∩ ϕ(x0 ) = ∅. By the upper semicontinuity of ϕ, U := {x ∈ X : ϕ(x) ⊂ Y \ V } is an open subset of X and x0 ∈ U . Thus U × V is an open neighborhood of (x0 , y0 ) disjoint from gr ϕ. −1 (x) = {x} × ϕ(x) is To verify (b) we first notice that πX is continuous and πX compact in X × Y for each x ∈ X. Let F ⊂ gr ϕ be a closed set. Then U := (X × Y ) \ F is an open subset of X × Y . −1 −1 Let x ∈ X satisfy πX (x) ⊂ U . By the compactness of πX (x), there exist open sets G1 ⊂ X and G2 ⊂ Y such that −1 πX (x) ⊂ G1 × G2 ⊂ U.

Since −1 {x ∈ X : πX (x) ⊂ G1 × G2 } = {x ∈ X : ϕ(x) ⊂ G2 } ∩ G1

is an open subset of X, −1 (x) ⊂ U } πX (F ) = πX ((X × Y ) \ U ) = X \ {x ∈ X : πX

is a closed set in X. Hence πX is perfect. As (c) is an immediate consequence of (b), the proof is finished. Lemma 10.48. Let X be a regular space. Then (a) {N : N ∈ N } is a network for each network N , (b) X is determined if and only if X admits a countable network.

342


Proof. Let N be a network for a regular space X. Since for each x ∈ X and open set U containing x there exists an open set V such that x ∈ V ⊂ V ⊂ U , assertion (a) follows. For the proof of (b), we start with the assumption that X is determined. Let f : Y → X be a continuous surjection from a separable metric space Y . If B is a countable base of open sets in Y , then it easily follows that {f (B) : B ∈ B} is a countable network for X. Assume now that X has a countable network N = {Mn : n ∈ N}. Let {0, 1}N be the Cantor set identified with the family of all subsets of N as in Subsection 10.3.A. We set \ \ Mic , σ ∈ {0, 1}N . Nσ := Mi ∩ i∈σ

i∈σ /

We claim that Nσ is either empty or a singleton for each σ ∈ {0, 1}N . Indeed, let σ ∈ {0, 1}N and x, y ∈ X be distinct points. We find a neighborhood V of x such that y ∈ / V and n ∈ N such that x ∈ Mn ⊂ V . Then, x ∈ / Nσ if σn = 0, whereas y∈ / Nσ if σn = 1. Hence Nσ cannot contain both points x, y. We set Y := {σ ∈ {0, 1}N : Nσ 6= ∅}. and define f : Y → X as the mapping determined by the property f (σ) ∈ Nσ . Obviously, Y is a separable metric space and f is onto. To finish the proof we have to verify the continuity of f . Let σ ∈ Y and an open neighborhood U of f (σ) be given. We find an index n ∈ N such that f (σ) ∈ Mn ⊂ U. Then the set U := {τ ∈ {0, 1}N : τn = 1} is a neighborhood of σ and f (U ) ⊂ V . Thus f is continuous at σ and the proof is complete. Lemma 10.49. A determined compact space is metrizable. Proof. Let X be a determined compact space. By Lemma 10.48, there exists a countable network N for X consisting of closed sets. For any pair N, N 0 ∈ N of disjoint sets we choose a continuous function f(N,N 0 ) ∈ C(X) such that f(N,N 0 )

( 0 = 1

on N, on N 0 .


343

Then {f(N,N 0 ) : N, N 0 ∈ N , N ∩ N 0 = ∅} is a countable family of continuous functions separating points of X. Thus X is metrizable by Lemma 10.45. Definition 10.50 (Hausdorff metric). Let X be a compact space with a compatible metric ρ that is bounded by 1. For a pair of sets A, B ⊂ X, let δ(A, B) := sup{dist(x, B) : x ∈ A}. Let K(X) denote the family of all compact subsets of X endowed with the following metric:   if A = B = ∅, 0 dH (A, B) = 1 if exactly one of A, B is empty,   δ(A, B) ∨ δ(B, A) if both A, B are nonempty. By Chapter II, Section 4.F in A.S. Kechris [262], dH is a metric on K(X) that turns it into a compact metric space. Moreover, it is compatible with the so-called Vietoris topology. Theorem 10.51. Let X be a compact convex set X with a determined boundary B. Then X is metrizable. Proof. Let N be a countable network of B (see Lemma 10.48) and let ε ∈ (0, 1) be fixed. For each finite family F ⊂ N , we set AF := {f ∈ BAc (X) : diam f (F ) ≤ ε, F ∈ F}. Let K([−1, 1]) denote the space of all compact subsets of [−1, 1] endowed with the Hausdorff metric dH (see Definition 10.50) and, for n ∈ N, let (K([−1, 1]))n be the product space of n copies of K([−1, 1]) with the maximum metric. If F = {F1 , . . . , Fn }, we define a mapping ϕF : AF → (K([−1, 1]))n as ϕF (f ) = (f (F1 ), . . . , f (Fn )),

f ∈ AF .

Since (K([−1, 1]))n is a compact metric space, ϕF (AF ) is its separable subspace. We select a countable set DF ⊂ AF such that ϕF (DF ) is dense in ϕF (AF ). Let [ D := {DF : F ⊂ N finite}. Step 1: For each f ∈ BAc (X) there exists a sequence {fn } of functions from D such that lim sup |fn (x) − f (x)| ≤ 2ε, x ∈ B. n→∞

344


Given a function f ∈ BAc (X) , we find a finite open cover U of X such that diam f (U ) ≤ ε for each U ∈ U. We set N 0 := {N ∈ N : there exists U ∈ U such that N ⊂ U }. Let {F1 , F2 , . . . } be an enumeration of N 0 . We notice that diam f (Fi ) ≤ ε for each i ∈ N. By setting F n := {F1 , . . . , Fn },

n ∈ N,

we get f ∈ AF n for each n ∈ N. Thus for each n ∈ N we may use density of DF n in ϕF n (AF n ) to choose a function fn ∈ DF n such that dH (f (Fi ), fn (Fi )) < ε,

i = 1, . . . , n.

(10.7)

Let x ∈ B be an arbitrary point. We find a natural number n ∈ N such that x ∈ Fn . Let k ∈ N with n ≤ k be given. By (10.7) and the definition of the Hausdorff metric on K([−1, 1]), there exists y ∈ Fn such that |f (x) − fk (y)| < ε. Since diameters of f (Fn ) and fk (Fn ) are smaller or equal than ε, |fk (x) − f (x)| ≤ |fk (x) − fk (y)| + |fk (y) − f (x)| ≤ 2ε. Hence lim supn→∞ |fn (x) − f (x)| ≤ 2ε for each x ∈ B, concluding the proof. Step 2: For each f ∈ BAc (X) there existP a sequence {fn } of functions from D and a sequence {λn } of positive numbers with ∞ n=1 λn = 1 such that ∞ X

λn |fn (x) − f (x)| ≤ 3ε,

x ∈ X.

n=1

Given f ∈ BAc (X) , by Step 1 there exists a sequence {fn } in D such that −2ε ≤ lim inf(fn (x) − f (x)) ≤ lim sup(fn (x) − f (x)) ≤ 2ε, n→∞

x ∈ B.

n→∞

Corollary 3.75 yields −2ε ≤ lim inf(fn (x) − f (x)) ≤ lim sup(fn (x) − f (x)) ≤ 2ε, n→∞

x ∈ X.

n→∞

In other words, lim sup |fn (x) − f (x)| ≤ 2ε,

x ∈ X.

n→∞

A usePof the Simons inequality 3.74 provides a sequence {λn } of positive numbers with ∞ n=1 λn = 1 such that ∞ X n=1

λn |fn (x) − f (x)| ≤ 3ε,

x ∈ X.


345

Step 3: For each f ∈ BAc (X) there Pnexist a finite family {f1 , . . . , fn } ⊂ D and a positive numbers {c1 , . . . , cn } with k=1 ck = 1 such that n X

ck |fk (x) − f (x)| ≤ 6ε,

x ∈ X.

k=1

If f ∈ BAc (X) , let {fk } and {λk } be as in Step 2. We find n ∈ N such that P −1 < 1 + ε and set λ := ∞ k=n λk satisfies (1 − λ) ck := (1 − λ)−1 λk ,

k = 1, . . . , n.

Then n X

ck |fk (x) − f (x)| = (1 − λ)−1

k=1

n X

λk |fk (x) − f (x)| ≤ 3ε(1 + ε) ≤ 6ε,

x ∈ X,

k=1

and Step 3 is proved. It follows that for every ε ∈ (0, 1) there exists a countable family Dε ⊂ BAc (X) such that BAc (X) ⊂ co Dε + 6εBAc (X) . This shows that BAc (X) is separable. Since Ac (X) separates points of X, X is metrizable by Lemma 10.45 and the proof is finished. Lemma 10.52. Let X be completely regular space such that • a countable set of continuous functions separates points of X, and X is distinguishable. Then X is separable metrizable. •

Proof. Let {gn : n ∈ N} be a family of continuous functions on βX that distinguishes X in βX. Let {fn : n ∈ N} be a family of continuous functions on X that separates points of X. We may assume that all functions fn are bounded. We extend the functions fn (and denote them likewise) to the whole βX. Let B be a countable base of open sets in R and let U := {fn−1 (B) : B ∈ B} ∪ {gn−1 (B) : B ∈ B}. We claim that {U ∩ X : U ∈ U} is a subbase of the topology of X. Indeed, let V ⊂ βX be an open neighborhood of x ∈ X. For each y ∈ βX we select sets Uy , Uy0 ∈ U such that x ∈ Uy , y ∈ Uy0 and Uy ∩ Uy0 = ∅. By the compactness of βX \ V there exist y1 , . . . , yn in βX \ V such that βX \ V ⊂

n [ i=1

Uy0 i .

346


Thus x∈

n \

Uyi ⊂ V,

i=1

and {U ∩ X : U ∈ U} is a subbase of X. As any regular space with a countable base is metrizable (see R. Engelking [169], Theorem 4.4.7) and clearly separable, the proof is finished.

Lemma 10.53. Let X be completely regular space. Then the following assertions are equivalent. (i) X is determined, (ii) X is K-countably determined and admits a countable set of continuous functions separating points of X. Proof. Assume first that X is determined, that is, there exist a separable metric space M and a continuous surjective mapping f : M → X. Then f ×f : M ×M → X ×X defined as f (m, n) := (f (m), f (n)),

(m, n) ∈ M × M,

is a continuous surjection. Thus (X × X) \ ∆ = (f × f ) (f × f )−1 ((X × X) \ ∆) , as a continuous image of a Lindelöf space, is Lindelöf. By Lemma 10.44, X admits a countable family of continuous functions separating points of X. Conversely, let {fn : n ∈ N} be a family of continuous functions separating points of X and let ϕ : M → 2X be a usco mapping of a separable metric space M onto X. Let πM : M ×X → M and πX : M ×X → X denotes the canonical projections. We select a countable family {gn : n ∈ N} of continuous functions on M that separates points of M . Then the family {gn ◦ πM : n ∈ N} ∪ {fn ◦ πX : n ∈ N} separates points of gr ϕ.

(10.8)


347

We claim that gr ϕ is distinguishable in β(gr ϕ). Indeed, we find a metric compactc of M and let ification M c π bM : β(gr ϕ) → M be the continuous extension of πM . To finish the reasoning of the claim we have to prove that gr ϕ = (b πM )−1 (b πM (gr ϕ)). (10.9) By Lemma 10.47(b), πM : gr ϕ → M is a perfect mapping. By Theorem A.37, c \ M. π bM (β(gr ϕ) \ gr ϕ) ⊂ M Hence (10.9) follows. Thus (10.8) along with (10.9) allows to use Lemma 10.52 and conclude that gr ϕ is a separable metrizable space. Hence X = πX (gr ϕ) is determined. Lemma 10.54. Let F be a countable family of bounded continuous functions on a topological space X. Then there exists a norm separable subspace G ⊂ C b (X) so that F ⊂ G, G contains constant functions and G is a lattice (with respect to the pointwise ordering). Proof. Given F as in the statement of the lemma, we may assume without loss of generality that 1 ∈ F. Then G := W(span F) − W(span F) is the desired space (recall that W(H) is the smallest min-stable cone generated by H; see Definition 3.10). Lemma 10.55. Let ϕ : X → Y be a continuous affine surjection of a compact convex set X onto a compact convex set Y . Let F 1 ⊂ C b (ext X) and G ⊂ Kc (Y ) satisfy the following property: for every f ∈ F 1 there exist bounded sequences {un }, {vn } in G such that sup{(un ◦ ϕ)|ext X : n ∈ N} = f = inf{(−vn ◦ ϕ)|ext X : n ∈ N}. Then there exists a family F 2 ⊂ C b (ext Y ) so that f1 ∈ F 1 if and only if there exists f2 ∈ F 2 satisfying f1 = f2 ◦ ϕ on ϕ−1 (ext Y ) ∩ ext X. Proof. Given F 1 and G as in the hypothesis, for each f1 ∈ F 1 we find sequences {un }, {vn } according to the assumption and set f2 (y) := sup un (y),

y ∈ ext Y.

n∈N

Since ext Y ⊂ ϕ(ext X) and sup un (ϕ(x)) = inf −vn (ϕ(x)), n∈N

n∈N

x ∈ ext X,

348


the function f2 satisfies f2 = sup un |ext Y = inf −vn |ext Y . n∈N

n∈N

Thus f2 is both upper and lower semicontinuous, and hence continuous on ext Y . Obviously, f2 is bounded. If x ∈ ext X ∩ ϕ−1 (ext Y ), then f2 (ϕ(x)) = sup un (ϕ(x)) = f1 (x). n∈N

Hence F 2 consisting of all such functions constructed for each f1 ∈ F 1 is the required family. Theorem 10.56. Let X be a compact convex set. Then the following assertions are equivalent. (i) X admits a continuous strictly convex function, (ii) ext X is a Baire subset of X, (iii) ext X is distinguishable in X, (iv) ext X is K-countably determined and countably many continuous functions on ext X separate points, (v) ext X is Lindelöf and countable many continuous functions on ext X separate points, (vi) X is metrizable. Proof. We will prove (i) ⇐⇒ (vi) =⇒ (ii) =⇒ (iii) =⇒ (iv) =⇒ (v) =⇒ (iii) and (iv) =⇒ (vi). For the proof of (i) =⇒ (vi), let f be a strictly convex continuous function on X. Then ϕ : X × X → [0, ∞) defined as 1 x + y , ϕ(x, y) := (f (x) + f (y)) − f 2 2

(x, y) ∈ X × X,

is a positive continuous function and ∞ n \ 1o ∆= (x, y) ∈ X × X : ϕ(x, y) < . n n=1

Hence ∆ is a Gδ subset of X × X and X is metrizable by Lemma 10.45. The implication (vi) =⇒ (i) follows from Theorem 3.42. Further, (vi) =⇒ (ii) follows from Theorem 3.42 and Proposition 3.43. If ext X is a Baire subset of X, then ext X is distinguishable in X by Lemma 10.46(e). Hence (ii) =⇒ (iii).


349

Assume now that (iii) is satisfied. By Lemma 10.46, ext X is a K-countably determined space. Define ϕ : X × X → X by the formula 1 ϕ(x, y) := (x + y), 2

(x, y) ∈ X × X.

It is easy to see that both ϕ−1 (ext X) and

ext X × ext X

are distinguishable in X × X. By Lemma 10.46(b), (ext X × ext X) \ ∆ = (ext X × ext X) ∩ ((X × X) \ ϕ−1 (ext X)) is a distinguishable set in X × X. It follows from Lemma 10.46(d) and Theorem A.111(e) that (ext X × ext X) \ ∆ is Lindelöf. By Lemma 10.44, ext X admits a countable family of continuous functions separating points of ext X. Thus (iv) holds. For the proof of (iv) =⇒ (v), we just use Theorem A.111(e) to conclude that ext X is Lindelöf. To close the chain of implications we have to prove (v) =⇒ (iii). To this end, assume that ext X is Lindelöf and F 1 is a countable family of continuous functions on ext X that separates points of ext X. Obviously, we may assume that all functions in F 1 are bounded. Using Lemma 10.54 we enlarge F 1 (if necessary) to get a norm separable subspace of C b (ext X) that contains constant functions and is a lattice with respect to the pointwise ordering. Let F 2 ⊂ F 1 be a countable norm dense set in F 1 . Given f ∈ F 2 , ext X f ∗ = f on ext X by Corollary 3.23(c). Since ext X is Lindelöf, for each f ∈ F 2 , according to Lemma A.54, there exist bounded sequences {fk }, {gk } in Kc (X) such that sup fk |ext X = f = inf (−gk |ext X ). k∈N

k∈N

Let G 1 consist of the elements of all these sequences for f running through F 2 . Claim. For every x1 ∈ ext X and x2 ∈ X \ ext X there exists g ∈ G 1 so that g(x1 ) 6= g(x2 ). Proof of the claim. We assume that g(x1 ) = g(x2 ) for each g ∈ G 1 . We find a Baire set A ⊂ X with x1 ∈ A ⊂ X \ {x2 } and a maximal measure µ ∈ Mx2 (X). We use Theorem 9.12, for maximal measures εx1 , µ, a Baire set A and a family G 1 , to get a continuous affine surjection ϕ : X → Y of X onto a metrizable compact convex set Y and a set G 2 ⊂ C(Y ) such that (a) ϕ] εx1 and ϕ] µ are maximal on Y ,

350


(b) ϕ−1 (ϕ(A)) = A, (c) for each g1 ∈ G 1 there exists g2 ∈ G 2 such that g1 = g2 ◦ ϕ. We denote yi := ϕ(xi ), i = 1, 2. By (b), y1 6= y2 and by (a), y1 ∈ ext Y . We claim that families F 1 , G 2 satisfy the assumptions of Lemma 10.55. Indeed, if f ∈ F 2 , we get the required sequences from the choice of G 1 and property (c). Since F 2 is norm dense in F 1 , the claim easily follows. Thus we may use Lemma 10.55 to get a family F 3 ⊂ C b (ext Y ) so that f1 ∈ F 1 if and only if there exists f3 ∈ F 3 satisfying f1 = f3 ◦ ϕ on ϕ−1 (ext Y ) ∩ ext X. Since F 1 separates points of ext X, F 3 separates points of ext Y . Further, F 3 is a subspace of C b (ext Y ) that contains constant functions and is, moreover, a lattice. We define B := {f ∈ C b (ext Y ) : ϕ] µ(f ) = f (y1 )}. Then F 3 ⊂ B. To see this, let f3 ∈ F 3 be given. We find f1 ∈ F 1 so that f1 = f3 ◦ ϕ on −1 ϕ (ext Y ) ∩ ext X. By the definition of G 1 , there exists a bounded sequence {un } in G 1 such that supn∈N un = f1 on ext X. Using (c), we select a sequence {b un } in G 2 with un = u bn ◦ ϕ, n ∈ N. Then (ϕ] µ)(f3 ) ≥ (ϕ] µ)(sup u bn ) ≥ sup (ϕ] µ)(b un ) n∈N

n∈N

= sup µ(un ) ≥ sup un (x2 ) n∈N

n∈N

= sup un (x1 ) = f1 (x1 ) n∈N

= f3 (y1 ). Analogously we get (ϕ] µ)(f3 ) ≤ f3 (y1 ). Since ϕ] µ is maximal, it is carried by ext Y . HenceSwe may find an increasing sequence {Kn } of compact sets in ext Y such that (ϕ] µ)( ∞ n=1 Kn ) = 1 and y1 ∈ K1 . Now we may use the Stone–Weierstrass theorem for the restriction of B to each Kn to conclude that B = C b (ext Y ). Indeed, let f ∈ C b (ext Y ) and ε > 0 be given. We find Kn so that (ϕ] µ)(Kn ) > 1 − ε and a function g ∈ F 3 so that sup |f (y) − g(y)| < ε. y∈Kn

Since F 3 is a lattice containing constant functions, we may assume that kgk ≤ kf k. Then Z |(ϕ] µ)(f ) − f (y1 )| ≤ |f (t) − g(t)| d(ϕ] µ)(t) Kn

+ 2kf k(ϕ] µ)(ext Y \ Kn ) + |(ϕ] µ)(g) − g(y1 )| + |g(y1 ) − f (y1 )| ≤ ε(2 + 2kf k).

10.6 Continuous affine images

351

Thus ϕ] µ = εy1 . But this is impossible, because ϕ] µ ∈ My2 (Y ) and y1 6= y2 . This concludes the proof of the claim. The claim yields that ext X is distinguishable in ext X by means of functions from G 1 . This concludes the proof of (v) =⇒ (iii). To finish the proof we verify (iv) =⇒ (vi). If (iv) holds, Lemma 10.52 yields that ext X is a separable metrizable space. By Theorem 10.51, X is metrizable.

10.6

Continuous affine images

In this section we study the following problem. Let ϕ : X → Y be a continuous affine mapping of a compact convex set X to a compact convex set Y . 1 • If ϕ(ext X) ⊂ ext Y , does it imply that ϕ (M1 ] max (X)) ⊂ Mmax (Y )? •

If ϕ(ext X) ⊂ ext Y and ϕ is injective on ext X, does it imply that ϕ] is injective on M1max (X)?

Theorem 10.57. Let ϕ : X → Y be a continuous affine mapping of a compact convex set X to a compact convex set Y and let ext Y be a Lindelöf space. (a) Then the following assertions are equivalent: (i) ϕ(ext X) ⊂ ext Y , (ii) ϕ] (M1max (X)) ⊂ M1max (Y ). (b) Further, the following assertions are equivalent: (i’) ϕ(ext X) ⊂ ext Y and ϕ is injective on ext X, (ii’) ϕ] (M1max (X)) ⊂ M1max (Y ) and ϕ] is injective on M1max (X). Proof. We first notice that the implications (ii) =⇒ (i) and (ii’) =⇒ (i’) are obvious. We start the proof of the converse implications by showing (i) =⇒ (ii). To this end, let µ ∈ M1max (X) be given. We fix an arbitrary closed set F ⊂ Y \ ext Y . Since ext Y is Lindelöf, there exists a countable family of cozero sets {Un : n ∈ N} in Y such that ∞ [ ext Y ⊂ Un ⊂ Y \ F. n=1

S∞

ϕ−1 (

Then G := µ(G) = 1. Thus

n=1 Un )

is an Fσ set. By the assumptions, ext X ⊂ G and hence (ϕ] µ)

∞ [

Un = µ(G) = 1,

n=1

and hence µ(F ) = 0. Thus (ϕ] µ)∗ (Y \ ext Y ) = 0, and ϕ] µ ∈ M1max (Y ) by virtue of Proposition 3.80.

352


We proceed with the proof of (i’) =⇒ (ii’). We start by proving ϕ(X \ ext X) ⊂ Y \ ext Y.

(10.10)

Indeed, given y ∈ ext Y ∩ ϕ(X), the set ϕ−1 (y) is a closed face. Since ϕ−1 (y) = co(ext ϕ−1 (y)) = co(ϕ−1 (y) ∩ ext X), the assumption yields that ϕ−1 (y) is a singleton. Hence (10.10) follows. Let µ ∈ M1max (X) be given. For any set F ⊂ X \ ext X, (10.10) gives ϕ(F ) ⊂ Y \ ext Y. This along with Proposition 3.80 and the first part of the proof yields (ϕ] µ)(ϕ(F )) = 0,

F ⊂ X \ ext X, F closed.

Hence µ(F ) ≤ µ(ϕ−1 (ϕ(F ))) = (ϕ] µ)(ϕ(F )) = 0,

F ⊂ X \ ext X, F closed,

and thus µ(ϕ−1 (ϕ(F ))) = µ(F ),

F ⊂ X, F closed.

(10.11)

If µ, ν ∈ M1max (X) are measures with ϕ] µ = ϕ] ν, then (10.11) yields µ(F ) = µ(ϕ−1 (ϕ(F ))) = (ϕ] µ)(ϕ(F )) = (ϕ] ν)(ϕ(F )) = ν(ϕ−1 (ϕ(F ))) = ν(F ) for any closed F ⊂ X. Hence µ = ν and ϕ] is injective on M1max (X). Remark 10.58. It can be easily verified that the mapping ϕ : X → Y is a homeomorphism of ext X onto ϕ(ext X) if ϕ(ext X) ⊂ ext Y and ϕ is injective on ext X. Indeed, since ϕ(ext X) ⊂ ext Y

and

ϕ(X \ ext X) ⊂ Y \ ext Y,

it is not difficult to realize that ϕ(F ∩ ext X) = ϕ(F ) ∩ ext Y for any F ⊂ X. Hence ϕ : ext X → ext Y is a closed mapping (that is, ϕ maps closed sets to closed sets), and thus a homeomorphism on ext X. Hence we obtain that ext X is a Lindelöf space if ext Y is Lindelöf and ϕ is a surjective mapping with the properties as above. Theorem 10.59. Let ϕ : X → Y be a continuous affine mapping of a compact convex set X to a simplex Y . (a) Then the following assertions are equivalent:


353

(i) ϕ(ext X) ⊂ ext Y , (ii) ϕ] (M1max (X)) ⊂ M1max (Y ), (iii) ϕ(F ) is a face for each closed face F ⊂ X, (iv) ϕ(F ) is a closed extremal set for each closed extremal F ⊂ X. (b) Further, the following assertions are equivalent: (i’) ϕ(ext X) ⊂ ext Y and ϕ is injective on ext X, (ii’) ϕ] (M1max (X)) ⊂ M1max (Y ) and ϕ] is injective on M1max (X), (iii’) ϕ is a homeomorphism onto ϕ(X). Proof. For the proof of (a), we first verify (i) =⇒ (ii). To this end, let µ be a maximal probability measure on X. To show that ϕ] µ is maximal on Y , we use Mokobodzki’s maximality test 3.58. Let g be a continuous convex function on Y . Since Y is a simplex, g ∗ is an affine function (see Theorem 6.5). By the assumption , g ∗ ◦ ϕ = (g ◦ ϕ)∗

on ext X.

By Proposition 3.88, g ∗ ◦ ϕ ≤ (g ◦ ϕ)∗ on X. On the other hand, given x ∈ X, there exists a measure λ ∈ Mx (X) such that λ(g ◦ ϕ) = (g ◦ ϕ)∗ (x) (see Lemma 3.22). Then ϕ] λ ∈ Mϕ(x) (Y ) and (g ◦ ϕ)∗ (x) = λ(g ◦ ϕ) = (ϕ] λ)(g) ≤ g ∗ (ϕ(x)). Hence g ∗ ◦ ϕ = (g ◦ ϕ)∗ on X. Thus the equality (ϕ] µ)(g) = µ(g ◦ ϕ) = µ((g ◦ ϕ)∗ ) = µ(g ∗ ◦ ϕ) = (ϕ] µ)(g ∗ ) shows that ϕ] µ is a maximal measure on Y . We proceed with the proof by showing (ii) =⇒ (iii). Let F ⊂ X be a closed face. Since ϕ(F ) is obviously convex, we need to check its extremality. Let ν ∈ M1max (Y ) satisfy r(ν) ∈ ϕ(F ). We find a point x ∈ F with ϕ(x) = r(ν) and select a measure µ ∈ M1max (X) such that r(µ) = x. Since F is a closed face, µ ∈ M1 (F ). Then ϕ] µ is carried by ϕ(F ) and by the assumption, ϕ] µ is maximal. Since r(ϕ] µ) = ϕ(r(µ)) = r(ν) and Y is a simplex, ϕ] µ = ν. Hence ν ∈ M1 (ϕ(F )). Since a closed set is extremal if and only if it is a union of closed faces (see Proposition 2.69), we get (iii) =⇒ (iv). We proceed with the proof of (iv) =⇒ (i). But this is immediate, because a set {x} is extremal if and only if x ∈ ext X. This concludes the proof of (a).

354


We start the proof of (b) by showing (i’) =⇒ (iii’). We know from (a) that ϕ(X) is a face of Y and hence a simplex. Since ext ϕ(X) = ϕ(X) ∩ ext Y , we may assume from now on that ϕ is a surjective mapping onto a simplex Y . By the assumption, ϕ(ext X) = ext Y

and

ϕ is injective on ext X.

Step 1: For any f ∈ Ac (X), the function g(y) := inf{f (x) : x ∈ ϕ−1 (y)},

y ∈ Y,

is lower semicontinuous and affine. It is easy to verify that g is lower semicontinuous and convex. Assume that g is not affine, that is, that there exist y1 , y2 ∈ Y , λ ∈ (0, 1) and ε > 0 such that λg(y1 ) + (1 − λ)g(y2 ) − g(λy1 + (1 − λ)y2 ) > ε > 0. By Proposition 4.9, we can find functions h1 , h2 ∈ Ac (Y ) so that h(yi ) − hi (yi ) < 31 ε and hi < g, i = 1, 2. Let h := h1 ∨ h2 and y := λy1 + (1 − λ)y2 . Since Y is a simplex, h∗ is affine (see Theorem 6.5) and hence the function f − ∗ h ◦ ϕ is lower semicontinuous affine and f − h∗ ◦ ϕ > 0 on ext X (since h < g and h∗ = h on ext Y ). By the Minimum principle 2.24, f − h∗ ◦ ϕ > 0 on X. On the other hand, 1 h∗ (y) > λh1 (y1 ) + (1 − λ)h2 (y2 ) > λg(y1 ) + (1 − λ)g(y2 ) − ε > g(y). 3 Thus there exists x ∈ ϕ−1 (y) such that f (x) < h∗ (y). Hence f (x) − h∗ (ϕ(x)) < 0, a contradiction. Step 2: The mapping ϕ is injective. Suppose that there exists a point y ∈ Y such that the fiber ϕ−1 (y) contains two different points, say x1 and x2 . Let f ∈ Ac (X) satisfy f (x1 ) > f (x2 ). Let g : Y → R be defined as above and let h := g ◦ ϕ − f . By the first step, h is upper semicontinuous and affine. Further, h = 0 on ext X by the assumption. By the Minimum principles 2.24 and 3.85, h = 0 on X. But h(x1 ) < 0, a contradiction. This finishes the proof of injectivity of ϕ, and thus ϕ is a homeomorphism. Since the remaining implications are obvious, the proof is finished. Theorem 10.60. Let F ⊂ H be function spaces on a compact space K, F be simplicial and Ac (H) = H. Then the following assertions are equivalent: (i) Ac (F) = H, (ii) ChF (K) = ChH (K).


355

Proof. If Ac (F) = H, Exercise 3.95 yields ChF (K) = ChAc (F ) (K) = ChH (K). Hence (i) =⇒ (ii). To prove (ii) =⇒ (i), suppose that ChF (K) = ChH (K). We know that Ac (F) ⊂ c A (H) = H. Let φ1 : K → S(H) and φ2 : K → S(Ac (F)) be the homeomorphic embedding from Definition 4.25. If ϕ : S(H) → S(Ac (F)) is the restriction mapping, then φ2 = ϕ ◦ φ1 . (10.12) By the assumption and Proposition 4.26(d), ϕ(ext S(H)) = ext S(Ac (F)) and ϕ is injective on ext S(H) by (10.12). Since S(Ac (F)) is a simplex (see Theorem 6.54), Theorem 10.59(b) implies that ϕ is a surjective homeomorphism. Hence Ac (F) = H, which finishes the proof. Notation 10.61. Let ϕ : X → Y be a continuous affine surjection of a compact convex set X onto a compact convex set Y . If f ∈ C(X), we denote as fϕ the function on Y defined as fϕ (y) := sup{f (x) : x ∈ ϕ−1 (y)},

y ∈ Y.

Lemma 10.62. Let ϕ be a continuous affine surjection of a compact convex set X onto a compact convex set Y and let f ∈ Ac (X). Then the following assertions hold: (a) fϕ ◦ ϕ ≥ f and fϕ is upper semicontinuous and concave. (b) fϕ = inf{h ∈ Ac (Y ) : h ◦ ϕ ≥ f }. Proof. Assertion (a) follows easily from the definition (see the proof of Theorem 10.59). To show (b), let h ∈ Ac (Y ) satisfy h ◦ ϕ ≥ f . Then fϕ (y) = max f (ϕ−1 (y)) ≤ h(y),

y ∈ Y.

Conversely, let y ∈ Y be fixed. Let p : Ac (X) → R be defined as p(g) = inf{h(y) : h ◦ ϕ ≥ g},

g ∈ Ac (X).

Then p is a sublinear functional on Ac (X) such that p(h ◦ ϕ) = h(y), h ∈ Ac (Y ). By the Hahn–Banach theorem we can find a linear functional s on Ac (X) such that s(f ) = p(f ) and s ≤ p on Ac (X). If g ∈ Ac (X) is negative, then s(g) ≤ p(g) ≤ 0. Further, s(1) ≤ p(1) and s(−1) ≤ p(−1). Since p(1) = 1, s(1) = p(1) = 1. Thus

356


s is a positive continuous functional on Ac (X), and hence there exists x ∈ X with s(g) = g(x), g ∈ Ac (X) (see Proposition 4.31(a)). For any h ∈ Ac (Y ), h(ϕ(x)) = s(h ◦ ϕ) ≤ p(h ◦ ϕ) = h(y). By using −h instead of h we get h(ϕ(x)) = h(y), and thus y = ϕ(x). Hence fϕ (y) ≥ f (x) = s(f ) = p(f ), and the proof is finished. Lemma 10.63. Let ϕ be a continuous affine surjection of a compact convex set X onto a compact convex set Y . Then the following conditions are equivalent: (i) ϕ is open (that is, ϕ maps open sets to open sets), (ii) fϕ is continuous on Y for any f ∈ Ac (X), (iii) for any f1 , . . . , fd ∈ Ac (X), the set U := ϕ({x ∈ X : f1 (x) > 0, . . . , fd (x) > 0} is open in Y . Proof. Let ϕ be open and f ∈ Ac (X). For given y ∈ Y and c < fϕ (y) we find x ∈ ϕ−1 (y) with f (x) > c. Then W := ϕ({z ∈ X : f (z) > c}) is an open set in Y containing y and fϕ > c on W . Thus fϕ is lower semicontinuous. By Lemma 10.62(a), fϕ is continuous. This proves (i) =⇒ (ii). To show (ii) =⇒ (iii), we set V := {t ∈ Rd : ti > 0 for all i = 1, . . . , d}, d X S := λ ∈ Rd : λi ≥ 0 for all i = 1, . . . , d, λi = 1 , i=1 −1

d

Cy := {(f1 (x), . . . , fd (x)) ∈ R : x ∈ ϕ

(y)},

y ∈ Y.

By the definition, y ∈ U if and only if Cy ∩ V 6= ∅. If y ∈ / U , a separation argument shows that there exists λ ∈ Rd such that λ · t ≤ 0 for t ∈ Cy and λ · t > 0 for t ∈ V . We observe that λ is nontrivial and λi ≥ 0 for all i = 1, . . . , d, so that we may assume that λ ∈ S. It follows that [ Y \U = {y ∈ Y : sup λ · t ≤ 0} λ∈S

=

[ y∈Y: λ∈S

=

[ y∈Y: λ∈S

t∈Cy

sup x∈ϕ−1 (y) d X i=1

λi fi

d X

λi fi (x) ≤ 0

i=1

ϕ

(y) ≤ 0 .


357

Hence Y \ U is a projection on the first coordinate of the set d X C := (y, λ) ∈ Y × S : λi fi ϕ (y) ≤ 0 . i=1

By the assumption, fϕ is continuous on Y for each f ∈ Ac (X). Since kfϕ − gϕ k ≤ kf − gk, the mapping (f, y) 7→ fϕ (y) is continuous on Ac (X) × Y . Hence C is compact and, consequently, U is open. For the proof of (iii) =⇒ (i) we notice that the sets {x ∈ X : f1 (x) > 0, . . . , fd (x) > 0},

f1 , . . . , fd ∈ Ac (X), d ∈ N,

form a base of the topology of X. Lemma 10.64. If X is a Bauer simplex, then the sets of the form {x ∈ ext X : f (x) > a}, f ∈ Ac (X), a ∈ R, form a base of open sets in ext X. T Proof. If X is a Bauer simplex, ext X is a compact set. Thus the sets ni=1 {x ∈ ext X : fi (x) > 0}, where n ∈ N and the functions fi are continuous on ext X, form a base of the topology of ext X. Given functions f1 , . . . , fn in Ac (X), let f ∈ Ac (X) be such that f1 ∧ · · · ∧ fn = f on ext X (see Theorem 6.37). Then n \

{x ∈ ext X : fi (x) > 0} = {x ∈ ext X : f (x) > 0},

i=1

which concludes the proof. Lemma 10.65. Let X be a compact convex set, Y be a simplex and ϕ : X → Y be a continuous affine surjective mapping satisfying ϕ(ext X) = ext Y . Then fϕ is strongly affine on Y for any f ∈ Ac (X). Proof. Given a function f ∈ Ac (X) and y ∈ Y , let δy be the unique maximal measure in My (Y ). Let x ∈ ϕ−1 (y) be such that f (x) = fϕ (y) and let µx ∈ Mx (X) be a maximal measure. By Theorem 10.59(a), ϕ] µx is a maximal measure on Y and represents y. Hence δy = ϕ] µx and fϕ (y) = f (x) = µx (f ) ≤ µx (fϕ ◦ ϕ) = (ϕ] µx )(fϕ ) = δy (fϕ ). Since fϕ is upper semicontinuous and concave by Lemma 10.62(a), Corollary 4.8 gives fϕ (y) ≥ δy (f ) for any y ∈ Y . Hence δy (f ) = fϕ (y) for y ∈ Y . By Corollary 6.12, fϕ is strongly affine.

358


Theorem 10.66. Let X, Y be Bauer simplices and ϕ : X → Y be a continuous affine surjective mapping satisfying ϕ(ext X) = ext Y . Then the following assertions are equivalent: (i) ϕ is open, (ii) ϕ : ext X → ext Y is open, (iii) fϕ |ext Y is continuous on ext Y for every f ∈ Ac (X). Proof. We start the proof by noticing that Lemma 10.65 yields that fϕ is strongly affine function for each f ∈ Ac (X). Since ext Y is a closed set, Corollary 5.32 gives that fϕ is continuous on X if and only if fϕ |ext X is continuous on ext X. Hence (i) ⇐⇒ (iii) by Lemma 10.63. Further, for any f ∈ Ac (X) and a ∈ R we have {y ∈ ext Y : fϕ (y) > a} = ϕ({x ∈ ext X : f (x) > a}).

(10.13)

Indeed, if fϕ (y) > a for some y ∈ ext Y , then max f (ϕ−1 (y)) > a. Since ϕ−1 (y) is a closed face in X (see Proposition 2.72(b)), there exists x ∈ ext ϕ−1 (y) = ϕ−1 (y) ∩ ext X such that f (x) = max f (ϕ−1 (y)) (cf. Proposition 2.64(b)). Hence y ∈ ϕ({x ∈ ext X : f (x) > a}). Since the reverse inclusion “⊃” is obvious, (10.13) follows. Thus (10.13) yields that fϕ |ext Y ∈ C(ext Y ) provided ϕ is open on ext X and f ∈ Ac (X). Conversely, if each fϕ |ext Y is continuous on ext Y , then ϕ is open on ext X due to Lemma 10.64. Hence (ii) ⇐⇒ (iii), which concludes the proof. Corollary 10.67. Let K, L be compact spaces and ϕ : K → L be a continuous surjection. Then ϕ is open if and only if ϕ] : M1 (K) → M1 (L) is open. Proof. Given ϕ : K → L as in the hypothesis, the mapping ϕ] : M1 (K) → M1 (L) is a continuous affine surjection of the Bauer simplex M1 (K) onto the Bauer simplex M1 (L) (see Proposition 6.38) satisfying the assumptions of Theorem 10.66. If we identify K and L with ext M1 (K) and ext M1 (L), respectively, then ϕ] = ϕ on ext M1 (K). Hence Theorem 10.66 yields the conclusion.

10.7 10.7.A

Several topological results on Choquet boundaries The Choquet boundary as a Baire space

Theorem 10.68. Let S be a function cone of continuous functions on a compact space K. Then ChS (K) is a Baire space.

10.7 Several topological results on Choquet boundaries

359

Proof. Let {Vn } be a sequence of open T∞sets in K such that each Vn ∩ChS (K) is dense in ChS (K). We need to show that n=1 Vn ∩ ChS (K) is dense in ChS (K). To this end, let V0 be an open set intersecting ChS (K). We select a point x0 ∈ V0 ∩ChS (K) and use Proposition 7.11(b) and Theorem 7.21 for the lower semicontinuous function cV0 to find f0 ∈ −S such that f0 (x0 ) > 0 and {x ∈ K : f0 (x) ≥ 0} ⊂ V0 . Inductively we construct functions fn ∈ −S and points xn ∈ ChS (K), n ∈ N, such that • f n+1 < fn , •

fn (xn ) > 0,

{x ∈ K : fn (x) ≥ 0} ⊂ Vn . Assume that the construction has been completed up to the n-th stage. By the inductive assumption and density of Vn+1 ∩ ChS (K) in ChS (K), there exists •

xn+1 ∈ {x ∈ K : fn (x) > 0} ∩ Vn+1 ∩ ChS (K). We apply Proposition 7.11(b) along with Theorem 7.21 to the function cVn+1 ∧ fn to find a function fn+1 ∈ −S with fn+1 (xn+1 ) > 0 and fn+1 < fn . This completes the construction. Let now f := infn∈N fn . Since the sequence {fn } is decreasing, f is an upper semicontinuous S-convex function. Moreover, by a compactness argument, {x ∈ K : f (x) ≥ 0} =

∞ \

{x ∈ K : fn (x) ≥ 0} 6= ∅.

n=0

By the Minimum principle 7.15(f), f attains its maximum at some point y ∈ ChS (K). Hence fn (y) ≥ 0, n ≥ 0, and thus y ∈ V0 ∩

∞ \

Vn ∩ ChS (K).

n=1

This finishes the proof.

10.7.B

Polish spaces as Choquet boundaries

Lemma 10.69. Let Y be a compact metric space and U be its open subset. Then for every ε > 0 there exists a family {gk : k ∈ N} of positive continuous functions on Y such that P∞ (a) k=1 gk = cU , (b) limk→∞ diam spt gk = 0, (c) diam spt gk < ε, k ∈ N.

360


Proof. Given an open set U in a compact metric space Y , we find a countable open cover V = {Vk : k ∈ N} of U such that • diam V < ε, k ∈ N, k •

limk→∞ diam Vk = 0,

dist(Vk , Y \ U ) > 0. S To achieve this, we find compact sets Fn , n ∈ N, such that U = ∞ n=1 Fn . For each n ∈ N we find a finite open cover of Fn consisting of sets of diameter less than 21n ε that have positive distance from Y \ U . By enumerating all these families into a single sequence we get the required family V. We use Theorem 5.1.9 in [169] to find a family {gn : n ∈ N} of positive continuous functions on U such that P∞ • n=1 gn = cU , •

•

for every n ∈ N there exists kn ∈ N such that {x ∈ U : gn (x) > 0} ⊂ Vkn ,

the family W := {gn−1 ((0, 1]) : n ∈ N} is locally finite, that is, for each x ∈ U there exists a neighborhood intersecting only finitely many sets of W. Due to the properties of the cover {Vk : k ∈ N} we can extend functions gn by 0 outside U and get continuous functions on Y . We finish the proof by realizing that {gn : n ∈ N} satisfies the requirements. To verify the only non-obvious property (b), we show that, for each k ∈ N, Vk intersects only finitely many sets from W. Since W is locally finite, for each x ∈ Vk we can find an open neighborhood Ux of x intersecting only finitely many sets from W. By the compactness of Vk , we choose finitely many points x1 , . . . , xj in Vk such that Vk ⊂ Ux1 ∪ · · · ∪ Uxj . Hence Vk intersects only finitely many elements of W. From this the property (b) follows and the proof is finished. •

Theorem 10.70. Let P be a Polish space. Then there exists a metrizable simplex X such that P is homeomorphic to ext X. Proof. Let P be a Polish space and Y be its metric compactification. We fix a compatible metric ρ on T Y . By [169], Theorem 4.3.24, P is a Gδ subset of Y , and hence we may write P = ∞ n=0 Gn , where Y = G0 ⊃ G1 ⊃ · · · are open subsets of Y . For each n ≥ 0 we use Lemma 10.69 to find continuous functions {hnk }∞ k=1 on Y and a sequence of strictly positive numbers {βnk }∞ converging to 0 such that k=1 P∞ n • = c , h Gn k=1 k diam spt hkn ≤ 2−n βkn , k ∈ N. We set pnk := hnk cY \Gn+1 , •

Then

∞ X ∞ X n=0 k=1

k ∈ N, n ≥ 0.

hnk = cY \P .

361


We take into consideration only nonzero functions pnk and inductively find points xnk , ykn ∈ spt hnk ∩ P such that the family {xnk , ykn : k ∈ N, n ≥ 0} consists of pairwise distinct elements. We remark that this is possible because spt hnk ∩ P is infinite, whenever pnk 6= 0. We define a mapping γ : M(Y ) → M(Y ) as γ(µ) := µ −

Z ∞ X ∞ X n=0 k=1

Y

pnk dµ

1 (εxn + εykn ). 2 k

(10.14)

We define a subspace M of M(Y ) as M := {γ(µ) : µ ∈ M(Y \ P )}. Claim 1. The space M is closed (in the w∗ -topology). Proof of Claim 1. It is enough to show that the unit ball BM is compact (see Theorem A.7). Since BM is metrizable, it suffices to check the sequential compactness of BM . To this end, let {µi } be a sequence of measures in M(Y \P ) such that the sequence {γ(µi )} is contained in BM . We have to extract a w∗ -convergent subsequence whose limit is in BM . To this end, we use an estimate kµi k ≤ kγ(µi )k, the compactness of BM (Y \Gn+1 ) , n ≥ 0, and the diagonal method to find a subsequence (denoted again n ∈ as {γ(µi )}) such that the sequence {µi |Y \Gn+1 }∞ i=1 converges to a measure ν M(Y \ Gn+1 ), n ≥ 0. We set ν :=

∞ X

ν n |Gn \Gn+1 .

n=0

P∞ P∞

We notice that ν = n=0 k=1 pnk ν n . Further, kνk ≤ 1. To show this, we first observe that ∞ X ∞ Z ∞ X ∞ Z X X f pnk dµi ≤ |f |pnk d|µi | ≤ kµi k ≤ 1, n=0 k=1

Y

n=0 k=1

i ∈ N,

Y

and pnk µi = hnk cY \Gn+1 µi = hnk µi |Y \Gn+1 → hnk ν n = pnk ν n ,

k ∈ N, n ≥ 0.

Hence, for every f ∈ BC(Y ) , we obtain ∞ ∞ X ∞ X X |ν(f )| = (ν n |Gn \Gn+1 )(f ) = n=0

n=0 k=1

Z

f hnk dν n ≤ 1.

Y

(HerePwe use the following fact: If limi→∞ cji = cj and then ∞ j=1 |cj | ≤ 1.)

P∞

j=1 |cji |

≤ 1 for i ∈ N,

362


Next we verify that γ(µi ) → γ(ν). To this end, we fix f ∈ C(Y ). For each λ ∈ M(Y \ P ) we denote Z 1 αkn (λ) := pnk (f − (f (xnk ) + f (ykn )) dλ, k ∈ N, n ≥ 0. 2 Y Then Z ∞ ∞ 1 XX γ(λ)(f ) = λ(f ) − pnk dλ (f (xnk ) + f (ykn )) 2 Y n=0 k=1 ∞ ∞ Z ∞ X ∞ Z X 1 XX ( pnk dλ)(f (xnk ) + f (ykn )) = pnk dλ − 2 Y Y =

n=0 k=1 ∞ X ∞ X

n=0 k=1

αkn (λ).

n=0 k=1

Let ε > 0 be given. We find δ > 0 such that |f (x) − f (y)| < ε provided ρ(x, y) < δ. Next we find N ∈ N such that 2−N < 21 δ and K ∈ N satisfying βkn < δ, n = 1, . . . , N , k ≥ K. Let A ⊂ N × N consists of all pairs (n, k), where n > N or n = 1, . . . , N and k > K. Then for any λ ∈ M(Y \ P ), the condition on diameters of the supports hnk yields Z N X K X X f dγ(λ) − αkn (λ) ≤ |αkn (λ)| Y

n=0 k=1

Z ≤ε

X

(n,k)∈A

pkn d|λ| ≤ εkλk.

Y (n,k)∈A

For fixed n ≥ 0, k ∈ N, Z 1 pnk (f − (f (xnk ) + f (ykn ))) dµi lim αkn (µi ) = lim i→∞ Y i→∞ 2 Z 1 = pnk (f − (f (xnk ) + f (ykn ))) dν n 2 Y Z 1 = pnk (f − (f (xnk ) + f (ykn ))) dν = αkn (ν). 2 Y P P∞ n n n Thus αk (µi ) → αk (ν) and since the series ∞ n=0 k=1 |αk (λ)| converges uniformly on BM(Y \P ) , an elementary limit procedure gives Z lim

i→∞ Y

f dγ(µi ) = lim

i→∞

∞ X ∞ X n=0 k=1

This concludes the proof of Claim 1.

αkn (µi ) =

∞ X ∞ X n=0 k=1

αkn (ν) =

Z f dγ(ν). Y

363


We proceed with the proof of the theorem by denoting π : M(Y ) → M(Y )/M the canonical quotient mapping and by setting X := π(M1 (Y )). Let ε : Y → M1 (Y ) denote the mapping assigning to each x ∈ Y the Dirac measure εx , and let ϕ := π ◦ ε. We notice the following fact. Claim 2. If x ∈ P , y ∈ Y and π(εx ) = π(εy ), then x = y. Proof of Claim 2. Assume that y ∈ Y \ {x} satisfies π(εy ) = π(εx ). Then εy − εx ∈ M , that is, there exists λ ∈ M(Y \ P ) such that ∞

εy − εx = γ(λ) = λ −

∞

1 XX 2 n=0 k=1

But the right-hand side has mass at most

1 2

Z Y

1 pnk dλ (εxnk + εykn ). 2

at each point of P , a contradiction.

It follows that ϕ : P → X is a homeomorphic embedding. Indeed, ϕ is injective by Claim 2 and, obviously, ϕ is continuous. Let F ⊂ Y be a closed set. Then for any x ∈ P and y ∈ F with ϕ(x) = ϕ(y), Claim 2 yields x = y. Hence ϕ(P ∩ F ) = ϕ(F ) ∩ ϕ(P ), and ϕ maps closed sets in P to closed sets in X. Hence ϕ is a homeomorphism. Claim 3. We have ext X ⊂ ϕ(P ). Proof of Claim 3. Let s ∈ ext X be given. Then π −1 (s) ∩ ext M1 (Y ) 6= ∅ by Proposition 2.72. Hence there exists x ∈ Y such that s = ϕ(x) (see Proposition 2.27). Assume that x ∈ / P , that is, x ∈ Gn \ Gn+1 for some n ≥ 0. Denote λ :=

Z ∞ X ∞ X n=0 k=1

pnk dεx

Y

1 (εxn + εykn ). 2 k

Then λ ∈ M1 (P ) and γ(εx ) = εx − λ. Hence 1 π(εx ) = π(λ) = 2

∞ X ∞ X

pnk (x)π(εxnk )

+

∞ X ∞ X

! pnk (x)π(εykn )

.

n=0 k=1

n=0 k=1

Since s is extreme, this implies ∞ X ∞ X

pnk (x)π(εxnk )

n=0 k=1

in other words,

∞ X ∞ X n=0 k=1

=

∞ X ∞ X

pnk (x)π(εykn ),

n=0 k=1

pnk (x)(εxnk − εykn ) ∈ M.

(10.15)

364


Since this measure does not charge Y \ P , (10.15) holds only in case ∞ X ∞ X

pnk (x)(εxnk − εykn ) = 0.

n=0 k=1

As all points xnk , ykn , where pnk is nonzero, are different, this is impossible. Hence x ∈ P and Claim 3 is proved. To show that X is a simplex we verify the following fact. Claim 4. If Λ ∈ (Ac (X))⊥ satisfies |Λ|(X \ ϕ(P )) = 0, then Λ = 0. Proof of Claim 4. To verify this, let Λ be as in the hypothesis. We find λ ∈ M(Y ) with |λ|(Y \ P ) = 0 such that Λ = ϕ] λ and pick a continuous function g on Y such that g ∈ M⊥ . Then G(s) := µ(g),

s ∈ X, µ ∈ π −1 (s),

is an affine continuous function on X (see Corollary 5.27). Thus 0 = Λ(G) = (ϕ] λ)(G) = λ(G ◦ ϕ) = λ(g). Hence λ ∈ (M⊥ )⊥ = M . Since λ is carried by P , we get λ = 0, and consequently Λ = 0. Hence X is a simplex by Proposition 6.9. To finish the proof we verify the following fact. Claim 5. We have ϕ(P ) ⊂ ext X. To verify this, assume that ϕ(x) ∈ / ext X for some x ∈ P . Then there exists a measure µ ∈ M1 (X) carried by ϕ(P ) that represents ϕ(x) and is different from εϕ(x) . Then εϕ(x) − µ ∈ (Ac (X))⊥ and |εϕ(x) − µ|(X \ ϕ(P )) = 0. Thus µ = εϕ(x) by Claim 4. This contradiction concludes the proof.

10.7.C

K-countably determined boundaries

Theorem 10.71. Let X be a compact convex set such that ext X is K-countably deterT mined. Then ext X ∈ Π2 (Bos(X)); that is, ext X can be written as ∞ (F ∪ Vn ), n n=1 where Fn ⊂ X are closed and Vn ⊂ X open.

10.8 Convex Baire-one functions

365

Proof. If ext X is K-countably determined, there exists a countable family {Kn : n ∈ N} of compact sets in X such that for each x ∈ X \ ext X, y ∈ ext X, there exists n ∈ N withSy ∈ Kn ⊂ X \ {x} (see Theorem A.111(e)). For each F ⊂ N finite we set AF := n∈F Kn and 1 . BF := x ∈ X : exists µ ∈ Mx (X) such that µ(AF ) ≥ 2 Since AF is closed and the barycentric mapping is continuous, BF is closed as well. Further, AF ⊂ BF . We claim that \ ext X = AF ∪ (X \ BF ). F ⊂N,F finite

To this end, choose x ∈ ext X. If x ∈ BF for some finite F ⊂ N, then there exists µ ∈ Mx (X) with µ(AF ) ≥ 21 . Since µ = εx , x ∈ AF . Hence x ∈ AF ∪ (X \ BF ) for every F ⊂ N finite. Conversely, let x ∈ / ext X. We choose a maximal measure µ ∈ Mx (X). By the property of the family {Kn : n ∈ N}, there exists a set N ⊂ N such that [ Kn ⊂ X \ {x}. ext X ⊂ n∈N

S Since µ( n∈N Kn ) = 1 (see Theorem 3.81), there exists a finite set F ⊂ N so that µ(AF ) ≥ 21 . Since x ∈ / AF , we get x ∈ BF \ AF and thus x ∈ / AF ∪ (X \ BF ). This concludes the proof.

10.8

Convex Baire-one functions

We recall that a function f on a topological space X is fragmented if for any F ⊂ X and ε > 0 there exists a nonempty open set U ⊂ X such that diam f (U ∩ F ) < ε. Lemma 10.72. If f is a fragmented convex function on a compact convex set X, then f is lower bounded. Proof. Assume that there exist points xn ∈ X, n ∈ N, such that f (xn ) → −∞. We consider S := {λ ∈ `1 : λ ≥ 0, kλk = 1} with the w∗ -topology. Then S is a compact convex set and the mapping ϕ : S → X defined as ∞ X ϕ(λ) := r λn εxn , λ = {λn } ∈ S, n=1

is continuous and affine. Hence f ◦ ϕ is convex and fragmented on S. By Theorem A.127, f ◦ ϕ is a Baire-one function on S. Thus Theorem 10.37 yields that f ◦ ϕ

366


is lower bounded. But f ◦ ϕ(en ) = f (xn ) → −∞, a contradiction (here {en } is the usual basis of `1 ). Lemma 10.73. Let f be a fragmented convex function on a compact convex set X and µ ∈ M+ (X) be a measure on X with kµk > 0. Then for every ε > 0 there exists a ν measure ν ∈ M+ (X) such that ν ≤ µ, kνk > 0 and ν(f ) ≥ kνk(f (r( kνk )) − ε). Proof. Let µ ∈ M+ (X) and ε > 0 be given. Set Y := co spt µ. By the assumption, there exists an open nonempty set U ⊂ Y so that diam f (U ) ≤ ε. Without loss of generality we may assume that inf f (U ) = 0 and sup f (U ) ≤ ε. By combining the Krein–Milman and Milman theorems 2.22 and 2.43, we get distinct points P y1 , . . . , ynP ∈ spt µ ∩ ext Y and coefficients λ1 , . . . , λn ∈ (0, 1) so that ni=1 λi = 1 and y := ni=1 λi yi ∈ U . We find a closed convex neighborhood V ⊂ U of y and disjoint open neighborhoods Ui of yi , i = 1, . . . , n, such that n X

λi z i ∈ V

whenever zi ∈ Ui .

(10.16)

i=1

Then µ(Ui ) > 0, i = 1, . . . , n. We set µi :=

µ|Ui , µ(Ui )

i = 1, . . . , n.

By convexity of f and (10.16), n X i=1

λi f (zi ) ≥ f

n X

λi zi ≥ 0,

zi ∈ Ui , i = 1, . . . , n.

(10.17)

i=1

Integrating (10.17) with respect to the probability measure µ1 ⊗ · · · ⊗ µn we obtain n X

λi µi (f ) ≥ 0.

i=1

P Let ν˜ := ni=1 λi µi and z be the barycenter of ν. ˜ From (10.16) we easily obtain that z ∈ V . Thus ν(f ˜ ) ≥ 0 ≥ sup f (U ) − ε ≥ f (z) − ε. We look for ν in the form ν = cν˜ with c > 0. To satisfy the requirement ν ≤ µ, we choose c ≤ min{µ(U1 ), . . . , µ(Un )}. Then the measure ν = cν˜ has all the required properties. Lemma 10.74. Let f be a fragmented convex function on a compact convex set X. P∞ P∞ Then n=1 λn f (x Pn ) ≥ f ( n=1 λn xn ) whenever numbers λn are positive, points xn are in X and ∞ n=1 λn = 1.

10.8 Convex Baire-one functions

367

Proof. Let P be the positive cone of `1 and S be its base, that is, P = {λ ∈ `1 : λ ≥ 0} and S = {λ ∈ P : kλk = 1} considered with the norm topology (see Notation 10.33). Without loss of generality we may assume that f ≥ 0 (see Lemma 10.72). Let λ0 ∈ S be given. We set αn := λ0n f (xn ),

n ∈ N.

Then {αn } ∈ P . We define g : P → R as ( 0, P g(λ) := kλkf ( ∞ n=1 Since λ 7→

λn kλk xn ),

λ = 0, λ 6= 0.

∞ X λn xn kλk n=1

is a continuous mapping of P \ {0} to X, g is a fragmented function on P \ {0}. Consequently, it is fragmented on P and thus g is a Baire-one function on P by Theorem A.127 (we recall that (P, k · k) is a complete metric space). Moreover, it is a sublinear function on P by the convexity of f . Let ϕ : {0, 1}N → R be defined as ϕ(σ) = g(σλ0 ),

σ ∈ {0, 1}N .

(Here σλ0 is the sequence {σn λ0n }.) Since σ 7→ σλ0 is continuous on {0, 1}N , ϕ is a Baire-one function on {0, 1}N . Set ∞ X N A := σ ∈ {0, 1} : ϕ(σ) ≤ σn αn . n=1

P∞

Since the mapping σ 7→ n=1 σn αn is continuous on {0, 1}N and ϕ is Baire-one, A is a Gδ set (see Theorem A.124). If σ ∈ {0, 1}N is finitely supported, say by a finite set J ⊂ N, then ϕ(σ) = g

X n∈J

∞ X X X 0 λ0n en ≤ λn g(en ) = λ0n f (xn ) = σn αn . n∈J

n∈J

n=1

Hence σ ∈ A and thus A is dense. The mapping σ 7→ N \ σ is a homeomorphism of {0, 1}N onto itself (see Subsection 10.3.A). Hence {N \ σ : σ ∈ A} is a dense Gδ set in {0, 1}N and we can find σ ∈ {0, 1}N such that both σ and N \ σ are in A. Then λ0 = σλ0 + (N \ σ)λ0 ,

368


and thus ϕ(1) = g(λ0 ) ≤ g(σλ0 ) + g((N \ σ)λ0 ) = ϕ(σ) + ϕ(N \ σ) ≤

∞ X

σn αn +

n=1

∞ X

(1 − σn )αn =

n=1

∞ X

αn .

n=1

In other words, 0

g(λ ) = f

∞ X

λ0n xn

n=1

≤

∞ X

λ0n f (xn ).

n=1

This finishes the proof. Theorem 10.75. Let f be a fragmented convex function on a compact convex set X. Then f is lower bounded and satisfies the subbarycentric formula. Proof. Let f be a function satisfying our assumptions. Then f is lower bounded by Lemma 10.72 and universally measurable by Proposition A.118 and Theorem A.121. Let µ ∈ M1 (X) and ε > 0 be given. We are going to construct a transfinite sequence {µα }0≤α≤ω1 of positive measures on X which satisfies •

µ0 = µ,

•

µα+1 ≤ µα and either kµα+1 k < kµα k or µα = 0, R X f d(µα − µα+1 ) ≥ kµα − µα+1 k(f (xα ) − ε), where xα is the barycenter of µα −µα+1 kµα −µα+1 k , P µα = β 0, we use Lemma 10.73 to µα and obtain a measure να such that 0 < να ≤ µα . By setting µα+1 := µα − να we finish the inductive step for an isolated ordinal number. If α is a limit ordinal number and µβ have been defined for every β < α, we define µα ∈ M+ (X) by setting µα (g) := limβ 0} , we obtain that the set {x ∈ K : f ∗ (x) > 0} is open for every f ∈ C(K). For a ∈ R, the equality {x ∈ K : f ∗ (x) > a} = {x ∈ K : (f − a)∗ (x) > 0} yields that f ∗ is lower semicontinuous for every f ∈ C(K). Since f ∗ is always upper semicontinuous, the upper envelope f ∗ is continuous for every f ∈ C(K). For the proof of the converse implication, assume that f ∗ is continuous for every f ∈ C(K). It is sufficient to show that, given f1 , . . . , fd ∈ C(K) and U := {µ ∈ M(H) : µ(f1 ) > 0, . . . , µ(fd ) > 0} , the set K \ r(U ) is closed. Set n o V : = (t1 , . . . , td ) ∈ Rd : t1 > 0, . . . , td > 0 , d n o X S : = (t1 , . . . , td ) ∈ Rd : t1 ≥ 0, . . . , td ≥ 0, ti = 1 , i=1

C(x) : =

n

o (µ(f1 ), . . . , µ(fd )) ∈ Rd : µ ∈ Mx (H) ,

x ∈ K.

10.9 Function spaces with continuous envelopes

377

Then x ∈ / r(U ) if and only if C(x) ∩ V = ∅. For every x ∈ K, the set C(x), as a continuous affine image of a compact convex set Mx (H), is convex and compact in Rd . Since V is a cone, the Hahn–Banach separation theorem yields that the intersection V ∩ C(x) is empty if and only if there exists s ∈ S such that sup {s · c : c ∈ C(x)} ≤ 0 ≤ inf {s · t : t ∈ V } . Hence Lemma 3.21 gives

[ K \ r(U ) = x∈K:

sup

[ x∈K:

[ x∈K: s∈S

µ

sup

d X

µ∈Mx (H)

s∈S

=

si µ(fi ) ≤ 0

µ∈Mx (H) i=1

s∈S

=

d X

d X

si fi ≤ 0

i=1

∗ si fi (x) ≤ 0 .

i=1

Thus K \ r(U ) = πK

(x, s) ∈ K × S :

d X

∗ si fi (x) ≤ 0 ,

i=1

where πK : K × S → K denotes the projection onto K. By the assumption and Lemma 3.18(c), the mapping (x, f ) 7→ f ∗ (x) is continuous on K × C(K). Hence (x, s) 7→

d X

∗ si fi (x)

i=1

is continuous on K × S. Thus

{(x, s) ∈ K × S :

d X

∗ si fi (x) ≤ 0}

i=1

is a compact set in K × S and K \ r(U ) is closed, which finishes the proof.

378

10.10


Exercises

Exercise 10.89. Find a boundary of a compact convex set disjoint from the set of extreme points. Hint. Take an uncountable set Γ and let X be the unit ball of `∞ (Γ) endowed with the w∗ -topology. Then ext X = {x ∈ X : |x(γ)| = 1, γ ∈ Γ}. If we set B := {x ∈ X : |x(γ)| ∈ {0, 1}, x with countable support}, then B is a boundary disjoint from ext X (use Proposition 4.36). Exercise 10.90. Let X be a compact subset of an infinite-dimensional Banach space E. Prove that there exists a sequence {xn } of linearly independent vectors in E converging to 0 such that X ⊂ co{xn : n ∈ N}. Hint. We first propose the following observation. If E1 , E2 are subspaces of E with E1 ∩ E2 = {0} and E2 = span{e1 , . . . , ek } where {e1 , . . . , ek } are linearly independent vectors, then the vectors a1 + e1 , . . . , ak + ek are linearly independent for any a1 , . . . , ak ∈ E1 . To construct the required sequence {xn }, we find inductively sets Kn and Fn as follows. Let K1 := X and F˜1 ⊂ X be a finite 2−2 -net for K1 . If E1 := span F˜1 , let E2 be a subspace of E such that its dimension equals cardinality of F˜1 and E2 ∩E1 = {0}. We select a basis {e1 , . . . , ek } of E2 and form the set F1 in such a way that the i-the element of F1 is a little perturbation of i-the element of F˜1 in the direction of ei , i = 1, . . . , k, so that F1 is still a 2−2 -net for K1 and consists of linearly independent vectors. Assuming that the construction has been completed for n ∈ N, let Kn+1 := (Kn − Fn ) ∩ 2−2n UE (here UE is the open unit ball of E). Then for any element x ∈ Kn there exists y ∈ Kn+1 and z ∈ Fn such that y = x − z. We choose a finite 2−2(n+1) net Fñ+1 for Kn+1 . If now E1 := span{F1 ∪ · · · ∪ Fn ∪ Fñ+1 }, let E2 be a subspace of E such that its dimension equals cardinality of Fñ+1 and E2 ∩ E1 = {0}. We produce Fn+1 from Fñ+1 as in the first step of the construction to achieve that F1 ∪ · · · ∪ Fn+1 consists of linearly independent vectors, Fn+1 is a 2−2(n+1) -net for Kn+1 and Fn+1 ⊂ 2−2n UE . If x ∈ K1 is arbitrary, we can find x2 ∈ K2 and y1 ∈ F1 such that x2 = x − y1 . Then we find x3 ∈ K3 and y2 ∈ F2 such that x3 = x2 − y2 , hence x = x3 + y1 + y2 . By proceeding with this construction we get points xn ∈ Kn , n ≥ 2, and yn ∈ Fn , n ∈ N, such that x = xn+1 + y1 + · · · + yn , n ∈ N. Since xn+1 ∈ Kn+1 ⊂ 2−2(n+1) BE , we get x=

∞ X n=1

yn =

∞ X n=1

2−n (2n yn ).

10.10 Exercises

379

The required sequence {xn } is obtained by putting the elements of 2F1 , 22 F2 , 23 F3 , . . . into a single sequence. Exercise 10.91. Let {xn } be a strongly linearly independent sequence of vectors in a Banach space E that converges to 0. Prove that X := co{xn : n ∈ N} is a Bauer simplex with ext X = {0} ∪ {xn : n ∈ N}. Hint. By Theorem 2.43, ext X ⊂ {xn : n ∈ N} = {0} ∪ {xn : n ∈ N}. Assume that an , bn , n = 0, 1, . . . , are positive numbers such that a0 +

∞ X n=1

and a0 0 +

∞ X

an = b0 +

∞ X

bn = 1

n=1

an xn = b0 0 +

n=1

∞ X

bn xn .

n=1

P∞

Then n=1 (an −bn )xn = 0, and thus the strong linear independency yields an = bn , n ∈ N. Consequently, a0 = b0 . From this we get that X is a simplex and ext X = {0} ∪ {xn : n ∈ N}. Hence X is a Bauer simplex. Exercise 10.92. Let X be a compact convex subset of a Banach space E. Prove that there exist compact convex Bauer simplices X1 , X2 in E such that X ⊂ co(X1 ∪X2 ). Hint. If X is in Rd , the assertion is trivial because then we can find a d-simplex S ⊂ Rd with X ⊂ S. Otherwise we use Exercise 10.90 to find a linearly independent sequence {xn } in E tending to 0 such that X ⊂ co{xn : n ∈ N}. Using Theorem 10.30 we find a partition N1 , N2 of N into two disjoint infinite sets such that the sequences {xn }n∈N1 and {xn }n∈N2 are strongly linearly independent. By Exercise 10.91, the sets Xi := co{xn : n ∈ Ni },

i = 1, 2,

are Bauer simplices and, clearly, X ⊂ co(X1 ∪ X2 ). Exercise 10.93. Let F1 , F2 be nonempty disjoint faces of a compact convex set X such that X = co(F1 ∪ F2 ). Then F2 is the complementary face of F1 . Hint. Let F ⊂ X be a face disjoint from F1 and let x ∈ F \ F2 . Then there exist α ∈ (0, 1) and xi ∈ Fi , i = 1, 2, such that x = αx1 + (1 − α)x2 . Then x1 , x2 ∈ F , a contradiction with F ∩ F1 = ∅.

380


Exercise 10.94. Let X, Y be compact convex sets and ϕ : ext X → Y be a mapping. Prove that the following conditions are equivalent: (i) There exists a unique continuous affine extension ϕ b : X → Y of ϕ. (ii) There exists a continuous affine extension ϕ b : X → Y of ϕ. c (iii) For every g ∈ A (Y ), the function g ◦ ϕ satisfies conditions (i1) and (i3) of Theorem 6.32. Hint. Obviously, (i) =⇒ (ii). If ϕ b : X → Y is a continuous affine extension of ϕ, g◦ϕ b is a continuous affine extension of g ◦ ϕ. Hence (i1) and (i3) of Theorem 6.32 are satisfied for g ◦ ϕ by Theorem 6.32. Hence (ii) =⇒ (iii). Assume that (iii) holds. By Theorem 6.32, for any g ∈ Ac (Y ) there exists a unique extension g[ ◦ ϕ ∈ Ac (X) of g ◦ ϕ. Let T : Ac (Y ) → Ac (X) be defined as g 7→ g[ ◦ ϕ,

g ∈ Ac (Y ).

By the Minimum principle 2.24, T is a linear positive mapping such that T 1 = 1. Let φ1 : X → S(Ac (X)) and φ2 : Y → S(Ac (Y )) be the affine homeomorphisms given by Proposition 4.31. Then the dual operator T ∗ maps S(Ac (X)) to S(Ac (Y )) and thus we may define ϕ(x) b := (φ2 )−1 (T ∗ (φ1 (x))),

x ∈ X.

Then ϕ b is a continuous affine mapping extending ϕ. Indeed, we get from g(ϕ(x)) b = (T ∗ φ1 (x))(g) = T g(x) = g[ ◦ ϕ(x),

g ∈ Ac (Y ),

that g(ϕ(x)) b = g[ ◦ ϕ(x) = g(ϕ(x)),

g ∈ Ac (Y ), x ∈ ext X.

Hence ϕ b is an extension of ϕ. To finish the proof we have to verify the uniqueness of ϕ. b Let ψ be another continuous affine extension of ϕ. Then for any g ∈ Ac (Y ), g ◦ ψ ∈ Ac (X) and extends g ◦ ϕ. By the minimum principle, g ◦ ψ = g ◦ ϕ. b Hence g ◦ ψ = g ◦ ϕ b for every g ∈ Ac (Y ), and thus ψ = ϕ. b Exercise 10.95. Let Σ be a σ-algebra on a set X and νn : Σ → R, n ∈ N, be signed measures on Σ. Assume that ν(A) := limn→∞ νn (A) exists for every A ∈ Σ. Prove that ν is countably additive on Σ. Hint. Obviously, ν is finitely additive on Σ. Let Ai , i ∈ N, be disjoint sets from Σ. For n ∈ N, let fn : {0, 1}N → R be defined as [ fn (σ) = νn Ai , σ ∈ {0, 1}N . i∈spt σ

Then each fn is countably additive and continuous on {0, 1}N .

10.10 Exercises

381

Indeed, the countable additivity follows from the fact that νn is countably additive. To show its continuity, let σ k → σ in {0, 1}N and ε > 0 be given. Let i0 ∈ N be such S∞ that |νn |( i=i0 Ai ) < ε. Since σ k → σ on each coordinate, there exists k0 ∈ N such that σ k |i0 = σ|i0 for each k ≥ k0 . Let N := {1, . . . , i0 }. For k ≥ k0 we have |fn (σ k ) − fn (σ)| ≤ |fn (σ k cN ) − fn (σcN )| + |fn (σ k cN\N )| + |fn (σcN\N )| ≤ |νn |

∞ [

∞ [ Ai + |νn | Ai ≤ 2ε.

i=i0

Hence fn is continuous. If f (σ) := ν

i=i0

[

Ai ,

σ ∈ {0, 1}N ,

i∈spt σ

then f is finitely additive and it is a pointwise limit of the sequence {fn }. Hence f has the Baire property in the restricted sense (see Subsection A.2.D) and thus f is continuous and countably additive by Theorem 10.26. Hence ν is countably additive. Exercise 10.96. Prove that a compact convex set X with ext X countable is metrizable. Hint. Since any countable space has a countable network, the assertion follows from Lemma 10.48 and Theorem 10.51. Exercise 10.97. There exists a continuous affine surjection ϕ of a simplex X onto a compact convex set Y and a measure µ ∈ M1max (X) such that •

ϕ(ext X) = ext Y and ϕ is injective on ext X,

•

ϕ] µ ∈ / M1max (Y ).

Hint. Let H1 be the function space on the compact space K1 from Example 3.82 and let K2 be the quotient of K1 where all points of [0, 1] × {0} are identified with the point (0, 0) (see [169], Section 2.4). We write q : K1 → K2 for the quotient mapping and take H2 := {f ∈ C(K2 ) : 2f (0, 0) = f (t, −1) + f (t, 1), t ∈ [0, 1]}. Let X, Y be the state space of H1 , H2 , respectively, and φ1 , φ2 be the respective embeddings. Then ext X = φ1 (K1 \([0, 1]×{0})) and ext Y = φ2 (K2 \{(0, 0)}). We denote by ϕ : X → Y the restriction of the adjoint operator to the operator h 7→ h ◦ q, h ∈ H2 . Then X is a simplex and φ] λ ∈ M1 (X) is maximal for any continuous measure λ ∈ M1 ([0, 1] × {0}), even though φ] λ is carried by a compact set disjoint from ext X. (We recall that λ is continuous if λ({x}) = 0 for each x ∈ X.)

382


Then ϕ(ext X) = ext Y and ϕ is even injective on ext X. On the other hand, if λ is any continuous probability measure on φ1 ([0, 1] × {0}), then λ is maximal on X, yet the measure ϕ] λ equals the Dirac measure at the point φ2 (0, 0), and hence ϕ] λ is not maximal. Exercise 10.98. There exists a continuous affine surjection ϕ of a metrizable simplex X onto a compact convex set Y such that • ϕ is injective on ext X, •

ϕ] (M1max (X)) ⊂ M1max (Y ) and ϕ] is injective on M1max (X),

•

ϕ is not injective on X.

Hint. Let K1 = {x1 , x2 , x3 , y1 , y2 , y3 } with the discrete topology and K2 be the quotient of K1 , if we identify y2 with x2 . Again, we denote by q : K1 → K2 the quotient mapping. Let H1 := {f ∈ C(K1 ) : 2f (x2 ) = f (x1 ) + f (x3 ), 2f (y2 ) = f (y1 ) + f (y3 )}

and

H2 := {f ∈ C(K2 ) : f (x1 ) + f (x3 ) = 2f (x2 ) = f (y1 ) + f (y3 )}. We take X, Y , φ1 , φ2 and ϕ : X → Y as above. Then X is a simplex, ext X = φ1 (K1 \ {x2 , y2 }), ext Y = φ2 (K2 \ {x2 }), ϕ : ext X → ext Y is a bijection, yet ϕ is not injective on X. Obviously, ϕ] maps injectively maximal measures to maximal measures. Exercise 10.99. There exists a continuous affine surjection ϕ of a simplex X onto a compact convex set Y such that • ϕ is injective on ext X, •

ϕ] (M1max (X)) ⊂ M1max (Y ),

•

ϕ] is not injective on M1max (X).

Hint. Let K1 = [0, 1] ∪ [2, 3] × {−1, 0, 1} endowed again with the “porcupine” topology as in Exercise 10.97 and let K2 be the quotient of K1 after identifying points (t + 2, 0) with (t, 0), t ∈ [0, 1]. Let H1 := {f ∈ C(K1 ) : 2f (t + i, 0) = f (t + i, −1) + f (t + i, 1), t ∈ [0, 1], i = 0, 2}, and H2 := {f ∈ C(K2 ) : 2f (t, 0) = f (t + i, −1) + f (t + i, 1), t ∈ [0, 1], i = 0, 2}, and let X, Y , φ1 , φ2 and ϕ be as above. Then ext X = φ1 (K1 \ (([0, 1] ∪ [2, 3]) × {0})), and ϕ maps injectively ext X onto ext Y .

ext Y = φ2 (K2 \ ([0, 1] × {0})),

10.10 Exercises

383

We claim that ϕ] (M1max (X)) ⊂ M1max (Y ). Indeed, a probability measure λ is maximal on X if and only if λ = (φ1 )] µ for some measure µ ∈ M1 (K1 ) whose restriction to [0, 1] ∪ [2, 3] × {0} is continuous. Similarly, any maximal probability measure on Y is of the form (φ2 )] µ for some measure µ ∈ M1 (K2 ) whose restriction to [0, 1] × {0} is continuous. From these observations the claim follows. Finally, if we take Lebesgue measure λ1 on [0, 1] × {0} and Lebesgue measure λ2 on [2, 3] × {0}, then ϕ] ((φ1 )] λ1 ) = ϕ] ((φ1 )] λ2 ). Hence ϕ] is not injective on M1max (X). Exercise 10.100. Let X be a compact convex set and r : M1 (X) → X be the barycentric mapping. If f ∈ C(X) is understood as an affine continuous function on M1 (X) via the formula µ 7→ µ(f ), µ ∈ M1 (X), show that fr = f ∗ (here fr is the function defined in Notation 10.61). Hint. Use Lemma 3.22. Exercise 10.101. If H is a Bauer simplicial space on K, then H is a CE-function space. Hint. The assertion follows immediately from Proposition 6.37. Exercise 10.102. If H is a CE-function space on a compact space K, prove that ChH (K) is closed. Hint. Let x ∈ K \ ChH (K). By Theorem 3.24, there is f ∈ C(K) such that f (x) < f ∗ (x). Then U := {y ∈ K : f ∗ (y) − f (y) > 0} is an open neighborhood of x and U ∩ ChH (K) = ∅. Exercise 10.103. Let H be a CE-function space on K. Prove that ChH (K) f ∗ is continuous for every f ∈ C(ChH (K)). Hint. Imitate the proof of Lemma 10.80. Exercise 10.104. Let K, L be compact spaces and ϕ : K → L be a continuous surjection. Prove that ϕ is open if and only if fϕ (y) := sup{f (x) : x ∈ ϕ−1 (y)},

y ∈ L,

is continuous on L for each f ∈ C(K). Hint. For necessity use the proof of Lemma 10.63. For sufficiency, asssume that ϕ is not open. Then there exist an open set U ⊂ K and x ∈ U such that ϕ(x) is not in the interior of ϕ(U ). Let f ∈ C(K) be positive such that f (x) > 0 and f = 0 on K \ U . Then the function fϕ is not continuous at ϕ(x).

384

10.11


Notes and comments

The idea of the proof of Theorem 10.2 goes back to R. Wittmann [474], where a more general result on the existence of a carrier in the sense of potential theory is established. A remarkable feature of the proof is that neither the Hahn–Banach theorem nor Bauer’s minimum principle are involved, thus Zorn’s lemma is not needed. A more general result than that of Theorem 10.2 is proved in H. Bauer [38]. A different approach yielding a very general existence result of the Shilov boundary containing results previously obtained can be found in R. Wittmann [475]. The relation to the Choquet boundary is studied there in detail. Wittmann’s method is based on the notion of so-called Shilov points. We also refer the reader to H. S. Bear [46]. The idea of I-envelopes can be traced to Section 5 in V. P. Fonf, J. Lindenstrauss and R. R. Phelps [179]. It was elaborated in V. P. Fonf and J. Lindenstrauss [178] where Theorem 10.7 is proved. Proposition 10.6 is taken from a paper by O. Kalenda [258] that along with [257] is devoted to a systematic study of I-envelopes. The notion of I-envelopes points to the following problem. Problem 10.105. Let E be a Banach space. Does there exist a topology τ (or even a locally convex topology τ ) on E ∗ such that I-env(A) = coτ (A) for any A ⊂ E ∗ ? It is not difficult to realize that I-env(A) is just the norm closed convex hull of A in the case when E ∗ is separable. Also, if E does not contain `1 , then it is not difficult to realize the existence of a desired locally convex topology on E ∗ . But in general the problem seems to be open. Theorem 10.8 is a consequence of a theorem by G. Rodé contained in [390]. For the case of extreme points it was proved by R. Haydon [220]. We refer the reader to papers by G. Godefroy [198] and [199] for more information related to this theorem. Theorem 10.9 is a particular version of R. C. James’ theorem characterizing weakly compact sets in Banach spaces (see [244] and [243]). The presented proof follows the same arguments as the proof of Theorem 3.55 in [173]. The notion of angelicity is due to D. H. Fremlin; it is explicitly defined, for example, in J. Bourgain, D. H. Fremlin and M. Talagrand [80], Definition 3A or in B. Cascales and G. Godefroy [98]. Theorem 10.11 is a simpler variant of Theorem 462B in [183]; it follows the paper by J. D. Pryce [379]. Theorem 10.12 was proved by J. Bourgain and M. Talagrand in [81]; a different proof, which we follow, appeared in S. S. Khurana [268]. Theorem 10.12 provides a positive answer in a particular case to the so-called Boundary problem formulated by G. Godefroy in [198]. Let E be a Banach space and B ⊂ BE ∗ be a boundary of E, that is, any element x ∈ E attains its norm on B. Is it true that any bounded σ(E, B)-compact set A in E is also weakly compact? Positive answers were known for many particular cases: A is convex (see Proposition 5.1(β) in M. De Wilde [136] and also Corollary 1 on p. 100 in K. Floret [176]);


385

B is w∗ -relatively sequentially compact in E ∗ (see B. Cascales and G. Vera [101], Corollary C); E = C(K), the space of continuous functions on a compact space K (see B. Cascales and G. Godefroy [98], Theorem 5); E = `1 (Γ) for a set Γ (see B. Cascales and R. Shvydkoy [100], Theorem 4.9]); E does not contain a subspace isomorphic to `1 ([0, 1]) (see B. Cascales, G. Manjabacas and G. Vera [99], Theorem D); E is an L1 -predual, (see J. Spurný [422], Theorem 1.1). This problem was solved in full generality by H. Pfitzner [373]. In the same paper he also proves that BE is angelic in the topology σ(E, B) for any boundary B ⊂ BE ∗ . For unbounded sets, an example constructed by W. Moors and E. Reznichenko [350], Section 4 (see also Section 4 in [422]) shows that the angelicity need not hold even for the topology σ(E, ext BE ∗ ). Several results on boundary topologies in L1 preduals can be found in W. B. Moors and J. Spurný [351]. Theorem 10.18 is proved in a paper A. J. Lazar [292] as a more general version of the classical Banach–Stone theorem (see for example Theorem 3.43 in [173]). The more precise description of isometries of spaces of affine continuous functions on simplices contained in Theorem 10.17 is given in the paper by T. S. S. R. K. Rao [385] and A. Curnock, J. Howroyd and N. C. Wong [128]. A. J. Ellis and W. S. So presented in [168] an example attributed to J. T. Chan showing that Theorem 10.18 does not hold for general compact convex set. An interesting question is whether a pair K1 , K2 of compact spaces is homeomorphic if the spaces C(K1 ) and C(K2 ) are isomorphic. It has turned out, as shown in a paper by A. Amir [14], that this is the case if there exists an isomorphism T : C(K1 ) → C(K2 ) such that kT k · kT −1 k < 2. H. B. Cohen showed in [122] that the bound 2 is the best possible. The case of C(K) spaces was further generalized for spaces of affine continuous functions on compact convex sets by C. H. Chu and H. B. Cohen [119]. They proved the following result: Let X, Y be compact convex sets such that their extreme points are weak peak points and let T : Ac (X) → Ac (Y ) be an isomorphism with kT k · kT −1 k < 2. If • X, Y are metrizable, or • ext X, ext Y are closed sets, then ext X is homeomorphic to ext Y . Another result of [119] is the following theorem: Let X, Y be simplices such that their extreme points are weak peak points and let T : Ac (X) → Ac (Y ) be an isomorphism with kT k · kT −1 k < 2. Then ext X is homeomorphic to ext Y . An example constructed in H. U. Hess [224] shows that the assumption on extreme points cannot be omitted. It seems to be an open problem whether the following assertion holds. Problem 10.106. Let X, Y be compact convex sets such that their extreme points are weak peak points and let T : Ac (X) → Ac (Y ) be an isomorphism with kT k·kT −1 k < 2. Is it true that ext X is homeomorphic to ext Y ?

386


The results of Section 10.3 are based mainly upon papers by J. P. R. Christensen. Theorem 10.24(a) appears in [114], part (b) is standard and we include it just for a comparison (see, for example, Proposition 272O in [181]). The first version of Theorem 10.26 can be found in [114]; we follow the improved variant of [116], Chapter 5, and [456], Theorem 2.3.13. Theorem 10.30 and Exercise 10.92 are taken from J. P. R. Christensen [115]. Theorems 10.31 and 10.37 on automatic boundedness can be found in J. Spurný [430] and J. Saint Raymond [407]. Theorem 10.32 can be found in J. P. R. Christensen [116], Theorem 5.8, and [114], Theorem 4. A result related to Theorem 10.32 is due to M. Talagrand who proved in [442] that an affine function f : [0, 1]N → R measurable with respect to the sets with the Baire property is continuous. (We note that [0, 1]N is affinely homeomorphic to (B`∞ , w∗ ).) We do not know the answer to the following problem. Problem 10.107. Let f : X → R be an affine function on a compact convex set X that has the Baire property. Is f necessarily bounded? Section 10.4 follows the paper [220] by R. Haydon. We refer the reader to S. A. Argyros, G. Godefroy and H. P. Rosenthal [19] for information on spaces not containing `1 . Section 10.5 contains results of several authors. Theorem 10.51 can be found in [350] with a slightly different proof; for the case of extreme points it was proved by H. H. Corson [127] and a different proof was given by R. Haydon [219]. The main Theorem 10.56 is an amalgamation of results from papers by M. Hervé [223], B. MacGibbon [333], J. E. Jayne [247] (see also [394], Section 5.10), G. Debs [141] and E. A. Reznichenko [387]. We remark that our proof of the result from [141] is different since it does not use the α-favorability of the set of extreme points of a compact convex set. We refer the reader to a paper by E. A. Reznichenko [387] where the following result (among others) was proved: Let X be a compact convex set such that •

X is a simplex, or

•

ext X is a Lindelöf space.

Then w(X) = w(ext X) = nw(ext X). (We recall that w(Y ) and nw(Y ) are the weight and network weight, respectively, of a topological space Y .) It is an open problem whether the conclusion of this theorem holds without additional assumptions on the set X. Problem 10.108. Let X be a compact convex set. Is it true that w(X) = w(ext X) = nw(ext X)? We also mention the following problem from B. MacGibbon [333].


387

Problem 10.109. Does there exist a nonmetrizable perfectly normal compact convex set? (We recall that a topological space is perfectly normal if it is normal and every open set is of type Fσ .) Descriptive structure of extreme and exposed points of convex sets was studied by many authors, we refer the reader to H. H. Corson [126], G. Choquet, H. H. Corson and V. Klee [112], J. E. Jayne and C. A. Rogers [248], P. Holický and V. Kom´ınek [233], P. Holický and M. Laczkovich [234] or P. Holický and T. Keleti [232]. The problem of transferring properties of a boundary of a compact convex set to the whole set is investigated in O. Kalenda and J. Spurný [260]. Theorem 10.57 is from M. Kaˇcena and J. Spurný [254]. It was stated in a more general version without a proof in S. Teleman [455]. However, Example 10.97 from C. J. K. Batty [32] shows that the result may fail in general. Theorem 10.59 and Theorem 10.60 are from D. A. Edwards and G. F. Vincent-Smith [161]; our proofs also use methods of E. A. Reznichenko [387]. An improvement can be found in D. A. Edwards [158], see also G. F. Vincent-Smith [461], A. de la Pradelle [134] and [135]. Theorem 10.66 and Corollary 10.67 are taken from J. Vesterstrøm [460]. Theorem 10.68 for the set of extreme points of a compact convex set is attributed in [5] to G. Choquet (unpublished result). The general result for Choquet boundaries is proved in D. A. Edwards [156] (see also Theorem I.5.13 in [5]). It can be found in G. Choquet [108], Theorem 27.9, that the set of extreme points of a compact convex set is even α-favorable. The result of Subsection 10.7.B can be found in G. Choquet [108], Theorem 29.9, and in R. Haydon [218]. Our exposition follows the paper [218] (see also [24], Section 3.6). P. R. Andeneas showed in [15] that every completely metrizable locally separable space is homeomorphic to ext X for a suitable simplex X. A related result is due to A. J. Lazar [294], who proved that any uncountable Polish space is homeomorphic to the set of extreme points of a closed convex body in `2 . A construction of simplices with a prescribed set of extreme points can be also found in P. J. Stacey [436]. A related result is due to D. Bensimon [51] and it reads as follows: Let M be a complete metric space of zero covering dimension. Then there exists a standard Choquet simplex X such that M is homeomorphic to ext X. Theorem 10.71 is taken from M. Talagrand [445]. We mention here another result of M. Talagrand [448]: There exists a simplex X such that •

ext X is K-analytic,

•

ext X is not contained in the smallest family of subsets of ext X containing compact sets and closed with respect to countable unions and intersections,

•

ext X is of type Kσδ in β(ext X),

388


•

there exist an open set U ⊂ ext X and a point ω ∈ ext X such that ext X = U ∪ {ω},

•

ext X \ ext X is discrete.

Section 10.8 follows the paper J. Saint Raymond [408]. Subsection 10.9.A is taken from papers A. Clausing and S. Papadapoulou [121], S. Papadopoulou [369] and [368] and R. C. O’Brien [365]. We also refer the reader ˚ Lima [302], A. Clausing and G. Mägerl [120] and to G. Debs [138] and [140], A. H. Höllein [237]. The results of Subsection 10.9.B follow the paper J. Vesterstrøm [460]. Exercise 10.90 is rather standard; we refer the reader to Lemma in J. P. R. Christensen [115] and H. H. Schaefer [409]. Exercise 10.92 is from [115]. Exercise 10.94 can be found in E. M. Alfsen [4], [2] and A. J. Lazar [292]. Exercise 10.95 is a scalar version of the Hahn–Vitali–Saks theorem (see, for example, N. Dunford and J. T. Schwartz [150], pp. 158–159). Examples of Exercises 10.97–10.99 are taken from [254].

Chapter 11

Continuous and measurable selectors

This chapter is devoted to continuous and Borel measurable selectors of multivalued mappings defined on compact convex sets. We start with the proof of the Lazar selection theorem 11.6 which is an affine analogue of the Michael selection theorem. Theorems 11.8 and 11.9 provide its further variants. First applications to affine retractions and extenders are contained in Theorem 11.13. Characterizations of separability of (Ac (X))∗ for the case of a simplex X is given in Theorem 11.14, the Michael selection theorem for compact spaces and the Dugundji theorem are presented as Theorems 11.17 and 11.18. Extension of Baire-α functions from compact subsets of extreme points is investigated in Theorem 11.23 as well as an application on extension of Baire-α functions from compact subsets of completely regular spaces. The last application shows that the mapping T from Definition 6.7 can be pointwise approximated by continuous mappings (see Theorem 11.27). The next section presents a general selection theorem and several of its applications. We start in Subsection 11.5.A with basic facts on measurability of multivalued mappings. The main selection result is contained in Theorem 11.35. Next, Subsection 11.5.C collects several applications of this result. Perhaps the most notable is Talagrand’s result on the existence of a Borel measurable mapping on a metrizable compact convex set X, that assigns to a point x ∈ X a maximal and simplicial measure m(x) representing x (see Theorem 11.41).

11.1

The Lazar selection theorem

Definition 11.1 (Affine mappings). A multivalued mapping ϕ : X → 2E from a convex set X to a vector space E is called affine if ϕ(x) is convex and αϕ(x) + (1 − α)ϕ(y) ⊂ ϕ(αx + (1 − α)y) whenever x, y ∈ X, α ∈ [0, 1]. Definition 11.2 (Semicontinuous mappings). A multivalued mapping ϕ : X → 2Y between topological spaces X and Y is termed lower semicontinuous if ϕ−1 (U ) := {x ∈ X : ϕ(x) ∩ U 6= ∅} is open in X for each open set U ⊂ Y . It is called upper semicontinuous if ϕ−1 (U ) := {x ∈ X : ϕ(x) ⊂ U } is open in X for each open set U ⊂ Y .

390

11 Continuous and measurable selectors d

Lemma 11.3. Let ϕ : X → 2R be a lower semicontinuous nonempty valued affine mapping on a simplex X and let U ⊂ Rd be a neighborhood of 0. Then there exists a continuous affine mapping h : X → Rd such that h(x) ∈ ϕ(x) + U for every x ∈ X. Proof. We proceed by induction on the dimension. If d = 1, let ϕ : X → 2R and U ⊂ R be as in the hypothesis. We find an open symmetric interval V ⊂ R such that 2V ⊂ U . Let g1 (x) := inf(ϕ(x) + V ),

x∈X

g2 (x) := sup(ϕ(x) + V ),

and

x ∈ X.

Since ϕ is lower semicontinuous and affine, g2 , −g1 are lower semicontinuous concave functions with g1 ≤ g2 . By the Edwards in-between theorem 6.6 there exists a continuous affine function h : X → R such that g1 ≤ h ≤ g2 . By the choice of V , h(x) ∈ ϕ(x) + U for every x ∈ X. d+1 Assume that the assertion holds for Rd . Let ϕ : X → 2R be a lower semicontinuous affine mapping and U be a neighborhood of 0. Let p and q be canonical projections of Rd+1 onto R and Rd , respectively. Let Up and Uq be open symmetric neighborhoods of 0 in R and Rd , respectively, such that Up × Uq ⊂ U . As p◦ϕ : X → 2R is lower semicontinuous and affine, we may apply the preceding argument to get a continuous affine function k : X → R such that k(x) ∈ (p ◦ ϕ)(x) + Up , Then the mapping from X to 2R

d+1

x ∈ X.

(11.1)

defined as

p−1 (k(x) + Up ),

x ∈ X,

is affine and lower semicontinuous. Moreover, ψ(x) := p−1 (k(x) + Up ) ∩ ϕ(x),

x ∈ X,

(11.2)

d+1

is a nonempty valued mapping from X to 2R . We claim that ψ is affine and lower semicontinuous. Since the intersection of a pair of affine mappings is again affine, we proceed to the lower semicontinuity of ψ. Let V ⊂ Rd+1 be an open nonempty set. Given a point x ∈ ψ−1 (V ), we find t ∈ ψ(x) ∩ V . Let W be an open ball around t such that W ⊂ p−1 (k(x) + Up ). Since the function k is continuous, we can find an open neighborhood G1 of x such that W ⊂ p−1 (k(y) + Up ) for each y ∈ G1 . As ϕ is lower semicontinuous and x ∈ ϕ−1 (W ), there exists an open set G2 containing x such that G2 ⊂ ϕ−1 (W ).

11.1 The Lazar selection theorem

391

We claim that G1 ∩ G2 ⊂ ψ−1 (V ). Indeed, if y ∈ G1 ∩ G2 , there is a point s ∈ ϕ(y) ∩ W . Since y ∈ G1 , s ∈ p−1 (k(y) + Up ). Then s ∈ p−1 (k(y) + Up ) ∩ ϕ(y) ∩ W ⊂ ψ(y) ∩ V, proving the claim. d Obviously, the mapping q ◦ ψ : X → 2R is also affine and lower semicontinuous. By the induction hypothesis, there exists an affine mapping l : X → Rd such that l(x) ∈ (q ◦ ψ)(x) + Uq ,

x ∈ X.

(11.3)

To finish the proof it is enough to verify that the mapping h : x 7→ (k(x), l(x)),

x ∈ X,

is the desired one. Obviously, h is continuous and affine. Let x ∈ X be given. By (11.3), l(x) = q(t) + uq for some t ∈ ψ(x) and uq ∈ Uq . As t ∈ ψ(x), p(t) = k(x) + up for some up ∈ Up . Thus (k(x), l(x)) = (p(t), q(t)) + (−up , uq ) = t + (−up , uq ) ∈ ϕ(x) + U. This finishes the inductive step as well as the proof. Lemma 11.4. Lex ϕ : X → 2E be an affine lower semicontinuous nonempty valued mapping from a simplex X to a locally convex space E and let U ⊂ E be a neighborhood of 0. Then there exists a continuous affine mapping h : X → E such that h(x) ∈ ϕ(x) + U for every x ∈ X. Proof. Let ϕ : X → 2E and U ⊂ E be as in the hypothesis. We may assume that U is an open convex balanced neighborhood of 0 in E. Since ϕ is lower semicontinuous, the family 1 n 1o , y ∈ E, Gy := ϕ−1 y − U = x ∈ X : y ∈ ϕ(x) + 2 2 is an open covering of X. We choose y1 , . . . , yd such that X⊂

d [

G yi .

(11.4)

i=1

By setting d n X 1 o d ψ(x) := λ = (λ1 , . . . , λn ) ∈ R : λi yi ∈ ϕ(x) + U , 2 i=1

we get an affine lower semicontinuous mapping with nonempty values.

x ∈ X,

392

11 Continuous and measurable selectors

Indeed, (11.4) yields that ψ is nonempty valued, and, obviously, ψ is affine. Its lower semicontinuity follows from the observation that d n X o 1 λi yi − U 6= ∅ . ψ(x) = λ ∈ Rd : ϕ(x) ∩ 2 i=1

P Let W ⊂ Rd be an open convex balanced neighborhood of 0 such that di=1 wi yi ∈ 1 4 U whenever w = (w1 , . . . , wd ) ∈ W . Lemma 11.3 provides an affine continuous mapping h0 : X → Rd such that h0 (x) = (h1 (x), . . . , hd (x)) ∈ ψ(x) + W,

x ∈ X.

Then the mapping h : X → E defined as h(x) :=

d X

hi (x)yi ,

x ∈ X,

i=1

is the desired continuous affine mapping. Indeed, given x ∈ X, let λ ∈ ψ(x) and w ∈ W be such that h0 (x) = λ + w. Then d X i=1

hi (x)yi =

d X i=1

λ i yi +

d X i=1

1 1 wi yi ∈ ϕ(x) + U + U ⊂ ϕ(x) + U. 2 4

This finishes the proof. Lemma 11.5. Let X be a simplex, E be a locally convex space, ϕ : X → 2E be a lower semicontinuous affine multivalued mapping, h : X → E be a continuous mapping and U ⊂ E be a convex balanced neighborhood of 0. Then ψ(x) := ϕ(x) ∩ (h(x) + U ),

x ∈ X,

is a lower semicontinuous affine mapping provided ψ(x) is nonempty for each x ∈ X. Proof. Let V be an open subset of E and x ∈ ψ−1 (V ). We choose t ∈ ϕ(x) ∩ (h(x) + U ) ∩ V . Let u ∈ U be such that t = h(x) + u. We find a balanced open convex neighborhood W of 0 such that t + W ⊂ V and u + 2W ⊂ U . Let G1 , G2 be open neighborhoods of x such that G1 ⊂ ϕ−1 (t + W ) and

h(y) − h(x) ∈ W for y ∈ G2 .

Let y ∈ G1 ∩ G2 . We pick s ∈ ϕ(y) ∩ (t + W ) and let w ∈ W satisfy s = t + w. Then s = t + w = h(y) + u + w + h(x) − h(y) ∈ h(y) + u + w + W ⊂ h(y) + u + 2W ⊂ h(y) + U

11.1 The Lazar selection theorem

393

and s ∈ ϕ(y) ∩ (t + W ) ⊂ ϕ(y) ∩ V. Hence s ∈ ϕ(y) ∩ (h(y) + U ) ∩ V and G1 ∩ G2 ⊂ ψ−1 (V ). This concludes the proof. Theorem 11.6 (Lazar). Let ϕ : X → 2E be a lower semicontinuous affine mapping from a simplex X to a Fréchet space E with nonempty closed values. Then there exists a continuous affine mapping h : X → E such that h(x) ∈ ϕ(x) for each x ∈ X. Proof. Given ϕ : X → 2E , let ρ be a translation invariant complete metric on E. We select a decreasing sequence {Un } of open convex balanced neighborhoods of 0 such that they form a base of neighborhoods of 0. We take p0n to be the Minkowski functional of the set Un and define pn := p01 + · · · + p0n . Then {pn } is an increasing sequence of pseudonorms on E that determines its topology. Moreover, if {yn } is a sequence in E which is Cauchy in every pseudonorm pn , it is convergent. For y ∈ E and r > 0 we set Un (y, r) := {z ∈ E : pn (y − z) < r}. By Lemma 11.4, there exists a continuous affine mapping h1 : X → E such that h1 (x) ∈ ϕ(x) + U1 (0, 2−1 ),

x ∈ X.

We inductively find continuous affine mappings hn : X → E such that hn+1 (x) ∈ ϕ(x) ∩ (hn (x) + Un (0, 2−n ) + Un (0, 2−n−1 ), x ∈ X, for each n ∈ N. For the inductive step, suppose that the mappings hk have been found for k = 1, . . . , n. Then we apply Lemma 11.4 to ϕ(x) ∩ (hn (x) + Un (0, 2−n )),

x ∈ X,

to get the desired mapping hn+1 (here we use Lemma 11.5). The completeness of E guarantees that the sequence {hn } converges uniformly to a function h : X → E. Since ϕ has closed values, h(x) ∈ ϕ(x) for each x ∈ X and the proof is complete. Theorem 11.7. Let X be a compact convex set. Then the following assertions are equivalent. (i) X is a simplex, (ii) any lower semicontinuous affine mapping ϕ : X → 2E with nonempty closed convex values in a Fréchet space E admits a continuous affine selection.

394


Proof. By Theorem 11.6, (i) =⇒ (ii). Conversely, assuming (ii), let f, −g ∈ C(X) be convex functions such that f ≤ g. Then ϕ : X → 2R defined as ϕ(x) := [f (x), g(x)],

x ∈ X,

is an affine lower semicontinuous function. According to the assumption, there exists a continuous affine function h ∈ Ac (X) such that f ≤ h ≤ g. By Theorem 6.6 and Theorem 6.56, X is a simplex. Theorem 11.8. Let F be a closed face of a simplex X, ϕ : X → 2E be an affine lower semicontinuous mapping to a Fréchet space E with nonempty closed convex values and f : F → E be a continuous affine mapping with f (x) ∈ ϕ(x), x ∈ F . Then there exists a continuous affine selection g : X → E from ϕ such that g = f on F . Proof. Given the objects as in the hypotheses, let ψ : X → 2E be defined as ( f (x), x ∈ F, ψ(x) := ϕ(x), x ∈ X \ F. A routine verification shows that ψ is a lower semicontinuous affine mapping and thus admits a continuous affine selection g. Obviously, g satisfies the required conditions. Corollary 11.9. Let ϕ : X → 2K be a lower semicontinuous affine mapping from a simplex X to a metrizable compact convex set K. Then ϕ admits a continuous affine selection. Proof. By Proposition 2.45, we may assume that K is a compact convex subset of `2 . The conclusion now follows from Theorem 11.6.

11.2

Applications of the Lazar selection theorem

Lemma 11.10. Let H be a function space on a compact space K and let F be a metrizable subset of K. Then coH F is metrizable as well. Proof. If F is metrizable, M1 (F ) is metrizable as well. By Proposition 8.18, the barycentric mapping from Proposition 3.37 r : M1 (F ) ∩ M(H) → coH F is surjective. Since a continuous image of a compact metrizable space is metrizable (see Lemma A.34), the proof is complete. Lemma 11.11. Let H be a simplicial function space on a compact space K. Let F ⊂ ChH (K) be a compact set and let C = coH F . Then

11.2 Applications of the Lazar selection theorem

395

(a) δx (F ) = 1 for each x ∈ C, (b) C is a Choquet set, (c) the mapping x 7→ δx , x ∈ C, is continuous. Proof. Let F be as in the hypothesis and x ∈ C. By Proposition 8.18, there exists a measure µ ∈ Mx (H) carried by F . Then µ is H-maximal and µ = δx as H is simplicial. Since (b) follows from Proposition 8.31, we proceed to (c). Let {xα } be a net of points in C converging to x and assume that δxα 9 δx . By passing to a subnet if necessary we may assume that δxα → µ where µ 6= δx . Then µ ∈ Mx (H) and, moreover, µ is maximal (see Corollary 3.60). By the simpliciality of H, µ = δx , a contradiction. This concludes the proof. Definition 11.12 (Retractions and extenders). If F is a subset of a topological space K, a continuous mapping ρ : K → F is a retraction if ρ(x) = x for x ∈ F . The set F is called a retract. Let F be a subset of compact space K and H be a subspace of C(K). Let G := {h|F : h ∈ H}. A mapping L:G→H is a called an extender if L is a linear operator, and for each f ∈ H, • Lf = f on F , • (Lf )(K) ⊂ co f (F ). Theorem 11.13. Let X be a simplex. (a) If F is a closed metrizable face of X, then there exists an affine retraction ρ : X → F (that is, ρ : X → F is a continuous affine mapping such that ρ(x) = x, x ∈ F ). (b) If F is a closed metrizable face of X, then there exists an extender L : Ac (F ) → Ac (X). (c) Let K ⊂ ext X be a metrizable compact set. Then there exists an extender L : C(K) → Ac (X). Proof. For the proof of (a), let ϕ : X → 2F be defined as ( x, x ∈ F, ϕ(x) := F, x ∈ X \ F. Then ϕ satisfies the assumptions of Corollary 11.9 and thus ϕ admits a continuous affine selector. This selector is the required affine retraction of X onto F . Assertion (b) immediately follows from (a), since the mapping h 7→ h ◦ ρ, is the required extender.

h ∈ Ac (F ),

396


For the proof of (c) we notice that F = co K is a closed metrizable face such that δx (K) = 1 for each x ∈ F and the mapping x 7→ δx , x ∈ F , is continuous (see Lemma 11.10 and Lemma 11.11). If ρ : X → F is a continuous affine retraction, the operator Lf (x) := δρ(x) (f ), x ∈ X, f ∈ C(K), is the required extender. This concludes the proof. Theorem 11.14. For a metrizable simplex X, the following assertions are equivalent: (i) ext X is uncountable, (ii) Ac (X) contains an isometric copy of C({0, 1}N ), (iii) Ac (X) contains an isometric copy of `1 , (iv) (Ac (X))∗ is nonseparable. Proof. Assume that ext X is uncountable. Since ext X is a Polish space (see Proposition 3.43), ext X contains a homeomorphic copy C of {0, 1}N . By Theorem 11.13(c), there exists an extender L : C(C) → Ac (X). Hence L(C(C)) is an isometric copy of C({0, 1}N ) contained in Ac (X). Thus (i) =⇒ (ii). Since C({0, 1}N ) contains an isometric copy of any separable Banach space (see [173], proof of Theorem 5.17), (ii) =⇒ (iii). Obviously, (iii) =⇒ (iv). Finally, assume that ext X is countable. Then Mbnd (X) is isometric to `1 (ext X) and the mapping π : Mbnd (X) → (Ac (X))∗ defined as π(µ) := µ|(Ac (X))∗ is easily seen to be an isometry (use Proposition 6.9). Hence (Ac (X))∗ is separable, which proves (iv) =⇒ (i). Theorem 11.15. Let H be a simplicial function space on a compact space K. Let F ⊂ K be a metrizable Choquet set and A := {h|F : h ∈ Ac (H)}. Then there exists an extender L : A → Ac (H). Proof. Assume that F is a Choquet set in K, X := S(Ac (H)) and φ : K → X is the embedding from Section 4.3. By Theorem 8.62, F is a P -set with respect to the function space Ac (H). As F is a Choquet set in K with respect to Ac (H) (see Lemma 8.16), by Theorem 8.39 there exists a closed face H ⊂ X such that φ(F ) = H ∩ φ(K). By Proposition 4.26(c) and Proposition 2.64, φ(F ) = H ∩ φ(K) ⊃ H ∩ ext X = ext H.

(11.5)

By the Milman theorem 2.43, co φ(F ) = H. By Lemma 11.10, H is metrizable.

11.2 Applications of the Lazar selection theorem

397

It follows from (11.5) that, if b h1 , b h2 ∈ Ac (X) satisfy b h1 = b h2 on φ(F ), then b h1 = c b h2 on H. Thus for any function h ∈ A there exists a unique function E1 h ∈ A (H) such that E1 h ◦ φ = h on F . Then E1 is a linear positive operator of norm 1. Since X is a simplex (see Theorem 6.54), Theorem 11.13 provides an extender E2 : Ac (H) → Ac (X). Then L : A → Ac (H) defined as Lh = (E2 ◦ E1 )(h) ◦ φ,

h ∈ A,

is the required extender. This concludes the proof. Theorem 11.16. Let H be a simplicial function space on a compact space K. Let F ⊂ ChH (K) be a metrizable compact set. Then there exists an extender L : C(F ) → Ac (H). Proof. Let X denote the simplex S(Ac (H)) and φ : K → X be the embedding from Section 4.3. If F ⊂ ChH (K) is a compact metrizable set, we use Theorem 11.13(c) to obtain an extender E : φ(F ) → Ac (X). Then Lf (x) := E(f ◦ φ−1 )(φ(x)),

x ∈ K, f ∈ C(F ),

is an extender from C(F ) into Ac (H). Theorem 11.17 (The Michael selection theorem). Let ϕ : K → 2E be a lower semicontinuous mapping from a compact space K to a Fréchet space E such that ϕ has nonempty closed convex values. Then ϕ admits a continuous selection. Proof. Given ϕ : K → 2E as above, let ϕ b : M1 (K) → 2E be defined as ( ϕ(x), µ = εx , x ∈ K, ϕ(µ) b := E otherwise. It is straightforward to verify that ϕ b is a lower semicontinuous affine mapping with nonempty closed convex values. Theorem 11.6 provides an affine continuous selection fb : M1 (K) → E for ϕ. b Then f (x) := fb(εx ),

x ∈ K,

is a continuous selection for ϕ. Theorem 11.18 (Dugundji). Let Y be a metrizable compact subset of a completely regular space X and f : Y → E be a continuous mapping to a locally convex space E. Assume that co f (Y ) is a compact subset of E (this is automatically satisfied if E is a Fréchet space). Then there exists a continuous mapping F : X → E such that F = f on Y and F (x) ⊂ co f (Y ), x ∈ X.

398


ˇ Proof. If X is as in the theorem, we consider its Cech–Stone compactification βX. 1 1 Then F := M (Y ) is a metrizable closed face of M (βX). Moreover, the function fb: F → E defined as Z µ 7→

f (y) dµ(y),

µ ∈ F,

Y

is continuous (we consider the Pettis integral). We remark that fb is well defined by the assumption on co f (Y ). By Corollary 11.13, there exists a continuous affine retraction r : M1 (βX) → F . Then F (x) := fb(r(εx )), x ∈ βX, is a continuous extension of f with the required properties. This concludes the proof.

Theorem 11.19 (Borsuk–Dugundji). Let X be a completely regular space and Y its metrizable compact subset. Then there exists a linear extender L : C(Y ) → C b (X). ˇ Proof. We consider the Cech–Stone compactification βX of X and let H := C(βX). c Then H = A (H) and ChH (βX) = βX. Thus the extender from Theorem 11.16 is the desired one.

11.3

The weak Dirichlet problem for Baire functions

Let H be a function space on a compact space K. We recall the inductive definition of functions of H-affine classes from Section 5.6. For any countable ordinal α and a family of functions H we define H0,b := H and having Hβ,b , β < α, already defined for an ordinal number α ∈ (0, ω1 ), we define H Sα,b to be the space of all pointwise limits of bounded sequences of functions from β 0. Proof. We select a countable base B of open sets in Y that is stable with respect to finite unions. We claim that [ {x ∈ X : diam ϕ(x) < ε} = ϕ−1 (B). (11.8) B∈B,diam B max fn (ϕ(x))

⇐⇒

ϕ(x) ⊂ fn−1 ((−∞, α)).

Hence ψn−1 (U ) ∈ Σ2 (S ). It follows from the Hahn–Banach theorem and density of {fn : n ∈ N} in Ac (Y ) that ∞ \ ψ(x) = ψn (x), x ∈ X. n=1

By Lemma 11.30, ψ is Σ2 (S )-upper semimeasurable.

406

11.5.B


Selection theorem

Let Y be a topological space and S ⊂ C(Y ) be a convex cone that separates points of Y and contains constant functions. Then for any compact K ⊂ Y , S K := {s|K : s ∈ S} is a function cone of continuous functions as defined in Chapter 7. Lemma 11.34. Let Y be a separable metrizable space and S ⊂ C(Y ) be a set. Then S is separable in the topology of the uniform convergence on compact subsets of Y . Proof. We fix a countable base {Vn : n ∈ N} of Y that is closed with respect to finite unions. For any n, m ∈ N with Vn ∩ Vm = ∅ we select a continuous function fn,m on Y such that ( 0 on Vm , fn,m = 1 on Vn . Let B consist of all products of finitely many just constructed functions and let A :=

n X

qi fi : qi ∈ Q, fi ∈ B, n ∈ N .

i=1

Then for any compact set K ⊂ Y , {f |K : f ∈ A} is dense in C(K) by the Stone– Weierstrass theorem A.29. We enumerate A as {gq : q ∈ N} and for any n, p, q ∈ N we find a function fnpq ∈ S such that 1 sup |fnpq (x) − gq (x)| < , p x∈Vn provided it exists. Otherwise we set fnpq := 0. We claim that the family {fnpq : n, p, q ∈ N} is dense in S in the topology of uniform convergence on compact sets. Indeed, let K ⊂ Y be compact, f ∈ S be arbitrary and ε > 0. We find p ∈ N with p2 < ε and select q ∈ N such that kf − gq kC(K) < p1 . Since the base {Vn : n ∈ N} is stable with respect to finite unions, an easy compactness argument provides n ∈ N such that supx∈Vn |f (x) − gq (x)| < p1 . Let fnpq ∈ S be the function chosen for the triple (n, p, q). Then kf − fnpq kC(K) ≤ sup |f (x) − fnpq (x)| < x∈Vn

2 < ε. p

This concludes the proof. Theorem 11.35. Let S be an algebra of sets in a set X, Y be a separable metrizable space and S ⊂ C(Y ) be a convex cone that separates points of Y and contains constant functions. Let ϕ : X → 2Y be a Σ2 (S )-measurable mapping with nonempty compact values. Then there exists a Σ2 (S )-measurable function f : X → Y such that f (x) ∈ ChS|ϕ(x) (ϕ(x)), x ∈ X.

11.5 Measurable selectors

407

Proof. We fix a compatible metric on Y . Using Lemma 11.34, we select a countable set Q := {vj : j ∈ N} ⊂ S that is dense in S in the topology of uniform convergence on compact sets. Let v0 ∈ S be an arbitrary strictly negative function on Y . We set Tj := {y ∈ Y : vj ≤ 0} and

Sj := {y ∈ Y : vj < 0},

j ≥ 0.

By induction, we construct mappings αn : X → N ∪ {0}, n ≥ 0, such that (a) αn is Σ2 (S )-measurable, n ≥ 0, (b) ϕ(x) ∩ Sαn (x) 6= ∅, x ∈ X and n ≥ 0, (c) vαn (x) > vαn−1 (x) on ϕ(x), x ∈ X and n ∈ N, (d) diam(ϕ(x) ∩ Tαn (x) ) < n1 , x ∈ X and n ∈ N. We start the construction by setting α0 (x) := 0, x ∈ X. Then the required property (a) is obvious and (b) is satisfied by the choice of v0 . Assume now that the construction has been completed for all k = 0, . . . , n − 1 for some n ∈ N. For each j ≥ 0 we define Aj := x ∈ X : ϕ(x) ∩ Sj 6= ∅ , Bj := x ∈ X : vαn−1 (x) < vj on ϕ(x) , 1 Cj := x ∈ X : diam(ϕ(x) ∩ Tj ) < . n It follows from Lemma 11.30 and Lemma 11.32 that both Aj and Cj are in Σ2 (S ). Further, the set [ Bj = ({x ∈ X : αn−1 (x) = m} ∩ {x ∈ X : vj − vm > 0 on ϕ(x)}) m≥0

=

[

−1 αn−1 (m) ∩ {x ∈ X : ϕ(x) ⊂ (vj − vm )−1 ((0, ∞))}

m≥0

is in Σ2 (S ) by the induction hypothesis and Σ2 (S )-upper semimeasurability of ϕ. Hence Dj := Aj ∩ Bj ∩ Cj ∈ Σ2 (S ), j ≥ 0. S We claim that X = ∞ j=0 Dj . Indeed, let x ∈ X be arbitrary. We denote u := vαn−1 (x) and use the induction hypothesis to realize that ϕ(x) ∩ {y ∈ Y : u(y) < 0} 6= ∅. By the Minimum principle 7.15(f) there exists y0 ∈ ChS|ϕ(x) (ϕ(x)) such that u(y0 ) < 1 ) such that U (y0 , r) ⊂ {y ∈ Y : u(y) < 0} and find a con0. We pick r ∈ (0, 2n tinuous function g on Y with g(y0 ) = u(y0 ) whose support is contained in U (y0 , r). Then for f := u ∨ g we get from Theorem 7.21 and Proposition 7.11(b) f (y0 ) = inf{s(y0 ) : s ∈ S, s > f on ϕ(x)}.

408


Since Q is dense in S with respect to the topology of uniform convergence on compact sets, we can find j ≥ 0 such that vj (y0 ) < 0

and

vj > f on ϕ(x).

(11.9)

It follows that ϕ(x) ∩ Tj ⊂ ϕ(x) ∩ spt g ⊂ U (y0 , r).

(11.10)

By (11.9) and (11.10), we get that x ∈ Dj as required. Using Lemma 11.31 we obtain a countable partition {Ej : j ≥ 0} of X consisting of sets from Σ2 (S ) such that Ej ⊂ Dj , j ≥ 0. We define αn : X → N ∪ {0} as αn (x) := m,

x ∈ Em .

Since αn is constant on each set from the partition {Ej : j ≥ 0}, αn is Σ2 (S )measurable. It follows from the inclusions Ej ⊂ Dj , j ≥ 0, that αn satisfies all the required conditions (b), (c), (d). This finishes the construction. We set ϕn (x) := Tαn (x) , x ∈ X, and use (b), (c) and (d) to deduce that f (x) := ϕ(x) ∩

∞ \

ϕn (x),

x ∈ X,

n=1

is a well-defined (single-valued) mapping from X to Y . Further, for x ∈ X we denote y := f (x) and show that y ∈ ChS|ϕ(x) (ϕ(x)). By (c), the function v(z) := sup vαn (x) (z), z ∈ ϕ(x), n∈N

is a lower semicontinuous S|ϕ(x) -concave function. Further, since {y} = ϕ(x) ∩ T∞ n=1 Tαn (x) , we get v(y) ≤ 0

and v > 0 on ϕ(x) \ {y}.

By Theorem 7.15(c), {y} is S|ϕ(x) -extremal, and thus y ∈ ChS|ϕ(x) (ϕ(x)). Since each mapping αn is Σ2 (S )-measurable, the mappings x 7→ ϕn (x), x ∈ X, are Σ2 (S )-upper semimeasurable. By Lemma 11.30, x 7→ f (x), x ∈ X, is Σ2 (S )-upper semimeasurable. Since f is single-valued, it is Σ2 (S )-measurable. This finishes the proof. Corollary 11.36. Let S be an algebra of sets in a set X, Y be a separable metrizable space and S ⊂ C(Y ) be a convex cone that separates points of Y and contains constant functions. Let ϕ : X → 2Y be an S -semimeasurable mapping with nonempty compact values. Then there exists a Σ2 (S )-measurable function f : X → Y such that f (x) ∈ ChS|ϕ(x) (ϕ(x)), x ∈ X. (11.11)


409

In particular, if X is a metrizable space and ϕ : X → 2Y with nonempty compact values is upper or lower semicontinuous, then there exists a mapping f ∈ Baf1 (X, Y ) satisfying (11.11) (see Definition 5.18). Proof. The first part follows from Lemma 11.29 and Theorem 11.35. If ϕ is upper or lower semicontinuous, then ϕ is Σ2 (Bas(X))-measurable, where Bas(X) is the algebra from Definition 5.13. An application of the first part yields a Σ2 (Bas(X))-measurable selection f satisfying (11.11). But this means that f is in Baf1 (X, Y ).

11.5.C

Applications of the selection theorem

We recall that, given a family of functions F, W(F) denotes the min-stable cone generated by F (see Definition 3.10). Theorem 11.37. Let E be a metrizable separable locally convex space and S be an algebra of sets in a set X. Let ϕ : X → 2E be a Σ2 (S )-semimeasurable mapping with nonempty compact convex values. Then there exists a Σ2 (S )-measurable mapping f : X → E such that f (x) ∈ ext ϕ(x),

x ∈ X.

Proof. We apply Corollary 11.36 to the separable metrizable space E and the cone S := W({x∗ + c : x∗ ∈ E ∗ , c ∈ R}). Then ChS|ϕ(x) (ϕ(x)) = ext ϕ(x) (see Proposition 7.6(c) and Exercise 7.63) and hence the assertion follows. Theorem 11.38. Let E be a separable metrizable locally convex space and S be ∗ ∗ an algebra of sets in a set X. Let ϕ : X → 2(E ,w ) be an S -lower semimeasurable mapping with nonempty convex w∗ -compact values. Then there exists a Σ2 (S )measurable mapping f : X → (E ∗ , w∗ ) such that f (x) ∈ ext ϕ(x),

x ∈ X.

Proof. Let {Un } be a basis of neighborhoods of 0 in E and let Yn be the polar of Un , that is, Yn := {x∗ ∈ E ∗ : |x∗ (x)| ≤ 1 for each x ∈ Un }, n ∈ N. We set

( ϕ−1 (Y1 ), n = 1, Xn := −1 −1 ϕ (Yn ) \ ϕ (Yn−1 ), n ≥ 2.

Then {Xn : n ∈ N} is a partition of X consisting of sets from S and ϕ|Xn : ∗ Xn → 2(Yn ,w ) is an S -lower semimeasurable mapping for each n ∈ N. For each n ∈ N, (Yn , w∗ ) is a compact metrizable space and we can apply Corollary 11.36 to

410


the cone W({x∗ + c : x∗ ∈ E ∗ , c ∈ R}) to obtain a Σ2 (S )-measurable selection fn : Xn → (Yn , w∗ ) such that fn (x) ∈ ext ϕ(x), x ∈ X. Then the mapping f (x) := fn (x),

x ∈ Xn , n ∈ N,

is the desired selection. This concludes the proof. Definition 11.39 (Convex cone S on M+ (X)). If E is a locally convex space, we consider the convex cone S := W({x∗ + c : x∗ ∈ E ∗ , c ∈ R}) on E. If X is a compact convex set in E, any function s ∈ W({x∗ + c : x∗ ∈ E ∗ , c ∈ R}) can be regarded as a continuous function on M+ (X) via the identification with µ 7→ µ(s), µ ∈ M+ (X). Hence we can view S as a convex cone of continuous functions on M+ (X) that separates points of M+ (X) and contains constant functions. If X ⊂ E is a compact convex set and x ∈ X, then the set Mx (X) of all probability Radon measures on X representing x is a compact convex set (in the w∗ -topology). Moreover, it is an ordered compact convex set (see Section 7.5) and the cone S|Mx (X) is dense in the cone −Lc (Mx (X)) (here Lc (Mx (X)) is the cone of all continuous affine isotone functions on Mx (X); see Definition 7.41). Hence by Theorem 7.54(d), ChS|Mx (X) (Mx (X)) = Ch−Lc (Mx (X)) (Mx (X)) = ext((Mx (X))max ). (Here (Mx (X))max is the set of all maximal elements in Mx (X).) Hence µ ∈ ext(Mx (X))max is both a maximal and an extreme point of Mx (X) (see Theorem 7.61). Theorem 11.40. Let S be an algebra in a set X and K be a metrizable compact convex subset of a locally convex space E. Let ϕ : X → 2K be an S -upper semimeasurable mapping with nonempty compact convex values and f : X → K be an S -measurable mapping with f (x) ∈ ϕ(x) for x ∈ X. Then there exists a Σ3 (S )-measurable mapping m : X → M1 (K) such that m(x) ∈ ext Mf (x) (ϕ(x)) max , x ∈ X. Proof. We recall that r : M1 (K) → K is the barycentric mapping. Let S be the cone on M+ (X) as above. 1 We define a multivalued mapping M : X → 2M (K) as M (x) := Mf (x) (ϕ(x)),

x ∈ X.

We claim that M is a Σ2 (S )-upper semimeasurable mapping.


411

Indeed, we notice that M (x) = M1 (ϕ(x)) ∩ r−1 (f (x)),

x ∈ X,

and the mappings M1 (x) := M1 (ϕ(x)),

M2 (x) := r−1 (f (x)),

x ∈ X,

are Σ2 (S )-upper semimeasurable. To verify this, let ε : K → M1 (K) be the homeomorphic embedding assigning to y ∈ K the Dirac measure εy at y. Then, by Corollary 2.28, M1 (x) = co{εy : y ∈ ϕ(x)} = co(ε(ϕ(x)), x ∈ X. Thus by Lemma 11.33, M1 is Σ2 (S )-upper semimeasurable. Concerning M2 , its Σ2 (S )-upper semimeasurability follows from the following equality (M2 )−1 (F ) = f −1 (r(F )) ∈ Σ2 (S ) valid for any closed set F ⊂ M1 (K) (we recall that F , and consequently r(F ), is a compact set). Hence Lemma 11.29 and Lemma 11.30 imply the Σ2 (S )-upper semimeasurability of M . Since ∆3 (S ) is an algebra of sets and Σ2 (∆3 (S )) = Σ3 (S ) by Proposition 5.4(e), Corollary 11.36 provides a Σ3 (S )-measurable selector m : X → M1 (K) such that m(x) ∈ ChS|M (x) (M (x)) = ext Mf (x) (ϕ(x)) max , x ∈ X. The proof is complete. Theorem 11.41. Let X be a metrizable compact convex set. Then there exists a mapping m ∈ Baf1 (X, M1 (X)) such that m(x) ∈ ext((Mx (X))max ),

x ∈ X.

Proof. We use Corollary 11.36, where Y = M1 (X), S is the convex cone from Definition 11.39 and ϕ : X → 2Y assigns to x ∈ X the set Mx (X). The role of the algebra S is played by the algebra Bas(X) generated by closed sets in X (see Definition 5.13). If r : Y → X is the barycentric mapping and F ⊂ Y is a closed set, ϕ−1 (F ) = r(F ) is closed in X. Hence ϕ is a Bas(X)-lower semimeasurable mapping. By Corollary 11.36, there exists a Σ2 (Bas(X))-measurable function m : X → Y such that m(x) ∈ ChS|ϕ(x) (ϕ(x)) = ext((Mx (X))max ), Hence m is the required mapping by Theorem 5.19.

x ∈ X.

412

11.6


Exercises

Exercise 11.42. Let K be the one-point compactification of an uncountable discrete set A and X := M1 (K). Then ext X is uncountable and Ac (X) does not contain `1 . Hint. Since ext X = {εx : x ∈ K}, ext X is uncountable. Further, Ac (X) is isometric to C(K) and C(K) is isomorphic to c0 (A) (we recall that c0 (A) denotes the space of all continuous functions on the discrete space A that vanish at infinity, see Subsection A.3.B). Thus it suffices to show that c0 (A) does not contain `1 . Let S : `1 → c0 (A) be an isomorphism and let B ⊂ A be countable such that it contains supports of all vectors Sen , n ∈ N (here {en : n ∈ N} denotes the usual basis in `1 ). Then any vector in S(`1 ) is supported by B, and thus S(`1 ) is an isomorphic copy of `1 in c0 , which is impossible. ˇ Exercise 11.43. Let βN be the Cech–Stone compactification of N and F := βN \ N. Prove that there exists no bounded linear operator E : C(F ) → C(βN) such that Ef = f on F for each f ∈ C(F ). Hint. Assume that E is such an operator and let P f := f − E(f |F ),

f ∈ C(βN).

If we identify `∞ with C(βN) and c0 with {f ∈ C(βN) : f = 0 on F }, then P is a bounded linear projection of `∞ onto c0 . But such a projection does not exist (see Theorem 5.15 in [173]). Exercise 11.44. Let X be a simplex such that ext X is countable. Then X is metrizable and every strongly affine function is in A1 (X). Hint. By Exercise 10.96, X is metrizable. If f is a strongly affine function on X, it is bounded by Lemma 4.5 and f |ext X is a Baire-one function due to Theorem A.124. If h ∈ A1 (X) is an extension of f |ext X provided by Theorem 6.49, f = h by the minimum principle (all maximal measures are carried by ext X). that ext X is uncountable. Then Exercise 11.45. Let X be a metrizable simplex such S for any α < ω1 there exists a function f ∈ Aα (X) \ β 0 such that for every real numbers c1 , . . . , cn we have n

X

C −1 k(c1 , . . . , cn )k∞ ≤ ck gk ≤ Ck(c1 , . . . , cn )k∞ ). k=1

Hint. Without loss of generality we may assume that 0 < f1 ≤ · · · ≤ fn ≤ · · · ≤ 1. Let x ∈ X and c, η > 0 be such that f (x) < c < c + η < lim sup f (y). y→x

P 1 Let εk > 0, k ∈ N, be chosen such that ∞ k=1 εk < 12 η. Inductively we find points xk ∈ X, indices nk ∈ N and neighborhoods Uk 3 x such that, for k ∈ N, (a) nk+1 > nk , (b) Uk+1 ⊂ Uk , (c) x1 = x and xk+1 ∈ Uk , (d) diam fnk (Uk ) < εk , (e) fnk+1 (xk+1 ) > c + η, (f) fnk (z) > f (z) − εk for z ∈ {x1 , x2 , . . . , xk }. We start the construction by setting x1 := x and finding n1 ∈ N such that fn1 satisfies (f) for z = x1 . Then we select a neighborhood U1 3 x with (d). Assume that the construction has been completed up to the k-th stage. Let xk+1 ∈ Uk be chosen in such a way that f (xk+1 ) > c + η. Then we find nk+1 > nk such that (e) and fnk+1 (z) > f (z)−εk+1 for z ∈ {x1 , x2 , . . . , xk+1 } are satisfied. We finish the construction by finding a neighborhood Uk+1 of x with (b) and diam fnk+1 (Uk+1 ) < εk+1 . Fix i ∈ N. If k < i, then (d) gives fnk+1 (xi+1 ) − fnk (xi+1 ) ≤ |fnk+1 (xi+1 ) − fnk+1 (x)| + |fnk+1 (x) − fnk (x)| + |fnk (x) − fnk (xi+1 )| < εk+1 + εk + εk . (11.12) For k > i, (f) yields fnk+1 (xi+1 ) − fnk (xi+1 ) ≤ f (xi+1 ) − fnk (xi+1 ) < εk .

(11.13)

11.6 Exercises

415

For k = i we have by (e) and (d) fni+1 (xi+1 ) − fni (xi+1 ) > c + η − (fni (x) + fni (xi+1 ) − fni (x)) > c + η − c − εi = η − ε i .

(11.14)

We claim that gk := fnk+1 − fnk , k ∈ N, form a sequence equivalent to the standard basis of c0 . To see this, let c1 , . . . , cn be real numbers and let i ∈ {1, . . . , n} be an index such that |ci | = k(c1 , . . . , cn )k∞ . Then, for any x ∈ X, n n X X ck gk (x) ≤ |ck |(fn

k+1

k=1

(x) − fnk (x))

k=1

≤ |ci |

n X

(fnk+1 (x) − fnk (x)) ≤ |ci |(f (x) − f1 (x))

k=1

≤ k(c1 , . . . , cn )k∞ . Hence

n

X

ck gk ≤ k(c1 , . . . , cn )k∞ . k=1

To show the estimate from below, assume that |ci | = ci . Then (11.12), (11.13) and (11.14) give n X

ck (fnk+1 (xi+1 ) − fnk (xi+1 ))

k=1

≥ ci (fni+1 (xi+1 ) − fni (xi+1 )) − |ci |

X

(fnk+1 (xi+1 ) − fnk (xi+1 ))

k∈{1,...,n}\{i}

≥ ci η − εi −

X

(εk+1 + 2εk )

k∈N\{i}

≥ ci η −

∞ X

3εk

k=1

3 ≥ ci η. 4 Hence

n

X

3

ck gk ≥ ηk(c1 , . . . , cn )k∞ 4 k=1

in this case.

416


If i ∈ {1, . . . , n} is such that −ci = k(c1 , . . . , cn )k∞ , then we apply the inequalities above to (−c1 , . . . , −cn ) and get n n

X

X

3 3

ck gk = (−ck )gk ≥ ηk(−c1 , . . . , −cn )k∞ = ηk(c1 , . . . , cn )k∞ . 4 4 k=1

k=1

Exercise 11.51. Let X be a metrizable compact convex set and f be an affine lower semicontinuous function on X that is not continuous. Then there exists a sequence {gn } of functions in Ac (X) that is equivalent to the standard basis of c0 . Hint. Without loss of generality we may assume that 0 < f ≤ 1 (see Lemma 4.20). By Proposition 4.12, Proposition A.53 and Lemma A.54 there exists an increasing sequence {fn } of affine continuous functions converging to f . By Exercise 11.50 we can extract a subsequence {fnk } such that gk = fnk+1 − fnk form a sequence equivalent to the standard basis of c0 . Exercise 11.52. Prove that the space c of convergent sequences endowed with the sup-norm is isomorphic to c0 . Hint. Consider the operator T : c → c0 which maps (x1 , x2 , . . . ) 7→ ( lim xn , x1 − lim xn , x2 − lim xn , . . . ), n→∞

n→∞

n→∞

(x1 , x2 , . . . ) ∈ c.

Exercise 11.53. Let X be an infinite-dimensional metrizable simplex. Then Ac (X) contains an isomorphic copy of c0 . Hint. Assume first that ext X is closed. Then ext X contains infinitely many distinct points xn , n ∈ N, and x such that xn → x. Let F := {x} ∪ {xn : n ∈ N}. Then C(F ) is isometric to the space c of convergent sequences which is isomorphic to c0 (see Exercise 11.52). By Theorem 11.13(c) there exists an extender L : C(F ) → Ac (X). Then T (C(F )) is isomorphic to c0 . If ext X is not closed, there exists a concave continuous function f such that f∗ is not continuous (see Theorem 6.37). Then f∗ is an affine (see Theorem 6.5) lower semicontinuous function that is not continuous. Hence Exercise 11.51 finishes the argument.

11.7

Notes and comments

The main selection Theorem 11.6, as well as its consequences Theorems 11.7, 11.8 and Corollary 11.9, is proved in A. J. Lazar [293]. Our proof follows the papers


417

Ch. Léger [299] and A. J. Lazar and J. Lindenstrauss [296], where a selection theorem for L1 -preduals is proved. Theorems 11.13, 11.15 and 11.16 are from A. J. Lazar [293]. Theorem 11.14 is a particular version of A. J. Lazar and J. Lindenstrauss [296], Theorem 2.3, and E. H. Lacey and P. D. Morris [291], Theorem 1.1. Theorem 11.17 is an affine version on simplices of the Michael selection theorem (see E. Michael [341] and [342]). Theorem 11.18 can be found in J. Dugundji [149] who proved there the following result by topological methods: If A is a closed subset of a metrizable space X and f : A → E is a continuous mapping to a locally convex space E, then there exists a continuous mapping F : X → E extending f such that F (X) ⊂ co f (A). R. Arens showed in [17] that X can be replaced by a paracompact space if E is a Banach space. Theorem 11.19 is a variant of results from J. Dugundji [149] and K. Borsuk [77]. An application to potential theory is given in J. Lukeˇs and J. Koláˇr [276] and D. Werner [465] (see also P. Harmand, D. Werner and W. Werner [216], Chapter II, pp. 96–98). Section 11.3 follows the paper J. Spurný [428] and the results of Section 11.4 can be found in J. Lukeˇs, J. Malý, I. Netuka, M. Smrˇcka and J. Spurný [319]. Section 11.5 follows ideas from an unpublished paper by G. Debs [137] (see also [139]). Theorem 11.41 is proved in M. Talagrand [443]. The existence of a Borel measurable selection assigning to a point x of a compact convex set X a maximal measure µ ∈ Mx (X) was established by M. Rao [384]. If X is a simplex, the mapping x 7→ δx has the property that T f (x) := δx (f ), x ∈ X, is Borel for every f ∈ C(X) (see Theorem 6.8). We do not know the answer to the following question. Problem 11.54. Let X be a compact convex set. Does there exist a mapping x 7→ µx , x ∈ X such that µx ∈ Mx (X) is maximal and x 7→ µx (f ) is Borel for each f ∈ C(X)? The question of the existence of Borel measurable selectors is a widely investigated area; we refer the reader for example to K. Kuratowski and A. Maitra [286], K. Kuratowski and C. Ryll-Nardzewski [287], Ch. Castaign and M. Valadier [102], D. H. Fremlin [180], R. W. Hansell [207] and [208], J. Kaniewski and R. Pol [261] and J. Spurný and M. Zelený [433]. Exercises 11.48 and 11.49 are proved in M. Capon [96]. The paper by M. Kaˇcena and J. Spurný [253] shows that there exists a metrizable simplex X such that T (B bα (X)) \ B bα (X) 6= ∅ for every α ∈ [0, ω0 ). If U is a bounded open set in Rd and H(U ) is the function space from Definition 13.26, Theorem 1.3 of [253] shows that the shift of classes ceases to exist for H(U ) at the second stage. Exercise 11.49 prompts a question whether the shift of finite classes is essential. This problem was answered in J. Spurný [420] by the following result.

418


There exist a metrizable simplex X and a strongly affine function f : X → R of the second Baire class that is not of the second affine class. Problem 11.55. Given n ≥ 3, does there exist a metrizable simplex X and a strongly affine function f ∈ B n (X) that is not of affine class n? Exercise 11.50 is a very particular version of techniques described in H. P. Rosenthal [398]; we also refer the reader to S. A. Argyros, G. Godefroy and H. P. Rosenthal [19].

Chapter 12

Constructions of function spaces

The goal of this chapter is to present methods of how to construct new function spaces from given ones. More precisely, we will deal with products of function spaces and inverse limits of function spaces. The key feature of both of these constructions is that they preserve simpliciality. We begin with products of function spaces, so after the basic definitions and auxiliary results we come to the results describing Choquet boundaries and maximal measures of products of function spaces (see Theorem 12.10 and 12.13). In order to show that the product preserves simpliciality, we need to prove in Subsection 12.1.C several technical results on peaked partitions of unity and approximations in product spaces. These results are essential for the proof of Theorem 12.21, showing that the product of simplicial spaces is simplicial. This method also opens a way for a construction of tensor products of compact convex sets which we indicate in Exercises 12.85–12.87. Inverse limits of function spaces and compact convex sets are studied in Subsections 12.2.B and 12.2.C. Again, the principal results are the preservation of simpliciality (see Theorem 12.34), its consequence to compact convex simplices (see Corollary 12.35) and a description of the Choquet boundary of the inverse limit (see Theorem 12.36). The general results are applied in the framework of compact convex sets to get in Theorem 12.40 and Corollary 12.42 that Ac (X) is an L∞,1+ε -space provided X is a simplex. Perhaps the most important result is Theorem 12.45 that asserts that any metrizable simplex is the inverse limit of a suitable sequence of finite-dimensional simplices. A much simpler method yields that any simplex is the inverse limit of a system of metrizable simplices (see Theorem 12.47). A fundamental object in the theory of metrizable simplices is the Poulsen simplex. Its construction and properties are presented in Subsection 12.3.A. Theorem 12.60 summarizes its remarkable properties. Further we show in Subsection 12.3.B the existence of a nonmetrizable simplex and a strongly affine function on it that is not determined by its values on the set of extreme points. Finally, a striking example due to M. Talagrand on functions of the second affine class is constructed in Subsection 12.3.C.

420

12.1 12.1.A

12 Constructions of function spaces

Products of function spaces Definitions and basic properties

Q If {Ei }i∈I is a family of topological spaces, i∈I Ei denotes their cartesian Q product with If J ⊂ I, the natural Q the usual product topology. Q Q projection of i∈I Ei onto E is denoted as π . If A ⊂ E and z ∈ i i J i∈J i∈I i∈I\J Ei , then n o Y πJz (A) := x ∈ Ei : (x, z) ∈ A . i∈J

If f :

Q

i∈I

Ei → R is a function and y ∈

Q

πJy (f )(x) := f (x, y),

i∈I\J

Ei , we define

x∈

Y

Ei .

i∈J

Q

Q In case f (x, y1 ) = f (x, y2 ) for any x ∈ i∈J Ei and y1 , y2 ∈ i∈I\J Ei , we write πJ (f ) instead of πJy (f ). If J = {j}, we write πj (f ) instead (f ). N of π{j}Q If fi : Ei → R are functions, i = 1, . . . , n, we write ni=1 fi : ni=1 Ei → R for the function ! n n O Y fi (x1 , . . . , xn ) := f1 (x1 ) · · · fn (xn ), (x1 , . . . , xn ) ∈ Ei . i=1

i=1

Definition 12.1 (Products of function Qspaces). Let Hi be a function space on a compact space K , i ∈ I. We set K := i i∈I Ki and define the algebraic tensor product J H as the linear span of the set i i∈I ( ) O Q hi ⊗ c i∈I\J Ki : J ⊂ I finite, hi ∈ Hi for i ∈ J . i∈J

N The multiaffine product i∈I Hi is defined as n o O Y Hi := f ∈ C(K) : πjy (f ) ∈ Hj for any j ∈ I, y ∈ Ki . i∈I

i∈I\{j}

S

We recall that {Jγ }γ∈Γ is a partition of a set I if γ∈Γ Jγ = I and Jα ∩ Jβ = ∅ for distinct and {Jγ }γ∈Γ is a partition of Q α, β ∈ Γ. If {Ki }i∈I are compact spaces Q Q I, then i∈I Ki is naturally homeomorphic to γ∈Γ ( i∈Jγ Ki ). This provides an Q Q Q identification of C( i∈I Ki ) with C( γ∈Γ ( i∈Jγ Ki )). Proposition 12.2. Let Hi be a function space on a compact space Ki , i ∈ I. Let {Jγ }γ∈Γ be a partition of I. Then the following assertions hold: J N (a) i∈I Hi ⊂ i∈I Hi ,

12.1 Products of function spaces

(b)

J

(c)

N

i∈I i∈I

Hi =

J

Hi =

N

421

J

i∈Jγ Hi ), N γ∈Γ ( i∈Jγ Hi ). γ∈Γ (

Proof. The proof follows by a straightforward verification. Unless stated otherwise, Hi are N function spaces on compact spaces Ki , i ∈ I, and H denotes the multiaffine product i∈I Hi . Notation 12.3. If J ⊂ I, we write HJ for the space of functions from H that depends only on coordinates from J; precisely, HJ := {h ∈ H : x, y ∈ K, πJ (x) = πJ (y) =⇒ h(x) = h(y)}. The space Hf of functions depending on finitely many coordinates is defined as Hf := {h ∈ H : exists J ⊂ I finite such that h ∈ HJ }. Q The function space πJ (H) on i∈J Ki is defined as n o Y Ki : f ⊗ cQi∈I\J Ki ∈ H . πJ (H) := f ∈ C i∈J

Proposition 12.4. The following assertions hold: (a) if I1 ⊂ I2 ⊂ I and h ∈ HI1 , then h ∈ HI2 , N (b) πJ (H) = i∈J Hi , (c) if h ∈ HJ , then h = πJ (h) ⊗ cQi∈I\J Ki , (d) if µ ∈ M+ (K) and h ∈ HJ , then µ(h) = ((πJ )] µ)(πJ (h)), (e) Hf is dense in H. Proof. We start the proof by noticing that assertions (a)–(c) are easy to verify from the definitions. If µ ∈ M+ (K) and h ∈ HJ , then ((πJ )] µ)(πJ (h)) = µ(πJ (h) ◦ πJ ) = µ(πJ (h) ⊗ cQi∈I\J Ki ) = µ(h). This proves (d). For the proof of (e), let h ∈ H and ε > 0 be given. By the compactness of K we can select a finite open cover {V1 , . . . , Vm } of K consisting of standard open basic sets such that diam h(Vi ) < ε, i = 1, . . . , m. We set J := {i ∈ I : there exists l ∈ {1, . . . , m} such that πi (Vl ) 6= Ki } Q and select y ∈ i∈I\J Ki . We define h0 : K → R as h0 (x) := h(πJ (x), y),

x ∈ K.

422


Then h0 ∈ Hf . If z ∈ K is given, we find Vl containing z. Then (πJ (z), y) ∈ Vl as well, and thus |h0 (z) − h(z)| = |h(πJ (z), y) − h(z)| < ε. Thus Hf is dense in H. Lemma 12.5. Let Hi be a function space on a compact space Ki , i = 1, 2. Then Ac (H1 ) Ac (H2 ) ⊂ Ac (H1 ⊗ H2 ). Proof. Let ai ∈ Ac (Hi ), i = 1, 2. We show that a1 ⊗ a2 ∈ Ac (H1 ⊗ H2 ). We assume first that both functions a1 , a2 are positive, and fix (x1 , x2 ) ∈ K1 × K2 and ε > 0. We find δ > 0 such that δ(a1 (x1 ) + a2 (x2 ) + δ) < ε, and choose functions hi ∈ Hi , i = 1, 2, such that, for i = 1, 2, we have ai ≤ hi

and hi (xi ) < ai (xi ) + δ.

(This is possible by Corollary 3.23(a).) Then h1 ⊗ h2 ∈ H1 ⊗ H2 , and a1 (x1 )a2 (x2 ) ≤ h1 (x1 )h2 (x2 ) < (a1 (x1 ) + δ)(a2 (x2 ) + δ) = a1 (x1 )a2 (x2 ) + δ(a1 (x1 ) + a2 (x2 ) + δ) < a1 (x1 )a2 (x2 ) + ε. Hence (a1 ⊗ a2 )∗ = a1 ⊗ a2 . Let now a1 be positive and a2 be arbitrary. Obviously, (a1 ⊗ c)∗ = a1 ⊗ c for any c ∈ R. Using the first step we get a1 ⊗ a2 ≤ (a1 ⊗ a2 )∗ = (a1 ⊗ (a2 + ka2 k − ka2 k))∗ = (a1 ⊗ (a2 + ka2 k) − a1 ⊗ ka2 k)∗ ≤ (a1 ⊗ (a2 + ka2 k))∗ + (a1 ⊗ (−ka2 k))∗ = a1 ⊗ (a2 + ka2 k) + (a1 ⊗ (−ka2 k)) = a1 ⊗ a2 . For the lower envelope we have (a1 ⊗ a2 )∗ = −(a1 ⊗ (−a2 ))∗ = −(a1 ⊗ (−a2 )) = a1 ⊗ a2 . Hence a1 ⊗ a2 ∈ Ac (H1 ⊗ H2 ). Finally, let a1 , a2 be arbitrary. Then a1 ⊗ a2 = (a1 + ka1 k) ⊗ a2 − ka1 k ⊗ a2 ∈ Ac (H1 ⊗ H2 ). This concludes the proof.


423

Proposition 12.6. The following assertions hold: (a) If x ∈ K, µ ∈ Mx (H) and J ⊂ I, then (πJ )] µ ∈ MπJ (x) (πJ (H)). In particular, (πi )] µ ∈ Mxi (Hi ), i ∈ I. N (b) If x = (xi )i∈I ∈ K and µi ∈ Mxi (Hi ), then i∈I µi ∈ Mx (H). N (c) Ac (H) ⊂ i∈I Ac (Hi ). (d) H = Ac (H) if and only if Hi = Ac (Hi ) for every i ∈ I. Proof. To verify (a), let x ∈ K, J ⊂ I and µ ∈ Mx (H) be given. For h ∈ πJ (H), we set h0 := h ⊗ cQi∈I\J Ki . Then h0 ∈ H, and hence ((πJ )] µ)(h) = µ(h ◦ πJ ) = µ(h0 ) = h0 (x) = h(πJ (x)). To prove (b), let µi ∈ Mxi (Hi ) for i ∈ I and h ∈ H. We denote µ := We first handle the case I = {1, . . . , n}. Then Z Z µ(h) = ··· h(y1 , . . . , yn ) dµn (yn ) · · · dµ1 (y1 ). K1

N

i∈I

µi .

Kn

Since the function yn 7→ h(y1 , . . . , yn ) is in Hn , Z h(y1 , . . . , yn ) dµn (yn ) = h(y1 , . . . , yn−1 , xn ),

(y1 , . . . , yn−1 ) ∈

Kn

n−1 Y

Ki .

i=1

Proceeding by induction, µ(h) = h(x1 , . . . , xn ). If I is an arbitrary index set, J ⊂ I is finite and g ∈ HJ , then by the previous part and Proposition 12.4(b),(d), O µ(g) = ((πJ )] µ)(πJ (g)) = ( µi )(πJ (g)) = πJ (g)(πJ (x)) = g(x). i∈J

Since Hf is dense in H (see Proposition 12.4(e)), µ ∈ MxQ (H). For the proof of (c), let h ∈ Ac (H), j ∈ I and y ∈ i∈I\{j} Ki be given. We have to show that πjy (h) ∈ AcN (Hj ). For any xj ∈ Kj and µj ∈ Mxj (Hj ), we set µ := µj ⊗ εy . Since µ = µj ⊗ i∈I\{j} εyi , we get µ ∈ M(xj ,y) (H) by the previous part. Hence πjy (h) = h(xj , y) = µ(h) = (µj ⊗ εy )(h) Z Z = h(s, t) d(µj ⊗ εy )(s, t) Q Kj

Z = Kj

Thus h ∈

N

i∈I

Ac (Hi ).

i∈I\{j}

Ki

h(s, y) dµj (s) = µj (πjy (h)).

424


To show (d), let Hi = Ac (Hi ) for each i ∈ I. Then by (c), H ⊂ Ac (H) ⊂

O

Ac (Hi ) =

i∈I

O

Hi = H.

i∈I

Conversely, let H = Ac (H) and j ∈ I beQgiven. We pick a function fj ∈ Ac (Hj ) and show that fj ∈ Hj . We set K 0 := i∈I\{j} Ki and f := fj ⊗ cK 0 . Then f ∈ Ac (H). Indeed, for x ∈ K and µ ∈ Mx (H) we get (πj )] µ ∈ Mxj (Hj ) by (a). Thus f (x) = fj (xj ) = ((πj )] µ)(fj ) = µ(fj ◦ πj ) = µ(f ). Hence f ∈ Ac (H) = H and fj = πj (f ) ∈ Hj . This concludes the proof.

12.1.B

Maximal measures and extremal sets

In this section, Hi is a function space on aN compact space Ki , i ∈ I (or i = 1, 2, respectively). IfQ H is the multiaffine product i∈I Hi , we denote by K the underlying compact space i∈I Ki . If H is a function Sspace on a compact space K and x ∈ K, as in Theorem 8.32 we denote Fx (H) := ν∈Mx (H) spt ν. Lemma 12.7. Let Hi be a function space on a compact space Ki , i = 1, 2, and let ϕ : K1 → K2 be a continuous mapping such that Fϕ(x) (H2 ) ⊂ ϕ(Fx (H1 )) for each x ∈ ChH1 (K1 ). Then ϕ] µ is H2 -maximal for any H1 -maximal measure µ ∈ M+ (K1 ). Proof. Let µ ∈ M+ (K1 ) be H1 -maximal. Then µ is carried by ChH1 (K1 ) by Proposition 3.64. For any function g ∈ C(K2 ) we know by Theorem 8.32 that g ◦ ϕ is constant on Fx (H1 ) for µ-almost all x ∈ K1 . Hence there exists an Fσ set A ⊂ K1 such that µ(K \ A) = 0 and g is constant on ϕ(Fx (H1 )) for all x ∈ A. By the assumption, g is constant on Fϕ(x) (H2 ) for all x ∈ A. Then (ϕ] µ)(K2 \ ϕ(A)) = 0 and g is constant on Fy (H2 ) for each y ∈ ϕ(A). Hence ϕ] µ is H2 -maximal by Theorem 8.32. 12.8. Let H = Proposition Q F (H ). i i∈I xi

N

i∈I

Hi . Let x = (xi )i∈I ∈ K. Then Fx (H) =

Proof. Let µ ∈ Fx (H) be given. By Lemma 12.6(a), (πi )] µ ∈ Mxi (Hi ) for any i ∈ I. By Lemma A.100(a), πi (spt µ) = spt(πi )] µ ⊂ Fxi (Hi ), Thus spt µ ⊂

Q

i∈I

Fxi (Hi ), which gives Fx (H) ⊂

Q

i∈I

i ∈ I. Fxi (Hi ).


425

N Conversely, let µi ∈ Mxi (Hi ) for i ∈ I. Then i∈I µi ∈ Mx (H) by Proposition 12.6(b), and thus Lemma A.100(b) yields Y O spt µi = spt( µi ) ⊂ Fx (H). i∈I

Q

i∈I

Fxi (Hi ) ⊂ Fx (H), which finishes the proof. N N Proposition 12.9. Let J ⊂ I, H = i∈I Hi and G := i∈J Hi . Q (a) Let E ⊂ K be a closed H-extremal set and y ∈ i∈I\J Ki . Then πJy (E) is either empty or G-extremal. Hence

i∈I

(b) If E ⊂ K is a closed H-extremal set, then πJ (E) is G-extremal. Q (c) If Ei ⊂ Ki are closed Hi -extremal, i ∈ I, then i∈I Ei is a closed H-extremal set. Proof. For the proof of (a), assume that πJy (E) is a nonempty set that is not Gextremal. Then there exist x ∈ πJy (E) and µ ∈ Mx (G) such that spt µ is not contained in πJy (E). We claim that the measure µ ⊗ εy ∈ M(x,y) (H). Indeed, by Propositions 12.6(b) and 12.2(c), O O µ ⊗ εy ∈ M(x,y) G ⊗ Hi = M(x,y) Hi . i∈I\J

i∈I

Hence µ ⊗ εy is a measure H-representing (x, y) ∈ E, but µ ⊗ εy is not carried by E. Hence we have arrived at a contradiction with the H-extremality of E. Q To show (b), let x ∈ πJ (E) and µ ∈ My x (G) be given. Then there exists y ∈ i∈I\J Ki such that (x, y) ∈ E. By (a), πJ (E) is G-extremal, and hence spt µ ⊂ y πJ (E) ⊂ πJ (E). We start the proof of (c) by showing the assertion for the case I = {1, 2}. Let Ei ⊂ Ki be closed Hi -extremal sets, i = 1, 2, and E := E1 × E2 . Let (x1 , x2 ) ∈ E and µ ∈ M(x1 ,x2 ) (H1 ⊗ H2 ). Since K \ E = ((K1 \ E1 ) × K2 ) ∪ (K1 × (K2 \ E2 )), we get µ(K \ E) ≤ µ((K1 \ E1 ) × K2 ) + µ(K1 × (K2 \ E2 )) = ((π1 )] µ)(K1 \ E1 ) + ((π2 )] )(K2 \ E2 ). By Proposition 12.6(a), (πi )] µ ∈ Mxi (Hi ) and thus ((πi )] µ)(KQ i \ Ei ) = 0, Ni = 1, 2. Hence µ(K \E) = 0 and E is H1 ⊗H2 -extremal. By induction, i∈I Ei is i∈I Hi extremal provided I is finite. Q Let now I be an arbitrary set and E := i∈I Ei . Assume that µ(K \ E) > 0 for some x ∈ E and µ ∈ Mx (H). We find f ∈ C(K) such that f = 0 on E and µ(f ) > 0. Let ε > 0 be arbitrary.

426


Q By Proposition 12.4(e), there exist a finite set J ⊂ I and g ∈ C( i∈J Ki ) such Q that, for K 0 := i∈I\J Ki , we have kf − g ⊗ cK 0 kC(K) < ε. Then (πJ )] µ ∈ MπJ (x) (πJ (H)) and, by the Q first part of the argument, πJ (x) is contained in the πJ (H)-extremal set EJ := i∈J Q Ei . Hence spt(πJ )] µ ⊂ Ej . Further, |g| < ε on EJ . Indeed, we pick t ∈ i∈I\J Ei . Then for any s ∈ EJ , (s, t) ∈ E. Hence |g(s)| = |(g ⊗ cK 0 )(s, t) − f (s, t)| < ε. Using this we obtain Z |µ(g ⊗ cK 0 )| = |((πJ )] µ)(g)| ≤

|g| d((πJ )] µ) < ε. EJ

Finally, 0 < |µ(f )| ≤ |µ(f − g ⊗ cK 0 )| + |µ(g ⊗ cK 0 )| < 2ε. Since ε is arbitrary, we get a contradiction. N Q Theorem 12.10. Let H = i∈I Hi . Then ChH (K) = i∈I ChHi (Ki ). Proof. The assertion follows from Proposition 12.9, because we have the following chain of equivalences for x = (xi )i∈I ∈ K: x ∈ ChH (K)

⇐⇒

{x} is H-extremal

⇐⇒

{xi } is Hi -extremal for each i ∈ I

⇐⇒

xi ∈ ChHi (Ki ) for each i ∈ I.

N N Proposition 12.11. Let J ⊂ I, H = i∈I Hi and G := i∈J Hi . (a) If µ ∈ M+ (K) is H-maximal, then (πJ )] µ is G-maximal. (b) If µ, ν ∈ M+ (K) with µ ≺H ν, then (πJ )] µ ≺G (πJ )] ν. Proof. Lemma 12.7 yields that for a verification of (a) it is enough to show that FπJ (x) (G) ⊂ πJ (Fx (H)) for any x ∈ K. But this follows from Proposition 12.8, which gives Y Y FπJ (x) (G) = Fxi (Hi ) = πJ Fxi (Hi ) = πJ (Fx (H)). i∈J

i∈I

Q To prove (b), let µ, ν ∈ M+ (K) be given. We write K 0 := i∈I\J Ki . For an arbitrary f ∈ −W(G), the function f ⊗ cK 0 is contained in −W(H). Hence ((πJ )] µ)(f ) = µ(f ⊗ cK 0 ) ≤ ν(f ⊗ cK 0 ) = ((πJ )] ν)(f ). Hence (πJ )] µ ≺G (πJ )] ν by Proposition 3.56.

427


N Proposition 12.12. Let H := H1 H2 and µ ∈ M+ (K1 × K2 ). If (πi )] µ is Hi maximal, i = 1, 2, then µ is H-maximal. In particular, if µi ∈ M+ (Ki ) is Hi -maximal, i = 1, 2, then µ1 ⊗µ2 is H-maximal. Proof. Let h ∈ H be arbitrary. We define T : K1 → C(K2 ) by assigning x1 7→ π2x1 (h). Then T (K1 ), as a compact subset of the Banach space C(K2 ), is norm separable. Let {hn : n ∈ N} ⊂ T (K1 ) be a dense countable set. Since (π2 )] µ is H2 -maximal, Theorem 8.32 yields that the set Fn := {x2 ∈ K2 : hn = hn (x2 ) on Fx2 (H2 )} T satisfies ((π2 )] µ)(K2 \ Fn ) = 0 for each n ∈ N. Then the set E2 := n∈N Fn is of full (π2 )] µ-measure and each hn is constant on Fx2 (H2 ) for every x2 ∈ E. Since {hn : n ∈ N} is dense in T (K1 ), we get h(x1 , y) = h(x1 , x2 ),

(x1 , x2 ) ∈ K1 × E2 , y ∈ Fx2 (H2 ).

(12.1)

Analogously we find a set E1 ⊂ K1 such that ((π1 )] µ)(K1 \ E1 ) = 0 and h(y, x2 ) = h(x1 , x2 ),

(x1 , x2 ) ∈ E1 × K2 , y ∈ Fx1 (H1 ).

(12.2)

Combining (12.1) and (12.2) we get h(x1 , x2 ) = h(y1 , y2 ),

(x1 , x2 ) ∈ E1 × E2 , y1 ∈ Fx1 (H1 ), y2 ∈ Fx2 (H2 ).

Since µ(K \ (E1 × E2 )) ≤ µ((K1 \ E1 ) × K2 ) + µ(K1 × (K2 \ E2 )) = ((π1 )] µ)(K1 \ E1 ) + ((π2 )] µ)(K2 \ E2 ) = 0, we get that h is constant on F(x1 ,x2 ) (H) = Fx1 (H1 ) × Fx2 (H2 ) for µ-almost all (x1 , x2 ) ∈ K1 ×K2 (here we use Proposition 12.8). By Theorem 8.32, µ is H-maximal. This concludes the proof. N + Theorem 12.13. Let H := i∈I Hi and µ ∈ M (K). If (πi )] µ is Hi -maximal, i ∈ I, then µ is H-maximal. N In particular, if µi ∈ M1 (Ki ) are Hi -maximal measures, i ∈ I, then i∈I µi is H-maximal. Proof. Let µ ∈ M+ (K) be as in the hypothesis. Step 1. We first prove the assertion by induction for finite index sets I. Assume that it holds for all finite sets with n elements and let I = J ∪ {i0 }, where J has n elements. For each i ∈ J, (πi )] ((πJ )] µ) = (πi )] µ

428


N is Hi -maximal. By Nthe inductive hypothesis, (πJ )] µ is i∈J Hi -maximal. By Proposition 12.12, µ is i∈J Hi ⊗ Hi0 -maximal. In other words, µ is H-maximal. Step 2. Let I be an arbitrary index set. We select an H-maximal measure ν with µ ≺H ν. By Proposition 12.11, for any finite set J ⊂ I, we have (πJ )] µ ≺πJ (H) (πJ )] ν. Since (πi )] ((πJ )] µ) = (πi )] µ for i ∈ J, the first part of the proof yields that (πJ )] µ is πJ (H)-maximal. Hence (πJ )] µ = (πJ )] ν. We get that µ(E) = ((πJ )] µ)(E) = ((πJ )] ν)(E) = ν(E) Q for any set E := i∈I Ei , where Ei ⊂ Ki is Borel, i ∈ I, and Ei 6= Ki only for indices in a finite set J. Hence, by Proposition A.99, µ = ν. We conclude that µ is H-maximal.

12.1.C

Partitions of unity and approximation in products of function spaces

Theorem 12.14. Let H be a simplicial space on a compact space K and −f, g be lower semicontinuous H-concave functions on K such that f ≤ g. Let F ⊂ K be a closed H-extremal set, h ∈ C(F ) satisfy f |F ≤ h ≤ g|F and h(x) = µ(h),

x ∈ F, µ ∈ Mx (H).

Then there exists a function a ∈ Ac (H) such that f ≤ a ≤ g and a = h on F . Proof. By setting ( h f := f 0

on F, on K \ F,

and

( h g := g 0

on F, on K \ F,

we obtain functions f 0 ∈ Kusc (H) and g 0 ∈ S lsc (H) with f 0 ≤ g 0 . By the Edwards in-between theorem 6.6 there exists a function a ∈ Ac (H) such that f 0 ≤ a ≤ g 0 . Then a satisfies the required properties. Definition 12.15 (Partition and peaked partition of unity). PA family {h1 , . . . , hn } of positive functions in H is called a partition of unity if ni=1 hi = 1. If moreover khi k = 1, i = 1, . . . , n, then {h1 , . . . , hn } is called a peaked partition of unity. Lemma 12.16. Assume that H is a simplicial function space on a compact space K, {f1 , . . . , fn }, {g1 , . . . , gn } are families of functions in Ac (H) and x1 , . . . , xp are pairwise distinct points in ChH (K). Let φ1 , . . . , φm be functions on ChH (K) and {αij : i = 1, . . . , n, j = 1, . . . , m} and {βjl : j = 1, . . . , m, l = 1, . . . , p} be families of real numbers such that


(a) fi |ChH (K) ≤

Pm

j=1 αij φj

429

≤ gi |ChH (K) , i = 1, . . . , n,

(b) φj (xl ) = βjl , j = 1, . . . , m, l = 1, . . . , p. Then there exists a family {ψ1 , . . . , ψm } of functions in Ac (H) such that P (c) fi ≤ m j=1 αij ψj ≤ gi on K, i = 1, . . . , n, (d) ψj (xl ) = βjl , j = 1, . . . , m, l = 1, . . . , p. Proof. We proceed by induction on m. Step 1. Assume that m = 1 and we are given fi |ChH (K) ≤ αi1 φ1 ≤ gi |ChH (K) , φ1 (xl ) = β1l ,

i = 1, . . . , n,

l = 1, . . . , p.

(12.3)

Without loss of generality we may assume that all numbers αi1 ≥ 0 (if αi1 < 0 for some i = 1, . . . , n, we would consider −gi |ChH (K) ≤ (−αi1 )φ1 ≤ −fi |ChH (K) for this index i in (12.3)). We set n 1 o fi : i ∈ {1, . . . , n} with αi1 6= 0 , A1 := αi1 o n 1 gi : i ∈ {1, . . . , n} with αi1 6= 0 . B 1 := αi1 By (12.3) and the Minimum principle 3.16, any function from A1 is on K smaller than any function from B 1 . We set ( max{f : f ∈ A1 } a1 := min{β1l : l = 1, . . . , p} ( min{g : g ∈ B 1 } b1 := max{β1l : l = 1, . . . , p}

if A1 = 6 ∅, if A1 = ∅, if B 1 = 6 ∅, if B 1 = ∅.

Then −a1 , b1 are continuous H-concave functions satisfying a1 ≤ b1

and

a1 (xl ) ≤ β1l ≤ b1 (xl ),

l = 1, . . . , p.

Using Theorem 12.14, we find a function ψ1 ∈ Ac (H) such that a1 ≤ ψ1 ≤ b1 and ψ1 (xl ) = β1l , l = 1, . . . , p. Obviously, fi ≤ αi1 ψ1 ≤ gi whenever αi1 6= 0. Otherwise fi |ChH (K) ≤ 0 ≤ gi |ChH (K) , and thus fi ≤ αi1 ψ1 ≤ gi again by the Minimum principle 3.16. This finishes the first step of the proof.

430


Step 2. Assume now that the assertion holds for m − 1 and we have given functions {φ1 , . . . , φm } satisfying the assumptions (a) and (b). We rewrite (a) as fi |ChH (K) −

m−1 X

αij φj ≤ αim φm ≤ gi |ChH (K) −

j=1

m−1 X

αij φj ,

i = 1, . . . , n.

j=1

As in the first step we may assume that αim ≥ 0 for every i = 1, . . . , n. Then we obtain for i ∈ {1, . . . , n} with αim 6= 0, m−1 m−1 X αij X αij 1 1 fi |ChH (K) − φj ≤ φm ≤ gi |ChH (K) − φj . αim αim αim αim j=1

(12.4)

j=1

Hence for any r, s with αrm 6= 0 6= αsm we obtain m−1 m−1 X αsj X αrj 1 1 gs |ChH (K) − φj . fr |ChH (K) − φj ≤ αrm αrm αsm αsm

(12.5)

j=1

j=1

By interchanging r with s we get from (12.5) that m−1 X αrj αsj 1 1 1 1 gs ≤ φj ≤ fs fr − − gr − αrm αsm αrm αsm αrm αsm

(12.6)

j=1

on ChH (K). For αim = 0 we get for the functions {φ1 , . . . , φm−1 } fi |ChH (K) ≤

m−1 X

αij φj ≤ gi |ChH (K) .

(12.7)

j=1

We apply the induction hypothesis to (12.6) and (12.7), where we consider the families 1 1 fr − gs ∪ {fi }, αrm αsm

1 1 gr − fs ∪ {gi }, αrm αsm

where r, s are such that αrm 6= 0 6= αsm and i satisfies αim = 0. It yields the existence of a family {ψ1 , . . . , ψm−1 } ⊂ Ac (H) such that ψj (xl ) = βjl ,

j = 1, . . . , m − 1, l = 1, . . . , p,

m−1 m−1 X αsj X αrj 1 1 gs − ψj , fr − ψj ≤ αrm αrm αsm αsm

(12.8)

αrm 6= 0 6= αsm , (12.9)

j=1

j=1

fi ≤

m−1 X i=1

αij ψj ≤ gi ,

αim = 0.

(12.10)


431

We set Am :=

m−1 X αij 1 fi − ψj : i ∈ {1, . . . , n} with αim 6= 0 , αim αim j=1

B m :=

m−1 X αij 1 gi − ψj : i ∈ {1, . . . , n} with αim 6= 0 . αim αim j=1

By (12.9), any function from Am is smaller than any define ( max{f : f ∈ Am } am := min{βml : l = 1, . . . , p} ( min{g : g ∈ B m } bm := max{βml : l = 1, . . . , p}

function from B m . Again we if Am = 6 ∅, if Am = ∅, if B m = 6 ∅, if B m = ∅.

Then −am , bm are continuous H-concave functions such that am ≤ bm and am (xl ) ≤ βml ≤ bm (xl ), l = 1, . . . , p (here we use (b), (12.4) and (12.8)). Using Theorem 12.14, we get a function ψm ∈ Ac (H) such that am ≤ ψm ≤ bm

and ψm (xl ) = βml ,

l = 1, . . . , p.

(12.11)

Then {ψ1 , . . . , ψm } is the required family of functions (for αim 6= 0 we get (c) from (12.11) and the definition of Am , B m , for αim = 0 we obtain (c) from (12.10)). This concludes the proof. Proposition 12.17. Let H be a simplicial function space on a compact space K, {f1 , . . . , fn } ⊂ Ac (H) and ε > 0. Assume that φ1 , . . . , φm are positive functions on ChH (K), {x1 , . . . , xm } is a family of distinct points in ChH (K) and {αij : i = 1, . . . , n, j = 1, . . . , m} is a family of real numbers such that Pm (a) j=1 φj = 1 on ChH (K), (b) for j, k = 1, . . . , m, we have ( 1, j = k, φj (xk ) = 0, j 6= k, Pm (c) |fi (x) − j=1 αij φj (x)| ≤ ε for x ∈ ChH (K) and i = 1, . . . , n. Then there exists a partition of unity {ψ1 , . . . , ψm } in Ac (H) such that (d) for j, k = 1, . . . , m, we have ( 1, j = k, ψj (xk ) = 0, j 6= k, Pm (e) |fi (x) − j=1 αij ψj (x)| ≤ ε for x ∈ K and i = 1, . . . , n.

432


Proof. Using (a) and (c) we get for any i = 1, . . . , n fi − ε − αim ≤

m X

αij φj − αim

j=1

=

m X

αij φj − αim

m−1 X

φj

j=1

j=1

=

m X

(αij − αim )φj

j=1

≤ fi + ε − αim on ChH (K). Positive functions φ1 , . . . , φm−1 satisfy P (f) fi −ε−αim ≤ m−1 j=1 (αij −αim )φj ≤ fi −αim +ε on ChH (K) for i = 1, . . . , n, Pm−1 (g) 0 ≤ j=1 φj ≤ 1, (h) 0 ≤ φj ≤ 1, j = 1, . . . , m − 1, (i) for j = 1, . . . , m − 1, l = 1, . . . , m, we have ( 1, j = l, φj (xl ) = 0, j= 6 l. Using Lemma 12.16 we find functions ψ1 , . . . , ψm−1 in Ac (H) such that (f)–(i) hold with φj replaced by ψj (in Lemma 12.16, we consider families {fi − αim − ε}ni=1 ∪ {0} ∪ {0}m−1 j=1 ,

{fi − αim + ε}ni=1 ∪ {1} ∪ {1}m−1 j=1

and n + m inequalities given by (f), (g) and (h)). We denote the properties of the functions ψj as (f∗ )–(i∗ ). We set ψm := 1 −

m−1 X

ψj .

j=1

It follows from (g∗ ) and (h∗ ) that ψj ≥ 0, j = 1, . . . , m, and thus {ψ1 , . . . , ψm } is a partition of unity. From (i∗ ) and ψm (xl ) = 1 −

m−1 X j=1

( 1, ψj (xl ) = 0,

m = l, m 6= l,

we obtain (d). Finally, by rewriting (f∗ ) we get (e). This concludes the proof.


433

Proposition 12.18. Let Hi be a function space on a compact space Ki , i = 1, 2, and let H1 be simplicial. Assume that {f1 , . . . , fn } is a family in Ac (H1 ) ⊗ H2 and ε > 0. Then there exist a partition of unity {ψ1 , . . . , ψm } in Ac (H1 ), {x1 , . . . , xm } ⊂ ChH1 (K1 ) and {gij : i = 1, . . . , n, j = 1, . . . , m} ⊂ H2 such that (a) for 1 ≤ j, l ≤ m we have ( 1, j = l, ψj (xl ) = 0, j 6= l, (b) kfi −

Pm

j=1 ψj

⊗ gij k < ε for i = 1, . . . , n.

Proof. Let (H2 )n stand for the cartesian product of n copies of H2 endowed with the maximum norm, that is, g = (g(1), . . . , g(n)) ∈ (H2 )n .

kgk∞ = max kg(i)kH2 , i=1,...,n

Let pi : (H2 )n → H2 , i = 1, . . . , n, be the projection on the i-th coordinate. Let f : K1 → (H2 )n be a continuous mapping defined as f (x) := (π2x (f1 ), . . . , π2x (fn )),

x ∈ K1 .

(The mapping f is continuous by Proposition A.35.) For each g ∈ (H2 )n we set ε Ug := x ∈ K1 : kg − f (x)k∞ < , 3

(12.12)

and choose g1 , . . . , gp ∈ (H2 )n such that K1 ⊂

p [

Ugj .

j=1

Let Vgj := Ugj ∩ ChH1 (K1 ),

j = 1, . . . , p.

We set V 0 := ∅ and inductively define families V j , j = 1, . . . , p, as follows. Let ( V j−1 ∪ {Vgj } if V j−1 ∪ {Vgj +1 , . . . , Vgp } does not cover ChH1 (K1 ), V j := V j−1 otherwise. After relabeling we can assume that V p = {Vg1 , . . . , Vgm } for some m ∈ {1, . . . , p}. The family V p is an open cover of ChH1 (K1 ) such that for any l = 1, . . . , m there exists [ x l ∈ V gl \ Vgj . j∈{1,...,m}\{l}

434


We denote C := {g1 , . . . , gp } − co{g1 , . . . , gm } and

D := C + U 0,

ε . 3

For any i = 1, . . . , n, the set pi (C) is a compact subset of H2 . By Theorem A.32 and the compactness of K2 , there exist points {ξi1 , . . . , ξiqi } in K2 and their open neighborhoods {Wi1 , . . . , Wiqi } such that K2 ⊂

qi [

Wir

and

r=1

ε diam h(Wir ) < , 3

r = 1, . . . , qi , h ∈ pi (C).

For any h ∈ pi (C) we find yh ∈ K2 and r ∈ {1, . . . , qi } such that yh ∈ Wir and |h(yh )| = khk. Hence khk −

ε < max |h(ξir )| ≤ khk, 3 r=1,...,qi

h ∈ pi (C).

Since pi (D) ⊂ pi (C) + U (0, 3ε ), we obtain khk −

2ε < max |h(ξir )| ≤ khk, r=1,...,qi 3

h ∈ pi (D).

(12.13)

We define functionals Γir : (H2 )n → R, i = 1, . . . , n, r = 1, . . . , qi , as Γir (g) := g(i)(ξir ),

g = (g(1), . . . , g(n)) ∈ (H2 )n .

Then equations (12.13) can be rewritten as khk∞ −

2ε < max |Γir (h)| ≤ khk∞ , i=1,...,n 3

h ∈ D.

(12.14)

r=1,...,qi

We define functions φj (x) on ChH1 (K1 ) as ( 1, j = min{l : x ∈ Vgl , l = 1, . . . , m}, φj (x) := 0, j 6= min{l : x ∈ Vgl , l = 1, . . . , m},

j = 1, . . . , m.

Then φj ≥ 0,

m X j=1

φj = 1,

( 1, φj (xl ) = 0,

j = l, j 6= l,

j, l = 1, . . . , m.

Further, for each x ∈ ChH1 (K1 ), there exists a unique index jk such that φjk (x) 6= 0. Hence, from (12.12) we get ε kf (x) − gjk k∞ < . 3


435

This can be rewritten as m X

ε

f (x) − φj (x)gj ∞ < , 3

x ∈ ChH1 (K1 ).

(12.15)

j=1

Since {Vg1 , . . . , Vgm } is a cover of ChH1 (K1 ), we see from (12.12) that ε f (x) ∈ {g1 , . . . , gm } + U (0, ), 3 Further

Pm

j=1 φj (x)gj

x ∈ ChH1 (K1 ).

∈ co{g1 , . . . , gm }, and thus f (x) −

m X

φj (x)gj ∈ D.

j=1

Using (12.14) and (12.15) we obtain   m m X X Γir (f (x)) −   = Γ f (x) − φ (x)g φ (x)Γ (g ) j j j ir j ir j=1 j=1 m X

ε φj (x)gj ∞ < , ≤ f (x) − 3

(12.16)

j=1

i = 1, . . . , n, r = 1, . . . , qi , x ∈ ChH1 (K1 ). We apply Proposition 12.17 to the family of functions {Γir ◦ f : i = 1, . . . , n, r = 1, . . . , qi } ⊂ Ac (H1 ) and to the family of numbers {Γir (gj ) : i = 1, . . . , n, r = 1, . . . , qi , j = 1, . . . , m}, and get a partition of unity {ψ1 , . . . , ψm } ⊂ Ac (H1 ) such that m X ε Γir (f (x)) − ≤ , ψ (x)Γ (g ) j ir j 3 j=1 i = 1, . . . , n, r = 1, . . . , qi , x ∈ ChH1 (K1 ), ( 1, j = l, ψj (xl ) = j, l = 1, . . . , m. 0, j= 6 l, Since {Ug1 , . . . , Ugp } is a cover of K1 , from (12.12) we infer f (x) ∈ {g1 , . . . , gp } + U 0,

ε , 3

x ∈ K1 .

(12.17)

(12.18)

436 Since


Pm

j=1 ψj (x)gj

∈ co{g1 , . . . , gm } for each x ∈ K1 , we obtain f (x) −

m X

ψj (x)gj ∈ D,

x ∈ K1 .

j=1

From (12.14) and (12.17) we get   m X 2ε   kf (x) − ψj (x)gj k∞ − < max Γir f (x) − ψj (x)gj i=1,...,n 3 j=1 j=1 r=1,...,qi m X ψj (x)Γir (gj ) = max Γir (f (x)) − i=1,...,n j=1 r=1,...,q m X

i

ε ≤ , 3 Hence

x ∈ K1 .

m X

f (x) − ψj (x)gj ∞ < ε,

x ∈ K1 .

(12.19)

j=1

To obtain the required family of functions {gij : i = 1, . . . , n, j = 1, . . . , m} in H2 , we define gij := pi (gj ), i = 1, . . . , n, j = 1, . . . , m. Property (a) then follows from (12.18) and (b) from (12.19). This finishes the proof.

12.1.D

Products of simplicial spaces

Proposition 12.19. Let Hi be function spaces on compact spaces Ki , i = 1, 2, and let H1 be simplicial. Then Ac (H1 ⊗ H2 ) = Ac (H1 ) ⊗ Ac (H2 ). Proof. It follows from Proposition 12.18 that any function in Ac (H1 ) ⊗ Ac (H2 ) can be uniformly approximated by functions from Ac (H1 ) Ac (H2 ). Using this fact, by Lemma 12.5 and Proposition 12.6(c) we obtain Ac (H1 ) ⊗ Ac (H2 ) ⊂ Ac (H1 ) Ac (H2 ) ⊂ Ac (H1 ⊗ H2 ) = Ac (H1 ⊗ H2 ) ⊂ Ac (H1 ) ⊗ Ac (H2 ), which finishes the proof. Theorem 12.20. Let Hi be simplicial function spaces on compact spaces Ki , i = 1, 2. Then H1 ⊗ H2 is simplicial.


437

Proof. Let H := H1 ⊗ H2 . We show that Ac (H) has the weak Riesz interpolation property (see Theorem 6.16). To this end, let a, b, c, d be functions from Ac (H) such that a ∨ b < c ∧ d. We choose ε > 0 such that c ∧ d − a ∨ b > ε. Since Ac (H) ⊂ Ac (H1 ) ⊗ Ac (H2 ) by Proposition 12.6(c), we may use Proposition 12.18 to find a partition of unity {ψ1 , . . . , ψm } ⊂ Ac (H1 ), points x1 , . . . , xm in ChH1 (K1 ) and a family of functions {aj , bj , cj , dj : j = 1, . . . , m} ⊂ Ac (H2 ) such that ( 1, j = l, ψj (xl ) = j, l = 1, . . . , m, (12.20) 0, j 6= l, and for the functions a0 :=

m X

ψj ⊗ aj ,

b0 :=

j=1 0

c :=

m X

m X

ψj ⊗ bj ,

j=1

ψj ⊗ cj ,

0

d :=

j=1

m X

(12.21) ψj ⊗ dj

j=1

we have a ∨ b < a0 ∨ b0 < c0 ∧ d0 < c ∧ d.

(12.22)

Then also π2x (a0 ) ∨ π2x (b0 ) < π2x (c0 ) ∧ π2x (d0 ),

x ∈ K1 .

(12.23)

For any j = 1, . . . , m we obtain from (12.20) and (12.21) that x

π2 j (a0 ) = aj ,

x

π2 j (b0 ) = bj ,

x

π2 j (c0 ) = cj ,

x

π2 j (d0 ) = dj .

From (12.23) it follows that aj ∨ bj < cj ∧ dj ,

j = 1, . . . , m.

Since Ac (H2 ) has the weak Riesz interpolation property (see Theorem 6.16), there exists a function hj ∈ Ac (H2 ) with aj ∨ bj < hj < cj ∧ dj , Then the function h :=

m X

j = 1, . . . , m.

(12.24)

ψj ⊗ hj

j=1

belongs to Ac (H1 ) ⊗ Ac (H2 ) = Ac (H) (see Proposition 12.19). Since the functions ψj are positive, (12.24) and (12.22) yield a ∨ b < h < c ∧ d. This concludes the proof.

438


Theorem 12.21. Let Hi be simplicial function spaces on compact spaces Ki , i ∈ I. Then N (a) i∈I Hi is simplicial, N (b) Ac (H) = i∈I Ac (Hi ), N Q (c) δx = i∈I δxi for x = (xi )i∈I ∈ i∈I Ki . Proof. Step 1. First we verify by induction that (a) and (b) hold for a finite set I. From Proposition 12.19 and Theorem 12.20, we know their validity in case I has two elements. Assume now that they hold for any index set containing n indices, and let I = J ∪ {i0 }, where J has n elements. Using induction hypothesis, Propositions 12.2(c) and 12.19, we get that the space O O Hi = H i ⊗ H i0 i∈I

i∈J

is simplicial, and Ac (H) = Ac

O

O Hi ⊗ Ac (Hi0 ) H i ⊗ H i 0 = Ac i∈J

i∈J

=

O

Ac (Hi ) ⊗ Ac (Hi0 ) =

i∈J

O

Ac (Hi ).

i∈I

NStep 2.c Let now Ic be an arbitrary index set and H := i∈I A (Hi ) = A (H). Let J ⊂ I be finite and ! O Ac (Hi ) h∈ i∈I

N

i∈I

Hi . We first show that

J

(that is, h depends only on coordinates from J). By the first part of the proof, O O πJ (h) ∈ Ac (Hi ) = Ac Hi . i∈J

i∈J

Using Lemma 12.5, Propositions 12.4(c) and 12.2(c) we get   O O h = πJ (h) ⊗ cQi∈I\J Ki ∈ Ac  Hi ⊗ Hi  = Ac (H). i∈J

i∈I\J

N N By Proposition 12.4(e), ( i∈I Ac (Hi ))f is dense in i∈I Ac (Hi ). It follows that !f O i∈I

c

A (Hi ) =

O i∈I

c

A (Hi )

⊂ Ac (H) = Ac (H).


439

Since the second inclusion follows from Proposition 12.6(c), we get the required N c c equality A (H) = i∈I A (Hi ). N c Let a, b, c, d ∈ Ac (H). Then a, b, c, d ∈ i∈I A (Hi ), and therefore Proposition 12.4(e) yields the existence of a finite set J ⊂ I and functions ! O 0 0 0 0 c a ,b ,c ,d ∈ A (Hi ) i∈I

J

such that a ∨ b < a0 ∨ b0 < c0 ∧ d0 < c ∧ d. Then πJ (a0 ) ∨ πJ (b0 ) < πJ (c0 ) ∧ πJ (d0 ). Since

N

i∈J

Hi is simplicial, there exists a function O O h0 ∈ Ac Hi = Ac (Hi ) i∈J

i∈J

such that πJ (a0 ) ∨ πJ (b0 ) < h0 < πJ (c0 ) ∧ πJ (d0 ). Then the function h := h0 ⊗ cQi∈I\J Ki belongs to

N

i∈I

Ac (Hi ) = Ac (H) and satisfies a ∨ b < h < c ∧ d.

This finishes the proof of (a) and (b). Q To verify (c), let x = (xi )N i∈I be a point in i∈I Ki . By Proposition 12.6(b) and Theorem 12.13, the measure i∈I δxi is H-maximal and H-represents x. Since H is simplicial, O δx = δ xi i∈I

as needed. This concludes the proof. Theorem 12.22. Let Hi be function spaces on compact spaces Ki , i ∈ I, such that N i∈I Hi is simplicial. Then Hi is simplicial for each i ∈ I. Proof. We Q fix j ∈ I and choose aj , bj , cj , dj ∈ Ac (Hj ) such that aj ∨ bj < cj ∧ dj . 0 Let K := i∈I\{j} Ki . Then the functions a := aj ⊗ cK 0 ,

b := bj ⊗ cK 0 ,

c := cj ⊗ cK 0 ,

are in Ac (H) (see Lemma 12.5), and a ∨ b < c ∧ d.

d := dj ⊗ cK 0

440


By the simpliciality of Ac (H) we find a function h ∈ Ac (H) with a∨b < h < c∧d. We select an arbitrary y ∈ K 0 . Proposition 12.6(c) yields πjy (h) ∈ Ac (Hj ). Then aj ∨ bj < πjy (h) < cj ∧ dj , and Ac (Hj ) has the weak Riesz interpolation property. Hence Hj is simplicial by Theorem 6.16.

12.2 12.2.A

Inverse limits of function spaces Admissible mappings

Definition 12.23 (Admissible mapping). Let Hi be a function space on a compact space Ki , i = 1, 2, and let ϕ : K1 → K2 be a continuous surjective mapping. It is said to be an admissible mapping of (K1 , H1 ) to (K2 , H2 ), if h2 ◦ ϕ ∈ H1 for any h2 ∈ H2 . If ϕ : (K1 , H1 ) → (K2 , H2 ) is an admissible mapping, H2 is naturally isometrically embedded in H1 via the mapping h2 7→ h2 ◦ ϕ. If S(Hi ) denote the respective state spaces of Hi , i = 1, 2, then the restriction mapping ψ : S(H1 ) → S(H2 ) is a continuous affine mapping. By the Hahn–Banach theorem, it is surjective. Another way of describing ψ is by the formula ψ(s1 )(h2 ) = s1 (h2 ◦ ϕ),

h2 ∈ H2 , s1 ∈ S(H1 ).

If φi : Ki → S(Hi ) denote the respective homeomorphic embedding of Ki into S(Hi ), i = 1, 2, we obtain that φ2 ◦ ϕ = ψ ◦ φ1 .

(12.25)

Proposition 12.24. Let ϕ : (K1 , H1 ) → (K2 , H2 ) be an admissible mapping. Then the following assertions hold. (a) If µ ∈ Mx (H1 ) for x ∈ K1 , then ϕ] µ ∈ Mϕ(x) (H2 ). (b) If g is a continuous H2 -convex function, then g ◦ ϕ is H1 -convex. (c) If µ ≺H1 ν for µ, ν ∈ M+ (K1 ), then ϕ] µ ≺H2 ϕ] ν. (d) ϕ(ChH1 (K1 )) ⊃ ChH2 (K2 ). (e) ϕ] (Mmax (H1 )) ⊃ Mmax (H2 ). Proof. To check (a), let x ∈ K1 and µ ∈ Mx (H1 ) be given. Then ϕ] µ(h2 ) = µ(h2 ◦ ϕ) = h2 (ϕ(x)), and thus ϕ] µ ∈ Mϕ(x) (H2 ).

h ∈ H2 ,

12.2 Inverse limits of function spaces

441

By an analogous straightforward argument we get (b). For the proof of (c), let µ ≺H1 ν for µ, ν ∈ M+ (K1 ) be given. For any g ∈ Kc (H2 ), we obtain ϕ] µ(g) = µ(g ◦ ϕ) ≤ ν(g ◦ ϕ) = ϕ] ν(g). Hence ϕ] µ ≺H2 ϕ] ν as required. Now let ψ : S(H1 ) → S(H2 ) be the restriction mapping. By Proposition 2.72(c), ψ(ext S(H1 )) ⊃ ext S(H2 ). In view of (12.25) and Proposition 4.26(d), −1 ChH2 (K2 ) = φ−1 2 (ext S(H2 )) ⊂ φ2 (ψ(φ1 (ChH1 (K1 )))) = ϕ(ChH1 (K1 )).

Analogously we proceed with the proof of (e). By Proposition 7.49, ψ(Mmax (S(H1 ))) ⊃ Mmax (S(H2 )). Since µ ∈ M+ (Ki ) is Hi -maximal if and only if (φi )] µ is Ac (S(Hi ))-maximal, i = 1, 2, (see Proposition 4.28(d)) and ψ] ◦ (φ1 )] = (φ2 )] ◦ ϕ] , we conclude that ϕ] (Mmax (H1 )) ⊃ Mmax (H2 ). Proposition 12.25. Let ϕ : (K1 , H1 ) → (K2 , H2 ) be an admissible mapping and let H2 be simplicial. Then the following assertions are equivalent: (i) ϕ(ChH1 (K1 )) = ChH2 (K2 ), (ii) (g ◦ ϕ)∗ = g ∗ ◦ ϕ for any continuous H2 -convex function g on K2 , (iii) ϕ] (Mmax (K1 )) = Mmax (K2 ). Proof. We start by showing (i) =⇒ (ii). Let g ∈ Kc (H2 ) be given. Proposition 12.24(a) and Lemma 3.22 imply (g ◦ ϕ)∗ ≤ g ∗ ◦ ϕ. On the other hand, g ∗ is H2 -affine by Theorem 6.5, and hence g ∗ ◦ ϕ is H1 -affine (use Proposition 12.24(a)). By the assumption, g ∗ ◦ ϕ = (g ◦ ϕ)∗ on ChH1 (K1 ), and hence the minimum principle of Proposition 3.88 gives g ∗ ◦ ϕ ≤ (g ◦ ϕ)∗ on K1 . Hence g ∗ ◦ ϕ = (g ◦ ϕ)∗ . To show (ii) =⇒ (iii), let µ ∈ Mmax (H1 ) be given. For any g ∈ Kc (H2 ), ϕ] µ(g) = µ(g ◦ ϕ) = µ((g ◦ ϕ)∗ ) = µ(g ∗ ◦ ϕ) = ϕ] µ(g ∗ ), and ϕ] µ is H2 -maximal by Theorem 3.58. Finally, (iii) =⇒ (i) is obvious.

442


12.2.B

Construction of inverse limits

Definition 12.26 (Inverse system and limit of function spaces). We assume that I is an up-directed set. Let (Ki , Hi ) be a function space, i ∈ I. We say that ((Ki , Hi ), pij )i,j∈I is an inverse system of function spaces if •

pij : (Kj , Hj ) → (Ki , Hi ) is an admissible mapping, i ≤ j,

•

pii is identity, i ∈ I,

•

pij ◦ pjk = pik , i ≤ j ≤ k.

If I = N, we say that the system is an inverse sequence. Let K := lim Ki be the inverse limit of the system (Ki , pij )i,j∈I , that is ←

K=

(xi )i∈I ∈

Y

Ki : pij (xj ) = xi , i ≤ j, i, j ∈ I

i∈I

and let πi : K → Ki be the i-th projection. The inverse limit lim((Ki , Hi ), pij )i,j∈I of this inverse system is the function space ←−

(K, H), where H :=

[

{h ◦ πi |K : h ∈ Hi }.

i∈I

If every Ki is nonempty, it follows from Theorem 3.2.13 in [169] that K is a nonempty compact space because each pij is surjective. Further, every πi is surjective and pij ◦ πj = πi , i ≤ j. Definition 12.27 (Inverse system of compact convex sets). Let I be an up-directed set and (Xi , pij )i,j∈I be a system of compact convex sets and mappings such that •

pij : Xj → Xi is a continuous affine surjection, i ≤ j,

•

pii is identity, i ∈ I,

•

pij ◦ pjk = pik , i ≤ j ≤ k.

Then ((Xi , Ac (Xi )), pij )i,j∈I is an inverse system of function spaces as defined in Definition 12.26. We call it an inverse family of compact convex sets and the compact convex set X := lim(Xi , pij )i,j∈I is its inverse limit. Sometimes we write briefly ←− X = lim Xi . ← Further, πi : X → Xi is an affine continuous surjection for each i ∈ I. Proposition 12.28. Let ((Ki , Hi ), pij )i,j∈I be an inverse system of function spaces and (K, H) := lim((Ki , Hi ), pij )i,j∈I be its inverse limit. Then ←−

(a) H is a well-defined function space, (b) every πi : (K, H) → (Ki , Hi ) is an admissible mapping.


443

Proof. To show that H is a function space, we first notice that H contains constant functions. If x, y ∈ K differs at a coordinate i ∈ I, let h ∈ Hi be such that h(xi ) 6= h(yi ). Then h ◦ πi ∈ H separates the points x and y. Let f ◦ πi , g ◦ πj ∈ H, where i, j ∈ I, f ∈ Hi and g ∈ Hj . We select k ∈ I such that i, j ≤ k. Then f ◦ πi + g ◦ πj = f ◦ pik ◦ πk + g ◦ pjk ◦ πk ∈ H, because f ◦ pik , g ◦ pjk ∈ Hk and Hk is a vector space. An analogous verification shows that cf ∈ H whenever c ∈ R and f ∈ H. Thus H is a vector space. Since (b) follows by the definition, the proof is finished. Proposition 12.29. If ((Ki , Hi ), pij )i,j∈I is an inverse system of function spaces and (K, H) is its inverse limit, we consider the family (S(Hi ), ψij )i,j∈I , where ψij : S(Hj ) → S(Hi ) are the restriction mappings. (a) The family (S(Hi ), ψij )i,j∈I is an inverse system of compact convex sets. (b) S(H) is affinely homeomorphic to lim S(Hi ). ←−

Proof. The system (S(Hi ), ψij )i,j∈I is inverse by a straightforward verification. For the proof of (b), let X := lim S(Hi ) be the limit of the system (S(Hi ), ψij )i,j∈I . Let ←

Φ : H → Ac (S(H)) be the mapping from Definition 4.25, that is, Φ(s)(h) = s(h),

s ∈ S(H), h ∈ H.

We define the required affine homeomorphism ϕ : X → S(H) as ϕ(s)(h ◦ πj ) = sj (h),

s = (si )i∈I ∈ X, h ∈ Hj for some j ∈ I.

Obviously, ϕ is a well-defined affine continuous mapping. We show that ϕ is injective. Let s, t ∈ X be distinct points, that is, they differ at a coordinate j ∈ I. Let h ∈ Hj be such that sj (h) 6= tj (h). Then ϕ(s)(h ◦ πj ) = sj (h) 6= tj (h) = ϕ(t)(h ◦ πj ), and ϕ(s) 6= ϕ(t). In order to prove surjectivity of ϕ it suffices to show that ϕ(X) ⊃ ext S(H) (use the Krein–Milman theorem 2.22). Assume that this is not the case and hence there exists t ∈ ext S(H) \ ϕ(X). Using Proposition 2.41 and density of Φ(H) in Ac (S(H)) (see Proposition 4.26(b)) we find a function h ∈ H such that Φ(h) > 0 on ϕ(X) and Φ(h)(t) < 0. Let h = h0 ◦ πj for some h0 ∈ Hj and j ∈ J. Then 0 ≤ Φ(h)(ϕ(s)) = sj (h0 ),

s ∈ X.

Hence sj (h0 ) ≥ 0 for each sj ∈ S(Hj ). It follows that h0 , and consequently h, is a positive function. Since Φ preserves order (see Proposition 4.26(e)), Φ(h) ≥ 0, which is a contradiction. This concludes the proof.

444


Proposition 12.30. Let (Xi , pij )i,j∈I be an inverse system of compact convex sets and let X := lim Xi be its limit. Let (X, H) be the inverse limit of the family ←

((Xi , Ac (Xi )), pij )i,j∈I , where every (Xi , Ac (Xi )) is regarded as a function space. Then (a) H = Ac (X), (b) S(H) is affinely homeomorphic to X. Proof. Given the objects as in the hypothesis, we notice that H = {f ◦ πi : f ∈ Ac (Xi ), i ∈ I} is contained in Ac (X). By Proposition 12.28(a), H contains constants and separates points of X. Thus Exercise 4.49 yields H = Ac (X) and (a) holds. By Proposition 4.31(a), X is affinely homeomorphic to S(Ac (X)) = S(H) = S(H). This concludes the proof. Theorem 12.31. Let ((Ki , Hi ), pij )i,j∈I be an inverse system of function spaces and (K, H) be its inverse limit. Let µ ∈ M1 (lim Ki ) be a measure such that (πi )] µ is ← Hi -maximal for any i ∈ I. Then µ is H-maximal. Proof. Let K := lim Ki and let ν ∈ M1 (K) satisfy µ ≺H ν. Proposition 12.28(b) ←

and Proposition 12.24(c) yield (πi )] µ ≺Hi (πi )] ν for each i ∈ I. Since (πi )] µ is Hi -maximal, (πi )] ν = (πi )] µ. Thus Lemma A.103 yields ν = lim(πi )] ν = lim(πi )] µ = µ. ←

←

This concludes the proof. Proposition 12.32. Let ((Ki , Hi ), pij )i,j∈I be an inverse system of function spaces and (K, H) be its inverse limit. (a) Let x = (xi )i∈I ∈ K and µ ∈ M1 (K). Then µ ∈ Mx (H) if and only if ((πi )] µ, pij )i,j∈I is an inverse system of measures (see Definition A.101) with (πi )] µ ∈ Mxi (Hi ) for i ∈ I. (b) Let x = (xi )i∈I ∈ K and let (µi , pij )i,j∈I be an inverse system of measures with µi ∈ Mxi (Hi ). Then µ := lim µi is an H-representing measure for x. ←−

Proof. For the proof of (a), we first assume that µ ∈ Mx (H). Propositions 12.28(b) and 12.24(a) yield that (πi )] µ ∈ Mxi (Hi ) for any i ∈ I. Lemma A.103 implies that ((πi )] µ, (pij )] )i,j∈I is an inverse system. Conversely, let ((πi )] µ, pij )i,j∈I be an inverse system of measures with (πi )] µ ∈ Mxi (Hi ) for i ∈ I. Let h ∈ Hj for some j ∈ I. Then µ(h ◦ πj ) = ((πj )] µ)(h) = h(xj ) = (h ◦ πj )(x). Thus µ ∈ Mx (H).


445

To check (b), Definition A.101 provides a measure µ ∈ M1 (K) such that (πi )] µ = µi , i ∈ I. By (a), µ ∈ Mx (H), and we are done.

Inverse limits of simplicial function spaces

12.2.C

We again consider an inverse system ((Ki , Hi ), pij )i,j∈I and its inverse limit (K, H). S Proposition 12.33. If every Hi is simplicial, then i∈I {h◦πi : h ∈ Ac (Hi )} is dense in Ac (H). Proof. Let a ∈ Ac (H) and ε > 0. By Corollary 3.23 and an easy compactness argument, there exist functions −f, g ∈ W(H) such that a − ε < f ≤ a ≤ g < a + ε. We write f = (f1 ◦ πi1 ) ∨ · · · ∨ (fn ◦ πin ), g = (g1 ◦ πj1 ) ∧ · · · ∧ (gm ◦ πjm ),

fk ∈ Hik , ik ∈ I, k = 1, . . . , n, gk ∈ Hjk , jk ∈ I, k = 1, . . . , m.

Let i ∈ I be greater than all the indices i1 , . . . , in , j1 , . . . , jm . Then (f1 ◦ pi1 i ) ∨ · · · ∨ (fn ◦ pin i ) ≤ (g1 ◦ pj1 i ) ∧ · · · ∧ (gm ◦ pjm i ), and the former function belongs to Kc (Hi ), the latter to S c (Hi ). Using the simpliciality of Ac (Hi ), we find a function h ∈ Ac (Hi ) such that (f1 ◦ pi1 i ) ∨ · · · ∨ (fn ◦ pin i ) ≤ h ≤ (g1 ◦ pj1 i ) ∧ · · · ∧ (gm ◦ pjm i ). Then a − ε < f ≤ h ◦ πi ≤ g < a + ε. Since πi is an admissible mapping, Proposition 12.24(b) yields h ◦ πi ∈ Ac (H). This concludes the proof. Theorem 12.34. If every Hi is simplicial, then H is simplicial. Proof. We show that Ac (H) has the weak Riesz interpolation property. Assume that a1 , . . . , a4 ∈ Ac (H) satisfy a1 ∨ a2 < a3 ∧ a4 . By Proposition 12.33 we may assume that ak = bk ◦ πik , k = 1, . . . , 4, where ik ∈ I and bk ∈ Ac (Hik ). Let i ∈ I be greater than indices i1 , . . . , i4 . As above we employ the simpliciality of Hi to find a function h ∈ Ac (Hi ) such that (b1 ◦ pi1 i ) ∨ (b2 ◦ pi2 i ) < h < (b3 ◦ pi3 i ) ∧ (b4 ◦ pi4 i ). Then h ◦ πi ∈ Ac (H) (use Proposition 12.24(b)) and a1 ∨ a2 < h ◦ πi < a3 ∧ a4 . This finishes the proof.

446


Corollary 12.35. Let (Xi , pij )i,j∈I be an inverse family of Choquet simplices. Then its inverse limit is a Choquet simplex. Proof. Let X := lim Xi be the inverse limit and let (X, H) be the inverse limit of ←

the family ((Xi , Ac (Xi )), pij )i,j∈I whose elements are regarded as function spaces. By Theorem 12.34, H is a simplicial function space. Proposition 12.30 yields H = Ac (X), and thus Ac (X) is simplicial as well. But this means exactly that X is a simplex. Theorem 12.36. Let x = (xi )i∈I be a point of K. (a) If xi ∈ ChHi (Ki ) for every i ∈ I, then x ∈ ChH (K). (b) Assume that each Hi is simplicial and pij (ChHj (Kj )) ⊂ ChHi (Ki ), i ≤ j. Then x ∈ ChH (K) if and only if xi ∈ ChHi (Ki ) for every i ∈ I. Proof. For the proof of (a), let x ∈ K satisfy xi ∈ ChHi (Ki ) for each i ∈ I. Let µ ∈ Mx (H) be given. By Proposition 12.32, ((πi )] µ, pij )i,j∈I is an inverse system of measures with (πi )] µ ∈ Mxi (Hi ) for i ∈ I. Then (πi )] µ = εxi for each i ∈ I and Lemma A.103 gives µ = lim(πi )] µ = lim εxi = εx . ←

←

Hence x ∈ ChH (K). To verify (b), let x ∈ ChH (K) be given and let the assumption of (b) be satisfied. We fix i ∈ I, and select a function f ∈ Kc (Hi ) along with ε > 0. Since x ∈ ChH (K), there exists h ∈ H such that f ◦ πi < h and (f ◦ πi )(x) < h(x) < (f ◦ πi )(x) + ε. Without loss of generality we may assume that h = g ◦πj for some h ∈ Hj and j ≥ i. Then (f ◦ pij )(xj ) = (f ◦ pij ◦ πj )(x) = (f ◦ πi )(x) < (h ◦ πj )(x) < (f ◦ pij )(xj ) + ε gives (f ◦ pij )(xj ) < h(xj ) < (f ◦ pij )(xj ) + ε. This along with Proposition 12.25 yields (f ∗ ◦ pij )(xj ) = (f ◦ pij )∗ (xj ) < h(xj ) < (f ◦ pij )(xj ) + ε. Hence f ∗ (xi ) < f (xi ) + ε. Since ε > 0 is arbitrary, f ∗ (xi ) = f (xi ) for any f ∈ Kc (Hi ). By Theorem 3.24, xi ∈ ChHi (Ki ).


12.2.D

447

Structure of simplices

∞ n Definition 12.37 (Admissible basis of `∞ n ). We recall that `n is the space R considered with the supremum norm and pointwise ordering (that is, x ≤ y if and only if y −x has positive coordinates). Let {e1 , . . . , en } be the canonical basis of `∞ n . A basis {a1 , . . . , an } of `∞ is called admissible if a = T e , i = 1, . . . , n, for some isometry i i n ∞ . An admissible basis is positive if the operator T above is positive. T : `∞ → ` n n

Lemma 12.38. (a) Let {a1 , . . . , an } be an admissible basis of `∞ n . Then there exists a permutation π : {1, . . . , n} → {1, . . . , n} and signs i ∈ {−1, 1}, i = 1, . . . , n, such that ai = i eπ(i) , i = 1, . . . , n. If, moreover, {a1 , . . . , an } is positive, all signs i are positive. ∞ (b) Let m > n, T : `∞ n → `m be an isometry and let {u1 , . . . , un } be an admissible ∞ basis of `n . Then there exist an admissible basis

{v1 , . . . , vm } ⊂ `∞ m and a family of numbers {ai,j : i = 1, . . . , n, j = n + 1, . . . , m} such that • •

P T ui = vi + m j=n+1 ai,j vj , i = 1, . . . , n, Pn i=1 |ai,j | ≤ 1, j = n + 1, . . . , m.

Further, for each k = n + 1, . . . , m, the space Fk := span{T u1 , . . . , T un , vn+1 , . . . , vk } is isometric to `∞ k . (c) If T in (b) is positive and {u1 , . . . , un } is an admissible positive basis, all numbers ai,j , i = 1, . . . , n, j = n + 1, . . . , m, and the basis {v1 , . . . , vm } can be chosen to be positive. ∞ (d) Let F ⊂ `∞ m be a subspace isometric to `n , n < m. Then there exist subspaces ∞ Fn+1 , . . . , Fm−1 of `∞ m such that F ⊂ Fn+1 ⊂ · · · ⊂ Fm−1 ⊂ `m and Fi is ∞ isometric to ì , i = n + 1, . . . , m − 1.

Proof. For the proof of (a), let ai := T ei , i = 1, . . . , n, for some isometry T : `∞ n → ∞ . Since e ∈ ext B ∞ , T e ∈ B ∞ , and `∞ . Let e stand for the vector (1, 1, . . . , 1) ∈ ` `n `n n n thus there exist signs i ∈ {−1, 1}, i = 1, . . . , n, such that T e = (1 , . . . , n ). ∞ Let S : `∞ n → `n be the isometry that maps ei to i ei , i = 1, . . . , n. Then ST is a positive isometry of `∞ n onto itself.

448


Since e = ST (e1 + · · · + en ),

kST ei k = 1,

and ST ei ≥ 0,

i = 1, . . . , n,

it follows that ST ei = eπ(i) , i = 1, . . . , n, for some permutation π : {1, . . . , n} → {1, . . . , n}. Hence T ei = i eπ(i) as needed. ∞ For the proof of (b), let T : `∞ n → `m be an isometry and {u1 , . . . , un } be an ∞ admissible basis of `n . We write T ui in coordinates with respect to the canonical basis {e1 , . . . , em } of `∞ m as T ui = (ci,1 , . . . , ci,m ),

i = 1, . . . , n.

For any choice of signs 1 , . . . , n we have n n n X

X X i ci,1 , . . . , i ci,m . i ui = 1 = T i=1

i=1

i=1

Pn

Hence i=1 |ci,j | ≤ 1 for each j = 1, . . . , m. For any i = 1, . . . , n, we can find an index π(i) ∈ {1, . . . , m} and i ∈ {−1, 1} ∞ such that i ci,π(i) = 1. Hence cl,π(i) = 0 for all l ∈ {1, . . . , n}\{i}. Let S : `∞ m → `m be the isometry that maps eπ(i) to i ei , i = 1, . . . , n, and the remaining vectors onto {en+1 , . . . , em }. P Then S defines the required admissible basis {v1 , . . . , vm }. We write T ui = vi + m j=n+1 ai,j vj , i = 1, . . . , n. Let k ∈ {n + 1, . . . , m} be fixed. It is easy to observe that the basis k k X X T u1 − a1,j vj , . . . , T un − an,j vj , vn+1 , . . . , vk j=n+1

j=n+1

is isometric to the usual basis (e1 , . . . , ek ) of `∞ k and spans Fk . This concludes the proof of (b). To verify (c), we follow the proof of (b) and notice that T ui are positive vectors. Hence the same argument as in (b) provides the required positive basis {v1 , . . . , vm } and numbers ai,j , i = 1, . . . , n and j = n + 1, . . . , m. ∞ To prove (d), let F ⊂ `∞ m be isometric to `n for some n < m. We find an ad∞ missible basis {v1 , . . . , vm } of `m from (b). Now, the spaces Fn+1 , . . . , Fm from (b) satisfy our requirements. This finishes the proof. Lemma 12.39. Let X be a compact convex set. c (a) Let {e1 , . . . , en } be the canonical basis of `∞ n , {f1 , . . . , fn } ⊂ A (X) be a peaked partition of unity and let E be the linear span of {f1 , . . . , fn }. Then the mapping T : E → `∞ n defined as T

n X

ai fi := (a1 , . . . , an ),

(a1 , . . . , an ) ∈ `∞ n ,

i=1

is a positive isometry satisfying T fi = ei , i = 1, . . . , n.


449

Further, the mapping ϕ : X → Rn defined as ϕ(x) := (f1 (x), . . . , fn (x)),

x ∈ X,

is an affine continuous surjection onto the (n − 1)-simplex co{e1 , . . . , en }. (b) Let E ⊂ Ac (X) be isometric to `∞ n and 1 ∈ E. Then there exist a positive isometry T : E → `∞ and a peaked partition of unity {f1 , . . . , fn } ⊂ Ac (X) n such that E = span{f1 , . . . , fn } and T fi = ei , i = 1, . . . , n. (c) Let {e1 , . . . , ep } ⊂ Ac (X) be a peaked partition of unity and Ep be the linear span of {e1 , . . . , ep }. If Ep+r ⊃ Ep is a subspace of Ac (X) isometric to `∞ p+r , then there exist positive numbers {a1,m , . . . , am,m },

m = p, . . . , p + r − 1,

and peaked partitions of unity {e1,m , . . . , em,m } ⊂ Ep+r ,

m = p + 1, . . . , p + r,

such that •

ei = ei,p+1 + ai,p ep+1,p+1 , i = 1, . . . , p,

•

ei,m = ei,m+1 + ai,m em+1,m+1 , i = 1, . . . , m, m = p + 1, . . . , p + r − 1, Pm i=1 ai,m = 1, m = p, . . . , p + r − 1.

•

Proof. For the proof of (a), let T and {f1 , . . . , fn } be as in the hypothesis. We find points {x1 , . . . , xn } such that fi (xi ) = 1, i = 1, . . . , n. Obviously, T is positive. For a given (a1 , . . . , an ) ∈ `∞ n , let i0 be an index satisfying |ai0 | = k(a1 , . . . , an )k. Then n n X X ai fi (x) ≤ |ai0 | fi (x) = k(a1 , . . . , an )k, i=1

On the other hand,

x ∈ X.

i=1 n X

ai fi (xi0 ) = ai0 ,

i=1

and hence T is an isometry. Obviously, T ei = fi for each i = 1, . . . , n. Further, ϕ(xi ) = ei , i = 1, . . . , n. For any x ∈ X, ϕ(x) =

n X

fi (x)ϕ(xi ),

i=1

and thus ϕ(X) ⊂ co{e1 , . . . , en }. This proves (a). To verify (b), let T : E → `∞ n be an isometry. Since 1 is an extreme point of BE , each coordinate of T 1 equals either 1 or −1. By composing T with an isometry of `∞ n

450


onto `∞ n if necessary we may assume that T 1 = (1, . . . , 1) =: e. Then T is positive and the functions fi := T −1 ei , i = 1, . . . , n, form the required partition of unity. Indeed, n n X X 1 = T −1 (e) = T −1 ei = fi , i=1

i=1

and all functions fi are positive. To prove (c), let Ep ⊂ Ep+r be given. Let T : Ep → `∞ p be a positive isometry given by (a). Using (b), we find a positive isometry S : Ep+r → `∞ p+r . Let I : Ep → −1 ∞ ∞ Ep+r denote the inclusion. Then SIT : `p → `p+r is an isometry into. Using ∞ Lemma 12.38(d) we find a space Fp+1 ⊂ `∞ p+r isometric with `p+1 such that ∞ SIT −1 (`∞ p ) ⊂ Fp+1 ⊂ `p+r .

Let J : Fp+1 → `∞ p+1 be a positive isometry. Since the set {T e1 , . . . , T ep } is the ∞ canonical basis of `p , using Lemma 12.38(c) we can find a positive admissible basis {v1 , . . . , vp+1 } of `∞ p+1 and positive numbers {a1 , . . . , ap } such that JSIei = vi + ai vp+1 ,

i = 1, . . . , p,

(12.26)

P and pi=1 ai ≤ 1. ∞ Let E := (JS)−1 (`∞ p+1 ) and V : E → `p+1 denote the restriction of JS. Since 1 ∈ E and V is a positive isometry, V 1 = (1, . . . , 1) ∈ `∞ p+1 . We set ei,p+1 := V −1 vi ,

i = 1, . . . , p + 1,

and ai,p := ai ,

i = 1, . . . , p.

Then {e1,p+1 , . . . , ep+1,p+1 } is a peaked partition of unity in Ep+r . Indeed, all the functions are obviously positive and have norm one. Since V 1 = (1, . . . , 1) =

p+1 X

vi ,

i=1

we obtain 1=

p+1 X

ei,p+1 .

i=1

We claim that

Pp

i=1 ai,p

= 1. Indeed, from (12.26) we get

ei = ei,p+1 + ai,p ep+1,p+1 , Hence 1=

p X i=1

ei,p+1 + ep+1,p+1

i = 1, . . . , p. p X i=1

ai,p .


451

If x ∈ X is a point where ep+1,p+1 (x) = 1, we obtain 1=

p X

ai,p .

i=1

Now the rest of the argument follows by an application of the this procedure to the family {e1,m , . . . , em,m } for all m = p + 1, . . . , p + r − 1. This concludes the proof. Theorem 12.40. Let X be a simplex and E ⊂ Ac (X) be a finite-dimensional space c isometric to `∞ n . Let {f1 , . . . , fk } ⊂ A (X) and ε > 0. Then there exist m ∈ N and a finite-dimensional subspace F ⊃ E of Ac (X) such that F is isometric to `∞ m and dist(fi , F ) < ε, i = 1, . . . , k. Proof. Given the objects as in the theorem, we first notice that without loss of generality we may assume that k = 1. Let f ∈ Ac (X), ε > 0 and E be given. Using the continuity of f and the compactness of X we can find a partition {U1 , . . . , Um } of ext X consisting of nonempty sets such that diam f (Uj ) < ε, j = 1, . . . , m. We select points xj ∈ Uj and set φj := cUj , αj := f (xj ), j = 1, . . . , m. Then m X f (x) − αj φj (x) ≤ ε,

x ∈ ext X.

j=1

By Lemma 12.17, there exists a peaked partition of unity {ψ1 , . . . , ψm } in Ac (X) such that m X f (x) − αj ψj (x) ≤ ε, x ∈ X. j=1

Then F := span{ψ1 , . . . , ψm } satisfies the required conditions by Lemma 12.39(a). Definition 12.41 (L∞ -spaces). We recall that a Banach space E is termed an L∞,λ space if for any finite-dimensional space A ⊂ E there exists a finite-dimensional subspace F ⊃ A of E whose Banach–Mazur distance from some `∞ n is smaller or equal than λ where n = dim F (that is, for every ε > 0 there exists an isomorphism −1 k < λ + ε). T : F → `∞ n such that kT kkT Corollary 12.42. If X is a simplex, then Ac (X) is an L∞,1+ε -space for every ε > 0. Proof. Let E ⊂ Ac (X) be a space of dimension n ∈ N and ε > 0. We choose ε0 ∈ (0, n) such that n ε0 < ε. n − ε0

452


Using Theorem 5.6(i) of [173], we find a basis {e1 , . . . , en } of E consisting of unit vectors and functionals {e∗1 , . . . , e∗n } ⊂ (Ac (X))∗ of norm 1 such that ( 1, i = j, ∗ ei (ej ) = 1 ≤ i, j ≤ n. 0, i 6= j, Using Theorem 12.40 we find a finite-dimensional space F 0 ⊂ Ac (X) isometric to `∞ m 0 for some m ∈ N such that dist(ei , F 0 ) < nε2 , i = 1, . . . , n. We select {f1 , . . . , fn } ⊂ 0 F 0 such that dist(ei , fi ) < nε2 , i = 1, . . . , n, and set G := span{f1 , . . . , fn }. If S : E → G is defined as n n X X ci ei := ci fi , c1 , . . . , cn ∈ R, S i=1

i=1

then S is an isomorphism satisfying ke − Sek ≤

n X

|ci |kei − fi k ≤

n ε0 ε0 X ∗ |e (e)| kek, ≤ i n2 n

e=

i=1

i=1

It follows that kS −1 k ≤

1 1−

ε0 n

n X

ci ei ∈ E.

i=1

.

(12.27)

Since G has dimension n, Theorem 5.6(ii) in [173] yields the existence of a projection P : F 0 → G with kP k ≤ n. We set H := ker P and F := E ⊕ H. Then F contains E and the operator T : F → F 0 defined as T f := Se + h,

f = e + h ∈ E ⊕ H,

satisfies by (12.27) 0 kf k − kT f k ≤ kf − T f k = kSe − ek ≤ ε kek n ε0 −1 ε0 1 ≤ kS kkSek ≤ kP (Se + h)k n n 1 − εn0 n kT f k. ≤ ε0 n − ε0 By our choice of ε0 , kf k − kT f k ≤ εkT f k and thus

kT k ≤ (1 − ε)−1 This finishes the proof.

and kT −1 k ≤ 1 + ε.


453

Remark 12.43. It is known that a Banach space E is an L∞,1+ε -space for every ε > 0 if and only if E ∗ is isometric to L1 (X, Σ, µ) for a suitable measure space (X, Σ, µ) (see p. 59 in [251]). Hence Corollary 12.42 provides another proof of Theorem 6.25. Lemma 12.44. Let E ⊂ Ac (X) be isometric to `∞ n and let 1 ∈ E. Then the set Sn := {e∗ ∈ E ∗ : e∗ ≥ 0, e∗ (1) = 1} is affinely homeomorphic to an (n − 1)-simplex. Proof. We start the proof by using Lemma 12.39(b) to find a peaked partition of unity {f1 , . . . , fn } ⊂ Ac (X) such that E = span{f1 , . . . , fn }. Let ψ : Sn → Rn be defined as ψ(e∗ ) := (e∗ (f1 ), . . . , e∗ (fn )), e∗ ∈ Sn . Then ψ is an affine homeomorphism of Sn to Rn . By Lemma 12.39(a) it is enough to show that ψ(Sn ) = ϕ(X) (ϕ is defined as in Lemma 12.39(a)). Let φ : X → Sn be the evaluation mapping defined as φ(x)(f ) := f (x), f ∈ E, x ∈ X. Obviously, for any e∗ ∈ Sn , ψ(e∗ ) has positive coordinates whose sum equals 1, hence ψ(Sn ) ⊂ ϕ(X). Conversely, if xi is a point of X with fi (xi ) = 1, then ψ(φ(xi )) = ei . Hence ψ(Sn ) ⊃ ext ϕ(X), and the conclusion follows. Theorem 12.45. Let X be a metrizable simplex. Then the following assertions hold. (a) There exists an increasing family {En } of finite-dimensional subspaces of Ac (X) S ∞ c such that E1 = span{1}, En is isometric to `∞ n and A (X) = n=1 En . (b) There exists an inverse sequence (Xn , pnm )n,m∈N such that each Xn is an (n−1)simplex and X = lim Xn . ←

Proof. For the proof of (a), let {fk : k ∈ N} be a dense subset of BAc (X) . We inductively use Theorem 12.40 to find an increasing sequence {nk } of natural numbers and spaces Enk ⊂ Ac (X) such that • n = 1 and E n1 = span{1}, 1 •

Enk is isometric to `∞ nk , k ∈ N,

•

Enk ⊂ Enk+1 , k ∈ N,

dist(fi , Enk ) < 2−k , i = 1, . . . , k, k ∈ N. By Lemma 12.39(a),(c), for each k ∈ N we can find spaces •

Enk ⊂ Enk +1 ⊂ · · · ⊂ Enk+1 −1 ⊂ Enk+1 such that each Ej is isometric to `∞ j , j = nk + 1, . . . , nk+1 − 1. Hence the proof of (a) is finished. To verify (b), let {En } be subspaces given by (a). We set Xn := Sn , n ∈ N, where Sn is the compact convex set from Lemma 12.44, and define pnm : Xm → Xn ,

454


n ≤ m, to be the restriction. By Lemma 12.44, the system (Xn , pnm )n,m∈N is an inverse sequence of (n − 1)-simplices. It is a routine verification to show that the mapping ψ : S(Ac (X)) → lim Xn defined as ←

ψ(s) := (s|En )n∈N ,

s ∈ S(Ac (X)),

is an affine homeomorphism. Since X is affinely homeomorphic to S(Ac (X)) (see Proposition 4.31), we are done. Corollary 12.46. For any metrizable simplex X,Sthere exists an increasing sequence {Fn } of finite-dimensional faces of X such that ∞ n=1 Fn is dense in X. Proof. Let {En } be a sequence of spaces in Ac (X) isometric to `∞ n such that E1 = c span{1} and their union is dense in A (X) (see Theorem 12.45). Let n ∈ N be fixed. By Lemma 12.39 there exists a peaked partition of unity {f1 , . . . , fn } such that En is its linear span and T : E → `∞ n defined as T fi := ei , i = 1, . . . , n, is a positive isometry. Let xi ∈ X be points such that fi (xi ) = 1,

i = 1, . . . , n.

Without loss of generality we may assume that all points xi are extreme. Then the set Fn := co{x1 , . . . , xn } is a finite-dimensional face of X. S We claim that {Fn } is the required sequence. Indeed, set F := ∞ n=1 Fn and assume the existence of ∞ [ Fn . x∈X\ n=1

Using the Hahn–Banach theorem we find a function f ∈ Ac (X) and δ > 0 such that 0 < f < 1 and m < m + δ < f (x),

where m := max f (F ).

Let n ∈ N and g ∈ En be such that 0 < g < 1 and kf − gk < 41 δ. If g = for some numbers a1 , . . . , an , then k(a1 , . . . , an )k`∞ = kgk. n Since

1 g ≤ δ + m on F, 4

for any i = 1, . . . , n we get 1 0 ≤ ai = g(xi ) ≤ m + δ. 4

Pn

i=1 ai fi

(12.28)

12.3 Several examples

455

By (12.28), kgk ≤ m + 41 δ. Hence 1 1 m + δ < f (x) ≤ δ + g(x) ≤ m + δ, 4 2 a contradiction. This finishes the proof. Theorem 12.47. Every simplex is the inverse limit of an inverse system of metrizable simplices. Proof. If X is a simplex, let E denote the family of all norm separable closed subspaces of Ac (X) that contain 1 and possess the weak Riesz interpolation property. By Theorem 9.12, this family is up-directed, that is, for any E1 , E2 ∈ E there exists S E ∈ E with E1 ∪ E2 ⊂ E. Moreover, {E : E ∈ E} is dense in Ac (X). For any E ∈ E, let XE := {e∗ ∈ E ∗ : e∗ ≥ 0, e∗ (1) = 1}. For any E, F ∈ E with E ⊂ F , let pE,F : XF → XE be the restriction mapping. Since each E ∈ E possesses the weak Riesz interpolation property, (XE , pE,F )E,F ∈E is an inverse system of metrizable simplices (as in Exercise 6.91). To finish the proof it is enough to realize that the mapping ψ : S(Ac (X)) → lim XE defined as ←

ψ(s) := (s|E )E∈E ,

s ∈ S(Ac (X)),

is an affine homeomorphism.

12.3 12.3.A

Several examples The Poulsen simplex

Lemma 12.48. Let X be a simplex in a locally convex space E and z ∈ E \ span X. Then co(X ∪ {z}) is a simplex and ext co(X ∪ {z}) = ext X ∪ {z}. Proof. We start the proof by noticing that z is an extreme point of co(X ∪ {z}). Further, any point x in co(X ∪{z}) can be uniquely decomposed as x = αy+(1−α)z for some y ∈ X and α ∈ [0, 1]. Hence there exists a unique maximal measure µ ∈ Mx (co(X ∪ {z})), namely µ = αδy + (1 − α)εz , where δy is the unique maximal measure for y on X. Thus co(X ∪ {z}) is a simplex and ext co(X ∪ {z}) = ext X ∪ {z}. Lemma 12.49. Let X be a simplex in a Banach space E. Then there exists a simplex Y ⊂ E ⊕ `2 such that X is a closed face of Y and ext Y is dense in Y .

456


Proof. Let X be a simplex in a Banach space E and let {en : n ∈ N} be the canonical basis of `2 . We identify x ∈ E with (x, 0) ∈ E ⊕ `2 and y ∈ `2 with (0, y) ∈ E ⊕ `2 . Let En := span{ek : k = 1, . . . , n} ⊂ `2 and let πn : E ⊕ `2 → E ⊕ En be the projection. Let Y0 := X. We construct inductively simplices Yn ⊂ E ⊕ `2 , points zn ∈ E ⊕ `2 and εn > 0, n ∈ N, such that (a) Yn = co(Yn−1 ∪ {zn }), n ∈ N, (b) Yn ⊂ E ⊕ En , n ∈ N, (c) ext Yn ⊂ ext Ym , n ≤ m, (d) πn (Ym ) = Yn and kπn (x) − xk ≤ 2−n+1 , x ∈ Ym , n ≤ m, (e) for each k ∈ N there exists n ≥ k such that dist(y, ext Yn ) < εk , y ∈ Yk , and εn+1 < 21 εk , (f) εn+1 ≤ εn , n ∈ N. To start the construction, we choose y1 ∈ Y0 and set ε1 := 2−1 , z1 := y1 + 2−1 e1 , Y1 := co(Y0 ∪ {z1 }). Assume now that the construction has been completed up to the n–th stage. We denote Mn (ε) := {x ∈ Yn : dist(x, {z1 , . . . , zn }) ≥ ε},

ε > 0.

Now we distinguish two cases. If Mn (εn ) 6= ∅, we choose yn+1 ∈ Mn (εn ) and set εn+1 = εn . Otherwise we find εn+1 ∈ (0, 12 εn ) such that Mn (εn+1 ) 6= ∅ and choose yn+1 ∈ Mn (εn+1 ). In both cases we set zn+1 := yn+1 + 2−n−1 en+1 , Yn+1 := co(Yn ∪ {zn+1 }). This finishes the construction. Properties (a)–(d) are obviously satisfied. Given k ∈ N, we find m ≥ k such that 2−m+3 < εk and notice that by (d) and the construction, kπm (zj ) − πm (zi )k ≥ εj − 2−m+2 ,

m ≤ i < j.

By the compactness of Ym we observe that infi,j≥m kπm (zj ) − πm (zi )k = 0, and thus there must exist j ∈ N with εj < εk . Hence we find n ≥ k such that the (n + 1)-th step of the construction follows the second case, which verifies property (e). S Let Y := ∞ n=1 Yn . Then Y is convex and compact (it follows from (d)).


457

Further, ext Yn ⊂ ext Y . Indeed, let x ∈ ext Yn equal αy1 + (1 − α)y2 for some y1 , y2 ∈ Y . Then x = απm (y1 ) + (1 − α)πm (y2 ),

n ≤ m,

and thus πm (y1 ) = πm (y2 ) = x for each m ≥ n. Hence y1 = y2 = x. S Thus ext Y is dense in Y by (e). It also follows from (d) that ∞ n=1 {f ◦ πn : f ∈ Ac (Yn )} is dense in Ac (Y ). Since each space in this union has the Riesz interpolation property, Ac (Y ) has the weak Riesz interpolation property and hence Y is a simplex (see Theorem 6.16). Since X is obviously a face of Y , the proof is finished. Corollary 12.50. There exists a simplex X ⊂ `2 such that ext X is dense in X. Proof. An immediate consequence of Theorem 12.49. Lemma 12.51. Le ϕ : X → Y be a continuous affine surjection of a compact convex set onto an n-simplex Y . Then ϕ is open. Proof. Let {e0 , . . . , en } be affinely independent vectors in some space Rd and ϕ : X → co{e0 , . . . , en } be a continuous affine surjection. Let U ⊂ X be a closed neighborhood of x ∈ X and let y = ϕ(x). For each ei we find xi ∈ X such that ϕ(xi ) = ei . Let α ∈ (0, 1) be such that αx + (1 − α)xi ∈ U , i = 0, . . . , n. Then V := co{αy + (1 − α)ei : i = 0, . . . , n} is a neighborhood of y contained in ϕ(U ). Hence ϕ is open. Lemma 12.52. Let F be a closed face of a metrizable simplex X and {e1 , . . . , en } ⊂ Ac (X) be a peaked partition of unity such that each ei attains its maximum on F . Then there exists an affine continuous retraction g : X → F such that ei = ei ◦ g, i = 1, . . . , n. Proof. Let {x1 , . . . , xn } ⊂ F be such that ei (xi ) = 1, i = 1, . . . , n. Let π : X → co{x1 , . . . , xn } be defined as π(x) :=

n X

ei (x)xi ,

x ∈ X.

i=1

Then ei = ei ◦ π for all i = 1, . . . , n. We define ϕ : X → 2F by the formula ϕ(x) := π −1 (π(x)) ∩ F,

x ∈ X.

458


Then ϕ is an affine mapping with nonempty closed convex values. Since the set co{x1 , . . . , xn } is an (n − 1)-simplex (see Lemma 6.68 and Exercise 6.85), the mapping π : F → co{x1 , . . . , xn } is open by Lemma 12.51. Hence for any open set U ⊂ X, {x ∈ X : ϕ(x) ∩ U 6= ∅} = {x ∈ X : π −1 (π(x)) ∩ F ∩ U 6= ∅} = π −1 (π(F ∩ U )), and thus ϕ is lower semicontinuous. If f (x) = x for x ∈ F , then f is a continuous affine mapping with f (x) ∈ ϕ(x), x ∈ F . By Theorem 11.8, there exists a continuous affine selection g : X → F such that g(x) = x on F . Then π ◦ g = π and hence ei ◦ g = ei ◦ π ◦ g = ei ◦ π = ei ,

i = 1, . . . , n.

This concludes the proof. Lemma 12.53. Let F be a closed proper face of a metrizable simplex X with ext X = X. Let {e1 , . . . , en } ⊂ Ac (X) be a peaked partition of unity, {f1 , . . . , fn+1 } ⊂ Ac (F ) be a partition of unity, and let {a1 , . . . , an } be a set of positive numbers such that ei |F = fi + ai fn+1 , i = 1, . . . , n. Then, for each ε > 0, there exists a peaked partition of unity {g1 , . . . , gn+1 } ⊂ Ac (X) such that • g | i F = fi , i = 1, . . . , n + 1, •

kei − (gi + ai gn+1 )k < ε, i = 1, . . . , n.

Proof. Let xi ∈ ext X be such that ei (xi ) = 1, i = 1, . . . , n. We find a point yn+1 ∈ ext X \ F such that n X ei (yn+1 ) − ei aj xj < ε,

i = 1, . . . , n.

(12.29)

j=1

This is possible because ext X \ F is dense in X (see Lemma 2.90). For j = 1, . . . , n we set yj := xj if xj ∈ / F , otherwise we find yj ∈ ext X \ F such that |ei (xj ) − ei (yj )| < ε,

i = 1, . . . , n.

(12.30)

Let H := co(F ∪ {y1 , . . . , yn+1 }). Then H is a closed face of X and by Lemma 12.52 there exists a continuous affine retraction ϕ : X → H such that ei ◦ ϕ = ei , i = 1, . . . , n. For i = 1, . . . , n + 1 we find gei ∈ Ac (H) defined by conditions gei = fi on F and ( 0, i 6= j, gei (yj ) = j = 1, . . . , n + 1, 1, i = j,


459

(use Theorem 6.6), and define gi := gei ◦ ϕ,

i = 1, . . . , n.

Then {g1 , . . . , gn+1 } is a peaked partition of unity and gi = fi on F . Since ei ◦ ϕ = ei and gi ◦ ϕ = gi , we get kei − (gi + ai gn+1 )k = kei ◦ ϕ − (gi ◦ ϕ + ai gn+1 ◦ ϕ)k =

max

j=1,...,n+1

|ei (yj ) − (gi + ai gn+1 )(yj )| < ε.

(Use (12.30) for j = 1, . . . , n and (12.29) for j = n + 1.) This finishes the proof. Theorem 12.54. Let X1 , X2 be metrizable simplices with ext Xi = Xi , i = 1, 2. Let Hi ⊂ Xi be a proper closed face, i = 1, 2, and let ϕ : H2 → H1 be an affine homeomorphism. Then ϕ can be extended to an affine homeomorphism of X2 onto X1 . Proof. Let {xn : n ∈ N} and {yn : n ∈ N} be dense subsets in the unit spheres of Ac (X1 ) and Ac (X2 ), respectively, such that x1 = y1 = 1. We construct inductively peaked partitions of unity {ej1,m , . . . , ejm,m } ⊂ Ac (X1 ),

j j {f1,m , . . . , fm,m } ⊂ Ac (X2 ),

m ≤ j, m ∈ N,

and sets of positive numbers {a1,m , . . . , am,m }, such that Pm (1) i=1 ai,m = 1, (2)

m ∈ N,

m ∈ N,

eji,m = eji,m+1 + ai,m ejm+1,m+1 , j j j fi,m = fi,m+1 + ai,m fm+1,m+1 ,

(3)

j eji,m (ϕ(t)) = fi,m (t),

(4)

−j keji,m − ej+1 i,m k < 2 , j j+1 kfi,m − fi,m k < 2−j ,

m ∈ N, 1 ≤ i ≤ m, j ≥ m + 1,

t ∈ H2 , m ∈ N, 1 ≤ i ≤ m, j ≥ m,

m ∈ N, 1 ≤ i ≤ m, j ≥ m.

m (5) If Em := span{em 1,m , . . . , em,m }, then for each n ∈ N there exists m ∈ N such that dist(xk , Em ) < 2−n , 1 ≤ k ≤ n.

460


m , . . . , f m }, then for each n ∈ N there exists m ∈ N such (6) If Fm := span{f1,m m,m that dist(yk , Fm ) < 2−n , 1 ≤ k ≤ n. 1 } := {1}. We start the construction by setting {e11,1 } := {1} and {f1,1 j j Assume that {ej1,m , . . . , ejm,m } and {f1,m , . . . , fm,m } have been constructed for 1 ≤ m ≤ j ≤ p such that (1), (3), (4) hold for m ≤ j ≤ p, (2) holds for m + 1 ≤ j ≤ p, and (5) and (6) hold for some n ∈ N. By Theorem 12.40, there exists a finite-dimensional space Ep+r ⊂ Ac (X1 ) isometric to `∞ p+r such that Ep ⊂ Ep+r and

dist(xk , Ep+r ) < 2−n−1 ,

1 ≤ k ≤ n + 1.

By Lemma 12.39, there exist sets of positive numbers {a1,m , . . . , am,m },

p ≤ m ≤ p + r − 1,

and peaked partitions of unity {g1,m , . . . , gm,m } ⊂ Ep+r , such that Pm •

• •

i=1 ai,m = 1, p ≤ m ≤ p + r − p ei,p = gi,p+1 + ai,p gp+1,p+1 , 1 ≤ i

p + 1 ≤ m ≤ p + r,

1, ≤ p,

gi,m = gi,m+1 + ai,m gm+1,m+1 , 1 ≤ i ≤ m, p + 1 ≤ m ≤ p + r − 1.

For every 1 ≤ m ≤ p + r, p + 1 ≤ j ≤ p + r and 1 ≤ i ≤ m we set ( epi,m , 1 ≤ m ≤ p, eji,m := gi,m , p + 1 ≤ m ≤ p + r. Then (1), (2) and (4) hold for the constructed functions eji,m and (5) holds for n + 1. We now transfer the constructed objects to Ac (X2 ). Since p+1 {ep+1 1,p+1 ◦ ϕ, . . . , ep+1,p+1 ◦ ϕ}

is a partition of unity on H2 and p fi,p (t) = epi,p (ϕ(t)) = gi,p+1 (ϕ(t)) + ai,p gp+1,p+1 (ϕ(t)) p+1 = ep+1 i,p+1 (ϕ(t)) + ai,p ep+1,p+1 (ϕ(t)),

t ∈ H2 , 1 ≤ i ≤ p,

Lemma 12.53 provides for any ε > 0 a peaked partition of unity {h1,p+1 , . . . , hp+1,p+1 } ⊂ Ac (X2 ) such that •

hi,p+1 (t) = ep+1 i,p+1 (ϕ(t)), t ∈ H2 , 1 ≤ i ≤ p + 1,

461

12.3 Several examples •

p kfi,p − (hi,p+1 + ai,p hp+1,p+1 )k < ε, 1 ≤ i ≤ p.

We take ε < 2−2(p+1) and define p+1 fi,p+1 := hi,p+1 ,

1 ≤ i ≤ p + 1,

p+1 p+1 p+1 fi,p := fi,p+1 + ai,p fp+1,p+1 ,

1 ≤ i ≤ p,

.. . p+1 p+1 p+1 fi,l := fi,l+1 + ai,l fl+1,l+1 ,

1 ≤ i ≤ l,

.. . p+1 p+1 p+1 f1,1 := f1,2 + a1,1 f2,2 .

Then p+1 p kfi,k − fi,k k < 2−p ,

1 ≤ k ≤ p, 1 ≤ i ≤ k.

Similarly we construct m m m m m {f1,m , . . . , fm,m }, . . . , {f1,2 , f2,2 }, {f1,1 },

p + 2 ≤ m ≤ p + r.

Then (4) holds for j ≤ p + r − 1. We interchange the roles of Ac (X1 ) and Ac (X2 ), that is, we find by Theorem 12.40 a space Fp+r+s ⊂ Ac (X2 ) isometric to `∞ p+r+s such that Fp+r ⊂ Fp+r+s and dist(yk , Fp+r+s ) < 2−(n+1) ,

1 ≤ k ≤ n + 1.

j Now we proceed with the construction as above and find vectors fi,m and then eji,m . Then (1)–(4) hold for relevant indices and (6) for n + 1. This completes the inductive construction. We set

ei,m := lim eji,m , j→∞

j fi,m := lim fi,m , j→∞

1 ≤ i ≤ m, m ∈ N.

It follows from (4), (5) and (6) that span{ei,m : 1 ≤ i ≤ m, m ∈ N}

and

span{fi,m : 1 ≤ i ≤ m, m ∈ N}

are dense in Ac (X1 ) and Ac (X2 ), respectively. Further, by (2), ei,m = ei,m+1 + ai,m em+1,m+1 , fi,m = fi,m+1 + ai,m fm+1,m+1 ,

m ∈ N, 1 ≤ i ≤ m,

(12.31)

and by (3) ei,m (ϕ(t)) = fi,m (t),

t ∈ H2 , m ∈ N, 1 ≤ i ≤ m.

(12.32)

462


Hence {e1,m , . . . , em,m } and

{f1,m , . . . , fm,m },

m ∈ N,

are peaked partitions of unity, in particular, they are linearly independent. It follows from (12.31) that the equalities T ei,m = fi,m ,

1 ≤ i ≤ m, m ∈ N,

j define an isometry T : Ac (X1 ) → Ac (X2 ). Since ej1,1 = 1 and f1,1 = 1 for each j ∈ N, we have e1,1 = 1 and f1,1 = 1, and thus T 1 = 1. Hence T is positive. Let φi : Xi → S(Ac (Xi )) be the affine homeomorphism from Proposition 4.31, i = 1, 2. As in the proof of Theorem 10.14, the adjoint operator T ∗ : (Ac (X2 )∗ →(Ac (X1 ))∗ defines an affine homeomorphism of S(Ac (X2 )) onto S(Ac (X1 )). Since

T e(t) = e(ϕ(t)),

t ∈ H2 , e ∈ Ac (X1 ),

the mapping ∗ φ−1 1 ◦ T ◦ φ2

is an affine homeomorphism of X2 onto X1 that extends ϕ. This finishes the proof. Corollary 12.55. Let X1 , X2 be metrizable simplices with ext Xi = Xi , i = 1, 2. Then X1 is affinely homeomorphic to X2 . Proof. We pick points xi ∈ ext Xi , i = 1, 2. Then the faces Fi := {xi }, i = 1, 2, are affinely homeomorphic and thus the conclusion follows from Theorem 12.54. Definition 12.56. (Poulsen simplex) A metrizable simplex, whose set of extreme points is dense, is called the Poulsen simplex. Lemma 12.57. Let F be a closed face of a simplex X and x1 , x2 be distinct points in the complementary face F 0 . Then there exists a positive function f ∈ Ac (X) such that f = 0 on F and f (x1 ) 6= f (x2 ). Proof. Given the objects as in the hypothesis, we find a function g ∈ Ac (X) such that 0 ≤ g < 21 and δ := g(x1 ) − g(x2 ) > 0. Then k := cF ∨ g is a convex upper semicontinuous function. We claim that k ∗ = k = g on F 0 . Indeed, let z ∈ F 0 be arbitrary. We find a measure ν ∈ Mz (X) such that ν(k) = ∗ k (z) (see Lemma 3.21). We find a maximal measure µ with ν ≺ µ and observe, using Proposition 3.56, that k ∗ (z) = ν(k) ≤ µ(k) ≤ k ∗ (z).


463

Since µ is maximal, Lemma 8.13 and Theorem 8.39 yield µ(F 0 ) = 1. Hence Z ∗ k(t) dµ(t) = µ(g) = g(z). k (z) = µ(k) = F0

We select a function h ∈ Ac (X) such that k ≤ h and h(xi ) < g(xi ) + 12 δ, i = 1, 2. By Theorem 6.6, there exists a function a ∈ Ac (X) such that k ≤ a ≤ 1 ∧ h. Then the function f := 1 − a satisfies our requirements. Proposition 12.58. Let F be a closed face of a simplex X and let A := {f ∈ Ac (X) : 0 ≤ f ≤ 1, f |F = 0}. Let ϕ : X → RA be defined as ϕ(x) := (f (x))f ∈A , x ∈ X, and let Y := ϕ(X). (a) The set Y is a simplex and ϕ(F ) is an extreme point of Y . (b) The mapping ϕ|F 0 is injective and ϕ(F 0 ) is the complementary face of ϕ(F ). (c) If X is the Poulsen simplex, then Y is the Poulsen simplex. Proof. For f ∈ A, let πf : Y → R be the restriction on the coordinate determined by f . Then πf ∈ Ac (Y ). It follows that for any function f ∈ Ac (X) constant on F there exists a function g ∈ Ac (Y ) such that f = g ◦ ϕ. For the proof of (a) we notice that ϕ(F ) = 0 is an extreme point of Y . Indeed, if αϕ(x)+(1−α)ϕ(y) = 0 for some x, y ∈ X and α ∈ (0, 1), then f (αx+(1−α)y) = 0 for each f ∈ A, and thus αx + (1 − α)y ∈ F (use Corollary 8.64). Hence x, y ∈ F and ϕ(x) = ϕ(y) = 0. Further, let g1 , −g2 ∈ Kc (Y ) be such that 0 ≤ g1 ≤ g2 ≤ 1. By Theorem 6.6, there exists a function f ∈ A such that g1 ◦ ϕ ≤ f ≤ g2 ◦ ϕ on X. Then g1 ≤ πf ≤ g2 and Y is a simplex by Theorem 6.6. To verify (b), we notice that ϕ is injective on F 0 by Lemma 12.57. The second assertion requires to show that (ϕ(F ))0 = ϕ(F 0 ). In other words, we need to prove that c∗ϕ(F ) (t) = 0

⇐⇒

t ∈ ϕ(F 0 ),

t ∈ Y.

464


Assume first that c∗ϕ(F ) (t) = 0, that is, t ∈ (ϕ(F ))0 . Let x ∈ X be such that ϕ(x) = t. For any ε > 0 there exists a function g ∈ Ac (Y ) such that g ≥ cϕ(F ) and g(t) < ε. Thus f := g ◦ ϕ ∈ Ac (X) satisfies f ≥ cF and f (x) < ε. Hence c∗F (x) = 0 and x ∈ F 0 . Thus t ∈ ϕ(F 0 ). Conversely, let x ∈ F 0 be given. For any ε > 0 we use Theorem 6.6 to find a function f ∈ Ac (X) such that f = 1 on F and f (x) < ε. For the function g ∈ Ac (Y ) satisfying f = g ◦ ϕ we get g ≥ cϕ(F ) and g(ϕ(x)) < ε. Thus c∗ϕ(F ) (ϕ(x)) = 0 and ϕ(x) ∈ (ϕ(F ))0 . Finally, let X be the Poulsen simplex. Since ϕ : F 0 → ϕ(F 0 ) is an affine bijection and (ϕ(F ))0 = ϕ(F 0 ), ϕ(ext X ∩ F 0 ) = ϕ(ext F 0 ) = ext ϕ(F 0 ) = ext Y ∩ (ϕ(F ))0 . Further, F as a closed proper face of X has empty interior (see Lemma 2.90), and thus ext X ∩ F 0 = ext F 0 is dense in X. Hence ext ϕ(F 0 ) is dense in Y and Y is the Poulsen simplex. This finishes the proof. Theorem 12.59. Let Fi , i = 1, 2, be closed faces of the Poulsen simplex and let Fi0 be their complementary faces, i = 1, 2. Then F10 is affinely homeomorphic to F20 . Proof. For each i = 1, 2, let ϕi : Xi → Yi be as in Proposition 12.58. Then the singletons {yi } := ϕi (Fi ), i = 1, 2, are obviously affinely homeomorphic, and thus by Theorem 12.54 there exists an affine homeomorphism ϕ : Y1 → Y2 sending ϕ1 (F1 ) onto ϕ2 (F2 ). By Proposition 12.58, ϕ(ϕ1 (F10 )) = ϕ({y1 }0 ) = {y2 }0 = ϕ2 (F20 ). 0 0 Thus ϕ−1 2 ◦ ϕ ◦ ϕ1 is an affine homeomorphism of F1 onto F2 .

Theorem 12.60 (Properties of the Poulsen simplex). Let S stand for the Poulsen simplex. (a) Any metrizable simplex is affinely homeomorphic to a face of S. (b) Let X be a metrizable simplex with the following properties: •

Any metrizable simplex is affinely homeomorphic to a face of X.

•

If F1 , F2 are faces of X with dim F1 = dim F2 < ∞, then there exists an affine homeomorphism ϕ of X onto itself such that ϕ(F1 ) = F2 .

Then X is affinely homeomorphic to S.


465

(c) If K ⊂ ext S is a compact set, then its interior (relative to ext S) is empty. (d) Every Polish space is homeomorphic to a closed subset of ext S. (e) If K1 , K2 are homeomorphic compact subsets of ext S, then there exists an affine homeomorphism ϕ of S onto itself such that ϕ(K1 ) = K2 . (f) The set ext S is arcwise connected (that is, for any points x1 , x2 ∈ ext S there exists a homeomorphic copy of [0, 1] contained in ext S that joins x1 with x2 ). Proof. To prove (a), let X be a metrizable simplex. By Proposition 2.45 we may assume that X is a norm compact subset of `2 . By Lemma 12.49, there exists a compact simplex Y ⊂ `2 ⊕ `2 such that X is a closed face of Y . Due to Corollary 12.55, Y is affinely homeomorphic to S. To verify (b), let X possess the prescribed property. By the assumption, X contains a face X0 affinely homeomorphic to the Poulsen simplex S. If F ⊂ X is an arbitrary finite-dimensional closed face, there exist a closed face of the same dimension F0 ⊂ X0 and an affine homeomorphism ϕ : X → X such that ϕ(F ) = F0 . Then ϕ(F ) = F0 ⊂ ext X0 ⊂ ext X. Hence F ⊂ ext X. Since finite-dimensional faces are dense in X by Corollary 12.46, X = ext X. Thus X is affinely homeomorphic to S by Corollary 12.55. To show (c), let K ⊂ ext S be a compact set. Assume that its interior relative to ext S is nonempty, that is, that there exists an open set U ⊂ S intersecting K such that U ∩ ext S ⊂ K. Since ext S is dense in S, U ∩ ext S is dense in U . Thus F := co K is a face in S (see Exercise 6.84) that is proper (any point in ext S \ K belongs to S \ F ). Since F contains U , we obtain a contradiction with Lemma 2.90. If P is a Polish space (a separable completely metrizable topological space), then there exists a metrizable simplex X such that ext X is homeomorphic to P (see Theorem 10.70). By (a), X can be realized as a closed face of S, which proves (d). If ϕ : K1 → K2 is a homeomorphism between compact subsets of ext S, we can extend it to an affine homeomorphism ψ : co K1 → co K2 . Indeed, co Ki , i = 1, 2, are closed faces with closed sets of extreme points, hence they are Bauer simplices (see Lemma 11.11). By Proposition 6.39, co Ki is affinely homeomorphic to M1 (Ki ), i = 1, 2, and the claim follows. Theorem 12.54 thus implies assertion (e). To prove (f), let x1 , x2 ∈ ext S be distinct points. By (d), ext S contains a homeomorphic copy l of [0, 1] joining a pair of points y1 , y2 ∈ ext S. By (e) we can find an affine homeomorphism ϕ : S → S such that ϕ(yi ) = xi , i = 1, 2. Hence ϕ(l) is a simple arc joining x1 and x2 .

12.3.B

A big simplicial space

Construction 12.61. Let H1 be a simplicial function space on a compact space K1 such that H1 = Ac (H1 ). Let K2 be the compact space constructed from K1 as in

466


Definition 6.13, where B = ChH1 (K1 ), Lb = [0, 1] and µb is Lebesgue measure on [0, 1] for each b ∈ B. If t ∈ [0, 1], we write (b, t) for a point in Lb . We consider the space K1 as a subset of K2 . Let p : K2 → K1 be the natural projection, that is, ( y, y ∈ K1 , p(y) = x, y ∈ Lx for some x ∈ B. Let H2 := {f ∈ C(K2 ) : f |K1 ∈ H1 , f (x) = µx (f ) for x ∈ ChH1 (K1 )}. Then H1 can be considered to be a subspace of H2 if we identify f ∈ H1 with f ◦ p ∈ H2 . Lemma 12.62. Let (K1 , H1 ) be a simplicial function space satisfying H1 = Ac (H1 ) and let (H2 , K2 ) be as in Construction 12.61. (a) Then H2 is a function space on the compact space K2 with H2 = Ac (H2 ) and ChH2 (K2 ) = K2 \ K1 . (b) The mapping p : K2 → K1 is admissible and p(ChH2 (K2 )) = ChH1 (K1 ). (c) If µ ∈ M1 (K2 ) is such that p] µ is continuous, then µ(K2 \ K1 ) = 0. (d) If µ ∈ M1 (K1 ) is H1 -maximal and continuous, then it is H2 -maximal. (e) The function space H2 is simplicial. (f) For any x ∈ K1 , the unique H2 -maximal measure H-representing x is continuous. Proof. We start the proof of (a) by noticing that K2 is a (Hausdorff) compact space and H2 is a function space. Indeed, 1 ∈ H2 , the space H2 separates points of K1 and any pair of points in K2 \ K1 not contained in the same Lb . Given x ∈ ChH1 (K1 ), let R1 g ∈ C([0, 1]) be a monotone function satisfying 0 g = 0. Then ( g on Lx , h := 0 otherwise, is in H2 and separates points of Lx . If f ∈ Ac (H2 ) is given, f |K1 ∈ Ac (H1 ) = H1 and f (x) = µx (f ) for any x ∈ ChH1 (K1 ). Thus f ∈ H2 . Obviously, any point of K1 has an H2 -representing measure distinct from the Dirac measure, and thus ChH2 (K2 ) ⊂ K2 \ K1 . On the other hand, it is easy to construct an H2 -exposing function for any point in K2 \ K1 . Hence ChH2 (K2 ) = K2 \ K1 . Since (b) is obvious, we proceed to the proof of (c). Let µ ∈ M1 (K2 ) be such that p] µ is continuous on K1 . Assume that µ(K2 \ K1 ) > 0. Since K2 \ K1 is a


467

disjoint union of open sets Lx , x ∈ ChH1 (K1 ), there exists x ∈ ChH1 (K1 ) such that µ(Lx ) > 0. Then p] µ({x}) = µ(Lx ) > 0, a contradiction with the assumption. To prove (d), let µ ∈ M1 (K1 ) be H1 -maximal and continuous. Let ν ∈ M1 (K2 ) be any measure with µ ≺H2 ν. By (b) and Proposition 12.24(c), µ ≺H1 p] ν. Since µ is H1 -maximal, µ = p] ν. By the continuity of µ and (c), ν is carried by K1 , and hence µ = ν. 1 The next step P∞ measure, P∞is to prove (e). Let µ ∈ M (K1 ) be a discrete H1 -maximal that is, µ = n=1 an εxn , where the numbers an are strictly positive, P∞ n=1 an = 1 and xn are distinct points in ChH1 (K1 ). We claim that ν := n=1 an µxn is the unique H2 -maximal measure with µ − ν ∈ H⊥ . 2 To see this, let ν 0 ∈ M1 (K2 ) be an H2 -maximal measure with µ − ν 0 ∈ H⊥ 2 . Then 0 is H -maximal (see Proposition 12.25). Thus p ν 0 = µ by p] ν 0 − µ ∈ H ⊥ and p ν 1 ] ] 1 Proposition 6.9. From this and from the fact that ν 0 does not charge anyPpoint in K1 (by Construction 12.61 and H2 -maximality of ν 0 ), we obtain that ν 0 = ∞ n=1 an λxn for some probability measures λxn on Lxn . Fix n ∈ N. If g ∈ C([0, 1]) is a continuous R1 function with 0 g = 0, then ( g on Lxn , h := 0 otherwise, is in H2 and ν 0 (h) = µ(h) = 0. Hence λxn is a multiple of µxn . Since they are both probability measures, λxn = µxn . Thus ν 0 = ν. It follows that any point x ∈ K1 has a unique H2 -maximal measure H2 -representing x. Indeed, let δx bePthe unique H1 -maximal measure H1 -representing x. We decompose δx = (δx )c + P∞ n=1 an εxn , where (δx )c is the continuous part of δx , the numbers an are positive, ∞ n=1 an ≤ 1 and the points xn are distinct elements of ChH1 (K1 ) (see Proposition A.78). Then we get that δ := (δx )c +

∞ X

a n µ xn

n=1

is the unique H2 -maximal measure H2 -representing x. To see this, let µ be an H2 -maximal measure with µ − δ ∈ H⊥ 2 . Then p] µ is H1 -maximal and p] µ − δx ∈ H⊥ 1 . Thus (δx )c +

∞ X

an εxn = p] µ = µ|K1 + p] (µK2 \K1 ).

n=1

By (c), p] (µ|K2 \K1 ) is discrete. Hence µ|K1 = (δx )c . Since ∞ X n=1

an µxn − µ|K2 \K1 ∈ H⊥ 2,

468


it follows that

∞ X

an εxn − µ|K2 \K1 ∈ H⊥ 2.

n=1

P∞

Hence µ|K2 \K1 = n=1 an µxn by the preceding argument. Thus µ = δ. Hence H2 is simplicial, which finishes the proof of (e). Finally, given x ∈ K1 , we repeat the argument from the proof of (e) to deduce that the unique H2 -maximal measure H2 -representing x is continuous. This concludes the proof. Construction 12.63. We inductively construct simplicial function spaces (Hα , Kα ) together with admissible mappings pβα : (Kα , Hα ) → (Kβ , Hβ ) and homeomorphic embeddings iβα : Kβ → Kα for all ordinals β ≤ α ≤ ω1 . Let K0 := {0} and H0 := C(K0 ). Suppose that α ∈ (0, ω1 ) and (Kβ , Hβ ) along with the respective mappings have been constructed for all β < α. If α is an isolated ordinal, say α = γ + 1, we use Construction 12.61 for the space (Kγ , Hγ ) to get (Kα , Hα ). Let pγα be the mapping p : Kα → Kγ from Construction 12.61 and iγα : Kγ → Kα be the identity mapping. Further we set pβα := pβγ ◦ pγα

and

iβα := iγα ◦ iβγ ,

β < α,

and pαα be the identity. If α is a limit ordinal, let (Kα , G α ) be the inverse limit of ((Kβ , Hβ ), pβγ )β,γ kj such that 0 yj+1 ∈ span{en,p : kj ≤ n < kj+1 } 0 and kyj+1 − ym k < ηj+1 .

475


After finishing this construction, we get the required sequence by normalizing the vectors yj0 , j ∈ N, and denoting them again as {yj }. Step 2. Assume that kyj k1 → 0. Then kyj k2 → 1, without loss of generality kyj k2 > 2−1 for all j ∈ N. For each j ∈ N we find σ j ∈ Σ such that gσj (yj ) > 2−1 .

(12.33)

We may assume that each σ j satisfies σ j (n) ≤ n + 1,

n ∈ N,

(12.34)

(otherwise we would replace σ j by τ j (n) := σ j (n) ∧ (n + 1), n ∈ N). Since kyj k1 → 0, for a fixed index r ∈ N we have lim

j→∞

∞ X r X

|e∗n,p (yj )| ≤ lim r j→∞

n=r p=1

∞ X

sup |e∗n,p (yj )| ≤ lim rkyj k1 = 0. j→∞

n=r 1≤p≤r

Hence we may achieve by choosing another subsequence (still denoted as {yj }) that kj ∞ X X n=kj p=1

1 |e∗n,p (yj 0 )| < , 4

j ∈ N, j 0 ≥ j.

(12.35)

We define a sequence σ ∈ Σ as follows: σ(n) := σ j (n) ∨ kj ,

where j is such that kj ≤ n < kj+1 .

By (12.34), σ is an increasing sequence. Using Step 1, for any j ∈ N we obtain kj+1 −1 kj −1

gσ (yj ) = gσj (yj ) −

X

X

e∗n,p (yj ).

n=kj p=σ j (n)

By combining (12.33) and (12.35), we obtain 1 gσ (yj ) ≥ , 4

j ∈ N.

Since gσ ∈ E ∗ , this is a contradiction with the requirement that {yj } weakly converge to 0. Step 3. Assume that there exists c > 0 such that kyj k1 > c for infinitely many indices j. We will prove that {yj } contains a subsequence that is equivalent to `1 basis (see Definition 10.38), which will yield a contradiction due to Theorem A.9.

476


Without loss of generality, let kyj k1 > c for all j ∈ N. Let c1 , . . . , cm be real numbers. Since e∗n,p

m X

cj yj = e∗n,p ci yi ,

ki ≤ n < ki+1 ,

j=1

we obtain m m ∞ m X X

X

X

cj yj ≥ cj yj 1 = sup e∗n,p cj yj j=1

n=1 1≤p≤n

j=1

=

m kj+1 X X−1

m X

kj+1 −1

|cj |

j=1

≥c

sup |e∗n,p (cj yj )|

1≤p≤n

j=1 n=kj

=

j=1

m X

X

sup |e∗n,p (yj )|

n=kj 1≤p≤n

|cj |.

j=1

This means that {yj } is equivalent to `1 -basis, which concludes the proof. Theorem 12.77. There exists a metrizable compact convex set X and a strongly affine function f ∈ B 2 (X) such that S • f ∈ / α p(z) − ε on V and a constant α > 0 such that p(z) , y ∈ ∂U \ V. lim inf s(x) > x→y, x∈U α ∗

Let u ∈ H + (Y ) satisfy u ≥ p on U c . If y ∈ ∂U ∩ V , then lim inf (αs(x) + u(x)) ≥ 0 + u(y) ≥ p(y) ≥ p(z) − ε,

x→y, x∈U

whereas for y ∈ ∂U \ V we have lim inf (αs(x) + u(x)) ≥ p(z) + 0.

x→y, x∈U

Then by the minimum principle (Theorem A.160), u + αs ≥ p(z) − ε, and passing to infimum over u (and using Theorem A.195) we obtain bpU c (xn ) = lim inf RpU c (xn ) ≥ lim inf(p(z) − αs(xn ) − ε) = p(z) − ε. lim inf R n→∞

n→∞

n→∞

Using the density of P − P in C(U ) (Proposition A.169) we obtain that {xn } is regular.

13.2 Boundary behavior of solutions

497

Remark 13.15. The converse to Lemma 13.14 is also true, see Exercise 13.132. Proposition 13.16. Let U ⊂ Y be a relatively compact open set. Then z ∈ ∂U is regular if and only if there is a barrier at z. Proof. Let s be a barrier for z. Then s is a barrier for each sequence {xn } of points of U converging to z. By Lemma 13.14, each such a sequence {xn } is regular. Hence z is regular. Conversely, let z ∈ ∂U be a regular point. We pick a potential p ∈ P such that V p < p on V for each regular set V . (Recall that pV denotes the Poisson modification. For the existence of such a potential, see Proposition A.206.) Let {Vn } be a sequence of regular sets such that z ∈ / Vn and U \ {z} ⊂

∞ [

Vn .

n=1

If we denote qn := pVn , we obtain that qn ∈ P, qn ≤ p, qn (z) = p(z) and qn < p on Vn . Set ∞ X bU c ). s := 2−n (p − R qn n=1 Uc

bq is positive and superharmonic on U . Since p(z) = qn (z), Then each function p − R n p, qn are continuous and z is regular, we have lim

x→z, x∈U

bqU c (x)) = 0, (p(x) − R n

lim

x→z, x∈U

s(x) = 0.

The passage to the sum is obvious because of uniform convergence. If y ∈ ∂U \ {z}, we find n ∈ N such that y ∈ Vn and obtain bU c (x)) ≥ 2−n (p(y) − qn (y)) > 0. lim inf s(x) ≥ 2−n lim inf (p(x) − R qn

x→y, x∈U

x→y, x∈U

Remark 13.17. If a function s ∈ H(U ) exposes z, then s|U is obviously a barrier at z. It is considerably more difficult to construct H(U )-exposing functions than barriers. For their existence, we use rather a general tool, see Proposition 13.103. The concept of barrier is mostly considered in a weaker form, see for example H. Bauer [40], §4.3.

13.2.A

Regular points for the Laplace equation

In this subsection, we deal with potential theory of the Laplace equation. We consider a bounded open set U in Rd . We already know that balls are regular sets; this we

498

13 Function spaces in potential theory and the Dirichlet problem

needed to verify that the regularity axiom of Bauer’s axiomatic theory holds for the Laplace equation. The famous Wiener criterion gives a necessary and sufficient condition for the regularity of a point z ∈ ∂U , namely, if d ≥ 3, z is regular if and only if ∞ X

αn cap(An \ U ) = ∞.

n=1

Here α > 1, An := {x ∈ Rd : αn ≤ |x − z|2−d ≤ αn+1 },

n ∈ N,

and cap denotes the Newtonian capacity (see [21], Theorem 7.7.2). A similar condition holds also for d = 2, see [21], Section 7.7. It follows that the problem of regularity reduces to the task how to measure the capacity of a set. Since the notion of capacity is not too easy to handle (and we do not plan to present it here), let us list also some geometric criteria giving conditions for regularity. These are less sharp but more transparent. Examples 13.18. (a) If z ∈ ∂U and there is a closed ball B ⊂ U c such that z ∈ ∂B, then z ∈ ∂ reg U ([21], Theorem 6.6.12). (b) If d = 2, z ∈ ∂U and z is an endpoint of a line-segment contained in U c , then z ∈ ∂ reg U ([21], Theorem 6.6.15(i)). (c) Cone criterion. If d ≥ 3, z ∈ ∂U , C is a closed cone with vertex z, P is a (d − 1)dimensional hyperplane containing the axis of C and C ∩ P ∩ B(z, r) ⊂ U c for some r > 0, then z ∈ ∂ reg U ([21], Theorem 6.6.15(ii)). (d) Punctured ball. This is the simplest example of an irregular set. If U := U (0, 1) \ {0}, f = 0 on S(0, 1) and f (0) = 1, then there exists no h ∈ H(U ) such that h|∂U = f . Indeed, according to the minimum principle (Theorem A.160) every such function h would satisfy h ≤ εN ,

ε>0

(N is the fundamental harmonic function, see Definition A.137), and passing to infimum we would obtain that h ≤ 0 in U . (e) Lebesgue spine. For the so-called Lebesgue spine q n o − 1 L := {0} ∪ (x1 , x2 , x3 ) ∈ R3 : x1 > 0, x22 + x23 ≤ e x1 , the origin is not a regular set of U := U (0, 1) \ L. Notice that both U and U c are connected and that ∂ reg U is not closed. Details can be found in [21], pp. 186–187.


499

Theorem 13.19. The set of all regular points for U is a Gδ set and the set of all irregular points is polar. Proof. The set of all regular points is ∂U ∩ b(U c ) and thus it is polar by the Kellogg property (Theorem A.215) and Gδ by Corollary A.207. See also [21], Corollary 6.8.4 and Theorem 6.6.8. Lemma 13.20. There exists a sequence {Bj }∞ j=0 of pairwise disjoint closed balls in d R such that all balls Bj are included in U (0, 31 ), ∞ [

Bj \

j=0

∞ [

Bj = {0}

(13.4)

j=0

and 0 is an irregular point of U := U (0,

1 3)

\

∞ [

Bj .

(13.5)

j=0

Proof. The property (13.4) is satisfied by a collection of pairwise disjoint balls if the sequence of centers converges to the origin. Now, we may assume that Y = U (0, 23 ). We denote V := U (0, 13 ), p(x) :=

4 9

− |x|2 ,

x ∈ Y,

q := p ∧ 31 . Then p, q ∈ P, p = q on V c and p(0) = 49 = q(0) + 91 . Consider a sequence {zk } of points of V such that zk → 0, and a sequence {ck } of strictly positive real numbers such that v(0)
0 are chosen small enough to guarantee that B(zj , rj ) ⊂ V , B(zj , rj ) are mutually disjoint, and v ≥ p on B(zj , rj ) (this is possible to achieve as v(y) ≥ ck N (y −zk )). Let U be defined by (13.5). The function c u := v + q is hyperharmonic on Y and majorizes p on Y \ U , so that RpU ≤ u on U . We have Z bpU c (0) ≤ lim inf u(y) ≤ lim inf − R u(x) dx ≤ u(0) < p(0), y→0

r→0+

and thus 0 is not a regular point for U .

B(0,r)

500


Proposition 13.21. There exists a bounded open set U ⊂ Rd (d ≥ 2) for which the set ∂ reg U of regular points is not an Fσ set. d Proof. Let {Bj }∞ j=0 be a sequence of pairwise disjoint closed balls in R such that all 1 balls Bj are included in U (0, 3 ), ∞ [

∞ [

Bj \

j=0

Bj = {0}

j=0

S ∗ ∞ and 0 is an irregular point of U (0, 31 ) \ ∞ j=0 Bj (Lemma 13.20). Further, let {Bj }j=0 be a sequence of pairwise disjoint closed balls in B0 such that ∞ [

Bj∗ \

j=1

Set F :=

∞ [

∞ [

Bj∗ = ∂B0∗ .

j=1

Bj ∪

j=1

∞ [

Bj∗ ∪ {0},

j=0

U := {Bj : j ≥ 1} ∪ {Bj∗ : j ≥ 1}. Then F is a compact set. By induction we will construct a sequence {Ki } of compact sets, a sequence {Hi } of closed balls, a sequence {zi } of points and a sequence {V i } of families of closed balls. We set K1 = F , H1 = B0∗ , z1 = 0 and V 1 = U. Let us suppose that these objects have been constructed for i < k. We select one of the largest balls in U k−1 ; this will be Ck . Consider a similarity Φk : Rd → Rd which maps B(0, 13 ) onto Ck . By setting Kk := (Kk−1 \ Ck ) ∪ Φk (F ), Hk := Φk (B0∗ ), zk := Φk ({0}), V k := (V k−1 \ {Ck }) ∪ {Φk (B) : B ∈ U} we finish the construction. We observe that {Kk } forms a decreasing family of compact sets and its intersection denoted by K is then a compact set. Further, given a number r > 0, each family V k contains only a finite number of balls of radius exceeding r. Thus, each ball which once occurs in V k will be in one of the next steps replaced by an isometric copy of F . However, the balls Hk do not fall into the families V k and remain in Fk for ever. We set U := U (0, 31 ) \ K. The points zk are all irregular for U , whereas all boundary points of the balls Hk are regular for U by the cone criterion (Example 13.18(c)). We observe that both regular points and irregular points are dense in ∂K and thus ∂reg U cannot be Fσ .


501

Now, we will study “negligibility” of irregular points. To this end, we need some preliminaries concerning polar sets. Lemma 13.22 (Evans). Let K ⊂ Rd be a compact polar set. Then there exists a Radon measure µ on K such that N µ = ∞ on K. Proof. Using a simple scaling argument we may assume that K is contained in an open ball V of diameter less than 1. This assumption is made in order to avoid problems with negative values of the logarithmic kernel in the two-dimensional case: now we can be sure that there exists α > 0 such that N (x − y) ≥ α for x, y ∈ V . Since K is polar, by Proposition A.214 there exists a Radon measure ν carried by V such that N ν = ∞ on K. With each z ∈ V \ K we associate the ball Uz := U z, 12 dist(z, K) . Choose zk ∈ V \ K such that {V ∩ Uzk : k ∈ N} is a locally finite covering of V \ K, and yk ∈ K such that S |zk − yk | = dist(zk , K). Find pairwise disjoint Borel sets Ek ⊂ V ∩ Uzk such that ∞ k=1 Ek = V \ K. If z ∈ Ek and x ∈ K, then |zk − yk | ≤ |zk − x| ≤ |zk − z| + |z − x| ≤ 21 |zk − yk | + |z − x|, and thus |zk − yk | ≤ 2|z − x|. We obtain |x − yk | ≤ |x − z| + |z − zk | + |zk − yk | ≤ |x − z| + 32 |zk − yk | ≤ 4|x − z|, and thus, for a suitable constant C ≥ 1, N (x − z) ≤ CN (x − yk ),

x ∈ K, z ∈ Ek .

(13.6)

Indeed, (13.6) is clear for d ≥ 3, in the two-dimensional case we estimate log 4 + 1 N (x − yk ). N (x − z) ≤ log 4 + N (x − yk ) ≤ α In addition, we can choose the constant C so that Z N (x − z) dν(z) ≤ C,

x ∈ K.

Vc

We define the measure µ as µ := ν|K +

∞ X

ν(Ek )εyk .

k=1

Then spt µ ⊂ K and µ(K) = ν(K) +

∞ X k=1

ν(Ek ) = ν(K) + ν(U \ K) = ν(U ) < ∞,

(13.7)

502


so that µ is a Radon measure. Let x ∈ K. Then by (13.6) and (13.7), Z ∞= N (x − z) dν(z) Rd

Z

Z

N (x − z) dν(z) +

N (x − z) dν(z) +

= Vc

K

k=1

Z ≤C +C

N (x − z) dν(z) + C K

∞ Z X

∞ X

N (x − z) dν(z)

Ek

ν(Ek )N (x − yk )

k=1

Z N (x − z) dµ(z) = C(1 + N µ(x)).

=C +C K

This concludes the proof. Theorem 13.23. There exists a positive harmonic function k on U such that limx→z k(x) = ∞ whenever z ∈ ∂ irr U . Proof. Using a simple scaling argument we may suppose that U is contained in an open ball V of diameter less than 1. By Theorem S∞ 13.19, there exists a sequence {Kn } of compact polar sets such that ∂ irr U = n=1 Kn . By the Evans lemma 13.22, there exist Radon measures µn on Kn such that N µn = ∞ on Kn and, since diamP U is small, N µn ≥ 0 on U . We may suppose that µn (Rd ) ≤ 2−n . Define µ := ∞ n=1 µn . Then µ is a Radon measure carried by ∂U . Putting k := N µ on U , we have k ∈ H + (U ). If z ∈ ∂ irr U , then z ∈ Kn for a suitable n ∈ N. Since N µn is lower semicontinuous, N µn (x) → ∞ as x → z. Hence k(x) → ∞ as x → z. Theorem 13.24. Let u be a lower bounded hyperharmonic function on U . If lim inf u(x) ≥ 0 x→z

whenever

z ∈ ∂ reg U,

then u ≥ 0 on U . In particular, if h is a bounded harmonic function on U for which limx→z h(x) = 0 for each z ∈ ∂ reg U , then h = 0 on U . Proof. Let ε > 0 and let k be the function from Theorem 13.23. Then lim

x→z, x∈U

(u + εk)(x) ≥ 0

whenever z ∈ ∂U . It follows from Theorem A.160 that u + εk ≥ 0 on U , hence u ≥ 0 on U .


13.2.B

503

Regular points for the heat equation

Similarly to the case of the Laplace equation, a necessary and sufficient condition for regularity analogous to the Wiener criterion is known for the heat equation. We will discuss it briefly in notes to this chapter. In Examples 13.18 we listed some criteria of regularity for the Laplace equation. Now, we are going to mention some particular criteria for the heat equation. We will tacitly use the barrier criterion of Proposition 13.16. Examples 13.25. (a) Let V ⊂ Rd be a bounded open set, T > 0 and U := V ×(0, T ). Suppose that ξ ∈ ∂V and τ ∈ (0, T ). We will show that (ξ, τ ) is regular for the heat equation if and only if ξ is regular for the Laplace equation. Let Y := U (0, R) be a ball in Rd containing V . We consider the harmonic structure of the Laplace equation on Y and the harmonic structure of the heat equation on Y ×R. bV c . Then p(x) = R2 − |x|2 is a potential on Y (see Example A.180). Set q := R p p Denote also ψ(t) := 1 + |t − τ |2 − 1, t ∈ R. Suppose first that ξ is regular for the Laplace equation. Then q is continuous at ξ, q ≤ p and q(ξ) = p(ξ). Set s(x, t) = p(x) − q(x) + 12 |x − ξ|2 + ψ(t),

(x, t) ∈ U .

(13.8)

We see that ∆s = −d in U and ∂s ∂t ≥ −1. Thus s|U is a barrier at (ξ, τ ) and (ξ, τ ) is therefore regular for U . (If V is regular for the Laplace equation, then s is S(U )exposing at (ξ, τ ).) If ξ is irregular for the Laplace equation, then q(ξ) < p(ξ) (otherwise p(x)−q(x)+ ∗ ∗ |x−ξ|2 would be a barrier for V ). Consider the operator T : H + (Y ) → H + (Y ×R) defined as T u(x, t) = u(x)et . Then T p is a heat potential on Y ×R (for the reasoning ∗ compare with Subsection A.8.E and Example A.180). Let u ∈ H + (Y ), u ≥ p bU c ≤ T q. Since on Y \ V . Then T u ≥ T p on (Y × R) \ U . It follows that R Tp T q(ξ, τ ) < T p(ξ, τ ), we see that (ξ, τ ) is not regular for the heat equation. (b) Let U be as above. Then each point (ξ, 0) ∈ V × {0} is a regular boundary point for the heat equation on U . The function u : (x, t) 7→

|x − ξ|2 + t, 2d

(x, t) ∈ U ,

is H(U )-exposing. (c) On the other hand, it is not difficult to show that no point of V × {T } is regular. Indeed, from the Harnack type inequality (Proposition A.151) we easily infer that there cannot exist a barrier at such a boundary point. (d) Let V the unit ball in Rd . There exists a strictly positive continuous function η which attains a strict maximum at 0 such that the point (0, η(0)) is not heat-regular

504


for the set {(x, t) ∈ Rd+1 : x ∈ V, 0 < t < η(x)}, see E. G. Effros and J. L. Kazdan [165]. (e) Let V , U be as in (a), W = V × (−1, 0)∪(0, 1) and ξ ∈ V . A similar reasoning as in (c) shows that (ξ, 0) is not regular. On the other hand, if u is as in (b) for t > 0 and u = 1 for t < 0, then u is a barrier for each sequence {(xn , tn )} of points of U converging to (ξ, 0). Hence such sequences converging to (ξ, 0) “from above” are regular.

13.3

Function spaces and cones in potential theory

Throughout this section, U will be a relatively compact subset of a Bauer harmonic space Y . For typical applications, U is open or compact. Definition 13.26. If U is open, recall that H(U ) denotes the function space of all functions continuous on U which are harmonic on U . Define also S(U ) as the convex cone of all continuous functions on U which are superharmonic on U . Definition 13.27. Now, we drop the assumption of openness. Suppose that U is as declared at the beginning of the section and A ⊃ U . We say that h : A → R is harmonically extendable on U if there exist an open set W ⊃ U and a harmonic function h˜ on W such that h = h˜ on U . Similarly we define superharmonic extendability. We define H0 (U ) as the function space of all functions continuous on U which are harmonically extendable on U ; also we define S0 (U ) as the convex cone of all continuous functions on U which are superharmonically extendable on U . Now (U is still not necessarily open), we define H(U ) := H0 (U ), S(U ) := S0 (U ). For U open, this gives obviously the same class of functions as in Definition 13.26. Lemma 13.28. We have H0 (U ) =

[

S0 (U ) =

[

{H(V )|U : V open, V ⊃ U }

and {S(V )|U : V open, V ⊃ U }.

Proof. The inequality “⊃” being obvious, let h ∈ H0 (U ). We find an open set W ⊃ U and a harmonic function h˜ on W such that h = h˜ on U . We use Tietze’s theorem to construct a continuous extension f ∈ C(Y ) of h, this means that h = f |U . We find positive functions u, w ∈ C(Y ) such that U = {x ∈ Y : u(x) = 0} and

W = {x ∈ Y : w(x) > 0}.

Let ˜ V := {x ∈ W : u(x) + |h(x) − f (x)| < w(x)}.

13.3 Function spaces and cones in potential theory

505

We observe that U ⊂ V and V ⊂ {x ∈ Y : u(x)≤w(x)} ⊂ {x ∈ Y : u(x)=0} ∪ {x ∈ Y : w(x)>0} = U ∪ W. Set

( h˜ g := h

on V ∩ W, on V \ W.

Since g = h˜ on V , the function g is obviously harmonic on V . Now we want to prove continuity on V . Since both functions h˜ and h are continuous and coincide on U , we have h˜ = h on W ∩ U . Thus g = h on U = (U ∩ W ) ∪ (U \ W ). The equality g = h˜ on V ∩ W shows the continuity of g at all points of V ∩ W and the continuity of h shows the continuity of g at all points of V \ W . Let z ∈ V ∩ ∂W . Then z ∈ U , lim g(x) = lim h(x) = h(z) = g(z), x→z, x∈U

x→z, x∈U

and |g(z) −

lim

x→z, x∈W

g(x)| = |f (z) −

lim

˜ h(x)| ≤

x→z, x∈W ∩V

lim sup

˜ |h(x) − f (x)|

x→z, x∈W ∩V

≤ lim w(x) = 0. x→z

This shows the continuity of g at z. The assertion concerning S0 (U ) is proved similarly. Remark 13.29. In literature, if U is finely open, S(U ) is defined as the function space of all functions continuous on U which are finely superharmonic on U . These two definitions are equivalent, see Exercise 13.155. Similar remark applies to the function space H(U ). Proposition 13.30. Suppose that U is open. Then Ac (S(U )) = Ac (H(U )) = H(U ). Proof. Obviously, H(U ) ⊂ Ac (H(U )) ⊂ Ac (S(U )). If h ∈ Ac (S(U )), then h ∈ C(U ). Let V ⊂ U be a regular open set with V ⊂ U and µVx be the harmonic measure, x ∈ V . Then µVx ∈ Mx (S(U )) and thus H V h = h on V . It follows that h is harmonic on V , and V being arbitrary, h is harmonic on U , so that h ∈ H(U ).

13.3.A

Function spaces and cones: Laplace equation

From now, we need to consider the case of Laplace equation and the general case separately. The reason is that our exposition for the Laplace equation is more selfcontained. Throughout this subsection, we will deal with the Laplace equation and a

506


bounded coanalytic subset U of Rd . This assumption makes it possible to approximate balayage on U c by balayage on compact subsets of U c (cf. Theorem A.197). Our main objective is to characterize the Choquet boundary of H(U ) and S(U ) and to show that both H(U ) and S(U ) are simplicial. We will also characterize the maximal measures. In order to realize this project, we need to evoke some deeper facts from potential theory. This is done in the course of the proof of the next lemma. Lemma 13.31. Let p ∈ P. Then there exists a sequence {pn } of potentials from P bpU c . In particular, pn % p on ∂ reg U . such that pn |U ∈ H0 (U ) and pn % R bU c . Proof. Denote u = R p Step 1. First, we suppose that U is open. By Theorem A.179, there exists a measure µ ∈ M(Y ) such that u is the Green potential Gµ. By Theorems A.181 and A.196, the measure µ is carried by U c . We use the Lusin theorem to find a sequence {Kn } of compact subsets of U c such that ∞ [ µ Y \ Kn = 0 n=1

and each (Gµ)|Kn is continuous. We denote µn := µ|Kn . Since µ − µn is a positive measure and (Gµ)|Kn = (Gµn )|Kn + (G(µ − µn ))|Kn , a continuous function is written as a sum of two lower semicontinuous functions, so we see that also (Gµn )|Kn is continuous. By the Evans–Vasilesco continuity principle (Theorem A.184), the functions pn := Gµn are continuous potentials. By the Lebesgue monotone convergence theorem, pn % p. Using Theorem A.181 again, we observe that pn are harmonic on U . This completes the proof for U open. Step 2. Now, we assume that U is coanalytic. Then by Theorem A.197 there exists an increasing sequence {Lk } of compact subsets of U c such that bpLk . u = sup R k∈N

bLk = Gνk . Applying There exist measures νk such that each νk is carried by Lk and R p ∞ the argument used in Step 1 to Lk , we find a sequence {Lk,m }m=1 of compact subsets of Lk such that c

bpLk = sup Gνk,m , R m∈N

where νk,m = νk |Lk,m ,

and the potentials Gνk,m are continuous. Now, we select k1 and m1 , for example k1 = m1 = 1, and set p1,1 = 12 Gνk1 ,m1 . Then p1,1 < 32 u and thus (by the continuity of Gνk1 ,m1 and the compactness of L1 ) we can find k2 > k1 and m2 > m1 such that 23 Gνk2 ,m2 > p1,1 on L1 and set


507

pi,2 = 23 Gνki ,m2 , i = 1, 2. Since νk1 ,m1 has support in L1 , by the comparison principle (Proposition A.182) we have p2,2 ≥ p1,1 on Y . At the n-th step we find kn > kn−1 and mn > mn−1 such that n n+1 Gνkn ,mn

≥ p1,n−1 ∨ · · · ∨ pn−1,n−1

(13.9)

on Ln−1 and observe by Proposition A.182 that (13.9) holds in fact on Y . We set n Gνki ,mn , i = 1, . . . , n. The final sequence will be {pn } with pn := pn,n , pi,n = n+1 n ∈ N. This sequence is obviously increasing. Since for each i ∈ N we have bpLki , u ≥ sup pn ≥ sup pi,n = R n∈N

n∈N

it follows that supn∈N pn = u. bU c (x) and thus pn (x) % u(x) = Now, if x ∈ ∂ reg U = U ∩ b(U c ), then p(x) = R p p(x). Lemma 13.32. We have ChH(U ) (U ) ⊂ ChS(U ) (U ) ⊂ ∂ reg U. Proof. Since H(U ) ⊂ S(U ) ∩ −S(U ), we have ChH(U ) (U ) ⊂ ChS(U ) (U ). / b(U c ). We find p ∈ P such that Let z ∈ ∂ irr U , so that by Theorem 13.5, z ∈ c bpU (z) < p(z). R Let s ∈ −S0 (U ), s < p. Then using Lemma 13.28 we find a relatively compact open set W ⊂ Y containing U and t ∈ −S(W ) such that t ≤ p on W and t = s on U . If u ∈ S + (Y ), u ≥ p on U c , then u ≥ t on ∂W and by the minimum principle c (Theorem A.160) we have u ≥ t on W . It follows that RpU ≥ t on W and passing to bpU c ≥ t on W . the lower semicontinuous regularization we observe that R Hence bU c (z) < p(z), sup{s(z) : s ∈ −S(U ), s ≤ p} ≤ R p and by Theorem 3.24, z ∈ / ChS(U ) (U ). Theorem 13.33. The convex cone S(U ) is simplicial. For any z ∈ U , the measure c εzU is the unique maximal S(U )-representing measure for z. In particular, ChS(U ) (U ) = ∂ reg U.

(13.10)

Proof. Given z ∈ U , let µ ∈ Mz (S(U )) be a maximal measure. By Theorem 7.26, µ is carried by the Choquet boundary ChS(U ) (U ). Let p ∈ P. Then by Lemma 13.31 bpU c and each there exists a sequence {pn } of potentials from P such that pn % R pn |U belongs to H(U ). From Lemma 13.32 we infer that pn → p on U ∩ b(U c ) ⊃ ChS(U ) (U ). We obtain c bpU c (z) = εU µ(p) = µ(sup pn ) = sup µ(pn ) = sup pn (z) = R z (p).

n∈N

n∈N

n∈N

508

13 Function spaces in potential theory and the Dirichlet problem c

By the density of P − P in C(U ) (Proposition A.169) we obtain that µ = εU z . Thus, c U S(U ) is simplicial and εz is the unique representing measure for z. Now, if z ∈ ∂ reg U and µ ∈ Mx (S(U )) is a maximal representing measure, then c µ = εU z = εz , thus z ∈ ChS(U ) (U ). Hence ∂ reg U ⊂ ChS(U ) (U ), which together with Lemma 13.32 proves (13.10). Theorem 13.34. We have Ac (H(U )) = H(U ) = S(U ) ∩ −S(U ). Proof. Obviously H(U ) ⊂ S(U )∩−S(U ). Suppose that h ∈ S(U )∩−S(U ). Choose ε > 0. Then there exist an open set W ⊃ U and functions s ∈ S(W ), t ∈ −S(W ), such that |s − h| < ε and |t − h| < ε on U . Let V = {x ∈ W : |s(x) − t(x)| < 2ε}. Then V is an open set containing U and |s − t| ≤ ε on V . Using the simpliciality of S(V ) and Theorem 7.31 we find f ∈ Ac (S(V )) such that t − ε ≤ f ≤ s + ε on V . By Proposition 13.30, Ac (S(V )) = H(V ), and hence f ∈ H0 (U ). It follows that h ∈ H(U ). The equality Ac (H(U )) = H(U ) follows from Bauer’s theorem 3.27. Theorem 13.35. The function space H(U ) is simplicial. For any z ∈ U , the measure c εzU is the unique maximal H(U )-representing measure for z. In particular, ChH(U ) (U ) = ∂ reg U. Proof. Once we know that H(U ) = Ac (S(U )) (Theorem 13.34), this follows from analogous assertions for the cone S(U ) (Theorem 13.33) by Theorem 7.38. (Alternatively, we can proceed as in the proof of Theorem 13.33.) Remark 13.36. By Theorems 13.35 and 13.33, the maximal representing measures for H(U ) and S(U ) coincide. This is no longer true for representing measures that are not maximal. For x ∈ U , the inclusion Mx (S(U )) ⊂ Mx (H(U )) is obvious. We are going to show that Mx (H(U )) \ Mx (S(U )) 6= ∅, if U is a ball and x is its center. We will consider the case d ≥ 3 only; for d = 2, we can produce a similar example using the logarithmic kernel. Let U := U (0, 1) and y = ( 12 , 0, . . . , 0). Then Z Z h(0) = − h dx ≥ 2−d − h dy = 2−d h(y) U

U (y, 21 )

for every h ∈ H+ (U ) (this is a weak version of the Harnack inequality, for a version with the sharp constant see [21], Theorem 1.4.1). Define s(x) := |x − y|2−d ∧ 22d ,

x ∈ U.


509

Then s ∈ S(U ), s(y) = 22d and s(0) = 2d−2 . According to Key lemma 3.21, there is a measure µ ∈ M0 (H(U )) such that µ(s) = s∗ (0) = inf {h(0) : h ∈ H(U ), h ≥ s} . We note that s∗ (0) ≥ 2−d s(y) = 2d > s(0), from which it follows that µ(s) = s∗ (0) > s(0). Hence µ ∈ / M0 (S(U )). Theorem 13.37. Suppose that f ∈ C(U ) and Hf = f on U . Then (a) f is H(U )-affine, (b) f ∈ H(U ). c

˜ ˜ (x) := εU Proof. Set Hf x (f ), x ∈ U . Then by assumptions, Hf = f in U . If / b(U c ) and thus x ∈ b(U ). This means that x is in the fine x ∈ ∂ irr U , then x ∈ closure of U and by the fine continuity of both functions (Proposition A.209) we see ˜ (x) = f (x). If x ∈ ∂ reg U , then simply Hf ˜ (x) = εx (f ) = f (x). It follows that Hf c U ˜ that Hf = f in U . Since the measures εx are the maximal H(U )-representing measures for x ∈ U , we obtain assertion (a) from Corollary 6.12. Then, (b) follows from Theorem 13.34. Proposition 13.38. Assume that U is open. If f ∈ C(∂U ), then Hf = sup{t|U : t ∈ −S(U ), t|∂U ≤ f }. Proof. Fix f ∈ C(∂U ) and extend f to U by setting f := ∞ on U . The set {−t|U : t ∈ −S(U ), t ≤ f } is a saturated family, and hence by Theorem A.165 g := sup{t ∈ −S(U ) : t ≤ f } is harmonic on U . Now f (z) = sup{h(z) : h ∈ H(U ), h ≤ f },

z ∈ ChH(U ) (U ),

by Theorem 3.24(v), and hence f = g on ChH(U ) (U ). Let u := g − Hf . Since ChH(U ) (U ) = ∂ reg U by Theorem 13.35 and g is lower semicontinuous, we have lim inf u(x) ≥ g(z) − f (z) = 0, x→z

z ∈ ∂ reg U,

and from Theorem 13.24 we infer that g ≥ Hf on U . The converse inequality is obvious.

510

13.3.B


Function spaces and cones in parabolic potential theory and harmonic spaces

In this subsection, Y will be again a general Bauer harmonic space and U be its relatively compact coanalytic subset. In this generality, the Choquet theory of H(U ) leads to the concept of essential solution. Our exposition is based on deep theorems by J. Bliedtner and W. Hansen which we took over without proofs. (See Theorem A.208 and the part in Appendix concerning essential base and essential balayage.) Lemma 13.39. Let p ∈ P. Then c bpβ(U c ) = sup{q ∈ P : q ≤ p, q| ∈ H0 (U )}. BpU = R U

In addition, there exists a sequence {pn } of potentials from P such that pn |U ∈ c H0 (U ) and pn % BpU . In particular, pn % p on ∂ess U . Proof. Recall that by Proposition 13.10, ∂ess U = U ∩ β(U c ). By Theorem A.227 c

BpU = sup{BpK : K compact, K ⊂ U c }. Let K ⊂ U c be compact. Then b(β(K)) = β(K). By Theorem A.208 and Proposition A.226, BpK = Rpβ(K) = sup{q ∈ P : q ≤ p, Rqβ(K) = q} = sup{q ∈ P : q ≤ p, BqK = q}. β(K)

If q ∈ P and q = BqK = Rq , then q is harmonic on K c ⊃ U (Theorem A.196) and thus c c BpU ≤ sup{q ∈ P : q ≤ BpU , q|U ∈ H0 (U )}. It remains to prove that the supremum can be achieved by an increasing sequence. For this we need that the family c

Q = {q ∈ P : q ≤ BpU , q|U ∈ H0 (U )} is up-directed and that there is a countable subset of Q with the same supremum. The latter fact follows from Lemma A.55, we will prove the former one. Let s, t ∈ Q, we want to show that there is q ∈ Q such that s ∨ t ≤ q. There exist open sets V, W ⊃ U such that s is harmonic on V and t is harmonic on W . Then g := s ∨ t is subharmonic on V ∩ W and the rest follows from Lemma A.171. Lemma 13.40. We have ChH(U ) (U ) ⊂ ChS(U ) (U ) ⊂ ∂ess U.


511

Proof. Since H(U ) ⊂ S(U ) ∩ −S(U ), we have ChH(U ) (U ) ⊂ ChS(U ) (U ). Hence we need to show that ChS(U ) (U ) ⊂ ∂ess U . Let z ∈ U \ ∂ess U and p ∈ P be such that c BpU (z) < p(z). Let s ∈ −S0 (U ) satisfy s < p. Then using the definition of S0 (U ) and Tietze’s theorem we find g ∈ C(Y ) and an open set W ⊂ Y such that g ≤ p, g = s on U and s is subharmonic on W . By Lemma A.171, there exists q ∈ P such that q|W ∈ H(W ) and g ≤ q ≤ p. By Lemma 13.39, c

g(z) ≤ q(z) ≤ BpU (z) < p(z). Hence sup{s(z) : s ∈ −S(U ), s ≤ p} < p(z) and z ∈ / ChS(U ) (U ) by Theorem 3.24. Theorem 13.41. The function space H(U ) and the convex cone S(U ) are simplicial β(U c ) is the unique and H(U ) = S(U ) ∩ −S(U ). For any z ∈ U , the measure εz H(U )-maximal H(U )-representing measure and the unique S(U )-maximal S(U )representing measure for z. In particular, ChH(U ) (U ) = ChS(U ) (U ) = ∂ess U.

(13.11)

Proof. The proof is almost the same as that of Theorems 13.33, 13.34 and and 13.35. Given z ∈ U , let µ ∈ Mz (S(U )) be a maximal measure. By Theorem 7.26, µ is carried by the Choquet boundary ChS(U ) (U ). Let p ∈ P. Then by Lemma 13.39 c there exists a sequence {pn } of potentials from P such that pn % BpU and each pn |U belongs to H(U ). Using Lemma 13.40 we see that pn → p on U ∩ β(U c ) ⊃ ChS(U ) (U ). We obtain c

c

µ(p) = µ(sup pn ) = sup µ(pn ) = sup pn (z) = BpU (z) = εzβ(U ) (p). n∈N

n∈N

n∈N

β(U c )

. By the density of P − P in C(U ) (Proposition A.169) we obtain that µ = εz β(U c ) Thus, S(U ) is simplicial and εz is the unique S(U )-maximal representing measure for z. It follows that ChS(U ) (U ) = {z ∈ U : εβ(U z

c)

= εz } = ∂ess U.

As in the proof of Theorems 13.34 and 13.35 we show that H(U ) = S(U ) ∩ −S(U ) and that the maximal representing measures and Choquet boundaries (when we compare H(U ) with S(U )) are the same.

512


Theorem 13.42. Suppose that f ∈ C(U ) and Df = f on U . Then (a) f is H(U )-affine, (b) f ∈ H(U ). Proof. By the continuity of f we have lim

x→z,x∈U

Df (x) = f (z),

By Lemma 13.9,

c

f (z) = εzβ(U ) (f ),

z ∈ U.

z ∈ U.

Then we obtain assertion (a) from Corollary 6.12 and (b) follows from Bauer’s theorem 3.27 (we use the fact that H(U ) = S(U ) ∩ −S(U )). Remark 13.43. Recall that in potential theory of the Laplace equation, the Choquet boundary coincides with ∂ reg U = U ∩ b(U c ). (Even for an arbitrary set A we have β(A) = b(A).) This is no longer true in potential theory of the heat equation as we will see from Example 13.45. But even if ChH(U ) (U ) = ∂ reg U , the harmonic measure U µU (x,t) and the maximal measure δ(x,t) may differ (and the example is even simpler), this we will see in the following Example 13.44. Example 13.44. Let U := (0, 1) × (0, 1) ∪ (1, 2) . For each (x, t) ∈ U ,

c

U µU (x,t) = ε(x,t) ,

and

β(U c )

U δ(x,t) = ε(x,t) .

In this example, since horizontal segments are totally thin (see Example A.218), β(U c ) = b(U c ) = V c and

c

with V := (0, 1) × (0, 2] c

U V U µU (x,t) = ε(x,t) 6= ε(x,t) = δ(x,t) ,

t > 1.

To demonstrate the difference, consider f ∈ C(∂U ) such that f > 0 on (0, 1) × {1} and f = 0 elsewhere. By Examples 13.25(e), Hf (x, t) > 0 for t > 1 close to 1. Using the Harnack-type inequality (Proposition A.151; together with a scaling and translating argument), we easily infer that Hf > 0 on (0, 1) × (1, 2). Thus µU (x,t) ((0, 1) × {1}) > 0,

(x, t) ∈ (0, 1) × (1, 2),

whereas by Theorem A.198 c

εV(x,t) ((0, 1) × {1}) = 0,

(x, t) ∈ (0, 1) × (1, 2).


513

Example 13.45. Let Y = R1+1 be considered with the structure of the heat equation, o n 1 U := (0, 1) × (0, 1) \ 1 − : n ∈ N n

and

V := (0, 1) × (0, 1).

Then ChH(U ) (U ) = ∂V \ (0, 1) × {1}

while

∂ reg U = ∂V.

Indeed, let [α] denote the integer part of a real number α. Given ξ ∈ (0, 1), set u(x, t) :=

1 |x − ξ|2 +t−1+ 1 1, 2 1−t − 2

(x, t) ∈ U.

Then u is a barrier at (ξ, 1), and thus (ξ, 1) is regular. On the other hand, using Example A.218 we infer that β(U c ) ⊂ b(b(U c )) ⊂ b(V c ) = V c \ (0, 1) × {1}).

13.3.C

Continuity properties of H(U )-concave functions

Throughout this subsection, U is a bounded open subset of Rd , d ≥ 2, considered with the structure of the Laplace equation. In this subsection, continuity properties of H(U )-concave functions are investigated. Recall that a function s : U → [−∞, ∞] is called H(U )-concave if for every x ∈ U and every µ ∈ Mx (H(U )), the integral µ(s) exists and µ(s) ≤ s(x). The following examples show that H(U )-concave (even H(U )-affine) functions are not necessarily continuous on U . Examples 13.46. (a) Let U be an open unit ball in Rd and f a bounded discontinuous Borel function on ∂U . Define c

h : x 7→ εU x (f ),

x ∈ U.

Then h is H(U )-affine, but not continuous at the points of ∂U where f is discontinuous. (b) If a set U is not regular, H(U )-affine functions may be discontinuous on U even for continuous boundary data. Consider, for example, U as the set obtained by removing the Lebesgue spine (see Example 13.18(e)) with cusp at 0 from the open unit ball in Rd , d ≥ 3. Define Z c h(x) := |y| dεU x ∈ U. x (y), ∂U

As before, h is H(U )-affine. Since all points in ∂U \ {0} are regular, there exists a sequence {xn } in U tending to 0 such that limn→∞ h(xn ) = 0. Since the probability c measure εU 0 does not charge 0, it is clear that h(0) > 0. We see that already the restriction of h to U ∪ {0} is not continuous.

514


Lemma 13.47. Let H be a function space on a compact space K, x, y ∈ K, µ ∈ Mx (H), ν ∈ My (H), and c > 0 such that µ ≤ cν. Then s(x) ≤ cs(y) whenever s is a positive H-concave function on K. Proof. If we define 1 1 σ := εx + ν − µ , c c then σ is a positive measure by assumption and, obviously, σ ∈ My (H). Hence σ(s) ≤ s(y), and therefore s(x) ≤ cσ(s) ≤ cs(y). Proposition 13.48. Let c > 0 and x, y ∈ U such that h(x) ≤ ch(y) for every h ∈ H + (U ). Then s(x) ≤ cs(y) for every H(U )-concave function s ≥ 0. Moreover, if U is connected, then, for every compact subset K of U and every x0 ∈ U , there exists c > 0 such that s ≤ cs(x0 ) on K for every H(U )-concave function s ≥ 0. In particular, every lower bounded H(U )-concave function which is finite at some point of U is locally bounded on U . c

Proof. Recall that the space H(U ) is simplicial and that, for every z ∈ U , εU z is a unique measure from Mz (H(U )) carried by ChH(U ) (U ) (see Theorem 13.35). Fix f ∈ C(∂U ), 0 ≤ f ≤ 1, η > 0, and a compact set L ⊂ ChH(U ) (U ) such that c Uc εU x + εy (U \ L) < η. By Theorems 6.27 and 13.35, there exists h ∈ H(U ), 0 ≤ h ≤ 1, such that h = f on L. Then c Uc εU x (f ) ≤ εx (h) + η = h(x) + η ≤ ch(y) + η c

c

U = cεU y (h) + η ≤ cεy (f ) + (c + 1)η, c

c

U hence εU x ≤ cεy . The first part now follows from Lemma 13.47. The second part follows easily from D. H. Armitage and S. J. Gardiner [21], Corollary 1.4.4. For the final part, we note that, for every H(U )-concave function s, the positive function s − inf s(U ) is also H(U )-concave.

Example 13.49. Let U be the open unit ball in Rd , d ≥ 3. There exists a positive function s ∈ S(U ) which is not H(U )-concave. Choose y ∈ U \ {0} and find c > 0 such that h(0) ≤ ch(y), h ∈ H + (U ). Set t(x) := |x|2−d − 1 for x ∈ Rd . Define s : x 7→ (c + 1)t(y) ∧ t(x),

x ∈ Rd .

Then s(0) = (c + 1)t(y) = (c + 1)s(y). By Proposition 13.48 we know that s is not H(U )-concave.


515

Theorem 13.50. Assume that U is connected. Then every locally lower bounded set F of H(U )-concave functions, which is upper bounded at some point of U , is locally bounded and locally equicontinuous on U . Proof. Assume that the set F is upper bounded at x0 ∈ U . Choose a compact set K in U containing x0 . It is easy to find an open connected set V containing K such that V ⊂ U . By adding a suitable constant, we may assume that s|V ≥ 0 for every s ∈ F. Then, by Proposition 13.48, F is bounded on K, say by a constant a. Fix η > 0. It follows from D. H. Armitage and S. J. Gardiner [21], Lemma 1.5.6, that there exists δ > 0 such that c

c

εVx ≤ (1 + η)εVy ,

x, y ∈ K, |x − y| < δ.

By Proposition 13.48, for every s ∈ F and for all x, y ∈ K satisfying |x − y| < δ, s(x) ≤ (1 + η)s(y) ≤ s(y) + ηa.

Corollary 13.51. Let U be bounded open connected set and s be a locally lower bounded H(U )-concave function, which is finite at some point of U . Then s is continuous on U .

13.3.D

Separation by functions from H(U )

Throughout this subsection, U is again a bounded open subset of Rd , d ≥ 2, considered with the structure of the Laplace equation. In view of the similarity with convex and concave functions on a compact convex set in Rd , the following question is quite natural: Given functions u and v on U := U (0, 1) ⊂ Rd , d ≥ 2, such that u is H(U )-concave, v is H(U )-convex and u ≤ v, does there exist a function h ∈ H(U ) satisfying u ≤ h ≤ v? We will show that, in contrast to the convex setting, the answer is negative. Define ϕ : t 7→ (1 − t)/(1 + t)d−1 , t ∈ [0, 1]. The function ϕ is decreasing and convex on [0, 1]. Given z ∈ ∂U , for the Poisson kernel with pole z we write Pz instead of P0,1 (·, z), so that Pz (x) = 1 − |x|2 |x − z|d , x ∈ U. We start with several lemmas. Lemma 13.52. Let h ∈ H + (U ) and h(0) = 1. Then h(x) ≥ ϕ(|x|) for every x ∈ U . Moreover, if y ∈ U , y 6= 0, and h(y) = ϕ(|y|), then h = P−y/|y| .

516


Proof. By the Riesz–Herglotz theorem (see D. H. Armitage and S. J. Gardiner [21], Theorem 1.3.8, or Theorem 14.18 below), there exists µ ∈ M1 (∂U ) such that Z Pz (x) dµ(z), x ∈ U. h(x) = ∂U

Let x ∈ U \ {0} and z ∈ ∂U . We have |x − z| ≤ |x| + 1 = x + x/|x| and the equality holds if and only if z = −x/|x|. Hence Pz (x) ≥ P−x/|x| (x) =

1 − |x|2 = ϕ(|x|), (1 + |x|)d

and therefore h(x) ≥ ϕ(|x|). Obviously, h(0) = ϕ(0). Also, for x 6= 0, Pz (x) = ϕ(|x|) holds if and only if z = −x/|x|. Suppose now that y ∈ U , y 6= 0, and h(y) = ϕ(|y|). Then Z Pz (y) − ϕ(|y|) dµ(z) = h(y) − ϕ(|y|) = 0. ∂U

Since Pz (y) − ϕ(|y|) > 0 for z ∈ ∂U \ {−y/|y|}, we get µ = ε−y/|y| . For z ∈ ∂U and r ∈ (0, 1) we set hz,r : x 7→ Pz (rx),

x ∈ U,

and we define U := {hz,r : z ∈ ∂U, r ∈ (0, 1)}. Lemma 13.53. Let u := inf U. Then u is an H(U )-concave function and u(x) = ϕ(|x|) for every x ∈ U . Proof. Let x ∈ U . We have hx,r (x) = Pz (rx) ≥ P−x/|x| (rx) =

1 − r2 |x|2 (1 + r|x|)d

for every z ∈ ∂U and rR ∈ (0, 1). Hence u(x) = ϕ(|x|). It follows that u ∈ C(U ) and, obviously, u(x) ≥ u dν whenever ν ∈ Mx H(U ) . For y ∈ U , y 6= 0, put 1 − ϕ(|y|) |x − y| vy : x 7→ ϕ(|y|) + , |y|

x ∈ U.

13.4 Dirichlet problem: solution methods

517

Lemma 13.54. Let y ∈ U \ {0}. The function vy is continuous and convex on U (in particular, vy is an H(U )-convex function), u ≤ vy on U , vy (0) = 1 and u(y) = vy (y). Proof. Define w : x 7→ 1 −

1 1 − ϕ(|y|) |x|, |y|

x ∈ U.

It is easy to see that w ≤ vy on U . If |y| ≤ |x| ≤ 1, then u(x) = ϕ(|x|) ≤ ϕ(|y|) ≤ vy (x) since ϕ is decreasing. If |x| ≤ |y|, then u(x) = ϕ(|x|) ≤ w(x) ≤ vy (x) since ϕ is convex, w(0) = u(0) and w(y) = u(y). Theorem 13.55. There exists a family U ⊂ H(U ) such that u := inf U is continuous on U and there exists a continuous convex function v on U such that u ≤ v and the inequalities u ≤ h ≤ v hold for no function h ∈ H(U ). Proof. The theorem is a consequence of previous Lemmas 13.52, 13.53 and 13.54. Corollary 13.56. There exist an H(U )-concave function u and an H(U )-convex function v on U such that u ≤ v on U but u and v cannot be separated by a function from H(U ).

13.4

Dirichlet problem: solution methods

Throughout this section, U will be a relatively compact open subset of Y . We consider various methods how to solve the Dirichlet problem. In fact, some of these methods at least look “constructive”, whereas other are only verification methods. Namely, given boundary data f ∈ C(∂U ) and a harmonic function h ∈ H (U ), these methods give criteria how to recognize when h = Hf . We must say that neither “constructive methods”, like balayage or Perron method, give hint how to solve the generalized Dirichlet problem in practice, because the task to find infimum of an infinite system of function is far from indicating a numerical algorithm. Fortunately, the issue of computational methods for the Laplace equation is well understood and there is a rich list of literature on this topic, which we do not pursue to extend. In spirit of the philosophy of potential theory, we remain being interested in subtle qualitative questions connected with irregularities, for which the variety of theoretical solution methods may help to find an appropriate approach.

518

13.4.A


PWB solution of the Dirichlet problem

Definition 13.57 (Upper and lower solutions, resolutive function). For an extended real-valued function f on ∂U , we denote ∗

U (f ) := {u ∈ H (U ) : u is bounded below on U and lim inf u(x) ≥ f (z) for each z ∈ ∂U }. x→z

Functions from the family U (f ) are called upper functions to f while functions from −U (−f ) are labelled as lower functions to f . We define the upper solution Hf and the lower solution Hf by Hf := inf {u : u ∈ U (f )}

and

Hf := −H(−f ).

According to the minimum principle for hyperharmonic functions A.160, Hf ≤ Hf on U . We say that f is resolutive if Hf = Hf on U and Hf is finite. It is clear that the operator H is sublinear, so that the class R(U ) of all real-valued resolutive functions is a linear space. If f is resolutive, it is customary to label the common value Hf = Hf as Hf . As we show later in Theorem 13.61, this is consistent with our earlier Definition 13.1. Proposition 13.58. Suppose that f1 , f2 , f are extended real-valued functions on ∂U and f = f1 + f2 on the set of all points where this sum makes sense. If f1 , f2 are resolutive, then f is resolutive and Hf = Hf1 + Hf2 . Proof. Let ui be upper functions to fi , i = 1, 2, and u := u1 + u2 . Choose z ∈ ∂U . If f (z) = f1 (z) + f2 (z), then obviously f (z) ≤ lim inf u(x). x→z, x∈U

(13.12)

Otherwise one of the functions fi , say f1 , satisfies fi (z) = ∞. Then lim inf u1 (x) = ∞.

x→z, x∈U

Since u2 is lower bounded, (13.12) holds. It follows that u is an upper function to f . The rest is obvious. 13.59. Let {fk } be a sequence of positive resolutive functions and f := Proposition P∞ f . Then k=1 k ∞ ∞ X X Hfk . Hf = Hf = Hfk = k=1

k=1


Proof. If we choose uk ∈ U (fk ), k ∈ N, then Hf ≤

∞ X

P∞

k=1 uk

519

∈ U (f ), so that

Hfk .

k=1

On the other hand, for each n ∈ N, n X

Hfk ≤ H

n X

fk ≤ Hf.

k=1

k=1

Theorem 13.60 (Wiener). Every continuous function f on ∂U is resolutive and the common value of Hf and Hf is Hf (in the sense of Definition 13.1). ∗

Proof. Let p ∈ P and consider u ∈ H + (Y ) such that u ≥ p on U c . Then obviously c u|U ∈ U (p) and thus Hp ≤ RpU = Hp on U . On the other hand, Hp is a lower function to p, so that Hp ≤ Hp. It follows that p is resolutive. Now the result follows from the density of P − P in C(∂U ) (Proposition A.169). Theorem 13.61 (Brelot). For any extended real-valued function f on ∂U and x ∈ U we have Z ∗ Hf (x) =

f dµx .

(13.13)

∂U

If an extended real-valued function f on ∂U is resolutive, then f is µx -integrable for every x ∈ U and the common value of Hf and Hf is Hf (in the sense of Definition 13.1). Conversely, if f is µx -integrable for every x from a dense subset of U , then f is resolutive. Proof. First, we claim that Hg = Hg = Hg if g is lower semicontinuous. Then g is lower bounded, so P that we may assume that g ≥ 0 and in this case we have the representation g = ∞ j=1 gj , where gj ≥ 0 are continuous functions (use Proposition A.53). By Theorem 13.60, the functions gj are resolutive and Hgj = Hgj . Thus by Proposition 13.59, ∞ X Hg = Hg = Hgj . j=1

Since

∞ X

Hgj = Hg

j=1

by the Lebesgue monotone convergence theorem, this verifies the claim for lower semicontinuous functions.

520


Now, suppose that f is an extended real-valued function f on ∂U and x ∈ U . If u ∈ U (f ), set g(z) := lim inf u(y), z ∈ ∂U. y→z

Then g is lower semicontinuous, u ∈ U (g) and g ≥ f . Hence Z Z ∗ g dµx = Hg(x) = Hg(x) ≤ u(x). f dµx ≤ ∂U

∂U

Passing to infimum over u we obtain Z ∗ f dµx ≤ Hf (x). ∂U

Conversely, let g ≥ f is lower semicontinuous. Then Z Hf (x) ≤ Hg(x) = g dµx , ∂U

and passing to the infimum over all lower semicontinuous majorants we obtain Z ∗ Hf (x) ≤ f dµx ∂U

(see Lemma A.90). This verifies (13.13). It is easy to deduce that resolutivity of f implies its µx -integrability, x ∈ U , and the coincidence Hf = Hf = Hf . Suppose now that f is µx -integrable for every x from a dense subset of U . We find a sequence {xn } forming a dense subset of U and a sequence {αn } of strictly positive numbers such that f is µxn -integrable for each n ∈ N and ∞ X

αn µxn (|f |) < ∞.

n=1

Set µ :=

∞ X

α n µ xn .

n=1

Then f is µ-integrable. First, if f is lower semicontinuous, then as above, Hf = limj→∞ Hfj where fj ∈ C(∂U ), j ∈ N, and fj % f . Since the limit Hf is finite on a dense set, by Doob’s convergence axiom, Hf is harmonic on U . Now, let f be an arbitrary µ-integrable function. By Lemma A.90, there exist decreasing sequences {gj } and {kj } of µ-integrable lower semicontinuous functions on ∂U such that −kj ≤ f ≤ gj on ∂U , j ∈ N, and µ(gj − f ) & 0,

µ(kj + f ) & 0.


521

We use the above reasoning to verify that for each j ∈ N, the functions gj and kj are resolutive and Hgj and Hkj are harmonic functions. Since −Hkj ≤ Hf ≤ Hf ≤ Hgj ,

(13.14)

using Doob’s convergence axiom again we obtain that the functions v := lim (−Hkj ) and j→∞

u := lim Hgj j→∞

are harmonic. We observe that µ(u) = µ(v), and thus u = v on a dense set. From the continuity of both u and v and (13.14) we deduce that v = Hf = Hf = u. Thus f is resolutive and the common value of Hf and Hf is Hf . Corollary 13.62. If R(U ) is equipped with the pointwise order, then it is a vector lattice and lattice supremum in R(U ) coincides with the pointwise supremum. Proof. This follows from the characterization of R(U ) in terms of an intersection of L1 -spaces. Definition 13.63 (Negligible set). We say that M ⊂ ∂U is negligible if µx (M ) = 0 for each x ∈ U . Proposition 13.64. A set M ⊂ ∂U is negligible if and only if there exists a dense set S ⊂ U such that µx (M ) = 0 for each x ∈ S. Proof. Suppose that S ⊂ U is dense and µx (M ) = 0 for each x ∈ S. Let f = cM . Then Hf (x) = 0 for each x ∈ S. Since U (f ) forms a saturated family, by Theorem A.165, Hf is harmonic in U . Thus, in view of Theorem 13.61, Hf = Hf = 0 in U . It follows that M is negligible. The converse implication is obvious. Lemma 13.65. Let f : ∂U → R be a resolutive function. Then there exist lower semicontinuous resolutive functions u and v on ∂U and a negligible set N ⊂ ∂U such that f = u − v on ∂U \ N and Hu, Hv are harmonic. Proof. We find a sequence {xn } of points of U and a sequence {αn } of strictly positive numbers such that {xn : n ∈ N} is dense in U and ∞ X

αn µxn (|f |) < ∞.

n=1

Set µ :=

∞ X n=1

α n µ xn .

522


Then f is µ-integrable. Hence there exists a sequence {fj } of continuous functions on ∂U such that ∞ X µ(|fj − f |) < ∞. j=1

It follows that the sum L1 (µxn ), n ∈ N. Then

P∞

j=1 fj

f (y) =

converges absolutely to f in each of the spaces

∞ X

fj+ (y)

j=1

−

∞ X

fj− (y) ∈ R

j=1

holds for each y ∈ ∂U \ N , where the exceptional set N is µxn -null for each n ∈ N. By Proposition 13.64, N is negligible. Using the Lebesgue monotone convergence theorem we obtain Hf = H

∞ X j=1

∞ ∞ ∞ X X X fj+ − H fj− = Hfj+ − Hfj− . j=1

j=1

j=1

P∞ P − + Now, the functions ∞ j=1 Hfj are harmonic by Doob’s convergence j=1 Hfj and axiom, as they are finite on {xn : n ∈ N}. Theorem 13.66. Let U ⊂ Y be relatively compact and open and f : ∂U → R be a resolutive function. Then Hf is harmonic in U . Proof. Let u, v be as in Lemma 13.65. Then Hf = Hu − Hv and thus Hf is harmonic. Remark 13.67. It may look that the harmonicity of Hf or even Hf follows easily from considering U (f ) as a saturated family. However, this approach works only if it is clear that U (f ) contains a superharmonic function. The situation is more transparent for the Laplace equation. Then we may assume that U is connected. If Hf is finite at one point, then there exists an upper function finite at this point and this upper function is already superharmonic. Definition 13.68 (PWB solution). If f is resolutive, then it follows from Theorem 13.61 that the common value of Hf and Hf is Hf which is a harmonic function on U (see Theorem 13.66). The procedure, how we obtained the solution, is called the Perron–Wiener–Brelot method of solving the Dirichlet problem and the solution obtained this way is called the Perron–Wiener–Brelot solution, for short the PWB solution, of the Dirichlet problem on U with boundary data f .

523


13.4.B

Cornea’s approach to the Dirichlet problem

Throughout this section, Y will be again a Bauer harmonic space and U ⊂ Y a relatively compact open set. Cornea’s approach yields a new method for setting and solving the Dirichlet problem by means of the so-called controlled convergence. It leads to the notion of resolutivity in Cornea’s sense, which actually turns out to be equivalent to the resolutivity introduced in Definition 13.57. Recall that the Perron–Wiener–Brelot solution h of the Dirichlet problem for a given boundary data f is based on the approximation of h by hyperharmonic functions from above and by hypoharmonic functions from below. However, for the Laplace equation it is possible to forget hyperharmonic and hypoharmonic functions in both PWB method and Cornea’s methods and to consider only harmonic upper and lower functions (see Theorem 13.81). Definition 13.69 (Controlled convergence). Let f be an extended real-valued function on ∂U , h and k be real-valued functions on U and k ≥ 0. We will say that h converges to f controlled by k, if for every z ∈ ∂U and every ε > 0 the following inequalities hold: ∞ 6= lim sup(h(x) − εk(x)) ≤ f (z) ≤ lim inf(h(x) + εk(x)) 6= −∞. x→z

x→z

Definition 13.70 (c-resolutive and C-resolutive functions, Cornea’s solutions of the Dirichlet problem). An extended real-valued function f on ∂U is called resolutive in Cornea’s sense (for short c-resolutive) if there are functions h ∈ H (U ) and k ∈ S + (U ) such that h converges to f controlled by k. The function h is then said to be a c-solution for f . We say that f is harmonically resolutive in Cornea’s sense (for short C-resolutive) if the control function k can be taken to be harmonic. The function h is then called a C-solution for f . We say that the set U is C-resolutive if each f ∈ C(∂U ) is C-resolutive. Note that in Theorem 13.74 below we will prove that for each c-resolutive (and the more C-resolutive) function f , the c-solution coincides with Hf . So, we will tacitly use this fact after proving Theorem 13.74. Remark 13.71. Suppose that h converges to f controlled by k. Consider the sets Gα := {x ∈ U : k(x) > α}. Then the sets Gα “potential-theoretically disappear” as α → ∞. However, for each fixed α we easily obtain lim

x→z, x∈U \Gα

h(x) = f (z),

z ∈ ∂U ∩ U \ Gα .

(13.15)

In this sense the c-solution generalizes the classical solution of the Dirichlet problem. Under the name “quasisolutions”, a general treatment of solutions of Dirichlet problem based on the limiting process as in (13.15) is developed in [320].

524


Lemma 13.72. Let f be a resolutive function on ∂U . Then there exists a superharmonic upper function to f . Proof. We may assume that f ≥ 0. Let {xn } be a sequence of points of U such that {xn : n ∈ N} is dense in U . Set fn = f ∧ n, n ≥ 0, and gn = fn − fn−1 for n ∈ N. Then gn are bounded and resolutive (see Corollary 13.62). Thus, for each n ∈ N there −j for each exists a bounded function n ∈ U (gn ) such that un (xj ) − Hgn (xj ) < 2 Pu ∗ + ∞ j = 1, . . . , n. Set u := n=1 un . Then u ∈ H (U ). Fix n ∈ N and z ∈ ∂U . Then u1 + · · · + un is an upper function to fn := g1 + · · · + gn and thus fn (z) ≤ lim inf(u1 (x) + · · · + un (x)) ≤ lim inf u(x). x→z

x→z

Letting n → ∞ we obtain f (z) on the left-hand side. Thus, u is an upper function to f . Given j ∈ N, we have X X X ui (xj ) < Hgi (xj ) + 2−i ≤ Hf (xj ) + 1, i>j

i>j

i>j

and thus u(xj ) ≤

X

ui (xj ) +

X

ui (xj ) < ∞.

i>j

i≤j

Thus, u is finite on a dense set, so that u is superharmonic. Proposition 13.73. Let f be a resolutive function on ∂U , kn ∈ U (f ), qn ∈ U (−f ) be such that each of the functions k :=

∞ X

(kn − Hf ) and

q :=

∞ X

(qn + Hf )

n=1

n=1

is finite on a dense subset of U . Then Hf converges to f controlled by k + q. Proof. Obviously, k + q is a positive superharmonic function. Put h := Hf and fix z ∈ ∂U , ε > 0 and n ∈ N such that εn ≥ 1. Then h + ε(k + q) ≥ h + εk ≥ h +

n

n

j=1

j=1

1X 1X (kj − h) = kj . n n

It follows that lim inf(h(x) + ε(k(x) + q(x))) ≥ x→z

n

n

j=1

j=1

X 1 1X lim inf lim inf kj (x). kj (x) ≥ x→z n x→z n

Since kj ∈ U (f ), the last sum is different from −∞ and lim inf kj (x) ≥ f (z), x→z

j = 1, . . . , n.


525

We conclude that f (z) ≤ lim inf(h(x) + ε(k(x) + q(x))) 6= −∞. x→z

The proof for lim sup is analogous and the assertion is proved. Theorem 13.74. Let f be an extended real-valued function on ∂U and h ∈ H (U ). Then the following assertions are equivalent: (i) f is resolutive and h = Hf , (ii) there exists s ∈ S + (U ) such that h converges to f controlled by s (so that f is c-resolutive). Proof. (i) =⇒ (ii): Suppose that f is resolutive and h = Hf . By Lemma 13.72, there exist u ∈ U (f ) ∩ S (U ) and v ∈ U (−f ) ∩ S (U ). Denote Z := {x ∈ U : u(x) + v(x) = ∞}. Then U \ Z is dense in U . Let {xn } be a sequence of points in U \ Z such that {xn : n ∈ N} is dense in U . Since U (f ) is min-stable, we easily find sequences {kn } of functions from U (f ) and {qn } of functions from U (−f ) such that kn ≤ u, qn ≤ v, kn (xj ) < h(xj ) + 2−n and qn (xj ) < −h(xj ) + 2−n for each n ∈ N and jP= 1, . . . , n. From Proposition 13.73 we obtain that h converges to f controlled by ∞ n=1 (kn + hn ). (ii) =⇒ (i): Let s be as in (ii) and V be a regular set with V ⊂ U . Then sV is harmonic on V , h + εsV ∈ U (f ) and −h + εsV ∈ U (−f ) for each ε > 0. It follows that h = Hf = Hf on V and h is harmonic on V . Since regular sets form a base of the topology on U , by Sheaf axiom of Definition A.156, h is harmonic. Thus f is resolutive and h = Hf . Theorem 13.75. Let f be a resolutive function. Then the inequalities lim inf Hf (x) ≤ f (z) ≤ lim sup Hf (x) x→z

x→z

hold for every z ∈ ∂U except for a negligible set. Proof. By Theorem 13.74, there exists k ∈ S + (U ) such that Hf converges to f controlled by k. Put n o N := z ∈ ∂U : lim inf k(x) = ∞ . x→z

Then N is a Borel set and for its characteristic function cN we obviously have δk ∈ U (cN ) whenever δ > 0. This shows that N is a negligible set. Fix z ∈ ∂U \ N . Then we have for every ε > 0 f (z) ≤ lim inf(Hf (x) + εk(x)) ≤ lim sup Hf (x) + ε lim inf k(x). x→z

x→z

x→z

526


Since lim infx→z k(x) < ∞, we have proved the inequality f (z) ≤ lim sup Hf (x). x→z

Replacing f by −f completes the proof. Corollary 13.76. If R(U ) is the class of all real-valued resolutive functions on ∂U , then f 7→ Hf , f ∈ R(U ), is an “injective” mapping in the following sense: If Hf = Hg, then the set {z ∈ ∂U : f (z) 6= g(z)} is negligible. Proof. It follows immediately from Theorem 13.75. Lemma 13.77. Let fj be extended real-valued functions on ∂U , hj and kj be harmonic functions on U , kj ≥ 0, j = 1, 2. Let hj converge to f controlled by kj , j = 1, 2. Let f be an extended real-valued function on ∂U such that f (z) = f1 (z) + f2 (z) at any point z ∈ ∂U at which the sum f1 (z)+f2 (z) is well defined. Then h := h1 +h2 converges to f controlled by k = k1 + k2 . Further, if λ ∈ R, λh1 converges to λf1 controlled by k1 . Proof. Let ε > 0 and z ∈ ∂U . Then we easily obtain ∞ 6= lim sup(h(x) − εk(x)) ≤ f (z) ≤ lim inf(h(x) + εk(x)) 6= −∞ x→z

x→z

(13.16)

if the sum f1 (z) + f2 (z) is well defined. If f1 (z) = −∞ and f2 (z) = ∞, then lim sup(h(x) − εk(x)) ≤ −∞ + lim sup(h2 (x) − εk2 (x)) = −∞ x→z

x→z

and lim inf(h(x) + εk(x)) ≥ lim inf(h1 (x) + εk1 (x)) + ∞ = ∞. x→z

x→z

The case f1 (z) = ∞ and f2 (z) = −∞ is analogous. We have obtained (13.16) for each z ∈ ∂U , and thus h converges to f controlled by k. Since the last part is obvious, the proof is complete. Lemma 13.78. Let {fn } be an increasing sequence of real-valued C-resolutive functions, fn % f . If Hf ∈ H (U ), then f is C-resolutive. Proof. Without loss of generality we may assume that fn ≥ 0, n ∈ N, and we set f0 := 0. Let {yj } be a sequence of points of U such that {yj : j ∈ N} is dense in U . Put hn := Hfn , n ≥ 0 and h := limn→∞ hn = Hf . Using the diagonal method we find an increasing sequence {ni } of natural numbers such that ∞ X i=1

i(hni+1 (yj ) − hni (yj )) < ∞,

j ∈ N.


527

We relabel the sequence {hni } as {hi }. There exist functions kn ∈ H + (U ) such that hn converges to fn controlled by kn . Choose numbers an > 0 such that ∞ X

an kn (yj ) < ∞,

j ∈ N,

n=1

P P∞ and put k := ∞ n=1 an kn , g := n=1 n(hn+1 − hn ). By Doob’s convergence axiom, k, g ∈ H + (U ). Clearly, h = lim hn = n→∞

∞ X

(hn+1 − hn ).

n=0

We claim that h converges to f controlled by k + g. Fix z ∈ ∂U and ε > 0. Since for every n ∈ N, fn (z) ≤ lim inf(hn (x) + εkn (x)) 6= −∞ x→z

and obviously hn + εkn ≤ h + ε(k + g), we have fn (z) ≤ lim inf(h(x) + ε(k(x) + g(x))) 6= −∞. x→z

Hence f (z) ≤ lim inf(h(x) + ε(k(x) + g(x))) 6= −∞. x→z

Fix a natural number m > 1 satisfying 1 ≤ εm. Then h − ε(k + g) =

m−1 X

(hn+1 − hn ) − εk +

(hn+1 − hn )

n=m

n=0

−ε

∞ X

m−1 X

n(hn+1 − hn ) − ε

n(hn+1 − hn )

n=m

n=1

≤ hm − εk +

∞ X

∞ X

(hn+1 − hn ) + 0 −

n=m

∞ X

(hn+1 − hn )

n=m

= hm − εk, which yields lim sup(h(x) − ε(k(x) + g(x))) ≤ lim sup(hm (x) − εk(x)). x→z

x→z

Since hm converges to fm controlled by km , hence also by k, we have ∞ 6= lim sup(hm (x) − εk(x)) ≤ fm (z) ≤ f (z). x→z

528


We conclude that ∞ 6= lim sup(h(x) − ε(k(x) + g(x))) ≤ f (z). x→z

It follows that h converges to f controlled by k + g, and hence f is C-resolutive. Lemma 13.79. Suppose that U is C-resolutive. Let M ⊂ ∂U be a negligible set and let f be an extended real-valued function on ∂U such that f = 0 on ∂U \ M . Then f is C-resolutive. Proof. Let g := ∞ on M and g := 0 on ∂U \ M . By Theorem 13.74, g is cresolutive and thus there exists a superharmonic function s ∈ S + (U ) such that the function s(z) ¯ := lim inf s(x), z ∈ ∂U , satisfies x→z

s(z) ¯ = ∞,

z ∈ M.

Then s ∈ U (s) ¯ and thus 0 ≤ H s¯ ≤ s on U . It follows that the function s¯ is lower semicontinuous and resolutive. Now, we find a sequence {fn } of continuous functions such that 0 ≤ f1 ≤ f2 ≤ · · · and fn % s. ¯ Since U is C-resolutive, all the functions fn are C-resolutive and by Lemma 13.78, s¯ is C-resolutive. Let h := H s¯ converges to s¯ controlled by a harmonic function k. Then ∞ = lim inf(h(x) + k(x)) ≤ lim sup(h(x) − k(x)) + lim inf(2k(x)), x→z

x→z

x→z

z ∈ M.

It follows that 0 converges to g (and thus also to f ) controlled by k. Theorem 13.80 (Cornea). Let U ⊂ Y be C-resolutive. Then an extended real-valued function f on ∂U is C-resolutive if and only if it is resolutive. Proof. Obviously, each C-resolutive function is resolutive. Conversely, let f be a resolutive function on ∂U . By Lemma 13.65, there exist lower semicontinuous resolutive functions u and v on ∂U and a negligible set N ⊂ ∂U such that f = u − v on ∂U \ N and Hu, Hv are harmonic. We may express functions u and v as limits of increasing sequences of continuous functions. From Lemma 13.78 we infer that u and v are C-resolutive. By Lemma 13.77 and Lemma 13.79, the function f is C-resolutive. Now, we will study the two most important concrete harmonic spaces from the point of view of C-resolutivity. Theorem 13.81. Let Y be a Bauer harmonic space associated with the Laplace equation and U be a relatively compact open subset of Y . Then U is C-resolutive.

529


Proof. Let f ∈ C(∂U ). Let h = Hf and k ∈ H + (U ) be a function such that limx→z k(x) = ∞ whenever z is an irregular point of U . Such a function exists by Theorem 13.23. For each ε > 0, we have f (z) = lim h(x) ≤ lim inf(h(x) + εk(x)),

z ∈ ∂ reg U,

f (z) ≤ ∞ = lim inf(h(x) + εk(x)),

z ∈ ∂ irr U.

x→z

x→z

x→z

Hence h + εk ∈ U (f ) ∩ H (U ). Similarly, −h + εk ∈ U (−f ) ∩ H (U ). Letting ε → 0 we obtain that f is C-resolutive. Example 13.82. Consider the heat equation in Rd+1 . Denote W = U (0, 1) × (−1, 2) and set U = W \ {(0, 0)}. Let f ∈ C(∂U ) be given by ( 1, (x, t) = (0, 0), f (x, t) := 0 otherwise. By the Harnack type inequality (Proposition A.151), there exists c > 0 such that k(0, 1) ≥ ck(0, 0),

k ∈ H (W ).

(13.17)

Suppose that u˜ ∈ U (f ) ∩ H (U ). Since (0, 0) is polar (Proposition A.217), there exists s ∈ S + (W ) such that s(0, 0) = ∞. Set ( ∞, (x, t) = (0, 0), un (x, t) := 1 u(x, ˜ t) + n s(x, t), (x, t) ∈ U. \ By Proposition A.189, u := inf un is a positive supercaloric function on W and u = u˜ on U . (In fact, we have shown how to prove that polar sets are removable for positive supercaloric functions.) Let h be the greatest subcaloric minorant of u on W , which is a caloric function on W (see Proposition A.166). We claim that h(x, t) = u(x, t) for t < 0. Indeed, let −1 < τ < 0 and ( 0, t ≤ τ, uτ (x, t) := u(x, t), t > τ. Then, by Corollary A.186, uτ is supercaloric and u − uτ is a subcaloric minorant of u. Hence h ≥ u − uτ on W, −1 < τ < 0, which proves the claim. Since u|U is an upper function to f , we have h(0, 0) = lim inf h(0, t) = lim inf u(0, t) ≥ f (0, 0) = 1. t→0−

t→0−

530


Then by (13.17), u(0, ˜ 1) ≥ h(0, 1) ≥ ch(0, 0) ≥ c. On the other hand, each strictly positive multiple of s|U is an upper function to f and thus Hf = 0. This shows that Hf cannot be obtained as infimum of caloric upper functions.

13.4.C

The Wiener solution

In this subsection, U will be a relatively compact open subset of a Bauer harmonic space Y . The Wiener method of solution of the generalized Dirichlet problem is the method based on inner stability of the Dirichlet problem. The issue of the Wiener solution has two aspects: •

Convergence of the Wiener method. We will prove that the Wiener approximations converge for any exhaustion.

•

Existence of special exhaustions. In general, the sets used for approximate solution are of the same type as the given set U . To have a reason to appreciate the Wiener method, we need to show that we can find exhaustions by “better sets”, where the Dirichlet problem is easier to solve or the solution is more canonical (to be convinced that the solution obtained by this way is the “right solution”).

Definition 13.83 (Exhaustion). A family V of subsets of a relatively compact set W is said to be an exhaustion of W if (a) For each V ∈ V, V ⊂ W , (b) for each compact set K ⊂ W there exists V ∈ V such that K ⊂ V . If V is an exhaustion of W , x ∈ U , and g ∈ C(W ), denote by limV ∈V H V g(x) a number λ which has the property that for any ε > 0 there exists a compact set K containing x such that V ∈ V,

V ⊃ K =⇒ |H V g(x) − λ| < ε.

Such a limit is clearly unique if it exists. Definition 13.84 (Wiener solution). Given an exhaustion V of U , we say that a function h ∈ H (U ) is a Wiener solution of the Dirichlet problem on U with boundary data f ∈ C(∂U ), if for any continuous extension g ∈ C(U ) of f we have h(x) = lim H V g(x), V ∈V

x ∈ U.

The existence and uniqueness is treated by the following theorem.


531

Theorem 13.85. Let V be an exhaustion of U , f ∈ C(∂U ) and g ∈ C(U ). Suppose that g|∂U = f . Then for each x ∈ U , lim H V g(x) = Hf (x).

V ∈V

In particular, the limit always exists and depends neither on g, nor on the exhaustion. Proof. Let p ∈ P and x ∈ U . By Theorems A.199 and A.195, c

Hp(x) = RpU (x) = inf{RpG (x) : G open , G ⊃ U c }. Given ε > 0, we find a closed (thus compact) set K ⊂ U such that c

RpK (x) < Hp(x) + ε. For any V ∈ V with K ⊂ V we then have c

c

c

bpU (x) ≤ R bpV (x) ≤ RpK (x) < Hp(x) + ε. Hp(x) = R From the density of P − P (Proposition A.169) we obtain the result. Proposition 13.86. For potential theory of the Laplace equation, there exists an exhaustion of U formed by regular sets. Proof. The proof can be found in [21], Corollary 6.6.13. We can indicate another approach. If K is a compact subset of U , choose a C ∞ -function F : Rd → [0, 1] so that F = 1 on K and F = 0 on U c and consider the sets Vα := {x ∈ U : F (x) > α} ,

α ∈ (0, 1).

By the Morse–Sard theorem (see, for example, S. Sternberg [437]) there exists β ∈ (0, 1) such that ∇F is nonvanishing at every point of ∂Vβ . Then Vβ is a regular set by Example 13.18(c) and K ⊂ Vβ ⊂ Vβ ⊂ U . Example 13.87 (Bauer). Now we consider the case of the heat equation. Let U := {(x, t) ∈ R2 : 0 < x2 + t2 < 2} and

K := {(x, t) ∈ R2 : x2 + t2 = 1}.

Then there does not exist a (calorically) regular set V such that K ⊂ V ⊂ V ⊂ U . Indeed, assume that such a set V exists. We find (ξ, τ ) ∈ ∂V ∩ U (0, 1) which minimizes τ in this set. Similarly to Example 13.25(c) we obtain that (ξ, τ ) ∈ ∂ irr V . Remark 13.88. We will see later that the Wiener construction has still a reasonable sense in harmonic spaces, in particular for the heat equation. There exists an exhaustion by sets, in which the solution of the Dirichlet problem is in some sense canonical (finely regular sets, Proposition 13.95, or Keldysh sets, Proposition 13.129). Another successful idea is to construct a solution using regular subsets of U by a method which does not require to treat an exhaustion, see Exercise 13.156.

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13.4.D


Fine Wiener solution

We first introduce finely regular sets as a class of not necessarily open sets for which something like classical solution of the Dirichlet problem is still available. By means of exhaustion of a general finely open set by finely regular sets we obtain a version of Wiener method for finely open sets. Surprisingly, this yields a new possibility of exhaustion also if the set U is open, but the harmonic space is “parabolic”. Definition 13.89 (Finely regular sets). We say that a relatively compact set U ⊂ Y is finely regular if b(U c ) = U c . In this case U is finely open and also β(U c ) = U c (see Proposition A.221). Also notice that a relatively compact open set U is finely regular if and only if U is regular. Theorem 13.90. Suppose that U is finely regular and f ∈ C(∂U ). Then there exists a unique H(U )-affine function h such that h = f on U \ U . Moreover, h = Hf on U , h is continuous at all points of U \ U and h is finely continuous in U . Proof. Since H(U ) is simplicial and for each x ∈ U , the H(U )-maximal representing c β(U c ) measure for x is δx = εx = εU x , any H(U )-affine function h which coincides with f on U \ U must satisfy h(x) = δx (f ) for x ∈ U . This shows the uniqueness. Let h be given by this formula. Then h is H(U )-affine by Theorem 6.8. Taking into account density of P −P in C(∂U ) (Proposition A.169), we observe that it is enough to verify the remaining properties for f = p|∂U , p ∈ P. c bpU c (x) for x ∈ U by Theorem A.195, and h(x) = Then h(x) = RpU (x) = R c bU c (x) for x ∈ U \ U ⊂ b(U c ) by the very definition of b(U c ). The RpU (x) = R p c U b function Rp is hyperharmonic and thus finely continuous. If z ∈ U \ U , then h is c upper semicontinuous at z as h = RpU on U , whereas h is lower semicontinuous at z c bpU on U . This shows the continuity of h at z. as h = R Theorem 13.91 (The Lusin–Menshov property). The fine topology τ has the following Lusin–Menshov property: Let F be a finely closed subset of Y and K a Kσ subset of Y disjoint from F . Then there exists a positive finely continuous and upper semicontinuous function ϕ on Y such that ϕ = 0 on F and ϕ > 0 on K. S Proof. Let {Kn } be an increasing sequence of compact subsets of Y , ∞ n=1 Kn = K and p a potential as in Proposition A.206. Using Theorems A.195 and A.205 we bpF < p on Y \ F . According to [66], Lemma obtain that b(F ) ⊂ F and RpF = R VI.2.5, there exists a sequence {Un } of open sets such that F ⊂ Un ⊂ Y and bpUn ≤ RpUn ≤ RpF + 1 R n Set ϕ :=

on Kn .

∞ X 1 bUn ). (p − R p 2n n=1


533

Then ϕ ≥ 0 is a finely continuous and upper semicontinuous function on Y , ϕ = 0 on F . Fix x ∈ K. There exists n ∈ N such that x ∈ Kn and RpF (x) + n1 < p(x). bpUn (x) < p(x) which yields ϕ(x) > 0. Hence R Corollary 13.92. Let K and U be subsets of Y , K ⊂ U , K compact and U finely open. Then there exists a Borel finely open set V such that K ⊂ V ⊂ V ⊂ U . Proof. We know from the Lusin–Menshov property 13.91 that there exists a finely continuous and upper semicontinuous function ϕ ≥ 0 such that ϕ = 0 on Y \ U

and

ϕ > 0 on K.

Let g ≥ 0 be a continuous function on Y such that K = {x ∈ Y : g(x) = 0}. Set V := {x ∈ Y : g(x) < ϕ(x)} . Then V is a Borel finely open set such that K ⊂ V ⊂ V ⊂ {x ∈ Y : g(x) ≤ ϕ(x)} ⊂ U. This completes the proof. Proposition 13.93. The fine interior of any compact subset of Y is a finely regular set. Proof. Let K ⊂ Y be a compact set and U its fine interior. Then U c is the fine closure of K c and thus U c = K c ∪ b(K c ) ⊂ b(U c ) ⊂ U c . This concludes the proof. Remark 13.94. The reasoning of the previous proposition shows that the assertion continues to hold even for finely closed sets (instead of compact ones). The following proposition shows that the fine solution of the Dirichlet problem is more efficient from the point of view of insertion of regular sets (recall that we appreciated the possibility to construct “regular exhaustions” for the Wiener method of Subsection 13.4.C). By Example 13.87, the possibility of ordinary regular exhaustion fails in the case of the heat equation. Proposition 13.95. Let U be a finely open set and K ⊂ U compact. Then there exists a finely regular set V such that K ⊂ V ⊂ V ⊂ U . Proof. The assertion is an immediate consequence of the Lusin–Menshov property in Corollary 13.92: There exists a finely open set G such that K ⊂ G ⊂ G ⊂ U . Since G is compact, the fine interior V of G is a finely regular set and K ⊂ G ⊂ V ⊂ V ⊂ G ⊂ U.

534


Corollary 13.96. Let U ⊂ Y be finely open and relatively compact. Then there exists a finely regular exhaustion V of U and Hf (x) = lim H V f (x), V ∈V

x ∈ U,

f ∈ C(∂U ).

Proof. This is only a recapitulation of Theorem 13.85 and Proposition 13.95.

13.4.E

PDE solutions in Sobolev spaces

In this subsection we will study the Dirichlet problem in the framework of the Laplace equation. The PDE methods indicated in the Appendix (Subsection A.6.A) yield a solution for Sobolev boundary data. In this subsection we show how to treat continuous Dirichlet data. The space W 1,2 (U )/W01,2 (U ) enables us to formulate some “new” variants of the Dirichlet problem. For example, if our domain has a part of “inner boundary” which is visible from two sides, we can impose distinct boundary data depending on which side is chosen. If we want to express these data as a function on the boundary, we must replace the topological boundary by something more complicated (mostly by the Martin boundary or another compactification). To have a better understanding, we would like to make some link between boundary data and an element of a space like W 1,2 (U )/W01,2 (U ). The theory of Sobolev spaces attaches boundary values or traces to Sobolev functions. This concept is well developed if the domain is smooth or at least satisfies the Lipschitz condition. However, now we observe a drawback: even for smooth domains, not every continuous function on the boundary is a trace of a Sobolev function. For example, if D is the unit disc in R2 , a function f is a trace of a W 1,2 (D)-function on the boundary of D if and only if ZZ |f (y) − f (x)|2 dS(x) dS(y) < ∞. (13.18) |y − x|2 ∂D×∂D (see A. Kufner, S. Fuˇc´ık and O. John [284], Theorems 6.8.13, 6.9.2). This condition is not implied by continuity. Indeed, it is easily seen that if fk is the real part of the function x 7→ (x1 + ix2 )k , then the integral as in (13.18) tends to infinity as k → ∞, although the norms of fk in the space C(∂D) remain bounded. Now it is easy to consider a countable number of pairwise disjoint arcs Ln on ∂D and a continuous function f ∈ C(∂D) which oscillates as n1 fkn on Ln (with kn large enough) such that the integral (13.18) diverges. To overcome this difficulty, a general continuous function f ∈ C(∂U ) is approximated by smooth functions (for example, by mollification). More precisely, we find a sequence {fn } of functions from C 1 (Rd ) such that fn converge uniformly on Rd to a continuous extension of f . Each fn |U is in W 1,2 (U ) and thus we can find hn ∈ W 1,2 (U ) as the unique harmonic function such that hn − fn |U ∈ W01,2 (U ).


535

Now we observe that the operator sending boundary data from W 1,2 (U ) to the variational solution from H (U ) described in Proposition A.143 is positive (this follows easily with the aid of testing by the negative part of the solution, cf. D. Gilbarg and N. S. Trudinger [193], Theorem 8.1), linear and preserves constants. Thus hn converge uniformly to some function h ∈ H (U ) and this h can be regarded as the solution of the given Dirichlet problem. We obtained the PDE solution (the abbreviation points to the theory of partial differential equations) of the Dirichlet problem, namely “a harmonic function h which is a limit of harmonic functions which are weak solutions of approximating Dirichlet problems”. This description of our main achievement is somewhat cumbersome. Therefore we want to call attention to one aspect of this approach to the Dirichlet problem which is not so widely known. Let us emphasize that in what follows, we will not construct a “new solution” but we will characterize the same PDE solution in terms which allow a more elegant description. In view of Tietze’s theorem, we may assume that our boundary data is a continuous function on U . The main new ingredient is the use of the space W01,2 (U ) + C0 (U ) where C0 (U ) is the space of continuous functions on U vanishing on ∂U . Lemma 13.97. Let U ⊂ Rd be a regular bounded open set and f ∈ C 1 (Rd ). If h is the classical solution of the Dirichlet problem on U with boundary data f and u is its solution in the sense of Proposition A.143, then h = u. Proof. We choose ε > 0. Let {vn } be a sequence of C ∞ c -functions such that vn → u−f in W01,2 (U ) and denote wn := f +vn −ε. Then the set Vn := {x ∈ U : wn (x) > h(x)} is open and Vn ⊂ U . Therefore h ∈ W 1,2 (Vn ) and we can use the energyminimizing property of h to show that Z Z |∇h|2 dx ≤ |∇wn |2 dx, Vn

Z

Vn

|∇h − ∇wn |2 dx

1/2

Vn

≤

Z

|∇h|2 dx

1/2

+

Z

Vn

≤2

1/2 |∇wn |2 dx

Vn

Z

|∇wn |2 dx

1/2

.

Vn

It follows that

Z

∇(wn − h)+ dx ≤ 4

U

Hence (u−h−ε)+

|∇wn |2 dx.

U

h)+

This means that (wn − Z

Z

∈

Wc1,2 (U ).

Letting n → ∞ we obtain that Z ∇(u − h − ε)+ dx ≤ 4 |∇u|2 dx.

U W01,2 (U ).

U

∈ Now, letting ε → 0 and considering also a symmetric estimate we obtain that u − h ∈ W01,2 (U ), and thus h − f ∈ W01,2 (U ). It follows that h is the solution in the sense of Proposition A.143 and by uniqueness, h = u.

536


Lemma 13.98. Let U ⊂ Rd be a bounded open set and h ∈ W01,2 (U ) + C0 (U ) be harmonic on U . Then h = 0. Proof. Suppose that h = u + g where u ∈ W01,2 (U ) and g ∈ C0 (U ). Notice that u is 1,2 (U ). Let {Vn } be a regular exhaustion of U . Set continuous on U and g ∈ Wloc ( H Vn u on Vn , un := u on U \ Vn , ( H Vn g on Vn , gn := g on U \ Vn . Here we use the notation H V f for the solution of the Dirichlet problem on V with boundary data f . For the definition of un and gn it is the classical solution, but by Lemma 13.97 also simultaneously the solution in the sense of (A.2) in Subsection A.6.A. Now, by the energy-minimizing property, we see that {un } is bounded in W01,2 (U ). It follows that (passing, if necessary, to a subsequence) that there exists a weak limit u∞ of un in W01,2 (U ), which is simultaneously a strong limit in L2 (U ) (by the Rellich–Kondrachov compact embedding theorem, see, for example, [334], Theorem 1.61) and a pointwise limit almost everywhere (at least for a suitable subsequence). It is easy to see that u∞ is weakly harmonic on U , thus u∞ = 0 a.e. The functions gn converge uniformly to 0 by the minimum principle, as sup |g|(∂Vn ) → 0. Since obviously h = gn + un , passing to the limit we conclude that h = 0. Theorem 13.99. Let U ⊂ Rd be a bounded open set and f ∈ C(U ). Then there exists a unique h ∈ H (U ) such that h − f ∈ W01,2 (U ) + C 0 (U ). Proof. The uniqueness follows from Lemma 13.98. For the existence, we consider a sequence {fn } of smooth functions on Rd such that |fn − f | < 2−n−2 on ∂U . We write h0 = f0 = 0. For each n ∈ N we find the unique harmonic function hn on U such that hn − fn ∈ W01,2 (U ) and decompose hn − fn − (hn−1 − fn−1 ) = un + gn , −n . For n ≥ 2 we have |f − f −n on where gn ∈ C ∞ n n−1 | < 2 c (U ) and kun kW 1,2 < 2 0

∂U and thus also |hn −hn−1 | < 2−n P on U . Therefore we can achieve that |gn | < 2n+1 on U . We observe that the series ∞ n=1 gn is uniformly convergent to a function P∞ 1,2 g ∈ C 0 (U ), n=1 un converges in W0 (U ) to a function u ∈ W01,2 (U ), and therefore h − f = u + g ∈ W01,2 (U ) + C 0 (U ).

13.5 Generalized Dirichlet problem and uniqueness questions

537

Remark 13.100. From now, we can define the PDE solution as the solution obtained in Theorem 13.99. It is an obvious feature of the space C 0 (U ) that the solution cannot depend on values of f off ∂U . Corollary 13.101. Let f ∈ C(∂U ). Suppose that the Dirichlet problem on U with boundary data f admits a classical solution h. Then h is the PDE solution of this Dirichlet problem.

13.5

Generalized Dirichlet problem and uniqueness questions

In this section, U will be a relatively compact open subset of Y . In Examples 13.18 we encountered sets for which the classical Dirichlet problem is not solvable for some continuous boundary data. The existence of irregular sets leads to a natural question of how to assign to those functions something like a solution in a reasonable way. This is the generalized Dirichlet problem. Keldysh operators are solution operators of the generalized Dirichlet problem. There is no problem with the existence of Keldysh operators. The operators H and D of Section 13.1 serve as examples. However, it is by no means clear whether there are different Keldysh operators on U , in particular whether H = D. Notice that the set H(U )|∂U of all “solvable functions” is, in the irregular case, a nowhere dense subset of C(∂U ). (Of course, any proper closed subspace of a Banach space is nowhere dense.) Therefore it may be somewhat surprising if the extension problem has a unique solution. We will show that the answer to the uniqueness question is “yes” in case of the Laplace equation. On the other hand, already the operators H and D may be different in case of the heat equation. This raises further interesting questions which we will address as well. Definition 13.102 (Keldysh operator). By a Keldysh operator A on U we mean a mapping A : C(∂U ) → H (U ) satisfying the following conditions: (a) A is linear and positive, (b) if there is a solution h ∈ H (U ) of the classical Dirichlet problem for a function f ∈ C(∂U ), then Af = h on U . The following proposition serves as a key tool. Proposition 13.103 (Keldysh lemma). Given x ∈ ChH(U ) (U ), there exists a function h ∈ H+ (U ) such that h(y) = 0 if and only if y = x. Proof. We know by Proposition 13.30 that H(U ) = Ac (H(U )). Combining Theorem 6.28 with Theorem 13.41 we obtain the result immediately. For the Laplace equation, we can use Theorem 13.35 instead of Theorem 13.41.

538

13.5.A


Lattice approach

In this subsection we fix a relatively compact open set U ⊂ Y and write H + = H + (U ). Since each Keldysh operator T on U is bounded and the family of all bounded harmonic functions is contained in H + −H + , we may use H + −H + as the target space for T . This enables us to exploit better lattice properties of H +−H + . Definition 13.104 (Riesz homomorphism). A vector lattice is said to be Dedekind complete if every nonempty upper bounded set has a least upper bound. Let E and F be ordered vector lattices. Then a mapping T : E → F is called a Riesz homomorphism if T (f ∨ g) = T (f ) ∨ T (g) for each f, g ∈ E. Definition 13.105 (Harmonic lattice). We consider H + −H + as an ordered vector space equipped with the natural order. This means that the relation u ≤ v is understood pointwise, but the supremum in H + −H + can differ from the pointwise supremum. Therefore, one has to be careful with the interpretation of the notation f ∧ g, f ∨ g in this subsection. Proposition 13.106. H + −H + is a Dedekind complete vector lattice. Proof. Let ∅ 6= F ⊂ H + −H + be an upper bounded set and k be its upper bound. We may assume that k ≥ 0. We set G := {k − h : h ∈ F}. Then G ⊂ H + . If u := inf G, then by Proposition A.189, u b is superharmonic. Thus, by Proposition A.166, there exists a greatest harmonic minorant u0 of u b. Then k − u0 ∈ H + −H + and it is the least upper bound of F. To complete the proof, we need also to show that given u, v ∈ H + −H + , the set {u, v} is upper bounded. We find u1 , u2 , v1 , v2 ∈ H + such that u = u1 −u2 and v = v1 −v2 . Then u1 +u2 +v1 +v2 is a common upper bound to u and v. Theorem 13.107. (a) The operator H is a Riesz homomorphism from the vector lattice R(U ) to H + −H + . The more this is a Riesz homomorphism from C(∂U ) to H + −H + . (Recall that R(U ) is a vector lattice in view of Corollary 13.62.) (b) If a Keldysh operator T : C(∂U ) → H + −H + is a Riesz homomorphism, then T = H. Proof. (a) Suppose that f, g ∈ R(U ) and h = Hf ∨ Hg. Choose u ∈ U (f ) and v ∈ U (g). Then h + (u − Hf ) + (v − Hg) ≥ max{u, v} and hence this is an upper function to f ∨ g. Passing to the infimum over u ∈ U (f ) and v ∈ U (g) we obtain that H(f ∨ g) ≤ h. Conversely, the function H(f ∨ g) is a common harmonic majorant of Hf and Hg and thus h ≤ H(f ∨ g). (b) We will use (a) and the following uniqueness argument: Since T = H on H(U )|∂U , then also T = H on W(H(U ))|∂U and by the Stone–Weierstrass theorem


539

(Theorem A.30), T = H on a dense subset of C(∂U ). Since T 1 = 1, T preserves uniform convergence and thus T = H on C(∂U ). Definition 13.108 (Extended Keldysh operators). A mapping T : `∞ (∂U ) → H (U ) is called an extended Keldysh operator, if T is linear, positive, and T |C(∂U ) is a Keldysh operator. Definition 13.109 (Envelopes). Let f ∈ `∞ (∂U ). We set o ^n D∗ f := h|U : h ∈ H(U ), h ≥ f on ∂U , o _n D∗ f := h|U : h ∈ H(U ), h ≤ f on ∂U , n o f ∗ := inf h|∂U : h ∈ H(U ), h ≥ f on ∂U , n o f∗ := sup h|∂U : h ∈ H(U ), h ≤ f on ∂U . (The lattice operations are with respect to the natural order in H + −H + .) Proposition 13.110. Let f ∈ `∞ (∂U ). Then D∗ f = Hf ∗ ,

D∗ f = Hf∗ .

Proof. Let g := inf{h ∈ H(U ) : h ≥ f on ∂U }. Then g is upper semicontinuous and thus f ∗ = g|∂U is resolutive. If h ∈ H(U ) and h ≥ f on ∂U , then h|U = H(h|∂U ) ≥ Hf ∗ . This proves ∗ D f ≥ Hf ∗ . The function D∗ f is the greatest harmonic minorant of g|U . If k is a harmonic minorant of g|U and h ∈ H(U ) majorizes f on ∂U , then k is a lower function to h|∂U . Hence k is a lower function to f ∗ . We have shown that D∗ f ≤ Hf ∗ . Since the second equality is analogous, the proof is complete. Theorem 13.111. (a) If S is a Keldysh operator on C(∂U ), then there exists an extended Keldysh operator T : `∞ (∂U ) → H (U ) such that S = T |C(∂U ) . (b) Let f ∈ `∞ (∂U ), h ∈ H (U ) and D∗ f ≤ h ≤ D∗ f . Then there exists an extended Keldysh operator T on `∞ (∂U ) such that T f = h. Proof. These are concrete applications of the abstract Hahn–Banach–Kantorovich theorem, which can be found for instance in [245], Corollary 2.5.8 and 2.5.9. Proposition 13.112. Let T be an extended Keldysh operator on `∞ (∂U ). Then D∗ f ≤ T f ≤ D∗ f, Proof. This is obvious.

f ∈ `∞ (∂U ).

540


Definition 13.113 (Keldysh functions). A function f ∈ `∞ (∂U ) is said to be a Keldysh function if T1 f = T2 f for each pair T1 , T2 of extended Keldysh operators. For f ∈ C(∂U ) this means that T1 f = T2 f for each pair T1 , T2 of Keldysh operators. Theorem 13.114. Let f ∈ `∞ (∂U ). Then f is a Keldysh function if and only if the set {z ∈ ∂U : f∗ (z) < f ∗ (z)} is negligible. Proof. Denote M = {z ∈ ∂U : f∗ (z) < f ∗ (z)}. If M is negligible, then Hf∗ = Hf ∗ , and thus by Propositions 13.110 and 13.112, each extended Keldysh operator maps f to D∗ f = D∗ f . Conversely, assume that M is not negligible. Then referring to Propositions 13.110 and 13.112 again, we obtain that there exists x ∈ U such that D∗ f (x) < D∗ f (x). By Theorem 13.111, there exist extended Keldysh operators T1 and T2 such that T1 f = D∗ f and T2 f = D∗ f . Hence f is not a Keldysh function.

13.5.B

Uniqueness for the Laplace equation

In this subsection we will prove uniqueness for the generalized Dirichlet problem in case of the Laplace equation. We will even show that condition (a) of Definition 13.102 may be weakened to the condition that the operator A is increasing, and hence linearity does not play an essential role. Theorem 13.115. There exists exactly one operator A : C(∂U ) → H (U ) satisfying (a’) Af1 ≤ Af2 , whenever f1 , f2 ∈ C(∂U ) satisfy f1 ≤ f2 , (b) A(h|∂U ) = h|U , whenever h ∈ H(U ) . Proof. We already know that the PWB solution f 7→ Hf , f ∈ C(∂U ), is an operator satisfying (a’) and (b) (cf. Corollary 13.60). Let A fulfil conditions (a’) and (b). Pick f ∈ C(∂U ). Our aim is to show that Af = Hf . Since Af and Hf are bounded harmonic functions on U , in view of Theorem 13.24 it suffices to verify that lim Af (x) = lim Hf (x) for any z ∈ ∂ reg U.

x→z

x→z

To this end, fix z ∈ ∂ reg U . Applying Theorem 13.35 and the Keldysh lemma (Proposition 13.103) we get a function h ∈ H(U ) such that h(z) = 0 and h > 0 elsewhere on U . Let ε > 0 and let a neighborhood V of z be chosen in such a way that f ≤ ε+f (z) on ∂U ∩ V . There exists β > 0 such that f ≤ ε + f (z) + βh on ∂U \ V . If g(x) := ε + f (z) + βh(x) for

x ∈ ∂U,

then f ≤ g on ∂U and Ag = ε + f (z) + βh on U.


541

Conditions (a’) and (b) imply Af ≤ Ag = ε + f (z) + βh on U. Hence lim sup Af (x) ≤ ε + f (z) + β lim h(x) = ε + f (z). x→z

x→z

An analogous reasoning yields lim inf Af (x) ≥ f (z) − ε. x→z

As ε > 0 is arbitrary, we conclude lim Af (x) = f (z) = lim Hf (x),

x→z

x→z

as needed. Corollary 13.116 (Keldysh theorem). There exists exactly one Keldysh operator on U. Proof. Every Keldysh operator A satisfies (a’) from Theorem 13.115. Corollary 13.117. For f ∈ C(∂U ) define Sf (x) := inf {s ∈ S(U ) : s ≥ f on ∂U } ,

x ∈ U.

Then Sf = Hf for every f ∈ C(∂U ). Proof. Given f ∈ C(∂U ), the restriction of the set {s ∈ S(U ) : s ≥ f on ∂U } on U forms a saturated family, and thus Sf ∈ H (U ). Obviously, the operator A : f 7→ Sf,

f ∈ C(∂U ),

satisfies (a’) and (b) from Theorem 13.115. Remark 13.118. An independent proof was given in Proposition 13.38. Corollary 13.119. Let U ⊂ Rd be a bounded open set and f ∈ C(U ). Then the PDE and PWB solutions of the Dirichlet problem on U with boundary data f coincide. Remark 13.120. In Section 13.4 we have already observed that there are several ways how to produce the solution operator H (for example, balayage method, PWB method, Wiener method, Cornea method). The Keldysh theorem offers the possibility to show that all these methods lead to the same result. The corollaries above show that this tool may be efficient. However, in some other situations, direct methods may be simpler. Also, a result obtained via the Keldysh theorem is not valid in the full generality of harmonic spaces.

542

13.5.C


Keldysh theorems in parabolic and axiomatic potential theories

In this section in general, U will be a relatively compact open subset of a Bauer harmonic space Y . We start, however, with an example concerning the heat equation. Example 13.121. Consider the heat equation on Rd+1 . Let U ⊂ Rd+1 be a bounded open set and (ξ, τ ) in U . Then the set N := {(x, t) ∈ ∂U : t > τ } is of µU (ξ,τ ) -measure zero. (The physical interpretation of this fact is that the solution of the Dirichlet problem does not depend on the future.) Indeed, if f ∈ C(∂U ) is such that f (x, t) = 0 for t ≤ τ and f (x, t) > 0 for t > τ , using Corollary A.186 we easily construct an upper function u to f such that u(ξ, τ ) = 0. This implies that Hf (ξ, τ ) = 0 and thus the set N = {(x, t) ∈ ∂U : f (x, t) > 0} is negligible. Consequently, if U ⊂ Rd × (−∞, T ), then the set {(x, t) ∈ ∂U : t = T } is negligible. Example 13.122. Recall that both the PWB operator H and the essential solution operator D are Keldysh operators on U . These operators can be really different, see Example 13.44. Definition 13.123 (Keldysh sets). We say that U is a Keldysh set if there is exactly one Keldysh operator on U . Remark 13.124. Example 13.122 shows that U can fail to be a Keldysh set. Definition 13.125 (Keldysh measures). If A is a Keldysh operator on U and x ∈ U , then the mapping f 7→ Af (x), f ∈ C(∂U ), is a positive Radon measure on ∂U which is called the associated Keldysh measure at x and denoted by αx . Lemma 13.126. Let A be a Keldysh operator on U and αx be its associated Keldysh c β(U c ) measure at x ∈ U . If w ∈ W(H(U )), then εx (w) ≤ αx (w) ≤ εU x (w). Proof. Let w = h1 ∧ · · · ∧ hn , where h1 , . . . , hn ∈ H(U ). Then, for each j ∈ {1, . . . , n}, hj = Ahj ≥ Aw, so that w ≥ Aw on U . By the maximality of β(U c ) β(U c ) β(U c ) εx (Theorem 13.41), εx ≺H(U ) αx , and thus εx (w) ≤ αx (w) (cf. Definition 3.19). On the other hand, since lim sup Aw(x) ≤ w(z), x→z, x∈U

Aw is a lower function to w and we have c

αx (w) = Aw(x) ≤ Hw(x) = εU x (w).


543

Proposition 13.127. If ∂ irr U is negligible, then ∂ reg U = ∂ess U . Proof. Let W = (β(U c ))c . Given p ∈ P, we claim that bb(U c ) = R bU c . R p p

(13.19)

∗

To this end, let u ∈ H + (Y ) satisfy u ≥ p on b(U c ). By Theorem 13.5, u ≥ p on ∂U ∩ b(U c ) = ∂ reg U , so that by the assumption u ≥ p holds µx -almost everywhere for each x ∈ U . Thus bpU c on U. u ≥ Hu ≥ Hp = R Taking infimum with respect to u and using Theorem A.195, we obtain bpb(U c ) = Rpb(U c ) ≥ R bpU c R

on U.

Since U is open, the converse inequality bb(U c ) ≤ R bU c R p p is obvious, so that (13.19) holds on U . By Lemma A.224, U is finely dense in W . By Proposition A.209, it follows that (13.19) holds on W , whereas (13.19) holds on W c ⊃ b(b(U c )). Indeed, W c = β(U c ) = b(β(U c )) ⊂ b(b(U c )) ⊂ b(U c ), c

bpb(U ) = p = R bpU c on W c . We have verified that (13.19) holds on Y for each hence R p ∈ P. If z ∈ b(U c ) and p ∈ P, then c

c

bU (z) = R bb(U ) (z), p(z) = R p p and thus z ∈ b(b(U c )). The converse inclusion is always true, so that b(U c ) = b(b(U c )). By Proposition A.223, b(U c ) = β(U c ). Using Proposition 13.10 and Theorem 13.5 we infer that ∂ reg U = ∂ess U . Theorem 13.128. The following conditions are equivalent: (i) U is a Keldysh set, (ii) ∂U \ ∂ess U is negligible, (iii) ∂ irr U is negligible, (iv) D = H on C(∂U ).

544


Proof. By Theorem 13.41, ∂ess U = ChH(U ) (U ). (i) =⇒ (ii): Suppose that U is a Keldysh set. By Theorem 3.42 and Proposition 3.43, there exists a function f ∈ C(U ) such that ChH(U ) (U ) = {x ∈ U : f∗ (x) = f ∗ (x)}, where f ∗ and f∗ are as in Definition 13.109. Since U is a Keldysh set, f is a Keldysh function and by Theorem 13.114, the set ∂U \ ∂ess U ⊂ {x ∈ ∂U : f∗ (x) 6= f ∗ (x)} is negligible. (ii) =⇒ (iii): This is obvious, as ∂ess U ⊂ ∂ reg U . (iii) =⇒ (ii): Suppose that ∂ irr U is negligible. Then by Proposition 13.127, ∂ reg U = ∂ess U , but using (iii) again we obtain that ∂U \ ∂ess U is negligible. (ii) =⇒ (iv): We will show that all functions from C(∂U ) are Keldysh functions. We choose f ∈ C(U ). By Theorem 3.24, {x ∈ ∂U : f∗ (x) 6= f ∗ (x)} ⊂ ∂U \ ChH(U ) (U ) = ∂U \ ∂ess U and this set is negligible by (ii). From Theorem 13.114 we obtain that f is a Keldysh function. (iv) =⇒ (i) Let A be a Keldysh operator. If (iv) holds, Lemma 13.126 yields that Dw ≤ Aw ≤ Hw = Dw for each w ∈ W(H(U )). By the Stone–Weierstrass theorem (Theorem A.30), A = H on C(∂U ). Recall that for the purpose of the Wiener solution, there does not exist a regular exhaustion in general (Example 13.87). Although not every set in parabolic potential theory is a Keldysh set, we will see that we can take profit from the fact that, from some point of view, there is “enough” of Keldysh sets. Proposition 13.129. Let K be a compact subset of U . Then there exists a Keldysh set V such that K ⊂ V ⊂ V ⊂ U . Proof. Fix a continuous function f on Y with compact support contained in U such that 0 ≤ f ≤ 1 and f = 1 on K. Let Us := {x ∈ Y : f (x) > s}, s ∈ (0, 1). Obviously, Us are open sets and K ⊂ Ub ⊂ Ub ⊂ Ua ⊂ U whenever 0 < a < b < 1. Let P 0 be a countable subset of P such that P 0 − P 0 is dense in C(U ) and S ⊂ Y be a dense set. Consider the countable family of functions {gp,y : p ∈ P 0 , y ∈ S}, where gp,y : (0, 1) → R is defined as bpUsc (y), gp,y (s) := R

s ∈ (0, 1).


545

The functions gp,y are increasing, and thus they are continuous except for a countable subset of (0, 1). Fix a level b ∈ (0, 1) such that all gp,y are continuous at b. If 0 < a < b, then (Ua )c ⊂ (Ub )c ⊂ β((Ub )c ) by Proposition A.223. It follows that c

c

bpUb (y) = gp,y (b) = sup gp,y (a) = sup R bpUac (y) ≤ R bpβ(Ub ) (y), R a 2c. Hence p∗ ≥ 2 on (− 21 , 12 ) × {1}. Let f be a continuous function on ∂U such that 0 ≤ f ≤ p∗ and f (0, 1) ≥ 2. By Theorem 13.111 there exists a Keldysh operator A such that

546


Ap = Hp∗ . Suppose that Ap ≤ p on U . Let {tk } be a sequence of real numbers such that tk < 2 and tk & 1. Then Hf (0, tk ) ≤ Hp∗ (0, tk ) = Ap(0, tk ) ≤ p(0, tk ). The barrier

( s(x, t) =

x2 2

+ t − 1,

1,

t > 1, t 0 such that ∞ k=1 λk pk ∞ converges uniformly on U and is finite on Y . Then p := k=1 λk pk is a potential by Proposition A.170. We find a sequence {zn } of points of U such that bU c (z) = lim R bU c (zn ). R p p n→∞

Fix k ∈ N. Then p − λk pk is a potential and thus bU c (zn ) − R bU c (zn ) = lim sup −R bU c lim sup λk R pk p p−λk pk (zn ) n→∞

n→∞

bU c bU c = − lim inf R p−λk pk (zn ) ≤ −Rp−λk pk (z) n→∞

bpU c (z) − R bpU c (z). = λk R k

13.6 Exercises

547

Hence, for any k ∈ N we have bU c (zn ) ≤ lim R bU c (zn ) + lim sup λk R bU c (zn ) − R bU c (zn ) lim sup λk R pk p pk p n→∞

n→∞

n→∞

b (z) + λk R b (z) − R bU c (z) = λk R bU c (z). ≤R p pk p pk Uc

Uc

Since the inequality c

c

bpU (zn ) ≥ R bpU (z) lim inf R k k n→∞

is trivial, we obtain bU c (zn ) = R bU c (z), lim R pk pk

n→∞

k ∈ N.

It follows that c

c

U εU z = lim εzn n→∞

in M(U ).

Exercise 13.134. Let U ⊂ Rd be a bounded open set. Then ∂ reg U , with respect to the Laplace equation, is dense in ∂U . Hint. Let z0 ∈ ∂U and r > 0. We find a point x ∈ U (z0 , r) \ U and a point z ∈ ∂U which minimizes |y −x| over y ∈ U . Then z ∈ ∂ reg U according to Example 13.18(a).

Exercise 13.135 (Semiregular sets, points and sequences). Let U ⊂ Y be relatively compact and open. A point z ∈ ∂U is termed semiregular if z is irregular and if there exists a limit limx→z, x∈U Hf (x) for any f ∈ C(∂U ). A sequence {xn } of points of U Uc converging to z ∈ ∂U is called semiregular if z ∈ ∂ irr U and µU xn → εz . We say that the set U is semiregular if the Perron–Wiener–Brelot solution Hf can be extended continuously to U for any function f ∈ C(∂U ). Prove that the set U is semiregular if and only if any point of ∂U is either regular or semiregular. Exercise 13.136. We use the terminology of Exercise 13.135. Let U ⊂ Y be relatively compact and open. Let z ∈ ∂U . The following statements are equivalent: (i) z is semiregular, (ii) there exists f ∈ C(∂U ) such that lim supx→z, x∈U Hf (x) < f (z), (iii) there exists a relative neighborhood of z in ∂U which is negligible, (iv) each sequence {xn } of points of U converging to z is semiregular, (v) each sequence {xn } of points of U converging to z violates limn→∞ µU xn = εz .

548


Hint. The implication (i) =⇒ (ii) is trivial. (ii) =⇒ (iii): We find ε > 0 and an open neighborhood V of z such that Hf < f (z) − 2ε on V ∩ U and f > f (z) − ε on V ∩ ∂U . Let g be a continuous function on ∂U with support in V such that 0 ≤ g ≤ ε and g(z) = ε. Then each lower function to f is also a lower function to f − g and thus Hg = Hf − H(f − g) = 0. It follows that {y ∈ ∂U : g(y) > 0} is negligible. (iii) =⇒ (iv): Let V be a regular open set containing z such that V ∩ ∂U is negli∗ gible and set W := U ∪ V . Choose p ∈ P and set g := p(1 − cV ). If u ∈ H + (Y ) is bpW c ≥ H U g such that u ≥ p on W c , then u is an upper function to g on U and thus R on U . Since V ∩ ∂U is negligible, we have H U g = H U p. Since the converse inbU c = R bW c on U , next, passing to the limes equality is easy, we obtain the equality R p p bW c is continuous at z, we easily conclude that inferior, also on ∂U ∩ V . Since R p Uc µU x → εz as x → z, x ∈ U . The implication (iv) =⇒ (v) is obvious. (v) =⇒ (ii): Let p be as in the hint to Exercise 13.133. Suppose that (ii) is false. Then there exists a sequence {xn } converging to z such that lim supn→∞ Hp(xn ) = p(z). Similarly as in Exercise 13.133 we deduce that µU xn → εz , so that (v) is false. Finally, (iv) =⇒ (i) is obvious. Exercise 13.137. Let U ⊂ Y be a relatively compact open set. For any x ∈ U , the harmonic measure µU x is carried by the set of all nonsemiregular points. The set of all semiregular points is negligible and open in ∂U . Hint. Use Exercise 13.136. Exercise 13.138. A relatively compact open set U is semiregular if and only if ∂ irr U is negligible and open in ∂U . Hint. Use Exercises 13.136 and 13.135. Exercise 13.139. Consider the Laplace equation on Rd . Let V ⊂ Rd be a regular set and U ⊂ V be an open set such that V \ U is polar. Then U is semiregular and the points of ∂U ∩ V are semiregular boundary points for U . Conversely, for each semiregular set U there exists a regular set V ⊃ U such that V \ U is polar. Hint. The first assertion follows from the fact that polar sets are negligible. For the second one, set V = U ∪ ∂ irr U . Then U = V . From Exercises 13.134 and 13.138 we deduce that V is open. By Theorem 13.19, V \ U is polar. By Propositions A.211 and A.203(c), b(U c ) = b(V c ) ∪ b(U c \ V c ) = b(V c ). Hence, using Theorem 13.5 we show that each boundary point of V is regular for V and thus V is regular. Exercise 13.140. Consider the Bauer harmonic space associated with the heat equation on Rd+1 . If U ⊂ Rd+1 is a cylinder of type V × (0, T ), where V ⊂ Rd is regular (a ball, for example), then U is semiregular. The Dirichlet problem for a cylinder is

13.6 Exercises

549

the most typical for the heat equation. Thus, semiregular sets appear already in potential theory of the Laplace equation, but for the heat equation they are of particular interest. Exercise 13.141. Consider potential theory of the Laplace equation on Rd . Let U ⊂ Rd be a bounded open set. (a) If z ∈ ∂ irr U , then lim r1−d Hd−1 (S(z, r) \ U ) = 0,

r→0+

where Hd−1 is the (d−1)-dimensional Hausdorff measure. (b) If d = 2 and z ∈ ∂ irr U , then there are arbitrarily small circles centered at z which are contained in U . Hint. Consult [21], Corollary 7.2.4, Theorem 7.3.9 and Theorem 7.5.1. Exercise 13.142. (a) For the Laplace equation: Let U ⊂ Rd be bounded and open. Suppose that f ∈ C(∂U ) and h ∈ H (U ) is bounded. Then h = Hf if and only if for each z ∈ ∂ reg U we have lim h(x) = f (z). x→z, x∈U

(b) Prove that an analogy to assertion (a) fails for the heat equation. (c) In general Bauer harmonic spaces: Let U ⊂ Y be relatively compact and open. Suppose that f ∈ C(∂U ) and h ∈ H (U ) is bounded. Then h = Hf if and only if for each z ∈ ∂U and regular sequence {xn } converging to z we have lim h(xn ) = f (z).

n→∞

Hint. (a) Use Theorem 13.23. (b) Let U be as in Example 13.44. Consider a function f ∈ C(∂U ) such that Hf 6= Df and h = Df . (c) For the “if” part, consider a sequence {fk } of continuous functions which is dense in the unit ball of C(∂U ). Choose x ∈ U and ε >P 0. Find upper functions uk to −k fk such that uk (x) − Hfk (x) < 2 ε. Then u := h + ∞ k=1 (uk − Hfk ) is an upper function to f . Indeed, let xn → z be such that lim u(xn ) = L := lim inf u(x).

n→∞

x→z, x∈U

We will show that L ≥ f (z). Passing to a subsequence we may assume that there exists a limit limn→∞ µxn in M1 (∂U ). If {xn } is regular, then L = lim u(xn ) ≥ lim h(xn ) = f (z). n→∞

n→∞

550


Otherwise there exists τ > 0 such that the set n o F := g ∈ C(∂U ) : kgk < 1, lim Hg(xn ) < g(z) − τ n→∞

is nonempty (and open). The set I of all indices k ∈ N with fk ∈ F is infinite. For each k ∈ I we have lim inf(uk (xn ) − Hfk (xn )) ≥ fk (z) − lim Hfk (xn ) > τ. n→∞

n→∞

It follows that this case is also safe as L = lim u(xn ) = ∞ > f (z). n→∞

We have verified that u is an upper function to f and hence Hf (x) ≤ h(x) + ε. Exercise 13.143 (Abstract Dirichlet problem). In this series of exercises we abandon the localizable structure of the Bauer harmonic space. For the Perron-like method on a single set U we extract what is really needed. We will show that this setting covers also the model of concave functions on a simplex, although the decomposition of the space to “interior part and boundary part” does not seem to be as natural there as in potential theoretic models. Let X be a locally compact space with a countable base and U a relatively compact subset of X. We assume that U(U ) is a convex cone of lower finite extended realvalued functions on U satisfying the following axioms: Minimum principle: If u ∈ U(U ), lim inf u(x) ≥ 0

x→z,x∈U

for each

z ∈ U \ U,

then u ≥ 0 on U . Completeness axiom: U+ (U ) is closed under countable sums. By analogy with the classical theory of harmonic functions we define for any extended real-valued function f on ∂U the upper class U (f ) as U (f ) := u ∈ U(U ) : u is lower bounded on U and lim inf u(x) ≥ f (z) for any z ∈ U \ U . x→z,x∈U

Define Hf and Hf by Hf := inf U (f ) and

Hf := −H(−f ).

By Minimum principle, Hf ≤ Hf on U . A function f is resolutive if Hf = Hf and if this common value, which we denote simply by Hf , is finite on U . The function Hf is called the Perron–Wiener–Brelot solution, briefly PWB solution of the Dirichlet problem for f on U . We denote by R(U ) the set of all real-valued resolutive functions on U . As the first task of this series of exercises, prove the following assertions.

13.6 Exercises

551

(a) If f1 , f2 , f are extended real-valued functions on ∂U , f1 , f2 are resolutive and f = f1 + f2 on the set of all points where this sum makes sense, then f is resolutive and Hf = Hf1 + Hf2 . P∞ (b) If {fk } is a sequence of positive resolutive functions and f := k=1 fk , then Hf = Hf . Hint. The part (a) is easy. For (b), use the method of the proof of Theorem 13.61. Exercise 13.144. The following models satisfy the axioms introduced in Exercise 13.143. (a) A sample example consists of a relatively compact open subset U of a Bauer ∗ harmonic space X, on which we consider as U(U ) the convex cone H (U ) of all hyperharmonic functions on U . See also Exercise 13.149. (b) U is a finely open subset of a Bauer harmonic space and U(U ) the family of all finely hyperharmonic functions, or, alternatively, the family all lower semicontinuous finely hyperharmonic functions. For definition, see Exercise 13.152. (c) X is a metrizable compact convex set in a locally convex space, U = X \ ext X and U(U ) is the family of all lower semicontinuous concave functions on U , see Exercises 13.150, 13.151. Exercise 13.145. Under the setting of Exercise 13.143, the set U is said to be resolutive (with respect to U(U )) if any function from C(∂U ) is resolutive. If U is resolutive and x ∈ U , then the Riesz representation theorem A.72 provides a unique Radon measure µx on ∂U such that Z f dµx for any f ∈ C(∂U ). Hf (x) = ∂U

We label µx as the harmonic measure at x. Let U be a resolutive set and let f be a lower semicontinuous function on ∂U . Prove that for any x ∈ U , µx (f ) = Hf (x). Hint. We may assume that f ≥ 0. Since the space X has a countable base, there exists an increasing sequence {fn } of functions from C(∂U ) such that fn % f . Using the Lebesgue monotone convergence theorem and Exercise 13.143(b) to the differences fn+1 − fn we obtain µx (f ) = lim µx (fn ) = lim Hfn (x) = Hf (x). n→∞

n→∞

Exercise 13.146. Under the setting of Exercise 13.143, let U be a coanalytic resolutive set (Exercise 13.145). Then the harmonic measure µx is carried by U \ U for any x ∈ U.

552


Hint. It is enough to show that µx (K) = 0 for any compact set K ⊂ U . Then the characteristic function cK is upper semicontinuous and cK = 0 on U \ U . In light of the preceding Exercise 13.145 we conclude that µx (K) = µx (cK ) = −H(−cK )(x) = 0.

Exercise 13.147. Under the setting of Exercise 13.143, let U be a coanalytic resolutive set (Exercise 13.145). Let f be an arbitrary extended real-valued function on ∂U . Then Z ∗ Hf (x) = f dµx for any x ∈ U. ∂U

Hint. Fix f and x ∈ U . For each function u from U (f ), let u b be defined on ∂U by u b(z) := lim inf u(y), y→z,y∈U

z ∈ ∂U.

Then u b is a lower semicontinuous function on ∂U and u ∈ U (b u). By Exercise 13.146, f ≤ u b holds µx -almost everywhere. Let S denote the family of all lower semicontinuous functions s on ∂U such that s ≥ f on U \ U . Then inf Hs : s ∈ S ≤ inf H u b : u ∈ U (f ) ≤ inf U (f ) = Hf ≤ inf Hs : s ∈ S . Having in mind Exercise 13.145, we get Hf (x) = inf Hs(x) : s ∈ S

= inf {µx (s) : s ∈ S } =

Z

∗

f dµx . ∂U

Exercise 13.148. Under the setting of Exercise 13.143, let U be a coanalytic resolutive set (Exercise 13.145). Let N ⊂ ∂U be a negligible set (this means, analogously as in Definition 13.63, that µx (N ) = 0 for each x ∈ U ). Let u ∈ U(U ) be lower bounded. Suppose that lim inf u(x) ≥ 0, z ∈ U \ (U ∪ N ). x→z, x∈U

Then u ≥ 0 on U . Hint. Pick a P point x ∈ U . Given ε > 0, find P∞a sequence {un } of upper functions to ∞ cN such that n=1 un (x) < ε. Then u + P n=1 un is an upper function to 0 and thus by Minimum principle axiom, 0 ≤ u + ∞ n=1 un , in particular u(x) > −ε.

13.6 Exercises

553

Exercise 13.149. Let U be an open relatively compact subset of a Bauer harmonic ∗ space. Then the family H (U ) satisfies both Minimum principle and Completeness axiom of Exercise 13.143. Moreover, the set U is resolutive (in the sense of Exercise 13.145). ∗

Hint. By Theorem A.161, the family H (U ) satisfies Completeness axiom while Minimum principle is guaranteed by Theorem A.160. The resolutivity follows from Theorem 13.60. Exercise 13.150. Let X be a compact convex set in a locally convex space and f be a lower semicontinuous concave function on X \ ext X. If lim infy→x f (y) ≥ 0 for every x ∈ ext X, then f ≥ 0 on X. Hint. If

 f (x), fb(x) := lim inf f (y), y→x

x ∈ X \ ext X, x ∈ ext X,

then fb is lower semicontinuous on X. Assume that m := min fb(X) < 0 and set M := {x ∈ X : fb(x) = m}. The set M is nonempty and closed. It is also extremal. Indeed, if λx + (1 − λ)y ∈ M , where x, y ∈ X and λ ∈ (0, 1), then fb = m on the open segment (x, y). Hence x, y ∈ X \ ext X by the assumption. We see that fb is concave on the closed segment [x, y], and therefore x, y ∈ M . Since M is extremal and M ∩ ext X = ∅, we arrive at a contradiction to Proposition 2.20. Exercise 13.151. Let X be a metrizable Choquet simplex, U = X \ ext X and U(U ) be the family of all lower semicontinuous concave function on U . Then the axioms of Exercise 13.143 are satisfied and any bounded continuous function on ext X is resolutive (in the sense of Exercise 13.145). Hint. For Minimum principle see Exercise 13.150. Let f ∈ C b (ext X) and δx be the unique maximal measure in Mx (X). Define T f : x 7→ δx (f ), x ∈ X. We show that T f ≥ Hf . Fix x ∈ X and find a sequence {Kn } of compact subsets of ext X such that δx (Kn ) → 1. If ( f on Kn , vn := kf k on X \ Kn , then vn is a lower semicontinuous concave function on X. The function un := (vn )∗ |X\ext X is an upper function to f and, moreover un (x) = δx (vn ) → δx (f ) = T f (x) (see Lemma 6.4). Hence Hf (x) = inf U (f )(x) ≤ inf{un (x) : n ∈ N} = T f (x). Analogously, Hf (x) ≥ T f (x). Since Hf ≤ Hf , we have T f = Hf = Hf .

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Exercise 13.152. Let Y be a Bauer harmonic space and U ⊂ Y be a finely open set. We say that a function u : U → (−∞, ∞] is finely hyperharmonic on U is for each x ∈ U and each relatively compact finely open set V with x ∈ V ⊂ V ⊂ U we have Z c u dεVx ≤ u(x) V

(in particular, we require the integral on the left-hand side to be well defined). Let ∗ ∗ H (U ) be the set of all finely hyperharmonic functions on U and H lsc (U ) be the set of all lower semicontinuous finely hyperharmonic functions on U . If U is relatively ∗ ∗ compact and coanalytic, then both H (U ) and H lsc (U ) satisfy the axioms of Exercise 13.143. Moreover, in both cases the set U is resolutive (in the sense of Exercise 13.145) and the PWB solution of the Dirichlet problem with continuous boundary data f is just H U f . Hint. Consult [320], Section 13.A. Exercise 13.153. Let U ⊂ Y be an open set and u be a superharmonic function on U . c Let V be a relatively compact set such that V ⊂ U and x ∈ V . Then εVx (u) ≤ u(x). Hint. As in the proof of Theorem A.198 we show that there exists a sequence {Vn } of c Vc relatively compact open subsets of Y such that x ∈ Vn ⊂ V n ⊂ U and εxn → εVx . c V By Theorem A.194, εxn (u) ≤ u(x). Hence we may pass to the limit and conclude c that εVx (u) ≤ u(x). Exercise 13.154. Let U ⊂ W be finely open subsets of a Bauer harmonic space Y and suppose that W \ U is semipolar. Let u be a lower semicontinuous function on W . Suppose that u is finely continuous (in the extended sense) and u|U is finely hyperharmonic on U (Exercise 13.152). Then u is finely hyperharmonic on W . Hint. Consider the largest finely open set V such that u is finely hyperharmonic on V ∩ W and prove that V c ⊂ b(V c ). Use Proposition A.223 to show that V c ⊂ β(U c ) ⊂ W c . For details we refer to [320], Theorem 12.20. Exercise 13.155. Recall the definition of fine hyperharmonicity (Exercise 13.152). Finely superharmonic means just finely hyperharmonic and finite on a finely dense set. A function h is called finely harmonic if h and −h are finely superharmonic. Let e ) the family of all U ⊂ Y be a relatively compact finely open set. We denote by S(U e ) the family of functions s ∈ C(U ) which are finely superharmonic on U and by H(U e ) = H(U ). all functions h ∈ C(U ) which are finely harmonic on U . Prove that H(U Then, using the result by J. Bliedtner and W. Hansen that e ) = H(U e ) + P| S(U U e ) = S(U ). (see [66], Proposition 9.1), prove that S(U

(13.20)


555

e ) and H(U ) ⊂ H(U e ) can be obtained from ExHint. The inclusions S(U ) ⊂ S(U c c e ercise 13.153. Set W = (β(U )) . If h ∈ H(U ), then by Exercise 13.154, h is finely harmonic on W . From Exercise 13.152 we infer that H W h = h. By Theorem 13.42, h ∈ H(U ). The inclusion S(U ) is then easy once we know the (deep) formula (13.20). Exercise 13.156. Let U ⊂ Y be a relatively compact open set and V be a countable base of topology on U consisting of regular sets V with V ⊂ U . Consider a sequence {Vn } of elements of V which repeats each set from V infinitely many times. Define the sequence of operators Tn : C(U ) → C(U ) recursively: T0 f = f, c

Tn f (x) = εVx (Tn−1 f ),

x ∈ U, n ∈ N.

Prove that (Tn f )|U → Hf for each f ∈ C(U ). Hint. Use the density of P − P in C(U ). If p ∈ C(U ), then the sequence {Tn p} decreases to the greatest harmonic minorant of p in U , which is Hp by Proposition 13.3. Exercise 13.157. If ∂ irr U is negligible, then for any x ∈ ∂ reg U there exists a function h ∈ H+ (U ) such that h(x) = 0 and h > 0 on U \ {x}. Hint. This follows from Proposition 13.127, Theorem 13.41 and Proposition 13.103. Exercise 13.158. Let U be a Keldysh set and f ∈ `∞ (∂U ). Then f is a Keldysh function if and only if the set of all discontinuity points of f is negligible. Hint. By Theorem 13.114, f is a Keldysh function if and only if the set M := {x ∈ ∂U : f∗ (x) < f ∗ (x)} is negligible. But M contains the set Z of all discontinuity points of f and M \ Z ⊂ ∂U \ ∂ess U . Since U is a Keldysh set, the set ∂U \ ∂ess U is negligible (Theorem 13.128).

13.7

Notes and comments

Potential theory has a long and interesting history. We give several references where further sources can be found: H. Bauer [41], [43], [41], M. Brelot [89], [88], [90], I. Netuka [362]; see also commentaries in D. H. Armitage and S. J. Gardiner [21], J. Bliedtner and W. Hansen [66], M. Brelot [87], C. Constantinescu and A. Cornea [123], J. L. Doob [145] and L. L. Helms [222]. Balayage is one of the central concepts of potential theory. Its physical background has to do with the phenomenon of electrostatic influence. Mathematically it appears

556


in the work of C. F. Gauss (1840). Méthode du balayage was used by H. Poincaré for a construction of the solution of the Dirichlet problem in 1890; see [375]. Ch. de la Vallée Poussin introduced the harmonic measure in [288], [289] and showed c that the balayage εU x (for a bounded open set U ) coincides with the harmonic measure U µx ; see also O. Frostman [188]. Since about 1940, balayages studied originally for measures by means of Green potentials, was systematically investigated by M. Brelot as a kind of a dual operation bA were on superharmonic functions. The notions of the réduite RsA and balayage R s introduced by M. Brelot [84]. This opened the way to a study of the Dirichlet problem for general sets, not restricting attention only to open domains. Balayage of measures as well as of hyperharmonic functions turns out to be one of the most efficient tools not only in classical theory, but also in abstract potential theory, such as the theory of harmonic spaces, of balayage spaces, H-cones etc. Readers are referred, for instance, to H. Bauer [40], [43], J. Bliedtner and W. Hansen [66], M. Brelot [87], N. Boboc, Gh. Bucur and A. Cornea [71], C. Constantinescu and A. Cornea [123], J. L. Doob [145], I. Netuka [362]. The notion of a regular boundary point was introduced by H. Lebesgue [298]. A c characterization of regular points by means of the equality εU x = εx (Theorem 13.5) goes back to O. Frostman [189]. Essential balayage was invented by J. Bliedtner and W. Hansen in connection with their studies of simplicial cones in potential theory [62], [66]. Essential solution is only our language used to make a parallel between two extreme cases of solution of the generalized Dirichlet problem and to describe corresponding regular points. The term “weak Dirichlet problem” is used in J. Bliedtner and W. Hansen [65] in this connection. The idea of a barrier goes back to H. Poincaré [375]. Explicitly this notion appears in H. Lebesgue [298] and the relation between a barrier and regularity in Proposition 13.16 is due to G. Bouligand [78]. Example 13.18(d) of an isolated irregular boundary point is given in S. Zaremba [478]; the striking Example 13.18(e) of the Lebesgue spine appears in H. Lebesgue [297]. Wiener’s criterion mentioned at the beginning of Subsection 13.2.A is due to N. Wiener [469]. O. D. Kellog [267] and F. Vasilesco [457] proved that the set of irregular points is of capacity zero (which here means polar) for the plane case. For the Laplace equation case (see Theorem 13.19) for R3 , the result is due to G. C. Evans [170]. For a detailed account of the early history of regular points, see F. Vasilesco [458]. Examples of bounded open sets U for which ∂ reg U is not an Fσ set were proposed by S. J. Gardiner, A. Cornea and W. Hansen in 2004 (private communication). The construction in Proposition 13.21 seems to be new. The result of Lemma 13.22 is due to G. C. Evans [171]. A similar result to that of Theorem 13.23 appears in M. G. Arsove [22]. A relation between regularity for the Laplace equation and the


557

heat equation (Example 13.25(a)) was established by I. Babuˇska and R. Výborný in [26]. For several results on regularity for the heat equation, see E. G. Effros and J. L. Kazdan [165], [166], C. L. Evans and R. F. Gariepy [172], W. Hansen [212] and I. Netuka [356]. The key result of Section 13.3 is Theorem 13.41, due to J. Bliedtner and W. Hansen [62], Theorem 3.3. Our exposition loosely follows J. Bliedtner and W. Hansen [66]. We focus our attention on a unified approach for all sets U , based on upper approximation. This covers the previous parallel treatment of compact sets and open (or, more generally, finely open) sets in the literature; on the other hand, it is more concrete than the dilation approach in [62] and [66]. We avoid fine potential theory tools there. Although local potential theory of harmonic spaces is less general than the nonlocal theory of balayage spaces, some statements are surprisingly more complicated in our local setting, due to different interpretation of domains of definition of harmonic functions. Thus, Lemma 13.28 and Theorem 13.34 seem to be new. For potential theory of the Laplace equation, we can use specific methods to simplify some arguments in the development of the theory of H(U ) and S(U ). This we present in Subsection 13.3.A, where we largely follow I. Netuka [358], but generalize the exposition to non-open sets. The idea of the example from Remark 13.36 is based on I. Netuka and J. Veselý [363]. Example 13.45 is from J. Köhn and M. Sieveking [275]. In this book, we focus our attention on cones of continuous functions. Nevertheless, in connection with potential theory, cones of lower semicontinuous functions are also sometimes investigated. In [63], J. Bliedtner and W. Hansen introduce the cone H∗ (U ) of all lower semicontinuous functions v on U , hyperharmonic on U and satisfying lim inf v(x) = v(z) for any z ∈ ∂U, x→z,x∈U

and show that the Choquet boundary for H∗ (U ) is the set of all non-semiregular boundary points of U . The cone H∗ (U ) is simplicial, but not geometrically simplicial in general (these notions thus can really differ for cones of discontinuous functions). Note that the cone H∗ (U ) is not convex. Results of Subsection 13.3.C are taken from W. Hansen and I. Netuka [214], where the topic is studied in more detail and in the framework of harmonic spaces. The relation between the notion of H(U )-concavity and the notion of superharmonicity (see Example 13.49) has been clarified in I. Netuka and J. Veselý [363]. In [214], W. Hansen and I. Netuka derived the following theorem for strong Brelot spaces. Theorem. Let U be a relatively compact open connected set and s a lower bounded function on U that is not identically ∞ . Then the following assertions are equivalent: (i) for every open set V with V ⊂ U , the restriction s|V belongs to the uniform closure of W(H(V )),

558


(ii) s = inf{h : h ≥ s, h harmonic on U } . Moreover, if s is an H(U )-concave function which is finite at some point of U , then the restriction s|U satisfies (ii). Theorem 13.54 and its Corollary 13.55 are taken from I. Netuka [361]. It is shown there that the separation is impossible in general even if we suppose that a strict inequality between H(U )-convex and H(U )-concave function is assumed. It should be mentioned that the inequality h(x) ≥ ϕ(|x|) from Lemma 13.52 is well known; see M. Brelot [87], p. 203. The material of Section 13.4 summarizes several methods of solution of the Dirichlet problem. This problem has a long and interesting history. Various methods for its solution were discovered before 1920 but none of them applied to the case of general domains (although nowadays we know that, for example, the balayage method is well adapted to work for general domains). The first method requiring no limitations on the shape of the boundary is due to O. Perron [372] and R. Remak [386]. In [468], N. Wiener defined a completely different type of solution and showed in [470] that his solution coincides with that of Perron, thus establishing the resolutivity of continuous functions (Theorem 13.60). The question of whether such a solution of the generalized Dirichlet problem is the only bounded harmonic function having preassigned values at each regular point was settled later (cf. Exercise 13.142). A general treatment of the Perron method in the case of classical potential theory is due to M. Brelot [83] who considers arbitrary boundary data and lower semicontinuous functions. His main result, the so-called Brelot resolutivity theorem (see Theorem 13.61), shows that resolutivity is equivalent to integrability with respect to harmonic measures. For the case of open sets in axiomatic theories of harmonic functions, Theorem 13.61 is due to H. Bauer [39]. It could be of interest to the reader to mention how M. Brelot was actually led to the question of resolutivity. In [470], N. Wiener notes that in the simplest cases of discontinuous boundary values, the Perron generalization of the Dirichlet problem is still satisfactory. He continues: However, we do not need to go much further to find an instance where the Perron generalization breaks down. For example, let G be the interior of a circle with boundary R. Let us establish a system of polar coördinates with the centre of G in pole. Let f (P ) be 1 for those points P of R with a θ-coördinate rational in terms of 2π, and let f (P ) be 0 elsewhere on R. Clearly any upper function of f is at least 1 and no lower function is greater than 0. Hence the Perron method yields us here no unique generalization of the Dirichlet problem. M. Brelot told of this example in 1973: All mathematicians (including himself) had shared N. Wiener’s belief that the Perron approach had been of no good for general discontinuous boundary data. This fact was frequently reproduced among specialists in potential theory (needless to say that N. Wiener was an authority in mathematical circles). It was one day in 1937, when M. Brelot was slowly walking to the Sorbonne to lecture on the Dirichlet problem and he also intended to mention Wiener’s example. Thinking on


559

upper functions (which are, of course (!), at least 1), he suddenly stopped his flow of ideas by saying to himself: Am I really able to establish this? It did not take long time to realize that the upper solution is in fact 0, and this moment opened the way to Brelot’s resolutivity investigations. The exposition of Subsection 13.4.B is based on papers [124] and [125] by A. Cornea. In Subsection 13.4.C we present the Wiener method and consider its generalizations. It was observed by H. Bauer ([40], p. 147; cf. Example 13.87) that in the case of potential theory of the heat equation, one cannot, in general, insert a regular set in between a compact set and an open set. Example 13.122 is taken from J. Lukeˇs [314] and Example 13.45 from the paper by J. Köhn and M. Sieveking [275]. In [190], B. Fuglede developed the so-called fine potential theory based on the notion of finely harmonic functions on finely open sets. In general harmonic spaces, first J. Bliedtner and W. Hansen [62], [64] considered cones of continuous finely superharmonic functions, and then J. Lukeˇs and J. Malý [318] proposed the foundations of “full” fine potential theory. This has been systematically studied by J. Lukeˇs, J. Malý and L. Zaj´ıcˇ ek in [320]. Notice that the minimum principle for Fuglede’s finely hyperharmonic functions is based on the so-called quasi-Lindelöf property of the fine topology, while in the “full” setting it is based on the Lusin–Menshov property 13.91 (cf. J. Lukeˇs [316] and J. Lukeˇs and J. Malý [318]). A part of this research concerning the fine Wiener method is presented in Subsection 13.4.D. The reader can consult a survey paper [191] by B. Fuglede. Solution methods of classical potential theory rely on special properties of the Laplace equation. Namely, the Poisson integral is mostly used for the solution of the Dirichlet problem on a ball. Alternatively, the method of integral equations uses knowledge of the fundamental solution. The theory of partial differential equations, however, provides methods which are more flexible without extra difficulty and cover a wide class of elliptic equations. For example, for an equation of the form − div A(x, ∇u) = 0,

(13.21)

where A : Ω×Rd → Rd is a continuous function satisfying some structure conditions, the existence of solutions can be proved by the following methods: •

• •

Riesz representation theorem in Hilbert spaces if A is the identity (so that we are again restricted to the Laplace equation); Lax–Milgram theorem if A is linear in the second variable; direct methods of the calculus of variations if (13.21) is the Euler–Lagrange equation of a functional of type Z Ju = F (x, ∇u) dx, U

which is the case if A(x, ξ) = ∇ξ F (x, ξ);

560 •


the theory of monotone operators in the general (possibly nonlinear and nonvariational) case. (Note that Banach’s contraction principle is powerful enough provided the structure is simple.)

All these methods lead to weak solutions of the equation (13.21). In Subsection 13.4.E, we illustrate this alternative PDE approach in the simple case of the Laplace equation. For an introduction to more general models and more detailed treatment of the ideas, we refer to D. Gilbarg and N. S. Trudinger [193], J. Heinonen, T. Kilpeläinen and O. Martio [221], J. Malý and W. P. Ziemer [334]. In our presentation we show how the PDE approach to continuous Dirichlet data can be viewed; we follow J. Malý and W. P. Ziemer [334], Section 2.3.9. A. F. Monna emphasized in 1938 and 1939 the fact that the methods of Perron and Wiener are only special constructions and investigated the uniqueness question of the solution of the Dirichlet problem from the functional analysis point of view (see [349] where relevant references and interesting comments on the subject may be found). He asked whether an operator of the Dirichlet problem (submitted to certain natural conditions) was uniquely determined. A similar question was investigated by M. V. Keldysh, who proved in [263] that there is exactly one positive linear operator sending continuous functions on ∂U into harmonic functions on U such that its value is the solution of the classical Dirichlet problem, provided it exists (see Corollary 13.116). An operator satisfying conditions (a’) and (b) from Theorem 13.115 is sometimes called a K-operator. In our presentation, the Keldysh lemma 13.103 is a statement from Choquet theory. For the classical potential theory, in view of Theorem 13.35, it provides a “high quality” barrier for any regular boundary point. For a characterization of all exposed sets, see J. Lukeˇs, T. Mocek and I. Netuka [321]. A hard analysis proof was given by M. V. Keldysh in [264]. This proof, based on the Wiener criterion of regularity, is very involved. A proof in the same spirit for the plane case can be found in P. C. Curtis, Jr. [129]. In fact, knowing the Keldysh lemma, the proof of the Keldysh theorem 13.116 is very easy. This was the approach of M. V. Keldysh in [263]. However, M. Brelot observed in [86] that a weakened form of the Keldysh lemma (which can be proved in a simpler way) is sufficient for a proof of the uniqueness of a Keldysh operator. He also realized that linearity plays no role; see Theorem 13.115. A modification of the definition of the Keldysh operator, the so-called Ninomiya operator, is investigated from the point of view of uniqueness by I. Netuka in [360]. The lattice approach in Subsection 13.5.A follows I. Netuka [359]. The envelopes ∗ D f and D∗ f in Definition 13.109 were considered by M. Brelot in [85]; see also I. Netuka [357], [359], H. and U. Schirmeier [410]. Extended Keldysh operators (Definition 13.108) are studied by I. Netuka in [357] and [359]. An elementary approach to the Keldysh theorem in classical potential theory is available in I. Netuka [358].


561

In [314], J. Lukeˇs noticed that the Keldysh theorem fails for potential theory for the heat equation; see Example 13.122 and J. Lukeˇs and I. Netuka [324]. The key result saying that ∂ irr U is negligible if and only if the complement of ChH(U ) U is negligible (Theorems 13.41 and 13.128) is a consequence of deep results of J. Bliedtner and W. Hansen [62]. Results in the spirit of Theorem 13.128 are established in H. and U. Schirmeier [410], J. Lukeˇs [314] and I. Netuka [357], [359]. The fact that a Keldysh set can be inserted between a compact and an open set (Proposition 13.129) was proved by J. Lukeˇs and I. Netuka [323]. Keldysh type operators satisfying additional conditions of Remark 13.130 are studied by W. Hansen in [213]. The construction from Example 13.131 seems to be new. The so-called principal solution of the Dirichlet problem considered in J. Lukeˇs [313] is a further example of a Keldysh operator. In this connection, we formulate the following problem. Problem 13.159. Determine the extreme points of the convex set K of all Keldysh operators on U . (The operators H and D of Section 13.1 are extreme points of K. What else can be said?) Concerning the exercises, the relation between regularity and barriers was generalized to filters by N. Boboc, C. Constantinescu and C. Cornea [72]; Exercise 13.132 concerning regular sequences is then a special case. The notion of a semiregular set goes back to H. Bauer [39]. Exercise 13.136 summarizes results of I. Netuka [355], J. Lukeˇs [315], J. Lukeˇs and J. Malý [317] and J. Bliedtner and W. Hansen [63]. Accumulation points of harmonic and balayaged measures were investigated in detail in [317]. Exercise 13.133 follows Köhn and Sieveking [275]. The behavior of the heat equation described in Exercise 13.140 motivated J. Köhn [273] to introduce an axiomatic system of potential theory built up on the notion of semiregular set. Exercise 13.142(c) is from J. Malý’s thesis (Prague, 1980). Exercises 13.143–13.149 generalize classical results of M. Brelot and H. Bauer (cf. above notes to Subsection 13.4.A) to a weak axiomatic system proposed by J. Malý in his thesis (Prague 1980); see also J. Lukeˇs, J. Malý and L. Zaj´ıcˇ ek [320] (in particular 11.C.1). Exercises 13.152–13.154 offer a short excursion into fine potential theory, following [320], parts of Sections 12 and 13. Exercise 13.155 clarifies the relations between the approximation approach and the fine approach to the introduction of function spaces or cones. This is based on ideas from Section VII.9 of [66]. The method of Exercise 13.156 goes back to H. A. Schwarz (1870), H. Poincaré and H. Lebesgue; cf. generalization and survey by J. Veselý [459]. Exercise 13.158 is due to I. Netuka [357]. At the end of these notes, we add a short historical remark concerning “abstract” harmonic spaces (cf. Section A.8). It seemed quite natural to develop an axiomatic

562


system of “harmonic functions” which would unify the classical theory of harmonic functions, the theory of caloric functions, or, more generally, solutions of some partial differential equations, and extend classical potential theory to these axiomatics. These theories were constructed around the 1950’s by G. Tautz, J. L. Doob, M. Brelot and H. Bauer. In the sixties N. Boboc, C. Constantinescu and A. Cornea, in a series of papers, built up a further more general theory, in fact, a localized version of Bauer’s axiomatic system, culminating by the appearance of the monograph [123]. For the purpose of our monograph, we decided to confine ourselves to a simplified version of Bauer’s axiomatic. Among many articles on axiomatic theories, we refer the reader to a survey paper [43] by H. Bauer.

Chapter 14

Applications

In classical analysis, there is a series of results where more complicated objects are expressed as a mixture (or an average) of simpler ones. This chapter shows that such simple objects are typically extreme points of an appropriate compact convex set and the mixture is an integral representation in the spirit of the Choquet theory. Usually, the extreme points are parametrized in a suitable way and representing measures are carried by the set of parameters rather than by the set of extreme points itself. The simplest example of such a representation is an expression of double stochastic matrices by means of permutation matrices (Theorem 14.16). In this finite-dimensional situation, Minkowski’s theorem 2.11 provides a sufficient tool. Using the Integral representation theorem 2.31, we present several results on representation of convex functions and concave functions on a one-dimensional interval (Theorem 14.4, Corollary 14.9, Corollary 14.12). Positive harmonic functions on a ball in Rd admit a representation by means of Poisson kernels (Theorem 14.18). This is the classical Riesz–Herglotz theorem. We note that this is a very special case of the Martin type representation extensively studied in potential theory. (Space limitations do not allow us to include this elegant part of mathematical analysis here.) In view of the relation between holomorphic functions and harmonic functions on a disc, Theorem 14.18 also offers a representation of holomorphic functions with positive real part. Nevertheless, we included this representation independently in Section 14.6. We believe that a reader could find it interesting to see how methods of complex analysis lead to the direct identification of extreme points (Theorem 14.36) without recourse to the Riesz–Herglotz theorem 14.18. In Section 14.5, elementary function theory combined with the Integral representation theorem 2.31 makes it possible to obtain a representation of typically real holomorphic functions. Extreme completely monotonic functions are identified in Section 14.7 and the classical Bernstein theorem on representability of completely monotonic functions as the Laplace transform of measures is proved (Theorem 14.43). For the special case of discrete abelian groups, we present a proof of the Bochner– Weil theorem saying that positively definite functions admit a representation by means of characters (Theorem 14.53). It seems to be of interest to show that an application of the Krein–Milman theorem turns out to be an efficient tool in establishing results having nothing to do with integral representation. This is illustrated in the proofs of the Lyapunov theorem on the range of nonatomic vector measures with values in Rd (Theorem 14.58) and the Stone–Weierstrass theorem (Theorem 14.61 and Theorem 14.62).

564

14 Applications

All the situations described above follow the same scenario: look at a suitable convex set, prove its compactness, identify the extreme points, show that they form a closed set. Then the Integral representation theorem does the job. A different situation appears in Section 14.11 where a representation of invariant measures is investigated (Theorem 14.70). The extreme points, that is, ergodic measures, do not necessarily form a closed set and the full strength of the Choquet representation theorem 3.45 is used.

14.1

Representation of convex functions

In this section, integral representation of convex functions will be deduced from the Krein–Milman theorem. Definition 14.1 (The space K). We denote by K the set of all positive convex functions f on [0, 1] normalized by the condition f (0) + f (1) = 1.

(14.1)

Obviously, K is a convex subset of the locally convex space R[0,1] of real-valued functions on [0, 1] endowed with the topology of pointwise convergence. We will frequently use that the evaluation functional g 7→ g(x),

g ∈ R[0,1] ,

is a continuous linear functional on R[0,1] , whenever x ∈ [0, 1]. Proposition 14.2. The set K is compact. Proof. Since F := {g ∈ R[0,1] : 0 ≤ g ≤ 1} is a compact space, we need to show that K is a closed subset of F. The inclusion is obvious. To see that it is closed, we observe that K is the intersection of the sets {g ∈ F : g(0) + g(1) = 1, g λx + (1 − λ)y ≤ λg(x) + (1 − λ)g(y)} over all x, y ∈ [0, 1] and λ ∈ (0, 1). Now, we are going to characterize the set of extreme points of K. To this end, for y ∈ [0, 1], define the functions ϕy and ψy on [0, 1] as follows: ϕ1 := c{1} ,

ψ0 := c{0} ,

ϕy : x 7→

(x − y)+ , 1−y

x ∈ [0, 1],

y ∈ [0, 1),

ψy : x 7→

(y − x)+ , y

x ∈ [0, 1],

y ∈ (0, 1].

and

14.1 Representation of convex functions

565

Proposition 14.3. If E0 := {ϕy : y ∈ [0, 1]} and E1 := {ψy : y ∈ [0, 1]}, then ext K = E0 ∪ E1 . Proof. It is easy to verify that, for each y ∈ [0, 1], the functions ϕy and ψy are extreme points of K. Suppose that k ∈ K \ (E0 ∪ E1 ). If we show that k = f + g for two positive convex functions that are not a real multiples of k, then k is not an extreme point of K. Indeed, we denote α := f (0)+f (1). Then α ∈ (0, 1), 1−α = g(0)+g(1) and we can write k as a nontrivial convex combination h=α

f g + (1 − α) . α 1−α

Now, we will distinguish several cases. If k is discontinuous but not of the form ϕ1 or ψ0 , then such a decomposition is easy. If k is affine and yet not in E0 ∪ E1 , then c := inf k > 0 and thus we can decompose as k = k(0)(1 − x) + k(1)x. From now we assume that k is continuous and not affine. Suppose that we can find a point a ∈ (0, 1) such that k is affine neither on [0, a] nor on [a, 1]. Let h be an affine function supporting k at a. If h is increasing, we can write k = f + g, where f , g are nonnegative convex, f = k on [0, a] and f = h on [a, 1]. If h is decreasing, we interchange the role of intervals [0, a] and [a, 1]. Anyway, f is not a real multiple of k. Now, if no such a point a as above exists, we use another rule for the choice of the distinguished point a, namely, now a will be the supremum of all points t from (0, 1) such that k is affine on [0, t]. Then 0 < a < 1 and k is affine both on [0, a] and [a, 1] (but not on [0, 1]). We can use the same trick with the supporting function as above, in addition, we will assume that h will touch k only at a. Taking into account that k∈ / E0 ∪ E1 we observe again that the function f as above is not a real multiple of k. Define the mapping Φ : [y, 0] 7→ ϕy ,

[y, 1] 7→ ψy ,

y ∈ [0, 1].

Obviously, Φ is a one-to-one continuous mappings of [0, 1] × {0, 1} onto E0 ∪ E1 . Let f ∈ K. Applying the Krein–Milman theorem with transfer 2.33 we find a Radon measure σ on [0, 1] × {0, 1} such that Z Z f (x) = ϕy (x) dσ(y, 0) + ψy (x) dσ(y, 1), x ∈ [0, 1]. (14.2) [0,1]

[0,1]

If we decompose σ = aε0,1 + bε1,0 + µ ⊗ ε0 + ν ⊗ ε1 , we can summarize our achievement as follows.

566

14 Applications

Theorem 14.4. Let k be a positive convex function on [0, 1]. Then there exist a ≥ 0, b ≥ 0 and Radon measures µ and ν on [0, 1] such that Z Z x−ξ ξ−x k(x) = ac{0} (x) + bc{1} (x) + dµ(ξ) + dν(ξ) ξ [0,x) 1 − ξ (x,1] for any x ∈ [0, 1]. Proof. We may assume that k is normalized by k(0) + k(1) = 1, which means that k ∈ K. Then the statement is a straightforward reformulation of (14.2). Proposition 14.5. The set K is not a Choquet simplex. Moreover, there is a unique maximal measure representing an element k ∈ K if and only if either k is affine on (0, 1) or inf k = 0. Proof. The function k can be uniquely written as a convex combination of the functions ϕ1 , ψ0 and a continuous function from K. Hence we may assume that k is continuous. If k is (continuous) affine, the only possibility of representation is a unique combination of ϕ0 and ψ1 . If inf k = 0, then there exist points a, b ∈ [0, 1] such that a ≤ b, k > 0 on [0, a) ∪ (b, 1] and k = 0 on [a, b]. Then we decompose k = k(1)

kc[0,a) kc(a,1] + k(0) , k(1) k(0)

which is a convex combination of monotone functions from K and this decomposition is unique. Hence we may assume that min k = 0 and k is monotone, say, increasing. If µ, ν are the measures from Theorem 14.4 for the function k, then µ = 0 and thus Z ξ−x k(x) = dν(ξ). ξ (x,1] Let η be a C 2 “test function” with support in (0, 1]. Using integration by parts and Fubini’s theorem we obtain Z 1 Z 1 Z ξ 0 η(ξ) η (x) dν(ξ) = dx dν(ξ) ξ ξ 0 0 0 Z 1 Z ξ ξ − x 00 η (x) dx dν(ξ) = ξ 0 0 Z 1 Z ξ − x 00 = η (x) dν(ξ) dx ξ 0 (x,1] Z 1 = k(x)η 00 (x) dx. 0

14.2 Representation of concave functions

567

This shows that the expression on the left-hand side does not depend on the measure ν, and using density of C 2 functions in C c ((0, 1]) we deduce that ν is unique. Now, we need to show that if k is not affine and inf k > 0, then the representation is not unique. Namely, for the representation we write first k = h + (k − h), where h is a supporting affine function such that 0 ≤ h ≤ k on [0, 1] and h is a positive multiple of ϕ0 or ψ1 . Then the representing measure for k − h does not charge ϕ0 and ψ1 (this would contradict the fact that h supports k). Since there exist at least two different supporting affine functions h1 , h2 which are distinct combinations of ϕ0 and ψ1 , the representation of k cannot be unique.

14.2

Representation of concave functions

Definition 14.6 (The set Sp ). We denote by Sp , 1 ≤ p ≤ ∞, the set of all positive concave functions f on (0, 1) satisfying kf kp ≤ 1. Define ( (1 − x)y, 0 ≤ y ≤ x ≤ 1, G(x, y) := x(1 − y), 0 ≤ x ≤ y ≤ 1. (this is in fact the Green function for the differential operator f 7→ −f 00 ). Theorem 14.7. For 1 ≤ p ≤ ∞, Sp is a compact convex set in C (0, 1) endowed with the topology of locally uniform convergence. Proof. First we observe that C((0, 1)) is metrizable (see Examples 1.44 in [403]). Therefore it is enough to consider a sequence {gk } and show that a subsequence is convergent in Sp . It is easy to see that the original sequence is bounded and thus the functions gk are equilipschitz on each compact subinterval of (0, 1). We exhaust (0, 1) by a sequence of compact intervals, use the Arzèla-Ascoli theorem A.32 and the diagonal method, and find a subsequence which converges uniformly to a function g in C((0, 1)) on every compact subinterval of (0, 1). The function g is concave and we can use the Lebesgue dominated convergence theorem (if p < ∞) or a trivial estimate (if p = ∞) to conclude that kgkp ≤ 1. Theorem 14.8. Extreme points of S1 are the zero function and the functions nx 1 − xo gy : x 7→ 2 min , , y 1−y

x ∈ (0, 1),

where y runs through [0, 1]. (Of course, g0 : x 7→ 2(1 − x), g1 : x 7→ 2x.)

568

14 Applications

Proof. Let y ∈ [0, 1]. To show that gy is an extreme point of S1 , suppose that gy + f , gy − f are in S1 for some f ∈ C((0, 1)). Then f is affine on (0, y] and [y, 1), as −gy + (gy − f ) and −gy + (gy + f ) are concave thereon. Trivially f is continuous and f (0+ ) = f (1− ) = 0. It follows that f is a multiple of gy , and thus also gy + f and gy − f are multiples of gy . The restriction on norm shows that f = 0. Conversely, let g be a nonzero extreme point of S1 . We want to prove that g = gy for some y ∈ [0, 1]. Find the greatest interval (0, a) on which g coincides with some multiple of x 7→ x. If none such interval exists, set a = 0. Similarly, find the greatest interval (b, 1) on which g coincides with some multiple of x 7→ 1 − x. If none such interval exists, set b = 1. If a = b, then it easily follows that g = ga (= gb ). Suppose that a < b. Fix c with a < c < b. Find an affine function h such that h ≥ g and h(c) = g(c). Set ( g(x) − h(1)x, x ≤ c, f0 (x) = h(x) − h(1)x, x > c, and ( g(x) − h(0)(1 − x), x ≥ c, f1 (x) = h(x) − h(0)(1 − x), x < c. Denote 1

Z λ0 =

f0 (x) dx, 0 1

Z λ1 =

f1 (x) dx. 0

Then f0 /λ0 , f1 /λ1 ∈ S1 and f0 + f1 = g. Hence Z λ0 + λ1 =

1

g(x) dx = 1 0

and g = λ0

f0 f1 + λ1 , λ0 λ1

that is, g is a convex combination of functions from S1 . Since g is an extreme point, it follows that f1 f0 = , g= λ0 λ1 in particular g(x) =

xh(1) , λ1

which contradicts the maximality of a.

x ∈ (0, c),


569

Corollary 14.9. Let f be a positive concave function on (0, 1). Then there exists a unique Radon measure µ on (0, 1) such that 1

Z

G(x, y) dµ(y) + (1 − x)f (0+ ) + xf (1− ),

f (x) =

x ∈ (0, 1),

0

and Z

1

Z y(1 − y) dµ(y) + f (0+ ) + f (1− ) = 2

1

f (x) dx. 0

0

Proof. We can assume that f is not the zero function and normalize f by kf k1 = 1. We rewrite the extreme points nx 1 − xo gy (x) = 2 min , y 1−y

(14.3)

alternatively as g0 : x 7→ 2(1 − x), g1 : x 7→ 2x and gy =

2G(·, y) , y(1 − y)

y ∈ (0, 1).

The Krein–Milman theorem with transfer 2.33 yields a probability Radon measure P on [0, 1] such that Z

Z gy (x) dP (y) = 2(1 − x) P ({0}) + 2x P ({1}) +

f (x) = [0,1]

0

1

2G(x, y) dP (y). y(1 − y)

(Due to the normalization, the representing measure on ext S1 does not charge the zero function.) By the Lebesgue dominated convergence theorem and (14.3) Z

Z gy (x) dP (y) = lim

lim

x→0+

x→1−

(0,1)

gy (x) dP (y) = 0. (0,1)

It follows that 2P ({0}) = f (0+ ),

2P ({1}) = f (1− ).

The measure µ with dµ 2 (y) = dP y(1 − y) has the required properties.

(14.4)

570

14 Applications

For the uniqueness, we pick a C 2 “test function” η with support in (0, 1). Using integration by parts, for each y ∈ (0, 1) we have Z

1

00

Z

Z

00

x(1 − y)η (x) dx +

G(x, y)η (x) dx = 0

y

1

(1 − x)yη 00 (x) dx

y

0

= y(1 − y)η 0 (y) −

y

Z

(1 − y)η 0 (x) dx

0

− y(1 − y)η 0 (y) +

Z

1

yη 0 (x) dx

y

= −(1 − y)η(y) − yη(y) = −η(y). (This calculation reflects that G is the Green function.) Hence using Fubini’s theorem we obtain Z 1 Z 1 Z 1 η(y) dµ(y) = − G(x, y)η 00 (x) dx dµ(y) 0

0

=−

0

Z 1 Z 0

Z =−

1

G(x, y)η 00 (x) dµ(y) dx

0 1

f (x)η 00 (x) dx.

0

This shows that the expression on the left-hand side does not depend on µ. Since any continuous function on (0, 1) with compact support can be uniformly approximated by C 2 functions on a suitable interval [a, b] ⊂ (0, 1), the measure µ is uniquely determined. R1 Remark 14.10. The Green potential u(x) = 0 G(x, y) dµ(y) solves in the sense of distributions the Dirichlet problem −u00 = µ, u(0) = u(1) = 0. Hence u0+ (0) − u0− (1) = µ((0, 1)) and if µ is a finite measure, then u is not only concave but also Lipschitz. Therefore to obtain the representation of general concave functions we need to consider for µ also infinite measures (but the measure P from (14.4) is a probability measure). Theorem 14.11. Extreme points of S∞ are the zero function and the functions

ga,b (x) :=

 x  a,

1−x ,  1−b

 1,

0 < x < a, b < x < 1, a ≤ x ≤ b,

where 0 ≤ a ≤ b ≤ 1.


571

Proof. Suppose that one of the function ga,b is expressed as a nontrivial convex combination of some functions f0 and f1 from S∞ . Then obviously f0 = f1 = 1 on [a, b]. Further, if a > 0, then f0 (0+ ) = f1 (0+ ) = 0, and if b < 1, then f0 (1− ) = f1 (1− ) = 0. Also, the functions f0 and f1 must be affine on (0, a] and on [b, 1). Then there is no other choice than f0 ≡ f1 ≡ ga,b . Conversely, let g be a nonzero extreme point of S∞ . Then obviously sup g = 1. Let [a, b] be the greatest (possibly degenerate) interval such that g = 1 on [a, b] ∩ (0, 1), b = 1 if g(1− ) = 1 and a = 0 if g(0+ ) = 1. Suppose that a > 0. We must show that g = xa on (0, a]. It is not difficult to deduce from the extremality of g that g(0+ ) = 0. Consider the function ( g(x) − xa , 0 < x < a, h(x) = 0, a ≤ x < 1, Ra and set α := 0 h. 0 (a) > 0, then there exists ε > 0 such that both g + εh, g − εh belong to S . If g− ∞ 0 (a) = 0. By the maximality of the interval Let us consider the most difficult case g− [a, b], g cannot be affine on any left neighborhood of a. Then by Theorem 14.8, h/α is not an extreme point of Z a n o S1 ((0, a)) := f concave on (0, a) : f ≥ 0, f ≤1 . 0

Thus we can express h as 21 (h0 + h1 ) where h0 and h1 are distinct functions of αS1 ((0, a)). The integral bound guarantees that h0 and h1 are linearly independent. We extend h0 and h1 to (0, 1) by setting hi = 0 on [a, 1), i = 0, 1. Since h0− (a) is finite, (hi )0− (a) is finite as well for i = 0, 1. Hence we can find a linear combination h2 of h0 and h1 such that (h2 )0− (a) = 0. By choosing a suitable ε > 0 we can achieve that g + εh2 ∈ S∞ , g − εh2 ∈ S∞ . This finishes the proof. Corollary 14.12. Let f be a positive concave function on (0, 1). Then there exists a Radon measure µ on T := {(s, t) ∈ [0, 1] × [0, 1] : s ≤ t} such that Z f (x) = µ([0, x] × [x, 1]) + T ∩((x,0]×[0,1])

Z + T ∩([0,1]×[0,x))

1−x 1−t

dµ(s, t),

x s

dµ(s, t) x ∈ (0, 1).

572

14 Applications

Proof. For f = 0 this is trivial. Otherwise, if f is normalized by kf k∞ = 1, we can use the Krein–Milman theorem with transfer 2.33 to find a representing measure µ on T providing the required representation. Theorem 14.13. Let 1 < p < ∞. Then extreme points of Sp are the zero function and all functions f ∈ K with kf kp = 1. Proof. Obviously, any nonzero extreme point has Lp -norm one. Conversely, let us assume that f is a function of norm one, which is expressed as a nontrivial convex combination of some functions f0 and f1 from Sp . Then due to the strict convexity of the Lp -norm, the norm of f is strictly smaller than one, which is a contradiction. Remark 14.14. The set S1 is simplex. To show that none of Sp with 1 < p < ∞ is simplicial, consider the equality 1 2 αp x

+ 12 αp (1 − x) = 12 αp ,

R1 R1 where αp is chosen so that 0 (αp x)p dx = 0 (αp (1−x))p dx = 1. Then the constant function 21 αp is represented as a convex combination of the functions αp x and 12 αp (1− x) as well as a convex combination of the functions 1 and 0. A similar argument works for S∞ with α∞ = 1.

14.3

Doubly stochastic matrices

Definition 14.15 (Doubly stochastic and permutation matrices). Let D be the set of all d × d doubly stochastic matrices, that is, matrices with positive entries in which the sum of the elements in each row and in each column is 1. A permutation matrix is a matrix obtained by permuting the rows of the identity matrix. It has exactly one 1 in each row and in each column, all other entries being 0. It is obvious that D is a convex set and that the set P of permutation matrices is a 2 subset of ext D. Considering D in a natural way as a subset of Rd , D is a compact set. Theorem 14.16. The equality ext D = P holds and every doubly stochastic matrix is a convex combination of permutation matrices. Proof. Suppose first that A = (ai,j )di,j=1 is a permutation matrix. Then all entries are 0 or 1 and it easily follows that A ∈ ext D. Conversely, consider A ∈ ext D and let the set S of all positions [i, j] such that 0 < ai,j < 1 be nonempty. We will show that there exist p, m ∈ N and sequences ∞ {ik }∞ k=1 , {jk }k=1 of indices from {1, 2, . . . , d} such that m > 1 and the following properties hold: (a) [ik , jk ], [ik , jk+1 ] ∈ S for each k ∈ N,

14.4 The Riesz–Herglotz theorem

573

(b) if k > p, then ik+m = ik and jk+m = jk , (c) if k > p and 1 ≤ q < m, then ik+q 6= ik and jk+q 6= jk . Having such sequences at our disposal, we define a matrix E = (ei,j )di,j=1 by

ei,j

  1 = −1   0

for i = ik , j = jk with k > p, for i = ik , j = jk+1 with k > p, otherwise.

The matrix E has the property that sum of its elements over each row and each column is zero. If we take ε > 0 so small that ε ≤ ai,j ≤ 1 − ε for each [i, j] ∈ S, then both the matrices A + εE and A − εE are doubly stochastic and thus A is not an extreme point of D, which is a contradiction. ∞ It remains to construct the sequences {ik }∞ k=1 , {jk }k=1 . We start with an arbitrary [i1 , j1 ] ∈ S. There must be at least one column index j2 6= j1 such that [i1 , j2 ] ∈ S, otherwise the sum over the i1 -th row would be ai1 ,j1 < 1. Similarly we find i2 6= i1 such that [i2 , j2 ] ∈ S. We proceed inductively and always find jk+1 6= jk such that [ik , jk+1 ] ∈ S and ik+1 6= ik such that [ik+1 , jk+1 ] ∈ S. We go on with the construction till we find p ∈ N when it happens that jp or ip is chosen for the second time. Then we continue with the construction more carefully. If this happens first with the column index, jp = jp−m , then we set ip = ip−m 6= ip−1 . If this happens first with the row index, ip = ip−m , then we set jp+1 = jp+1−m 6= jp . At all the next steps we follow the rule of period m. It is clear that the sequences constructed according to this algorithm satisfy the properties (a), (b) and (c). Once we know the equality ext D = P, Theorem 2.11 concludes the proof. Remark 14.17. It is easy to show that the set D is a Choquet simplex in the case d = 2 whereas it is not if d ≥ 3. Namely, the arithmetic mean of all permutation matrices with positive determinant is the matrix with all entries n1 , as well as the arithmetic mean of all permutation matrices with negative determinant.

14.4

The Riesz–Herglotz theorem

For ρ > 0, let U (0, ρ) denote the open ball in Rd of dimension d ≥ 2 with center 0 and radius ρ. We will write U instead of U (0, 1) and ∂U for its boundary. Denote by e ) the convex set of all positive harmonic functions on U equal to 1 at the origin. H(U e ) as a subset of C(U ) endowed with the topology of locally uniform We consider H(U e ) is a compact set. convergence. By [21], Lemma 1.5.6, H(U Let S stand for the surface area measure on ∂U .

574

14 Applications

We recall that the function P := P0,1 (cf. Definition A.131) P : (x, z) 7→

1 − |x|2 , |x − z|d

z ∈ ∂U, x ∈ U,

is called the Poisson kernel of U . Denote by Px and P z the functions Px : y 7→ P (x, y),

y ∈ ∂U,

P z : y 7→ P (y, z),

y ∈ U.

and

Differentiation under the integral sign shows that the Poisson integral Z P µ(x) :=

Px dµ,

x ∈ U,

∂U

is a positive harmonic function on U whenever µ is a Radon measure on ∂U . We write σ := (S(∂U ))−1 S and P f instead of P (f σ) if f is an S-integrable function on ∂U . e ). Then there exists a unique Radon Theorem 14.18 (Riesz–Herglotz). Let h ∈ H(U 1 measure µ ∈ M (∂U ) such that Z h(x) =

Px dµ,

x ∈ U.

∂U

Proof. For r ∈ (0, 1) and x ∈ U (0, 1/r) let hr (x) := h(rx). Then the function hr is harmonic on U (0, 1/r), hence hr = P hr on U by Remark A.133. If µr := hr σ, then Z µr (∂U ) =

hr dσ = hr (0) = 1 ∂U

by the mean value theorem (Theorem A.173). By the compactness of M1 (∂U ), there exists a measure µ ∈ M1 (∂U ) and a sequence {rn } so that rn % 1 and µrn → µ. Then, for any x ∈ U , we have Z h(x) = lim h(rn x) = lim Px hrn dσ = lim (P µrn )(x) n→∞ n→∞ ∂U n→∞ Z Z = lim Px dµrn = Px dµ = P µ(x). n→∞ ∂U

∂U

To prove uniqueness, suppose that h = P µ = P ν for some µ, ν ∈ M1 (∂U ). Pick

14.5 Typically real holomorphic functions

575

r ∈ (0, 1) and f ∈ C(∂U ). Then Z Z Z f dµr = f (z)hr (z) dσ(z) = f (z)h(rz) dσ(z) ∂U ∂U ∂U Z Z f (z) Prz (y) dν(y) dσ(z) = ∂U ∂U Z Z = f (z) Pry (z) dν(y) dσ(z) ∂U ∂U Z Z Z P f (ry) dν(y). Pry (z)f (z) dσ(z) dν(y) = = ∂U

∂U

∂U

Setting r = rn and letting n → ∞, we get Z

Z f dµrn →

P f (rn y) → f (y) and ∂U

from which it follows that

R ∂U

f dµ =

R ∂U

f dµ, ∂U

f dν. Hence µ = ν, finishing the proof.

e ) of extreme points of H(U e ) coincides with the set Corollary 14.19. The set ext H(U z {P : z ∈ ∂U }. Proof. Let D := {P z : z ∈ ∂U }. If ψ : z 7→ P z ,

z ∈ ∂U,

then ψ is continuous, injective and ψ(∂U ) = D. Consequently, D is compact and ψ maps ∂U homeomorphically onto D. It follows from the Riesz–Herglotz theorem 14.18 combined with the Milman thee ) ⊂ D = D. Since ext H(U e ) 6= ∅ by the Krein–Milman orem 2.43 that ext H(U z e theorem 2.22, P ∈ ext H(U ) for a point z ∈ ∂U . Given y ∈ ∂U , there is a linear isometry T : Rd → Rd with P z = P y ◦ T . Since T preserves harmonicity, P y e ), and hence necessarily belongs to ext H(U e ), {P z : z ∈ ∂U } ⊂ ext H(U finishing the proof.

14.5

Typically real holomorphic functions

Definition 14.20 (Typically real holomorphic functions). Let U := {z ∈ C : |z| < 1} and A(U ) be the set of all holomorphic functions on U . A function f ∈ A(U ) is said to be typically real, if f (z) is real if and only if z ∈ U is real. The set of all typically real functions will be denoted by T (U ).

576

14 Applications

Here is an example of functions from T (U ). Fix t ∈ [−1, 1] and set ht : z 7→

z , 1 + 2tz + z 2

z ∈ U.

√ Obviously ht ∈ A(U ), because 1 + 2tz + z 2 = 0 implies z = −t ± + 1 − t2 , hence |z| = 1. If z ∈ U and ht (z) is real, we have ht (z) = ht (z), which yields the equality z − z = |z|2 (z − z), or z = z. The importance of functions ht is illustrated by the following result. Theorem 14.21. Let f ∈ T (U ). Then there exist real numbers a, b and a Radon probability measure µ on [−1, 1] such that Z ht (z) dµ(t), z ∈ U. f (z) = a + b [−1,1]

We will prove this result using the Krein–Milman theorem. Define T 0 (U ) := g ∈ T (U ) : g(0) = 0, g 0 (0) = 1 . Clearly, T 0 (U ) is a convex subset of A(U ). If f ∈ T (U ), then f 0 (0) ∈ R, hence there are real numbers a, b and g ∈ T 0 (U ) such that f = a + bg. In what follows we will use the notation U + := {z ∈ U : Im z > 0}

and

U − := {z ∈ U : Im z < 0} .

The following simple result will be useful. Lemma 14.22. Let f ∈ A(U ), f (0) = 0, f 0 (0) = 1. Then f ∈ T 0 (U ) if and only if Im z · Im f (z) > 0,

z ∈ U, Im z 6= 0.

(14.5)

Proof. Suppose that f ∈ T 0 (U ). Since f 0 (0) = 1, we have 1t Im f (it) → 1 for t → 0+ . Hence there exists z0 ∈ U + such that Im f (z0 ) > 0. In fact, Im f > 0 on U + , since otherwise Im f would have zero there, which is impossible by the definition of a typically real function. Similarly Im f < 0 on U − and (14.5) holds. The converse implication is obvious. Lemma 14.23. If g ∈ T 0 (U ) and 0 < r < 1, then max {|g(z)| : |z| ≤ r} ≤

r . (1 − r)2 2

) = −2i Im(ρw), Proof. Fix 0 < ρ < 1 and θ ∈ R and denote w := eiθ . Since ρ(1−w w we have ρ(1 − w2 )g(ρw) Re = 2 Im(ρw) · Im g(ρw) ≥ 0 w


577

by Lemma 14.22. Since g 0 (0) = 1, the function z 7→ Re

ρ(1 − z 2 )g(ρz) z

,

z ∈ U \ {0} ,

has a harmonic extension to U and is positive on U by Proposition A.135, being positive on ∂U . Defining f (z) :=

(1 − z 2 )g(z) , z

z ∈ U \ {0} ,

and f (0) = 1, we have f ∈ A(U ) and Re f ≥ 0. In fact, Re f > 0 on U by the strong minimum principle for harmonic functions (see Proposition A.134). Define ϕ(z) :=

1+z , 1−z

z ∈ U.

Then ϕ ∈ A(U ), ϕ(0) = 1, ϕ is injective, ϕ(U ) = (0, ∞) × R and ϕ−1 (w) =

w−1 , w+1

w ∈ ϕ(U );

cf. R. B. Burckel [95], Exercise 2.5, p. 43. The function F := ϕ−1 ◦ f has the following properties: F ∈ A(U ), F (0) = 0, F (U ) ⊂ U. Consequently, |F (z)| ≤ |z| for every z ∈ U by Schwarz’s lemma (see R. B. Burckel [95], Theorem 6.1). It follows |(1 − z 2 )g(z) − z| ≤ |z||(1 − z 2 )g(z) + z|,

z ∈ U.

The left-hand side of the inequality is minorized by |1 − z 2 ||g(z)| − |z| and the righthand side is majorized by |z|(|1 − z 2 ||g(z)| + |z|). This yields (1 − |z|2 )|g(z)|(1 − |z|) ≤ |1 − z 2 ||g(z)|(1 − |z|) ≤ |z|(1 + |z|) and |g(z)| ≤

|z| , (1 − |z|)2

z ∈ U.

The result now follows by the maximum modulus principle; see R. B. Burckel [95], Exercise 5.7 and Corollary 5.9, p. 127 and 128. Now we are going to characterize the set of extreme points of T 0 (U ). To this end, fix β > 2 and g ∈ T 0 (U ) and notice that the functions ( (β − z −1 − z)g(z), z ∈ U \ {0} , f1 (z) := −1, z = 0,

578

14 Applications

and ( (β + z −1 + z)g(z), f2 (z) := 1,

z ∈ U \ {0} , z = 0,

belong to T (U ). We will show this for f1 , the proof for f2 follows the same lines. We can suppose that g is holomorphic on a neighborhood of U , otherwise we would consider the function g(rz)r−1 and let r → 1+ . If z is real, −1 ≤ z ≤ 1, then f1 (z) is real, so that Im f1 (z) = 0. For z = eiθ , θ real, Im f1 (eiθ ) = (β − 2 cos θ) Im g(eiθ ). Since Im g(z) > 0 if Im z > 0, we have Im f1 (eiθ ) ≥ 0 for θ ∈ [−π, π], thus the harmonic function Im f1 is positive on U + by Proposition A.135. In fact, Im f1 is strictly positive there since g(z) 6= 0 if Im z > 0. Similarly, Im f1 < 0 on U − . This shows that f1 ∈ T (U ). Lemma 14.24. We have ext T 0 (U ) ⊂ {ht : t ∈ [−1, 1]}. Proof. Suppose that g ∈ ext T 0 (U ) and t = − 41 g 00 (0). We will show that t ∈ [−1, 1] and g = ht . Fix α > 2|t| + 2 and define   1 (1 − (z −1 + z + 2t)g(z)), z ∈ U \ {0} , f (z) := α 0, z = 0. Then f ∈ A(U ) and f 0 (0) = 0. We have (g + f )(z) =

1 1 1 + (α − 2t − − z)g(z), α α z

z ∈ U \ {0} ,

hence the preceding result (with β = α − 2t > 2) shows that the function g + f is typically real. Similarly, g − f is typically real since α + 2t > 2. Also, (g + f )(0) = (g − f )(0) = 0 and (g + f )0 (0) = 1,

(g − f )0 (0) = 1.

Hence g + f, g − f ∈ T 0 (U ). The assumption g ∈ ext T 0 (U ) implies that f = 0, thus g(z)(1 + 2tz + z 2 ) = z for any z ∈ U \ {0}. We see that 1 + 2tz + z 2 6= 0 on U , which easily implies t ∈ [−1, 1] and g = ht .


579

In order to prove that any function ht , t ∈ [−1, 1], is an extreme point of T 0 (U ), we consider the space A(U ) endowed with the topology of locally uniform convergence. It is known that a subset of A(U ) is compact if and only if it is closed and bounded (see W. Rudin [403], Sec. 1.45). Note that, for k ∈ N and z ∈ U , the mapping f 7→ f (k) (z) is a continuous linear functional on A(U ). Lemma 14.25. The set T 0 (U ) is compact. Proof. Suppose that the sequence {gn } of functions from T 0 (U ) converges locally uniformly on U to a function g. Obviously, g ∈ A(U ), g(0) = 0 and g 0 (0) = 1. Since Im z · Im gn (z) ≥ 0 for z ∈ U by Lemma 14.22, we have Im gn ≥ 0, and therefore Im g ≥ 0 on U + . Actually, since g 0 (0) = 1, Im g > 0 there by the strong minimum principle (Proposition A.134). A similar reasoning shows that Im g < 0 on U − . We conclude that g ∈ T 0 (U ) and T 0 (U ) is closed. By Lemma 14.23, T 0 (U ) is bounded. We see that T 0 (U ) is compact. Theorem 14.26. The set {ht : t ∈ [−1, 1]} is compact and is equal to ext T 0 (U ). Proof. The mapping Φ : t 7→ ht , t ∈ [−1, 1], is obviously continuous and injective, thus the set {ht : t ∈ [−1, 1]} is compact. We have already shown that it contains ext T 0 (U ). Fix s ∈ [−1, 1] and define 00

000

Fs (f ) := Re(12sf (0) + f (0)),

f ∈ A(U ).

Then Fs , as a continuous linear functional on A(U ), attains by Bauer’s concave minimum principle 2.24 its minimum on T 0 (U ) at an extreme point. For t ∈ [−1, 1], a simple calculation shows that 00

ht (0) = −4t

and

000

ht (0) = 24t2 − 6.

Hence Fs (ht ) = 24t2 − 48st − 6 > 24s2 − 48s2 − 6 = Fs (hs ), provided that t ∈ [−1, 1] \ {s}. It follows that hs ∈ ext T 0 (U ), and consequently ext T 0 (U ) = {ht : t ∈ [−1, 1]}. Proof of Theorem 14.21. We may (and do) suppose that f ∈ T0 (U ). Using the Krein–Milman theorem with transfer 2.33 (see also Remark 2.32(c)), we obtain the required representation Z f (z) = ht (z) dµ(z). [−1,1]

580

14.6

14 Applications

Holomorphic functions with positive real part

Definition 14.27 (Holomorphic functions with positive real part). Let U := {z ∈ C : |z| < 1}, R := {z ∈ C : Re z > 0} and T := {z ∈ C : |z| = 1} and let A(U ) be the set of all holomorphic functions on U . Denote P(U ) := {f ∈ A(U ) : f (0) = 1, Re f ≥ 0}. It is known that the function ϕ : z 7→

1+z , 1−z

z ∈ U,

(14.6)

is an injective holomorphic mapping of U onto R, so that ϕ ∈ P(U ). Therefore, if η ∈ T and 1 + ηz hη : z 7→ , z ∈ U, (14.7) 1 − ηz then also hη belongs to P(U ). The importance of the functions hη is illustrated by the following result. Theorem 14.28. Let f ∈ P(U ). Then there exists a Radon probability measure µ on T such that Z 1 + ηz f (z) = dµ(η), z ∈ U. 1 − ηz T We will prove this representation theorem using the Krein–Milman theorem with transfer 2.33. At first, we establish several auxiliary results. Lemma 14.29. If f ∈ P(U ) and 0 < r < 1, then max{|f (z)| : |z| ≤ r} ≤

1+r . 1−r

Proof. Note that the inverse of the function ϕ from (14.6) is given by ϕ−1 (w) =

w−1 , w+1

w ∈ R.

Let f ∈ P(U ). Then f (0) = 1 and Re f > 0 on U in view of the strong minimum principle for harmonic functions (see Proposition A.134). Hence f (U ) ⊂ R. Define F := ϕ−1 ◦ f . Then F ∈ A(U ), F(0) = 0 and F (U ) ⊂ U . By Schwarz’s lemma (see R. B. Burckel [95], Theorem 6.1), |F (z)| ≤ |z|,

z ∈ U.

14.6 Holomorphic functions with positive real part

It follows that

|f (z)| − 1 f (z) − 1 ≤ |z|, ≤ |f (z)| + 1 f (z) + 1

581

z ∈ U.

If |z| ≤ r, then |f (z)| ≤

1+r . 1−r

We will consider the space A(U ) endowed with the topology of locally uniform convergence. It is known that a subset of A(U ) is compact if and only if it is closed and bounded (see W. Rudin [403], Sec. 1.45). Lemma 14.30. The set P(U ) is a compact convex subset of A(U ). If M := {hη : η ∈ T }, then the sets M and co M are compact. Proof. The set P(U ) is obviously convex and closed. It is bounded by Lemma 14.29, hence compact. Define Φ : η 7→ hη , η ∈ T. (14.8) Then Φ : T → A(U ) is an injective mapping and Φ(T ) = M . If 0 < r < 1, η, ξ ∈ T and |z| ≤ r, then |hη (z) − hξ (z)| =

2|z||η − ξ| 2r |η − ξ|. ≤ |(1 − ηz)(1 − ξz)| (1 − r)2

Hence Φ is continuous and M is compact. Since the topology of A(U ) is induced by a complete translation invariant metric, co M is compact (see W. Rudin [403], p. 72). S be the set of all sequences {αn }∞ Definition 14.31 (The dual space of A(U )). Let n=0 p n of complex numbers satisfying lim supn→∞ |αn | < 1. Proposition 14.32. Let {αn }∞ n=0 ∈ S. For f ∈ A(U ) of the form f (z) :=

∞ X

cn z n ,

z ∈ U,

n=0

define Af :=

∞ X

αn cn .

n=0

Then A : f 7→ Af is a continuous linear functional on A(U ).

(14.9)

582

14 Applications

Proof. The functional A is obviously linear. Put p r := lim sup n |αn | n→∞

and define

∞ X

a : z 7→

1 z ∈ U. r

αn z n ,

n=0

Fix ρ ∈ (r, 1) and define ψ(t) := ρeit ,

t ∈ [0, 2π].

Then for f of the form (14.9) we have Z ∞ dz X 1 = αn cn = Af. a 1/z f (z) 2πi ψ z n=0

It follows that if gk ∈ A(U ) and gk → g locally uniformly on U , then gk → g uniformly on the closure of the set ρU and Agk → Ag as k → ∞. Hence A is continuous. Theorem 14.33. Let A be a continuous linear functional on A(U ). Then there exists a unique sequence {αn }∞ n=0 ∈ S such that Af :=

∞ X

αn cn ,

(14.10)

n=0

whenever f ∈ A(U ) is as in (14.9). Proof. Write id : z 7→ z, z ∈ C, and for n = 0, 1, . . . define αn := A(id)n . Suppose ∞ that therep exists a subsequence {βk }∞ k=0 of {αn }n=0 such that βk 6= 0 for every k and limk→∞ k |βk | ≥ 1. Define f (z) :=

∞ X

βk−1 z k ,

z ∈ U.

k=0

Then f ∈ A(U ) and, by the linearity and the continuity of A, Af =

∞ X

∞ X A (id)k βk−1 = βk βk−1 = ∞,

k=0

k=0

which is a contradiction. We conclude that lim supn→∞ ∈ S. If f is the function from (14.9) and m ∈ N, then A

m X n=0

m X cn (id)n = αn cn n=0

p n

|αn | < 1, hence {αn }∞ n=0

583


and (14.10) holds by the continuity of A. Obviously, the sequence {αn }∞ n=0 is unin quely determined by A (consider f := (id) in (14.10)). Remark 14.34. For a characterization of the dual space of the space of holomorphic functions on a general open set in C, see D. H. Luecking and L. A. Rubel [312]. Proposition 14.35. Suppose that the functions p and q, ∞ X

p(z) := 1 + 2

pn z n

q(z) := 1 + 2

and

n=1

∞ X

qn z n ,

z ∈ U,

n=1

belong to P(U ). Define pq : z 7→ 1 + 2

∞ X

p n qn z n ,

z ∈ U.

n=1

Then pq ∈ P(U ). Proof. Obviously, pq ∈ A(U ) and pq (0) = 1. Fix z ∈ U , ρ ∈ (|z|, 1) and define ψ(t) := ρeit , t ∈ [0, 2π]. We have Z Z ∞ ∞ dw X X dw 1 1 n −n qm w m 1+2 pn z w p(z/w)q(w) 1+2 = 2πi ψ w 2πi ψ w m=1 n=1 Z X ∞ ∞ dw X 1 =1+4 pn qn z n pn qn z n 4 =1+ 2πi ψ w n=1

n=1

and 1 2πi

Z p(z/w)q(w) ψ

1 = 2πi

Z

dw w

1+2

ψ

∞ X

n

pn z w

−n

n=1

We arrive at 1 2πi

1+2

∞ X

qm ρ2m w−m

m=1

dw w

= 1.

∞

Z p(z/w) Re q(w) ψ

X dw =1+2 pn qn z n . w n=1

The integral on the left-hand side is equal to Z 2π 1 dt p zρ−1 e−it Re q ρeit iρeit it 2πi 0 ρe

(14.11)

584

14 Applications

and its real part is equal to Z 2π 1 Re p zρ−1 e−it Re q ρeit dt ≥ 0. 2π 0

(14.12)

We conclude from (14.11) and (14.12) that ∞ X 0 ≤ Re 1 + 2 pn qn z n = Re pq (z). n=1

Theorem 14.36. The equality ext P(U ) = M holds. Proof. We first prove that ext P(U ) ⊂ M . Recall that M is closed and co M is compact, hence it is sufficient to show that co M = P(U ). Indeed, by the Milman theorem 2.43, we then have ext P(U ) ⊂ M = M . Obviously, co M ⊂ P(U ). We will show that the assumption P(U ) \Pco M 6= ∅ leads to a contradiction. ∞ n Let g ∈ P(U ) \ co M , g(z) := n=0 dn z for z ∈ U . By the Hahn–Banach theorem, there are a continuous linear functional B on A(U ) and β ∈ R such that Re B(f ) > β > Re B(g) whenever

f ∈ co M.

By Theorem 14.33, there exists a sequence {βn }∞ n=0 ∈ S such that for every f ∈ A(U ) of the form (14.9) we have Bf =

∞ X

βn cn .

n=0

Put α0 := Re β0 − β

and

αn := βn

for

n ≥ 1.

Then {αn }∞ n=0 ∈ S and the equality Af :=

∞ X

αn cn ,

f ∈ A(U ),

n=0

for f of the form (14.9) defines a continuous linear functional on A(U ). We have Re Af > 0 > Re Ag

for any f ∈ co M.

If η ∈ T , then the function hη from (14.7) satisfies hη (z) = 1 + 2

∞ X n=1

ηnzn,

z ∈ U,

(14.13)


hence Re Ahη = α0 + 2 Re

∞ X

585

αn η n > 0.

n=1

Since the function η 7→ Re Ahη is strictly positive on T , the strong minimum principle for harmonic functions of Proposition A.134 yields α0 + 2 Re

∞ X

αn z n > 0

for any

z ∈ U.

n=1

In particular, α0 > 0 and the function ∞ X αn

zn,

z ∈ U,

belongs to P(U ) because Re a > 0 on U . Recall that ∞ X n 1 g(z) = 1 + 2 2 dn z ,

z ∈ U.

a : z 7→ 1 + 2

n=1

α0

n=1

By Proposition 14.35, the function ag : z 7→ 1 +

∞ X

dn

n=1

αn n z , α0

z ∈ U,

(14.14)

p belongs to P(U ), thus Re(ag) ≥ 0 on U . Since lim supn→∞ n |αn | < 1, the function ag can be defined by the series in (14.14) on a disc with center 0 and radius greater than 1. Then we have Re(ag)(1) ≥ 0 and Re Ag =

∞ X n=1

∞ X αn αn dn = α0 1 + dn = α0 Re(ag)(1) ≥ 0. α0 n=1

This contradicts to (14.13), hence we proved that ext P(U ) ⊂ M . By the Krein–Milman theorem 2.22, ext P(U ) 6= ∅. If h ∈ ext P(U ), then h = hη for a suitable η ∈ T . For ξ ∈ T , hξ is obtained from hη by a rotation. Therefore hξ ∈ ext P(U ). The equality ext P(U ) = M is proved. Proof of Theorem 14.28. Let f ∈ P(U ). Using the Krein–Milman theorem with transfer 2.33 (see Remark 2.32(c)), we obtain the desired representation Z f (z) = hη (z) dµ(η). T

586

14 Applications

Remark 14.37. If h is a positive harmonic function on U and h(0) = 1, then there exists a uniquely determined harmonic function e h on U such that the function f := h+ie h is holomorphic on U and f (0) = 1. Thus Theorem 14.28 gives a representation of normalized positive harmonic functions. More specifically, Z Z 1 + ηz 1 − |z|2 h(z) = Re dµ(η) = dµ(η). 2 1 − ηz T T |η − z| Thus, for dimension d = 2, the Riesz–Herglotz theorem 14.18 is a consequence of Theorem 14.28. Conversely, knowing the Riesz–Herglotz theorem, the representation from Theorem 14.28 is obvious. Nevertheless, we find interesting to explain the way how the extreme points of P(U ) may be identified without recourse to the Riesz–Herglotz theorem, and how the representation follows from the Krein–Milman theorem.

14.7

Completely monotonic functions

Definition 14.38 (Completely monotonic functions). For a real-valued function f on [0, ∞) and α > 0 denote ∆α f : x 7→ f (x + α) − f (x),

x ∈ [0, ∞).

Let E := R[0,∞) be equipped with the product topology. Then E is a locally convex space (with the pointwise convergence topology) and ∆α : E → E is a linear operator. A real-valued function f on [0, ∞) is said to be completely monotonic if f ≥ 0 on [0, ∞) and (−1)n ∆αn . . . ∆α1 f ≥ 0 on [0, ∞) (14.15) holds for all αj > 0 and n ∈ N. Let CM denote the set of all completely monotonic functions on [0, ∞). Example 14.39. If µ is a finite Radon measure on [0, ∞), the function Z F : x 7→ e−tx dµ(t), x ∈ [0, ∞), [0,∞)

is completely monotonic. Indeed, the following computation (−1)n ∆αn . . . ∆α1 F (x) =

Z

n Y

1 − e−tαj e−tx dµ(t) ≥ 0,

[0,∞) j=1

yields the assertion. Lemma 14.40. Any completely monotonic function is decreasing and convex.

14.7 Completely monotonic functions

587

Proof. The monotonicity of f follows from (14.15) for n = 1 and convexity from (14.15) for n = 2. Definition 14.41 (The space K). Define K := {f ∈ E : f is completely monotonic, f (0) = 1}. For λ ∈ [0, ∞] define the functions fλ on [0, ∞) as ( e−λx , λ ∈ [0, ∞), fλ (x) = c{0} (x), λ = ∞. Proposition 14.42. The set K is a compact convex subset of E and ext K = {fλ : λ ∈ [0, ∞]}. Moreover, the set ext K is closed. Proof. Denote Λ := {fλ : λ ∈ [0, ∞]}. It is easy to check that the sum and a positive multiple of completely monotonic functions is again in CM. Hence, K is convex. It follows from Lemma 14.40 that 0 ≤ f ≤ 1 for any f ∈ K. Therefore, K ⊂ [0, 1][0,∞) ⊂ E. Since [0, 1][0,∞) is compact and K is obviously a closed subset of [0, 1][0,∞) , K is compact. Assume that ϕ ∈ ext K. When ϕ(z) = 1 for some z ∈ [0, ∞), taking into account that ϕ is decreasing and convex, we see that ϕ = f0 . So suppose that ϕ(t) < 1 for all t ∈ (0, ∞). Fix s > 0 such that ϕ(s) > 0 and define functions hs : t 7→

ϕ(t) − ϕ(t + s) 1 − ϕ(s)

and

gs : t 7→

ϕ(t + s) , ϕ(s)

t > 0.

Since hs (0) = gs (0) = 1 and the function t 7→ −∆s hs (t) = hs (s) − hs (t + s) is completely monotonic, hs , gs ∈ K. Since ϕ(t) = 1 − ϕ(s) hs (t) + ϕ(s)gs (t), t > 0, ϕ ∈ ext K implies that ϕ(t) = hs (t) = gs (t),

t > 0.

(14.16)

Hence, ϕ(t)ϕ(s) = ϕ(s + t)

(14.17)

588

14 Applications

for any pair s, t > 0 with ϕ(s) > 0. By a symmetry reason, (14.17) is also true if ϕ(t) > 0. Finally, if ϕ(s) = ϕ(t) = 0, then ϕ(s + t) = 0 as well by monotonicity. Completely monotonic functions are convex, and therefore continuous on (0, ∞). As a continuous solution of (14.17), we get ϕ(t) = e−λt ,

t > 0,

unless ϕ = 0 on (0, ∞). The exponent λ cannot be negative as completely monotonic functions are decreasing. From the above discussion it follows that ϕ = fλ for some λ ∈ [0, ∞], and thus ext K ⊂ Λ. It remains to show that all functions fλ , λ ∈ [0, ∞], are extreme points of K. Obviously, f0 , f∞ ∈ ext K, but the convex hull of {f0 , f∞ } does not equal K. Since ext K ⊂ Λ, the Krein–Milman theorem 2.22 provides λ ∈ (0, ∞) such that fλ ∈ ext K. For a fixed b > 0, consider the mapping Tb : f (x) 7→ f (bx), x ∈ [0, ∞). Then Tb is an affine homeomorphism of K onto K, and thus Tb carries every extreme point of K into an extreme point of K. Hence, Tb (fλ ) = fλb is an extreme point of K. Since this holds for every b > 0, we see that all functions from Λ are extreme. As the set Λ is obviously closed in K, the set ext K is closed. Theorem 14.43 (Bernstein). Let f be a real-valued function on [0, ∞). Then the following assertions are equivalent: (i) f ∈ K, (ii) there exist a unique Radon measure µ ∈ M+ ([0, ∞)) and ω > 0 such that kµk + ω = 1 and Z f (x) = e−xt dµ(t) + ωc{0} (x), x ∈ [0, ∞), (14.18) [0,∞)

(iii) f (0) = 1, f is a C ∞ -function on (0, ∞) and (−1)n f (n) ≥ 0 on (0, ∞) for each n ≥ 0.

(14.19)

Proof. (i) =⇒ (ii): We use the Krein–Milman theorem with transfer 2.33 to obtain the sought representation Z f (x) = fλ (x) dµ(λ) + ωc{0} (x), x ∈ [0, ∞). [0,∞]

Concerning the uniqueness, let µ, µ0 and ω, ω 0 be two representations. Since x → 7 R −xλ 0 . Let A be e dµ is a continuous function on [0, ∞), we obtain ω = ω [0,∞)

14.8 Positive definite functions on discrete groups

589

the linear span of the functions fλ , λ ∈ [0, ∞). Then A is an algebra and by the Stone–Weierstrass theorem A.29, A is dense in the space of all continuous functions on [0, ∞] (here we understand [0, ∞] as the one-point compactification of [0, ∞)). It follows that the representation (14.18) of a completely monotonic function is unique. (ii) =⇒ (iii): This follows from the fact that the measure µ is finite and therefore we can differentiate the integral with respect to t under the integral sign. (iii) =⇒ (i): Let CCM be the set of all C ∞ functions on (0, ∞) satisfying (14.19). From the mean value theorem we easily infer that each mapping f 7→ −∆α f maps CCM into itself. By iteration we see that (−1)n ∆αn . . . ∆α1 f ∈ CCM for any f ∈ CCM. Hence the functions satisfying (iii) are completely monotonic.

14.8

Positive definite functions on discrete groups

Definition 14.44 (Positive definite functions). Let G be a discrete abelian group (written additively). A function f : G → C is said to be positive definite if for all natural numbers n, all λ1 , . . . , λn ∈ C and x1 , . . . , xn ∈ G we have n X

λj λk f (xj − xk ) ≥ 0.

(14.20)

j,k=1

The aim of this section is to show that every positive definite function admits an integral representation in terms of very special positive definite functions, the socalled characters defined below. Denote by Md (G) the set of all finite linear combinations of Dirac measures on G. So to say that µ belongs to Md (G) means that there are complex numbers λ1 , . . . , λn and elements x1 , . . . , xn ∈ G such that µ=

n X

λj εxj .

(14.21)

j=1

With µ of the form (14.21), we associate the measure µ e ∈ Md (G) by µ e :=

n X

λj ε−xj .

j=1

Definition 14.45 (Convolutions). Given x, y ∈ G, we define εx ? εy := εx+y and, by linearity, µ ? ν is defined for all µ, ν ∈ Md (G) in an obvious way. Clearly, the convolution so defined is commutative and associative. Also, µ] ?ν =µ e ? νe. For f : G → C and µ ∈ Md (G) define µ ? f by Z (µ ? f )(x) := f (x − y) dµ(y), x ∈ G. G

Given y ∈ G, we put fy := εy ? f , so that fy (x) = f (x − y) for all x ∈ G.

590

14 Applications

It is worth mentioning that for µ as in (14.21) and f : G → C, the condition (µ ? µ e)(f ) ≥ 0 is nothing else than (14.20). Notice also that, for µ ∈ Md (G) and f : G → C, we have Z Z f (y) d(µ ? µ e)(y). f (−y) d(µ ? µ e)(y) = (µ ? µ e) ? f (0) = G

G

Consequently, f is positive definite if and only if (µ ? µ e ? f )(0) ≥ 0 whenever µ ∈ Md (G). Fix a positive definite function f on G, c ∈ C, x ∈ G and put µ := ε0 + cεx . Then 0≤ µ?µ e (f ) = (1 + |c|2 )f (0) + cf (x) + cf (−x). The choice c = 0 gives f (0) ≥ 0, the choice c = 1 and c = i yields 2f (0) + f (x) + f (−x) ≥ 0 and 2f (0) + i(f (x) − f (−x)) ≥ 0, respectively. Thus both numbers f (x) + f (−x) and i(f (x) − f (−x)) are real, which shows that f (−x) = f (x). If f (x) 6= 0, the choice c = −|f (x)|/f (x) then gives |f (x)| ≤ f (0). In particular, f (0) = 0 implies that f vanishes identically. Definition 14.46 (Characters). A character of G is a homomorphism γ : G → T where T denotes the (multiplicative) group of complex numbers of modulus one. In other words, γ is a character if |γ(x)| = 1 and γ(x + y) = γ(x)γ(y) whenever x, y ∈ G. It follows that γ(0) = 1 and 1 = γ(0) = γ(x − x) = γ(x)γ(−x), thus γ(−x) = γ(x). If µ is of the form (14.21) and γ is a character, then (µ ? µ e)(γ) =

n X

λj λk γ(xj − xk ) =

j,k=1

n X j=1

n X λj γ(xj ) λk γ(xk ) ≥ 0. k=1

We conclude that every character is a positive definite function on G. Let P denote the set of all positive definite functions f on G such that f (0) = 1 b be the set of all characters of G. Obviously, P is a convex set in the vector and G b ⊂ P. space of complex functions on G and G Notice first that for a positive definite function f and for ν ∈ Md (G), the function g := ν ? νe ? f is positive definite. Indeed, for an arbitrary µ ∈ Md (G) we have µ?µ e?g =µ?µ e ? ν ? νe ? f = (µ ? ν) ? (µ] ? ν) ? f, so that (µ ? µ e ? g)(0) ≥ 0. b Lemma 14.47. We have ext P ⊂ G.

14.8 Positive definite functions on discrete groups

591

Proof. Suppose that f ∈ ext P , y ∈ G and λ ∈ C. Define ν := ε0 + λεy , κ := ε0 − λεy and denote g := ν ? νe ? f, h := κ ? κ e ? f. Recall that g, h are positive definite functions and g = (1 + |λ|2 )f + λfy + λf−y ,

h = (1 + |λ|2 )f − λfy − λf−y .

Suppose that g(0) = 0. Since |g| ≤ g(0), the function λfy + λf−y is a real multiple of f . Similarly for the case h(0) = 0. If g(0) 6= 0 and h(0) 6= 0, we have f= and

g g(0)

∈ P,

h h(0)

g(0) g h(0) h + 2 2 1 + |λ| g(0) 1 + |λ| h(0)

∈ P . We know that g(0) ≥ 0, h(0) ≥ 0 and, obviously, g(0) + h(0) = (1 + |λ|2 )f (0) = 1 + |λ|2 .

Thus f is a convex combination of functions from P , and therefore both g and h are real multiples of f because f ∈ ext P . The same is true for g − h. We conclude that there exists a number a(λ) ∈ R such that λfy + λf−y = a(λ)f. Taking λ = 1 and λ = i we arrive at f−y = af , where a = 21 (a(1) + ia(i)). Hence f (x + y) = af (x), x ∈ G. The choice x = 0 shows that a = f (y). We see that f (x + y) = f (x)f (y) whenever x, y ∈ G. Pick z ∈ G. Since 1 = f (0) = f (z − z) = f (z)f (−z) = f (z)f (z) = |f (z)|2 , b we get f ∈ G. Definition 14.48 (The space S). Consider P as a subset of the topological vector space S of complex functions on G endowed with the topology of pointwise convergence. Obviously, for every x ∈ G, the mapping g 7→ g(x), g ∈ S, is a continuous linear b are closed subsets of {h ∈ S : |h| ≤ 1}, hence P and functional on S. Both P and G b are compact. G Lemma 14.49. For x ∈ G define x b : γ 7→ γ(x),

b γ ∈ G.

Then A := {b x : x ∈ G} b is a dense subset of the space C(G).

592

14 Applications

b A separates points of G b and 1 ∈ A Proof. Obviously, A is a linear subspace of C(G), (consider x b for x = 0). Further, the conjugate of every function from A belongs to A b Since x since γ(−x) = γ(x) for every x ∈ G and every γ ∈ G. byb = x[ + y whenever x, y ∈ G, A is an algebra. By the Stone–Weierstrass theorem A.29, A is a dense b subset of C(G). b define Lemma 14.50. For ν ∈ M+ (G) Z ν f : x 7→ γ(x) dν(γ),

x ∈ G.

b G

Then f ν is positive definite. b If ν ∈ M1 (G), b Proof. The assertion is obvious if ν is a molecular measure on G. ν then ν is a limit of a net of molecular measures. Hence f is a pointwise limit of positive definite functions, so f ν is positive definite. b Lemma 14.51. For every f ∈ P there exists a unique Radon measure ν ∈ M1 (G) ν such that f = f . b (see Lemma 14.47) and that G b is a closed subset of Proof. We know that ext P ⊂ G P . By the Integral representation theorem 2.31, given f ∈ P , there exists a Radon b such that measure ν ∈ M1 (G) Z f (x) = γ(x) dν(γ) b G

for each x ∈ G. The density result from Lemma 14.49 shows that ν is uniquely determined. b Corollary 14.52. We have ext P = G. Proof. Apply the result of Exercise 2.109. Now we are in a position to state the representation theorem. b is a bijection of M+ (G) b onto Theorem 14.53. The mapping ν 7→ f ν , ν ∈ M+ (G), the set of all positive definite functions. Proof. The assertion is an immediate consequence of Lemmas 14.50 and 14.51. Example 14.54. Consider the additive group Z of integers, so that complex functions on G = Z are simply complex sequences indexed by Z. Fix a character γ on Z and 1 . Since denote z := γ(1) n γ(n) = γ(1)

and

γ(−n) = γ(n)

14.9 Range of vector measures

593

for every natural number n, we have γ(k) = z −k , k ∈ Z. Since |z| = 1, we can b with the unit circle T. Given a complex Radon measure µ on T, we define identify Z the k-th Fourier coefficient of µ by Z z −k dµ(z), k ∈ Z. µ b(k) := T

Our result in the situation of the group G = Z reads as follows: A sequence {ak }∞ k=−∞ is positive definite if and only if there exists a Radon measure µ on T such that ak = µ b(k) for every k ∈ Z. Thus positive definite functions on Z are exactly sequences of Fourier coefficients.

14.9

Range of vector measures

In this section we will prove the so-called Lyapunov theorem on the range of nonatomic vector measures. The proof presented is a nice application of the Krein– Milman theorem 2.22. We will use the following notation: Σ will be a σ-algebra on a set X and ν a signed measure on Σ. The symbol |ν| stands for the variation of the measure ν. Recall that ν is said to be nonatomic, if every set M ∈ Σ such that |ν|(M ) > 0 contains a set Q ∈ Σ with 0 < |ν|(Q) < |ν|(M ). Lemma 14.55. Let ν be a finite measure on Σ and P := {g ∈ L∞ (ν) : 0 ≤ g ≤ 1} . Then P is a w∗ -compact convex subset of L∞ (ν) and ext P = {cM : M ∈ Σ} . Proof. Obviously, P is a convex set and cM ∈ ext P whenever M ∈ Σ. Suppose that g ∈ P is not a characteristic function of any set of Σ. Then there exist a ∈ (0, 1) and a set Q ∈ Σ such that ν(Q) > 0 and a≤g ≤1−a

on Q.

Define g1 := g + acQ and g2 := g − acQ . Then g1 , g2 ∈ P , g = 21 (g1 + g2 ) and g1 6= g, hence g ∈ / ext P . Since g ∈ P if and only if Z Z f g dν ≤ f dν 0≤ X

L1 (ν), P

X

is a w∗ -closed subset of the closed unit ball in L∞ (ν),

for every positive f ∈ where L∞ (ν) is considered as the dual of L1 (ν). Consequently, P is w∗ -compact by the Banach–Alaoglu theorem.

594

14 Applications

Lemma 14.56. Let ν be a finite measure on Σ, fj be elements of L1 (ν), j = 1, . . . , d , Z Z f1 g dν, . . . , fd g dν ∈ Rd , g ∈ L∞ (ν), T : g 7→ X

X

and N := ker T . Then T : L∞ (ν) → Rd is w∗ -continuous. If ν is nonatomic, then P ⊂ N + ext P . Proof. It follows immediately from the definition of the w∗ -topology that T is w∗ continuous. Consequently, N is a w∗ -closed subset of L∞ (ν). Suppose that ν is nonatomic and fix g ∈ P . Then K := P ∩ (g − N ) is a nonempty w∗ -compact convex subset of L∞ (ν). By the Krein–Milman theorem 2.22, there exists f ∈ ext K ⊂ K ⊂ P . We are going to show that f is the characteristic function of a set of Σ. Then f ∈ ext P by Lemma 14.55 and g = f + h for a suitable h ∈ N , since f ∈ g − N . We see that g ∈ N + ext P , which will finish the proof. So suppose that f ∈ ext K is not the characteristic function of any set of Σ. Then there exist a ∈ (0, 1) and Q ∈ Σ such that ν(Q) > 0 and a≤f ≤1−a

on Q.

Fix a natural number n > d. Since ν is nonatomic, there is a partition of Q into disjoint sets Q1 , . . . , Qn ∈ Σ, each of them of strictly positive measure. Denote by B the d × n matrix with entries Z fj dν, j = 1, . . . , d, k = 1, . . . , n, Qk

fix a non-trivial solution b = (b1 , . . . , bn ) ∈ Rn of the system Bx = 0 and define h :=

n X

bk cQk .

k=1

Then Z fj h dν = X

n X k=1

Z bk

fj dν = 0,

j = 1, . . . , d,

Qk

hence h ∈ N . Set h0 :=

ah , khk∞

g1 := f + h0 ,

g2 := f − h0 .

Then g1 , g2 ∈ P and h0 ∈ N . Since, by the definition of K, f = g − g0 for a suitable g0 ∈ N and g0 − h0 , g0 + h0 ∈ N , we have g1 , g2 ∈ g − N , so that g1 , g2 ∈ K. Since f = 12 (g1 + g2 ) and h0 is a nonzero element of L∞ (ν), f ∈ / ext K, which is a contradiction.

14.10 The Stone–Weierstrass approximation theorem

595

Corollary 14.57. If ν is nonatomic, then ext P is w∗ -dense in P . Proof. Let V be a w∗ -neighborhood of 0 in L∞ (ν), so that there are a natural number d and f1 , . . . , fd ∈ L1 (ν) such that Z ∞ V = g ∈ L (ν) : gfj dν ≤ 1, 1 ≤ j ≤ d . X

If N has the same meaning as in Lemma 14.56, then P ⊂ N + ext P ⊂ V + ext P , hence P is a subset of the w∗ -closure of ext P . Theorem 14.58. Let µ1 , . . . , µd be nonatomic signed measures on Σ and µ : M 7→ (µ1 (M ), . . . , µd (M )) ∈ Rd ,

M ∈ Σ.

Then µ(Σ) is a compact convex set in Rd . Proof. Let ν := |µ1 | + · · · + |µd |. Then the measure ν is obviously nonatomic, and finite by W. Rudin [402], Theorem 6.4. By the Radon–Nikodym theorem, there exist real-valued functions fj ∈ L1 (ν) such that µj = fj ν, j = 1, . . . , d. Let, as above, P := {g ∈ L∞ (ν) : 0 ≤ g ≤ 1} , T be defined as in Lemma 14.56 and N := ker T . By Lemma 14.55, µ(Σ) = T (ext P ). Lemma 14.56 yields T (P ) ⊂ T (N + ext P ) = T (ext P ) ⊂ T (P ). Thus µ(Σ) = T (P ). We see that µ(Σ) is a w∗ -continuous linear image of a w∗ compact convex set. Hence µ(Σ) is a compact convex set in Rd .

14.10

The Stone–Weierstrass approximation theorem

The aim of this section is to show how the Krein–Milman theorem 2.22 opens a way to a proof of the algebra version of the Stone–Weierstrass theorem. Also a lattice version of the Stone–Weierstrass theorem can be easily deduced using the Choquet theory. We start with the algebra version of the Stone–Weierstrass theorem. If K is a compact space and C(K) is a real or a complex space, we recall that A is an algebra if it is a linear space and f g ∈ A whenever f, g ∈ A. If C(K) is a real space, A ⊂ C(K) is said to be a lattice if A is a subspace of C(K) and f ∨ g, f ∧ g ∈ A whenever f, g ∈ A. Lemma 14.59. Let C(K) be a real or complex space and A ⊂ C(K) be an algebra such that f ∈ A implies f ∈ A, 1 ∈ A and A separates points of K. If s, t are different points of K, then there exists g ∈ A such that 0 < g < 1 and g(s) 6= g(t).

596

14 Applications

Proof. Obviously, there exists f ∈ A such that f (s) = 0 and f (t) 6= 0. Then the function ff , h := 1 + kf k2 belongs to A, 0 ≤ h < 1 and h(s) 6= h(t). Now, the function g := h +

1 1 − sup h(K) 2

has the required properties. Lemma 14.60. Let C(K) be a real or complex space, A ⊂ C(K) be an algebra and n o X := µ ∈ A⊥ : kµk ≤ 1 . (14.22) Let µ ∈ ext X and g ∈ A be a real-valued function such that 0 < g < 1. Then g is constant on spt µ. Proof. The case A⊥ = {0} is trivial. Otherwise we have kµk = 1. Define κ := gµ,

ν := (1 − g)µ.

Since 0 < g < 1, g and 1 − g are in A and A is an algebra, the measures κ and ν are nontrivial and belong to X. Notice that κ ν µ = kκk + kνk kκk kνk and

Z

Z kκk + kνk =

(1 − g) d|µ| = |µ|(K) = 1.

g d|µ| + K

K

By assumption, µ is an extreme point of X. Consequently, µ = κ/kκk, and therefore gµ = κ = kκkµ. Thus, g = kκk µ-almost everywhere. Hence g = kκk on spt µ because g is continuous. Theorem 14.61 (Stone–Weierstrass). Let K be a compact space, C(K) be a real or complex space and A ⊂ C(K) be an algebra such that A contains the constant functions, separates points of K and such that f ∈ A implies f ∈ A. Then A is dense in C(K). Proof. By the Hahn–Banach theorem it is sufficient to show that A⊥ = {0}. Assume that A⊥ 6= {0} and define X by (14.22). Obviously, X is a convex set. By Banach– Alaoglu’s theorem, X is w∗ -compact. By the Krein–Milman theorem 2.22, there is an extreme point µ of X. Obviously, kµk = 1. It follows from Lemmas 14.59 and

14.11 Invariant and ergodic measures

597

14.60 that spt µ is a singleton. Then µ = λεx for a suitable x ∈ X and λ ∈ C of modulus one. Now, µ ∈ A⊥ and 1 ∈ A implies Z 0= 1 dµ = λεx (1) = λ 6= 0, K

which is a contradiction. The proof is complete. Using the Choquet theory we also present a proof of the lattice version of the Stone– Weierstrass theorem. Theorem 14.62 (Stone–Weierstrass). Let K be a compact space and L a linear subspace of the real space C(K). If L is a lattice containing the constant functions and separating points of K, then L is dense in C(K). Proof. Obviously, L is a function space. We will show that ChL (K) = K. To this end, pick x ∈ K, ν ∈ Mx (L) and h ∈ L. Since the function ϕ : t 7→ |h(t) − h(x)|,

t ∈ K,

belongs to L, we get Z

Z |h(t) − h(x)| dν(t).

ϕ dν =

0 = ϕ(x) = K

K

Hence spt ν ⊂ {t ∈ K : h(t) = h(x)} . Since L separates points of K, \ spt ν ⊂ {t ∈ K : h(t) = h(x)} = {x} . h∈L

Hence ν = εx , so that x ∈ ChL (K). It follows that Ac (L) = C(K) (cf. Bauer’s characterization of the Choquet boundary 3.24 and Corollary 3.23(a)). Select now f ∈ C(K) = Ac (L) ⊂ Kc (L) and ε > 0. By Proposition 3.55, there exists h ∈ −W(L) = L such that f − ε ≤ h ≤ f . This implies f ∈ L and finishes the proof.

14.11

Invariant and ergodic measures

In the preceding sections, the Krein–Milman theorem was applied in order to establish an integral representation in various situations. Until now, in the cases considered, the set of extreme points was closed, so that the full strength of the Choquet representation theorem was not needed. In the case of the present section, the set of extreme points turns out to be nonclosed in general, hence the Krein–Milman theorem does not suffice for the integral representation theorem proved below.

598

14 Applications

Definition 14.63 (T -invariant measures and functions). Let K be a nonempty compact space and T be a nonempty family of continuous maps from K to itself. A Radon measure µ ∈ M+ (K) is called T -invariant if µ satisfies µ(B) = µ(S −1 B) for every Borel set B ⊂ K and every S ∈ T . In other words, for the image measure S] µ we have S] µ = µ. It is easy to see that T -invariant measures form a convex cone. If µ is a T -invariant measure, a Borel measurable function f on K is said to be µ-invariant if f =f ◦S

µ-almost everywhere on K for every S ∈ T .

A Borel set B ⊂ K is called µ-invariant if µ(B4S −1 B) = 0,

S ∈T.

Hence B is µ-invariant if and only if cB is a µ-invariant function. The following result will be useful for a description of extreme T -invariant measures. Lemma 14.64. Let µ be a T -invariant measure on K and f be a positive Borel measurable function on K. Then the measure ν := f µ is T -invariant if and only if f is µ-invariant. Proof. Let f be T -invariant, B ⊂ K be a Borel set and S ∈ T . Then Z Z Z −1 ν(S B) = f dµ = f ◦ S dµ = f d(S] µ) S −1 B S −1 B B Z = f dµ = ν(B). B

For the converse, fix S ∈ T and suppose that S] ν = ν. For a given α ∈ R, let A := {x ∈ K : f (x) ≤ α}, A1 := S −1 A \ A and A2 := A \ S −1 A. Then f − α > 0 on A1 , so Z ν(A1 ) − αµ(A1 ) =

(f − α)dµ ≥ 0 A1

and the equality holds if and only if µ(A1 ) = 0. Obviously, Z ν(A2 ) = f dµ ≤ αµ(A2 ). A2


599

Note that ν(A1 ) = ν(S −1 A) − ν(S −1 A ∩ A) = ν(A) − ν(S −1 A ∩ A) = ν(A2 ) and, similarly, µ(A1 ) = µ(A2 ). Now ν(A1 ) ≥ αµ(A1 ) = αµ(A2 ) ≥ ν(A2 ) = ν(A1 ), which yields ν(A1 ) − αµ(A1 ) = 0, hence 0 = µ(A1 ) = µ(A2 ). We conclude that the sets A and S −1 A = {y ∈ K : f ◦ S (y) ≤ α} differ only by a set of µ-measure zero. Now we will use the following equality for real-valued functions ϕ and ψ on K: [ {x ∈ K : ϕ(x) > ψ(x)} = {x ∈ K : ϕ(x) > r > ψ(x)} r∈Q

=

[

{x ∈ K : ψ(x) ≤ r} \ {x ∈ K : ϕ(x) ≤ r} .

r∈Q

Choosing ϕ = f , ψ = f ◦ S, we obtain f ≤ f ◦ S µ-almost everywhere. The choice ϕ = f ◦ S and ψ = f gives f ≥ f ◦ S µ-almost everywhere, finishing the proof. Lemma 14.65. If µ, ν are T -invariant measures, then µ ∧ ν is T -invariant. Proof. Given T -invariant measures µ, ν, we can write µ ∧ ν = (f ∧ g)(µ + ν), where f, g are positive Borel function satisfying µ = f (µ + ν) and ν = g(µ + ν). By Lemma 14.64, both f and g are µ + ν-invariant. Given S ∈ T , let A := {x ∈ K : f (x) < g(x)}. A straightforward verification shows that (µ + ν)(B4S −1 B) = 0, B is either A or K \ A. Hence (f ∧ g) ◦ S = f ∧ g

(µ + ν)-almost everywhere

for all S ∈ T . It follows that f ∧ g is (µ + ν)-invariant and thus µ ∧ ν is T -invariant by Lemma 14.64. Definition 14.66 (Ergodic measures). A T -invariant measure µ ∈ M1 (K) is called ergodic if the only µ-invariant sets B are trivial, that is, if they satisfy µ(B) = 0 or µ(B) = 1. (Obviously, any Borel set B with µ(B) ∈ {0, 1} is µ-invariant.) Let X denote the set of T -invariant probability measures on K considered as a subset of M(K) endowed with the w∗ -topology. Obviously, X is a convex set. Since Z Z o \n X= µ ∈ M1 (K) : f dµ − f ◦ S dµ = 0 K

K

for f running through C(K) and S through T , X is a compact set.

600

14 Applications

Let us remark that X may be empty in general (see the simple Example 14.67 below). The Markov–Kakutani theorem A.3 shows that X is nonempty provided that T is a commuting family of continuous maps. The theorem is applied as follows: the set C is M1 (K) with the w∗ -topology, the mapping TS : C → C is for S ∈ T defined by TS µ = S] µ, µ ∈ M1 (K), and T = {TS : S ∈ T }. Example 14.67. Set K := {x, y} (equipped with the discrete topology) and T := {S1 , S2 } where S1 (x) = S1 (y) = x

and

S2 (x) = S2 (y) = y.

Suppose that µ is an S1 -invariant measure. Then µ S1−1 {x} = µ(K) = 1, hence µ({x}) = 1. We conclude that µ = εx . Similarly we deduce that µ = εy . Therefore, the set of T -invariant measures is empty. Proposition 14.68. Given a T -invariant probability measure µ, the following assertions are equivalent: (i) µ ∈ ext X, (ii) µ is ergodic, (iii) every positive Borel measurable T -invariant function on K is constant µ-almost everywhere. Proof. (i) =⇒ (ii): Assume that the measure µ is not ergodic. Fix a µ-invariant set A such that 0 < µ(A) < 1. Define −1 µ1 := µ(A) µ|A

and

−1 µ2 := µ(K \ A) µ|K\A .

Then µ1 , µ2 ∈ M1 (K), µ1 6= µ and µ = µ(A)µ1 + 1 − µ(A) µ2 .

(14.23)

Choose S ∈ T and a Borel set B ⊂ K. Since (S −1 B) ∩ (A4S −1 A) = (S −1 B ∩ A)4(S −1 B ∩ S −1 A) and µ(A4S −1 A) = 0, we have µ(S −1 B ∩ S −1 A) = µ (S −1 B) ∩ A , hence µ|A (B) = µ(B ∩ A) = µ(S −1 (B ∩ A)) = µ(S −1 B ∩ S −1 A) = µ (S −1 B) ∩ A

= µ|A (S −1 B). It follows that µ1 is T -invariant and, by (14.23), also µ2 is T -invariant. The equality (14.23) shows that µ ∈ / ext X, proving the implication (i) =⇒ (ii).


601

(ii) =⇒ (iii): Assume that the measure µ is ergodic and f is a positive Borel µ-invariant function. Then f −1 (U ) is µ-invariant for any U ⊂ R Borel, and thus µ(f −1 (U )) ∈ {0, 1}. If {Un : n ∈ N} is a countable base of open sets in R, the set [ L = K \ {f −1 (Un ) : µ(f −1 (Un )) = 0} satisfies µ(L) = 0. Further, f is constant on K \ L. Indeed, if it were not true and f attains two different values y1 , y2 ∈ R on K \ L, we could separate them by sets Un1 , Un2 from the base and obtain a contradiction with our choice of L. Let (iii) hold and let µ, µ1 , µ2 ∈ X be such that µ = 12 (µ1 + µ2 ). Let fj , j = 1, 2, be the Radon–Nikodym derivative of µj with respect to µ. By Lemma 14.64, fj is T -invariant, hence constant µj -almost everywhere. Since µj ∈ M1 (K), fj = 1 µj almost everywhere, so that µ = µ1 = µ2 and µ ∈ ext X proving the implication (iii) =⇒ (i). Proposition 14.69. The set X is a simplex. Proof. We consider the space span X with the order inherited from M(K). Let φ : span X → (Ac (X))∗ be the evaluation mapping (that is, φ(µ)(F ) = F (µ), µ ∈ X, F ∈ Ac (X)). We claim that φ preserves order, that is, φ(µ) is a positive functional if and only if µ is a positive measure. Any function f ∈ C(K) provides an element fb ∈ Ac (X) via the identification fb(µ) = µ(f ), µ ∈ X. (We remark that it is easy to see from Lemma 2.34 that the space {fb : f ∈ C(K)} is dense in Ac (X).) If µ ∈ span X is a positive measure, then µ = cν for some c ≥ 0 and ν ∈ X. Thus φ(µ)(F ) ≥ 0 for any F ∈ Ac (X) positive. Conversely, let φ(µ) be a positive functional on Ac (X). Then for any f ∈ C(K) positive, fb ≥ 0 on X, and thus µ(f ) = φ(µ)(fb) ≥ 0. Hence µ is a positive measure. It follows from Lemma 14.65, Proposition 4.31(d) and Proposition A.15 that the ordered vector space (Ac (X))∗ is a lattice. By Theorem 6.54, X is a simplex. Now, the Choquet theorem 3.45 shows that every T -invariant measure has a representation as an “integral average” of ergodic measures. Theorem 14.70. Let K be a metrizable compact convex set and X be as in Definition 14.66. For every µ ∈ X there exists a unique Radon measure m ∈ M1 (X) such that m is carried by the set of ergodic measures and Z µ(f ) = ν(f ) dm(ν), f ∈ C(K). (14.24) X

Proof. Let µ ∈ X be fixed. Since C(K) is separable, the w∗ -topology on X is metrizable. By the Choquet representation theorem 3.45, Propositions 14.69 and 14.68, there exists a unique probability measure m on X carried by the set of ergodic measures such that Z F (µ) = F (ν) dm(ν), X

602

14 Applications

whenever F is a w∗ -continuous linear functional on M(K). Since, for a fixed f ∈ C(K), the mapping fb : λ 7→ λ(f ), λ ∈ M(K), is linear and w∗ -continuous on M(K), we get the assertion. As already mentioned, the set of ergodic measures need not be closed, as the following example reproduced from R. R. Phelps [374] shows. Example 14.71. Let I := [0, 1], let J be the unit circle represented as R(mod 1). Consider K := I × J with the product topology and T consisting of the single mapping S : (x, y) 7→ (x, x + y), (x, y) ∈ K. For n ∈ N denote by µn the Radon measure on K which gives mass 1/n to the n points (1/n, k/n), 0 ≤ k ≤ n − 1. Then µn is an ergodic measure. Indeed, let B ⊂ K be a Borel set and Kn := (1/n, k/n) : 0 ≤ k ≤ n − 1 . Then B is µn -invariant if and only if either B ∩ Kn = ∅ or B ∩ Kn = Kn , hence µn is an ergodic measure. The sequence {µn } converges to the uniform distribution µ on {0} × J (the Haar measure on J). However, µ is not ergodic since every Borel set B ⊂ K is µ-invariant. Remark 14.72. We indicate an instructive example of a compact convex set of T invariant measures which is a Bauer simplex. Let L be a compact metrizable space and K be the product space LN . Let T be the set of all transformations of K of the form (xj ) 7→ (xσ (j)), where σ is a permutation of N that changes only finitely many elements. In this situation, if µ ∈ M 1 (K) is a T -invariant measure, then N µ is ergodic, 1 if and only if there exists a measure λ ∈ M (L) such that µ = i∈N λi , where λi = λ for all i ∈ N. (This is the result of E. Hewitt and L. J. Savage [225]; a probabilistic interpretation is mentioned in V. P. Fonf, J. Lindenstrauss and R. R. Phelps [179], p. 616.) As before, let us denote by X the compact convex set of T -invariant N probability measures on K. Then the mapping λ 7→ i∈N λi , λi = λ for i ∈ N, is a homeomorphism of the compact set M1 (L) onto the set of ergodic measures, that is, by Proposition 14.68, onto ext X. Hence X is a metrizable Bauer simplex. Remark 14.73. However, T -invariant measures may also give rise to the Poulsen simplex. An example of such a situation is outlined in V. P. Fonf, J. Lindenstrauss and R. R. Phelps [179], p. 618: K is the product space {0, 1}Z and T consists of the natural shift mapping of K. Then the compact convex set X of T -invariant probability measures is the Poulsen simplex. We note that both examples mentioned above fit into a general framework studied in E. Glasner and B. Weiss [194]. They proved, in particular, for the case of a countable discrete group G and a compact metric space

603

14.12 Exercises

K := {0, 1}G , the following dichotomy result: the set of G-invariant measures on K is either a Bauer simplex or the Poulsen simplex. Actually, the former case occurs if and only if G has the so-called property T of Kazhdan. For further results and literature on simplices of invariant measures we refer to T. Downarowicz [148].

14.12

Exercises

Exercise 14.74. Let µ ∈ M1 ([0, 1]) and Z Z x−ξ ξ−x 1 1 dµ(ξ) + dµ(ξ), k : x 7→ 2 [0,x) 1 − ξ 2 (x,1] ξ

x ∈ [0, 1].

Prove that k ∈ K (see Definition 14.1). Hint. The proof follows by a routine verification. Exercise 14.75. If f ∈ P(U ) and n ∈ N, then |f (n) (z)| ≤

2n! , (1 − |z|)n+1

z ∈ U.

Hint. The function f 7→ |f (n) (z)| is convex and continuous on A(U ). By Bauer’s concave minimum principle (see Corollary 2.24) and Theorem 14.36, max |f (n) (z)| : f ∈ P(U ) = max |g (n) (z)| : g ∈ M ≤

2n! . (1 − |z|)n+1

Exercise 14.76. Let W be an infinite subset of U , g ∈ A(U ), g (n) (0) 6= 0 for any n = 0, 1, . . . . For w ∈ W define gw : z 7→ g(wz),

z ∈ U.

Then the linear hull of {gw : w ∈ W } is dense in A(U ). Hint. Assume that there exists a nontrivial continuous linear functional A on A(U ) annihilating the linear hull of {gw : w ∈ W }. It follows from Theorem 14.33 that there exists {αn }∞ n=0 ∈ S such that (14.10) holds for f of the form (14.9). For w ∈ W we have ∞ X g (n) (0) n n gw (z) = w z , z ∈ U, n! n=0

hence Agw =

∞ X n=0

αn

g (n) (0) n w = 0. n!

(14.25)

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14 Applications

Since lim supn→∞

p n |αn | < 1, the function z 7→

∞ X g (n) (0) n=0

n!

wn z n

is holomorphic on a neighborhood of U . By (14.25), it has an infinite number of zeros in U . Consequently, g (n) (0) = 0 for any n = 0, 1, . . . , n! which implies A = 0, a contradiction. αn

Exercise 14.77. Let K be a compact space and A a linear subspace of the real space C(K). If A is an algebra, then A is a lattice . (Hence, if moreover A contains the constant functions and separates points of K, then the result of Theorem 14.61 is an immediate consequence of Theorem 14.62.) Hint. Let f ∈ A, f 6= 0, and let {Pn } be a sequence of polynomials such that Pn (0) = 0 and Pn (x) → |x| uniformly on [−1, 1]. It is easy to verify that the sequence {kf kPn kff k } converges uniformly to |f | on K. Hence |f | ∈ A. Exercise 14.78. Let K be a compact space and L a subspace of the real space C(K) containing the constant functions. If L is a lattice, then L is an algebra. First hint. Fix f ∈ L, f 6= 0, and prove that f 2 ∈ L. For this, put a := kf k and denote by A the set of all functions ϕ ∈ C([−a, a]) such that ϕ ◦ f ∈ L. Then A is a closed linear subspace of C([−a, a]), A is a lattice containing the constant functions and separating points of [−a, a] (consider the identity function). By Theorem 14.62, A = C([−a, a]). In particular, the function ϕ : t 7→ t2 , t ∈ [−a, a], belongs to A, thus f 2 ∈ L. Second hint. Fix f ∈ L, f 6= 0, and prove again that f 2 ∈ L. Put a := kf k. Given ε > 0, there are affine functions ϕ1 , . . . , ϕn on [−a, a] such that 2 t − (ϕ1 ∨ · · · ∨ ϕn )(t) ≤ ε, whenever t ∈ [−a, a] (a polygonal line inscribed to the graph of the function t 7→ t2 on [−a, a]). Since ϕj ◦ f ∈ L, j = 1, . . . , n, and L is obviously a lattice, f 2 is a uniform limit of functions of L, thus f 2 ∈ L. Exercise 14.79. Let f be a real-valued function on [0, ∞). Prove that f is completely monotonic if and only if n X n (−1)k f (x + kδ) ≥ 0 k k=1

for any n ≥ 0, x ∈ [0, ∞) and δ ≥ 0.


605

Hint. The assertion is merely a reworded definition of completely monotonic functions. Exercise 14.80. Let X, Y be Banach spaces and T be an isometric isomorphism of X onto Y . Then T is an extreme point of the unit ball BL(X,Y ) of the space L(X, Y ) of bounded operators from X to Y . Hint. The dual mapping T ∗ maps the set ext BY ∗ onto the set ext BX ∗ . Suppose that T = 12 (T1 +T2 ) for T1 , T2 ∈ BL(X,Y ) . If f ∈ ext BY ∗ , then T ∗ f ∈ ext BX ∗ , and since T ∗ f = 12 (T1∗ f + T2∗ f ), we have T ∗ f = T1∗ f = T2∗ f . This yields f (T1 x − T2 x) = 0 for every x ∈ X and every f ∈ ext BY ∗ . By the Krein–Milman theorem 2.22, f (T1 x − T2 x) = 0 for every x ∈ X and every f ∈ BY ∗ , which implies T1 x = T2 x for every x ∈ X, and therefore T1 = T2 .

14.13

Notes and comments

The exposition of Section 14.1 follows R. M. Rakestraw [383], where also the case of uniqueness of measures µ and ν of Theorem 14.4 is discussed. For the following integral representation result, see L. Hörmander [239], p. 24: Let f be a bounded continuous convex function on [0, 1], ( (x − 1)y, 0 ≤ y ≤ x ≤ 1, G(x, y) = x(y − 1), 0 ≤ x ≤ y ≤ 1. Then there exists a uniquely determined Radon measure µ on (0, 1) such that Z f (x) = f (1) + (x − 1)f (0) + G(x, y) dµ(y), x ∈ [0, 1]. [0,1]

In fact, µ turns out to be the second distributive derivative of f . The same result (formulated in terms of one-dimensional potential theory) can be found in J. L. Doob [145], p. 260. A representation theorem similar to that of Theorem 14.4 is proved in W. Blaschke und G. Pick [60] using tools of classical analysis. The proof of main Theorem 14.16 on a representation of doubly stochastic matrices was inspired by a hint to Exercise 5.15 (p. 60) in C. Berg, J. P. R. Christensen and P. Ressel [54] where matrices are considered as chessboards. Another proof can be based on M. Hall Jr. [205] (Theorem 5.1.9), and a similar proof is given in K. Jacobs [242]. A still different proof is presented in A. W. Roberts and D. E. Varberg [388] (Sec. 63, Theorem A). The result on the representation of doubly stochastic matrices goes back to G. Birkhoff [57]. The history of representation of doubly stochastic matrices is described in the paper by S. King and R. Shiflett [269] where related results on infinite doubly stochastic matrices and operators are reviewed. In particular, doubly stochastic measures are considered. Let I = [0, 1] and let λ stand

606

14 Applications

for Lebesgue measure on I. Recall, that a probability measure µ defined on I × I is said to be doubly stochastic if µ(A × I) = µ(I × A) = λ(A) whenever A ⊂ I is a Borel set. A description of the extreme points of the convex set M of doubly stochastic measures was independently discovered by R. G. Douglas [147] and J. Lindenstrauss [305]. Their result is expressed in terms of the space of functions S := {(x, y) 7→ f (x) + g(y), (x, y) ∈ I × I : f, g ∈ L1 (λ)} and reads as follows: A measure µ ∈ M is an extreme point of M if and only if S is a dense subset of L1 (µ) (cf. Theorem 7.60). e ) in Corollary 14.19 without recourse to the The description of the set ext H(U Riesz–Herglotz theorem was established in D. H. Armitage [20] or in J. Lukeˇs and I. Netuka [325], giving an answer to the question of R. R. Phelps who asks for a e ) ⊂ {P z : z ∈ ∂U } in [374] simple and elementary proof of the inclusion ext H(U e ) obtained in (1st ed., p. 118 and 2nd ed., p. 103). The characterization of ext H(U [325] uses standard facts from abstract analysis combined with a minimum of very basic results on harmonic functions. In [325] the reader can also find historical notes and comments concerning the Riesz–Herglotz theorem. The representation theorem 14.21 for typically real function goes back to M. S. Robertson [389]. The proof presented here is patterned after G. A. Edgar [154]. For a different approach (and a proof of uniqueness of the representation), see G. Schober [411], Theorem 2.17. The exposition on holomorphic functions with positive real part presented in Section 14.6 follows the paper [279] of R. A. Kortram. The origin of Theorem 14.43 appeared in Bernstein’s paper [55]. Several proofs of this theorem came afterwards; see, for example, D. V. Widder [467]. But it was G. Choquet who draws attention in [104] to the use of the Krein–Milman theorem for the proof. Another proof following Choquet’s idea can be found in R. R. Phelps [374]. Note that there exists a closely related class of functions, namely totally monotone functions. To derive a similar representation of totally monotone functions, G. Choquet used a theory of well-capped cones; for details, see G. Choquet [108], Theorem 32.6. The Bochner type theorem 14.53 for a discrete group is treated, for instance, in G. Choquet [107] and R. Becker [47]. The classical Bochner theorem reads as follows: If f : R → C is a continuous positive definite function, f (0) = 1, then there exists a unique Radon measure µ ∈ M1 (R) such that Z f (x) = eixy dµ(y), x ∈ R. R

For various generalizations of the classical Bochner theorem (where the group is R), consult G. Choquet [108], volume II, or R. R. Phelps [374], 2nd edition, pp. 103–105, or G. B. Folland [177], pp. 76–121, where further references can be found. Note that


607

the general case usually requires an extension of the representation theorems introducing, for instance, the notion of universal cap of a cone. However, in [110], G. Choquet gives a proof of a general Bochner–Weil theorem on abelian locally compact groups based on a reduction to the discrete case presented here. Theorem 14.58 was proved by A. A. Lyapunov in [300]. The idea to use the Krein– Milman theorem for the proof of the Lyapunov convexity theorem 14.58 goes back to J. Lindenstrauss [306]. Lemma 14.56 is taken from R. B. Holmes [238]. Let us remark that the assertion of the Lyapunov convexity theorem holds only in finite-dimensional spaces. The proof of the Stone–Weierstrass theorem 14.61 follows L. de Branges [133]. For a generalization, see W. Rudin [403], Chap. 5. The proof of Theorem 14.62 was inspired by a paper of J. Bliedtner [61]. The exposition of Section 14.11 is based on Phelps’ book [374] where a more general situation is considered; see also the survey paper [179] by V. P. Fonf, J. Lindenstrauss and R. R. Phelps, or G. Choquet [108] (volumes II and III). A discussion on various definitions of ergodic measures is contained there and uniqueness of the representation (14.24) is established. Also, further references related to the subject are to be found there. The hints of Exercise 14.78 are taken from a short paper of C. Dellacherie [142]; cf. also a paper by G. Nöbeling and H. Bauer [364]. Exercise 14.80 is taken from a paper by D. Milman [345]. In general, an operator T ∈ ext BL(X,Y ) need not be an isometry. For a partial converse and further discussions, see N. M. Roy [401]. For a finite-dimensional Banach space X, every extreme point of BL(X) is an isometry if and only if X is a Hilbert space; see M. A. Navarro [354].

Chapter A

Appendix

This chapter presents a survey of notions and facts needed throughout the book. As was already mentioned in the introduction, our key references are W. Rudin [403] and M. Fabian, P. Habala, P. Hájek, V. Montesinos Santaluc´ıa, J. Pelant and V. Zizler [173] for functional analysis, L. Asimow and A. J. Ellis [24] for ordered vector spaces, D. H. Fremlin [182], [181] and [183] for measure theory, R. Engelking [169] and K. Kuratowski [285] for topology, A. S. Kechris [262] and C. A. Rogers and J. E. Jayne [394] for descriptive set theory, D. H. Armitage and S. J. Gardiner [21] for classical potential theory and J. Bliedtner and W. Hansen [66] for abstract potential theory. Section A.5 also uses the paper G. Koumoullis [280].

A.1 A.1.A

Functional analysis Locally convex spaces

We recall that a real linear topological space E is locally convex if there exists a base of neighborhoods of 0 consisting of convex sets. Alternatively, E is a locally convex space if there exists a family P of pseudonorms that generates the topology of E (see [403], Theorem 1.36 and Theorem 1.37). A base of neighborhoods of 0 thus can be chosen in such a way that each of its elements U is closed, convex and balanced, that is, U is convex and λU ⊂ U for each |λ| ≤ 1. If E is a vector space and M is a set of linear mappings of E to R that separates points of E, the locally convex topology on E generated by M is denoted as σ(E, M ). We recall that a locally convex space is Fréchet if its topology is induced by a complete translation invariant metric (see Definition 1.8 in [403]). A set A ⊂ E is bounded if, given an open set containing 0, there exists λ > 0 such that A ⊂ λU . Any compact set in E is bounded (see [403], Theorem 1.15) and bounded sets can be characterized by the following property: whenever {xn } is a sequence in A and λn → 0, then λn xn → 0 (see [403], Theorem 1.30). We recall that a nonempty set C in a vector space is a convex cone if C is stable with respect to addition and multiplication by positive scalars. It is said to be proper if C ∩ (−C) = {0}. The following geometric versions of the Hahn–Banach theorem are used throughout the book. Theorem A.1 (Geometric Hahn–Banach theorem). Let A, B be nonempty disjoint convex sets in a locally convex space E.

A.1 Functional analysis

609

(a) If A is open, then there exists f ∈ E ∗ and c ∈ R such that f (a) < c ≤ f (b),

a ∈ A, b ∈ B.

(b) If A is compact and B is closed, then there exists f ∈ E ∗ , c1 , c2 ∈ R such that f (a) < c1 < c2 < f (b),

a ∈ A, b ∈ B.

Proof. See [403], Theorem 3.4. Theorem A.2 (Mazur). Let A be a convex subset of a locally convex space. Then A is closed if and only if it is weakly closed. Proof. See Theorem 3.12 in [403]. We also recall the following fixed point theorem. Theorem A.3. Let X be a nonempty compact convex set in a locally convex space Y and T be a commuting family of continuous affine mappings carrying X into itself. Then there exists a point x ∈ X such that T x = x for every T ∈ T . Proof. See [403], Theorem 5.23.

A.1.B

Banach spaces

If E is a Banach space we recall that E is canonically isometrically embedded to its second dual E ∗∗ via the evaluation mapping x 7→ εx defined as εx (x∗ ) = x∗ (x), x∗ ∈ E ∗ . Theorem A.4 (Goldstine’s lemma). The closed unit ball BE is w∗ -dense in BE ∗∗ . Proof. See [173], Theorem 3.27. Theorem A.5 (Banach–Alaoglu). The closed unit ball BE ∗ in E ∗ is a w∗ -compact subset of E ∗ . It is metrizable in the w∗ -topology if and only if E is separable. Proof. See [173], Theorem 3.21 and [173], Proposition 3.24. If A is a subset of a Banach space E, we recall that the annihilator A⊥ is defined as A⊥ := {x∗ ∈ E ∗ : x∗ = 0 on A}. Analogously, for B ⊂ E ∗ we denote B⊥ := {x ∈ E : b(x) = 0 for each b ∈ B}. We recall the following standard descriptions of dual spaces for subspaces.

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Proposition A.6. Let F be a closed subspace of a Banach space E. (a) For f ∗ ∈ F ∗ , let e∗ be its Hahn–Banach extension and let T f ∗ := e∗ + F ⊥ . Then T is an isometric isomorphisms of F ∗ onto E ∗ /F ⊥ . (b) If q : E → E/F is the quotient mapping, let T : (E/F )∗ → F ⊥ be defined as T y ∗ := y ∗ ◦ q. Then T is an isometric isomorphism of (E/F )∗ onto F ⊥ . (c) Let q : E ∗ → F ∗ be the restriction mapping and T : F ∗∗ → E ∗∗ be defined as T f ∗∗ (e∗ ) = f ∗∗ (qe∗ ), e∗ ∈ E ∗ . Then T is an isometric isomorphism of F ∗∗ onto F ⊥⊥ , which is moreover a (w∗ –w∗ )-homeomorphism. Proof. The assertions (a) and (b) can be found as Theorem 4.9 of [403], and (c) follows from (a) and (b). The verification of the fact that T is a (w∗ –w∗ )-homeomorphism is straightforward. It easily follows from the duality theory that a linear functional f on E ∗ is w∗ continuous if and only if f = εx for some x ∈ E. The following result shows that the w∗ -continuity of f can be checked only on bounded subsets of E ∗ provided E is a Banach space. Theorem A.7 (Banach–Dieudonné). Let E be a Banach space. A convex set A ⊂ E ∗ is w∗ -closed if and only if A ∩ nBE ∗ is w∗ -closed for every n ∈ N. Proof. See [173], Theorem 4.44. Theorem A.8 (Bishop–Phelps). If E is a Banach space, then the set of functionals in E ∗ that attain their norm is norm dense in E ∗ . Proof. See [173], Theorem 3.54. Theorem A.9 (Schur). If {xn } is a sequence in `1 weakly converging to 0, then xn → 0 in norm. Proof. See [173], Theorem 5.19.

A.1.C

Ordered Banach spaces and lattices

Definition A.10 (Ordered vector spaces). Let E be a (real) vector space and C be a proper convex cone in E satisfying E = C − C. By setting x≤y

if and only if

y − x ∈ C,

we get a partial order on E. It has the following properties • x ≤ x for every x ∈ E, •

if x, y ∈ E, x ≤ y and y ≤ x, then x = y,

•

if x, y, z ∈ E, x ≤ y and y ≤ z, then x ≤ z,

A.1 Functional analysis •

if x, y ∈ E, x ≤ y, then x + z ≤ y + z for all z ∈ E,

•

if x, y ∈ E, x ≤ y, then λx ≤ λy for all λ ≥ 0,

•

if x, y ∈ E, x ≤ y, then −y ≤ −x.

611

We say that E is an ordered vector space with the positive cone E + := C. Proposition A.11. For an ordered vector space E the following properties are equivalent: (i) (Riesz decomposition property): If b1 , b2 , a ∈ E + and 0 ≤ a ≤ b1 + b2 , then there exist a1 , a2 ∈ E + such that a = a1 + a2 and a1 ≤ b1 , a2 ≤ b2 . (ii) (Riesz refinement property): If a1 , a2 , b1 , b2 ∈ E + and a1 + a2 = b1 + b2 , then there exist cij ∈ E + , i, j = 1, 2, such that ai = ci1 + ci2 , i = 1, 2,

and

bj = c1j + c2j , j = 1, 2.

(iii) (Riesz interpolation property): If a1 , a2 , b1 , b2 ∈ E and ai ≤ bj for each i, j = 1, 2, then there exists c ∈ E such that ai ≤ c ≤ bj for each i, j = 1, 2. Pn + (i’) If 0 ≤ a ≤ P i=1 bi with b1 , . . . , bn positive, then there exist a1 , . . . , an ∈ E n such that a = i=1 ai and ai ≤ bi , i = 1, . . . , n. P P (ii’) If ni=1 ai = m 1 , . . . , bm positive, then there exist j=1 bj for a1 , . . . , an , bP Pn {cij : m 1 ≤ i ≤ n, 1 ≤ j ≤ m} such that ai = j=1 cij , 1 ≤ i ≤ n, and bj = i=1 cij , 1 ≤ j ≤ m. (iii’) If a1 , . . . , an , b1 , . . . , bm are in E and ai ≤ bj for each i = 1, . . . , n and j = 1, . . . , m, then there exists c ∈ E with ai ≤ c ≤ bj , i = 1, . . . , n, j = 1, . . . , m. Proof. We first show by induction the equivalences (i) ⇐⇒ (i’), (ii) ⇐⇒ (ii’) and Pn+1 . . , bn+1 positive. Then there exist (iii) ⇐⇒ (iii’). Let 0 ≤ a ≤ i=1 bi for b1 , . P c, an+1 ∈ E with a = c+an+1 such that 0 ≤ c ≤ ni=1 bi and 0 ≤ an+1 ≤ Pbnn+1 . The inductive assumption yields the existence of a1 , . . . , an positive with a = i=1 ai and ai ≤ bi , i = 1, . . . , n. Hence (i) =⇒ (i’). To show (ii) =⇒ (ii’), we first prove the assertion inductively for the case of a pair a1 , a2 . Let a1 , a2 and b1 , . . . , bm+1 be such that a1 , a2 ≤ bj , 1 ≤ j ≤ m + 1. Using the inductive assumption we find c0 with a1 , a2 ≤ c0 ≤ bj , 1 ≤ j ≤ m, and then we select c such that a1 , a2 ≤ c ≤ c0 , bm+1 . We proceed analogously in the general case a1 , . . . , an . To verify (iii) =⇒ (iii’)P we use a similar reasoning as in the previous First we Pn case.P m handle the case a1 + a2 = j=1 bj and then we pass to the case i=1 ai = m j=1 bj . Now we prove (i) =⇒ (ii) =⇒ (iii) =⇒ (i). (i) =⇒ (ii): Given a1 +a2 = b1 +b2 with positive terms, we have 0 ≤ a1 ≤ b1 +b2 . By (i) there exist c11 , c12 ∈ E + such that c11 ≤ b1 , c12 ≤ b2 and a1 = c11 + c12 . Now it suffices to put c21 := b1 − c11 and c22 := b2 − c12 .

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(ii) =⇒ (iii): If ai ≤ bj for each i, j = 1, 2, then (b1 − a1 ) + (b2 − a2 ) = (b1 − a2 ) + (b2 − a1 ) and each member of this equality is positive. By (ii), there exist cij ∈ E + , i, j = 1, 2, such that b1 − a1 = c11 + c12 , b2 − a2 = c21 + c22 , b1 − a2 = c11 + c21 ,

b2 − a1 = c12 + c22 .

We set c := a1 + c12 . Then c = a2 + c21 ,

b1 = c + c11 ,

b2 = c + c22 ,

and thus (iii) follows. (iii) =⇒ (i): Let b1 , b2 , a ∈ E + with 0 ≤ a ≤ b1 + b2 . Then 0, a − b2 ≤ a, b1 . Assuming (iii), there exists c ∈ E such that 0, a − b2 ≤ c ≤ a, b1 . By setting a1 := c and a2 := a − c, we finish the proof. Definition A.12 (Supremum and infimum). If (E, ≤) is a partially ordered set and M a nonempty subset of E, then z is called an upper bound of M if m ≤ z for all m ∈ M , and z is called a least upper bound of M , or supremum of M , if z is an upper bound of M and z ≤ y for any upper bound y of M . Any subset of E has at most one supremum. W The supremum of M will be denoted by M . The supremum of a two-point set {x, y} will be denoted by x ∨ y. TheVnotion of greatest lower bound or infimum is defined in a similar way with notation M and x ∧ y. Definition A.13 (Vector lattices). A partially ordered set (E, ≤) is called a lattice if every two-point subset of E has a supremum and an infimum. The ordered vector space E is called a vector lattice if it is a lattice. Given x ∈ E, we write x+ := x ∨ 0, x− := (−x) ∨ 0 = −(x ∧ 0) and |x| := x ∨ (−x). Lemma A.14. Let E be an ordered vector space and assume that a, b ∈ E are such that a ∧ b exists. Then (a) for any c ∈ E, (a + c) ∧ (b + c) exists and equals (a ∧ b) + c, (b) if λ ≥ 0, then λa ∧ λb exists and equals λ(a ∧ b), (c) (−a) ∨ (−b) exists and equals −(a ∧ b).

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613

Proof. The proof follows by a straightforward verification; see also Proposition 2.5.1 in [24]. Proposition A.15. Let E be an ordered vector space. Then the following assertions are equivalent: (i) E is a vector lattice, (ii) a ∧ b exists for each a, b ∈ E + , (iii) a ∨ 0 exists for all a ∈ E. Proof. Obviously, (i) =⇒ (ii) and (i) =⇒ (iii). If a, b ∈ E are arbitrary, we can find c ∈ E such that a + c, b + c ∈ E + (we recall that E = E + − E + ). Then a ∧ b = (a + c) ∧ (b + c) − c and a ∨ b = −((−a) ∧ (−b)). Hence (ii) =⇒ (i). Now assume (iii) and choose a, b ∈ E. By (iii), (a − b) ∨ 0 exists. Then (a − b) ∨ 0 + b is the supremum a ∨ b. So (iii) =⇒ (i). Lemma A.16. Let E be a vector lattice. Then (a) a + b = a ∨ b + a ∧ b, (b) a = a+ − a− , (c) if a = b − c for b, c positive, then b ≥ a+ , c ≥ a− , (d) a+ + a− = |a| = a+ ∨ a− , (e) a+ ∧ a− = 0. Proof. We follow the proof of Proposition 2.5.3 in [24]. For the proof of (a), a − (a ∨ b) = a + (−a ∧ −b) = 0 ∧ (a − b) = 0 ∧ (a − b) + b − b = (b ∧ a) − b. Hence (b) follows with b = 0 in (a) and (c) is obvious from the definition of a+ , a− . Since |a| + a = (a ∨ −a) + a = (2a) ∨ 0, we get |a| = 2a+ − (a+ − a− ) = a+ + a− . Hence |a| ≥ 0 and thus |a| = |a| ∨ 0 = (a ∨ −a) ∨ 0 = (a ∨ 0) ∨ (−a ∨ 0) = a+ ∨ a− yields (d). Finally, (e) follows from (a) applied to a+ and a− . Proposition A.17. If an ordered vector space E is a lattice, then it has the Riesz decomposition property.

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A Appendix

Proof. Let 0 ≤ a ≤ b1 + b2 with b1 , b2 ≥ 0 be given. We set a1 := a ∧ b1 and a2 := a − a1 . Then a1 ≤ b1 . Since a − b2 ≤ a, b1 , we get a − b2 ≤ a ∧ b1 , and thus a2 = a − a ∧ b1 ≤ b2 . Thus E has the Riesz decomposition property. Definition A.18 (Ordered locally convex spaces and their duals). A locally convex space E is an ordered locally convex space if E is ordered by a proper closed convex cone E + . If E is a normed linear space or even a Banach space, we call it an ordered normed linear space or an ordered Banach space. A functional x∗ ∈ E ∗ is positive if x∗ (x) ≥ 0 for all x ∈ E + . If (E ∗ )+ denotes the proper convex cone of all positive elements x∗ ∈ E ∗ , then (E ∗ , w∗ ) is also an ordered locally convex space provided E ∗ = (E ∗ )+ − (E ∗ )+ . If E is an ordered normed linear space, we note that (E ∗ )+ is w∗ -closed and thus norm closed. Hence E ∗ is an ordered Banach space. Examples A.19. (a) A function space H with the pointwise ordering serves as an example of an ordered normed linear space. (In the case of the pointwise ordering considered on families of functions we sometimes also use the term natural ordering.) (b) The dual space H∗ of a function space H with the ordering defined as ϕ ≤ ψ if ψ(h) − ϕ(h) ≥ 0 for all h ∈ H+ is an ordered Banach space. Indeed, if ϕ ∈ H∗ is given, we extend it to a measure µ ∈ (C(K))∗ . If µ = µ+ −µ− is the decomposition of µ into its positive and negative part, we set ϕ+ := µ+ |H and ϕ− := µ− |H . Then ϕ+ , ϕ− are positive functionals on H and ϕ = ϕ+ − ϕ− . (c) The spaces Ac (X) (X a compact convex set), C(K) (K a compact space), C c (X), C 0 (X) (X a σ-compact locally compact space) and their duals are ordered Banach spaces. Lemma A.20. If E is an ordered locally convex space, then x ∈ E is positive if and only if x∗ (x) ≥ 0 for all positive x∗ ∈ E ∗ . Proof. If x ∈ / E + , the Hahn–Banach theorem provides x∗ ∈ E ∗ such that x∗ (x) < ∗ + inf x (E ). Since E + is a convex cone, inf x∗ (E + ) = 0 and thus x∗ is positive and x∗ (x) < 0. Lemma A.21. Let E be an ordered vector space. Then every additive and positively homogeneous functional ϕ on E + can be uniquely extended to a linear functional on E. Proof. The proof is straightforward. Suppose that x1 − x2 = y1 − y2 , where x1 , x2 , y1 , y2 ∈ E + . Since x1 + y2 = y1 + x2 , we have ϕ(x1 ) + ϕ(y2 ) = ϕ(y1 ) + ϕ(x2 ), or ϕ(x1 ) − ϕ(x2 ) = ϕ(y1 ) − ϕ(y2 ), so we can extend ϕ to E by defining ϕ(x) := ϕ(x1 ) − ϕ(x2 ) whenever x ∈ E, x = x1 − x2 and x1 , x2 ∈ E + . Then ϕ is obviously additive and positive homogeneous on E. Since ϕ(−x) = −ϕ(x) for x ∈ E, ϕ is linear.

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615

Proposition A.22. Let E be an ordered Banach space. Then there exists α > 0 such that for every x ∈ E we can find x1 , x2 ≥ 0 satisfying x = x1 − x2 and kx1 k + kx2 k ≤ αkxk. Proof. The proof is a variant of the proof of the open mapping theorem; see Theorem 2.1.2 of [24]. Remark A.23. In all cases of ordered normed linear spaces that we encounter, the existence of α > 0 from Proposition A.22 is easily shown. Theorem A.24. Let E be an ordered Banach space with the Riesz decomposition property such that E ∗ = (E ∗ )+ − (E ∗ )+ . Then E ∗ is a lattice. Proof. Let x∗ , y ∗ ∈ E ∗ be positive. By Proposition A.15 it is enough to check that x∗ ∧ y ∗ exists. We define w∗ (x) := inf{x∗ (y) + y ∗ (z) : x = y + z, y, z ∈ E + },

x ∈ E+.

Then w∗ (x1 + x2 ) ≤ w∗ (x1 ) + w∗ (x2 ) for all x1 , x2 ∈ E + and w∗ (λx) = λw∗ (x) for all x ∈ E + and λ ≥ 0. It follows from the Riesz refinement property that w∗ (x1 + x2 ) = w∗ (x1 ) + w∗ (x2 ), x1 , x2 ∈ E + . Using Lemma A.21 we extend w∗ to the whole space E. Then w∗ is positive on + E and we claim that w∗ is continuous. Indeed, let α > 0 be the constant from Proposition A.22. Given x ∈ E, let x = x1 − x2 , where x1 , x2 ∈ E + and kx1 k + kx2 k ≤ αkxk. Then |w∗ (x)| ≤ w∗ (x1 ) + w∗ (x2 ) ≤ x∗ (x1 ) + y ∗ (x2 ) ≤ α(kx∗ k ∨ ky ∗ k)kxk. If z ∗ ≤ x∗ , y ∗ , x ≥ 0 and x = x1 + x2 , x1 , x2 ≥ 0, then z ∗ (x) = z ∗ (x1 + x2 ) ≤ x∗ (x1 ) + y ∗ (x2 ). Hence z ∗ ≤ w∗ and w∗ = x∗ ∧ y ∗ .

A.2

Topology

We consider all topological spaces to be Hausdorff (the only exception are topologies on extreme points considered in Chapter 9). We recall that Tychonoff spaces are just completely regular spaces (see [169], Section 1.5). A topological space X is metrizable if there exists a metric on X that induces the same family of open sets. A topological space is Polish if it is separable and completely metrizable. It is Lindelöf if any open cover has a countable subcover. We recall that a topological space X is σcompact if it can be written as a countable union of compact sets; it is locally compact if any point of X has a neighborhood whose closure is compact. A topological space

616

A Appendix

X is normal if for a given pair of disjoint closed sets F1 , F2 in X there exist disjoint open sets U1 , U2 in X such that Fi ⊂ Ui , i = 1, 2. We note that any σ-compact locally compact or metrizable space is normal (see [169], Sections 3.3, 3.8 and 4.1). A set A in a topological space X is perfect if A is closed and has no isolated points. Theorem A.25 (Urysohn’s lemma). If X is a normal space and F1 , F2 ⊂ X are nonempty disjoint closed sets, then there exists a continuous function f : X → [0, 1] such that f = 0 on F1 and f = 1 on F2 . Proof. See [169], Theorem 1.5.11. Theorem A.26 (Tietze’s theorem). If X is a normal space and f is a continuous function on a closed set F ⊂ X, then there exists a continuous function h on X such that f = h on F and h(X) ⊂ co f (F ). Proof. See [169], Theorem 2.1.8. Notation A.27 (The topology of pointwise convergence). If X is a topological space and RX is the space of all real-valued functions on X, we write τX for the topology of pointwise convergence on X (see [169], Section 2.3, for its properties).

A.2.A

ˇ Compact spaces and Cech–Stone compactification

If K is a compact space and ∼ is a closed equivalence relation on K (that is, the classes of equivalence are closed), then there exists a unique (up to a homeomorphism) compact space K/ ∼ and a mapping q : K → K/ ∼ such that x ∼ y if and only if q(x) = q(y) (see [169], Theorem 3.2.11). Given a family {Ki : i ∈ I} of compact Q spaces, by the Tychonoff theorem (see [169], Theorem 3.2.4) the product space i∈I Ki with the product topology is comQ pact. We write πi : i∈I Ki → Ki for the projection on the i-th coordinate. We say that a relation ≤ up-directs a set I, if •

i ≤ j and j ≤ k, then i ≤ k,

•

i ≤ i for any i ∈ I,

•

for any i, j ∈ I there exists k ∈ I such that i ≤ k, j ≤ k.

We will need the notion of inverse limit of compact spaces. We recall that (Ki , pij )i,j∈I is an inverse system of compact spaces Ki , i ∈ I, if I is a set up-directed by ≤, •

pij : Kj → Ki is a continuous surjection, i, j ∈ I,

•

pii is identity on Ki , i ∈ I,

•

pij ◦ pjk = pik for any i ≤ j ≤ k.

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617

The set

(xi )i∈I ∈

Y

Ki : pij (xi ) = xj , j ≤ i, j, i ∈ I

i∈I

is the inverse limit of the system (Ki , pij )i,j∈I and is denoted as lim(Ki , pij )i,j∈I . ← Sometimes we write only lim Ki . ← By [169], Theorem 3.2.13, the inverse limit is compact and nonempty provided the spaces Ki are nonempty. Also, the projections πi : lim(Ki , pij )i,j∈I → Ki ←

are onto by [169], Corollary 3.2.15. We recall several known results for compact spaces and start with the following well-known observation. Proposition A.28. Let (K, τ ) be a compact space and σ be a topology on K weaker than τ . Then σ = τ . Proof. We recall that, by our convention, all topologies are Hausdorff and thus the assertion follows from [169], Corollary 3.1.14. Theorem A.29 (The algebra version of the Stone–Weierstrass theorem). Let K be a compact space and let A be a closed subalgebra of the complex or real space C(K) that contains constant functions, is self-adjoint (that is, if f ∈ A then f ∈ A) and separates points of K. Then A = C(K). Proof. See Theorem 281E and Theorem 281G of [181]. Theorem A.30 (The lattice version of the Stone–Weierstrass theorem). Let K be a compact space and let A be a closed sublattice of the real space C(K) that contains constant functions and separates points of K. Then A = C(K). Proof. See Theorem 281A in [181]. Proposition A.31. Let F ⊂ C(K) be a min-stable convex cone containing constant functions and separating points of K. Then F − F is dense in C(K). Proof. Indeed, if f1 , f2 , f3 , f4 ∈ F, then the equality (f1 − f2 ) ∧ (f3 − f4 ) = ((f1 + f4 ) ∧ (f3 + f2 )) − (f2 + f4 ) shows that F − F is a lattice. Obviously, F − F is a vector space and F ⊂ F − F. Hence the conclusion follows from Theorem A.30. Theorem A.32 (Arzelà–Ascoli). If K is a compact space and F ⊂ C(K), then F is relatively compact if and only if F is bounded and equicontinuous (that is, for each x ∈ K and ε > 0 there exists a neighborhood U of x such that |f (y) − f (x)| < ε for each f ∈ F and y ∈ U ).

618

A Appendix

Proof. See [169], Theorem 3.4.20. Proposition A.33. A compact space K is metrizable if and only if C(K) is separable. Proof. See [173], Lemma 3.23. Lemma A.34. A continuous image of a compact metrizable space is a compact metrizable space. Proof. If ϕ : K → L is a continuous mapping of a metrizable compact space K onto L, then L is compact and C(L) is isometrically embedded in C(K) via the mapping g 7→ g ◦ ϕ, g ∈ C(L). Thus C(L) is separable which yields metrizability of L by Proposition A.33. Proposition A.35. Let K1 , K2 be compact spaces. Then the mapping T : C(K1 × K2 ) → C(K1 , C(K2 )) defined as (T f (x1 ))(x2 ) := f (x1 , x2 ),

x1 ∈ K1 , x2 ∈ K2 ,

is an isometric isomorphism (here C(K1 , C(K2 )) denotes the space of all continuous mappings on K1 with values in C(K2 )). Proof. It is easy to check that T f is indeed a continuous mapping from K1 to C(K2 ). Conversely, given a continuous mapping g : K1 → C(K2 ), the function f (x1 , x2 ) := (g(x1 ))(x2 ),

(x1 , x2 ) ∈ K1 × K2 ,

is easily seen to be a continuous function on K1 × K2 and T f = g. Since sup{|f (x1 , x2 )| : (x1 , x2 ) ∈ K1 × K2 } = sup kT f (x1 )k = kT f k, x1 ∈K1

the conclusion follows. If X is a completely regular space, there exists a unique (up to a homeomorphism) compactification βX of X such that any bounded continuous function f on X can be ˇ extended to a continuous function on βX. The space βX is called the Cech–Stone compactification of X (see [169], Section 3.6). Theorem A.36. Let ϕ : X → Y be a continuous mapping of a completely regular space X to a completely regular space Y and cY be a compactification of Y . Then there exists a unique continuous mapping ϕ b : βX → cY extending ϕ. Proof. See [169], Corollary 3.6.6. A mapping ϕ : X → Y between topological space is perfect if ϕ is continuous, maps closed sets to closed sets and ϕ−1 (y) is compact for each y ∈ Y .

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619

Theorem A.37. Let ϕ : X → Y be a perfect mapping of a completely regular space X to a completely regular space Y . If cY is a compactification of Y , then the extension ϕ b : βX → cY satisfies ϕ(βX b \ X) ⊂ cY \ Y . Proof. See [169], Theorem 3.7.16. Proposition A.38. Let ϕ : X → Y be a continuous surjection of a compact space X onto a compact space Y . Then there exists a minimal (with respect to inclusion) compact set X 0 ⊂ X such that ϕ(X 0 ) = Y . Proof. If F := {F ⊂ X : F closed, ϕ(F ) = Y } is ordered by inclusion (that is, F1 ≤ F2 if F1 ⊃ F2 ), then it is easy to see by Zorn’s lemma that there exists a maximal element X 0 ∈ F. Definition A.39 (Up and down-directed families of sets and functions). A family F of extended real-valued functions on a set X is down-directed if for every f1 , f2 ∈ F there exists f3 ∈ F with f3 ≤ f1 ∧ f2 . A family F of sets is down-directed if {cF : F ∈ F} is down-directed. Analogously we define up-directed families of functions and sets. Proposition A.40. Let ϕ : X → Y be a perfect mapping of a completely regular space X into a completely regular T T space Y . If F is a down-directed family of closed sets in X, then ϕ( F ∈F F ) = F ∈F ϕ(F ). T Proof. To show the nontrivial inclusion, let y ∈ F ∈F ϕ(F ). Then {ϕ−1 (y) ∩ F : F ∈ F} is a down-directed family T of nonempty compact sets, and thus their intersection contains a point x. Then x ∈ F ∈F F and ϕ(x) = y.

A.2.B

Baire and Borel sets

Definition A.41 (Zero and cozero sets). A subset A of a topological space is called a zero set (or a functionally closed set) if A = f −1 (0) for some continuous function f on X. It is said to be a cozero set (or functionally open set) if X \ A is a zero set. Definition A.42 (Fσ and Gδ sets). A subset A of a topological space is an Fσ set if A is a countable union of closed sets. A complement of an Fσ set is a Gδ set. Proposition A.43. Let X be a topological space. (a) The family of all zero sets is stable with respect to finite unions and countable intersections.

620

A Appendix

(b) If X is normal and A ⊂ X, then A is a zero set if and only if A is a closed Gδ set. (c) If X is metrizable, any closed set is a zero set. Proof. For the proof of (a) see [169], p. 42, (b) can be found as Corollary 1.5.12 in [169] and (c) follows from Corollary 4.1.12 in [169]. Definition A.44 (Baire and Borel sets). If X is a topological space, the σ-algebra generated by all open sets is the σ-algebra of Borel sets. The σ-algebra of Baire sets is generated by all cozero sets. Hence, in the case of a metrizable space X, these families coincide (see Proposition A.43(c)). Definition A.45 (Baire and Borel functions). If X, Y are topological spaces, then f : X → Y is a Borel measurable mapping (or Borel mapping) if f −1 (U ) is Borel for each U ⊂ Y open. If Y = R we speak about Borel functions. We say that f is a Baire measurable mapping if f −1 (U ) is a Baire set for every U ⊂ Y open. In the case when Y = R, it follows from Corollary 5.22 that every Baire measurable function is of some Baire class as defined in Definition 5.20 and thus we usually speak about Baire functions. Proposition A.46. Let K be a compact space and let F be a vector space of bounded functions on K satisfying • F contains all bounded upper semicontinuous functions (see Definition A.49), F is closed with respect to pointwise limits of bounded monotone sequences. Then F contains all bounded Borel functions. •

Proof. Set B := {B ⊂ K Borel : cB ∈ F}. It follows from the assumptions that B is a Dynkin system (see Definition A.67). Since B contains all open sets, B contains all Borel sets by Proposition A.68. If f is an arbitrary bounded Borel function, we may assume that 0 ≤ f ≤ 1. For k n n ∈ N, let An,k := {x ∈ K : k−1 2n < f (x) ≤ 2n }, k = 1, . . . , 2 , and fn :=

n X k−1 k=1

2n

cAn,k .

By the first part, fn ∈ F and fn % f . Hence f ∈ F. Proposition A.47. Let K be a compact space and let F be a vector space of bounded functions on K satisfying • F contains all continuous functions, F is closed with respect to pointwise limits of bounded monotone sequences. Then F contains all bounded Baire functions. •

A.2 Topology

621

Proof. Similarly to above, let B := {B ⊂ K Baire : cB ∈ F}. Then B is a Dynkin system and it is easy to see that B contains all cozero sets. Indeed, if U ⊂ K is a cozero set, using Tietze’s theorem we construct fn ∈ C(K), n ∈ N, such that 0 ≤ fn ≤ 1 and fn % cU . Hence cU ∈ F and U ∈ B. Again by Proposition A.68, B contains all Baire sets and we may finish the proof as in Proposition A.46. Proposition A.48. Let A be a Baire set in a normal space X. Then there exist continuous functions fn , n ∈ N, on X with values in [0, 1] such that A is contained in the σ-algebra generated by the family {fn−1 ((0, 1]) : n ∈ N}. Proof. Let A denote the family of all Baire sets satisfying the condition. Then A contains all cozero sets by definition and is obviously closed with respect to taking complements and countable unions. Thus A contains all Baire sets.

A.2.C

Semicontinuous functions

Definition A.49 (Semicontinuous functions). We recall that a function f : X → (−∞, ∞] on a topological space is lower semicontinuous if {x ∈ X : f (x) > a} is open for each a ∈ R. A function f is upper semicontinuous if −f is lower semicontinuous. We collect several characterizations of semicontinuous functions suitable for our purposes. Proposition A.50. For a lower bounded function f : X → (−∞, ∞] on a completely regular space the following assertions are equivalent: (i) f is lower semicontinuous, (ii) f = sup{g ∈ C(X) : g ≤ f }, (iii) f (x) ≤ lim infi∈I f (xi ) for every net {xi }i∈I converging to x. Proof. Proof of (i) ⇐⇒ (iii) is elementary and (i) ⇐⇒ (ii) follows from [169], Exercise 1.7.15. Obviously, lower semicontinuous functions on a completely regular space form a convex cone that is closed with respect to taking minimum of finite families, suprema of arbitrary families and the cone of real-valued lower semicontinuous functions is closed with respect to uniform convergence (see [169], Exercise 1.7.14).

622

A Appendix

Definition A.51 (Lower semicontinuous regularization). If f : X → (−∞, ∞] is a lower bounded function on a topological space X, its lower semicontinuous regularization fb is defined as fb(x) := f (x) ∧ lim inf f (y), y→x

x ∈ X.

It is not difficult to realize that for any completely regular space X we have fb = sup{g ∈ C(X) : g ≤ f }. Proposition A.52. A lower semicontinuous function on a nonempty compact space is lower bounded and attains its infimum. Proof. If f is lower semicontinuous on K, inf f (K) > −∞ because f > −∞ on K and K is compact. Thus ∞ \ n=1

x ∈ K : f (x) ≤ inf f (K) +

1 n

is nonempty by the compactness of K and hence f attains its infimum. Proposition A.53. Let f : X → (−∞, ∞] be a lower bounded function on a normal space X. Then the following assertions are equivalent: (i) There exists an increasing sequence {fn } of continuous function on X such that fn % f . (ii) The set {x ∈ X : f (x) > c} is an open Fσ set in X for each c ∈ R. Proof. The implication (i) =⇒ (ii) is straightforward. For the proof of the converse implication, we assume that f is positive. For any q ∈ Q∩(0, ∞), set Uq := {x ∈ X : f (x) > q}. Using Urysohn’s lemma A.25 we find a sequence of positive continuous functions {fn,q } such that fn,q % qcUq . Then f = sup{fn,q : n ∈ N, q ∈ Q ∩ (0, ∞)}, and we finish the proof by creating an increasing sequence of continuous functions from the latter family. Lemma A.54. Let X be a Lindelöf topological space and f a pointwise limit of a decreasing sequence {hn } of continuous functions on X. Let F be a down-directed family of continuous functions on X such that f = inf{h ∈ F : h ≥ f }. Then there is a decreasing sequence {fn } of functions from F which converges pointwise to f .

A.2 Topology

623

Proof. Without loss of generality, we may assume that {hn } is a strictly decreasing sequence. We fix n ∈ N. For each x in X, there exist a function gx ∈ F and an open neighborhood U (x) of x such that f ≤ gx and gx < hn on U (x). The Lindelöf property ensures the existence of a countable family {U (xm )} chosen from {U (x)}x∈X which covers X. We relabel the corresponding functions {gxm } as {gn,m }∞ m=1 . Then we have inf{gn,m : m ∈ N} < hn . We enumerate the functions {gn,m : n, m ∈ N} as a single sequence {gn }. Then f = inf{gn : n ∈ N}. Since F is down-directed, we may inductively find a sequence {fn } such that g1 = f1 and fn+1 ≤ gn+1 ∧ fn . Then {fn } has the desired properties.

Lemma A.55. Let X be a topological space with a countable base of open sets and F be a family of lower semicontinuous functions on X. Then there exists a countable family F 0 ⊂ F such that sup F = sup F 0 . Proof. For a number q ∈ Q, we use the existence of a countable base to find a countable family F q ⊂ F such that [

{x ∈ X : f (x) > q} =

f ∈F

[

{x ∈ X : f (x) > q}.

f ∈F q

S Let F 0 := q∈Q F q . If x ∈ X, r ∈ R and there exists f ∈ F with f (x) > r, let q ∈ Q satisfy r < q < f (x). Then a function g ∈ F q with g(x) > q witnesses that the function sup F 0 is greater than r at the point x. Thus sup F 0 = sup F. Theorem A.56. Let X be a normal space and f, −g be real-valued upper semicontinuous functions on X such that f ≤ g. Then there exists a continuous function h on X such that f ≤ h ≤ g. Proof. See [169], Exercise 1.7.15.

A.2.D

Baire spaces and sets with the Baire property

Definition A.57 (Baire T∞ spaces and hereditarily Baire spaces). A topological space X is a Baire space if n=1 Gn is dense provided Gn , n ∈ N, are dense open sets. It is called hereditarily Baire (or strong Baire) if any closed subset of X is a Baire space. Theorem A.58. If X is a locally compact or completely metrizable space, then X is hereditarily Baire. Proof. See Section 3.9 and Theorem 4.3.26 of [169].

624

A Appendix

Definition A.59 (Meager sets and sets with the Baire property). A set A ⊂ X is meager if A can be covered by countably many nowhere dense sets. It is residual (sometimes comeager), if X \ A is meager. The set A is said to have the Baire property, if A is contained in the σ-algebra generated by Borel sets and meager sets. It has the Baire property in the restricted sense if A ∩ F has the Baire property in F for every F ⊂ X. A function f : X → R has the Baire property (or the Baire property in the restricted sense) if f −1 (U ) has the Baire property (or the Baire property in the restricted sense) for any open set U ⊂ R. Proposition A.60. Any Borel set has the Baire property in the restricted sense. Proof. See §11, VI in [285]. Proposition A.61. Let f : X → R have the Baire property on a topological space X. Then there exists a comeager set Y ⊂ X such that f |Y is continuous. Proof. See §28, II in [285].

A.3 A.3.A

Measure theory Measure spaces

We recall that a triple (X, Σ, µ) is a measure space if Σ is a σ-algebra of sets in X and µ : Σ → [0, ∞] is a σ-additive function. The measure µ is finite if µ(X) < ∞. (We use the term finite instead of totally finite of Section 211 in [181] since we work in much less complicated measure spaces.) We write Lp (X, Σ, µ) or simply Lp (µ) for the usual Lebesgue spaces (see [181], Chapter 24). The measure space (X, Σ, µ) is complete if the following condition holds: if A ⊂ X, N ∈ Σ, A ⊂ N and µ(N ) = 0, then A ∈ Σ. Of course, then µ(A) = 0. If (X, Σ, µ) is a measure space, by adding all sets with the above property we obtain the completion (X, Σ, µ) of (X, Σ, µ) (see [181], Section 212). If (X, Σ, µ) is a measure space and A ∈ Σ satisfies µ(X \ A) = 0, we say that µ is carried by A. If (X, Σ, µ) is a measure space and µ is carried by a set X 0 ∈ Σ, we consider the measure space (X 0 , Σ0 , µ0 ), where Σ0 := {X 0 ∩ A : A ∈ Σ} and µ0 (A0 ) := µ(A), where A ∈ Σ satisfies X 0 ∩ A = A0 . We write µ|X 0 for the measure µ0 . Conversely, if (X 0 , Σ0 , µ0 ) is a measure space and X ⊃ X 0 , we consider the measure space (X, Σ, µ), where Σ := {A ⊂ X : X 0 ∩ A ∈ Σ0 } and µ(A) = µ0 (X 0 ∩ A) for any A ∈ Σ. By this procedure we may consider the measure µ to be defined on the whole set X.

A.3 Measure theory

625

If (P) is a property of a measure space (X, Σ, µ), where no confusion can arise we sometimes say that the measure µ has property (P). If (X, Σ, µ) is a measure space and Y is a topological space, a mapping f : X → Y is µ-measurable (briefly measurable) if f −1 (U ) ∈ Σ for each U ⊂ Y open. If Σ is a σ-algebra of sets in X and µ : Σ → R is σ-additive, it is called a signed measure (analogously we define a complex measure). Any signed measure can be decomposed as µ = µ+ − µ− , where µ+ , µ− is the positive and negative part of µ, respectively. Moreover, the measures µ+ , µ− are finite (see [181], Corollary 231F). We write |µ| for the variation of µ, that is, ∞ X |µ|(A) := sup{ |µ(Ai )| : {Ai } is a partition of A},

A ∈ Σ.

n=1

Then |µ|(A) = µ+ (A) + µ− (A) for any A ∈ Σ (see Section 231 of [181]). (Similarly we define the variation of a complex measure µ.) If (X, Σ, µ) is S a measure space, µ is σ-finite if there exist sets Xn ∈ Σ, n ∈ N, such that X = n∈N Xn and µ(Xn ) < ∞. Definition A.62 (Inner and outer measure). If (X, Σ, µ) is a measure space, it induces an inner measure µ∗ on the family 2X of all sets in X by the formula µ∗ (B) := sup{µ(A) : A ∈ Σ, A ⊂ B},

B ⊂ X.

The outer measure µ∗ is defined analogously. Obviously, µ is complete if and only if any set A ⊂ X with µ∗ (A) = µ∗ (A) < ∞ is contained in Σ. Definition A.63 (Absolutely continuous and singular measures). Suppose that (X, Σ, µ) is a measure space and ν : Σ → (−∞, ∞] is a σ-additive function. We say that ν is absolutely continuous with respect to µ (written as ν µ) if ν(A) = 0 whenever A ∈ Σ and µ(A) = 0. We say that ν is singular with respect to µ (denoted by ν ⊥ µ) if there exists a set A ∈ Σ such that µ(A) = 0 and ν(B) = 0 whenever B ∈ Σ and B ⊂ X \ A. We refer the reader to Section 232 of [181] for more information on this subject. Theorem A.64 (The Radon–Nikodym theorem). Let (X, Σ, µ) be a σ-finite measure space and ν : Σ → R be a σ-additive function. If ν is absolutely continuous with R 1 respect to µ, then there exists a function f ∈ L (µ) such that ν(A) = A f dµ for any A ∈ Σ. Proof. See [181], Corollary 232F. Proposition A.65 (Lebesgue decomposition). Let (X, Σ, µ) be a σ-finite measure space and ν : Σ → R be σ-additive. Then ν can be uniquely expressed as ν = νac +νs where νac µ and νs ⊥ µ.

626

A Appendix

Proof. See [181], Proposition 232I. A.66. Let (X, Σ, µ) be a measure space and An ∈ Σ, n ∈ N, satisfy Theorem P∞ µ(A n ) < ∞. Then the set {n ∈ N : x ∈ An } is finite for µ-almost all n=1 x ∈ X. Proof. See Lemma 273A of [181]. Definition A.67 (Dynkin system). A system D of subsets of a set X is called a Dynkin system if X ∈ D, X \ D ∈ D for each D ∈ D and if D is closed with respect to the formation of countable unions of pairwise disjoint elements of D. Proposition A.68. Let D be a Dynkin system of subsets of a set X and F ⊂ D a system which is closed with respect to finite intersections. Then D contains the σalgebra generated by F. Proof. See Theorem 136B of [182]. Definition A.69 (Upper and lower integral). If (X, Σ, µ) is a measure space and f : X → [−∞, ∞] is a function, the upper integral of f is defined as Z ∗ Z f dµ := inf g dµ : g is µ-integrable, f ≤ g µ-almost everywhere . X

X

The lower integral is defined analogously. We refer the reader to Proposition 133J of [182] for more information on this subject.

A.3.B

Radon measures on locally compact σ-compact spaces

Definition A.70 (Radon measures). If X is a locally compact σ-compact topological space and Σ is a σ-algebra of sets in X, a σ-additive function µ : Σ → [0, ∞] is called a Radon measure, if (a) Σ contains all Borel sets in X, (b) µ(K) < ∞ for each K ⊂ X compact, (c) µ(A) = sup{µ(K) : K ⊂ A compact} for any A ∈ Σ, (d) the measure space (X, Σ, µ) is complete. (Sometimes we say that (X, Σ, µ) is a Radon measure space.) We remark that our definition of a Radon measure coincides with [183], Definition 411H. A signed (or complex) measure µ is Radon if the total variation |µ| of µ is a Radon measure. If x ∈ X, we write εx for the Dirac measure at x. Obviously, εx is a Radon measure.

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627

Definition A.71 (Universally measurable sets and functions). If X is a locally compact σ-compact space, a set A ⊂ X is universally measurable, if A is µ-measurable for any Radon measure µ on X (cf. [183], Definition 434E). A mapping ϕ from X to a topological space Y is universally measurable if ϕ−1 (U ) is universally measurable for any U ⊂ Y open. If Y = R, we speak about universally measurable functions. Theorem A.72 (Riesz representation theorem for positive functionals). Let X be a locally compact σ-compact space and let C c (X) stand for the space of all real-valued continuous functions with compact support. Let T : C c (X) → R be a positive functional (that is, T f ≥ 0 whenever f ≥ 0). Then there exists a unique Radon measure R µ on X such that T f = X f dµ, f ∈ C c (X). Proof. See [183], Theorem 436J. If X is a locally compact space and f : X → R is continuous, we say that f vanishes at infinity if for any ε > 0 there exists a compact F ⊂ X such that |f | < ε on X \ F . Clearly, the space C 0 (X) of all continuous functions vanishing at infinity is a Banach space when endowed with the supremum norm. It is easy to see that C 0 (X) is an ordered Banach space that is a lattice, and thus (C 0 (X))∗ is a lattice by Theorem A.24. We recall that a measure ν defined on the σ-algebra of all Borel sets in a topological space X is said to be inner regular with respect to compact sets if ν(B) = sup{ν(F ) : F ⊂ B compact} for any B ⊂ X Borel. The following proposition shows the relation of Radon measures with measures defined on Borel sets. Proposition A.73. Let X be a locally compact σ-compact space and B stand for the σ-algebra of all Borel sets in X. (a) If µ is a Radon measure on X, the measure ν(A) := µ(A), A ⊂ X Borel, is inner regular with respect to compact sets and it is finite on compact sets. (b) Conversely, if (X, B, ν) is a measure space such that ν is inner regular with respect to compact sets and finite on compact sets, then there exists a unique Radon measure space (X, Σ, µ) such that µ(B) = ν(B) for any B ∈ B. Proof. The first assertion is obvious, the second follows from Proposition 416F in [183]. Remark A.74 (Inner and outer regularity of Radon measures). Let X be a locally compact σ-compact space and µ be a measure defined on Borel sets in X that is finite on compact sets. It is not difficult to realize that µ is inner regular with respect to compact sets if and only if it is outer regular with respect to open sets (that is, µ(A) = inf{µ(G) : G ⊃ A open} for A ⊂ X Borel). S Indeed, let Xn ⊂ X, n ∈ N, be compact sets with X = ∞ n=1 Xn . If µ is inner regular with respect to compact sets, A ⊂ X is Borel and ε > 0, let Kn ⊂ Xn \ A be

628

A Appendix

compact sets such that µ((Xn \ A) \ Kn ) < 2−n ε, n ∈ N. Let Un ⊂ X be open such that µ(Un \ Xn ) < 2−n ε, n ∈ N. Then U :=

∞ [

Un ∩ (X \ Kn )

n=1

is an open set containing A such that µ(U \ A) < 2ε. Conversely, let µ be outer regular with respect to open sets, A ⊂ X be Borel and ε > 0. If µ(A) = ∞ and c > 0, let n ∈ N be such that µ(A ∩ Xn ) > c. If U ⊂ X open satisfy µ(U \ (Xn \ A)) < ε, then K := Xn \ U is compact and µ(K) > µ(A ∩ Xn ) − ε > c − ε. If µ(A) < ∞, we can find n ∈ N with µ(A ∩ Xn ) > µ(A) − ε and proceed as above. Hence it follows that the definition of Radon measures in Definition A.70 coincides on locally compact σ-compact spaces with the definition of Radon measures from Theorem 2.14 of [402]. Theorem A.75 (Riesz representation theorem for (C 0 (X))∗ ). Let X be a locally compact σ-compact space and let C 0 (X) denote the Banach space of continuous functions vanishing at infinity. Then for anyR T ∈ (C 0 (X))∗ there exists a unique signed Radon measure µ on X such that T f = X f dµ, f ∈ C 0 (X). Proof. The assertion is essentially Proposition 437I of [183] (or, by Remark A.74, we can use Theorem 6.19 from [402]). We only explain how to get the complete measure representing T . Given T ∈ (C 0 (X))∗ , we use the aforementioned Proposition 437I of [183] to find a signed measure ν defined on the σ-algebra B of all Borel sets in X such that •

•

if ν + and ν − are the positive and negative parts of ν, respectively, then they are inner regular with respect to compact sets, R T f = X f dν for any f ∈ C 0 (X).

Let (X, Σ+ , µ+ ) and (X, Σ− , µ− ) be the unique Radon measures extending ν + and ν − , respectively (see Proposition A.73). Let Σ := Σ+R ∩ Σ− and µ := µ+ − µ− on Σ. Then (X, Σ, µ) is a Radon measure space and T f = X f dµ for any f ∈ C 0 (X). Theorem A.76 (Lusin). Let X be a locally compact σ-compact space and µ be a Radon measure on X. Then for each µ-measurable real-valued function f and ε > 0 there exists a compact set F ⊂ X such that µ(X \ F ) < ε and f |F is continuous. Proof. See Theorem 418J in [183]. Definition A.77 (Discrete, molecular and continuous Radon measures). A Radon measure µ on a locally compact σ-compact space X is called discrete if there exists a set S ⊂ X such that µ(X \ S) = 0 and µ({x}) > 0 for each x ∈ S. Hence

A.3 Measure theory

629

a finite Radon measure µ is discrete if and only if there exists a sequence of positive numbers {λj } and points xj ∈ X such that ∞ X

λj < ∞ and

µ=

j=1

∞ X

λ j εxj .

j=1

Pk A Radon measure µ is molecular if µ = j=1 λj εxj , where x1 , . . . , xk ∈ X, Pk λ1 , . . . , λk are positive and j=1 λj = 1. A Radon measure µ is called continuous if µ({x}) = 0 for each x ∈ X. Proposition A.78. Every Radon measure µ on a locally compact σ-compact space X can be uniquely expressed in the form µ = µd + µc , where µd is discrete and µc continuous. Proof. Set S := {x ∈ X : µ({x}) > 0} (the set S is countable) and define µd := µ|S

and µc := µ|X\S

to get the desired decomposition. Definition A.79 (Support of a Radon measure). If µ is a Radon measure on a locally compact σ-compact space X, then its support (denoted as spt µ) is defined as [ spt µ := X \ {G : G ⊂ X open, µ(G) = 0}. Definition A.80 (Radon measures as a convex cone). If X is a locally compact σcompact space and µ1 , µ2 are Radon measures on X, µ1 + µ2 is the unique Radon measure ν satisfying ν(f ) = µ1 (f ) + µ2 (f ) for any f ∈ C 0 (X). Hence ν(B) = µ1 (B) + µ2 (B) for any Borel set B ⊂ X. In what follows, we consider Radon measures as a cone. Lemma A.81. Let µ and ν be Radon measures on a locally compact σ-compact space X. Then spt(µ + ν) = spt µ ∪ spt ν. Proof. If µ, ν are Radon measures on X and if G ⊂ X is an open set, then (µ + ν)(G) = 0 if and only if µ(G) = ν(G) = 0. From this, the assertion easily follows. Definition A.82 (Signed Radon measures as an ordered Banach space). If X is a locally compact σ-compact space, we write M(X), M+ (X), and M1 (X) for the set of all signed, positive finite, and probability Radon measures on X, respectively. As above, the identification of M(X) with (C 0 (X))∗ given by Theorem A.75 provides

630

A Appendix

a way to understand M(X) as a Banach space with the norm kµk = |µ|(X) for any µ ∈ M(X) (see Proposition 437I of [183]). Then µ ∈ M(X) is a limit of a sequence {µn } in M(X), if and only if |µ − µn |(A) → 0 for every Borel set A in X. Further, if we define an order “≤” on M(X) by µ≤ν

if

µ(A) ≤ ν(A) for any A ⊂ X Borel,

then M(X) becomes an ordered RBanach space. Moreover, the mapping T : M(X) → (C 0 (X))∗ given by T µ(f ) = X f dµ, f ∈ C 0 (X), is an isometric isomorphism preserving order. (We recall that (C 0 (X))∗ is an ordered Banach space as defined in Subsection A.1.C that is even a lattice.) Hence it follows from Theorem A.24 that M(X) is a lattice. Further, if µ ∈ M(X), then the positive part µ+ of µ is precisely the supremum µ ∨ 0 in the ordered Banach space M(X). Hence the cone of all positive finite Radon measures coincides with the cone of all positive elements in (C 0 (X))∗ . There is a strong relation between the order and the norm on M(X), namely kµk = k|µ|k, µ ∈ M(X), and kµ + νk = kµk + kνk for µ, ν ∈ M+ (X). Remark A.83. The space C(K) of continuous functions on a compact space K, the space C 0 (X) of continuous functions vanishing at infinity on a locally compact space X and M(X) are examples of Banach lattices (see Section 5 in [251]). Theorem A.84 (Lebesgue monotone convergence theorem for nets). Let µ be a Radon measure on a compact space K and F a nonempty up-directed set of lower semicontinuous functions on K. Then Z Z (sup F) dµ = sup f dµ : f ∈ F . K

K

Proof. See [183], Corollary 414B. Theorem A.85. Let X be a locally compact σ-compact space and M(X) be identified with (C 0 (X))∗ as in Definition A.82. Consider M(X) to be endowed with the w∗ topology. (a) BM(X) and M1 (X) are compact convex sets. (b) If f is lower semicontinuous function on X, then µ 7→ µ(f ), µ ∈ M1 (X), is a lower semicontinuous function on M1 (X). (c) If X is metrizable, BM(X) and M1 (X) are metrizable as well. Proof. The dual unit ball BM(X) is compact by Theorem A.5 and M1 (X) is compact because M1 (X) = {µ ∈ BM(X) : µ ≥ 0, µ(1) = 1}. Further, (b) follows from Theorem A.84 and (c) from Theorem A.5 and Proposition A.33.

A.3 Measure theory

631

Convention A.86. If X is a locally compact σ-compact space, we consider M(X) to be endowed with the w∗ -topology. Thus all topological notions concerning Radon measures, if not stated otherwise, are considered with respect to the w∗ -topology. The following result enables us to consider bounded universally measurable functions as elements of (C 0 (X))∗∗ . Theorem A.87. Let X be a locally compact σ-compact space and U be the space of all bounded universally measurable functions on X. Let U be endowed with the supremum norm andR pointwise order. Then the mapping S : U → (C 0 (X))∗∗ defined as Sf (µ) := X f dµ, µ ∈ (C 0 (X))∗ , is an isometric isomorphism of U into (C 0 (X))∗∗ that preserves the order. Proof. See Proposition 437I in [183]. The following lemma can be traced in [73], Lemma 1.2, and in [66], Lemma I.1.9. Lemma A.88. Let K be a compact space K and F be a nonempty convex set of lower semicontinuous functions on K. Then the following conditions are equivalent: (i) there exists a function f in F such that f > 0, (ii) for any nonzero Radon measure µ on K there exists a function f in F such that µ(f ) > 0. Proof. The implication (i) =⇒ (ii) is easy: If f ∈ F is a strictly positive function and µ a nonzero Radon measure on K, we get Z Z f dµ ≥ min f (K) dµ = µ(K) min f (K) > 0. K

K

Conversely, denote H := {h ∈ C(K) : there exist f ∈ F and λ ≥ 0 such that h ≤ λf } and G := {g ∈ C(K) : g > 0}. Our aim is to prove that H ∩ G 6= ∅. To do this, assume that H and G are disjoint sets in order to arrive at a contradiction. Obviously G and H are nonempty convex sets and G is open in C(K). According to the geometric version of the Hahn–Banach theorem A.1, there exist a nonzero linear functional Φ ∈ (C(K))∗ and α ∈ R such that sup Φ(H) ≤ α ≤ inf Φ(G). Obviously, α = 0, since H and G are cones. Therefore Φ is positive. By the Riesz representation theorem A.72, there exists a Radon measure µ ∈ M+ (K) such that Φ(ϕ) = µ(ϕ) for any ϕ ∈ C(K). The measure µ is nontrivial. Let f ∈ F. Since f = sup{h ∈ H : h ≤ f } and µ(h) = 0 for each h ∈ H, it follows that µ(f ) ≤ 0, which is a contradiction. Hence the sets H and G are not disjoint and the proof is complete.

632

A Appendix

Lemma A.89. If X is locally compact σ-compact space, µ ∈ M+ (X) and A ⊂ X is a Baire set, then µ(A) = sup{µ(F ) : F ⊂ A, F closed Gδ } = inf{µ(U ) : U ⊃ A, U open Fσ }.

(A.1)

Proof. It is easy to show that the family of those Baire subsets of X that satisfy (A.1) is a σ-algebra, and that it contains each cozero set. Hence it contains all Baire sets. Lemma A.90. Let X be a locally compact σ-compact space and µ be a Radon measure on X. If f : X → [−∞, ∞] is a function, then Z ∗ f dµ = inf{µ(g) : g ≥ f, g lower semicontinuous}. X

R∗

Proof. If X f dµ = ∞, it is enough to take g = ∞. We assume that h ≥ f µalmost everywhere and it is µ-integrable. Let ε > 0. Using Theorem A.76, we find an increasing sequence {Kn } of compact sets in X such that h|Kn is continuous S∞ and µ(X \ n=1 Kn ) = 0. For n ∈ N, we find open sets Gn ⊃ Kn such that R |h| dµ < 2−n ε. Let hn be a continuous function on X such that |hn | ≤ |h| Gn \Kn on X, hn = h on Kn and hn = 0 on X \ Gn . Then g1 := supn∈N hn is a lower semicontinuous function on X such that g1 ≥ h µ-almost everywhere. Let N be a set with µ(N ) = 0 such that h ≥ f and T∞g1 ≥ h on X \ N . Let {Un } be a decreasing sequence of open sets such that N ⊂ n=1 Un and P∞ −n µ(Un ) < ε2 , n ∈ N. Then g2 := n=1 cUn is lower semicontinuous on X and g := g1 + g2 is a lower semicontinuous function satisfying g ≥ f . Since Z Z n Z X (h1 ∨ · · · ∨ hn ) dµ ≤ h dµ + |h| dµ, n ∈ N, X

Kn

i=1

Gi \Ki

we get Z µ(g) = µ(g2 ) + lim

n→∞ X

Z (h1 ∨ · · · ∨ hn ) dµ ≤

h dµ + 2ε. X

This finishes the proof.

A.3.C

Images, products and inverse limits of Radon measures

Definition A.91 (Image of a measure). Let (X, Σ, µ) be a measure space, Y be a set and ϕ : X → Y be a mapping. Let T := {T ⊂ Y : ϕ−1 (T ) ∈ Σ} and let ν be a set function on T defined as ν(T ) := µ(ϕ−1 (T )),

T ∈T.

Then (Y, T , ν) is a measure space and ν is called the image of the measure µ (under the mapping ϕ). We write ϕ] µ for the measure ν.

A.3 Measure theory

633

Proposition A.92. Let (X, Σ, µ), Y and T be as in Definition A.91. Then a T measurable function g on Y is ϕ] µ-integrable if and only if g ◦ ϕ is µ-integrable. In this case, Z Z g d(ϕ] µ) = g ◦ ϕ dµ. Y

X

Proof. See Theorem 235I of [181]. Theorem A.93. Let X, Y be locally compact σ-compact spaces and ϕ : X → Y be continuous. If µ is a Radon measure on X, then ϕ] µ is a Radon measure on Y . Proof. This is a special case of [183], Theorem 418I. Theorem A.94. Let ϕ : K → L be a continuous surjection of a compact space K onto a compact space L. Then ϕ] : M(K) → M(L) is a continuous (with respect to the w∗ -topologies) surjection that maps M1 (K) onto M1 (L). Proof. For the surjectivity of ϕ] , see Theorem 418L in [183]. The verification of the w∗ -continuity is straightforward. Definition A.95 (Product of Radon measures). If {(Ki , Σi , µi ) : i ∈ I} is a family of compact spaces Qwith Radon probability Q measures, Q then there exists a unique product measure µ on i∈I Ki such that µ( i∈I Ei ) = i∈I µi (Ei ), whenever Ei are Borel subsets of Ki such that Ei 6= Ki for finitely many indices (see Proposition 417D in [183]). By Theorem 417Q in [183], µ can be uniquely extended to a Radon measure N µ i∈I i , which we call the Radon product measure. Theorem A.96 (Associative law). Let {(Ki , ΣN i , µi ) : i ∈ I} be a family of compact spaces with Radon probability measures andN i∈I µi be their Radon product measure. If {Ij }j∈J is a partition of I and µj = i∈Ij µi is the Radon product measure Q Q Q on K j := i∈Ij Ki , j ∈ J, then the natural bijection ϕ : i∈I Ki → j∈J K j N N defined as ϕ((xi )i∈I ) = ((xi )i∈Ij )j∈J identifies i∈I µi with j∈J µj . Proof. See Theorem 417J in [183]. Theorem A.97 (Fubini’s theorem). Let (K1 , Σ1 , µ1 ) and (K2 , Σ2 , µ2 ) be Radon measure Nspaces on compact spaces K1 and K2 , respectively. If f : K1 × K2 → R is a µ1 µ2 -integrable function, then Z Z Z f (x, y) d(µ1 ⊗ µ2 )(x, y) = f (x, y) dµ2 (y) dµ1 (x) K1 ×K2

K1

Z

K2

Z

= K2

Proof. See Theorem 417H in [183].

K1

f (x, y) dµ1 (x) dµ2 (y).

634

A Appendix

Q Proposition A.98. Let Ki , i ∈ I, be compact spaces, K := i∈I Ki , µ be a Radon measure Q on K and ε > 0. Then for any µ-measurable set A ⊂ K, there exists a set U := i∈I Ui , where Ui ⊂ Ki are open and Ui 6= Ki for only finitely many indices, such that µ((A \ U ) ∪ (U \ A)) < ε. Q Proof. Let U be the family of all finite unions of sets of the form i∈I Ui , where Q Ui ⊂ Ki are open and Ui 6= Ki for finitely many i ∈ I. If Fb ⊂ i∈I Ki is compact and Vb ⊃ Fb is open, we can find an element U ∈ U with Fb ⊂ U ⊂ Vb . Given A ⊂ K µ-measurable, by regularity of µ there are F ⊂ A ⊂ V such that F is closed, V open and µ(V \ F ) < 12 ε. By the above reasoning we choose U ∈ U such that F ⊂ U ⊂ V and get the sought set.

Q Proposition A.99. If two Radon measures on i∈I Ki coincide on the cylinder sets Q i∈I Ui , where Ui ⊂ Ki are open and Ui 6= Ki for finitely many i ∈ I, then they are equal. Proof. The assertion is a consequence of Proposition A.98.

Lemma A.100. Let K :=

Q

i∈I

Ki be the product of compact spaces Ki , i ∈ I. Then

(a) spt(πj )] µ = πj (spt µ) for any µ ∈ M+ (K) and j ∈ I, (b) spt

N

i∈I

µi =

Q

i∈I

spt µi , where µi ∈ M1 (Ki ) for i ∈ I.

Proof. For the proof of (a), let Gj := Kj \ πj (spt µ). Then Y

0 = µ(Gj ×

Ki ) = ((πj )] µ)(Gj ),

i∈I\{j}

and hence spt(πj )] µ ⊂ πj (spt µ). Conversely, let Gj := Kj \ spt(πj )] µ. Then 0 = ((πj )] µ)(Gj ) = µ(Gj ×

Y i∈I\{j}

which yields πj (spt µ) ⊂ spt(πj )] µ.

Ki ),

635

A.3 Measure theory

N Q Q To verify (b), we notice that spt i∈I µi ⊂ i∈I spt µQ i by (a). Let x ∈ i∈I spt µi and G be a basic open neighborhood of x, that is, G = i∈I Gi , where Gi ⊂ Ki are open and Gi 6= Ki only for indices i in a finite set J ⊂ I. Then µ(G) =

Y

µi (Gi ) > 0.

i∈J

Hence µ(G) > 0 for any open set G 3 x, in other words, x ∈ spt

N

i∈I

µi .

Definition A.101 (Inverse limits of Radon measures). Let (Ki , pij )i,j∈I be an inverse system of compact spaces and (Ki , Σi , µi )i∈I be Radon probability spaces such that (pij )] µj = µi , i ≤ j. By Proposition 418M of [183], there exists a Radon probability measure µ on K = lim(Ki , pij )i,j∈I such that (πi )] µ = µi for each i ∈ I. ← Lemma A.102 below shows that µ is uniquely determined. We call the family ((Ki , Σi , µi )i∈I , (pij )i,j∈I ) the inverse system of measures and µ the inverse limit of this system. Sometimes we write (µi , pij )i,j∈I for the inverse system and lim µi for the inverse limit. ←

Lemma A.102. Let ((Ki , Σi , µi )i∈I , (pij )i,j∈I ) be an inverse system of Radon probability spaces. Then there is a unique Radon probability measure µ on K := lim Ki ←

such that (πi )] µ = µi for all i ∈ I. Proof. Assume that Then we can regard them as measures on Qµ, ν are such measures. Q the product space i∈I Ki . Let U := i∈I Ui ∩ K, where Ui ⊂ Ki are open and the set Ui differs from Ki for only finitely many indices i ∈ I. Let J ⊂ I be the set of those indices. We find l ∈ I that is greater than all indices in J and define Ul :=

\

p−1 jl (Uj ).

j∈J

Then Ul is an open set in Kl and U = πl−1 (Ul ). Hence µ(U ) = µ(πl−1 (Ul )) = ((πl )] µ)(Ul ) = ((πl )] ν)(Ul ) = ν(πl−1 (Ul )) = ν(U ). By Proposition A.99, µ = ν. Lemma A.103. Let (Ki , pij )i,j∈I be an inverse system of compact spaces and let µ be a Radon probability measure on lim Ki . Then ((πi )] µ, pij )i,j∈I is an inverse system ←

of measures and µ = lim(πi )] µ. ←

636

A Appendix

Proof. To check that ((πi )] µ, pij )i,j∈I is an inverse system is easy and the fact µ = lim µi follows from Lemma A.102. ←

Lemma A.104. Let (Ki , pij )i,j∈I be an inverse system of compact spaces and let µ be a continuous Radon probability measure on lim Ki . Then for each δ > 0 there exists ←

i ∈ I such that ((πj )] µ)({x}) < δ for each j ≥ i and x ∈ Kj . Proof. For each i ∈ I, let Li := {x ∈ Ki : ((πi )] µ)({x}) ≥ δ}. If we assume the contrary, then the set J := {i ∈ I : Li 6= ∅} is cofinal (that is, for every i ∈ I there exists j ∈ J such that i ≤ j). Obviously, each set Li is finite and, moreover, pij (Lj ) ⊂ Li ,

i ≤ j,

i, j ∈ J.

Hence we can find x := (xi )i∈I ∈ K such that xi ∈ Li for each i ∈ J. Then \ {x} = πi−1 (xi ) i∈J

and δ ≤ ((πi )] µ)({xi }) = µ(π −1 (xi )),

i ∈ J.

Thus µ({x}) ≥ δ, a contradiction.

A.3.D

Kernels and disintegration of measures

Definition A.105 (Kernels of measures). Let X be a locally compact σ-compact space and C 0 (X), as before, be the space of continuous functions vanishing at infinity. We say that a mapping T : X × C 0 (X) → R is a kernel if (a) x 7→ T (x, f ) is a Borel function for each f ∈ C 0 (X), (b) there exists C > 0 such that f 7→ T (x, f ) is a functional on C 0 (X) of norm at most C for all x ∈ X. By (b), a kernel T determines for any x ∈ X a signed Radon measure Tx ∈ M(X) such that T (x, f ) = Tx (f ) for any f ∈ C 0 (X) and |Tx |(X) ≤ C (see Theorem A.75). Moreover, the mapping x 7→ Tx has the property that x 7→ Tx (f ) is a Borel function for each f ∈ C 0 (X). Conversely, if x 7→ Tx is a mapping from X to M(X) such that x 7→ Tx (f ) is a Borel function for each f ∈ C 0 (X) and kTx k ≤ C, x ∈ X, for some C > 0, then the mapping (x, f ) 7→ Tx (f ) is a kernel as defined above.

A.4 Descriptive set theory

637

We can extend the domain of the mapping T to all bounded universally measurable functions on X by defining Z f (t) dTx (t), x ∈ X. T f (x) := X

We use the same letter T for the mapping assigning a bounded universally measurable function f ∈ U(X) the function T f . We can further consider the mapping T to be defined on the space M(X) by the formula Z T µ(f ) := Tx (f ) dµ(x), f ∈ C 0 (X). X

By Theorem A.75, T µ is a well-defined signed Radon measure. Again we use the letter T for the mapping T : M(X) → M(X). We remark that T : M(X) → M(X) is positive if and only if T : U(X) → `∞ (X) is positive and this is the case if and only if Tx ≥ 0 for all x ∈ X. Theorem A.106 (Disintegration theorem). Let K, L be compact spaces and µ be a Radon probability measure on K × L. Let π : K × L → K be the projection. Then there exists a family {Tx }x∈K of Radon probability measures on L such that Z Z Z f dµ = f (x, y) dTx (y) d((π)] µ)(x), f ∈ L1 (µ). K×L

K

L

Proof. This follows from Proposition 452O in [183] applied to the projection π. Theorem A.107 (Disintegration theorem for metrizable spaces). Let X, Y be metrizable compact spaces, ϕ : X → Y be a continuous mapping, µ ∈ M1 (X) and ν = ϕ] µ. Then there exists a mapping T : y 7→ Ty , y ∈ Y , such that, for y ∈ Y , • T ∈ M1 (X), y • • •

spt Ty ⊂ ϕ−1 (y), y 7→ Ty (f ) is Borel for each f ∈ C(X), R R X f dµ = Y Ty (f ) dν, f ∈ C(X).

Proof. See [79], p. 58.

A.4

Descriptive set theory

If X, Y are topological spaces, a multivalued mapping ϕ : X → 2Y is a mapping whose values are subsets of Y . If A ⊂ X, its image ϕ(A) is defined as ϕ(A) := S ϕ(x). A multivalued mapping is called upper semicontinuous if x∈A ϕ−1 (U ) := {x ∈ X : ϕ(x) ⊂ U } is open in X for each U ⊂ Y open.

638

A Appendix

If ϕ : X → 2Y is upper semicontinuous and has nonempty compact values, then ϕ is sometimes called a usco mapping. Proposition A.108. If a topological space is the image of a Lindelöf or compact space under a usco mapping, then it is Lindelöf or compact, respectively. Proof. See Section 2.7 of [394]. We recall that NN stands for the usual product space endowed with the pointwise topology and N

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