An introduction to the theory of functional equations and inequalities

Marek Kuczma

An Introduction to the Theory of Functional Equations and Inequalities Cauchy’s Equation and Jensen’s Inequality Second Edition Edited by Attila Gilányi

Birkhäuser Basel · Boston · Berlin

Editor: Attila Gilányi Institute of Mathematics University of Debrecen P.O. Box 12 4010 Debrecen Hungary e-mail: [email protected]

2000 Mathematical Subject Classification: 39B05, 39B22, 39B32, 39B52, 39B62, 39B82, 26A51, 26B25

´ The first edition was published in 1985 by Uniwersytet Slaski (Katowicach) (Silesian University of Katowice) and Pánstwowe Wydawnictwo Naukowe (Polish Scientific Publishers) ´ © Uniwersytet Slaski and Pánstwowe Wydawnictwo Naukowe

Library of Congress Control Number: 2008939524 Bibliographic information published by Die Deutsche Bibliothek Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data is available in the Internet at .

ISBN 978-3-7643-8748-8 Birkhäuser Verlag AG, Basel – Boston – Berlin This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in other ways, and storage in data banks. For any kind of use permission of the copyright owner must be obtained. © 2009 Birkhäuser Verlag AG Basel · Boston · Berlin P.O. Box 133, CH-4010 Basel, Switzerland Part of Springer Science+Business Media Printed on acid-free paper produced of chlorine-free pulp. TCF ∞ Printed in Germany ISBN 978-3-7643-8748-8

e-ISBN 978-3-7643-8749-5

987654321

www.birkhauser.ch

Preface to the Second Edition The first edition of Marek Kuczma’s book An Introduction to the Theory of Functional Equations and Inequalities was published more than 20 years ago. Since then it has been considered as one of the most important monographs on functional equations, inequalities and related topics. As J´ anos Acz´el wrote in Mathematical Reviews “. . . this is a very useful book and a primary reference not only for those working in functional equations, but mainly for those in other fields of mathematics and its applications who look for a result on the Cauchy equation and/or the Jensen inequality.” Based on the considerably high demand for the book, which has even increased after the first edition was sold out several years ago, we have decided to prepare its second edition. It corresponds to the first one and keeps its structure and organization almost everywhere. The few changes which were made are always marked by footnotes. Several colleagues helped us in the preparation of the second edition. We cordially thank Roman Ger for his advice and help during the whole publication process, Karol Baron and Zolt´ an Boros for their conscientious proofreading, and Szabolcs Baj´ ak for typing and continuously correcting the manuscript. We are grateful to Eszter Gselmann, Fruzsina M´esz´aros, Gy¨ongyv´er P´eter and P´ al Burai for typesetting several chapters, and we would like to thank the publisher, Birkh¨ auser, for undertaking and helping with the publication. The new edition of Marek Kuczma’s book is paying tribute to the memory of the highly respected teacher, the excellent mathematician and one of the most outstanding researchers of functional equations and inequalities.

Debrecen, October 2008 Attila Gil´ anyi

Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii

Part I Preliminaries 1 Set 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

Theory Axioms of Set Theory . . . . . Ordered sets . . . . . . . . . . . Ordinal numbers . . . . . . . . Sets of ordinal numbers . . . . Cardinality of ordinal numbers Transfinite induction . . . . . . The Zermelo theorem . . . . . Lemma of Kuratowski-Zorn . .

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19 23 25 28 32 35 37 39 42

3 Measure Theory 3.1 Outer and inner measure . . . . 3.2 Linear transforms . . . . . . . . 3.3 Saturated non-measurable sets 3.4 Lusin sets . . . . . . . . . . . . 3.5 Outer density . . . . . . . . . . 3.6 Some lemmas . . . . . . . . . .

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2 Topology 2.1 Category . . . . . . . . . . . 2.2 Baire property . . . . . . . 2.3 Borel sets . . . . . . . . . . 2.4 The space z . . . . . . . . . 2.5 Analytic sets . . . . . . . . 2.6 Operation A . . . . . . . . . 2.7 Theorem of Marczewski . . 2.8 Cantor-Bendixson theorem 2.9 Theorem of S. Piccard . . .

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3.7 3.8

Theorem of Steinhaus . . . . . . . . . . . . . . . . . . . . . . . . . . . Non-measurable sets . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4 Algebra 4.1 Linear independence and dependence . . 4.2 Bases . . . . . . . . . . . . . . . . . . . 4.3 Homomorphisms . . . . . . . . . . . . . 4.4 Cones . . . . . . . . . . . . . . . . . . . 4.5 Groups and semigroups . . . . . . . . . 4.6 Partitions of groups . . . . . . . . . . . 4.7 Rings and fields . . . . . . . . . . . . . . 4.8 Algebraic independence and dependence 4.9 Algebraic and transcendental elements . 4.10 Algebraic bases . . . . . . . . . . . . . . 4.11 Simple extensions of fields . . . . . . . . 4.12 Isomorphism of fields and rings . . . . .

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75 78 83 87 89 95 98 101 103 105 106 108

Part II Cauchy’s Functional Equation and Jensen’s Inequality 5 Additive Functions and Convex Functions 5.1 Convex sets . . . . . . . . . . . . . . 5.2 Additive functions . . . . . . . . . . 5.3 Convex functions . . . . . . . . . . . 5.4 Homogeneity fields . . . . . . . . . . 5.5 Additive functions on product spaces 5.6 Additive functions on C . . . . . . .

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117 128 130 137 138 139

6 Elementary Properties of Convex Functions 6.1 Convex functions on rational lines . . 6.2 Local boundedness of convex functions 6.3 The lower hull of a convex functions . 6.4 Theorem of Bernstein-Doetsch . . . .

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7 Continuous Convex Functions 7.1 The basic theorem . . . . . . . . . . 7.2 Compositions and inverses . . . . . . 7.3 Differences quotients . . . . . . . . . 7.4 Differentiation . . . . . . . . . . . . 7.5 Differential conditions of convexity . 7.6 Functions of several variables . . . . 7.7 Derivatives of a function . . . . . . . 7.8 Derivatives of convex functions . . . 7.9 Differentiability of convex functions . 7.10 Sequences of convex functions . . . .

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161 162 164 168 171 174 177 180 188 192

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Contents

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8 Inequalities 8.1 Jensen inequality . . . . . . . . . . . . . . . . . 8.2 Jensen-Steffensen inequalities . . . . . . . . . . 8.3 Inequalities for means . . . . . . . . . . . . . . 8.4 Hardy-Littlewood-P´ olya majorization principle 8.5 Lim’s inequality . . . . . . . . . . . . . . . . . . 8.6 Hadamard inequality . . . . . . . . . . . . . . . 8.7 Petrovi´c inequality . . . . . . . . . . . . . . . . 8.8 Mulholland’s inequality . . . . . . . . . . . . . 8.9 The general inequality of convexity . . . . . . .

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197 201 208 211 214 215 217 218 223

9 Boundedness and Continuity of Convex Functions and Additive 9.1 The classes A,B,C . . . . . . . . . . . . . . . . . . . . . . 9.2 Conservative operations . . . . . . . . . . . . . . . . . . . 9.3 Simple conditions . . . . . . . . . . . . . . . . . . . . . . . 9.4 Measurability of convex functions . . . . . . . . . . . . . . 9.5 Plane curves . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 Skew curves . . . . . . . . . . . . . . . . . . . . . . . . . . 9.7 Boundedness below . . . . . . . . . . . . . . . . . . . . . . 9.8 Restrictions of convex functions and additive functions . .

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227 229 231 241 242 244 246 251

10 The Classes A, B, C 10.1 A Hahn-Banach theorem 10.2 The class B . . . . . . . 10.3 The class C . . . . . . . 10.4 The class A . . . . . . . 10.5 Set-theoretic operations 10.6 The classes D . . . . . . 10.7 The classes AC and BC

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257 260 266 267 269 271 276

11 Properties of Hamel Bases 11.1 General properties . . . 11.2 Measure . . . . . . . . . 11.3 Topological properties . 11.4 Burstin bases . . . . . . 11.5 Erd˝ os sets . . . . . . . . 11.6 Lusin sets . . . . . . . . 11.7 Perfect sets . . . . . . . 11.8 The operations R and U

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12 Further Properties of Additive Functions and 12.1 Graphs . . . . . . . . . . . . . . . . . . . 12.2 Additive functions . . . . . . . . . . . . 12.3 Convex functions . . . . . . . . . . . . . 12.4 Big graph . . . . . . . . . . . . . . . . . 12.5 Invertible additive functions . . . . . . . 12.6 Level sets . . . . . . . . . . . . . . . . . 12.7 Partitions . . . . . . . . . . . . . . . . . 12.8 Monotonicity . . . . . . . . . . . . . . .

Convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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305 308 313 316 322 327 330 335

Part III Related Topics 13 Related Equations 13.1 The remaining Cauchy equations 13.2 Jensen equation . . . . . . . . . . 13.3 Pexider equations . . . . . . . . . 13.4 Multiadditive functions . . . . . 13.5 Cauchy equation on an interval . 13.6 The restricted Cauchy equation . 13.7 Hossz´ u equation . . . . . . . . . 13.8 Mikusi´ nski equation . . . . . . . 13.9 An alternative equation . . . . . 13.10The general linear equation . . .

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343 351 355 363 367 369 374 376 380 382

14 Derivations and Automorphisms 14.1 Derivations . . . . . . . . . . . . . . 14.2 Extensions of derivations . . . . . . . 14.3 Relations between additive functions 14.4 Automorphisms of R . . . . . . . . . 14.5 Automorphisms of C . . . . . . . . . 14.6 Non-trivial endomorphisms of C . .

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391 394 399 402 403 406

15 Convex Functions of Higher Orders 15.1 The difference operator . . . . . . . . . . . 15.2 Divided differences . . . . . . . . . . . . . . 15.3 Convex functions of higher order . . . . . . 15.4 Local boundedness of p-convex functions . . 15.5 Operation H . . . . . . . . . . . . . . . . . 15.6 Continuous p-convex functions . . . . . . . 15.7 Continuous p-convex functions. Case N = 1 15.8 Differentiability of p-convex functions . . . 15.9 Polynomial functions . . . . . . . . . . . . .

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415 421 429 432 435 439 442 444 446

Contents

16 Subadditive Functions 16.1 General properties . . . . . . . 16.2 Boundedness. Continuity . . . . 16.3 Differentiability . . . . . . . . . 16.4 Sublinear functions . . . . . . . 16.5 Norm . . . . . . . . . . . . . . 16.6 Infinitary subadditive functions

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455 458 465 471 473 475

17 Nearly Additive Functions and Nearly Convex Functions 17.1 Approximately additive functions . . . . . . . . . . . 17.2 Approximately multiadditive functions . . . . . . . . 17.3 Functions with bounded differences . . . . . . . . . . 17.4 Approximately convex functions . . . . . . . . . . . 17.5 Set ideals . . . . . . . . . . . . . . . . . . . . . . . . 17.6 Almost additive functions . . . . . . . . . . . . . . . 17.7 Almost polynomial functions . . . . . . . . . . . . . 17.8 Almost convex functions . . . . . . . . . . . . . . . . 17.9 Almost subadditive functions . . . . . . . . . . . . .

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483 485 486 490 498 505 510 515 524

18 Extensions of Homomorphisms 18.1 Commutative divisible groups . . 18.2 The simplest case of S generating 18.3 A generalization . . . . . . . . . 18.4 Further extension theorems . . . 18.5 Cauchy equation on a cylinder . 18.6 Cauchy nucleus . . . . . . . . . . 18.7 Theorem of Ger . . . . . . . . . . 18.8 Inverse additive functions . . . . 18.9 Concluding remarks . . . . . . .

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Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571 Indices Index of Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587 Subject Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 Index of Names . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593

Introduction The present book is based on the course given by the author at the Silesian University in the academic year 1974/75, entitled Additive Functions and Convex Functions. Writing it, we have used excellent notes taken by Professor K. Baron. It may be objected whether an exposition devoted entirely to a single equation (Cauchy’s Functional Equation) and a single inequality (Jensen’s Inequality) deserves the name An introduction to the Theory of Functional Equations and Inequalities. However, the Cauchy equation plays such a prominent role in the theory of functional equations that the title seemed appropriate. Every adept of the theory of functional equations should be acquainted with the theory of the Cauchy equation. And a systematic exposition of the latter is still lacking in the mathematical literature, the results being scattered over particular papers and books. We hope that the present book will fill this gap. The properties of convex functions (i.e., functions fulfilling the Jensen inequality) resemble so closely those of additive functions (i.e., functions satisfying the Cauchy equation) that it seemed quite appropriate to speak about the two classes of functions together. Even in such a large book it was impossible to cover the whole material pertinent to the theory of the Cauchy equation and Jensen’s inequality. The exercises at the end of each chapter and various bibliographical hints will help the reader to pursue further his studies of the subject if he feels interested in further developments of the theory. In the theory of convex functions we have concentrated ourselves rather on this part of the theory which does not require regularity assumptions about the functions considered. Continuous convex functions are only discussed very briefly in Chapter 7. The emphasis in the book lies on the theory. There are essentially no examples or applications. We hope that the importance and usefulness of convex functions and additive functions is clear to everybody and requires no advertising. However, many examples of applications of the Cauchy equation may be found, in particular, in books Acz´el [5] and Dhombres [68]. Concerning convex functions, numerous examples are scattered throughout almost the whole literature on mathematical analysis, but especially the reader is referred to special books on convex functions quoted in 5.3. We have restricted ourselves to consider additive functions and convex functions defined in (the whole or subregions of) N -dimensional euclidean space RN . This gives the exposition greater uniformity. However, considerable parts of the theory presented

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Introduction

can be extended to more general spaces (Banach spaces, topological linear spaces). Such an approach may be found in some other books (Dhombres [68], Roberts-Varberg [267]). Only occasionally we consider some functional equations on groups or related algebraic structures. We assume that the reader has a basic knowledge of the calculus, theory of Lebesgue’s measure and integral, algebra, topology and set theory. However, for the convenience of the reader, in the first part of the book we present such fragments of those theories which are often left out from the university courses devoted to them. Also, some parts which are usually included in the university courses of these subjects are also very shortly treated here in order to fix the notation and terminology. In the notation we have tried to follow what is generally used in the mathematical literature1 . The cardinality of a set A is denoted by card A. The word countable or denumerable refers to sets whose cardinality is exactly ℵ0 . The topological closure and interior of A are denoted by cl A and int A. Some special letters are used to denote particular sets of numbers. And so N denotes the set of positive integers, whereas Z denotes the set of all integers. Q stands for the set of all rational numbers, R for the set of all real numbers, and C for the set of all complex numbers. The letter N is reserved to denote the dimension of the underlying space. The end of every proof is marked by the sign . Other symbols are introduced in the text, and for the convenience of the reader they are gathered in an index at the end of the volume. The book is divided in chapters, every chapter is divided into sections. When referring to an earlier formula, we use a three digit notation: (X.Y.Z) means formula Z in section Y in Chapter X. The same rule applies also to the numbering of theorems and lemmas. When quoting a section, we use a two digit notation: X.Y means section Y in Chapter X. The same rule applies also to exercises at the end of each chapter. The book is also divided in three parts, but this fact has no reflection in the numeration. Many colleagues from Poland and abroad have helped us with bibliographical hints and otherwise. We do not endeavour to mention all their names, but nonetheless we would like to thank them sincerely at this place. But at least two names must be mentioned: Professor R. Ger, and above all, Professor K. Baron, whose help was especially substantial, and to whom our debt of gratitude is particularly great. We thank also the authorities of the Silesian University in Katowice, which agreed to publish this book. We hope that the mathematical community of the world will find it useful. Katowice, July 1979 Marek Kuczma

1 The notation in the second edition has been slightly changed. The following sentences are modified accordingly.

Part I

Preliminaries

Chapter 1

Set Theory 1.1 Axioms of Set Theory The present book is based on the Zermelo-Fraenkel system of axioms of the Set Theory augmented by the axiom of choice. The axiom of choice plays a fundamental role in the entire book. The mere existence of discontinuous additive functions and discontinuous convex functions depends on that axiom1 . Therefore the axiom of choice will equally be treated with the remaining axioms of the set theory and no special mention will be made whenever it is used. The primitive notions of the set theory are: set, belongs to (∈), and being a relation type (τ ) [ατ A, R means α is a relation type of A, R; cf. Axiom 8]. The eight axioms read as follows. Axiom 1.1.1. Axiom of Extension. Two sets are equal if and only if they have the same elements: A = B if and only if (x ∈ A) ⇔ (x ∈ B). Axiom 1.1.2. Axiom of Empty Set. There exists a set ∅ which does not contain any element: For every x, x ∈ / ∅.  Axiom 1.1.3. Axiom of Unions. For every collection2 A of sets there exists a set A which contains exactly those elements that belong to at least one set from A:    x∈ A ⇔ (there exists an A ∈ A such that x ∈ A). Axiom 1.1.4. Axiom of Powers. For every set A there exists a collection P(A) of sets which consists exactly of all the subsets of A:   B ∈ P(A) ⇔ (B ⊂ A). 1

R. M. Solovay has shown (Solovay [292]) that a model of mathematics (without axiom of choice) is possible in which all subsets of R (and consequently also all functions f : R → R) are Lebesgue measurable. 2 The word collection is, of course, a synonym of set.

4

Chapter 1. Set Theory

Axiom 1.1.5. Axiom of Infinity. There exists a collection A of sets which contains the empty set ∅ and for every X ∈ A there exists a Y ∈ A consisting of all the elements of X and X itself: ∅ ∈ A and for every X ∈ A there exists a Y ∈ A such that (x ∈ Y ) ⇔ (x ∈ X) or (x = X). Axiom 1.1.6. Axiom of Choice. The cartesian product of a non-empty family of nonempty sets is non-empty: If A = ∅ and for every A ∈ A, A = ∅, then

× A = ∅.

A∈A

Axiom 1.1.7. Axiom of Replacement. Let ψ(x, y) be a two-place propositional formula such that, for every x, there exists exactly one y such that ψ(x, y) holds. Then for every set A there exists a set B which contains those and only those y for which ψ(x, y), x ∈ A: If for every x there exists a z such that ψ(x, y) ⇔ y = z, then for every set A there exists a set B such that (y ∈ B) ⇔ there exists an x ∈ A such that ψ(x, y). Roughly speaking, if to every x there corresponds (according to ψ) a unique y, and if x runs over a set, then the corresponding y’s run over a set. Before stating the last axiom, we must introduce certain notions. Let A be a set and R ⊂ A2 a relation in A. A couple A, R is called a relation system. Two relation systems A, R and B, S are said to be isomorphic iff there exists a one-toone function f from A onto B (a bijection) such that for every a, b ∈ A we have aRb if and only if f (a)Sf (b). Axiom 1.1.8. Axiom of Relation Types. To every relation system A, R there corresponds an object α such that ατ A, R and if ατ A, R and βτ B, S, then α = β if and only if A, R and B, S are isomorphic. The Axiom of Replacement implies the following statement. Axiom of Specification. For every set A and for every propositional formula Φ(x) there exists the set {x ∈ A | Φ(x)} consisting of exactly those x ∈ A for which Φ(x) holds true: x ∈ {x ∈ A | Φ(x)} ⇔ x ∈ A ∧ Φ(x). It is enough to take ψ(x, y) ⇔ Φ(x) ∧ y = x. The Axiom of Empty Set can be replaced by the weaker Axiom of Existence. There exists a set. The empty set can be defined as (A being an existing set) ∅ = {x ∈ A | x = x}.

1.2. Ordered sets

5

If we take into account the definition of the cartesian product of an arbitrary collection of sets, we can reformulate the Axiom of Choice as follows: For  every non-empty collection A of non-empty sets there exists a function w : A → A (the choice function) such that w(A) ∈ A for every A ∈ A. The Axiom of Choice is usually used in this form. The Axiom of Extension implies the uniqueness of sets whose existence is guaranteed by the remaining Axioms 2–7. The Axiom of Relation Types can be omitted. The whole set theory can be built without a use of this axiom. The ordinal numbers (as well as cardinal numbers) must then be defined otherwise. (Cf., e.g., Halmos [130]). From the Axioms 1.1.1–1.1.8 all the set theory can be built (cf. KuratowskiMostovski [198], Halmos [130], Rasiowa [262]). We assume that the reader is familiar with it. However, in the sequel we outline the theory of ordinal numbers, as the latter is often omitted in the university courses of the set theory.

1.2 Ordered sets Let A be a set, and  ⊂ A2 a relation which is reflexive, antisymmetric and transitive: (i) a  a, (ii) (a  b) ∧ (b  a) ⇒ (a = b), (iii) (a  b) ∧ (b  c) ⇒ (a  c). Such a relation  is called an order3 in A and the couple (A, ) is called an ordered set. Clearly, every ordered set is a relation system in the sense of 1.1. The strict relation < is defined as (a < b) ⇔ (a  b) ∧ (a = b). Instead of a  b, a < b, we shall often write b  a, b > a. If, besides (i), (ii) and (iii), also the trichotomy law holds: (iv) For every a, b ∈ A, we have either a < b, or b < a or a = b, then the set A is called linearly ordered or a chain. Let (A, ) be an ordered set. An element a ∈ A is called maximal [minimal ] iff there is no b ∈ A strictly greater [smaller] than a. In other words, a is maximal iff (b ∈ A) ∧ (a  b) ⇒ (b = a). [a is minimal iff (b ∈ A) ∧ (b  a) ⇒ (b = a)]. 3

In earlier texts order is often called a partial order, the word order being reserved for what is here called a linear order. This is due to the fact that, for arbitrary a, b ∈ A, we are often unable to decide whether a  b or b  a. An illustrative example is the power set P(A) of a set A with the order relation defined as the inclusion: a  b ⇔ a ⊂ b.

6

Chapter 1. Set Theory

One ordered set may have several (or none) maximal [minimal] elements. If a ∈ A is a maximal [minimal] element, then there may exist in A elements b which are not comparable with a, i.e., for which neither a  b, nor b  a holds. An element a ∈ A is called the greatest [smallest] (or the last [least]) element iff x  a [a  x] holds for every x ∈ A. The last [least] element, if it exists, is unique. An element a ∈ A is called the upper bound of a set E ⊂ A iff x  a holds for every x ∈ E. It is not required that a ∈ E, but it is possible. There may exist several (or none) upper bounds of a set E ⊂ A. If (B, ) is another ordered set, then we say that (A, ) and (B, ) are similar   and write (A, ) ∼ (B, ≺) iff there exists a one-to-one order-preserving mapping f from A onto B. The relation of similarity is an isomorphism of relation systems (A, ) and (B, ) as defined in 1.1:   (a  b) ⇔ f (a)  f (b) An ordered set, every non-empty subset of which has the smallest element, is called a well-ordered set, and the corresponding order is called a well-order . We have the following Theorem 1.2.1. Every well-ordered set is linearly ordered. Proof. This follows from the fact that for any a, b ∈ A, the pair {a, b} ⊂ A has the smallest element.  Any finite linearly ordered set is well ordered and two such sets are similar if and only if they have the same number of elements. (The proof of these facts is left to the reader.) The set (N, ), where  stands for the usual inequality between numbers, is well ordered.

1.3 Ordinal numbers Let (A, ) be a well-ordered set. Any set P ⊂ A such that if x ∈ P and y  x, then y ∈ P , is called an initial segment of A. Theorem 1.3.1. If P is an initial segment of a well-ordered set A4 , and P = A, then there exists in A an x such that P = P (x) = {y ∈ A | y < x} . Proof. The set A \ P = ∅ has the smallest element x. We will show that P = P (x). Let y ∈ P . If we had x  y, then we would have x ∈ P , which contradicts the condition x ∈ A \ P . Thus y < x and y ∈ P (x). Consequently P ⊂ P (x). If y ∈ P (x), then y < x, and since x is smallest in A \ P , we must have y ∈ P . Consequently P (x) ⊂ P . Thus P and P (x) have the same elements, and so they are equal: P = P (x).  Instead of saying: A is the first component of the ordered set (A, ), we often say simply: A is an ordered set.

4

1.3. Ordinal numbers

7

The formulas, valid for arbitrary a, b ∈ A, are left to the reader as exercises:   1. (a  b) ⇔ P (a) ⊂ P (b) ,   2. (a = b) ⇔ P (a) = P (b) , 3. For arbitrary well-ordered set (A, ) put P = {P (x) | x ∈ A}. The ordered set (P, ⊂) is well ordered and is similar to (A, ). The relation types of well-ordered sets are called the ordinal numbers. If (A, ) is a well-ordered set and if ατ (A, ), then we write α = A. The N is denoted by ω. If (A, ) is a finite (well-ordered) set consisting of n ∈ N elements, then we assume A = n. In particular, ∅ = 0. If (A, ) is a well-ordered set, then, by the Axiom of Infinity there exists a set B which contains all the elements of A and A itself. We order B by assuming additionally that A  a for any a ∈ A. The ordinal number B is denoted by A + 1. If an ordinal number cannot be written as α + 1 with another ordinal number α, then it is called a limit number . An example of a limit number is ω. If α and β are ordinal numbers, say α = A and β = B, then we say that α < β iff the set A is similar to an initial segment of B different from B. Theorem 1.3.2. For any ordinal number α, it is not true that α < α. Proof. For an indirect proof, suppose that α < α, i.e., A is similar to its initial segment different from A. Let f be a similarity function. Put B = {x ∈ A | f (x) < x} . By Theorem 1.3.1 there exist an a ∈ A such that A ∼ P (a). Hence f (a) < a, i.e., a ∈ B. So B = ∅, and B ⊂ A, and hence there exists the smallest element, say b, in B. Then f (b) < b, (1.3.1)   and since f is order-preserving, f f (b) < f (b). This means that f (b) ∈ B, which, by (1.3.1) contradicts the condition that b is smallest in B.  Theorem 1.3.3. If α < β and β < γ, then α < γ. Proof. Let A, B, C be well ordered sets such that A = α, B = β and C = γ. Moreover, let b ∈ B and c ∈ C be such that A ∼ P (b) and B ∼ P (c), the similarity functions being f and g, respectively. Then it is easily checked that g ◦ f is a one-to-one, order-preserving mapping of A onto an initial segment of C, different from C.  Theorem 1.3.4. If α < β, then it is not true that β < α. Proof. This follows from Theorems 1.3.3 and 1.3.2.



Theorems 1.3.2 and 1.3.3 imply that the inequality  defined for ordinal numbers as follows: α  β iff either α < β, or α = β, is an order in the sense of 1.2.

8

Chapter 1. Set Theory

1.4 Sets of ordinal numbers We start with a lemma. Lemma 1.4.1. If (A, ) and (B, ) are similar well-ordered sets5 and f is a similarity function, then f maps initial segments of A onto initial segments of B. Proof. Let f be a one-to-one order preserving mapping of A onto B. Then f −1 is a one-to-one order preserving mapping of B onto A. Let P be an initial segment of A. The thing to prove is that f (P ) is an initial segment of B. Suppose that there exist b1 < b2 , b2 ∈ f (P ) and b1 ∈ / f (P ). Put ai = f −1 (bi ), i = 1, 2. Then a1 < a2 , a2 ∈ P and a1 ∈ / P , which is impossible since P is an initial segment of A.  Corollary 1.4.1. No two different initial segments of a well-ordered set are similar to each other. Proof. Let (A, ) be a well-ordered set, and P1 = P2 initial segments of A, P1 ∼ P2 . Since P1 = P2 , at least one of these segments, say P1 , must be different from A, and hence of the form P (a) with an a ∈ A. If a ∈ P2 , then P1 is an initial segment of (P2 , ). Indeed, if x ∈ P1 and y < x, then y < a and hence y ∈ P (a) = P1 . And if a∈ / P2 , then P2 is an initial segment of (P1 , ). Indeed, then P2 = A and hence of the form P (b) with a b ∈ A. If x ∈ P2 and y < x, then y < b and y ∈ P2 . Thus one of the sets P1 , P2 is similar to its initial segment different from this set, which contradicts Theorem 1.3.2.  Theorem 1.4.1. Of any two well-ordered sets, one is similar to an initial segment of the other. Proof. Let (A, ) and (B, ) be two well-ordered sets. Define the set Z, Z = {x ∈ A | there exists a y ∈ B such that P (x) ∼ P (y)} . By Corollary 1.4.1, such a y is unique. Thus we may define a function f : Z → B by putting f (x) = y iff P (x) ∼ P (y). Again by Corollary 1.4.1 f is one-to-one. The set Z is an initial segment of A. For if x ∈ Z, then P (x) ∼ P (y) for a y ∈ B. If x < x, then P (x ) ⊂ P (x). Let g, mapping P (x) onto P (y), be a similarity function. Then g maps P (x ) onto an initial segment P of P (y) and hence P is an initial segment of B. Since y ∈ / P (y), also y ∈ / P , and thus P = B. So there exists a  y  ∈ B such that P = P (y  ). Henceg establishes a similarity between P(x ) and  P (y ).    Thus x ∈ Z and P (y ) = P f (x ) is an initial segment of P (y) = P f (x) , whence f (x ) < f (x). So as a by-product we have obtained the fact that f is order-preserving. Similarly, f (Z) is an initial segment of B. For if y ∈ f (Z), then P (y) ∼ P (x) for an x ∈ Z. Let h be a similarity mapping. If y  < y, then h maps P (y  ) onto an initial segment P (x ) of P (x), whence y  = f (x ), x ∈ Z and y  ∈ f (Z). 5 It would be enough to postulate that one of these sets is well ordered. It follows then by the similarity that the other is well ordered, too. (Cf. Exercise 1.6)

1.4. Sets of ordinal numbers

9

We have already shown that Z ∼ f (Z). To complete the proof it is enough to show that either Z = A or f (Z) = B. Suppose that Z = A and f (Z) = B. Then there exist a ∈ A and b ∈ B such that Z = P (a) and f (Z) = P (b). Thus P (a) ∼ P (b) and a ∈ Z, which is incompatible with Z = P (a).  Theorem 1.4.1 implies the trichotomy law for ordinal numbers: For any ordinal numbers α, β either α < β, or β < α, or α = β. Let us note also the following Theorem 1.4.2. For every ordinal number α, there exists the set Γ(α) of all ordinal numbers β < α. Proof. Let (A, ) be a well-ordered set such that A = α. Let for x ∈ A, ψ(x, y) be the propositional formula yτ P (x), i.e., y = P (x). By the Axiom of Replacement there exists a set B such that y ∈ B ⇔ there exists an x ∈ A such that yτ P (x). But this, in turn, is equivalent to the fact that y < α. Thus B is the required set.  Once we know that Γ(α) is a set, the formula y = P (x) for x ∈ A defines a function from A onto Γ(α). This function clearly is one-to-one and order-preserving. It follows that Γ(α) ∼ A and consequently Γ(α) is well ordered and Γ(α) = α.

(1.4.1)

Actually, the fact that Γ(α) is well ordered is a particular case of the following statement. Theorem 1.4.3. Any set of ordinal numbers is well ordered by the inequality . Proof. The thing to prove is that if A = ∅ is a set of ordinal numbers, then there exists the smallest element in A. Take an α ∈ A. If α is smallest in A, there is nothing more to prove. If not, the set A ∩ Γ(α) ⊂ Γ(α) = ∅. Since Γ(α) is well ordered, there exists the smallest element β in A ∩ Γ(α). For an arbitrary γ ∈ A we have either γ  α, or γ < α. In the first case, since β ∈ Γ(α), we have β < α  γ, whence β < γ. In the other case γ ∈ Γ(α) ∩ A, so β  γ. Thus β is the smallest element in A.  Next we have Lemma 1.4.2. For every ordinal number α we have α < α + 1. ∗

Proof. Let α = A. Then α + 1 = A , where A∗ is the set consisting of all the elements of A and A itself ordered so that A  a for all a ∈ A. Hence P (A) = A in A∗ . The function f : A → A∗ defined as f (a) = a for a ∈ A establishes the similarity of A with P (A) in A∗ . Thus A is similar to an initial segment of A∗ different from A∗ , i.e., ∗ A
10

Chapter 1. Set Theory

Proof. Let A be a set of ordinal numbers. By the Axiom of Replacement there exists the collection of sets {Γ(β)}β∈A , and by the Axiom of Unions there exists the set B=



Γ(β).

β∈A

Then, for every β ∈ A,   the set Γ(β) clearly is an initial segment of B. Hence B  Γ(β) = β cf. (1.4.1) . In view of Lemma 1.4.2 the ordinal number B + 1 has the desired property.  Corollary 1.4.2. There does not exist the set of all ordinal numbers. Corollary 1.4.3. For every set of ordinal numbers there exists the smallest ordinal number which does not belong to the given set. Proof. Let A be a set of ordinal numbers and let α be a number greater than any β ∈ A. If α is not the smallest number with this property, then the set Γ(α)\A ⊂ Γ(α) is non-empty, and consequently it has the smallest element γ. It is already seen that γ is the smallest ordinal number which does not belong to A. 

1.5 Cardinality of ordinal numbers If α is an ordinal number, then by definition any two well-ordered sets of type α are similar, i.e., there exists a one-to-one mapping from one set onto the other. Consequently these sets have the same cardinality. Consequently to any ordinal number α we may assign a cardinal number, the common cardinality of all well-ordered sets of type α. This cardinal  numberis called the cardinality of α and is denoted by α. In particular we have cf. (1.4.1) α = card Γ(α). It is easy to check that for arbitrary ordinal numbers α, β we have α  β ⇒ α  β,

(1.5.1)

Lemma 1.5.1. There exists the set A which contains all ordinal numbers α with α  ℵ0 . Proof. If α  ℵ0 for an ordinal number α, then there exists a one-to-one mapping   f : Γ(α) → N. If for a, b ∈ f Γ(α) we put   (a  b) ⇔ f −1 (a)  f −1 (b) ,     the set f Γ(α) ,  will become a well-ordered set similar to Γ(α) (f being the     similarity function), and hence f Γ(α) ,  = α. Thus for any ordinal number α with α  ℵ0 there exists a well-ordered set (B, ) such that B ⊂ N, and B = α. The converse is also true. If (B, ) is a well-ordered set of a type α, and B ⊂ N, then α  ℵ0 . Consequently we may describe ordinal numbers α such that α  ℵ0 as order

1.5. Cardinality of ordinal numbers

11

types of well-ordered subsets of N. (However, the order in these sets may be different from the natural order in N.) Let P be the power set of N (Axiom 4) and, for P ∈ P, let RP be the power set of P × P . Every element R of RP is a relation in P . There exists the set (Axiom of Specification) B = {(P, R) | P ∈ P, R ∈ RP } and the set C = {(P, R) ∈ B | R is a well order} . By the Axiom of Replacement there exists the set of the types of all (P, R) ∈ C. This is the desired set.  The same argument shows that for any cardinal number m there exists the set of all ordinal numbers α such that α  m. This set is denoted in the sequel by M (m). Lemma 1.5.2. For every cardinal number m, we have   M (m) = Γ M (m) .

(1.5.2)

Proof. There exists an ordinal number β which is greater than any α ∈ M (m). First we show that M (m) is an initial segment of Γ(β). Let γ ∈ M (m) and ξ < γ. Then by (1.5.1) ξ  γ  m, i.e., ξ ∈ M (m). So either M (m) = Γ(β), or there exists an η ∈ Γ(β) such that M (m) = {ξ ∈ Γ(β) | ξ < η} = Γ(η). At any case M (m) = Γ(α) for an ordinal number α. Since α = Γ(α) = M (m), we obtain hence (1.5.2).  Theorem 1.5.1. For every cardinal number m, we have card M (m) > m.

(1.5.3)

Proof. Put α = M (m) so that M (m) = Γ (α). Suppose that (1.5.3) does not hold, i.e., α  m. This means that α ∈ M (m), i.e., α ∈ Γ (α), which is impossible.  Remark 1.5.1. Here we have made use of the trichotomy law for the cardinals, which, however, cannot be proved at this stage. What we actually prove here is that the inequality α  m is impossible. This is sufficient to prove the remaining theorems of the present chapter, and then the trichotomy law for the cardinals follows from Theorem 1.4.1 and 1.7.1, and hence also condition (1.5.3). We define Ω to be the order type of the set M (ℵ0 ), and ℵ1 to be the cardinality of M (ℵ0 ). Thus Ω = M (ℵ0 ), ℵ1 = Ω = card M (ℵ0 ). By Theorem 1.5.1 ℵ1 > ℵ0 . Moreover Theorem 1.5.2. There is no cardinal number strictly between ℵ0 and ℵ1 . In other words, ℵ1 is the next cardinal number after ℵ0 .

12

Chapter 1. Set Theory

Proof. The thing to show is that if for a cardinal number m we have m < ℵ1 , then m  ℵ0 . Let m < ℵ1 and let X be a set such that card X = m. Since card X = m < ℵ1 = card M (ℵ0 ), there exists a subset Y of M (ℵ0 ) with the same cardinality as X: there exists Y ⊂ M (ℵ0 ) : card Y = card X = m. Let α = Y . Thus Y ∼ Γ (α), and m = card Γ (α). Of course,  α < Ω, for  otherwise we would have α  Ω, i.e., m  ℵ1 . Thus α ∈ Γ (Ω) = M (ℵ0 ) cf. (1.5.2) , which means that m = α  ℵ0 .  On the basis of Axioms 1–8 it is impossible to compare ℵ1 with the power of continuum c. The conjecture that ℵ1 = c is the contents of the celebrated Continuum hypothesis: ℵ1 = c. P. J. Cohen [48] showed that the continuum hypothesis is independent of Axioms 1–8. In the present book, unless explicitely stated otherwise, we do not assume the continuum hypothesis. Note that, as a result of (1.5.2), Γ (Ω) = M (ℵ0 ). Consequently, for every ordinal number α < Ω we have α  ℵ0 .

1.6 Transfinite induction If α is an ordinal number, then any function defined on the set Γ (α) is called  a transfinite sequence of type α. The values of this function are denoted by xβ the value at the point β ∈ Γ (α) and the sequence itself by {xβ }β<α . Theorem 1.6.1. Let X be a set, {xβ }β<α a transfinite sequence containing all the elements of X, Φ a propositional formula defined for x ∈ X, and α an ordinal number. If the following conditions are fulfilled (i) Φ (x0 ); (ii) If β = γ + 1 < α and Φ (xγ ), then Φ (xβ ); (iii) If β < α is a limit number and Φ (xγ ) for every γ < β, then Φ (xβ ); then Φ (x) for every x ∈ X 6 . Proof. Let E = {β ∈ Γ (α) | Φ (xβ )}, and let ξ be the smallest ordinal number in Γ (α) which does not belong to E. By (i) ξ > 0. If ξ = γ + 1, then γ ∈ E, and consequently Φ (xγ ) holds. Then by (ii) Φ (xξ ). If ξ is a limit number, then Φ (xγ ) holds for every γ < β, whence by (iii) Φ (xξ ). In both cases we arrive at ξ ∈ E. The contradiction obtained shows that there is not such a ξ in Γ (α), or, in other words, E = Γ (α). Thus Φ (xβ ) for all β < α, and since the sequence {xβ }β<α contains all the elements of X, we have Φ (x) for all x ∈ X.  Theorem 1.6.1 extends the well-known induction principle for N. The next theorem, known as definitions by transfinite induction, also is an extension of the corresponding theorem for natural numbers. 6

Condition (ii) could be omitted, as it is contained in (iii), where β need not be a limit number. Similarly (i). Thus (i)–(iii) jointly can be replaced  by (iv) If β < α, and Φ (xγ ) for every γ < β, then Φ xβ .

1.6. Transfinite induction

13

Let F (α, X), where α is an ordinal number and X is a set, be the collection of all transfinite sequences of types β < α and with values in X: f ∈ F (α, X) ⇔ f : Γ (β) → X for some β < α. Theorem 1.6.2. For every ordinal number α, for every set X, and for every function h : F (α, X) → X, there exists exactly one transfinite sequence f : Γ (α) → X such that   f (β) = h f | Γ(β) for β < α. The symbol f | Γ (β) denotes the restriction of f to the set Γ (β). Proof. First we prove the uniqueness. Suppose that there exist two such functions f, g. Let Φ be the propositional formula for β ∈ Γ (α) Φ (β) ⇔ [f (β) = g (β)] . For β = 0 we have f (0) = h (f | ∅) = h (g | ∅)  = g (0),so Φ (0)  holds true. Let Φ (γ) for γ be less than a β < α. Then f (β) = h f | Γ(β) = h g | Γ(β) = g (β), i.e., Φ (β). By Theorem 1.6.1 Φ (β) for all β ∈ Γ (α), i.e., f = g. To prove the existence, consider the set E of all those β  α for which there exists a transfinite sequence(by the first  part of the proof necessarily unique) fβ : Γ (β) → X such that fβ (γ) = h fβ | Γ(γ) for all γ < β. If δ  γ, δ, γ ∈ E, then the sequence fγ | Γ(δ) fulfils for ξ < δ fγ | Γ (δ) (ξ) = fγ (ξ) = h fγ | Γ(ξ) = h fγ | Γ (δ) | Γ (ξ) so that, by the uniqueness, fγ | Γ (δ) = fδ . Suppose that there exists a β  α, β / E. We may assume that β is the smallest such number, i.e., γ ∈ E for γ < β  ∈ for otherwise we could choose the smallest such number in Γ (β) . Let β be a limit number. Then for every ξ < β we have ξ + 1 < β. Put for ξ < β f (ξ) = fξ+1 (ξ) . Then f : Γ (β) → X is a transfinite sequence of type β and for ξ < β we have     f (ξ) = fξ+1 (ξ) = h fξ+1 | Γ(ξ) = h f | Γ(ξ) . And if β = γ + 1, then we put for ξ < γ f (ξ) = fξ+1 (ξ) and

  f (γ) = h f | Γ(γ) . (1.6.1)    Then we have f (ξ) = h f | Γ(ξ) for ξ < γ as in the preceding case and f (γ) = h f | Γ(γ) by (1.6.1). Thus β ∈ E, which shows that all β  α belong to E. 

14

Chapter 1. Set Theory

1.7 The Zermelo theorem We start with a lemma. Lemma 1.7.1. Let X be a set. If there a transfinite sequence f : Γ (α) → X (of  exists  a certain type α) such that X = f Γ(α) , then the set X can be well ordered. Proof. For x ∈ X put

Ax = {β ∈ Γ (α) | f (β) = x}

and let ξx be the smallest element in Ax . Define the order in X (x  y) ⇔ (ξx  ξy ) . Relation  is a well-order in X.



Theorem 1.7.1. [Zermelo [326]] Every set can be well ordered. Proof. It is enough to show that there exists a transfinite sequence whose range is   X. Let w : P (X) \ ∅ → X be a choice function (cf. 1.1). Let card X = m and α = M (m). Define the transfinite sequence f : Γ (α) → X by    f (β) = w X \ {f (γ)} , β < α. γ<β

By Theorem 1.6.2 the sequence {xβ }β<α is unambiguously defined. It remains to show that the range of f is X. Put  B= {f (β)} . β∈Γ(α)

Clearly B ⊂ X. Moreover card B = card Γ (α) = α. Hence m = card X  card B = α = card M (m). This shows that f cannot be defined for all β ∈ Γ (α), i.e., for a certain β < α we must have X \ f (γ) = ∅. Thus f is of type β and its range γ∈Γ(β)

is X.



In this proof of Zermelo theorem we have used the Axiom of Choice. This could not be avoided, the Axiom of Choice had to be used in the proof. Actually the Zermelo theorem is equivalent to the Axiom of Choice, as results from the following Theorem 1.7.2. If every set can be well ordered, then for every set there exists a choice function. Proof. Let X be an arbitrary set, and let  be a relation which well orders X. Put for every A ⊂ X, A = ∅ w (A) = the smallest element in A.   Then w : P (X) \ ∅ → X is a choice function.



1.8. Lemma of Kuratowski-Zorn

15

1.8 Lemma of Kuratowski-Zorn In applications the Axiom of Choice is usually used in one of its equivalent forms. We have encountered one of such forms in the preceding section. It was the Zermelo theorem. But perhaps the most famous statement equivalent to the Axiom of Choice is the following Theorem 1.8.1. [Lemma of Kuratowski-Zorn7] Let (X, ) be an ordered set, X = ∅, in which every chain has an upper bound. Then for every x0 ∈ X there exists in X a maximal element x such that x  x0 . Proof. Let card X = m, M (m) = α. For an indirect proof suppose that for every y  x0 the set {z ∈ X | z > y} is non-empty. Define the transfinite sequence {yβ }β<α as  the upper bound of {xγ }γ<β , if it exists, yβ = (1.8.1) x0 otherwise , and define the transfinite sequence {xβ }β<α by   xβ = w {z ∈ X | z > yβ } ,

(1.8.2)

  where w : P (X) \ ∅ → X is a choice function. Clearly, by (1.8.1), yβ  x0 for every β < α so that the set occurring in (1.8.2) is non-empty. So the sequence {xβ }β<α is well defined. This sequence is increasing. To show this consider the propositional formula Φ(β) (β < α) meaning: “For β < α, if γ < ξ  β, then xγ < xξ ”. If Φ(γ) for γ < β, then {xδ }δ<β is a chain, and consequently yβ  xδ , δ < β. Hence also xβ > xδ for δ < β, i.e., Φ(β) holds. By Theorem 1.6.1 we have Φ(β) for all β < α. Now put  {xβ } . B= β<α

We have B ⊂ X, whence card B  card X = m, whereas card B = card Γ (α) = α = card M (m) > m. This shows that for some β < α we must have {z ∈ X | z > yβ } = ∅, i.e., yβ is a maximal element in X, and clearly yβ  x0 .  Theorem 1.8.2. The Lemma of Kuratowski-Zorn implies the Axiom of Choice.  Proof. Let A be any collection of non-empty sets, and put B = A. Let X be the family of those sets F ⊂ B for which the intersection F ∩ A contains at most one point for every A ∈  A. (X, ⊂) is an ordered set, ∅ ∈ X. If C ⊂ X is a chain, then  C ∈ X. In fact, if C ∩ A for an A ∈ A contains two different elements, say x, y, 7

Kuratowski [197], Zorn [327].

16

Chapter 1. Set Theory

then there exist sets Dx , Dy ∈ C such that x ∈ Dx ∩ A and y ∈ Dy ∩ A. But since C is a chain, one of the sets Dx , Dy is contained in the other, say Dx ⊂ Dy . But then x, y ∈ Dy and Dy ∩ A contains more than one point. By the Lemma of Kuratowski-Zorn there exists in X a maximal element F0 . We will show that F0 ∩ A = ∅ for every A ∈ A. If we had F0 ∩ A0 = ∅ for an A0 ∈ A, then for an x0 ∈ A0 we might define a set F ∗ = F0 ∪ {x0 }. Clearly F ∗ ∈ X and F ∗ is larger than F0 , which is impossible, since F0 is maximal in X. Thus F0 ∩ A is a singleton for every A ∈ A, and we can define a function w : A → B by w (A) = A ∩ F0 . w is a choice function.



Another equivalent formulation of the Axiom of Choice is the following one. Theorem 1.8.3. For every non-empty family A of disjoint non-empty sets, there exists a set which has exactly one element in common with every set A ∈ A.  Proof. Let w : A → A be a choice function. The set w (A) has the desired properties.  We shall show also that Theorem 1.8.3 implies the Axiom of Choice.   Let A be any non-empty family of non-empty sets. Consider the subsets of A × A of the form A × {A} for A ∈ A. Let B be the collection of all such subsets. The sets from B     are disjoint, since for A1 = A2 , we have A1 × {A1 } ∩ A2 × {A2 } = ∅.  Let F be  a set that has exactly one element in common with every B ∈ B. If B ∩ A × {A} =    x, {A} , put w(A) = x. This formula defines a choice function w : A → A. P. J. Cohen [48] proved that the Axiom of Choice is independent of the remaining Axioms 1–8. A detailed discussion of The Axiom of Choice is found in Jech [154].

Exercises 1. Prove that for any two sets A, B there exists the set A ∩ B containing those and only those elements which are common to A and B. 2. The ordered pair a, b is defined as {a, {a, b}}, where {a, b} is the unordered pair consisting of a and b. Prove that a, b = c, d if and only if a = c and b = d. 3. Prove that every ordered set (A, ) is similar to a set (C, ⊂), where C is a collection of sets. 4. Let X, Y be arbitrary sets, and let F be the family of all functions whose domain is contained in X and range is contained in Y . For f, g ∈ F write f  g iff domain f ⊂ domain g and g | domain f = f . Prove that (F , ) is an ordered set in which every chain has an upper bound. 5. Let (A, ) and (B, ) be well-ordered sets and order the cartesian product A × B lexicographically. (I.e., a, b  c, d iff a < b or a = b and c  d.) Prove that (A × B, ) is well ordered. 6. If (A, ) is well ordered, then so is also every ordered set similar to (A, ).

1.8. Lemma of Kuratowski-Zorn

17

7. Prove that if (A, ) and (B, ) are two similar well-ordered sets, then there is exactly one similarity function f : A → B. 8. Prove that if (X, ) is a well-ordered set, then every subset of X is similar either to X, or to an initial segment of X. 9. A subset A of an ordered set X is called cofinal in X iff for every x ∈ X there exists an a ∈ A with a  x. Prove that every linearly ordered set has a cofinal well-ordered subset. 10. Let {xβ }β<α be a transfinite sequence of type α, where α is a limit ordinal number, whose terms are ordinal numbers. The smallest ordinal number greater than every xβ , β < α, is called the limit of {xβ }β<α and denoted by lim xβ . If β<α

f and g are two increasing transfinite sequences of  ordinal  numbers, α is a limit ordinal and lim f (β) = ξ, then lim g (δ) = lim g f (β) . β<α

δ<ξ

β<α

11. An ordinal number ξ is cofinal with a limit ordinal number α iff there exists an increasing sequence {xβ }β<α of ordinal numbers such that ξ = lim xβ . Prove β<α

that ξ is cofinal with α if and only if Γ (ξ) contains a cofinal subset of type α. 12. Prove the following version of the transfinite induction principle: Let (X, ) be a well-ordered set, x0 the smallest element in X, and Φ a propositional formula defined on X. If the following conditions are fulfilled (i) Φ (x0 ), (ii) If Φ (y) for y ∈ P (x), then Φ (x), x ∈ X, then Φ (x) for all x ∈ X. 13. Prove that if (A, ) is a well-ordered set similar to its subset A0 , then the similarity function f : A → A0 satisfies f (x)  x for all x ∈ A. 14. Prove that in every family E of sets there exists a maximal subfamily E0 of mutually disjoint sets.

Chapter 2

Topology 2.1 Category In 2.1–2.2 X is a topological space, so, e.g., X may be a metric space, or, in particular, RN . A set A ⊂ X is called nowhere dense iff int cl A = ∅. A set A ⊂ X is said to be of the first category iff A is a countable union of nowhere dense sets: A=

∞ 

Ai , int cl Ai = ∅, i ∈ N .

(2.1.1)

i=1

A set A ⊂ X which is not of the first category is said to be of the second category. A set A ⊂ X is called residual iff A = X \ A is of the first category. Theorem 2.1.1. If A ⊂ X is of the first category, then there exists a set B ∈ Fσ such that A ⊂ B, and B is of the first category. Proof. A has a representation (2.1.1). Put B=

∞ 

cl Ai .

i=1

The sets cl Ai are nowhere dense, just like Ai : int cl cl Ai = int cl Ai = ∅ . B is an Fσ set, being a countable union of closed sets, and B is of the first category, being a countable union of nowhere dense sets. Obviously A ⊂ B.  Theorem 2.1.2. If A ⊂ X is residual, then there exists a set B ∈ Gδ such that B ⊂ A, and B is residual. Proof. There exists (by Theorem 2.1.1) a set C ∈ Fσ such that A ⊂ C, and C is of the first category. The set B = C  = X \ C has the desired properties. 

20

Chapter 2. Topology

If we assume, moreover, that the space X is complete, then we may assert that the set B in Theorem 2.1.2 is of the second category. Otherwise X = B ∪ C would be1 of the first category contrary to a celebrated Baire theorem that a complete space is of the second category. The following theorem yields a slight generalization of Theorem 2.1.2: Theorem 2.1.3. If A ⊂ H ⊂ X , H ∈ Gδ , and H \ A is of the first category, then there exists a set B ⊂ A such that B ∈ Gδ and H \ B is of the first category. Proof. 2 Put A1 = A ∪ (X \ H). The set A1 is residual3 (X \ A1 ⊂ H \ A), so by Theorem 2.1.2 there exists a residual set B1 ⊂ A1 , B1 ∈ Gδ . Put B = H ∩ B1 . Since B1 ⊂ A∪(X \H), we have B ⊂ H ∩[A∪(X \H)] = (H ∩A)∪[H ∩(X \H)] = H ∩A = A, and clearly B ∈ Gδ as an intersection of Gδ sets. It remains to show that H \ B is of the first category. Now, H \ B = H \ (H ∩ B1 ) = H \ (X ∩ B1 ) ⊂ X \ B1 , which is a set of the first category. Consequently also H \ B is of the first category.  The following lemma will be needed later. Lemma 2.1.1. If F ∈ Fσ and int F = ∅, then F is of the first category. Proof. F =

∞ 

Fi , where Fi are closed. If, for a certain i ∈ N, we had int Fi = ∅,

i=1

then we would have also int F = ∅, since int Fi ⊂ int F . Consequently int cl Fi = int Fi = ∅ for all i ∈ N, and thus the sets Fi are nowhere dense. Hence F is of the first category.  We say that a set A ⊂ X is of the first category at a point x ∈ X iff there exists a neighbourhood Ux of x (in X) such that the set A ∩ Ux is of the first category. And we say that a set A ⊂ X is of the second category at a point x ∈ X iff it is not of the first category at x, i.e., iff A ∩ Ux is of the second category for every neighbourhood Ux of x. Put D(A) = {x ∈ X | A is of the II category at x} . We have the following calculation rules for D(A): D(A ∪ B) = D(A) ∪ D(B) ,

(2.1.2)

A ⊂ B ⇒ D(A) ⊂ D(B) .

(2.1.3)

Simple proofs of formulas (2.1.2) and (2.1.3) are left to the reader (cf. Kuratowski [196]). 1

Cf. Exercise 2.4. Theorem 2.1.3 follows also from Theorem 2.1.2 on replacing X by H and observing that every Gδ set in H is also a Gδ set in X. 3 Cf. Exercise 2.4. 2

2.1. Category

21

Theorem 2.1.4. If the space X is separable4 and A ∩ D(A) = ∅ then A is of the first category. Proof. Since X is separable, it has a countable base of neighbourhoods. Let {Bn }n∈N be such a base. For every  x ∈ A there exists aneighbourhood Ux of x such that A∩Ux is of the first category since A ∩ D(A) = ∅ , and there exists an nx ∈ N such that x ∈ Bnx ⊂ Ux . The set A ∩ Bnx ⊂ A ∩ Ux also is of the first category. Since we have A=



A ∩ Bnx ,

(2.1.4)

x∈A

and there are only at most countably many sets Bnx , the union in (2.1.4) is at most countable, and consequently (cf. Exercise 2.4) the set A is of the first category.  Theorem 2.1.5. For every set A ⊂ X, the set D(A) is closed. Proof. We shall show that the set X \ D(A) is open. Take an x ∈ D(A). Then there exists an open neighbourhood Ux of x such that A ∩ Ux is of the first category. Now, since Ux is a neighbourhood of every point from Ux , we have Ux ⊂ X \ D(A). Consequently x is an inner point of X \ D(A), which shows that X \ D(A) is open, and so D(A) is closed.  Theorem 2.1.6. If the space X is separable, then for every set A ⊂ X, the set A\D(A) is of the first category. Proof. By (2.1.3) we have D[A \ D(A)] ⊂ D(A) , whence [A \ D(A)] ∩ D[A \ D(A)] ⊂ [A \ D(A)] ∩ D(A) = ∅ . The theorem results now from Theorem 2.1.4.



Now we are going to investigate category problems in product spaces. Let Y be an arbitrary space. For every set A ⊂ X × Y and every x0 ∈ X we write A[x0 ] = {y ∈ Y | (x0 , y) ∈ A} .

(2.1.5)

Lemma 2.1.2. Let Y be a separable5 topological space, and let A ⊂ X ×Y be a nowhere dense set. Then there exists a set P ⊂ X of the  first category, such that for every x ∈ P  = X \ P the set A[x] defined by (2.1.5) is nowhere dense. 4

Theorem 2.1.4 (as well as Theorem 2.1.6) is generally valid, without assuming that X is separable, but the proof in the general case is much more difficult (cf., e.g., Kuratowski [196]). Since in the sequel the results of the present chapter will be used for X = RN , which is separable, we restrict ourselves to separable X only. Separable means here having an at most countable neighbourhood base. 5 Here, as well as in Theorems 2.1.7 and 2.1.8, the assumption that Y is separable is essential.

22

Chapter 2. Topology

Proof. Since Y is separable, it has a countable base of neighbourhoods. Let {Bn }n∈N be such a base. Put (2.1.6) En = {x ∈ X | {x} × Bn ⊂ cl[A ∩ ({x} × Y )]} .    {x} × Bn ⊂ cl[A ∩ ({x} × Y )] ⊂ cl[A ∩ ({x} × Y )] ⊂ Hence En × Bn = x∈En x∈E x∈X   n  ⊂ [cl A ∩ cl({x} × Y )] = cl A ∩ cl({x} × Y ) = cl A ∩ ({x} × Y ) = x∈X

cl A ∩ (X × Y ) = cl A so that

x∈X

x∈X

En × Bn ⊂ cl A .

(2.1.7)

Since A is nowhere dense, it follows from (2.1.7) that En × Bn is nowhere dense. Suppose that En is not nowhere dense. Then there exists a non-empty open set U ⊂ cl En . Then (cf. Exercise 2.1) U × Bn ⊂ cl En × Bn ⊂ cl En × cl Bn = cl(En × Bn ) so that U × Bn ⊂ int cl(En × Bn ), which is impossible, since En × Bn is nowhere dense. Thus, for every n ∈ N, the set En is nowhere dense. Put ∞  P = En . n=1

The set P is of the first category (cf. Exercise 2.4). Suppose that for an x ∈ X the set A[x] is not nowhere dense. Then cl A[x] contains an open set, and hence a set from the base. So there exists an n ∈ N such that Bn ⊂ cl A[x], whence {x} × Bn ⊂ {x} × cl A[x] = cl {x} × A[x] = cl[A ∩ ({x} × Y )]. By (2.1.6) x ∈ En ⊂ P . Hence, for every x ∈ P  , the set A[x] is nowhere dense.  Theorem 2.1.7. Let Y be a separable topological space, and let A ⊂ X × Y be a set of the first category. Then there exists a set P ⊂ X, of the  first category, such that for every x ∈ P  = X \ P the set A[x] defined by (2.1.5) is of the first category. Proof. We have A =

∞ 

An , where the sets An ⊂ X × Y are nowhere dense. By

n=1

Lemma 2.1.2, for every n ∈ N, there exists a set Pn ⊂ X, of the first category, such that for every x ∈ Pn the set An [x] is nowhere dense. Put P =

∞ 

Pn .

n=1

So P is a set of the first category, and for every x ∈ P  ⊂ Pn the set An [x] is nowhere dense (n ∈ N), whence ∞  An [x] (2.1.8) A[x] = n=1

is of the first category. To see that (2.1.8) is true observe that y ∈ A[x] ⇔ (x, y) ∈ A ⇔ there exists an n ∈ N such that (x, y) ∈ An ⇔ there exists an n ∈ N such that ∞  y ∈ An [x] ⇔ y ∈ An [x].  n=1

2.2. Baire property

23

Theorem 2.1.8. Let Y be a separable topological space, and let A ⊂ X, B ⊂ Y . Then the set A × B ⊂ X × Y is of the first category if and only if at least one of the sets A, B is of the first category. Proof. Suppose that A × B is of the first category, and B is of the second category. By Theorem 2.1.7 there exists a set P ⊂ X, of the first category, such that for every x ∈ P  the set (A × B)[x] is of the first category. But for x ∈ A the set (A × B)[x] = B is of the second category. Hence A ⊂ P , and thus A is of the first category. To prove the converse implication, we first prove that if A is nowhere dense in X, then A× Y is nowhere dense in X × Y . We have cl(A× Y ) = cl A× cl Y = cl A× Y , whence int cl(A × Y ) = int(cl A × Y ) = int cl A × int Y = ∅ × Y = ∅. Thus A × Y is nowhere dense. ∞  Now suppose that the set A is of the first category. Then A = An , where An are nowhere dense, whence A × B ⊂ A × Y =

∞  



An × Y =

n=1 ∞ 

set An × Y, n ∈ N, is nowhere dense, whence

∞ 

n=1

(An × Y ) . Every

n=1

(An × Y ) is of the first category,

n=1

and so is also A × B. If the set B is of the first category, then so is A × B. The proof is similar.



2.2 Baire property A set A ⊂ X is said to have the Baire property iff A = (G ∪ P ) \ R ,

(2.2.1)

where the set G is open, and the sets P, R are of the first category. It follows directly from the definition that every open set and every set of the first category have the Baire property. Theorem 2.2.1. The family of the sets with the Baire property is a σ-algebra. Proof. If An , n ∈ N, are sets with the Baire property, then there exist open sets Gn , n ∈ N, and sets of the first category Pn, Rn , n ∈ N, such that An = (Gn ∪ Pn ) \ Rn , n ∈ N . Then

∞ 

An ⊂

n=1

and

∞  n=1

Gn ∪

∞ 

Gn ∪

n=1 ∞  n=1

∞ 

Pn

n=1

∞ ∞    Pn \ An ⊂ Rn n=1

n=1

24

Chapter 2. Topology

which shows that ∞  n=1

and

∞ 

An =

∞  n=1

Gn ∪

∞ 

∞ ∞ ∞      

Pn \ Gn ∪ Pn \ An

n=1

n=1

n=1

n=1

An is a set with the Baire property.

n=1

Now suppose that A, with a representation (2.2.1), is a set with the Baire property. Then A = (G ∩ P  ) ∪ R . Put H = int G , C = R ∪ (G \ H) , D = P \ R. Then A = (H ∪ C) \ D .

(2.2.2)

In fact, (H ∪ C) \ D = [H ∪ R ∪ (G \ H)] ∩ (P ∩ R ) = [H ∪ R ∪ (G ∩ H  )] ∩ (P  ∪ R) = (H ∪ R ∪ G ) ∩ (H ∪ R ∪ H  ) ∩ (P  ∪ R) . Now, H ∪ R ∪ H  = X, and H ∪ R ∪ G = R ∪ G , since H = int G ⊂ G . Hence (H ∪C)\D = (R∪G )∩(P  ∪R) = (R∩P  )∪R∪(G ∩P  )∪(G ∩R) = (G ∩P  )∪R = A since R ∩ P  ⊂ R and G ∩ R ⊂ R. This proves (2.2.2). Now, the set H is open, D ⊂ P is of the first category, and G \ H is a closed set without inner points, and so it is nowhere dense, and hence of the first category. Thus C also is of the first category, and it follows from (2.2.2) that A has the Baire property.  Theorem 2.2.2. If the space X is separable6 , then for every set A ⊂ X there exists a set B with the Baire property such that A ⊂ B and for every set Z with the Baire property containing A the set B \ Z is of the first category. Proof. By Theorem 2.1.6 the set A \ D(A) is of the first category, so there exists (Theorem 2.1.1) a set C ∈ Fσ , of the first category such that A \ D(A) ⊂ C. Put B = D(A) ∪ C. The set B has the Baire property, since both, D(A) (being closed; cf. Theorems 2.1.5 and 2.3.1) and C (being of the  first category)   have the Baire property (now use Theorem 2.2.1), and evidently A = A∩D(A) ∪ A\D(A) ⊂ D(A)∪C = B.   Let Z ⊃ A be an arbitrary set with the Baire property. Then cf. (2.1.3)       B \ Z = D(A) \ Z ∪ (C \ Z) ⊂ D(Z) \ Z ∪ (C \ Z) ⊂ D(Z) \ Z ∪ C is of the first category (cf. Exercise 2.5).



6 The assumption that X is separable is not essential and can be omitted. Cf. the footnote 4 on p. 21.

2.3. Borel sets

25

Now let Y be a separable topological space. We preserve notation (2.1.5). Theorem 2.2.3. Let Y be a separable topological space. If the set A ⊂ X × Y has the Baire property, then there exists a set Q ⊂ X, of the first category, such that for every x ∈ Q = X \ Q the set A[x] has the Baire property. Proof. We have A = (G ∪ P ) \ R, where G is open, and P, R are of the first category. By Theorem 2.1.7 there exist sets Q1 ⊂ X and Q2 ⊂ X such that for every x ∈ Q1 the set P [x] is of the first category, and for every x ∈ Q2 the set R[x] is of the first category. Put Q = Q1 ∪ Q2 . For x ∈ Q the sets P [x] and R[x] both are of the first category (in Y ), and Q itself is of the first category (in X). Now, we have   A[x] = G[x] ∪ P [x] \ R[x] . For every x ∈ X the set G[x] is open (in Y ), therefore for every x ∈ Q the set A[x] has the Baire property.  There are far reaching analogies between the sets of the first category and the sets of Lebesgue measure zero (cf. Oxtoby [250]). In this context the sets with the Baire property play the role of measurable sets. These analogies will become even more evident in the sequel of this book.

2.3 Borel sets Let B(X) be the family of all Borel sets in a metric space X, i.e., the smallest σalgebra of subsets of X containing the open sets. Theorem 2.3.1. Every Borel set has the Baire property. Proof. By Theorem 2.2.1 the family of all sets (in X) with the Baire property is a σ-algebra, and it contains the open sets, so it must contain the smallest σ-algebra containing the open sets, i.e., B(X).  Now we describe a construction of B(X). Let G(X) and F (X) be the families of all open and all closed sets in X, respectively. We define (cf. Theorem 1.6.2) two transfinite sequences {Aα }α<Ω and {Mα }α<Ω of set classes in X. We put A0 = G(X) , M0 = F (X) ,     Mξ σ , Mα = Aξ δ , α < Ω , Aα = ξ<α

ξ<α

where, for any collection ζ of sets, ζσ denotes the collection of all the countable unions of sets from ζ, and ζδ denotes the collection of all the countable intersections of sets from ζ. So A1 = Fσ , M1 = Gδ . Lemma 2.3.1. For every ordinal number β < α < Ω (i) Aβ ∪ Mβ ⊂ Aα ∩ Mα ; (ii) if Z ∈ Aα , then Z  ∈ Mα , and conversely; (iii) Aα ∪ Mα ⊂ B(X).

26

Chapter 2. Topology

Proof. (i) The proof will be by transfinite induction (Theorem 1.6.1) with respect to α. The smallest possible value of α is α = 1, because for α = 0 there is no β < α. If α = 1, then the only β < α is β = 0, and (i) becomes G(X) ∪ F(X) ⊂ Fσ ∩ Gδ , which is known to be true. Now take arbitrary ordinal numbers β < α < Ω. We must distinguish two cases. 1. α = β + 1. Then   Mξ σ = Aα . Mβ ⊂ (Mβ )σ ⊂ ξ<α

Further,



Aξ ⊂

ξ<β

whence also Mβ =



Aξ



Aξ ,

ξ<α

 δ

⊂



ξ<β

Aξ

 δ

= Mα .

ξ<α

Hence Mβ ⊂ Aα ∩ Mα . Similarly it is proved that Aβ ⊂ Aα ∩ Mα . Thus (i) holds. 2. α > β + 1. For the inductive proof assume that (∗) for every γ < α, if η < γ, then Aη ∪ Mη ⊂ Aγ ∩ Mγ . Put γ = β + 1. Then β < γ < α. By (∗) Aβ ∪ Mβ ⊂ Aγ ∩ Mγ . Hence Aβ ∪ Mβ ⊂



       Aξ ∩ Mξ ⊂ Aξ δ ∩ Mξ σ = Mα ∩ Aα .

ξ<α

ξ<α

ξ<α

ξ<α

Thus (i) holds in this case, too. (ii) The proof is again by transfinite induction with respect to α. For α = 0 (ii) is true in virtue of the well-known property of the open and closed sets. Suppose that for all β < α < Ω we have (∗∗) if Z ∈ Aβ , then Z  ∈ Mβ , and conversely. Take a Z ∈ Aα . Then ∞  Z= En , where, for every n ∈ N, En ∈



n=1

Mξ , i.e., for every n ∈ N there exists a ξn < α such

ξ<α

that En ∈ Mξn . Hence, by (∗∗), En ∈ Aξn , n ∈ N, and Z =

∞  n=1

En ∈



Aξ

 δ

ξ<α

The proof of the converse implication is analogous.

= Mα .

2.3. Borel sets

27

(iii) This results from the fact that B(X) is a σ-algebra (again a proof by transfinite induction is necessary, which is left to the reader).  Theorem 2.3.2. We have 

Aα =

α<Ω



Mα = B(X) .

α<Ω

Proof. For  every α < Ω we  have α +  1 < Ω, whence by Lemma 2.3.1 (i) Aα ⊂ Mα+1 ⊂ Mα , and Aα ⊂ Mα . Similarly, for every α < Ω we have α<Ω  α<Ω  α<Ω  Mα ⊂ Aα+1 ⊂ Aα , and Mα ⊂ Aα . Consequently α<Ω

α<Ω

α<Ω



Aα =

α<Ω



Mα ,

α<Ω

  and by Lemma 2.3.1 (iii) Aα = Mα ⊂ B(X). So it is enough to show that α<Ω α<Ω  Aα is a σ-algebra, for then it must contain the smallest σ-algebra containing α<Ω

G(X) = A0 , i.e., it must contain B(X).  Aα . Then, for every n ∈ N, there exists an Take a sequence of sets An ∈ α<Ω

αn < Ω such that An ∈ Aαn . By Theorem 1.4.4 there exists an ordinal number α greater than every number αn . This has been constructed as α = B + 1, where, in ∞  the present case B = Γ(αn ). We have for every n ∈ N , Γ(αn ) = αn < Ω, whence n=1

card Γ(αn) = αn  ℵ0 . Hence also card B  ℵ0 , and consequently α  ℵ0 . Therefore α ∈ M (ℵ0 ) = Γ(Ω), i.e., α < Ω. Thus we have αn < α < Ω for every n ∈ N. By Lemma 2.3.1 (i) Aαn ⊂ Mα for every n ∈ N, whence An ∈ Mα for every n ∈ N. ∞      Hence An ∈ (Mα )σ ⊂ Mξ σ = Aα+1 ⊂ Aα . n=1 α<Ω ξ<α+1 Aα . Then A ∈ Aβ for a certain β < Ω. By Lemma 2.3.1 Now take a set A ∈ α<Ω   (ii) A ∈ Mβ ⊂ Mα = Aα . Consequently Aα is a σ-algebra.  α<Ω

α<Ω

α<Ω

Theorem 2.3.3. B(X) is the smallest class K of subsets of X with the properties: (i) F (X) ⊂ K, ∞ ∞  An ∈ K and An ∈ K. (ii) if An ∈ K for n ∈ N, then n=1

n=1

Proof. Clearly B(X) has properties (i) and (ii), so K ⊂ B(X) . To prove the converse inclusion, we will show that for every α < Ω Aα ∪ M α ⊂ K .

(2.3.1)

28

Chapter 2. Topology

For α = 0 (2.3.1) is true by (i) and by the fact that G(X) ⊂ Fσ ⊂ K. Assume that for a certain α < Ω we have Aξ ∩ Mξ ⊂ K for every ξ < α. Then also     Aξ ∪ Mξ ⊂ K , ξ<α

and by (ii)



Aξ

ξ<α

 δ

ξ<α

∪



Mξ

 σ

⊂ K.

ξ<α

Thus (2.3.1) holds. The inclusion B(X) ⊂ K results now from (2.3.1) in view of Theorem 2.3.2.  Theorem 2.3.4. If the space X is separable, then card B(X)  c. Proof. In a separable space X we have card G(X)  c (since every open set may be represented as a union of base neighbourhoods, and there exists a countable base), and hence also card F (X)  c. An easy proof by transfinite induction shows that for every α < Ω we have card Aα  c

In the union



card Mα  c .  Aα we have ℵ1  c summands, so card Aα  c. The theorem and

α<Ω

α<Ω

results now from Theorem 2.3.2.



As an easy consequence of Theorem 2.3.4 we obtain that in R there are sets with the Baire property which are not Borel sets. Take a first category set C ⊂ R with card C = c (for instance the Cantor set). Then all subsets of C are of the first category, and consequently they have the Baire property, whereas card P(C) > card C = c, so among subsets of C there must exist some which are not Borel sets, as card B(R) = c.

2.4 The space z Let Xi (with metrics i , respectively), i ∈ N, be metric spaces and put ∞

X = × Xi . i=1

Thus the elements of X are infinite sequences {xi }, with xi ∈ Xi , i ∈ N. Define the metric  in X putting for x = {xi } , y = {yi } , x, y ∈ X, (x, y) =

∞

1 i (x, y) . i 1 +  (x, y) 2 i i=1

(2.4.1)

Lemma 2.4.1. If xn = {xni } ∈ X , n ∈ N, then the sequence {xn } converges: lim xn = x0 = {x0i } ∈ X

n→∞

if and only if lim xni = x0i for every i ∈ N. n→∞

(2.4.2)

2.4. The space z

29

Proof. If (2.4.2) holds, then lim (xn , x0 ) = 0 .

n→∞

By (2.4.1) (xn , x0 ) 

(2.4.3)

1 i (xni , x0i ) , whence 2i 1 + i (xni , x0i ) lim i (xni , x0i ) = 0

n→∞

(2.4.4)

for every i ∈ N. Conversely, suppose that (2.4.4) holds for every i ∈ N. We have (xn , x0 ) 

m ∞ m

1 i (xni , x0i ) 1 1 i (xni , x0i ) 1 + + m = i 1 +  (x , x ) i i 1 +  (x , x ) 2 2 2 2 i ni 0i i ni 0i i=1 i=m+1 i=1

for every m ∈ N. By (2.4.4) we get hence lim sup (xn , x0 )  2−m , whence, on letting n→∞

m → ∞, we obtain (2.4.3).



Theorem 2.4.1. If all spaces Xi are separable, then so is also X. Proof. For every i ∈ N there exists a countable set Yi ⊂ Xi , dense in Xi . Fix an x = {xi } ∈ X and consider the set Y ⊂ X consisting of the points {y1 , x2 , x3 , . . .} , {y1 , y2 , x3 , . . .} , {y1 , y2 , y3 , . . .} ,

(2.4.5)

............... , yi ∈ Yi . The cardinality of Y equals that of

∞ 

∞

× Yi , and thus Y is countable. Take

n=1 i=1

an x = {xi } ∈ X. For every i ∈ N there exists a sequence {yni } ⊂ Yi of points of Yi such that (2.4.6) lim yni = xi . n→∞

Replacing in (2.4.5) in the n-th row yi by yni we obtain a sequence of points yn ∈ Y and it follows from (2.4.6) and Lemma 2.4.1 that lim yn = x .

n→∞

Thus Y is dense in X, and consequently X is separable.



Theorem 2.4.2. If all spaces Xi are complete, then so is also X. Proof. Let {xn } ⊂ X , xn = {xni } , be an arbitrary Cauchy sequence in X. By (2.4.1) (xn , xn+k ) 

1 i (xni , xn+k,i ) 2i 1 + i (xni , xn+k,i)

30

Chapter 2. Topology

for all n , k , i ∈ N, whence it follows that for every i ∈ N the sequence {xni } ⊂ Xi is a Cauchy sequence in Xi . Since Xi is complete, there exists an xi ∈ Xi such that lim xni = xi .

n→∞

Hence, by Lemma 2.4.1, lim xn = x ,

n→∞

where x = {xi } ∈ X. Thus the space X is complete.



Now take every Xi to be N with the metric i (n, k) = |n − k|. In the sequel the resulting product space X will be always denoted by z. Since the space N is separable (being countable; N itself is a countable and dense subset of N) and complete (there are no Cauchy sequences in N besides stationary ones), from Theorems 2.4.1 and 2.4.2 results Theorem 2.4.3. The space z is complete and separable. The generality of the space z may be seen from the following Theorem 2.4.4. Every complete and separable metric space X is a continuous image of z. Proof. Let X be a complete and separable metric space. Then    1 X= , K x, 2

(2.4.7)

x∈X

where K(x, r) is the open ball centered  at x with  the radius r. Denoting by d(A) the diameter of the set A ⊂ X, we have d K(x, 12 )  1 for every x ∈ X. Since X is separable, by the Lindel¨ of theorem we can choose from (2.4.7) a countable cover: ∞    X= Kn(1) , d Kn(1)  1 , n=1

and also X=

∞ 

  cl Kn(1) , d cl Kn(1)  1 .

n=1

We construct a family of sets {Dn1 ...nm } , m, n1 , . . . , nm ∈ N, as follows. We put (1) Dn = cl Kn . If we have already defined the sets Dn1 ...nm for all n1 , . . . , nm ∈ N and for a certain m ∈ N, we have    1 Dn1 ...nm ⊂ (2.4.8) K x, m . 2 x∈Dn1 ...nm

Again we choose from (2.4.8) a countable cover Dn1 ...nm ⊂

∞  n=1

  Kn(m) , d Kn(m) 

1 2m−1

,

2.4. The space z

31

and we put , nm+1 ∈ N . Dn1 ...nm ,nm+1 = Dn1 ...nm ∩ cl Kn(m) m+1 Thus we have defined inductively a family of sets {Dn1 ...nm } , m, n1 , . . . , nm ∈ N, with the properties (i) Every set Dn1 ...nm is closed; 1 (ii) d(Dn1 ...nm )  m−1 ; 2 (iii) Dn1 ...nm ,nm+1 ⊂ Dn1 ...nm , m, n1 , . . . , nm , nm+1 ∈ N. Now take a z = {ni } ∈ z and put f (z) =

∞ 

Dn1 ...nm .

(2.4.9)

m=1

By Cantor’s theorem (2.4.9) represents a single point in X, consequently formula (2.4.9) defines a function f : z → X. For every m ∈ N the sets Dn1 ...nm , n1 . . . nm ∈ N, form a cover of X. Consequently for every x ∈ X and for every m ∈ N there exists a set Dn1 ...nm such that x ∈ Dn1 ...nm . The sequence {ni } represents a point z ∈ z and we have f (z) = x. Thus f is onto: f (z) = X . It remains to show that f is continuous. Fix an ε > 0 and choose a p ∈ N so that 1 1 < ε. Put δ = p+1 . Take z  , z  ∈ z, z  = {ni }, z  = {ni }. (z  , z  ) < δ means 2p−1 2 ∞

1 |ni − ni | 1 < p+1 . i 1 + |n − n | 2 2 i i i=1

(2.4.10)

Since ni and ni are integers, ni = ni implies 1 |ni − ni |  . 1 + |ni − ni | 2 Thus it follows from (2.4.10) that ni = ni = ni for i = 1, . . . , p. But f (z  ) =

∞  m=1

Dn1 ...nm , f (z  ) =

∞ 

Dn1 ...nm

m=1

and by (iii) above f (z  ), f (z  ) ∈ Dn1 ...np . By (ii) this last set has the diameter at 1 most p−1 < ε, so the distance between f (z  ) and f (z  ) is less than ε. This proves 2 that f is uniformly continuous in z. 

32

Chapter 2. Topology

2.5 Analytic sets Let X be a separable and complete metric space. A set A ⊂ X is called analytic iff there exists a continuous function f : z → X such that f (z) = A. Moreover, we assume that the empty set is analytic. It is of the form f (z) with the empty function f . The class of all analytic sets in X is denoted by A(X). Theorem 2.5.1. Every closed set in X is analytic, i.e., F (X) ⊂ A(X). Proof. If F ⊂ X is a closed set, then F considered as a metric space (with the same metric as in X) is separable, since the separability is a hereditary property. F is also complete. In fact let xn ∈ F be a Cauchy sequence. Then {xn } converges in X, since X is complete: lim xn = x ∈ X. n→∞

But then x ∈ F , since F is closed. Thus {xn } converges in F : lim xn = x ∈ F.

n→∞



The theorem results now from Theorem 2.4.4.

Theorem 2.5.2. Let X, Y be separable and complete metric spaces, A ⊂ X an analytic set, and g : A → Y a continuous function. Then the set g(A) ⊂ Y also is analytic. Proof. There exists a continuous function f : z → X such that f (z) = A. The function h = g ◦ f (the composite of g and f ) is continuous, h : z → Y , and h(z) = g(A).  Theorem 2.5.3. The cartesian product of a finite or countable number of analytic sets is an analytic set. Proof. Let Ai , i ∈ I, where I = {1, . . . , n} or I = N, be in A(Xi ), where Xi are separable complete spaces. Then × Ai ⊂ × Xi , and × Xi is separable and complete. i∈I

i∈I

i∈I

(If I = N, this follows from Theorems 2.4.1 and 2.4.2, and if I is finite, the proof is similar). Thus there exist continuous functions fi : z → Xi such that fi (z) = Ai , i ∈ I. The function f : z → × Xi defined as f = (f1 , f2 . . .) is continuous (if I = N, this i∈I

follows from Lemma 2.4.1, and if I is finite, the proof is similar) and f (z) = × Ai . Thus × Ai ∈ A( × Xi ). i∈I

i∈I

i∈I



Theorem 2.5.4. The union of a finite or countable number of analytic sets is an analytic set. Proof. We need only consider countable unions, since a finite union can always be written as a countable one. Let X be a separable and complete metric space, and Ai ⊂ X , i ∈ N, analytic sets. So there exist continuous functions fi : z → X such that fi (z) = Ai , i ∈ N. We define f : z × N → X putting f (z, n) = fn (z). The

2.5. Analytic sets

33

function f is continuous. To see this, take a sequence yk = (zk , nk ) ∈ z × N , k ∈ N, convergent to a y0 = (z0 , n0 ) ∈ z × N: lim yk = y0 .

k→∞

Then lim zk = z0 and lim nk = n0 , i.e., nk = n0 for k sufficiently large. For such k k→∞

k→∞

we have f (yk ) = f (zk , n0 ) = fn0 (zk ) → fn0 (z0 ) = f (z0 , n0 ) = f (y0 ) , since fn0 is continuous. Now, z×N is complete and separable, being a product of complete and separable spaces, and hence, by Theorem 2.4.4 is an analytic set. Moreover we have ∞  f (z × N) = Ai . i=1

In fact, if x ∈ f (z × N), then x = f (z, n) for certain z ∈ z , n ∈ N, i.e., x = fn (z), ∞ ∞   whence x ∈ fn (z) = An and x ∈ Ai . And if x ∈ Ai , then there exists an n ∈ N i=1

i=1

such that x ∈ An = fn (z), i.e., there exists a z ∈ z such that x = fn (z) = f (z, n) ∈ f (z × N). The theorem results now from Theorem 2.5.2.  Theorem 2.5.5. The intersection of a finite or countable number of analytic sets is an analytic set. Proof. Again we need only consider countable intersections, and we may assume that the intersection in question is non-empty, for the empty set is always analytic. Let X be a separable and complete metric space, and Ai ⊂ X , i ∈ N, analytic sets. So there ∞ exist continuous functions fi : z → X such that fi (z) = Ai , i ∈ N. Put Z = × Zi , i=1

where Zi = z for every i ∈ N. By Theorems 2.4.1, 2.4.2 and 2.4.3 Z is a separable and complete space. Further define Q ⊂ Z as Q = {z = {zi } ∈ Z | fi (zi ) is independent of i}   the sequence fi (zi ) is a constant . First we show that Q is a closed subset of Z, and hence, by Theorem 2.5.1, is analytic. (n) (0) Take a sequence z (n) ∈ Q , lim z (n) = z (0) . Let z (n) = {zi }, z (0) = {zi }, n→∞

(k)

zi

(n)

∈ z for k = 0, 1, 2, . . .. By Lemma 2.4.1 lim zi n→∞

(0)

(n)

(0)

= zi

for every i ∈ N. Thus (n)

fi (zi ) = lim fi (zi ) is independent of i, since every sequence fi (zi ) is constant. n→∞

Hence z (0) ∈ Q. We define a function f : Q → X as f (z) = fi (zi ) (independent of i !) for z ∈ Q. Take again sequence z (n) ∈ Q converging to z (0) ∈ Q, as above. Then f (z (n) ) = (n) (0) fi (zi ) → fi (zi ) = f (z (0) ), since the functions fi are continuous. Now, ∞  f (Q) = Ai . i=1

34

Chapter 2. Topology

For, if x ∈ f (Q), then x = fi (zi ) for a z in Z and all i ∈ N, and hence x ∈ fi (z) = Ai ∞ ∞ for all i ∈ N, i.e., x ∈ Ai . And if x ∈ Ai , then x ∈ Ai = fi (z) for all i ∈ N, i.e., i=1

i=1

there exist zi ∈ z , i ∈ N, such that x = fi (zi ) , i ∈ N. Putting z = {zi } ∈ Q we get x = f (z) ∈ f (Q). The theorem results now from Theorem 2.5.2.  Theorem 2.5.6. If X is a complete and separable metric space, then B(X) ⊂ A(X), i.e., every Borel set is analytic.  Proof. The class A(X) fulfils conditions (i) and (ii) of Theorem 2.3.3. (i) results  from Theorem 2.5.1, and (ii) from Theorems 2.5.4 and 2.5.5 , and so it must contain the smallest class fulfilling these conditions, i.e., B(X).  Finally we give here two characterizations of the class A(X). The first of them will not be proved (cf., e.g., Kuratowski [196], where the condition is taken as the definition of the analytic sets.) Theorem 2.5.7. Let X be a complete and separable metric space. A set A ⊂ X is analytic if and only if there exist a Borel set B ⊂ X and a continuous function f : B → X such that f (B) = A. Let us note that the sufficiency results easily from Theorems 2.5.6 and 2.5.2. Theorem 2.5.8. Let X be a complete and separable metric space. The family A(X) is the smallest class Z of subsets of X which fulfils the following three conditions (i) F (X) ⊂ Z; ∞ ∞  (ii) If Ai ∈ Z for i ∈ N, then also Ai and Ai ∈ Z; i=1

i=1

(iii) If Z ∈ Z, and f : Z → X is a continuous function, then f (Z) ∈ Z. Proof. The class A(X) fulfils conditions (i)–(iii) in virtue of Theorems 2.5.1, 2.5.4, 2.5.5 and 2.5.2. So Z ⊂ A(X). Conversely, let Z ⊂ P(X) be a collection of sets such that conditions (i)–(iii) are fulfilled. By Theorem 2.3.3 B(X) ⊂ Z and hence, by (iii), Z contains also all the continuous images of the Borel subsets of X. But by Theorem 2.5.7 this is exactly the class of analytic subsets of X. So A(X) ⊂ Z.  However, the class A(X) is not a σ-algebra. The complement of an analytic set need not be analytic1 . It can be generally proved that if A and A both are analytic, then A is a Borel set2 . The notion of an analytic set is due to M. Souslin and N. Lusin (cf. Lusin [210]). Expositions of the theory of analytic sets may be found, e.g., in Kuratowski [196], Kuratowski-Mostowski [198], Na˘ımark [236], Sierpi´ nski [284]. 1 Souslin [293], Lusin-Sierpi´ nski [212], Hurewicz [149]. Cf. also Kuratowski-Mostowski [198], p. 443, Kuratowski [196], p. 408. 2 Souslin [293], Cf. also Kuratowski-Mostowski [198], p. 440, Kuratowski [196] p. 395. Cf. also Exercise E2.10.

2.6. Operation A

35

2.6 Operation A Let S = {Sn1 ...nm } , m, n1 , . . . , nm ∈ N, be a family of sets. (To every z = {ni } ∈ z there corresponds a sequence Sz = {Sn1 , Sn1 n2 , Sn1 n2 n3 , . . .} of elements of S, and S= Sz ). We define the operation A (cf. Lusin-Souslin [213]) for such a family as  z∈z ( denotes the union extended over all z = {ni } ∈ z) z

AS =

∞   z

Sn1 ...nm .

m=1

If, for every m, n1 , . . . , nm , nm+1 ∈ N, we have Sn1 ...nm nm+1 ⊂ Sn1 ...nm , the family S is called a regular system. If S is any family as described above, then putting m  ∗ Sn1 ...nm = Sn1 ...ni , m, n1 , . . . , nm ∈ N , i=1

we obtain a regular system S ∗ = {Sn∗1 ...nm } , m, n1 , . . . , nm ∈ N, such that AS ∗ = AS . So when studying the operation A we may always assume, without a loss of generality, that the system S in question is regular. Theorem 2.6.1. Every analytic set is the result of the operation A performed on a regular system of closed sets. Proof. Let X be a separable and complete metric space and A ⊂ X an analytic set. Thus there exists a continuous function f : z → X such that A = f (z). Define the sets zn1 ...nm ⊂ z by zn1 ...nm = { z = {ki } ∈ z | k1 = n1 , . . . , km = nm } , m, n1 , . . . , nm ∈ N . It follows from Lemma 2.4.1 that the sets zn1 ...nm , m, n1 , . . . , nm ∈ N, are closed, and clearly zn1 ...nm ,nm+1 ⊂ zn1 ...nm , m, n1 , . . . , nm , nm+1 ∈ N . (2.6.1) Now put Sn1 ...nm = cl f (zn1 ...nm ) , m, n1 , . . . , nm ∈ N . The family S = {Sn1 ...nm } , m, n1 , . . . , nm ∈ N , is, by (2.6.1), a regular system of closed sets. If z = {ni } ∈ z is fixed, then, for every m ∈ N , z ∈ zn1 ...nm , whence, for every m ∈ N , f (z) ∈ Sn1 ...nm so that f (z) ∈

∞  m=1

Sn1 ...nm .

(2.6.2)

36

Chapter 2. Topology

But the diameter d(zn1 ...nm ) 

1 , whence, by the continuity of f , 2m

lim d(Sn1 ...nm ) = lim d [cl f (zn1 ...nm )] = 0 .

m→∞

m→∞

So the intersection in (2.6.2), as an intersection of a descending sequence of closed sets with diameters tending to zero in a complete space, represents, by Cantor’s theorem, a single point. Thus ∞  f (z) = Sn1 ...nm . m=1

Hence AS =

∞   z

Sn1 ...nm =



f (z) = f (z) = A.



z

m=1

Theorem 2.6.2. The result of the operation A performed on a family of analytic sets is an analytic set. Proof. Let X be a separable and complete metric space and An1 ...nm ⊂ X, m, n1 , . . . , nm ∈ N analytic sets. The sets zn1 ...nm , m, n1 , . . . , nm ∈ N , defined in the proof of Theorem 2.6.1, are analytic in virtue of Theorem 2.5.1. Hence also, by Theorem 2.5.3, the sets An1 ...nm ×zn1 ...nm are analytic subsets of X ×z, and consequently, by Theorem ∞  2.5.4, so are also the sets (for every fixed m ∈ N) (An1 ...nm × zn1 ...nm ) and the set (cf. Theorem 2.5.5)

∞

n1 ,...,nm =1

∞ 

(An1 ...nm × zn1 ...nm ). The projection

m=1 n1 ,...,nm =1

p:X ×z→X

p(x, z) = x ∞  is a continuous function, so by Theorem 2.5.2 p

∞ 

m=1 n1 ,...,nm =1

[An1 ...nm × zn1 ...nm ]



is an analytic set. But p

∞  

∞ 

∞    [An1 ...nm × zn1 ...nm ] = An1 ...nm = A{An1 ...nm } . (2.6.3) z

m=1 n1 ,...,nm =1

To prove (2.6.3), take an x ∈ p

∞ 

m=1

 [An1 ...nm × zn1 ...nm ] . Then there exists

∞ 

m=1 n1 ,...,nm =1 ∞ 

a z = {ki } ∈ z such that (x, z) ∈

n1 ,...,nm =1

[An1 ...nm × zn1 ...nm ] for every m ∈ N. This

means that x ∈ An1 ...nm for all m ∈ N, where ni = ki for i ∈ N and consequently x∈

∞  m=1

Ak1 ...km ⊂

∞   z

An1 ...nm .

m=1

This proves an inclusion in (2.6.3). The proof of the converse inclusion is analogous. 

2.7. Theorem of Marczewski

37

2.7 Theorem of Marczewski The theorem of Marczewski (Marczewski [219]) assures that some important regularity properties of sets are preserved by the operation A. We start with two lemmas. Lemma 2.7.1. If {Sn1 ...nm }, m, n1 , . . . , nm ∈ N , is a regular system, then for every k1 , . . . , kj ∈ N ∞ ∞    z

l=1

Sk1 ...kj ln1 ...nm =

∞   z

m=1

Sk1 ...kj n1 ...nm .

m=1

The easy proof of Lemma 2.7.1 is left to the reader. Lemma 2.7.2. If {Sn1 ...nm } , m, n1 , . . . , nm ∈ N , is a regular system, then for every set A ∞ ∞ ∞        A\ Sn1 ...nm \ Sn1 ...nm ⊂ Sn1 ...nm l , z

z

m=1

m=0

l=1

where for m = 0 we assume Sn1 ...nm = A. Proof. For an indirect proof suppose that there exists an x ∈ A \ such that x ∈ m ∈ N ∪ {0}:



∞ 

z m=0



Sn1 ...nm \

∞  l=1

∞  z m=1

Sn1 ...nm

 Sn1 ...nm l . This means that for every z ∈ z and

if x ∈ Sn1 ...nm , then there exists an l ∈ N such that x ∈ Sn1 ...nm l .

(∗)

We have x ∈ A, so using (∗) for m = 0 we infer that there exists an l1 ∈ N such that x ∈ Sl1 . Using (∗) for m = 1 and n1 = l1 , we infer that there exists an l2 ∈ N such that x ∈ Sl1 l2 . Proceeding so further we get the existence of a z0 = {li } ∈ z such ∞  that, for every m ∈ N , x ∈ Sl1 ...lm . Hence x ∈ Sn1 ...nm , which contradicts our z m=1

supposition.



A family Σ of subsets of a set (space) X is called an M-algebra (Marczewski algebra) iff it is a σ-algebra and for every set A ⊂ X there exists a set B ∈ Σ , A ⊂ B, such that if Z ∈ Σ and A ⊂ Z, then every set Y ⊂ B \ Z belongs to Σ. By Theorems 2.2.1 and 2.2.2 the class of the sets with the Baire property (in separable metric space X) is an M -algebra. Similarly, as will be shown later, the class of all Lebesgue measurable sets in RN is an M -algebra. Theorem 2.7.1 (Theorem of Marczewski). If Σ is an M -algebra, then the result of the operation A performed on a regular system of sets from Σ belongs to Σ. Proof. Let Sn1 ...nm ∈ Σ , m, n1 , . . . , nm ∈ N, form a regular system, and put A = A{Sn1 ...nm } .

38

Chapter 2. Topology

For the set A choose a set B ∈ Σ such that A ⊂ B and if A ⊂ Z ∈ Σ, then every ∞  Y ⊂ B \ Z belongs to Σ. Similarly, to every set Sk1 ...kj n1 ...nm , k1 , . . . , kj ∈ N, z m=1

choose a set Bk1 ...kj ∈ Σ according to the same pattern. We may assume that Bn1 ...nm ⊂ Sn1 ...nm , n1 , . . . , nm ∈ N ,

(2.7.1)

for otherwise we could replace Bn1 ...nm by Bn1 ...nm ∩ Sn1 ...nm which would not spoil the remaining properties of Bn1 ...nm . By Lemma 2.7.2 we have B\A=B\

∞  

Sn1 ...nm ⊂ B \

z m=1

∞  

Bn1 ...nm ⊂

z m=1

∞ 

Bn1 ...nm \

m=0

∞ 

Bn1 ...nm l ,

l=1

where for m = 0 we put Bn1 ...nm = B. Now, by Lemma 2.7.1, ∞  

Sk1 ...kj , n1 ...nm =

z m=1

∞   ∞ 

Sk1 ...kj ln1 ...nm ⊂

l=1 z m=1

∞ 

Bk1 ...kj l ∈ Σ ,

l=1

since Σ is a σ-algebra. Hence, by the choice of Bk1 ...kj , ∞ 



(B \ A) ∩ Bk1 ...kj \

Bk1 ...kj l ∈ Σ .

(2.7.2)

l=1

Since by (2.7.1) and Lemma 2.7.2 B\A = B\

∞  

Sn1 ...nm ⊂ B \

z m=1

∞  

Bn1 ...nm ⊂

∞  

z m=1

we have B \ A = (B \ A) ∩

z m=0

z m=0

Let [. . .] stand for (B \ A) ∩ Bn1 ...nm \ ∞   z

m=0

Bn1 ...nm l ,

l=1

∞ ∞   

Bn1 ...nm \ Bn1 ...nm l .



B\A=

∞ 

Bn1 ...nm \

[. . .] =

∞  l=1

l=1



Bn1 ...nm l . Then

∞   m=0 z

[. . .] =

∞ 

∞ 

[. . .] .

m=0 n1 ...nm =1

The last union contains only countably many summands, therefore, since Σ is a σalgebra, we get by (2.7.2) B \ A ∈ Σ. Since A ⊂ B, we have A = B \ (B \ A) ∈ Σ.  From Theorem 2.7.1, Theorem 2.6.1 and the fact that the class of all sets with the Baire property is an M -algebra, we obtain Theorem 2.7.2. Every analytic set has the Baire property. In Part II of the present book we will encounter many sets without the Baire property, and hence non-analytic.

2.8. Cantor-Bendixson theorem

39

2.8 Cantor-Bendixson theorem The Cantor-Bendixson theorem belongs to standard part of any university course on the topology of metric spaces. But we prove it here because of its application in the theory of analytic sets. Let X be a metric space and A ⊂ X. A point x ∈ X is called a point of accumulation of A iff for every neighbourhood U of x we have card (U ∩ A) > 1. The set of all the accumulation points of A is denoted by Ad . A point x ∈ A is called a point of condensation of A iff for every neighbourhood U of x we have card (U ∩ A) > ℵ0 . The set of all the condensation points of A is denoted by A• . A set A ⊂ X is called perfect iff A = Ad . The reader will easily verify the following properties of the operations • and d : A• ⊂ Ad ⊂ cl A , (A ∪ B)• = A• ∪ B • , A ⊂ B ⇒ A• ⊂ B • , A• = cl A• , for every set collection K ,



Ad ⊂

A∈K

 

d A ,

A∈K

(A ∪ B) = A ∪ B , d

d

d

Add ⊂ Ad , cl A = A ∪ Ad . Theorem 2.8.1.

If X is a separable metric space, then for every set A, card (A \ A• )  ℵ0 .

Proof. There exists a countable base of neighbourhoods {Gn }n∈N in X. Take an x ∈ A \ A• . Then there exists a neighbourhood U of x such that card (U ∩ A)  ℵ0 . Further, there exists an nx ∈ N such that x ∈ Gnx ⊂ U , and clearly we have card (Gnx ∩ A)  ℵ0 . Now we have  A \ A• ⊂ (Gnx ∩ A) . (2.8.1) x∈A\A•

The union in (2.8.1) is at most countable, many  since there are at most countably distinct sets Gnx , and hence the set (Gnx ∩ A), and so also A \ A• , is at most x∈A\A•

countable. Corollary 2.8.1. If X is a separable metric space and A ⊂ X, then (A \ A• )• = ∅. Corollary 2.8.2. If X is a separable metric space and A ⊂ X, then A•• = A• . Proof. We have, since A• is closed A•• ⊂ cl A• = A• .



40

Chapter 2. Topology

On the other hand A ⊂ A• ∪ (A \ A• ), whence A• ⊂ A•• ∪ (A \ A• )• = A•• .



Theorem 2.8.2 (Theorem of Cantor-Bendixson). If X is a separable metric space, then X = A ∪ B, where A = Ad and card B  ℵ0 . Proof. Let and put A =



K = { C ⊂ X | C ⊂ Cd } , K , B = X \ A. Then A ⊂ Ad , whence Ad = A ∪ Ad = cl A and (cl A)d = Ad ∪ Add = Ad = cl A .

Thus cl A ∈ K and cl A ⊂ A. This means that A ∪ Ad ⊂ A, i.e., Ad ⊂ A. Since A ⊂ Ad , we obtain hence A = Ad , i.e., A is perfect. Now, B • = B •• ⊂ B •d , i.e., B • ∈ K and B • ⊂ A. By the definition of B we have B∩B • = ∅ and B = B\B • . Hence by Theorem 2.8.1 card B = card (B\B • )  ℵ0 .  Any subset D of X may be considered as a space, so D = A∪B with A = Ad and card B  ℵ0 . But in the formula A = Ad the accumulation points of A are taken with respect to the relative topology of D. (These are points x such that card (D∩U ∩A) > 1 for every neighbourhood U of x).  But since every accumulation point of A in D is an accumulation point of A in X card (D ∩ U ∩ A) > 1 implies card (U ∩ A) > 1 , Ad in D is contained in Ad in X. Consequently D = A ∪ B with A ⊂ Ad and card B  ℵ0 . Theorem 2.8.3. If X is a complete and separable metric space and A ∈ A(X), then either card A  ℵ0 , or card A = c. Remark. Theorem 2.8.3 says that the analytic sets realize the continuum hypothesis. Every uncountable analytic set has the power of continuum. Proof. Let A ∈ A(X) and card A > ℵ0 . There exists a continuous function f : z → X such that A = f (z). Consider the family of sets f −1 ({y}) for y ∈ A. These sets clearly are non-empty and disjoint. By the Axiom of Choice (cf. Theorem 1.8.3) there exists a set Z ⊂ z which has exactly one point in common with every set f −1 ({y}) for y ∈ A. Consequently there is a one-to-one correspondence between the points of A and the points of Z. Thus card Z  card A > ℵ0 . Moreover, the function f | Z is one-to-one. By Theorem 2.8.2 Z = D ∪ B with D ⊂ Dd and card B  ℵ0 .

(2.8.2)

It follows that card D > ℵ0 and the function f | D is one-to-one. Take x0 , x2 ∈ D , x0 = x2 . There exist open balls K0 , K2 ⊂ z such that xi ∈ 1 Ki , d(Ki ) < , i = 0, 2, and f (cl K0 ) ∩ f (cl K2 ) = ∅, since f is continuous. Let I 2 be the family of all sequences I = {in } , in = 0, 2. For every I = {in } ∈ I we shall

2.8. Cantor-Bendixson theorem

41

define a sequence of points {xi1 , xi1 i2 , . . . xi1 ...in , . . .} and a sequence of open balls {Ki1 , Ki1 i2 , . . . Ki1 ...in , . . .} in such a manner that xi1 ...xn ∈ Ki1 ...in , xi1 ...xn ∈ D , d(Ki1 ...in ) <

1 , 2n

f (cl Ki1 ...in ) ∩ f (cl Kj1 ...jn ) = ∅ for (i1 , . . . , in ) = (j1 , . . . , jn ) ,

(2.8.3) (2.8.4) (2.8.5)

where jk = 0, 2. Suppose that we have already defined points xi1 ...in and balls Ki1 ...in for a certain n ∈ N, so that conditions (2.8.3), (2.8.4), (2.8.5) are fulfilled. Since every  neighbourhood U of any x ∈ D contains points of D different from x cf. (2.8.2) , we may find points xi1 ...in 0 and xi1 ...in 2 in D, different from xi1 ...in , and open balls Ki1 ...in 0 and Ki1 ...in 2 such that (in+1 = 0, 2) xi1 ...in in+1 ∈ Ki1 ...in in+1 , xi1 ...in in+1 ∈ D , 1 d(Ki1 ...in in+1 ) < n+1 , 2 f (cl Ki1 ...in 0 ) ∩ f (cl Ki1 ...in 2 ) = ∅ . Moreover, we may choose the balls Ki1 ...in in+1 ⊂ Ki1 ...in . In this manner we have inductively defined, for every I = {in } ∈ I, a sequence of points {xi1 ...in }n∈N and a sequence of balls {Ki1 ...in }n∈N so that conditions (2.8.3), (2.8.4), (2.8.5) are fulfilled and Ki1 ...in+1 ⊂ Ki1 ...in . (2.8.6) Thus, by (2.8.4), lim d(cl Ki1 ...in ) = 0, moreover, (2.8.6) implies that also n→∞

cl Ki1 ...in+1 ⊂ cl Ki1 ...in . By Cantor’s theorem the set

∞ n=1

∞ 

cl Ki1 ...in consists exactly of a single point

cl Ki1 ...in = xI , I = {in } .

n=1

Put E=



{xI } .

I∈I

For I1 = I2 , I1 , I2 ∈ I, we havexI1 = xI2 , since if i11 = i21 , . . . , i1n−1 = i2n−1 , i1n = i2n for an n ∈ N I1 = {i1n }, I2 = {i2n } , then by (2.8.5) f (cl Ki11 ...i1n )∩f (cl Ki21 ...i2n ) = ∅, and also f (Ki11 ...i1n ) ∩ f (Ki21 ...i2n ) = ∅, whence f (xI1 ) = f (xI2 ). Thus f | E is one-to-one, and since card I = c, also card E = c and card f (E) = c. Since E ⊂ z , f (E) ⊂ f (z) = A. Thus card A  c. On the other hand card A = card f (z)  card z = c. So necessarily card A = c. 

42

Chapter 2. Topology

∞ i  n of the n 3 n=1 Cantor set (cf. Exercise 2.11), and the correspondence is one-to-one. Actually it can be shown that every uncountable analytic set contains topologically the Cantor set (i.e., contains a subset homeomorphic with the Cantor set). Cf. Exercise 2.12.

We may assign to every f (xI ) ∈ f (E) , I = {in }, the point

2.9 Theorem of S. Piccard In the year 1942 S. Piccard published a theorem to the effect that if A, B ⊂ R are sets of the second category with the Baire property, then the algebraic sum A + B of A and B contains an interval (cf. [256]3 ). This theorem was later generalized in various ways (cf., e.g., Kominek [172], Orlicz-Ciesielski [247], Sander [273] and also Sm´ıtal-Snoha [291]) Some of these generalizations exceed the scope of this book. Here we will present one of such generalizations based on Dubikajits-Ferens-Ger-Kuczma [72]. Let X be a topological group (not necessary commutative), i.e., a set X endowed with a topology (e.g., a metric space) and with a binary operation + (i.e., + is a function + : X 2 → X) such that x + (y + z) = (x + y) + z

(2.9.1)

for arbitrary x, y, z ∈ X, there exists an element 0 ∈ X such that x+ 0 = x,

(2.9.2)

and for every x ∈ X there exists an element (−x) ∈ X such that x + (−x) = 0 .

(2.9.3)

x + (−y) is usually written as x − y. The commutativity of the operation +: x+y =y+x

(2.9.4)

is, in general, not assumed. Moreover, the functions f1 : X 2 → X and f2 : X → X f1 (x, y) = x + y , f2 (x) = −x , are continuous. Then obviously, for every a ∈ X, the functions Ta : X → X and Sa : X → X Ta (x) = a + x, Sa (x) = x + a (2.9.5) are homeomorphisms. For any two sets A , B ⊂ X we write A + B = {x = a + b | a ∈ A, b ∈ B }, and for a set A ⊂ X and an element b ∈ X we write A + b = {x = a + b | a ∈ A}, b + A = {x = b + a | a ∈ A}. Theorem 2.9.1. Let (X, +) be a topological group, and A, B ⊂ X two sets of the second category with the Baire property. Then int (A + B) = ∅ . 3

Added in the

2nd

edition by K. Baron.

(2.9.6)

2.9. Theorem of S. Piccard

43

Proof. X itself is of the second category, since it contains sets of the second category. Then every non-empty open subset of X is of the second category. In fact, suppose that G ⊂ X is non-empty, open and of the first category. Fix an a ∈ G. Then, for every x ∈ X, the set G − a+ x is of the first category and open since functions (2.9.5) are homeomorphisms and x ∈ G − a + x. Thus X is of the first category at every point x ∈ X. By Theorem 2.1.4 X is of the first category, which is impossible. The sets A, B have the Baire property, i.e., there exist non-empty open sets G, H (if G, H were empty, the sets A, B would be of the first category) and first category sets P , Q , R , S such that A = (G ∪ P ) \ R , B = (H ∪ Q) \ S . Hence it follows that the sets G \ A ⊂ R and B \ H ⊂ S are of the first category. Fix elements g ∈ G  and h ∈ H and put G0 = −g + G, H0 = H − h , A0 = −g + A , B0 = B − h. Then since (2.9.5) are homeomorphisms the set G0 \ A0 = (−g + G) \ (−g + A) = −g + (G \ A) is of the first category, and similarly the set H0 \ B0 = (H \ B) − h is of the first category. Put U = G0 ∩ H0 . U is open, since so are G0 and H0 , and non-empty, since 0 ∈ U . Consequently U is of the second category. For every t ∈ U write Ut = U ∩ (t − U ) . Ut is open and non-empty (t ∈ Ut ) for every t ∈ U . The set (for every t ∈ U ) Ut \ A0 ⊂ U \ A0 ⊂ G0 \ A0 is of the first category. Similarly the set Ut \ (t − B0 ) ⊂ (t − U ) \ (t − B0 ) = t − (U \ B0 ) ⊂ t − (H0 \ B0 ) is of the first category. Since Ut ⊂ [A0 ∩ (t − B0 )] ∪ (Ut \ A0 ) ∪ [Ut \ (t − B0 )] , the set A0 ∩ (t − B0 ) is of the second category (for otherwise Ut would be of the first category) and hence non-empty. So, for every t ∈ U , there exists an at ∈ A0 ∩(t− B0 ), i.e., at ∈ A0 and t − at ∈ B0 . This means that there exists a bt ∈ B0 such that at + bt = t, i.e., t ∈ A0 + B0 . Now, A0 + B0 = (−g + A) + (B − h) , so U ⊂ (−g + A) + (B − h) , and g+U +h⊂A+B. Thus the set A + B contains a non-empty open set g + U + h, i.e., (2.9.6) holds. 

44

Chapter 2. Topology

Exercises 1. Let X, Y be topological spaces. Let A ⊂ X and B ⊂ Y be arbitrary sets. Prove that cl (A × B) = cl A × cl B and int (A × B) = int A × int B 2. Let X, Y be topological spaces, and f : X → Y a homeomorphism. Prove that for any A ⊂ X cl f (A) = f (cl A) , int f (A) = f (int A) , and hence the sets A and f (A) are both of the same category. 3. Let X, Y be topological spaces, X compact, and let f : X → Y be a one-to-one continuous function. Prove that f is a homeomorphism. 4. Prove that any subset of a first category set is of the first category, and the union of a finite or countable collection of first category sets is of the first category. 5. Let X be a separable4 topological space. Prove that each of the following properties is equivalent to the fact that A ⊂ X has the Baire property: (i) there exists a residual set B such that A is both open and closed in B; (ii) there exist sets B ∈ Gδ and R of the first category such that A = B ∪ R; (iii) there exist sets C ∈ Fσ and Q of the first category such that A = C \ Q; (iv) the set D(A) ∩ D(A ) is nowhere dense; (v) the set D(A) \ A is of the first category. 6. Prove that every set E with the Baire property has a decomposition E = A ∪ B, where A is of the first category, and B is a Borel set. 7. Let X, Y be topological spaces, Y separable. Prove that if a set A ⊂ X × Y is residual (with respect to X × Y ), then there exists a residual set C ⊂ X such that for every x ∈ C the set A[x] is residual. 8. Let X be a topological space. We say that sets A, B ⊂ X are Borel separable iff there exists a set D ∈ B(X) such that A ⊂ D ⊂ B  . Let Pn , Qn ⊂ X , n ∈ N, ∞ ∞   be arbitrary sets, P = Pn , Q = Qn . Show that if for every n, m ∈ N the n=1

n=1

sets Pn and Qm are Borel separable, then so are also P and Q. [Hint: Choose Dnm ∈ B(X) such that Pn ⊂ Dnm ⊂ Qm . The set D=

∞ ∞ 

Dnm ∈ B(X)

n=1 m=1

separates P and Q.] 9. Let X be a separable complete metric space, and let A, B ⊂ X be disjoint analytic sets. Then A, B are Borel separable. (Lusin separation theorem; Lusin [211]). [Hint: Choose continuous functions f , g : z → X such that A = f (z) , B = g(z). For every m1 , . . . , mj ∈ N put zm1 ...mj = { z = {ni } ∈ z | n1 = m1 , . . . , nj = mj } , j ∈ N . 4

The assumption that X is separable is not essential, and can be omitted.

2.9. Theorem of S. Piccard

45

Suppose that A, B are not Borel separable, and, using Exercise 2.8, find mi ∈ N , i ∈ N, such that f (zm1 ...mj ) and g(zm1 ...mj ) are not Borel separable, j ∈ N. Put z = {mi } ∈ z. Then  f (z),

g(z) > 0 ( being the distance in X), and d f (zm1 ...mj ) + d g(zm1 ...mj ) <  f (z), g(z) for every j ∈ N. Hence cl f (zm1 ...mj ) ∩ g(zm1 ...mj ) = ∅ so that f (zm1 ...mj ) and g(zm1 ...mj ) are Borel separable.] 10. Let X be a separable complete metric space and A ⊂ X an analytic set. Prove that if A is analytic, then A ∈ B(X). [Hint: Use Lusin separation theorem for A and B = A .] 11. The Cantor set C ⊂ R [Cantor discontinuum] is the set   ∞

in , in = 0, 2; n ∈ N . C = x ∈ [0, 1] | x = 3n n=1 Prove that C is a nowhere dense perfect set. 12. Let X be a separable complete metric space, and A ⊂ X an uncountable analytic ∞ i  n set. Put for ξ = ∈ C h(ξ) = f (xI ), where I = {in } (cf. the proof of n 3 n=1 Theorem 2.8.3) and f : z → X is a continuous function such that f (z) = A. Prove that h : C → A is a homeomorphism. 13. Prove that C + C = [0, 2]. [Hint: Consider the set   ∞

jn , jn = 0, 1; n ∈ N D = x ∈ [0, 1] | x = 3n n=1 and show that D + D = [0, 1].]

Chapter 3

Measure Theory 3.1 Outer and inner measure We assume that the reader is acquainted with the theory of Lebesgue measure and integral. We consider sets situated in RN with a fixed positive integer N . The Lebesgue measure in RN is denoted by m. The σ-algebra of all Lebesgue measurable subsets of RN is denoted by L. There are various possible approaches to the theory of Lebesgue measure (cf., e.g., Halmos [129], Riesz [266], Natanson [237], Carath´eodory [40], L  ojasiewicz [208], Hartman-Mikusi´ nski [138], etc.). We do not presuppose any of those approaches. At any case the reader should be familiar with the following theorem, which we quote here without proof. Theorem 3.1.1. For any set A ⊂ RN the following conditions are equivalent: (i) A ∈ L. (ii) For every ε > 0 there exist open sets G and U such that A ⊂ G , G \ A ⊂ U , m(U ) < ε . (iii) There exist sets H ∈ Gδ and B ∈ L such that A ⊂ H , A = H \ B , m(B) = 0 . (iv) For every ε > 0 there exist a closed set F and an open set V such that F ⊂ A , A \ F ⊂ V , m(V ) < ε . (v) There exist sets E ∈ Fσ and C ∈ L such that E ⊂ A , A = E ∪ C , m(C) = 0 . Theorem 3.1.1 will be our main tool in the sequel of this chapter.

48

Chapter 3. Measure Theory

Let A ⊂ RN be an arbitrary set. We define the outer measure1 and the inner measure of A (denoted in the sequel by me (A) and mi (A), respectively) as me (A) = inf m(G), mi (A) = G⊃A G open

sup m(F ) .

F ⊂A F closed

(3.1.1)

The outer and inner measure are defined for all sets A ⊂ RN . It follows directly from definition (3.1.1) that me (∅) = mi (∅) = 0 . The connection between the outer measure, inner measure and the measure is expressed by the following theorems: Theorem 3.1.2. If A ∈ L, then me (A) = mi (A) = m(A) .

(3.1.2)

Proof. If A ∈ L, then by Theorem 3.1.1 there exist, for every ε > 0, a closed set F and open sets G, U, V such that F ⊂ A ⊂ G, G \ A ⊂ U , A\ F ⊂ V , m(U ) < ε , m(V ) < ε . Hence G = A ∪ (G \ A) and A = F ∪ (A \ F ), and since G \ A ∈ L , A \ F ∈ L , m(G \ A)  m(U ) < ε , m(A \ F )  m(V ) < ε, we have m(G)  m(A) + ε , m(A)  m(F ) + ε , i.e., m(G) − ε  m(A)  m(F ) + ε .

(3.1.3)

Since ε > 0 has been arbitrary, relation (3.1.3) shows that m(A) = inf m(G) = G⊃A G open

sup m(F ) .

F ⊂A F closed

which in view of (3.1.1) is equivalent to (3.1.2).



Theorem 3.1.3. If mi (A) = me (A) < ∞ ,

(3.1.4)

then A ∈ L and (3.1.2) holds. Proof. Relation (3.1.4) implies that there exist, for every ε > 0, a closed set F and an open set G such that F ⊂ A ⊂ G and ε ε m(G) −  me (A) = mi (A)  m(F ) + . 2 2 Hence m(G \ F ) = m(G) − m(F )  ε, and clearly the set G \ F is open, G \ A ⊂ G \ F , A \ F ⊂ G \ F . Now the theorem follows from Theorem 3.1.1 if we take U = V = G \ F . Relation (3.1.2) is a consequence of Theorem 3.1.2.  1

If the Lebesgue measure is introduced via an outer measure, then the reader will easily verify that me defined by (3.1.1) coincides with the previous notion.

3.1. Outer and inner measure

49

The next four theorems give the relation between the outer [inner] measure of the union of a countable collection of sets and those of particular sets. ∞  Theorem 3.1.4. If An ⊂ RN , n ∈ N, are arbitrary sets, and A ⊂ An , then n=1

me (A) 

∞

me (An ) .

(3.1.5)

n=1

In particular me

∞ 

∞  An  me (An ) .

n=1

(3.1.6)

n=1

Proof. Fix an ε > 0. According to definition (3.1.1), for every n ∈ N we can find an open set Gn such that An ⊂ Gn , m(Gn )  me (An ) + ∞ 

ε . 2n

(3.1.7)

Gn also is an open set, and hence measurable. For measurable sets (and with me

n=1

replaced by m) formulas (3.1.5) and (3.1.6) are true. Since A ⊂ (3.1.1) me (A)  m(

∞ 

∞ 

Gn , we have by

n=1

Gn ), and hence

n=1

me (A)  m

∞ 

∞ ∞

 Gn  m(Gn )  me (An ) + ε ,

n=1

n=1

(3.1.8)

n=1

by (3.1.7). Letting in (3.1.8) ε → 0 we obtain (3.1.5). (3.1.6) is a particular case of (3.1.5).  Relations (3.1.5) and (3.1.6) (as well as similar relations occurring in Theorems 3.1.5–3.1.7) are true also for finite unions. It is enough to put An = ∅ for n > m. Taking, in particular, m = 1 and A1 = B, we obtain from (3.1.5) the monotonicity of the outer measure: If A ⊂ B, then me (A)  me (B) . The Lebesgue measure, as is well known, is σ-additive: if An ∈ L , n ∈ N, are pairwise disjoint: Ai ∩ Aj = ∅ for i = j , i, j ∈ N , then m

∞  n=1

∞  An = m(An ) . n=1

For the outer measure we have the following weaker version of this fact.

50

Chapter 3. Measure Theory

Theorem 3.1.5. If An ∈ L , n ∈ N, are pairwise disjoint measurable sets, and A ⊂ RN is arbitrary, then ∞ ∞    me A ∩ An = me (A ∩ An ) . (3.1.9) n=1

n=1 ∞ 

Proof. Fix an ε > 0. We can find an open set G such that A ∩

An ⊂ G and

n=1 ∞    An . m(G) − ε  me A ∩

(3.1.10)

n=1

The sets G ∩ An , n ∈ N, are pairwise disjoint measurable sets, A ∩ An ⊂ A ∩ An ∩ ∞  Ai ⊂ G ∩ An , whence by Theorems 3.1.4 and 3.1.2 i=1

me (A ∩ An )  me (G ∩ An ) = m(G ∩ An ) . Since

∞ 

∞ 

(G ∩ An ) = G ∩

n=1

(3.1.11)

An ⊂ G, we get by (3.1.10) and (3.1.11)

n=1

∞ ∞      me A ∩ An  m(G) − ε  m (G ∩ An ) − ε n=1

n=1

=

∞

m(G ∩ An ) − ε 

n=1

∞

me (A ∩ An ) − ε . (3.1.12)

n=1

When ε → 0 we obtain (3.1.9) from (3.1.12) and Theorem 3.1.4.



For A2 = A1 = RN \ A1 , An = ∅ for n  3, we obtain from Theorem 3.1.5 the so-called Carath´eodory condition me (A) = me (A ∩ A1 ) + me (A ∩ A1 ) for all A ⊂ RN , which is equivalent to the measurability of the set A1 , and in the Carath´eodory theory of the Lebesgue measure is even taken as the definition of the measurability of A1 . The following Theorems 3.1.6 and 3.1.7 are counterparts of Theorems 3.1.4 and 3.1.5 for inner measure. Theorem 3.1.6. If An ⊂ RN , n ∈ N, are pairwise disjoint, and A ⊂ RN is an ∞  arbitrary set such that An ⊂ A, then n=1

mi (A) 

∞

mi (An ) .

(3.1.13)

n=1

In particular, mi

∞  n=1

∞  An  mi (An ) . n=1

(3.1.14)

3.1. Outer and inner measure

51

Proof. For every n ∈ N take a real number an < mi (An ) and a closed set Fn ⊂ An such that m(Fn ) > an . (3.1.15) Take an arbitrary k ∈ N. The set

k 

Fn is closed and

n=1

k 

Fn ⊂ A. By the definition

n=1

of the inner measure we have mi (A)  m

k 

k k

 Fn = m(Fn ) > an ,

n=1

n=1

(3.1.16)

n=1

since the sets Fn are disjoint, just like An . Now let a1 → mi (A1 ), . . . , ak → mi (Ak ). Thus we get from (3.1.16) k

mi (A)  mi (An ) . n=1

Letting k → ∞ we obtain hence (3.1.13). (3.1.14) is a particular case of (3.1.13).  Taking in (3.1.13) A1 = B , An = ∅ for n  2, we obtain the monotonicity of the inner measure: If B ⊂ A, then mi (B)  mi (A) . Theorem 3.1.7. If An ⊂ RN , n ∈ N, are pairwise disjoint measurable sets, and A ⊂ RN is arbitrary, then ∞ ∞    An = mi (A ∩ An ) . mi A ∩ n=1

(3.1.17)

n=1 ∞ 

Proof. Fix a real number a < mi (A∩

An ). We can find a closed set F ⊂ A∩

n=1

∞ 

An

n=1

such that a < m(F ) .

(3.1.18)

The sets F ∩ An n ∈ N, are pairwise disjoint measurable sets. Since F ⊂ F ⊂F∩

∞ 

An =

n=1

∞ 

∞ 

An , also

n=1

(F ∩ An ). Moreover, we have F ∩ An ⊂ A ∩ An , n ∈ N. Hence

n=1

by the monotonicity of the inner measure m(F )  m

∞ 

∞ ∞ ∞

 (F ∩ An ) = m(F ∩ An ) = mi (F ∩ An )  mi (A ∩ An ) .

n=1

n=1

n=1

n=1

(3.1.19) Relations (3.1.18) and (3.1.19) imply a<

∞

n=1

mi (A ∩ An ) .

52

Chapter 3. Measure Theory

Letting a → mi (A ∩

∞ 

An ), we obtain

n=1

 mi

A∩



∞ 

An



n=1

∞

mi (A ∩ An ) ,

n=1

which together with Theorem 3.1.6 yields equality (3.1.17).



Before proceeding further let us note that for an arbitrary set A ⊂ RN we have mi (A)  me (A) .

(3.1.20)

This follows directly from definition (3.1.1), but one can also argue as follows: For an arbitrary ε > 0 there exists a closed set F ⊂ A such that mi (A)  m(F ) + ε . By the monotonicity of the outer measure we have m(F ) = me (F )  me (A) . Hence mi (A)  me (A) + ε, and on letting ε → 0 we obtain (3.1.20). No we prove Lemma 3.1.1. For every set A ⊂ RN there exists a measurable set B such that A ⊂ B and mi (B \ A) = 0. Proof. Consider first the case where me (A) < ∞. For every n ∈ N there exists an open set Gn such that A ⊂ Gn and m(Gn )  me (A) + Put B=

∞ 

1 . 2n

Gn .

(3.1.21)

(3.1.22)

n=1

For an indirect proof suppose that mi (B \ A) = 2η > 0. Then there exists a closed set F ⊂ B \ A such that m(F )  η. For every n ∈ N the set Gn \ F is open, A ⊂ Gn \ F , whence me (A)  m(Gn \ F ) = m(Gn ) − m(F )  m(Gn ) − η . Hence, by (3.1.21), m(Gn )  m(Gn ) + or

1 −η, 2n

1 −η  0, 2n whence, on letting n → ∞, we obtain a contradiction.

3.1. Outer and inner measure

53

Now let me (A) be arbitrary. Put A1 = K(0, 1) , An = K(0, n) \ K(0, n − 1) for n = 2, 3, 4, . . . , where K(0, n) is the open ball centered at the origin and with ∞  radius n. The sets An are measurable, pairwise disjoint, and An = RN . Moreover, n=1

m(An ) < ∞ for n ∈ N, whence also me (A ∩ An )  m(An ) < ∞ for n ∈ N. By the first part of the proof we can find, for every n ∈ N, a measurable set Bn such that ∞  A ∩ An ⊂ Bn and mi [Bn \ (A ∩ An )] = 0. Put B = (Bn ∩ An ). Then B ∈ L , A = A∩R

N

= A∩

∞ 

∞ 

An =

n=1

(A ∩ An ) ⊂

n=1

B\A=

∞ 

∞ 

n=1

(Bn ∩ An ) = B. Moreover,

n=1

(Bn ∩ An ) \ A =

n=1

∞ 

[(Bn ∩ An ) \ A] .

n=1

Now, (Bn ∩An )\A = (Bn ∩An )\(A∩An ) ⊂ Bn \(A∩An ), whence mi [(Bn ∩ An ) \ A]  mi [Bn \ (A ∩ An )] = 0, and ultimately mi [(Bn ∩ An ) \ A] = 0 , n ∈ N. The sets Bn ∩ An are measurable and pairwise disjoint (since An are pairwise disjoint). By Theorem 3.1.7 mi (B \ A) = mi

∞  

∞   



(Bn ∩ An ) \ A = mi (Bn ∩ An ) ∩ A

n=1

=

∞

n=1

∞



mi (Bn ∩ An ) ∩ A = mi (Bn ∩ An ) \ A = 0 .



n=1

n=1

So the set B has the required properties.



Lemma 3.1.2. For every set A ⊂ RN there exists a measurable set B such that A ⊂ B and m(B) = me (A) Proof. For every n ∈ N there exists an open set Gn such that A ⊂ Gn and (3.1.21) holds. Define B by (3.1.22). Then B ⊂ Gn , n ∈ N, whence by (3.1.21) m(B)  1 m(Gn )  me (A) + n , n ∈ N. As n → ∞, we obtain hence m(B)  me (A). On the 2  other hand, A ⊂ B, whence me (A)  me (B) = m(B). Lemma 3.1.3. For every set A ⊂ RN there exists a measurable set B such that A ⊂ B and for every measurable set Z such that A ⊂ Z we have m(B \ Z) = 0. Proof. Let B be the set occurring in Lemma 3.1.1. For every Z ∈ L, A ⊂ Z, we have B \ Z ⊂ B \ A, whence by Lemma 3.1.1 mi (B \ Z) = 0. Since B \ Z is measurable, this implies in virtue of Theorem 3.1.2 m(B \ Z) = 0.  The Lebesgue measure is complete, i.e., if Y ⊂ B \Z, then Y ∈ L and m(Y ) = 0. Thus Lemma 3.1.3 together with Theorems 2.7.1 and 2.6.1 imply the following Theorem 3.1.8. Every analytic set is Lebesgue measurable.

54

Chapter 3. Measure Theory

3.2 Linear transforms As is well known, the Lebesgue measure in RN is invariant under translations. We are going to show that this important property of m is shared also  by the outer and inner  measure. More exactly, we are going to prove the following cf. L  ojasiewicz [208] Theorem 3.2.1. Let f : RN → RN be linear transform f (x) = Lx + b , x ∈ RN ,

(3.2.1)

where L is an N × N non-singular square matrix, and b ∈ RN . Then, for every set A ⊂ RN , we have     me f (A) = |det L| me (A) , mi f (A) = |det L| mi (A) . (3.2.2) Proof. First suppose that the set A is closed2 . Function (3.2.1) is a homeomorphism, so f (A) also is closed, and thus both A and f (A), are measurable. For such an A formula (3.2.2) takes the form   m f (A) = |det L| m(A) . (3.2.3) Let χA be the characteristic function of the set A:  1 if x ∈ A, χA (x) = 0 if x ∈  A. Then χA ◦ f = χf −1 (A) (◦ denotes the functional composition). Hence, by the known rules of the integral calculus,    m(A) = χA dx = (χA ◦ f ) |det L| dx = χf −1 (A) |det L| dx RN



= |det L|

RN



RN

 χf −1 (A) dx = |det L| m f −1 (A) ,

RN

whence (3.2.3) follows on replacing A by f (A). Now let A ⊂ RN be Take an  arbitrary.    arbitrary closed  set F ⊂ A. Then  f (F ) ⊂ f (A), whence mi f (F )  mi f (A) . By (3.2.3) mi f (F ) = m f (F ) = |det L| m(F ), whence   |det L| m(F )  mi f (A) . (3.2.4) Taking in (3.2.4) the supremum over all closed sets F ⊂ A, we obtain according to (3.1.1)   |det L| mi (A)  mi f (A) . (3.2.5) Now take an arbitrary closed set F0 ⊂ f (A). Then the set F = f −1 (F0 ) is also closed, and we have f (F ) = F0 . By (3.1.3) m(F0 ) = |det L| m(F ) , 2

The same argument applies also if A is open.

3.2. Linear transforms

55

and m(F )  mi (A), since F ⊂ A. Thus m(F0 )  |det L| mi (A) .

(3.2.6)

Taking in (3.2.6) the supremum over all closed sets F0 ⊂ f (A), we obtain   mi f (A)  |det L| mi (A) , which together with (3.2.5) proves the second relation in (3.2.2). The proof of the first relation in (3.2.2) is analogous.  Let us note that formulas (3.2.2) remain valid also for singular matrices L. N Then  the set f (A)  is contained in an (N − 1)-dimensional subspace of R , whence mi f (A) = me f (A) = 0. Taking in Theorem 3.2.1 L = E, the unit matrix of order N , we obtain the invariance of me and mi under translations. Corollary 3.2.1. Let f : RN → RN be transform (3.2.1), where L is an N × N nonsingular square matrix, and b ∈ RN . If a set A ⊂ RN is measurable, then also f (A) is measurable, and relation (3.2.3) holds. Proof. Consider first the case where m(A) < ∞. By Theorem 3.1.2 me (A) = mi (A) = m(A), whence by Theorem 3.2.1     me f (A) = |det L| m(A) , mi f (A) = |det L| m(A) (3.2.7)     so that me f (A) = mi f (A) < ∞. By Theorem 3.1.3 f (A) ∈ L, and (3.2.3) results from (3.2.7). If m(A) = ∞, then there exist pairwise disjoint measurable sets An such that ∞  A = An , m(An ) < ∞ , n ∈ N (cf. the proof of Lemma 3.1.1). Then f (A) = ∞ 

n=1

f (An ), and f (An ) ∈ L , n ∈ N, by the first part of the proof. Hence also f (A) ∈ L,

n=1

and by Theorem 3.2.1     m f (A) = mi f (A) = |det L| mi (A) = |det L| m(A). 

Thus (3.2.3) holds. Corollary 3.2.2. Let f : RN → RN be the transform f (x) = ax + b , x ∈ RN , where a ∈ R \ {0} , b ∈ RN . Then, for every set A ⊂ RN , we have   N me f (A) = me (aA + b) = |a| me (A) ,   mi f (A) = mi (aA + b) = |a|N mi (A) , and, if A ∈ L, then also f (A) ∈ L, and   N m f (A) = m(aA + b) = |a| m(A).

(3.2.8)

56

Chapter 3. Measure Theory

Proof. Take L = aE, the diagonal matrix with all the elements of the main diagonal equal a. Then det L = aN = 0, and function (3.2.1) reduces to (3.2.8). Now the corollary results from Theorem 3.2.1 and Corollary 3.2.1. 

3.3 Saturated non-measurable sets A set A ⊂ RN is called saturated non-measurable (Halperin [132]) iff mi (A) = mi (A ) = 0 .

(3.3.1)

It follows directly from definition (3.3.1) that if a set A is saturated non-measurable, then so is also its complement A . To justify the attribute non-measurable, let us note that a saturated non-measurable set actually is not Lebesgue measurable. Supposing the contrary, we would get by (3.3.1) m(A) = m(A ) = 0, whence m(RN ) = m(A ∪ A ) = m(A) + m(A ) = 0, a contradiction. An insight into the structure of saturated non-measurable sets may be gained from the following Theorem 3.3.1. Let A ⊂ RN be an arbitrary set. Then the following conditions are equivalent: (i) mi (A ) = 0; (ii) for every measurable set E of positive measure we have A ∩ E = ∅; (iii) for every measurable set E we have me (A ∩ E) = m(E); (iv) for every open interval3 I ⊂ RN we have me (A ∩ I) = m(I). Proof. First we prove that condition (i) implies (ii). Let a set A ⊂ RN fulfil (i), and let E ∈ L be an arbitrary set such that m(E) > 0. If we had A ∩ E = ∅, then E ⊂ A , whence mi (A )  mi (E) = m(E) > 0 , which is incompatible with (i). Now assume that A ⊂ RN fulfils (ii). Let E ⊂ RN be an arbitrary measurable set. Clearly me (A ∩ E)  me (E) = m(E) . It remains to exclude the strict inequality. For an indirect proof suppose that for a certain E ∈ L we have me (A ∩ E) < m(E) . By Lemma 3.1.2 there exists a set B ∈ L such that A ∩ E ⊂ B and m(B) = me (A ∩ E) < m(E). Hence also m(B ∩ E)  m(B) < m(E) . Now, by (3.3.2) m(B ∩ E) < ∞, whence, again by (3.3.2) m [E \ (B ∩ E)] = m(E) − m(B ∩ E) > 0 . 3

An open interval in RN is a cartesian product of N one-dimensional open intervals.

(3.3.2)

3.3. Saturated non-measurable sets

57

Since A ∩ E ⊂ B ∩ E, we have E \ (B ∩ E) ⊂ E \ (A ∩ E) ⊂ A . Thus we have constructed a measurable set of positive measure viz. E \ (B ∩ E) disjoint with A. This contradicts (ii). Condition (iv) is a particular case of condition (iii), so (iii) ⇒ (iv). It remains to show that (v) ⇒ (i). Let condition (iv) be fulfilled, and put Jn = (−n, n)N , n ∈ N. Thus Jn ⊂ RN ∞  are finite intervals, Jn ⊂ Jn+1 , n ∈ N , Jn = RN . Suppose that mi (A ) > 0. n=1

Then there exists a closed set F ⊂ A such that m(F ) > 0. We have F = F ∩ RN = F ∩

∞  n=1

Jn =

∞ 

(F ∩ Jn ) ,

n=1

whence m(F ) = lim m(F ∩ Jn ) . n→∞

Thus m(F ∩ Jn ) > 0 for n sufficiently large. Let J denote an arbitrary interval from the sequence {Jn } such that m(F ∩ J) > 0. We have A ∩ J = (A ∩ J ∩ F ) ∪ (A ∩ J ∩ F  ) = A ∩ J ∩ F  ⊂ J ∩ F  , since A ∩ J ∩ F = ∅. Hence by (iv) m(J) = me (A ∩ J)  me (J ∩ F  ) = m(J ∩ F  ) = m [J \ (J ∩ F )] = m(J) − m(J ∩ F ) < m(J) , 

a contradiction.

It is clear from the proof that in condition (iv) open intervals may be replaced by closed intervals. Let us note that conditions of Theorem 3.3.1 do not imply that the set A is nonmeasurable. They are fulfilled, e.g., for A = RN . But for saturated non-measurable sets we can derive from Theorem 3.3.1 the following result, which to a certain extent justifies the attribute saturated. Theorem 3.3.2. Let A ⊂ RN be a saturated non-measurable set. Then, for every measurable set B of positive measure, the set A ∩ B is non-measurable. Proof. We have A ∩ B ⊂ A, whence mi (A ∩ B)  mi (A) = 0. Thus mi (A ∩ B) = 0, whereas by Theorem 3.3.1 me (A ∩ B) = m(B) > 0. The theorem now follows from Theorem 3.1.2.  We defer the construction of a saturated non-measurable set A ⊂ R to the last section. Many examples of saturated non-measurable sets A ⊂ RN will occur in the second part of this book. However, in the sequel we assume that there exist saturated non-measurable sets, the assumption which will be justified only later. Let us note that if A is a saturated non-measurable set, then by Theorem 3.3.1 me (A) = me (A ∩ RN ) = m(RN ) = ∞ .

58

Chapter 3. Measure Theory

Theorem 3.3.3. Every set B ⊂ RN of positive outer measure contains a non-measurable set. Proof. If B is non-measurable, then B itself is good. Now suppose that B ∈ L so that m(B) = me (B) > 0, and let A be a saturated non-measurable set. Then the set A ∩ B ⊂ B is non-measurable in virtue of Theorem 3.3.2.  The saturated non-measurable set A constructed in 3.8 will also have the property: (∗) If B ⊂ A or B ⊂ A , and B has the Baire property, then B is of the first category. Property (∗) is a topological analogue of the saturated non-measurability (Kuczma [189]). Making use of (∗) we will prove topological analogues of Theorems 3.3.1, 3.3.2, 3.3.3. Theorem 3.3.4. If A ⊂ RN has property (∗), and B ⊂ RN is a set of the second category and with the Baire property, then the set A ∩ B is of the second category and without the Baire property. Proof. B = (A ∩ B) ∪ (A ∩ B). If A ∩ B had the Baire property, then also A ∩ B = B \ (A ∩ B) would have the Baire property, and since A ∩ B ⊂ A , we would get by (∗) that A ∩ B is of the first category. Similarly, A ∩ B ⊂ A would be of the first category, and hence also B would be of the first category. If A ∩ B were of the first category, then it would have the Baire property, and as we have just seen, this leads to a contradiction. Consequently A ∩ B is of the second category and without the Baire property.  Corollary 3.3.1. If a set A ⊂ RN has property (∗), then for every set B of the second category and with the Baire property we have A ∩ B = ∅. Theorem 3.3.5. If A ⊂ RN has property (∗), and B ⊂ RN is of the second category and with the Baire property, then D(A ∩ B) = D(B). Proof. The inclusion D(A ∩ B) ⊂ D(B) follows from A ∩ B ⊂ B. Now let x ∈ D(B). For every open neighbourhood U of x the set U ∩ B is of the second category. The set U ∩ B has the Baire property, since B has it, and U , being open, also has the Baire property. By Theorem 3.3.4 the set U ∩ B ∩ A is of the second category, which proves that x ∈ D(A ∩ B). Consequently D(B) ⊂ D(A ∩ B), which completes the proof.  Taking in Theorem 3.3.5 B = RN , we obtain Corollary 3.3.2. If a set A ⊂ RN has property (∗), then D(A) = RN . Theorem 3.3.6. Every set B ⊂ RN of the second category contains a subset without the Baire property. Proof. If B is without the Baire property, then B itself is good. Otherwise, let A ⊂ RN be a set with property (∗). By Theorem 3.3.4 A ∩ B ⊂ B is a set without the Baire property. 

3.4. Lusin sets

59

3.4 Lusin sets An uncountable set A ⊂ RN is called a Lusin set (Lusin [210]) iff it intersects every nowhere dense perfect set in RN at at most countably many points. In other words, A is a Lusin set iff card A > ℵ0 and for every nowhere dense perfect set F ⊂ RN we have card (A ∩ F )  ℵ0 . It follows directly from the definition that an uncountable subset of a Lusin set is a Lusin set. The existence of Lusin sets was shown, under the assumption of the continuum hypothesis, by Lusin [210]. This will follow also from theorems in 11.6. We will need the following Lemma 3.4.1. Let E ⊂ RN be an arbitrary set, and let f : E → RM , M ∈ N, be a continuous function. There exists a decomposition E =A∪B,

(3.4.1)

where A ∩ B = ∅ , A is of the first category, and f (B) is measurable and of measure zero. Proof. If E is finite, the lemma is obvious. So in the sequel we assume that E is infinite. Let {pk }k∈N be a sequence of points of E, dense in E. By the continuity of f we may choose positive numbers δnk , n, k ∈ N, such that for x ∈ K(pk , δnk ) (the open ball centered at pk and with the radius δnk ) we have   1 . (3.4.2) f (x) ∈ K f (pk ), k 2 n Put En = E \

∞ 

K(pk , δnk ) , A =

k=1

∞ 

En , B = E \ A . Then (3.4.1) holds, and

n=1

A ∩ B = ∅. Suppose that y ∈ int cl En . So there exists an open ball K ⊂ RN such that y ∈ K ⊂ cl En . There exists a k ∈ N such that pk ∈ K, since the sequence {pk } is dense in E. Hence K(pk , δnk ) ∩ K = ∅. On the other hand, K(pk , δnk ) ∩ En = ∅, so K(pk , δnk ) ∩ K cannot contain any point of En . But this contradicts the fact that K ⊂ cl En . Thus int cl En = ∅, which means that  En are nowhere dense, and A is of the first category. It remains to show that m f (B) = 0.   ∞  1 ⊂ RM , and let c denote the M -dimensional K f (pk ), k Let Un = 2 n k=1 measure of the unit ball in RM . By Corollary 3.2.2        1 1 1 c =m K(0, 1) + f (pk ) = k M m K(0, 1) = k M , m K f (pk ), k 2 n 2k n (2 n) (2 n) whence by Theorem 3.1.4    ∞ ∞

c 1 1 c m(Un )  = . m K f (pk ), k = M M k M 2 n (2 n) n 2 −1 k=1

k=1

60

Chapter 3. Measure Theory

If an x ∈ B, then x ∈ En for n ∈ N, i.e., x ∈ K(pk , δnk ) for a certain kn ∈ N. Hence by (3.4.2) f (x) ∈ Un for every n ∈ N, which means that f (B) ⊂ Un for every n ∈ N. Hence   1 c me f (B)  m(Un )  M M n 2 −1   for all n ∈ N. Letting n → ∞ we obtain hence me f (B) = 0, i.e., f (B) is measurable and of measure zero.  An interesting property of Lusin sets is expressed by the following theorem of Sierpi´ nski [281]. Theorem 3.4.1. If E ⊂ RN is a Lusin set, and f : E → RM a continuous function, then f (E) is Lebesgue measurable and of measure zero. Proof. By Lemma 3.4.1 the set E has a decomposition (3.4.1), where A is of the first ∞    category, and m f (B) = 0. Thus A = En , where int cl En = ∅ for n ∈ N. By n=1

Theorem 2.8.2 cl En = Fn ∪Pn , where Fn is perfect in cl En , and card Pn  ℵ0 , n ∈ N. But cl En is closed, so Fn is perfect. Moreover, Fn ⊂ cl En , and cl cl En = cl En so that int cl cl En = int cl En = ∅. This means that cl En is nowhere dense, and hence also Fn is nowhere dense. Since A ⊂ E , A itself is a Lusin set4 , whence card(A ∩ Fn )  ℵ0 for n ∈ N. We have ∞ ∞   A⊂ cl En = (Fn ∪ Pn ) , n=1

n=1

∞ 

[(A ∩ Fn ) ∪ (A ∩ Pn )]. The sets occurring under the union sign are at most   countable, so also card A  ℵ0 , and card f (A)  card A  ℵ0 , whence m f (A) = 0. Now f (E) = f (A) ∪ f (B), and being a union of measurable sets of measure zero, f (E) itself is a measurable set of measure zero.  so A =

n=1

Taking M = N and f (x) = x for x ∈ E, we obtain from Theorem 3.4.1 Corollary 3.4.1. Every Lusin set is Lebesgue measurable and of measure zero. Theorem 3.4.1 and Corollary 3.4.1 express the fact that from the point of view of measure theory Lusin sets are very small. The situation is different from the point of view of topology (Sierpi´ nski [282]). Theorem 3.4.2. If E ⊂ RN is a Lusin set, then E is of the second category and without the Baire property. Proof. Suppose that E is of the first category. Then E =

∞ 

En , where the sets En

n=1

are nowhere dense. Since E, being a Lusin set, is uncountable, at least one of the sets En , say Em , is uncountable. By Theorem 2.8.2 Em = A ∪ B, where A is perfect in 4

If A is uncountable. If card A  ℵ0 , then so the more card(A ∩ Fn )  ℵ0 .

3.5. Outer density

61

Em and card B  ℵ0 . The fact that A is perfect in Em means that every point of A is its accumulation point (this fact is independent of Em ), in other words A ⊂ Ad ,

(3.4.3)

and every accumulation point of A, belonging to Em , belongs also to A: Ad ∩Em ⊂ A. Put F = cl A = A ∪ Ad . Hence F d = Ad ∪ Add = Ad , and by (3.4.3) F = A ∪ Ad ⊂ Ad ∪ Ad = Ad = F d . On the other hand, F d ⊂ F , since F is closed. Consequently F = F d , and F is perfect. Further, since A ⊂ Em , A is nowhere dense, and also F = cl A is nowhere dense. Em ⊂ E is a Lusin set, so card(Em ∩ F )  ℵ0 . Hence also card(Em ∩ A)  ℵ0 , since A ⊂ F , and since Em ∩ A = A, card A  ℵ0 . Therefore also Em = A ∪ B is at most countable. This contradiction shows that E cannot be of the first category. So E is of the second category, and suppose that E has the Baire property: E = (G ∪ P ) \ R , where G is open, and P, R are of the first category. G = ∅, since otherwise E would be of the first category. By Theorem 2.1.1 there exists a set B ∈ Fσ , of the first category, such that R ⊂ B. Put A = G \ B. The set A is analytic, since it is a Borel set, and we have A ⊂ E. Clearly A is of the second category, and hence uncountable. There exists (cf. Exercise 2.13) a homeomorphism f : C → A, where C is the Cantor set. C is perfect and nowhere dense (Exercise 2.11), so also F = f (C) ⊂ A ⊂ E is perfect and nowhere dense, since f is a homeomorphism. Therefore card(E ∩F )  ℵ0 . But E ∩ F = F , so card F  ℵ0 . But f , being a homeomorphism, is one-to-one, so card F = card f (C) = card C = c. This contradiction shows that E cannot have the Baire property. 

3.5 Outer density Let A ⊂ RN be an arbitrary set. A point x ∈ RN is called a point of outer density of A iff me (A ∩ Qr ) lim = 1, r→0 m(Qr ) where Qr denotes the closed cube with the edge of length r centered at x: N  r r , x = (x1 , . . . , xN ) . Qr = × xi − , xi + 2 2 i=1

If A ∈ L, we omit the attribute outer and say simply that x is a density point of A. The famous Lebesgue density theorem says that if A ∈ L, then almost every point x ∈ A is a density point of A. For an arbitrary set A, let A∗ denote the set of the points of outer density of A, which reduces to the set of density points of A whenever A is measurable. The Lebesgue density theorem may be expressed as follows: If A ∈ L, then m(A \ A∗ ) = 0 . (3.5.1)

62

Chapter 3. Measure Theory

Formula (3.5.1) can be extended to arbitrary sets A ⊂ RN . We will prove this below making use of the Vitali covering theorem, with which, as we assume, the reader is familiar. The proof given here is due to Z. Opial (cf. L  ojasiewicz [208]). Theorem 3.5.1. Almost every point of an arbitrary set A ⊂ RN is a point of outer density of A. Proof. Let E = A \ A∗ be the set of those x ∈ A which are not points of outer density of A. The thing to prove is that m(E) = 0. First assume that A is bounded. For every cube Qr we have A ∩ Qr ⊂ Qr , whence me (A ∩ Qr )  m(Qr ). Thus for x ∈ E we have me (A ∩ Qr ) < 1. lim inf r→0 m(Qr ) Let H(α) , α ∈ (0, 1), be the set of those x ∈ A for which lim inf r→0

me (A ∩ Qr ) < α. m(Qr )

   1 H 1 − , it is enough to prove that m H(α) = 0 for every α ∈ (0, 1). k k=2 Fix an α ∈ (0, 1), and write H = H(α). Take an ε > 0. There exists an open set G such that H ⊂ G and m(G)  me (H) + ε . (3.5.2) Since E =

∞ 

Moreover, since H ⊂ A is bounded, we have me (H) < ∞. Let Q be the family of all closed cubes Q ⊂ G such that me (A ∩ Q) < α. m(Q)

(3.5.3)

By definition Q is a Vitali cover of H. By the Vitali covering theorem there exists a finite or countably infinite sequence of pairwise disjoint cubes Qn ∈ Q such that    m H\ (3.5.4) Qn = 0 . n

Since H=

         H ∩ Qn ∪ H \ A ∩ Qn ∪ H \ Qn ⊂ Qn ,

 n

n

n

n

we have by (3.5.4) and by Theorem 3.1.5   A ∩ Qn = me (A ∩ Qn ) . me (H)  me n

n

But all Qn ∈ Q, and so fulfil (3.5.3). Moreover, Qn ⊂ G, whence also

 n

Qn ⊂ G.

Thus by (3.5.2) me (H)  α

n

m(Qn ) = α m

 n

   Qn  α m(G)  α me (H) + ε ,

3.6. Some lemmas

63

whence, on letting ε → 0, we obtain me (H)  α me (H), whence me (H) = 0. In view of the inequalities 0  mi (H)  me (H) this implies by Theorem 3.1.3 that H ∈ L and m(H) = 0. ∞  Now let A be unbounded. Then A can be represented as a union A = An of n=1

disjoint bounded sets. Since An ⊂ A, we have me (An ∩ Qr )  me (A ∩ Qr )  m(Qr ) for every n ∈ N and r > 0. Thus, if an x is a point of outer density of an An , it must be also a point of outer density of A. In other words, A∗n ⊂ A∗ , n ∈ N. Hence, for every n ∈ N , An \ A∗ ⊂ An \ A∗n , and ∗

E = A\A =

∞  n=1



∞  

∗

An \ A =

∗

An \ A



⊂

n=1

∞  

 An \ A∗n .

(3.5.5)

n=1

By the first part of the proof m(An \ A∗n ) = 0 for n ∈ N. Thus all the sets in (3.5.5) are of measure zero, and, in particular, m(E) = 0.  The classical Lebesgue density theorem is a particular case of Theorem 3.5.1 above.

3.6 Some lemmas In this section we prove some unrelated theorems which will be useful in later parts of this book. Theorem 3.6.1 is known as Sm´ıtal’s lemma (Kuczma-Sm´ıtal [190], Kuczma [185]), although similar ideas are found also in earlier works (Ostrowski [248], Halmos [129], Erd˝ os [76]; R. A. Rosenbaum [269] mentions an equivalent theorem). Lemma 3.6.1 is Zygmund’s lemma (cf. L  ojasiewicz [208]). Theorem 3.6.4 is found in Borel [36]. Theorem 3.6.1. Let B, D ⊂ RN be such that me (B) > 0 and D is dense in RN . Put A = B + D. Then mi (A ) = 0. Proof. By Theorem 3.5.1 there exists an x0 ∈ B which is a point of outer density of B. Thus, for every c ∈ (0, 1), there exists a δ > 0 such that, if Qr is the cube centered at x0 and with the edge r, then r < δ implies me (B ∩ Qr ) > c m(Qr ) . Put D0 = x0 + D. Then every x ∈ D0 may be written as x = x0 + y , y ∈ D. By Corollary 3.2.2 me [(B ∩ Qr ) + y] = me (B ∩ Qr ) and m(Qr + y) = m(Qr ). Thus for any x = x0 + y ∈ D0 and for any cube Q centered at x and with the edge r < δ we have me [(B + y) ∩ Q] > c m(Q). Since A = B + D = (B + y), we have y∈D

me (A ∩ Q)  me [(B + y) ∩ Q], whence me (A ∩ Q) > c m(Q) for every cube Q centered at a point of D0 and with the edge r < δ.

(3.6.1)

64

Chapter 3. Measure Theory

Take an arbitrary open interval5 I ⊂ RN . Let Q be the family of cubes Q ⊂ I centered at points of D0 and with edges less than δ. The family Q is a Vitali cover of I, since the set D0 is, like D, dense in RN . By the Vitali covering theorem6 there exists a (finite of infinite) sequence of pairwise disjoint cubes Qn ∈ Q such that    m I \ Qn = 0 .

(3.6.2)

n

Then we have7 by Theorem 3.1.5 and by (3.6.1)      

m(I)  me (I ∩ A) = me A ∩ Q n + me A ∩ I \ Q n 

= me A ∩





Qn =

n

n

me (A ∩ Qn )  c

n

n

m(Qn ) = c m(I) ,

n

  since m(I) = m( Qn ) + m(I \ Qn ), and by (3.6.2). Hence, letting c → 1, we get n

n

m(I) = m(I ∩ A) .

(3.6.3)

Relation (3.6.3) is valid for an arbitrary open interval I ⊂ RN . The theorem now follows from Theorem 3.3.1.  The following theorem (Kuczma [189]) is a topological analogue of Theorem 3.6.1. Theorem 3.6.2. Let X be a separeble8 topological group, and let B, D ⊂ X be such that B is of the second category, and D is dense in X. Put A = B + D. Then the set A = X \ A does not contain any subset of the second category and with the Baire property. Proof. It is enough to show that A intersects every set of the second category and with the Baire property in X. Let E ⊂ X be such a set. Then E = (G ∪ P ) \ R, where G is open, and P, R are of the first category. Since E is of the second category, we must have G = ∅. Since B is of the second category, D(B) = ∅ by Theorem 2.1.4. Let x0 ∈ D(B). There exists a d ∈ D such that x0 + d ∈ G. Hence the set G − d is a neighbourhood of x0 . Consequently B ∩ (G − d) is of the second category, and so is also its translation [B ∩ (G − d)] + d. (Translation is a homeomorphism). But [B ∩ (G − d)] + d = (B + d) ∩ G ⊂ (B + D) ∩ G = A ∩ G. Thus A ∩ G is of the second category, and so is also (A∩G)\ R. On the other hand, (A∩G)\ R = A∩(G\ R) ⊂ A∩[(G ∪ P ) \ R] = A∩E. Consequently A ∩ E also is of the second category, and, in particular, A ∩ E = ∅.  5 6 7 8

Cf. 3.3. The proof given in Kuczma-Sm´ıtal [190]does not usethe Vitali theorem Since Qn ⊂ I for all n, we have I ∩ A ∩ Qn = A ∩ Qn . n

n

The assumption that X is separable is not essential and may be omitted.

3.6. Some lemmas

65

Now let f : (a, b) → R be a function of a single real variable, and let D+ f (x) denote the upper right Dini derivative of f at x: D+ f (x) = lim sup h→0+0

f (x + h) − f (x) . h

Lemma 3.6.1. Let f : (a, b) → R be a continuous function, and let Z ⊂ (a, b) be a set such that int f (Z) = ∅. If D + f (x) > 0 for x ∈ (a, b) \ Z, then f is increasing9 . Proof. For an indirect proof suppose that there exist c, d ∈ (a, b) such that c < d, and f (d) < f (c) .   Take an arbitrary y ∈ f (d), f (c) . There exist x ∈ (c, d) such that f (x) = y. Put A = { x ∈ (c, d) | f (x) = y } . By the continuity of f the set A is closed10 , and since A ⊂ (c, d) , A is bounded. Thus A is compact. Hence x0 = sup A ∈ A ⊂ (c, d) . For x ∈ (x0 , d) we have f (x) < y = f (x0 ). Thus for x ∈ (x0 , d) f (x) − f (x0 ) f (x) − y = < 0. x − x0 x − x0 Hence D+f (x0 )  0, whence   x0 ∈ Z and y ∈ f (Z). Since y was arbitrary in f (d), f (c) this shows that f (d), f (c) ⊂ f (Z), contrary to the assumption that int f (Z) = ∅. Thus f must be increasing in (a, b).  Theorem 3.6.3. Let f : (a, b) → R be a continuous function, and let Z ⊂ (a, b) be a countable set. If D + f (x)  0 for x ∈ (a, b) \ Z, then f is increasing. x Proof. For every n ∈ N put gn (x) = f (x) + . Then n D+ gn (x) = D+ f (x) +

1 > 0 for x ∈ (a, b) \ Z . n

Since card gn (Z)  card Z  ℵ0 , we have int gn (Z) = ∅ for n ∈ N. By Lemma 3.6.1 all functions gn are increasing, and hence also f (x) = lim gn (x) n→∞



is increasing.

9 In this book increasing resp. decreasing means weakly increasing resp. weakly decreasing, i.e., x < y implies f (x)  f (y) resp. f (x)  f (y). 10 The set A is closed in (c, d). But since f (d) < y < f (c),

A = { x ∈ [c, d] | f (x) = y } , and so A is also closed in [c, d], and hence closed.

66

Chapter 3. Measure Theory

Now fix integers k , p , q , p  1 , q  2 , 0  k  q − 1, and put C = { c ∈ N | 0  c  q − 1 , c = k } ,  Ap =

(3.6.4)

∞

ci , ci ∈ C for i = 1 , . . . , p − 1, cp = k , i q i=1  (3.6.5) 0  ci  q − 1 for i > p .

x ∈ [0, 1] | x =

Lemma 3.6.2. Set (3.6.5) is measurable and m(Ap ) = (q − 1)p−1 q −p .

(3.6.6)

Proof. For every c1 . . . , cp−1 ∈ C we define the interval P (c1 , . . . , cp−1 ) as  ⎧p−1  ci  ci k p−1 k+1 ⎪ ⎪ if k < q − 1 , + , + ⎪ ⎪ q p i=1 q i qp ⎨ i=1 q i P (c1 , . . . , cp−1 ) =  p−1 ⎪ ⎪ ⎪  ci  ci k p−2 cp−1 + 1 ⎪ ⎩ if k = q − 1 . + , + i q p i=1 q i q p−1 i=1 q Then



Ap =

P (c1 , . . . , cp−1 ) .

(3.6.7)

c1 ,...,cp−1 ∈C

The length of each interval P (c1 , . . . , cp−1 ) is q −p , their number is (q − 1)p−1 , and the different intervals are disjoint. Thus (3.6.6) is a direct consequence of (3.6.7).  Theorem 3.6.4. Let q  3 and let  A=

∞

ci x ∈ [0, 1] | x = , ci ∈ C i q i=1

 .

Then A ∈ L and m(A) = 0. Proof. Put A = [0, 1] \ A. Then A =

∞ 

Ap ,

p=1

where Ap are given by (3.6.5). The sets Ap are pairwise disjoint and measurable, thus A ∈ L and by Lemma 3.6.2 m(A ) =

∞

p=1

m(Ap ) =

∞

(q − 1)p−1 p=1

qp

= 1.

Hence also A ∈ L and m(A) = 1 − m(A ) = 0. Corollary 3.6.1. The Cantor set C (cf. Exercise 2.11) has measure zero.



3.7. Theorem of Steinhaus

67

3.7 Theorem of Steinhaus The celebrated theorem of Steinhaus is the measure theoretical analogue of the theorem of S. Piccard (Theorem 2.9.1). Originally Steinhaus proved it (Steinhaus [297]) for measurable subsets of R, but then it was extended to more general settings (cf., e.g., Kurepa [200], Kuczma-Kuczma [194], Sander [274], M. E. Kuczma [192], [193], Paganoni [251]). Here we will present two different proofs of Theorem 3.7.1. The first one (Kuczma-Kuczma [194]) is based on an idea of H. Kestelman [169] and will use the following Lemma 3.7.1. If A, B, C ⊂ RN are measurable sets, m(A) < ∞, then |m(A ∩ B) − m(A ∩ C)|  m(B ÷ C) , where B ÷ C = (B \ C) ∪ (C \ B) is the symmetric difference of the sets B and C. Proof. We have m(A∩B)−m(A∩C) = m(A∩B ∩C)+m(A∩B ∩C  )−m(A∩B ∩C)−m(A∩B  ∩C) , whence |m(A ∩ B) − m(A ∩ C)|  m(A ∩ B ∩ C  ) + m(A ∩ B  ∩ C)  m(B ∩ C  ) + m(B  ∩ C) = m(B ÷ C) .



Theorem 3.7.1 (Theorem of Steinhaus). Let A, B ⊂ RN be arbitrary sets such that mi (A) > 0, mi (B) > 0. Then int(A + B) = ∅. Proof. Every set A such that mi (A) > 0 contains a compact subset A0 such that m(A0 ) > 0. So there exist compact sets A0 ⊂ A and B0 ⊂ B , m(A0 ) > 0 , m(B0 ) > 0. Since A0 +B0 ⊂ A+B, it is enough to prove the theorem for compact A, B of positive measure. Assume that A, B ⊂ RN are compact, m(A) > 0 , m(B) > 0. For every t ∈ RN put Bt = t − B. We define a function ω : RN → R as  ω(t) = m(A ∩ Bt ) =

χA (x)χBt (x) dx ,

(3.7.1)

RN

where χ denotes the characteristic function (cf. 3.2). First we prove that the function ω is continuous. Fix a t ∈ RN and take an ε > 0. By Theorem 3.1.1 there exists an open set Gt 1 such that Bt ⊂ Gt and m(Gt \ Bt ) < ε. Thus Bt ∩ Gt = ∅, and both these sets 2 being closed, their distance δ is a positive number. So for h ∈ RN with11 |h| < δ we

68

Chapter 3. Measure Theory

have Bt+h ⊂ Gt , whence m(Bt+h \ Bt )  m(Gt \ Bt ) <

1 ε. 2

Similarly, since Bt+h = Bt + h, we have by Corollary 3.2.2 m(Bt+h ) = m(Bt ), and m(Bt \ Bt+h )  m(Gt \ Bt+h ) = m(Gt ) − m(Bt+h ) 1 = m(Gt ) − m(Bt ) = m(Gt \ Bt ) < ε . 2 Hence m(Bt+h ÷ Bt ) = m(Bt+h \ Bt ) + m(Bt \ Bt+h ) < ε for |h| < δ . For such h we have by Lemma 3.7.1 |ω(t + h) − ω(t)| = |m(A ∩ Bt+h ) − m(A ∩ Bt )|  m(Bt+h ÷ Bt ) < ε . Thus ω is continuous at t, and t ∈ RN being arbitrary, ω is continuous in RN . Next we have by (3.7.1)          ω(t) dt = χA (x) χBt (x) dt dx = χA (x) χB (t − x) dt dx . RN

RN

RN

RN

RN

Making in the inner integral the change of variables u = t − x, we get hence      ω(t) dt = χA (x) χB (u) du dx RN

RN

=





RN

RN

χB (u) du





 χA (x) dx = m(B) m(A) > 0 .

RN

This shows that ω ≡ 0 in RN , and by (3.7.1) ω(t)  0 in RN . Consequently there exists a t0 ∈ RN such that ω(t0 ) > 0, and, by the continuity of ω, there  exists an open neighbourhood U of t0 such that ω(t) > 0 for t ∈ U . Thus cf. (3.7.1) m(A ∩ Bt ) > 0 for t ∈ U , in particular, A ∩ Bt = ∅ for t ∈ U . Thus for every t ∈ U there exists an a ∈ A such that a ∈ Bt , i.e., a = t − b with a b ∈ B. In other words, t = a + b ∈ A + B, whence U ⊂ A + B and int(A + B) = ∅.  The other proof is due to J. H. B. Kemperman [165]. It is based on the following Lemma 3.7.2. Let A, B ⊂ RN be measurable sets, and let A∗ , B ∗ be the sets of their density points. If a ∈ A∗ and b ∈ B ∗ , then there exist positive numbers δ and η such that for x, y ∈ RN such that |x − a| < δ , |y − b| < δ we have m [(A − x) ∩ (y − B)]  η . 11

(3.7.2)

|h| denotes the Euclidean norm of h. If h = (h1 , . . . , hN ), then |h| = (

N 

i=1

h2i )1/2 .

3.7. Theorem of Steinhaus

69

Proof. The reader will easily verify that 0 is a density point of the sets A − a and 1 B − b. Fix an ε , 0 < ε < . There exists a cube Q centered at the origin and such 2 that m(C)  (1 − ε) m(Q) , m(D)  (1 − ε) m(Q) , (3.7.3) where C = Q ∩ (A − a) , D = Q ∩ (B − b). We have C ⊂ A − a , D ⊂ B − b, whence C + a ⊂ A , D + b ⊂ B and C + a − x ⊂ A − x , y − (D + b) ⊂ y − B. Hence Q ∩ (C + a − x) ∩ (y − D − b) ⊂ (C + a − x) ∩ (y − D − b) ⊂ (A − x) ∩ (y − B) , and m [(A − x) ∩ (y − B)]  m [Q ∩ (C + a − x) ∩ (y − D − b)] .

(3.7.4)

Put Cx = C + a − x , Dy = y − D − b . Since (Cx ∩ Dy ) ∪ Cx ∪ Dy = RN , we have m(Q)  m(Q ∩ Cx ∩ Dy ) + m(Q ∩ Cx ) + m(Q ∩ Dy ) , and by (3.7.4) m [(A − x) ∩ (y − B)]  m(Q ∩ Cx ∩ Dy )  m(Q) − m(Q ∩ Cx ) − m(Q ∩ Dy ) . (3.7.5) Further, (Q ∩ Cx ) ∪ Cx = Q ∪ Cx = Q ∪ (Q ∩ Cx ), and similarly (Q ∩ Dy ) ∪ Dy = Q ∪ (Q ∩ Dy ), whence m(Q∩Cx )+m(Cx ) = m(Q)+m(Q ∩Cx ) , m(Q∩Dy )+m(Dy ) = m(Q)+m(Q ∩Dy ) , the sets in question being disjoint and measurable, whence by (3.7.3), since by Corollary 3.2.2 m(Cx ) = m(C) and m(Dy ) = m(D), m(Q ∩ Cx ) = m(Q) − m(Cx ) + m(Q ∩ Cx ) = m(Q) − m(C) + m(Q ∩ Cx )  εm(Q) + m(Q ∩ Cx ) , m(Q ∩ Dy ) = m(Q) − m(Dy ) + m(Q ∩ Dy ) = m(Q) − m(D) + m(Q ∩ Dy )  εm(Q) + m(Q ∩ Dy ) . The cube Q is symmetric with respect to the origin, i.e., −Q = Q. Thus we have C ⊂ Q , D ⊂ Q , −D ⊂ −Q ⊂ Q. Let Qx and Qy be the cubes concentric with Q and such that edge Qx = edge Q + 2 |a − x| , edge Qy = edge Q + 2 |b − y|. Then Q ∩ Cx ⊂ Qx \ Q , Q ∩ Dy ⊂ Qy \ Q, whence m(Q ∩ Cx )  m(Qx \ Q) = m(Qx ) − m(Q) , m(Q ∩ Dy )  m(Qy \ Q) = m(Qy ) − m(Q), and m(Q ∩ Cx )  εm(Q) + m(Qx ) − m(Q) , m(Q ∩ Dy )  εm(Q) + m(Qy ) − m(Q) .

70

Chapter 3. Measure Theory

The functions ϕ(x) = m(Qx ) − m(Q) , ψ(y) = m(Qy ) − m(Q) approach zero as x → a , y → b. We have by (3.7.5) m [(A − x) ∩ (y − B)]  (1 − 2ε)m(Q) − ϕ(x) − ψ(y) .

(3.7.6)

1 1 1 (1 − 2ε)m(Q). There exists a δ > 0 such that ϕ(x)  η and ψ(y)  η 2 2 2 whenever |x − a| < δ and |y − b| < δ. For such x, y (3.7.2) follows from (3.7.6). 

Put η =

Theorem 3.7.2. Let A, B ⊂ RN be measurable sets of positive measure, and let A∗ , B ∗ be sets of their density points. Then the set (A∩A∗ )+(B ∩B ∗ ) is open and non-empty. Proof. By the Lebesgue density theorem A ∩ A∗ = ∅ , B ∩ B ∗ = ∅, so also (A ∩ A∗ ) + (B ∩ B ∗ ) = ∅. Take a d ∈ (A ∩ A∗ ) + (B ∩ B ∗ ). Thus d = a + b, where, in particular, a ∈ A∗ , b ∈ B ∗ . Choose η and δ according to Lemma 3.7.2 and take a t ∈ RN such that |d − t| < δ. Put x = a, y = t− a. Then |x − a| = 0 < δ, and |y − b| = |t − a − b| = |t − d| < δ. By Lemma 3.7.2 m [(A − x) ∩ (y − B)]  η .

(3.7.7)

By Lebesgue density theorem the sets A and A ∩ A∗ differ only by a set of measure zero, and similarly the sets B and B∩B ∗ differ only by a set of measure zero. Therefore also the sets A ∩ (t − B) and (A ∩ A∗ ) ∩ [t − (B ∩ B ∗ )] differ only by a set of measure zero, and hence they have equal measure. Thus (3.7.7) implies in virtue of Corollary 3.2.2   m (A ∩ A∗ ) ∩ [t − (B ∩ B ∗ )] = m [A ∩ (t − B)] = m [(A − a) ∩ (t − B − a)] = m [(A − x) ∩ (y − B)]  η . Hence (A ∩ A∗ ) ∩ [t − (B ∩ B ∗ )] = ∅, i.e., there exists an a∗ ∈ A ∩ A∗ such that a∗ ∈ t− (B ∩B ∗ ), or a∗ = t− b∗ with a b∗ ∈ B ∩B ∗ . Thus t = a∗ + b∗ ∈ (A∩A∗ )+ (B ∩B ∗ ). The set { t ∈ RN | |d − t| < δ }

(3.7.8)

is a neighbourhood of d. The argument presented above shows that the set (3.7.8) is  contained in (A ∩ A∗ ) + (B ∩ B ∗ ). Hence (A ∩ A∗ ) + (B ∩ B ∗ ) is open. Theorem 3.7.1 is an easy consequence of Theorem 3.7.2. In fact, let A, B ⊂ RN be arbitrary sets such that mi (A) > 0 , mi (B) > 0. Similarly as previously we may assume that A, B are measurable and of positive measure. By Theorem 3.7.2 the set (A ∩ A∗ ) + (B ∩ B ∗ ) is open and non-empty, and clearly (A ∩ A∗ ) + (B ∩ B ∗ ) ⊂ A + B, and hence (A ∩ A∗ ) + (B ∩ B ∗ ) ⊂ int(A + B). Thus int(A + B) = ∅.

3.8. Non-measurable sets

71

3.8 Non-measurable sets We conclude this chapter with a construction, following P. R. Halmos [129], of an example of a non-measurable set and of a saturated non-measurable set. Lemma 3.8.1. Let ξ ∈ R be an irrational number. Then the sets D = { x ∈ R | x = p + qξ , p, q ∈ Z } , D1 = { x ∈ R | x = p + qξ , q ∈ Z , p ∈ 2Z, } , D2 = { x ∈ R | x = p + qξ , q ∈ Z , p ∈ 2Z + 1 } are dense in R. Proof. For every n ∈ N there exists a unique pn ∈ Z such that pn + nξ ∈ (0, 1). Fix 1 a non-empty open interval I ⊂ R. There exists a k ∈ N such that m(I) > . Choose k k + 1 distinct integers n1 , . . . , nk+1 ∈ N and corresponding pn1 , . . . , pnk+1 ∈ Z such that xi = pni + ni ξ ∈ (0, 1), i = 1, . . . , k + 1. The points xi are distinct, since ξ is irrational. So among the points xi there are at least two, say xr and xs , such that 1 |xr − xs | < . Thus for some l ∈ Z we have k l(xr − xs ) ∈ I . But l(xr − xs ) = l(pnr − pns ) + l(nr − ns )ξ ∈ D. Thus D ∩ I = ∅ for every non-empty open interval I ⊂ R, i.e., D is dense R. The proof for D1 is similar. It is enough to replace the interval (0, 1) by (0, 2). The proof for D2 follows from the equality D2 = D1 + 1.  Now we proceed to construct a non-measurable set. For any x, y ∈ R we write x ∼ y iff x − y ∈ D. The relation ∼ is an equivalence relation, so R may be split into a family of mutually disjoint equivalence classes; two numbers x, y belong to the same class if and only if x ∼ y. By Theorem 1.8.3 there exists a set B ⊂ R which has exactly one point in common with every such class. So any two different members b1 , b2 of B are in different classes, whence b1 − b2 ∈ D. In other words b1 − b2 ∈ D , b1 , b2 ∈ B, means b1 = b2 . Hence (B − B) ∩ D = {0}. Suppose that mi (B) > 0. By Corollary 3.2.2 mi (−B) = mi (B) > 0. By Theorem 3.7.1 int(B − B) = int [B + (−B)] = ∅. But then the set B − B must contain points of D other than 0, since by Lemma 3.7.1 D is dense in R. But as we have just seen, this is not the case. Consequently mi (B) = 0 .

(3.8.1)

Now suppose that me (B) = 0. By Corollary 3.2.2 me (B +d) = 0 for every d ∈ D. Since 0  mi (B + d)  me (B + d) = 0, this actually means (cf. Theorem 3.1.3) that all the sets B + d , d ∈ D, are measurable and of measure zero. Every x ∈ R belongs to an equivalence class with respect to ∼. If b is the only element of B lying in the same class, then x ∼ b, i.e., d = x − b ∈ D. Hence

72

Chapter 3. Measure Theory

x = b + d ∈ B + d. Consequently 

(B + d) = R .

(3.8.2)

d∈D

The union in (3.8.2) is countable (since D is countable), and thus R, as a union of countably many sets of measure zero, would be of measure zero, which is not the case. Thus we must have me (B) > 0 .

(3.8.3)

Let us note that the same argument shows that the set B cannot be of the first category, and consequently it is of the second category. By (3.8.1), (3.8.3), and Theorem 3.1.2, the set B is not Lebesgue measurable. Now we define a set A ⊂ R as A = B + D1 ,

(3.8.4)

where B is the non-measurable set constructed above, and D1 is defined in Lemma 3.7.1. Let x ∈ A − A and suppose that x ∈ D2 . Since x ∈ A − A, there exist b1 , b2 ∈ B and d1 , d2 ∈ D1 such that x = (b1 + d1 ) − (b2 + d2 ) = (b1 − b2 ) + (d1 − d2 ). Obviously D1 ⊂ D and D2 ⊂ D, whence d1 − d2 ∈ D, and, as we have just supposed, x ∈ D2 , whence x − (d1 − d2 ) ∈ D and b1 − b2 = x − (d1 − d2 ) ∈ D. But, as pointed out above, this implies that b1 = b2 , whence x = d1 − d2 . Now, d1 − d2 ∈ D1 , whereas x ∈ D2 , and evidently D1 ∩ D2 = ∅. This contradiction shows that (A − A) ∩ D2 = ∅ .

(3.8.5)

Theorem 3.8.1. The set A defined by (3.8.4) is saturated non-measurable. Proof. Relation (3.8.5) and Lemma 3.8.1 imply that int(A−A) = ∅, whence it follows from Theorem 3.7.1 that mi (A) = 0. The relation mi (A ) = 0 results from Theorem 3.6.1 in view of (3.8.3) and of Lemma 3.8.1.  Theorem 3.8.2. The set A defined by (3.8.4) has property (∗) from 3.3. Proof. Suppose that a set E ⊂ A is of the second category and with the Baire property. Then also −E is of the second category and with the Baire property, since the function f (x) = −x is a homeomorphism. We have E − E = E + (−E), and so by Theorem 2.9.1 int(E − E) = ∅. Since E ⊂ A, we have E − E ⊂ A − A, whence also int(A − A) = ∅, which contradicts (3.8.5). Thus A cannot contain any set of the second category and with the Baire property. Neither A can contain such a set, as results from Theorem 3.6.2, in view of (3.8.4), of the fact that (as pointed above) B is of the second category, and of Lemma 3.8.1. 

3.8. Non-measurable sets

73

Exercises 1. Show that if A ⊂ RN is a set such that mi (A) > 0, then there exists a compact set F ⊂ A such that 0 < m(F ) < ∞. 2. Let A1 ⊂ A2 ⊂ RN be arbitrary sets, mi (A2 ) < ∞. Show that there exist measurable sets K1 ⊂ A1 , K2 ⊂ A2 such that K1 ⊂ K2 , m(K1 ) = mi (A1 ), m(K2 ) = mi (A2 ). 3. Let An ⊂ RN , n ∈ N, be arbitrary sets such that An ⊂ An+1 , n ∈ N. Show that  ∞  me An = lim me (An ) . n=1

n→∞

4. Let An ⊂ RN , n ∈ N, be arbitrary sets such that An+1 ⊂ An , n ∈ N, mi (A1 ) < ∞. Show that  ∞  mi An = lim mi (An ) . n=1

n→∞

5. Show by an example that there exist sets A1 , A2 ⊂ RN such that A1 ∩ A2 = ∅ and mi (A1 ) + mi (A2 ) < mi (A1 ∪ A2 )  me (A1 ∪ A2 ) < me (A1 ) + me (A2 ). 6. For every set A ⊂ RN let A∗ denote the set of all points of outer density of A. Show that if A ⊂ RN is a saturated non-measurable set, and E ⊂ RN is a measurable set, then (E ∩ A)∗ = E ∗ . 7. Show that there exist disjoint measurable sets A, B ⊂ RN which are not Borel separable (cf. Exercise 2.8). [Hint: Take A to be a non-Borel subset of the Cantor set C , and B = C \ A]. 8. Show that there exists a decomposition RN = A∪B such that A, B ∈ L, A∩B = ∅, m(A) = 0, and B is of the first category. [Hint: Let D be a countable dense subset of RN . Thus m(D) = 0. For every n ∈ N there exists an open set Gn such that D ⊂ Gn , m(Gn ) < 2−n . Take ∞ A= Gn ]. n=1

9. Show that if f : (a, b) → R is a continuous real function of a single real variable, and f  (x) = 0 in (a, b) except for at most countably many points, then f is constant in (a, b). 10. Let A ⊂ R be symmetric with respect to zero (i.e., −A = A) and such that A + A = A. Show that either A = R, or A ∈ L and m(A) = 0, or A is saturated non-measurable. 11. Let A be a collection of mutually disjoint measurable subsets of RN of positive measure. Show that card A  ℵ0 . [Hint: Let Kr = K(0, r) denote the open ball centered at the origin and with radius r. Given a c > 0, let A(c) = { A ∈ A | m(A)  c }, and let Ar = { A ∈ A | m(A ∩ Kr ) > 0 }. Consider first the case where there exists an r > 0 such that ∞   A (1/n). In A ⊂ Kr . Show that, for every c > 0 , card A(c) < ℵ0 and A = n=1

the general case show that, for every r > 0, card Ar  ℵ0 and A =

∞ 

r=1

Ar ].

Chapter 4

Algebra 4.1 Linear independence and dependence Let F be a field (cf. 4.7), and let L be a set endowed with two operations: the addition of elements of L, and the multiplication of elements of L by elements of F such   that (L, +) is a commutative group i.e., fulfils conditions (2.9.1)–(2.9.4); cf. 4.5 , and moreover 1x = x for every x ∈ L, α(βx) = (αβ)x for every α, β ∈ F, x ∈ L, (α + β)x = αx + βx, α(x + y) = αx + αy for every α, β ∈ F, x, y ∈ L. Then the quadruple (L; F ; +; · ) is called a linear space over the field F . Often we say that L is a linear space (over F ), without making specific reference to the operations + and · . It follows from the group property of L that there exists an element 0 ∈ L such that x + 0 = 0 + x = x for every x ∈ L, and for every x ∈ L there exists an element −x ∈ L such that (−x) + x = x + (−x) = 0. Instead of writing x + (−y), we write simply x − y. Elements x1 , . . . , xn ∈ L are called linearly dependent (over F ) iff there exist α1 , . . . , αn ∈ F , not all zero, such that1 α1 x1 +· · ·+αn xn = 0. Elements x1 , . . . , xn ∈ L are called linearly independent (over F ) iff they are not linearly dependent, i.e., iff the equality α1 x1 + · · · + αn xn = 0 implies that α1 = · · · = αn = 0. It follows that a singleton x ∈ L is linearly dependent if and only if x = 0, and if a system x1 , . . . , xn ∈ L is linearly dependent, then so is also every system containing x1 , . . . , xn . (In particular, any linearly independent system cannot contain zero.) For if there exist α1 , . . . , αn ∈ F , not all zero, such that α1 x1 + · · · + αn xn = 0, 1 Every expression α x + · · · + α x , x ∈ L, α ∈ F , i = 1, . . . , n, is called the linear combination n n 1 1 i i of x1 , . . . , xn with the coefficients α1 , . . . , αn .

76

Chapter 4. Algebra

then for any y1 , . . . , yk ∈ L we have α1 x1 + · · · + αn xn + 0y1 + · · · + 0yk = 0, and α1 , . . . , αn , 0, . . . , 0 ∈ F are not all zeros. This last property gives rise to the following definition. A set B ⊂ L is called linearly independent (over F ) iff the elements of every finite subset of B are linearly independent. A set B ⊂ L is called linearly dependent (over F ) iff it is not linearly independent, i.e., iff B contains a finite subset whose elements are linearly dependent. An important example of a linear space is (RN ; R; +; · ) with customary addition of elements of RN and customary multiplication of elements of RN by real numbers. Actually, for every field F ⊂ R, the quadruple (RN ; F ; +; · ) is a linear space. In the sequel of this book an important role will be played by the linear space (RN ; Q; +; · ). If (L; F ; +; · ) is a linear space, and L0 ⊂ L is such that (L0 ; F ; +; · ) (with the same addition and multiplication by scalars) itself is a linear space, then L0 is called a subspace of L. If A ⊂ L is an arbitrary non-empty set, then by E(A) we denote the linear subspace of L spanned by A, i.e., the set2 E(A) = {x ∈ L | x = α1 a1 + · · · + αn an , α1 , . . . , αn ∈ F, a1 , . . . , an ∈ A, n ∈ N} of all the finite linear combinations of elements of A (with coefficients from F ). In other sources the set E(A) is also denoted by Span A, Lin A. The reader will easily check the following properties of the operation E. Here (L; F ; +; · ) is a linear space. (i) For every A ⊂ L we have A ⊂ E(A) ⊂ L. (ii) If A ⊂ B ⊂ L, then E(A) ⊂ E(B). (iii) E(L) = L. Further properties are contained in the following lemmas. Lemma 4.1.1. If (L; F ; +; · ) is a linear space and A ⊂ L is a linearly independent set, then every x ∈ E(A) has a representation, unique up to the terms with coefficients zero, x = α1 a1 + · · · + αn an , αi ∈ F, ai ∈ A, i = 1, . . . , n.

(4.1.1)

Proof. It follows from the definition of E(A) that every x ∈ E(A) has a representation (4.1.1). Suppose that x has also another representation x = β1 b1 + · · · + βm bm , βi ∈ F, bi ∈ A, i = 1, . . . , m.

(4.1.2)

We may assume that m = n and ai = bi , i = 1, . . . , n, adding in (4.1.1) and (4.1.2) the lacking terms with coefficients zero, and changing, if necessary, the numeration of ai ’s and bi ’s. Hence α1 a1 + · · · + αn an = β1 a1 + · · · + βn an , 2 We will write E (A) instead of E(A) if a question may arise which field is involved. From Chapter F 5 on E(A) will always mean EQ (A).

4.1. Linear independence and dependence

77

or (α1 − β1 )a1 + · · · + (αn − βn )an = 0.

(4.1.3)

Since A is linearly independent and {a1 , . . . , an } ⊂ A, (4.1.3) implies that αi − βi = 0, i.e., αi = βi , i = 1, . . . , n. Thus representations (4.1.1) and (4.1.2) may differ only by the added terms with coefficients zero.  Lemma 4.1.2. Let (L; F ; +; · ) be a linear space. If A ⊂ L is linearly independent, and a ∈ L\A, then the set A ∪ {a} is linearly independent if and only if a ∈ L\E(A). Proof. Suppose that A ∪ {a} is linearly dependent. Then there exists a finite set {a1 , . . . , an } ⊂ A ∪ {a} and αi ∈ F , i = 1, . . . , n, such that α1 a1 + · · · + αn an = 0. If we had ai ∈ A for i = 1, . . . , n, then A itself would be linearly dependent. So one of ai ’s, say a1 , equals a. If we had α1 = 0, then we are back in the previous situation. So α1 = 0 and α2 αn a = a1 = − a 2 − · · · − an , α1 α1 which means that a ∈ E(A). Now, if a ∈ E(A), then a = α1 a1 + · · · + αn an , αi ∈ F, ai ∈ A, i = 1, . . . , n, and a − α1 a1 − · · · − αn an = 0. Not all coefficients in the above linear combination are zeros (the coefficient of a is 1). Consequently the elements of the finite set {a, a1 , . . . , an } ⊂ A ∪ {a} are linearly dependent, and hence also A ∪ {a} is linearly dependent.  Lemma 4.1.3. Let A ⊂ RN be a finite or countable set. Then the set E(A) = EQ (A) is countable. Proof. For every n ∈ N, let Hn be the collection of all finite subsets K of A such that card K = n. If n > card A, we assume Hn = ∅. For K ∈ Hn the set E(K) may be identified with Qn , and hence card E(K) = ℵ0 . On the other hand, we have card Hn  (card A)n , or card Hn = 0 whenever Hn = ∅, so every Hn is finite or ∞  countable, and so is also H = Hn . For every set K ∈ H there exists an n ∈ N n=1

such that K ∈ Hn , whence card E(K)  ℵ0 , as observed above. The reader will easily check the formula  E(A) = E(K). (4.1.4) K∈H

Thus (4.1.4) is a finite or countable union of countable sets, and therefore E(A) is countable. 

78

Chapter 4. Algebra

Let (L; F ; +; · ) be a linear space over a field F ⊂ R. A set C ⊂ L is called a cone iff x + y ∈ C for all x, y ∈ C, and αx ∈ C for all x ∈ C, α ∈ F, α  0. For an arbitrary set A ⊂ L the symbol E + (A) will denote the cone spanned by A, i.e., the set E + (A) = { x ∈ L | x = α1 a1 + · · · + αn an , a1 , . . . , an ∈ A, α1 , . . . , αn ∈ F, α1  0, . . . , αn  0, n ∈ N} of all the finite linear combinations of elements of A with non-negative coefficients from F .

4.2 Bases Let (L; F ; +; · ) be a linear space. A set B ⊂ L is called a base of L iff (i) B is linearly independent; (ii) E(B) = L. Similarly, if C ⊂ L is a cone, then a set B ⊂ C is called a cone-base of C if B fulfils (i) and (iii) E + (B) = C. As we shall see later, only very few cones have a base, whereas we have the following Lemma 4.2.1. Let (L; F ; +; · ) be a linear space, and let A ⊂ L be a linearly independent set. Let C ⊂ L be an arbitrary set such that A ⊂ C. Then there exists a linearly independent set B such that A ⊂ B ⊂ C and E(B) = E(C). Proof. Let R be the family of sets R = { D ⊂ L | A ⊂ D ⊂ C and D is linearly independent }.  A ∈ R, so R = ∅. (R, ⊂) is an ordered set. Let Z ⊂ R be any chain, and put E = Z. Then E is linearly independent. In fact, let {a1 , . . . , an } ⊂ E be an arbitrary finite set. Then, since Z is a chain, there exists a set D0 ∈ Z such that {a1 , . . . , an } ⊂ D0 , and consequently a1 , . . . , an are linearly independent. Clearly A ⊂ E ⊂ C so that E ∈ R. Thus E is an upper bound of Z in R. By Theorem 1.8.1 there exists in R a maximal element B. Thus B is linearly independent, A ⊂ B ⊂ C. Hence E(B) ⊂ E(C). Suppose that E(B) = E(C). If we had C ⊂ E(B), then E(C) ⊂ E E(B) = E(B), and E(C) = E(B), which we have supposed untrue. Consequently there exists an a ∈ C \E(B) ⊂ L\B. By Lemma 4.1.2 the set B ∪ {a} is linearly independent, and, since a ∈ C, we have A ⊂ B ∪ {a} ⊂ C. Thus B ∪ {a} ∈ R and is strictly larger than B, which contradicts the fact that B is maximal in R. Consequently we must have E(B) = E(C).  Taking C = L we have hence Theorem 4.2.1. If (L; F ; +; · ) is a linear space, and A ⊂ L is a linearly independent set, then there exists a base B of L such that A ⊂ B.

4.2. Bases

79

Corollary 4.2.1. Every linear space L = {0} has a base. To see this take in Theorem 4.2.1 A = {a} with an arbitrary a ∈ L, a = 0. Two further important corollaries of Lemma 4.2.1 run as follows. Corollary 4.2.2. Let L = {0} be a linear space. Then every set C ⊂ L such that E(C) = L contains a base of L. Corollary 4.2.3. If (L; F ; +; · ) is a linear space, and L0 ⊂ L is a linear subspace of L, then for every base B0 of L0 there exists a base B of L such that B0 ⊂ B. In other words, every base of a subspace can be extended to a base of the entire space. In proving Lemma 4.2.1 we have used Theorem 1.8.1, and so the Axiom of Choice. As has been shown by J. D. Halpern [133], the statement in Lemma 4.2.1 is actually equivalent to the Axiom of Choice (cf. also L¨ auchli [205]). There is no uniqueness attached to bases of a linear space. One linear space has many different bases. But there is an important relation between all those bases. Theorem 4.2.2. Let (L; F ; +; · ) be a linear space. Then any two bases of L are equipollent. Proof. Let B ⊂ L and C ⊂ L be two bases of L. Consider the family Φ of all functions ϕ whose domain Dϕ is contained in B, range Rϕ is contained in C, the set Rϕ ∪ (B \ Dϕ ) is linearly independent3 , and ϕ is one-to-one. For ϕ1 , ϕ2 ∈ Φ we introduce the relation  : ϕ1  ϕ2 iff Dϕ1 ⊂ Dϕ2 and ϕ2 | Dϕ1 = ϕ1 (cf. Exercise 1.4). (Φ, ) is an ordered set. Fix an x0 ∈ B. We  have C \ E(B  \ {x0 }) = ∅. In fact, if we had C ⊂ E(B \ {x0 }), then L = E(C) ⊂ E E(B \ {x0 }) = E(B \ {x0 }) ⊂ E(B) = L, whence E(B \ {x0 }) = L. This implies that x0 ∈ E(B \ {x0 }), whence by Lemma 4.1.2 the set B = (B \ {x0 }) ∪ {x0 } would be linearly dependent. Consequently there exists a y0 ∈ C \ E(B \ {x0 }). We define a transform ϕ : {x0 } → {y0 } by putting ϕ(x0 ) = y0 . Then Dϕ = {x0 } ⊂ B, Rϕ = {y0 } ⊂ C, and Rϕ ∪ (B \ Dϕ ) = {y0 } ∪ (B \ {x0 }) is linearly independent in virtue of Lemma 4.1.2. Clearly ϕ is one-to-one. Thus ϕ ∈ Φ, i.e., Φ = ∅.  Now let Z ⊂ Φ be any chain. Put D = Dϕ and define a function ϕ0 : D → L ϕ∈Z

by putting ϕ0 (x) = ϕ(x) if x ∈ Dϕ , ϕ ∈ Z. This definition is correct, since if x ∈ Dϕ1 ∩ Dϕ2 , ϕ1 , ϕ2 ∈ Z, then either ϕ1  ϕ2 , or ϕ2  ϕ1 . Let, e.g., ϕ1  ϕ2 . This means that Dϕ1 ⊂ Dϕ2 and ϕ2 | Dϕ1 = ϕ1 so that ϕ1 (x) = ϕ2 (x).  Rϕ . Since for Clearly ϕ  ϕ0 for every ϕ ∈ Z. Further we have Rϕ0 = ϕ∈Z

ϕ ∈ Z we have Dϕ ⊂ B, Rϕ ⊂ C, also Dϕ0 ⊂ B, Rϕ0 ⊂ C. We are going to show that the set Rϕ0 ∪ (B \ Dϕ0 ) is linearly independent. Supposing the contrary, let {a1 , . . . , an } ⊂ Rϕ0 ∪(B\Dϕ0 ) be linearly dependent. There exists a k, 0  k  n, such that {a1 , . . . , ak } ⊂ Rϕ0 , {ak+1 , . . . , an } ⊂ B \ Dϕ0 . (If necessary, we may change the numeration of a1 , . . . , an ). Note that ϕ1  ϕ2 implies 3

In the present proof we understand under this fact also that Rϕ ∩ (B \ Dϕ ) = ∅.

80

Chapter 4. Algebra

Rϕ1 ⊂ Rϕ2 . Since Z is a chain, there exists a ϕ ∈ Z such that a1 , . . . , ak ∈ Rϕ . Since Dϕ ⊂ Dϕ0 , we have B \ Dϕ0 ⊂ B \ Dϕ , and ak+1 , . . . , an ∈ B \ Dϕ . Hence {a1 , . . . , an } ⊂ Rϕ ∪ (B \ Dϕ ), which is incompatible with the linear independence of the latter. It is easy to check that ϕ0 is one-to-one. Consequently ϕ0 ∈ Φ and is an upper bound of Z in Φ. By Theorem 1.8.1 there exists in Φ a maximal element ϕ. ˆ We will show that Dϕˆ = B. Suppose that Dϕˆ = B. If we had Rϕˆ = C, then taking an x0 ∈ B \ Dϕˆ = ∅ we would get that C ∪ {x0 } = Rϕˆ ∪ {x0 } would be linearly independent, which, according to Lemma 4.1.2, is impossible, since x0 ∈ E(C) = L. So Rϕˆ = C. There exists a y0 ∈ C \ Rϕˆ . We must distinguish two cases.   1. y0 ∈ / E Rϕˆ ∪ (B \ Dϕˆ ) . Take an x0 ∈ B \ Dϕˆ . Define a function ψ : Dϕˆ ∪ {x0 } → L by putting ψ(x) = ϕ(x) ˆ for x ∈ Dϕˆ , ψ(x0 ) = y0 . Then Dψ = Dϕˆ ∪ {x0 } ⊂ B, Rψ = Rϕˆ ∪ {y0 } ⊂ C. Further, since Dϕˆ ⊂ Dψ , Rψ ∪ (B \ Dψ ) ⊂ Rϕˆ ∪ {y0 } ∪ (B \ Dϕˆ ) = [Rϕˆ ∪ (B \ Dϕˆ )] ∪ {y0 }. By Lemma 4.1.2 the set [Rϕˆ ∪ (B \ Dϕˆ )] ∪ {y0 } is linearly independent, and hence also its subset Rψ ∪ (B \ Dψ ) is linearly independent. Since y0 ∈ / Rϕˆ , the function ψ is one-to-one. Hence ψ ∈ Φ and is strictly larger than ϕ, ˆ which contradicts the maximality of ϕ. ˆ   2. y0 ∈ E Rϕˆ ∪ (B \ Dϕˆ ) . Then y0 has a representation y0 =

k

i=1

αi ti +

n

αi si ,

(4.2.1)

i=k+1

αi ∈ F, i = 1, . . . , n; ti ∈ Rϕˆ , i = 1, . . . , k; si ∈ B \ Dϕˆ , i = k + 1, . . . , n. If we had αk+1 = · · · = αn = 0, then we would get by (4.2.1) y0 −

k

αi ti = 0, y0 ∈ C, ti ∈ Rϕˆ ⊂ C, i = 1, . . . , k,

i=1

i.e., C would be linearly dependent. So there must exist an i0 , k + 1  i0  n, such that αi0 = 0. Put x0 = si0 ∈ B \ Dϕˆ , and define a function ψ : Dϕˆ ∪ {x0 } → L as above ψ(x) = ϕ(x) ˆ for x ∈ Dϕˆ , ψ(x0 ) = y0 . As in case 1 Dψ ⊂ B, Rψ ⊂ C, ψ is one-to-one. Suppose that the set Rψ ∪ (B \ Dψ ) is linearly dependent. But Rϕˆ ∪ (B \ Dψ ) ⊂ Rϕˆ ∪ (B \ Dϕˆ ), and since the latter set is linearly independent, also its subset Rϕˆ ∪ (B \ Dψ ) is linearly independent. Since Rψ = Rϕˆ ∪ {y0 }, we have Rψ ∪ (B \ Dψ ) = [Rϕˆ ∪ (B \ Dψ )] ∪ {y0 },

4.2. Bases

81

whence by Lemma 4.1.2

  y0 ∈ E Rϕˆ ∪ (B \ Dψ ) . (4.2.2)   But x0 = si0 ∈ Dψ and x0 ∈ / Rϕˆ in representation (4.2.1) only ti ’s belong to Rϕˆ , so x0 ∈ / Rϕˆ ∪ (B \ Dψ ). By (4.2.2) y0 has a representation y0 =

m

γi bi ,

(4.2.3)

i=1

where bi ∈ Rϕˆ ∪ (B \ Dψ ), and so bi = x0 for i = 1, . . . , m. Thus the term αi0 x0 occurs in representation (4.2.1), but not in representation (4.2.3), which contradicts Lemma 4.1.1, since αi0 = 0. / Consequently the set Rψ ∪ (B \ Dψ ) is linearly independent, and since y0 ∈ Rϕˆ , the function ψ is one-to-one. Hence ψ ∈ Φ and is strictly larger that ϕ, ˆ which contradicts the maximality of ϕ. ˆ Thus we have proved that Dϕˆ = B. Thus there exists a one-to-one function ϕˆ : B → C, which proves that card B  card C. Since the role of B and C is symmetric, we have also card C  card B. Hence card B = card C 

which means that B and C are equipollent. Let (L; F ; +; · ) be a linear space, L = {0}. We define the dimension of L as dim L = card B,

(4.2.4)

where B is an arbitrary base of L. Theorem 4.2.2 shows that definition (4.2.4) is correct (is independent of the choice of the base B). We assume that if L = {0}, then dim L = 0 (for every field F ). The following lemma is an analogue of Lemma 4.1.3 for sets of higher cardinality. Lemma 4.2.2. card A.

4

Let A ⊂ RN be a set such that card A > ℵ0 . Then card EQ (A) =

Proof. Write E(A) = EQ (A), and card A = n. Let B ⊂ A be a base of E(A) so that E(B) = E(A) (cf. Corollary 4.2.2). Put m = card B. We have m > ℵ0 , for m  ℵ0 would imply by Lemma 4.1.3 card A  card E(A) = card E(B) = ℵ0 . For every n ∈ N ∞  let Hn be the collection of all subsets K of B such that card K = n. Put H = Hn . n=1

For every n ∈ N we have card Hn  mn = m, whence card H  ℵ0 m = m. On the other hand, for every K ∈ H we have card E(K) = ℵ0 . Since  E(K), E(B) = K∈H

we have card E(B)  m ℵ0 = m, and card E(A) = card E(B)  m. Since B ⊂ A ⊂ E(A), we get hence m  n  m, i.e., m = n. Hence card E(A) = m = n = card A.  4 Tarski

[314]

82

Chapter 4. Algebra

We will also need the following Lemma 4.2.3. Let {h1 , . . . , hN } ⊂ RN be a base of the space (RN ; R; +; · ). Then the set EQ ({h1 , . . . , hN }) is dense in RN . Proof. Take an x ∈ RN and an ε > 0. x may be written as x = x1 h1 + · · · + xN hN , xi ∈ R, i = 1, . . . , N. To every xi , i = 1, . . . , N , there exists an αi ∈ Q such that |xi − αi | < ε/N |hi |. Put y = α1 h1 + · · · + αN hN ∈ EQ ({h1 , . . . , hN }). Then x − y = (x1 − α1 )h1 + · · · + (xN − αN )hN , whence |x − y| 

N

|xi − αi | |hi | < ε.

i=1

Thus arbitrarily close to every x ∈ RN there is a y ∈ EQ ({h1 , . . . , hN }), which means that the latter set is dense in RN .  Lemma 4.2.4. Let L ⊂ RN be such that (L;Q;+; ·) is a linear subspace of (RN ;Q;+; ·). If me (L) > 0, or L is of the second category, then L is dense in RN . Proof. Put M = ER (L). Then me (M ) > 0 or M is of the second category, since L ⊂ M . Hence M = RN , for lower dimensional subspaces of RN over R (k-dimensional hyperplanes, k < N ) are of measure zero and nowhere dense, and hence of the first category. By Corollary 4.2.2 L contains a base B of RN over R. By Lemma 4.2.3 EQ (B) is dense in RN , and so is also L, since EQ (B) ⊂ EQ (L) = L.  Every base of the space (RN ; Q; +; · ) will be referred to as a Hamel5 basis6 . Theorem 4.2.3. Every Hamel basis has the power of continuum. Proof. Let H ⊂ RN be a Hamel basis, and let card H = n. By Lemma 4.2.2 and Lemma 4.1.3 n = card H = card EQ (H) = card RN = c Thus card H = c.



Consequently the dimension of the space (RN ; Q; +; · ) is c, whereas, as is well known, the space (RN ; R; +; · ) has dimension N . 5

After G. Hamel, who in a celebrated paper Hamel [134] first proved the existence of a base of the space (R; Q; +; · ). 6 Some authors use the term Hamel basis to denote any base of a linear space, as defined above, in distinction from other types of bases (e.g., a Schauder basis, etc.), where topology is involved. This use will not be followed in the present book.

4.3. Homomorphisms

83

4.3 Homomorphisms Let (L; F ; +; · ) and (M ; F ; +; · ) be linear spaces (over the same field F ). A mapping f : L → M is called a homomorphism iff f (x + y) = f (x) + f (y)

(4.3.1)

f (αx) = αf (x)

(4.3.2)

for arbitrary x, y ∈ L, and for arbitrary x ∈ L, α ∈ F . It follows from (4.3.1) by induction that   n n xi = f (xi ) f i=1

(4.3.3)

i=1

for arbitrary xi ∈ L, i = 1, . . . , n, and n ∈ N. The general construction of homomorphisms f : L → M is described by the following Theorem 4.3.1. Let (L; F ; +; · ) and (M ; F ; +; · ) be linear spaces, and let B ⊂ L be a base of L. Then for every function g : B → M there exists a unique homomorphism f : L → M such that f | B = g. Proof. Take an arbitrary x ∈ L. By Lemma 4.1.1 x has a representation (unique up to terms with coefficients zero) x=

n

αi bi , αi ∈ F, bi ∈ B, i = 1, . . . , n.

(4.3.4)

i=1

For such an x we define f (x) as7 f (x) =

n

αi g(bi ).

(4.3.5)

i=1

Thus the function f has been unambiguously defined in the whole of L. If x ∈ B, then representation (4.3.4) takes the form x = x, whence by (4.3.5) f (x) = g(x). Thus f | B = g. Now, besides point (4.3.4) take also a y ∈ L. y has a representation y=

n

βi bi , βi ∈ F, i = 1, . . . , n,

i=1 7

Note that adding in (4.3.4) terms with coefficients zero does not change x, nor f (x).

(4.3.6)

84

Chapter 4. Algebra

because, adding in representations (4.3.4) and (4.3.6) terms with coefficients zero, we may assume that the number of summands is the same in (4.3.4) and (4.3.6), and (4.3.4) and (4.3.6) contain the same bi ’s. Hence x+y =

n

(αi + βi )bi .

i=1

By (4.3.5) f (x + y) =

n

(αi + βi )g(bi ) =

i=1

n

αi g(bi ) +

i=1

n

βi g(bi ) = f (x) + f (y).

i=1

Thus f satisfies (4.3.1). Take an arbitrary α ∈ F . By (4.3.4) αx =

n

ααi bi ,

i=1

where ααi ∈ F . According to definition (4.3.5) f (αx) =

n

ααi g(bi ) = α

i=1

n

αi g(bi ) = αf (x).

i=1

Thus f satisfies also (4.3.2) and, consequently, is a homomorphism. Thus the desired homomorphism exists. To prove the uniqueness, let f : L → M be an arbitrary homomorphism such that f (b) = g(b) for b ∈ B.

(4.3.7)

Take an x ∈ L with a representation (4.3.4). We have by (4.3.3), (4.3.2) and (4.3.7) f (x) = f

n  i=1

n n n

 αi bi = f (αi bi ) = αi f (bi ) = αi g(bi ). i=1

i=1

i=1

Thus f must be given by (4.3.5), and so the function defined by (4.3.5) is the unique homomorphism f : L → M such that f | B = g.  Theorem 4.3.1 says that every function g : B → M can be extended onto L to a homomorphism f : L → M , and the extension is unique. We prove also the following theorem about the extensions of homomorphisms. Theorem 4.3.2. Let (L; F ; +; · ) and (M ; F ; +; · ) be linear spaces, and let L0 be a linear subspace of L. Then every homomorphism f0 : L0 → M can be extended onto L to a homomorphism f : L → M .

4.3. Homomorphisms

85

Proof. Let f0 : L0 → M be a homomorphism, and let B0 ⊂ L0 be a base of L0 . By Corollary 4.2.3 there exists a base B of L such that B0 ⊂ B. Put g(b) = f0 (b) for b ∈ B0 ,

(4.3.8)

and define the function g : B → M arbitrarily on B \ B0 . By Theorem 4.3.1 g can be extended onto L to a homomorphism f : L → M . Take an x ∈ L0 . Then x has a representation n

x= αi bi , αi ∈ F, bi ∈ B0 . i=1

We get by (4.3.8), since f and f0 are homomorphisms, f (x) = f

n 

n n n n

   αi bi = αi f (bi ) = αi g(bi ) = αi f0 (bi ) = f0 αi bi = f0 (x).

i=1

i=1

i=1

i=1

i=1

Thus f | L0 = f0 , i.e., f is an extension of f0 .



The next theorem is slightly more sophisticated. If f : L → M is a homomorphism, the kernel of f is defined as Ker f = f −1 (0) = { x ∈ L | f (x) = 0 }. Lemma 4.3.1. Let (L; F ; +; · ) and (M ; F ; +; · ) be linear spaces, and let f : L → M be a homomorphism. Then Ker f is a linear subspace of L. Proof. The thing to show is that for x, y ∈ Ker f and α, β ∈ F we have αx + βy ∈ Ker f . Now, x, y ∈ Ker f means f (x) = f (y) = 0, whence by (4.3.1) and (4.3.2) f (αx + βy) = f (αx) + f (βy) = αf (x) + βf (y) = 0, i.e., αx + βy ∈ Ker f .



Theorem 4.3.3. Let (L; F ; +; · ) and (M ; F ; +; · ) be linear spaces, and let L0 be a linear subspace of L. Let B0 be a base of L0 and B a base of L such that B0 ⊂ B. If dim M  card(B \ B0 ), then there exists a homomorphism f : L → M such that Ker f = L0 . Proof. Let C ⊂ M be a base of M . We have card(B \ B0 )  dim M = card C, so there exists a one-to-one mapping g : B \ B0 → C. Put g(b) = 0 for b ∈ B0 .

(4.3.9)

Thus g is a mapping from B into M , and by Theorem 4.3.1 there exists a homomorphism f : L → M such that f | B = g.

86

Chapter 4. Algebra

Now take an x ∈ Ker f ⊂ L, i.e., f (x) = 0. Let x have a representation x=

k

n

αi ti +

i=1

αi si ,

(4.3.10)

i=k+1

where αi ∈ F , i = 1, . . . , n; ti ∈ B0 , i = 1, . . . , k; si ∈ B \ B0 , i = k + 1, . . . , n. Then 0 = f (x) =

k

n

αi g(ti ) +

i=1

αi g(si ).

i=k+1

By (4.3.9) n

αi g(si ) = 0.

(4.3.11)

i=k+1

But since g | B\B0 is one-to-one, g(si ) are different members of C, and since the latter is linearly independent, elements g(sk+1 ), . . . , g(sn ) also are linearly independent. Thus (4.3.11) implies αi = 0, i = k + 1, . . . , n, whence (4.3.10) reduces to x=

k

αi ti , αi ∈ F, ti ∈ B0 , i = 1, . . . , k.

(4.3.12)

i=1

Hence x ∈ E(B0 ) = L0 , i.e., Ker f ⊂ L0 . On the other hand, let x ∈ L0 , and let x have a representation (4.3.12). Then f (x) =

k

i=1

αi f (ti ) =

k

αi g(ti ) = 0

i=1

by (4.3.9). Thus x ∈ Ker f , whence L0 ⊂ Ker f . Consequently L0 = Ker f .



Theorem 4.3.4. Let (L; F ; +; · ) and (M ; F ; +; · ) be linear spaces, and let L0 be a linear subspace of L. Let B0 be a base of L0 and B a base of L such that B0 ⊂ B. If dim M = card(B \ B0 ), then there exists a homomorphism f : L → M such that Ker f = L0 , f (L) = M . Proof. We argue as in the proof of Theorem 4.3.3. Let C ⊂ M be a base of M . Because of the equality card(B \ B0 ) = dim M = card C we may find a g : B \ B0 → C which is one-to-one and onto. If we define g on B0 by (4.3.9), then we can extend it onto L to a homomorphism f : L → M .

4.4. Cones

87

The relation Ker f = L0 is shown in the proof of Theorem 4.3.3. Clearly f (L) ⊂ M . Now take a y ∈ M . It has a representation y=

n

βi ci , βi ∈ F, ci ∈ C, i = 1, . . . , n.

(4.3.13)

i=1

Since g | B \ B0 is onto C, there exist bi ∈ B \ B0 , i = 1, . . . , n, such that g(bi ) = ci , i = 1, . . . , n. Put x=

n

(4.3.14)

βi bi ∈ E(B \ B0 ) ⊂ E(B) = L.

i=1

Then by (4.3.14) and (4.3.13), since f is a homomorphism, f (x) = f

n 



βi b i =

i=1

n

i=1

βi f (bi ) =

n

βi g(bi ) =

i=1

n

βi ci = y.

i=1

Thus y ∈ f (L), whence M ⊂ f (L), and ultimately f (L) = M . Thus f has the desired properties. 

4.4 Cones Let (L; F ; +; · ) and (M ; F ; +; · ) be linear spaces (over the same field F ⊂ R), and let C ⊂ L be a cone. Let a function f : C → M fulfil condition  (4.3.1) for arbitrary x, y ∈ C and hence also (4.3.3) for arbitrary x1 , . . . , xn ∈ C and (4.3.2) for arbitrary x ∈ C, α ∈ F , α  0. Such a function is called additive and positively homogenous. In the present section we are going to investigate the possibility of extending an additive and positively homogenous function onto L to a homomorphism from L into M (Kuczma [179]). Lemma 4.4.1. Let (L; F ; +; · ) and (M ; F ; +; · ) be linear spaces over a field F ⊂ R, and let C ⊂ L be a cone such that E(C) = L.

(4.4.1)

Further, let B ⊂ C be a base of L. Then any two additive and positively homogenous functions from C into M which coincide on B are identical on C. Proof. Let f : C → M and g : C → M be additive and positively homogenous functions such that f (b) = g(b) for b ∈ B. (4.4.2) Take an arbitrary x ∈ C. Then x has a representation x=

n

i=1

αi bi , αi ∈ F, bi ∈ B, i = 1, . . . , n.

(4.4.3)

88

Chapter 4. Algebra

Setting in (4.3.1) x = y = 0 we get f (0) = 2f (0), whence f (0) = 0, and similarly g(0) = 0. So f (0) = g(0). If x = 0, then some αi ’s in (4.4.3) must be different from zero, and hence α0 = max |αi | > 0. We put  = 1/2α0 . Then 1 + αi ∈ F and i

1 + αi > 0 for i = 1, . . . , n, and consequently n

bi + x =

n

i=1

(1 + αi )bi ∈ C,

i=1

since bi ∈ C, i = 1, . . . , n. Further   n n f bi + x = f (bi ) + f (x), f

 n

i=1



(1 + αi )bi

=

i=1

i=1 n

(1 + αi )f (bi ).

(4.4.4) (4.4.5)

i=1

Relations (4.4.4) and (4.4.5) imply f (x) =

n

αi f (bi ).

i=1

Similarly, g(x) =

n

αi g(bi ).

i=1

Condition (4.4.2) now implies that f (x) = g(x).



Theorem 4.4.1. Let (L; F ; +; · ) and (M ; F ; +; · ) be linear spaces over a field F ⊂ R, and let C ⊂ L be a cone fulfilling condition (4.4.1). Then every additive and positively homogenous function f : C → M can be uniquely extended onto L to a homomorphism f0 : L → M . Proof. In virtue of Corollary 4.2.2 and of condition (4.4.1) there exists a base B of L such that B ⊂ C. By Theorem 4.3.1 the function f | B can be uniquely extended onto L to a homomorphism f0 : L → M . Since f and f0 | C both are additive and positively homogenous, and they coincide on B, we have f0 | C = f by Lemma 4.4.1. The uniqueness of the extension also results from Lemma 4.4.1.  Condition (4.4.1) can be released, if we renounce from the uniqueness of the extension. Theorem 4.4.2. Let (L; F ; +; · ) and (M ; F ; +; · ) be linear spaces over a field F ⊂ R, and let C ⊂ L be a cone. Then every additive and positively homogenous function f : C → M can be extended onto L to a homomorphism f0 : L → M . Proof. By Theorem 4.4.1 there exists a unique homomorphism f1 : E(C) → M such that f1 | C = f . By Theorem 4.3.2 f1 can be extended onto L to a homomorphism f0 : L → M . f0 is the required extension of f . 

4.5. Groups and semigroups

89

4.5 Groups and semigroups As was pointed out in 2.9, a group (G, + ) is a non-empty set G endowed with a binary inner operation8 + satisfying condition (2.9.1), (2.9.2) and (2.9.3). If, moreover, (G, + ) satisfies also condition (2.9.4), then it is said to be commutative, or abelian. A semigroup (S, + ) is a non-empty set S endowed with a binary inner operation + satisfying conditions (2.9.1). Of course, every group is, in particular, a semigroup. If (G, + ) is a group, and ∅ = H ⊂ G, and if (H, + ) (with the same operation +) is a group resp. semigroup, then (H, + ) is called a subgroup resp. subsemigroup of (G, + ). If (G, + ) is a group resp. semigroup, then the n-fold sum (n ∈ N) x + · · · + x is usually denoted by nx. Often, when speaking about a group (G, + ), we omit the reference to the operation in G, and say simply: the group G. Lemma 4.5.1. Let (G, + ) be a group, and ∅ = H ⊂ G. In order that (H, + ) be a group (a subgroup of G) it is necessary and sufficient that x−y ∈H

(4.5.1)

for every x, y ∈ H. Proof. Suppose that condition (4.5.1) is satisfied for every x, y ∈ H. The associativity of the operation + in H (i.e., condition (2.9.1) for all x, y, z ∈ H) results from H ⊂ G and from the fact that (2.9.1) holds for all x, y, z ∈ G. Taking an x ∈ H and y = x, we get from (4.5.1) 0 ∈ H, i.e., (2.9.2) is fulfilled in H. Finally, taking in (4.5.1) x = 0, we obtain that −y ∈ H for every y ∈ H. This means that every x ∈ H has in H the inverse element −x, i.e., condition (2.9.3) is fulfilled, too. Now, for arbitrary u, v ∈ H, also −v ∈ H, as has been just pointed out, and taking in (4.5.1) x = u, y = −v, we obtain u + v ∈ H, which means that + restricted to H is an inner operation. Conversely, if (H, + ) is a subgroup of (G, + ), then (4.5.1) results from (2.9.3) and from the fact that + is an inner operation in H.  Lemma 4.5.2. Let (G, + ) be a semigroup, and ∅ = S ⊂ G. In order that (S, + ) be a semigroup (a subsemigroup of G) it is necessary and sufficient that9 x+y ∈S

(4.5.2)

for every x, y ∈ S. Proof. Condition (4.5.2) guarantees that + is an inner operation in S. The associativity of + in S results from that in G, as in the proof of Lemma 4.5.1. Conversely, if (S, + ) is a semigroup, then (4.5.2) results from the fact that + is an inner operation in S.  8

Usually the operation in a group G is denoted by + only in the case of commutative groups. For non-commutative groups the multiplicative notation is employed. However, because of the connection with additive functions, in the present book we use the additive notation also in non-commutative cases. An operation in a set A is called binary iff it is defined for pairs of elements from A, and a binary operation in a set A is called inner, if the result of the operation performed on two elements from A is again an element of A. 9 Condition (4.5.2) can equivalently be written as S + S ⊂ S.

90

Chapter 4. Algebra

Lemma 4.5.3. Let (G, + ) be a group, and (H, + ) a subgroup of (G, + ). Then (H  = G \ H) H = −H, H  = −H  , H + H  = H  + H = H  . Proof. The relation H = −H results from the fact that (H, + ) is a group. Take an x ∈ H  . If we had −x ∈ H, then we would have x = −(−x) ∈ −H = H, a contradiction. Thus −x ∈ H  and −H  ⊂ H  . Hence H  = −(−H  ) ⊂ −H  and so −H  = H  . If x ∈ H and y ∈ H  , then x + y ∈ H  , since otherwise we would have y = (−x) + (x + y) ∈ −H + H = H + H = H. Similarly, y + x ∈ H  , since otherwise we would have y = (y + x) − x ∈ H − H = H + H = H. Thus H + H  ⊂ H  and H  + H ⊂ H  , whence10 H  ⊂ −H + H  = H + H  and H  ⊂ H  − H = H  + H. So H + H  = H  + H = H .  Let (G, + ) be an arbitrary group, and (H, + ) a subgroup of (G, + ), H = G. We say that the index of H is 2 iff x−y ∈ H for every x, y ∈ H  (i.e., iff H  −H  ⊂ H). Lemma 4.5.4. Let (G, + ) be a group, and (H, + ) a subgroup of (G, + ), H = G. Then the index of H is 2 if and only if H  + H  ⊂ H.

(4.5.3)

Proof. Let the index of H be 2. Take arbitrary x, y ∈ H  , and suppose that x+y ∈ H  . Then x = (x + y) − y ∈ H  − H  ⊂ H, since the index of H is 2. This contradiction shows that x + y ∈ H, i.e., (4.5.3) holds. Now assume that (4.5.3) holds. By Lemma 4.5.3 −H  = H  so that by (4.5.3)  H − H  = H  + H  ⊂ H  and the index of H is 2.  Corollary 4.5.1. Let (G, + ) be a group, and (H, + ) a subgroup of (G, + ), H = G. Then the index of H is 2 if and only if H  + H  = H. Proof. By Lemma 4.5.3, H + H  ⊂ H  , whence H ⊂ H  − H  = H  + H  . Hence H  + H  = H  if and only if (4.5.3) holds.  As an example we mention the additive groups of integers (Z, + ) and of even integers (2Z, + ), where + denotes the usual addition of numbers. (2Z, + ) is a subgroup of (Z, + ), and its index is 2. On the other hand, (Q, + ) and (R, + ) have no subgroups of index 2 (cf. Corollary 4.5.2 below). Let (G, + ) be a group, and n ∈ N. We say that the division by n is performable in G iff for every y ∈ G there exists a unique x ∈ G such that nx = y. In such a case this unique x is denoted by y/n. If the division by n is performable in G for every n ∈ N, then the group (G, + ) is called divisible. The additive groups (Q, + ), (R, + ), (C, + ), (RN , + ) (where + denotes the usual addition, as defined in the corresponding sets) are all divisible. 10 If A, B, C ⊂ G are arbitrary sets such that A + B ⊂ C, then A ⊂ A + 0 ⊂ A + B − B ⊂ C − B, and, similarly, B ⊂ 0 + B ⊂ −A + A + B ⊂ −A + C. On the other hand, in general it is not true that if A + B = C, then A = C − B.

4.5. Groups and semigroups

91

Lemma 4.5.5. Let (G, + ) be a group in which the division by 2 is performable. Then G has no subgroup of index 2. Proof. Every x ∈ G can be written as x = 2(x/2) ∈ 2G. So G ⊂ 2G. The converse inclusion is evident so that G = 2G. Now suppose that H = G is a subgroup of index 2. Then 2H ⊂ H, since (H, + ) is a group, and 2H  ⊂ H  + H  ⊂ H by (4.5.3). Now, G = 2G = 2(H ∪ H  ) = 2H ∪ 2H  ⊂ H, whereas H ⊂ G, since (H, + ) is a subgroup of (G, + ). Thus H = G, contrary to the supposition.  Corollary 4.5.2. No divisible group has a subgroup of index 2. Corollary 4.5.3. The group (RN , + ) has no subgroup of index 2. Let (G, + ) be a group, and (H, + ) a subgroup of (G, + ). We say that the subgroup (H, + ) of (G, + ) is normal iff −y + x+ y ∈ H for every x ∈ H, y ∈ G, i.e., iff −y + H + y ⊂ H

(4.5.4)

for every y ∈ G. Condition (4.5.4) can also be formulated as −y + H + y = H

(4.5.5)

for every y ∈ G. In fact, since in (4.5.4) y ∈ G is arbitrary, we may replace y by −y, arriving at y + H − y ⊂ H, i.e., H ⊂ −y + H + y (for every y in G), which together with (4.5.4) yields (4.5.5). Let (H, + ) be a group, and S ⊂ H an arbitrary set. We say that a subgroup (G, + ) of (H, + ) is generated by S iff (G, + ) is the smallest subgroup of (H, + ) containing S, i.e., for every subgroup (K, + ) of (H, + ) such that S ⊂ K we have G ⊂ K.

(4.5.6)

(In this connection cf. Exercise 4.8.) Note that a set S ⊂ H can generate only one ˜ + ) is another subgroup of (H, + ) generated by S, then subgroup of (H, + ). If (G, ˜ and G ˜ ⊂ G, which yields G ˜ = G. we must have, according to (4.5.6), G ⊂ G Theorem 4.5.1. Let (H, + ) be a commutative group, and (S, + ) a subsemigroup of (H, + ). Then the group G generated by S is G = S − S.

(4.5.7)

Proof. Suppose that (4.5.7) holds. For every x, y ∈ G there exist u, v, w, z ∈ S such that x = u − v, y = w − z. Hence, since S ⊂ H, and (H, + ) is commutative x − y = (u − v) − (w − z) = (u + z) − (v + w) ∈ S − S = G, since by Lemma 4.5.2 u + z ∈ S, v + w ∈ S. It follows by Lemma 4.5.1 that (G, + ) is a subgroup of (H, + ). Now take an arbitrary x ∈ S. By Lemma 4.5.2 2x ∈ S, and hence, by (4.5.7), x = 2x − x ∈ S − S = G.

92

Chapter 4. Algebra

Thus S ⊂ G. Let (K, + ) be an arbitrary subgroup of (H, + ) such that S ⊂ K. For every x, y ∈ S we have x − y ∈ K, i.e., S − S ⊂ K. By (4.5.7) we get hence (4.5.6). Consequently (G, + ) is the group generated by S.  Lemma 4.5.6. Let (G, + ) be a group, and let (S, + ) be a subsemigroup of (G, + ) such that for every x ∈ G, x = 0, either x ∈ S, or −x ∈ S (or both). Then S generates G and (4.5.7) holds. Proof. Take an arbitrary x ∈ G, x = 0. If x ∈ S, then also 2x ∈ S (since S is a semigroup), and x = 2x − x ∈ S − S. (4.5.8) If x ∈ / S, then −x ∈ S and −2x = 2(−x) ∈ S. Hence x = (−x) − (−2x) ∈ S − S.

(4.5.9)

Finally, with an arbitrary y ∈ S 0 = y − y ∈ S − S.

(4.5.10)

Relations (4.5.8)–(4.5.10) imply that G ⊂ S − S. On the other hand S − S ⊂ G, since S ⊂ G and (G, + ) is a group. Hence we get (4.5.7). If (K, + ) is an arbitrary subgroup of (H, + ) such that S ⊂ K, then G = S − S ⊂ K. Thus S generates G.  Let (S, + ) be a semigroup. We say that the left cancellation law holds in S iff z + x = z + y implies x = y for arbitrary x, y, z ∈ S. We say that the right cancellation law holds in S iff x + z = y + z implies x = y for arbitrary x, y, z ∈ S. We say that S is cancellative iff both, left and right cancellation laws hold in S. We say that S is left reversible iff (x + S) ∩ (y + S) = ∅ for arbitrary x, y ∈ S.  We quote here without proof the following Theorem of Øre [246] cf. also Rees [265], Dubreil [73], and also Clifford-Preston [47] . Theorem 4.5.2. Let (S, + ) be a semigroup. There exists a group (G, + ) such that (S, + ) is a subsemigroup of (G, + ), S generates G, and (4.5.7) holds, if and only if S is left reversible and cancellative. Let (G, + ) be a group, and let (H, + ) be a subgroup of (G, + ). We may introduce in G the equivalence relation11 x ∼ y iff − x + y ∈ H. 11

(4.5.11)

The verification of the fact that (4.5.11) actually is an equivalence relation is left to the reader.

4.5. Groups and semigroups

93

The equivalence classes with respect to relation (4.5.11) are called the left cosets12 with respect to H, and the coset generated by an x ∈ G is denoted by [x]: [x] = { y ∈ G | x ∼ y } = { y ∈ G | −x + y ∈ H } = { y ∈ G | y ∈ x + H } = x + H. (4.5.12) In particular, we have [0] = H. (4.5.13) The set of all left cosets in G with respect to H will be denoted by G/H. If (H, + ) is a normal subgroup of (G, + ), then in the set G/H we may define a binary operation as follows: [x] + [y] = [x + y].

(4.5.14)

We must show that definition (4.5.14) does not depend on the choice of x and y, i.e., that if [x] = [u] and [y] = [v], then [x + y] = [u + v]. But [x] = [u] and [y] = [v] means x ∼ u and y ∼ v, i.e., −x + u ∈ H, −y + v ∈ H. Put a = −x + u ∈ H. Thus u = a + x, and −(x + y) + (u + v) = −y − x + u + v = −y + a + y + (−y + v) ∈ −y + H + y + H = H + H ⊂ H. This means that x + y ∼ u + v, i.e., [x + y] = [u + v], which was to be shown. Theorem 4.5.3. Let (G, + ) be a group, and (H, + ) a normal subgroup of (G, + ). With the operation + defined by (4.5.14), (G/H, + ) is a group. Proof. If a, b, c ∈ G/H, then there exist x, y, z ∈ G such that a = [x], b = [y], c = [z]. Then (a + b) + c = ([x] + [y]) + [z] = [x + y] + [z] = [(x + y) + z] = [x + (y + z)] = [x] + [y + z] = [x] + ([y] + [z]) = a + (b + c). Consequently condition (2.9.1) is satisfied. Let a ∈ G/H be arbitrary, and x ∈ G be such that a = [x]. Then a + [0] = [x] + [0] = [x + 0] = [x] = a.   Thus [0] = H cf. (4.5.13) is the neutral element of (G/H, + ), which shows that condition (2.9.2) is satisfied. Finally, for every a ∈ G/H, we have (with x ∈ G such that a = [x]) a + [−x] = [x] + [−x] = [x − x] = [0], which shows that −a = [−x] is the inverse element to a. Consequently, also condition (2.9.3) is satisfied.  12

The right cosets are the equivalence classes with respect to the relation x ≈ y iff x − y ∈ H.

The left cosets and right cosets coincide if and only if the subgroup H is normal.

94

Chapter 4. Algebra

Let (G, + ) and (H, + ) be two arbitrary groups.13 A mapping f : G → H is called a homomorphism14 iff f (x + y) = f (x) + f (y)

(4.5.15)

holds for all x, y ∈ G. The set Ker f = f −1 (0) = { x ∈ G | f (x) = 0 } is called the kernel of f . Lemma 4.5.7. Let (G, + ) and (H, + ) be groups, and let f : G → H be a homomorphism. Put K = Ker f . Then (K, + ) is a normal subgroup of (G, + ).  Proof. Putting in (4.5.15) x = y = 0 we get f (0) = 0, whence putting in (4.5.15)  y = −x f (−x) = −f (x). Hence, replacing in (4.5.15) y by −y, f (x − y) = f (x) − f (y).

(4.5.16)

Now, if x, y ∈ K, then by (4.5.16) f (x − y) = f (x) − f (y) = 0 − 0 = 0, i.e., x − y ∈ K. In virtue of Lemma 4.5.1, (K, + ) is a subgroup of (G, + ). Take arbitrary x ∈ K, y ∈ G. Then f (−y + x + y) = f (−y) + f (x) + f (y) = f (−y) + f (y) = −f (y) + f (y) = 0. Thus −y + x + y ∈ K, which means that K is normal.



Lemma 4.5.8. Let (G, + ) and (H, + ) be groups, and let f : G → H be a homomorphism. Let K = Ker f . Then for every x ∈ G [x] = { y ∈ G | f (y) = f (x) },

(4.5.17)

where [x] denotes the elements of G/K containing x. Proof. Let y ∈ [x]. This means that x ∼ y, i.e., −x+y ∈ K. In other words f (−x+y) = 0. Now, since f is a homomorphism   f (y) = f x + (−x + y) = f (x) + f (−x + y) = f (x). (4.5.18) Thus y ∈ { y ∈ G | f (y) = f (x) }. Conversely, let f (y) = f (x). Then from (4.5.18) we get f (−x + y) = 0, whence −x + y ∈ K, i.e., x ∼ y and y ∈ [x]. Hence we get (4.5.17).  13

In particular, (H, + ) need not be a subgroup of (G, + ), and so + in G and in H may denote two quite unrelated operations. 14 The word homomorphism is used to denote several distinct notions: homomorphism of linear spaces, homomorphism of groups (and later of rings and fields). In fact, all these notions are particular cases of one more general notion: homomorphism of algebraic structures. We presume that the reader is familiar with this situation, and we give here the definitions only for the sake of completeness. No ambiguities can possibly arise.

4.6. Partitions of groups

95

Corollary 4.5.4. Let (G, + ) and (H, + ) be groups, and let f : G → H be a homomorphism. Let K = Ker f . Then, for arbitrary [x], [y] ∈ G/K, [x] = [y] if and only if f (x) = f (y). Let (G, + ) be a group, and let (H, + ) be an arbitrary subgroup of (G, + ). Let the mapping π : G → G/H be defined by π(x) = [x], x ∈ G

(4.5.19)

Mapping (4.5.19) is called the canonical projection, or (in the case, where the subgroup H is normal), the canonical homomorphism from G onto G/H. This last name is due to the following fact. Lemma 4.5.9. Let (G, + ) be a group, and let (H, + ) be a normal subgroup of (G, + ). Then the mapping π : G → G/H, defined by (4.5.19), is a homomorphism. Proof. Take arbitrary x, y ∈ G. By (4.5.19) and (4.5.14) π(x + y) = [x + y] = [x] + [y] = π(x) + π(y), which shows that π is a homomorphism.



Let (G, + ) be a group, and let  (H,  + ) be an arbitrary subgroup of it. Every mapping ξ : G/H → G such that π ξ(a) = a for every a ∈ G/H is called a lifting 15 from G/H into G. If we take an arbitrary set T ⊂ G which has exactly one element in common with every class in G/H, then the mapping ξ : G/H → G defined by ξ(a) = T ∩ a, a ∈ G/H,

(4.5.20)

is a lifting from G/H into G. Conversely, if ξ : G/H → G is a lifting, then the set T = ξ(G/H) has exactly one element in common with every class in G/H, and has property (4.5.20); the verification of these facts is left to the reader.

4.6 Partitions of groups Let A and T be arbitrary sets. Every family {At }t∈T of subsets of A satisfying the conditions  At = A (4.6.1) t∈T

and

At ∩ At = ∅ for t = t , t , t ∈ T,

(4.6.2)

is called a partition of A. A partition {Bs }s∈S of a set B ⊂ A is called a subpartition of a partition {At }t∈T of A iff the following condition is fulfilled: (i) For every s ∈ S there exists a t ∈ T such that Bs ⊂ At . It is easily seen that, by (4.6.2), this t ∈ T is unique. Hence we have the following Lemma 4.6.1. Let B ⊂ A be arbitrary non-empty sets, and let a partition {Bs }s∈S of B be a subpartition of a partition {At }t∈T of A. If, for any s ∈ S and t ∈ T , we have Bs ∩ At = ∅, then actually Bs ⊂ At . 15

There is no analogue of Lemma 4.5.9 for liftings.

96

Chapter 4. Algebra

Proof. By (i) there exists a t ∈ T such that Bs ⊂ At . Then evidently At ∩ At = ∅, whence by (4.6.2) t = t and At = At . Consequently Bs ⊂ At .  Let ∅ = B ⊂ A. A partition {Bs }s∈S of B which is a subpartition of a partition {At }t∈T of A is called semiselective iff the following condition is fulfilled: (ii) For every t ∈ T and s, s ∈ S, if Bs ⊂ At and Bs ⊂ At , then Bs = Bs . A partition {Bs }s∈S of B which is a subpartition of a partition {At }t∈T of A is called selective iff it is semiselective and (iii) For every t ∈ T there exists an s ∈ S such that Bs ⊂ At . Again, as a result of (ii), this s ∈ S is unique. Let (G, + ) be an arbitrary group, and let (H, + ) be a subgroup of (G, + ). Then clearly G/H (the family of the left cosets with respect to H) is a partition16 of G. The following three theorems are due to A. Grz¸a´slewicz [123]. Theorem 4.6.1. Let (G, + ) be a group, and (H, + ) its subgroup. Let (G0 , + ) be another subgroup of (G, + ), and let (H0 , + ) be a subgroup of (H, + ). Then H/H0 is a subpartition of G/G0 if and only if H0 ⊂ G0 . Proof. Let H0 ⊂ G0 . Let [x] be a member of H/H0 , and x a member of G/G0 generated by the same x ∈ H ⊂ G. Thus u ∈ [x] means −x+u ∈ H0 . But then also −x+u ∈ G0 , which means that u ∈ x. Consequently [x] ⊂ x, i.e., condition (i) is satisfied. Now suppose that for every x ∈ H there exists a y ∈ G such that [x] ⊂ y. In particular, [0] ⊂ y for a certain y ∈ G. But then 0, which clearly belongs to [0], belongs also to y, which means that 0 and y are equivalent (with respect to G0 ), or y = 0. Hence H0 = [0] ⊂ 0 = G0 , which ends the proof.  Theorem 4.6.2. Let (G, + ) be a group, and (H, + ) its subgroup. Let (G0 , + ) be another subgroup of (G, + ), and let (H0 , + ) be a subgroup of (H, + ). Then H/H0 is a semiselective subpartition of G/G0 if and only if H0 = H ∩ G0 .

(4.6.3)

Proof. Suppose that H/H0 is a semiselective subpartition of G/G0 . Similarly as in the proof of Theorem 4.6.1, the elements of H/H0 will be denoted by [ · ], whereas the elements of G/G0 will be denoted by  · . We clearly have H0 ⊂ H, and by Theorem 4.6.1 H0 ⊂ G0 , whence H0 ⊂ H ∩ G0 . (4.6.4) Now take an arbitrary x ∈ H ∩ G0 . Then, in particular, x ∈ H, so x ∈ [x], and, on the other hand x ∈ G0 = 0. Thus [x]  ∩ 0 = ∅,  and, by Lemma 4.6.1, [x] ⊂ 0. Further, [0] = H0 ⊂ H ∩ G0 ⊂ G0 = 0 cf. (4.6.4) . Condition (ii) then implies that [x] = [0] = H0 , i.e., x ∈ H0 . Hence H ∩ G0 ⊂ H0 , which together with (4.6.4) implies (4.6.3). However, G/H cannot be indexed as {[x]}x∈G , for then it could happen that [x] = [x ] for x = x . We may take as T an arbitrary set which has exactly one element in common with every class contained in G/H, and writing At = [t] consider G/H as {At }t∈T .

16

4.6. Partitions of groups

97

Now suppose that condition (4.6.3) holds. Hence H0 ⊂ G0 , and according to Theorem 4.6.1 H/H0 is a subpartition of G/G0 . Now take arbitrary x, y ∈ H and z ∈ G such that [x] ⊂ z and [y] ⊂ z. In order to prove (ii)  we need to show that then [x] = [y]. But [x] = x+H0 and [y] = y +H0 cf. (4.5.12) . Thus x+H0 ⊂ z and y + H0 ⊂ z. Since (H0 , + ) is a group, 0 ∈ H0 , and so, in particular, x ∈ z, and y ∈ z, which means that x and y are equivalent with respect to G0 , i.e., −x+y ∈ G0 . Since x, y ∈ H, also −x + y ∈ H, and we get by (4.6.3) −x + y ∈ H0 . Thus x and y are equivalent with respect to H0 , i.e., [x] = [y].  Theorem 4.6.3. Let (G, + ) be a group, and (H, + ) its subgroup. Let (G0 , + ) be another subgroup of (G, + ), and let (H0 , + ) be a subgroup of (H, + ). Then H/H0 is a selective subpartition of G/G0 if and only if relation (4.6.3) and the relation G = H + G0

(4.6.5)

hold. Proof. Let H/H0 be a selective subpartition of G/G0 . Then, in particular, H/H0 is a semiselective subpartition of G/G0 , and in virtue of Theorem 4.6.2 condition (4.6.3) holds. Since H ⊂ G, G0 ⊂ G, and (G, + ) is a group, we have H + G0 ⊂ G + G = G.

(4.6.6)

In order to prove the converse inclusion, let z be an arbitrary element of G. We continue to employ the notation [ · ],  · , to denote elements of H/H0 and G/G0 , respectively. Thus we have z ∈ z. H/H0 is a selective subpartition of G/G0 , therefore by (iii) there exists an x ∈ H such that [x] ⊂ z. Since x ∈ [x], we have x ∈ z, i.e., x and z are equivalent with respect to G0 , which means that −x + z ∈ G0 . Put y = −x+z so that y ∈ G0 . We have z = x+y ∈ H +G0 . This shows that G ⊂ H +G0 , which together with (4.6.6) implies (4.6.5). Conversely, suppose that conditions (4.6.3) and (4.6.5) are fulfilled. By Theorem 4.6.2 H/H0 is a semiselective subpartition of G/G0 . Take an arbitrary class z ∈ G/G0 . Then z ∈ G, and by (4.6.5) there exist x ∈ H and y ∈ G0 such that z = x + y, whence −x + z =  y ∈ G0 , or −z + x ∈ −G0 = G0 (since G0 is a group). Thus x ∈ z + G0 = z cf. (4.5.12) . Hence x ∈ [x] ∩ z, whence [x] ∩ z = ∅. By Lemma 4.6.1 [x] ⊂ z, which means that for an arbitrary class z from G/G0 there exists a class [x] from H/H0 such that [x] ⊂ z. Consequently condition (iii) is fulfilled and H/H0 is a selective subpartition of G/G0 .  Example 4.6.1. Let G = Z + 2iZ be the set of all complex numbers p + iq such that p is an integer (p ∈ Z), and q is an even integer (q ∈ 2Z). Let G0 = 2Z + 2iZ be the set of all complex numbers p + iq such that both, p and q, are even integers (p, q ∈ 2Z). Let H = Z be the set of integers and let H0 = 2Z be the set of even integers. Endowed with the operation + of the usual addition of complex or real numbers, all these sets become groups. These groups fulfil conditions (4.6.3) and (4.6.5).

98

Chapter 4. Algebra

4.7 Rings and fields A set R = ∅ endowed with two  inner binary operations + and · is called a ring iff (R, + ) is a commutative group i.e., fulfils conditions (2.9.1)–(2.9.4) , and, moreover, (xy)z = x(yz)

(4.7.1)

(x + y)z = xz + yz, z(x + y) = zx + zy

(4.7.2)

for arbitrary x, y, z ∈ R, and

for arbitrary x, y, z ∈ R. If, moreover, xy = yx

(4.7.3)

for arbitrary x, y ∈ R, then the ring (R; +, · ) is called commutative. A commutative17 ring (F ; +, · ) is called a field iff (F \ {0}, · ) is a (commutative) group. In every field18 F there exists an element 1 ∈ F such that 1x = x1 = x

(4.7.4)

for every x ∈ F , and to every x ∈ F , x = 0, there exists an element x−1 ∈ F \ {0} such that xx−1 = x−1 x = 1. (4.7.5) Let (R; +, · ) be a ring. Every expression p(x1 , . . . , xm ) =

n



ai1 ...im xi11 . . . ximm , ai1 ...im ∈ R,

(4.7.6)

k=0 i1 +···+im =k ij ∈N∪{0}

is called a polynomial in variables x1 , . . . , xm . Every such polynomial can be considered as a function from Rm into R, for if we replace x1 , . . . , xm by concrete elements of R, and then calculate expression (4.7.6), then p(x1 , . . . , xm ) will become a concrete element of R. The collection of all polynomials (4.7.6) is denoted by R[x1 , . . . , xm ]. For m = 1 (4.7.6) becomes simply n

ak xk , (4.7.7) p(x) = k=0

and the collection of all such polynomials is denoted by R[x] . Let R be a subring of a ring P and let a ∈ P . We denote by R[a] the set of the values of all polynomials (4.7.7) for x = a, i.e., R[a] = { y ∈ P | y = p(a) =

n

ak ak , p ∈ R[x] },

(4.7.8)

k=0 17 Sometimes the commutativity of the operation · in a field is not required, but in the present book we are going to consider such fields only in which the operation · is commutative. 18 Sometimes, where there is no need to underline the operations in R resp. F , we say simply a ring R, or a field F , instead of a ring (R; +, · ), or a field (F ; +, · ).

4.7. Rings and fields

99

and similarly for polynomials in more variables. It is easily seen that if (R; +, · ) is a ring, then also (R[x]; + , · ) and (R[x1 , . . . , xm ]; + , · ), with the addition and multiplication of polynomials defined in the usual way, are rings. Similarly, if a is an element of P , then19 (R[a];  + , · ) is a ring.  If all ak in (4.7.7) or, more generally, all ai1 ...im in (4.7.6) are zeros, we write p p = 0 means that at least one of the coefficients in (4.7.7)  = 0. The condition  resp. (4.7.6) is different from zero. If p is a polynomial (4.7.7), then the polynomial p (x) =

n

kak xk−1

k=1

is called the derivative of the polynomial p. If in (4.7.7) an = 0, then   the number n is called the degree of p. Polynomials of degree zero p(x) = a0 = 0 are elements of R. For p = 0 we assume that its degree is −1. A structure intermediate between a ring and a field is an integral domain. An integral domain (R; +, · ) is a commutative ring with an element 1 ∈ R fulfilling (4.7.4) for all x ∈ R, and without divisors of zero, i.e., fulfilling, for arbitrary x, y ∈ R, the condition xy = 0 implies that either x = 0, or y = 0. (4.7.9) Every field is an integral domain. (If xy = 0, but x = 0, then x−1 xy = 0, whence y = 0). However, in a field, or in an integral domain, or even in a group, it may happen that for an element u = 0 we have u + u = 0, or, more generally, for a certain k ∈ N ku = 0.

(4.7.10)

If (4.7.10) holds, but lu = 0 for l = 1, . . . , k − 1, then we say that u is an element of order k. If such a k does not exist, we say that u is an element of order zero. In an integral domain (and so also in a field) all elements u = 0 have the same order. In fact, if u fulfils (4.7.10), and v = 0 is arbitrary, then u(kv) = (ku)v = 0, whence by (4.7.9), since u = 0, we get kv = 0. The same argument shows also that if lu = 0 for an l ∈ N, then also necessarily lv = 0 (for otherwise we would have lu = 0), and so u and v have the same order. The common order k of all u = 0 in an integral domain (or in a field) (R; +, · ) is called the characteristic of R: char R = k. If all elements u = 0 in R have order zero, then we say that R is of characteristic zero. 19 The operations in R[a] are those existing in P , since by (4.7.8) R[a] ⊂ P . So (R[a]; + , · ) is a subring of (P ; + , · ).

100

Chapter 4. Algebra

If (R; +, · ) is an integral domain, then we may associate with R a field (F ; +, · ), x the so called field of fractions of R. The elements of F are the expressions , x, y ∈ R, y u x = iff xv = yu. The operations (the y = 0, where it is to be understood that y v addition and multiplication) for such expressions are defined as for usual fractions   elements of Q, which is the field of fractions of the integral domain (Z; +, · ) , and the rules of calculations are the same. We do not go into details here. Let (F ; +, · ) be a field. It is not difficult to check that then, for every m ∈ N, the ring F [x1 , . . . , xm ] is an integral domain. The field of fractions of F [x1 , . . . , xm ] is denoted by F (x1 , . . . , xm ), and its elements are called rational functions (over F ). Thus a rational function is an expression of the form w(x1 , . . . , xm ) =

p(x1 , . . . , xm ) , p, q ∈ F [x1 , . . . , xm ], q = 0. q(x1 , . . . , xm )

(4.7.11)

(Of course, the polynomials p and q need not depend effectively on all variables x1 , . . . , xm ). Again, if we substitute for x1 , . . . , xm concrete elements a1 , . . . , am ∈ F such that q(a1 , . . . , am ) = 0, and calculate expression (4.7.11) with such a substitution, then we obtain an element of F : w(a1 , . . . , am ) =

p(a1 , . . . , am ) ∈ F. q(a1 , . . . , am )

Thus a rational function w may be considered as a function from a subset of F m into F . Lemma 4.7.1. Let (K; +, · ) be a field, and ∅ = F ⊂ K. Then (F ; +, · ) is a field if and only if for every x, y ∈ F x − y ∈ F, (4.7.12) and, if y = 0,

xy −1 ∈ F.

(4.7.13)

Proof. Suppose that conditions (4.7.12) and (4.7.13) are fulfilled. By Lemma 4.5.1 (F, + ) is a group, and, of course, (F, + ) is commutative, since (K, + ) is a commutative group. Conditions (4.7.1)–(4.7.3) hold for all x, y ∈ F , since they hold in K and F ⊂ K. It remains to check that (F \ {0}, · ) is a group. But this follows again from Lemma 4.5.1, in virtue of condition (4.7.13). Conversely, if (F ; +, · ) is a field, then conditions (4.7.12) and (4.7.13) are evidently fulfilled. (They result also from Lemma 4.5.1.)  Theorem 4.7.1. Let K = ∅ be an arbitrary non-empty collection of subfields of a field. Then K0 = K is a field. Proof. If x, y ∈ K0 , then x, y ∈ K for every K ∈ K. Hence also x − y ∈ K and, if y = 0, xy −1 ∈ K, for every K ∈ K. Consequently x − y ∈ K0 and xy −1 ∈ K0 . The theorem results now from Lemma 4.7.1. 

4.8. Algebraic independence and dependence

101

Let (K; +, · ) be a field, and let (F ; +, · ), F ⊂ K, be a subfield of (K; +, · ). Let A ⊂ K be an arbitrary set. Let K be the collection of all fields contained in K (subfields of K) and containing F ∪ A. By Theorem 4.7.1 the intersection K0 of all fields in K (K = ∅, since K ∈ K) is a field containing F ∪ A, and clearly is the smallest field with this property. In the sequel this field K0 will be denoted20 by F (A) and called the extension of F by the set A. If A = {a} is a singleton, then instead of F ({a}) we write simply F (a) and call the extension F (a) simple extension. Similarly, the extension of F by a finite set {a1 , . . . , am } will be denoted by F (a1 , . . . , am ). As is easily seen we have F (a1 , . . . , am ) = F (a1 )(a2 ) . . . (am ).

(4.7.14)

We assume that the reader is familiar with the main properties of rings, fields, and polynomials, and here we present only not-so-known properties and definitions, and also we recall some known facts and notions in order to refresh the reader’s memory and fix the terminology and notation.

4.8 Algebraic independence and dependence Let F , K be fields, F ⊂ K. The elements a1 , . . . , am ∈ K are called algebraically dependent (over F ) iff there exists a polynomial p ∈ F [x1 , . . . , xm ], p = 0, such that p(a1 , . . . , am ) = 0. If a1 , . . . , am are not algebraically dependent, they are called algebraically independent (over F ). Then p(a1 , . . . , am ) = 0 for every polynomial p ∈ F [x1 , . . . , xm ], p = 0. If a1 , . . . , am ∈ F , then they are algebraically dependent over F . One can take the polynomial p(x1 , . . . , xm ) =

m

(xi − ai ) ∈ F [x1 , . . . , xm ].

i=1

Of course, p = 0, and p(a1 , . . . , am ) = 0. The reader will verify that every part (every sub-system) of an algebraically independent system is algebraically independent. This gives rise to the following definition. An arbitrary set A ⊂ K is called algebraically independent (over F ) iff every finite system {a1 , . . . , am } ⊂ A is algebraically independent, i.e., iff p(a1 , . . . , am ) = 0 for every a1 , . . . , am ∈ A, m ∈ N, and for every p ∈ F [x1 , . . . , xm ], p = 0. A set A ⊂ K is called algebraically dependent (over F ) iff it is not algebraically independent (over F ), i.e., iff there exist a1 , . . . , am ∈ A and p ∈ F [x1 , . . . , xm ], p = 0, such that p(a1 , . . . , am ) = 0. The notion of the algebraic independence [dependence] extends 20 The operations in F (A) are those originally existing in K. Also in Theorem 4.7.1 it is understood that the operations in all the fields K ∈ K, and in K0 , are the same. In the sequel of this book, whenever two fields F and K occur, and F ⊂ K, it is understood that (F ; +, · ) is a subfield of (K; +, · ), i.e., the operations in F are the same as in K.

102

Chapter 4. Algebra

that of the linear independence [dependence], which is the special case, where the polynomial p is linear homogenous: p(x1 , . . . , xm ) =

m

αi xi , αi ∈ F, i = 1, . . . , m.

i=1

Let F ⊂ K be fields, and let S ⊂ K be an arbitrary non-empty set. Consider the set21 ∞    {w(t1 , . . . , tm )}, (4.8.1) F˜ = m=1 w∈F (x1 ,...,xm ) ti

where the third union extends over all t1 , . . . , tm ∈ S such that w(t1 , . . . , tm ) has sense22 . Lemma 4.8.1. Let F ⊂ K be fields, and let ∅ = S ⊂ K. Then with notation (4.8.1), F˜ = F (S). Proof. Let x ∈ F˜ . Then there exist an m ∈ N, a w ∈ F (x1 , . . . , xm ) and t1 , . . . , tm ∈ S such that x = w(t1 , . . . , tm ). (4.8.2) But since the coefficients in w belong to F ⊂ F (S), and t1 , . . . , tm ∈ S ⊂ F (S), and rational operations23 do not lead us out of a field, this implies that x ∈ F (S). Thus F˜ ⊂ F (S). Now observe that (F˜ ; +, · ) is a field. Indeed, if x, y ∈ F˜ , then x is given by (4.8.2) and y = u(s1 , . . . , sn ), with some m, n ∈ N, w ∈ F (x1 , . . . , xm ), u ∈ F (x1 , . . . , xn ), and t1 , . . . , tm , s1 , . . . , sn ∈ S. Moreover, if y = 0, then u = 0. Define rational functions v, z ∈ F (x1 , . . . , xm , xm+1 , . . . , xm+n ) as v(x1 , . . . , xm+n ) = w(x1 , . . . , xm ) − u(xm+1 , . . . , xm+n ), and (if u = 0) z(x1 , . . . , xm+n ) =

w(x1 , . . . , xm ) . u(xm+1 , . . . , xm+n )

Then x − y = v(t1 , . . . , tm , s1 , . . . , sn ) ∈ F˜ , and (if y = 0)

xy −1 = z(t1 , . . . , tm , s1 , . . . , sn ) ∈ F˜ .

Since F˜ ⊂ F (S), this implies in virtue of Lemma 4.7.1 that F˜ is a field. 21 Of course, since F ⊂ K, we have F [x , . . . , x ] ⊂ K[x , . . . , x ] and F (x , . . . , x ) ⊂ m m m 1 1 1 K(x1 , . . . , xm ). 22 I.e., the denominator of w(t , . . . , t ) does not vanish. If the set S is algebraically independent m 1 over F , then w(t1 , . . . , tm ) has sense for every t1 , . . . , tm ∈ S, and the third union in (4.8.1) extends over all t1 , . . . , tm ∈ S. 23 Addition, subtraction, multiplication, and division (by an element different from zero).

4.9. Algebraic and transcendental elements

103

The constant polynomials pa (x) = a, and q(x) = 1, as well as the polynomial r(x) = x, belong to F [x] for every a ∈ F . Hence the rational functions wa (x) =

r(x) pa (x) and u(x) = q(x) q(x)

belong to F (x). Now, for every a ∈ F and arbitrary t ∈ S we have a = wa (t) ∈ F˜ , t = u(t) ∈ F˜ , i.e., F ⊂ F˜ , S ⊂ F˜ . Thus F˜ is a field containing F ∪ S, and consequently it must contain the smallest field with these properties, i.e., F (S). Thus F (S) ⊂ F˜ , and consequently F˜ = F (S).  Corollary 4.8.1. Let F ⊂ K be fields, and let a1 , . . . , an ∈ K. Then for every b ∈ F (a1 , . . . , an ) there exists a w ∈ F (x1 , . . . , xn ) such that b = w(a1 , . . . , an ). Proof. By Lemma 4.8.1 b = w0 (t1 , . . . , tk ), where w0 ∈ F (x1 , . . . , xk ), and t1 , . . . , tk ∈ {a1 , . . . , an }. Every such w0 can be considered as a rational function w ∈ F (x1 , . . . , xn ) such that w(a1 , . . . , an ) = w0 (t1 , . . . , tk ).  Lemma 4.8.2. Let F ⊂ K be fields, and let ∅ = S ⊂ K. Let a1 , . . . , an ∈ F (S). Then there exists a finite set S1 ⊂ S such that a1 , . . . , an ∈ F (S1 ). Proof. By Lemma 4.8.1 there exist rational functions w1 ∈ F (x1 , . . . , xm1 ), . . . , wn ∈ F (x1 , . . . , xmn ), m1 , . . . , mn ∈ N, and points t11 , . . . , t1m1 , . . . , tn1 , . . . , tnmn ∈ S such that ai = wi (ti1 , . . . , timi ), i = 1, . . . , n. Then, again by Lemma 4.8.1, a1 , . . . , an ∈ F (S1 ), where S1 = { t11 , . . . , t1m1 , . . . , tn1 , . . . , tnmn } is a finite subset of S.



4.9 Algebraic and transcendental elements Let F ⊂ K be fields. An element t ∈ K is called algebraic (over F ) iff the singleton {t} forms an algebraically dependent system over F . If a t ∈ K is not algebraic (over F ), then it is called transcendental (over F ). Alternatively, a t ∈ K is algebraic (over F ) iff there exists a non-trivial (= 0) polynomial p ∈ F [x] such that p(t) = 0. This follows directly from the definition of the algebraic dependence. The algebraic closure of F (in K) is the set algcl F = { t ∈ K | t is algebraic over F }.

104

Chapter 4. Algebra

It can be shown that algcl F is a field (a subfield of K; cf. Exercise 4.15). The algebraic closure depends not only on F , but also on K. For example, algcl Q in C is larger than algcl Q in R. E.g., the imaginary unit i is an element of algcl Q in C, but not of algcl Q in R. As is well known, we have card algcl Q = ℵ0 so that R\algcl Q = ∅. If t ∈ K is algebraic over F , then among all the polynomials p ∈ F [x], p = 0, such that p(t) = 0 there are polynomials of the smallest degree. (This is a consequence of the fact that N is well ordered.) If p0 is such a polynomial and p0 (x) =

n

ak xk , an = 0,

k=0

1 p0 (x) also is such a polynomial, and has the coefficient of xn equal one. For an a given algebraic t ∈ K such a polynomial (of the smallest possible degree, with the coefficient of x in the highest power equal one and such that its value at the point x = t is zero) is unique, and is called the minimal polynomial of t. The degree of the minimal polynomial of t is called the degree of t. Algebraic elements of degree 1 are simply all the elements of F . In fact, if t ∈ F , then the polynomial p(x) = x − t belongs to F [x], p(t) = 0, and clearly p is the minimal polynomial of t. On the other hand, if t is algebraic of degree 1, then there exists a polynomial p(x) = x − c ∈ F [x] such that p(t) = 0. Thus t = c, and p ∈ F [x] implies that c ∈ F . Consequently t ∈ F . It may be shown that a p ∈ F [x] is the minimal polynomial of an algebraic   element t ∈ K if and only if p(t) = 0, p is irreducible24 in F [x] , and the coefficient of x in the highest power in p is one (cf. Exercise 4.11). then

Lemma 4.9.1. Let F ⊂ K be fields, and let a set S ⊂ K be algebraically independent over F . Let a ∈ K \ S. Then S ∪ {a} is algebraically independent if and only if a∈ / algcl F (S). Proof. Suppose that a ∈ / algcl F (S). Take an arbitrary finite system {a1 , . . . , am } ⊂ S ∪ {a}, and for an indirect proof suppose that the system {a1 , . . . , am } is algebraically dependent. Note that the system {a1 . . . , am } must contain a, for otherwise S would be algebraically dependent. Let, e.g., am = a. There exists a polynomial p ∈ F [x1 , . . . , xm ], p = 0, such that p(a1 , . . . , am ) = 0. Arrange the polynomial p according to the increasing powers of xm : p(x1 , . . . , xm ) =

n

pi (x1 , . . . , xm−1 )xim , pi ∈ F [x1 , . . . , xm−1 ], i = 0, . . . , n.

i=0

Thus

n

pi (a1 , . . . , am−1 )ai = 0.

(4.9.1)

i=0

  A polynomial p ∈ F [x] is called reducible in F [x], or over F iff there exist polynomials u, v ∈ F [x], of degrees smaller than that of p, and such that p = uv. If p is not reducible, it is said to be irreducible. 24

4.10. Algebraic bases

105

Since a is transcendental over F (S), (4.9.1) implies that pi (a1 , . . . , am−1 ) = 0, i = 0, . . . , n, and at least one of pi ’s is non-trivial, for otherwise we would have p = 0. So the system {a1 , . . . , am−1 } ⊂ S is algebraically dependent, which is incompatible with the fact that S is algebraically independent. Now assume that S ∪ {a} is algebraically independent, and for an indirect proof suppose that a ∈ algcl F (S). Consequently there exists a polynomial p ∈ F (S)[x], p = 0, such that p(a) = 0. Let p(x) =

n

ai xi , ai ∈ F (S), i = 0, . . . , n.

(4.9.2)

i=0

By Lemma 4.8.2 there exist t1 , . . . , tm ∈ S such that ai ∈ F (t1 , . . . , tm ), i = 0, . . . , n. According to Lemma 4.8.1 there exist rational functions w0 , . . . , wn ∈ F (x1 , . . . , xm ) such that25 ai = wi (t1 , . . . , tm ), i = 0, . . . , n. (4.9.3) Write every wi as wi =

Li , M, Li ∈ F [x1 , . . . , xm ], i = 0, . . . , n. M

(4.9.4)

(We can find a common denominator M for all wi , i = 0, . . . , n.) Not all Li are zero, for otherwise, by (4.9.2), (4.9.3) and (4.9.4), we would have p = 0. Inserting (4.9.3) and (4.9.4) into (4.9.2) we obtain n

Li (t1 , . . . , tm )ai = 0.

(4.9.5)

i=0

Define a polynomial P ∈ F [x1 , . . . , xm , xm+1 ] as P (x1 , . . . , xm , xm+1 ) =

n

Li (x1 , . . . , xm )xim+1 .

i=0

Then by (4.9.5) P (t1 , . . . , tm , a) = 0, i.e., the system {t1 , . . . , tm , a} ⊂ S ∪ {a} is algebraically dependent, which is incompatible with the algebraic independence of S ∪ {a}. 

4.10 Algebraic bases Let F ⊂ K be fields. A set S ⊂ K is called an algebraic base of K over F iff it fulfils the two conditions: (i) S is algebraically independent over F ; (ii) algcl F (S) = K. 25 We may assume all w belong to F (x , . . . , x ) (with the same m), because a rational function m 1 i of a smaller number of variables may always be regarded as a rational function of larger number of variables, which does not depend effectively on all the variables.

106

Chapter 4. Algebra

In order to prove the existence of algebraic bases, we prove first the following Lemma 4.10.1. Let F ⊂ K be fields. A set S ⊂ K is an algebraic base of K over F if and only if it is a maximal26 algebraically independent subset of K. Proof. Let S ⊂ K be an algebraic base of K over F . Suppose that there exists an algebraically independent set S1 ⊂ K such that S ⊂ S1 and S = S1 . Then there exists an a ∈ S1 \ S. The set S ∪ {a} ⊂ S1 is algebraically independent (as a subset of an algebraically independent set), and so, by Lemma 4.9.1, a ∈ / algcl F (S) = K, which is impossible. Conversely, suppose that S is a maximal algebraically independent subset of K, but is not an algebraic base of K over F . Being algebraically independent, S satisfies (i) above, so necessarily we must have algcl F (S) = K. Thus there exists an a ∈ K \ algcl F (S), and then, by Lemma 4.9.1, the set S1 = S ∪ {a} is algebraically independent, S ⊂ S1 , S = S1 . Thus S cannot be maximal.  Theorem 4.10.1. Let F ⊂ K be fields, and let K = algcl F . Then there exists an algebraic base of K over F . Proof. Let R be the family R = { S ⊂ K | S is algebraically independent over F }. Any singleton {t}, where t ∈ K \ algcl F (so that t is transcendental over F ) belongs to R, and so R = ∅. (R, ⊂ )is an ordered set. If Z ⊂ R is a chain, then, as the reader will easily verify, S0 = Z ∈ R is an upper bound for Z. By Theorem 1.8.1 R contains a maximal element Smax . By Lemma 4.10.1, Smax is an algebraic base of K over F . 

4.11 Simple extensions of fields In this section we prove some lemmas concerning simple extensions of a field. Let F ⊂ K be fields. Lemma 4.11.1. If F ⊂ K are fields, and a ∈ K, then  F (a) = {w(a)}, where the union extends over w ∈ F (x), whose denominator does not vanish at the point a. This is a particular case of Lemma 4.8.1. It is enough to observe that if w ∈ F (x1 . . . , xm ), then w(x, . . . , x) ∈ F (x). Corollary 4.11.1. If F ⊂ K are fields, and a ∈ K is transcendental over F , then  {w(a)}. F (a) = w∈F (x) 26

I.e., there does not exist an algebraically independent set S1 ⊂ K such that S ⊂ S1 and S = S1 .

4.11. Simple extensions of fields

107

Lemma 4.11.2. If F ⊂ K are fields, and a ∈ K is algebraic over F , then  {p(a)}. F (a) = F [a] = p∈F [x]

Proof. The inclusion F [a] ⊂ F (a) is clear. Now we show that F [a] is a field. Of course, F [a] ⊂ K. If b, c ∈ F [a], b = p(a), c = q(a), p, q ∈ F [x], then also r = p − q ∈ F [x], and b−c = p(a)−q(a) = r(a) ∈ F [a]. If, moreover, c = 0 and m ∈ F [x] is the minimal polynomial of a, then27 (q, m) = 1. It follows that there exist polynomials r, s ∈ F [x] such that (cf. Exercise 4.10) r(x) q(x) + s(x) m(x) = 1.

(4.11.1)

Putting in (4.11.1) x = a we obtain cr(a) = 1, i.e., c−1 = r(a) ∈ F [a], whence also bc−1 = p(a) r(a) ∈ F [a].  By Lemma 4.7.1 F [a] is a field.  Clearly F ⊂ F[a] any d ∈ F is the value at a of the constant polynomial p(x) = d , and a ∈ F [a] a is the value at a of the polynomial p(x) = x . F [a] must contain the smallest field with these properties, i.e., F (a) ⊂ F [a]. Hence F [a] = F (a).  If F ⊂ K are fields, then we can, in particular, add elements of K, and multiply elements of K by elements of F . Thus we may consider K as a linear space28 over F : (K; F ; +; · ). Lemma 4.11.3. If F ⊂ K are fields, and a ∈ K is algebraic over F, of a degree k,  then the system {1, a, a2 , . . . , ak−1 } forms a base of the linear space F (a); F ; +; · . Proof. If the system {1, a, . . . , ak−1 } were linearly dependent, then there would exist α0 , α1 , . . . , αk−1 ∈ F , not all zero, and such that α0 + α1 a + · · · + αk−1 ak−1 = 0.

(4.11.2)

Put p(x) = α0 + α1 x + · · · + αk−1 xk−1 . We have p ∈ F [x], p = 0. Relation (4.11.2) means that p(a) = 0, and since the degree of p is at most k − 1, the degree of a cannot exceed k − 1. This shows that the set B = {1, a, . . . , ak−1 } is linearly independent. Now take an arbitrary b ∈ F (a), b = 0. By Lemma 4.11.2 b ∈ F [a], i.e., there exists a p ∈ F [x] such that b = p(a). Let q ∈ F [x] be the minimal polynomial of a. There exist polynomials u, r ∈ F [x] such that p(x) = u(x) q(x) + r(x), 27 (q, m) denotes the greatest common divisor of q and m, i.e., the polynomial u ∈ F [x], normalized so that the coefficient of x in the highest power is 1, and such that u divides both, q and m, and if a polynomial v ∈ F [x] divides q and m, then v divides also u. If we had u = 1, then we would have q = ut and m = uv with certain t, v ∈ F [x]. Since q(a) = c = 0, we get u(a) = 0, and hence, since m(a) = 0, we get v(a) = 0. But since u = 1, the degree of v is smaller than that of m, and m would not be minimal. 28 The reader will verify that all the conditions from the definition of a linear space (cf. 4.1) are fulfilled.

108

Chapter 4. Algebra

and the degree of r  k−1, or else r = 0. The latter case is impossible, since otherwise we would have for x = a b = p(a) = u(a) q(a) = 0. Thus r(x) = α0 + α1 x + · · · + αn xn , αn = 0, n  k − 1, and b = p(a) = r(a) = α0 + α1 a + · · · + αn an ∈ E(B). Of course, 0 ∈ E(B) (take all the coefficients 0), and so F (a) ⊂ E(B). The converse inclusion is obvious, whence E(B) = F (a), and, as has already been  shown, B is linearly independent. This means that B is a base of F (a); F ; +; · .  Lemma 4.11.4. Let F ⊂ K be fields, let a ∈ K be algebraic over F , and let q ∈ F [x] be its minimal polynomial. If p1 , p2 ∈ F [x] and p1 (a) = p2 (a), then q divides p1 − p2 .   Proof. Let r = p1 − p2 so that r(a) = 0 , and let u = (q, r). Since q is a minimal polynomial (and hence irreducible), we must have either u = 1, or u = q. u = 1 is impossible, for otherwise there would exist s, t ∈ F [x] such that s(x) q(x)+t(x) r(x) = 1, and for x = a the left-hand side becomes zero. So u = q, and hence q divides r. 

4.12 Isomorphism of fields and rings Let (R1 ; +, · ) and (R2 ; +, · ) be rings. A mapping ϕ : R1 → R2 is called a homomorphism (of R1 into R2 ) iff ϕ(x + y) = ϕ(x) + ϕ(y)

(4.12.1)

ϕ(xy) = ϕ(x) ϕ(y)

(4.12.2)

for every x, y ∈ R1 , and for every x, y ∈ R1 . If, moreover, ϕ is one-to-one, then ϕ is called a monomorphism. If ϕ is onto   ϕ(R1 ) = R2 , then ϕ is called an epimorphism. A homomorphism which is a monomorphism and an epimorphism is called an isomorphism. A homomorphism ϕ : R → R (R2 = R1 = R) is called an endomorphism. An isomorphism ϕ : R → R is called an automorphism. R1 and R2 may be in particular (and the same terminology is applied) fields. An isomorphism between two algebraic structures states their complete similarity (from the algebraic point of view). All algebraic properties of one structure are also shared by the other. Note that if in (4.12.1) we put x = y = 0, we get ϕ(0) = 0,

(4.12.3)

and if in (4.12.2) we put x = y, and apply the mathematical induction, then we get ϕ(xi ) = ϕ(x)i , i ∈ N,

(4.12.4)

4.12. Isomorphism of fields and rings

109

for every homomorphism ϕ. Further, if ϕ is a monomorphism, and thus one-to-one, then by (4.12.3) ϕ(x) = 0 for x = 0, and setting in (4.12.2) x = y = 1 we obtain ϕ(1) = 1.

(4.12.5)

Let F1 , F2 be fields, and ϕ : F1 → F2 a homomorphism. We define a mapping Iϕ : F1 [x] → F2 [x]. If p(x) =

n

αi xi , αi ∈ F1 , i = 0, . . . , n,

(4.12.6)

i=0

then we put Iϕ (p)(x) =

n

ϕ(αi )xi .

(4.12.7)

i=0

Thus Iϕ (p) ∈ F2 [x]. We have the following Lemma 4.12.1. Let F1 , F2 be fields, and ϕ : F1 → F2 a homomorphism. Then the mapping Iϕ : F1 [x] → F2 [x] defined by (4.12.6) and (4.12.7) also is a homomorphism. If ϕ is an isomorphism, then Iϕ also is an isomorphism. Proof. Let q ∈ F1 [x], q(x) =

m

βj xj , βj ∈ F1 , j = 0, . . . , m,

j=0

be an arbitrary polynomial from F1 [x]. Consider polynomial (4.12.6). We may assume that, e.g., m  n. Then (p + q)(x) =

m n

(αj + βj )xj + αj xj , j=0

j=m+1

whence Iϕ (p + q)(x) =

m

ϕ(αj + βj )xj +

j=0

=

=

m

j=0 n

n

ϕ(αj )xj

j=m+1

ϕ(αj )xj + ϕ(αj )xj +

j=0

Similarly (pq)(x) =

m

j=0 m

ϕ(βj )xj +

n

j=m+1

ϕ(βj )xj = Iϕ (p)(x) + Iϕ (q)(x).

j=0



m+n

k=0

ϕ(αj )xj

i+j=k 0in 0jm

 αi βj xk ,

110

Chapter 4. Algebra

whence Iϕ (pq)(x) =

m+n

k=0

=

n 

m+n

    ϕ αi βj xk = ϕ(αi )ϕ(βj ) xk i+j=k 0in 0jm

ϕ(αi ) xi

i=0

k=0

m 

i+j=k 0in 0jm

 ϕ(βj ) xj = Iϕ (p)(x)Iϕ (q)(x).

j=0

Thus Iϕ is a homomorphism. Now let ϕ be an isomorphism. If r ∈ F2 [x], r(x) =

k

γi xi , γi ∈ F2 , i = 0, . . . , k,

i=0

then there is a unique p ∈ F1 [x] such that Iϕ (p) = r, viz. p(x) =

k

ϕ−1 (γi )xi .

i=0

It follows that Iϕ is one-to-one and onto. Consequently Iϕ is an isomorphism.



Lemma 4.12.2. Let F1 , F2 be fields, and ϕ : F1 → F2 a homomorphism. Then either ϕ = 0 in F1 , or ϕ is one-to-one, and hence a monomorphism. Proof. Suppose that ϕ is not one-to-one. Then there exist x1 , x2 ∈ F1 , x1 = x2 , such that ϕ(x1 ) = ϕ(x2 ). Put y = x1 − x2 = 0. Then x1 = y + x2 , whence by (4.12.1) ϕ(x1 ) = ϕ(y) + ϕ(x2 ), and ϕ(y) = 0. Thus we have by (4.12.2) for every x ∈ F1     x x = ϕ(y) ϕ = 0. ϕ(x) = ϕ y y y 

Consequently ϕ = 0 in F1 .

Now we are going to investigate the possibility of extending an isomorphism of fields to an isomorphism of their simple extensions. Let F1 , F2 , K1 , K2 be fields, F1 ⊂ K1 , F2 ⊂ K2 , and let ϕ : F1 → F2 be an isomorphism. We say that elements a ∈ K1 and A ∈ K2 are ϕ-conjugate iff either they are both transcendental over F1 and F2 , respectively, or they are both algebraic, and their minimal polynomials q and Q, respectively, are related in the following fashion Q = Iϕ (q).

(4.12.8)

If F ⊂ K are fields, then the identity map e(x) = x for x ∈ F

(4.12.9)

4.12. Isomorphism of fields and rings

111

is an automorphism of F . If elements a, A ∈ K are e-conjugate, then we say shortly that they are conjugate. Thus a, A ∈ K are conjugate iff either they are both transcendental over F , or they are both algebraic over F and have the same minimal polynomial. Two elements of F are conjugate if and only if they are equal. The “if” part is trivial. Now assume that a, A ∈ F are conjugate, and let q be their common minimal polynomial. Then q must have the form q(x) = x − c, whence, since q(a) = q(A) = 0, we get a = c and A = c. Thus a = A. Now we prove the following Theorem 4.12.1. Let F1 , F2 , K1 , K2 be fields, F1 ⊂ K1 , F2 ⊂ K2 , and let ϕ : F1 → F2 be an isomorphism. Let a ∈ K1 , A ∈ K2 , a = 0, A = 0. Then there exists an isomorphism Φ : F1 (a) → F2 (A) such that Φ(a) = A, Φ | F1 = ϕ, if and only if a and A are ϕ-conjugate. Proof. First we assume that the required isomorphism Φ exists. Let a be algebraic over F1 , and let q be its minimal polynomial. Then q is irreducible, and since, by Lemma 4.12.1, Iϕ is an isomorphism, the polynomial Q = Iϕ (q) also is irreducible (cf. Exercise 4.16). By (4.12.5) the coefficient of x in the highest power in Q is 1 (just like in q). If we show that Q(A) = 0, then it follows that A is algebraic over F2 , and Q is its minimal polynomial. (4.12.8) holds by the definition of Q. If we assume A to be algebraic, we argue in the same way (replacing Iϕ by Iϕ−1 ), since by Lemma 4.12.1 Iϕ is an isomorphism. Let n

αi xi . q(x) = i=0

Then Q(x) = Iϕ (q)(x) =

n

ϕ(αi )xi ,

i=0

whence, using (4.12.1), (4.12.2), (4.12.3), (4.12.4) and the facts that Φ(a) = A, Φ | F1 = ϕ, we obtain Q(A) =

n

ϕ(αi )Ai =

i=0

n

i=0

Φ(αi )Φ(a)i = Φ

n 

   αi ai = Φ q(a) = Φ(0) = 0.

i=0

Now assume that a and A are ϕ-conjugate, and let a and A be algebraic over F1 and F2 , respectively, their minimal polynomials being q and Q, respectively. Since by Lemma 4.11.2 F1 (a) = F1 [a], for every b ∈ F1 (a) there exists a polynomial p ∈ F1 [x] such that b = p(a). (4.12.10) For such a b we put Φ(b) = Iϕ (p)(A). (4.12.11)  To see that  this definition is correct i.e., it is independent of the choice of p fulfilling (4.12.10) suppose that p1 (a) = p2 (a) for some p1 , p2 ∈ F1 [x]. Then, by Lemma 4.11.4,

112

Chapter 4. Algebra

p1 − p2 = uq with a polynomial u ∈ F1 [x], whence, by Lemma 4.12.1 and (4.12.8), Iϕ (p1 )(A) − Iϕ (p2 )(A) = Iϕ (p1 − p2 )(A) = Iϕ (u)(A)Iϕ (q)(A) = Iϕ (u)(A)Q(A) = 0. Thus Φ is unambiguously defined. Take arbitrary u, v ∈ F1 (a) = F1 [a]. There exist polynomials r, s ∈ F1 [x] such that u = r(a), v = s(a), whence u + v = r(a) + s(a) = (r + s)(a), uv = r(a) s(a) = (rs)(a). By Lemma 4.12.1 Φ(u + v) = Iϕ (r + s)(A) = Iϕ (r)(A) + Iϕ (s)(A) = Φ(u) + Φ(v), and Φ(uv) = Iϕ (rs)(A) = Iϕ (r)(A)Iϕ (s)(A) = Φ(u)Φ(v). Thus Φ is a homomorphism. Take an arbitrary C ∈ F2 (A). By Lemma 4.11.2 there exists a polynomial P ∈ F2 [x] such that C = P (A). Put p = Iϕ−1 (P ) and c = p(a) ∈ F1 [a] = F1 (a). Then Φ(c) = Iϕ (p)(A) = P (A) = C. Consequently Φ is an epimorphism, and so we cannot have Φ = 0 in F1 (a). By Lemma 4.12.2 Φ is a monomorphism, and hence an isomorphism. For polynomial (4.12.9) we have e(a) = a, whence Φ(a) = Iϕ (e)(A) = A, whereas for the polynomial p(x) = α ∈ F1 we have Iϕ (p) = ϕ(α), whence Φ(α) = Iϕ (p)(A) = ϕ(α), i.e., Φ | F1 = ϕ. Thus Φ has all the required properties.  If a and A are ϕ-conjugate and transcendental, then by Corollary 4.11.1 F1 (a) = {w(a)}, and for every b ∈ F1 (a) there exist polynomials r, s ∈ F1 [x] such that w∈F1 (x)

b=

r(a) . s(a)

(4.12.12)

Iϕ (r)(A) . Iϕ (s)(A)

(4.12.13)

For such a b we put Φ(b) =

Again we must check that this definition is correct. Suppose that also b = u(a)/v(a) with u, v ∈ F1 [x]. Hence (rv − su)(a) = 0, rv − su ∈ F1 [x], and since a is a transcendental over F1 , we must have rv − su = 0, whence by Lemma 4.12.1 Iϕ (r)Iϕ (v) = Iϕ (s)Iϕ (u), and Iϕ (u)(A) Iϕ (r)(A) = . Iϕ (v)(A) Iϕ (s)(A) Thus Φ is well defined by (4.12.13). Similarly as in the case, where a and A are algebraic, it can be checked that Φ is an isomorphism.

4.12. Isomorphism of fields and rings

We have a =

113

e(a) , whence 1 Φ(a) =

Iϕ (e)(A) = A, Iϕ (1)(A)

and for every α ∈ F1 , taking p(x) = α, we have α = Φ(α) =

p(a) , whence 1

Iϕ (p)(A) = ϕ(α). Iϕ (1)(A)

Thus Φ has all the required properties.



Hence we derive the following result, which will be needed later. Theorem 4.12.2. Let F ⊂ K be fields, and let a, b, A, B ∈ K, abAB = 0. There exists an isomorphism ϕ : F (a, b) → F (A, B) such that ϕ(a) = A, ϕ(b) = B, ϕ | F = e if and only if a and A are conjugate and for a certain isomorphism ϕˆ : F (a) → F (A) such that ϕ(a) ˆ = A, ϕˆ | F = e, the elements b and B are ϕ-conjugate. ˆ Proof. The theorem follows by applying twice Theorem 4.12.1 according to (4.7.14). In the proof of the necessity one has to put ϕˆ = ϕ | F (a). 

Exercises 1. Let (L; F ; +; · ) be a linear space. Show that L0 ⊂ L is a linear subspace if and only if E(L0 ) = L0 . Let, moreover, F ⊂ R. Show that C ⊂ L is a cone if and only if E + (C) = C. 2. Let (L; F ; +; · ) be a linear space, let F ⊂ R, and let C ⊂ L be a cone such that C ∩ (−C) = {0}. Show that the relation  defined as x  y iff y − x ∈ C 3. 4. 5.

6.

is an order in L. Let (L; F ; +; · ) be a linear space, and let a set B ⊂ L be linearly independent. Suppose that B = B1 ∪ B2 , B1 ∩ B2 = ∅. Show that E(B1 ) ∩ E(B2 ) = {0}. Let (L; F ; +; · ) and (M ; F ; +; · ) be linear spaces, and let f : L →M be a homomorphism. Show that for every set A ⊂ L we have f E(A) = E f (A) . Let (L; F ; +; · ) and (M ; F ; +; · ) be linear spaces, and let f : L → M be a homomorphism. Show that f (L) is a linear subspace of M , and dim Ker f + dim f (L) = dim L. [Hint: Let B0 be a base of Ker f , and B a base of L such that B0 ⊂ B. Prove that f | B \ B0 is one-to-one and that f (B \ B0 ) is a base of f (L).] Let (G, + ) be a group, and (S, + ) its subsemigroup. Show that if S is finite, then (S, + ) is a subgroup of (G, + ) . [Hint: Let S = {a1 , . . . , an }. Take arbitrary a, b ∈ S. Then {a1 + a, . . . , an + a} is a set of n distinct elements of S, so one of them must equal b. Therefore the equation x + a = b has a unique solution x ∈ S.]

114

Chapter 4. Algebra

7. Let (G, + ) be a group, and (H, + ) its subgroup. Show that the index of H is 2 if and only if there exists a y0 ∈ G \ H such that x + y0 ∈ H for every x ∈ G \ H (Grz¸a´slewicz-Pow¸azka-Tabor [124]). [Hint: If u, v ∈ G \ H, then u + y0 ∈ H, v + y0 ∈ H, and u − v = (u + y0 ) − (v + y0 ) ∈ H.] 8. Let (H, + ) be a group. Show that for every family of subgroups of (H, + ) their intersection is a subgroup of (H, + ), and hence for every set S ⊂ H there exists the subgroup of (H, + ) generated by S. 9. Let (R; +, · ) be a ring. A principal ideal of R is every set of the form Ia = { x ∈ R | x = pa, p ∈ R }, where a ∈ R. Show that29 Ia + Ib = I(a,b) for arbitrary a, b ∈ R such that (a, b) exists. 10. Let (R; +, · ) be a ring, and let a, b ∈ R. Show that if (a, b) exists, then there exist r, s ∈ R such that (a, b) = ra + sb. [Hint: Use Exercise 4.9.] 11. Let F ⊂ K be fields, and let a ∈ K be algebraic over F . Show that p ∈ F [x] is the minimal polynomial of a if and only if p is irreducible (over F ), p(a) = 0, and the coefficient of x in the highest power in p is one. [Hint: If p is irreducible, and there exists a q ∈ F [x], of a degree smaller than that of p, such that q(a) = 0, then consider (p, q) ∈ F [x]. It follows that p must be reducible.] 12. Show that if F1 ⊂ F2 ⊂ K are fields, and (F2 ; F1 ; +; · ) has a finite dimension, then F2 ⊂ algcl F1 . [Hint: Let dim F2 = k ∈ N. For every a ∈ F2 the elements 1, a, . . . , ak are linearly dependent.] 13. Show that if F1 ⊂ F2 ⊂ F3 are fields and the dimensions of both the spaces (F2 ; F1 ; +; · ) and (F3 ; F2 ; +; · ) are finite, then also the dimension of the space (F3 ; F1 ; +; · ) is finite. 14. Show that if F ⊂ K are fields, and a, b ∈ K are algebraic over F , then F (a, b) ⊂ algcl F . 15. Show that if F ⊂ K are fields, then algcl F is a field. [Hint: For every a, b ∈ algcl F , consider the field F (a, b) ⊂ algcl F .] 16. Let F1 and F2 be fields, and let ϕ : F1 → F2 be an isomorphism. Show that a polynomial p ∈ F1 [x] is reducible if and only if Iϕ (p) ∈ F2 [x] is reducible. 17. A field F is called algebraically closed iff every polynomial p ∈ F [x] of a degree greater than one is reducible (over F ). Show that if a field F is algebraically closed, then for every field K ⊃ F we have algcl F = F . 18. Let F1 and F2 be fields, and let ϕ : F1 → F2 be an isomorphism. Show that F1 is algebraically closed if and only if F2 is algebraically closed. 19. Let (K; +, · ) be a field, let (F ; +, · ) be a subfield of (K; +, · ), and let S ⊂ K be algebraically independent over F . Show that if S1 ⊂ S and a ∈ S \ S1 , then a is transcendental over F (S1 ). 29

(a, b) denotes the greatest common divisor of a and b.

Part II

Cauchy’s Functional Equation and Jensen’s Inequality

Chapter 5

Additive Functions and Convex Functions 5.1 Convex sets Let C ⊂ R be an arbitrary set. A non-empty set A ⊂ RN is called C-convex iff λx + (1 − λ) y ∈ A for all x, y ∈ A and λ ∈ C ∩ [0, 1] .

(5.1.1)

In order to avoid the tiresome necessity of considering exceptional cases, we note the convention that the empty set ∅ is not C-convex for any C (cf. also Eggleston [74]). C-convex sets have been thoroughly investigated by J. W. Green–W. Gustin [120]. Some particular sets C will play a prominent role in the sequel, and we shall apply a special terminology in such cases. If C = R, then a set A fulfilling (5.1.1) is  1 called shortly convex 1 . If C = Q, then A is called Q-convex. If C = , then A is 2 called J-convex. Let D be the set of diadic numbers:  k D = x ∈ R | x = n , k ∈ Z, n ∈ N ∪ {0} . 2 Lemma 5.1.1. A set A ⊂ RN is J-convex if and only if it is D-convex. Proof. Both conditions of convexity imply that A = ∅. The “if” part of the lemma 1 ∈ D. Now suppose that A is J-convex. We have to show that is trivial, since 2 k λx + (1 − λ) y ∈ A for all x, y ∈ A and for every λ of the form λ = n , 0  k  2 1

The theory of convex sets is well developed and there exist many monographs of this subject. Cf., e.g., Bonnesen–Fenchel [35], Eggleston [74]. When λ runs over [0, 1], the point λx + (1 − λ) y runs over the segment joining x and y. Thus the geometrical meaning of convexity is: a non-empty set A ⊂ RN is convex iff together with arbitrary two its points it contains also the segment joining these points.

118

Chapter 5. Additive Functions and Convex Functions

2n , n ∈ N ∪ {0}. This is certainly true for n = 0 and n = 1. Now assume it to be true k 1 for an n  1 and consider a λ = n+1 . If k = 2n , then λ = , and we are back in the 2 2 case n = 1. So let k < 2n (if k > 2n we need only interchange the roles of x and y). Then k 2n − k 1 λx + (1 − λ) y = n+1 x + n+1 y + y. (5.1.2) 2 2 2 k 2n − k By the induction hypothesis the point n x + y belongs to A. But then (5.1.2) 2 2n may be written as 1 1 λx + (1 − λ) y = z + y, 2 2 and this is a point of A by the J-convexity of A.  Lemma 5.1.2. Suppose that we are given n + 1 non-negative real numbers α1 , . . . , αn+1 , n  2, such that n+1

αi = 1 (5.1.3) i=1

and αn+1  αi , i = 1, . . . , n.

(5.1.4)

Then the numbers α1 , . . . , αn can be partitioned into two non-empty disjoint classes 1 such that the sum of numbers in each class does not exceed . 2 Proof. Let us divide the numbers α1 , . . . , αn into two non-empty disjoint classes A and B in an arbitrary fashion. Suppose that the sum of numbers in one class, say B, 1 is greater than . Let 2

s= αi , αj = max αi . αi ∈A

αi ∈B

1 Observe that αj cannot be the only member of B, for otherwise we would have αj > 2 1 and by (5.1.4) αn+1 > , which is incompatible with (5.1.3). The sum of the elements 2 1 1 in B is, according to (5.1.3), 1 − αn+1 − s > , whence s + αn+1 < , and by (5.1.4), 2 2 1 s + αj < . Thus we may transfer αj from class B to class A, keeping still the sum of 2 1 the elements in A under . Continuing this procedure, after a finite number of steps 2 we arrive at the desired decomposition.  1 ∈ C, and let A ⊂ RN be a C-convex 2 set. Then for every n ∈ N, every a1 , . . . , an ∈ A, and every α1 , . . . , αn ∈ C ∩ [0, 1]

Lemma 5.1.3. Let C ⊂ R be a ring2 such that 2

It would be sufficient to assume that (C, +) is a group. But in Theorem 5.1.3 the assumption that (C; +, ·) is a ring is essential, and anyhow the only important cases are C = R, C = Q, and C = D.

5.1. Convex sets n 

such that

119

αi = 1 we have

i=1

α1 a1 + · · · + αn an ∈ A.

(5.1.5)

Proof. For n = 1 (5.1.5) is trivial, for n = 2 (5.1.5) is equivalent to (5.1.1). Assume (5.1.5) to be true for an n ∈ N, n  2, and consider a combination x0 = α1 a1 + · · · + αn+1 an+1 , where a1 , . . . , an+1 ∈ A, α1 , . . . , αn+1 ∈ C ∩ [0, 1], and (5.1.3) holds. Taking into account Lemma 5.1.2, we can change the numeration of a1 , . . . , an+1 (and, correspondingly, that of α1 , . . . , αn+1 ) in such a manner that (5.1.4) holds, and, for a certain k ∈ N, 1  k  n − 1, we have s1 =

k

i=1

αi 

1 , 2

s2 =

n

i=k+1

αi 

1 . 2

Put  1 − s1 an+1 , x = α1 a1 + · · · + αk ak + 0 · ak+1 + · · · + 0 · an−1 + 2   1 − s2 an+1 . y = 0 · a2 + · · · + 0 · ak + αk+1 ak+1 + · · · + αn an + 2 

1 ∈ C, By the induction hypothesis we have 2x ∈ A, 2y ∈ A, and hence, since 2 1  x0 = (2x + 2y) ∈ A. 2 The following two results are almost self-evident, so we leave the task of proving them to the reader. Theorem 5.1.1. Let a ∈ R, b ∈ RN , and let C ⊂ R be an arbitrary set. If a set A ⊂ RN is C-convex, then also the set aA + b is C-convex. Lemma 5.1.4. Let H ⊂ RN be a k-dimensional hyperplane, 1  k  N − 1. If a set A ⊂ RN is convex, then also the perpendicular projection of A onto H is convex. The next result, although equally simple, has a fundamental importance. Theorem 5.1.2. Let C ⊂ R be an arbitrary set, and let A be a non-empty collection A is either empty or of C-convex subsets of RN . Then the set A0 = A = A∈A

C-convex. Proof. Suppose that A0 = ∅, and take arbitrary x, y ∈ A0 , λ ∈ C ∩ [0, 1]. Then x, y ∈ A for every A ∈ A, whence also λx + (1 − λ) y ∈ A for every A ∈ A. Therefore λx + (1 − λ) y ∈ A0 , which shows that A0 is C-convex. 

120

Chapter 5. Additive Functions and Convex Functions

Let A ⊂ RN and C ⊂ R be arbitrary sets, A = ∅. Taking as A the family of all C-convex sets B ⊂ RN such that A ⊂ B (A = ∅, since RN ∈ A), we infer from N Theorem 5.1.2 that there exists the smallest C-convex subset of R containing A ( A = ∅, since A ⊂ A). This set is called the C-convex hull of A. For any non-empty set A ⊂ RN the convex hull (C = R) of A will be denoted by conv A  , the Q-convex hull (C = Q) of A will be denoted by Q (A), the J-convex  1  of A will be denoted by J (A). hull C = 2 Theorem 5.1.3. Let A ⊂ RN be an arbitrary set, A = ∅, and let C ⊂ R be a ring. Then the C-convex hull of A is the set B given by   n n

N B= x∈R |x= αi ai , αi ∈ C ∩ [0, 1] , ai ∈ A, i = 1, . . . , n, αi = 1; n ∈ N . i=1

i=1

(5.1.6) Proof. Let C (A) denote the C-convex hull of A. It is easily checked that the set B given by (5.1.6) is C-convex, whence C (A) ⊂ B. The inclusion B ⊂ C (A) results from Lemma 5.1.3. Hence C (A) = B.  Hence we derive a number of results, which will be needed only later. Lemma 5.1.5. For an arbitrary non-empty set A ⊂ RN and for an arbitrary a ∈ R we have Q (aA) = aQ (A) . (5.1.7) Proof. If x ∈ Q (aA), then by Theorem 5.1.3 there exist n ∈ N, α1 , . . . , αn ∈ Q ∩ [0, 1] n  and a1 , . . . , an ∈ A such that αi = 1 and i=1

x = α1 (aa1 ) + · · · + αn (aan ) = a (α1 a1 + · · · + αn an ) ∈ aQ (A) .   Conversely, if x ∈ aQ (A), then with suitable αi ∈ Q ∩ [0, 1] and ai ∈ A, i = 1, . . . , n x = a (α1 a1 + · · · + αn an ) = α1 (aa1 ) + · · · + αn (aan ) ∈ Q (aA) . 

Consequently (5.1.7) holds.

Corollary 5.1.1. If a set A ⊂ RN , A = ∅, is symmetric with respect to zero: A = −A, then so is also Q (A). Proof. By Lemma 5.1.5 −Q (A) = Q (−A) = Q (A).



Lemma 5.1.6. For arbitrary non-empty sets A, B ⊂ RN we have Q (A + B) = Q (A) + Q (B) .

(5.1.8)

5.1. Convex sets

121

Proof. By Theorem 5.1.3 if an x ∈ Q (A + B) then there exist n ∈ N, α1 , . . . , αn ∈ n  Q ∩ [0, 1], αi = 1, and a1 , . . . , an ∈ A, b1 , . . . , bn ∈ B such that i=1

x=

n

αi (ai + bi ) =

i=1

n

αi ai +

i=1

n

αi bi ∈ Q (A) + Q (B) .

i=1

On the other hand, if x ∈ Q (A) + Q (B), then x=

n

αi ai +

i=1

m

βj b j ,

j=1

with αi , βj ∈ Q, ai ∈ A, bj ∈ B, i = 1, . . . , n; j = 1, . . . , m,

n 

αi = 1,

i=1

m 

βi = 1. Let

j=1

d ∈ N be a common denominator of all αi ’s and βj ’s. Then x may be written as x=

d

k=1

γk ak +

d

k=1

γk bk =

d

γk (ak + bk ) ∈ Q (A + B) ,

k=1

where γk = 1/d, k = 1, . . . , d, and ak , bk equal to suitable ai , bj .



If a, b ∈ RN are arbitrary points, we write Q (a, b) instead of Q ({a, b}), and call Q (a, b) the rational segment joining the points a, b. It is easily seen from Theorem 5.1.3 that ! " Q(a, b) = x ∈ RN | x = λa + (1 − λ) b, λ ∈ Q ∩ [0, 1] , (5.1.9) for if in the combination x = α1 a1 + · · ·+ αn an several ai ’s equal a, and the remaining equal b, we can make the suitable reduction, denoting the sum of the coefficients of ai = a by λ. And if one of a, b is lacking in the combination, we can always add it with the coefficient zero. The set (a = b) ! " l(a, b) = x ∈ RN | x = λa + (1 − λ) b, λ ∈ Q (5.1.10) is called the rational line passing through the points a and b. If we take b = 0, (5.1.10) becomes " ! l(a, 0) = x ∈ RN | x = λa, λ ∈ Q . This is the general form of a rational line passing through the origin. Note that a = 0 (i.e., a is linearly independent over Q) so that we have l (a, 0) = E ({a}). Similarly, if a, b are linearly independent over Q, then the set E ({a, b}) is called the rational plane passing through the origin, a, and b. Generally, for arbitrary k ∈ N, if the points a1 , . . . , ak ∈ RN are linearly independent over Q, then the set E ({a1 , . . . , ak }) is called a k-dimensional rational hyperplane through the origin and the points a1 , . . . , ak . Note that for any k ∈ N it may happen that a k-dimensional rational hyperplane is entirely contained in a less dimensional (real) hyperplane. In particular, in R there exist kdimensional rational hyperplanes for every k ∈ N.

122

Chapter 5. Additive Functions and Convex Functions

One of the simplest examples of a convex set is a simplex. Suppose that we are given k + 1 (k  N ) points p1 , . . . , pk+1 ∈ RN which do not lie entirely on a (k − 1)dimensional (real) hyperplane. The convex hull S of the set {p1 , . . . , pk+1 } is called the simplex with vertices p1 , . . . , pk+1 , and we say, that the dimension of S is k. By Theorem 5.1.3 (cf. the remarks following formula (5.1.9)) S = conv {p1 , . . . , pk+1 }   k+1 k+1

αi pi , αi ∈ [0, 1] , i = 1, . . . , k + 1; αi = 1 . = x ∈ RN | x = i=1

i=1

Thus one-dimensional simplex is the segment joining the points p1 and p2 , the two dimensional simplex is the triangle (including its interior) with the vertices p1 , p2 , p3 , and so on. Lemma 5.1.7. Suppose that the points p1 , . . . , pN +1 ∈ RN do not lie entirely on an (N − 1)-dimensional hyperplane. Then the set   N +1 N +1

◦ N S = x∈R |x= αi pi , αi ∈ (0, 1) , i = 1, . . . , N + 1; αi = 1 i=1

i=1

is open. Proof. Take an x0 ∈ S ◦ , x0 =

N +1

α0i pi , α0i ∈ (0, 1) , i = 1, . . . , N + 1;

i=1

N +1

α0i = 1.

i=1

Consider the equalities 

x = α1 p1 + · · · + αN +1 pN +1 , 1 = α1 + · · · + αN +1 .

(5.1.11)

We may consider (5.1.11) as a system of N + 1 linear equations with N + 1 unknowns α1 , . . . , αN +1 . Since the points p1 , . . . , pN +1 do not lie on an (N − 1)-dimensional hyperplane, the determinant of the system is different from zero, and thus (5.1.11) is a Cramer system. Thus for every x ∈ RN (5.1.11) has a unique solution (α1 , . . . , αN +1 ), this solution  depends on x in a continuous manner. For x = x0 the solution is and α01 , . . . , α0N +1 , α0i ∈ (0, 1) , i = 1, . . . , N + 1. Therefore if x remains in a small neighbourhood U of x0 , the resulting α’s remain in (0, 1), and thus x ∈ S ◦ . Consequently U ⊂ S ◦ , which shows that S ◦ is open.  Corollary 5.1.2. If S ⊂ RN is an N -dimensional simplex, then int S = ∅. Proof. S being N -dimensional means that S = conv {p1 , . . . , pN +1 }, where the points p1 , . . . , pN +1 do not lie entirely on an (N − 1)-dimensional hyperplane. By Lemma 5.1.7 the set S ◦ is open, and clearly S ◦ ⊂ S, whence S ◦ ⊂ int S. Thus int S = ∅. 

5.1. Convex sets

123

Remark 5.1.1. If G ⊂ RN is a non-empty open set, then for every x ∈ G there exists an N -dimensional simplex S ⊂ G such that x ∈ int S. In fact, let {e1 , . . . , eN } be the usual orthonormal base of RN (over R). Consider the points pi = x + rei ,

i = 1, . . . , N,

pN +1 = x − r

N

ei ,

(5.1.12)

i=1

where r > 0 is so small that all points (5.1.12) lie in an open ball K centered at x and contained in G. Since K is convex, also the set S = conv {p1 , . . . , pN +1 } is contained in K, and hence in G. It is easy to check that points (5.1.12) do not lie on an (N − 1)-dimensional hyperplane, and thus S is an N -dimensional simplex. Since N +1 x= αpi , where α = 1/ (N + 1), x ∈ S ◦ ⊂ int S. i=1

Theorem 5.1.4. Let A ⊂ RN be a convex set, which does not lie entirely on an (N − 1)dimensional hyperplane. Then int A = ∅. Proof. Since A does not lie on an (N − 1)-dimensional hyperplane, it must contain N + 1 points p1 , . . . , pN +1 which do not lie on an (N − 1)-dimensional hyperplane. Let S be the N -dimensional simplex with the vertices p1 , . . . , pN +1 . Since A is convex, S ⊂ A, and by Corollary 5.1.2 we obtain int A = ∅.  Lemma 5.1.8. Let A ⊂ RN be a convex set, and let a ∈ int A, b ∈ A. Then, for every λ ∈ (0, 1), x = λa + (1 − λ) b ∈ int A. (5.1.13) Proof. Since a ∈ int A, there exists an r > 0 such that the open ball K = K (a, r) centered at a and with the radius r is contained in A: K ⊂ A.

(5.1.14)

Fix a λ ∈ (0, 1), and consider x given by (5.1.13). Let U = K (x, λr) be the open ball centered at x and with the radius λr. Take a y ∈ U . Then |y − x| < λr, i.e., |y − λa − (1 − λ) b| < λr, or # # # y − (1 − λ) b # # − a## < r. # λ By (5.1.14) y − (1 − λ) b ∈ A. λ Now we have y=λ

(5.1.15)

y − (1 − λ) b + (1 − λ) b, λ

which, in virtue of (5.1.15) and of the convexity of A, implies that y ∈ A. Hence U ⊂ A, which means that x is n interior point of A. 

124

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Corollary 5.1.3. Let A ⊂ RN be a convex set, and let a ∈ int A, b ∈ cl A. Then, for every λ ∈ (0, 1), (5.1.13) holds. Proof. Choose r > 0 so that K = K (a, r) fulfils (5.1.14). Fix a λ ∈ (0, 1), and consider points (5.1.13). We may find a y ∈ A such that |y − b| < r Put z =a−

|b − x| . |a − x|

|a − x| (y − b) . |b − x|

(5.1.16)

(5.1.17)

Then by (5.1.16) |z − a| =

|a − x| |y − b| < r, |b − x|

whence z ∈ K, and by (5.1.14) z ∈ A so that actually z ∈ int A. Now, since x = λa + (1 − λ) b = λ (a − b) + b = (1 − λ) (b − a) + a, we have λ=

|a − x| |b − x| , 1−λ= . |b − a| |b − a|

Thus (5.1.17) may be written as z =a−

1−λ (y − b) , λ

whence λa + (1 − λ) b = λz + (1 − λ) y, i.e., x = λz + (1 − λ) y. By Lemma 5.1.8 x ∈ int A.



Hence we derive some consequences pointing at a certain regularity of convex sets. Theorem 5.1.5. Let A ⊂ RN be a convex set such that int A = ∅. Then the set int A is convex3 . Proof. This is an immediate consequence of Lemma 5.1.8.



Theorem 5.1.6. Let A ⊂ RN be a convex set such that int A = ∅. Then clint A = cl A. Proof. Since int A ⊂ A, we have clint A ⊂ cl A. Now let b ∈ cl A, and choose an arbitrary a ∈ int A. By Corollary 5.1.3 we have (5.1.13) for every λ ∈ (0, 1). As λ → 0, the point x tends to b, whence b ∈ cl int A. Consequently cl A ⊂ cl int A, whence cl int A = cl A.  3

Cf. also Exercise 5.4 for a dual theorem.

5.1. Convex sets

125

A dual theorem is also true (cf. Exercise 5.5). Let H ⊂ RN be an (N − 1)-dimensional hyperplane. Then H separates RN into 1 2 two open half-spaces. These half-spaces will be denoted by πH and πH . So RN = 1 2 πH ∪ πH ∪ H, and the sets occurring in this decomposition are pairwise disjoint. Let A ⊂ RN , A = ∅, be an arbitrary set. An (N − 1)-dimensional hyperplane H is called a support hyperplane of A iff (i) H ∩ cl A = ∅; 1 2 (ii) A ∩ πH = ∅, or A ∩ πH = ∅. In other words, H is a support hyperplane of A iff H ∩ cl A = ∅ and A lies entirely on one side of H. Lemma 5.1.9. Let A ⊂ RN be a convex set such that int A = ∅. An (N − 1)dimensional hyperplane H is a support hyperplane of A if and only if H ∩ cl A = ∅ and H ∩ int A = ∅. Proof. Suppose that H is a support hyperplane of A. Then H satisfies (i)  2conditions  1 2 2 and (ii). Let, e.g., A∩πH = ∅. Then A ⊂ πH ∪H, whence int A ⊂ int πH ∪ H = πH , and H ∩ int A = ∅. Now suppose that H fulfils (i) and H ∩ int A = ∅. Put B = int A, and suppose 1 2 1 2 that B ∩πH = ∅ and B ∩πH = ∅. Then there exist points a ∈ B ∩πH and b ∈ B ∩πH . By Theorem 5.1.5 the set B is convex, so the whole segment joining a and b lies in B. But this segment must intersect H, whence we would get H ∩ B = ∅. Consequently, 1 2 1 either πH ∩ int A = ∅, or πH ∩ int A = ∅. Suppose that, e.g., πH ∩ int A = ∅. Then 2 2 int A ⊂ πH ∪ H, and since H ∩ int A = ∅, int A ⊂ πH . Hence, using Theorem 5.1.6, 2 2 1 we get A ⊂ cl A = clint A ⊂ cl πH = πH ∪ H. Hence A ∩ πH = ∅, and H fulfils (ii). Consequently H is support hyperplane of A.  Lemma 5.1.10. Let D ⊂ RN be a convex and open set, and let L0 be a linear subspace of RN (over R) of a dimension r, 0  r < N , such that L0 ∩ D = ∅. Then there exists a linear subspace L of RN , of the dimension N − 1, and such that L0 ⊂ L and L ∩ D = ∅. Proof. Let L be a linear subspace of RN , of the greatest possible dimension, such that L0 ⊂ L and L ∩ D = ∅. Let dim L = s so that r  s < N . The thing to prove is that s = N − 1. Let M be a linear subspace of RN perpendicular to L of the dimension N −s. Let B be the perpendicular projection of D onto M . The set B is convex (cf. Lemma 5.1.4) and is open in the relative topology of M . The perpendicular projection of L onto M is the point 0. Let l ⊂ M be a straight line passing through 0. If l∩B = ∅, then the linear subspace L1 of RN spanned by L∪l is disjoint with D, contains L0 , and dim L1 = s+1, which contradicts the fact that the dimension of L is maximal. Thus l ∩ B = ∅ for every line l in M passing through the origin. Of course, 0 ∈ / B, since L ∩ D = ∅. If N − s = 1, the lemma is proved. So suppose that N − s  2. Let P ⊂ M be a plane through the origin (a two-dimensional linear subspace of RN ). The set A = B ∩ P is not empty and hence, by Theorem 5.1.2, A is convex. A is also open in the relative topology of P . We have l ∩ A = ∅ for every line l ⊂ P passing through

126

Chapter 5. Additive Functions and Convex Functions

the origin. Moreover, 0 = A.

(5.1.18)

All rays starting from the origin and passing through the points of A form a sector C in P with vertex at 0. We have A ⊂ C.

(5.1.19)

Let α be the angle of C at the vertex. If α < π, then we may draw a line l in P such that l ∩ C = {0}, whence l ∩ A = ∅, which is impossible. Suppose that α = π. Then C is a half-plane determined by a line l ⊂ P passing through zero. But A ∩ l = ∅, so A must have a point on l, and being open in P , must have points on both the sides of l, which is incompatible with (5.1.19). Consequently α > π. Then C contains a half-plane P1 . Let P2 be the other half-plane of P . Since α > π, C ∩ P2 = ∅, and on every ray (from 0) contained in C ∩ P2 there are points of A. Choose an a ∈ A ∩ P2 . Draw a line l in P through a and 0. Then l ∩P1 is a ray in C, whence A∩(l ∩ P1 ) = ∅. Choose a b ∈ A ∩ P1 . Then the whole segment joining a and b is contained in A, since A is convex, hence 0 ∈ A, which contradicts (5.1.18). Consequently we must have N − s = 1, i.e., s = N − 1.  Now we are able to prove one of the fundamental results in the theory of convex sets. Theorem 5.1.7. Let A ⊂ RN be a convex set. Through every point of the frontier of A there passes a support hyperplane of A. Proof. Suppose that A lies on an (N − 1)-dimensional hyperplane H: A ⊂ H. Then H fulfils conditions (i) and (ii), and hence it is a support hyperplane of A. Now suppose that A does not lie on any (N − 1)-dimensional hyperplane. Then by Theorem 5.1.4 the set int A is non-empty, and hence, by Theorem 5.1.5, int A is a convex open set. Let p be a frontier point of A so that p ∈ cl A, but p ∈ / int A. Put D = int A − p. According to Theorem 5.1.1 D is a convex and open set, 0 ∈ / D. The set L0 = {0} is a 0-dimensional linear subspace of RN , and L0 ∩ D = ∅, since 0∈ / D. By Lemma 5.1.10 there exists an (N − 1)-dimensional linear subspace L of RN such that 0 ∈ L (which is true about every subspace of RN ), and L ∩ D = ∅. Then H = L + p is an (N − 1)-dimensional hyperplane, p ∈ H so that H ∩ cl A = ∅, and H ∩ int A = H ∩ (D + p) = ∅. By Lemma 5.1.9 H is a support hyperplane of A.  There is no uniqueness attached to the constructions in Lemma 5.1.10 and Theorem 5.1.7, and there is no uniqueness attached to support hyperplanes. A convex set may have many support hyperplanes passing through the same point. E.g., in the case N = 2 consider the set ! " A = (x, y) ∈ R2 | y > |x| . (5.1.20) It is easy to check that the set (5.1.20) is convex (cf. also Exercise 7.1), the origin is its frontier point, and every line y = ax with |a|  1 is a support line4 of A passing through the origin. 4 In the case where N = 2 (and thus (N − 1)-dimensional hyperplanes are simply lines) we say support line instead of support hyperplane.

5.1. Convex sets

127

The existence of support hyperplanes is a very characteristic feature of convex sets. It may be proved that if A ⊂ RN is a closed and bounded set such that int A = ∅, and A has a support hyperplane at every frontier point, then A is convex (Eggleston [74]). Lemma 5.1.11. Let A ⊂ RN be convex and closed set, and let x ∈ / A. Then there exists a support hyperplane H of A which separates x from A. Proof. Let r = d (x, A) > 0 be the distance from x to A, and let K be a closed ball centered at x and with a radius greater than r. Then d (x, A) = inf |x − y| = y∈A

inf

y∈A∩K

|x − y| ,

and since the set A ∩ K is compact (being closed and bounded), there is a point z ∈ A such that d (x, A) = |x − z| . (5.1.21) Let H be the (N − 1)-dimensional hyperplane passing through z and perpendicular to the segment xz. Let π denote that of the two open half-spaces into which H divides RN which contains x. Suppose that there exists a u ∈ A∩π. Then the angle Θ = ∠xzu is acute so that cos Θ > 0. (5.1.22) Consider the point v = z + λ (u − z) = λu + (1 − λ) z, with a λ ∈ (0, 1). Since z and u belong to A, and A is convex, also v belongs to A. Now, we have v−z = λ (u − z), whence x−v = (x − z)−(v − z) = (x − z)−λ (u − z), and |x − v|2 = |x − z|2 + λ2 |u − z|2 − 2λ (x − z) (u − z) 2

2

= |x − z| + λ2 |u − z| − 2λ |x − z| |u − z| cos Θ 2

= |x − z| + λ |u − z| (λ |u − z| − 2 |x − z| cos Θ) . If λ ∈ (0, 1) is small enough, we have by (5.1.22) λ |u − z| < 2 |x − z| cos Θ, whence |x − v| < |x − z|. But this contradicts (5.1.21), since v ∈ A. Consequently we must have A ∩ π = ∅, which means that H is a support hyperplane of A and that H separates x from A.  Theorem 5.1.8. Let A ⊂ RN , A = RN , be a convex and closed set5 . Then A is equal to the intersection of all the closed half-spaces containing A, determined by all the support hyperplanes of A. Proof. Let H be the collection of all the support hyperplanes of A. Since A = RN the frontier of A is non-empty, and in virtue of Theorem 5.1.7 also H = ∅. For every 5

Theorem remains true also for A = RN if we agree that



∅ = RN

128

Chapter 5. Additive Functions and Convex Functions

H ∈ H let ΠH denote that of the two closed subspaces into which H divides RN which contains A: A ⊂ ΠH for H ∈ H. (5.1.23) The thing to prove is that



A= By (5.1.23) we have A ⊂

ΠH .

(5.1.24)

H∈H



ΠH . Take an x ∈ / A. By Lemma 5.1.11 there exists an H0 ∈ H such that x ∈ / ΠH0 , whence x ∈ / ΠH . This shows that ΠH ⊂ A. H∈H

H∈H

Hence (5.1.24) follows.

H∈H



5.2 Additive functions A function6 f : RN → R is called additive iff it satisfies Cauchy’s functional equation f (x + y) = f (x) + f (y)

(5.2.1)

for all x, y ∈ RN . For N = 1 equation (5.2.1) was first treated by A. M. Legendre [206] and C. F. Gauss [96], but A. L. Cauchy [41] first found its general continuous solution, and the equation has been named after him. Concerning equation (5.2.1) and the vast literature of the subject, cf. also Acz´el [5]. It follows from (5.2.1) by induction that if f : RN → R is additive, then   n n

xi = f (xi ) (5.2.2) f i=1

i=1

for every n ∈ N and for every x1 , . . . , xn ∈ RN . Lemma 5.2.1. If f1 : RN → R and f2 : RN → R are additive functions, then, for every a, b ∈ R, the function f = af1 + bf2 is additive. This results immediately from (5.2.1). The following theorem gives a fundamental property of solutions of equation (5.2.1). Theorem 5.2.1. If f : RN → R satisfies equation (5.2.1), then f (λx) = λf (x)

(5.2.3)

for every x ∈ RN and λ ∈ Q. Proof. For x = y = 0 we get from (5.2.1) f (0) = 0, 6

(5.2.4)

So in the present book we consider only finite-valued additive functions. Additive functions with infinite values have been considered by I. Halperin [131]. He showed that the only such functions are f = +∞ and f = −∞ (cf. also 16.6).

5.2. Additive functions

129

whence, setting in (5.2.1) y = −x, we obtain 0 = f (0) = f (x − x) = f (x) + f (−x) , i.e., f (−x) = −f (x) .

(5.2.5)

Thus the function f is odd. Taking in (5.2.2) x1 = · · · = xn = x we obtain (5.2.3) for all λ ∈ N. According to (5.2.4) and (5.2.5), (5.2.3) holds for all λ ∈ Z. k Now, an arbitrary λ ∈ Q may be written as λ = , where k ∈ Z, m ∈ N. Hence m kx = m (λx) and so, by what already has been proved,   kf (x) = f (kx) = f m(λx) = mf (λx), 

whence (5.2.3) follows. If N = 1, then we get from (5.2.3) with c = f (1) f (λx) = cλ for λ ∈ Q, whence, if f is continuous, f (x) = cx for λ ∈ R.

(5.2.6)

Formula (5.2.6) gives the general continuous additive functions f : R → R. For continuous additive functions f : RN → R, N > 1, the corresponding formula will be found in 5.5. N Theorem 5.2.1 says  Nthat every  additive function f : R → R is a homomorphism from the space R ; Q; +; · into (R; Q; +; ·)(cf. 4.3). Thus as an immediate consequence of Theorem 4.3.1 we obtain the following   Theorem 5.2.2. Let H be an arbitrary Hamel basis of the space RN ; Q; +; · . Then for every function g : H → R there exists a unique additive function f : RN → R such that f | H = g. Theorem 5.2.2 gives the general construction of all the additive functions f : RN → R. In fact, every additive f may be obtained as the unique additive extension of a certain function g : H → R, viz. g = f | H. For many years the existence of discontinuous additive functions was an open problem. Mathematicians could neither prove that every additive function is continuous (which is false), nor exhibit an example of a discontinuous additive function (as we will see later, it was a hopeless task, since there do not exist effective examples of discontinuous additive functions). It was G. Hamel who first succeeded in proving that there exist discontinuous additive functions. In Hamel [134] he proved Theorem 5.2.2 above (for N = 1), whence the existence of discontinuous additive functions follows easily. In fact, we obtain from Theorem 5.2.2 Corollary 5.2.1. Let H be an arbitrary Hamel basis of the space (R; Q; +; ·), and let g : H → R be an arbitrary function. Let f : R → R be the unique additive extension of g. The function f is continuous if and only if g(x)/x = const for x ∈ H.

130

Chapter 5. Additive Functions and Convex Functions

Proof. If f is continuous, then it has form (5.2.6), and in particular, for x ∈ H, we must have g(x) = f (x) = cx, i.e., g(x)/x = c = const. If g (x) = cx for x ∈ H, then the function cx is an additive extension of g, and by the uniqueness of such an extension f must be given by (5.2.6). Thus f is continuous.  Now, it is enough to take any function g : H → R such that g(x)/x = const on H, and as an additive extension of g we obtain a discontinuous additive function. This argument would have to be modified for N > 1, but we can also argue otherwise. It follows from the proof of Theorem 4.3.1 that the values of an additive extension f : RN → R of a function g : H → R are given as linear combinations of the values of g (at suitable points of H) with rational coefficients. Hence, if g takes on only rational values, then also all values of f will be rational, and thus f cannot be continuous. We shall often make use of this observation, therefore we will formulate it as   Corollary 5.2.2. Let H be an arbitrary Hamel basis of the space  RN ; Q; +; · , and let g : H → R be a function with rational values only g(H) ⊂ Q , g = 0. Then the additive extension f : RN → R of g is discontinuous. Hence we infer the existence of discontinuous additive functions f : RN → R for arbitrary N ∈ N. Discontinuous additive functions are sometimes called Hamel functions. They exhibit many pathological properties, as will be seen later in this book. Since such functions are undesirable in applications, we will try to find possibly weak conditions which assure the continuity of an additive function.

5.3 Convex functions Let D ⊂ RN be a convex and open7 set. A function f : D → R is called convex 8 iff it satisfies Jensen’s functional inequality   f (x) + f (y) x+y  (5.3.1) f 2 2 for all x, y ∈ D. If the inequality in (5.3.1) for x = y is sharp, f is called strictly convex. If a function f : D → R (where D ⊂ RN is a convex and open set) satisfies the inequality   f (x) + f (y) x+y f  (5.3.2) 2 2 In this book we consider convex functions only on convex and open subsets of RN , and we consider only finite-valued functions. Continuous convex functions defined on subsets of RN which are convex, but not necessarily open, and also infinite-valued convex functions, are discussed in detail in Rockafellar [268]. Adjoining to D also some of its frontier points would seriously complicate the presentation, but the essentials would remain the same. (Let us note also that convex functions are often considered in more general spaces than RN ; cf., e.g., Roberts–Varberg [267]). 8 Some authors call functions fulfilling (5.3.1) J-convex or Jensen-convex, reserving the name convex for functions fulfilling (5.3.7) for all real λ ∈ [0, 1]. Since in the present book we are primarily interested in functions fulfilling Jensen’s inequality (5.3.1), we call such functions simply convex, as many authors do. Functions fulfilling (5.3.7) for all real λ ∈ [0, 1] will be referred to as continuous and convex. As we shall see later, they are necessarily continuous. 7

5.3. Convex functions

131

for all x, y ∈ D, then f is called concave. If the inequality for x = y is sharp, f is called strictly concave. It is easily seen that a function f : D → R is concave if and only if the function −f is convex. So the properties of concave functions can easily be obtained from those of convex functions. Every theorem about convex functions can be “translated” into a theorem about concave functions, and vice versa. Therefore concave functions will not be separately treated in this book. Convex functions were introduced (for N = 1) by J. L. W. V. Jensen [155], [156], although functions satisfying similar conditions were already treated by O. H¨ older [143], J. Hadamard [127] and O. Stolz [298]. Basic properties of convex functions in the one-dimensional case were proved by Jensen himself and by F. Bernstein - G. Doetsch [30]. Generalizations to higher dimensions were made by H. Blumberg [33] and E. Mohr [227]. In his fundamental paper J. L. W. V. Jensen [156] wrote: “Il semble que la notion de fonction convexe est a peu pr`es aussi fondamentale que celles-ci fonction positive, fonction croissante. Si je ne me trompe pas en ceci, la notion devra trouver sa place dans les expositions ´el´ementaires de la th´eorie des fonctions r´eelles.” (It seems to me that the notion of convex function is just as fundamental as positive function or increasing function. If I am not mistaken in this, the notion ought to find its place in elementary expositions of the theory of real functions). And he was certainly not mistaken. Convex functions are more or less extensively treated in various textbooks on calculus (cf., e.g., Bourbaki [37], Haupt–Aumann [139]). The whole monographs9 (e.g., Rockafellar [268], Roberts–Varberg [267], Popoviciou [260]; cf. also Krasnosel’ski˘ı–Ruticki˘ı [175]) and expositions( Beckenbach [22], Moldovan [230], Kuczma [181]) have been devoted to them. They have been given many generalizations (Beckenbach [21], Beckenbach–Bing [25], Popoviciou [259], Moldovan [228], [229], [230], Kemperman [167], Ger [104], De´ ak [65], Czerwik [51], Ponstein [258], Vertge˘ın– Rubinˇste˘ın [317], Mitrinovi´c [226], Guerraggio–Paganoni [126], Sander [275]; cf. also Chapter 15). They play extremely important role in many branches of mathematics. In the present book we deal with seemingly less important aspects of the theory of convex functions. However, the importance of considering pathological examples of non-measurable functions lies not in investigating their properties, but in establishing that such examples may occur and finding what conditions are essential and cannot be released when we want to avoid such functions. Continuous convex functions have very beautiful properties (cf. Chapter 7). But in order to enjoy them we must know, what (possibly weak) conditions should be imposed on the functions considered in order to guarantee that they are actually continuous. Properties of convex functions are strikingly similar to those of additive functions. Actually we have by Theorem 5.2.1 for any additive function f : RN → R   1 f (x) + f (y) x+y = f (x + y) = , f 2 2 2 i.e., every additive function is convex 10 . 9

These monographs deal mainly with continuous convex functions. As well as concave. The converse is not true: if a function f : RN → R is at the same time convex and concave, then it differs from an additive function by a constant. Cf. 13.2 for details. 10

132

Chapter 5. Additive Functions and Convex Functions

Many other convex functions are well known; for N = 1 the functions f (x) = x2 , f (x) = exp x, f (x) = |x| are all convex on any open interval I ⊂ R (also on I = R). In Chapter 7 we will see how to find examples of convex functions in RN . Some examples of convex functions will also be given at the end of this section. We start with some trivial results. Theorem 5.3.1. The linear combination, with non-negative real coefficients, of convex functions is a convex function. Theorem 5.3.111 is quite obvious and requires no proof. Let us note that, as a result of Theorem 5.3.1, the sum of a finite number of convex functions is a convex function, as is also the product of a convex function by a non-negative constant. On the other hand, the product of (even non-negative) convex functions need not be convex (the function f (x) = x2 exp x on R is a counter-example). But we have (cf. also Corollary 7.2.1) Theorem 5.3.2. The square of a non-negative convex function is a convex function. Also Theorem 5.3.2 is almost self-evident and will be given no proof, similarly as the following Theorem 5.3.3. The limit of a convergent sequence of convex functions is a convex function. Corollary 5.3.1. The sum of a convergent series of convex functions is a convex function. Theorem 5.3.4. Let D ⊂ RN be a convex and open set, and  let f1 , f2 :D → R be convex functions. Then the function f given by f (x) = max f1 (x), f2 (x) , x ∈ D, is convex. Proof. Take arbitrary x, y ∈ D. We have   f1 (x) + f1 (y) f (x) + f (y) x+y   , f1 2 2 2 and

 f2

x+y 2

 

f2 (x) + f2 (y) f (x) + f (y)  2 2

Whence also  f

x+y 2

i.e., f is convex.



     f (x) + f (y) x+y x+y = max f1 , f2  , 2 2 2 

11 Theorem 5.2.1 says that the family of all convex functions f : D → R forms a cone in the space of all functions f : D → R. Cf. Kemperman [168] concerning further research in this direction.

5.3. Convex functions

133

Lemma 5.3.1. Let D ⊂ RN be a convex and open set. If f : D → R is a convex function, then for every n ∈ N and for every x1 , . . . , xn ∈ D   f (x1 ) + · · · + f (xn ) x1 + · · · + xn  . (5.3.3) f n n Proof. It follows from (5.3.1) by induction that for every p ∈ N and for every x1 , . . . , xn ∈ D   2p 2p 1 1 f xi  p f (xi ). (5.3.4) 2p i=1 2 i=1 Now fix an n ∈ N, and choose a p ∈ N such that n < 2p . Take arbitrary x1 , . . . , xn ∈ D, and put n 1 xk = xi for k = n + 1, . . . , 2p . (5.3.5) n i=1 Since D is convex, points (5.3.5) belong to D in virtue of Lemma 5.1.3. We have p

n 2 1 1 x = xi , i 2p i=1 n i=1

whence by (5.3.4)     n 2p 2p 1 1 1 xi = f xi  p f (xi ) f n i=1 2p i=1 2 i=1 $ n %  n 1 1 p . f (xi ) + (2 − n) f xi = p 2 i=1 n i=1 From (5.3.6) we obtain

 nf

which yields (5.3.3).

1 xi n i=1 n



(5.3.6)

1 f (xi ) , n i=1 n





Theorem 5.3.5. Let D ⊂ RN be a convex and open set. If f : D → R is a convex function, then for every x, y ∈ D and for every λ ∈ Q ∩ [0, 1] we have   f λx + (1 − λ)y  λf (x) + (1 − λ)f (y). (5.3.7) k , n ∈ N, 0 < k < n. Put x1 = · · · = xk = x, xk+1 = · · · = xn = y. n By (5.3.3)   kf (x) + (n − k) f (y) kx + (n − k) y  , f n n which is the same as (5.3.7). If λ = 0 or 1, (5.3.7) is trivial. 

Proof. Let λ =

If f : D → R is convex and continuous, then (5.3.7) holds for all real λ ∈ [0, 1]. The converse theorem is also true (cf. 7.1).

134

Chapter 5. Additive Functions and Convex Functions

Lemma 5.3.2. Let D ⊂ RN be a convex set, and let F ⊂ R be a field. If a function12 f : D → [−∞, ∞) satisfies (5.3.7) for all λ ∈ F ∩ [0, 1], then   n n

λi xi  λi f (xi ) (5.3.8) f i=1

i=1

for every n ∈ N, x1 , . . . , xn ∈ D, and λ1 , . . . , λn ∈ F ∩ [0, 1] such that

n 

λi = 1.

i=1

Proof. The proof runs by induction. For n = 2 (5.3.8) is identical with (5.3.7). Now suppose (5.3.8) to be true for an n ∈ N. Take arbitrary x1 , . . . , xn+1 ∈ D and λ1 , . . . , λn+1 ∈ F ∩[0, 1] such that λ1 + · · ·+ λn+1 = 1. If λ1 = · · · = λn = 0, λn+1 = 1, then (5.3.8) is trivial. If λ1 + · · · + λn = 0, then we get by (5.3.7)  λ1 x1 + · · · + λn xn f λi xi = f λi + λn+1 xn+1 λ1 + · · · + λn i=1 i=1    n 

λn λ1  λi f x1 + · · · + xn + λn+1 f (xn+1 ). λ1 + · · · + λn λ1 + · · · + λn i=1 n+1







n

(5.3.9)

By the induction hypothesis   λn λ1 x1 + · · · + xn f λ1 + · · · + λn λ1 + · · · + λn λ1 λn  f (x1 ) + · · · + f (xn ), λ1 + · · · + λn λ1 + · · · + λn and we obtain from (5.3.9) f

n+1

i=1

 λi xi



n+1

λi f (xi ),

i=1

i.e., (5.3.8) for n + 1. Thus (5.3.8) is generally true.



Thus inequality (5.3.8) with arbitrary x1 , . . . , xn ∈ D is valid for every convex function f : D → R with arbitrary λ1 , . . . , λn ∈ Q ∩ [0, 1], adding up to 1, and for every continuous convex function f : D → R with arbitrary λ1 , . . . , λn ∈ [0, 1], adding up to 1. (Cf. Theorem 5.3.5 and the subsequent remark). Theorem 5.3.5 is an analogue of Theorem 5.2.1. Theorem 5.2.2 has no analogue for convex functions. We do not know a way to construct all convex functions f : D → R. Below we give a few examples of convex functions which are not additive. Example 5.3.1. Let g : R → R be convex function (possibly continuous), and let a : RN → R be an additive function. Then the function f : RN → R:   f (x) = g a(x) (5.3.10) 12

Thus we admit that f may assume the value −∞. The lemma is also true for f : D → (−∞, ∞].

5.3. Convex functions

135

is convex. In fact,            g a(x) + g a(y) x+y a(x) + a(y) f (x) + f (y) x+y =g a =g  = . f 2 2 2 2 2 It would be tempting to conjecture that every convex function f : RN → R can be written in form (5.3.10), where a is additive, and g is a continuous convex function. But this is not the case, as will become apparent from the following Example 5.3.2. Let H be a Hamel basis for RN , and let, for every h ∈ H, a convex function (possibly continuous) gh : R → R be given such that gh (0) = 0.

(5.3.11)

We define a function f : RN → R as follows. If x ∈ RN , then x has an expansion x=

n

βi hi , βi ∈ Q, hi ∈ H, i = 1, . . . , n.

(5.3.12)

i=1

For such an x we put f (x) =

n

ghi (βi ).

(5.3.13)

i=1

The function f thus defined is convex. In fact, take an arbitrary y ∈ RN . y has an expansion n

y= γi hi , γi ∈ Q, i = 1, . . . , n. (5.3.14) i=1

We may assume that the same hi ’s occur in (5.3.12) and (5.3.14), adding in both expressions the lacking hi ’s with the coefficients 0. Due to condition (5.3.11) this does not affect formula (5.3.13). According to the definition of f f (y) =

n

ghi (γi ),

i=1

and, since by (5.3.12) and (5.3.14) x + y βi + γi = hi , 2 2 i=1 n

we have  f

x+y 2

 =

n

i=1

 g hi

βi + γi 2

 

n

f (x) + f (y) ghi (βi ) + ghi (γi ) = . 2 2 i=1

Now fix h1 , h2 ∈ H, and take gh1 (x) = x2 , gh2 (x) = x4 , gh (x) = 0 for h ∈ H \ {h1 , h2 }. Suppose that function (5.3.13) can be written in form (5.3.10) with a

136

Chapter 5. Additive Functions and Convex Functions

continuous g and additive a. Then for x = λh1 , λ ∈ Q, we have according to Theorem  5.2.1 a1 = a (h1 ) f (x) = f (λh1 ) = g (a (λh1 )) = g (λa (h1 )) = g (λa1 ) , whereas from (5.3.13) f (x) = f (λh1 ) = gh1 (λ) = λ2 . Hence g (λa1 ) = λ2 for all λ ∈ Q. Note that a1 cannot be zero, for otherwise we would obtain λ2 = g (0) for all λ ∈ Q, which is clearly impossible. Therefore g (x) = x2 /a21 for all x = λa1 , λ ∈ Q, and since g is continuous, this implies that g (x) = x2 /a21 for all real x.   Similarly, we have for x = λh2 , λ ∈ Q a2 = a (h2 ) :     f (x) = f (λh2 ) = g a(λh2 ) = g λa(h2 ) = g(λa2 ), and by (5.3.13) f (x) = f (λh2 ) = gh2 (λ) = λ4 , whence it follows, as in the preceding case, g (x) = x4 /a42 . Consequently x2 x4 = a21 a42 for all real x, which is impossible. But also formula (5.3.13) with continuous gh does not yield the general form of convex functions f : RN → R. To show this, we consider Example 5.3.3. Let H be a Hamel basis for RN . For x ∈ RN with expansion (5.3.12) we define f (x) as13 n

f (x) = exp βi . (5.3.15) i=1

Adding to (5.3.12) further hi ’s with the coefficients 0 does not affect expression (5.3.15). Thus if we take a y ∈ RN , we may assume that y has an expansion (5.3.14), and we have, since the function exp is convex,     n n n

βi + γi 1 f (x) + f (y) x+y = exp  exp . f βi + exp γi = 2 2 2 2 i=1 i=1 i=1 Consequently our function f is convex. If we suppose that f can be written in form (5.3.13) with continuous gh ’s, then14 for x = λh, λ ∈ Q, h ∈ H, we get by (5.3.13) f (x) = gh (λ), whereas by (5.3.15) f (x) = eλ . Hence gh (λ) = eλ for λ ∈ Q, and for all h ∈ H. Hence, for x = λh1 + μh2 , λ, μ ∈ Q, h1 , h2 ∈ H, we get by (5.3.13) f (x) = eλ +eμ , whereas by (5.3.15) f (x) = eλ+μ . Thus eλ+μ = eλ +eμ for all λ, μ ∈ Q, which is impossible. 13

It is easily seen that expression (5.3.15) is a particular case of (5.3.10). We could argue simpler: (5.3.15) yields f (0) = 1, whereas from (5.3.13) f (0) = 0. But this n  ghi (βi ), where c ∈ R is argument fails if instead of (5.3.13) we consider the formula f (x) = c + 14

a constant.

i=1

5.4. Homogeneity fields

137

5.4 Homogeneity fields Let f : RN → R be an additive function. Then, by Theorem 5.2.1, f satisfies relation (5.2.3) for all x ∈ RN and all λ ∈ Q. But it may happen that f satisfies (5.2.3) (for all x ∈ RN ) also for some λ ∈ / Q. Let " ! Hf = λ ∈ R | f (λx) = λf (x) for all x ∈ RN . The set Hf is called the homogeneity field of the function f . The name is well motivated, as may be seen from the following theorem, due to J. R¨ atz [263]. Theorem 5.4.1. Let f : RN → R be an additive function. Then the set Hf is a field. Proof. Let λ, μ ∈ Hf . Then, for an arbitrary x ∈ RN ,   f (λ − μ)x = f (λx − μx) = f (λx) − f (μx) = λf (x) − μf (x) = (λ − μ)f (x). Consequently λ − μ ∈ Hf . Similarly, if, moreover, μ = 0     λ λ x , λf (x) = f (λx) = f μ x = μ f μ μ 

whence f

 λ λ x = f (x). μ μ

Thus λ/μ ∈ Hf , and in virtue of Lemma 4.7.1 Hf is a field.



The following result (R¨ atz [263]) is in a certain sense converse to Theorem 5.4.1. Theorem 5.4.2. Let F ⊂ R be an arbitrary field. Then there exists an additive function f : RN → R such that Hf = F . Proof. Consider the linear space   N R ; F ; +; · ,

(5.4.1)

and let B be a base of (5.4.1) (cf. Corollary 4.2.1). Fix an arbitrary c ∈ R, c = 0, and let g : B → R be the constant function g (x) = c for x ∈ B.

(5.4.2)

According to Theorem 4.3.1 there exists a homomorphism f from (5.4.1) into (R; F ; +; ·) such that f |B = g. Since f is homomorphism, f is an additive function, and satisfies f (λx) = λf (x) (5.4.3) for all x ∈ RN and λ ∈ F . Consequently F ⊂ Hf . Now take an arbitrary x ∈ RN . x can be written in the form x=

n

i=1

λi bi , λi ∈ F, bi ∈ B, i = 1, . . . , n.

138

Chapter 5. Additive Functions and Convex Functions

By (5.4.3) and (5.4.2), since f is additive and f | B = g, f (x) =

n

i=1

Thus

λi f (bi ) =

n

λi g(bi ) = c

n

i=1

λi ∈ cF.

i=1

  f RN ⊂ cF.

(5.4.4)

Take arbitrary λ ∈ Hf and b0 ∈ B. We have f (λb0 ) = λf (b0 ) = λg(b0 ) = λc.

(5.4.5)

On the other hand, since λb0 ∈ RN , we infer from (5.4.4) that there exists an α ∈ F such that f (λb0 ) = αc. (5.4.6) Since c = 0, relations (5.4.5) and (5.4.6) yield λ = α ∈ F . Hence Hf ⊂ F , and ultimately we obtain that Hf = F .  Similar investigations for convex functions have been carried out by R. Ger [113]. However, the results are not so complete as those for additive functions. Therefore we shall not go into details here.

5.5 Additive functions on product spaces Consider RN as a product of lower dimensional spaces: RN = R p × Rq

(5.5.1)

with p ∈ N, q ∈ N, p + q = N . Every x ∈ RN can be represented as x = (xp , xq ), with xp ∈ Rp , xq ∈ Rq , and if y ∈ RN , y = (yp , yq ), yp ∈ Rp , yq ∈ Rq , then x + y = (xp , xq ) + (yp , yq ) = (xp + yp , xq + yq ) .

(5.5.2)

With these notations we have the following (Acz´el [5]) Theorem 5.5.1. If f : RN → R is an additive function, and RN has decomposition (5.5.1), then there exist additive functions fp : Rp → R and fq : Rq → R such that f (x) = f (xp , xq ) = fp (xp ) + fq (xq ).

(5.5.3)

Proof. Put fp (xp ) = f (xp , 0), fq (xq ) = f (0, xq ). Then by (5.5.2) x = (xp , xq ) = (xp , 0) + (0, xq ), and f (x) = f (xp , xq ) = f (xp , 0) + f (0, xq ) = fp (xp ) + fq (xq ), i.e., we obtain formula (5.5.3). Further, we have for arbitrary xp , yp ∈ Rp and xq , y q ∈ Rq   fp (xp +yp ) = f (xp +yp , 0) = f (xp , 0)+(yp , 0) = f (xp , 0)+f (yp , 0) = fp (xp )+fp (yp ),

5.6. Additive functions on C

139

and   fq (xq +yq ) = f (0, xq +yq ) = f (0, x0 )+(0, yq ) = f (0, xq )+f (0, yq ) = fq (xq )+fq (yq ), i.e., the functions fp and fq are additive. This completes the proof15 .



Applying Theorem 5.5.1 several times we arrive at the representation f (x) = f (x1 , . . . , xN ) =

N

fi (xi ),

(5.5.4)

i=1

valid for additive functions f : RN → R, where fi : R → R, i = 1, . . . , N , are additive functions of a single variable. Hence we derive the following result. Theorem 5.5.2. If f : RN → R is continuous additive function, then there exists a c ∈ RN such that f (x) = cx, (5.5.5) where cx =

N 

  ci xi c = (c1 , . . . , cN ) , x = (x1 , . . . , xN ) denotes the scalar product.

i=1

Proof. As we saw in 5.2, the theorem is true for N = 1. If N > 1, we apply formula (5.5.4). Fix an i, 1  i  N , and put in (5.5.4) x1 = · · · = xi−1 = xi+1 = · · · = xN = 0. Since fj (0) = 0, j = 1, . . . , N (cf. Theorem 5.2.1), we get fi (xi ) = f (0, . . . , 0, xi , 0, . . . , 0), whence it follows that the function fi is continuous. Consequently all fi in (5.5.4) are continuous additive functions of a single variable, and thus there exist constants ci ∈ R, i = 1, . . . , N , such that fi (xi ) = ci xi , i = 1, . . . , N. Put c = (c1 , . . . , cN ). Then (5.5.5) is a consequence of (5.5.4) and (5.5.6).

(5.5.6) 

Thus it has been rather easy to find the general form of continuous additive functions f : RN → R. For discontinuous additive functions no such formula exists. As we shall see later (Chapter 9), such functions are non-measurable, and consequently they cannot be effectively displayed. They can be obtained only by a use of a Hamel basis, and thus, implicitely, the Axiom of Choice.

5.6 Additive functions on C In this section we consider functions f : C → C which are additive, i.e., satisfy the functional equation f (z1 + z2 ) = f (z1 ) + f (z2 ) (5.6.1) for all z1 , z2 ∈ C. 15 The same argument works for homomorphisms on products of groups. Concerning homomorphisms on products of semigroups, cf. Kuczma [182], Szymiczek [311], Martin [221].

140

Chapter 5. Additive Functions and Convex Functions

If we abstract from the analytic structure of C, C may be considered as R2 : C = R × R, and z ∈ C may be considered as z = (Re z, Im z). An arbitrary function f : C → C may be written as f (z) = f1 (z) + if2 (z), (5.6.2) where f1 : C → R and f2 : C → R are the functions f1 (z) = Re f (z), f2 (z) = Im f (z).

(5.6.3)

If f is additive, then by (5.6.1) and (5.6.3), for arbitrary z1 , z2 ∈ C f1 (z1 + z2 ) = Ref (z1 + z2 ) = Re[f (z1 ) + f (z2 )] = Ref (z1 ) + Ref (z2 ) = f1 (z1 ) + f1 (z2 ), and f2 (z1 + z2 ) = Imf (z1 + z2 ) = Im[f (z1 ) + f (z2 )] = Imf (z1 ) + Imf (z2 ) = f2 (z1 ) + f2 (z2 ),   i.e., the functions f1 and f2 which may be considered as functions from R2 into R are additive. Thus we obtain from Theorem 5.5.1 according to (5.6.2) Theorem 5.6.1. If f : C → C is additive, then there exist additive functions fkj : R → R, k, j = 1, 2, such that f (z) = f11 (Re z) + f12 (Im z) + if21 (Re z) + if22 (Im z).

(5.6.4)

Similar is the situation with the continuous solutions of (5.6.1). Theorem 5.6.2. If f : C → C is a continuous additive function, then there exist complex constants c1 , c2 such that f (z) = c1 z + c2 z¯,

(5.6.5)

where z¯ denotes the complex conjugate of z. Proof. The continuity of f implies the continuity of each function fkj occurring in (5.6.4), and so there exist real constants ckj , k, j = 1, 2, such that fkj (x) = ckj x, k, j = 1, 2. Hence f (z) = c11 Re z + c12 Im z + ic21 Re z + ic22 Im z = (c11 + ic21 ) Re z + (c12 + ic22 ) Im z. Put a = c11 + ic21 ∈ C, b = c12 + ic22 ∈ C. Then f (z) = a Re z + b Im z.

(5.6.6)

5.6. Additive functions on C

141

Since i2 = −1, relation (5.6.6) can be written as f (z) = a Re z − (bi) i Im z = and with c1 =

a − bi a + bi a − bi a + bi Re z + Re z − i Im z + i Im z, 2 2 2 2

1 1 (a − bi), c2 = (a + bi), we obtain 2 2 f (z) = c1 (Re z + i Im z) + c2 (Re z − i Im z) , 

i.e., (5.6.5).

Functions (5.6.5) are not analytic, since z¯ is not an analytic function of z. For analytic additive functions we have the following result. Theorem 5.6.3. If f : C → C is an analytic additive function, then there exists a complex constant c such that f (z) = cz.

(5.6.7)

Proof. f , being analytic, is differentiable. Differentiating relation (5.6.1) with respect to z1 we obtain f  (z1 + z2 ) = f  (z1 ). (5.6.8) Relation (5.6.8), valid for arbitrary z1 , z2 ∈ C, says that the function f  is constant, i.e., f  (z) = c, z ∈ C, with a c ∈ C. Hence f (z) = cz + b,

(5.6.9)

with a, b ∈ C. Inserting (5.6.9) into (5.6.1) we obtain b = 0, whence (5.6.7) follows. 

Exercises 1. Let C ⊂ R be a set such that C ∩ (0, 1) = ∅. Show that if a set A ⊂ RN is C-convex, then there exists a set D ⊂ R dense in (0, 1) and such that A is D-convex. [Hint: Let D = {λ ∈ + (1 − λ) y ∈ A for every x, y ∈ A}. If λ ∈ C ∩  [0, 1] |λx n (0, 1), then αnk = 1 − λk ∈ D for every k, n ∈ N. Moreover, 0 < αnk − αn+1,k < λk for n, k ∈ N. Let (a, b) ⊂ (0, 1) be a non-empty interval. If k ∈ N is such that λk < min (1 − b, b − a), then there exists an n ∈ N, such that αnk ∈ (a, b).] 2. Let C ⊂ R be a set such C ∩ (0, 1) = ∅. Show that if A ⊂ RN is C-convex, then cl A is convex. 3. For arbitrary a ∈ RN and A ⊂ RN let d (a, A) denote the Euclidean distance of a from A, and, given an ε > 0, let ! " Aε = x ∈ RN | d (x, A) < ε . Show that if A ⊂ RN is convex, then so is also Aε for every ε > 0.

142

Chapter 5. Additive Functions and Convex Functions

4. Show that if A ⊂ RN is convex, then so is also cl A. [Hint: Use Exercise 4.3.] 5. Let A ⊂ RN be a convex set such that int A = ∅. Show that int cl A = int A. [Hint: Let y ∈ int A, x ∈ cl A \ int A. Using Corollary 5.1.3 show that y + λ (x − y) ∈ / cl A for λ > 1, whence x ∈ / int cl A.] 6. Let f : R → R be an additive function, which is not one-to-one. Show that, for every y ∈ f (R), the set f −1 (y) is dense in R. (Sm´ıtal [286].) 7. Let f : R → R be an additive function such that f (R) = R. Show that f either is one-to-one, or has the Darboux property (the intermediate value property; Sm´ıtal [286]). 8. Let f : R →  R, f = 0 be an additive function. Show that, for every y ∈ f (R), mi f −1 (y) = 0. (Sm´ıtal [286]). 9. Let D ⊂ RN be a convex and open set. Show that a function f : D → R is convex if and only if the set {(x, y) ∈ D × R | y > f (x)} is J-convex. 10. Show that if an additive function f : R2 → R is continuous with respect to either variable, then it is jointly continuous in R2 . 11. Show that if an additive function f : R2 → R is unbounded on a square [a, b] × [a, b] ⊂ R2 , then for every fixed y ∈ R, either f (x, y), or f (y, x) is unbounded on [a, b]. 12. Let f : R2 → R be an additive function. Show that if f (x, y) (as a function of x) is unbounded on an interval I ⊂ R for a y ∈ R, then it is unbounded on I for every y ∈ R.

Chapter 6

Elementary Properties of Convex Functions 6.1 Convex functions on rational lines In this chapter we discuss some properties of convex functions connected with their boundedness and continuity. We start with the following Lemma 6.1.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. Then f (x) − f (x − nd) f (x) − f (x − md) f (x + md) − f (x) f (x + nd) − f (x)    n m m n (6.1.1) for every x ∈ D, d ∈ RN and m, n ∈ N such that 0 < m < n and x ± nd ∈ D. Proof. We use Lemma 5.3.1. Take in (5.3.3) x1 = · · · = xm = x + nd, xm+1 = · · · = xn = x. We obtain   mf (x + nd) + (n − m) f (x) m (x + nd) + (n − m) x  , f n n i.e., nf (x + md)  mf (x + nd) + (n − m) f (x) , and n [f (x + md) − f (x)]  m [f (x + nd) − f (x)] .

(6.1.2)

Hence the last inequality results. The first inequality in (6.1.1) results from (6.1.2) if we replace d by −d. For the proof of the middle inequality in (6.1.1) use (5.3.1) with x, y replaced by x + md, x − md, respectively. We obtain f (x) 

f (x + md) + f (x − md) , 2

whence f (x) − f (x − md)  f (x + md) − f (x) , and the middle inequality in (6.1.1) results on dividing both the sides by m.



144

Chapter 6. Elementary Properties of Convex Functions

Lemma 6.1.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. If f is bounded above on a set A ⊂ D, then it is also bounded above (by the same constant) on Q (A). Proof. Suppose that for a certain constant M ∈ R we have f (t)  M for t ∈ A.

(6.1.3)

Take an arbitrary x ∈ Q (A). By Theorem 5.1.3 there exist an n ∈ N, t1 , . . . , tn ∈ A, and λ1 , . . . , λn ∈ Q ∩ [0, 1] such that n

λi = 1,

(6.1.4)

i=1

and x=

n

λi ti .

(6.1.5)

i=1

By Lemma 5.3.2 and Theorem 5.3.5 we obtain from (6.1.5), (6.1.3), and (6.1.4) f (x) 

n

i=1

λi f (ti ) 

n

λi M = M

i=1

This shows that f is bounded above by M on Q (A).

n

λi = M.

i=1



Corollary 6.1.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. If f is bounded above on a set A ⊂ D, then it is also bounded above (by the same constant) on J (A). Proof. This could be proved by the same argument as Lemma 6.1.2, using Lemma 1 5.1.1, but we will derive this directly from Lemma 6.1.2. Since ∈ Q, the set Q (A) 2 is J-convex, and hence J (A) ⊂ Q (A). So our corollary is an immediate consequence of Lemma 6.1.2.  Corollary 6.1.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. For arbitrary x, y ∈ D, the function f is bounded above on the rational segment Q(x, y). Proof. This follows from Lemma 6.1.2 in view of the fact that every function is bounded on every finite set.  Lemma 6.1.3. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. For arbitrary x, y ∈ D, the function f is bounded below on the rational segment Q (x, y).

6.1. Convex functions on rational lines

145

Proof. Take arbitrary x, y ∈ D, and t ∈ Q (x, y). According to (5.1.9) there exists a λ ∈ Q ∩ [0, 1] such that t = λx + (1 − λ) y. (6.1.6) Moreover, by Corollary 6.1.2, there exists a constant M such that f (s)  M for s ∈ Q (x, y) . Put u =

(6.1.7)

1 (x + y), and v = 2u − t. By (6.1.6) 2 v = x + y − λx − (1 − λ) y = λy + (1 − λ) x ∈ Q (x, y) ,

whence by (6.1.7) f (v)  M. Since u =

(6.1.8)

1 (v + t), and f is convex, we have 2 f (u) 

1 [f (v) + f (t)] , 2

whence by (6.1.8) f (t)  2f (u) − f (v)  2f (u) − M. Thus f is bounded below on Q (x, y) by the constant1 2f (u) − M .



Theorem 6.1.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. For arbitrary a, b ∈ D the function f | Q (a, b) is uniformly continuous. Proof. Fix a, b ∈ D. We may assume that a = b. Since D is open, there exists a positive number ρ ∈ Q such that a = a − ρ (b − a) ∈ D and b = b + ρ (b − a) ∈ D.

(6.1.9)

By Corollary 6.1.2 and Lemma 6.1.3 there exist real constants K  M such that

First we show that

K  f (x)  M for x ∈ Q (a , b ) .

(6.1.10)

Q (a, b) ⊂ Q (a , b ) .

(6.1.11)

In fact, if t ∈ Q (a, b), then, according to (5.1.9), there exists a λ ∈ Q ∩ [0, 1] such that ρ+λ t = λa + (1 − λ) b. Put μ = . It is easily seen that μ ∈ Q ∩ (0, 1). Moreover, by 1 + 2ρ (6.1.9), μa +(1 − μ) b =

1+ρ−λ ρ+λ [a − ρ (b − a)]+ [b + ρ (b − a)] = λa+(1 − λ) b = t. 1 + 2ρ 1 + 2ρ

Consequently t ∈ Q (a , b ), which proves inclusion (6.1.11). 1

This constant depends on x and y, but not on a particular t ∈ Q (x, y).

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Chapter 6. Elementary Properties of Convex Functions

  M −K M −K + 1 the integral part of + Given an arbitrary ε > 0, put n = ε ε  1 so that n ∈ N and M −K < n. (6.1.12) ε ρ Further put δ = |b − a|. Take arbitrary x , x ∈ Q (a, b) such that n 

|x − x | < δ.

(6.1.13)

Then there exist α, β ∈ Q ∩ [0, 1] such that x = αa + (1 − α) b, x = βa + (1 − β) b, whence

x − x = (β − α) (b − a) , ρ and by (6.1.13) |β − α| |b − a| < δ = |b − a|, and finally n n |β − α| < ρ. Consider the numbers μ1,2 =

(6.1.14)

ρ + α ± n (β − α) ∈ Q. 1 + 2ρ

By (6.1.14) 0

α α + 2ρ < μ1,2 < 1 1 + 2ρ 1 + 2ρ

so that μ1,2 ∈ Q ∩ [0, 1]. Now, μ1,2 a + (1 − μ1,2 ) b = [α ± n (β − α)] a + [1 − α ∓ n (β − α)] b = αa + (1 − α) b ± n (β − α) (b − a) = x ± n (x − x ) . Consequently x ± n (x − x ) ∈ Q (a , b ) ⊂ D. Taking in Lemma 6.1.1 x = x , d = x − x , and m = 1, we get by (6.1.10) and (6.1.11) f (x ) − f (x − nd) M −K f (x + nd) − f (x ) K −M   f (x ) − f (x )   , n n n n and by (6.1.12)

M −K < ε. n This shows that f is uniformly continuous on Q (a, b). |f (x ) − f (x )| 



Every uniformly continuous function on a set A ⊂ RN can be uniquely extended onto cl A to a continuous function (cf. Exercise 6.1), so we obtain from Theorem 6.1.1 the following

6.1. Convex functions on rational lines

147

Theorem 6.1.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. For arbitrary a, b ∈ D there exists a unique continuous function2 gab : ab → R such that gab | Q (a, b) = f | Q (a, b). The function gab satisfies   gab (x) + gab (y) x+y  (6.1.15) gab 2 2 for every x, y ∈ ab. Proof. The set cl Q (a, b) is convex (cf. Exercise 5.2), so ab ⊂ cl Q (a, b). On the other hand, Q (a, b) ⊂ ab, whence cl Q (a, b) ⊂ cl ab = ab. Thus ab = cl Q (a, b). The existence of a unique continuous extension gab of f | Q (a, b) onto ab results from Theorem 6.1.1. It remains to prove formula (6.1.15). Take arbitrary x, y ∈ ab = cl Q (a, b). There exist two sequences of points 1 xn , yn ∈ Q (a, b), n ∈ N, such that lim xn = x, lim yn = y. Then lim (xn + yn ) = n→∞ n→∞ n→∞ 2 1 (x + y), and, since f is convex, gab is continuous, and gab | Q (a, b) = f | Q (a, b), 2       f (xn ) + f (yn ) x+y xn + yn xn + yn = lim gab = lim f  lim gab n→∞ n→∞ n→∞ 2 2 2 2 gab (xn ) + gab (yn ) gab (x) + gab (y) = . = lim n→∞ 2 2 

This proves relation (6.1.15).

Remark 6.1.1. We do not say that gab is convex, since gab is not defined on an open subset of RN . Let l = l (a, b) be the rational line passing through a and b (cf. 5.1), and put l0 = l ∩ D. Similarly, let L be the (real) line passing through a and b, and let L0 = L ∩ D. With this notation we have Theorem 6.1.3. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. For every a, b ∈ D there exists a unique continuous function Gab : L0 → R such that Gab | l0 = f | l0 . The function Gab satisfies inequality (6.1.15) for every x, y ∈ L0 . Proof. Let an , bn ∈ l0 , n ∈ N, be two monotonic sequences of points converging to the two ends of L0 . As is easy to check, we have Q (an , bn ) ⊂ l0 , an bn ⊂ L0 for every ∞ ∞   n ∈ N, and Q (an , bn ) = l0 , an , bn = L0 . Put gn = gan bn (Theorem 6.1.2). If n=1

n=1

for some m, n ∈ N we have am bm ⊂ an bn , then g | am bm is a continuous function on am bm which restricted to Q (am , bm ) coincides with f . But gm is the unique function with these properties, whence gn | am bm = gm . Consequently we my define a function Gab on L0 in the following way: Gab (x) = gn (x) if x ∈ an bn , n ∈ N. 2 ab

= cl Q (a, b) = conv (a, b) denotes the closed segment joining the points a and b.

(6.1.16)

148

Chapter 6. Elementary Properties of Convex Functions

Inequality (6.1.15) for Gab results from Theorem 6.1.2 in virtue of the fact that for every x, y ∈ L0 there exists an n ∈ N such that x, y ∈ an bn . The continuity of Gab results from that of gn , n ∈ N. Finally, if G : L0 → R is an arbitrary continuous function such that G | l0 = f | l0 , then Gn = G | an bn is continuous, and Gn | Q (an , bn ) = f | Q (an , bn ) for every n ∈ N, whence it follows that Gn = gn , n ∈ N. Thus G must be given by (6.1.16), which proves the uniqueness of Gab .  With the same notation we have Theorem 6.1.4. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. Then, for arbitrary a, b ∈ D, the function f | l0 is continuous. This is an immediate consequence of Theorem 6.1.3. However, f itself need not be continuous in D. In order to find conditions ensuring the continuity of f in D, we are going to investigate, in the next section, the problem of the boundedness of f on open subsets of D.

6.2 Local boundedness of convex functions Let D ⊂ RN . A function f : D → R is called locally bounded [locally bounded above, locally bounded below ] at a point x0 ∈ D iff there exists a neighbourhood U ⊂ D of x0 such that the function f is bounded [bounded above, bounded below] on U . The following three theorems refer to the local boundedness of convex functions. Theorem 6.2.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. If f is locally bounded above at a point x0 ∈ D, then it is locally bounded above at every point x ∈ D. Proof. Take an arbitrary x ∈ D. We may assume that x = x0 , for otherwise f is locally bounded above at x by hypothesis. There exists a positive number ρ ∈ Q such that the point y = x − ρ (x0 − x) (6.2.1) belongs to D. Let λ =

ρ ∈ Q ∩ (0, 1). We have by (6.2.1) 1+ρ x = λx0 + (1 − λ) y.

(6.2.2)

By hypothesis there exists an r > 0 such that the open ball K = K (x0 , r) centered at x0 and with the radius r, is contained in D, and f is bounded above on K f (t)  M for t ∈ K,

(6.2.3)

where M is a real constant. Let U = K (x, λr) be the open ball centered at x and with the radius λr. Take an arbitrary u ∈ U , and put t=

u − (1 − λ) y . λ

(6.2.4)

6.2. Local boundedness of convex functions

149

Since u ∈ U , we have |u − x| < λr, whence by (6.2.2) |u − λx0 − (1 − λ) y| < λr, and # # # u − (1 − λ) y # # # < r, − x 0 # # λ i.e., t − x0 < r, which means that t ∈ K. Since by (6.2.4) u = λt + (1 − λ) y,

(6.2.5)

and y ∈ D, t ∈ K ⊂ D, and the set D is convex, we have u ∈ D, whence U ⊂ D. Moreover, by (6.2.5), Theorem 5.3.5, and (6.2.3)   f (u)  λf (t) + (1 − λ) f (y)  λM + (1 − λ) f (y)  max M, f (y) .   Thus f is bounded above on U by the constant max M, f (y) .  Theorem 6.2.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. If f is locally bounded below at a point x0 ∈ D, then it is locally bounded below at every point x ∈ D. Proof. Take an arbitrary x ∈ D, x = x0 . There exists a positive number ρ ∈ Q such that the point y = x0 − ρ (x − x0 ) ρ ∈ Q ∩ (0, 1) we have belongs to D. With λ = 1+ρ x0 = λx + (1 − λ) y.

(6.2.6)

By hypothesis there exists an r > 0, such that the open ball K = K (x0 , r) centered at x0 and with the radius r, is contained in D and f is bounded below on K f (t)  M for t ∈ K,

(6.2.7)

where M is a real Moreover, we may assume that r is so small that the  constant. r open ball U = K x, is contained in D. λ Now take an arbitrary u ∈ U , and put t = λu + (1 − λ) y.

(6.2.8)

r Then by (6.2.6) and (6.2.8) |t − x0 | = λ |u − x| < λ = r, which means that t ∈ K. λ By Theorem 5.3.5 we have f (t)  λf (u) + (1 − λ) f (y) , whence by (6.2.7) f (u) 

1 1−λ 1 1−λ f (t) − f (y)  M − f (y) = λ λ λ λ

 1+

1 ρ



1 M − f (y) . ρ

Thus f is bounded below on U by constant, which depends on the constants M , f (y), ρ, but not on a particular u ∈ U . Hence f is locally bounded below at x. 

150

Chapter 6. Elementary Properties of Convex Functions

Theorem 6.2.3. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. If f is locally bounded above at a point x0 ∈ D, then it is locally bounded at every point x ∈ D. Proof. By Theorem 6.2.1 f is locally bounded above at every point x ∈ D, and so if we show that f is locally bounded below at x0 , then it will follow by Theorem 6.2.2 that it is locally bounded below, and hence locally bounded, at every point x ∈ D. Let K ⊂ D be an open ball around x0 such that f is bounded above on K, i.e., (6.2.3) holds with a real constant M . 1 Take an arbitrary u ∈ K, and put t = 2x0 − u so that x0 = (u + t). We have 2 t − x0 = − (u − x0 ), whence |t − x0 | = |u − x0 |, which shows that t ∈ K. We have 1 f (x0 )  [f (u) + f (t)], whence by (6.2.3) 2 f (u)  2f (x0 ) − f (t)  2f (x0 ) − M. Thus f is bounded below on K by the constant 2f (x0 ) − M .



6.3 The lower hull of a convex functions Throughout this section the symbol K (x, r) denotes the open ball in RN centered at x and with the radius r. Let D ⊂ RN be an  open set, and let f : D → R be function. For x ∈ D and r > 0 sufficiently small such that K(x, r) ⊂ D we define ϕx (r) as ϕx (r) = inf f. K(x,r)

If r1 < r2 and K(x, r2 ) ⊂ D, we have K(x, r1 ) ⊂ K(x, r2 ), whence ϕx (r1 ) =

inf

K(x,r1 )

f

inf

K(x,r2 )

f = ϕx (r2 ) .

Thus the function ϕx is decreasing. Consequently there exists the limit mf (x) = lim ϕx (r) = lim r→0+

inf f.

r→0+ K(x,r)

(6.3.1)

Function mf : D → [−∞, +∞) is called the lower hull of f , and its value at an x ∈ D is called the infimum of f at x. Similarly we can define the function Mf : D → (−∞, +∞] Mf (x) = lim sup f, (6.3.2) r→0+ K(x,r)

which is called an upper hull of f , and its value at an x ∈ D is called the supremum of f at x. If the set D and the function f are convex, and f is locally bounded below t a point x0 ∈ D, then, by Theorem 6.2.2, f is locally bounded below at every point x ∈ D and the infimum of f at every x ∈ D is finite. Thus, in such a case, the function mf is finite: mf : D → R. In the other case f is locally unbounded below at every

6.3. The lower hull of a convex functions

151

point x ∈ D, which means that mf (x) = −∞ for every x ∈ D. Similarly, it follows from Theorem 6.2.1 that either Mf (x) is finite for every x ∈ D: Mf : D → R, or Mf (x) = +∞ for every x ∈ D. Theorem 6.3.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. If mf = −∞, then the function mf : D → R is continuous and convex. Proof. For every z ∈ D and ε > 0 there exists a δ > 0 such that ϕz (r)  lim ϕz (ρ) − ε = mf (z) − ε for r < δ, ρ→0+

(6.3.3)

whence for t ∈ K (z, r), r < δ, f (t)  inf f = ϕz (r)  mf (z) − ε.

(6.3.4)

K(z,r)

1 (x + y), fix an ε > 0, and choose a δ > 0 2 according to (6.3.3). Fix a positive r < δ. Then (6.3.4) holds for t ∈ K (z, r). Further, we can find points u ∈ K (x, r) and v ∈ K (y, r) such that (we assume that r has been chosen so small that K (x, r) , K (y, r) ⊂ D) Now take arbitrary x, y ∈ D, put z =

f (u)  inf f + ε = ϕx (r) + ε  lim ϕx (ρ) + ε = mf (x) + ε,

(6.3.5)

f (v)  inf f + ε = ϕy (r) + ε  lim ϕy (ρ) + ε = mf (y) + ε,

(6.3.6)

ρ→0+

K(x,r)

and ρ→0+

K(y,r)

since the functions ϕx , ϕy are decreasing. We have # # # # # #u + v #u + v x + y ## 1 1 1 1 # # # # 2 − z # = # 2 − 2 #  2 |u − x| + 2 |v − y| < 2 r + 2 r = r, which means that

1 (u + v) ∈ K (z, r). Thus we get by (6.3.4) 2     u+v x+y  mf − ε. f 2 2

From (6.3.7), (6.3.5) and (6.3.6) we obtain, since the function f is convex,     u+v f (u) + f (v) mf (x) + mf (y) x+y mf −εf   + ε, 2 2 2 2 i.e.,

 mf

x+y 2

 

mf (x) + mf (y) + 2ε. 2

Letting ε → 0, we obtain hence the convexity of mf .

(6.3.7)

152

Chapter 6. Elementary Properties of Convex Functions

Now we turn to the proof of the continuity of mf . Fix an x0 ∈ D. Let {e1 , . . . , eN } be an orthonormal base of RN over R so that we have, in particular, |ei | = 1, i = 1, . . . , N.

(6.3.8)

For every x ∈ RN we can write x − x0 uniquely in the form x − x0 = λ1 e1 + · · · + λN eN ,

(6.3.9)

where λi ∈ R, i = 1, . . . , N . For every k, 1  k  N , let Mk denote the set of those points x ∈ D for which in expansion (6.3.9) at least N − k of the coefficients λi are zeros. By induction on k we will prove that mf | Mk is continuous at x0 . Let k = 1. Put Li = { x ∈ D | x = x0 + λei , λ ∈ R } , i = 1, . . . , N. We have M1 =

N 

Li ,

i=1

so in order to prove that mf | M1 is continuous at x0 it is enough to prove that mf | Li is continuous at x0 for every i = 1, . . . , N . Fix an i, 1  i  N . Since x0 ∈ Li , we can find a λ ∈ R, λ = 0 such that 1 1 u = x0 + λei and v = x0 − λei both belong to D. Hence x0 = u + v ∈ Q (u, v). 2 2 By Theorem 6.1.1 mf | Q (u, v) is continuous, so given an ε > 0 we can find a δ > 0 such that ε |mf (t) − mf (x0 )| < (6.3.10) for t ∈ Q (u, v) ∩ K (x0 , δ) . 2 Moreover, we may assume that δ < |λ|. On the other hand, we have by (6.3.4) for y ∈ K (x0 , δ), provided that δ has been chosen small enough, ε f (y)  mf (x0 ) − . 2 If z ∈ K (x0 , δ), and r > 0 is so small that K (z, r) ⊂ K (x0 , δ), we obtain hence ε inf f  mf (x0 ) − , 2 K(z,r) and on letting r → 0+ we get (for arbitrary z ∈ K (x0 , δ)) mf (z)  mf (x0 ) −

ε > mf (x0 ) − ε. 2

(6.3.11)

Now suppose that z ∈ K (x0 , δ) ∩ Li , and take an arbitrary r > 0 such that K (z, r) ⊂ K (x0 , δ). Since z ∈ Li , there exists a κ ∈ R such that z = x0 + κei , and since z ∈ K (x0 , δ), we have |κ| |ei | = |z − x0 | < δ, i.e., by (6.3.8), |κ| < δ < |λ|. Hence κ+λ 1 1κ = + ∈ (0, 1) . 2λ 2 2λ

6.3. The lower hull of a convex functions

153

Consequently there exists a μ ∈ Q ∩ (0, 1) such that # # # # #μ − κ + λ # < r . # 2λ # 2 |λ|

(6.3.12)

Put t = μu + (1 − μ) v = x0 + (2μλ − λ) ei . Thus t ∈ Q (u, v). Moreover by (6.3.8) and (6.3.12) # # # κ + λ ## # |t − z| = |(2μλ − λ − κ) ei | = |2μλ − λ − κ| = 2 |λ| #μ − < r, 2λ # whence t ∈ K (z, r) ⊂ K (x0 , δ). By (6.3.10) ε mf (t) < mf (x0 ) + . 2

(6.3.13)

Let Ut be an arbitrary ball centered at t and such that Ut ⊂ K (z, r). There exists an ε s ∈ Ut such that f (s)  mf (t) + , or by (6.3.13) 2 f (s) < mf (x0 ) + ε.

(6.3.14)

Consequently every ball K (z, r) ⊂ K (x0 , δ) contains a point s such that (6.3.14) holds. Hence inf f  mf (x0 ) + ε, K(z,r)

and on letting r → 0+ we obtain mf (z)  mf (x0 ) + ε.

(6.3.15)

Relations (6.3.11) and (6.3.15) imply that |mf (z) − mf (x0 )|  ε for z ∈ K (x0 , δ) ∩ Li . Consequently the function mf | Li is continuous at x0 , and hence also the function mf | M1 is continuous at x0 . Now assume that for a k ∈ N, 2  k  N , mf | Mk−1 is continuous at x0 . Thus, given an ε > 0, there exists a δ > 0 such that for all y ∈ Mk−1 ∩ K (x0 , δ) we have |mf (y) − mf (x0 )| <

1 ε. 3

(6.3.16)

  1 Now take an x ∈ Mk ∩ K x0 , δ . Then, possibly after a suitable renumbering of 2 ei ’s, x − x0 = λ1 e1 + · · · + λk ek with some λi ∈ R, i = 1, . . . , k. Put 1 u = x0 − λ1 e1 + λ3 e3 + · · · + λk ek , v = x0 + λ2 e2 + λ3 e3 + · · · + λk ek . 2

154

Chapter 6. Elementary Properties of Convex Functions

1 (x + u). Moreover, 2 1  1  1 |u − x0 | = λ21 + λ23 + ···+ λ2k 2  λ21 + λ22 + λ23 + ···+ λ2k 2 = |x − x0 | < δ < δ, 2 1  2 1  1 1 2 λ2 + λ23 + ··· + λ2k |v − x0 | =  λ21 + λ22 + λ23 + ··· + λ2k 2 = |x − x0 | < δ < δ, 4 2

Then u, v ∈ Mk−1 and v =

which means that u, v ∈ K (x0 , δ). By (6.3.16) |mf (u) − mf (x0 )| <

1 ε, 3

|mf (v) − mf (x0 )| <

1 ε, 3

(6.3.17)

and, since the function mf is convex,   1 u+x  [mf (u) + mf (x)] , mf (v) = mf 2 2 i.e., mf (x)  2 mf (v) − mf (u) . Hence we obtain in view of (6.3.17)     1 1 mf (x) > 2 mf (x0 ) − ε − mf (x0 ) + ε = mf (x0 ) − ε. 3 3

(6.3.18)

Now put w = x0 + 2λ2 e2 + λ3 e3 + · · · + λk ek , z = x0 + 2λ1 e1 + λ3 e3 + · · · + λk ek . 1 Then w, z ∈ Mk−1 , x = (w + z). Moreover, |w − x0 |  2 |x − x0 | < δ, |z − x0 |  2 2 |x − x0 | < δ so that w, z ∈ K (x0 , δ). By (6.3.16) |mf (w) − mf (x0 )| <

1 ε, 3

|mf (z) − mf (x0 )| <

1 ε, 3

(6.3.19)

and, since the function mf is convex,   1 z+w  [mf (z) + mf (w)] . mf (x) = mf 2 2 Hence we obtain in view of (6.3.19)   1 1 1 1 mf (x0 ) + ε + mf (x0 ) + ε = mf (x0 )+ ε < mf (x0 )+ε. (6.3.20) mf (x) < 2 3 3 3 Relations (6.3.18) and (6.3.20) imply that |mf (x) − mf (x0 )| < ε.

(6.3.21)

Relation (6.3.21) holds for every x ∈ Mk ∩ K (x0 , δ), which means that the function mf | Mk is continuous at x0 . Induction shows that mf | Mk is continuous at x0 for k = 1, . . . , N . But MN = RN , so mf | MN = mf . Thus the function mf is continuous at x0 . Since x0 ∈ D has been arbitrary, the function mf is continuous in D. 

6.4. Theorem of Bernstein-Doetsch

155

6.4 Theorem of Bernstein-Doetsch Let D ⊂ RN be an open set, and let f : D → R be a function. It follows directly from the definition of the lower hull of f that mf (x)  f (x) for x ∈ D.

(6.4.1)

But if the function f and the set D are convex, we have more precise informations. Theorem 6.4.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. If at a point ξ ∈ D we have f (ξ) = mf (ξ), then f is not locally bounded at ξ. Proof. If mf (ξ) = −∞, then f is not locally bounded below at ξ, as results from the definition of mf . So let mf (ξ) > −∞. Let U ⊂ D be an arbitrary neighbourhood of ξ. Take an arbitrary M > 0. According to (6.4.1) p = f (ξ) − mf (ξ) > 0.

(6.4.2)

Thus there exists an n ∈ N such that 1 (n − 1) p + mf (ξ) > M. 2

(6.4.3)

Further, there exists an h ∈ RN such that ξ ± ih ∈ U , i = 1, . . . , n, and mf (ξ) −

1 1 p < f (ξ − h) < mf (ξ) + p. 2 2

(6.4.4)

Take in Lemma 5.3.1 x1 = · · · = xn−1 = ξ − h, xn = ξ + (n − 1) h. We obtain   (n − 1) f (ξ − h) + f (ξ + (n − 1) h) (n − 1) (ξ − h) + ξ + (n − 1) h  . f (ξ) = f n n Hence f (ξ + (n − 1) h)  nf (ξ) − (n − 1) f (ξ − h) = n [f (ξ) − f (ξ − h)] + f (ξ − h) . (6.4.5) 1 By (6.4.2) and (6.4.4) f (ξ − h) < f (ξ) − p, so we obtain by (6.4.4), (6.4.5) and 2 (6.4.3) f (ξ + (n − 1) h) >

n 1 1 p + mf (ξ) − p = (n − 1) p + mf (ξ) > M. 2 2 2

Thus for every neighbourhood U of ξ and for every M > 0 there exists a point t ∈ U [viz. t = ξ + (n − 1) h] such that f (t) > M . It follows that f is not locally bounded above at ξ.  Hence we derive the following Theorem of Bernstein-Doetsch [30]. Theorem 6.4.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. If f is locally bounded above at a point of D, then it is continuous in D.

156

Chapter 6. Elementary Properties of Convex Functions

Proof. Let f be locally bounded above at a point of D. By Theorem 6.2.3 f is locally bounded at every point of D, whence it follows in virtue of Theorem 6.4.1 that f (x) = mf (x) for x ∈ D. Since f is locally bounded, we have mf = −∞, whence by Theorem 6.3.1 mf is continuous in D. Consequently f is continuous in D.  Because of the basic importance of the Theorem of Bernstein-Doetsch in the theory of convex functions, we present below two other proofs, which do not rely on the difficult Theorem 6.3.1. Second proof of Theorem 6.4.2. If f is locally bounded above at a point of D, then, by Theorem 6.2.3, it is locally bounded at every point of D. Consequently formulas (6.3.1) and (6.3.2) define finite numbers for every x ∈ D. Evidently mf (x)  f (x)  Mf (x)

(6.4.6)

for every x ∈ D. Now take an arbitrary x ∈ D. There exists a sequence of points xn ∈ D, n ∈ N, such that lim xn = x, lim f (xn ) = mf (x) , (6.4.7) n→∞

n→∞

and the sequence of points zn ∈ D, n ∈ N, such that lim zn = x,

n→∞

lim f (zn ) = Mf (x) .

n→∞

(6.4.8)

Put yn = 2zn − xn , n ∈ N. We have by (6.4.7) and (6.4.8) lim yn = x.

n→∞

Moreover zn =

(6.4.9)

1 (xn + yn ), n ∈ N, whence by the convexity of f 2 f (zn ) 

1 [f (xn ) + f (yn )] , 2

or, f (yn )  2f (zn ) − f (xn ) . Hence, on letting n → ∞ we obtain in virtue of (6.4.7) and (6.4.8) lim inf f (yn )  2Mf (x) − mf (x) ,

(6.4.10)

lim inf f (yn )  Mf (x) .

(6.4.11)

n→∞

whereas by (6.4.9) n→∞

Relations (6.4.10) and (6.4.11) yield the inequality Mf (x)  mf (x), which together with (6.4.6) implies that Mf (x) = mf (x). Thus (cf. Exercise 6.8) f is continuous at x, and so, since x ∈ D has been arbitrary, f is continuous in D. 

6.4. Theorem of Bernstein-Doetsch

157

Third proof of Theorem 6.4.2. Let f be locally bounded above at a point of D. Take an arbitrary x0 ∈ D. By Theorem 6.2.3 f is locally bounded at x0 , thus there exists positive constants M and δ such that K (x0 , δ) ⊂ D and |f (t)|  M for t ∈ K (x0 , δ) .

(6.4.12)

For an arbitrary x ∈ K (x0 , δ) we have |x − x0 | < δ, and consequently we can find a λ ∈ Q ∩ (0, 1) such that |x − x0 | |x − x0 | <λ<2 . (6.4.13) δ δ Further put x − x0 x − x0 y = x0 + , z = x0 − . (6.4.14) λ λ Hence, in view of (6.4.13), |y − x0 | =

|x − x0 | < δ, λ

|z − x0 | =

|x − x0 | < δ, λ

which means that y, z ∈ K (x0 , δ). Moreover, by (6.4.14) x = λy + (1 − λ) x0 ,

x0 =

1 λ x+ z. λ+1 λ+1

By Theorem 5.3.5, since f is convex f (x)  λf (y) + (1 − λ) f (x0 ) ,

f (x0 ) 

λ 1 f (x) + f (z) , λ+1 λ+1

i.e., f (x) − f (x0 )  λ [f (y) − f (x0 )] ,

f (x) − f (x0 )  λ [f (x0 ) − f (z)] .

Hence we obtain by (6.4.12), since x0 , y, z ∈ K (x0 , δ), −2M λ  f (x) − f (x0 )  2M λ, and taking into account (6.4.13) we get −

4M 4M |x − x0 | < f (x) − f (x0 ) < |x − x0 | , δ δ

or

4M |x − x0 | . (6.4.15) δ Relation (6.4.15) proves the continuity of f at x0 , whence, since x0 ∈ D has been arbitrary, the continuity of f in D follows.  |f (x) − f (x0 )| <

Since every open set is a neighbourhood of any of its points, we get hence Corollary 6.4.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. If f is bounded above on a non-empty open set U ⊂ D, then it is continuous in D.

158

Chapter 6. Elementary Properties of Convex Functions

Let us observe, that the local (or even global) boundedness below does not imply the continuity of a convex function. For example, the function exp is continuous and convex on R, and let a : RN → R be a discontinuous additive function. Then the function (cf. Example 5.3.1) f : RN → R: f (x) = exp a(x) is convex, discontinuous, and f (x)  0 for x ∈ RN . If a function is continuous at a point, then it is locally bounded at this point. Thus the Theorem of Bernstein-Doetsch implies the following Theorem 6.4.3. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. Then either f is continuous in D, or f is totally discontinuous3 in D. In the next chapter we are going to discuss the main properties of continuous convex functions.

Exercises 1. Let A ⊂ RN be an arbitrary set, and let f : A → R be a function uniformly continuous on A. Prove that there exists a unique continuous function F : cl A → R such that F |A = f . 2. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. Let x ∈ D and y ∈ RN be such that x + y ∈ D. Prove that  y  lim f x + n = f (x) . n→∞ 2 3. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. Let 0, a, b, a + b ∈ D. Prove that the set ! " A = x ∈ RN | x = λ1 a + λ2 b, λ1 , λ2 ∈ Q ∩ [0, 1] is contained in D and that the function f is bounded on A. [Hint: Clearly f is bounded above on A, and bounded below on Q (0, a + b). For any x ∈ A, x = λ1 a + λ2 b, λ1 , λ2 ∈ Q ∩ (0, 1) consider the points y = λ2 a + λ1 b 1 and z = (x + y).] 2 4. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. Let 0, a, b, a+b ∈ D, and let λn , μn ∈ Q∩[0, 1], n ∈ N, be such that lim λn = λ0 , lim μn = μ0 . Prove that if λ0 , μ0 ∈ Q ∩ [0, 1], then

n→∞

n→∞

lim f (λn a + μn b) = f (λ0 a + μ0 b) .

n→∞ 3

I.e., discontinuous at every point of D.

(∗)

6.4. Theorem of Bernstein-Doetsch

159

5. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. Let 0, a, b ∈ D, and let λn , μn ∈ Q, n ∈ N, be such that lim λn = λ0 , n→∞ lim μn = μ0 , and λ0 a + μ0 b ∈ D, λn a + μn b ∈ D, n ∈ N. Prove that if n→∞

λ0 , μ0 ∈ Q, then (∗) holds. 6. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. Let 0, a, b ∈ D. Prove that if a, b are linearly independent over R, then the function f |E (a, b) ∩ D is continuous. 7. Show that the conclusion of Exercise 6.6 may be invalid if a, b are linearly dependent over R. [Hint: Take incommensurable a, b ∈ R, and let f : R → R be an additive function such that f (a) = 1, f (b) = 0. Consider a sequence xn = pn a + qn b, pn , qn ∈ Z, n ∈ N, such that lim xn = 0]. n→∞

8. Let D ⊂ RN be an open set, and let f : D → R be a function. Prove that the condition Mf (x) = mf (x) (cf. (6.3.1) and (6.3.2)) is necessary and sufficient for the continuity of f at x, x ∈ D.

Chapter 7

Continuous Convex Functions 7.1 The basic theorem Let D ⊂ RN be a convex and open set. In 5.3 we saw that a convex function f : D → R fulfills the inequality   f λx + (1 − λ)y  λf (x) + (1 − λ)f (y) (7.1.1) for all x, y ∈ D and all λ ∈ Q ∩ [0, 1]. It was also pointed out that if, moreover, f is continuous, then inequality (7.1.1) holds actually for all real λ ∈ [0, 1]. The following lemma is analogous to Lemma 6.1.2 Lemma 7.1.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a function fulfilling inequality (7.1.1) for all λ ∈ [0, 1]. If the function f is bounded above on a set A ⊂ D, then it is also bounded above (by the same constant ) on the set conv(A). Proof. Suppose that f (t)  M

for t ∈ A ,

(7.1.2)

with a certain real constant M . Take an arbitrary x ∈ conv (A). By Theorem 5.1.3 n  there exist an n ∈ N, t1 , . . . , tn ∈ A, and λ1 , . . . λn ∈ [0, 1] such that λi = 1 and i=1

x=

n

λi ti .

(7.1.3)

i=1

Relation (7.1.3) implies in virtue of Lemma 5.3.2 that f (x) 

i=1

n

λi f (ti ) 

i=1

n

λi M = M

i=1

λi = M

n

where we have used also (7.1.2). Consequently f is also bounded above by M at an arbitrary point of conv (A). 

162

Chapter 7. Continuous Convex Functions

Theorem 7.1.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a function. The function f is convex and continuous if and only if it satisfies inequality (7.1.1) for all x, y ∈ D and all λ ∈ [0, 1]. Proof. If the function f is convex and continuous, then clearly it satisfies (7.1.1) for all x, y ∈ D and all λ ∈ [0, 1]. Now suppose that a function f : D → R satisfies 1 inequality (7.1.1) for all x, y ∈ D and λ ∈ [0, 1]. Putting λ = we see that f is 2 convex. Take an arbitrary x0 ∈ D. According to the Remark in 5.1, theres exist an N -dimensional simplex S ⊂ D such that x0 ∈ int(S). Let p1 , . . . , pN +1 be the vertices of S: S = conv {p1 , . . . , pN +1 } Since the set A = {p1 , . . . , pN +1 } is finite, f is bounded above on A. By Lemma 7.1.1 f is bounded above on S, and thus on a neighborhood of x0 . In other words, f is locally bounded above at x0 . By Theorem 6.4.2 f is continuous.  Condition (7.1.1) has a very simple geometric interpretation. When λ runs over the interval [0, 1], the points λx + (1 − λ) y fill the segment  joining x and y, and  thus the points λx + (1 − λ)y, λf (x) + (1 − λ)f (y) = λ x, f (x) + (1 − λ) y, f (y) fill the       segment joining the points x, f (x) and y, f (y) in RN +1 , i.e., the chord joining the corresponding points of the graph of f . Condition (7.1.1) says that the points of  the graph of f between x, f (x) and y, f (y) , for arbitrary x, y ∈ D, lie below the points of the corresponding chord. In many sources (where the authors are interested exclusively in continuous convex functions) condition (7.1.1) is taken as the definition of convexity. Then discontinuous convex functions are automatically eliminated. In the present book, however, a convex function is any function satisfying the Jensen-inequality (5.3.1). If we want the function in question to be continuous, we shall always distinctly say so. Contrary to discontinuous convex functions, which behave in a rather pathological manner, continuous convex functions display a nice behaviour. Continuous convex functions are very thoroughly treated, e.g., in Rockafellar [268], and Roberts-Varberg [267]. In the sequel of this chapter we are going to present only a few properties of continuous convex functions. Our exposition is based on Rockafellar [268]. 7.2–7.5 are devoted particularly to continuous convex functions of a single real variable. The results presented there have no simple extension to the case of several variables. In the rest if this chapter we deal with continuous convex functions of an arbitrary number of variables.

7.2 Compositions and inverses In the present section we investigate the convexity of compositions of, and of inverse functions to, convex and concave functions. Note that continuity of the functions in question is never explicitely used, however, a monotonic function an interval is locally bounded at every point of this interval, and hence a monotonicity condition on a convex or concave function authomatically implies its continuity. Only the functions f and ϕ in Theorem 7.2.1 below need not be continuous.

7.2. Compositions and inverses

163

Theorem 7.2.1. Let d ⊂ RN be a convex and open set, and let J ⊂ R be an open interval1 . Let f : D → J, and g : J → R, be functions, and let ϕ : D → R be the composition ϕ = g ◦ f = g (f ). Then: 1. 2. 3. 4.

if if if if

f f f f

is is is is

convex, and g is convex and increasing, then ϕ is convex; concave, and g is concave and increasing, then ϕ is concave; convex, and g is concave and decreasing, then ϕ is concave; concave, and g is convex and decreasing, then ϕ is convex.

Proof. Let f be convex and g convex and increasing. We have for arbitrary x, y ∈ D   f (x) + f (y) x+y  f 2 2 whence

     f (x) + f (y) g (f (x)) + g (f (y)) x+y g  , g f 2 2 2

i.e.,

 ϕ

x+y 2

 

ϕ (x) + ϕ (y) 2

Thus ϕ is convex. The remaining cases are dealt with similarly.



Note that if f is additive, then we do not need the monotonicity condition on g (cf. Example 5.3.1) The function g : R → R given by  xp for x  0, (7.2.1) g(x) = 0 for x < 0, is convex and increasing whenever p > 1 (cf. Theorem 7.5.2). Hence we obtain from Theorem 7.2.1 Corollary 7.2.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a nonnegative convex function. If p > 1, then the function ϕ : D → R, ϕ (x) = [f (x)]p , x ∈ D, is convex. Theorem 7.2.2. Let J ⊂ R be an open interval, and let f : J → R be a strictly monotonic function. Let f −1 : f (J) → J be the function inverse to f . Then: 1. 2. 3. 4.

if if if if

f f f f

is is is is

convex and increasing, then f −1 is concave; convex and decreasing, then f −1 is convex; concave and increasing, then f −1 is convex; concave and decreasing, then f −1 is concave.

By an open interval we understand in this book every interval of the form (a, b), where −∞  a < b  +∞

1

164

Chapter 7. Continuous Convex Functions

Proof. Let f be convex and increasing. Take arbitrary x, y ∈ f (J), and put u = f −1 (x), v = f −1 (y). By the convexity of f   f (u) + f (v) x+y u+v  = f 2 2 2 whence, since f −1 is increasing, just like f , u+v  f −1 2 or



x+y 2

f −1 (x) + f −1 (y)  f −1 2





,

x+y 2

 .

Thus f −1 is concave. The remaining cases are dealt similarly.



7.3 Differences quotients We start with the following result. Theorem 7.3.1. Let J ⊂ R be an open interval, and let f : D → R be a function. Each of the following conditions (postulated for every x1 , x2 , x3 ∈ J, x1 < x2 < x3 ) is necessary and sufficient for the function f to be continuous and convex: (x3 − x1 ) f (x2 )  (x2 − x1 ) f (x3 ) + (x3 − x2 ) f (x1 ) ;

(7.3.1)

f (x3 ) − f (x1 ) f (x2 ) − f (x1 )  ; x2 − x1 x3 − x1

(7.3.2)

f (x3 ) − f (x1 ) f (x3 ) − f (x2 )  . x3 − x1 x3 − x2

(7.3.3)

Proof. First we show that the three conditions (7.3.1), (7.3.2), (7.3.3) are equivalent to each other. Take arbitrary x1 , x2 , x3 ∈ J, x1 < x2 < x3 . Suppose that a function f : J → R fulfills (7.3.1). Then we get subtracting the term (x3 − x1 ) f (x1 ) from both the sides of (7.3.1) (x3 − x1 ) [f (x2 ) − f (x1 )]  (x2 − x1 ) [f (x3 ) − f (x1 )]

(7.3.4)

whence (7.3.2) follows on dividing by (x2 − x1 ) (x3 − x1 ) > 0. Conversely, (7.3.4) results from (7.3.2), and adding the term (x3 − x1 )f (x1 ) to both sides we obtain (7.3.1). Thus (7.3.1) and (7.3.2) are equivalent. Similarly, adding to both sides of (7.3.1) the term (x3 − x2 )f (x3 ) we obtain (x3 − x1 ) f (x2 ) + (x3 − x2 ) f (x3 )  (x3 − x1 ) f (x3 ) + (x3 − x2 ) f (x1 ) ,

(7.3.5)

whence (x3 − x2 ) [f (x3 ) − f (x1 )]  (x3 − x1 ) [f (x3 ) − f (x2 )] ,

(7.3.6)

7.3. Differences quotients

165

and (7.3.3) results on dividing by (x3 − x2 ) (x3 − x1 ) > 0. Conversely, (7.3.3) implies (7.3.6) and (7.3.5), and hence (7.3.1) follows by subtracting the term (x3 − x2 ) f (x3 ) from both the sides of (7.3.5). Thus (7.3.1) and (7.3.3) are equivalent. Hence it follows that all the three conditions (7.3.1), (7.3.2), (7.3.3) are equivalent to each other. Now let f : J → R be a continuous and convex function. By Theorem 7.1.1   f λx + (1 − λ)y  λf (x) + (1 − λ)f (y) (7.3.7) holds for arbitrary x, y ∈ J and λ ∈ [0, 1]. Take arbitrary x1 , x2 , x3 ∈ J, x1 < x2 < x3 , x2 − x1 and put in (7.3.7) x = x1 , y = x3 , λ = . Since x1 < x2 < x3 , we have x3 − x1 x2 − x1 0 < x3 − x2 < x3 − x1 , and consequently λ ∈ (0, 1). Then 1 − λ = , and x3 − x1 λx + (1 − λ) y =

x3 − x2 x2 − x1 x1 + x3 = x2 . x3 − x1 x3 − x1

Thus (7.3.7) yields f (x2 ) 

x3 − x2 x2 − x1 f (x1 ) + f (x3 ) , x3 − x1 x3 − x1

and (7.3.1) follows on multiplying both the sides by x3 − x1 > 0. Now assume that a function f : J → R fulfills (7.3.1) for arbitrary x1 , x2 , x3 ∈ J, x1 < x2 < x3 . Take arbitrary x, y ∈ J, x = y, and an arbitrary λ ∈ (0, 1). One of x, y is smaller than the other, e.g., x < y. Put in (7.3.1) x1 = x, x3 = y, x2 = λx + (1 − λ) y. We have x2 − x1 = λx + (1 − λ) y − x = (1 − λ) (y − x) > 0, and x3 − x2 = y − λx − (1 − λ) y = λ (y − x) > 0 so that x1 < x2 < x3 . (7.3.1) yields   (y − x)f λx + (1 − λ)y  (1 − λ)(y − x)f (y) + λ(y − x)f (x). Dividing both the sides by y − x > 0 we obtain (7.3.7). If x = y, or if λ = 0, or λ = 1, then (7.3.7) is trivially fulfilled. So f satisfies (7.3.7) for all x, y ∈ J and all λ ∈ [0, 1]. By Theorem 7.1.1 f is continuous and convex.  Now let J ⊂ R be an open interval, and let f : J → R be a function. We define a function I (x, h) for x ∈ J and h ∈ R such that h = 0, x + h ∈ J: I (x, h) =

f (x + h) − f (x) . h

(7.3.8)

Theorem 7.3.2. Let J ⊂ R be an open interval, and let f : J → R be a continuous and convex function. Then the corresponding function I defined by (7.3.8) is increasing with respect to either variable.

166

Chapter 7. Continuous Convex Functions

Proof. Fix an arbitrary x ∈ J, and take h1 , h2 ∈ R such that h1 < h2 , x + h1 ∈ J, x + h2 ∈ J. If 0 < h1 < h2 , put in (7.3.2) x1 = x, x2 = x + h1 , x3 = x + h2 . Then we get I (x, h1 )  I (x, h2 ). If h1 < h2 < 0, then we put in (7.3.3) x1 = x + h1 , x2 = x + h2 , x3 = x. We obtain f (x) − f (x + h1 ) f (x) − f (x + h2 )  , −h1 −h2 i.e.,

f (x + h2 ) − f (x) f (x + h1 ) − f (x)  , h1 h2

and so I (x, h1 )  I (x, h2 ). Finally, if h1 < 0 < h2 , then in the inequality f (x2 ) − f (x1 ) f (x3 ) − f (x2 )  , x2 − x1 x3 − x2 resulting from (7.3.2) and (7.3.3), we put x1 = x+h1 , x2 = x, x3 = x+h2 , and obtain again I (x, h1 )  I (x, h2 ). Consequently the function I is increasing with respect to h. Now take arbitrary x1 , x2 ∈ J, x1 < x2 , and an h = 0 such that x1 +h, x2 +h ∈ J. Using the monotonicity (already proved) of I with respect to the second variable, we have f (x2 + h) − f (x1 ) f (x1 + h) − f (x1 ) I (x1 , h) =  , h h + (x2 − x1 ) and    f (x2 + h) + − h − (x2 − x1 ) − f (x2 + h) f (x2 + h) − f (x1 ) = h + (x2 − x1 ) −h − (x2 − x1 )   f (x2 + h) + (−h) − f (x2 + h)  −h f (x2 + h) − f (x2 ) = I(x2 , h). = h This means that I is increasing also with respect to the first variable.



The converse is only partially true. Theorem 7.3.3. Let J ⊂ R be an open interval, and let f : J → R be a function. Let I be defined by (7.3.8). If for h > 0 the function I is increasing with respect to h, then f is continuous and convex. If I is increasing with respect to x, then the function is convex, but need not be continuous. Proof. Suppose that I is increasing with respect to h. Take arbitrary x1 , x2 , x3 ∈ J, x1 < x2 < x3 , and put x = x1 , h1 = x2 − x1 , h2 = x3 − x1 > h1 . Then I (x, h1 )  I (x, h2 ), i.e., f (x2 ) − f (x1 ) f (x3 ) − f (x1 )  . x2 − x1 x3 − x1

7.3. Differences quotients

167

Consequently f satisfies condition (7.3.2) of Theorem 7.3.1, and thus it is continuous and convex. Now let I be increasing with respect to x. Take arbitrary x, y ∈ J, x = y. Thus 1 one of x, y is smaller than the other; let, e.g., x < y. Put h = (y − x), x1 = x, 2 1 x2 = x + (y − x) = x1 + h > x1 . Hence I (x1 , h)  I (x2 , h), i.e., 2     x+y x+y − f (x) f (y) − f f 2 2  , 1 1 (y − x) (y − x) 2 2 or, after multiplication by

1 (y − x) > 0, and rearranging the remaining terms, 2   x+y 2f  f (x) + f (y) . (7.3.9) 2

If x = y, (7.3.9) is trivially fulfilled. (7.3.9) is equivalent to the Jensen inequality (5.3.1). Thus the function f is convex. The monotonicity of I with respect to x does not imply the continuity of f . For example, if f : R → R is a discontinuous additive function, then the function I (x, h) =

f (x + h) − f (x) f (h) = h h

does not depend on x, and thus, as a function of x, it is constant, and hence increasing. But f is discontinuous.  On the other hand, we may observe that the convexity of f alone does not imply the monotonicity if I with respect to x. Let g : R → R be a discontinuous additive 2 function, and let f : R → R be defined by f (x) = [g (x)] , x ∈ R. Thus f is convex (cf. Example 5.3.1). Fix an h such that g (h) = 0. We have I (x, h) = Hence

2 1 2 g (x) g (h) + [g (h)] . h h

h g (x) = 2g (h)

  1 2 I (x, h) − [g (h)] . h

If I were increasing with respect to x, then the function g would be monotonic, and hence continuous. Consequently I cannot be increasing with respect to x. As particular case of Theorem 7.3.2 we get Theorem 7.3.4. Let J ⊂ R be an open interval, and let f : J → R be a continuous and convex function. Then, for every fixed h > 0, the function Δh f (x) = f (x + h) − f (x) (defined for x ∈ J such that x + h ∈ J) is increasing. Proof. This follows from Theorem 7.3.2 and the equality Δh f (x) = hI (x, h)



168

Chapter 7. Continuous Convex Functions

Theorem 7.3.5. Let J ⊂ R be an open interval, J = (a, b), and let f : J → R be a continuous and convex function. Then either of f is monotonic in J, or there exist a point x0 ∈ J such that f is decreasing in (a, x0 ), and increasing in (x0 , b). Proof. If f is decreasing in J, there is nothing more to prove. Otherwise there exist u, v ∈ J such that u < v and f (u) < f (v). Then we have for every x ∈ (u, v) x = λu + (1 − λ) v with certain λ ∈ (0, 1), and hence, by Theorem 7.1.1, f (x)  λf (u) + (1 − λ) f (v) < λf (v) + (1 − λ) f (v) . Thus there exist arbitrary close points u, v ∈ J such that u < v and f (u) < f (v). Define a set A ⊂ J: A = {u ∈ J | For every ε > 0 there exist a v ∈ J such that u < v < u + ε and f (u) < f (v)}. As we have already seen, A = ∅. Put2 x0 = inf A. By what has already been established, f is decreasing in (a, x0 ). Now take arbitrary x, y ∈ J, x0  x  y. By the definition of x0 there exist a u ∈ A such that x0  u  x, and by the definition of A there exist a v ∈ J such that u < v < y and f (u) < f (v). By Theorem 7.3.2 0<

f (v) − f (u) f (y) − f (u) f (y) − f (x)   . v−u y−u y−x

Consequently f (x) < f (y), which proves that f is increasing in (x0 , b).



7.4 Differentiation According to Theorem 7.3.2, if f : J → R (J – an open interval) is convex and continuous, then for every fixed x ∈ J the differences quotient I (x, h) is an increasing function of h. Consequently it has finite one-sided limits as h tends to zero from the right, and from the left. But these limits are one-sided derivatives of f at x:   f+ (x) = lim I (x, h) , f− (x) = lim I (x, h) h→0+

h→0−

(7.4.1)

Thus we have Theorem 7.4.1. Let J ⊂ R be an open interval, and let f : J → R be a continuous  and convex function. Then at every point x ∈ J there exist the right derivative f+ (x),  and the right derivative f− (x), and we have for every x, y ∈ J, x < y,     f− (x)  f+ (x)  f− (y)  f+ (y) .

(7.4.2)

2 It may happen that x = a, but, as we prove further, the function f is increasing in (x , b). So, if 0 0 x0 = a, the function f is increasing in (x0 , b) = (a, b) = J, and the theorem is true also in this case.

7.4. Differentiation

169

Moreover,

lim f  (t) t→x+ + lim f  t→x− +

  = lim f− (t) = f+ (x) ,

(7.4.3)

  (t) = lim f− (t) = f− (x) .

(7.4.4)

t→x+

t→x−

Proof. The existence of derivatives (7.4.1) results from the monotonicity of I as a function of h, as has been pointed out above. Moreover, we have for every x, y ∈ J, x < y, and for every h such that x − h ∈ J, x + h ∈ J, 0 < h < y − x, I (x, −h)  I (x, h)  I (y, −h)  I (y, h) .

(7.4.5)

Only the inequality I (x, h)  I (y, −h) requires motivation, the remaining ones result from the inequality −h < h and from Theorem 7.3.2. Now, f (y − h) − f (y) f ((y − h) + h) − f (y − h) = −h h f (x + h) − f (x) = I (x, h)  h

I (y, −h) =

since y − h > x, and the function I is monotonic with respect to the first variable (Theorem 7.3.2). Letting in (7.4.5) h → 0+ we obtain inequalities (7.4.2).   Inequalities (7.4.2) show that the functions f+ and f− are increasing, and hence at every point x ∈ J they have one-sided limits. Since f is continuous, the function I is continuous with respect to x. Moreover, because of the monotonicity of I with respect to h we have for every t ∈ J and every fixed h > 0 such that t + x ∈ J  f+ (t)  I (t, h) .

(7.4.6)

Now fix an x ∈ J and an h > 0 such that x + h ∈ J. By (7.4.2) and (7.4.6) we have for every t > x such that t + x ∈ J    f+ (x)  f− (t)  f+ (t)  I (t, h) .

(7.4.7)

Letting in (7.4.7) t → x+ we obtain    f+ (x)  lim f− (t)  lim f+ (t)  I (x, h) , t→x+

t→x+

whence, as h → 0+,     f+ (x)  lim f− (t)  lim f+ (t)  f+ (x) . t→x+

t→x+

(7.4.8)

Inequalities (7.4.8) yield (7.4.3). Relation (7.4.4) may be established in a similar manner.  Corollary 7.4.1. Let J ⊂ R be an open interval, and let f : J → R be a continuous   and convex function. Then the functions f+ , f− : J → R are increasing.

170

Chapter 7. Continuous Convex Functions

Theorem 7.4.2. Let J ⊂ R be an open interval, and let f : J → R be a continuous and convex function. Then the following three conditions are equivalent for every x ∈ J: (i) f is differentiable at x;  (ii) f+ is continuous at x;  is continuous at x. (iii) f−    Proof. Condition (i) means that f+ (x) = f− (x), condition (ii) that lim f+ (t) = t→x+

   lim f− (t), and condition (iii) that lim f− (t) = lim f− (t). Relations (7.4.3) and

t→x−

t→x−

t→x−

(7.4.4) show that all the three conditions are equivalent.



Theorem 7.4.3. Let J ⊂ R be an open interval, and let f : J → R be a continuous and convex function. Then f is differentiable in J except at most countably many points. If J0 ⊂ J is the set of the points of the differentiability of f , then the function f  : J0 → R is increasing and continuous.  Proof. The function f+ , being monotonic, may be discontinuous at at most countably many points. In view of Theorem 7.4.2 we obtain hence the first part of our assertion.  The second results from the fact that for x ∈ J0 we have f  (x) = f+ (x), and by the  Theorem 7.4.2 f+ is continuous at every point of J0 . 

In sequel we say that f is twice differentiable at a point x0 ∈ J iff x0 ∈ J0 (the set of the points of differentiability of f ) and the limit f  (x0 ) = y→x lim y∈J0

f  (y) − f  (x0 ) y − x0

(7.4.9)

exists. Limit (7.4.9) is equal to f  (x0 ) = lim

y→x0

  f+ (y) − f+ (x0 ) , y − x0

(7.4.10)

 is monotonic, it follows whenever the latter limit exists. But since the function f+ from the famous Theorem of Lebesgue (cf., e.g., L  ojasiewicz [208]) that limit (7.4.10) exists almost everywhere in J. Moreover, the difference quotient in (7.4.10) is non negative (since f+ is increasing), and consequently so is also f  (x0 ) if it exist. Thus we have

Theorem 7.4.4. Let J ⊂ R be an open interval, and let f : J → R be a continuous and convex function. Then f is twice differentiable almost everywhere in J. Whenever it exist, f  (x) > 0.

7.5. Differential conditions of convexity

171

Since f  (defined almost everywhere in J) is monotonic, it is measurable, and bounded on every compact subinterval of J. Thus the expression3 x

f  (t)dt

(7.4.11)

x0

is meaningful for every x0 , x ∈ J. We have the following Theorem 7.4.5. Let J ⊂ R be an open interval, and let f : J → R be a continuous and convex function. Then, for every x0 , x ∈ J, we have x f (x) = f (x0 ) +

f  (t)dt.

(7.4.12)

x0

Proof. Put

x F (x) = f (x0 ) +

f  (t)dt.

x0 

The function F is differentiable whenever f is continuous, and thus everywhere in J with the exception of an at most countable set. If x ∈ J is a point of the continuity of f  , then we have F  (x) = f  (x), i.e., (F − f ) (x) = 0. By Theorem 3.6.3 (cf. also Exercise 3.9) F − f = const. Since at the point x0 we have F − f = 0, the constant must be zero, and F = f . Hence we obtain (7.4.12).  As an immediate consequence of Theorem 7.4.5 we obtain Theorem 7.4.6. Let J ⊂ R be an open interval, and let f : J → R be a continuous and convex function. The function f is absolutely continuous in J.

7.5 Differential conditions of convexity The convexity of differentiable function can be inferred from the behaviour of its derivatives. The theorems in this section are in some respect converse to those of 7.4. Theorem 7.5.1. Let J ⊂ R be an open interval, and let f : J → R be a differentiable function. The function f is convex if and only if the function f  is increasing in J. Proof. The “only if” part results from Corollary 7.4.1. now assume that f  is increasing in J, and take arbitrary x, y ∈ J, x < y. We have by the mean-value theorem     x+y y−x x+y  = f (v) y − = f  (v) f (y) − f , 2 2 2     x+y y−x x+y − f (x) = f  (u) − x = f  (u) , f 2 2 2 3

The integral in (7.4.11) and (7.4.12) are Lebesgue integrals, but everything remains unchanged if &x  f+ (t)dt we replace them by the Riemann integral x0

172

Chapter 7. Continuous Convex Functions

where u, v ∈ J are points such that x
x+y < v < y. 2

Hence f  (u)  f  (v), i.e.,     x+y x+y − f (x)  f (y) − f , f 2 2 

and f

x+y 2

 

f (x) + f (y) 2

(7.5.1)

Relation (7.5.1) has been so far established for x < y, but due to the symmetry of (7.5.1) with respect to x and y we actually obtain (7.5.1) for x = y. For x = y (7.5.1) is obvious.  Theorem 7.5.2. Let J ⊂ R be an open interval, and let f : J → R be a twice differentiable function. The function f is convex if and only if the function f  is non-negative in J. Proof. This is a direct consequence of Theorem 7.5.1 above.



It follows from Theorem 7.5.2 that the function (7.2.1) (p > 1), and the function exp are convex in R, the fact that has already been used earlier in this book. For every function f : J → R and for every x ∈ J the expressions (finite or not) D+ f (x) = lim sup I (x, h) ,

D− f (x) = lim sup I (x, h) ,

h→0+

d+ f (x) = lim inf I (x, h) , h→0+

h→0−

d− f (x) = lim inf I (x, h) , h→0−

are meaningful. They are called the Dini derivatives of f at x. Thus D+ f , D− f , d+ f and d− f are functions from J into [−∞, ∞]. Lemma 7.5.1. Let J ⊂ R be an open interval, and let f : J → R be a continuous function. If one of the Dini derivatives of f is finite and increasing in J, then f is absolutely continuous in J. Proof. Assume that the upper right Dini derivative D+ f of f is finite and increasing in J. Thus D+ f is measurable and bounded on every compact subinterval of J. &y Consequently for every x, y ∈ J the integral |D+ f (t)| dt is meaningful. In order to x

prove the lemma it is enough to show that for every x, y ∈ J, x < y, we have y |f (y) − f (x)|  x

# + # #D f (t)# dt.

(7.5.2)

7.5. Differential conditions of convexity

173

Fix an x ∈ J, and put y ϕ (y) = |f (y) − f (x)| ,

ψ (y) =

# + # #D f (t)# dt.

x

We have

# #  ψ (y) = #D+ f (y)#

(7.5.3)

at every point y ∈ J at which D+ f , and hence also |D+ f |, is continuous. Take arbitrary y ∈ J such that D+ f is continuous at y, and take an arbitrary sequence of points yn ∈ J, yn > y, n ∈ N, such that lim yn = y

n→∞

and lim

n→∞

f (yn ) − f (y) = D+ f (y). yn − y

Since ϕ(yn ) − ϕ(y) = |f (yn ) − f (y)| − |f (y) − f (x)|  |f (yn ) − f (y)|, we get hence d+ ϕ (y)  lim inf n→∞

# ϕ (yn ) − ϕ (y) ## +  D f (y)# . yn − y

(7.5.4)

By (7.5.3) and (7.5.4) # # D+ (ψ − ϕ) (y) = ψ  (y) − d+ ϕ (y)  #D+ f (y)# − d+ ϕ (y)  0. Thus D+ (ψ − ϕ)  0 at all points of continuity of D+ f , and since the latter function is increasing, D+ (ψ − ϕ)  in J except at at most countably many points. By Theorem 3.6.3 the function ψ − ϕ is increasing. For x = y we have ϕ (x) = ψ (x) = 0, and so ψ (y)  ϕ (y) for y > x. But this is just relation (7.5.2).  Theorem 7.5.3. Let J ⊂ R be an open interval, and let f : J → R be a continuous function. If one of the Dini derivatives of f is finite and increasing in J, then f is convex. Proof. Assume that the upper right Dini derivative D + f of f is finite and increasing in J. By Lemma 7.5.1 f is absolutely continuous in J, and hence it is differentiable almost everywhere in J, and we have x f (x) = f (x0 ) +

x



D+ f (t) dt,

f (t) dt = f (x0 ) + x0

(7.5.5)

x0

where x0 ∈ J is an arbitrary fixed point. Take arbitrary x, y ∈ J, x < y. By (7.5.5)  f (y) − f

x+y 2





y D+ f (t)dt,

= 1 2 (x+y)

f

x+y 2

1 2 (x+y) 

 − f (x) =

D+ f (t)dt. x

(7.5.6)

174

Chapter 7. Continuous Convex Functions

Since D+ f is increasing, and x <

x+y < y, we have 2 1 2 (x+y) 

y D+ f (t)dt 

D+ f (t)dt, x

1 2 (x+y)

whence by (7.5.6)  f (y) − f

x+y 2



 f

x+y 2

 − f (x).

Hence we obtain (7.5.1), and in the same way as in the proof of Theorem 7.5.1 it follows that f is convex. 

7.6 Functions of several variables In the present section we consider arbitrary families {fα }α∈A of continuous and convex functions. Lemma 7.6.1. Let D ⊂ RN be a convex and open set, and let A = ∅ be an arbitrary set. Suppose that for every α ∈ A we are given a continuous and convex function fα : D → R. Then the function f : D → (−∞, ∞), f (x) = sup fα (x) satisfies the α

inequality

  f λx + (1 − λ) y  λf (x) + (1 − λ) f (y)

(7.6.1)

for all x, y ∈ D and λ ∈ [0, 1].

  Proof. Take arbitrary x, y ∈ D, λ ∈ [0, 1], and a ∈ R such that a < f λx + (1 − λ)y . There exist an α ∈ A such that   a < fα λx + (1 − λ)y Hence by Theorem 7.1.1   a < fα λx + (1 − λ)y  λfα (x) + (1 − λ)fα (y)  λf (x) + (1 − λ)f (y).   Letting a tend to f λx + (1 − λ)y , we obtain (7.6.1).



Lemma 7.6.2. Let D ⊂ RN be a convex and open set, and let Δ ⊂ D be a dense subset of D, and let A = ∅ be an arbitrary set. Suppose that for every α ∈ A the function fα : D → R is continuous and convex, and sup fα (x) < ∞ α

forx ∈ Δ,

inf fα (x0 ) > −∞ α

(7.6.2) (7.6.3)

for an x0 ∈ D. Then the functions fα , α ∈ A are jointly bounded on every compact set contained in D.

7.6. Functions of several variables

175

Proof. Put f (x) = sup fα (x), α

x ∈ D.

By (7.6.2) f (x) < ∞ for x ∈ Δ and by Lemma 7.6.1 f satisfies (7.6.1) for all x, y ∈ D, λ ∈ [0, 1]. Take an arbitrary x ∈ D. According to the Remark in 5.1 x is an interior point of an N -dimensional simplex contained in D. Very small changes of the vertices of this simplex do not affect this fact, so we may assume that the vertices of the simplex belong to Δ. Consequently there exist points p1 , . . . , pN +1 ∈ Δ, and numbers λ1 , . . . , λN +1 ∈ [0, 1] such that x=

N +1

λi pi ,

i=1

N +1

λi = 1.

i=1

Hence by Lemma 5.3.2 f (x) 

N +1

λi f (pi ) < ∞.

i=1

Consequently f (x) < ∞ for all x ∈ D, and in virtue of Lemma 7.6.1, and Theorem 7.1.1 f is continuous and convex function. Being continuous, f is bounded above on every compact set contained in D, and so is also every function fα  f , α ∈ A. Now write L = inf fα (x0 ) > −∞ (7.6.4) α   cf. (7.6.3) , and let K be a closed ball centered at x0 and with a radius r > 0 such that K ⊂ D. The ball K is a compact set, so by the first part of the proof theres exist a constant M such that fα (x)  M

for x ∈ K and α ∈ A.

(7.6.5)

Take an arbitrary x ∈ D, x = x0 , and put y = x0 −

r (x − x0 ) |x − x0 |

Then |y − x0 | = r so that y ∈ K. Moreover, x0 = λx + (1 − λ) y with λ=

r ∈ (0, 1) . r + |x − x0 |

For every α ∈ A we have by (7.6.6) fα (x)  λfα (x) + (1 − λ)fα (y), whence by (7.6.4) and (7.6.5) fα (x) 

1−λ 1 1 1−λ 1 fα (x0 ) − fα (y)  L − M  (L − M ) , λ λ λ λ λ

(7.6.6) (7.6.7)

176

Chapter 7. Continuous Convex Functions

and by (7.6.7)

L−M (r + |x − x0 |) , α ∈ A . (7.6.8) r So far estimation (7.6.8) has been derived for all x ∈ D, x = x0 . For x0 we have according to (7.6.4), for all α ∈ A, fα (x) 

fα (x0 )  L  L − M =

L−M (r + |x − x0 |) . r

Thus (7.6.8) is valid for all x ∈ D. The expression on the right-hand side of (7.6.8) is a continuous function of x, and thus it is bounded below on every compact set  contained in D, and in virtue of (7.6.8) so is also every function fα , α ∈ A. Theorem 7.6.1. Let D ⊂ RN be a convex and open set, and let Δ ⊂ D be a dense subset of D and let A = ∅ be an arbitrary set. Suppose that for every α ∈ A the function fα : D → Ris continuous and convex and that relations (7.6.2) and (7.6.3) hold with an x0 ∈ D. Then all functions fα , α ∈ A, fulfil a Lipschitz condition with a common constant on every compact set contained in D.  Proof. Take  a compact C ⊂ D, C = ∅, and letd (C, D ) be the distance of the sets C and D we assume d (C, D ) = ∞ if D = ∅ . Since C and D are disjoint closed sets, d (C, D ) > 0, and so we can find an r,

0 < r < d (C, D ) .

(7.6.9)

Further, let d (x, C) denote the distance of the point x from C and put " ! K = x ∈ RN | d (x, C)  r . K is a compact set, and by (7.6.9) K ⊂ D. By Lemma 7.6.2 there exist constant Q and M such that −∞ < Q  fα (x)  M < ∞

for all x ∈ K, α ∈ A.

(7.6.10)

Now take arbitrary x, y ∈ C, x = y, and put z=y−

r (x − y) . |x − y|

We have |z − y| = r, whence z ∈ K; moreover x ∈ C ⊂ K. Since y = λx + (1 − λ) z with λ=

r , r + |x − y|

we have by (7.6.10) for every α ∈ A fα (y)  λfα (x) + (1 − λ) fα (z) = fα (x) − (1 − λ) [fα (x) − fα (z)]  fα (x) − (1 − λ) (Q − M ) = fα (x) + (1 − λ) (M − Q) ,

(7.6.11)

7.7. Derivatives of a function

177

and by (7.6.11) M −Q |x − y| (M − Q)  |x − y| . r + |x − y| r (7.6.12) Since the role of x and y has been symmetric, we may interchange x and y arriving at M −Q fα (x) − fα (y)  |x − y| . (7.6.13) r Relations (7.6.12) and (7.6.13) together yield fα (y) − fα (x)  (1 − λ) (Q − M ) =

|fα (x) − fα (y)| 

M −Q |x − y| . r

(7.6.14)

Relation (7.6.14), so far obtained for x = y, y ∈ C and all α ∈ A, is evidently valid also for x = y. Condition (7.6.14) means that all functions fα , α ∈ A fulfill on C the Lipschitz condition with the same constant (M − Q) /r.  Taking as A a singleton, we obtain as a particular case of Theorem 7.6.1 Theorem 7.6.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a continuous and convex function. The function fulfils a Lipschitz condition on every compact set contained in D. Let us note that Theorem 7.4.6 results also from Theorem 7.6.2 above.

7.7 Derivatives of a function For functions of several variables we have three different notions of the differentiability at a point, in general not equivalent to each other4 . The simplest fact is that the function in question, say f , has at a point x all the partial derivatives ∂f /∂ξi . This fact refers only to the behaviour of f may be quite wild5 . Therefore we need stronger conditions. A function f : RN → R is called linear if the function f0 defined by f0 (x) = f (x) − f (0) is additive and homogeneous: f0 (λx) = λf0 (x)

(7.7.1)

for all λ ∈ R. If (7.7.1) holds only for positive λ ∈ R, the function f0 (independently of whether it is additive, or not) is called positively homogeneous. If f : RN → R is linear, then for the corresponding function f0 we have by (7.7.1)     f0 λx + (1 − λ)y = f0 (λx) + f0 (1 − λ)y = λf0 (x) + (1 − λ)f0 (y) 4 In the one-dimensional space (N = 1) the difference between the three notions disappears and they are all equivalent to each other. But since for N=1 the problem of the differentiability of continuous convex functions are related questions have been thoroughly discussed in 4–5, here we may restrict ourselves to N > 1. 5 However, if we know that the partial derivatives exist at every point of an open set D, and are continuous in D, then it follows already that the function in question has a Stolz differential (cf. below) at every point of D.

178

Chapter 7. Continuous Convex Functions

for every x, y ∈ RN , λ ∈ [0, 1]. Thus f0 we have by (7.7.1), and by Theorem 7.1.1 it is continuous. Now it follows from Theorem 5.5.2 that f0 (x) = cx with a certain c ∈ RN (cx is the scalar product), and f (x) = cx + b,

x ∈ RN ,

(7.7.2)

where b = f (0). Conversely, every function f : RN → R of form (7.7.2) is linear. Let D ⊂ RN be an open set, and let f : D → R be a function. We say that f is differentiable at a point x ∈ D iff for every y ∈ RN there exist the limit f (x + λy) − f (x) , λ→0+ λ

fy (x) = lim

(7.7.3)

and fy (x) is a linear function of y. It follows from (7.7.3) that we always have f0 (x) = 0

(7.7.4)

Lemma 7.7.1. Let D ⊂ RN be an open set, and let f : D → R be a function. If f is differentiable at a point x ∈ D, then f has all partial derivatives at x. Proof. Let ei = (0, . . . , 0, 1, 0, . . . , 0) (1 at i-th place), i = 1, . . . , N , be the usual orthonormal base of RN (over R). There exist the limits (i = 1, . . . , N ) f (x + λei ) − f (x) = fe i (x) λ→0+ λ lim

lim

λ→0−

f (x + λei ) − f (x) f (x − λei ) − f (x)  = − lim = −f−e (x) . i λ→0+ λ λ

It follows from the linearity of the function f. (x) and from condition (7.7.4) that  f. (x) is homogeneous so that in particular −f−e (x) = fe i (x). Consequently there i exists the limit f (x + λei ) − f (x) ∂f = lim (x) , λ→0 λ ∂ξi i = 1, . . . , N . Thus f has all its partial derivatives at x.



Conversely, the function f : R2 → R which is one whenever ξ1 , ξ2 > 0 (x = (ξ1 , ξ2 )), and 0 otherwise, has at zero the partial derivatives equal to zero, but is not differentiable at zero, since limit (7.7.3) does not exist for x = 0 and y such that f (y) = 1. But also this notion of differentiability is not sufficient in order to rend the function in question sufficiently smooth around the point at which it is differentiable. For example, the function f : R2 → R defined as  1 if ξ2 = ξ12 , ξ1 > 0, (7.7.5) f (x) = f (ξ1 , ξ2 ) = 0 otherwise,

7.7. Derivatives of a function

179

  is differentiable at zero fy (0) = 0 for all y ∈ RN but is not continuous at the point x = 0. So we introduce one more notion. Let D ⊂ RN be an open set, and let f : D → R be a function. We say that f has a Stolz differential at the point x ∈ D iff there exists an x∗ ∈ RN such that f (z) = f (x) + x∗ (z − x) + r(x, z)

forz ∈ D,

(7.7.6)

where the function r : R2N → R fulfills the condition lim

z→x

r(x, z) = 0.6 |z − x|

(7.7.7)

If it exists, x∗ in (7.7.6) is called the Stolz differential or the gradient of f at x, and is denoted by ∇f (x) . In the above example function (7.7.5) has no Stolz differential at zero. In the opposite direction we have the following Lemma 7.7.2. Let D ⊂ RN be an open set, and f : D → R be a function. If f has a Stolz differential ∇f (x) at a point x ∈ D, then it is differentiable at x, and we have   ∂f ∂f ∇f (x) = ,..., , (7.7.8) ∂ξ1 (x) ∂ξN (x) and

  fy (x) = ∇f (x) y.

(7.7.9)

Proof. Take an arbitrary y ∈ RN , y = 0. Putting in (7.7.6) z = x + λy, λ > 0, we have by (7.7.7)   f (x + λy) − f (x) − ∇f (x) λy lim = 0, λ→0+ λ|y| i.e.,

   1 f (x + λy) − f (x)  − ∇f (x) λy = 0. λ→0+ |y| λ lim

Hence we obtain (7.7.9). According to (7.7.4), (7.7.9) holds for y = 0. It follows from (7.7.9) that fy (x) is a linear function of y. (7.7.8) results from (7.7.9) on putting y = ei , i = 1, . . . , N (compare the proof of Lemma 7.7.1).  In the sequel we will need one more notion. Let D ⊂ RN be an open set, and let f : D → R be a function. Any x∗ ∈ RN such that f (z)  f (x) + x∗ (z − x)

for z ∈ D

(7.7.10)

is called a subgradient of f at x. The set of all subgradients of f at x will be denoted by ∂f (x). Of course, it may happen that ∂f (x) = ∅. 6 Of course, every function f can be written in form (7.7.6) with every x∗ ∈ RN . It is enough to put r (x, z) = f (z) − f (x) − x∗ (z − x). Here the problem lies in that x∗ should be chosen in such a manner that the corresponding function r should satisfy (7.7.7).

180

Chapter 7. Continuous Convex Functions

The simple examples given above show that a function of several variables may have all partial derivatives at a point, but be non-differentiable at this point, and may be differentiable at a point, but have no Stolz differential at this point. However, the considerations of next section aim at showing that in the case of a continuous convex function all the three conditions (the existence of partial derivatives, the differentiability, the existence of a Stolz differential) are equivalent to each other.

7.8 Derivatives of convex functions In this section we will be concerned with continuous and convex functions. Lemma 7.8.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex and continuous function. At every point x ∈ D the derivative fy (x) exist for every y ∈ RN . Moreover, f. (x) is a convex, continuous and positively homogeneous function and, for every y ∈ RN ,  −fy (x)  f−y (x). (7.8.1) Proof. Fix an x ∈ D and y ∈ RN , and put J = {λ ∈ R | x + λy ∈ D}. J is an open interval containing zero. Further put g(λ) = f (x + λy) − f (x),

λ∈J.

(7.8.2)

The function g : J → R defined by (7.8.2) clearly is continuous, and we have for arbitrary λ, μ ∈ J       λ+μ λ+μ (x + λy) + (x + μy) g =f x+ y − f (x) = f − f (x) 2 2 2 f (x + λy) + f (x + μy)  − f (x) 2 [f (x + λy) − f (x)] + [f (x + μy) − f (x)] g (λ) + g (μ) = = 2 2 so that g is convex. By Theorem 7.4.1 there exist the derivative  g+ (0) = lim

λ→0+

g (λ) f (x + λy) − f (x) = lim = fy (x) . λ→0+ λ λ

Now, we have for arbitrary u, v ∈ RN , μ ∈ [0, 1] and λ > 0 sufficiently small 1 [f (x + λ [μu + (1 − μ) v]) − f (x)] λ 

1  f μ (x + λu) + (1 − μ) (x + λv) − f (x) = λ 1  [μf (x + λu) + (1 − μ) (x + λv) − f (x)] λ f (x + λu) − f (x) f (x + λv) − f (x) =μ + (1 − μ) , λ λ

7.8. Derivatives of convex functions

181

and as λ → 0+ we obtain in the limit  fμu+(1−μ)v (x)  μfu (x) + (1 − μ) fv (x) .

By Theorem 7.1.1 the function f. (x) is convex and continuous. Further, we have for arbitrary positive μ ∈ R and λ > 0 sufficiently small f (x + λμy) − f (x) f (x + λμy) − f (x) =μ , λ λμ hence, letting λ → 0+, we obtain  fμy (x) = μfy (x),

i.e., f. (x) is positively homogeneous. Finally we have for arbitrary y ∈ RN by (7.7.4) and by the convexity of f. (x) 0 = f0 (x) = f y−y (x)  2

1   fy (x) + f−y (x) , 2 

whence (7.8.1) follows.

Lemma 7.8.2. 7 Let D ⊂ RN be a convex and open set, and let f : D → R be a convex and continuous function. For every x ∈ D, we have ∂f (x) = ∅ Proof. Consider the set " ! A = (x, y) ∈ RN +1 | x ∈ D, y > f (x)   The set A is open and convex (cf. Exercise 7.1)and the points x, f (x) ∈ RN +1 are its frontier points. Take an x ∈ D. According toTheorem  5.1.7 there exist a support hyperplane H of A passing through the point x, f (x) . The hyperplane H has an equation of the form   c(z − x) + α y − f (x) = 0, (7.8.3) where c ∈ RN , α ∈ R, and (z, y) is the current point of RN +1 . If we had α = 0, then every point (x, y) with an arbitrary y ∈ R, would satisfy (7.8.3), and taking y > f (x) we would get that H ∩ A = ∅, and so A being open, would have points on both sides of H, which is impossible. Consequently α = 0. So we may write (7.8.3) in the form y = f (x) −

c (z − x). α

The two halfspaces into which H divides RN +1 are determined by the inequalities y > f (x) − 7

c (z − x) α

and

The converse result is also true. Cf. Exercise 7.8.

y < f (x) −

c (z − x) . α

182

Chapter 7. Continuous Convex Functions

c For arbitrary y > f (x) the point (x, y) ∈ A and clearly we have y > f (x) − (x − x) α so that the set A must be contained in the half-space determined by the inequality y > f (x) −

c (z − x). α

Take an arbitrary z ∈ D, and an arbitrary t > f (z). Then (z, t) ∈ A, whence t > f (x) −

c (z − x) . α

Letting t → f (z), we obtain hence c (z − x) , α c and this is valid for all z ∈ D, which shows that − ∈ ∂f (x), and ∂f (x) = ∅. α f (z)  f (x) −



Lemma 7.8.3. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex and continuous function. For every x∗ ∈ RN we have x∗ ∈ ∂f (x) if and only if fy  x∗ y

(7.8.4)

for all y ∈ RN . Proof. By the definition of a subgradient, x∗ ∈ ∂f (x) iff (7.7.10) holds. Fix a λ > 0 1 and put y = (z − x). Then (7.7.10) goes into λ f (x + λy)  f (x) + x∗ λy.

(7.8.5)

Conversely (7.7.10) results from (7.8.5) on putting z = x + λy. thus x∗ ∈ ∂f (x) iff (7.8.5) holds for all y ∈ RN and λ > 0 such that x + λy ∈ D. Now, if x∗ ∈ ∂f (x), then we have by (7.8.5) f (x + λy) − f (x)  x∗ y λ for all y ∈ RN and λ > 0 such that x + λy ∈ D, whence on passing to the limit as λ → 0+ we obtain (7.8.4). Conversely, let (7.8.4) hold. Function (7.8.2) is continuous and convex, whence by Theorem 7.3.2  g(λ)  g+ (0) λ for λ > 0, i.e., by (7.8.4) f (x + λy) − f (x)  fy (x)  x∗ y. λ Hence we obtain (7.8.5), i.e., x∗ ∈ ∂f (x).



7.8. Derivatives of convex functions

183

Lemma 7.8.4. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. Then, for every x ∈ D and y ∈ RN , fy (x) =

sup

x∗ y.

x∗ ∈∂f (x)

Proof. In virtue of Lemma 7.8.3 it is enough to show that there exists an x∗ ∈ ∂f (x) such that fy (x) = x∗ y (7.8.6) (x ∈ D and y ∈ RN fixed). If y = 0, (7.8.6) results from (7.7.4) and Lemma 7.8.2. So let y = 0. By a simple transformation of the coordinate system we can make y = e1 . We define a function ϕ : RN → R putting ϕ(z) = fz (x). By Lemma 7.8.1 ϕ is convex, continuous and positively homogeneous. Next we define a function ψ : RN −1 → R putting ψ(ζ2 , . . . , ζN ) = ϕ(1, ζ2 , . . . , ζN ) It is easily seen that also the function ψ is convex and continuous. For every z = (ζ1 , . . . , ζN ) ∈ RN we write z˜ = (ζ2 , . . . , ζN ) ∈ RN −1 . By Lemma 7.8.2 there exists a z0∗ ∈ ∂f (0) (so that, in particular, z0∗ ∈ RN −1 ). By the definition of a subgradient we have ψ(˜ z ) − ψ(0)  z0∗ z˜ (7.8.7) for every z ∈ RN . Now let H ⊂ RN be the hyperplane " ! H = z = (ζ1 , . . . , ζN ) ∈ RN | ζ1 = 1 . For z ∈ H we have ϕ(z) = ψ(˜ z ) whence by (7.8.7) z ) − ψ(0)  z0∗ z˜, ϕ(z) − ϕ(e1 ) = ψ(˜

z ∈H.

(7.8.8)

Put z ∗ = (ϕ (e1 ) , z0∗ ) ∈ Rn . For z ∈ H we have by (7.8.8) ϕ(z)  z ∗ z,

z ∈H.

(7.8.9)

In particular, according to the definition of z ∗ ϕ(e1 ) = z ∗ e1 .

(7.8.10)

Let P be the half-space " ! P = z = (ζ1 , . . . , ζN ) RN | ζ1 > 0 . For every z ∈ P there exist a zˆ ∈ H and a λ > 0 such that z = λˆ z . Hence we get by (7.8.9) and by the positive homogeneity of ϕ ϕ(z) = ϕ(λˆ z ) = λϕ(ˆ z )  λz ∗ zˆ = z ∗ z,

z∈P.

(7.8.11)

184

Chapter 7. Continuous Convex Functions

Now take an arbitrary z ∈ RN and consider the point z¯ = λz + (1 − λ)e1

(7.8.12)

with λ ∈ R. When λ → 0, point (7.8.12) approaches e1 ∈ P = int P so we may find a λ ∈ (0, 1) such that for point (7.8.12) we have z¯ ∈ P . Suppose that we have fixed a suitable λ ∈ (0, 1) and z¯ ∈ P given by (7.8.12) and that ϕ(z) < z ∗ z.

(7.8.13)

Then we have by (7.8.12), (7.8.10) and by the convexity of ϕ   ϕ(¯ z )  λϕ(z) + (1 − λ)ϕ(e1 ) < λz ∗ z + (1 − λ)z ∗ e1 = z ∗ λz + (1 − λ)e1 = z ∗ z¯, which contradicts (7.8.11), since z¯ ∈ P . The contradiction obtained shows that supposition (7.8.13) was false and that we actually have ϕ(z)  z ∗ z for all z ∈ RN . By Lemma 7.8.3 z ∗ ∈ ∂f (x). (7.8.10) means that we have (7.8.6) with x∗ = z ∗ and y = e1 .  Lemma 7.8.5. Let f : RN → R and g : RN → R be additive functions. If f (x)  g (x) for all x ∈ RN , then f = g. Proof. By Theorem 5.2.1 f and g are odd. Suppose that for an x ∈ RN we have f (x) < g(x). Then g(−x) − f (−x) = −g(x) + f (x) = − [g(x) − f (x)] < 0 i.e., g(−x) < f (−x), contrary to the assumption. Thus we must have f (x) = g(x) for all x ∈ RN .  Theorem 7.8.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex and continuous function. If at a point x ∈ D the function f has Stolz differential ∇f (x), then ∂f (x) = {∇f (x)}. Conversely, if ∂f (x) is a singleton, ∂f (x) = {x∗ }, then f has at x Stolz differential ∇f (x), and ∇f (x) = x∗ . Proof. Assume that at a certainpoint x ∈ D the function f has Stolz differential ∇f (x). By Lemma 7.7.2 ∇f (x) y = fy (x). Take any x∗ ∈ ∂f (x). By Lemma 7.8.3 fy (x)  x∗ y, or   ∇f (x) y  x∗ y (7.8.14) the sides of (7.8.14) are additive functions of for all y ∈ RN . The expressions  on both  y, whence by Lemma 7.8.5 ∇f (x) y = x∗ y for all y ∈ RN . But this is only possible if x∗ = ∇f (x). Consequently ∂f (x) = {∇f (x)}. Conversely, assume ∂f (x) to be a singleton, ∂f (x) = {x∗ }. We define a function p : (D − x) → R by p (y) = f (x + y) − f (x) − x∗ y.

7.8. Derivatives of convex functions

185

Since x∗ is a subgradient, p  0 in D − x, and moreover, as may be easily checked, we have for every u, v ∈ D − x and μ ∈ [0, 1]   p μu + (1 − μ) v  μp (u) + (1 − μ) p (v) , whence it follows in virtue of Theorem 7.1.1 that the function p is convex and continuous. Also the function g : R → R defined for a fixed, but arbitrary, y ∈ RN by (7.8.2), is convex and continuous. Thus we have by Theorem 7.3.2 g (λ) f (x + λy) − f (x) g(λ) =  lim = fy (x) λ→0+ λ λ λ for every y ∈ RN and λ > 0 such that x + λy ∈ D. Taking λ = 1 we obtain hence by Lemma 7.8.3 p (y) = f (x + y) − f (x) − x∗ y  fy (x) − x∗  0 (7.8.15) for every y ∈ D − x. Since p (0) = 0, (7.8.15) can be written as p (y)  p (0) + 0 (y − 0) , which shows that 0 ∈ ∂p (0). And if a y ∗ ∈ ∂p (0), then we have for all y ∈ D − x p (y)  p (0) + y ∗ (y − 0) = y ∗ y, or

f (x + y)  f (x) + (x∗ + y ∗ ) y.

This means that x∗ + y ∗ ∈ ∂f (x) = {x∗ }, whence x∗ + y ∗ = x∗ , and y ∗ = 0. Consequently ∂p (0) = {0}, and by Lemma 7.8.4 pu (0) = 0 for every u ∈ RN . Now write p (λu) , λ > 0. h (λ; u) = λ Since p(0) = 0, we have h(λ; u) = and

p(0 + λu) − p(0) p(λu) − p(0) = , λ λ lim h(λ; u) = pu (0) = 0

(7.8.16)

λ→0+

The set D − x is open, and 0 ∈ D − x, so by the Remark in 5.1 there exists a simplex S = conv{p1 , . . . , pN +1 } such that S ⊂ D − x and 0 ∈ int S. Consequently there exists a closed ball K centered at the origin and with a radius r > 0, contained in S. Take an arbitrary u ∈ K. Then N +1 by Theorem 5.1.3 there exist λ1 , . . . , λN +1 ∈ [0, 1] such that λi = 1 and i=1

u=

N +1

i=1

λi pi .

186

Chapter 7. Continuous Convex Functions

Hence we have, using the convexity and continuity if p, and Lemma 5.3.2, for λ ∈ (0, 1),  N +1 N +1 +1

p(λpi ) N

1 1 h(λ; u) = p(λu) = p λi (λpi )  λi h(λ; pi ), (7.8.17)  λ λ λ i=1 i=1 i=1 since for λ ∈ (0, 1) we have λpi = λpi + (1 − λ) 0 ∈ S ⊂ D − x, and since λi ∈ [0, 1], i = 1, . . . , N + 1. Since evidently h  0, relations (7.8.16) and (7.8.17) show that lim h (λ; u) = 0 uniformly on K. λ→0

Now take an arbitrary y ∈ D − x. There exist a u ∈ K and a λ > 0 such that |u| = r, and y = λu. Hence we have for arbitrary ε > 0 p(λu) 1 p(y) = = h(λ; u) < ε |y| λr r

provided that λ is small enough. In other words f (x + y) − f (x) − x∗ y p(y) = <ε |y| |y| provided that |y| is small enough, which means that x∗ is Stolz differential of f at x.  Lemma 7.8.6. Let f : RN → R be a convex and positively homogeneous function, and let b1 , . . . , bN be a base of RN over R. If f (−bi ) = −f (bi ) ,

i = 1, . . . , N ,

(7.8.18)

then f is linear. Proof. Since f is positively homogeneous, we have f (0) = f (2 · 0) = 2f (0), whence f (0) = 0

(7.8.19)

Further, for arbitrary λ < 0 we have in view of (7.8.18) for i = 1, . . . , N f (λbi ) = f (− |λ| bi ) = |λ| f (−bi ) = − |λ| f (bi ) = λf (bi ), which together with (7.8.19) and the positive homogeneity of f shows that for λ ∈ R and i = 1, . . . , N.

f (λbi ) = λf (bi )

(7.8.20)

We also have, since f is positively homogeneous and convex,   f (x) + f (y) x+y 2 = f (x) + f (y), f (x + y) = 2f 2 2 whence by introduction

 f

n

 xi

i=1

for every n ∈ N and x1 , · · · , xn ∈ RN .



n

i=1

f (xi )

(7.8.21)

7.8. Derivatives of convex functions

187

It follows from (7.8.19) and (7.8.21) that for x ∈ RN 0 = f (0) = f (x − x)  f (x) + f (−x) and so −f (−x)  f (x)

for x ∈ RN .

(7.8.22)

Now take an arbitrary x ∈ RN . There exist λ1 , · · · , λN ∈ R such that x=

N

λi bi .

i=1

By (7.8.20), (7.8.21) and (7.8.22) N

i=1

λi f (bi ) =

N

i=1

= −f

 f (λi bi )  f 

N

 λi bi

= f (x)  −f (−x)

i=1

N

 (−λi )bi

−

i=1

N

f (−λi bi ) =

i=1

N

λi f (bi ),

i=1

whence it follows, in particular, that f is odd: f (−x) = −f (x)

for x ∈ RN .

Thus, for arbitrary x, y ∈ RN , f (x + y) = −f (−x − y)  − [f (−x) + f (−y)] = f (x) + f (y) , which together with (7.8.21) yields that f is additive. Similarly, the argument leading to (7.8.20) shows that f is homogeneous. Thus f is linear.  Theorem 7.8.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex and continuous function. For every x ∈ D the following three conditions are equivalent: (i) f has a Stolz differential at x; (ii) f is differentiable at x; (iii) f has at x all partial derivatives. Proof. The implications (i) ⇒ (ii) ⇒ (iii) result from Lemmas 7.7.1 and 7.7.2. Now we are going to prove the converse implications. Assume that f satisfies (iii). By Lemma 7.8.1 the derivatives fy (x) exist for every y ∈ RN , moreover the function f. (x) is convex and positively homogeneous. Let e1 , . . . , eN be the usual orthonormal base of RN over R. We have for i = 1, . . . , N fe i (x) = lim

λ→0+

f (x + λei ) − f (x) ∂f = (x) , λ ∂ξi

188

Chapter 7. Continuous Convex Functions

and  (x) = lim f−e i

λ→0+

f (x − λei ) − f (x) ∂f =− (x) , λ ∂ξi

 (x) = −fe i (x), i = 1, . . . , N . By Lemma 7.8.6 the function f. (x) is linear, Hence f−e i which means that f is differentiable at x. Now assume that f satisfies (ii). In view of (7.7.4) there exist a y ∗ ∈ RN such that fy (x) = y ∗ y. (7.8.23)

Take an arbitrary x∗ ∈ ∂f (x). By (7.8.23) and Lemma 7.8.3 y ∗ y  x∗ y for every y ∈ RN , whence by Lemma 7.8.5 y ∗ y = x∗ y for all y ∈ RN , i.e., y ∗ = x∗ . Thus ∂f (x) = {y ∗ } is a singleton, and by Theorem 7.8.1 f satisfies (i). 

7.9 Differentiability of convex functions In this section we extend Theorem 7.4.3 to the case of functions of several variables. We start with some lemmas. Lemma 7.9.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex and continuous function. Let xn ∈ D and yn ∈ RN , n ∈ N, be convergent sequences: lim xn = x0 ∈ D, lim yn = y0 ∈ RN . Then n→∞

n→∞

lim sup fy n (xn )  fy 0 (x0 ).

(7.9.1)

n→∞

Proof. Since f is continuous and convex, fy 0 (x0 ) exist. Take an arbitrary μ  fy 0 (x0 ). Then f (x0 + λy0 ) − f (x0 ) <μ (7.9.2) λ for λ > 0 sufficiently small. On the other hand, for every fixed n ∈ N, the differences quotient [f (xn + λyn ) − f (xn )] /λ is an increasing function of λ (cf. Theorem 7.3.2), whence for λ > 0 f (xn + λyn ) − f (xn )  fy n (xn ). (7.9.3) λ Moreover, (7.9.4) lim [f (xn + λyn ) − f (xn )] = f (x0 + λy0 ) − f (x0 ) , n→∞

since f is continuous. Let us find a λ > 0 such that (7.9.2) holds. We have by (7.9.3), (7.9.4), and (7.9.2) lim sup fy n (xn )  lim n→∞

n→∞

f (x0 + λy0 ) − f (x0 ) f (xn + λyn ) − f (xn ) = < μ. λ λ

Letting μ → fy 0 (x0 ), we obtain hence (7.9.1).



Corollary 7.9.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex and continuous function. Then the function f  : D×RN → R is upper semicontinuous.

7.9. Differentiability of convex functions

189

Lemma 7.9.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex and continuous function. Then, for every x ∈ D and y ∈ RN ,  (x). lim inf fy (z) = f−y z→x

(7.9.5)

  fy , whence we obtain according to Lemma 7.9.1 Proof. By Lemma 7.8.1 −f−y    lim inf fy (z)  lim inf (−f−y )(z) = − lim sup f−y (z)  −f−y (x). z→x

z→x

(7.9.6)

z→x

Fix x ∈ D, y ∈ RN , and let J = {λ ∈ R | x + λy ∈ D}.

(7.9.7)

J is an open interval, and we define a function g : J → R by g (λ) = f (x + λy) .

(7.9.8)

g is a convex and continuous function. For arbitrary λ ∈ J we have f (x + λy + μy) − f (x + λy) μ→0+ μ  g (λ + μ) − g (λ) = lim = g+ (λ) . μ→0+ μ

fy (x + λy) = lim

(7.9.9)

and f (x + λy − μy) − f (x + λy) μ g (λ − μ) − g (λ)  = lim = −g− (λ) . μ→0+ μ

 (x + λy) = lim f−y

μ→0+

(7.9.10)

By Theorem 7.4.1 we obtain in virtue of (7.9.9) and (7.9.10)    lim fy (x + λy) = lim g+ (λ) = g− (0) = −f−y (x) .

λ→0−

λ→0−

(7.9.11) 

Relations (7.9.6) and (7.9.11) imply (7.9.5).

Lemma 7.9.3. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex and continuous function. Let a y ∈ RN be fixed, and let ! "  Y0 = x ∈ D | f−y (x) = −fy (x) . (7.9.12)  . Y0 is dense Then the set Y0 is the set of the points of continuity of the function f−y N in D, D \ Y0 is of measure zero and of the first category in R . More exactly

D \ Y0 =

∞ 

Sk ,

(7.9.13)

k=1

where the sets Sk , k ∈ N, are closed in D, and at most countable on every straight line with the direction y.

190

Chapter 7. Continuous Convex Functions

Proof. First assume that y = 0. Take an x ∈ Y0 . Then by Lemmas 7.9.2 and 7.9.1  (x) = fy (x)  lim sup fy (z) lim inf fy (z) = −f−y z→x

z→x

whence it follows that the limit lim

z→x

fy

(z) exist, and

lim fy (z) = fy (x) .

z→x

Thus the function fy is continuous at x. Conversely, let fy be continuous at an x ∈ D. We have by Lemma 7.8.1 and Lemma 7.9.2  fy (x)  −f−y (x) = lim inf fy (z) = lim fy (z) = fy (x) , z→x

z→x

 (x) = −fy (x), and x ∈ Y0 . Consequently Y0 is the set of points of whence f−y continuity of the function fy . Now define a function h : D → R putting  (x) , h (x) = fy (x) + f−y

x ∈ D.

By Lemma 7.8.1, h  0 in D, and we have Y0 = {x ∈ D|h (x) = 0}. Thus relation (7.9.13) holds with  1 , k ∈ N. (7.9.14) Sk = x ∈ D | h(x)  k Since by Corollary 7.9.1 h is upper semicontinuous in D, the sets (7.9.14) are closed in D. The straight line Lx passing through a point x ∈ RN and with the direction y can be written as ! " Lx = z ∈ RN | z = x + λy, λ ∈ R . Define the interval J by (7.9.7), and the function g : J → R by (7.9.8). Take an x ∈ RN and an arbitrary z ∈ D ∩ Lx (if D ∩ Lx = ∅). Then z = x + λy with λ ∈ J, and we have by (7.9.9) and (7.9.10)  fy (z) = g+ (λ),   (λ) − g− (λ), and Hence h(z) = g+  1 Sk ∩ Lx = z ∈ D ∩ Lx | h(z)  k

  f−y (z) = −g− (λ)



1 . k (7.9.15) Set (7.9.15) is at most countable, since it is contained in the set of the points of nondiffentiability (cf. Theorem (7.4.3) of the continuous and convex function g : J → R. Sets (7.9.14) being closed in D, are Lebesgue measurable. Let H be an (N − 1)dimensional hyperplane perpendicular to y. Then  m (Sk ) = m (Sk ∩ Lx ) dx = 0, k ∈ N , (7.9.16) H

=



 z = x + λy ∈ D | g+ (λ) − g− (λ) 

7.9. Differentiability of convex functions

191

since m (Sk ∩ Lx ) = 0 for every x ∈ RN , the set (7.9.15) being at most countable. It follows by (7.9.13) that m(D \ Y0 ) = 0. This last property implies, in particular, that the set Y0 is dense in D. Similarly, (7.9.16) implies that int Sk = ∅, k ∈ N. The fact that Sk is closed in D means that Sk = D ∩ Fk , where Fk is closed in RN . Thus both sets, D and Fk , belong to the class Fσ , and consequently so does also Sk . By Lemma 2.1.1 Sk is of the first category, k ∈ N, and by (7.9.13) also D \ Y0 is of the first category.  If y = 0, then by (7.7.4) Y0 = D and in this case the theorem is evident. Theorem 7.9.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex and continuous function. Let D0 = {x ∈ D | f is differentiable at x} . Then the set D0 is dense in D, D \ D0 is of measure zero and of first category, and the function ∇f : D0 → RN is continuous in D0 . Proof. Put  ∂f Yj = x ∈ D | (x) exists ∂ξj

   = x ∈ D | fe j (x) = −f−e (x) , j

j = 1, . . . , N,

where e1 , . . . , eN is the usual orthonormal base of RN (over R). By Theorem 7.8.2 D0 =

N 

Yj .

Hence

j=1

D \ D0 =

N 

(D \ Yj ) .

j=1

The properties of D \ D0 result now from Lemma 7.9.3. Also by Lemma 7.9.3 it follows that all the partial derivatives of f are continuous in D0 , and hence, in view  of Theorem 7.8.2 and Lemma 7.7.2 cf., in particular, formula (7.7.8) , also ∇f is continuous in D0 .  Corollary 7.9.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex and continuous function. Then f is differentiable almost everywhere in D. Corollary 7.9.3. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex and differentiable function. Then f is of class C 1 in D. We prove yet the following generalization of Theorem 7.5.2. Theorem 7.9.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a twice differentiable8 function. The function f is convex if and only if the matrix  2  ∂ f (x) (7.9.17) ∂ξi ∂ξj is positive semi-definite for every x ∈ D. I.e., f is differentiable in D and fy is differentiable in D for every y ∈ RN (in the sense of the definition in 7.7).

8

192

Chapter 7. Continuous Convex Functions

Proof. For every x ∈ D and y ∈ RN define the interval J = Jxy by (7.9.7) and the function g = gxy : J → R by (7.9.8). Observe that g is twice differentiable in J, and f is convex in D if and only if g = gxy is convex in Jxy for every choice of x ∈ D and y ∈ RN . In fact, if f is convex, then it is easy to check that gxy is convex for every x ∈ D, y ∈ RN . Conversely, let gxy be convex for every x ∈ D, y ∈ RN , and take an arbitrary u, v ∈ D. Put x = u, y = v − u. Then       u+v 1 gxy (0) + gxy (1) f (u) + f (v) 1 f = f u + (v − u) = gxy  = . 2 2 2 2 2 So f is convex. By Theorem 7.5.2 the function g is convex if and only if g   0 in J. Therefore d2 f (x + λy)  0 for every choice of x ∈ D and y ∈ RN , f is convex if and only if dλ2 and for every λ from the corresponding interval Jx y. In other words, f is convex if d2 f (x + λy) |x+λy=u  0 for every u ∈ D and y ∈ RN . But and only if dλ2

∂2f d2 f (x + λy) |x+λy=u = (u)ηi ηj , 2 dλ ∂ξi ∂ξj i=1 j=1 N

N

(7.9.18)

where y = (η1 , . . . , ηN ). But the condition that (7.9.18) is non-negative for every u ∈ D and y ∈ RN is equivalent to the fact that the matrix (7.9.17) is positive semi-definite. 

7.10 Sequences of convex functions First let us note the following variant of Theorem 5.3.3. Theorem 7.10.1. Let D ⊂ RN be a convex and open set, and let fn : D → R, n ∈ N, be a sequence of convex and continuous functions. If the sequence {fn } converges in D to a finite function f , then f is convex and continuous. Proof. Fix an arbitrary x, y ∈ D and λ ∈ [0, 1]. Then by Theorem 7.1.1   fn λx + (1 − λ) y  λfn (x) + (1 − λ) fn (y) for all n ∈ N. As n → ∞, we obtain hence   f λx + (1 − λ)y  λf (x) + (1 − λ)f (y).

(7.10.1)

Relation (7.10.1) holds for arbitrary x, y ∈ D and λ ∈ [0, 1], whence, again by Theorem 7.1.1, the function f is convex and continuous.  Theorem 7.10.2. Let D ⊂ RN be a convex and open set, and let Δ ⊂ D be a dense subset of D. Let fn : D → R, n ∈ N, be a sequence of convex and continuous functions. If the sequence {fn (x)} converges (to a finite limit ) for every x ∈ Δ, then the sequence {fn } converges uniformly on every compact set contained in D.

7.10. Sequences of convex functions

193

Proof. Fix a compact C ⊂ D, and an ε > 0. By Theorem 7.6.1 there exist a constant L > 0 such that |fn (x) − fn (y)|  L |x − y| (7.10.2) for all x, y ∈ C and all n ∈ N. For every x ∈ Δ, let Bx be the open ball centered at x and with the radius ε/3L. Since Δ is dense in D, we have  Bx . C⊂D⊂ x∈Δ

Since C is compact, there exist x1 , . . . , xm ∈ Δ such that C⊂

m 

Bxi .

(7.10.3)

i=1

Since the set X = {x1 , . . . , xm } is finite, the sequence fn converges uniformly on X. Therefore there exist an n0 ∈ N such that for every indices p, q > n0 we have ε for x ∈ X . (7.10.4) |fp (x) − fq (x)| < 3 Take an arbitrary u ∈ C. By (7.10.3) there exists an xj ∈ X such that ε |u − xj | < . (7.10.5) 3L Thus we have by (7.10.2), (7.10.5) and (7.10.4) for p, q > n0 |fp (u) − fq (u)|  |fp (u) − fp (xj )| + |fp (xj ) − fq (xj )| + |fq (xj ) − fq (u)|  L |u − xj | + |fp (xj ) − fq (xj )| + L |u − xj | < ε. Thus the sequence {fn } satisfies on C the uniform Cauchy condition, and therefore converges uniformly on C.  Corollary 7.10.1. Let D ⊂ RN be a convex and open set, and let f, fn : D → R, n ∈ N, be convex and continuous functions. If lim sup fn (x)  f (x) n→∞

for x ∈ D,

(7.10.6)

then to every compact C ⊂ D and every ε > 0 there exists an index n0 ∈ N such that for every index n > n0 and every x ∈ C (7.10.7) fn (x)  f (x) + ε.  Proof. Put gn (x) = max fn (x), f (x) , x ∈ D, n ∈ N. The functions gn : D → R are continuous, and by Theorem 5.3.4 they are convex. Relation (7.10.6) implies that 

lim gn (x) = f (x)

n→∞

for x ∈ D .

Consequently, by Theorem 7.10.2, given a compact C ⊂ D and an ε > 0, we can find an n0 ∈ N such that for n > no , n ∈ N, gn (x)  f (x) + ε.   Since fn (x)  max fn (x), f (x) = gn (x), we obtain hence (7.10.7).



194

Chapter 7. Continuous Convex Functions

Corollary 7.10.2. Let D ⊂ RN be a convex and open set, and let Δ ⊂ D be a dense subset of D. Let fn : D → R, n ∈ N, be a sequence of convex and continuous functions. If the sequence {fn (x)} is bounded for every x ∈ Δ, then it is possible to choose from the sequence {fn } a subsequence uniformly convergent on every compact set contained in D. Proof. Since the space RN is separable, there exist a countable set Δ   0 ⊂ Δ dense in D. Using the diagonal method of choice cf., e.g., L  ojasiewicz [208] , we can choose from {fn } a subsequence convergent at every point of Δ0 . The corollary results now from Theorem 7.10.2.  Theorem 7.10.3. Let D ⊂ RN be a convex and open set, and let fn : D → R, n ∈ N, be a sequence of convex and differentiable functions, convergent in D: lim fn = f.

n→∞

If the function f : D → R is differentiable in D, then lim ∇fn = ∇f

n→∞

uniformly on every compact set contained in D. Proof. It is enough to show that for every y ∈ RN , y = 0,     lim ∇fn (x) y = ∇f (x) y n→∞

 uniformly on every compact set contained in D. For y = ei , i = 1, . . . , N , we obtain hence the  suitable convergence of partial derivatives of f , and thus, by formula (7.7.8), of ∇fn . For an indirect proof suppose that there exist a compact C ⊂ D, a y ∈ RN \ {0}, and an ε > 0 such that for infinitely many n ∈ N there exist points xn ∈ C such that     | ∇fn (xn ) y − ∇f (xn ) y|  ε, i.e., either or



   ∇fn (xn ) y  ∇f (xn ) y + ε,



   ∇fn (xn ) y  ∇f (xn ) y − ε,

(7.10.8) (7.10.9)

which is equivalent to     ∇fn (xn ) (−y)  ∇f (xn ) (−y) + ε, Of the two possibilities, (7.10.8) and (7.10.9), at least one must occur for infinitely many n. We may assume that (7.10.8) occurs for infinitely many n, for (7.10.9) may be reduced to (7.10.8) on replacing y by −y. Since C is compact, we may assume that the sequence {xn } converges to an x0 ∈ C: lim xn = x0

n→∞

7.10. Sequences of convex functions

195

and that(7.10.8) holds for all n ∈ N, for otherwise we replace {fn } and {xn } by suitable subsequences. The point x0 + λy belongs to D for small λ > 0, and hence xn + λy ∈ D for small λ > 0 and large n ∈ N. By Theorem 7.10.2 lim f (xn ) = f (x0 ),

(7.10.10)

lim f (xn + λy) = f (x0 + λy)

(7.10.11)

n→∞

and n→∞

  cf., e.g., L  ojasiewicz [208] . Thus we have by Lemma 7.7.2 and Theorem 7.3.2 for λ > 0 and n ∈ N such that xn + λy ∈ D 

 fn (xn + λy) − fn (xn ) ∇fn (xn ) y = (fn )y (xn )  , λ

and letting n → ∞, we obtain hence in view of (7.10.10) and (7.10.11)   f (x0 + λy) − f (x0 ) . lim inf ∇fn (xn ) y  n→∞ λ On the other hand, we have by Corollary 7.9.3 and by (7.10.8)       ∇f (x0 ) y + ε = lim ∇f (xn ) y + ε  lim inf ∇fn (xn ) y. n→∞

n→∞

(7.10.12)

(7.10.13)

Relations (7.10.12) and (7.10.13) yield   f (x0 + λy) − f (x0 ) ∇f (x0 ) + ε  . λ

(7.10.14)

for sufficiently small λ > 0. When λ → 0+, (7.10.14) goes into (cf. 7.7.9)     ∇f (x0 ) y + ε  fy (x0 ) = ∇f (x0 ) y which is a contradiction with ε > 0.



Exercises 1. Let D ⊂ RN be a convex and open set. Show that a function f : D → R is convex and continuous if and only if the set ! " (x, y) ∈ RN +1 | x ∈ D, y > f (x) is convex. 2. Let f : RN → R be a convex function. Show that if f is bounded above on RN , then it is constant. 3. Let f : RN → R be a convex and positively homogeneous function. Show that f is continuous.

196

Chapter 7. Continuous Convex Functions

4. Let J ⊂ R be an open interval, and let f : J → R be a continuous function. Suppose that the derivative f  (x) exists for x ∈ J0 ⊂ J such that the set J \ J0 is at most countable. Show that f is convex if and only if f  is an increasing function on J0 . 5. Let J ⊂ R be an open interval, and let J0 ⊂ J be a set such that the set J \ J0 is at most countable. Show that there exists a convex and continuous function f : J → R such that the derivative f  exists if and only if x ∈ J0 . 6. Let J ⊂ R be an open interval, and let f : J → R and g : J → R be non-negative, increasing, convex functions. Show that the product f g is a convex function. 7. Let J ⊂ R be an open interval, and let f : J → R be a convex and continuous

  (x) , f+ (x) . function. Prove that, for every x ∈ J, we have ∂f (x) = f− 8. Let D ⊂ RN be a convex and open set, and let f : D → R be a function. Show that if ∂f (x) = ∅ for every x ∈ D, then f is convex and continuous. [Hint: Take arbitrary u, v ∈ D, λ ∈ [0, 1], and let x = λu+(1 − λ) v, x∗ ∈ ∂f (x). Then f (x) = λ [f (x) + x∗ (u − x)] + (1 − λ) [f (x) + x∗ (v − x)]  λf (u) + (1 − λ) f (v).] 9. Prove the following extension of Lemma 7.9.1: Let D ⊂ RN be a convex and open set, and let fn : D → R, n ∈ N, be a sequence of convex and continuous functions. Further, let xn ∈ D and y ∈ RN , n ∈ N, be convergent sequences: lim xn = x0 ∈ D,

n→∞

lim yn = y0 ∈ RN .

n→∞

If the sequence {fn } converges in D to a finite function f : D → R lim fn = f,

n→∞

then

lim sup (fn )yn (xn )  fy 0 (x0 ) . n→∞

Chapter 8

Inequalities 8.1 Jensen inequality Since the convex functions are defined by a functional inequality, it is not surprising that this notion will lead to a number of interesting and important inequalities. Some inequalities connected with the notion of convexity will be presented in this chapter. Nowadays the theory of inequalities is a well-developed mathematical subject with scores of results which have been gathered in various books and monographs devoted to this topic (e.g., Hardy-Littlewood-P´ olya [136], Mitrinovi´c [226], BeckenbachBellman [24]). A number of interesting results about inequalities (a.o. about inequalities connected with the notion of convexity) can be found in various articles published in Publikacije Elektrotehniˇckog Fakulteta Univerzita u Beogradu. An ample bibliography of the subject can also be found in those sources. Here we will not endeavour to give complete references related to every particular inequality considered. The reader is referred to the above mentioned books and articles. Also, we do not claim to have exhausted the topic (cf. also Roberts-Varberg [267]). Simply we present here a number of inequalities connected with the notion of convexity which seem to us particulary interesting and important. Jensen’s inequality is inequality (5.3.1). Also direct consequences of (5.3.1), viz. (5.3.3), (5.3.7) and (5.3.13) are referred to as Jensen’s inequalities. Now we are going to derive further inequalities of this type, all known under the name of Jensen’s inequalities. Theorem 8.1.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a continuous and convex function. Then for every n ∈ N, x1 , . . . , xn ∈ D, and for every n  non-negative real numbers q1 , . . . , qn with qi > 0, we have i=1

⎛ n

⎞ qi xi

⎜ i=1 f⎜ n ⎝ 

i=1

qi

⎟ ⎟ ⎠

n 

qi f (xi )

i=1

n 

i=1

. qi

(8.1.1)

198

Chapter 8. Inequalities

Proof. Inequality (8.1.1) results from (5.3.13) on setting λi = qi /

n 



qj .

j=1

Theorem 8.1.2. Let (X, M, μ) be a measure space such that μ (X) = 1, and let J ⊂ R be an open interval. Let p : X → J be an integrable function, and let f : J → R be a continuous and convex function. Then ⎛ ⎞   f ⎝ pdμ⎠  (f ◦ p) dμ. (8.1.2) X

X

&

Proof. (Zygmund [328]). Put t =

pdμ. Clearly t ∈ J. Take a k ∈ ∂f (t). Then

X

f (y) − f (t)  k (y − t)

(8.1.3)

for all y ∈ J. Replacing y by p (x), and integrating over x ∈ X, we get from (8.1.3) ⎛ ⎞   (f ◦ p) dμ − f (t)  k ⎝ pdμ − t⎠ = 0, X

X



which is equivalent to (8.1.2).

Theorem 8.1.3. Let (X, M, μ) be a measure space, and let J ⊂ R be an open interval. Let p : X → J and & q : X → [0, ∞] be integrable functions such that the product pq is integrable and qdμ > 0, and let f : J → R be a continuous and convex function. X

Then

⎛&

pqdμ

f ⎝X&

qdμ

&

⎞

⎠X

X

q (f ◦ p) dμ & . qdμ

(8.1.4)

X

Proof. Define the new measure ν : M → [0, ∞] putting for A ∈ M & qdμ A & ν (A) = . qdμ X

In particular, we have ν (X) = 1, and dν = &

q dμ. qdμ

(8.1.5)

X

By Theorem 8.1.2

⎛ f⎝



⎞ pdν ⎠ 

X

whence by (8.1.5) we obtain (8.1.4).

 (f ◦ p) dν, X



8.1. Jensen inequality

199

N Theorem 8.1.4. Let X ⊂ & R be an open set, and let Λ  0 be an additive function of intervals in X, 0 < dΛ < ∞. Let J ⊂ R be an open interval, let p : X → J X

be a function integrable (in the sense of Lebesgue-Stieltjes) with respect to Λ, and let f : J → R be a continuous and convex function. Then ⎛& ⎞ & pdΛ (f ◦ p) dΛ ⎠X & f ⎝X& . (8.1.6) dΛ dΛ X

X

Proof. Inequality (8.1.6) is a particular case of (8.1.2), where μ (A) =

&

dΛ/

A

&

dΛ. 

X

From the above theorems we can derive some classical inequalities. Corollary 8.1.1. Let n ∈ N, and let a1 , . . . , an , b1 , . . . , bn be arbitrary real numbers. Then  n 2  n  n 

2 2 (8.1.7) ai b i  ai bi . i=1

i=1

i=1

Proof. Assume first that ai = 0 for i = 1, . . . , n. Put in Theorem 8.1.1 D = R, f (x) = x2 , xi = abii , qi = a2i , i = 1, . . . , n. Then (8.1.1) becomes (8.1.7). If all ai ’s are zeros, (8.1.7) is trivial. If some ai ’s are zeros, we can rearrange sequences {ai } and {bi } so that ai = 0 for i = 1, . . . , m, ai = 0 for i = m + 1, . . . , n, 0 < m < n. Then, on account of what has already been proved 

n

2 ai b i

 =

i=1

 =

m

2 

ai b i

i=1 n



 a2i

i=1

m

 a2i

i=1 m



b2i





i=1

m

i=1

n

 b2i



a2i

i=1

n

 b2i

,

i=1



and (8.1.7) holds true. Inequality (8.1.7) is the famous Cauchy-Buniakowski-Schwarz inequality.

Corollary 8.1.2. Let (X, M, μ) be a measure space, and let a, b : X → [−∞, ∞] be integrable functions. Then ⎛ ⎝



X

⎞2

⎛

abdμ⎠  ⎝



X

⎞⎛ a2 dμ⎠ ⎝



⎞ b2 dμ⎠ .

(8.1.8)

X

Proof. Since b is integrable, it is equivalent to a finite function (i.e., b = b∗ a.e. in X, where b∗ is finite), and thus we may assume that b itself is finite: b : X → R. Assume first that a = 0 in X. Set in Theorem 8.1.3 J = R, f (x) = x2 , p (x) = b (x) /a (x) 2 and q (x) = a (x) . Then (8.1.4) becomes (8.1.8).

200

Chapter 8. Inequalities

If a = 0 a.e. in X, inequality (8.1.8) is trivial. Otherwise put Y = { x ∈ X | a (x) = 0 } . By what has already been proved ⎞2 ⎛ ⎞2 ⎛ ⎞⎛ ⎞ ⎛     ⎝ abdμ⎠ = ⎝ abdμ⎠  ⎝ a2 dμ⎠ ⎝ b2 dμ⎠ X

Y

⎛ =⎝



Y

Y

⎞⎛ ⎞ ⎛ ⎞⎛ ⎞    a2 dμ⎠ ⎝ b2 dμ⎠  ⎝ a2 dμ⎠ ⎝ b2 dμ⎠ ,

X

Y

X

X



and (8.1.8) holds true.

Corollary 8.1.3. Let n ∈ N, and let x1 , . . . , xn , q1 , . . . , qn be non-negative real numbers. Then p  n p−1 n  n

qi xi  qi qi xpi for p > 1 and p < 0 , (8.1.9) i=1

i=1

i=1

and if, moreover, xi > 0 for i = 1, . . . , n, then  n p  n p−1 n

qi xi  qi qi xpi i=1

i=1

for 0 < p < 1.

(8.1.10)

i=1

Proof. If all qi ’s are zeros, (8.1.9) and (8.1.10) are trivial, so we assume that

n 

qi > 0.

i=1

Let p > 1 or p < 0, and assume that xi > 0 for i = 1, . . . , n. Set in Theorem 8.1.1 N = 1, D = (0, ∞), f (x) = xp , and (8.1.1) becomes (8.1.9). If some xi ’s are zeros, we argue as in the proof of Corollary 8.1.1. If 0 < p < 1 and all xi ’s are positive, we take in Theorem 8.1.1 N = 1, D = (0, ∞), and f (x) = −xp , and then we multiply the resulting inequality by (−1), arriving thus at (8.1.10).  Corollary 8.1.4. Let n ∈ N, and let x1 , . . . , xn be non-negative real numbers. Then p  n n

xi  np−1 xpi for p > 1 and p < 0 , (8.1.11) i=1

i=1

and if, moreover, xi > 0 for i = 1, . . . , n, then p  n n

xi  np−1 xpi i=1

for 0 < p < 1.

(8.1.12)

i=1

Proof. We obtain (8.1.11) and (8.1.12) from (8.1.9) and (8.1.10) taking qi = 1, i = 1, . . . , n. 

8.2. Jensen-Steffensen inequalities

201

Corollary 8.1.5. Let n ∈ N, let a1 , . . . , an , b1 , . . . , bn be arbitrary non-negative num1 1 bers. Let p > 1 and let q be given by + = 1. Then p q 1/p  1/q  n n n

ai b i  api bqi . (8.1.13) i=1

i=1

i=1

Proof. Assume first that ai > 0, bi > 0 for i = 1, . . . , n. Taking in Theorem 8.1.1 N = 1/p 1, D = (0, ∞), f (x) = xp , qi = bqi , xi = (api /bqi ) , i = 1, . . . , n, we obtain (8.1.13). If some ai ’s and/or some bi ’s are zeros, we argue as in the proof of Corollary 8.1.1.  Inequality (8.1.13) is the famous H¨ older inequality. Corollary 8.1.6. Let (X, M, μ) be a measure space, and let a, b : X → [−∞, ∞] be functions integrable in p-th and q-th power, respectively, where p > 1, p1 + 1q = 1. Then the function ab is integrable, and  1/p  1/q  p q |ab| dμ  |a| dμ |b| dμ . (8.1.14) X

X

X

Proof. The functions a and b are equivalent to finite functions, so we may assume that a and b are themselves finite. Assume first that a = 0 and b = 0 in X. Setting in  p q 1/p q Theorem 8.1.3 J = (0, ∞), p (x) = |a (x)| / |b (x)| , q (x) = |b (x)| , f (x) = xp , we obtain inequality (8.1.14). The integrability of the product ab results from (8.1.14). If a, b are allowed to assume the value zero, we argue as in the proof of Corollary 8.1.2. Here again ab is integrable, since by the above the function ab restricted to the set { x ∈ X | a (x) b (x) = 0 } is integrable. 

8.2 Jensen-Steffensen inequalities Now we are going to generalize the result of the preceding section (cf., e.g., Mitrinovi´c [225], [226], Boas [34]). We start with a lemma. Lemma 8.2.1. Let J = [a, b] be a closed real interval, and let p, λ : J → R be functions such that p is non-negative and monotonic in J and the Riemann-Stieltjes integral &b pdλ exists1 . Then, if p is decreasing a

b p (a) inf [λ (c) − λ (a)] 

pdλ  p (a) sup [λ (c) − λ (a)] ,

acb

a 1

(8.2.1)

acb

The Stieltjes integrals in this section denote Riemann-Stieltjes integrals. The Riemann-Stieltjes & integral ab pdλ certainly exists if one of the functions p, λ is continuous in J, and the other has a finite variation in J. In the case of Lemma 8.2.1 the function p (being monotonic) is of the finite variation, so for the existence of the integral in question it is sufficient (but not necessary) that the function λ is continuous. Note that if the function λ is increasing, Lemma 8.2.1 is trivial, and Theorem 8.2.1 is a particular case of Theorem 8.1.4. The main interest of inequality (8.2.3) below lies in the fact that λ need not be a monotonic function. Similarly, inequality (8.2.6) below is more general than (8.1.1) in that now qi need not be non-negative.

202

Chapter 8. Inequalities

and if p is increasing, b p (b) inf [λ (b) − λ (c)] 

pdλ  p (b) sup [λ (b) − λ (c)] .

acb

(8.2.2)

acb

a

Proof. Assume p to be decreasing. Then b

b p (x) d [λ (x) − λ (a)]

p (x) dλ (x) = a

a

b = [λ (b) − λ (a)] p (b) −

[λ (x) − λ (a)] dp (x) a

 p (b) sup [λ (c) − λ (a)] + [p (a) − p (b)] sup [λ (c) − λ (a)] acb

acb

= p (a) sup [λ (c) − λ (a)] . acb

This is the right inequality in (8.2.1). The left inequality in (8.2.1), and, in the case where p is increasing, inequalities (8.2.2) are proved similarly.  Theorem 8.2.1. Let J = [a, b] be a closed real interval, and let p, λ : J → R be functions such that λ (a)  λ (x)  λ (b) in J, λ (a) < λ (b), p is monotonic in J, &b and pdλ exists. Let D ⊂ R be an open interval such that p ([a, b]) ⊂ D, and let a

f : D → R be a continuous and convex function. Then ⎛

⎞

&b

&b pdλ (f ◦ p) dλ ⎟ ⎜ ⎟ ⎜ a a . f⎜ ⎟ ⎝ λ (b) − λ (a) ⎠ λ (b) − λ (a)

(8.2.3)

Proof. We assume that the function p is decreasing; in the case of an increasing p the proof is similar. Put &b pdλ a t= . (8.2.4) λ (b) − λ (a) We have b

b pdλ − p (b) [λ (b) − λ (a)] =

a

b pdλ − p (b)

a

b [p (x) − p (b)] dλ (x) .

dλ = a

a

8.2. Jensen-Steffensen inequalities

203

By Lemma 8.2.1 b [p (x) − p (b)] dλ (x)  [p (a) − p (b)] sup [λ (c) − λ (a)] acb

a

= [p (a) − p (b)] [λ (b) − λ (a)] . Hence

b pdλ  p (a) [λ (b) − λ (a)] . a

Similarly

b pdλ  p (b) [λ (b) − λ (a)] . a

The above inequalities show that p (b)  t  p (a). Take a k ∈ ∂f (t). Then f (y) − f (t)  k (y − t) for all y ∈ D. Put r (y) = f (y) − f (t) − k (y − t) ,

y ∈ D.

It is easily seen that r : D → R is a non-negative convex function, and r (t) = 0. It follows from Theorem 7.3.5 that r decreases in [p (b) , t] and increases in [t, p (a)]. Next put R (x) = r (p (x)), T ∈ p−1 (t). Then R  0, and R decreases in [a, T ], R increases in [T, b]. By Lemma 8.2.1 b

T Rdλ =

a

b Rdλ  R (a) inf [λ (c) − λ (a)]+R (b) inf [λ (b) − λ (c)] = 0.

Rdλ+ a

acb

acb

T

Next, by (8.2.4), b 0

b (r ◦ p) dλ =

Rdλ = a

b

a

[f (p (x)) − f (t) − kp (x) + kt] dλ a

b (f ◦ p) dλ − f (t) [λ (b) − λ (a)] − k

= a

 b

  pdλ − t λ(b) − λ(a)

a

b (f ◦ p) dλ − f (t) [λ (b) − λ (a)] .

= a

Hence (8.2.3) results.



204

Chapter 8. Inequalities

Theorem 8.2.2. Let J = [a, b] be a closed real interval, and let p : J → R and q : J → R be a monotonic and an integrable function, respectively, such that x 0

b qdt 

a

b qdt

x ∈ J,

for

a

qdt > 0. a

Let D ⊂ R be an open interval such that p ([a, b]) ⊂ D, and let f : D → R be a continuous and convex function. Then ⎛ ⎞ &b &b (f ◦ p) qdt ⎜ pqdt ⎟ ⎜ ⎟ a . (8.2.5) f ⎜ ab ⎟ &b ⎝& ⎠ qdt qdt a

a

Proof. We take in Theorem 8.2.1 λ (x) =

&x

q (t) dt, and (8.2.5) results from (8.2.3).

a



Theorem 8.2.3. Let n ∈ N, and let a < x1  · · ·  xn < b be real numbers. Further, let q1 , . . . , qn be real numbers such that for every k = 1, . . . , n 0

n

qi 

i=k

n

qi ,

i=1

n

qi > 0.

i=1

If f : (a, b) → R is a continuous and convex function, then ⎞ ⎛ n n  qi xi q f (xi ) ⎟ i=1 i ⎜ i=1 ⎟ ⎜ . f⎝  n n ⎠  qi qi i=1

(8.2.6)

i=1

Proof. Take arbitrary points ξ1 , . . . , ξn such that 0 < ξ1 < · · · < ξn < 1, and let p be an arbitrary continuous increasing function from [0, 1] onto [a, b] such that p (ξi ) = xi , i = 1, . . . , n. Define the function λ : [0, 1] → R by

λ (x) = qi . ξi <x



Then inequality (8.2.3) goes into (8.2.6).

Originally inequalities (8.2.3) and (8.2.5) were obtained (Steffensen [296]) from the Steffensen inequality (Steffensen [295]) b

b f (t) dt 

b−λ

a+λ 

f (t) g (t) dt  a

b

f (t) dt, where λ = a

g (t) dt, a

8.2. Jensen-Steffensen inequalities

205

valid for arbitrary integrable functions f, g : (a, b) → R such that f is decreasing and 0  g  1 in (a, b). The generalizations presented here are due to R. P. Boas [34]. Theorem 8.2.5 and 8.2.6 are also related to some results of Z. Ciesielski [45]. Theorem 8.2.4. Let J = [a, b] be a closed real interval, let p : J → R be a continuous, &b non-negative, decreasing function, let λ : J → R be a function such that pdλ exists, a

and let μ : J → R be a function. Assume that the following conditions are fulfilled: λ (a)  λ (x)  λ (b) + λ∗

for x ∈ J,

where λ∗  0, and 0 < μ (b) − μ (a) ,

λ (b) − λ (a) + λ∗  μ (b) − μ (a) .

Let D ⊂ R be an open interval such that 0, p (a) ∈ D, and let f : D → R be a continuous and convex function such that f (0)  0. Then ⎛ ⎞ &b &b pdλ (f ◦ p) dλ ⎜ ⎟ ⎜ ⎟ a a f⎜ . (8.2.7) ⎟ ⎝ μ (b) − μ (a) ⎠ μ (b) − μ (a) Proof. We shall distinguish a few cases. &b I. λ∗ = 0. Put A = pdλ. For y0 = λ (b) − λ (a) we have by Theorem 8.2.1 a



b (f ◦ p) dλ  y0 f a

We will prove that the function F (y) = yf

A y0

 .

(8.2.8)

  A y

is decreasing. With the exception of

at most countably many points we have (cf. Theorem 7.4.3)     A A A − f . F  (y) = f y y y Since f is continuous and convex and f (0)  0, we have by Theorem 7.3.2 for u > 0, u ∈ D, f (u) f (u) − f (0)    f− (u) . u u−0 Thus F  (y)  0 (for y > 0 such that A/y ∈ D and F  (y) exists). By Theorem 3.6.3 F is decreasing. Consequently for y1 = μ (b) − μ (a)  λ (b) − λ (a) = y0 we have by (8.2.8)     b A A  y1 f , (f ◦ p) dλ  y0 f y0 y1 a

which is equivalent to (8.2.7).

206

Chapter 8. Inequalities

II. λ∗ > 0 and p (b) = 0. We put  0 for x ∈ [a, b) , δ(x) = and 1 for x = b,

ω(x) = λ(x) + λ∗ δ(x),

x ∈ J.

Then ω(a)  ω(x)  ω(b) for x ∈ J, and

ω(b) − ω(a) = λ(b) + λ∗ − λ(a)  μ(b) − μ(a).

By the first part of the proof ⎛

&b

⎞

&b

(f ◦ p) dω ⎟ ⎜ ⎟ ⎜ a a  . f⎜ ⎟ ⎝ μ (b) − μ (a) ⎠ μ (b) − μ (a) But

b

b pdω =

a

pdω

b

∗

pdλ + λ

b pdδ =

a

a

∗

(8.2.9)

b

pdλ + λ p (b) = a

pdλ, a

and b

b (f ◦ p) dω =

a

∗

b

(f ◦ p) dλ + λ a

b (f ◦ p) dδ =

a

∗

b

(f ◦ p) dλ + λ f (0)  a

(f ◦ p) dλ, a

and so (8.2.7) results from (8.2.9). III. λ∗ > 0 and p (b) > 0. Then we extend the functions p, λ, μ onto an interval [a, c], where c > b, so that p is continuous, non-negative, decreasing in [a, c], p (c) = 0, and λ(x) = λ(b), μ(x) = μ(b) for x ∈ [b, c] . By the second part of the proof ⎛ &c

⎞

&c

(f ◦ p) dλ ⎜ ⎟ a a ⎜ ⎟ f⎝ .  μ (c) − μ (a) ⎠ μ (c) − μ (a) pdλ

(8.2.10)

But μ (c) − μ (a) = μ (b) − μ (a), c

b pdλ =

a

c (f ◦ p) dλ =

pdλ, a

So (8.2.7) results from (8.2.10).

b

a

(f ◦ p) dλ. a



8.2. Jensen-Steffensen inequalities

207

Theorem 8.2.5. Let J = [a, b] be a closed real interval, let p : J → R be a continuous, non-negative, decreasing function, and let q : J → R be an integrable function such that x b qdt  0 for x ∈ J, |q| dt > 0. a

a

Let D ⊂ R be an open interval such that 0, p (a) ∈ D, and let f : D → R be a continuous and convex function such that f (0)  0. Then ⎛

⎞

&b

&b

⎜ pqdt ⎟ ⎜ ⎟ f ⎜ ab ⎟ ⎝& ⎠ |q| dt

(f ◦ p) qdt

a

&b

a

.

(8.2.11)

|q| dt

a

Proof. We take in Theorem 8.2.4 x

x qdt,

λx =

|q| dt,

μx =

a

∗

b (|q| − q) dt,

λ =

a

a



and (8.2.7) becomes (8.2.11).

Theorem 8.2.6. Let n ∈ N, let x1 , . . . , xn , a, b be real numbers such that a < 0  x1  · · ·  xn < b, and let q1 , . . . , qn be real numbers such that for k = 1, . . . , n n

qi  0,

n

|qi | > 0.

i=1

i=k

If f : (a, b) → R is a continuous and convex function such that f (0)  0, then ⎛ n

⎜ i=1 f⎜ n ⎝ 

⎞ qi xi |qi |

⎟ ⎟ ⎠

n 

qi f (xi )

i=1

i=1

n 

.

(8.2.12)

|qi |

i=1

Proof. We take arbitrary ξ1 , . . . , ξn such that 0 < ξn < · · · < ξ1 < 1, and let p be an arbitrary continuous, non-negative, decreasing function from [0, 1] onto [0, b] such that p (ξi ) = xi , i = 1, . . . , n. Define the function λ, μ : [0, 1] → R and the number λ∗ by n

λ (x) = qi , μ (x) = |qi | , λ∗ = (|qi | − qi ) . ξi <x

ξi <x

Then inequality (8.2.7) goes over into (8.2.12).

i=1



208

Chapter 8. Inequalities

8.3 Inequalities for means Let F be a strictly monotonic function on an interval J ⊂ R, let w1 , . . . , wn be non n   negative real numbers wi > 0 , and let a1 , . . . , an ∈ F (J) be arbitrary points. i=1

The expression

⎛ n

wi F −1 (ai )

⎜ i=1 Mn (F ; a; w) = F ⎜ n ⎝ 

wi

⎞ ⎟ ⎟ ⎠

(8.3.1)

i=1

is called a weighted quasiarithmetic mean of a1 , . . . , an with the weights w1 , . . . , wn . In (8.3.1) and in the sequel a stands for (a1 , . . . , an ), and w stands for (w1 , . . . , wn ). Theorem 8.3.1. Let J ⊂ R be an open interval, and let the functions F, G : J → R be strictly monotonic, F (J) = G (J). Let the function G−1 ◦ F be convex2 . Then, for every n ∈ N, every a1 , . . . , an ∈ F (J), and every non-negative w1 , . . . , wn such that n  wi > 0, we have i=1

Mn (F ; a; w)  Mn (G; a; w)

(8.3.2)

whenever G is increasing. If G is decreasing, inequality (8.3.2) gets reversed. Proof. We take in (8.1.1) f = G−1 ◦ F , qi = wi , xi = F −1 (ai ), i = 1, . . . , n. We get ⎛ ⎞ n n  −1 wi F (ai ) w G−1 (ai ) ⎜ i=1 ⎟ i=1 i −1 ⎜ ⎟ . (8.3.3) G ◦F ⎝ n n ⎠   wi wi i=1

i=1

Applying to both the sides of (8.3.3) the function G we obtain (8.3.2) if G is increasing, and the reversed inequality if G is decreasing.  Remark. If the function G−1 ◦ F is concave, then inequality (8.3.2) holds whenever G is decreasing, and gets reversed whenever G is increasing. The well-known inequality Hn  Gn  An between the arithmetic mean An =

1 (a1 + · · · + an ) , n

(8.3.4)

the geometric mean 1/n

Gn = (a1 · · · an )

,

(8.3.5)

and the harmonic mean   −1 −1 Hn = n a−1 , 1 + · · · + an

(8.3.6)

(a1 , . . . , an > 0) can be obtained from Theorem 8.3.1, since each mean (8.3.4), (8.3.5), (8.3.6) is a quasiarithmetic mean. Actually, we can prove a more general result. 2

Since the function G−1 ◦ F is monotonic its convexity implies the continuity.

8.3. Inequalities for means

209

For every positive a1 , . . . , an and non-negative w1 , . . . , wn such that

n 

wi > 0,

i=1

and for every r ∈ R, r = 0, we put Mnr [a; w] =

 n

wi ari /

i=1

n

1/r wi

.

(8.3.7)

i=1

For r = 0 we assume  Mn0 [a; w] = exp

n

wi log ai /

i=1

n

 wi

=

i=1

n -

wi /

ai

n  j=1

wj

.

(8.3.8)

i=1

We define also Mnr for r = +∞ and r = −∞: Mn∞ [a; w] = max ai , i

Mn−∞ [a; w] = min ai .

(8.3.9)

i

In particular, we have for means (8.3.4), (8.3.5), (8.3.6) An = Mn1 [a; 1] ,

Gn = Mn0 [a; 1] ,

Hn = Mn−1 [a; 1] ,

where 1 means (1, . . . , 1). Theorem 8.3.2. With fixed n ∈ N, a = (a1 , . . . , an ) and w = (w1 , . . . , wn ) (ai > n  0, wi  0, i = 1, . . . , n; wi > 0) the function ϕ (r) = Mnr [a; w] is increasing on [−∞, ∞].

i=1

Proof. Inequalities ϕ (−∞)  ϕ (r)  ϕ (∞) for −∞ < r < ∞ are obvious. Now the theorem results from Theorem 8.3.1. For r = 0 (8.3.7) coincides with (8.3.1) for F (x) = x1/r , whereas (8.3.8) coincides with (8.3.1) for F (x) = ex . Taking r, s ∈ R, r < s, we need only use Theorem 8.3.1 for F and G as follows. 1. If r = 0, s = 0, then F (x) = x1/r , G (x) = x1/s , whence G−1 ◦ F (x) = xs/r . If 0 < r < s or r < 0 < s, then G is increasing and G−1 ◦ F is convex, and if r < s < 0, then G is decreasing and G−1 ◦ F is concave, so at any case ϕ (r)  ϕ (s). 2. r = 0 < s. Then F (x) = ex , G (x) = x1/s . The function G−1 ◦ F (x) = esx is convex and G is increasing, so ϕ (0)  ϕ (s). 3. r < 0 = s. Then F (x) = x1/r , G (x) = ex . The function G−1 ◦ F (x) = 1r log x is convex since r < 0, and the function G is increasing, so ϕ (r)  ϕ (0).  We will prove one more generalization of the inequality between means (8.3.4) and (8.3.5). Let F be a strictly monotonic function on an interval D ⊂ R,  let n ∈ N, let I be a subset of {1, . . . , n}, and let a1 , . . . , an ∈ F (D), w1 , . . . , wn  0, wi > 0. Let H : F (D) → R be a continuous function. Put  α (H; F ; a; w; I) =

i∈I

 wi

i∈I

⎛

⎞ wi F −1 (ai ) ⎜ i∈I  ⎟ H ◦F ⎝ ⎠. wi i∈I

210

Chapter 8. Inequalities

With this notation, and under the conditions specified, we have Theorem 8.3.3. If the function F is continuous, the function H ◦ F is convex, and I, J are disjoint subsets of {1, . . . , n}, then for fixed a and w the function α (F ; I) = α (H; F ; a; w; I) fulfils α (F ; I ∪ J)  α (F ; I) + α (F ; J) .

(8.3.10)

If the function H ◦ F is concave, the inequality gets reversed.    wi , q2 = wi , x1 = wi F −1 (ai ) / Proof. Take in (8.1.1) n = 2, f = H ◦F , q1 = i∈I i∈J i∈I    / wi , x2 = wi F −1 (ai ) / wi . Then i∈I

i∈J

i∈J

⎛ 

⎞ wi F −1 (ai ) ) + q2 H ◦ F (x2 ) ⎜ i∈I∪J  ⎟ q1 H ◦ F (x1 H ◦F ⎝ , ⎠ wi wi i∈I∪J

(8.3.11)

i∈I∪J

whence, after multiplying by



wi we obtain (8.3.10). If H ◦ F is concave, then we

i∈I∪J

obtain (8.3.11) with H ◦ F replaced by −H ◦ F , and the reversed inequality (8.3.10)  results on multiplying by − wi .  i∈I∪J

Now suppose that F, G : D → R are strictly monotonic continuous functions such that F (D) ∩ G (D) = ∅, H : F (D) ∪ G (D) → R is continuous, H ◦ F is convex, H ◦ G is concave. By Theorem 8.3.3 α (F ; I ∪ J) − α (G; I ∪ J)  α (F ; I) − α (G; I) + α (F ; J) − α (G; J) .

(8.3.12)

Take in particular I = {1, . . . , n − 1}, J = {n}. Then α (F ; J) = α (G; J) = k  wi we get from (8.3.12) wn H (an ), and with notation (8.3.1) and Wk = i=1

Wn {H (Mn [F ; a; w]) − H (Mn [G; a; w])}  Wn−1 {H (Mn−1 [F ; a; w]) − H (Mn−1 [G; a; w])} . Specifying further w = 1, H (x) = G (x) = x, F (x) = ex , we get hence with notation (8.3.4), (8.3.5) n (Gn − An )  (n − 1) (Gn−1 − An−1 ), or n (An − Gn )  (n − 1) (An−1 − Gn−1 ) .

(8.3.13)

Since A1 = G1 for every a1 > 0, inequality (8.3.13) implies that An  Gn for n ∈ N. From (8.3.13) one can obtain the inequality An − Gn 

√ 1 √ 2 max ai − aj . i,j=1,...,n n

(8.3.14)

8.4. Hardy-Littlewood-P´ olya majorization principle

211

Inequalities (8.3.13), (8.3.14) are due to R. Rado3 . Similarly one can derive the following inequalities of T. Popoviciu4  n  n−1   .  .  2/n aj ai An 1 An−1 An  max  , and + . i,j=1,...,n 2 Gn Gn−1 Gn ai aj One can also show that if lim an = ∞, then lim n (An − Gn ) = ∞; and if lim an = n→∞ n→∞ n→∞ ∞  2 α < ∞, then lim n (An − Gn ) < ∞ if and only if (an − α) < ∞. n→∞

n=1

8.4 Hardy-Littlewood-P´olya majorization principle The Hardy-Littlewood-P´olya majorization principle

5

reads as follows.

Theorem 8.4.1. Let J ⊂ R be an interval , let xi , yi ∈ J, i = 1, . . . , n, be real numbers such that x1  · · ·  xn , y1  · · ·  yn , (8.4.1) 6

k

xi 

i=1

k

yi ,

k = 1, . . . , n − 1,

n

i=1

xi =

i=1

n

yi .

(8.4.2)

i=1

If f : J → R is a continuous and convex function, then n

f (xi ) 

i=1

n

f (yi ) .

(8.4.3)

i=1

Conversely, if for some xi , yi ∈ J, i = 1, . . . , n, such that (8.4.1) holds, inequality (8.4.3) is fulfilled for every continuous and convex function, then relations (8.4.2) hold. Theorem 8.4.1 is a particular case of the following theorem of L. Fuchs [93]. Theorem 8.4.2. Let J ⊂ R be an interval, let xi , yi ∈ J, i = 1, . . . , n, be real numbers fulfilling (8.4.1), and let p1 , . . . , pn ∈ R be such that k

i=1

pi xi 

k

pi y i ,

k = 1, . . . , n − 1,

i=1

n

pi xi =

i=1

n

pi y i .

(8.4.4)

i=1

If f : J → R is a continuous and convex function, then n

i=1 3

pi f (xi ) 

n

pi f (yi ) .

(8.4.5)

i=1

Cf. Mitrinovi´c [226] Cf. Mitrinovi´c [226] 5 Hardy-Littlewood-P´ olya [135]. Sometimes Theorem 8.4.1 is inscribed to J. Karamata. (Karamata [164]. Cf., e.g., Beckenbach-Bellman [24]). 6 Here and in the sequel, if the interval J is not open, the expression “f : J → R is a continuous and convex function” should be understood: “f : J → R is continuous, and f0 = f | int J, f0 : int J → R is a convex function”. 4

212

Chapter 8. Inequalities

Conversely, if pi  0 for i = 1, . . . , n, and if for some xi , yi ∈ J, i = 1, . . . , n, such that (8.4.1) holds, inequality (8.4.5) is fulfilled for every continuous and convex function f : J → R, then relations (8.4.4) hold. The following theorem of M. Tomi´c [315] is very similar to Theorem 8.4.2. Theorem 8.4.3. Let J ⊂ R be an interval, let xi , yi ∈ J, i = 1, . . . , n, be real numbers fulfilling (8.4.1), and let p1 , . . . , pn ∈ R be such that k

pi xi 

i=1

k

pi y i ,

k = 1, . . . , n.

(8.4.6)

i=1

If f : J → R is a continuous, convex and increasing function, then inequality (8.4.5) holds. Conversely, if pi  0 for i = 1, . . . , n, and if for some xi , yi ∈ J, i = 1, . . . , n, such that (8.4.1) holds inequality (8.4.5) is fulfilled for every continuous convex, and increasing function f : J → R, then relations (8.4.6) hold. We will prove Theorems 8.4.2 and 8.4.3 simultaneously. Proof. I. Sufficiency. Adding the same term to both sides of (8.4.5) does not affect the inequality. Therefore we may assume that xi = yi for i = 1, . . . , n. Put f (xi ) − f (yi ) , i = 1, . . . , n. Di = xi − yi We have by Theorem 7.3.2 and by (8.4.1) Di =

f (xi ) − f (yi ) f (xi+1 ) − f (yi ) f (xi+1 ) − f (yi+1 )   = Di+1 . xi − yi xi+1 − yi xi+1 − yi+1

(8.4.7)

Further, put X0 = Y0 = 0,

Xk =

k

pi xi ,

Yk =

i=1

k

pi y i ,

k = 1, . . . , n.

i=1

By (8.4.4) and (8.4.7), or by (8.4.6), (8.4.7) and the monotonicity of f , n−1

(Xk − Yk ) (Dk − Dk+1 ) + (Xn − Yn ) Dn  0,

k=1

or

n−1

Xk (Dk − Dk+1 ) + Xn Dn 

k=1

n−1

Yk (Dk − Dk+1 ) + Yn Dn .

(8.4.8)

k=1

But n−1

Xk (Dk − Dk+1 ) + Xn Dn =

k=1

=

n

k=1 n

k=1

Xk Dk −

n−1

Xk Dk+1 =

k=1

(Xk − Xk−1 ) Dk =

n

Xk Dk −

k=1 n

k=1

pk xk Dk ,

n

Xk−1 Dk

k=1

(8.4.9)

8.4. Hardy-Littlewood-P´ olya majorization principle

213

and similarly n−1

n

Yk (Dk − Dk+1 ) + Yn Dn =

k=1

pk y k Dk .

(8.4.10)

k=1

Thus we get from (8.4.8), (8.4.9) and (8.4.10) n

pk xk Dk 

k=1

n

pk y k Dk ,

n

i.e.,

k=1

pk (xk − yk ) Dk  0,

k=1

which is equivalent to (8.4.5). II. Necessity. Suppose that pi  0 for i = 1, . . . , n, and that (8.4.5) holds for some xi , yi ∈ J, i = 1, . . . , n, and for every continuous and convex [and increasing] function f : J → R. Taking f (x) = x we obtain from (8.4.5) n

pi xi 

i=1

n

pi y i ,

(8.4.11)

i=1

i.e., (8.4.6) for k = n. Moreover, if there is no restriction on the monotonicity of f , we take f (x) = −x to obtain −

n

pi xi  −

i=1

n

pi y i .

(8.4.12)

i=1

Relations (8.4.11) and (8.4.12) together yield the last equality in (8.4.4). Now take an arbitrary k, 1  k < n, and f (x) = max (0, x − yk ). Thus f (x)  0, f (x)  x − yk , and, since pi  0 (i = 1, . . . , n), we have pi f (x)  pi (x − yk ) .

(8.4.13)

Putting in (8.4.13) x = xi , i = 1, . . . , k, and summing up over i = 1, . . . , k, we get by (8.4.5) k

pi (xi − yk ) 

i=1

k

pi f (xi ) 

i=1

n

pi f (xi ) 

i=1

n

pi f (yi ) .

(8.4.14)

i=1

Now, by (8.4.1), since f (x) = 0 for x  yk , n

pi f (yi ) =

i=1

k

i=1

pi f (yi ) =

k

pi (yi − yk ) .

(8.4.15)

i=1

Relations (8.4.14) and (8.4.15) yield k

i=1

pi (xi − yk ) 

k

pi (yi − yk ) ,

k = 1, . . . , n − 1,

i=1

whence the first n − 1 inequalities in (8.4.4) follow.



214

Chapter 8. Inequalities

8.5 Lim’s inequality From Theorem 8.4.1 we derive the following result: Theorem 8.5.1. Let a  0, b  0, c  a + b be real numbers, and let f : [0, ∞) → R be a continuous and convex function. Then f (a) + f (b + c)  f (a + b) + f (c) .

(8.5.1)

Proof. We use Theorem 8.4.1 with n = 2, x1 = c, x2 = a + b, y1 = b + c, y2 = a, and (8.5.1) follows.  Lim’s inequality (Lim [207]) is the special case f (x) = xr , r  1. Lim’s inequality can also be generalized as follows (cf. Stankovi´c-Lackovi´c [294]). Theorem 8.5.2. Let m ∈ N, and let Qi  0, i = 1, . . . , m, be real numbers. Let Q  m  max Qi and Q0 = Qi − nQ for a certain non-negative integer n. If 0  Q0  Q,

i=1,...,m

i=1

and f : [0, ∞) → R is a continuous and convex function such that f (0) = 0, then nf (Q) + f (Q0 ) 

m

f (Qi ) .

(8.5.2)

i=1

Proof. We assume that Q0 > 0. The case, where Q0 = 0, will be covered by Theorem 8.5.3. Then m

nQ < nQ + Q0 = Qi  mQ, i=1

whence n < m. We may assume that Qi ’s are numbered in such a way that Q1  · · ·  Qm , and we put in Theorem 8.4.1 (with n replaced by m) y1 = · · · = yn = Q, yn+1 = Q0 , yi = 0 for i = n + 2, . . . , m; xi = Qi , i = 1, . . . , m. Then nf (Q) + f (Q0 ) =

m

i=1

f (yi ) 

m

i=1

f (xi ) =

m

f (Qi ) ,

i=1



i.e., we obtain (8.5.2).

In order to obtain (8.5.1) from (8.5.2) (in such an argument f must be assumed to vanish at the origin) we put m = 2, n = 1, Q1 = a + b, Q2 = c, Q = b + c, Q0 = a. Remark. If n is not an integer, Theorem 8.5.2 fails to hold. Take, for example, m = 2, n = 32 , Q1 = Q2 = Q = 1, Q0 = 12 , f (x) = x2 . Then all the remaining assumptions of Theorem 8.5.2 are fulfilled, and (8.5.2) turns into the false inequality 3 1 +  2. 2 4

8.6. Hadamard inequality

215

But for such a case we have the following Theorem 8.5.3. Let m ∈ N, and let Qi  0, i = 1, . . . , m, be real numbers. Let m  Q  max Qi , and let s ∈ R be such that sQ = Qi . If f : [0, ∞) → R is a i=1,...,m

i=1

continuous and convex function such that f (0) = 0, then m

sf (Q) 

f (Qi ) .

(8.5.3)

i=1

Proof. If Q = 0, then all Qi ’s are zeros, and (8.5.3) becomes 0  0, which is true. Now assume that Q > 0. For every λ ∈ [0, 1] we have f (λx) = f (λx + (1 − λ) 0)  λf (x) + (1 − λ) f (0) = λf (x) . 

Hence f (Qi ) = f and

Qi Q Q

m



1 f (Qi )  Q i=1

Qi f (Q) , Q

 

m

i = 1, . . . , m,

 Qi

f (Q) = sf (Q) ,

i=1



i.e., we obtain (8.5.3).

8.6 Hadamard inequality We start with the following, more general theorem of Lackovi´c-Stankovi´c [203]. Theorem 8.6.1. Let J = [a, b] be a closed real interval, let p : J → R be a non-negative &b integrable function such that pdx > 0, let f : J → R be a continuous and convex a

function, let c ∈ int J, and let k ∈ ∂f (c). Then &b

&b

xp (x) dx

f (c) + k a b &

− ck  p (x) dx

a

p (x) f (x) dx

a

&b

p (x) dx

a

&c

&b

&b

a

c

a

f (a) p (x) dx + f (b) p (x) dx + k 

&b

$

&c

&b

a

c

xp (x) dx − k a p (x) dx + b

% p (x) dx .

p (x) dx

a

(8.6.1) Proof. Put g (x) = f (x) − kx,

x ∈ J.

(8.6.2)

216

Chapter 8. Inequalities

Then g : J → R is a continuous and convex function, 0 ∈ ∂g (c), and thus g (x) − g (c)  0 in J. Since p  0 we obtain hence b

b p (x) g (x) dx 

a

p (x) g (c) dx.

(8.6.3)

a

Relations (8.6.3) and (8.6.2) yield b

b p (x) f (x) dx − k

a

Dividing by

b xp (x) dx  f (c)

a

&b

b p (x) dx − kc

a

p (x) dx. a

p (x) dx > 0 we obtain hence the left inequality in (8.6.1).

a

It follows from Theorem 7.3.5 that g (x) decreases in [a, c] and increases in [c, b] so that g (a) − g (x)  0 in [a, c], and g (b) − g (x)  0 in [c, b]. Hence b

c p (x) g (a) dx 

a

b

and thus

p (x) g (b) dx 

p (x) g (x) dx, a

b

c

c

b

b p (x) g (b) dx 

p (x) g (a) dx + a

p (x) g (x) dx, c

c

p (x) g (x) dx. a

By (8.6.2) we get hence c

c p (x) dx − ka

f (a) a

b c

b  &b

p (x) dx c

b p (x) f (x) dx − k

a

Dividing by

p (x) dx − kb

p (x) dx + f (b) a

b

xp (x) dx. a

p (x) dx we obtain hence the right inequality in (8.6.1).



a

1 (a + b), p (x) = 1 for x ∈ J, we obtain from (8.6.1) the Hadamard 2 inequality (Hadamard [127]; cf. also Vasi´c-Lackovi´c [316]). Setting c =

Theorem 8.6.2. Let J = [a, b] be a closed real interval, and let f : J → R be a continuous and convex function. Then  f

a+b 2



1  b−a

b f (x) dx  a

f (a) + f (b) . 2

8.7. Petrovi´c inequality

217

8.7 Petrovi´c inequality The following inequality is due to M. Petrovi´c [253]. Theorem 8.7.1. Let [0, a) ⊂ R, 0 < a  ∞, and let f : [0, a) → R be a continuous and convex function. Then for every n ∈ N and every x1 , . . . , xn ∈ [0, a) such that x1 + · · · + xn ∈ [0, a) we have f (x1 + · · · + xn )  f (x1 ) + · · · + f (xn ) − (n − 1)f (0). Proof I. For n = 2 we obtain from (8.1.1)   p q px + qy  f (x) + f (y) f p+q p+q p+q

(8.7.1)

(8.7.2)

for every p, q  0 such that p + q > 0. Assume that x1 , x2 ∈ [0, a) are such that x1 + x2 ∈ (0, a), and set in (8.7.2) x = x1 + x2 , y = 0, p = x1 , q = x2 . Then (8.7.2) goes into x1 x2 f (x1 )  f (x1 + x2 ) + f (0) . (8.7.3) x1 + x2 x1 + x2 Interchanging in (8.7.3) x1 and x2 , we obtain f (x2 ) 

x2 x1 f (x1 + x2 ) + f (0) . x1 + x2 x1 + x2

(8.7.4)

Adding (8.7.3) and (8.7.4) we get f (x1 ) + f (x2 )  f (x1 + x2 ) + f (0) , which is (8.7.1) for n = 2. If x1 = x2 = 0, (8.7.5) is trivial. Further the proof runs by induction.

(8.7.5) 

We will give also another proof of Theorem 8.7.1. Proof II. Put g (x1 , . . . , xn ) = f (x1 + · · · + xn ) − f (x1 ) − · · · − f (xn ) .

(8.7.6)

Fix a j, 1  j  n. Differentiating (8.7.6) with respect to xj , we get by Theorem 7.4.3 ∂g (x1 , . . . , xn ) = f  (x1 + · · · + xn ) − f  (xj )  0 ∂xj except for at most countably many values of xj . By Theorem 3.6.3 g is increasing with respect to each variable. Hence, in particular, for x1 , . . . , xn ∈ [0, a) such that x1 + · · · + xn ∈ [0, a) g (x1 , . . . , xn )  g (0, . . . , 0) = (1 − n) f (0) . Relations (8.7.6) and (8.7.7) yield inequality (8.7.1).

(8.7.7) 

218

Chapter 8. Inequalities

8.8 Mulholland’s inequality The results in this section are due to H. P. Mulholland [234]. Consider the two sets of conditions: (i) The function ϕ : [0, ∞) → [0, ∞) is continuous, strictly increasing, convex, ϕ (0) = 0, and log ϕ is a convex function of log x (i.e., the function F : R → R, F (t) = log ϕ (et ) is convex). (ii) There exists a function γ : R → R, continuous, convex and increasing and such that ϕ(x) = x exp [γ (log x)] for x > 0, ϕ (0) = 0. (8.8.1) Lemma 8.8.1. Conditions (i) and (ii) are equivalent. Proof. I. Assume (i) to be fulfilled. Put γ(t) = log

  ϕ (et ) = log ϕ et − t. t e

(8.8.2)

It is clear that γ is defined by (8.8.2) for t ∈ R, and it is continuous and convex. For x > 0 the function ϕ(x) − ϕ(0) ϕ(x) = x x−0 is increasing in virtue of Theorem 7.3.2, and thus ϕ (et ) /et , and hence also γ (t) is an increasing function of t. Formula (8.8.1) results from (8.8.2). II. Assume (ii) to be fulfilled. Then we have (8.8.1) with γ given by (8.8.2). Hence   F (t) = log ϕ et = γ (t) + t is a convex function. Moreover, since γ is increasing, we have for 0 < x < 1 the inequality γ (log x)  γ (0), whence by (8.8.1) 0 < ϕ (x)  x exp γ (0) , whence lim ϕ (x) = 0. Thus if we put ϕ (0) = 0, ϕ will be defined and continuous in x→0

[0, ∞), since γ is continuous in R. It follows from (8.8.1) that ϕ is strictly increasing. Further, we have for x > 0 1  D+ ϕ (x) = eγ(log x) + xeγ(log x) γ+ (log x) , x and thus D+ ϕ is an increasing function. By Theorem 7.5.3 ϕ is convex.  Lemma 8.8.2. Let ϕ satisfy condition (i), and put f (x) = ϕ (x) /x, x > 0. Let n ∈ N, and let s1 , . . . , sn be positive real numbers. Put " ! Q = x = (x1 , . . . , xn ) ∈ RN | 0  xi  si , i = 1, . . . , n \ {(0, . . . , 0)} ,

8.8. Mulholland’s inequality

219

and define the function Φ : Q → R by Φ (x1 , . . . , xn ) =

n

 −1

xi f (si ) /ϕ

n

i=1

 ϕ (xi ) .

i=1

Then Φ admits the maximum at the point (s1 , . . . , sn ). Proof. By Theorem 7.3.2 f is an increasing function. We have on suitable subsets of Q (where the derivative in question exists), and for every j = 1, . . . , n,  n  n

∂ ∂ ∂ log Φ = log xi f (si ) − log ψ ϕ (xi ) ∂xj ∂xj ∂xj i=1 i=1  n 

∂ f (sj ) − log ψ ϕ (xi ) , =  n ∂xj i=1 xi f (si ) i=1

where ψ = ϕ−1 is concave in virtue of Theorem 7.2.2. For the function F (t) = log ϕ (et ) we have F −1 (t) = log ψ (et ), and again by Theorem 7.2.2, since F is convex   and increasing, F −1 is concave, and thus its derivative F −1 is decreasing on the set where it exists. But  −1  ψ  (u) u. F (log u) = ψ (u) Consequently ∂ log ψ ∂xj

 n

 ϕ (xi )

i=1

 

n 



ϕ (xi )  ϕ (xj ) ψ ϕ (xi ) i=1  n     ϕ (xi )  ψ n ϕ (xj ) i=1  ϕ (xi )  = n n  i=1 ϕ (xi ) ψ ϕ (xi ) ψ

 i=1 n 

=

i=1

i=1







ϕ (xj ) f (xj ) ψ (ϕ (xj )) ϕ (xj ) ϕ (xj ) = =  . n n n  ψ (ϕ (xj ))  ϕ (xi ) xj ϕ (xi ) xi f (xi ) i=1

i=1

i=1

Thus n 

∂ f (xj ) f (sj ) −  = log Φ   n n ∂xj xi f (si ) xi f (xi ) i=1

i=1

i=1

xi f (xi ) f (sj ) − n 

n 

i=1 n  

xi f (si )

i=1

xi f (si ) f (xj )  xi f (xi )

.

i=1

Let xi = xi (t) ,

i = 1, . . . , n,

(8.8.3)

220

Chapter 8. Inequalities

be a regular curve passing inside Q. Then

d log Φ (x1 (t) , . . . , xn (t)) = dt j=1 n



 ∂ log Φ xj ∂xj

n  n  



j=1 i=1

 xi f (xi ) f (sj ) − xi f (si ) f (xj ) xj

n 

xi f (si )

i=1

n  

xi f (xi )



.

i=1

We have n n



 xi f (xi )f (sj ) − xi f (si )f (xj ) xj

j=1 i=1

=

1ijn

=



xi f (xi )f (sj )xj − xi f (si )f (xj )xj + xj f (xj )f (si )xi − xj f (sj )f (xi )xi



f (xi )f (sj ) − f (xj )f (si )



1ijn

 xi xj − xj xi =



Aij ,

1ijn

where    Aij = f (xi ) f (sj ) − f (xj ) f (si ) xi xj − xj xi ,

i, j = 1, . . . , n.

Now take the functions xi (t) of the special form xi (t) = ki t, ki  0, or xi (t) = si . In order to evaluate the sign of Aij , we must distinguish four cases. 1. xi (t) = ki t, xj (t) = kj t. Then xi xj = xj xi = ki kj t, whence Aij = 0. 2. xi (t) = si , xj (t) = sj . Then xi = xj = 0, and Aij = 0. 3. xi (t) = ki t, xj (t) = sj . Then Aij = f (sj ) (f (xi ) − f (si )) (−ki sj )  0. (Note that xi  si and f is increasing.) 4. xi (t) = si , xj (t) = kj t. Then Aij = f (si ) (f (sj ) − f (xj )) si kj  0.   d log Φ x1 (t), . . . , xn (t) is non-negative whenever it exists, and we could dt   carry out this argument for D+ log Φ x1 (t), . . . , xn (t) . By Theorem 3.6.3 log Φ, and hence also Φ, is increasing along every line (8.8.3), where every xi (t) is either ki t, ki  0, or si . Since every point in Q different from (s1 , . . . , sn ) can be joint with (s1 , . . . , sn ) by segments of lines of this form, the value of Φ at (s1 , . . . , sn ) exceeds or equals any other value of Φ in Q.  Ultimately

8.8. Mulholland’s inequality

221

Theorem 8.8.1. Let a function ϕ : [0, ∞) → [0, ∞) fulfil condition (i), let n ∈ N, let ai , bi , i = 1, . . . , n, be non-negative real numbers. Then  n  n  n   

−1 −1 −1 ϕ ϕ (ai + bi )  ϕ ϕ (ai ) + ϕ ϕ (bi ) . (8.8.4) i=1

i=1

i=1

Proof. It is enough to prove (8.8.4) for ai > 0, bi > 0, i = 1, . . . , n. By the continuity of ϕ (8.8.4) then results for arbitrary non-negative ai , bi , i = 1, . . . , n. Put in Lemma 8.8.2 xi = ai , si = ai + bi , i = 1, . . . , n. Then n 

n  ai f (ai + bi ) (ai + bi ) f (ai + bi ) i=1 i=1 n n  .   ϕ−1 ϕ (ai ) ϕ−1 ϕ (ai + bi ) i=1

(8.8.5)

i=1

Similarly, putting xi = bi , si = ai + bi , i = 1, . . . , n, we get n 

n  bi f (ai + bi ) (ai + bi ) f (ai + bi ) i=1 n   i=1  n .   −1 −1 ϕ ϕ (bi ) ϕ ϕ (ai + bi ) i=1

 Multiplying (8.8.5) by ϕ−1

(8.8.6)

i=1

 n   n   −1 ϕ (ai ) and (8.8.6) by ϕ ϕ (bi ) , and adding

i=1

i=1

together the resulting expressions, we obtain n 

 n  n  % (ai + bi )f (ai + bi ) $

 ϕ−1 (ai + bi )f (ai + bi )  i=1  n ϕ(ai ) + ϕ−1 ϕ(bi ) ,  −1 i=1 i=1 i=1 ϕ ϕ(ai + bi )

n

i=1



whence (8.8.4) results. Taking in Theorem 8.8.1 ϕ (x) = xr , r  1, we obtain

Corollary 8.8.1. Let n ∈ N, let ai , bi , i = 1, . . . , n, be non-negative real numbers, and let r  1. Then / / / 0 n 0 n 0 n

0 0 0 r r r r r 1 1 (ai + bi )  ai + 1 bri . (8.8.7) i=1

i=1

i=1

Inequality (8.8.7) is the famous Minkowski inequality. Theorem 8.8.2. Let a function ϕ : [0, ∞) → [0, ∞) fulfil condition (i), and let the function f (x) = ϕ (x) /x be strictly increasing. Put g = f −1 and ψ (x) = xg (x) for x > 0, ψ (0) = 0. Let n ∈ N, and let ai , bi , i = 1, . . . , n, be non-negative real numbers such that 0  ϕ (ai )  ψ (bi ) , i = 1, . . . , n. (8.8.8)

222

Chapter 8. Inequalities

Then

n

 −1

ai b i  ϕ

i=1

n

 ϕ (ai ) ψ

−1

 n

i=1

 ψ (bi ) .

(8.8.9)

i=1

  Proof. Taking in Lemma 8.8.2 xi = ai , si = ϕ−1 ψ(bi ) , i = 1, . . . , n (cf. (8.8.8)), we have n n   ai f (si ) si f (si ) i=1 i=1 n n  . (8.8.10)   ϕ−1 ϕ (ai ) ϕ−1 ϕ (si ) i=1

i=1

Put A=

n

i=1

ϕ (si ) =

n

n 

ψ (bi ) ,



i=1

B=

ϕ−1

i=1

n 

si f (si )

ϕ (si )

A  n =  = −1 .  ϕ (A) ϕ (si ) ϕ−1 ϕ (si ) i=1

n 

i=1

i=1

(8.8.11) Then

ϕ−1 (A) = A/B, 

whence A=ϕ

A B

 =

A f B



A B



 ,

and f

A B

 = B,

i.e., A = f −1 (B) = g (B) , B

and ψ (B) = Bg (B) = A.

Hence B = ψ −1 (A) = ψ −1

 n

 ψ (bi ) .

(8.8.12)

i=1

Relations (8.8.10), (8.8.11) and (8.8.12) yield  n  n   n

−1 −1 ai f (si )  ϕ ϕ (ai ) ψ ψ (bi ) . i=1

i=1

(8.8.13)

i=1

Now, ϕ (si ) = ψ (bi ) ,

i = 1, . . . , n.

(8.8.14)

On the other hand, ϕ (si ) = si f (si ) = g [f (si )] f (si ) = ψ [f (si )] ,

i = 1, . . . , n.

(8.8.15)

Relations (8.8.14) and (8.8.15) yield ψ (bi ) = ψ [f (si )] ,

i = 1, . . . , n,

and, since ψ is one-to-one, bi = f (si ) , (8.8.16) inserted into (8.8.13) yield (8.8.9).

i = 1, . . . , n.

(8.8.16) 

8.9. The general inequality of convexity

223

8.9 The general inequality of convexity Let E be an arbitrary set, and let P be a family of functions f : E → [0, ∞) such that f1 + f2 ∈ P and λf ∈ P whenever f, f1 , f2 ∈ P and λ  0 (thus P is a cone in the space of all functions f : E → R). Let M : P → [0, ∞] be a function such that M (0) = 0 and M (λf ) = λM (f ) M (f + g)  M (f ) + M (g) f  g ⇒ M (f )  M (g)

for λ > 0, f ∈ P, for f, g ∈ P,

(8.9.1) (8.9.2)

for f, g ∈ P,

(8.9.3)

where f  g means f (x)  g (x) for all x ∈ E. Thus M is positively homogeneous, subadditive, and increasing. E.g., if (E, M, μ) is a measure space, we&may take P to be the family of all integrable functions f : E → [0, ∞), and M (f ) = f dμ. Another E

example: E = {1, . . . , N } a finite set, P = RN + , where ! " N RN + = x = (x1 , . . . , xN ) ∈ R | x1  0, . . . , xN  0 , and M any standard norm in RN .

Theorem 8.9.1. Let ϕ : RN + → [0, ∞) be a continuous and positively homogeneous function such that ϕ (x1 , . . . , xN ) > 0 whenever x1 > 0, . . . , xN > 0, and the set ! " K = x = (x1 , . . . , xN ) ∈ RN + | ϕ (x1 , . . . , xN )  1 is convex. If f1 , . . . , fN ∈ P are such that M (fi ) < ∞ for i = 1, . . . , N , and ϕ (f1 , . . . , fN ) ∈ P , then     M ϕ (f1 , . . . , fN )  ϕ M (f1 ) , . . . , M (fN ) . (8.9.4) Proof. By Theorem 5.1.8 the set K is equal to the intersection of all the closed halfspaces containing K, determined by all the support hyperplanes of K. Thus K is the set of points (x1 , . . . , xN ) ∈ RN such that ατ1 x1 + · · · + ατN xN − βτ  0 for all τ from a certain set T of indices. If x1 > 0, . . . , xN > 0, then ϕ (λx1 , . . . , λxN ) = λϕ (x1 , . . . , xN )  1 for λ sufficiently large, and thus (λx1 , . . . , λxN ) ∈ K for λ sufficiently large. In other words, λ (ατ1 x1 + · · · + ατN xN ) − βτ  0 for λ sufficiently large, whence ατi  0 for i = 1, . . . , N and all τ ∈ T . If for a τ ∈ T we had βτ  0, then the point 0 = (0, . . . , 0) would lie on the suitable support hyperplane of K. If we draw a line from 0, passing through a point x = (x1 , . . . , xN ) such that x1 > 0, . . . , xN > 0 (note that K ⊂ RN + ) and crossing the support hyperplane in

224

Chapter 8. Inequalities

question, then for large λ the points λx would lie on the other side of the hyperplane, and thus do not belong to K, which is impossible. Consequently βτ > 0 for all τ ∈ T . Let " ! +1 | xN +1  ϕ(x1 , . . . , xN ) . C = (x1 , . . . , xN , xN +1 ) ∈ RN + If xN +1 = 0, then (x1 , . . . , xN , xN +1 ) ∈ C for every (x1 , . . . , xN ) ∈ RN + , and also 0 = βτ xN +1  ατ1 x1 + · · · + ατN xN for every (x1 , . . . , xN ) ∈ RN homogeneityof ϕ, the + . If xN +1 > 0, then, by the positive  x1 xN , which, inequality xN +1  ϕ (x1 , . . . , xN ) is equivalent to 1  ϕ ,..., xN +1 xN +1 in turn, is equivalent to βτ xN +1  ατ1 x1 + · · · + ατN xN

for all τ ∈ T.

(8.9.5)

Thus condition (8.9.5) is necessary and sufficient for (x1 , . . . , xN , xN +1 ) ∈ C. Now take arbitrary f1 , . . . , fN ∈ P such that M (fi ) < ∞ for i = 1, . . . , N , and let x ∈ E be arbitrary. We have    f1 (x), . . . , fN (x), ϕ f1 (x), . . . , fN (x) ∈ C, whence   βτ ϕ f1 (x), . . . , fN (x)  ατ1 f1 (x) + · · · + ατN fN (x)

for all τ ∈ T.

(8.9.6)

Relation (8.9.6) holds for all x ∈ E, consequently βτ ϕ (f1 , . . . , fN )  ατ1 f1 + · · · + ατN fN (for all τ ∈ T ), and by (8.9.1), (8.9.2) and (8.9.3)     βτ M ϕ (f1 , . . . , fN ) = M βτ ϕ (f1 , . . . , fN )  M (ατ1 f1 + · · · + ατN fN )  ατ1 M (f1 ) + · · · + ατN M (fN ) for all τ ∈ T . But this means that    M (f1 ), . . . , M (fN ), M ϕ(f1 , . . . , fN ) ∈ C, which is equivalent to (8.9.4).



β Example 8.9.1. Let N = 2, and let ϕ (x1 , x2 ) = xα 1 x2 , where α > 0, β > 0, α + β = 1. −α/β Then ϕ (x1 , x2 ) = 1 means x2 = x1 , and consequently the set K is the part of −α/β the first quadrant lying above the hiperbola x2 = x1 , i.e., it is a convex set. From inequality (8.9.4) we obtain a generalization of the H¨ older inequality (8.1.13)   α β M f1α f2β  M (f1 ) M (f2 ) . (8.9.7)

8.9. The general inequality of convexity

225

r  1/r 1/r Example 8.9.2. Let N = 2, and let ϕ (x1 , x2 ) = x1 + x2 , where r  1. Then r  1/r ϕ (x1 , x2 ) = 1 means x2 = 1 − x1 , and consequently the set K is the part of the r  1/r first quadrant lying above the curve x2 = 1 − x1 , i.e., it is a convex set. From inequality (8.9.4) we obtain r   r  1/r 1/r 1/r 1/r  M (f1 ) + M (f2 ) . M f1 + f2

(8.9.8)

Replacing in (8.9.8) fj by fjr , j = 1, 2, we obtain a generalization of the Minkowski inequality (8.8.7) r 1/r

M [(f1 + f2 ) ]

 M (f1r )

1/r

1/r

+ M (f2r )

.

Exercises 1. Let D ⊂ RN be a convex and open set, and let f : D → (0, ∞) be a positive function. Show that if the function log f is convex, then f also is convex. 2. Let f : [0, ∞) → R be a continuous and concave function such that f (0)  0. Show that for every n ∈ N and x1 , . . . , xn ∈ [0, ∞) we have f (x1 + · · · + xn )  f (x1 ) + · · · + f (xn ). 3. Let f : [0, ∞) → R be a continuous and convex function. Derive inequality (8.7.1) from Theorem 8.5.3. 4. Derive inequality (8.1.13) from (8.9.7). n  qi = 1, then 5. Prove that if n ∈ N, xi  0, qi > 0 for i = 1, . . . , n, and i=1 n i=1

xqi i 

n

qi xi .

i=1

6. Under conditions of Exercise 8.5, show that, for every p  1, p  n n

qi xi  qi xpi . i=1

i=1

7. Let n ∈ N, xi  0, yi  0, i = 1, . . . , n. Show that n 1/n  n 1/n  n 1/n (xi + yi )  xi + yi i=1

i=1

i=1

(cf. Roberts-Varberg [267]). [Hint: Note that 1/n  n 1/n  n 2n 2n 1/n 1/n - xi - yi + ( i=1 yi ) ( i=1 xi ) = + , 2n 1/n x + yi x + yi ( i=1 (xi + yi )) i=1 i i=1 i and use Exercise 8.5.]

226

Chapter 8. Inequalities

8. Let J ⊂ R be an open interval, and let f : J → R be a continuous function. Prove that f is convex if and only if 1 f (x)  2h

h f (x + t) dt −h

for every x ∈ J and for every h > 0 such that x − h, x + h ∈ J. (Beckenbach [23], [22].)

Chapter 9

Boundedness and Continuity of Convex Functions and Additive Functions 9.1 The classes A,B,C The Theorem of Bernstein–Doetsch (cf., in particular, Corollary 6.4.1) says that if D ⊂ RN is a convex and open set, f : D → R is a convex function, T ⊂ D is open and non-empty, and f is bounded above on T , then f is continuous in D. Are there other sets T with this property? What are possibly weak conditions which assure the continuity of a convex function, or of an additive function? In this and in the next chapter we will deal with such questions. In order to simplify the notation we introduce the following classes of sets (GerKuczma [115]). A

= { T ⊂ RN | every convex function f : D → R, where T ⊂ D ⊂ RN and D is convex and open, bounded above on T is continuous in D}, B = { T ⊂ RN | every additive function f : RN → R bounded above on T is continuous}, C = { T ⊂ RN | every additive function f : RN → R bounded on T is continuous}. If we want to stress that the classes A, B, C refer to a particular space RN , we will write AN , BN , CN instead of A, B, C. Since every additive function is convex (cf. 5.3), it follows directly from the definition that A ⊂ B ⊂ C. (9.1.1) In the next chapter it will be proved (Theorem 10.2.2) that actually A = B.

(9.1.2)

Relation (9.1.2) implies that the class A is independent of the choice of the convex and open set D occurring in the definition. Relation (9.1.2) implies also that there

228

Chapter 9. Boundedness and Continuity

are no special conditions guaranteeing that a set T ∈ B which would not guarantee at the same time that also T ∈ A. In the sequel of this chapter we will freely use relation (9.1.2), deferring its proof to 10.2. On the other hand, the inclusion B ⊂ C is strict, i.e., the class C is strictly larger than B. Namely we have the following Theorem 9.1.1. C \ B = ∅. Proof. Let f0 : RN → R be a discontinuous additive function, and put T = { x ∈ RN | f0 (x)  0 }.

(9.1.3)

Then T ∈ / B, since there exists a discontinuous additive function (viz. f0 ) bounded above on T . We will show that T ∈ C. Let f : RN → R be an arbitrary additive function which is bounded on T . Thus there exists a constant M > 0 such that |f (x)|  M

for x ∈ T.

(9.1.4)

Take an arbitrary x0 ∈ RN . By Theorem 5.2.1 f (kx0 ) = kf (x0 ),

(9.1.5)

f0 (kx0 ) = kf0 (x0 )

(9.1.6)

and

for every k ∈ Z. By (9.1.6), depending on the sign of f0 (x0 ), kx0 ∈ T for k = 1, 2, 3, . . . , or for k = −1, −2, −3, . . . . Making use of (9.1.5) and (9.1.4) we get hence |kf (x0 )|  M for k = 1, 2, 3, . . . , or for k = −1, −2, −3, . . . . This is possible if and only if f (x0 ) = 0. Consequently f (x0 ) = 0 for arbitrary x0 ∈ RN , i.e., f = 0, and hence it is continuous. Thus every additive function f : RN → R bounded on T is continuous, i.e., T ∈ C. Hence T ∈ C \ B.  Let us also note the following Theorem 9.1.2. Let f : RN → R be an additive function bounded below on a set T ∈ B. Then f is continuous. Proof. If f is an additive function bounded below on T , then −f is an additive function bounded above on T . Since T ∈ B, this implies that −f is continuous, and hence also f is continuous. 

9.2. Conservative operations

229

9.2 Conservative operations We recall that P(X) denotes the collection of all subsets of X (cf. Axiom 1.1.4). A function F : P(RN ) → P(RN ) is called an A-conservative operation iff, for every set T ⊂ RN and every additive function f : RN → R, if f is bounded above on T , then f is bounded above also on F (T ). A function F : P(RN ) → P(RN ) is called a C-conservative operation iff, for every set T ⊂ RN and every additive function f : RN → R, if f is bounded on T , then f is bounded also on F (T ). Lemma 9.2.1. Every A-conservative operation is also C-conservative. Proof. Let F : P(RN ) → P(RN ) be an A-conservative operation, let T ⊂ RN , and let f : RN → R be an additive function bounded on the set T . Then both the additive functions, f and −f , are bounded above on T , and hence also on F (T ). Consequently f is bounded on F (T ), which proves that F is C-conservative.  But, as we shall see later, there exist C-conservative operations which are not A-conservative. Theorem 9.2.1. If F : P(RN ) → P(RN ) is an A-conservative operation and, for a certain T ⊂ RN , F (T ) ∈ A, then also T ∈ A. Proof. Let f : RN → R be an arbitrary additive function bounded  above on T . Then f is also bounded above on F (T ), and since F (T ) ∈ A = B cf. (9.1.2) , f is continuous. Consequently T ∈ B = A.  Theorem 9.2.2. If F : P(RN ) → P(RN ) is a C-conservative operation and, for a certain T ⊂ RN , F (T ) ∈ C, then also T ∈ C. Proof. Let f : RN → R be an arbitrary additive function bounded on T . Then f is bounded on F (T ), and since F (T ) ∈ C, f is continuous. Consequently T ∈ C.  Theorem 9.2.3. The composition of A-conservative [C-conservative] operations is Aconservative [C-conservative]. Proof. We will prove Theorem 9.2.3 for A-conservative operations only. The proof for C-conservative operations is similar. Let F1 , F2 : P(RN ) → P(RN ) be A-conservative operations, let T ⊂ RN , and let N f : R → R be an arbitrary additive function bounded above on T . Then f is bounded   above on F1 (T ), and hence also on F2 F1 (T ) . Thus F2 ◦ F1 is A-conservative.  By induction this generalizes to an arbitrary finite number of operations F1 , . . . , Fn . Trivially we have Theorem 9.2.4. Any operation F : P(RN ) → P(RN ) such that F (T ) ⊂ T for every set T ⊂ RN is A-conservative. Corollary 9.2.1. If A ⊂ B ⊂ RN and A ∈ A [A ∈ C], then also B ∈ A [B ∈ C]. Proof. By Theorem 9.2.4 and Lemma 9.2.1 operation F (T ) = A ∩ T is A-conservative and C-conservative, and F (B) = A. Theorems 9.2.1 and 9.2.2 complete the proof. 

230

Chapter 9. Boundedness and Continuity

But a real interest lies in operations F such that F (T ) is larger than T . Below we exhibit a number of such conservative operations. For every n ∈ N we define Sn : P(RN ) → P(RN ) by Sn (A) = A + · · · + A (n summands). The operations J and Q are defined in 5.1. Theorem 9.2.5. The operations J, Q and Sn for every n ∈ N are A-conservative. Proof. Let T ⊂ RN , and let f : RN → R be an additive (and hence convex) function bounded above on T f (x)  M for x ∈ T. (9.2.1) By Lemma 6.1.2 and Corollary 6.1.1 f is bounded above on Q(T ) and J(T ), which shows that the operations J and Q are A-conservative. Take a y ∈ Sn (T ). Then y = t1 + · · · + tn , where ti ∈ T for i = 1, . . . , n. Hence f (y) = f (t1 ) + · · · + f (tn )  M + · · · + M = nM. Thus f is bounded above on Sn (T ) by the constant nM , which shows that the operation Sn is A-conservative.  Now we define the operations R and U R(T ) = T − T

(9.2.2)

U (T ) = { x ∈ RN | x = t1 +(t2 −t3 ), t1 , t2 , t3 ∈ T } = T +(T −T ) = T +R(T ). (9.2.3) Theorem 9.2.6. The operations R and U are C-conservative. Proof. Let T ⊂ RN , and let f : RN → R be an arbitrary additive function bounded on T |f (x)|  M for x ∈ T. (9.2.4) If y ∈ R(T ), then there exist t1 , t2 ∈ T such that y = t1 − t2 , whence |f (y)| = |f (t1 ) − f (t2 )|  |f (t1 )| + |f (t2 )|  2M. If y ∈ U (T ), then there exist t1 , t2 , t3 ∈ T such that y = t1 + (t2 − t3 ), whence |f (y)| = |f (t1 ) + f (t2 ) − f (t3 )|  |f (t1 )| + |f (t2 )| + |f (t3 )|  3M. Thus f is bounded on R(T ) and on U (T ), which shows that the operations R and U are C-conservative.  On the other hand, R and U are not A-conservative. Let f0 : RN → R be a discontinuous additive function, and define the set T ⊂ RN by (9.1.3). Then f0 is bounded above on T . Let x0 ∈ RN be such that f0 (x0 ) < 0. Then we have (9.1.6) for all k ∈ Z, whence kx0 ∈ T for k ∈ N. Hence −kx0 = x0 − (k + 1)x0 ∈ T − T = R(T )

9.3. Simple conditions

231

for every k ∈ N, and lim f0 (−kx0 ) = lim

k→∞

k→∞



 − kf0 (x0 ) = ∞.

Consequently f0 is not bounded above on R(T ). Similarly it can be shown that f0 is not bounded above on U (T ). It follows that the operations R and U are not Aconservative. Let us note also the following two simple results. Theorem 9.2.7. The operation F : P(RN ) → P(RN ) defined by F (T ) = αT,

(9.2.5)

where α is a fixed rational number, is C-conservative. If α  0, then F is A-conservative. Proof. Let T ⊂ RN , and let f : RN → R be an arbitrary additive function fulfilling (9.2.4). If y ∈ αT , then y = αt for a certain t ∈ T . Hence by Theorem 5.2.1 |f (y)| = |f (αt)| = |αf (t)| = |α| |f (t)|  |α|M. Thus f is bounded on αT , which means that operation (9.2.5) is C-conservative. If f fulfils (9.2.1) and α  0, then f (y) = αf (t)  αM, and f is bounded above on αT . Consequently in this case operation (9.2.5) is Aconservative.  Theorem 9.2.8. The operation F : P(RN ) → P(RN ) defined by F (T ) = T + t,

(9.2.6)

where t ∈ RN is fixed, is A-conservative. Proof. Let T ⊂ RN , and let f : RN → R be an arbitrary additive function fulfilling (9.2.1). If y ∈ T + t, then y = x + t, where x ∈ T . Thus by (9.2.1) f (y) = f (x) + f (t)  M + f (t). This means that f is bounded above on T + t, and operation (9.2.6) is A-conservative. 

9.3 Simple conditions In this section we give some sufficient conditions and some necessary conditions for a set T ⊂ RN to belong to the class A or C. Lemma 9.3.1. If T ⊂ RN and int T = ∅, then T ∈ A.

232

Chapter 9. Boundedness and Continuity

Proof. Corollary 6.4.1 says that int T ∈ A. The operation F (T ) = int T fulfils the condition of Theorem 9.2.4, and hence is A-conservative. In virtue of Theorem 9.2.1 we have T ∈ A.  Theorem 9.3.1. [Theorem of Ostrowski]. If T ⊂ RN and mi (T ) > 0, then T ∈ A. Proof. By the Theorem of Steinhaus (Theorem 3.7.1) int S2 (T ) = ∅, whence by Lemma 9.3.1 S2 (T ) ∈ A. Theorems 9.2.5 and 9.2.1 now imply that T ∈ A.  For N = 1 Theorem 9.3.1 was proved by A. Ostrowski [248]. The generalizations to higher dimensions are due to A. Cs´asz´ar [50] and S. Marcus [218]. The topological analogue of the Theorem of Ostrowski is the following theorem of M. R. Mehdi [224] (cf. also Kominek [172]). Theorem 9.3.2. [Theorem of Mehdi]. If T ⊂ RN contains a second category set with the Baire property, then T ∈ A. Proof. By the Theorem of S. Piccard (Theorem 2.9.1) int S2 (T ) = ∅, and the proof runs as that of Theorem 9.3.1.  From Theorems 9.3.1 and 9.3.2 and from Theorems 9.2.5 and 9.2.1 we obtain the following Theorem 9.3.3. Let T ⊂ RN . Each of the following conditions is sufficient for T to belong to A:   (i)n mi Sn (T ) > 0. (ii)n Sn (T ) contains a set of the second category and with the Baire property.   (iii) mi J(T ) > 0. (iv) (v)

J(T ) contains a set of the second category and with the Baire property.   mi Q(T ) > 0.

(vi)

Q(T ) contains a set of the second category and with the Baire property.

Condition (i)2 is due to S. Kurepa [199]; (i)n for arbitrary n to R. Ger [97] (cf. also Kemperman [165], Marcus [218]). Condition (iii) is found with R. Ger–M. Kuczma [115] (cf. also Kuczma [176]), condition (v) with R. Ger [97]. From Theorem 9.3.3 we obtain immediately (cf. Exercise 2.11 and Exercise 2.13) Corollary 9.3.1. C ∈ A = A1 . Concerning the mutual relations of the conditions occurring in Theorem 9.3.3, we have the following Lemma 9.3.2. With the notation of Theorem 9.3.3, (i)n ⇒ (i)n+1 , (i)n ⇒ (iii), (iii) ⇒ (v).

n ∈ N, n ∈ N,

9.3. Simple conditions

233

Similarly (ii)n ⇒ (ii)n+1 , n ∈ N, n ∈ N, (ii)n ⇒ (iv), (iv) ⇒ (vi). None of these implications can be reversed. Proof. Fix an n ∈ N, T ⊂ RN , and take a t0 ∈ T . Clearly we have Sn (T ) + t0 ⊂ Sn+1 (T ).

(9.3.1)

Relation (9.3.1) proves the implication (i)n ⇒ (i)n+1 and (ii)n ⇒ (ii)n+1 . 1 Sm (T ). Now, to a given n ∈ N, choose a p so that m = 2p > n. Take an x ∈ m There exist t1 , . . . , tm ∈ T such that

1 1 ti . ti = m i=1 m i=1 m

x=

m

By Theorem 5.1.3 x ∈ J(T ), which shows that 1 Sm (T ) ⊂ J(T ). m

(9.3.2)

By what we have already proved, (i)n ⇒ (i)m and (ii)n ⇒ (ii)m . Inclusion (9.3.2) proves now the implication (i)n ⇒ (iii) and (ii)n ⇒ (iv). Since every Q-convex set is, in particular, J-convex, we have J(T ) ⊂ Q(T ) for every T ⊂ RN . Hence we obtain the implications (iii) ⇒ (v) and (iv) ⇒ (vi). Now we are going to show, by suitable counterexamples, that none of the implications in Theorem 9.3.3 can be reversed. These counterexamples are due to R. Ger [97]. Fix an n ∈ N, and consider the set T0 ⊂ R:  T0 =

x ∈ [0, 1] | x =

∞

i=1

εi , (n + 2)i

 εi = 0 or 1, i ∈ N .

Then for every x ∈ Sn (T0 ) the expansion of x with the base (n + 2) does not contain the digit (n + 1). By Theorem 3.6.4 the set Sn (T0 ) has measure zero, and it is easy to check that Sn (T0 ) is nowhere dense. On the other hand, Sn+1 (T0 ) contains all the numbers from the interval [0, 1]. The set T = T0 × RN −1 ⊂ RN has similar properties: Sn (T ) is nowhere dense and of measure zero, whereas Sn+1 (T ) contains the strip [0, 1] × RN −1 . Consequently conditions (i)n+1 and (ii)n+1 are fulfilled, whereas (i)n and (ii)n are not. Now let H ⊂ RN be a Hamel basis, and put T = QH = {x ∈ RN | x = λh, λ ∈ Q, h ∈ H}. Suppose that for a certain n ∈ N condition (i)n or (ii)n is fulfilled. By Theorem 3.7.1 or 2.9.1

int S2n (T ) = int Sn (T ) + Sn (T ) = ∅.

234

Chapter 9. Boundedness and Continuity

Every x ∈ int S2n (T ) (except, at most, x = 0) may be written in the form x = λ1 (x)h1 (x) + · · · + λk(x) (x)hk(x) (x),

(9.3.3)

where λi (x) ∈ Q, λi (x) = 0, hi (x) ∈ H, i = 1, . . . , k(x), k(x) ∈ N. Moreover, k(x)  2n,

(9.3.4)

since x ∈ S2n (T ), and due to the uniqueness of expansion (9.3.3) (Lemma 4.1.1). Write m = sup k(x). By (9.3.4) there exists an x0 ∈ int S2n (T ) such that x∈int S2n (T )

k(x0 ) = m. Let x0 = λ1 (x0 )h1 (x0 ) + · · · + λm (x0 )hm (x0 ),

λi (x0 ) = 0, i = 1, . . . , m,

be the Hamel expansion of x0 , and take an h0 ∈ H, h0 = hi (x0 ), i = 1, . . . , m. We have x¯ = x0 + λh0 ∈ int S2n (T ) for sufficiently small positive λ ∈ Q. For such a λ we have x ¯ = λ1 (x0 )h1 (x0 ) + · · · + λm (x0 )hm (x0 ) + λh0 , and x¯ ∈ int S2n (T ), whence k(¯ x) = m + 1, which contradicts the maximality of m. Consequently none of conditions (i)n , (ii)n , n ∈ N, can be fulfilled. In order to show that conditions (iii) and (iv) are fulfilled, we will prove that J(T ) = RN .

(9.3.5)

Take an arbitrary x ∈ RN . x has an expansion (9.3.3), where λi (x) ∈ Q, hi (x) ∈ H, i = 1, . . . , k(x). Let p ∈ N be chosen so that m = 2p > n, take arbitrary hj ∈ H, j = k(x) + 1, . . . , m, so that hj = hi (x), i = 1, . . . , k(x); j = k(x) + 1, . . . , m, and put λj = 0, j = k(x) + 1, . . . , m. Then mx = mλ1 (x)h1 (x) + mλk(x) (x)hk(x) (x) + λk(x)+1 hk(x)+1 + · · · + λm hm ∈ Sm (T ), 1 Sm (T ), and by (9.3.2) x ∈ J(T ). Consequently RN ⊂ J(T ), and since m the converse inclusion is obvious, we obtain (9.3.5). Now, again, let H ⊂ RN be an arbitrary Hamel basis, and put

whence x ∈

T = { x ∈ RN | x =

n

λi hi , λi ∈ Z, hi ∈ H, i = 1, . . . , n; n ∈ N }.

i=1

The set J(T ) contains those x ∈ RN whose coefficients in the Hamel expansions are diadic numbers. A sum of such x’s again has such a form, whence J(T ) + J(T ) ⊂ J(T ). Consequently if either of conditions (iii), (iv) were fulfilled, we would have int J(T ) = ∅. Let x ∈ int J(T ) have a Hamel expansion x = λ1 h1 + · · · + λk hk ,

(9.3.6)

9.3. Simple conditions

235

λi ∈ D, hi ∈ H, i = 1, . . . , k, and take an h0 ∈ H, h0 = hi , i = 1, . . . , k. The point x ¯ = λ1 h1 + · · · + λk hk + 3−n h0 does not belong to J(T ) (since 3−n ∈ / D), and, if n is large enough, belongs to int J(T ). The contradiction obtained shows that neither (iii), nor (iv) is fulfilled. On the other hand, we have clearly Q(T ) = RN . Consequently conditions (v) and (vi) are fulfilled.  We are going to show yet that neither of the conditions (vii) mi

∞ 

 Sn (T ) > 0,

n=1

(viii)

∞ 

Sn (T ) contains a set of the second category and with the Baire property,

n=1

does imply that T ∈ C. Let H ⊂ RN be a Hamel basis, and put T = { x ∈ RN | x = λh, λ ∈ Q ∩ [−1, 1], h ∈ H }. First we show that T ∈ / C. Let g : H → R be given by g(h) = 1, h ∈ H, and let f : RN → R be the additive extension of g (Theorem 5.2.2). By Corollary 5.2.2 f is discontinuous, and by Theorem 5.2.1 we have for x ∈ T , x = λh, |f (x)| = |f (λh)| = |λf (h)| = |λg(h)| = |λ|  1. Consequently f is bounded on T , and hence T ∈ / C. N Now take an arbitrary x ∈ R . x has a Hamel expansion (9.3.6), which may be written as x = λ11 h1 + · · · + λ1p1 h1 + · · · + λk1 hk + · · · + λkpk hk , where λij ∈ Q ∩ [−1, 1], λi1 + · · · + λipi = λi , i = 1, . . . , k. Consequently x ∈ Sm (T ), ∞ ∞   where m = p1 + · · · + pk . Hence RN ⊂ Sn , i.e., Sn = RN , which shows that n=1

n=1

conditions (vii) and (viii) are fulfilled. Also this example is due to R. Ger [97]. Theorem 9.3.4. None of the conditions occurring in Theorem 9.3.3 is necessary for the set T to belong to A. Proof. None of conditions (i)n , (ii)n , n ∈ N, (iii), (iv) is necessary, as results from Lemma 9.3.2. It remains to exhibit an example of a set T ∈ A which does not fulfil (v) nor (vi). Because the set we are going to construct is of a more general interest, we will employ a special letter (V0 ) to denote it. Let H ⊂ RN be a Hamel basis. By Theorem 1.7.1 H can be well ordered, and so we can arrange the elements of H into a transfinite sequence of the type γ (where γ is the smallest ordinal with γ¯ = c): H = {hα }α<γ .

236

Chapter 9. Boundedness and Continuity

Now every element x ∈ RN , except for x = 0, can be uniquely written in the form x = λ1 hα1 + · · · + λn hαn ,

(9.3.7)

where n ∈ N, λi ∈ Q, λi = 0, i = 1, . . . , n, α1 < · · · < αn . We define V0 as the set of those x ∈ RN which in expansion (9.3.7) have λn > 0 (the last non-zero coefficient is positive). Additionally we assume that 0 ∈ V0 . It is easily seen from this definition that V0 is a cone in the linear space (RN ; Q; +; ·); in particular, V0 + V0 ⊂ V0 . (9.3.8) Moreover, V0 is Q-convex, and so Q(V0 ) = V0 . If the set T = V0 fulfilled either of conditions (v), (vi), then we would have by (9.3.8) int V0 = ∅. Suppose that an x with expansion (9.3.7) belongs to int V0 . Then, for sufficiently small positive λ ∈ Q, we have x − λhαn +1 ∈ int V0 . On the other hand for such λ we have x − λhαn +1 ∈ / V0 , since the coefficient of hαn +1 is negative. This contradiction shows that T = V0 fulfils neither (v), nor (vi). In order to prove that V0 ∈ A consider an arbitrary additive function f : RN → R bounded above on V0 . Take an arbitrary h ∈ H. There exists an α < γ such that h = hα . The points λhα + hα+1 belong to V0 for every λ ∈ Q, and f (λhα + hα+1 ) = λf (hα ) + f (hα+1 ). When λ ranges over Q this expression should remain bounded above, which is possible only if f (hα ) = 0. Consequently f | H = 0, whence by the uniqueness of the extension f = 0, i.e., f is continuous. This shows that V0 ∈ A.  For the class C we have the following Theorem 9.3.5. Let T ⊂ RN . Each of the following conditions is sufficient for T to belong to C:   (ix) mi R(T ) > 0. (x) R(T ) contains a set of the second category and with the Baire property.   (xi) measurablei U (T ) > 0. (xii) U (T ) contains a set of the second category and with the Baire property. Theorem 9.3.5 results from Theorems 9.3.1 and 9.3.2 and from Theorems 9.2.6 and 9.2.2. Again, none of the above conditions is necessary for a set T to belong to C, as may be seen from the example of the set T = Q · H considered in the proof of Lemma 9.3.2. Note that for this particular T we have −T = T so that R(T ) = S2 (T ) and U (T ) = S3 (T ). Now we pass to some necessary conditions for a set to belong to C. Theorem 9.3.6. Let T ⊂ RN . If T ∈ C, then T contains a Hamel basis. Proof. Suppose that T does not contain a Hamel basis. It follows from Corollary 4.2.2 that then E(T ) = RN . (9.3.9)

9.3. Simple conditions

237

Let B ⊂ T be a base of E(T ) (Corollary 4.2.2), and let H be a Hamel basis of RN such that B ⊂ H (Corollary 4.2.3). By (9.3.9) we have H \ B = ∅. Define a function g : H → Q putting  0 for x ∈ B, g(x) = 1 for x ∈ H \ B, and let f : RN → R be the additive extension of g (Theorem 5.2.2). By Corollary 5.2.2 f is discontinuous (f = 0, since f | H \ B = g | H \ B = 1), and, as is easy to see, f vanishes on E(B) = E(T ). Thus, in particular, f is bounded on T ⊂ E(T ), which shows that T ∈ / C.  From relation (9.1.1) and Theorem 9.3.6 we obtain Corollary 9.3.2. Let T ⊂ RN . If T ∈ A, then T contains a Hamel basis. The converse to Theorem 9.3.6 or to Corollary 9.3.2 is not true. Any Hamel basis yields an example of a set containing a Hamel basis, but belonging neither to A, nor to C (cf. 11.1). But we have the following (Kominek [173]) Theorem 9.3.7. Let T ⊂ RN . If T is analytic and contains a Hamel basis, then T ∈ C. Proof. For every λ1 , . . . , λn ∈ R the function g : RnN → RN g(x1 , . . . , xn ) = λ1 x1 + · · · + λn xn is continuous. By Theorem 2.5.3 the set T n = T × · · · × T (n-fold product) is analytic, and hence, by Theorem 2.5.2 the set g(T n ) = λ1 T + · · · + λn T is analytic, and hence, by Theorem 3.1.8, Lebesgue measurable. Let H ⊂ T be a Hamel basis. We have RN =

∞ 



(λ1 T + · · · + λn T ).

(9.3.10)

n=1 λ1 ,...,λn ∈Q

In fact, if x ∈ RN , then

x = α1 h1 + · · · + αk hk ,

where αi ∈ Q, hi ∈ H ⊂ T , i = 1, . . . , k. Hence x ∈ α1 T + · · · + αk T , which is one of the summands in (9.3.10). Consequently (9.3.10) holds. As pointed out above, the summands in (9.3.10) are measurable. If they all had measure zero, also RN would be of measure zero, since the union in (9.3.10) is countable. Consequently there exist λ1 , . . . , λn ∈ Q such that T0 = λ1 T + · · · + λn T has a positive measure: m(T0 ) = mi (T0 ) > 0.

238

Chapter 9. Boundedness and Continuity

Let f : RN → R be an arbitrary additive function bounded on T : |f (x)|  M

for x ∈ T.

Then we have for x ∈ T0 |f (x)|  |λ1 |M + · · · + |λn |M  (max |λi |)M. i

Thus f is bounded on T0 . By the Theorem of Ostrowski (Theorem 9.3.1) and by (9.1.1) T0 ∈ C, which implies that f is continuous. Consequently T ∈ C.  From Theorem 9.3.6 and Corollary 9.3.1 we obtain yet Corollary 9.3.3. The Cantor set C contains a Hamel basis for R. Next we prove

    Lemma 9.3.3. Let T ⊂ RN . If m J(T ) = 0, then m E + (T ) = 0. Proof. We will distinguish two cases. I. 0 ∈ J(T ). Fix arbitrarily λ1 , . . . , λn ∈ Q ∩ [0, ∞). We may write λi = ai /b,

i = 1, . . . , n,

(9.3.11)

where ai ∈ N ∪ {0}, b ∈ N. (We may assume that all λi ’s have been written with a common denominator). There exists a p ∈ N such that a1 + · · · + an < 2p . Take an arbitrary x such that x ∈ λ1 J(T ) + · · · + λn J(T ). (9.3.12) Then x = λ1 t1 + · · · + λn tn =

1 (a1 t1 + · · · + an tn ), b

ti ∈ J(T ),

i = 1, . . . , n,

or x=

1 2p s1 + · · · + s2p [(t1 + · · · + t1 ) + · · · + (tn + · · · + tn ) + 0 + · · · + 0] = , b b 2p

where s i = t1

for i = 1, . . . , a1 ,

....................................... s i = tn si = 0

for i = for i =

n−1

k=1 n

ak + 1, . . . ,

n

ak ,

k=1

ak + 1, . . . , 2p .

k=1

Since J(T ) is J-convex, we have by Lemma 5.1.3 2−p (s1 + · · · + s2p ) ∈ J(T ), and hence 2p λ1 J(T ) + · · · + λn J(T ) ⊂ J(T ). b   Since m J(T ) = 0, we get hence m[λ1 J(T ) + · · · + λn J(T )] = 0.

(9.3.13)

9.3. Simple conditions

239

Relation (9.3.13) is valid for arbitrary λ1 , . . . , λn ∈ Q ∩ [0, ∞) (n ∈ N). We have ∞    E + J(T ) =



[λ1 J(T ) + · · · + λn J(T )],

(9.3.14)

n=1 λ1 ,...,λn ∈Q∩[0,∞)

     whence m E + J(T) =0. The union in (9.3.14) is countable. Since T ⊂ J(T ), we   have E + (T ) ⊂ E + J(T ) , whence m E + (T ) = 0. II. 0 ∈ / J(T ). Take an a ∈ J(T ) and put T0 = J(T )− a. The set T0 is J-convex: if 1 1 x, y ∈ T0 , then x = x −a, y = y  −a with x , y  ∈ J(T ), whence (x+y) = (x +y  )−  2  2  a ∈ J(T ) − a = T0 . Hence J(T0 ) = T0 . By Corollary 3.2.2 m J(T ) = m J(T ) = 0. 0  +  Clearly 0 ∈ T0 = J(T0 ). By Part I of the proof m E (T0 ) = 0. We will show that     +  E (T0 ) + λa . (9.3.15) E + J(T ) ⊂ Let x ∈ E

+

λ∈Q∩[0,∞)



 J(T ) . Then x may be written as x = λ1 t1 + · · · + λn tn + λ0 a,

where n ∈ N, λ0 , . . . , λn ∈ Q ∩ [0, ∞), t1 , . . . , tn ∈ J(T ). Hence x = λ1 (t1 − a) + · · · + λn (tn − a) + (λ0 + λ1 + · · · + λn )a. Now, λi ∈ Q ∩ [0, ∞), i = 1, . . . , n; λ = λ0 + λ1 + · · · + λn ∈ Q ∩ [0, ∞); ti − a ∈ J(T ) − a = T0 , i = 1, . . . , n. Hence   +  x ∈ E + (T0 ) + λa ⊂ E (T0 ) + λa , λ∈Q∩[0,∞)

which proves (9.3.15).      Relation (9.3.15) yields m E + J(T ) = 0, whence also m E + (T ) = 0.     Corollary 9.3.4. Let T ⊂ RN . If m J(T ) = 0, then also m Q(T ) = 0.



This follows from the inclusion J(T ) ⊂ Q(T ) ⊂ E + (T ) and from Lemma 9.3.3.     Lemma 9.3.4. Let T ⊂ RN . If m Sn (T ) = 0 for all n ∈ N, then also m J(T ) = 0. Proof. Take an x ∈ J(T ). Then, by Theorem 5.1.3 x = λ1 t1 + · · · + λn tn , n  where n ∈ N, ti ∈ T , λi ∈ D, i = 1, . . . , n, λi = 1. λi ’s may be written in form i=1

(9.3.11), where b = 2 for a certain p ∈ N ∪ {0}, and ai ∈ N, i = 1, . . . , n, p

n 

ai = b.

i=1

By a similar argument as in the proof of Lemma 9.3.3 we get x ∈ 2−p S2p (T ), whence J(T ) ⊂

∞ 

2−p S2p (T ),

p=0

and the lemma follows.



240

Chapter 9. Boundedness and Continuity

     Lemma 9.3.5. Let T ⊂ RN . If m R J(T ) = 0, then also m E(T ) = 0. Proof. Again we distinguish two cases. I. 0 ∈ J(T ). Fix arbitrarily λ1 , . . . , λn ∈ Q. They may be written in form (9.3.11), with ai ∈ Z, i = 1, . . . , n ; b ∈ N. Moreover, they may be numbered in such a way that ai  0 for i = 1, . . . , q ; ai < 0 for i = q + 1, . . . , n. There exists a p ∈ N such that q n  

2p > max ai , |ai | . i=1

i=q+1

Take an arbitrary x fulfilling (9.3.12). Then x = y1 − y2 , where y1 =

1 (t1 + · · · + t1 ) + · · · + (tq + · · · + tq ) + 0 + · · · + 0 , b

1 (tq+1 + · · · + tq+1 ) + · · · + (tn + · · · + tn ) + 0 + · · · + 0 , b 2p 2p J(T )− whence (similarly as in the proof of Lemma 9.3.3) y1 , y2 ∈ J(T ) and x ∈ b b p

2   J(T ) = R J(T ) . Hence it follows that b y2 =

λ1 J(T ) + · · · + λn J(T ) ⊂ and since

∞    E J(T ) =



 2p  R J(T ) , b

[λ1 J(T ) + · · · + λn J(T )],

n=1 λ1 ,...,λn ∈Q

  we obtain hence m E(T ) = 0. II. 0 ∈ / J(T ). Take an a ∈ J(T ) and put T0 = J(T ) − a. Then J(T0 ) = T0 , and     R J(T ) = J(T ) − J(T ) = [J(T ) − a] − [J(T ) − a] = R[J(T ) − a] = R J(T0 ) .   By the first part of the proof m E(T0 ) = 0. As in the proof of Lemma 9.3.3 we can show that     E J(T ) ⊂ E(T0 ) + λa . λ∈Q

     Hence m E J(T ) = 0, whence also m E(T ) = 0.     Theorem 9.3.8. Let T ⊂ RN . If T ∈ C, then me R J(T ) > 0.      Proof. Suppose the contrary. Then m R J(T ) = 0, and by Lemma 9.3.5 m E(T ) = 0. Hence E(T ) = RN , and T cannot contain a Hamel basis. By Theorem 9.3.6 T ∈ / C.  A full characterization of the classes A and C (necessary and sufficient conditions for a set T ⊂ RN to belong to A resp. C) will be given in the next chapter.

9.4. Measurability of convex functions

241

9.4 Measurability of convex functions We start with the following result. Theorem 9.4.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. Further, let T ⊂ D be a Lebesgue measurable set of positive measure, and let g : T → R be a measurable function. If f (x)  g(x)

for

x ∈ T,

(9.4.1)

then the function f is continuous. Proof. Put Tk = {x ∈ T | k − 1  g(x) < k},

k ∈ Z.

(9.4.2)

Since the function g is measurable, all sets Tk are measurable. Moreover, the sets Tk ∞  are pairwise disjoint and Tk = T . Hence k=−∞ ∞

m(Tk ) = m(T ) > 0.

k=−∞

Thus there exists a k0 ∈ Z such that m(Tk0 ) > 0.

(9.4.3)

By (9.4.1) and (9.4.2) f (x)  g(x) < k0

for x ∈ Tk0 ,   and thus f is bounded above on a set of positive measure cf. (9.4.3) . By the Theorem of Ostrowski (Theorem 9.3.1) f is continuous.  Taking in Theorem 9.4.1 T = D and g = f , we obtain the following (Sierpi´ nski [277], Blumberg [33], Bonnesen-Fenchel [35]). Theorem 9.4.2. [Theorem of Sierpi´ nski]. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. If f is measurable, then it is continuous. Since every additive function is convex, we obtain as a particular case of Theorem 9.4.2 the following result, which we will formulate as a separate theorem. Theorem 9.4.3. [Theorem of Fr´echet]. Every measurable additive function f : RN → R is continuous. In the case N = 1, Theorem 9.4.3 was first proved by M. Fr´echet [90], [91]. Various other proofs were then supplied by W. Sierpi´ nski [279], [280], S. Banach [19], M. Kac [160], A. Alexiewicz-W. Orlicz [13] and T. Figiel [80]. Concerning other measures than the Lebesgue measure cf. Kuczma-Sm´ıtal [190], Paganoni [251]; cf. also Fischer-Slodkowski [85].

242

Chapter 9. Boundedness and Continuity

Theorem 9.4.2 explains why all examples of discontinuous convex functions are non-effective. In fact, all such functions are necessarily non-measurable, and nonmeasurable functions exist only under the assumption of the Axiom of Choice. Such functions are highly pathological (cf. also Chapter 12), and have no practical importance. Therefore, when dealing with convex functions, it is of importance to know possibly weak conditions which would guarantee that the functions dealt with are continuous. The theorems in the present chapter furnish such conditions.

9.5 Plane curves We have seen that in R there exist sets of measure zero and of the first category which nevertheless belong to the class C1 . (The Cantor set C is such a set; cf. Corollary 9.3.1). Do there exist sets with similar properties in spaces of higher dimensions? In the present section we answer this question in the affirmative for N = 2. The case of an arbitrary N > 1 will be dealt with in 9.6. The results in the present section are due to M. Kuczma [184]. Suppose that N > 1, let D ⊂ RN −1 be an arbitrary set, and let ϕ : D → R be an arbitrary function. By Gr(ϕ) we denote the graph of ϕ, i.e., the set Gr(ϕ) = { (x, y) ∈ RN | x ∈ D, y = ϕ(x) }. If the function ϕ is continuous, then the set Gr(ϕ) ⊂ RN is of measure zero and nowhere dense. Theorem 9.5.1. Let D ⊂ RN −1 be an arbitrary set, and let ϕ : D → R be given as ϕ(x) = c + f (x)

for x ∈ D,

(9.5.1)

where f : RN −1 → R is an additive function, and c ∈ R is a constant. Then Gr(ϕ) ∈ / CN . Proof. Assume first that D = RN −1 . Suppose that Gr(ϕ) ∈ CN . Let c = (0, . . . , 0, c) ∈ RN . We have Gr(f ) = Gr(ϕ) − c. By Theorems 9.2.8 and 9.2.2 we get hence Gr(f ) ∈ CN . Take arbitrary p1 , . . . , pn ∈ Gr(f ), and arbitrary λ1 , . . . , λn ∈ Q. We may write pi = (xi , yi ), where xi ∈ RN −1 , yi = f (xi ), i = 1, . . . , n. Hence by Theorem 5.2.1 λ1 y1 + · · · + λn yn = λ1 f (x1 ) + · · · + λn f (xn ) = f (λ1 x1 + · · · + λn xn ). (9.5.2)   Relation (9.5.2) means that the point λ1 p1 + · · · + λn pn ∈ Gr(f ). Hence1 E Gr(f ) ⊂ Gr(f ) = RN . By Theorem 9.3.6 Gr(f ) ∈ / CN . This contradiction shows that also Gr(ϕ) ∈ / CN . Now, if D = RN −1 , then take the extension ϕ0 of ϕ onto RN −1 : ϕ0 (x) = c + f (x)

for x ∈ RN −1 .

We have Gr(ϕ) ⊂ Gr(ϕ0 ), and, as we have already shown, Gr(ϕ0 ) ∈ / CN . By Corollary 9.2.1 also Gr(ϕ) ∈ / CN .  1

Actually we have the equality, since the converse inclusion is trivial.

9.5. Plane curves

243

Now we restrict ourselves to the case N = 2, so x is a single real variable. If D ⊂ R is an interval, and a function ϕ of form (9.5.1) is continuous, then also f is continuous on D, and hence bounded on every compact subinterval of D. By Lemma 9.3.1 f is continuous in R, and hence, by Theorem 5.5.2, of the form f (x) = ax with a certain a ∈ R. Thus (9.5.1) reduces to ϕ(x) = c + ax

for x ∈ D.

(9.5.3)

Theorem 9.5.2. Let D ⊂ R be an interval, and let ϕ : D → R be a continuous function which is not of form (9.5.3) for any constants a, c ∈ R. Then Gr(ϕ) ∈ A2 . Proof. There exists an interval [α, β] ⊂ D such that ϕ is not of form (9.5.3) for all x ∈ [α, 3 = ϕ(x + α) − ϕ(α) for x ∈ [0, β − α]. We have Gr(ϕ) =  β]. Put  ϕ(x) Gr(ϕ) 3 − α, ϕ(α) , and thus, by Theorems 9.2.8 and 9.2.2, it is enough to prove that Gr(ϕ) 3 ∈ A2 . Note that also the function ϕ 3 is not of form (9.5.3) for x ∈ I = [0, β − α]. Put φ(x, t) = ϕ(t) 3 + ϕ(x 3 − t) for x ∈ I, t ∈ [0, x]. Suppose that for every x ∈ I and for all t, s ∈ [0, x] we have φ(x, t) = φ(x, s), i.e., ϕ(t) 3 + ϕ(x 3 − t) = ϕ(s) 3 + ϕ(x 3 − s). Setting, in particular, t = x, we obtain  ϕ(x) 3 + ϕ(0) 3 = ϕ(s) 3 + ϕ(x 3 − s), or, with u = s, v = x − s, since ϕ(0) 3 =0 , ϕ(u 3 + v) = ϕ(u) 3 + ϕ(v). 3

(9.5.4)

1 I. In 13.5 we will show that then 2 ϕ 3 can be extended onto R to an additive function, and thus, being continuous, is of 1 1 1 the form ϕ(x) 3 = ax for x ∈ I. Now take an arbitrary x ∈ I. Then x ∈ I, whence 2 2 2   1 1 1 ϕ 3 x = ax. Setting in (9.5.4) u = v = x we get 2 2 2     1 1 1 1 x +ϕ 3 x = ax + ax = ax. ϕ(x) 3 =ϕ 3 2 2 2 2

Relation (9.5.4) holds, in particular, for all u, v ∈

Thus ϕ 3 has form (9.5.3) (with c = 0) for all x ∈ I, contrary to the supposition. Consequently we may assume that there exists an x0 ∈ I such that M=

sup φ(x0 , t) > t∈[0, x0 ]

inf

t∈[0, x0 ]

φ(x0 , t) = m,

and due to the continuity of ϕ 3 we may even assume that x0 ∈ int I. Since φ(x0 , t) is a continuous function of t, there exist in the interval [0, x0 ] points t0 and T0 such that φ(x0 , t0 ) = m, φ(x0 , T0 ) = M. Put d = M − m. Again by the continuity of φ (now with respect to x) we can find an x1 ∈ (x0 , β − α) such that 1 φ(x, t0 ) < m + d, 3

1 φ(x, T0 ) > M − d 3

for x ∈ [x0 , x1 ].

(9.5.5)

244

Chapter 9. Boundedness and Continuity

  1 1 Now take a p = (x, y) ∈ R = (x0 , x1 ) × m + d, M − d = ∅. It follows from 3 3 (9.5.5) and from the Darboux property (intermediate value property) of φ(x, t) (as  a function of t) that there exists a t between t0 and T0 and so in [0, x0 ] such that φ(x, t) = y.

(9.5.6)

Put s = x − t. Then, by (9.5.6) and by the definition of φ, x = t + s,

y = ϕ(t) 3 + ϕ(s). 3

(9.5.7)

Relation (9.5.7) shows that p ∈ Gr(ϕ) 3 + Gr(ϕ). 3 Consequently R ⊂ Gr(ϕ) 3 + Gr(ϕ) 3 so that int[Gr(ϕ) 3 + Gr(ϕ)] 3 = ∅. By Theorem 9.3.3 Gr(ϕ) 3 ∈ A2 , whence also Gr(ϕ) ∈ A2 .  Theorem 9.5.2 has been extended by R. Ger [102], [103] to higher dimensions, but he had to make stronger assumptions concerning the regularity of ϕ. In the next section we will present one of these generalizations (Ger [103]). At any case, this result implies, in particular, that in the space RN (N > 1) there exist sets belonging to AN which are nowhere dense and of the N -dimensional Lebesgue measure zero (cf. also Exercise 10.6).

9.6 Skew curves Also in this section we assume that N > 1. Let D ⊂ R be an interval, and let ϕ : D → RN be a C 1 function such that2 ϕ(0) = 0. The function ϕ can be written as ϕ = (ϕ1 , . . . , ϕN ), and the functions ϕi : D → R, i = 1, . . . , N , are of class C 1 in D. The function ϕ describes a curve in RN . We assume that this curve does not lie entirely on an (N − 1)-dimensional hyperplane. Analytically this means that there do not exist constants c1 , . . . , cN ∈ R such that |c1 | + · · · + |cN | > 0 and c1 ϕ1 + · · · + cN ϕN = 0

in D.

(9.6.1)

In other words, the functions ϕ1 , . . . , ϕN are linearly independent over R. Then also the functions ϕ1 , . . . , ϕN are linearly independent (over R), for if we had c1 ϕ1 + · · · + cN ϕN = 0

in D

(9.6.2)

with certain c1 , . . . , cn ∈ R, then, integrating relation (9.6.2), we would get (9.6.1). We may form the determinant d : DN → R: #  # # ϕ1 (x1 ), . . . , ϕ1 (xN ) # # # # # .. .. (9.6.3) d(x1 , . . . , xN ) = # # . . # # # ϕ (x1 ), . . . , ϕ (xN ) # N N 2

This assumption is not essential, since it can always be realized by a suitable shift.

9.6. Skew curves

245

and the linear independence of ϕ1 , . . . , ϕN implies that d(3 x1 , . . . , x 3N ) = 0 for some x 31 , . . . , x 3N ∈ D. Write p3 = (3 x1 , . . . , x 3N ), then d(3 p) = 0. The continuity of the derivatives of ϕ implies that there exists a neighbourhood U of p3 such that d(p) = 0

for p ∈ U

  p = (x1 , . . . , xN ) . Define the function φ : U → RN by φ(x1 , . . . , xN ) =

N 

ϕ1 (xi ), . . . ,

i=1

N

 ϕN (xi ) .

i=1

The derivative of φ is an N × N square matrix with the entries

∂φi = ϕi (xj ), ∂xj

i, j = 1, . . . , N . Hence det φ (x1 , . . . , xN ) = d(x1 , . . . , xN ), whence det φ (p) = 0 in U . Consequently there exist intervals (αi , βi ) ⊂ D, i = 1, . . . , N , such that x 3i ∈ (αi , βi ), i = 1, . . . , N , and φ is a diffeomorphism of V = N

× (αi , βi ) onto a suitable domain in RN , and V ⊂ U .

i=1

For any A ⊂ D put HA = { p ∈ RN | p = ϕ(x), x ∈ A }.

Theorem 9.6.1. With the above notation, assume that Z ⊂ D is a set of positive one-dimensional Lebesgue measure and that x 3i , i = 1, . . . , N , are density points of Z. Then HZ ∈ AN . Proof. Put Zi = Z ∩ (αi , βi ),

i = 1, . . . , N.

Since x 3i , i = 1, . . . , N , are density points of Z, the one dimensional Lebesgue measure   N of Zi is positive for i = 1, . . . , N , whence also m × Zi > 0 (m stands here for the i=1

N -dimensional Lebesgue measure). Further, we have  m(HZ1 + · · · + HZN ) = | det φ (p)| dp > 0, N

× Zi

i=1

N

since det φ = 0 in × Zi ⊂ V ⊂ U . On the other hand, i=1

HZ1 + · · · + HZN ⊂ HZ + · · · + HZ = SN (HZ )   so that m SN (HZ ) > 0. By Theorem 9.3.3 HZ ∈ AN .



246

Chapter 9. Boundedness and Continuity

The condition that x 3i , i = 1, . . . , N , are density points of Z can be replaced by the condition, which is less general but easier to verify that the one-dimensional measure of Z ∩(α, β) is positive for every interval (α, β) ⊂ D. Another possible variant is that d(x1 , . . . , xN ) = 0 for almost every (x1 , . . . , xN ) ∈ DN such that xi = xj for i = j. Then x 31 , . . . , x 3N can always be chosen so that d(3 x1 , . . . , x 3N ) = 0 and x 31 , . . . , x 3N are density points of Z. Taking these remarks into account, we may formulate the following corollary to Theorem 9.6.1. Theorem 9.6.2. Let D ⊂ R be an interval, and let ϕ : D → RN be a function of class C 1 defining in RN a curve which does not lie entirely on an (N − 1)-dimensional hyperplane. Let Z ⊂ D be a measurable set of positive one-dimensional measure. If either of the conditions is fulfilled: (i) The one-dimensional measure of Z ∩ (α, β) is positive for every interval (α, β) ⊂ D; (ii) The determinant (9.6.3) is non-zero for almost every (x1 , . . . , xN ) ∈ DN such that xi = xj for i = j; then HZ ∈ AN . The set HZ is very small. It may even be much smaller than the curve y = ϕ(x), which is of N -dimensional measure zero and nowhere dense. Thus in every space RN there exist very small sets belonging to AN . As an example consider the curve ϕ1 (x) = x, . . . , ϕi (x) = xi , . . . , ϕN (x) = xN ,   x ∈ D, an interval. We have cf. (9.6.3) for x1 , . . . , xN ∈ D such that xi =  xj for i = j # # # # # 1 # # ,..., 1 ## 1 ,..., 1 # # # # x1 # 2x1 ,..., 2xN ## ,..., xN ## # # d(x1 , . . . , xN ) = # # = N! # # = 0, .. .. .. .. # # # # . . . . # # # # # xN −1 , . . . , xN −1 # # N xN −1 , . . . , N xN −1 # 1 1 N N since the last determinant is Van der Monde’s determinant. The condition (ii) of Theorem 9.6.2 is fulfilled. For any set Z ⊂ D of positive one-dimensional measure we have HZ ∈ AN .

9.7 Boundedness below In the definition of the class B the boundedness above may be replaced by the boundedness below. In fact, if f : RN → R is an additive function bounded below on a set T ∈ B, then −f is an additive function bounded above on T . Thus −f , and hence also f , is continuous. But in the definition of the class A the boundedness above cannot be replaced by the boundedness below. There exist discontinuous convex functions even globally bounded below (cf. the end of 6.4). But Theorem 6.2.2 suggests that one might look

9.7. Boundedness below

247

for conditions that would imply that a convex function is locally bounded below at every point of its domain of definition. In this connection J. Sm´ıtal [289] introduced the classes D(D) which may be defined as follows. Let D ⊂ RN be a convex and open set. D(D) is the class of all sets T ⊂ D with the property that every convex function f : D → R which is bounded below on T is locally bounded below at every point of D: D(D) = { T ⊂ D |

every convex function f : D → R bounded below on T is locally bounded below at every point of D}.

Later, in 10.6, we will show that the classes D(D) actually depend on D. By Theorem 6.2.2 every open set G ⊂ D belongs to D(D). Now we will prove the following Lemma 9.7.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. Let T ⊂ D be a set fulfilling one of the following conditions: (i) T ∈ L and m(T ) > 0; (ii) T is of the second category and with the Baire property. Then3 mf (x) < K  f (x) for x ∈ T

(9.7.1)

cannot hold for any real constant K. Proof. Suppose that (9.7.1) holds with a certain constant K ∈ R. If (i) holds, then take s to be an arbitrary density point of T . If (ii) holds, then T = (G ∪ P ) \ R,

(9.7.2)

where G = ∅ is open, and the sets P, R are of the first category. Note that then for every set U ⊂ G the set U \ T is of the first category. Let s be an arbitrary point of the set G \ R ⊂ T . Fix an ε > 0, and, for any x ∈ RN and r > 0, let K(x, r) denote the open ball centered at x and with the radius r. Put S = K(s, 2ε). If condition (i) is fulfilled, then we may choose ε so small that for every open ball U ⊂ S centered at s we have m(U ∩ T ) >

1 m(U ), 2

(9.7.3)

and, moreover, then we assume that ε is so small  S ⊂ D. If condition (ii) is fulfilled,  that S ⊂ G whence also S ⊂ D; cf. (9.7.2) so that for every set U ⊂ S the set U \ T is of the first category. 3

Concerning the definition of mf cf. 6.3.

248

Chapter 9. Boundedness and Continuity

Put

    2ε 1 1 Sn = n S + 1 − n s = K s, n ⊂ S, 2 2 2

n ∈ N,

(9.7.4)

and T n = T ∩ Sn ,

n ∈ N.

(9.7.5)

There exists an L ∈ R such that mf (s) < L < K, and, by the definition of mf , for every n ∈ N there exists a point xn ∈ D such that |xn − s| <

ε − 1)

2(2n

and

f (xn ) < L.

(9.7.6)

Further put for n ∈ N   1 1 n Un = 2 Sn+1 + (1 − 2 )xn = S + 2 1 − n+1 s + (1 − 2n )xn 2 2 1 1 1 1 = S + 2n (s − xn ) − s + xn = S + s + (2n − 1)(s − xn ) 2 2 2 2 = K(s, ε) + (2n − 1)(s − xn ) ⊂ S, n

n

since by (9.7.6) |(2n − 1)(s − xn )| <

ε , and 2

Bn = 2n Tn+1 + (1 − 2n )xn ⊂ Un ⊂ S. We have by (9.7.5) Un \Bn = 2n (Sn+1 \Tn+1 )+(1−2n )xn = 2n (Sn+1 \T )+(1−2n)xn . If, in particular, condition (ii) is fulfilled, then Sn+1 \ T is of the first category by (9.7.4), and hence also Un \ Bn is of the first category (Un \ Bn = g(Sn+1 \ T ), where g(x) = 2n x+(1−2n)xn is a homeomorphism) for every n ∈ N. Note also that since Un are open balls with a fixed radius ε, whose centers lie not further apart from   sε than ε , there is an open set U0 ⊂ Un for all n ∈ N. We may take, e.g., U0 = K s, . 2 2 Put also ∞ ∞   Qn = Bi , Vn = Ui , (9.7.7) i=n

Q=

∞ 

i=n

Qn ,

V =

n=1

∞ 

Vn

(9.7.8)

n=1

so that we have Qn+1 ⊂ Qn ,

Vn+1 ⊂ Vn

for n ∈ N.

(9.7.9)

Hence it follows that U0 ⊂ Vn for all n ∈ N, and hence U0 ⊂ V . Since U0 is open (and non-empty), it is of the second category4, and also V is of the second category. 4

Cf. the proof of Theorem 2.9.1.

9.7. Boundedness below

249

Now, if (i) is fulfilled, then we argue as follows. All Un are open balls with a fixed radius ε, so there is a positive number c such that m(Un ) = c for all n ∈ N. Hence m(Vn )  c for all n ∈ N, and by (9.7.8) and (9.7.9) m(V ) = lim m(Vn )  c. n→∞

(9.7.10)

Moreover, by (9.7.3), m(Tn ) = m(Sn ∩ T ) >

1 m(Sn ), 2

(9.7.11)

and by (9.7.11) m(Bn ) = 2nN m(Tn+1 ) > 2nN −1 m(Sn+1 ) = whence also m(Qn ) >

1 c, 2

1 1 m(2n Sn+1 ) = c, 2 2

n ∈ N,

n ∈ N,

and by (9.7.8) and (9.7.9) m(Q) = lim m(Qn )  n→∞

1 c. 2

Hence Q = ∅.

(9.7.12)

If condition (ii) is fulfilled, then we argue as follows. For every n ∈ N the set Un \ Bn is of the first category, as pointed out above. Since Bn ⊂ Un , this means that Un = Bn ∪ (Un \ Bn ), and Vn = Qn ∪

∞ 

n ∈ N,

(Ui \ Bi ),

n ∈ N.

(9.7.13)

i=n

Thus Vn \ Qn ⊂

∞ 

(Ui \ Bi ) is a set of the first category for every n ∈ N. Similarly,

i=n

we obtain from (9.7.13) and (9.7.8) V ⊂Q∪

∞ 

(Ui \ Bi ).

(9.7.14)

i=1

Since V is of the second category, (9.7.14) implies that also Q is of the second category, whence (9.7.12) results. Thus (9.7.12) is valid in either case. Now take an arbitrary x ∈ Q ⊂ Qm , m ∈ N. There exists an index n  m such that x ∈ Bn . Therefore x = 2n t + (1 − 2n )xn

250

Chapter 9. Boundedness and Continuity

with a t ∈ T . Hence

  1 1 t = n x + 1 − n xn , 2 2

and f (t) 

  1 1 f (xn ). f (x) + 1 − 2n 2n

Hence it follows (note that 1 − 2n < 0) that f (x)  2n f (t) + (1 − 2n )f (xn )  2n K + (1 − 2n )L = L + 2n (K − L). Letting m → ∞ (whence also n → ∞) we obtain hence, since K − L > 0, f (x) = ∞, which contradicts the assumption f : D → R. The contradiction obtained shows that (9.7.1) is impossible.  Part (i) of the above Lemma, similarly as Theorem 9.7.1 below, is due to A. Cs´asz´ar [50]. Theorem 9.7.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. Let T ⊂ D be a measurable set such that m(T ) > 0. Then for no measurable function g : T → R we may have mf (x) < g(x)  f (x)

for x ∈ T.

(9.7.15)

Proof. Suppose that (9.7.15) holds for a certain measurable set T ⊂ D of positive measure, and for a certain measurable function g : T → R. Put for λ ∈ Q Tλ = { x ∈ T | mf (x) < λ  g(x) }. Tλ , λ ∈ Q, are measurable sets, and T =



Tλ .

(9.7.16)

λ∈Q

If all Tλ had measure zero, we would get from (9.7.16) m(T ) = 0. Consequently there exists a λ0 ∈ Q such that m(Tλ0 ) > 0. For x ∈ Tλ0 we have by (9.7.15) mf (x) < λ0  f (x), contrary to Lemma 9.7.1.



We also prove the following Theorem of Hukuhara5 . Theorem 9.7.2. Let D ⊂ RN be a convex and open set, and let T ⊂ D be a set fulfilling one of conditions (i), (ii) of Lemma 9.7.1. Then T ∈ D(D). 5

M. Hukuhara [148] proved part (i) of Theorem 9.7.2 in the case N = 1. This result has then been extended to higher dimensions by S. Marcus [218] and A. Cs´ asz´ ar [50].

9.8. Restrictions of convex functions and additive functions

251

Proof. Suppose the contrary. Let f : D → R be a convex function bounded below on T : f (x)  K for x ∈ T, (9.7.17) but locally unbounded below at a point of D. By Theorem 6.2.2 f is locally unbounded below at every point of D so that mf (x) = −∞

for x ∈ D.

(9.7.18)

Relations (9.7.17) and (9.7.18) imply (9.7.1), contrary to Lemma 9.7.1.



Of course, in Theorem 9.7.2 and Lemma 9.7.1 condition (i) may be replaced by the condition mi (T ) > 0 , (i’) since by the definition of the inner measure (3.1) condition (i’) implies that there exists a measurable set T0 ⊂ T fulfilling (i), and (9.7.17) resp. (9.7.1) holds for x ∈ T0 as well. A more thorough study of the classes D(D) is deferred to 10.6.

9.8 Restrictions of convex functions and additive functions Theorem 6.1.1 says that for every convex function f : D → R and for every a, b ∈ D the restriction f | Q(a, b) is continuous. We may ask what can be inferred from the fact that for a certain set T ⊂ D the restriction f | T is continuous. This gives rise to the following definitions (cf. Kominek-Kominek [171]). N ∈ N being fixed, let AC denote the class of sets T ⊂ RN such that if D is a convex and open set such that T ⊂ D, and f : D → R is a convex function such that the restriction f | T is continuous, then f is continuous in D: AC = { T ⊂ RN | every convex function f : D → R, where D is a convex and open and T ⊂ D ⊂ RN , such that the restriction f | T is continuous, is continuous}. Similarly, BC is the class of sets T ⊂ RN such that if f : RN → R is an additive function such that the restriction f | T is continuous, then f is continuous in RN : BC = { T ⊂ RN | every additive function f : RN → R such that the restriction f | T is continuous, is continuous}. Let us note that the definition of the class AC makes the latter independent of D. The classes AC and BC will be studied also in the next chapter. In particular, the relation AC = BC does not hold. But directly from the definition it follows, since every additive function is convex, that AC ⊂ BC . Before proving analogues of Theorems 9.3.1 and 9.3.2 we will prove some other criteria for a set T ⊂ RN to belong to AC or BC .

252

Chapter 9. Boundedness and Continuity

We start with a lemma which will be proved in Chapter 11 (cf. Theorem 11.8.1). The operation R has been defined in 9.2. For any operation F : P(RN ) → P(RN ) we define the iterates F n , n ∈ N, as follows   F 1 = F, F n+1 (A) = F F n (A) , n ∈ N. N Lemma  n 9.8.1.  Let H ⊂ R be a Hamel basis. Then there exists an n ∈ N such that me R (H) > 0.

Theorem 9.8.1 below is due to F. B. Jones [158]. Theorem 9.8.1. Let T ⊂ RN be an analytic set containing a Hamel basis. Then T ∈ BC . Proof. Let f : RN → R be an additive function such that f | T is continuous. Put     Tk = { x ∈ T | |f (x)| < k } = T ∩ f −1 (−k, k) = (f | T )−1 (−k, k) , k ∈ N.   Since f | T is continuous, (f | T )−1 (−k, k) is open (in T ). Hence there exists an open set Gk ⊂ RN such that Tk = T ∩ Gk . It follows that Tk is analytic (Theorems 2.5.5 and 2.5.6) for every k ∈ N. ∞  We have Tk ⊂ Tk+1 , k ∈ N, and T = Tk . We want to show that k=1

R(T ) =

∞ 

R(Tk ).

(9.8.1)

k=1

If x ∈ R(T ), then x = u − v, where u, v ∈ T . There exists a k ∈ N such that u, v ∈ Tk , ∞ ∞   whence x ∈ R(Tk ) ⊂ R(Tk ). Thus R(T ) ⊂ R(Tk ). The converse inclusion is k=1

k=1

obvious. Obviously, we have also for arbitrary A, B ⊂ RN A ⊂ B ⇒ R(A) ⊂ R(B).

(9.8.2)

Repeating the above argument and making use of (9.8.1), we prove by induction that Rn (T ) =

∞ 

Rn (Tk ),

n ∈ N.

(9.8.3)

k=1

If A ⊂ RN is an analytic set, then by Theorem 2.5.3 A × A also is analytic, and since the function g : A × A → RN defined as g(x, y) = x − y is continuous, also the set R(A) = g(A × A) is analytic (Theorem 2.5.2). Hence it follows that all the sets Rn (Tk ), n, k ∈ N, are analytic.  Let H ⊂ T be a Hamel basis. By Lemma 9.8.1 there exists  an n∈ N such that me Rn (H) > 0. By (9.8.2) Rn (H) ⊂ Rn (T ), whence also me Rn (T ) > 0. Relation (9.8.3) implies (cf. Exercise 3.3) that     me Rn (T ) = lim me Rn (Tk ) , k→∞

9.8. Restrictions of convex functions and additive functions

253

  whence it follows that there exists a k ∈ N such that me Rn (Tk ) > 0. But Rn (Tk ) is an analytic set, and therefore measurable (Theorem 3.1.8). Thus   m Rn (Tk ) > 0. (9.8.4) We have |f (x)| < k for x ∈ Tk . Now (9.8.4) implies in virtue of Theorem 9.3.1 that Rn (Tk ) ∈ C and by Theorems 9.2.3, 9.2.6 and 9.2.2 Tk ∈ C. f is bounded on Tk , and hence it is continuous.  Before stating the next theorem we need some more definitions. If a, b ∈ RN , a = (a1 , . . . , aN ), b = (b1 , . . . , bN ), then we say that a < b resp. a  b iff ai < bi resp. ai  bi for i = 1, . . . , N . If a < b, then the open interval (a, b) and the closed interval [a, b] are defined as in the one-dimensional case: (a, b) = { x ∈ RN | a < x < b },

[a, b] = { x ∈ RN | a  x  b }.

If N = 2, then the interval (open or closed) is a rectangle (open or closed), its lower left and upper right vertices being the endpoints of the interval (the sides of the rectangle are parallel to the coordinate axes). The following theorem is due to Z. Kominek [173]. Theorem 9.8.2. Let T ⊂ RN be a set such that for certain a, b ∈ RN , a < b, and for a certain n ∈ N, either {a} ∪ (a, b) ⊂

1 Sn (T ) ⊂ [a, b], n

(9.8.5)

(a, b) ∪ {b} ⊂

1 Sn (T ) ⊂ [a, b]. n

(9.8.6)

or Then T ∈ AC .

Proof. We assume that relation (9.8.5) holds. If (9.8.6) is fulfilled, the proof is analogous. 1 Since T ⊂ Sn (T ), we have T ⊂ [a, b]. Relation (9.8.5) implies that a ∈ n 1 Sn (T ), i.e., n a=

1 (t1 + · · · + tn ), n

ti ∈ T,

i = 1, . . . , n.

(9.8.7)

Thus ti ∈ [a, b], in particular, ti  a, i = 1, . . . , n. If, for some i, we had ti = a, we n 1  would get from (9.8.7) ti = a. Consequently all ti = a, which shows that a ∈ T . n i=1 Now let D ⊂ RN be a convex and open set such that T ⊂ D, and let f : D → R be a convex function such that f | T is continuous. Let, for any ε > 0, K(a, ε) be the open ball centered at a and with the radius ε. We will show that there exists an ε > 0 such that f is bounded above on (a, b) ∩ K(a, ε). Hence it will follow by Lemma 9.3.1 that f is continuous in D, i.e., T ∈ AC .

254

Chapter 9. Boundedness and Continuity

  1 Supposing the contrary, for every p ∈ N we can find a point xp ∈ (a, b)∩K a, p such that f (xp ) > p. It follows that lim xp = a,

p→∞

By (9.8.5) xp ∈ xp =

lim f (xp ) = ∞.

(9.8.8)

p→∞

1 Sn (T ), p ∈ N, i.e., n

1 (tp1 + · · · + tpn ), n

tpi ∈ T,

i = 1, . . . , n;

p ∈ N.

(9.8.9)

We have tpi ∈ [a, b], p ∈ N, i = 1, . . . , n. It is easily seen that (9.8.8) implies lim tpi = a

p→∞

for i = 1, . . . , n.

(9.8.10)

Since f | T is continuous, a ∈ T , and tpi ∈ T for p ∈ N, i = 1, . . . , n, (9.8.10) implies that lim f (tpi ) = f (a), i = 1, . . . , n. (9.8.11) p→∞

By Lemma 5.3.1 and by (9.8.9) f (xp ) 

1 [f (tp1 ) + · · · + f (tpn )], n

p ∈ N,

i = 1, . . . , n.

(9.8.12)

Letting p → ∞, we obtain from (9.8.8), (9.8.11) and (9.8.12) ∞  f (a), which is a contradiction and completes the proof.



1 Since for the Cantor set C we have S2 (C) = [0, 1] (cf. Exercise 2.13), we obtain 2 from Theorem 9.8.2 Corollary 9.8.1. C ∈ AC . The following two theorems are due to B. Kominek-Z. Kominek [171] and Z. Kominek [173]. Theorem 9.8.3. Let T ⊂ RN be a set such that mi (T ) > 0. Then T ∈ AC . Proof. The condition mi (T ) > 0 implies that there exists a compact set C ⊂ T such that m(C) > 0. Let D ⊂ RN be a convex and open set such that T ⊂ D, and let f : D → R be a convex function such that the restriction f | T is continuous. Hence also the restriction f | C = f | T | C is continuous, and hence bounded. Thus f is bounded above on C, and by Theorem 9.3.1 f is continuous.  Theorem 9.8.4. Let T ⊂ RN be a set of the second category and with the Baire property. Then T ∈ AC .

9.8. Restrictions of convex functions and additive functions

255

Proof. The conditions on T imply that T = (G ∪ P ) \ R, where G is a non-empty open set, and P, R are of the first category. We may find a, b ∈ G \ R such that a < b and [a, b] ⊂ G. Put T0 = [a, b] \ R ⊂ [a, b]. Then also

T0 + T0 ⊂ [a, b]. 2 Since a ∈ G \ R, we have a ∈ T0 , and hence a∈

T0 + T0 . 2

(9.8.13)

(9.8.14)

N x+y ∈ Now take an x ∈ (a,b). There exists a y ∈ R such

that

(a, b) and x−y ∈ (a, b) since (a, b) is open , and hence y ∈ (a, b) − x ∩ x − (a, b) . Consequently the last set is non-empty, and being the intersection of two open sets, it is itself open. It follows that it is of the second category (cf. the proof of Theorem 2.9.1). Since the set (R − x) ∪ (x − R) is of the first category, we have !



" ! " (a, b) − x ∩ x − (a, b) \ (R − x) ∪ (x − R) = ∅. (9.8.15)

Further,

!



" ! " (a, b) − x ∩ x − (a, b) \ (R − x) ∪ (x − R) !

"

" ! = (a, b) − x \ (R − x) ∩ x − (a, b) \ (x − R) !

" !

" = (a, b) \ R − x ∩ x − (a, b) \ R . (9.8.16)

Write shortly W = (a, b) \ R. (9.8.15) and (9.8.16) yield (W − x) ∩ (x − W ) = ∅. Hence also the translation



W ∩ (2x − W ) = (W − x) ∩ (x − W ) + x = ∅.

Thus there exists a u ∈ W ∩ (2x − W ). In other words, u ∈ W and u = 2x − v, where 1 1 1 v ∈ W . Hence x = (u + v), where u, v ∈ W , i.e., x ∈ (W + W ) ⊂ (T0 + T0 ). 2 2 2 Since x has been arbitrary in (a, b), we obtain hence (a, b) ⊂

T 0 + T0 . 2

(9.8.17)

Relations (9.8.13), (9.8.14) and (9.8.17) imply in virtue of Theorem 9.8.2 that T0 ∈ AC . Since T0 ⊂ T , we infer hence that T ∈ AC .  Theorems 9.8.3 and 9.8.4 will be obtained again on another way in 11.5. Finally let us note that it is not known if the conditions of Theorem 9.8.1 imply that T ∈ AC .

256

Chapter 9. Boundedness and Continuity

Exercises 1. Show that Theorem 9.2.7 fails to hold if α is not assumed to be rational (Kuczma [184]). √ Hint: Let H be an arbitrary Hamel basis of R containing 1 and 2. Let g : √ H → R be an arbitrary function such that g(1) = 0, √ g( 2) = 1. Let f : R → R be the additive extension of g. Take T = Q and α = 2. 2. Show by suitable examples that none of the conditions occurring in Theorem 9.3.5 is necessary for T ∈ C. 3. Show that conditions (iii) and (iv) occurring in Theorem 9.3.3 are equivalent; similarly, conditions (v) and (vi) are equivalent. 4. Let D ⊂ RN be a convex and open set, let f : D → R be a convex function, and let W : R → R be a polynomial, W = const. Show that if the function W (f ) is continuous, then also f is continuous. 5. Let A ⊂ RN −1 be an arbitrary set, and let ϕ : A → R be a continuous function. Show that Gr(ϕ) is a set of N -dimensional measure zero. Hint: It is enough to consider the case, where A is bounded. Let G ⊂ RN −1 be a bounded open set such that A ⊂ G. Fix an ε > 0. Let Q be the family of closed cubes Q ⊂ G such that sup f − inf f < ε. Use Vitali theorem. A∩Q

A∩Q

6. Show that every subset of a circle with positive linear measure belongs to A2 . 7. A function f : X → Y (where X, Y are separable metric spaces) is said to satisfy the Baire condition iff for every Borel subset B of Y the set f −1 (B) satisfies the Baire condition. This is equivalent to the condition that there exists a residual subset T of X such that f | T is continuous (cf. Kuratowski [196]). Show that every convex function which satisfies the Baire condition is continuous.

Chapter 10

The Classes A, B, C 10.1 A Hahn-Banach theorem The main objective of the present chapter is to prove equality (9.1.2), and to give some characterizations of the classes A, C introduced in the preceding chapter. Thus we do not assume equality (9.1.2) as valid, it will be proved in the sequel1 . To this aim we will need a rational version of the Hahn-Banach theorem, which we will presently prove. We start with a definition. Let A ⊂ RN be a set. We say that A is Q-radial at a point a ∈ A iff for every y ∈ RN , y = 0, there exists an εy > 0 such that a + λy ∈ A for every λ ∈ Q ∩ (0, εy ). Lemma 10.1.1. Let D ⊂ RN be a set Q-convex and Q-radial at a point x0 ∈ D, and let L ⊂ RN be a linear space (over Q), x0 ∈ L. Let f : D → R be a function fulfilling the inequality   f λx + (1 − λ)y  λf (x) + (1 − λ)f (y) (10.1.1) for all x, y ∈ D and all λ ∈ Q ∩ [0, 1], and let g : L → R be a homomorphism2 . Let z∈ / L, and put L1 = E L ∪ {z} . If g(x)  f (x)

for x ∈ D ∩ L,

(10.1.2)

then there exists a homomorphism G : L1 → R such that G(x)  f (x)

for x ∈ D ∩ L1

(10.1.3)

and G | L = g. Proof. For every x, y ∈ L and every λ, μ ∈ Q ∩ (0, ∞) such that x + μz ∈ D and y − λz ∈ D we have by (10.1.2) and (10.1.1)   λ μ λ μ g(x) + g(y) = g x+ y λ+μ λ+μ λ+μ λ+μ 1

Theorem 10.2.2. The results of Chapter 9 (where equality (9.1.2) was postulated) will not be used in 10.1–10.2 of the present chapter, except for Theorem 9.2.8, in whose proof relation (9.1.2) was not used. 2 From the space (L ; Q ; + ; ·) into (R ; Q ; + ; ·).

Chapter 10. The Classes A, B, C

258

 μ λ x+ y f λ+μ λ+μ   μ λ (x + μz) + (y − λz) =f λ+μ λ+μ λ μ  f (x + μz) + f (y − λz). λ+μ λ+μ 

Hence

g(x) − f (x + μz) f (y − λz) − g(y)  , μ λ

and α = sup U

g(x) − f (x + μz) f (y − λz) − g(y)  inf = β, V μ λ

(10.1.4)

where U = {(x, μ) ∈ L × Q | μ > 0, x + μz ∈ D}, V = {(y, λ) ∈ L × Q | λ > 0, y − λz ∈ D}. Since D is Q-radial at x0 , we have (x0 , μ) ∈ U for μ ∈ Q ∩ (0, εz ) and (x0 , λ) ∈ V for λ ∈ Q ∩ (0, ε−z ). Consequently U = ∅ and V = ∅. Observe also that (10.1.4) implies that −∞ < α  β < +∞. In particular, [α, β] = ∅. Choose any c ∈ [α, β]. Every t ∈ L1 may be uniquely written as t = x + λz,

(10.1.5)

where x ∈ L and λ ∈ Q. For such a t define G(t) = g(x) − cλ. It is easily seen that G : L1 → R is a homomorphism and that G | L = g. Now take an arbitrary t ∈ D ∩ L1 . t may be represented in form (10.1.5). Consider three cases: I. λ = 0. Then t = x, and by (10.1.2) G(t) = g(x)  f (x) = f (t). Consequently (10.1.3) holds. II. λ > 0. Since t ∈ D, we have (x, λ) ∈ U . Since c  α, this implies g(x) − f (x + λz)  c, λ or g(x) − f (x + λz)  cλ, i.e., G(t) = g(x) − cλ  f (x + λz) = f (t). Again we obtain (10.1.3).

10.1. A Hahn-Banach theorem

259

III. λ < 0. Since t ∈ D, we have (x, −λ) ∈ V . Since c  β, this implies   f x − (−λ)z − g(x) c , −λ or −cλ  f (x + λz) − g(x), i.e., G(t) = g(x) − cλ  f (x + λz) = f (t). 

Thus (10.1.3) holds in this case, too.

Hence we derive the rational version of the Hahn-Banach theorem (cf. M. E. Kuczma [191], Berz [31]). Theorem 10.1.1. Let D ⊂ RN be a set Q-convex and Q-radial at a point x0 ∈ D, and let L ⊂ RN be a linear space (over Q), x0 ∈ L. Let f : D → R be a function fulfilling inequality (10.1.1) for all x, y ∈ D and all λ ∈ Q ∩ [0, 1], and let g : L → R be a homomorphism fulfilling (10.1.2). Then there exists a homomorphism (i.e., and additive function) G : RN → R such that G | L = g and (10.1.3) holds in D. Proof. Let R be the family of all couples (X, A), where X is a linear space (over Q), L ⊂ X ⊂ RN , and A : X → R is a homomorphism such that A | L = g and A(x)  f (x)

for x ∈ D ∩ X.

(10.1.6)

(L, g) ∈ R, so R = ∅. We introduce the order in R in the usual manner: for (X1 , A1 ), (X2 , A2 ) ∈ R we agree that (X1 , A1 )  (X2 , A2 ) iff X1 ⊂ X2 and A2 | X1 = A1 . If I ⊂ R is any chain, then put  Y = X, (X,A)∈I

and define B : Y → R putting B(y) = A(y) if y ∈ X and (X, A) ∈ I. The couple (Y, B) is an upper bound of I in R. In fact, if x, y ∈ Y and α ∈ Q, then there exists an (X, A) ∈ I such that x, y ∈ X. Then also x + y ∈ X ⊂ Y and αx ∈ X ⊂ Y , N which shows that Y is a linear space  (over Q).NSince L ⊂ X ⊂ R for all X such that (X, A) ∈ I, also L ⊂ Y = X ⊂ R . Similarly it is shown that B is a (X,A)∈I

homomorphism fulfilling (10.1.6) and such that B | L = g. By Theorem 1.8.1 (Lemma of Kuratowski-Zorn) in R there exists a maximal element (Z, G). The only thing we need to show is that Z = RN . Supposing the contrary, let z ∈ RN \ Z. By Lemma 10.1.1 there exists a homomorphism G∗ : E(Z ∪ {z}) → R such that G∗ | Z = G, whence G∗ | L = G | L = g, and   G∗ (x)  f (x) for x ∈ D ∩ E Z ∪ {z} .         Consequently E Z ∪ {z} , G∗ ∈ R, and clearly (Z, G) < E Z ∪ {z} , G∗ , which contradicts the maximality of (Z, G). Consequently we must have Z = RN . 

Chapter 10. The Classes A, B, C

260

Theorem 10.1.2. Let D ⊂ RN be a set which is symmetric with respect to 0 (i.e., −D = D), and let f : D → R be an even function: for x ∈ D.

f (−x) = f (x)

(10.1.7)

Further, let L ⊂ RN be a linear space (over Q), and let g : L → R be a homomorphism fulfilling (10.1.2). Then |g(x)|  f (x)

for x ∈ D ∩ L.

(10.1.8)

Proof. Take an arbitrary x ∈ D ∩ L. We have by (10.1.2) and (10.1.7) g(−x)  f (−x) = f (x), whence g(x) = −g(−x)  −f (x). This together with (10.1.2) yields −f (x)  g(x)  f (x), 

which is equivalent to (10.1.8).

10.2 The class B First we prove the following lemma. Lemma 10.2.1. Let T ⊂ R be a set Q-convex and Q-radial at a point x0 ∈ T . Then T ∈ B if and only if int T = ∅. Proof. If int T = ∅, then every additive function bounded above on T is continuous, as results from Corollary 6.4.1. Consequently T ∈ B. In order to prove the necessity of the condition int T = ∅ we will show that if int T = ∅, then T ∈ / B. At first assume that x0 = 0. Since int T = ∅ and T is Q-radial at zero, there exist x, y ∈ R such that 0 < x < y, x ∈ / T , y ∈ T . Put S = {(ξ, η) ∈ Q × Q | (1 + ξ)x + ηy ∈ T, η > 0}. Observe that (−1, 1) ∈ S, so S = ∅. Write s = sup (ξ, η)∈S

ξ . η

Since T is Q-radial at 0, there exists an ε > 0 (ε = ε−y ) such that −αy ∈ T for α ∈ Q ∩ (0, ε). We will show that 1 s . (10.2.1) ε ξ 1 Supposing the contrary, we would be able to find (ξ, η) ∈ S such that > . Then η ε necessarily ξ > 0, and we have η 0 < < ε. ξ

10.2. The class B

261

η Hence − y ∈ T , and, by the definition of S, (1 + ξ)x + ηy ∈ T . Since T is Q-convex, ξ this implies that  

ξ η 1 (1 + ξ)x + ηy + − y ∈ T, x= 1+ξ 1+ξ ξ contrary to the choice of x ∈ / T . The contradiction obtained proves (10.2.1). Since (−1, 1) ∈ S, we have s  −1. (10.2.2)   Put L = E {x, y} . Observe that x and y are linearly independent over Q. Supposing the contrary, let αx + βy = 0 for some α, β ∈ Q, α2 + β 2 > 0. Since x, y = 0, we must have α, β = 0, αβ < 0. And since 0 < x < y, we must have |β| < |α|. Since 0, y ∈ T and T is Q-convex,   β β x=− y+ 1+ 0 ∈ T. α α This contradiction shows that x and y are linearly independent over Q, and consequently form a base for L. By Theorem 4.3.1 there exists a unique homomorphism g : L → R such that g(x) = 1, g(y) = −s. We propose to show that g(z)  1

for z ∈ T ∩ L.

(10.2.3)

Suppose that there exists a z ∈ T ∩ L such that g(z) > 1. We have z = λx + μy with some λ, μ ∈ Q. Hence g(z) = λg(x) + μg(y) = λ − μs, and g(z) > 1 means μs < λ − 1.

(10.2.4)

We must distinguish three cases. I. μ > 0. Then (λ − 1, μ) ∈ S, and the inequality (10.2.4) yields s<

λ−1 , μ

which contradicts the definition of s. II. μ = 0. Then (10.2.4) turns into λ > 1. Moreover, λx = λx + μy = z ∈ T . Hence also   1 1 x = (λx) + 1 − 0 ∈ T, λ λ a contradiction.

Chapter 10. The Classes A, B, C

262

III. μ < 0. Then, by (10.2.4), 1−λ λ−1 = . μ −μ

s>

Consequently there exists a pair (ξ, η) ∈ S such that

1−λ ξ > . This means that η −μ

λη > η + μξ. Put κ=

(10.2.5)

λη − μ(1 + ξ) . η−μ

By (10.2.5) λη − μ(1 + ξ) > η − μ, whence κ > 1. Since (ξ, η) ∈ S, we have (1 + ξ)x + ηy ∈ T. Since z ∈ T , we have Hence also u= and v=

λx + μy ∈ T.   1 1 1 (λx + μy) = (λx + μy) + 1 − 0 ∈ T, κ κ κ

 

1 1 1 (1 + ξ)x + ηy = [(1 + ξ)x + ηy] + 1 − 0 ∈ T. κ κ κ

Finally, x=

−μ η u+ v ∈ T, η−μ η−μ

a contradiction. Thus (10.2.3) has been proved. By Theorem 10.1.1 (where N = 1, f (x) = 1) there exists an additive function G : R → R such that G | L = g and G(z)  1

for z ∈ T.

(10.2.6)

The function G is discontinuous. If G were continuous, then we would have G(z) = cz for z ∈ R with a certain real c, whence in particular G(x)/x = G(y)/y. But then 1 G(x) G(y) s = = =− , x x y y y > 1, which is a contradiction with (10.2.2). x  Thus  there exists a discontinuous additive function bounded above on T cf. (10.2.6) . Consequently T ∈ / B.

whence −s =

10.2. The class B

263

Now assume that x0 = 0. Then the set T −x0 is Q-convex and Q-radial at 0 and, similarly as T , has empty interior. By what has already been proved, T − x0 ∈ / B. Consequently there exists a discontinuous additive function G : R → R bounded above on T − x0 . By Theorem 9.2.8 G is also bounded above on T . This means that T ∈ / B, which completes the proof of the lemma.  The above Lemma 10.2.1, similarly as all the theorems in the present section, is due to M. E. Kuczma [191]. Theorem 10.2.1. Let T ⊂ RN be a set Q-convex and Q-radial at a point x0 . Then T ∈ B if and only if int T = ∅. Proof. The sufficiency results from Corollary 6.4.1 similarly as in the proof of Lemma 10.2.1. Thus we need only prove that if int T = ∅, then T ∈ / B. First assume that x0 = 0. Let e1 , . . . , eN be a base of RN over R. For i = 1, . . . , N , let Ti = {x ∈ R | xei ∈ T } ⊂ R. First we show that if int Ti = ∅ for i = 1, . . . , N , then int T = ∅. So assume that int Ti = ∅ for i = 1, . . . , N . Then there exist open intervals Ji ⊂ Ti \ {0}, i = 1, . . . , N . Put ! G = x ∈ RN | x = α1 x1 e1 + · · · + αN xN eN , αi ∈ Q ∩ (0, 1), xi ∈ Ji , i = 1, . . . , N,

N

" αi = 1 .

i=1

The set G = ∅ is open. For let x ∈ G, y ∈ RN . x may be written as x = α1 x1 e1 + · · · + αN xN eN , αi ∈ Q ∩ (0, 1), xi ∈ Ji , i = 1, . . . , N,

N

αi = 1.

i=1

(10.2.7) The points x1 e1 , . . . , xN eN are linearly independent over R (just like e1 , . . . , eN ), and since dim RN = N (over R), {x1 e1 , . . . , xN eN } is a base of RN over R. Hence y may be written as y = β1 x1 e1 + · · · + βN xN eN ,   Hence cf. (10.2.7)

βi ∈ R,

i = 1, . . . , N.

y = x + (β1 − α1 )x1 e1 + · · · + (βN − αN )xN eN , or

    β1 − α1 βN − αN y = α1 x1 + x1 e1 + · · · + αN xN + xN eN . α1 αN

In other words y = α1 y1 e1 + · · · + αN yN eN ,

Chapter 10. The Classes A, B, C

264

where yi = xi +

βi − αi xi , αi

i = 1, . . . , N.

If y is sufficiently close to x, then |βi − αi | are arbitrarily small for i = 1, . . . , N , and yi ∈ Ji for i = 1, . . . , N . Hence y ∈ G. Consequently G is open. Since T is Q-convex, we have by Theorem 5.1.3 G ⊂ Q(T ) = T . This shows that int T = ∅. Thus if int T = ∅, there exists an i, 1  i  N , such that int Ti = ∅. Clearly Ti is Q-convex and Q-radial at zero (in R). By Lemma 10.2.1 there exists a discontinuous additive function g : R → R bounded above on Ti . Let L = Rei , and define G : L → R by G(xei ) = g(x). It is easily checked that G is a homomorphism bounded above on T ∩ L. By Theorem 10.1.1 G may be extended onto RN to an additive function A : RN → R bounded above on T . Since A | L = G is discontinuous, also A is discontinuous. Consequently T ∈ / B. The case where x0 = 0 may be reduced to that where x0 = 0 like in the proof of Lemma 10.2.1.  The following example (cf. M. E. Kuczma [191]) shows that the assumption that T is Q-radial at a point x0 is essential. Example 10.2.1. Let H ⊂ RN be a Hamel basis and let V0 be the cone defined in the proof of Theorem 9.3.4. We have E + (V0 ) = V0 , whence also Q(V0 ) = V0 , i.e., V0 is Q-convex. Let f : RN → R be an additive function bounded above on V0 : f (x) < M

for x ∈ V0 .

(10.2.8)

Take an arbitrary h ∈ H. There is an α < γ such that h = hα . Then λh + hα+1 ∈ V0 for every λ ∈ Q. Hence by (10.2.8) λf (h) + f (hα+1 ) = f (λh + hα+1 )  M and letting λ range over whole Q we obtain f (h) = 0. Consequently f | H = 0, and since both, 0 and f , are additive extensions of f | H onto RN , we get by Theorem 5.2.2 f = 0. Consequently f is continuous, and thus V0 ∈ B. On the other hand, int V0 = ∅, since arbitrarily close to any x ∈ RN there are points which do not belong to V0 . In fact, if x has form x = λ1 hα1 + · · · + λn hαn , where n ∈ N, λi ∈ Q, hαi ∈ H, i = 1, . . . , n, α1 < · · · < αn , then the points y = x − λhαn +1 ,

λ ∈ Q ∩ (0, ∞),

do not belong to V0 and approach x when λ tends to zero. The same argument shows also that V0 is not Q-radial at any point.

10.2. The class B

265

Lemma 10.2.2. Let D ⊂ RN be a convex and open set, let f : D → R be a convex function, and let M ∈ R be a constant such that M > inf f . Then the set D

T = {x ∈ D | f (x) < M } is Q-convex and Q-radial at every point x ∈ T . Proof. If x, y ∈ T ⊂ D and λ ∈ Q ∩ [0, 1], then λx + (1 − λ)y ∈ D and by Theorem 5.3.5   f λx + (1 − λ)y  λf (x) + (1 − λ)f (y) < λM + (1 − λ)M = M. Consequently λx + (1 − λ)y ∈ T . This means that T is Q-convex. If x ∈ T and y ∈ RN , then by Theorem 6.1.1 lim f (x + λy) = f (x),

λ→0 λ∈Q

since for λ ∈ Q ∩ [0, 1] x + λy = (1 − λ)x + λ(x + y) ∈ Q(x, x + y). Thus there exists an ε > 0 such that for λ ∈ Q ∩ (0, ε) we have f (x + λy) < M , and x + λy ∈ T . This means that T is Q-radial at x.  Now we prove the theorem which has been our main aim. Theorem 10.2.2. A = B. Proof. The inclusion A ⊂ B is obvious, so we need only prove that B ⊂ A. Take a set T ⊂ RN such that T ∈ / A. We must show that T ∈ / B. Since T ∈ / A, there exist a convex and open set D ⊂ RN and a discontinuous convex function f : D → R such that T ⊂ D and f is bounded above on T : for x ∈ T.

(10.2.9)

T0 = {x ∈ D | f (x) < M }.

(10.2.10)

f (x) < M Put

Thus T ⊂ T0 . By Lemma 10.2.2 T0 is Q-convex and Q-radial at every its point. If we had int T0 = ∅, the function f would be continuous by Corollary 6.4.1. Thus int T0 = ∅. By Theorem 10.2.1 T0 ∈ / B, and hence also T ∈ / B.  Incidentally we can observe that at the same time we have obtained a new proof N of the fact that Q is an A-conservative  operation. In fact, if f : R → R is an additive function bounded above on a set T i.e., (10.2.9) holds , then the set T0 defined by (10.2.10) is Q-convex and contains T , and hence also contains the smallest set with these properties, i.e., Q(T ): Q(T ) ⊂ T0 .   Moreover, f is bounded above on T0 cf. (10.2.10) , and hence also on Q(T ).

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266

10.3 The class C The results in this section are due to J. Sm´ıtal [287] and J. Mo´scicki [231]. We start with some lemmas. Lemma 10.3.1. Let T ⊂ RN be a set Q-radial at zero. Then T contains a Hamel basis. Proof. Put L = E(T ). Suppose that L = RN . Then there exists a y ∈ RN \ L. Since T is Q-radial at 0, there exists a positive λ ∈ Q such that λy ∈ T ⊂ E(T ) = L. Hence 1 also y = (λy) ∈ L, a contradiction with y ∈ RN \ L. By Corollary 4.2.2 there exists λ a Hamel basis contained in T .  Lemma 10.3.2. Let T ⊂ RN be a set which is Q-convex and symmetric with respect to zero. Then T contains a Hamel basis if and only if T is Q-radial at zero. Proof. If T is Q-radial at zero, then it contains a Hamel basis in virtue of Lemma 10.3.1. Conversely, let H ⊂ T be a Hamel basis. Take an arbitrary y ∈ RN . y may be written as y = α1 h1 + · · · + αn hn , where αi ∈ Q, αi > 0, hi ∈ H ∪ (−H) ⊂ T ∪ (−T ) = T , since by the symmetry −T = T . Put α = α1 + · · · + αn > 0. Then y = λ1 h1 + · · · + λn hn , α y where λi ∈ Q, λi > 0, λ1 + · · · + λn = 1. Hence (cf. Lemma 5.1.3) ∈ T , since T is α y Q-convex. By the symmetry also − ∈ T , whence also α 1  y  y  + − ∈ T, 0= 2 α α and y λy = αλ + (1 − αλ) 0 ∈ T α   1 . This means that T is Q-radial at zero.  for λ ∈ Q ∩ 0, α Theorem 10.3.1. Let T ⊂ RN be a set which is Q-convex and symmetric with respect to zero. Then T ∈ C if and only if int T = ∅. Proof. The sufficiency results from Lemma 9.3.1 and from (9.1.1). To prove the necessity, assume that int T = ∅. We must distinguish two cases. I. T is Q-radial at zero. Then by Theorem 10.2.1 T ∈ / B, i.e., there exists a discontinuous additive function g : RN → R bounded above on T : g(x)  M



for x ∈ T.

Hence it follows by Theorem 10.1.2 that we take f (x) = M |g(x)|  M i.e., g is bounded on T . Consequently T ∈ / C.

for x ∈ T,



10.4. The class A

267

II. T is not Q-radial at zero. By Lemma 10.3.2 T does not contain any Hamel basis, and by Theorem 9.3.6 T ∈ / C.  Theorem 10.3.2. Let T ⊂ RN be an arbitrary set. The following conditions are equivalent: (i) T ∈ C,   (ii) int Q R(T ) = ∅,   (iii) int R Q(T ) = ∅,   (iv) int Q T ∪ (−T ) = ∅.     Proof. Assume (i). The sets Q R(T ) and Q T ∪ (−T ) obviously are Q-convex and, by Corollary 5.1.1, symmetric with respect to zero. Since T ⊂ T ∪ (−T ) ⊂ Q T ∪   (−T ) , (i) implies that Q T ∪ (−T ) ∈ C. Take a t ∈ T . Then T = (T − t) +  t, whence  by Theorems 9.2.8and 9.2.2 T − t ∈ C. Since T − t ⊂ T − T = R(T ) ⊂ Q R(T ) , we get hence Q R(T ) ∈ C. By Theorem 10.3.1 conditions (ii) and (iv) are fulfilled. Now assume (ii). By Lemmas 5.1.6 and 5.1.5       Q R(T ) = Q(T − T ) = Q T + (−T ) = Q(T ) + Q(−T ) = Q(T ) − Q(T ) = R Q(T ) . Thus (ii) implies (iii). Now assume (iii). By Theorems 9.2.5, 9.2.6, 9.2.3, 9.2.2 and by Lemma 9.3.1 T ∈ C, i.e., (i) is fulfilled. Finally assume (iv). If f : RN → R is an additive function bounded on T : |f (x)|  M

for x ∈ T,

(10.3.1)

then for x ∈ −T we have |f (x)| = | − f (x)| = |f (−x)|  M by (10.3.1). Consequently F : P(RN ) → P(RN ) given by F (T ) = T ∪(−T ) is a C-conservative operation. Hence, like above, it follows that T ∈ C, i.e., (i) is fulfilled. 

10.4 The class A Theorem 10.3.2 gives a characterization of the class C. In this section we prove a similar characterization for the class A = B. The results are due to J. Sm´ıtal [288] and J. Mo´scicki [231]. Theorem 10.4.1. Let T ⊂ RN be an arbitrary set. The following conditions are equivalent: (i) T ∈ A, (ii) For every set A ⊂ RN , Q-radial at a point, int Q(T + A) = ∅,

(iii) For every set A ⊂ RN , Q-radial at a point, int Q(T ) + Q(A) = ∅, (iv) For every set A ⊂ RN , Q-radial at a point, int Q(T ∪ A) = ∅.

Chapter 10. The Classes A, B, C

268

Proof. Assume (i). The sets Q(T + A) and Q(T ∪ A) obviously are Q-convex and Q-radial at a point. We have T ⊂ Q(T ∪ A), whence Q(T ∪ A) ∈ A. Similarly, with an a ∈ A, we have by Theorems 9.2.8 and 9.2.2 T + a ∈ A, and since T + a ⊂ T + A ⊂ Q(T + A), we get hence Q(T + A) ∈ A. By Theorems 10.2.1 and 10.2.2 conditions (ii) and (iv) are fulfilled. Now assume (ii). By Lemma 5.1.6 Q(T + A) = Q(T ) + Q(A), whence (iii) follows. Finally assume that either (iii) or (iv) is fulfilled. Let f : RN → R be an arbitrary additive function bounded above on T : f (x) < M

for x ∈ T.

Put A = {x ∈ RN | f (x) < M }. By Lemma 10.2.2 the set A is Q-convex and Q-radial at every point. Moreover T ⊂ A, whence T ∪A = A and Q(T ) ⊂ Q(A) = A so that Q(T )+Q(A) ⊂ A+A = S2 (A). Thus condition (iii) resp. (iv) implies that int S2 (A) = ∅ resp. int A = ∅. By Theorem 9.3.3 resp. Lemma 9.3.1 A ∈ A. Since the function f is bounded above on A, f is continuous. Consequently T ∈ B. By Theorem 10.2.2 T ∈ A.  We may observe that in conditions (ii), (iii), (iv) of Theorem 10.4.1 we may restrict ourselves to sets A which are Q-radial at a point and such that T ⊂ A. Hence we get the following Theorem 10.4.2. Let T ⊂ RN be an arbitrary set. Then T ∈ A if and only if for every set A which is Q-radial at a point and such that T ⊂ A we have int Q(A) = ∅. The proof of Theorem 10.4.2 does not differ from that of Theorem 10.4.1, since the set A constructed there has the property that T ⊂ A. Similarly, in the “if” part of Theorem 10.4.2 it would be enough to assume that A is Q-radial at every point. In Theorem 10.4.1 we can replace A by −A, since A is Q-radial at a point if and only if −A is Q-radial at a point. It is less obvious that we can replace T by −T . Theorem 10.4.3. Let T ⊂ RN be an arbitrary set. The following conditions are equivalent: (i) T ∈ A, (ii) For every set A, Q-radial at a point, int Q(A − T ) = ∅,

(iii) For every set A, Q-radial at a point, int Q(A) − Q(T ) = ∅. Proof. Let condition (i) be fulfilled. If A is an arbitrary set Q-radial at a point, then the set −A also is Q-radial at a point. By Theorem 10.4.1 int Q(−A + T ) = ∅.   But Q(−A + T ) = Q − (A − T ) = −Q(A − T ) (cf. Lemma 5.1.5). Hence also int Q(A − T ) = ∅, i.e., condition (ii) is fulfilled. By Lemmas 5.1.5 and 5.1.6 Q(A − T ) = Q(A) − Q(T ), so (ii) implies (iii).

10.5. Set-theoretic operations

269

Now suppose that condition (iii) is fulfilled, and let A ⊂ RN be an arbitrary set Q-radial at a point. Then also −A is Q-radial at a point, and by (iii) int Q(−A) −

Q(T ) = ∅. But

Q(−A) − Q(T ) = − Q(A) + Q(T ) ,

so int Q(A) + Q(T ) = ∅. By Theorem 10.4.1 T ∈ A. 

10.5 Set-theoretic operations The theorems of 9.3 say that every reasonably large set belongs to the class A. Consequently the sets which do not belong to A may be described as “small”. Consequently it is rather surprising that the union of two “small” sets may be “large”. Theorem 10.5.1. There exist sets T1 , T2 ⊂ RN such that Ti ∈ / C, i = 1, 2, T1 is countable and T1 ∪ T2 ∈ A. Proof. Let H ⊂ RN be an arbitrary Hamel basis, and write H = H 1 ∪ H2 , where H1 is countable, H1 ∩ H2 = ∅. Put T1 = E(H1 ),

T2 = E(H2 ).

Since H is linearly independent (over Q), we have for arbitrary h1 ∈ H1 , h2 ∈ H2 h1 ∈ / T2 ,

h2 ∈ / T1 .

Consequently T1 = RN , T2 = RN . Suppose that T1 contains a Hamel basis H0 . Then RN = E(H0 ) ⊂ E(T1 ) = T1 = RN , a contradiction. Similarly we show that also T2 does not contain a Hamel basis. By Theorem 9.3.6 T1 ∈ / C and T2 ∈ / C. Moreover, since H1 is countable, so is also T1 , in virtue of Lemma 4.1.3. Write T = T1 ∪ T2 , and let f : RN → R be an arbitrary additive function bounded above on T . For every h ∈ H we have λh ∈ T for every λ ∈ Q. Hence λf (h) = f (λh) remains bounded when λ varies over the whole Q, which is impossible unless f (h) = 0. Consequently f | H = 0. Zero is an additive extension of f | H onto RN , and so is also f , whence by Theorem 5.2.2 f = 0 in RN . Consequently f is continuous. This means that T ∈ B. By Theorem 10.2.2 T ∈ A.  Now we are going to deal with cartesian products. Write N = p + q, p, q ∈ N, and consider RN as the product RN = Rp × Rq . Theorem 10.5.2. Let T ⊂ Rp and S ⊂ Rq be arbitrary sets, and put W = T ×S ⊂ RN . Then W ∈ AN [W ∈ CN ] if and only if T ∈ Ap , S ∈ Aq [T ∈ Cp , S ∈ Cq ]. Proof. Assume that, e.g., T ∈ / Ap . Then there exists a discontinuous additive function g : Rp → R bounded above on T . For every x ∈ RN write x = (xp , xq ), where xp ∈ Rp , xq ∈ Rq . Put f (x) = g(xp ).

Chapter 10. The Classes A, B, C

270

The function f is additive, discontinuous, and bounded above on W . This shows that W ∈ / BN = AN (cf. Theorem 10.2.2). Conversely, let W ∈ / AN = BN . Then there exists a discontinuous additive function f : RN → R bounded above on W . By Theorem 5.5.1 f (x) = fp (xp ) + fq (xq ),

(10.5.1)

where fp : Rp → R and fq : Rq → R are additive functions. Since f is discontinuous, at least one of the functions fp and fq , say fp , must be discontinuous. Fix an xq ∈ S. Then, as xp varies over T , fp (xp ) = f (x) − fq (xq ),

x = (xp , xq ),

remains bounded above. Consequently T ∈ / Bp = Ap . The proof for the class C is analogous.



Now consider an arbitrary set W ⊂ RN . For every xp ∈ Rp and xq ∈ Rq write Wp [xq ] = {xp ∈ Rp | (xp , xq ) ∈ W }, Wq [xp ] = {xq ∈ Rq | (xp , xq ) ∈ W }, (cf. 2.1), and Wp = {xp ∈ Rp | There exists an xq ∈ Rq such that (xp , xq ) ∈ W }, Wq = {xq ∈ Rq | There exists an xp ∈ Rp such that (xp , xq ) ∈ W }. The sets Wp and Wq are the projections of W onto Rp and Rq , respectively, the sets Wp [xq ] and Wq [xp ] are sections of W . / CN and the set Wq is bounded. Theorem 10.5.3. Let W ⊂ RN be a set such that W ∈ If there exists an x ¯p ∈ Rp such that Wq [¯ xp ] ∈ Cq , then Wp ∈ / Cp . Proof. There exists a discontinuous additive function f : RN → R bounded on W . f can be written in form (10.5.1). Thus xp ), fq (xq ) = f (x) − fp (¯

x = (¯ xp , xq ).

xp ], fq (xq ) remains bounded. Since Wq [¯ xp ] ∈ Cq , fq is conWhen xq varies over Wq [¯ tinuous. By Theorem 5.5.2 fq (xq ) = cxq with a certain c ∈ Rq . Hence fq is bounded on bounded sets, in particular on Wq . Varying x = (xp , xq ) in such a manner that xp runs over the whole Wp and x ∈ W , we see from (10.5.1) that fp (xp ) remains bounded. Moreover, fp is discontinuous, since f is discontinuous and fq is continuous. Consequently Wp ∈ / Cp .  The assumption that Wq is bounded cannot be dropped, as may be seen from the following example.

10.6. The classes D

271

Example 10.5.1. Let p = q = 1, N = 2, and let f1 : R → R be a discontinuous additive function. Put W = {x = (x1 , x2 ) ∈ R2 | f1 (x1 ) < x2 < f1 (x1 ) + 1}.   For every x1 ∈ R the set W2 [x1 ] is the open interval f1 (x1 ), f1 (x1 ) + 1 and consequently (Lemma 9.3.1) belongs to C1 . Also the set W ∈ / C2 , since the additive function f (x) = x2 − f1 (x1 ) is discontinuous and bounded on W : 0 < f (x) < 1

for x ∈ W.

But the set W1 = R ∈ C1 . / CN , then either all Theorem 10.5.3 says that if the set Wq is bounded and W ∈ sections Wp [xq ] are not in Cp , or all sections Wq [xp ] are not in Cq . (Of course, here the assumption of the boundedness of Wq could be replaced by that of the boundedness of Wp ). One could expect that if W ∈ CN , then all (or, in some sense, almost all) sections Wp [xq ] and Wq [xp ] are in Cp resp. Cq . In particular, the following statement would be an analogue of the Fubini theorem: If W ∈ CN , then {xp ∈ Rp | Wq [xp ] ∈ Cq } ∈ Cp . However, this is not the case, as may be seen from the following example. Example 10.5.2. Let p = q = 1, N = 2, and let I ⊂ R be a non-trivial finite closed interval. Further, let ϕ : I → R be non-linear and one-to-one continuous function [e.g., ϕ(x) = x3 for x ∈ I]. Put W = Gr(ϕ). By Theorem 9.5.2 W ∈ A2 ⊂ C2 . Also the sets W1 = I and W2 = ϕ(I) are bounded. But for all x ∈ R the sets W1 [x] and W2 [x] contain each at most one point, and thus are not in C1 .

10.6 The classes D The classes D(D) have been defined in 9.7. The results in this section are due to J. Sm´ıtal [289]. Our knowledge of the classes D(D) is much narrower than that of the classes A, B, C. Here we give only some sufficient conditions for T ∈ D(D). (Other such sufficient conditions are contained in Theorem 9.7.2). No characterization of the classes D(D) is known. Theorem 9.7.2 suggests a likeness between the classes D(D) and A. In fact, we have the following Theorem 10.6.1. Let D ⊂ RN be a convex and open set. We have D(D) ⊂ A.

(10.6.1)

Proof. Let T ∈ D(D), and let f : RN → R be an arbitrary additive function bounded above on T : f (x)  M for x ∈ T.

Chapter 10. The Classes A, B, C

272

Put g = −f | D. Then g : D → R is a convex function bounded below on T : g(x)  −M

for x ∈ T.

Since T ∈ D(D), this means that g is locally bounded below at every point of D, and hence f is locally bounded above at every point of D. By the Theorem of BernsteinDoetsch (Theorem 6.4.2) f is continuous. Hence T ∈ B = A.  But actually the inclusion in (10.6.1) cannot be replaced by the equality, as may be seen from the following example. Example 10.6.1. Let N = 1, and let D ⊂ R be an open interval such that 0 ∈ D. There exists non-trivial compact interval S ⊂ D symmetric with respect to zero (−S = S). By Lemma 9.3.1 S ∈ C. By Theorem 9.3.6 S contains a Hamel basis H. Let B be a countable subset of H. Put T = {x ∈ D | If x = α1 h1 + · · · + αn hn , αi ∈ Q, hi ∈ H, i = 1, . . . , n, / (0, 1) whenever hi ∈ B}. then αi ∈ Take an arbitrary x ∈ D. Then x = α1 h1 + · · · + αn hn ,

αi ∈ Q, hi ∈ H, i = 1, . . . , n.

Choose hn+1 , hn+2 ∈ H \ B different from h1 , . . . , hn . Since hn+1 , hn+2 ∈ H, they are linearly independent over   Q, i.e., incommensurable. It follows from Lemma 3.8.1 that the set E {hn+1 , hn+2 } is dense in R. Thus we can choose β1 , . . . , βn , γ1 , γ2 ∈ Q in such a manner that βi > max(1, αi ), i = 1, . . . , n, n  and γ1 hn+1 + γ2 hn+2 is close to (βi − αi )hi . Put i=1

y = β1 h1 + · · · + βn hn − γ1 hn+1 − γ2 hn+2 , where γ1 , γ2 have been chosen so that y − x ∈ D and y = x + (y − x) ∈ D. By the choice of βi we have also y ∈ T , x − y ∈ T . Hence x = y + (x − y) ∈ T + T = S2 (T ). Consequently D ⊂ S2 (T ), and hence by Theorem 9.3.3 T ∈ A. On the other hand, T ∈ / D(D). To show this, arrange B into a sequence B = {bk }k∈N , and for every k ∈ N, let gk : R → R be the function ⎧ 1 ⎪ ⎪ −2kt for t < , ⎪ ⎨ 2 gk (t) = ⎪ ⎪ ⎪ ⎩2k(t − 1) for t  1 . 2 Thus gk is a broken linear function, convex in R, and with the properties   1 = −k, gk (t)  0 gk for t ∈ / (0, 1). gk (0) = gk (1) = 0, 2

(10.6.2)

10.6. The classes D

273

Now we define a function f : D → R as follows. If x ∈ D, then x may be written as x = α1 bk1 + · · · + αn bkn + β1 h1 + · · · + βm hm , where αi , βj ∈ Q, bki ∈ B, hj ∈ H \ B, i = 1, . . . , n; j = 1, . . . , m. For such an x put f (x) =

n

gki (αi ).

i=1

The function f is convex (cf. Example 5.3.2, where we must take gbk = gk for bk ∈ B, gh = 0 for h ∈ H \ B). For x ∈ T we have for the corresponding coefficients αi ∈ / (0, 1), whence gki (αi )  0, and consequently also f (x)  0. Thus f is bounded below on T . 1 We have bk ∈ B ⊂ S, k ∈ N. Since 0 ∈ S, also bk ∈ S for k ∈ N. Since S 2   1 1 bk bk is compact, we can choose from the sequence a subsequence 2 2 n n∈N k∈N convergent to a point of S: 1 bkn = x¯ ∈ S ⊂ D. n→∞ 2 lim

According to the definition of f we have by (10.6.2)     1 1 bkn = gkn = −kn f 2 2 

so that lim f

n→∞

1 bk 2 n

 = −∞.

This shows that f is not locally bounded below at x¯. Consequently T ∈ / D(D). Example 10.6.1 shows also that the condition int S2 (T ) = ∅ is not sufficient for T ∈ D(D). Lemma 10.6.1. Let D ⊂ RN be a convex and open set such that 0 ∈ D, and let T ⊂ D be an arbitrary set. If there exists a point a ∈ D such that for every set A which is Q-convex and Q-radial at zero (0 ∈ A) a ∈ int(T − A), then T ∈ D(D). Proof. Let f : D → R be an arbitrary convex function bounded below on T : f (x) > −M

for x ∈ T,

(10.6.3)

and we may assume that M > 0 and M > |f (0)|. Put B = {x ∈ D | f (x) < M }.

(10.6.4)

By Lemma 10.2.2 the set B is Q-convex and Q-radial at zero. Take a λ ∈ Q ∩ (1, 2) such that λa ∈ D. Write   1 B. A= 1− λ

Chapter 10. The Classes A, B, C

274

It is easily seen that also A is Q-convex and Q-radial at zero. By the conditions of the theorem a ∈ int(T − A), whence λa ∈ int λ(T − A). Let G = D ∩ int λ(T − A). Then G is a non-empty (λa ∈ G) open set. Take an arbitrary z ∈ G. Then z ∈ λ(T − A), which means that there exist a t ∈ T and an x ∈ A such that z = λt − λx. Hence z = λt − (λ − 1) b, where b ∈ B ⊂ D. Consequently   1 1 t= z+ 1− b. λ λ By Theorem 5.3.5

  1 1 f (t)  f (z) + 1 − f (b), λ λ

whence by (10.6.3) and (10.6.4) f (z)  λf (t) − (λ − 1)f (b)  −λM − (λ − 1)M = (−2λ + 1)M > −3M.

(10.6.5)

Relation (10.6.5) says that f is bounded below on G. By Theorem 6.2.2 f is locally bounded below in D, i.e., D ∈ D(D).  It may be shown by a suitable example that the condition occurring in Lemma 10.6.1 is not necessary for T ∈ D(D) (cf. Sm´ıtal [289]). Now, using Lemma 10.6.1, we prove a theorem which shows that the classes D(D) actually depend on D. Theorem 10.6.2. Let D1 ⊂ D2 ⊂ RN be convex and open sets such that 0 ∈ D1 and D1 = D2 . Then there exists a set T ⊂ D1 such that T ∈ D(D2 ) \ D(D1 ). Proof. Let H ⊂ RN be a Hamel basis. Since the linear spaces ER (H) and EQ (H) spanned by H over R and Q, respectively, evidently fulfil EQ (H) ⊂ ER (H) ⊂ RN , and EQ (H) = RN we have ER (H) = RN . Thus, by Corollary 4.2.2, H must contain a basis of RN over R. Let {b1 , . . . , bN } ⊂ H be such a basis. Fix an h0 ∈ H \{b1, . . . , bN }. Let B denote the frontier of D1 , and let f : D1 → R be a bounded below convex function such that lim f (x) = +∞. (10.6.6) x→B

(One can take, e.g., f (x) =

1 , (x, B)

where  denotes the Euclidean distance in RN .) Define g0 : H → R putting g0 (h0 ) = 1,

g0 (h) = 0

for h ∈ H \ {h0 }.

(10.6.7)

By Theorem 5.2.2 the function g0 can be extended to an additive function g : RN → R, and by Corollary 5.2.2 g is discontinuous, and hence locally unbounded below at any point of RN . (Otherwise −g would be an additive function locally bounded above at

10.6. The classes D

275

a point of RN , and hence, by Theorem 6.4.2, continuous, and thus also g would be continuous.) We define the convex function F : D1 → R by F = f + g. Thus F is locally unbounded below at any point of RN . Consequently the set

does not belong to D(D1 ):

T = {x ∈ D1 | F (x) > 0}

(10.6.8)

T ∈ / D(D1 ).

(10.6.9)

T ∈ D(D2 ).

(10.6.10)

We are going to show that To this aim take an a ∈ B ∩ D2 and an arbitrary set A which is Q-convex and Q-radial at zero, 0 ∈ A. Consequently there exists an ε > 0 such that for every β ∈ Q with |β|  ε we have βbi ∈ A, i = 1, . . . , N, and clearly we can choose ε rational. Let   C = Q {±εbi }i=1,...,N . The set cl C is convex (cf. Exercise 5.2). Since for every i = 1, . . . , N the rational segment Q(−εbi , εbi ) is contained in C, the real segment −εbi , εbi is contained in cl C. Hence it follows easily that 0 ∈ int cl C. Consequently there exists an open ball S centered at the origin and contained in cl C: S ⊂ cl C.

(10.6.11)

Since a ∈ D2 , and D2 is open, there exists a small ball around a contained in D2 . In other words, there exists a λ ∈ Q ∩ (0, 1) such that a + λS ⊂ D2 .  Take an arbitrary u ∈ (a + λS) \ D1 . Note that (a + λS) \ D1 = ∅, since a is a frontier point of D1 . We have (u + λS) ∩ D1 = ∅, since a ∈ B, and since by (10.6.11) λS ⊂ cl C, also (u + cl C) ∩ D1 = ∅ and (u + C) ∩ D1 = ∅. Consequently there exists a c ∈ C such that u + c ∈ D1 . Put s = inf { ξ ∈ Q ∩ [0, 1] | u + ξc ∈ D1 }.

(10.6.12)

Since ξ = 1 belongs to the set on the right-hand side of (10.6.12) and D1 is open, we have s < 1, and since D1 is convex, for every ξ ∈ (s, 1] ∩ Q we have u + ξc ∈ D1 , whence F (u + ξc) = f (u + ξc) + g(u) + ξg(c).

Chapter 10. The Classes A, B, C

276

  Now, c ∈ C ⊂ EQ {b1 , . . . , bN } , so there exist α1 , . . . , αN ∈ Q such that c = α1 b1 + · · · + αN bN . Since bi = h0 for i = 1, . . . , N , we obtain hence by (10.6.7) g(c) = α1 g0 (b1 ) + · · · + αN g0 (bN ) = 0, and F (u + ξc) = f (u + ξc) + g(u). Since u ∈ / D1 and u + ξc ∈ D1 , and by (10.6.6) f becomes infinite at the frontier of D1 , there exists an η ∈ Q ∩ (s, 1) such that F (u + ηc) > 0. According to (10.6.8) this means that u + ηc ∈ T . Now, ± εbi ∈ A, i = 1, . . . , N , whence   C = Q { ± εbi }i=1,...,N ⊂ Q(A) = A, since A is Q-convex. Consequently c ∈ A, and ηc = ηc + (1 − η)0 ∈ A. Thus u = (u + ηc) − ηc ∈ T − A. Since u could have been arbitrary in (a + λS) \ D1 , we infer that (a + λS) \ D1 ⊂ T − A. (10.6.13) Similarly, for arbitrary u ∈ (a + λS) ∩ D1 we have (u + C) \ D1 = ∅, and so there exists a c ∈ C such that u + c ∈ / D1 . Let s = sup { ξ ∈ Q ∩ [0, 1] | u + ξc ∈ D1 }. Since u ∈ D1 and D1 is open, we have s > 0. By the convexity of D1 u + ξc ∈ D1 for every ξ ∈ Q ∩ (0, s), and F (u + ξc) = f (u + ξc) + g(u) + ξg(c) = f (u + ξc) + g(u). Thus we can find an η > 0 such that F (u + ηc) > 0. Similarly as above we show that u ∈ T − A, whence (a + λS) ∩ D1 ⊂ T − A. (10.6.14) Relations (10.6.13) and (10.6.14) prove that a + λS ⊂ T − A, i.e., a ∈ int(T − A). By Lemma 10.6.1 we obtain hence (10.6.10). Relations (10.6.9) and (10.6.10) show that T ∈ D(D2 ) \ D(D1 ). 

10.7 The classes AC and BC The classes AC and BC have been defined in 9.8. We do not know any characterization of these classes, but we do know that they differ from the classes A and B. The relations between the classes A, B, AC and BC have been established by B. KominekZ. Kominek [171] and Z. Kominek [173]. As pointed out in 9.8 it follows directly from the definition that AC ⊂ BC . (10.7.1) Now we will prove Theorem 10.7.1. AC ⊂ A.

10.7. The classes AC and BC

277

Proof. The inclusion asserted will be proved if we show that if a set T ⊂ RN does not belong to A, then also T ∈ / AC . Let T ⊂ RN be an arbitrary set such that T ∈ / A. This means that there exists a convex and open set D ⊂ RN , T ⊂ D, and a discontinuous convex function f : D → R such that f is bounded above on T : f (x)  M

for x ∈ T.

Put T0 = {x ∈ D | f (x)  M }, and define the function F : D → R:  F (x) =

M for x ∈ T0 , f (x) for x ∈ D \ T0 .

Clearly T ⊂ T0 ⊂ D and F | T is continuous (being constant). To prove that F is convex, take arbitrary x, y ∈ D. We must distinguish some cases. 1 1. x, y ∈ D \ T0 and (x + y) ∈ D \ T0 . Then 2     f (x) + f (y) F (x) + F (y) x+y x+y =f  = . F 2 2 2 2 1 (x + y) ∈ T0 . Then 2   M +M f (x) + f (y) F (x) + F (y) x+y =M = < = . F 2 2 2 2

2. x, y ∈ D \ T0 , but

1 3. x, y ∈ T0 . By Lemma 10.2.2 the set T0 is Q-convex, and thus (x + y) ∈ T0 . 2 Hence   M +M F (x) + F (y) x+y =M = = . F 2 2 2 4. x ∈ T0 , y ∈ D \ T0 ,  F

x+y 2



1 (x + y) ∈ T0 . Then 2

=M =

M +M F (x) + M F (x) + f (y) F (x) + F (y) = < = . 2 2 2 2

1 (x + y) ∈ D \ T0 . Then 2     x+y f (x) + f (y) M + f (y) F (x) + F (y) x+y =f   = . F 2 2 2 2 2

5. x ∈ T0 , y ∈ D \ T0 ,

The cases where x ∈ D \ T0 , y ∈ T0 , are analogous to 4 and 5.

Chapter 10. The Classes A, B, C

278

Thus F is convex. Suppose that F is continuous in D. Then the set T0 is closed (in D), and hence Lebesgue measurable. Since f is a discontinuous convex function bounded above on T0 , we infer hence in virtue of Theorem 9.3.1 that m(T0 ) = 0. Consequently m(D \ T0 ) > 0. On D \ T0 we have f = F , so f | D \ T0 is continuous. By Theorem 9.8.3 f is continuous, a contradiction. Hence F is discontinuous. This shows that there exists a discontinuous convex function F such that the restriction F | T is continuous. Consequently T ∈ / AC .  Now we exhibit two examples which show that no other inclusion can be asserted. First we show that the inclusion in Theorem 10.7.1 is strict. Example 10.7.1. Let H ⊂ RN be a Hamel basis. Put g(h) = 1 for h ∈ H, and let f : RN → R be the additive extension of g (cf. Theorem 5.2.2). By Corollary 5.2.2 f is discontinuous. Moreover, all the values of f are rational. For every λ ∈ Q put Tλ = {x ∈ RN | f (x) = λ}. Let {λn }n∈N be a sequence of all rational numbers. We define a new sequence {μn }n∈N as follows μ1 = λ1 , μ3 = λ1 ,

μ2 = λ2 , μ4 = λ2 ,

μ5 = λ3 ,

μ6 = λ1 , μ7 = λ2 , μ8 = λ3 , μ9 = λ4 , .......................................... Every λ ∈ Q occurs in the sequence {μn }n∈N infinitely many times. Let " ! Un = Tμn ∩ x ∈ RN | 22n−1 < |x| < 22n+1 , n ∈ N, and T = {0} ∪

∞ 

Un .

n=1

It is easily seen that the restriction f | T is continuous (f is constant on every Un , and the sets Un are separated). Since f is discontinuous, this shows that T ∈ / BC .

(10.7.2)

N  take an arbitrary x ∈ R , x = 0. Let n ∈ N be such that f (x) = λn . Then  Now 1 1 1 x = f (x) = λn . f 2 2 2

Let l be the straight line passing through the origin and x. Since card l = c, whereas f (l) ⊂ Q is at most countable, f | l cannot be one-to-one. Thus there exist a, b ∈ l, a = b, such that f (a) = f (b), whence f (a − b) =f (a) −   f (b)  = 0. For every 1 1 1 α ∈ Q we have x + α(a − b) ∈ l and f x + α(a − b) = f x + αf (a − b) = 2 2 2   1 1 f x = λn . Consequently the set T 21 λn is dense on l. 2 2

10.7. The classes AC and BC

279

It follows from the definition of {μn } and {Un } that there exists a k ∈ N such that

1 λn for all t ∈ Uk . 2 Take a y ∈ Uk ∩l ∩{x ∈ RN | 22k < |x| < 22k+1 } ⊂ Uk ⊂ T lying on the same semiline 1 1 of l as x, and put z = x − y. Obviously f (z) = f (x) − f (y) = λn − λn = λn . 2 2 Moreover, z ∈ l, whence 22k−1 < |z| < 22k+1 , whence z ∈ Uk ⊂ T . Consequently x = y + z ∈ T + T . Evidently also 0 ∈ T + T , whence T + T = RN . By Theorem 9.3.3 T ∈ A, and by Theorem 10.2.2 T ∈ B. Thus by (10.7.2) T ∈ B \ BC = A \ BC , whence in virtue of (10.7.1) T ∈ A \ AC . 0 < |x| < 22k−1

and

f (t) =

Example 10.7.1 shows also that the condition int S2 (T ) = ∅ is not sufficient for T ∈ AC nor T ∈ BC . Example 10.7.2. For every x ∈ RN we write x = (ξ1 , . . . , ξN ), ξi ∈ R, i = 1, . . . , N , and let e1 = (1, 0, . . . , 0), . . . , eN = (0, . . . , 0, 1) be the usual orthonormal base of RN over R. Put V = {x ∈ RN | ξ1  0, . . . , ξN  0, ξ1 + · · · + ξN  1} and VQ = {x ∈ RN | x ∈ V and ξ1 ∈ Q, . . . , ξN ∈ Q}. We have VQ ⊂ V , moreover VQ is dense in V . Also e1 , . . . , eN ∈ V and int V = ∅, whence by Lemma 9.3.1 V ∈ A, and by Corollary 9.3.2 V contains a Hamel basis. Hence it follows that E(V ) = RN . Since e1 , . . . , eN are linearly independent over R, they are also linearly independent over Q. By Lemma 4.2.1 there exists a Hamel basis H of RN such that e1 , . . . , eN ∈ H ⊂ V. (10.7.3) Define g0 : H → R by g0 (h) = 1 for h ∈ H, and let g : RN → R be the additive extension of g0 (cf. Theorem 5.2.2). Let T = {x ∈ RN | 0  g(x)  1}. By Corollary 5.2.2 g is discontinuous, and since g is bounded on T , we obtain T ∈ / C.

(10.7.4)

Moreover, g | H = g0 = 1, whence H ⊂ T . Also, if x ∈ VQ , then x = ξ1 e1 + · · ·+ ξN eN with ξi ∈ Q ∩ [0, ∞), ξ1 + · · · + ξN  1. Hence by (10.7.3) g(x) = ξ1 g(e1 ) + · · · + ξN g(eN ) = ξ1 g0 (e1 ) + · · · + ξN g0 (eN ) = ξ1 + · · · + ξN ∈ [0, 1]. Thus VQ ⊂ T . Let f : RN → R be an arbitrary additive function such that the restriction f | T is continuous. Put ci = f (ei ), i = 1, . . . , N , and c = (c1 , . . . , cN ) ∈ RN . For x ∈ VQ , x = (ξ1 , . . . , ξN ), we have f (x) = f (ξ1 e1 + · · · + ξN eN ) = ξ1 f (e1 ) + · · · + ξN f (eN ) = ξ1 c1 + · · · + ξN cN = cx.

Chapter 10. The Classes A, B, C

280

Take an arbitrary h ∈ H ⊂ V . Since VQ is dense in V , there exist xn ∈ VQ , n ∈ N, such that lim xn = h. (10.7.5) n→∞

We have VQ ∪ H ⊂ T and f | T is continuous, so (10.7.5) implies that lim f (xn ) = f (h),

n→∞

i.e., f (h) = lim cxn = c lim xn = ch. n→∞

n→∞

Consequently for h ∈ H we have f (h) = ch. By the uniqueness part of Theorem 5.2.2 f (x) = cx in RN , i.e., f is continuous. Consequently T ∈ BC and by (10.7.4) T ∈ BC \ C. By (9.1.1) T ∈ BC \ B and T ∈ BC \ A, whence in virtue of Theorem 10.7.1 T ∈ BC \ AC . Gathering together the conclusions following from the above examples, we have the following Theorem 10.7.2. BC \ C = ∅.

A \ AC = ∅, B \ BC = ∅, BC \ B = ∅, BC \ AC = ∅ and

We conclude this section with the following Theorem 10.7.3. Let T ⊂ RN be an arbitrary set. If T ∈ AC , then T contains a Hamel basis. Proof. This is a consequence of Theorem 10.7.1 and Corollary 9.3.2.



Investigations similar to those contained in 10.5, but for classes AC and BC , were carried out by Z. Kominek [174].

Exercises 1. Let T ⊂ RN be an arbitrary set. Show that if T ∈ A, then also −T ∈ A. 2. Let T ⊂ RN be an arbitrary set. Show that if T ∈ A, then also αT + b ∈ A for arbitrary α ∈ Q \ {0}, b ∈ RN . 3. Let T ⊂ RN be an arbitrary set. Show that if T ∈ C, then also αT + b ∈ C for arbitrary α ∈ R \ {0}, b ∈ RN . 4. Let T ⊂ RN be an arbitrary set. Show that if T ∈ A, then also αT + b ∈ A for arbitrary α ∈ R \ {0}, b ∈ RN . (Generalization of Exercise 10.2). 5. Let T ⊂ RN be an analytic set such that E(T ) = RN . Show that, for every α ∈ R \ {0} and b ∈ RN , we have E(αT + b) = RN . [Hint: Use Exercise 10.3]. 6. Let C ⊂ R be the Cantor set, and let CN be the N -fold cartesian product of C. Show that CN ∈ AN . (Note that CN is nowhere dense in RN , and of N -dimensional measure zero.) 7. Let D ⊂ RN be a convex and open set, and let T ⊂ D be an arbitrary set. Show that if T ∈ D(D), or T ∈ BC , then T contains a Hamel basis. 8. Let T ⊂ RN be the set constructed in Example 10.7.1. Show that T does not contain an analytic subset containing a Hamel basis.

Chapter 11

Properties of Hamel Bases 11.1 General properties

  We recall that a Hamel basis is any base of the linear space RN ; Q; +; · . We have constructed Hamel bases already many times in this book. Theorem 4.2.1 (cf., in particular, Corollary 4.2.1) asserts that there exist Hamel bases. More exactly (Lemma 4.2.1), for every set A ⊂ C ⊂ RN such that A is linearly independent over Q, and E(C) = RN , there exists a Hamel basis H of RN such that A ⊂ H ⊂ C. In particular, every set belonging to any of the classes A = B, C, D(D), AC , BC contains a Hamel basis (Theorems 9.3.6 and 10.7.3 and Exercise 10.7). On the other hand, we have the following Theorem 11.1.1. No Hamel basis belongs to any of the classes A = B, C, D(D), AC , BC . Proof. Let H ⊂ RN be a Hamel basis. Let the function g : H → R be defined by g(h) = 1

for h ∈ H,

and let f : RN → R be the additive extension of g (Theorem 5.2.2). By Corollary 5.2.2 f is discontinuous. f | H = g = 1 is bounded and continuous, which shows that H does not belong to C, nor to BC , and hence H ∈ / A = B ⊂ C, H ∈ / D(D) ⊂ A and H∈ / AC ⊂ BC .  Theorem 4.2.3 says that every Hamel basis has the power of continuum. Consequently the linear space RN over Q is infinitely (continuum) dimensional. Every additive function f : RN → R can be prescribed arbitrarily on a Hamel basis (Theorem 5.2.2), and then uniquely extended onto RN . Now we prove the following (Kuczma [179]) Theorem 11.1.2. Let C ⊂ RN be a cone, and suppose that C contains a cone-base1 . Then C ∈ / B. 1

Cf. 4.2.

282

Chapter 11. Properties of Hamel Bases

Proof. Let B ⊂ C be a cone-base of C. Since B is linearly independent over Q, there exists a Hamel basis H of RN such that B ⊂ H. Define the function g : H → R by g(h) = −1

for h ∈ H,

and let f : RN → R be the additive extension of g. By Corollary 5.2.2 f is discontinuous. C = E + (B), consequently every x ∈ C can be written as x=

n

αi ∈ Q ∩ [0, ∞),

αi bi ,

bi ∈ B,

i = 1, . . . , n.

i=1

For such an x we have f (x) =

n

i=1

αi f (bi ) =

n

αi g(bi ) = −

i=1

n

αi  0.

i=1

Thus f is bounded above on C, which shows that C ∈ / B.



Theorem 11.1.2 justifies the claim made in 4.2 that only very small cones may have a cone-base. In particular, we obtain from Theorem 11.1.2 and Lemma 9.3.1 the following result of J. Acz´el and P. Erd˝ os [9] Corollary 11.1.1. There does not exist a cone-base of [0, ∞) ⊂ R (over Q). Let H ⊂ RN be a Hamel basis. Then of E + (H). By Theorem  + H is a cone-base + + 11.1.2 E (H) ∈ / B. But evidently R E (H) = E (H) − E + (H) = E(H) = RN . Hence we obtain (Erd˝ os [76]). Corollary 11.1.2. There exists a set T ⊂ RN which does not belong to the class B and such that R(T ) = RN . Corollary 11.1.3. The condition int R(T ) = ∅ is not sufficient for T ∈ B. This, together with Theorem 9.3.5, yields another proof of Theorem 9.1.1 (B = C).

11.2 Measure Theorem 11.2.1 below is due to W. Sierpi´ nski [276] (cf. also Marcus [216], Abian [1], Kuczma [189]). Theorem 11.2.1. Let H ⊂ RN be a Hamel basis. Then mi (H) = 0. Proof. This results from Theorems 11.1.1 and 9.3.1. Corollary 11.2.1. If H ⊂ R set, then m(A) = 0.

N



is a Hamel basis, and A ⊂ H is a Lebesgue measurable

In particular, every measurable Hamel basis has measure zero. It is worthwhile to observe that there exist measurable Hamel bases. In Chapter 9 we saw that in every space RN there exist measurable sets T ∈ A of measure zero. By Corollary 9.3.2 every such set contains a Hamel basis, which necessarily is Lebesgue measurable and of measure zero. Now let H ⊂ RN be a Hamel basis, and let H0 ⊂ H be a finite or countable subset of H. We have (Sierpi´ nski [276])

11.2. Measure

283

Theorem 11.2.2. If H ⊂ RN is a Hamel basis, and H0 ⊂ H is a finite or countable set, then the set E(H \H0 ) is saturated non-measurable and has property (∗) from 3.3. Proof. First suppose that E(H \ H0 ) is Lebesgue measurable and of measure zero, or E(H \ H0 ) is of the first category. Take an arbitrary x ∈ RN . Then x=

n

αi hi ,

αi ∈ Q,

hi ∈ H,

i = 1, . . . , n.

i=1

Let hi ∈ H \ H0 for i = 1, . . . , k, and hi ∈ H0 for i = k + 1, . . . , n. Then u=

k

αi hi ∈ E(H \ H0 ),

i=1

v=

n

αi hi ∈ E(H0 ),

i=k+1

(u = 0 if k = 0, v = 0 if k = n), and x = u + v. This means that RN = E(H \ H0 ) + E(H0 ), or  RN = E(H \ H0 ) + v. (11.2.1) v∈E(H0 )

By Lemma 4.1.3 the union in (11.2.1) is countable. Thus our supposition leads to the conclusion that RN itself is  of measure  zero, resp. of the first category, which is not the case. Consequently me E(H \ H0 ) > 0, and E(H \ H0 ) is of the second category. Next observe that every proper linear subspace of RN over R (a k-dimensional hyperplane, where k < N ) is of measure zero and of the first category (even nowhere dense). Since E(H \H0 ) = EQ (H \H0 ) ⊂ ER (H \H0 ), we must have ER (H \H0 ) = RN . By Lemma 4.2.1 there exists a base B = {b1 , . . . , bN } ⊂ H \ H0 of RN over R. The set H1 = H0 ∪ B is finite or countable, we may replace H0 by H1 in the  and therefore  argument above. Consequently me E(H \ H1 ) > 0, and E(H \ H1 ) is of the second category. In virtue of Lemma 4.2.3 the set E(B) is dense in RN . As is easily seen (cf. the argument above), we have E(H \ H0 ) = E(H \ H1 ) + E(B).   By Theorems 3.6.1 and 3.6.2 mi RN \ E(H \ H0 ) = 0 and every setT ⊂ RN \ E(H \  H0 ) with the Baire property is of the first category. Suppose that mi E(H \ H0 ) > 0, or E(H \ H0 ) contains a set of the second category and with the Baire property. By Theorem 4.3.3 there exists an additive function f : RN → R such that Ker f = E(H \H0 ). Since B ⊂ H \H0 , we have E(B) ⊂ E(H \H0 ), whence it follows that Ker f is dense in RN . If f were continuous, we would obtain hence E(H \H0 ) = Ker f = RN , which is impossible. Consequently f must be discontinuous. But f is zero (and hence bounded) on E(H \ H0 ), and thus, 9.3.1 resp. 9.3.2 it is continuous.  by Theorem  This contradiction shows that mi E(H \ H0 ) = 0 and E(H \ H0 ) cannot contain a set of the second category and with the Baire property. Consequently E(H \ H0 ) is saturated non-measurable and has property (∗) from 3.3. 

284

Chapter 11. Properties of Hamel Bases

Theorem 11.2.2 shows that the existence of non-measurable sets can be derived directly (without a use of the Axiom of Choice or of an equivalent statement) from the existence of a Hamel basis. Of course, the Axiom of Choice is inherent in the existence of a Hamel basis, but once we know that a Hamel basis exists, then we are able to construct non-measurable sets by elementary devices, without making a further appeal to the Axiom of Choice (Sierpi´ nski [276]). The next theorem (Kurepa [199]) shows a certain irregularity of the operation “+”. Theorem 11.2.3. There exist measurable sets A, B ⊂ RN such that the set A + B is non-measurable. Proof. Let H ⊂ RN be a measurable Hamel basis. Then also all sets λH with λ ∈ Q are measurable. We have ∞     RN = λ1 H + · · · + λn H (11.2.2) n=1 λ1 ,...,λn ∈Q

  cf. the proof of Theorem 9.3.7, in particular formula (9.3.10) . Put g(h) = 1 for h ∈ H, and let f : RN → R be the additive extension of g (Theorem 5.2.2). By Corollary 5.2.2 f is discontinuous. Fix λ1 , . . . , λn ∈ Q, and let μ = max(|λ1 |, . . . , |λn |). Every x ∈ λ1 H + · · · + λn H has a form x = λ1 h1 + · · · + λn hn , with h1 , . . . , hn ∈ H. For such an x we have f (x)

= =

λ1 f (h1 ) + · · · + λn f (hn ) = λ1 g(h1 ) + · · · + λn g(hn ) λ1 + · · · + λn  |λ1 | + · · · + |λn |  nμ.

/ B. Thus f is bounded above (by nμ) on λ1 H + · · · + λn H, whence λ1 H + · · · + λn H ∈ It follows in virtue of Theorem 9.3.1 that mi (λ1 H + · · · + λn H) = 0.

(11.2.3)

Consequently all summands in (11.2.2) have inner measure zero. But since the union in (11.2.2) is countable, they cannot all have the outer measure zero. Let n0 ∈ N be the minimal positive integer such that there exist λ1 , . . . , λn0 ∈ Q such that me (λ1 H + · · · + λn0 H) > 0. (11.2.4) By what has been said at the beginning of the proof n0 > 1. It follows from (11.2.3), (11.2.4) and Theorem 3.1.2 that the set λ1 H + · · · + λn0 H is non-measurable, whereas by the minimal property of n0 and by Theorem 3.1.3 the set λ1 H + · · · + λn0 −1 H is measurable (and of measure zero). Also the set λn0 H is measurable. Now, we have λ1 H + · · · + λn0 H = (λ1 H + · · · + λn0 −1 H) + (λn0 H).

(11.2.5)

Our theorem results now from (11.2.5) with A = λ1 H + · · · + λn0 −1 H and B = λn0 H. 

11.3. Topological properties

285

11.3 Topological properties Theorems 11.3.1, 11.3.2 and Corollary 11.3.1 (Sierpi´ nski [282]; cf. also Marcus [216]) are topological analogues of Theorems 11.2.1 and 11.2.3 and of Corollary 11.2.1. Theorem 11.3.1 results from Theorems 11.1.1 and 9.3.2, whereas the proof of Theorem 11.3.2 is identical as that of Theorem 11.2.3. Theorem 11.3.1. Every subset T of a Hamel basis such that T has the Baire property, is of the first category. Corollary 11.3.1. If a Hamel basis H ⊂ RN has the Baire property, then H is of the first category. Again we may observe that there exist Hamel bases H ⊂ RN which are of the first category, or even nowhere dense (and hence have the Baire property), since in every space RN there exist sets T ∈ A (and consequently containing a Hamel basis; cf. Corollary 9.3.2) which are nowhere dense in RN . Theorem 11.3.2. There exist sets A, B ⊂ RN with the Baire property and such that A + B does not have the Baire property. As an immediate consequence of Theorems 11.1.1 and either Theorem 9.3.7 or Theorem 9.8.1 we obtain the following (Sierpi´ nski [276], Jones [158]) Theorem 11.3.3. No Hamel basis is an analytic set. From Theorem 11.3.3 and Theorem 2.5.6 we immediately get (Sierpi´ nski [276]) Corollary 11.3.2. No Hamel basis is a Borel set. Also the topological part of Theorem 11.1.2 belongs in this section.

11.4 Burstin bases In 11.2–11.3 we saw that there exist very small Hamel bases: of measure zero and nowhere dense. Now we will try to answer the question: how large a Hamel basis can be. Theorems 11.1.1, 11.2.1 and 11.3.1 suggest that not too large. However, as we will presently see, there exist quite large Hamel bases. Let X ⊂ RN . A Hamel basis H ⊂ X is called a Burstin basis relative to X iff H intersects every uncountable Borel subset of X. A Burstin basis relative to RN is simply called a Burstin basis. (After C. Burstin, who in Burstin [39] first considered similar constructions. Cf. also Abian [1], Kuczma [189]).   Theorem 11.4.1. If the set X ⊂ RN is of the full measure i.e., m(RN \ X) = 0 , then every Burstin basis relative to X is saturated non-measurable. Proof. Let H ⊂ X be a Burstin basis relative to X, and suppose that H is not saturated non-measurable. In view of Theorem 11.2.1 this means that mi (RN \H) > 0. Thus there exists a closed set F ⊂ RN \ H with m(F ) > 0. Since H ⊂ X, we have

RN \H = (RN \X)∪(X\H), whence F = F ∩(RN \H) = F ∩(RN \X) ∪ F ∩(X\H) , and

F ∩ (X \ H) = F \ F ∩ (RN \ X) . (11.4.1)

286

Chapter 11. Properties of Hamel Bases

Since m(RN \ X) = 0, also the set F ∩ (RN \ X) is measurable and of measure zero.  F is measurable, being closed. Evidently F ∩ RN \ X ⊂ F , thus (11.4.1) implies



m F ∩ (X \ H) = m(F ) − m F ∩ (RN \ X) = m(F ) > 0. Consequently there exists a closed set K ⊂ F ∩ (X \ H) with m(K) > 0. Hence it follows that K is uncountable. (Otherwise we would have m(K) = 0). K, being closed, is a Borel set, and we have K ⊂ X. Thus by the definition of a Burstin basis we have H ∩ K = ∅, which contradicts the relation K ⊂ X \ H.  A topological analogue of Theorem 11.4.1 is the following Theorem 11.4.2. If the set X ⊂ RN is residual, then every Burstin basis relative to X fulfils condition (∗) from 3.3. Proof. Let H ⊂ X be a Burstin basis relative to X, and suppose that H does not fulfil (∗). In view of Theorem 11.3.1 this means that the set RN \ H contains a subset E of the second category and with the Baire property. Thus E = (G ∪ P ) \ R, where G is open and non-empty (since E is of the second category), and the sets P, R are of the first category. G, being open, is in particular of class Gδ and

G \ (G ∩ E) = G \ E = G ∩ E  = G ∩ (G ∪ P ) ∪ R ⊂ G ∩ R ⊂ R, whence G \ (G ∩ E) is of the first category. By Theorem 2.1.3 there exists a set F ⊂ G ∩ E ⊂ E such that F ∈ Gδ and G \ F is of the first category, whence it follows that F is of the second category. We have F ⊂ E ⊂ RN \ H, whence  



F = F ∩ RN \ H = F ∩ (X \ H) ∪ F ∩ (RN \ X) , N whence (11.4.1) follows.

Since X isNresidual,  R \ X is of the first category, and hence also F \ F ∩ (X \ H) = F ∩ R \ X is of the first category. Again by Theorem 2.1.3 there exists a set K ⊂ F ∩ (X \ H) such that K ∈ Gδ and F \ K is of the first category. Consequently K is of the second category, and thus uncountable. Since K belongs to Gδ , it is a Borel set, and we have K ⊂ X. Consequently H ∩ K = ∅, which contradicts the relation K ⊂ X \ H. 

Taking X = RN we obtain from Theorems 11.4.1 and 11.4.2 Corollary 11.4.1. Every Burstin basis is saturated non-measurable and has property (∗) from 3.3. We see that a Burstin basis yields an example of a comparatively “large” Hamel basis. The obvious question arises as to the existence of a Burstin basis. Presently we will answer this question. Theorem 11.4.3. Let X ⊂ RN be a Borel set such that E(X) = RN . Then there exists a Burstin basis relative to X.

11.4. Burstin bases

287

Proof. Let F be the family of all uncountable Borel subsets of X. Note that if we had card X  ℵ0 , then by Lemma 4.1.3 we would have card RN = card E(X) = ℵ0 . Consequently X must be uncountable, and thus X ∈ F. Consequently F = ∅. By Theorem 2.3.4 we get, since F ⊂ B(RN ), card F  card B(RN ) = c.

(11.4.2)

On the other hand, for every x ∈ X the set X \ {x} is an!uncountable Borel subset of " X, and hence belongs to F . Thus F contains the family X \ {x} x∈X . Consequently card F  card X = c

(11.4.3)

(cf. Theorem 2.8.3). Relations (11.4.2) and (11.4.3) show that card F = c. Let (cf. 1.5) A = {α ∈ M (c) | α = c}, and let γ be the smallest element of A. It follows from Theorem 1.7.1 that A = ∅, and thus γ exists in virtue of Theorem 1.4.3. Thus γ = c, but for every ordinal number α < γ we have α < c. Since card Γ(γ) = γ = c = card F , there exists a one-to-one mapping f : Γ(γ) → F which is onto. For every α ∈ Γ(γ) write Fα = f (α) ∈ F. Thus we have arranged F into a transfinite sequence F = {Fα }α<γ of type γ. Now we define atransfinite sequence {bα }α<γ of elements of X, and for every α < γ we put Bα = {bβ }. As b0 we take an arbitrary element of F0 with the only restriction b0 = 0:

β<α

  b0 = w F0 \ {0} ,

(11.4.4)

where w : P(X) \ {∅} → X is a choice function (cf. 1.1). Having already defined bβ for β < α < γ, we take   bα = w Fα \ E(Bα ) . (11.4.5) Since card Bα = α < c, we have by Lemmas 4.1.3 and 4.2.2 card E(Bα ) < c, whereas by Theorem 2.8.3 card Fα = c. Consequently Fα \ E(Bα ) = ∅, and bα is well defined. By Theorem 1.6.2 there exists a transfinite sequence {bα }α<γ fulfilling (11.4.4) and   (11.4.5) for every α < γ. Let B = {bα } = Bα . We will prove by transfinite α<γ

α<γ

induction that B is linearly independent over Q.  First we prove that for every α < γ the set Bα∗ = Bα ∪ {bα } = {bβ } is βα

linearly independent. For α = 0 we have B0∗ = {b0 }, which is linearly independent, since b0  = 0. Assuming that Bβ∗ are linearly independent forβ < α, we check that the set Bβ∗ also is linearly independent. In fact, let C ⊂ Bβ∗ be any finite set, β<α

β<α

C = {bβ1 , . . . , bβk }, where βi < α for i = 1, . . . , k. Put β = max(β1 , . . . , βk ) < α. We have βi  β for i = 1, . . . , k, and since Bβ∗  ⊂ Bβ∗  for β   β  , we have Bβ∗i ⊂ Bβ∗ for i = 1, . . . , k, and so bβi ∈ Bβ∗i ⊂ Bβ∗ for i = 1, . . . , k. Hence C ⊂ Bβ∗ . Since β < α, the

288

Chapter 11. Properties of Hamel Bases

set Bβ∗ is linearly independent, and consequently also C is linearly independent. This  ∗ shows that the set Bβ is linearly independent.  ∗ β<α Now, Bβ = Bα , and by (11.4.5) bα ∈ / E(Bα ). By Lemma 4.1.2 the set β<α

Bα∗ = Bα ∪ {bα } is linearly independent. Thus by Theorem 1.6.1 the sets Bα∗ are independent for all α < γ. By  linearly the same argument as above the set B = Bα∗ is linearly independent. α<γ

By Lemma 4.2.1 there exists a Hamel basis H such that B ⊂ H ⊂ X. Let F ⊂ X be an uncountable Borelset. Then F ∈ F, and hence there exists an α < γ such that F = Fα . Then bα ∈ F cf. (11.4.5) , and bα ∈ B ⊂ H, whence bα ∈ F ∩ H, and F ∩ H = ∅. This proves that H is a Burstin basis relative to X.  Taking in Theorem 11.4.3 X = RN , we obtain Corollary 11.4.2. There exists a Burstin basis. Let D ⊂ RN be a dense and countable set. Since D is countable, m(D) = 0, and hence for every n ∈ N there exists an open set Gn such that D ⊂ Gn and ∞ m(Gn ) < n−1 . Put A = Gn , B = RN \ A. Then A ∈ Gδ , and hence B ∈ Fσ . Since n=1

D ⊂ A, A is dense and int B = ∅. By Lemma 2.1.1 B is of the first category, i.e., A is residual. (In particular, A is of the second category and with the Baire property, whence by Theorem 9.3.2 A ∈ A). We have m(A)  m(Gn ) < n−1 for n ∈ N, whence m(A) = 0. Hence B is of the full measure. (In particular, m(B) = ∞ > 0, whence by Theorem 9.3.1 B ∈ A). Thus A and B are uncountable Borel sets, and by Corollary 9.3.2 E(A) = E(B) = RN . By Theorem 11.4.3 A contains a Burstin basis H1 relative to A, and B contains a Burstin basis H2 relative to B. By Theorem 11.4.1 H2 is saturated non-measurable, whereas by Theorem 11.4.2 H1 has property (∗) from 3.3. Since m(A) = 0, also H1 has measure zero, and since B is of the first category, so is also H2 . Hence we obtain Corollary 11.4.3. There exists a Hamel basis which is of measure zero and has property (∗) from 3.3, and there exists a Hamel basis which is of the first category and is saturated non-measurable.

11.5 Erd˝ os sets Let H ⊂ RN be an arbitrary Hamel basis. With every such a basis we associate the set Z(H) defined as follows (Erd˝ os [76]) Z(H) = {x ∈ RN | x = κ1 h1 + · · · + κn hn , κi ∈ Z, hi ∈ H, i = 1, . . . , n; n ∈ N}. (11.5.1) The set Z(H) will be referred to as the Erd˝ os set associated with the basis H (Kuczma [188]). Before proving more sophisticated properties of Erd˝ os sets let us first observe some simple facts.

11.5. Erd˝ os sets

289

Lemma 11.5.1. For every Hamel basis H ⊂ RN we have Z(H) + Z(H) = Z(H).

(11.5.2)

Proof. If x and y are linear combinations with integral coefficients of elements of H, then so is also x + y, which shows that Z(H) + Z(H) ⊂ Z(H).   On the other hand, always 0 ∈ Z(H) take in (11.5.1) all κi = 0 , whence Z(H) = Z(H) + 0 ⊂ Z(H) + Z(H).

(11.5.3)

(11.5.4) 

(11.5.3) and (11.5.4) yield (11.5.2).

Lemma 11.5.2. For every Hamel basis H ⊂ RN and for every x ∈ RN there exists a k ∈ N such that kx ∈ Z(H). Proof. There exist λ1 , . . . , λn ∈ Q and h1 , . . . , hn ∈ H such that x = λ1 h1 + · · · + λn hn . Let k be a common multiple of the denominators of λ1 , . . . , λn . Then kλ1 , . . . , kλn ∈ Z, and so by (11.5.1) kx = kλ1 h1 + · · · + kλn hn ∈ Z(H).



Corollary 11.5.1. For every Hamel basis H ⊂ RN we have RN =

∞  1 Z(H). n n=1

(11.5.5)

Proof. By Lemma 11.5.2 for every x ∈ RN there exists a k ∈ N such that x∈ This shows that RN ⊂

∞  1 1 Z(H) ⊂ Z(H). k n n=1

∞ 1  Z(H). The converse inclusion is obvious. n=1 n



Lemma 11.5.3. For every Hamel basis H ⊂ RN the set Z(H) is dense in RN . Proof. Take an arbitrary x ∈ RN . By Lemma 11.5.2 there exist  k, l ∈ Z  such that √ √ l√ kx ∈ Z(H) and 2 lx ∈ Z(H). Hence also pkx + ql 2 x = k p + q 2 x ∈ Z(H) k for every p, q ∈ Z. But by Lemma 3.8.1 the set  l √ t∈R|t=p+q 2, p, q ∈ Z k

290

Chapter 11. Properties of Hamel Bases

is dense in R, so there exist sequences {pn }n∈N and {qn }n∈N , pn , qn ∈ Z for n ∈ N, such that   l √ 1 lim pn + qn 2 = . n→∞ k k   l √ Put xn = pn + qn 2 x, n ∈ N. Then kxn ∈ Z(H) for n ∈ N and lim kxn = x. n→∞ k N Consequently Z(H) is dense in R .  Theorem 11.5.1. For every Hamel basis H ⊂ RN the set Z(H) is saturated nonmeasurable and has property (∗) from 3.3. 1 Proof. Take an h0 ∈ H. Then, for every x ∈ Z(H), we have h0 + x ∈ / Z(H), i.e., 2 1 1 Z(H) is disjoint with h0 + Z(H). By Lemma 11.5.3 the set h0 + Z(H) is dense 2 2 in RN , consequently int Z(H) = ∅. (11.5.6)   It follows from (11.5.2), (11.5.6), and Theorems 2.9.1 and 3.7.1 that mi Z(H) = 0 and Z(H) does not contain   any subset of the second category and with the Baire property. If we had m Z(H) = 0, or if Z(H) were of the first category, then according N to (11.5.5)   the same would be true about R , which is not the case. Consequently me Z(H) > 0 and Z(H) is of the second category. Theorems 3.6.1 and 3.6.2 now imply in virtue of relation (11.5.2) and Lemma  11.5.3 that mi RN \ Z(H) = 0 and RN \ Z(H) does not contain any subset of the second category and with the Baire property. Hence the theorem follows.  Independently of how small Hamel basis H we start from, Theorem 11.5.1 asserts that the corresponding Erd˝ os set is always large. This fact is also reflected in the relation Z(H) ∈ A. (11.5.7) Relation (11.5.7) (Ger [98]) has already been mentioned in the proof of Lemma 9.3.1, but we will prove here a stronger result (Kuczma [188]; cf. also Kuczma [180]). Lemma 11.5.4. For every Hamel basis H ⊂ RN and every non-empty open set G ⊂ RN we have G ∩ Z(H) ∈ A. (11.5.8) Proof. First consider the case where 0 ∈ G. Let f : RN → R be a discontinuous additive function. Since RN = E(H) = EQ (H) ⊂ ER (H) ⊂ RN , we have ER (H) = RN , and thus, by Corollary 4.2.2, H contains a base {b1 , . . . , bN } of RN over R. Every x ∈ RN can be uniquely represented as x = x1 b1 + · · · + xN bN , and x1 , . . . , xN depend on x in a continuous manner. We have, since f is additive, f (x) = f (x1 b1 ) + · · · + f (xN bN ) = ϕ1 (x1 ) + · · · + ϕN (xN ),

11.5. Erd˝ os sets

291

where ϕi (t) = f (tbi ). If all functions ϕ1 , . . . , ϕN were continuous, so would also be f , so there exists an i0 , 1  i0  N , such that ϕi0 is discontinuous. In the sequel we suppress the index i0 and write ϕ instead of ϕi0 and b instead of bi0 . Note that b ∈ H. It is easily seen that ϕ : R → R is an additive function. If we had ϕ(t) = ϕ(1)t, ϕ would be continuous, consequently there exists a u ∈ R such that ϕ(u) = ϕ(1)u. Write ϕ(u) = ϕ(1)u + c (11.5.9) so that c = 0. It follows from Theorem 5.2.1 that u is irrational. By Lemma 11.5.2 there exists a k ∈ N such that kub ∈ Z(H), and clearly ku is also irrational. For every p, q ∈ Z we have pb + qkub ∈ Z(H). Write D = {α ∈ R | α = p + qku, p, q ∈ Z}. By Lemma 3.8.1 the set D is dense in R, so there exists a sequence {αn }n∈N of elements of D such that αn = 0 for n ∈ N, (i) lim αn = 0.

(ii)

n→∞

We have also αn b ∈ Z(H)

for n ∈ N.

(iii)

Every αn can be written as pn + qn ku, with pn , qn ∈ Z, n ∈ N. By (ii) the sequence {αn } is bounded. If the sequence {qn } were bounded, also {pn } would have to be bounded, and thus the coefficients pn and qn would assume only a finite number of different values. Thus also the sequence {αn } could only have a finite number of different terms, which is incompatible with (i) and (ii). Thus the sequence {qn } must be unbounded. Note that properties (i)–(iii) of {αn } remain unchanged if we replace the sequence {αn } by {−αn }, or by a subsequence. Consequently we may assume that the sequence {αn } has been chosen in such a manner that lim c qn = +∞. (11.5.10) n→∞

Now, we have by Theorem 5.2.1 and by (11.5.9) f (αn b) = f (pn b + qn kub) = f (pn b) + f (qn kub) = pn f (b) + qn kf (ub) = pn ϕ(1) + qn kϕ(u) = (pn + qn ku)ϕ(1) + qn kc = αn ϕ(1) + qn kc, whence by (ii) and (11.5.10) lim f (αn b) = +∞.

n→∞

(11.5.11)

By (ii) αn b ∈ G for n sufficiently large, and thus in view of (iii) αn b ∈ G ∩ Z(H) for n sufficiently large. Relation (11.5.11) shows now that f is unbounded above on G ∩ Z(H).

292

Chapter 11. Properties of Hamel Bases

Consequently every discontinuous additive function f : RN → R is unbounded above on G ∩ Z(H). Hence it follows that G ∩ Z(H) ∈ B. By Theorem 10.2.2 we obtain hence (11.5.8). Now let 0 ∈ / G, and take an h ∈ G ∩ Z(H). (Such an h exists in virtue of Lemma 11.5.3). G − h is a non-empty open set and 0 ∈ G − h, so by the first part of the proof we have (G − h) ∩ Z(H) ∈ A. Next   (G − h) ∩ Z(H) + h = G ∩ Z(H) + h = G ∩ Z(H), for2 by Lemma 11.5.1 Z(H) ⊂ Z(H)+ h ⊂ Z(H)+ Z(H) ⊂ Z(H), and so Z(H)+ h = Z(H). Consequently (G − h) ∩ Z(H) = G ∩ Z(H) − h. By Theorems 9.2.8 and 9.2.1 we obtain hence (11.5.8).  Note that we obtain (11.5.7) from (11.5.8) taking G = RN . Lemma 11.5.5. Let H ⊂ RN be an arbitrary Hamel basis, and let f : RN → R be a discontinuous additive function. Then, for every M ∈ R, the set AM = {x ∈ Z(H) | f (x) > M }

(11.5.12)

is saturated non-measurable and has property (∗) from 3.3. Proof. It follows from (11.5.7) that there exists an x0 ∈ Z(H) such that d = f (x0 ) > 0. Write Bk = { x ∈ Z(H) | kd < f (x)  (k + 1)d }, k ∈ Z. First we prove that Bk+1 = Bk + x0 ,

k ∈ Z.

(11.5.13)

In fact, if x ∈ Bk + x0 , then x = t + x0 , where t ∈ Bk . Thus t ∈ Z(H) and kd < f (t)  (k + 1)d. Then by (11.5.2) x = t + x0 ∈ Z(H) + Z(H) = Z(H), and f (x) = f (t) + f (x0 ) = f (t) + d so that (k + 1)d < f (x)  (k + 2)d. Consequently x ∈ Bk+1 , and Bk + x0 ⊂ Bk+1 . Conversely, if x ∈ Bk+1 , then x ∈ Z(H) and (k + 1)d < f (x)  (k + 2)d. Write t = x − x0 . Then, since evidently −Z(H) = Z(H), we have t ∈ Z(H)−Z(H) = Z(H)+Z(H) = Z(H), and f (t) = f (x)−f (x0 ) = f (x)−d so that kd < f (t)  (k + 1)d. Consequently t ∈ Bk and x = t + x0 ∈ Bk + x0 , and Bk+1 ⊂ Bk + x0 . This proves (11.5.13). Relation (11.5.13) shows that all the sets Bk are congruent to each other under translation, and hence they all have the same measure and category. Clearly +∞ 

Bk = Z(H).

(11.5.14)

k=−∞

If the sets Bk had measure zero, or if they all were of the first category, then by (11.5.14) the same would be true about Z(H), which contradicts Theorem 11.5.1. Consequently me (Bk ) > 0 for k ∈ Z, and all Bk are of the second category, k ∈ Z. 2 If h ∈ Z(H), then also −h ∈ Z(H), whence Z(H) − h ⊂ Z(H) + Z(H) = Z(H), and Z(H) ⊂ Z(H) + h.

11.5. Erd˝ os sets

293

For k  M/d we have Bk ⊂ AM . It follows that, for every M ∈ R, me (AM ) > 0 and AM is of the second category. Suppose that for a non-empty open set G ⊂ RN we have G ∩ AM = ∅. This means that f (x)  M for x ∈ G ∩ Z(H). Lemma 11.5.4 would then imply that f is continuous, which is not the case. This shows that, for every M ∈ R, the set AM is dense in RN . Take arbitrary x, y ∈ A 12 M . Then x ∈ Z(H) and y ∈ Z(H), whence x + y ∈ 1 1 Z(H) + Z(H) = Z(H) (cf. Lemma 11.5.1), and f (x) > M and f (y) > M , whence 2 2 1 1 f (x + y) = f (x) + f (y) > M + M = M . Consequently x + y ∈ AM , and 2 2 A 12 M + A 12 M ⊂ AM .

(11.5.15)

  By Theorems 3.6.1 and 3.6.2 mi RN \ (A 12 M + A 12 M ) = 0 and the set RN \ (A 12 M + A 12 M ) does not contain any subset of the second category and with the Baire property. It follows from (11.5.15) that also mi (RN \ AM ) = 0 and the set RN \ AM does not contain any subset of the second category with the Baire property. According to definition (11.5.12) we have AM ⊂ Z(H). In virtue of Theorem 11.5.1 mi (AM ) = 0 and AM does not contain any subset of the second category with the Baire property. Consequently AM is saturated non-measurable and has property (∗) from 3.3.  Now we can improve on Lemma 11.5.4. Theorem 11.5.2. Let H ⊂ RN be an arbitrary Hamel basis, and let A ⊂ RN be an arbitrary measurable set of positive measure. Then A ∩ Z(H) ∈ A.

(11.5.16)

Proof. Let f : RN → R be a discontinuous additive function, and let the sets AM be defined by (11.5.12). By Lemma 11.5.5 and Theorem 3.3.1 we have A ∩ AM = ∅ for every M ∈ R, which means that f is unbounded above on A ∩ Z(H). Thus every discontinuous additive function f : RN → R is unbounded above on A ∩ Z(H). This means that A ∩ Z(H) ∈ B. By Theorem 10.2.2 we obtain (11.5.16).  Theorem 11.5.3. Let H ⊂ RN be an arbitrary Hamel basis, and let A ⊂ RN be an arbitrary set of the second category and with the Baire property. Then we have (11.5.16). Proof. Let f : RN → R be a discontinuous additive function, and let the sets AM be defined by (11.5.12). By Lemma 11.5.5 and Corollary 3.3.1 we have A ∩ AM = ∅ for every M ∈ R, and we argue as in the proof of Theorem 11.5.2.  Theorem 11.5.4. Let H ⊂ RN be an arbitrary Hamel basis, and let A ⊂ RN be an arbitrary measurable set of positive measure. Then A ∩ Z(H) ∈ AC .

(11.5.17)

294

Chapter 11. Properties of Hamel Bases

Proof. Let D ⊂ RN be a convex and open set such that A ∩ Z(H) ⊂ D, and let f : D → R be an arbitrary convex function such that the restriction f | A ∩ Z(H) is continuous. By the Lebesgue density theorem3 (cf. also Theorem 3.5.1) m(A ∩ A∗ ) = m(A) > 0. By Theorems 11.5.1 and 3.3.1 A ∩ A∗ ∩ Z(H) = ∅. Thus there exists a point a ∈ A ∩ Z(H) which is a density point of A. Further, because of the continuity of f | A ∩ Z(H), there exists a neighbourhood U of a in RN such that the function f is bounded on U ∩ A ∩ Z(H). Since a is a density point of A and U is a neighbourhood of a, m(U ∩ A) > 0. By Theorem 11.5.2 U ∩ A ∩ Z(H) ∈ A, whence it follows that f is continuous. This proves (11.5.17).  Theorem 11.5.5. Let H ⊂ RN be an arbitrary Hamel basis, and let A ⊂ RN be an arbitrary set of the second category and with the Baire property. Then we have (11.5.17). Proof. Let D ⊂ RN be a convex and open set such that A ∩ Z(H) ⊂ D, and let f : D → R be an arbitrary convex function such that the restriction f | A ∩ Z(H) is continuous. By Theorems 2.1.5, 2.3.1 and 2.2.1 the set4 A ∩ D(A) has the Baire property. Moreover, by Theorem 2.1.6 the set A\ D(A) is of the first category, whence A ∩ D(A) is of the second category, in view of the relation A = A ∩ D(A) ∪ A \

D(A) . By Theorem 11.5.1 and Corollary 3.3.1 we have A ∩ D(A) ∩ Z(H) = ∅. Thus there exists a point a ∈ A ∩ Z(H) at which the set A is locally of the second category. Further, there exists a neighbourhood U of a in RN such that the function f is bounded on U ∩ A ∩ Z(H). Since A is of the second category at a and U is a neighbourhood of a, the set U ∩ A is of the second category, and since U is open, U ∩ A has the Baire property (cf. Theorems 2.3.1 and 2.2.1). By Theorem 11.5.3 U ∩ A ∩ Z(H) ∈ A, whence it follows that f is continuous. This proves (11.5.17).  Taking in (11.5.17) A = RN we obtain from Theorem 11.5.4 or 11.5.5, for an arbitrary Hamel basis H ⊂ RN , Z(H) ∈ AC . Let us observe that Theorems 9.3.1, 9.3.2, 9.8.3 and 9.8.4 are immediate consequences of Theorems 11.5.2, 11.5.3, 11.5.4 and 11.5.5, respectively. Another example of a set T ⊂ R such that A ∩ T ∈ A for every measurable set A ⊂ R of positive measure can be found in Kuczma-Sm´ıtal [190].

11.6 Lusin sets In 11.4 we saw that there exist “large” Hamel bases. In 11.5 we encountered the Erd˝os set Z(H) associated with the Hamel basis H, which is “large” independently of what Hamel basis H we start with. The fact that in the definition of Z(H) there occur positive as well as negative coefficients is essential, as we will presently see. In this section we meet very small Hamel bases. 3 4

A∗ is the set of the density points of A; cf. 3.5. D(A) is the set of those points at which A is locally of the second category; cf. 2.1.

11.6. Lusin sets

295

Lusin sets were defined in 3.4. W. Sierpi´ nski [282] proved, assuming the continuum hypothesis (cf. 1.5) that there exists a Hamel basis which is a Lusin set. This is a simple corollary from the following theorem of P. Erd˝ os [76]. Theorem 11.6.1. Under assumption of the continuum hypothesis there exists a Hamel basis H ⊂ RN such that E + (H) is a Lusin set. Proof. Let F be the collection of all nowhere dense perfect sets in RN . By Theorem 2.3.4 card F  c. If N > 1, then every closed segment in RN is a nowhere dense perfect set, and there are continuum many closed segments in RN , whence card F  c. If N = 1, then every translation C + x, x ∈ R, of the Cantor set C is a nowhere dense perfect set, and there are continuum many such translations. We argue further as in the preceding case and arrive at the inequality card F  c. Thus finally we get card F = c. Under assumption of the continuum hypothesis we have (cf. 1.5) card Γ(Ω) = Ω = ℵ1 = c, so there exist functions, one-to-one and onto, f : Γ(Ω) → F and g : Γ(Ω) → RN . For α ∈ Γ(Ω) we put Fα = f (α), xα = g(α) so that F = {Fα }α<Ω , RN = {xα }α<Ω . Put for α < Ω  F (α) = Fβ . βα

Since the sets Fβ are nowhere dense, and there are only countably many β  α, the sets F (α) are of the first category. We will define a transfinite  sequence {hα }α<Ω ofα linearly independent elements of RN , and we write Hα = {hβ } (H0 = ∅), and E = E(Hα ), α < Ω. Since there β<α

are at most countably many β < α (< Ω), the sets Hα are at most countable, and hence, by Lemma 4.1.3, the sets E α are countable (α < Ω). Suppose that hβ have already been defined for β < α < Ω.  If α = α0, no hβ have yetNbeenαdefined so that 0 H0 = ∅, and we assume E = {0} . Since E is countable, R \ E = ∅, so there exists the smallest δ < Ω such that xδ ∈ / E α . We seek elements u, v ∈ RN such that (i) u − v = xδ ; (ii) For every h ∈ E α \ {0} the set {u} ∪ {v} ∪ {h} is linearly independent (over Q); (iii) For every h ∈ E α and  ,  ∈ Q  u +  v + h ∈ F (α) implies  +  = 0. Inserting from (i) u = v + xδ , we may express (ii) and (iii) as follows: (iv) For every h ∈ E α and σ  , σ  ∈ Q such that |σ  | + |σ  | > 0 (σ  + σ  ) v + σ  xδ + h = 0; (v) For every h ∈ E α and  ,  ∈ Q such that  +  = 0 / F (α) . ( +  ) v +  xδ + h ∈

296

Chapter 11. Properties of Hamel Bases

If for some σ  , σ  ∈ Q, v ∈ RN , and h ∈ E α we have (σ  + σ  ) v + σ  xδ + h = 0, then σ  + σ  = 0, since otherwise we would have σ  xδ + h = 0, whence xδ ∈ E α , which is untrue. Thus (iv) may be expressed as v = − or



v∈ /

σ , σ ∈Q σ + σ =0

σ  xδ + h , σ  + σ 

  σ  xδ + E α −  . σ + σ 

(11.6.1)

Similarly, (v) may be expressed as v∈ /



 F (α) −  xδ − h .  +  α

(11.6.2)

 ,  ∈Q h∈E  +  =0

Every set under the union sign in (11.6.1) is countable, since E α is countable, and hence the countable union of such sets is countable, and so of the first category. Similarly, every set under the union sign in (11.6.2) is of the first category, just like F (α) , and hence countable union of such sets is of the first category. Therefore the set    σ  xδ + E α   F α −  xδ − h −  ∪ Bα = σ + σ   +      α σ , σ ∈Q σ + σ =0

 ,  ∈Q h∈E  +  =0

is of the first category, and therefore RN \ Bα = ∅. Let w : P(RN ) \ {∅} → RN be a choice function. If we put v = w(RN \ Bα ),

u = v + xδ ,

then u, v so defined will satisfy conditions (i)–(iii). Next we define hα = u, hα+1 = v.

(11.6.3)

In this way, by transfinite induction  (cf. Theorem 1.6.2), we have definedNa sequence {h } of linearly independent compare condition (ii) elements of R . Let H = α α<Ω   {hα } = Hα . Then the set H is linearly independent5 . We will show that α<Ω

α<Ω

E(H) = RN .

(11.6.4)

Supposing (11.6.4) untrue, there exists an x ∈ RN \ E(H). Then x = xγ for a certain γ < Ω. Since Hα ⊂ H for every α < Ω, we have xγ ∈ / E(Hα ) = E α for every 5

If H were linearly dependent, then there would exist an α < Ω such that Hα , and hence also Hα+2 (note that Hα ⊂Hα+2 ) is linearly dependent. But Hα+2 = Hα ∪ {hα } ∪ {hα+1 } = {u} ∪ {v} ∪ Hα ⊂  {u} ∪ {v} ∪ E α cf. (11.6.3) . Thus {u} ∪ {v} ∪ E α would be linearly dependent, which contradicts (ii).

11.6. Lusin sets

297

α < Ω. Since in the induction step described above the index δ was minimal such that xδ ∈ / E α , we must have δ  γ every time. But there are at most countably many δ  γ < Ω, whereas the induction involves uncountably many steps at every step we define only two elements of the sequence {hα }α<Ω , and the whole sequence has Ω = c elements, and at every step we must take different δ, since by (i) xδ = hα − hα+1 ∈  E(Hα+2 ) = E α+2 . The contradiction obtained shows that (11.6.4) actually holds. Thus the set H is linearly independent and fulfils (11.6.4). Consequently H is a Hamel basis. It remains to show that E + (H) is a Lusin set. Note that H ⊂ E + (H), and since, by Theorem 4.2.3 H is uncountable, also E + (H) is uncountable. Let F ⊂ RN be a nowhere dense perfect set. Then there exists an α < Ω such that F = Fα ⊂ F (α) . Let x ∈ E + (H) ∩ F (α) . Then x = 1 hα1 + · · · + n hαn ,

(11.6.5)

where i are rational and positive, and hαi ∈ H, i = 1, . . . , n. We may arrange hαi in such a manner that α1 < · · · < αn . Suppose that αn > α. hαn is either u or v in our construction, and possibly adding in (11.6.5) the suitable term with the coefficient zero, we may assume that hαn−1 = u, hαn = v. But then hα1 , . . . , hαn−2 ∈ Hαn , and h = 1 hα1 + · · · + n−2 hαn−2 ∈ E(Hαn ) = E αn . We have x = h + n−1 u + n v ∈ F (α) ⊂ F (αn ) , whence by (iii) n−1 + n = 0. But n−1 , n are non-negative, and at least one of them is strictly positive. Hence n−1 + n > 0. This contradiction shows that αn  α. But there are at most countably many 1 , . . . , n ∈ Q and α1 < · · · < αn  α < Ω. Consequently the set E + (H) ∩ F (α) is at most countable, and hence also the set E + (H) ∩ Fα = E + (H) ∩ F is at most countable. Consequently E + (H) is a Lusin set.  Corollary 11.6.1. Under assumption of the continuum hypothesis there exists a Hamel basis H ⊂ RN which is a Lusin set. Proof. Let H be the basis whose existence is ensured by Theorem 11.6.1. Then E + (H) is a Lusin set, H ⊂ E + (H), and by Theorem 4.2.3 H is uncountable. Consequently (cf. 3.4) H also is a Lusin set.  By a slight modification of the above construction we obtain the following (Sm´ıtal [286]) Lemma 11.6.1. Under assumption of the continuum hypothesis, there exists a linearly independent set H (1) ⊂ RN such that E(H (1) ) is a Lusin set. Proof. Let F = {Fα }α<Ω , {xα }α<Ω , and F (α) have the same meaning as in the proof of Theorem 11.6.1. By the transfinite induction we define a transfinite sequence  (1) {hα }α<Ω of linearly independent elements of RN . Write Hα = {hβ } for α < Ω. β<α

Suppose that hβ have already been defined for β < α < Ω. We seek a u ∈ RN with the properties

298

Chapter 11. Properties of Hamel Bases (1)

(vi) {u} ∪ Hα is linearly independent; (1) (vii) For every h ∈ E(Hα ) and  ∈ Q u + h ∈ F (α) implies  = 0. (1)

By Lemma 4.1.2 condition (vi) will be fulfilled if u ∈ / E(Hα ), and by Lemma 4.1.3   (1) (1) E(Hα ) is at most countable we assume E(H0 ) = {0} , and hence of the first category. Similarly, condition (vii) means u∈ /





∈Q h∈E(H (1) ) α  =0

F (α) − h . 

(11.6.6)

Again, the set occurring in (11.6.6) is of the first category. Thus we may take  % $   F (α) − h N (1) u = w R \ E(Hα ) ∪ ,  (1) ∈Q h∈E(H ) α  =0

where w : P(RN ) \ {∅} → RN is a choice function. Then we put hα = u. Thus we have defined by transfinite induction a sequence {hα }α<Ω , and the set   (1) (1) H = {hα } = Hα is linearly independent. But this time it is not true that α<Ω

α<Ω

E(H (1) ) = RN , so H (1) is not a Hamel basis. Actually E(H (1) ) is a Lusin set. To see this, first observe that card H (1) = Ω = c, so H (1) , and hence also (1) E(H ), is uncountable. Next, suppose that for a certain α < Ω x ∈ E(H (1) ) ∩ Fα ⊂ E(H (1) ) ∩ F (α) .

(11.6.7)

Then we have (11.6.5), where i ∈ Q, i = 0, hαi ∈ H (1) , i = 1, . . . , n; α1 < · · · < αn . (1) Suppose that αn > α, hαn is a u from our construction, and hα1 , . . . , hαn−1 ∈ Hαn , (1) whence h = 1 hα1 + · · · + n−1 hαn−1 ∈ E(Hαn ). Hence x = h + n u, n = 0, whence / F (α) ⊂ F (αn ) , a contradiction with (11.6.7). By the by (vii) x ∈ / F (αn ) , and also x ∈ same argument as in the proof of Theorem 11.6.1 we infer that the set E(H (1) ) ∩ Fα is at most countable for every α < Ω.  Also the next theorem is due to J. Sm´ıtal [286]. Theorem 11.6.2. Under assumption of the continuum hypothesis, there exists a Hamel basis H ⊂ RN which admits a decomposition H = H (1) ∪H (2) , H (1) ∩H (2) = ∅, where H (1) and H (2) have the power of continuum, and E(H (1) ) is a Lusin set. Proof. Let H (1) be the set occurring in Lemma 11.6.1. By Theorem 4.2.1 there exists a Hamel basis H in RN such that H (1) ⊂ H. Put H (2) = H \ H (1) . By Lemma 11.6.1 E(H (1) ) is a Lusin set, and card H (1) = c, as has been pointed out in the proof of Lemma 11.6.1. Suppose that card H (2) < c. Then, by the continuum hypothesis, card H (2)  ℵ0 , and by Lemma 4.1.3 card E(H (2) )  ℵ0 . Every x ∈ RN =

11.7. Perfect sets

299

E(H) = E(H (1) ) + E(H (2) ) can be written as x = u + v, where u ∈ E(H (1) ), v ∈ E(H (2) ). Consequently    E(H (1) ) + v . (11.6.8) RN = v∈E(H (2) )

  By Corollary 3.4.1 E(H (1) ) is of measure zero. Consequently also E(H (1) ) + v is of measure zero for every v ∈ E(H (2) ). Since the union in (11.6.8) is at most countable, it follows that the set on the right-hand side of (11.6.8) is of measure zero. The contradiction obtained shows that card H (2) = c. 

11.7 Perfect sets Corollary 11.3.2 says that no Hamel basis is a Borel set. The question arises as to whether a Hamel basis may contain an uncountable Borel set6 . In the case of a Burstin basis the answer is “no” (Jones [158]). Theorem 11.7.1. Let H ⊂ RN be a Burstin basis. Then H contains no uncountable Borel subset. Proof. Suppose that B is an uncountable Borel set and B ⊂ H. Then 2B also is an uncountable Borel set, and consequently H ∩ 2B = ∅. Let y ∈ H ∩ 2B. In particular y ∈ 2B, i.e., y = 2x, where x ∈ B ⊂ H. Thus x ∈ H and 2x = y ∈ H, which is impossible, since x and 2x are linearly dependent (over Q).  But the situation as described in Theorem 11.7.1 is not general. Following F. B. Jones [158] we will prove Theorem 11.7.2. There exists a Hamel basis H ⊂ RN which contains a perfect set. Proof. Let {λn }n∈N be a sequence of all rational numbers such that λ1 = 0. Choose arbitrarily x0 , y0 ∈ RN (x0 = y0 ). Then the continuous functions fij (x, y) = λi x + λj y,

i, j = 1, 2,

have for x = x0 and y = y0 values different from x0 and y0 except when fij (x0 , y0 ) = x0 or y0 . There exist small closed balls K11 and K12 , of diameter less than 1 and such that x0 ∈ K11 , y0 ∈ K12 , K11 ∩ K12 = ∅, and / K11 ∪ K12 fij (x, y) ∈ for x ∈ K11 , y ∈ K12 , except when fij (x, y) = x or y. By a similar argument we can find closed balls K21 , K22 , K23 , K24 , pairwise disjoint, of diameter less than 12 , and such that K21 ∪K22 ⊂ K11 , K23 ∪K24 ⊂ K12 , and / λi x + λj y + λk z + λm u ∈

4 

K2ν ,

i, j, k, m = 1, 2, 3, 4,

ν=1 6 Or, what ammounts to the same, whether a Hamel basis may contain an uncountable analytic subset. In fact, every Borel set is analytic, and every uncountable analytic set contains an uncountable Borel subset (a homeomorphic image of the Cantor set C; cf. the proof of Theorem 2.8.3 and Exercise 2.12).

300

Chapter 11. Properties of Hamel Bases

for all x ∈ K21 , y ∈ K22 , z ∈ K23 , u ∈ K24 , except when the combination is x, y, z, or u. In the n-th step we find closed balls Kn1 , . . . , Kn,2n , pairwise disjoint, of diam1 eter less than , and such that Kn1 ∪ Kn2 ⊂ Kn−1,1 , . . . , . . . , Kn,2n−1 ∪ Kn,2n −1 ⊂ n Kn−1,2n−1 , and n

λi1 x1 + · · · + λi2n x

2n

∈ /

2 

i1 , . . . , i2n = 1, . . . , 2n ,

Knν ,

(11.7.1)

ν=1

for all x1 ∈ Kn1 , . . . , x2n ∈ Kn,2n , except when the combination is one of xν , ν = 1, . . . , 2n . In this way we have defined a sequence {Knν }, n ∈ N, ν = 1, . . . , 2n , of closed balls with the properties described above. Put n

F =

2 ∞  

Knν .

(11.7.2)

n=1 ν=1

It is easily seen that F is a perfect set. We will show that F is linearly independent over Q. Supposing the contrary, let {x1 , . . . , xp } ⊂ F be a finite linearly dependent system (xi = xj for i = j). Then one of x’s, say xp , is a linear combination of the remaining ones: xp = λi1 x1 + · · · + λip−1 xp−1 . (11.7.3) Choose an n ∈ N so that 2n  p, 2n 

max

κ=1,...,p−1

iκ , and

1 < min |xμ − xν |. By n μ,ν=1,...,p μ =ν

(11.7.2) we have x1 , . . . , xp−1 ∈

n 2 

Knν , and since the diameters of the balls Knν

ν=1

1 , every xk lies in a different ball. Choose arbitrarily points xk in the n remaining balls (one in each ball). Then (11.7.3) may be written as

are less than

xp = λi1 x1 + · · · + λip−1 xp−1 + λ1 xp + · · · + λ1 x2n . (11.7.4)  We may choose the points so that none of them coincides with xp . All the coefficients λi in (11.7.4) fulfil the condition 1  i  2n . By (11.7.1)



n

xp ∈ /

2  ν=1

n

Knν ,

whereas by (11.7.2)

xp ∈ F ⊂

2 

Knν .

ν=1

This contradiction shows that F is linearly independent (over Q). By Theorem 4.2.1 there exists a Hamel basis H ⊂ RN such that F ⊂ H. H is the required basis.  It follows easily from Corollary 11.2.1 and Theorem 11.3.1 that if a Hamel basis contains a perfect set F , then F is nowhere dense and of measure zero.

11.8. The operations R and U

301

11.8 The operations R and U The operations R and U were introduced in 9.2: R(T ) = T − T,

U (T ) = T + R(T ).

The iterates of the operations R and U are defined by the recurrence: R1 (T ) = R(T ), U 1 (T ) = U (T ),     Rn+1 (T ) = R Rn (T ) , U n+1 (T ) = U U n (T ) ,

n ∈ N.

Since J(T ), Q(T ) and Sn (T ) for all n ∈ N are contained in E + (T ), it follows from Theorem 11.6.1 that (under assumption of the continuum hypothesis) there exists a Hamel basis H such that J(H), Q(H) and Sn (H) for all n ∈ N are measurable sets of measure zero. For the iterates of the operations R and U we have, of course, the following Lemma 11.8.1. For every Hamel basis H ⊂ RN and every n ∈ N we have     mi Rn (H) = mi U n (H) = 0. n n Proof. By Theorems 9.2.3 and  n 9.2.6 the operations  n R and U are C-conservative for every n ∈ N. If we had mi R (H) > 0 or mi U (H) > 0, then by Theorem 9.3.1 we would get Rn (H) ∈ C or U n (H) ∈ C, whence, by Theorem 9.2.2, H ∈ C, which contradicts Theorem 11.1.1. 

But, contrary to the case of the operations J, Q, Sn , it cannot happen that Rn (H) and U n (H) are of measure zero for all n ∈ N. The following two theorems are due to F. B. Jones [158]. Theorem 11.8.1. For every Hamel basis H ⊂ RN there exists an n ∈ N such that   me Rn (H) > 0. Proof. For every non-empty set T ⊂ RN we have 0 ∈ R(T ), whence 0 ∈ Rn (H) for all n ∈ N. Hence Rn (H) = Rn (H) − 0 ⊂ Rn (H) − Rn (H) = Rn+1 (H), i.e., Rn (H) ⊂ Rn+1 (H),

n ∈ N.

(11.8.1)

Take an x ∈ Z(H) (cf. 11.5) and an a ∈ H. x may be written as x=

k

ε i hi =

i=1

k

εi (hi − a) + la,

(11.8.2)

i=1

where εi = ±1, hi ∈ H (i = 1, . . . , k), l =

k 

εi ∈ Z. (Not all hi ’s need be distinct).

i=1

We will show that

k

i=1

εi (hi − a) ∈ Rk (H).

(11.8.3)

302

Chapter 11. Properties of Hamel Bases

The proof runs by induction on k. For k = 1 we have h1 − a ∈ H − H = R(H), and −(h1 − a) = a − h1 ∈ H − H = R(H). Assuming (11.8.3) true for a k ∈ N, we have k+1

εi (hi − a) =

i=1

k

εi (hi − a) − εk+1 (a − hk+1 ).

i=1

By the induction hypothesis we have (11.8.3), whereas by what we have just shown and by (11.8.1) we have εk+1 (a − h1 ) ∈ R(H) ⊂ Rk (H). Hence

k+1

  εi (hi − a) ∈ Rk (H) − Rk (H) = R Rk (H) = Rk+1 (H).

i=1

Thus the validity of (11.8.3) has been established for every k ∈ N. Put ∞  M= Rk (H).

(11.8.4)

k=1

By (11.8.2) and (11.8.3)  

 M + la .

x ∈ M + la ⊂

l∈Z

Since x has been arbitrary in Z(H), we get hence    Z(H) ⊂ M + la .

(11.8.5)

l∈Z

If M were of measure zero, also all sets M + la would be of measure zero, whence  by (11.8.5) we would get m Z(H) = 0, which is not the case (Theorem 11.5.1). Consequently me (M ) > 0, whence by (11.8.4) me Rn (H) > 0 for a certain n ∈ N.  Theorem 11.8.2. For every Hamel basis H ⊂ RN there exists an n ∈ N such that   me U n (H) > 0. Proof. We have, since 0 ∈ Rn (T ) for every n ∈ N and every non-empty set T ⊂ RN ,     U n (H) = U n (H) + 0 ⊂ U n (H) + R U n (H) = U U n (H) = U n+1 (H), (11.8.6) and H = H + 0 ⊂ H + R(H) = U (H) ⊂ U n (H). Put M=

∞  k=1

U k (H).

(11.8.7)

11.8. The operations R and U

303

If x, y, z ∈ M , then by (11.8.6) there exists a k ∈ N such that x, y, z ∈ U k (H), whence     x + (y − z) ∈ U k (H) + R U k (H) = U U k (H) = U k+1 (H) ⊂ M. (11.8.8) Take an x ∈ Z(H) and an a ∈ M . Then x=

k

ε i hi =

i=1

k

εi (hi − a) + la,

i=1

where εi = ±1, hi ∈ H(i = 1, . . . , k), l =

k 

εi ∈ Z. We have for every k ∈ N

i=1

a+

k

εi (hi − a) ∈ M.

(11.8.9)

i=1

The proof runs by induction. For k = 1 we have a + ε1 (h1 − a) ∈ M   by (11.8.8), since a ∈ M and h1 ∈ H ⊂ M cf. (11.8.7) . Assuming (11.8.9) true for a k ∈ N, we have a+

k+1

εi (hi − a) = a +

i=1

k

εi (hi − a) + εk+1 (hk+1 − a) ∈ M

i=1

by (11.8.9) and (11.8.8). Thus (11.8.9) is true for all k ∈ N. Hence a+x=a+

k

εi (hi − a) + la ∈ M + la,

i=1

whence x ∈ M + (l − 1) a, and Z(H) ⊂



(M + la).

l∈Z

  Hence me (M ) > 0 follows as in the proof of Theorem 11.8.1, whence me U n (H) > 0 for a certain n ∈ N.  The following results can be proved quite similarly as Lemma 11.8.1 and Theorems 11.8.1 and 11.8.2: Lemma 11.8.2. For every Hamel basis H ⊂ RN , every n ∈ N, and every set A ⊂ RN with the Baire property, if A ⊂ Rn (H) or A ⊂ U n (H), then A is of the first category. Theorem 11.8.3. For every Hamel basis H ⊂ RN there exists an n ∈ N such that Rn (H) is of the second category. Theorem 11.8.4. For every Hamel basis H ⊂ RN there exists an n ∈ N such that U n (H) is of the second category.

304

Chapter 11. Properties of Hamel Bases

We conclude this chapter with an open problem7 (Acz´el-Erd˝ os [10]) concerning Hamel bases: Problem. – Does there exist a Hamel basis H ⊂ R such that h−1 ∈ H for every h ∈ H? – Does there exist a Hamel basis H ⊂ R such that hn ∈ H for every h ∈ H and n ∈ N? On the other hand it is known that there exists no Hamel basis H ⊂ R such that ab ∈ H whenever a, b ∈ H. (Cf. Exercise 14.7).

Exercises 1. Let H ⊂ RN be a Hamel basis. Show that aH is a Hamel basis for every a ∈ R \ {0}. 2. Let H ⊂ RN be a Hamel basis, and let h ∈ H. Show that H ∩ (H + h) = ∅. 3. Let H ⊂ RN be a Hamel basis. Show that there is a b ∈ RN such that the set H + b contains no Hamel basis.

Hint: Let b ∈ −H. Show that b ∈ / E(H + b) . 4. Show that the conclusion of Exercise 10.5 fails to hold if we drop the analiticity assumption on T . 5. Let H ⊂ RN be an arbitrary Hamel basis. Show that D(RN \ H) = RN and ∗ (RN \ H) = RN (cf. 2.1 and 3.5). 6. Show that under assumption of the continuum  hypothesis there exists a Hamel basis H ⊂ RN such that card E + (H) ∩ G  ℵ0 for every set G ⊂ RN such that G ∈ Gδ and m(G) = 0. 7. Show that under assumption of the continuum hypothesis there exists a Hamel basis H ⊂ RN such that the set E + (H) is of the first category. Hint: Use Exercise 11.6 and the decomposition RN = A ∪ B, where A ∈ Gδ and m(A) = 0, B ∈ Fσ and B is of the first category; cf. the end of 11.4. 8. Let H ⊂ RN be an arbitrary Hamel basis. Show that the set Z(H) does not contain an analytic subset containing a Hamel basis. (Consequently Theorems 9.3.7 and 9.8.1 cannot be reversed).   Hint: Show that for any A ⊂ Z(H) and n ∈ N we have Rn (A) ⊂ Rn Z(H) = Z(H).

7

Added in proof. Recently solved in the positive (under assumption of the continuum hypothesis) by J. Sm´ıtal [290].

Chapter 12

Further Properties of Additive Functions and Convex Functions 12.1 Graphs Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. Let mf be the lower hull of f (cf. 6.3). By Theorem 6.3.1 either mf (x) = −∞ for all x ∈ D, or mf : D → R is a continuous and convex function. Let Gr(f ) = {(x, y) ∈ RN +1 | x ∈ D, y = f (x)} be the graph of f (cf. 9.5). We have Theorem 12.1.1. If D ⊂ RN is a convex and open set and f : D → R is a discontinuous convex function, then the set Gr(f ) is dense in the set Af = {(x, y) ∈ RN +1 | x ∈ D, y > mf (x)}. Proof. Let K ⊂ D be an arbitrary closed ball, and J = [c, d] ⊂ R an arbitrary closed interval such that mf (x) < c for x ∈ K. (12.1.1) (Both, K and J, assumed non-degenerated.) Note that if mf (x) ≡ −∞, then (12.1.1) certainly is fulfilled, and if mf is a continuous function, then it is bounded above on the compact set K. Every point of the set Af can be included in a set K × J, where K and J are as described above. Thus it is enough to show that (K × J) ∩ Gr(f ) = ∅.

(12.1.2)

Since f is discontinuous, it is not bounded above on K. Thus there exists an x1 ∈ K such that f (x1 ) > d. (12.1.3) By (12.1.1) there exists an x2 ∈ K such that f (x2 ) < c.

(12.1.4)

306

Chapter 12. Further Properties of Additive Functions and Convex Functions

Let L = Q(x1 , x2 ) be the rational segment from x1 to x2 , and l = cl L = x1 x2 = {z ∈ RN | z = λx1 + (1 − λ)x2 , λ ∈ [0, 1]} be the closed segment joining x1 and x2 . We have L ⊂ l ⊂ K. (12.1.5) By Theorem 6.1.2 there exists a continuous function g : l → R such that g | L = f . Since x1 , x2 ∈ L, we have by (12.1.3) and (12.1.4) g(x1 ) > d,

g(x2 ) < c.

Since g is continuous, there exists a non-degenerated segment S ⊂ l such that g(S) ⊂ [c, d]. Clearly S ∩ L = ∅. Hence there exists an x0 ∈ L such that f (x0 ) = g(x0 ) ∈ J.   By (12.1.5) x0 ∈ K, and thus the point x0 , f (x0 ) belongs to the set on the left-hand side of (12.1.2).  If f : RN → R is a discontinuous additive function, then f is not locally bounded below at any point of RN . (Otherwise the additive function −f would be locally bounded above at a point, and hence continuous.) Therefore mf = −∞. Since in this case D = RN , the set Af becomes the whole space RN +1 . Thus, as a particular case of Theorem 12.1.1, we obtain Theorem 12.1.2. If f : RN → R is a discontinuous additive function, then the set Gr(f ) is dense in RN +1 . Theorem 12.1.2 can also be reformulated as follows. Theorem 12.1.3. If f : RN → R is a discontinuous additive function, then for every (non-degenerated ) interval J ⊂ R the set f −1 (J) is dense in RN . In connection with Theorems 10.3.2 and 10.4.1 here two examples  we will exhibit  (Mo´scicki [231]) showing that the conditions int Q T ∪ (−T ) = ∅ in Theorem 10.3.2 and int Q(T ∪A) = ∅ in Theorem 10.4.1 cannot be replaced by int[Q(T )∪Q(−T )] = ∅ and int[Q(T ) ∪ Q(A)] = ∅, respectively. Example 12.1.1. Let g : RN → R be an arbitrary discontinuous additive function, and put T = {x ∈ RN | g(x) < −1}. By Lemma 10.2.2 the set T is Q-convex: Q(T ) = T,

(12.1.6)

and, moreover, we have T ∈ C. To see this, let f : R → R be an arbitrary additive function bounded on T : |f (x)|  M for x ∈ T. N

Take an arbitrary x ∈ RN . We must distinguish two cases. I. x ∈ T . Then, for k ∈ N, we have g(kx) = kg(x) < −k  −1, whence kx ∈ T , and k |f (x)| = |f (kx)|  M for k ∈ N. But this implies f (x) = 0. Thus f | T = 0.

12.1. Graphs

307

II. x ∈ / T . Then g(x)  −1. Take an arbitrary y ∈ T . Then g(y − x) = g(y) − g(x) < −1 + for k ∈ N sufficiently large k(y − x) ∈ T . Therefore (by I.)   1 = 0. Hence, f k(y − x) = 0, i.e., k f (y) − f (x) = 0. But f (y) = 0, since y ∈ T , thus −kf (x) = 0, whence f (x) = 0. Consequently f (x) = 0 for all x ∈ RN , i.e., f is continuous. Now, by Corollary 5.1.1 and by (12.1.6)

Q(T ) ∪ Q(−T ) = Q(T ) ∪ − Q(T ) = T ∪ (−T ), and since −T = {x ∈ RN | g(x) > 1}, we get Q(T ) ∪ Q(−T ) = {x ∈ RN | |g(x)| > 1}. (12.1.7)

 Let G = int Q(T ) ∪ Q(−T ) , and suppose that G = ∅. By Theorem 12.1.3 g −1 [−1, 1] ∩ G = ∅, i.e., there exists a point x0 ∈ G ⊂ Q(T ) ∪ Q(−T ) such that |g(x0 )|  1. But this contradicts (12.1.7). Example 12.1.1 shows that the condition T ∈ C is not equivalent to int Q(T ) ∪

Q(−T ) = ∅. Example 12.1.2. Let H ⊂ RN be a Burstin basis (cf. 11.4). H may be arranged into a transfinite sequence H = {hα }α<γ , where γ is the smallest ordinal number such that γ = c. Every x ∈ RN may be written as x = λ1 hα1 + · · · + λn hαn ,

(12.1.8)

where λi ∈ Q, i = 1, . . . , n, and α1 < · · · < αn < γ. We put T = V0 = {x ∈ RN | λn > 0} ∪ {0}, where we assume (12.1.8) (cf. 10.2). In 10.2 it was shown that  that x has expansion  T is Q-convex i.e., fulfils (12.1.6) and that T ∈ B = A. Let g : H → R be given by g(h) = 1 for h ∈ H, and let f : RN → R be the additive extension of g (Theorem 5.2.2). By Corollary 5.2.2 f is discontinuous. Put A = {x ∈ RN | f (x) < 0}. By Lemma 10.2.2 the set A is Q-convex and Q-radial at every point. Suppose that there exists a non-empty open ball K ⊂ Q(T ) ∪ Q(A) = T ∪ A. Since H is a Burstin basis, H is dense in RN . Hence    K + h α = RN . 0<α<γ

Thus there exists an α > 0 such that h0 ∈ K + hα , i.e., h0 − hα ∈ K ⊂ T ∪ A. But f (h0 − hα ) = f (h0 ) − f (hα ) = g(h0 ) − g(hα ) = 1 − 1 = 0, so h0 − hα ∈ / A. Also, the point h0 − hα has expansion (12.1.8) with n = 2, α = 0, α = α, λ = 1, λ2 = −1, so 1 2 1

h0 − hα ∈ / T . This contradiction shows that int Q(T ) ∪ Q(A) = ∅, i.e., the condition

T ∈ B = A is not equivalent to the condition that int Q(T ) ∪ Q(A) = ∅ for every set A which is Q-convex and Q-radial at a point.

308

Chapter 12. Further Properties of Additive Functions and Convex Functions

12.2 Additive functions The contents of Theorem 12.1.2 can be expressed as follows: If f : RN → R is a discontinuous additive function, then for every non-empty and open set G ⊂ RN and every (non-degenerated) interval J ⊂ R (G × J) ∩ Gr(f ) = ∅.

(12.2.1)

The question arises whether in (12.2.1) G and J may be replaced by more general sets. In this section we will show that in (12.2.1) G may be an arbitrary Lebesgue measurable set of positive measure, or an arbitrary set of the second category and with the Baire property (Ostrowski [248], Kuczma [185], [189]). As expressed in Theorem 12.1.3, condition (12.2.1) with arbitrary non-empty and open set G is equivalent to the statement that f −1 (J) is dense in RN . We shall prove that (12.2.1) holds for an arbitrary set G of positive measure or of the second category with the Baire property when we shall have proved that f −1 (J) is saturated non-measurable and has property (∗) from 3.3. In other words this will mean that on an arbitrary set G of positive measure or of the second category with the Baire property f assumes values from every interval J ⊂ R, i.e., that for every such set G the set f (G) is dense in R. Theorem 12.2.1. If f : RN → R is a discontinuous additive function, then for every (non-degenerated ) interval J ⊂ R, J = R, the set f −1 (J) is saturated non-measurable and has property (∗) from 3.3. Proof. If 1 J = c, d , −∞  c < d < +∞, then f is bounded above on f −1 (J) by d. If J = c, d , −∞ < c < d  +∞, then f is bounded below on f −1 (J) by c, and then the discontinuous additive function −f is bounded above on f −1 (J) by  −c. At any case f −1 (J) ∈ / B = A, whence by Theorems 9.3.1 and 9.3.2 mi f −1 (J) = 0 and f −1 (J) does not contain any subset of the second category with the Baire property. It remains to show that also the complement RN \ f −1 (J) of f −1 (J) has the same properties, and this will be shown when we prove it for an arbitrary subinterval of J. Thus in the sequel we may assume that J = [c, d], where d/c > 0 and q = d/c ∈ Q. By Theorem 5.2.1 we have f (qx) = qf (x) for every x ∈ RN , whence f −1 (q n J) = q n f −1 (J) But

∞ 

for n ∈ N.

q n J is an infinite interval I = [c, ∞) (if 0 < c < d) or I = (−∞, d] (if

n=1

c < d < 0). Hence ∞  n=1

n −1

q f

(J) = f

−1

∞ 

 q n J = f −1 (I).

(12.2.2)

n=1

If f −1 (J) were of measure zero or of the first category, then by (12.2.2) so would be also f −1 (I), which means that RN \ f −1 (I) would be of positive (infinite) measure or of the second category with the Baire property. But on RN \ f −1 (I) the function f 1

c, d denotes any of the intervals (c, d) (open), [c, d] (closed), (c, d] and [c, d) (semiclosed).

12.2. Additive functions

309

  takes valuesoutside of I, i.e., itis bounded above by c when I = [c, ∞) , or bounded below by d when I = (−∞, d] , which contradicts the fact that f is discontinuous.   Thus me f −1 (J) > 0 and f −1 (J) is of the second category. This is valid for arbitrary non-degenerated interval J ⊂ R.     c+d d−c Suppose that 0 < c < d. Put J1 = c, , J2 = 0, , J3 = 2 2   c+d , d . We will show that 2 f −1 (J) = f −1 (J1 ) + f −1 (J2 ).

(12.2.3)

Let x ∈ f −1 (J). Then there exists a u ∈ J such that f (x) = u. Either u ∈ J1 or u ∈ J3 . If u ∈ J1 , then note that f (0) = 0 ∈ J2 , whence 0 ∈ f −1 (J2 ), whereas x ∈ f −1 (J1 ). Then x = x + 0 ∈ f −1 (J1 ) + f −1 (J2 ). Now suppose  that c + d c + d , d− = ∅. u ∈ J3 . If u < d, then by Theorem 12.1.3 f −1 u − 2 2 −1 1−q x we have according to Theorem 5.2.1, since If u = d, then for w = 2 −1 1−q 1 − q −1 d−c 1 − q −1 ∈ Q and q −1 d = c, f (w) = f (x) = d = so that 2 2 2 2    c + d c +d c + d c + d f −1 u − , d− = ∅, too. Take a y ∈ f −1 u − , d− 2 2 2     2 c+d d−c c+d −1 , d− ⊂ 0, = J2 . Thus y ∈ f (J2 ). and put v = f (y) ∈ u − 2 2 2 Let z = x − y. Then f (z) = f (x) − f (y) = u − v ∈ J1 so that z ∈ f −1 (J1 ). Now, x = z + y ∈ f −1 (J1 ) + f −1 (J2 ). Thus we have obtained f −1 (J) ⊂ f −1 (J1 ) + f −1 (J2 ). −1

f and

(12.2.4)

Conversely, let x ∈ f −1 (J1 ) + f −1 (J2 ). Then x = z + y, where z ∈ f −1 (J1 ), y ∈ (J2 ) so that f (z) ∈ J1 , f (y) ∈ J2 . Hence f (x) = f (z) + f (y) ∈ J, i.e., x ∈ f −1 (J), f −1 (J1 ) + f −1 (J2 ) ⊂ f −1 (J).

(12.2.5)

Relations (12.2.4) and (12.2.5) prove (12.2.3). Now, as we have already shown, f −1 (J1 ) is of the second category and of positive outer measure, whereas by Theorem 12.1.3 f −1 (J2 ) is dense in RN . By Theorems 3.6.1 and 3.6.2 the set RN \f −1 (J) has zero inner measure and does not contain any second category subset with the Baire property, whence the theorem follows.  The next natural question is whether in relation (12.2.1) also J may be an arbitrary set of positive measure or of the second category with the Baire property. In other words, whether a discontinuous additive function on every set of positive measure assumes values from every set of positive measure (resp. the corresponding question for the category). The answer to this question clearly is in the negative. It was pointed out in 5.2 that there exist discontinuous additive functions f : RN → R

310

Chapter 12. Further Properties of Additive Functions and Convex Functions

which admit only rational values. For such an f relation (12.2.1) with J being the set of irrational numbers is invalid even with G = RN . But a result of A. Ostrowski [248] gives us a certain insight into the situation. We recall that if f : RN → R is an additive function, then the kernel of f (cf. 4.3) is the set Ker f = f −1 (0) = {x ∈ RN | f (x) = 0}. Lemma 12.2.1. Let N = 1, and let f : R → R be an additive function. The set Ker f is dense in R if and only if f is non-invertible. Proof. If Ker f is dense in R, then it contains infinitely many points, i.e., f (x) = 0 for several x, and consequently f cannot be invertible. If f is non-invertible, then there exist x1 , x2 ∈ R such that x1 = x2 and f (x1 ) = f (x2 ). Put x0 = x1 − x2 = 0. We have f (x0 ) = f (x1 ) − f (x2 ) = 0, whence, for every λ ∈ Q, we obtain by Theorem 5.2.1 f (λx0 ) = λf (x0 ) = 0. Consequently Qx0 ⊂ Ker f , and Qx0 is dense in R, since x0 = 0, and consequently so is also Ker f .  For N > 1 Lemma 12.2.1 is no longer valid. If x0 ∈ RN , x0 = 0, then the set Qx0 is not dense in RN . Qx0 is a linear subspace (over Q) of RN , and {x0 } is its base. By Theorem 4.2.1 there exists a Hamel basis H of RN such that x0 ∈ H. We have by Theorem 4.2.3 card(H \ {x0 }) = card H = c = dim R. By Theorem 4.3.3 there exists an additive function f : RN → R such that Ker f = Qx0 . This f clearly is non-invertible2 (since Ker f is infinite), but Ker f is not dense in RN . Some sufficient conditions for the set Ker f to be dense in RN can be inferred from Lemma 4.3.1 and Lemmas 4.2.3 and 4.2.4. Now we pass to the announced theorem of A. Ostrowski3. Theorem 12.2.2. Let f : RN → R be an additive function4 such that Ker f is dense in RN . Then either on every set of positive measure f assumes values from every set of positive measure, or there exists a decomposition RN = A1 ∪ A2 ,   such that m(A1 ) = 0 and m f (A2 ) = 0.

(12.2.6)

Proof. RN is separable, so there exists a countable set D ⊂ Ker f dense in Ker f , and hence in RN . Since D ⊂ Ker f , we have f (x) = 0 for every x ∈ D. Now suppose that there exist sets A ⊂ RN of positive (N -dimensional) measure, and B ⊂ R of positive (one-dimensional) measure such that f (A) ∩ B = ∅. 2

(12.2.7)

f is also discontinuous, since otherwise Ker f would have to be closed, which is not the case with Qx0 . An example of continuous non-invertible f : RN → R with the kernel non-dense in RN would be easier. E.g., if N = 2, and x = (x1 , x2 ), then the function f (x) = x1 + x2 yields such an example. 3 A. Ostrowski [248] proved this theorem in the one-dimensional case (N = 1) under the condition that f is non-invertible. But Lemma 12.2.1 ensures that in this case the latter condition is equivalent to the condition that Ker f is dense in R. 4 There is no explicit assumption that f is discontinuous. If f is continuous, then Ker f is closed, and, being dense in RN , equals the whole RN . Hence f = 0. For such an f the theorem is trivially true, viz. (12.2.6) holds with A1 = ∅, A2 = RN .

12.2. Additive functions



Let C = A + D =

311

(A + d), and C  = RN \ C. Every set A + d is measurable and

d∈D

the union is countable, so C is measurable. By Theorem 3.6.1 mi (C  ) = 0, and thus, since C is measurable, m(C  ) = 0. (12.2.8) Let t ∈ f (A). Then t = f (x) for a certain x ∈ A. Take a d ∈ D. Then x + d ∈ C, and t = f (x) = f (x) + f (d) = f (x + d) ∈ f (C). Thus f (A) ⊂ f (C). Conversely, let t ∈ f (C). Then t = f (x) for a certain x ∈ C, i.e., x = a + d, where a ∈ A, d ∈ D. Hence t = f (x) = f (a + d) = f (a) + f (d) = f (a) ∈ f (A). Thus f (C) ⊂ f (A), and finally we obtain f (A) = f (C), whence by (12.2.7) f (C) ∩ B = ∅.

(12.2.9)

Put P =



λC  ,

M=

λ∈Q



λB,

P  = RN \ P,

M  = R \ M.

λ∈Q

By (12.2.8) m(λC  ) = 0 for every λ ∈ Q, whence m(P ) = 0.

(12.2.10)

Further, with B + = B ∩ (0, ∞), B − = B ∩ (−∞, 0) and B 0 = B ∩ {0}, we have B = B + ∪ B − ∪ B 0 . The sets B + , B − and B 0 are measurable, m(B 0 ) = 0, whence 0 < m(B) = m(B + ) + m(B − ). Hence at least one of the sets B + and B − must have positive measure. Let, e.g., m(B + ) > 0. Let L : (0, ∞) → R be the logarithmic function: L(x) = log x. The functions −L and L−1 (the exponential function) are continuous and convex, and hence, by Theorem 7.4.6 are absolutely continuous. Therefore also L is absolutely continuous. Thus if one of the sets E, L(E) is measurable, so is also the other,  andif one of E, L(E) has measure zero, so has also the other. It follows that m L(B + ) > 0. Now, with Q+ = Q ∩ (0, ∞), Q− = Q ∩ (−∞, 0), we have L

  λ∈Q+

  

L(B + ) + L(λ) = L(B + ) + L(Q+ ). λB + = L(λB + ) = λ∈Q+

λ∈Q+

The above set is measurable (as a countable  union of measurable sets)  and by Theorem 3.6.1 its complement has measure zero since L(Q+ ) is dense in R . But R\L

  λ∈Q+

    λB + = L (0, ∞) \ λB + , λ∈Q+

(12.2.11)

312

Chapter 12. Further Properties of Additive Functions and Convex Functions

since L is one-to-one. The set occurring in (12.2.11) has measure zero, whence, since L−1 is absolutely continuous,    m (0, ∞) \ (12.2.12) λB + = 0. Further, we have

 λ∈Q−

λB + = −



λ∈Q+

λB + , whence by (12.2.12)

λ∈Q+

   λB + = 0. m (−∞, 0) \

(12.2.13)

λ∈Q−

Now,

 λ∈Q−

λB + ∪



λB + ⊂

λ∈Q+



λB + ⊂

λ∈Q



λB = M,

λ∈Q

whence by (12.2.12) and (12.2.13) m(M  ) = 0.

(12.2.14)

Suppose that there exists a y = 0 such that y ∈ f (P  ) ∩ M . Since y ∈ M, y = 0, there exists a λ ∈ Q, λ = 0, such that y ∈ λB, i.e., y/λ ∈ B. Since y ∈ f (P  ), there exists an x, x ∈ P , (12.2.15) such that y = f (x). Therefore, by Theorem 5.2.1, x 1 y = f (x) = f , λ λ λ and (12.2.9) implies that y/λ ∈ / f (C), i.e., x/λ ∈ / C. Thus x/λ ∈ C  and x ∈ λC  ⊂ P , which contradicts (12.2.15). Thus the set f (P  ) ∩ M may contain at most one element, viz. 0. Consequently f (P  ) ⊂ M  ∪ {0}, which together with (12.2.14) shows that   m f (P  ) = 0.

(12.2.16)

Now, in order to obtain decomposition (12.2.6), we take A1 = P, A2 = P  . Relations (12.2.10) and (12.2.16) show that this decomposition has the required properties.  By the same argument one can also prove the topological analogue of Theorem 12.2.2. Theorem 12.2.3. Let f : RN → R be an additive function such that Ker f is dense in RN . Then either on every set of the second category and with the Baire property f assumes values from every set of the second category with the Baire property, or there exists a decomposition (12.2.6) such that A1 is of the first category (in the topology of RN ), and f (A2 ) is of the first category (in the topology of R). The invariance of the Baire property and of the category under L will now result from the fact that L is a homeomorphism. Theorem 3.6.1 must everywhere be replaced by Theorem 3.6.2.

12.3. Convex functions

313

12.3 Convex functions The next arising question is how far the results of 12.2 generalize to the case of convex functions. Unfortunately, we are unable to answer this question. Below we prove a result (Theorem 12.3.1) due to A. Ostrowski [249], which is a very weak analogue of Theorem 12.2.1. Lemma 12.3.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a discontinuous convex function. Further, let J ⊂ R be an interval with a finite upper   bound. Then mi f −1 (J) = 0. Proof. Let d ∈ R be such that t  d for t ∈ J. Then f is bounded above (by d) on f −1 (J), whence the lemma results in view of Theorem 9.3.1 and of the fact that f is discontinuous.  Theorem 12.3.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a discontinuous convex function. Let the interval J = [c, d] ⊂ R be such that inf f (x) = inf mf (x) < c < d < ∞.

x∈D

x∈D

(12.3.1)

  Then me f −1 (J) > 0. Proof. By (12.3.1) there exists an x0 ∈ D such that f (x0 ) < c.

(12.3.2)

For every x ∈ D, x = x0 , let l(x) denote the closed segment joining x0 and x: l(x) = {z ∈ D | z = x0 + t(x − x0 ), t ∈ [0, 1]}, and let L(x) = Q(x0 , x) denote the rational segment between x0 and x: L(x) = {z ∈ D | z = x0 + t(x − x0 ), t ∈ Q ∩ [0, 1]}. By Theorem 6.1.2 for every x ∈ D, x = x0 , there exists a continuous function gx : l(x) → R such that gx | L(x) = f | L(x). Put   C = {x ∈ D | f (x) > d} = f −1 (d, ∞) .   The complement C  = D \ C of C is the set C  = f −1 (−∞, d] . If C had measure zero, then C  would have a positive measure, which, according to Lemma 12.3.1, is impossible. Consequently me (C) > 0. (12.3.3) For every x ∈ C we have gx (x) = f (x) > d, whereas by (12.3.2) gx (x0 ) = f (x0 ) < c. Consequently we have gx (z) ∈ [c, d] for z from a certain non-degenerated segment Sx ⊂ l(x). In other words, there exists a (non-degenerated) segment Sx ⊂ l(x) such that   Sx ⊂ gx−1 [c, d] .

314

Chapter 12. Further Properties of Additive Functions and Convex Functions

1 (Sx − x0 ). Thus I(x) also is a non-degenerated segment con|x − x0 |   1 l(x) − x0 . Let |I(x)| denote the length of I(x). Consequently tained in |x − x0 | 0 < |I(x)|  1. Let #  # 1 1 < |I(x)|  , k ∈ N. Ck = x ∈ C ## k+1 k Clearly ∞  Ck = C, (12.3.4) Put I(x) =

k=1

and it follows by (12.3.3) and (12.3.4) that there exists a k0 ∈ N such that me (Ck0 ) > 0.

(12.3.5)

i x − x0 , i = 0, . . . , n+1. n + 1 |x − x0 |   1 i , and |x − x0 | xi = x0 + (x − x0 ) − x0 ∈ l(x) − x0 , We have |xi+1 − xi | = n+1 n+1   1 1 l(x) − x0 . Since for x ∈ Ck0 we have |I(x)| > , for every such whence xi ∈ |x − x0 | n x there must exist an i, 0  i  n + 1, such that xi ∈ I(x). Let Write n = k0 +1, and consider the points xi = xi (x) =

Ei = {x ∈ Ck0 | xi (x) ∈ I(x)},

i = 0, . . . , n + 1.

Again n+1 

Ei = Ck0 ,

(12.3.6)

i=0

and it follows by (12.3.5) and (12.3.6) that there exists an i0 ∈ { 0, . . . , n + 1 } such that me (Ei0 ) > 0. (12.3.7) Let x ∈ Ei0 . Then xi0 (x) ∈ I(x), i.e., i0 x − x0 1 ∈ (Sx − x0 ), n + 1 |x − x0 | |x − x0 | or x0 + and

But x0 +

  i0 (x − x0 ) ∈ Sx ⊂ gx−1 [c, d] , n+1

 gx x0 +

 i0 (x − x0 ) ∈ [c, d]. n+1

i0 (x − x0 ) ∈ L(x), whence n+1     i0 i0 (x − x0 ) = gx x0 + (x − x0 ) ∈ [c, d]. f x0 + n+1 n+1

12.3. Convex functions

315

In other words, for x ∈ Ei0 , x0 + Write E ∗ = x0 +

  i0 (x − x0 ) ∈ f −1 [c, d] . n+1

(12.3.8)

i0 (Ei0 − x0 ). By (12.3.7) (cf. Corollary 3.2.2) n+1

and (12.3.8) shows that

me (E ∗ ) > 0,

(12.3.9)

  E ∗ ⊂ f −1 [c, d] = f −1 (J).

(12.3.10)

Now, the theorem results from (12.3.9) and (12.3.10).



From Lemma 12.3.1 and Theorem 12.3.1 we obtain Corollary 12.3.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a discontinuous convex function. Then, for every finite and non-degenerated interval J ⊂ R such that inf J > inf mf (x) = inf f (x) (12.3.11) x∈D

the set f

−1

x∈D

(J) is non-measurable.

Proof. Since J is non-degenerated, it contains a non-degenerated interval [c, d], and by (12.3.11) condition (12.3.1) is fulfilled. Hence, by Theorem 12.3.1,      me f −1 (J)  me f −1 [c, d] > 0. (12.3.12) On the other hand, since J is finite, it has a finite upper bound. Thus by Lemma 12.3.1   mi f −1 (J) = 0. (12.3.13) Relations (12.3.12) and (12.3.13) imply, in virtue of Theorem 3.1.2, that f −1 (J) is non-measurable.  Theorem 12.3.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a discontinuous convex function. Then, for every non-degenerated interval J ⊂ R fulfilling (12.3.11) the set f −1 (J) is non-measurable. Proof. If sup J < ∞, then J is finite, and the theorem results from Corollary 12.3.1. So we may restrict ourselves to the case where J = (c, ∞), or J = [c, ∞) with a certain c ∈ R. Put J0 = R \ J, and c0 = inf mf (x). x∈D

  Then c0 < c < ∞. Choose a γ ∈ (c0 , c). By Corollary 12.3.1 the set f −1 (γ, c) is non-measurable, whence      me f −1 (J0 )  me f −1 (γ, c) > 0.

316

Chapter 12. Further Properties of Additive Functions and Convex Functions

On the other hand, by Lemma 12.3.1,   mi f −1 (J0 ) = 0. Consequently f −1 (J0 ) is non-measurable. We have f −1 (J0 ) = f −1 (R) \ f −1 (J) = D \ f −1 (J). Since D is measurable, and f −1 (J0 ) is not, the set f −1 (J) also must be non-measurable.  The attribute “non-measurable” in Theorem 12.3.2 cannot be replaced by “saturated non-measurable”. Clearly, f −1 (J) ⊂ D, and there and open D ⊂  exist convex   RN such that m(RN \D) > 0. For such D we have 0 < m RN \D  mi RN \f −1 (J) . But if J ⊂ R is an interval fulfilling (12.3.11), write d = sup J and D0 = {x ∈ D | mf (x)  d}. If x ∈ f −1 (J), then mf (x)  f (x)  d, whence x ∈ D0 . Thus f −1 (J) ⊂ D0 . It is an open problem whether   mi D0 \ f −1 (J) = 0 for every non-degenerated interval J ⊂ R fulfilling (12.3.11). By exactly the same argument as that employed in the respective proofs, we obtain also topological analogues of Lemma 12.3.1 and Theorems 12.3.1 and 12.3.2. Lemma 12.3.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a discontinuous convex function. Further, let J ⊂ R be an interval with a finite upper bound. Then the set f −1 (J) does not contain any set of the second category and with the Baire property. Theorem 12.3.3. Let D ⊂ RN be a convex and open set, and let f : D → R be a discontinuous convex function. Let J = [c, d] ⊂ R be an interval fulfilling (12.3.1). Then the set f −1 (J) is of the second category. Theorem 12.3.4. Let D ⊂ RN be a convex and open set, and let f : D → R be a discontinuous convex function. Then, for every non-degenerated interval J ⊂ R fulfilling (12.3.11) the set f −1 (J) does not have the Baire property.

12.4 Big graph Let f : RN → R be a discontinuous additive function. In 12.1 we have shown that the set Gr(f ) is dense in RN . Nothing more can be said in general about Gr(f ). In 5.2 we have seen that there exist (necessarily discontinuous) additive functions f : RN → R, f = 0, which assume only rational values: f (RN ) ⊂ Q. Take arbitrary λ ∈ Q and an arbitrary x ∈ RN such that f (x) = 0. Then f (x) ∈ Q, whence  also  q = λ/f (x) ∈  Q.By Theorem 5.2.1 f (qx) = qf (x) = λ. Consequently Q ⊂ f RN and we get f RN = Q. Thus   there exist (necessarily discontinuous) additive functions f : RN → R such that f RN is countable. For such an f we have    RN × {y} . Gr(f ) ⊂ { (x, y) ∈ RN +1 | x ∈ RN , y ∈ f (RN ) } = y∈f (RN )

12.4. Big graph

317

Every set RN × {y} is of (N +1-dimensional) measure zero, and nowhere dense (in the topology of RN +1 ). Hence Gr(f ) also is of measure zero, and of the first category. Moreover, Gr(f ) is disconnected, since for arbitrary y0 ∈ / f (RN ) we have    

Gr(f ) = Gr(f ) ∩ RN × (−∞, y0 ] ∪ Gr(f ) ∩ RN × [y0 , ∞) ,     and the sets Gr(f ) ∩ RN × (−∞, y0 ] and Gr(f ) ∩ RN × [y0 , ∞) are non-empty5 , disjoint, and closed in the space Gr(f ).   N N is countable cardf (RN ) =  Every additive function f : R → R such that f R ℵ0 will henceforth be referred to as a function with small graph. We may summarize what we have already established in the following theorems. Theorem 12.4.1. There exist additive functions f : RN → R with small graphs. Every such function is discontinuous. Theorem 12.4.2. If f : RN → R is an additive function with small graph, then the set Gr(f ) is of measure zero and of the first category (in RN +1 ), and is not connected. Now, we want to show that there exist additive functions f : RN → R which have particularly big graphs. Let π : RN +1 → RN be the projection: if p ∈ RN +1 and p = (x, y), x ∈ RN , y ∈ R, then π(p) = x. An additive function f : RN → R is called a function with big graph iff for every Borel set F ⊂ RN +1 such that6 card π(F ) = c we have F ∩ Gr(f ) = ∅. (12.4.1) We have7 Theorem 12.4.3. There exist additive functions f : RN → R with big graphs. Every such function is discontinuous. Proof. Let F be the family of all Borel sets F ⊂ RN +1 such that card π(F ) = c. Since F ⊂ B RN +1 , we get by Theorem 2.3.4 card F  c. On the other hand, F contains all open balls in RN +1 , and the family of all such balls has the power of continuum. Hence card F  c, whence card F = c. (12.4.2) Let γ be the smallest ordinal such that γ = c (cf. the proof of Theorem 11.4.3). Hence also card Γ(γ) = c, and by (12.4.2) there exists a one-to-one mapping F : Γ(γ) → F (onto). Instead of F (α) we write Fα . Thus F = {Fα }α<γ . Now we define a  transfinite sequence {(xα , yα )} of points of RN +1 . For every α < γ we write Xα = {xβ } and Eα = E(Xα ), except for α = 0, where we assume β<γ

  is countable, then there exists a y0 ∈ f RN such that y0 = 0. Thus y0 = f (x0 ) for an If f R     x0 ∈ RN . For every λ ∈ Q we have λy0 = λf (x0 ) = f (λx0 ) ∈ f RN , whence Qy0 ⊂ f RN . Hence  N f R is dense in R. 6 It ammounts to the same to require that card π(F ) > ℵ . In fact, every Borel set is analytic, and 0 the function π is continuous, hence the set π(F ) also is analytic (Theorems 2.5.6 and 2.5.2). Thus, by Theorem 2.8.3, the conditions card π(F ) = c and card π(F ) > ℵ0 are equivalent. 7 Cf. Jones [157]. 5



 N

318

Chapter 12. Further Properties of Additive Functions and Convex Functions

E0 = {0}. We want to construct the sequence {(xα , yα )}α<γ so  thatfor every α < γ the set {xα } ∪ Xα is linearly independent (over Q). Let w : P RN +1 \ {∅} → RN +1 be a choice function. Suppose that we have already defined (xβ , yβ ) for β < α < γ, (if α = 0, nothing has been defined yet) so that the sets {xβ } ∪ Xβ are linearly independent. We put   (xα , yα ) = w Fα \ [Eα × R] . (12.4.3) Note that card Xα = α < c, since α < γ, and by Lemmas 4.1.3 and 4.2.2 we have card Eα < c, whereas card π(Fα ) = c. Consequently Fα \ [Eα × R] = ∅. Suppose that the set Xα is linearly dependent. Then there exist xβ1 , . . . , xβn ∈ Xα such that the system {xβ1 , . . . , xβn } is linearly dependent. Let β = max(β1 , . . . , βn ) < α. Then {xβ1 , . . . , xβn } ⊂ Xβ ∪ {xβ }, which contradicts the fact that {xβ } ∪ Xβ is linearly independent (β < α). According to (12.4.3) we have xα ∈ / Eα , whence it follows by Lemma 4.1.2 that the set {xα } ∪ Xα is linearly independent. (If α = 0, then we get from (12.4.3) x0 = 0, and thus {x0 } is linearly independent). By Theorem 1.6.2 there exists a transfinite sequence {(xα , yα )}α<γ fulfilling (12.4.3) for every α < γ and such that {xα } ∪ Xα (and hence also Xα ) is linearly independent for every α < γ. Put   X= Xα = {xα }. α<γ

α<γ

The same argument as above shows that the set X is linearly independent. By Theorem 4.2.1 there exists a Hamel basis H ⊂ RN such that X ⊂ H. Define the function g : H → R putting  yα if x = xα , α < γ, g(x) = 0 if x ∈ H \ X. Let f : RN → R be the additive extension of g (Theorem 5.2.2). Thus (xα , yα ) ∈ Gr(f )

for every α < γ.

(12.4.4)

Let F ⊂ RN +1 be a Borel set such that card π(F ) = c. Thus F ∈ F, and there exists an α < γ such that F = Fα . By (12.4.3) (xα , yα ) ∈ Fα = F , whence by (12.4.4) (xα , yα ) ∈ F ∩ Gr(f ). Hence we obtain (12.4.1). Consequently f is a function with big graph. If f were continuous, then by Theorem 5.5.2 f would have the form f (x) = cx with a c ∈ RN . Hence Gr(f ) = {(x, y) ∈ RN +1 | x ∈ RN , y = cx}.

(12.4.5)

But set (12.4.5) does not have property (12.4.1). Consequently f is discontinuous.  Let us also note the following N Lemma  N  12.4.1. Let f : R → R be an additive function with big graph. Then f R = R.

12.4. Big graph

319

Proof. Let y ∈ R be arbitrary. The set F = RN × {y} ⊂ RN +1 is closed, and hence Borel, and π(F ) = RN , so card π(F ) = c. Thus we must have (12.4.1). In other words, there exists a point (x0 , y0 ) ∈ F ∩ Gr(f ). Since (x0 , y0 ) ∈ F , we must have y0 = y. Since (x0 , y0 ) ∈ Gr(f ), we must have f (x0 ) = y0 = y. Consequently y ∈ f RN , whence the lemma follows.  Lemma 12.4.1 shows, in particular, that a function with big graph cannot be at the same time a function with small graph. This will follow also from further theorems. A big graph has none of the properties specified in Theorem 12.4.2 for small graphs. Namely, we have (cf. also Marczewski [220], Marcus [215]) Theorem 12.4.4. If f : RN → R is an additive function with big graph, then the set Gr(f ) is saturated non-measurable, and has property (∗) from 3.3. Proof. Let (x1 , y1 ), (x2 , y2 ) ∈ Gr(f ). This means that y1 = f (x1 ), y2 = f (x2 ). Hence y1 + y2 = f (x1 ) + f (x2 ) = f (x1 + x2 ), i.e., (x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 ) ∈ Gr(f ). Consequently8 Gr(f ) + Gr(f ) ⊂ Gr(f ).

(12.4.6)

N On the other hand, int Gr(f ) = ∅, since  on every vertical line l = {x0 } × R, x0 ∈ R , there is only one point of Gr(f ), viz. x0 , f (x0 ) . By Theorem 3.7.1   mi Gr(f ) = 0. (12.4.7)

Similarly, by Theorem 2.9.1 we infer that Gr(f ) does not contain any second category set with the Baire property.   Now suppose that mi RN +1 \ Gr(f ) > 0. By the definition of the inner measure there must exist a closed (and hence a Borel) set F ⊂ RN +1 \ Gr(f )

(12.4.8)

m(F ) > 0.

(12.4.9)

such that If the set π(F ) were countable, then F would be contained in a countable union of vertical lines: ∞  ln . (12.4.10) F ⊂ n=1

In R every line ln is of measure zero, and hence (12.4.10) would imply that m(F ) = 0, contrary to (12.4.9). Therefore card π(F ) > ℵ0 , whence by Theorem 2.8.3 N +1

8 Actually, in (12.4.6) we have the equality. In fact, since f (0) = 0, we have (0, 0) ∈ Gr(f ). Now, for an arbitrary (x, y) ∈ Gr(f ) we have (x, y) = (x, y) + (0, 0) ∈ Gr(f ) + Gr(f ), whence Gr(f ) ⊂ Gr(f ) + Gr(f ). This together with (12.4.6) yields Gr(f ) = Gr(f ) + Gr(f ).

320

Chapter 12. Further Properties of Additive Functions and Convex Functions

card π(F ) = c. Thus we must have (12.4.1), which is incompatible with (12.4.8). Therefore we must have   mi RN +1 \ Gr(f ) = 0. (12.4.11) Relations (12.4.7) and (12.4.11) show that Gr(f ) is saturated non-measurable. Similarly, suppose that there exists a second category set B with the Baire property such that B ⊂ RN +1 \ Gr(f ). (12.4.12) Then B = (G ∪ P ) \ R, where G is open and non-empty, and P, R are of the first category. Put C = G\R ⊂ B. Hence G \ C ⊂ R is of the first category, and by Theorem 2.1.3 there exists a set F ∈ Gδ such that F ⊂ C ⊂ B, (12.4.13) and the set G \ F is of the first category. Consequently F must be of the second category. If we had card π(F )  ℵ0 , then F would be contained in a countable union of vertical lines, i.e., we would have (12.4.10). In RN +1 every line is nowhere dense, and hence the sets in (12.4.10) would be of the first category, which is impossible. Consequently card π(F ) = c, and we obtain (12.4.1), which contradicts (12.4.12) and (12.4.13). Thus neither Gr(f ), nor RN +1 \ Gr(f ), contains a second category set with the Baire property, which means that Gr(f ) has property (∗).  Theorem 12.4.4 shows, in particular, that of f is non-measurable and  the graph  does not have the Baire property, whence me Gr(f ) > 0 and Gr(f ) is of the second category. This is a very surprising property   of a graph of a function. The existence of functions f : R → R such that me Gr(f ) > 0 was first shown (on another way) by W. Sierpi´ nski [278]. N Let us also note  that  everyvertical line {x} × R, x ∈ R , intersects Gr(f ) at exactly one point viz. x, f (x)  , and consequently this intersection has measure zero. Nevertheless, me Gr(f ) > 0. This shows that in the Fubini theorem to the effect that if A ∈ L, and for almost every vertical line l we have m(A ∩ l) = 0, then m(A) = 0, the assumption of the measurability of A is essential. Theorem 12.4.5. If f : RN → R is an additive function with big graph, then the set Gr(f ) is connected. Proof. For an indirect proof suppose that Gr(f ) is not connected. Then Gr(f ) = A0 ∪ B0 ,

(12.4.14)

where the sets A0 and B0 are separated: (A0 ∩ cl B0 ) ∪ (B0 ∩ cl A0 ) = ∅.

(12.4.15)

Write A = cl A0 ,

B = cl B0 ,

F = A ∩ B.

(12.4.16)

12.4. Big graph

321

Suppose that there exists a point p ∈ F ∩ Gr(f ). By (12.4.14) p belongs either to A0 , or to B0 , and by (12.4.16) p belongs to A and to B. Hence p ∈ (A0 ∩ B) ∪ (B0 ∩ A), which is impossible in view of (12.4.15). Consequently F ∩ Gr(f ) = ∅,

(12.4.17)

which implies, since F is closed (and hence Borel) that (cf. Theorem 2.8.3) card π(F )  ℵ0 .

(12.4.18)

Note also that A ∪ B = cl A0 ∪ cl B0 = cl(A0 ∪ B0 ) = cl Gr(f ), whence by Theorem 12.1.2 A ∪ B = RN +1 . (12.4.19) Relation (12.4.19) implies that A = RN +1 \ A ⊂ B and B  = RN +1 \ B ⊂ A, whence A ∩ B  = ∅. Hence A \ F = A \ B = A ∩ B = B, i.e., A \ F is open, and A \ F ⊂ int A. On the other hand, if p ∈ F ∩ int A, then there exists a neighbourhood U of p such that U ⊂ A. But since p ∈ B = cl B0 , we have U ∩ B0 = ∅, i.e., A ∩ B0 = ∅, which is incompatible with (12.4.15). Thus F ∩ int A = ∅ and int A ⊂ A \ F . Hence int A = A \ F . A similar argument shows also that int B = B \ F . So int A = A \ F,

int B = B \ F.

(12.4.20)

Let l ⊂ RN +1 be any vertical line such that l ∩ F = ∅. Then, by (12.4.19) and (12.4.20), l = (l ∩ int A) ∪ (l ∩ int B). Sets (12.4.20) are clearly disjoint, and hence also the sets l ∩ int A and l ∩ int B are disjoint, and are open in l. Since l is connected, we must have either l ∩ int A = ∅, or l ∩ int B = ∅. Now let l ⊂ RN +1 be an arbitrary vertical line. We will show that one of the sets (l\F )∩int A and (l\F )∩int B is empty. Supposing the contrary, let p ∈ (l\F )∩int A ⊂ l ∩ int A, and q ∈ (l \ F ) ∩ int B ⊂ l ∩ int B. There exist neighbourhoods U and V of p and q, respectively, such that U ⊂ int A and V ⊂ int B. By (12.4.18)

π(U ) ∩ π(V ) \ π(F ) = ∅.  Observe that p and q lie on the same vertical line l, whence π(p) = π(q) and π(U ) ∩ π(V ) = ∅. Since the function π is open, the sets π(U ) and π(V ) are open, and hence  π(U ) ∩ π(V ) is a non-empty open set, and thus of the power of continuum. Let

x0 ∈ π(U ) ∩ π(V ) \ π(F ), and let l0 = {x0 } × R be the vertical line through x0 . Then l0 ∩ U = ∅, l0 ∩ V = ∅, l0 ∩ F = ∅. Hence l0 ∩ int A = ∅,

l0 ∩ int B = ∅,

which, as we have already shown, is impossible.

l0 ∩ F = ∅,

322

Chapter 12. Further Properties of Additive Functions and Convex Functions

Consequently, by (12.4.20), for every vertical line l either l ∩ int A or l ∩ int B is empty. This means that the sets π(int A) and π(int B) are disjoint. These sets are N also open (since the function π is open), and since R is connected, there must exist N an x ∈ R \ π(intA) ∪ π(int B) . Let l = {x} × R be the vertical line through x. Then cf.   (12.4.20) l ∩ [int A ∪ int B] = l ∩ [(A ∪ B) \ F ] = ∅, i.e., l ⊂ F . We have x, f (x) ∈ l ∩ Gr(f ) ⊂ F ∩ Gr(f ), which contradicts (12.4.17). The contradiction obtained shows that Gr(f ) is connected.  The connectedness of the graph of an additive function was also studied by F. B. Jones [157], F. Obreanu [245], S. Marcus [215] and W. Kulpa [195]. Since every function f : R → R without the Darboux property must have a disconnected graph, we get from Theorem 12.4.5 Corollary 12.4.1. If f : R → R is an additive function with big graph, then f has the Darboux property. Thus an additive function f : R → R with big graph yields also an example of a discontinuous function with the Darboux property. The Darboux property of additive functions was also considered by J. Sm´ıtal [286], S. Marcus [215], A. M. Bruckner–J. G. Ceder–Max Weiss [38]. Cf. also Exercise 5.7.

12.5 Invertible additive functions As is generally known, and as we have repeatedly seen before, discontinuous additive functions have many pathological properties. Therefore it is often believed that such functions cannot have any nice property. In particular, one could believe that a discontinuous additive function cannot be invertible. But this is not the case (Marczewski [220], Sm´ıtal [286], Makai [214]), as we are presently going to show. We start with a rather well-known lemma, which is a particular case of a much more general fact, and which we prove here for the sake of completeness. Lemma 12.5.1. Let f : RN → R be an additive function. Then f is invertible if and only if Ker f = {0}. (12.5.1) Proof. If f is invertible, then there may exist only one x ∈ RN such that f (x) = 0, which is of course x = 0. Thus (12.5.1) holds. Conversely, assume (12.5.1), and let x1 , x2 ∈ RN be such that f (x1 ) = f (x2 ). Then f (x1 − x2 ) = f (x1 ) − f (x2 ) = 0, i.e., x1 − x2 ∈ Ker f , and by (12.5.1), x1 − x2 = 0. Thus x1 = x2 . This shows that f is one-to-one, and hence invertible.  Theorem 12.5.1. There exist discontinuous, invertible additive functions f : RN → R. Proof. We shall distinguish two cases: N = 1 or N > 1.

12.5. Invertible additive functions

323

At first consider the case N = 1. Let H be a Hamel basis of the space (R; Q; +; ·). Let h1 and h2 be two fixed but different points of H. Define the function g : H → H putting g(h1 ) = h2 , g(h2 ) = h1 , g(h) = h whenever h = h1 and h = h2 . Let f : R → R be the additive extension of g (cf. Theorem 5.2.2). It follows from Corollary 5.2.1 that f is discontinuous, since otherwise we would have h2 /h1 = h1 /h2 = 1. Observe that f is invertible. In fact, since g is one-to-one and takes the values in H, we have (12.5.1). By Lemma 12.5.1 the function f is invertible. Now consider the case N > 1. Let H ⊂ RN and B ⊂ R be Hamel bases of N R and R, respectively. Since card H = card B (cf. Theorem 4.2.3), there exists a bijection g : H → B. Let f : RN → R be the additive extension of g (cf. Theorem 5.2.2). Of course, f is invertible, because g is one-to-one and takes the values in B. Moreover, f is discontinuous, since otherwise we would have (cf. Theorem 5.2.2) f (x) = cx for every x ∈ RN with a constant c ∈ RN , but such a function is not invertible.



Note that if N > 1, then there do not exist invertible continuous additive functions f : RN → R. (So the existence of discontinuous invertible additive functions can again be considered as a pathological property.) But if N = 1, functions f (x) = cx with c = 0 are all continuous and invertible. Actually, in the case N = 1 we can prove more (Makai [214]). Let X be an arbitrary set. A function f : X → X fulfilling for every x ∈ X the relation

f f (x) = x (12.5.2) is called involutory or an involution (cf. Kuczma [177]). Lemma 12.5.2. Let X be an arbitrary set, and let f : X → X be an involution. Then f is invertible and onto: f (X) = X.

Proof. If f (x1 ) = f (x2 ) for some x1 , x2 ∈ X, then by (12.5.2) x1 = f f (x1 ) = f f (x2 ) = x2 . Thus f is one-to-one, and hence invertible. Further, we have

X = f f (X) ⊂ f (X) ⊂ X, whence f (X) = X.



Condition (12.5.2) can be expressed as f −1 = f . Theorem 12.5.2. Let B ⊂ R be a non-empty finite or countable set. Then there exists a discontinuous additive function f : R → R fulfilling (12.5.2) for all x ∈ R and such that f (x) = x for x ∈ B.

324

Chapter 12. Further Properties of Additive Functions and Convex Functions

Proof. Let H ⊂ R be a Hamel basis. Every x ∈ B can be written in the form (x) (x)

(x) x = λ1 h1 + · · · + λ(x) nx hnx , (x)

(x)

(x)

(12.5.3)

(x)

where λ1 , . . . , λnx ∈ Q, h1 , . . . , hnx ∈ H. Put H0 =

nx   ! (x) " hi . x∈B i=1

Since B is finite or countable, so is also H0 . Hence H \ H0 is infinite (of the power of continuum). There exists a decomposition H \ H 0 = H1 ∪ H 2 , where H1 ∩ H2 = ∅ and card H1 = card H2 (= c). Hence there exists a one-to-one mapping g0 : H1 → H2 (onto). Thus the function g0−1 is defined on H2 and maps H2 onto H1 . Now we define the function g : H → H as follows: ⎧ ⎪ g (h) for h ∈ H1 , ⎪ ⎨ 0 −1 g(h) = g0 (h) for h ∈ H2 , (12.5.4) ⎪ ⎪ ⎩ h for h ∈ H0 . Let f : R → R be the additive extension of g. For h ∈ H0 we have g(h)/h = 1, and for h ∈ H1 we have g(h) = g0 (h) ∈ H2 so that g(h) = h and g(h)/h = 1. By Corollary 5.2.1 f is discontinuous.



For arbitrary h ∈ H1 we have g(h) = g0 (h) ∈ H2 , whence g g(h) = g0−1 g0 (h) = h.



Similarly, for h ∈ H2 we have g(h) = g0−1 (h) ∈ H 1 and g g(h) = g0 g0−1 (h) = h. Obviously we have g g(h) = h for h ∈ H0 . Thus g g(h) = h for all h ∈ H. Now take an arbitrary x ∈ R. Then x = λ1 h1 + · · · + λn hn ,

λi ∈ Q,

hi ∈ H,

i = 1, . . . , n,

whence f (x) = λ1 f (h1 ) + · · · + λn f (hn ) = λ1 g(h1 ) + · · · + λn g(hn ), and, since g maps H onto H,





f f (x) = λ1 f g(h1 ) + · · · + λn f g(hn )



= λ1 g g(h1 ) + · · · + λn g g(hn ) = λ1 h1 + · · · + λn hn = x. Thus f fulfils (12.5.2).

12.5. Invertible additive functions

325 (x)

Q,

Now take an arbitrary x ∈ B. Then x has a representation (12.5.3) with λi ∈ H0 , i = 1, . . . , nx . Hence by (12.5.4)  (x)  (x)  (x)  f (x) = λ1 f h1 + · · · + λ(x) nx f hnx  (x)  (x)  (x)  (x) (x) (x) (x) = λ1 g h1 + · · · + λ(x) nx g hnx = λ1 h1 + · · · + λnx hnx = x.

∈

(x) hi

Thus f (x) = x for x ∈ B.



It can be shown that for the function f constructed above the set {x ∈ R | f (x) = x} is always larger than B. Generally, if f : R → R is an additive function, then so is also the function ϕ : R → R given by ϕ(x) = f (x) − x. Consequently, in virtue of Lemma 4.3.1, the set {x ∈ R | f (x) = x} = Ker ϕ is a linear subspace of R. In the proof of Theorem 12.5.2 the function f has been constructed in such a way that f (H) = H (12.5.5) for a certain Hamel basis H ⊂ R. But there exist involutory additive functions such that (12.5.5) does not hold for any Hamel basis H ⊂ R. Such is, e.g., the function9 f (x) = −x. If H ⊂ R is an arbitrary Hamel basis and h ∈ H, then f (h) = −h ∈ /H (note that h and −h are linearly dependent). Consequently for this f we even have H ∩ f (H) = ∅ for every Hamel basis H ⊂ R. On the other hand, if f : R → R is an involutory additive function, and H ⊂ R is a Hamel basis, then f (H) also is a Hamel basis. In order to show this we prove two lemmas. Lemma 12.5.3. Let f : RN → R be an additive function. Then, for every linearly independent set A ⊂ RN the set f (A) also is linearly independent if and only if f is invertible. Proof. Let f be invertible. Take an arbitrary finite system {y1 , . . . , yn } ⊂ f (A), an arbitrary linear combination y = λ1 y1 + · · · + λn yn , λ1 , . . . , λn ∈ Q, of y1 , . . . , yn , and suppose that y = 0. There exist a1 , . . . , an ∈ A such that yi = f (ai ), i = 1, . . . , n. Hence 0 = y = λ1 f (a1 ) + · · · + λn f (an ) = f (λ1 a1 + · · · + λn an ), and by Lemma 12.5.1 λ1 a1 + · · · + λn an = 0. Since A is linearly independent, and {a1 , . . . , an } ⊂ A, we must have λ1 = · · · = λn = 0, which proves that y1 , . . . , yn are linearly independent. Hence also f (A) is linearly independent. Now suppose that f (A) is linearly independent for every linearly independent set A ⊂ RN . Then, in particular, f (x) = 0 for x = 0. It follows that Ker f = {0}. By Lemma 12.5.1 f is invertible.  9

This function is continuous. We have been unable to find an example of a discontinuous involutory additive function f : R → R such that f (H) \ H = ∅ for every Hamel basis H ⊂ R.

326

Chapter 12. Further Properties of Additive Functions and Convex Functions

  Lemma 12.5.4. Let f : RN → R be an additive function. Then E f (A) = R for every set A ⊂ RN such that E(A) = RN if and only if   (12.5.6) f RN = R. Proof. Assume that we have (12.5.6). Take an arbitrary y ∈ R and an arbitrary set A ⊂ RN such that E(A) = RN . By (12.5.6) there exists an x ∈ RN = E(A) such that y = f (x). Further, there exist an n ∈ N, λ1 , . . . , λn ∈ Q, and a1 , . . . , an ∈ A such that x = λ1 a1 + · · · + λn an . Hence

  y = λ1 f (a1 ) + · · · + λn f (an ) ∈ E f (A) .

  Thus E f (A) = R.   Now suppose that E f (A) = R for every set A ⊂ RN such that E(A) = RN . Let N H⊂ RNbe a Hamel basis, and take  an arbitrary y ∈ R. We have E(H) = R , whence E f (H) = R, and y ∈ E f (H) . Consequently there exist an n ∈ N, λ1 , . . . , λn ∈ Q, and y1 , . . . , yn ∈ f (H) such that y = λ1 y1 + · · · + λn yn . Further, there exist h1 , . . . , hn ∈ H such that yi = f (hi ), i = 1, . . . , n. Hence y = λ1 f (h1 ) + · · · + λn f (hn ) = f (λ1 h1 + · · · + λn hn ) ∈ f (RN ).  Thus R ⊂ f (RN ). The converse inclusion is trivial, whence (12.5.6) follows.     Remark 12.5.1. This proof can be simplified by using the relation E f (A) = f E(A) (cf. Exercise 4.4). Assuming (12.5.6) we have for every set A ⊂ RN such that E(A) = RN     E f (A) = f E(A) = f (RN ) = R.   In the other direction observe that E(RN ) = RN , whence E f (RN ) = R and     f (RN ) = f E(RN ) = E f (RN ) = R. Theorem 12.5.3. Let f : RN → R be an invertible additive function fulfilling condition (12.5.6). Then, for every Hamel basis H ⊂ RN , the set f (H) ⊂ R is a Hamel basis. Proof. Results from Lemmas 12.5.3 and 12.5.4.



Corollary 12.5.1. Let f : R → R be an additive function fulfilling condition (12.5.2) for all x ∈ R. Then, for every Hamel basis H ⊂ R, the set f (H) ⊂ R is a Hamel basis. Proof. Results from Lemma 12.5.2 and Theorem 12.5.3.



12.6. Level sets

327

12.6 Level sets In Theorem 12.2.1 the restriction on the interval J to be non-degenerated is essential. If J is degenerated (i.e., reduces to a point), the set f −1 (J) need not be large. E.g., if f : RN → R is an invertible additive function, then for every t ∈ R the set f −1 (t) contains at most one point. Of course, if t ∈ / f (RN ), then f −1 (t) = ∅. For the sake of convenience in the sequel we exclude such cases. Thus, if f : RN → R is an arbitrary function, then under a level set of f we understand any set of the form L = f −1 (t), where t ∈ f (RN ). If f = 0, then the only level set of f is Ker f = f −1 (0) = RN . We will pay no attention to this trivial case, and so we will usually assume that the function in question is not identically zero. In the present section we are going to study the level sets of additive functions. We start with the following theorem of S. Ruziewicz [271]. Theorem 12.6.1. Let f : RN → R be an arbitrary additive function. Then any two level sets of f are congruent under translation.   Proof. Let L = f −1 (t) and M = f −1 (s), t, s ∈ f RN , be arbitrary two level sets of f . Take arbitrary u ∈ L and v ∈ M . Then f (u) = t, f (v) = s, and f (u − v) = f (u) − f (v) = t − s. Now let x ∈ M be arbitrary. Then   f x + (u − v) = f (x) + f (u − v) = s + (t − s) = t, whence x + (u − v) ∈ L. Thus M + (u − v) ⊂ L. Now take an arbitrary x ∈ L, and put y = x − (u − v). Then f (y) = f (x) − f (u − v) = t − (t − s) = s, i.e., y ∈ M . On the other hand, x = y +(u−v) ∈ M +(u−v), whence L ⊂ M +(u−v). Consequently L = M + (u − v), which shows that L is a translation of M .



Theorem 12.6.1 shows that all level sets of an additive function are congruent to each other, and thus have analogical properties. Thus in the sequel, in order to establish a property of all level sets of an additive function f , we will  establish this property for a particular level set, e.g., for L = Ker f = f −1 (0). Note that, since f (0) = 0, always 0 ∈ f (RN ) so that f −1 (0) is a level set of f . The next two theorems are due to J. Sm´ıtal [286]. Theorem 12.6.2. Let f : RN → R be an arbitrary additive function. Then, for every level set L of f , we have either card L = 1, or card L  ℵ0 . Proof. Let L = Ker f , and suppose that card L > 1. Then there exists an x ∈ L such that x = 0. By Lemma 4.3.1 L is a linear subspace of RN , whence λx ∈ L for every λ ∈ Q. In other words Qx ⊂ L. But, since x = 0, card Qx = ℵ0 . Hence card L  ℵ0 . 

328

Chapter 12. Further Properties of Additive Functions and Convex Functions

Of course, if f : RN → R is an invertible additive function (cf. Theorem 12.5.1), then we have card L = 1 for every level set L of f . On the other hand, we have the following complement to Theorem 12.6.2. Note that, for every level set L of an additive function f : RN → R, we have L ⊂ RN , whence card L  c. Theorem 12.6.3. For every cardinal number m such that ℵ0  m  c there exists an additive function f : RN → R such that f (RN ) = R and card L = m for every level set L of f . Proof. Let H ⊂ RN be a Hamel basis. According to Theorem 4.2.3 card H = c, so there exists a set H1 ⊂ H such that card H1 = m, card(H \ H1 ) = c. By Lemmas 4.1.3 and 4.2.2 card E(H1 ) = m. Now, H1 as a subset of H, is linearly independent, so H1 is a base of E(H1 ). We have card(H \ H1 ) = c = dim R. By Theorem 4.3.4 there exists an additive function f : RN → R such that f (RN ) = R and Ker f = E(H1 ). This function has the desired properties.  Now we turn to measure-theoretical and topological properties of the level sets of additive functions (Sm´ıtal [286], Erd˝ os-Marcus [77], Halperin [132], Marcus [217]). Theorem 12.6.4. Let f : RN → R be an additive function, f = 0. Then, for every level set L of f , we have mi (L) = 0 and L does not contain any second category set with the Baire property. Proof. Let L = Ker f , and suppose that mi (L) > 0 or L contains a second category set with the Baire property. By Theorems 9.3.1, 9.3.2 and Corollary 9.2.1 we have L ∈ A. By Corollary 9.3.2 there exists a Hamel basis H ⊂ L. By Lemma 4.3.1 we have E(L) = L. Thus RN = E(H) ⊂ E(L) = L ⊂ RN i.e., Ker f = L = RN , which means that f = 0, contrary to the assumption.



The following theorem yields more information. Theorem 12.6.5. Let f : RN → R be an additive function, f = 0. Then, for every level set L of f , we have m(L) = 0, or L is saturated non-measurable. Similarly, either L is of the first category, or has property (∗) from 3.3. Proof. Let L = Ker f . By Lemma 4.3.1 L + L = L.

(12.6.1)

N If m  e (L) > 0, then by Lemma 4.2.4 L is dense in R , whence by Theorem 3.6.1 mi RN \ L = 0. Since by Theorem 12.6.4 also mi (L) = 0, L is saturated nonmeasurable. Similarly, if L is of the second category, then L is dense in RN according to Lemma 4.2.4. It follows from (12.6.1) and from Theorem 3.6.2 that the set RN \ L does not contain any second category set with the Baire property. Theorem 12.6.4 asserts that L has the same property. This means that L has property (∗). 

12.6. Level sets

329

Again, if f is an invertible additive function, then the level sets of f consist of a single point each, and consequently are of measure zero and of the first category. The following two theorems assert that the other possibility in Theorem 12.6.5 also can happen. Theorem 12.6.6. Let f : RN → R be an additive function with small graph. Then every level set of f is saturated non-measurable and has property (∗) from 3.3. Proof. Two different level sets of f are disjoint, and we have  RN = f −1 (t).

(12.6.2)

t∈f (RN )

If the sets f −1 (t) were of measure zero, or of the first category, then so would be also RN , since the union in (12.6.2) is countable. The theorem now results from Theorem 12.6.5.  Theorem 12.6.7. Let f : RN → R be an additive function with big graph. Then every level set of f is saturated non-measurable and has property (∗) from 3.3.   Proof. Let L = Ker f , and suppose that m(L) = 0. Then m RN \ L = ∞, and there exists a closed set F0 ⊂ RN \ L such that m(F0 ) > 0.

(12.6.3)

Similarly, suppose that L is of the first category. Then the set RN \ L is residual, and by Theorem 2.1.2 there exists a set F0 ∈ Gδ , F0 ⊂ RN \ L, and such that F0 is residual.

(12.6.4)

In both cases, by (12.6.3) or (12.6.4), we have card F0 = c. Now let F = F0 × {0} ⊂ RN +1 . The set F is a Borel set (since F0 is a Borel set), and π(F ) = F0 , whence card π(F ) = c. Thus we must have (12.4.1). This means that there exists a point (x, y) ∈ F ∩ Gr(f ). Hence x ∈ F0 ⊂ RN \ L, and f (x) = 0 so that x ∈ Ker f = L. Consequently x ∈ L ∩ (RN \ L), which is absurd. The theorem results now from Theorem 12.6.5.  It follows from Theorem 12.6.7 that there exists an additive function f : RN → R such that f (RN ) = R (cf. Lemma 12.4.1) so that card f (RN ) = c, and the level sets of f are saturated non-measurable and have property (∗). Similarly, it follows from Theorem 12.6.6 that there exists an additive function f : RN → R such that card f (RN ) = ℵ0 and the level sets of f are saturated non-measurable and have property (∗). The following theorem (Halperin [132], Sm´ıtal [286], Erd˝ os-Marcus [77]) yields a joint generalization of these statements. Theorem 12.6.8. For every cardinal number m such that ℵ0  m  c there exists an additive function f : RN → R such that card f (RN ) = m and the level sets of f are saturated non-measurable and have property (∗) from 3.3.

330

Chapter 12. Further Properties of Additive Functions and Convex Functions

Proof. Let H ⊂ R be a Hamel basis of R. By Theorem 4.2.3 card H = c, so there exists a set H1 ⊂ H such that card H1 = m, card(H \ H1 ) = c. By Lemma 4.2.2 card E(H1 ) = m. H \ H1 is linearly independent set, so it is a base of E(H \ H1 ). We have H \ (H \ H1 ) = H1 , whence card[H \ (H \ H1 )] = card H1 = m = dim E(H1 ). By Theorem 4.3.4 there exists an additive function f1 : R → R such that f1 (R) = E(H1 ), Ker f1 = E(H \ H1 ). N Let f2 : R N → R be anN additive function with big graph, and put f (x) = f1 f2 (x) , x ∈ R . Then f : R → R is an additive function and (cf. Lemma 12.4.1)

  f RN = f1 f2 (RN ) = f1 (R) = E(H1 ).    Consequently card f (RN ) = m. Now, Ker f = f −1 (0) = f2−1 f1−1 (0) = f2−1 E(H \    H1 ) . Take an arbitrary t ∈ E(H \ H1 ). Then f2−1 (t) ⊂ f2−1 E(H \ H1 ) = Ker f . By Theorem 12.6.7 the set f2−1 (t) is saturated non-measurable and has property (∗), so,  −1 in particular, me f2 (t) > 0 and f2−1 (t) is of the second category. Consequently also Ker f has the same properties. By Theorem 12.6.5 Ker f is saturated non-measurable and has property (∗).  As we have pointed out before, every invertible additive function yields an example of an additive function with level sets of measure zero. The question arises whether there exist additive functions with uncountable level sets of measure zero. Following J. Sm´ıtal [286] we will answer this question in the affirmative, assuming the continuum hypothesis. Theorem 12.6.9. Under assumption of the hypothesis there exists an ad  continuum ditive function f : RN → R such that f RN = R and the level sets of f have the power of continuum and measure zero. Proof. By Theorem 11.6.2 there exists a Hamel basis H ⊂ RN which admits a decomposition H = H1 ∪ H2 , H1 ∩ H2 = ∅, such that card H1 = card H2 = c and E(H1 ) is a Lusin set. H1 is a base of E(H1 ), and card(H \ H1 ) = card H2 = c = dim R. By Theorem 4.3.4 there exists an additive function f : RN → R such that f (RN ) = R, Ker f = E(H1 ). Since E(H1 ) is a Lusin set, it is uncountable, and by the continuum   hypothesis card Ker f = card E(H1 ) = c. By Corollary 3.4.1 m(Ker f ) = m E(H1 ) = 0.  If N  2, then for every continuous additive function f : RN → R the level sets have the power of continuum and measure zero. But the function f constructed in the proof of Theorem 12.6.9 is discontinuous. In fact, if f were continuous, then the set Ker f = f −1 (0) would be closed. But actually, according to Theorem 3.4.2, Ker f , being a Lusin set, does not have the Baire property.

12.7 Partitions Let f : RN → R be an additive function. Then the level sets of f are pairwise disjoint, fulfil (12.6.2), and, according to Theorem 12.6.1, they are congruent under translation. Thus every additive function f : RN → R furnishes a partition (12.6.2)

12.7. Partitions

331

of RN into disjoint congruent sets, and the number of constituents of the partition is N equal to card f (RN ). Partitions  of R were studied by Burstin [39], Ruziewicz [271], Hahn-Rosenthal [128], §8.3.3 , Sierpi´ nski [283], Halperin [132], Erd˝ os-Marcus [77], Sm´ıtal [286]. Directly from Theorem 12.6.8 we obtain the following (Halperin [132], Sm´ıtal [286], Ruziewicz [271]) Theorem 12.7.1. For every cardinal number m such that ℵ0  m  c there exists a partition of RN into m pairwise disjoint saturated non-measurable sets with property (∗) from 3.3, congruent under translation. In this context the following question arises: if A is a family of pairwise disjoint sets congruent under translation and such that  (12.7.1) A = RN , does there exist an additive function f : RN → R such that the sets A ∈ A are the level sets of f ? The following theorem (Kuczma [180]) answers this question. Observe that, since the sets from A are pairwise disjoint and fulfil (12.7.1), there exists a unique set A0 ∈ A such that 0 ∈ A0 . Theorem 12.7.2. Let A be a family of pairwise disjoint sets which are congruent under translation and fulfil (12.7.1). Let A0 ∈ A be such that 0 ∈ A0 . Then there exists an additive function f : RN → R such that the sets A ∈ A are the level sets of f if and only if the set A0 is a linear subspace (over Q) of RN . Proof. Suppose that such an f exists. We have f (0) = 0, consequently A0 = Ker f , and by Lemma 4.3.1 A0 is a linear subspace of RN . Now suppose that the sets from A are pairwise disjoint, congruent under translation, fulfil (12.7.1), and A0 is a linear subspace of RN . Since the sets A ∈ A are congruent under translation, for every A ∈ A there exists a dA ∈ RN such that A = A0 + dA .

(12.7.2)

Let B ⊂ A0 be a base of A0 . By Corollary 4.2.3 there exists a Hamel basis H of RN such that B ⊂ H. We have card(H \ B)  card H = c = dim R. By Theorem 4.3.3 there exists an additive function f : RN → R such that Ker f = A0 . Put tA = f (dA ) for A ∈ A. Now take an A ∈ A. If x ∈ A, then by (12.7.2) there exists an x0 ∈ A0 such that x = x0 + dA , whence f (x) = f (x0 ) + f (dA ) = f (dA ) = tA . Thus x ∈ f −1 (tA ) and A ⊂ f −1 (tA ). Take an x ∈ f −1 (tA ). Thus f (x) = tA = f (dA ) and f (x − dA ) = f (x) − f (dA ) = 0 , i.e., x − dA ∈ Ker f = A0 and x ∈ A0 + dA = A. Consequently f −1 (tA ) ⊂ A, and finally we get A = f −1 (tA ). Thus every set A ∈ A is a level set of f . 

332

Chapter 12. Further Properties of Additive Functions and Convex Functions

Example 12.7.1. Let H ⊂ RN be an arbitrary Hamel basis, and let Z(H) be the associated Erd˝os set (cf. 11.5). Define the relation ∼ in RN × RN as follows: x ∼ y ⇔ x − y ∈ Z(H).

(12.7.3)

Since Z(H) fulfils (11.5.2), ∼ is an equivalence relation. Consequently RN can be partitioned into Let A be the collection of  disjoint classes of equivalent elements.  these classes. Note that we have A = RN /Z(H). Then the sets from A are pairwise disjoint and fulfil (12.7.1). Take arbitrary A, B ∈ A, u ∈ A, v ∈ B, and put d = v − u. Take an x ∈ A. Then x ∼ u, i.e., by (12.7.3), x − u ∈ Z(H). Now let y = x + d = x + v − u = v + (x − u), whence y − v = x − u ∈ Z(H), and by (12.7.3) y ∼ v. Hence y ∈ B, i.e., x + d ∈ B. This shows that A + d ⊂ B. Now take an arbitrary y ∈ B. Thus y ∼ v, i.e., y − v ∈ Z(H). Let x = y − d. Then x = y − (v − u) = u + (y − v), and x − u = y − v ∈ Z(H). Consequently x ∼ u, whence x ∈ A, and y = x + d ∈ A + d. Hence B ⊂ A + d, and finally B = A + d. We have shown that any two sets from A can be obtained from each other by translation, i.e., the sets from A are congruent under translation. Observe that 0 ∈ Z(H), whence A0 = Z(H). The set Z(H) is not a linear subspace of RN , consequently the sets from A are not the level sets of an additive function. Now we are going to deal with the one-dimensional case. Let I ⊂ R be an interval. Then every partition (12.7.1) of R (N = 1) generates a partition of I:  

 A∩I .

I=

A∈A

But now the sets A ∩ I need not be congruent under translation. We will say that two sets A, B ⊂ I are congruent by decomposition iff there exist non-negative real numbers t, s such that t + s = |I| and



(A + t) ∩ I ∪ (A − s) ∩ I = B, where |I| denotes the length of I. Lemma 12.7.1. Let f : RN → R be an additive function. If x0 ∈ Ker f and x0 = 0, then x0 is a period of f . Proof. We have, for arbitrary x ∈ RN , f (x + x0 ) = f (x) + f (x0 ) = f (x), i.e., x0 is a period of f .



Lemma 12.7.2. Let f : R → R be a non-invertible additive function. Then f is microperiodic10 . 10

I.e., f has arbitrarily small positive periods.

12.7. Partitions

333

Proof. Since f is non-invertible, the set Ker f is dense in R according to Lemma 12.2.1. Consequently there exists an arbitrarily small positive x0 ∈ Ker f . By Lemma 12.7.1 every such x0 is a period of f .  Now we will prove a theorem (cf. Lusin-Sierpi´ nski [212], Halperin [132]) on a partition of the unit interval. Theorem 12.7.3. Let I = [0, 1] be the unit interval. For every cardinal number m such that ℵ0  m  c there exists a partition  I= A (12.7.4) of I into m sets which are congruent by decomposition and such that mi (A) = 0, me (A) = 1 for every A ∈ A. Also, if A ∈ A, then neither A, nor I \ A, contains a second category subset with the Baire property. Proof. There exists an additive function f˜ : R → R such that card f˜(R) = m and the level sets of f˜ are saturated non-measurable and have property (∗) from 3.3 (Theorem 12.6.8). Consequently there exists a z ∈ Ker f˜ such that z = 0. Define the function f : R → R by f (x) = f˜(zx). Then f is an additive function, f (R) = f˜(zR) = f˜(R) so that card f (R) = m, and Ker f = z −1 Ker f˜ is saturated non-measurable and has property (∗) so that all level sets of f are saturated non-measurable and have property (∗). Moreover, f (1) = 0, whence, by Lemma 12.7.1, 1 is a period of f : f (x + 1) = f (x)

for x ∈ R.

(12.7.5)

For every t ∈ f (R) put At = f −1 (t), and let A = {I ∩ At }t∈f (R) . Thus card A = m, and since R = At , we have t∈f (R)

I=



(I ∩ At ).

(12.7.6)

t∈f (R)

Formula (12.7.6) yields a partition of I into m pairwise disjoint sets. Since the sets At are saturated non-measurable, we have by Theorem 3.3.1 me (I ∩ At ) = |I| = 1 for every t ∈ f (R), whereas evidently mi (I ∩ At )  mi (At ) = 0, whence mi (I ∩ At ) = 0 for t ∈ f (R). Also, it follows from (∗) that neither I ∩ At ⊂ At , nor I \ At ⊂ R \ At can contain a second category subset with the Baire property, t ∈ f (R). It remains to show that the sets from A are congruent by decomposition. Take an arbitrary t ∈ f (R) and x ∈ At = f −1 (t). By (12.7.5) x + 1 ∈ At , whence At + 1 ⊂ At . Similarly, if x ∈ At + 1, then x = y + 1, where y ∈ At . Thus f (x) = f (y + 1) = f (y) = t, i.e., x ∈ f −1 (t) = At and At + 1 ⊂ At . Hence At + 1 = At

for t ∈ f (R).

(12.7.7)

Now take arbitrary t, s ∈ f (R), t = s. The sets At and As are congruent under translation (Theorem 12.6.1), i.e., there exists a d ∈ R such that At + d = As .

(12.7.8)

334

Chapter 12. Further Properties of Additive Functions and Convex Functions

Write d = [d] + r, where [d] is the integral part of d, and 0  r < 1. By (12.7.8) and (12.7.7) At + r = As , (12.7.9) and since t = s, whence At = As , we must have actually 0 < r < 1. Subtracting 1 from both sides of (12.7.9) we obtain by (12.7.7) At − (1 − r) = As .

(12.7.10)

By (12.7.9) (At ∩ I) + r = (At + r) ∩ (I + r) = (At + r) ∩ [r, r + 1] = As ∩ [r, r + 1], and



(At ∩ I) + r ∩ I = As ∩ [r, r + 1] ∩ I = As ∩ [r, 1].

(12.7.11)

Similarly, by (12.7.10)       (At ∩I)−(1−r) = At −(1−r) ∩ I −(1−r) = At −(1−r) ∩[r−1, r] = As ∩[r−1, r], whence

(At ∩ I) − (1 − r) ∩ I = As ∩ [r − 1, r] ∩ I = As ∩ [0, r].

(12.7.12)

From (12.7.11) and (12.7.12) we obtain 

 

 (At ∩ I) + r ∩ I ∪ (At ∩ I) − (1 − r) ∩ I = As ∩ I. Consequently the sets At ∩ I and As ∩ I are congruent by decomposition.



It is clear that in the above theorem the closed interval [0, 1] may be replaced by the open interval (0, 1), or by the half-open interval (0, 1] or [0, 1), or by any other interval I ⊂ R (with the obvious modification in the formulation of the theorem). As pointed out in 10.5 the sets from class A may be considered as “large”. Therefore also the following theorem (Kuczma [184]) may be of interest. Theorem 12.7.4. There exists a partition of RN into c pairwise disjoint, congruent (under translation) sets belonging to the class A.  Proof. Consider the partition described in Example 12.7.1. By Lemma 11.5.4 cf. formula (11.5.7) Z(H) ∈ A, whence, by Theorem 9.2.8, every translation of Z(H) belongs to A. It remains to show that card A = c. Let D be the set of those x ∈ RN which have expansions x = α1 h1 + · · · + αn hn ,

n ∈ N, hi ∈ H, i = 1, . . . , n,

(12.7.13)

1 with αi ∈ Q ∩ (0, 1). Thus H ⊂ D, whence (cf. Theorem 4.2.3) c = card H = 2   1 card H  card D  c, and card D = c. If y, z ∈ Z(H) + d for a d ∈ D, then 2

12.8. Monotonicity

335

y = u + d, z = v + d, where u, v ∈ Z(H), whence y − z = u − v ∈ Z(H). Thus y ∼ z. Conversely, if y ∈ Z(H) + d and z ∼ y, then z − y ∈ Z(H),

y = u + d, u ∈ Z(H). Hence z = (z − y) + y = (z − y) + (u + d) = (z − y) + u + d. By (11.5.2) we have (z − y) + u ∈ Z(H) + Z(H) = Z(H), and thus z ∈ Z(H) + d. Thus Z(H) + d is a class of equivalent elements of RN , whence Z(H) + d ∈ A for every d ∈ D. We must show that for d1 = d2 , d1 , d2 ∈ D, also Z(H) + d1 = Z(H) + d2 . This will imply that card A  card D = c, and since evidently card A  c, we will get hence card A = c. First observe that −Z(H) = Z(H), whence by (11.5.2) Z(H) − Z(H) = Z(H). Now suppose that for certain d1 , d2 ∈ D, d1 = d2 , we have Z(H) + d1 = Z(H) + d2 . Then d1 − d2 ∈ Z(H) − Z(H) = Z(H). (12.7.14) The point x = d1 − d2 has an expansion (12.7.13), where αi ∈ Q ∩ (−1, 1), and, since d1 = d2 , not all αi are zero. Thus x ∈ / Z(H). This contradicts (12.7.14). 

12.8 Monotonicity Now we are going to deal with quite a different problem. If D ⊂ R is an open interval and f : D → R is a convex function, and if f is monotonic in D, then f is bounded on every closed interval I ⊂ D, and hence it is continuous (cf. also Corollary 12.8.1 below). This fact can be extended to higher dimensions (Bereanu [28]). Here we are going to present one of such extensions (Kuczma [180]). Let I ⊂ R be an interval, and let ϕ : I → R be a function. Let ω be a positive real number. The function ϕ is called ω-increasing (Anastassiadis [14]) iff ϕ(x + ω)  ϕ(x) holds for all x ∈ I such that x + ω ∈ I. Similarly, ϕ is called ω-decreasing iff ϕ(x + ω)  ϕ(x) holds for all x ∈ I such that x + ω ∈ I. Let Ω ⊂ (0, ∞) be a set of positive numbers. The function ϕ is called Ω-increasing [Ω-decreasing] iff it is ω-increasing [ω-decreasing] for every ω ∈ Ω. A function which is either Ω-increasing, or Ω-decreasing, is called Ω-monotonic. There exist sets Ω ⊂ (0, ∞) and functions ϕ : I → R which are Ω-monotonic, but not monotonic in the customary sense. To see this consider Example 12.7.1 (N = 1). Since, as was shown in the proof of Theorem 12.7.4, card A = c, there exists a oneto-one mapping ψ : A → R (onto). Define the function ϕ : R → R putting ϕ(x) = ψ(A)

if x ∈ A ∈ A.

Let Ω = Z(H) ∩ (0, ∞).

(12.8.1)

For every ω ∈ Ω ⊂ Z(H) and every x ∈ R we have: if x ∈ A ∈ A, then x + ω ∈ A, whence ϕ(x + ω) = ϕ(x). Thus ϕ is ω-monotonic, and hence Ω-monotonic. But ϕ is not monotonic, since for every t ∈ ϕ(R) we have ϕ−1 (t) = A for a certain A ∈ A, and these sets, being translations of Z(H), are non-measurable (cf. Theorem 11.5.1). Consequently also the function ϕ is non-measurable.

336

Chapter 12. Further Properties of Additive Functions and Convex Functions

Lemma 12.8.1. Let I ⊂ R be an interval, let Ω1 , Ω2 ⊂ (0, ∞) be arbitrary sets, and let ϕ : I → R be a function which is Ω1 -increasing and Ω2 -increasing [Ω1 -decreasing and Ω2 -decreasing]. Then ϕ is (Ω1 + Ω2 )-increasing [(Ω1 + Ω2 )-decreasing]. Proof. Take an x ∈ I and an ω ∈ Ω1 +Ω2 such that x+ω ∈ I. Then ω = ω1 +ω2 , ω1 ∈ Ω1 , ω2 ∈ Ω2 , and x + ω1 ∈ I. Hence, if ϕ is Ω1 -increasing and Ω2 -increasing,   ϕ(x + ω) = ϕ (x + ω1 ) + ω2  ϕ(x + ω1 )  ϕ(x), i.e., ϕ is ω-increasing. If ϕ is Ω1 -decreasing and Ω2 -decreasing, the inequalities are reversed.  Lemma 12.8.2. Let I ⊂ R be an interval, let Ω ⊂ (0, ∞) be a set, and let ϕ : I → R be an Ω-monotonic function. For every δ ∈ (0, ∞) put Ωδ = Ω ∩ (0, δ). If int(Ωδ + Ωδ ) = ∅

for δ ∈ (0, δ0 )

(12.8.2)

and for a δ0 ∈ (0, ∞), then ϕ is monotonic. Proof. Suppose that ϕis Ω-increasing, and take arbitrary x, y ∈ I, x < y. Choose a  1 δ < min δ0 , (y − x) . By (12.8.2) there exists an interval (α, β) ⊂ (Ω+Ω)∩(0, 2δ). 2 By Lemma 12.8.1 ϕ is ω-increasing for every ω ∈ (α, β). We have β < 2δ < y − x. Note that by (12.8.2) Ωδ = ∅ for every δ  > 0. Hence there exists an ω ∈ Ω such that 0 < ω < β−α < y−x−α. Consider the interval J = (y−x−β, y−x−α) ⊂ (0, ∞). Since the length of J is β −α, there exists a k ∈ N such that kω ∈ J, i.e., y −x−kω ∈ (α, β). Hence     ϕ(y) = ϕ (y − kω) + kω  ϕ(y − kω) = ϕ x + (y − x − kω)  ϕ(x). (Note that y −kω ∈ (x+α, x+β) and kω > 0, whence x < y−kω < y and y −kω ∈ I.) Consequently ϕ is increasing. If ϕ is Ω-decreasing, the proof is the same, except that the inequalities must be reversed.  Condition (12.8.2) is certainly fulfilled if mi (Ωδ ) > 0 for δ ∈ (0, δ0 ) (Theorem 3.7.1) or if, for δ ∈ (0, δ0 ), Ωδ contains a second category set with the Baire property (Theorem 2.9.1). But, e.g., for the set Ω given by (12.8.1) condition (12.8.2) is not fulfilled. In fact, Ωδ ⊂ Ω ⊂ Z(H), whence by (11.5.2) Ωδ + Ωδ ⊂ Z(H) + Z(H) = Z(H), and int(Ωδ + Ωδ ) ⊂ int Z(H) = ∅. But we will prove the following Theorem 12.8.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function. Let x0 ∈ D, and let a1 , . . . , aN ∈ RN be linearly independent over R. For every i = 1, . . . , N put di = sup{t > 0 | x0 + tai ∈ D}, and let the functions ϕi : [0, di ) → R be given by ϕi (t) = f (x0 + tai ),

i = 1, . . . , N.

Suppose that εi ∈ (0, di ), i = 1, . . . , N , and that we are given sets Ωi ⊂ (0, εi ) such that Ωi ∈ A1 , i = 1, . . . , N . If, for every i = 1, . . . , N , the function ϕi is Ωi -monotonic in the interval [0, εi ] ⊂ [0, di ), then f is continuous.

12.8. Monotonicity

337

Proof. Write bi = x0 + εi ai ∈ D, i = 1, . . . , N . Let Ti = {x ∈ D | x = x0 + tai , t ∈ Ωi }

(12.8.3)

if ϕi is Ωi -decreasing, and Ti = {x ∈ D | x = x0 + (εi − t)ai = bi − tai , t ∈ Ωi }

(12.8.4)

if ϕi is Ωi -increasing, i = 1, . . . , N . Put T =

N 

Ti .

i=1

We will show that f is bounded above on T by M = max{f (x0 ), f (b1 ), . . . , f (bN )}. Take an arbitrary x ∈ T . There exists an i, 1  i  N , such that x ∈ Ti . Suppose that Ti is given by (12.8.3) so that ϕi is Ωi -decreasing. Then x = x0 + tai with a t ∈ Ωi , so ϕi (t)  ϕi (0). In other words, f (x) = f (x0 + tai ) = ϕi (t)  ϕi (0) = f (x0 )  M. Now suppose that Ti is given by (12.8.4) so that ϕi is Ωi -increasing. Then x = bi − tai with a t ∈ Ωi . Since Ωi ⊂ (0, εi ), we have εi − t ∈ (0, εi ) ⊂ [0, di ), and ϕi (εi − t)  ϕi (εi ). In other words,   f (x) = f x0 + (εi − t)ai = ϕi (εi − t)  ϕi (εi ) = f (x0 + εi ai ) = f (bi )  M. Thus f is bounded above on T . It remains to show that T ∈ AN . Let g : RN → R be an arbitrary additive function bounded above on T : g(x)  K

for x ∈ T .

Put γi (t) = g(x0 + tai ) − g(x0 ) = g(tai ), i = 1, . . . , N. Thus for every i = 1, . . . , N, γi : R → R is an additive function. Suppose that Ti is given by (12.8.3) and t ∈ Ωi . Then x0 + tai ∈ Ti and γi (t)  K − g(x0 ), i.e., γi is bounded above on Ωi , and since Ωi ∈ A1 = B1 (cf. Theorem 10.2.2), γi is continuous. Now suppose that Ti is given by (12.8.4) and t ∈ εi − Ωi . Then t = εi − t , where t ∈ Ωi and x0 + tai = x0 + (εi − t )ai ∈ Ti , whence γi (t)  K − g(x0 ), i.e., γi is bounded above on εi − Ωi . By Theorems 10.2.2, 9.2.8 and 9.2.1 we have εi − Ωi ∈ B1 . Hence γi is continuous. Thus all the functions γi , i = 1, . . . , N , are continuous. By Theorem 5.5.2 there exist constants c1 , . . . , cN ∈ R such that γi (t) = ci t for i = 1, . . . , N . An arbitrary x ∈ RN can be written as x = x1 a1 + · · · + xN aN ,

xi ∈ R, i = 1, . . . , N ,

338

Chapter 12. Further Properties of Additive Functions and Convex Functions

since a1 , . . . , aN are linearly independent over R. Then g(x) = g(x1 a1 ) + · · · + g(xN aN ) = γ1 (x1 ) + · · · + γN (xN ) =

N

ci xi ,

i=1

whence it follows that g is continuous. Thus T ∈ BN = AN (Theorem 10.2.2).



Since every function monotonic in an interval (0, ε) is (0, ε)-monotonic, and by Lemma 9.3.1 open intervals belong to A1 , it follows that (with the notation as in Theorem 12.8.1), if every function ϕi is monotonic in a right vicinity of zero, then f is continuous. In the particular case N = 1 we obtain hence Corollary 12.8.1. Let D ⊂ R be an open interval, and let f : D → R be a convex function. If f is monotonic on an interval I ⊂ D, then f is continuous. The following theorem is in a sense converse to Theorem 12.8.1. Theorem 12.8.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a continuous convex function. Let x0 ∈ D and a ∈ RN , a = 0. Let d = sup{t > 0 | x0 + ta ∈ D}, and let ϕ : [0, d) → R be given by ϕ(t) = f (x0 + at) . Then the function ϕ is monotonic in an interval (0, ε) ⊂ (0, d). Proof. Let ψ = ϕ | (0, d). Then ψ : (0, d) → R is a continuous convex function of a single real variable. By Theorem 7.3.5 either ψ is monotonic in (0, d), or there exists an ε ∈ (0, d) such that ψ is decreasing in (0, ε). The theorem follows. 

Exercises 1. Let f : RN → R be a discontinuous additive function, and let A ⊂ RN be a set which is either measurable of positive measure, or of the second category and with the Baire property. Show that cl f (A) = R. 2. Let f : RN → R be a discontinuous additive function, and let w : R → R be an arbitrary function. Show that for every interval I ⊂ R we have     mi f −1 w−1 w(R) \ w(I) = 0. 3. Let f : R → R be an additive function, let A ⊂ R \ {0} be a set which is either measurable of positive measure, or of the second category and with the Baire property. Suppose that f satisfies the condition   1 f (x)f > c for x ∈ A x with a certain c > 0. Show that f is continuous11 . 11

Acz´el [7]; cf. also Kannappan [161]. The problem whether this result remains valid also when c = 0 remains still open (Rothberger [270]; cf. also Benz [26]). Added in proof: This has recently been solved by G. M. Bergman [29].

12.8. Monotonicity

339



4.

5.

6.

7.

8.

9.

Hint: Let g : A → R be given by g(x) = x+x−1 . On g(A) we have f (t) = f (x)+ √ −1 −1 1/2 f (x−1 ) with >2 c

an x ∈ A, and |f (t)| = |f (x)| + |f (x )|  2|f (x)f (x )| (cf. 8.3). Show that in Lemma 12.3.1 the assumption that J has a finite upper bound is essential. function. Consider the Hint: Let g : RN → R be a discontinuous additive

function f : RN → R given by f (x) = |x| + |g(x)|. Let D ⊂ RN be a convex and open set, and let f : D → R be a discontin uous convex function. Show that for every c ∈ R we have mi f −1 (c, ∞) ∩  

(−∞, c) = 0. m−1 f Let f : RN → R be a function with big graph, and let A ⊂ RN be a set which is either measurable of positive measure or of the second category with the Baire property. Show that f (A) = R. Let f : RN → R be an additive function, and let H ⊂ RN be a Hamel basis. Show that if f (H) = {0} and card f (H)  ℵ0 , then f is a function with small graph.

Hint: Use Exercise 4.4 and Lemma 4.1.3. Show that there exists a non-constant periodic convex function f : R → R which is not additive. Hint: Let

g : R → R, g = 0, be a non-invertible additive function, and take f = |g|. Show by a suitable counter-example that the following extension of the Egorov theorem is invalid (Kuczma [189]): Let A ⊂ R be a measurable set with m(A) < ∞, and let f : A × (0, 1) → R be a function such that for every fixed t ∈ (0, 1) the function f (x, t) as a function of x is measurable. Suppose that there exists a function fˆ : A → R such that ˆ lim f (x, t) = f(x) for almost every x ∈ A. Then, for every ε > 0 and η > 0, t→0

there exists a closed set F ⊂ A and a δ > 0 such that m(A \ F ) < η and ˆ |f (x, t) − f (x)| < ε uniformly on F × (0, δ). Hint: Let g : R → R be an invertible discontinuous additive function. Take

A = [0, 1], and f (x, t) = 1 when (x, t) ∈ Gr(g), f (x, t) = 0 when (x, t) ∈ / Gr(g). 10. Show that there exists an additive function f : RN → R such that f (RN ) = R, and the level sets of f are saturated non-measurable, have property (∗) from 3.3, and have the power of continuum (Sm´ıtal [286]). Hint: Let f1 : R → R be an additive function such that f1 (R) = R and card Ker f1 = c, and let f2 : RN → R be an additive function with big graph. Put f = f1 ◦ f2 .

Part III

Related Topics

Chapter 13

Related Equations 13.1 The remaining Cauchy equations The following functional equations are also referred to as Cauchy’s equations (Cauchy [41]; cf. also Acz´el [5]) f (x + y) = f (x)f (y),

(13.1.1)

f (xy) = f (x) + f (y),

(13.1.2)

f (xy) = f (x)f (y).

(13.1.3)

Equation (13.1.1) will be considered for functions f : RN → R, whereas equations (13.1.2) and (13.1.3) will be considered only for functions of a real variable, because of the operation of multiplication occurring in the argument. Lemma 13.1.1. Let f : RN → R be a solution of equation (13.1.1). Then either f = 0, or f > 0 in RN . Proof. Suppose that there exists an x0 ∈ RN such that f (x0 ) = 0. Then, for arbitrary x ∈ RN ,   f (x) = f (x − x0 ) + x0 = f (x − x0 )f (x0 ) = 0, i.e., f = 0. Now suppose that f = 0. By what has just been shown, f is never zero in RN . For an arbitrary x ∈ RN we have 1 1 1 f (x) = f ( x + x) = [f ( x)]2  0, 2 2 2 and since f (x) = 0, we obtain f (x) > 0. Thus f is positive in RN .



Theorem 13.1.1. Let f : R → R be a solution of equation (13.1.1). Then either f = 0, or there exists an additive function g : RN → R such that N

f = exp g.

(13.1.4)

Proof. The function f = 0 clearly is a solution of (13.1.1). So suppose that f = 0. By Lemma 13.1.1 f is positive in RN . Put g = log f . Then, by (13.1.1), g : RN → R is an additive function, and the theorem follows. 

344

Chapter 13. Related Equations

On the other hand, it is easily seen that every function of form (13.1.4), where g : RN → R is additive, actually satisfies (13.1.1). Lemma 13.1.2. Let D ⊂ R be such that xy ∈ D for every x, y ∈ D, and 0 ∈ D. The only solution f : D → R of equation (13.1.2) is f = 0. Proof. Setting in (13.1.2) y = 0 we obtain f (0) = f (x) + f (0), whence f (x) = 0 for every x ∈ D. Of course, f = 0 is a solution of (13.1.2).  But we can obtain non-trivial solutions of (13.1.2) if we exclude 0 from the domain of definition of f . Thus we have Theorem 13.1.2. Let D = (0, ∞), or D = (−∞, 0) ∪ (0, ∞). If a function f : D → R satisfies equation (13.1.2), then there exists an additive function g : R → R such that   f (x) = g log |x| for x ∈ D. (13.1.5) Proof. For t ∈ R put g(t) = f (et ) so that g : R → R. We have by (13.1.2) for arbitrary u, v ∈ R g(u + v) = f (eu+v ) = f (eu ev ) = f (eu ) + f (ev ) = g(u) + g(v),     i.e., g is additive. Moreover, f (x) = f |x| = g log |x| for x > 0. For x = 1 we obtain hence f (1) = g(0) = 0. Now, if D = (−∞, 0) ∪ (0, ∞), set in (13.1.2) x = y = −1. Then 0 = f (1) = f (−1) + f (−1) = 2f (−1), whence f (−1) = 0. For x < 0, y = −1, equation (13.1.2) yields   f |x| = f (−x) = f (x) + f (−1) = f (x). Hence we obtain (13.1.5) for x < 0. So (13.1.5) is generally valid.



On the other hand, it is easily seen that every function of form (13.1.5), where g : R → R is additive, actually satisfies (13.1.2). Equation (13.1.3) is most complicated of all the three equations. We will solve it for functions f : D → R, where D is one of the sets (0, 1), [0, 1), (−1, 1), (−1, 0) ∪ (0, 1), (1, ∞), (0, ∞), [0, ∞), (−∞, 0) ∪ (0, ∞), R. (13.1.6) Lemma 13.1.3. Let D be one of sets (13.1.6). If f : D → R, f = 0, is a solution of (13.1.3), then f (1) = 1 if 1 ∈ D; f (−1) = 1 or −1 if −1 ∈ D; f (0) = 0 or 1 if 0 ∈ D. If 0 ∈ D and f (0) = 1, then f (x) = 1 for all x ∈ D. Proof. Setting in (13.1.3) in turn x = y = 1; x = y = −1; x = y = 0, we obtain

2 f (1) = f (1)f (1) = f (1) , whence f (1) = 0 or 1; if f (1) = 0, then we have for arbitrary x ∈ D f (x) = f (1)f (x) = 0,

13.1. The remaining Cauchy equations

345

contrary to the assumption that f = 0. So f (1) = 1. Next,

2 1 = f (1) = f (−1)f (−1) = f (−1) , whence f (−1) = 1 or −1. And

2 f (0) = f (0)f (0) = f (0) , whence f (0) = 0 or 1. If f (0) = 1, then we have for every x ∈ D f (x) = f (0)f (x) = f (0) = 1, 

and the proof is complete. Lemma 13.1.4.1 If D is one of the sets (0, 1),

[0, 1),

(1, ∞),

and if f0 : D → R, f0 = 0, is a solution of (13.1.3), then the function f : D ∪ {1} ∪  −1

→ R given by D \ {0} ⎧ ⎪ ⎨f0 (x) f (x) = 1 ⎪

−1 ⎩ f0 (x−1 )

if x ∈ D , if x = 1 ,  −1 if x ∈ D \ {0}

(13.1.7)

 −1 satisfies equation (13.1.3) in D ∪ {1} ∪ D \ {0} and f0 = f | D. Proof. First note that f0 (x) = 0 for x ∈ D, x = 0. In fact, suppose that f0 (x0 ) = 0 for an x0 ∈ D, x0 = 0. Suppose, for the argument’s sake, that x0 ∈ (−1, 1) ∩ D. Take an arbitrary x ∈ D such that |x| < |x0 |. Then |x/x0 | < 1 and x/x0 ∈ D. Hence     x x = f0 (x0 )f0 = 0. f0 (x) = f0 x0 x0 x0 Now take an arbitrary x ∈ D. We have from (13.1.3) by induction

n for n ∈ N. f0 (xn ) = f0 (x)

(13.1.8)

But for sufficiently large n ∈ N we have |xn | < |x0 |. Thus f0 (xn ) = 0, and by (13.1.8) f0 (x) = 0. Consequently f0 = 0 in D, contrary to the assumption. Thus definition (13.1.7) is meaningful. The relation f0 = f |D results directly from (13.1.7). It remains to show that f given by (13.1.7) satisfies (13.1.3) for x, y ∈  −1 D ∪ {1} ∪ D \ {0} . 1 In the following part of this section, some corrections were made in the 2nd edition, based on M. Kuczma, Note on multiplicative functions, Ann. Math. Sil. 3(15) (1990), 45–50, by K. Baron.

346

Chapter 13. Related Equations

We must distinguish several cases. I. x, y ∈ D. Then also xy ∈ D, and by (13.1.7) and (13.1.3) for f0 f (xy) = f0 (xy) = f0 (x)f0 (y) = f (x)f (y). II. x arbitrary, y = 1. Then, by (13.1.7), f (y) = 1. Hence f (xy) = f (x) = f (x)f (y). Of course, if x = 1 and y is arbitrary, then we obtain hence (13.1.3) in virtue of the commutativity of the operation of multiplication.  −1  −1 . Then also xy ∈ D \ {0} , and by (13.1.7) and (13.1.3) III. x, y ∈ D \ {0}





−1

−1 −1 −1 −1 −1 −1 f0 (y ) = f0 (x )f0 (y −1 ) = f0 ((xy)−1 ) = for f0 f (x)f (y) = f0 (x ) f (xy).  −1

−1 IV. x ∈ D, y ∈ D \ {0} , xy ∈ D. Then by (13.1.7) f (y) = f0 (y −1 ) and y −1 ∈ D. We have     f0 (xy)f0 y −1 = f0 xyy −1 = f0 (x), whence

 −1  = f0 (xy) = f (xy). f (x)f (y) = f0 (x) f0 y −1

 −1

−1 V. x ∈ D, y ∈ D \ {0} , xy = 1. Then y −1 = x, whence f (y) = f0 (y −1 ) =

−1 and f0 (x)

f (x)f (y) = f0 (x) f0 (x) n−1 = 1 = f (xy).  −1  −1 VI. x ∈ D, y ∈ D \ {0} , xy ∈ D \ {0} . Then x−1 y −1 = (xy)−1 ∈ D, and

−1

−1

−1  −1 −1  −1

−1 f0 (x−1 y −1 ) f0 (x) = f0 (x−1 y −1 )f0 (x) = f0 x y x = f0 (y −1 ) , whence

 −1  −1 −1  −1  = f0 x y = f (xy). f (x)f (y) = f0 (x) f0 y −1

 −1 Of course, if x ∈ D \ {0} and y ∈ D, then we obtain (13.1.3) in virtue of the commutativity of the operation of the multiplication.  The next lemma shows that we may restrict (13.1.6) to the following sets (0, ∞),

[0, ∞),

(−∞, 0) ∪ (0, ∞),

R.

(13.1.9)

Lemma 13.1.5. If D0 is one of sets (13.1.6), and f0 : D0 → R satisfies equation (13.1.3), then there exist a set D of form (13.1.9) and a function f : D → R satisfying (13.1.3) such that D0 ⊂ D and f0 = f | D0 .

13.1. The remaining Cauchy equations

347

Proof. Clearly, we may assume that f0 = 0 and by Lemma 13.1.4 it is enough to consider sets D0 of the form (−1, 1),

(−1, 0) ∪ (0, 1)

only. Like in the proof of Lemma 13.1.4, we show that f (x) = 0 for x ∈ D0 , x = 0. For arbitrary x, y ∈ D0 \ {0} we have −x, −y ∈ D0 \ {0} and by (13.1.3) f0 (−x) f0 (y) = f0 (−xy) = f0 (x) f0 (−y), whence

f0 (−x) f0 (−y) = . f0 (x) f0 (y)

(13.1.10)

Relation (13.1.10) means that ε=

f0 (−x) , f0 (x)

x ∈ D0 \ {0},

(13.1.11)

is constant. Setting y = −x in (13.1.10), we obtain ε2 = 1 .

(13.1.12)

Relation (13.1.11) may be written as f0 (−x) = εf0 (x)

(13.1.13)

for x ∈ D0 \ {0}. If 0 ∈ D0 , then according to Lemma 13.1.3 f (0) is either 0 or 1; in the latter case f0 (x) = 1 for all x ∈ D0 . Thus, (13.1.13) is fulfilled also for x = 0. Consequently, (13.1.13) holds in the whole of D0 .

Now, we define the function f : D0 ∪{1}∪(D0 \ {0})−1 ∪{−1} → R by (13.1.7) with D replaced by D0 and f (−1) = ε , (13.1.14) where ε is defined by (13.1.11). It remains to verify that f satisfies equation (13.1.3) −1 in D0 ∪ (D0 \ {0}) ∪ {1} ∪ {−1}. Beside the cases dealt with in the proof of Lemma 13.1.4, we need to consider also the cases where one of x, y, xy is −1. If x = −1, y ∈ D0 , then (13.1.3) results from (13.1.7), (13.1.13) and (13.1.14). If x = −1, y = 1, then we obtain (13.1.3) from (13.1.7) and (13.1.14). If x = y = −1, then (13.1.7), (13.1.14) and (13.1.12) imply (13.1.3). If x = −1, y ∈ (D0 \ {0})−1 , −1 then also −y ∈ (D0 \ {0}) and by (13.1.7), (13.1.13), (13.1.12) and (13.1.14)

−1

−1

−1 = εf0 (y −1 ) = ε f0 (y −1 ) = f (x) f (y) . f (xy) = f (−y) = f0 (−y −1 ) If x ∈ D0 , y ∈ (D0 \ {0})−1 , xy = −1, then y −1 = −x and

−1

−1

−1 = f0 (x) f0 (−x) = f0 (x) εf0 (x) = ε = f (xy) f (x) f (y) = f0 (x) f0 (y −1 ) in virtue of (13.1.7), (13.1.13), (13.1.12) and (13.1.14). The cases where y = −1, or x ∈ (D0 \ {0})−1 , y ∈ D0 , xy = −1, follow from those already considered in view of the commutativity of the multiplication of real numbers. 

348

Chapter 13. Related Equations

Theorem 13.1.3. Let D be one of the sets (13.1.6). If a function f : D → R satisfies equation (13.1.3), then there exists an additive function g : R → R such that f has one of the following forms: f = 0, f = 1,

(13.1.15) 



 exp g log |x| f (x) = 0 ⎧   ⎪ ⎨exp g log |x| f (x) = 0 ⎪   ⎩ − exp g log |x|

(13.1.16) for x = 0, x ∈ D, for x = 0 (if 0 ∈ D),

(13.1.17)

for x > 0, x ∈ D, for x = 0 (if 0 ∈ D), for x < 0, x ∈ D.

(13.1.18)

Proof. By Lemma 13.1.5 it is enough to consider sets D of form (13.1.9). Let x > 0, x ∈ D. Then by (13.1.3) √ 2 f (x) = f ( x)  0. (13.1.19) Suppose that there exists an x0 ∈ D, x0 = 0, such that f (x0 ) = 0. Take an arbitrary x ∈ D. Then x/x0 ∈ D and f (x) = f

    x x = f (x0 )f = 0. x0 x0 x0

Thus f = 0 in D, and we obtain (13.1.15). Now suppose that f (x) = 0 for x ∈ D \{0}. By Lemma 13.1.3 f (0) = 0 or 1. If f (0) = 1, then f = 1 in D in virtue of Lemma 13.1.3, and we obtain (13.1.16). So let f (0) = 0. By (13.1.19) f (x) > 0 for x > 0. Let f1 (x) = log f (x) for x > 0. Then f1 : (0, ∞) → R satisfies equation (13.1.2), whence by Theorem 13.1.2 there exists an additive function g : R → R such that f1 (x) = g(log x). Again by Lemma 13.1.3 f (−1) = 1 or − 1 (if −1 ∈ D). If −1 ∈ D and f (−1) = 1, then we have for arbitrary x ∈ D f (−x) = f (−1)f (x) = f (x), i.e., f is even. In this case we obtain (13.1.17). If −1 ∈ D and f (−1) = −1, then for arbitrary x ∈ D f (−x) = f (−1)f (x) = −f (x), i.e., f is odd. Hence we obtain (13.1.18).



On the other hand, it is easy to check that every function of form (13.1.15), (13.1.16), (13.1.17), or (13.1.18), where g : R → R is additive, actually satisfies equation (13.1.3). Now we are going to determine the continuous solutions of (13.1.1), (13.1.2) and (13.1.3).

13.1. The remaining Cauchy equations

349

Theorem 13.1.4. A function f : RN → R is a continuous solution of equation (13.1.1) if and only if either f = 0, or f (x) = ecx ,

x ∈ RN ,

(13.1.20)

with a certain constant c ∈ RN . Proof. Let f = 0. Since f satisfies (13.1.1), according to Theorem 13.1.1 f has form (13.1.4), where g : RN → R is an additive function. Since f is continuous, g = log f also is continuous. By Theorem 5.5.2 g(x) = cx with a certain c ∈ RN . Formula (13.1.20) results now from (13.1.4). It is obvious that f = 0 and every f of form (13.1.20) is a continuous solution of (13.1.1).  Theorem 13.1.5. Let D = (0, ∞) or D = (−∞, 0) ∪ (0, ∞). A function f : D → R is a continuous solution of equation (13.1.2) in D if and only if f (x) = c log |x|,

x ∈ D,

(13.1.21)

with a certain constant c ∈ R.

  Proof. By Theorem 13.1.2 f has form (13.1.5), where g(t) = f et is an additive function. Since f is continuous, g also is continuous, and hence, by Theorem 5.5.2, g(t) = ct with a certain c ∈ R. Formula (13.1.21) results now from (13.1.5). It is obvious that every function of form (13.1.21) is a continuous solution of (13.1.2).  Theorem 13.1.6. Let D be one of sets (13.1.6). A function f : D → R is a continuous solution of equation (13.1.3) if and only if either f = 0, or f = 1, or f has one of the following forms f (x) = |x|c , x ∈ D, (13.1.22) f (x) = |x|c sgn x,

x ∈ D,

(13.1.23)

with a certain c ∈ R. If 0 ∈ D, then c > 0. Proof. By Theorem 13.1.3 either f = 0, or f = 1, or f has form (13.1.17) or (13.1.18),   where g : R → R is an additive function. We have g(t) = log f et , and since f is continuous, so is also g, and consequently g(t) = ct with a certain c ∈ R. Formulas (13.1.22) and (13.1.23) result now from (13.1.17) and (13.1.18). Suppose that 0 ∈ D. If we had c = 0, then (13.1.22) yields f (x) = 1 for x = 0, and by the continuity we must have f (0) = 1. Thus we obtain the solution f = 1, already listed. Formula (13.1.23) with c = 0 yields  f (x) = 1 for x > 0 and f (x) = −1 for x < 0, and thus f cannot  be continuous. If D ⊂ [0, ∞), then both formulas (13.1.22) and (13.1.23) coincide . Similarly, if c < 0 then f given by (13.1.22) or (13.1.23) satisfies lim f (x) = ∞, and x→0+

thus f cannot be continuous at zero. It is obvious that every one of the specified functions is a continuous solution of (13.1.3). 

350

Chapter 13. Related Equations

Remark 13.1.1. If 0 ∈ D, and f is assumed continuous only in D \ {0}, then besides the solutions listed in Theorem 13.1.6 we obtain also  |x|c for x = 0, x ∈ D, f (x) = (13.1.24) 0 for x = 0, 

and f (x) =

|x|c sgn x for x = 0, x ∈ D, 0 for x = 0,

(13.1.25)

where c  0. This follows from the proof of Theorem 13.1.6 and from Lemma 13.1.3. Simple conditions ensuring the continuity of solutions of equations (13.1.1)– (13.1.3) may be inferred from (13.1.4), (13.1.5), (13.1.17) and (13.1.18). Below we formulate some such conditions. Theorem 13.1.7. Let f : RN → R be a solution of equation (13.1.1). If f is measurable, or bounded above on a set T ∈ A, or bounded below by a positive constant on a set T ∈ A, then f is continuous. Proof. If f = 0, then it is already continuous. If f = 0, then, according to Theorem 13.1.1, f is given by (13.1.4), where g : RN → R is an additive function. If f is measurable, then so is also g = log f , and hence, by Theorem 9.4.2, g is continuous. Then by (13.1.4), f is continuous, too. If f is bounded above, or f is bounded below by a positive constant, on a set T ⊂ RN , then g is bounded above resp. bounded below on T . If T ∈ A, then g is continuous, and hence also f is continuous.  Theorem 13.1.8. Let D = (0, ∞), or D = (−∞, 0) ∪ (0, ∞), and let f : D → R be a solution of equation (13.1.2). If f is measurable, or bounded above or below on a set T ⊂ D such that mi (T ) > 0 or T is of the second category and with the Baire property, then f is continuous. Proof. By Theorem 13.1.2 f has form(13.1.5), where g : R → R is an additive  function. (13.1.5) implies that g(t) = f et , so if f is measurable, then so is also g. By Theorem 9.4.2 g is then continuous, so also f is continuous. Suppose that f is bounded above or below on a set T ⊂ D. It follows from (13.1.5) that f is even, so f is also bounded (above or below) on the set T0 = (0, ∞) ∩ T ∪ (−T ) , and is bounded (above or below) on the set log T0 = {t ∈ R | et ∈ T0 }. If mi (T ) > 0, then also mi (T0 ) > 0, and since the function log is absolutely continuous (Theorem 7.4.6), also mi (log T0 ) > 0. If T is of the second category with the Baire property, then so is also T0 and log T0 , since the function log is a homeomorphism. By Theorems 9.3.1 and 9.3.2 g is continuous, and hence also f is continuous in D.  For solutions of equation (13.1.3) the situation is more complicated. For example, if 0 ∈ D, then functions (13.1.24) and (13.1.25) (c  0) are measurable, bounded on suitable open sets, satisfy equation (13.1.3), but are not continuous at zero. However, we have the following theorem, which may be proved similarly as Theorem 13.1.7 and 13.1.8, and therefore its proof is left to the reader. Theorem 13.1.9. Let D be one of sets (13.1.6), and let f : D → R be a solution of equation (13.1.3). If f is measurable, then it is continuous in D \ {0}.

13.2. Jensen equation

351

13.2 Jensen equation The equation resulting on replacing in the Jensen inequality (5.3.1) the sign of inequality by that of equality   f (x) + f (y) x+y = (13.2.1) f 2 2 is known as the Jensen equation (cf. Acz´el [5]). We will consider equation (13.2.1) for functions f : D → R, where D ⊂ RN is a convex set. If D were also open, then f satisfying (13.2.1) would be convex, and thus all the results established in the previous chapters of this book for convex functions would apply. But most of these results become invalid when the set D is no longer open. For example, let N = 1, D = [−1, 1], and let f : D → R be defined as  x2 if x ∈ (−1, 1), f (x) = 2 if |x| = 1. Such an f satisfies inequality (5.3.1) in D = [−1, 1], is measurable and bounded in D, but is discontinuous at x = 1 and x = −1. So we treat equation (13.2.1) in a more general setting. Lemma 13.2.1. Let D ⊂ RN be a convex set such that 0 ∈ D, and let f : D → R be a solution of equation (13.2.1) such that f (0) = 0.

(13.2.2)

Then, for every x ∈ D and n ∈ N, f

x 1 = n f (x). n 2 2

(13.2.3)

1 (x + 0) ∈ D, and by (13.2.1) and (13.2.2) 2   x x+0 f =f 2 2 f (x) f (x) + f (0) = . (13.2.4) = 2 2

Proof. Take an x ∈ D. Since D is convex,

Thus (13.2.3) holds for n = 1. Assuming it true for an n ∈ N, we have   x 1 1 0 ∈ D, = x + 1 − 2n+1 2n+1 2n+1 and by (13.2.3) (for n) and (13.2.4)  x  1 1 x = n+1 f (x). f n+1 = n f 2 2 2 2 Induction completes the proof.



352

Chapter 13. Related Equations

Lemma 13.2.2. Let D ⊂ RN be a convex set such that int D =  ∅, and let f : D → R be a solution of equation (13.2.1). Fix an x0 ∈ int D, and define the function f0 : (D − x0 ) → R by f0 (x) = f (x0 + x) − f (x0 ). (13.2.5) Then there exists a unique function f1 : RN → R satisfying equation (13.2.1) in RN and such that f1 (x) = f0 (x) for x ∈ D − x0 . (13.2.6) Proof. Function (13.2.5) is defined for x ∈ D − x0 . First we verify that f0 satisfies equation (13.2.1) in D − x0 . For every x, y ∈ D − x0 , we have x0 + x, x0 + y ∈ D, and by (13.2.5) and (13.2.1)       x+y x+y x0 + x + x0 + y f0 = f x0 + − f (x0 ) = f − f (x0 ) 2 2 2 1 1 = f (x0 + x) + f (x0 + y) − f (x0 ) 2 2

1

1 = f (x0 + x) − f (x0 ) + f (x0 + y) − f (x0 ) 2 2

1 = f0 (x) + f0 (y) . 2 Also, it is easily seen that 0 ∈ D − x0 and by (13.2.5) f0 (0) = 0.

(13.2.7)

Now put D0 = D − x0 and Dn = 2 n D0 ,

n ∈ N.

x x If x ∈ Dn , then n ∈ D0 . D0 is convex, just like D, and 0 ∈ D0 , whence n+1 = 2 2  1x + 0 ∈ D0 , and x ∈ 2n+1 D0 = Dn+1 . Thus 2 2n Dn ⊂ Dn+1 ,

n ∈ N ∪ {0}.

(13.2.8)

Also, 0 ∈ int D0 , since x0 ∈ int D. For every x ∈ RN we have lim x/2n = 0, whence n→∞

it follows that there exists an n ∈ N ∪ {0} such that x/2n ∈ D0 , whence x ∈ Dn . Hence ∞  D n = RN . (13.2.9) n=0

Define the function f1 : R

→ R as follows: x f1 (x) = 2n f0 n if x ∈ Dn , n ∈ N ∪ {0}. 2 N

(13.2.10)

First we must check whether definition (13.2.10) is correct. If x ∈ Dn = 2n D0 , then x/2n ∈ D0 = D − x0 , and the expression f0 (x/2n ) is meaningful. If x ∈ Dn and

13.2. Jensen equation

353

x ∈ Dm for n = m, say n < m, then by (13.2.7) and Lemma 13.2.1  x  x x 1 2m f0 m = 2m m−n f0 n = 2n f0 n , 2 2 2 2 and thus (13.2.10) is unambiguous. By (13.2.9) the function f1 is defined in the whole of RN . We must verify that it satisfies equation (13.2.1) in RN . Take arbitrary x, y ∈ RN . By (13.2.9) and (13.2.8) there exists an n ∈ N ∪ {0} x+y 1x y  such that x, y ∈ Dn . Hence x/2n , y/2n ∈ D0 = D − x0 , and n+1 = + n ∈ n 2 2 2 2 x+y ∈ Dn . Now, D0 , whence 2     x+y x+y n = 2 f0 f1 2 2n+1   x  1  y  n 1 f0 n + f0 n =2 2 2 2 2     x y  f1 (x) + f1 (y) 1 n 2 f0 n + 2 n f0 n . = = 2 2 2 2 Relation (13.2.6) results from (13.2.10) for n = 0. To prove the uniqueness, suppose that a function f2 : RN → R satisfies equation (13.2.1) in RN and fulfils the condition f2 (x) = f0 (x) for x ∈ D − x0 .

(13.2.11) x By (13.2.11) and (13.2.7) f2 (0) = f0 (0) = 0, and hence, by Lemma 13.2.1, f2 n = 2 1 N N f2 (x), x ∈ R , n ∈ N ∪ {0}. Take an arbitrary x ∈ R . By (13.2.9) there exists an 2n n ∈ N ∪ {0} such that x ∈ Dn = 2n D0 , whence x/2n ∈ D0 = D − x0 . Thus we have by (13.2.11) and (13.2.10) x x f2 (x) = 2n f2 n = 2n f0 n = f1 (x). 2 2 Consequently f2 = f1 in RN . Lemma 13.2.3. Let a function f : R (13.2.2). Then f is additive.

 N

→ R satisfy equation (13.2.1) and relation

Proof. We have by Lemma 13.2.1 for arbitrary x, y ∈ RN   f (x) + f (y) x+y =2 = f (x) + f (y), f (x + y) = 2f 2 2 

i.e., f is additive.

Theorem 13.2.1. Let D ⊂ R be a convex set such that int D = ∅, and let f : D → R be a solution of equation (13.2.1). Then there exist an additive function g : RN → R and a constant a ∈ R such that N

f (x) = g(x) + a for x ∈ D.

(13.2.12)

354

Chapter 13. Related Equations

Proof. Fix an x0 ∈ int D and define the function f0 : (D − x0 ) → R by (13.2.5). By Lemma 13.2.2 there exists a function f1 : RN → R satisfying equation (13.2.1) and condition (13.2.6). Hence by (13.2.7) f1 (0) = f0 (0) = 0. By Lemma 13.2.3 f1 is additive. For arbitrary x ∈ D we have x − x0 ∈ D − x0 , whence by (13.2.5) and (13.2.6)   f (x) = f x0 + (x − x0 ) = f0 (x − x0 ) + f (x0 ) = f1 (x − x0 ) + f (x0 ). Since f1 is additive, we get hence f (x) = f1 (x) − f1 (x0 ) + f (x0 ).

(13.2.13)

Put g = f1 , a = f (x0 ) − f1 (x0 ). Relation (13.2.12) results now from (13.2.13).



It is easily seen that every function of form (13.2.12), where g : RN → R is additive, actually satisfies equation (13.2.1). Theorem 13.2.2. Let D ⊂ RN be a convex set such that int D = ∅. A function f : D → R is a continuous solution of equation (13.2.1) if and only if f (x) = cx + a,

x ∈ D,

(13.2.14)

with certain constants c ∈ RN , a ∈ R. Proof. By Theorem 13.2.1 we have (13.2.12) with an additive g : RN → R. We have g = f − a in D, and since f is continuous, also g is continuous in int D and hence (Theorem 6.4.3) in RN . By Theorem 5.5.2 g(x) = cx with a certain constant c ∈ RN . Thus (13.2.14) results from (13.2.12). It is obvious that every function of form (13.2.14) is a continuous solution of (13.2.1).  Even if D is convex and open, then, using the fact that f fulfilling (13.2.1) is convex, we can infer on its continuity only from the boundedness above on a suitable set. If D is not open, then we cannot do even this, as it has been pointed out at the beginning of this section. But from Theorem 13.2.1 we obtain the following result. Theorem 13.2.3. Let D ⊂ RN be a convex set such that int D = ∅, and let f : D → R be a solution of equation (13.2.1). If f is measurable, or is bounded above, or below, on a set T ∈ A, then f is continuous. Proof. By Theorem 13.2.1 we have (13.2.12), and if f is measurable, then so is also g | int D. The function g | int D is convex (Theorem 5.1.5), and hence, by Theorem 9.4.1 is continuous. By Theorem 6.4.3 g is continuous, and thus, by (13.2.12), f is continuous. If f is bounded above, or below, on a set T ⊂ D, then so is also g. Since T ∈ A, this implies that g is continuous (if g is bounded below on T , then −g is bounded above on T ), and hence, by (13.2.12), f is continuous. 

13.3. Pexider equations

355

13.3 Pexider equations These are the equations (Pexider [254], Acz´el [5]) f (x + y) = g(x) + h(y),

(13.3.1)

f (x + y) = g(x)h(y), f (xy) = g(x) + h(y),

(13.3.2) (13.3.3)

f (xy) = g(x)h(y),

(13.3.4)

with three unknown functions f, g, h. Equation (13.3.1) has a simple interpretation: the value of f at x + y is the sum of two values, of which one depends only on x, and the other only on y (separation of variables). Similar interpretations can also be given for equations (13.3.2)–(13.3.4). Equations (13.3.1)–(13.3.4) can easily be reduced to the Cauchy equations (5.2.1) and (13.1.1)–(13.1.3). Below we give theorems to this effect. Theorem 13.3.1. Let f : RN → R , g : RN → R, and h : RN → R satisfy equation (13.3.1). Then there exist an additive function f0 : RN → R and constants a, b ∈ R such that f = f0 + a + b, g = f0 + a, h = f0 + b. (13.3.5) Proof. Put a = g(0), b = h(0), f0 = f − a − b. Setting in (13.3.1) y = 0 we obtain f (x) = g(x) + b, x ∈ RN , or f0 + a + b = g + b, i.e., g = f0 + a. Similarly, setting in (13.3.1) x = 0 and writing x instead of y, we obtain f (x) = a + h(x), x ∈ RN , or f0 + a + b = a + h, i.e., h = f0 + b. The formula f = f0 + a + b results from the definition of f0 . It remains to check that f0 : RN → R is additive. We have by (13.3.5) and (13.3.1) f0 (x + y) = f (x + y) − a − b = g(x) − a + h(y) − b = f0 (x) + f0 (y). 

Consequently f0 is additive.

On the other hand, it is easily seen that every triple (f, g, h) of form (13.3.5), where f0 : RN → R is additive, and a, b ∈ R are arbitrary constants, actually satisfies equation (13.3.1). Before we give the general solution of equation (13.3.2), note that every triple as follows: ⎧ ⎧ ⎪ ⎪ ⎨f = 0, ⎨f = 0, (13.3.6) g = 0, g arbitrary, ⎪ ⎪ ⎩ ⎩ h arbitrary, h = 0, is a solution of (13.3.2). Theorem 13.3.2. Let f : RN → R , g : RN → R, and h : RN → R satisfy equation (13.3.2). Then either f, g, h have one of forms (13.3.6), or there exist a function f0 : RN → R satisfying equation (13.1.1) and constants a, b ∈ R \ {0} such that f = abf0 ,

g = af0 ,

h = bf0 .

(13.3.7)

356

Chapter 13. Related Equations

Proof. If f (0) = 0, then setting in (13.3.2) x = y = 0 we obtain that either g(0) = 0, or h(0) = 0. If g(0) = 0, then setting in (13.3.2) x = 0 we obtain f = 0. Similarly, if h(0) = 0, then setting in (13.3.2) y = 0 we obtain that f = 0. Thus f (0) = 0 implies that f = 0 in RN . If there existed x, y ∈ RN such that g(x) = 0 and h(y) = 0, then we would have by (13.3.2) f (x + y) = g(x)h(y) = 0, which is impossible. Consequently at least one of the functions g, h must be identically zero. Thus f, g, h must have one of forms (13.3.6).  So let f (0) = 0 as may be seen from (13.3.2)  = 0. Then also g(0) = 0 and h(0) for x = y = 0 . Put a = g(0), b = h(0), f0 = a−1 b−1 f . Then, setting in (13.3.2) y = 0 we obtain f = bg, or abf0 = bg, i.e., g = af0 . Similarly setting in (13.3.2) x = 0 we obtain f = ah, or abf0 = ah, i.e., h = bf0 . The formula f = abf0 results from the definition of f0 . Now, for arbitrary x, y ∈ RN , f0 (x + y) = a−1 b−1 f (x + y) = a−1 b−1 g(x)h(y) = f0 (x)f0 (y), i.e., f0 satisfies (13.1.1).



On the other hand, it is easily seen that every triple of form (13.3.7), where f0 : RN → R satisfies (13.1.1), and a, b ∈ R \ {0} are arbitrary constants, actually satisfies equation (13.3.2). Theorem 13.3.3. Let D = (0, ∞), or D = [0, ∞), or D = (−∞, 0)∪(0, ∞), or D = R, and let f : D → R, g : D → R, and h : D → R satisfy equation (13.3.3). Then there exist a function f0 : D → R satisfying equation (13.1.2) and constants a, b ∈ R such that (13.3.5) holds2 . Proof. Put a = g(1), b = h(1), f0 = f − a − b. Setting in (13.3.3) in turn y = 1 and x = 1 we obtain (13.3.5). Then f0 (xy) = f (xy) − a − b = g(x) − a + h(y) − b = f0 (x) + f0 (y), i.e., f0 satisfies (13.1.2).



On the other hand, it is easily seen that every triple of form (13.3.5), where f0 : D → R satisfies (13.1.2), and a, b ∈ R are arbitrary constants, actually satisfies equation (13.3.3). Lemma 13.3.1. Let D be one of sets (13.1.6), and let f : D → R, g : D → R and h : D → R satisfy equation (13.3.4). If there exists a point x0 ∈ D, x0 = 0, at which one of the functions f, g, h is zero, then f, g, h have one of forms (13.3.6) in D. Proof. We will give the proof for D = R. If D = R, then simply some parts of the proof can be omitted.  4  4 If f (x0 ) = 0, then g |x0 | h |x0 | sgn x0 = f (x0 ) = 0, whence either  4  4 g |x0 | = 0 or h |x0 | sgn x0 = 0. So we may assume that g(x0 ) = 0 or h(x0 ) =   0. If x0 < 0, then g(−x0 )h(−x0 ) = f x20 = g(x0 )h(x0 ) = 0, whence either g(−x0 ) = 2 If 0 ∈ D, then, according to Lemma 13.1.2, necessarily f = 0. Then f = a + b = const, g = a = 0 const, and h = b = const.

13.3. Pexider equations

357

0, or h(−x0 ) = 0. So we may assume that x0 > 0. For the argument’s sake we assume that x0 ∈ (0, 1). If x0 ∈ [1, ∞) the proof     is similar.   x x x = g(x0 )h =g h(x0 ) = 0. For x ∈ (0, x0 ) we have f (x) = f x0 x0 x0 x0 So f is zero in (0, x0 ).   For every x ∈ (0, x0 ) also x2 ∈ (0, x0 ), whence 0 = f x2 = g(x)h(x) and either g(x) = 0, or h(x) = 0. We will show that f (x) = 0 and either g(x) = 0 or h(x) = 0 (13.3.8)   2−n , n ∈ N∪{0}. We have already it for every x ∈ 0, x0  for n = 0. Suppose  proved this   2−(n+1) 2−n 2 . Then x ∈ 0, x0 , true for an n ∈ N∪{0}, and take an arbitrary x ∈ 0, x0 and   0 = f x2 = g(x)h(x),   2−(n+1) whence either g(x) = 0 or h(x) = 0. Now take a y ∈ x, x0 . Then       x x x = g(y)h =g h(y) = 0. f (x) = f y y y y   2−(n+1) . Induction shows that (13.3.8) Consequently (13.3.8) holds for x ∈ 0, x0   2−n holds in every interval 0, x0 , n ∈ N ∪ {0}. Since every x ∈ (0, 1) belongs to a   for an n ∈ N ∪ {0}, (13.3.8) holds for all x ∈ (0, 1). certain interval 0, x2−n 0   x = Now take an x ∈ (1, ∞) and a y ∈ (0, 1). Then x/y > 1 and f (x) = f y y     x x g(y)h = g h(y) = 0. Thus f is zero in (1, ∞). For arbitrary x ∈ (1, ∞) y y   also x2 ∈ (1, ∞), and 0 = f x2 = g(x)h(x), whence either g(x) = 0, or h(x) = 0. Consequently we have (13.3.8) for every x ∈ (0, 1) ∪ (1, ∞). Now take an arbitrary x ∈ (0, 1) ∪ (1, ∞). Then     f (1) = f xx−1 = g(x)h x−1 = 0. Hence 0 = f (1) = g(1)h(1), and either g(1) = 0, or h(1) = 0. Thus (13.3.8) holds for x ∈ (0, ∞). Now, for any x ∈ (0, ∞), we have f (0) = f (0x) = g(0)h(x) = g(x)h(0) = 0, whence 0 = f (0) = g(0)h(0), and either g(0) = 0, or h(0) = 0. Thus (13.3.8) holds for x ∈ [0, ∞).   Take an x < 0. Then x2 ∈ (0, ∞), and 0 = f x2 = g(x)h(x), 4 whence  4 either  4 g(x) = 0, or h(x) = 0. Further, |x| ∈ (0, ∞), whence f (x) = g |x| h − |x| =  4  4  g − |x| h |x| = 0. Thus (13.3.8) holds for every x ∈ R. If there existed x, y ∈ R such that g(x) = 0 and h(y) = 0, then we would have f (xy) = g(x)h(y) = 0, contrary to (13.3.8). Consequently one of the functions g, h must be identically zero, and we obtain (13.3.6). 

358

Chapter 13. Related Equations

Theorem 13.3.4. Let D be one of sets (13.1.6), and let f : D → R, g : D → R, h : D → R satisfy equation (13.3.4). Then either f, g, h have one of forms (13.3.6), or there exist a function f0 : D → R satisfying equation (13.1.3) and constants a, b ∈ R \ {0} such that (13.3.7) holds in D. Proof. Suppose that f, g, h have not one of forms (13.3.6). By Lemma 13.3.1 f (x) = 0, g(x) = 0, h(x) = 0 for all x ∈ D \ {0}. If 1 ∈ D, then we put a = g(1), b = h(1), f0 = a−1 b−1 f , and the proof runs as that of Theorem 13.3.3. The case where 1 ∈ / D, is more difficult. We will give here a proof for D = (−1, 1). For other D of form (13.1.6) the proof is similar. By (13.3.4) we have f (xy) = g(x)h(y) = g(y)h(x), whence for x ∈ (−1, 0) ∪ (0, 1) = D \ {0} h(x) h(y) = . g(x) g(y) Thus h(x)/g(x) = c = const, and x ∈ (−1, 0) ∪ (0, 1).

h(x) = cg(x),

(13.3.9)

Equation (13.3.4) takes now the form f (xy) = cg(x)g(y),

x, y ∈ (−1, 0) ∪ (0, 1).

(13.3.10)

Now we show that each of the functions f, g has a constant sign in (0, 1). For x ∈ (0, 1) we have by (13.3.10) √  2 f (x) = c g x , and so sgn f (x) = sgn c = const. If there existed x, y ∈ (0, 1) such that g(x)g(y) < 0, then we would have f (xy) = −c|g(x)g(y)|, (13.3.11) whence sgn c = sgn f (xy) = − sgn c, and sgn c = 0, which is impossible. Consequently also g has  a constant sign in (0, 1). Similarly we prove  that g has a constant sign in (−1, 0) possibly different from the sign of g in (0, 1) . If there existed x, y ∈ (−1, 0) such that g(x)g(y) < 0, then xy ∈ (0, 1), sgn f (xy) = sgn c, and we obtain (13.3.11), which again leads us to a contradiction. Put ϕ0 (t) = log|f (et )|, γ0 (t) = log|g (et )|, λ = log|c|, t ∈ (−∞, 0). Then, for arbitrary u, v ∈ (−∞, 0) ϕ0 (u + v) = log |f (eu+v )| = log |f (eu ev )| = log |cg (eu ) g (ev )| = log |g (eu )| + log |g (ev )| + log |c| = γ0 (u) + γ0 (v) + λ, or, with ϕ = ϕ0 + λ, γ = γ0 + λ, ϕ(u + v) = γ(u) + γ(v). Hence

 γ(u) + γ(v) = ϕ(u + v) = γ

i.e.,

 γ

u+v 2

u+v 2

 =



 +γ

u+v 2

γ(u) + γ(v) . 2

(13.3.12) 

 = 2γ

u+v 2

 ,

13.3. Pexider equations

359

By Theorem 13.2.1 there exist an additive function α : R → R and a constant μ ∈ R such that γ(t) = α(t) + μ for t ∈ (−∞, 0). Hence, by (13.3.12), for t ∈ (−∞, 0),       t t t ϕ(t) = γ +γ = 2γ = α(t) + 2μ. 2 2 2 Thus, with λ1 = μ − λ, λ2 = 2μ − λ, ϕ0 (t) = α(t) + λ2 , and

  |f (et )| = exp α(t) + λ2 ,

γ0 (t) = α(t) + λ1 ,

t ∈ (−∞, 0),

  |g(et )| = exp α(t) + λ1 ,

t ∈ (−∞, 0),

or with c1 = exp λ1 , c2 = exp λ2 , |f (et )| = c2 exp α(t),

|g(et )| = c1 exp α(t), t ∈ (−∞, 0).   Writing c1 = c1 sgn g, c2 = c2 sgn f the sign in (0, 1) , we have f (x) = c2 exp α(log x),

g(x) = c1 exp α(log x),

x ∈ (0, 1).

(13.3.13)

The function f1 (x) = exp α(log x), x ∈ (0, 1), satisfies equation (13.1.3): f1 (xy) = exp α(log xy) = exp α(log x + log y)





= exp α(log x) + α(log y) = exp α(log x) exp α(log y) = f1 (x)f1 (y). Hence, by (13.3.13) and (13.3.9), for x ∈ (0, 1), f (x) = c2 f1 (x),

g(x) = c1 f1 (x),

h(x) = cc1 f1 (x).

(13.3.14)

If we write a = c1 , b = cc1 , and insert (13.3.14) into (13.3.4), we obtain c2 f1 (xy) = f (xy) = g(x)h(y) = af1 (x)bf1 (y) = abf1 (xy), whence c2 = ab. So we obtain from (13.3.14) for x ∈ (0, 1) f (x) = abf1 (x),

g(x) = af1 (x),

h(x) = bf1 (x).

(13.3.15)

Now, for x ∈ (−1, 0) we have by (13.3.10) and (13.3.15)

2   2   2   c g(x) = f x2 = c g |x| = ca2 f1 |x| , whence

  g(x) = εaf1 |x| ,

x ∈ (−1, 0),

(13.3.16)

360

Chapter 13. Related Equations

  where  ε = +1 or −1 remember that g has a constant sign in (−1, 0)! . With a y ∈ |x|, 1 we get hence according to (13.3.10), (13.3.15) and the fact that f1 satisfies (13.1.3)       x x |x| 2 f (x) = f y = cg(y)g = cεa f1 (y)f1 y y y     (13.3.17) = cεa2 f1 |x| = εabf1 |x| . Similarly, for x ∈ (−1, 0), we have by (13.3.9), and (13.3.16)     h(x) = cg(x) = cεaf1 |x| = εbf1 |x| .

(13.3.18)

If g(0) = 0,

(13.3.19)

f (0) = 0,

(13.3.20)

then by (13.3.4) and, taking an x ∈ (0, 1), we obtain from (13.3.4) 0 = f (0) = f (x0) = g(x)h(0), whence h(0) = 0, (13.3.21) since g(x) = 0 for x = 0. If g(0) = 0, then by (13.3.4), with an x ∈ (0, 1), g(x)h(0) = f (0) = g(0)h(x) = 0, whence also f (0) = 0 and h(0) = 0. Let a = g(0) = 0, By (13.3.4)

b = h(0) = 0.

f (0) = g(0)h(0) = a b .

(13.3.22) (13.3.23)

Now take an arbitrary x ∈ (0, 1). Then by (13.3.22), (13.3.23) and (13.3.15) a b = f (0) = g(0)h(x) = a bf1 (x), whence

b = const. b Since f1 satisfies (13.1.3), we have for arbitrary x, y ∈ (0, 1) f1 (x) =

b = f1 (xy) = f1 (x)f1 (y) = b

(13.3.24)

  2 b , b

whence b /b = 1 (since b /b = 0), and b = b.

(13.3.25)

Next, by (13.3.25), (13.3.23), (13.3.4), (13.3.15), (13.3.22), again (13.3.25), and (13.3.24) a b = a b = f (0) = g(x)h(0) = abf1 (x) = ab = ab,

13.3. Pexider equations

361

whence a = a,

(13.3.26)

a b = ab.

(13.3.27)

and

Now put

⎧ ⎪ ⎨f1 (x) f0 (x) = η ⎪ ⎩   εf1 |x|

for x ∈ (0, 1), for x = 0, for x ∈ (−1, 0),

(13.3.28)

where η = 0 if g(0) = 0, and η = 1 if g(0) = 0 (note that in the latter case f1 = 1). We will show that function (13.3.28) satisfies equation (13.1.3) in (−1, 1). If x > 0, y > 0, then xy > 0, and f0 (xy) = f1 (xy) = f1 (x)f1 (y) = f0 (x)f0 (y). If x < 0, y < 0, then xy > 0, and f0 (xy) = f1 (xy) = f1 (|xy|) = f1 (|x|) f1 (|y|) = ε2 f1 (|x|) f1 (|y|) = [εf1 (|x|)] [εf1 (|y|)] = f0 (x)f0 (y). If x > 0, y < 0, then xy < 0, and f0 (xy) = εf1 (|xy|) = εf1 (x|y|) = f1 (x)εf1 (|y|) = f0 (x)f0 (y). If x < 0, y > 0, the argument is analogous. If x = y = 0, then xy = 0, and f0 (xy) = f0 (0) = η = η 2 = f0 (0)f0 (0) = f0 (x)f0 (y). In order to consider the cases, where one of x, y is zero, we will distinguish two cases. I. g(0) = 0. Then η = 0. If x = 0, then we have for arbitrary y ∈ (−1, 1) f0 (xy) = f0 (0) = 0 = 0f0 (y) = f0 (x)f0 (y). If y = 0, the argument is analogous. II. g(0) = 0. Then η = 1, g = a and ε = 1. If x = 0, y = 0, then xy = 0, and f0 (xy) = 1 = f0 (x)f0 (y). If x = 0, y = 0, the argument is analogous. Now, by (13.3.15)–(13.3.23) and (13.3.25)–(13.3.27) we have (13.3.7) for all x ∈ (−1, 1) = D. The proof is completed.  On the other hand, it is easily seen that every triple of form (13.3.7), where f0 : D → R satisfies (13.1.3), and a, b ∈ R \ {0} are arbitrary constants, actually satisfies equation (13.3.4). Now we will determine the continuous solutions of equations (13.3.1)–(13.3.4). Theorem 13.3.5. Functions f : RN → R , g : RN → R and h : RN → R are a continuous solution of equation (13.3.1) if and only if there exist constants a, b ∈ R and c ∈ RN such that f (x) = cx + a + b,

g(x) = cx + a,

h(x) = cx + b,

x ∈ RN .

(13.3.29)

Proof. The functions f, g, h have form (13.3.5). Hence if any one of f, g, h is continuous, then so is also f0 , and by Theorem 5.5.2 f0 (x) = cx with a certain c ∈ RN . (13.3.29) results now from (13.3.5). It is obvious that every triple of form (13.3.29), where a, b ∈ R and c ∈ RN are arbitrary constants, represents a continuous solution of equation (13.3.1). 

362

Chapter 13. Related Equations

An analogous theorem is not true for equation (13.3.2). The functions ⎧ ⎧ ⎪ ⎪ ⎨f = 0, ⎨f = 0, (13.3.30) g = 0, g arbitrary continuous, ⎪ ⎪ ⎩ ⎩ h arbitrary continuous, h = 0, are continuous solutions of (13.3.2), and nothing more can be said about g resp. h in (13.3.30). But we have the following Theorem 13.3.6. Functions f : RN → R , g : RN → R and h : RN → R are a continuous solution of equation (13.3.2) if either they have one of the forms (13.3.30), or there exist constants a, b ∈ R \ {0} and c ∈ RN such that h(x) = becx, x ∈ RN .  Proof. This results from Theorems 13.3.2 and 13.1.4. Note that the case, where   f0 = 0, is contained in (13.3.30). f (x) = abecx,

g(x) = aecx ,

Theorem 13.3.7. Let D = (0, ∞), or D = [0, ∞), or D = (−∞, 0)∪(0, ∞), or D = R. Functions f : D → R, g : D → R and h : D → R are a continuous solution of equation (13.3.3) if and only if there exist constants a, b, c ∈ R such that f (x) = a + b,

g(x) = a,

h(x) = b,

x ∈ D,

whenever 0 ∈ D, or f (x) = c log |x| + a + b,

g(x) = c log |x| + a,

h(x) = c log |x| + b,

x ∈ D,

whenever 0 ∈ / D. Proof. This results from Theorems 13.3.3 and 13.1.5, and Lemma 13.1.2.



Theorem 13.3.8. Let D be one of sets (13.1.6). Functions f : D → R, g : D → R and h : D → R are a continuous solution of equation (13.3.4) if and only if either they have one of forms (13.3.30), or there exist constants a, b ∈ R \ {0} and c ∈ R (moreover, c > 0 if 0 ∈ D) such that for x ∈ D f (x) = ab,

g(x) = a,

h(x) = b,

or f (x) = ab|x|c ,

g(x) = a|x|c ,

h(x) = b|x|c ,

or f (x) = ab|x|c sgn x,

g(x) = a|x|c sgn x,

h(x) = b|x|c sgn x.  Proof. This results from  Theorems 13.3.4 and 13.1.6. The case, where f0 = 0, is contained in (13.3.30).  Remark 13.3.1. If f, g, h are assumed to be continuous only in D \ {0}, then the condition c > 0 disappears, and the value of f, g, h at zero (unless f, g, h are constant) is zero.

13.4. Multiadditive functions

363

Simple proofs of the following theorems are left to the reader. Theorem 13.3.9. Let f : RN → R , g : RN → R and h : RN → R satisfy equation (13.3.1). If one of the functions f, g, h is measurable, or bounded above, or below, on a set T ∈ A, then f, g, h are continuous. Theorem 13.3.10. Let f : RN → R, g : RN → R and h : RN → R satisfy equation (13.3.2). If f = 0, and one of the functions f, g, h is measurable, or bounded on a set T ∈ C, then f, g, h are continuous. Theorem 13.3.11. Let D = (0, ∞) or D = (−∞, 0) ∪ (0, ∞), and let f : D → R, g : D → R and h : D → R satisfy equation (13.3.3). If one of the functions f, g, h is measurable, or bounded above, or below, on a set T ⊂ D such that mi (T ) > 0, or T is of the second category and with the Baire property, then f, g, h are continuous. Theorem 13.3.12. Let D be one of sets (13.1.6), and let f : D → R, g : D → R and h : D → R satisfy equation (13.3.4). If f = 0, and one of the functions f, g, h is measurable, then f, g, h are continuous in D \ {0}.

13.4 Multiadditive functions Let p ∈ N. A function f : RpN → R is called p-additive, iff, for every i, 1  i  p, and for every x1 , . . . , xp , yi ∈ RN f (x1 , . . . , xi−1 , xi + yi , xi+1 , . . . , xp ) = f (x1 , . . . , xp ) + f (x1 , . . . , xi−1 , yi , xi+1 , . . . , xp ), (13.4.1) i.e., f is additive in each of its variables xi ∈ RN , i = 1, . . . , p. A 2-additive function is called biadditive. An example of a p-additive function is furnished by the product f1 (x1 ) · · · fp (xp ), where fi : RN → R, i = 1, . . . , p, are additive. At first let us note the following obvious Lemma 13.4.1. Let f1 , f2 : RpN → R be p-additive functions, and let a, b ∈ R be arbitrary constants. Then the function f = af1 + bf2 also is p-additive. Lemma 13.4.2. Let f : RpN → R be a p-additive function, and let (i1 , . . . , ip ) be an arbitrary permutation of (1, . . . , p). Define the function g : RpN → R by g(x1 , . . . , xp ) = f (xi1 , . . . , xip ). Then g is p-additive. Keeping the p − 1 variables x1 , . . . , xi−1 , xi+1 , . . . , xp fixed, we obtain from Theorem 5.2.1 Theorem 13.4.1. If f : RpN → R is p-additive, then for every x1 , . . . , xp ∈ RN , every λ ∈ Q, and every i, 1  i  p, f (x1 , . . . , xi−1 , λxi , xi+1 , . . . , xp ) = λf (x1 , . . . , xp ).

(13.4.2)

364

Chapter 13. Related Equations

Hence we obtain Lemma 13.4.3. If f : RpN → R is a p-additive function, and f is bounded in RpN , then f = 0. Proof. Suppose that there exists a point (x1 , . . . , xp ) ∈ RpN such that f (x1 , . . . , xp ) = 0. Then we have by Theorem 13.4.1 f (kx1 , x2 , . . . , xp ) = kf (x1 , . . . , xp ) for all k ∈ Z, and thus f cannot be bounded in RpN .



A multiadditive function cannot, in general, be expressed by additive functions, but its general construction can be found with the aid of Hamel basis (Acz´el [4]). Theorem 13.4.2. Let H ⊂ RN be a Hamel basis, and let g : H p → R be an arbitrary function. Then there exists a unique p-additive function f : RpN → R such that f | H p = g. Proof. Suppose that f : RpN → R is a p-additive function such that f | H p = g. Take arbitrary x1 , . . . , xp ∈ RN . Every xi has a representation xi =

ni

λij hij ,

i = 1, . . . , p,

(13.4.3)

j=1

where λij ∈ Q, hij ∈ H, i = 1, . . . , p; j = 1, . . . , ni . Then by (13.4.1) and (13.4.2) f (x1 , . . . , xp ) =

=

n1 n2

···

np

j1 =1 j2 =1

jp =1

n1 n2

np

···

j1 =1 j2 =1

jp =1

 

p -

 λiji

i=1 p -

 λiji

  f h1j1 , . . . , hpjp

(13.4.4)

  g h1j1 , . . . , hpjp .

i=1

Consequently, if such an f exists, it must have form (13.4.4), and hence it is unique. On the other hand, it is easily checked that a function f : RpN → R given by (13.4.4) is p-additive. If x1 = h1 , . . . , xp = hp are elements of H, then in (13.4.3) every ni = 1, and λij = λi1 = 1, i = 1, . . . , p. Hence by (13.4.4) f (h1 , . . . , hp ) =

p -

λi1 g(h1 , . . . , hp ) = g(h1 , . . . , hp ).

i=1

Thus f | H p = g.



It is obvious that the above theorem gives the general construction of p-additive functions, since every p-additive function f : RpN → R is the unique p-additive extension of g = f | H p .

13.4. Multiadditive functions

365

Corollary 13.4.1. Let H ⊂ RN be a Hamel basis, and let g : H p → R be a function. Further, let f : RpN → R be the p-additive extension of g. Then f is symmetric [antisymmetric] if and only if g is symmetric [antisymmetric]. p Proof. If f is symmetric [antisymmetric]  then so is also g= f | H . Now suppose  that g is symmetric [antisymmetric]. Then g hi1 ji1 , . . . , hip jip = εg h1j1 , . . . , hpjp , where (i1 , . . . , ip ) is an arbitrary permutation of (1, . . . , p), and ε = +1 if g is symmetric or g is antisymmetric and (i1 , . . . , ip ) is an even permutation, and ε = −1 if g is antisymmetric and the permutation (i1 , . . . , ip ) is odd. But then, according to (13.4.4), n

ni1 ip

  ··· f xi1 , . . . , xip =

=

ji1 =1

jip =1

n1

np

···

j1 =1





jp =1



p -

p -

g(hi1 ji1 , . . . , hip jip )

λiκ jiκ

κ=1



λiji

  εg h1j1 , . . . , hpjp

i=1

= εf (x1 , . . . , xp ), 

which means that f is also symmetric [antisymmetric].

Theorem 13.4.3. Let f : RpN → R be a continuous p-additive function. Then there exist constants cj1 ...jp ∈ R, j1 , . . . , jp = 1, . . . , N , such that f (x1 , . . . , xp ) =

N

j1 =1

···

N

cj1 ...jp x1j1 · · · xpjp ,

(13.4.5)

jp =1

where xi = (xi1 , . . . , xiN ), i = 1, . . . , p. Proof. We will prove (13.4.5) by induction on p. For p = 1 (13.4.5) holds in virtue of Theorem 5.5.2. Suppose (13.4.5) true for a p ∈ N, and let f : R(p+1)N → R be a continuous (p + 1)-additive function. Keep the p variables x1 , . . . , xp fixed. Then the function f (x1 , . . . , xp , y) as a function of y is additive and continuous, whence by Theorem 5.5.2 there exists a constant c = (c1 , . . . , cN ) ∈ RN such that f (x1 , . . . , xp , y) = cy =

N

ci y i

(13.4.6)

i=1

where y = (y1 , . . . , yN ). Of course c, and hence every ci , depends on x1 , . . . , xp : ci = ci (x1 , . . . , xp ),

i = 1, . . . , N.

(13.4.7)

Take y = (0, . . . , 0, 1, 0, . . . , 0), where 1 stands at i-th place. Then we get by (13.4.6) and (13.4.7) ci (x1 , . . . , xp ) = f (x1 , . . . , xp , y). (13.4.8)

366

Chapter 13. Related Equations

Relation (13.4.8) shows, since f is (p + 1)-additive, that ci is p-additive. By the induction hypothesis ci (x1 , . . . , xp ) =

N

···

j1 =1

N

cij1 ...jp x1j1 · · · xpjp ,

jp =1

with certain constants cij1 ...jp , whence by (13.4.6) f (x1 , . . . , xp , xp+1 ) =

N N

jp+1 =1 j1 =1

···

N

cjp+1 j1 ...jp x1j1 · · · xpjp xp+1,jp+1 .

jp =1

Putting cj1 ...jp jp+1 = cjp+1 j1 ...jp , we obtain hence (13.4.5) for p + 1.



In the proof of Theorem 13.4.3 only the continuity of f with respect to each variable separately was used. Therefore the theorem remains valid if f is supposed separately continuous with respect to each variable. Corollary 13.4.2. Let f : RpN → R be a p-additive function. If f is separately continuous with respect to each variable, then f is continuous. Proof. If f is separately continuous with respect to each variable, then the proof of Theorem 13.4.3 shows that f has form (13.4.5), and hence it is continuous.  Theorem 13.4.4. Let f : RpN → R be a measurable p-additive function. Then f is continuous. Proof. We will prove, by induction on p, that f has form (13.4.5). For p = 1 this follows from Theorems 9.4.3 and 5.5.2. Assume it true for a p ∈ N, and let f : R(p+1)N → pN R be a measurable (p + 1)-additive function.  pNThen,  for almost all (x1 , . . . , xp ) ∈ R , pN say for (x1 , . . . , xp ) ∈ A ⊂ R , where m R \ A = 0, the function f (x1 , . . . , xp , y), as a function of y, is a measurable additive function, and therefore (13.4.6) holds with a certain c = (c1 , . . . , cN ) ∈ RN depending on x1 , . . . , xp . For almost all (x1 , . . . , xp−1 ) ∈ R(p−1)N the set

B x1 , . . . , xp−1 = {xp ∈ RN | (x1 , . . . , xp−1 , xp ) ∈ A}     N has full

measure i.e., m R \ B = 0 . Take x1 , . . . , xNp−1 such that the set B x1 , . . . , xp−1 has full measure, and take an arbitrary xp ∈ R . The set xp − B x1 , . . . , xp−1 also has full measure, and consequently also the set



 (13.4.9) B x1 , . . . , xp−1 ∩ xp − B x1 , . . . , xp−1 has full measure. Thus there exists an x in set (13.4.9), which means that x and  xp − x both belong to B x1 , . . . , xp−1 , or (x1 , . . . , xp−1 , x ) ∈ A and (x1 , . . . , xp−1 , xp − x ) ∈ A.

13.5. Cauchy equation on an interval

367

Therefore, for arbitrary y = (y1 , . . . , yN ) ∈ RN , f (x1 , . . . , xp−1 , x , y) =

N

ci (x1 , . . . , xp−1 , x )yi ,

i=1

and 

f (x1 , . . . , xp−1 , xp − x , y) =

N

ci (x1 , . . . , xp−1 , xp − x )yi .

i=1

If we put ci (x1 , . . . , xp−1 , xp ) = ci (x1 , . . . , xp−1 , x ) + ci (x1 , . . . , xp−1 , xp − x ), i = 1, . . . , N, we will have by (13.4.1) f (x1 , . . . , xp−1 , xp , y) = f (x1 , . . . , xp−1 , x , y) + f (x1 , . . . , xp−1 , xp − x , y) =

N

ci (x1 , . . . , xp−1 , xp )yi

i=1

for all y ∈ RN , all xp ∈ RN , and almost all (x1 , . . . , xp−1 ) ∈ R(p−1)N . Repeating this procedure p times we obtain formula (13.4.6) with (13.4.7) valid for all y ∈ RN and all (x1 , . . . , xp ) ∈ RpN . Further the proof runs as that of Theorem 13.4.3.  Similarly one can prove Theorem 13.4.5. Let f : RpN → R be a p-additive function, bounded above, or below, on a set T ⊂ RpN , where either mi (T ) > 0, or T is of the second category and with the Baire property. Then f is continuous.

13.5 Cauchy equation on an interval Now let J ⊂ I ⊂ R be (non-degenerated) intervals, and suppose that J + J ⊂ I. Suppose that we are given a function f : I → R such that f (x + y) = f (x) + f (y) for x, y ∈ J.

(13.5.1)

Following J. Acz´el [3] we will prove the following Theorem 13.5.1. Let J ⊂ I ⊂ R be intervals such that J +J ⊂ I, and let f : I → R be a function satisfying equation (13.5.1). Then there exist an additive function g : R → R and a constant a ∈ R such that  g(x) + a for x ∈ J, f (x) = (13.5.2) g(x) + 2a for x ∈ J + J.

368

Chapter 13. Related Equations

1 (x + y) ∈ J. By (13.5.1) 2       x+y x+y x+y f (x) + f (y) = f (x + y) = f +f = 2f , 2 2 2

Proof. Take arbitrary x, y ∈ J. Then also

whence (13.2.1) follows. Consequently f | J satisfies the Jensen equation (13.2.1). By Theorem 13.2.1 there exist an additive function g : R → R and a constant a ∈ R such that f (x) = g(x) + a for x ∈ J. Now, if x ∈ J + J, then there exist u, v ∈ J such that x = u + v. Hence we get by what has already been shown f (x) = f (u + v) = f (u) + f (v) = g(u) + a + g(v) + a = g(u + v) + 2a = g(x) + 2a. Thus formula (13.5.2) has been completely shown.



Theorem 13.5.2. Let J ⊂ I ⊂ R be intervals such that J + J ⊂ I, and let f : I → R be a function satisfying equation (13.5.1). If J ∩ 2J = ∅, then there exists an additive function g : R → R such that f (x) = g(x)

for x ∈ J ∪ (J + J).

(13.5.3)

If J ∩ 2J = ∅, then there exist an additive function g : R → R and a constant a ∈ R such that (13.5.2) holds. Proof. First let us note that 2J = J + J.

(13.5.4)

In fact, the inclusion 2J ⊂ J + J is evident. On the other hand, if x ∈ J + J, then 1 there exist u, v ∈ J such that x = u + v. Then also w = (u + v) ∈ J, since J, being 2 an interval, is convex. Moreover, x = 2w ∈ 2J. Thus J + J ⊂ 2J, and we obtain (13.5.4).  Now, if J ∩ 2J = ∅, then (13.5.2) results from Theorem 13.5.1. Note  that in this case every function of form (13.5.2) actually satisfies equation (13.5.1). If there exists an x0 ∈ J ∩ 2J, then by Theorem 13.5.1 and (13.5.4) f (x0 ) = g(x0 ) + a and f (x0 ) = g(x0 ) + 2a. Hence g(x0 ) + a = g(x0 ) + 2a, and a = 0. Formula (13.5.3) results now from  (13.5.2). Again, every function of form (13.5.3) actually satisfies eqution (13.5.1) .  Theorem 13.5.3. Let J ⊂ R be an interval such that 0 ∈ cl J, and let I = J + J = 2J. If a function f : I → R satisfies equation (13.5.1), then f can be uniquely extended onto R to an additive function. 1 Proof. Since 0 ∈ cl J and J is convex, with every x ∈ J also x ∈ J, whence 2   1 x=2 x ∈ 2J. Thus J ⊂ 2J so that J ∩ 2J = J = ∅, and J ∪ (J + J) = J ∪ 2J = 2 2J = I. The function g occurring in Theorem 13.5.2 is the desired extension. The uniqueness of the extension results from the fact that, since J is non-degenerated,

13.6. The restricted Cauchy equation

369

int J = ∅, whence by Lemma 9.3.1 J ∈ A, and by Corollary 9.3.2 J contains a Hamel basis H. Thus any additive extension of f must also be an additive extension of f | H., and by Theorem 5.2.2 such an extension is unique.  The particular case J = I = (0, ∞) was dealt with by J. Acz´el–P. Erd˝ os [9].   The form in J ∪ (J + J) of the continuous solutions of equation (13.5.1), and conditions ensuring the continuity on J ∪ (J + J) of an f satisfying (13.5.1) will be easily found by the reader. The results of the present section for the case, where the interval J is open, will be extended to higher dimensions in the next section.

13.6 The restricted Cauchy equation Now we are going to consider the case (cf. Dar´oczy- Losonczi [59], and also Gola¸ bLosonczi [119], Dar´oczy-Gy˝ory [58], Sz´ekelyhidi [309], Lajk´ o [204]), where f is a function satisfying the equation f (x + y) = f (x) + f (y)

(13.6.1)

only for (x, y) ∈ G, where G ⊂ R2N is a certain set. As usually K(x, r) ⊂ RN will denote the open ball centered at x and with the radius r. Lemma 13.6.1. Let g1 : RN → R and g2 : RN → R be additive functions, and let a, b ∈ R be arbitrary constants. If U ⊂ RN is a non-empty open set and g1 (x) + a = g2 (x) + b for x ∈ U, then g1 = g2 and a = b. Proof. For x ∈ U we have g1 (x) − g2 (x) = b − a = const. The function g1 −g2 is additive, and is constant on the open set U . Thus U is contained in a level set of g1 − g2 . By Theorem 12.6.4 g1 − g2 = 0, whence g1 = g2 and a = b.  Lemma 13.6.2. Let U = K(0, r) ⊂ RN with a certain r > 0, and let D ⊂ RN be a set such that U + U = K(0, 2r) ⊂ D. If a function f : D → R satisfies equation (13.6.1) for all x, y ∈ U , then there exists a unique additive function g : RN → R such that f (x) = g(x) for x ∈ U + U . Proof. First observe that we have U = K(0, r) ⊂ K(0, 2r) = U + U . x+y ∈ U , and by (13.6.1) Take arbitrary x, y ∈ U . Then 2       x+y x+y x+y +f = 2f , f (x) + f (y) = f (x + y) = f 2 2 2

370

Chapter 13. Related Equations

i.e., f satisfies in U the Jensen equation (13.2.1). By Theorem 13.2.1 there exist an additive function g : RN → R and a constant a ∈ R such that f (x) = g(x) + a

for x ∈ U.

(13.6.2)

Take an x ∈ U + U . Then x = u + v, where u, v ∈ U , and by (13.6.1) and (13.6.2) f (x) = f (u+v) = f (u)+f (v) = g(u)+a+g(v)+a = g(u+v)+2a = g(x)+2a. (13.6.3) By (13.6.1) f (0) = 0, and similarly we have g(0) = 0, whence by (13.6.3) 0 = f (0) = g(0) + 2a = 2a, and a = 0. Thus we get from (13.6.3) f (x) = g(x) for x ∈ U + U . The uniqueness results from Lemma 13.6.1.  Now let G ⊂ R2N be an arbitrary set. Every point p ∈ R2N can be written as p = (x, y), where x, y ∈ RN . For every G ⊂ R2N we define sets G1 , G2 , G3 ⊂ RN as follows: G1 = {x ∈ RN | There exists a y such that (x, y) ∈ G}, G2 = {y ∈ RN | There exists an x such that (x, y) ∈ G}, G3 = {z ∈ RN | z = x + y, (x, y) ∈ G}. Thus the sets G1 and G2 are the projections of G onto the suitable spaces. For every p = (x, y) ∈ R2N and r > 0 let M (p, r) = K(x, r) × K(y, r). Every M (p, r) is an open subset of R2N . Let G ⊂ R2N , and let D ⊂ RN be a set such that G1 ∪ G2 ∪ G3 ⊂ D.

(13.6.4)

We say that a function f : D → R satisfies equation (13.6.1) on G iff (13.6.1) holds for every x, y such that (x, y) ∈ G. Lemma 13.6.3. Let G = M (p, r) with a certain p ∈ R2N and a certain r > 0, and let D ⊂ RN be a set fulfilling (13.6.4). If a function f : D → R satisfies equation (13.6.1) on G, then there exist a unique additive function g : RN → R and unique constants a, b ∈ R such that ⎧ ⎪ for x ∈ G1 , ⎨f (x) = g(x) + a (13.6.5) f (x) = g(x) + b for x ∈ G2 , ⎪ ⎩ f (x) = g(x) + a + b for x ∈ G3 . Proof. Let p = (u, v). We have G1 = K(u, r), G2 = K(v, r) and G3 = G1 + G2 = K(u + v, 2r).

(13.6.6)

13.6. The restricted Cauchy equation

371

In fact, if z ∈ G3 , then z = x + y, where (x, y) ∈ G. Then x ∈ G1 , y ∈ G2 , and z = x + y ∈ G1 #+ G# 2 = K(u + v, 2r). If z ∈ K(u + v, 2r), then z = u + v + x, where #1 # 1 1 |x| < 2r. Then ## x## < r so that u + x ∈ K(u, r) = G1 and v + x ∈ K(v, r) = G2 . 2  2 2    1 1 Consequently z = u + x + v + x ∈ K(u, r) + K(v, r) = G1 + G2 . Hence we 2 2 obtain (13.6.6). Now take arbitrary x, y ∈ K(0, r) and write x = u + x ∈ K(u, r), y  = v + y ∈ K(v, r). Thus (x , y  ) ∈ K(u, r) × K(v, r) = G. So we get by (13.6.1) f (x + y  ) = f (x ) + f (y  ) = f (u + x) + f (v + y).

(13.6.7)

We have also (u, y  ) ∈ K(u, r) × K(v, r) = G, whence by (13.6.1) f (u + y  ) = f (u) + f (y  ) = f (u) + f (v + y).

(13.6.8)

Similarly, (x , v) ∈ K(u, r) × K(v, r) = G, whence by (13.6.1) f (x + v) = f (x ) + f (v) = f (u + x) + f (v).

(13.6.9)

Relations (13.6.7), (13.6.8), (13.6.9) yield f (u + v + x + y) = f (x + y  ) = f (x + v) − f (v) + f (u + y  ) − f (u)

(13.6.10)

= f (u + v + x) − f (v) + f (u + v + y) − f (u). Obviously, since (u, v) ∈ G, f (u + v) = f (u) + f (v).

(13.6.11)

Subtracting (13.6.11) from (13.6.10) we obtain f (u + v + x+ y)− f (u +v) = f (u + v + x)− f (u + v)+ f (u +v +y)− f (u +v). (13.6.12) Relations (13.6.6) and (13.6.4) show that K(0,r)  ⊂ K(0,r)+  K(0,r) = K(0,2r) = G3 − (u + v) ⊂ D − (u + v). Define the function f0 : D − (u + v) → R by f0 (x) = f (u + v + x) − f (u + v).

(13.6.13)

Relation (13.6.12) shows that f0 satisfies equation (13.6.1) for x, y ∈ K(0, r). By Lemma 13.6.2 there exists an additive function g : RN → R such that f0 (x) = g(x) for x ∈ K(0, r). Now, if x ∈ G1 = K(u, r), then x − u ∈ K(0, r). We have, since x ∈ K(u, r) and v ∈ K(v, r), and thus (x, v) ∈ K(u, r) × K(v, r) = G, f (x + v) = f (x) + f (v), i.e.,  f u + v + (x − u) = f (x) + f (v), and by (13.6.13) and (13.6.11)   f (x) = f (x + v) − f (v) = f u + v + (x − u) − f (v) = f0 (x − u) + f (u + v) − f (v) = g(x − u) + f (u) = g(x) − g(u) + f (u).

372

Chapter 13. Related Equations

With a = f (u) − g(u) we obtain hence the first relation in (13.6.5). The second is established similarly. The third results from the first and the second: if x ∈ G3 , then x = t + s, where (t, s) ∈ G. Hence t ∈ G1 and s ∈ G2 , and f (t) = g(t) + a, f (s) = g(s) + b. Since (t, s) ∈ G, we have by (13.6.1) f (x) = f (t + s) = f (t) + f (s) = g(t) + a + g(s) + b = g(t + s) + a + b = g(x) + a + b. 

The uniqueness results from Lemma 13.6.1. From Lemma 13.6.3 we derive the following result (Dar´oczy-Losonczi [59]).

Theorem 13.6.1. Let G ⊂ R2N be an open and connected set, and let D ⊂ RN be a set fulfilling (13.6.4). If a function f : D → R satisfies equation (13.6.1) on G, then there exist a unique additive function g : RN → R and constants a, b ∈ R such that (13.6.5) holds. Proof. For every x ∈ G there exists a set M (x, rx ) ⊂ G. The family {M (x, rx )}x∈G forms an open cover of G. Since R2N is separable, it follows from Lindel¨ of’s theorem that thereexists a countable family M = {M (xn , rn )}n∈N ⊂ {M (x, rx )}x∈G such that G ⊂ M. Since the converse inclusion is trivial, we must have  G= M. Mβ have already been Now we re-arrange M. Let M0 = M (x1 , r1 ).Suppose that   defined for β < α < Ω in such a manner that β<ξ Mβ ∩ Mξ = ∅ for all ξ < α. If     ∩ Mβ = G, we end the induction here. If Mβ = G, and if we had M β β<α β<α β<α   M∈M\{Mβ }β<α M = ∅, then G would be a union of two disjoint open sets ⎛ G=⎝



⎞

⎛

Mβ ⎠ ∪ ⎝



⎞ M⎠ ,

M∈M\{Mβ }β<α

β<α

which contradicts the connectedness of G. Thus there exists a set Mα such that ⎛ ⎞  ⎝ Mβ ⎠ ∩ Mα = ∅. (13.6.14) β<α

We cannot have



Mβ = G for all α < Ω, since then we would have card M  ℵ1 >

β<α

ℵ0 . So there exists a γ < Ω such that 

Mβ = G,

β<γ

and we stop there. By Theorem 1.6.2 there exists a transfinite sequence {Mβ }β<γ , where γ < Ω (so that γ = ℵ0 ) fulfilling (13.6.14) for all α < γ.

13.6. The restricted Cauchy equation

373

By Lemma 13.6.3 for every β < γ there exists an additive function g : RN → R and constants aβ , bβ ∈ R such that f (x) = gβ (x) + aβ for x ∈ [Mβ ]1 , f (x) = gβ (x) + bβ for x ∈ [Mβ ]2 ,

(13.6.15)

f (x) = gβ (x) + aβ + bβ for x ∈ [Mβ ]3 . We will prove that gβ = g0 ,

aβ = a 0 ,

bβ = b0

(13.6.16)

for all β < γ. For β = 0 this is certainly true. Assume (13.6.16) for β < α < γ. It follows by (13.6.14) that there exists a β < α such that Mβ ∩ Mα = ∅. Thus Mβ ∩ Mα is a non-empty open set, and so is also the set [Mβ ]1 ∩ [Mα ]1 . By the induction hypothesis we have (13.6.16), whence by (13.6.15), for x ∈ [Mβ ]1 ∩ [Mα ]1 g0 (x) + a0 = gβ (x) + aβ = gα (x) + aα . By Lemma 13.6.1 gα = g0 and aα = a0 . The relation bα = b0 is shown similarly. By Theorem 1.6.1 (13.6.16) for all β < γ.

holds   Mβ 1 = [Mβ ]1 we have Put g = g0 , a = a0 , b = b0 . For x ∈ G1 = β<γ

β<γ

by (13.6.15)

and  (13.6.16) f (x) = g(x) + a. Similarly f (x) = g(x) + b for x ∈ G2 = Mβ 2 = [Mβ ]2 . If x ∈ G3 , then x = t + s, where (t, s) ∈ G. Then t ∈ G1 and β<γ

β<γ

s ∈ G2 . Consequently f (t) = g(t) + a, f (s) = g(s) + b, whence f (x) = f (t + s) = f (t) + f (s) = g(t) + a + g(s) + b = g(t + s) + a + b = g(x) + a + b. The uniqueness of g, a, b results from Lemma 13.6.1.



Corollary 13.6.1. Let G ⊂ R2N be an open and connected set such that G1 ∩ G2 = ∅, and let D ⊂ RN be a set fulfilling (13.6.4). If a function f : D → R satisfies equation (13.6.1) on G, then there exist a unique additive function g : RN → R and a constant a ∈ R such that f (x) = g(x) + a

for x ∈ G1 ∪ G2 ,

f (x) = g(x) + 2a

for x ∈ G3 .

Proof. This results from Theorem 13.6.1 and Lemma 13.6.1.



Corollary 13.6.2. Let G ⊂ R2N be an open and connected set such that G1 ∩G2 ∩G3 = ∅, and let D ⊂ RN be a set fulfilling (13.6.4). If a function f : D → R satisfies equation (13.6.1) on G, then there exist a unique additive function g : RN → R such that f (x) = g(x) for x ∈ G1 ∪ G2 ∪ G3 . Proof. This results from Theorem 13.6.1 and Lemma 13.6.1.



Note that if (0, 0) ∈ G, then 0 ∈ G1 ∩ G2 ∩ G3 . Thus we obtain from Corollary 13.6.2 Corollary 13.6.3. Let D = K(0, 2r) ⊂ RN , with a certain r > 0. If a function f : D → R satisfies equation (13.6.1) for all x, y ∈ K(0, r), then f can be uniquely extended onto RN to an additive function.

374

Chapter 13. Related Equations

The next theorem(Dar´oczy-Losonczi [59]) refers to the case N = 1. Fix an r > 0. Let I = [0, r], and let H ⊂ R2 be the set H = {p = (x, y) ∈ R2 | x ∈ I, y ∈ I, and x + y ∈ I}. Theorem 13.6.2. Let f : I → R be a function satisfying equation (13.6.1) on H. Then there exists a unique additive function g : R → R such that f = g | I.   1 Proof. Let J = 0, r so that I = 2J = J + J. Take arbitrary x, y ∈ J. Then 2 x, y ∈ I and x + y ∈ I, whence (x, y) ∈ H, and (13.6.1) holds. Thus f satisfies (13.6.1) for x, y ∈ J. By Theorem 13.5.3 f can be uniquely extended onto R to an additive function g : R → R. The function g has all the properties specified in the theorem.  Theorem 13.6.2 with r = 1 is particularly important in the probability theory. The variables x, y are then to be interpreted as probabilities, and the conditions of the theorem say that in the case where the sum of probabilities is again a probability, then the value of f for such a sum is the sum of the values of f for the corresponding probabilities. It is clear that in Theorem 13.6.2 the closed interval I = [0, r] may be replaced by any one of the intervals (0, r), [0, r) and (0, r]. Equation (13.6.1) postulated not for all x, y from the space considered, but only for x, y satisfying a certain additional condition (in the case of Theorem 13.6.1 the condition was (x, y) ∈ G) is called the Cauchy equation on a restricted domain, or a restricted Cauchy equation or a conditional Cauchy equation. A review of results and references pertinent to restricted Cauchy equations may be found in Kuczma [187], Dhombres-Ger [69], [70], Ger [108]. In the present book we only touch upon such problems. A complete treatment of these questions is found in the book Dhombres [68].

13.7 Hossz´ u equation The functional equation3 (Blanuˇsa [32], Dar´oczy [56], [57], Davison [60],[61],[62], ´ Davison-Redlin [63], Feny˝ o [78], Glowacki-Kuczma [117], Lajk´ o [204], Swiatak [300], [301], [304]) f (x + y − xy) + f (xy) = f (x) + f (y) (13.7.1) is referred to as the Hossz´ u equation. Its connection with the theory of additive functions is expressed by the following (Dar´ oczy [57]; cf. also Blanuˇsa [32], Davison ´ [60], Swiatak [304]) Theorem 13.7.1. Let a function f : R → R satisfy equation (13.7.1). Then there exist an additive function g : R → R and a constant a ∈ R such that f (x) = g(x) + a. 3

(13.7.2)

This equation was mentioned for the first time by M. Hossz´ u at the International Symposium on Functional Equations held in Zakopane (Poland) in October 1967; cf. Hossz´ u [147].

13.7. Hossz´ u equation

375

Proof. Define the function G : R2 → R by G(x, y) = f (x) + f (y) − f (xy).

(13.7.3)

It is easily seen from (13.7.3) that we have for arbitrary u, v, w ∈ R G(uv, w) + G(u, v) = G(u, vw) + G(v, w).

(13.7.4)

By (13.7.1) G(x, y) = f (x + y − xy). Inserting this into (13.7.4) yields f (uv + w − uvw) + f (u + v − uv) = f (u + vw − uvw) + f (v + w − vw). Setting in (13.7.5) w = v −1 , we obtain the relation     f uv + v −1 − u + f (u + v − uv) = f (1) + f v + v −1 − 1

(13.7.5)

(13.7.6)

valid for all u ∈ R and v ∈ R \ {0}. Now take arbitrary x, y ∈ R such that x + y > 0. We are looking for u, v ∈ R, v = 0, fulfilling the equations uv + v −1 − u = x + 1,

u + v − uv = y + 1.

(13.7.7)

Adding equations (13.7.7) we obtain v + v −1 = x + y + 2, i.e., v 2 − (x + y + 2)v + 1 = 0.

(13.7.8)

Now, (x + y + 2)2 − 4 = x2 + y 2 + 2xy + 4x + 4y = (x + y)(x + y + 4) > 0. Thus there exist two distinct real values of v fulfilling (13.7.8), and at least one of them must be different from zero4 . Let v0 be this value. We cannot have v0 = 1, since v0 + v0−1 = x + y + 2 > 2. y + 1 − v0 , then u0 , v0 satisfy the system of equations (13.7.7). Now, if u0 = 1 − v0 Define the function g : R → R as g(x) = f (x + 1) − f (1).

(13.7.9)

We will show that g is additive. Take arbitrary x, y ∈ R such that x + y > 0. Let u, v satisfy (13.7.7), v = 0. Then we get from (13.7.6) f (x + 1) + f (y + 1) = f (1) + f (x + y + 1), i.e., after subtracting from both sides 2f (1) f (x + y + 1) − f (1) = f (x + 1) − f (1) + f (y + 1) − f (1), whence by (13.7.9) g(x + y) = g(x) + g(y). 4

Actually both roots of (13.7.8) are different from zero, since their product is 1.

(13.7.10)

376

Chapter 13. Related Equations

If x + y  0, then we can find a t ∈ R such that t + x > 0 and t + x + y > 0. Then g(t + x) = g(t) + g(x), g(t + x + y) = g(t + x) + g(y) = g(t) + g(x) + g(y), g(t + x + y) = g(t) + g(x + y). Thus g(t) + g(x + y) = g(t) + g(x) + g(y), whence again we obtain (13.7.10). Consequently g fulfills (13.7.10) for all x, y ∈ R, which means that g is additive. By (13.7.9) f (x) = g(x − 1) + f (1) = g(x) − g(1) + f (1), whence we obtain (13.7.2) with a = f (1) − g(1).



On the other hand, it is easily seen that every function of form (13.7.2), where g : R → R is additive and a ∈ R is an arbitrary constant, actually satisfies equation (13.7.1). Theorem 13.7.1 remains true also for functions f : F → Γ, where (F ; +; ·) is an arbitrary field with card F  5, and (Γ, +) is a commutative group. (An additive function is then understood as a function g : F → Γ satisfying equation (13.7.10) for all x, y ∈ F ; the constant a in (13.7.2) must be taken from Γ). The proof is then more difficult (cf. Davison [61]). Equation (13.7.1) was also studied on other structures (Davison [60],[62], Glo´ wacki-Kuczma [117], Swiatak [304]). In particular, on Z equation (13.7.1) has solutions which are not of form (13.7.2). For example, the function  1 if n is even, f (n) = 0 if n is odd is such a solution. By Theorem 13.7.1 equation (13.7.1) has exactly the same solutions as the Jensen equation (13.2.1) (cf. Theorem 13.2.1). Thus Theorem 13.7.1 is often formulated as follows. Corollary 13.7.1. For functions f : R → R the Hossz´ u equation and the Jensen equation are equivalent.

13.8 Mikusi´ nski equation At the Symposium on Functional and Differential Equations held in Zawoja (Poland) in October 1971, J. Mikusi´ nski mentioned the equation

f (x + y) f (x + y) − f (y) − f (x) = 0, (13.8.1) which since has been named after him (cf. Dubikajtis-Ferens-Ger-Kuczma [72], BaronGer [20], Ger [105], [107]; cf. also Kannappan-Kuczma [162], Ger [109] concerning

13.8. Mikusi´ nski equation

377

more general equations of this type). Our primary aim is to show that every function f : RN → R fulfilling (13.8.1) is additive. But we will carry out our considerations for functions f : X → Y , where (X, +) and (Y, +) are arbitrary (not necessarily commutative) groups (cf. 4.5). If the range of f is a ring without divisors of zero (e.g., a field or an integral domain), then equation (13.8.1) can be written as if f (x + y) = 0, then f (x + y) = f (x) + f (y).

(13.8.2)

Equation (13.8.2) is meaningful for functions f with the range in a group. Lemma 13.8.1. Let (X, +) and (Y, +) be arbitrary groups, and let f : X → Y satisfy (13.8.2). Write K = f −1 (0) = {x ∈ X | f (x) = 0}. Then (K, +) is a subgroup of (X, +). Proof. If we had f (0) = 0, then we would get by (13.8.2) f (0) = f (0) + f (0), whence f (0) = 0, a contradiction. Consequently 0 ∈ K so that K = ∅. Suppose that for an x ∈ K we have −x ∈ / K, i.e., f (−x) = 0. Hence by (13.8.2) f (−x) = f (−2x + x) = f (−2x) + f (x) = f (−2x),

(13.8.3)

since x ∈ K means f (x) = 0. By (13.8.3) f (−2x) = f (−x) = 0, whence by (13.8.2) f (−2x) = f (−x) + f (−x).

(13.8.4)

By (13.8.3) and (13.8.4) f (−x) = f (−x) + f (−x), whence f (−x) = 0, contrary to the supposition. Now take arbitrary x, y ∈ K, and suppose that x + y ∈ / K. Then f (x + y) = 0, and by (13.8.2) f (x + y) = f (x) + f (y). (13.8.5) But f (x) = f (y) = 0, whence f (x + y) = 0, contrary to the supposition. For arbitrary x, y ∈ K we have −y ∈ K, and x − y = x + (−y) ∈ K. By Lemma 4.5.1 (K, +) is a subgroup of (X, +).  Theorem 13.8.1. Let (X, +) and (Y, +) be arbitrary groups, and let a function f : X → Y satisfy equation (13.8.2). Then either f satisfies (13.8.5) for all x, y ∈ X (i.e., f is a homomorphism), or there exist a subgroup (K, +) of (X, +), of index 2, and a constant c ∈ Y \ {0} such that  0 for x ∈ K, f (x) = (13.8.6) c for x ∈ / K. Proof. 5 Suppose that f : X → Y satisfies (13.8.2), but does not satisfy (13.8.5). Then there exist u, v ∈ X such that f (u + v) = f (u) + f (v). 5

Ger [107].

(13.8.7)

378

Chapter 13. Related Equations

We will prove that for every x ∈ X either f (u + v + x) = 0, or f (v + x) = 0.

(13.8.8)

Suppose that (13.8.8) does not hold, i.e., there exists a z ∈ X such that f (u + v + z) = 0 and f (v + z) = 0.

(13.8.9)

Hence, by (13.8.2), f (u + v + z) = f (u) + f (v + z) = f (u) + f (v) + f (z).

(13.8.10)

On the other hand, we have by (13.8.9) and (13.8.2) f (u + v + z) = f (u + v) + f (z).

(13.8.11)

Now, (13.8.10) and (13.8.11) yield f (u + v) = f (u) + f (v), contrary to (13.8.7). Now put K = f −1 (0) and take an arbitrary x ∈ X. By (13.8.8) either u + v + x ∈ K, or v + x ∈ K, i.e., either x ∈ −v − u + K, or x ∈ −v + K. In other words, X ⊂ (−v − u + K) ∪ (−v + K) = −v − u + K ∪ (u + K) . This implies that X = X + u + v ⊂ K ∪ (u + K), and since the converse inclusion is obvious, we have K ∪ (u + K) = X.

(13.8.12)

Put c = f (u) ∈ Y , and take an arbitrary x ∈ X \ K. By (13.8.12) x ∈ u + K, i.e., there exists a y ∈ K such that x = u + y. The condition x ∈ X \ K means f (u + y) = f (x) = 0, whereas the condition y ∈ K means f (y) = 0. Thus by (13.8.2) f (x) = f (u + y) = f (u) + f (y) = f (u) = c. Hence we obtain (13.8.6) in view of the definition of K. We also have c = f (x) = 0, whence c ∈ Y \ {0}. By Lemma 13.8.1 (K, +) is a subgroup of (X, +). Take arbitrary x, y ∈ K  = X \ K. By (13.8.6) f (x) = f (y) = c. If we had f (x + y) = 0, then we would get by (13.8.6) and (13.8.2) c = f (x + y) = f (x) + f (y) = 2c, whence c = 0, which is not the case. Consequently f (x + y) = 0, i.e., x + y ∈ K. This shows that K  + K  ⊂ K. By Lemma 4.5.4 the index of (K, +) is 2.  Corollary 13.8.1. Let (X, +) and (Y, +) be arbitrary groups, and let f : X → Y satisfy equation (13.8.2) but not (13.8.5). Put K = f −1 (0). Then there exists a u ∈ X such that (13.8.12) holds. Clearly, if f satisfies (13.8.5), then it satisfies also (13.8.2). If f has form (13.8.6), where (K, +) is a subgroup of (X, +) of index 2, then f satisfies (13.8.2), too. In fact,  let x, y ∈ X be arbitrary. If x, y ∈ K, then x + y ∈ K, since (K, +) is group a  subgroup of (X, +) , and f (x + y) = 0 = 0 + 0 = f (x) + f (y), and we have (13.8.2). If x ∈ K, y ∈ / K, then by Lemma 4.5.3 x + y ∈ / K, and by (13.8.6) f (x + y) = c, f (y) = c. Thus f (x + y) = c = 0 + c = f (x) + f (y),

13.8. Mikusi´ nski equation

379

and (13.8.2) is fulfilled. Similarly, if x ∈ / K, y ∈ K, then by Lemma 4.5.3 x + y ∈ / K, and f (x + y) = c = c + 0 = f (x) + f (y). Finally, if x, y ∈ / K, then by Lemma 4.5.4 x + y ∈ K, whence f (x + y) = 0, and again (13.8.2) is fulfilled. Note that if 2c = 0, then function (13.8.6) does not satisfy (13.8.5) for x, y ∈ / K, since otherwise we would have, since by Lemma 4.5.4 x + y ∈ K, 0 = f (x + y) = f (x) + f (y) = 2c, a contradiction. Since by Corollary 4.5.3 the group (RN , +) has no subgroup of index 2, we obtain from Theory 13.8.1 Theorem 13.8.2. If a function f : RN → R satisfies equation (13.8.1), then it is additive. We will prove also the following Theorem 13.8.3. Let (X, +) be a second category topological group6 such that for every neighbourhood V of zero  nV = X. (13.8.13) n∈N

Let (Y, +) be an arbitrary group. If a function f : X → Y satisfies (13.8.2) and the set K = f −1 (0) has the Baire property, then f satisfies (13.8.5) for all x, y ∈ X. Proof. If K is of the first category, then so are the sets x+K. Suppose that f does not satisfy (13.8.5). Then, by Corollary 13.8.1, there exists a u ∈ X such that (13.8.12) holds. (13.8.12) implies that K  = X \ K ⊂ u + K. If K were of the first category, then so would be also K  , and hence also X = K ∪ K  would be of the first category, contrary to the assumption. Thus K is a set of the second category and with the Baire property. By Theorem 2.9.1 int(K +K) = ∅. But, since by Lemma 13.8.1 (K, +) is a group, we have K +K ⊂ K, whence int K = ∅. Let x ∈ int K. Then the set V = int K − x is a neighbourhood of zero, and, by Lemmas 13.8.1 and 4.5.3, int K − x ⊂ K − K = K + K = K. Thus V ⊂ K, and hence also nV ⊂ K for every n ∈ N. By (13.8.13)  X= nV ⊂ K, n∈N

and thus K = X, which is incompatible with the fact that (K, +) is a subgroup of (X, +) of index 2.    Finally let us observe that equation (13.8.1) especially in form (13.8.2) can be regarded as a conditional Cauchy equation, the condition being f (x + y) = 0. 6 This means that (X, +) is a topological group (cf. 2.9) and the set X in the topological space X is of the second category (i.e., cannot be represented as a countable union of nowhere dense sets).

380

Chapter 13. Related Equations

13.9 An alternative equation If the function f : RN → R is additive, then it satisfies also

2

2 f (x + y) = f (x) + f (y) .

The converse implication is less trivial. (13.9.1) yields immediately only

either f (x + y) = f (x) + f (y), or f (x + y) = − f (x) + f (y) .

(13.9.1)

(13.9.2)

Since equation (13.9.2) has the form of an alternative, it is often referred to (as are also similar functional equations) as an alternative functional equation (cf. Hossz´ u ´ ´ [145], Vincze [318], [319], Swiatak [303], Swiatak–Hossz´ u [307], Fischer [81], Fischer– ´ Musz´ely [83, 84], Kuczma [186]; more general equations are dealt with in Swiatak ´ [299], [302], [305], Swiatak–Hossz´ u [306], Vincze [320], [321]. Cf. also Kuczma [187] concerning a review of results and further references). In this sense also the Mikusi´ nski functional equation (13.8.1) and related equations (Kannappan–Kuczma [162], Ger [109]) are alternative functional equations. Alternative functional equations may be regarded as functional equations on restricted domains (conditional equations). Similarly as in the case of Mikusi´ nski’s functional equation (13.8.1) we will carry out our considerations in more general structures. If (X, +) is a semigroup, and (Y ; +, ·) is an integral domain (cf. Chapter 4), then equation (13.9.1) for functions f : X → Y can be written as (13.9.2). It is in this form that we are going to deal with equation (13.9.1). Lemma 13.9.1. Let (X, +) be a semigroup, let (Y ; +, ·) be an integral domain, and let a function f : X → Y satisfy equation (13.9.1) and the condition f (2x) = 2 f (x)

(13.9.3)

for all x ∈ X. Then f satisfies equation (13.8.5) for all x, y ∈ X.

 Proof. Suppose that for a certain x, y ∈ X (13.8.5) does not hold. By (13.9.1) or, equivalently, (13.9.2) f (x + y) = −f (x) − f (y). (13.9.4) We have



f (x + x + y) = f x + (x + y) = e1 f (x) + f (x + y) = −e1 f (y), and by (13.9.3)



f (x + x + y) = f (2x + y) = e2 f (2x) + f (y) = e2 2f (x) + f (y) , where e1 , e2 = ±1. Hence 2f (x) + f (y) = e3 f (y),

(13.9.5)

2f (y) + f (x) = e4 f (x),

(13.9.6)

where e3 = ±1. Similarly

13.9. An alternative equation

381

where e4 = ±1. If e3 = e4 = 1, then we get by (13.9.5) and (13.9.6) 2f (x) = 2f (y) = 0, i.e.,

2 f (x) + f (y) = 0. (13.9.7) If e3 = −1, then (13.9.7) results from (13.9.5), and if e4 = −1, then (13.9.7) results from (13.9.6). Thus (13.9.7) holds in every case. (13.9.7) means that

f (x) + f (y) = − f (x) + f (y) = −f (x) − f (y), and by (13.9.4) we get f (x + y) = f (x) + f (y), 

contrary to the supposition.

Lemma 13.9.2. Let (X, +) be a semigroup, let (Y ; +, ·) be an integral domain, and let f : X → Y satisfy equation (13.9.1). If f (2x0 ) = 2f (x0 ) for a certain x0 ∈ X, then d0 = f (x0 ) fulfils the conditions 2d0 = 0,

4d0 = 0,

(13.9.8)

and 

6d0 = 0.

(13.9.9)

f (2x0 ) = −2f (x0 ),

(13.9.10)

 Proof. We have by (13.9.1) or, equivalently, (13.9.2)

whence



f (3x0 ) = f (2x0 + x0 ) = e1 f (2x0 ) + f (x0 ) = e1 − 2f (x0 ) + f (x0 ) = −e1 f (x0 ) = −e1 d0 , where e1 = ±1. Hence



f (4x0 ) = f (3x0 + x0 ) = e2 f (3x0 ) + f (x0 ) = e2 (−e1 + 1)d0 ,

and, on the other hand

f (4x0 ) = f (2x0 + 2x0 ) = e3 f (2x0 ) + f (2x0 ) = −4e3 d0 , where e2 , e3 = ±1. Hence

(k1 − k2 )d0 = 0,

where k1 = e2 (−e1 + 1), k2 = −4e3 . Thus the only possible values of k1 − k2 are ±2, ±4, ±6. If we had 4d0 = 0, then we would have 2d0 = −2d0 , and by (13.9.10) f (2x0 ) = −2f (x0 ) = −2d0 = 2d0 = 2f (x0 ), contrary to the assumption. Thus 2d0 = 0 also is impossible, for otherwise we would have 4d0 = 2(2d0 ) = 0. Hence we get (13.9.8). Consequently we must have (13.9.9), or −6d0 = 0, which also implies (13.9.9). 

382

Chapter 13. Related Equations

Theorem 13.9.1. Let (X, +) be a semigroup, and let (Y ; +, ·) be an integral domain of characteristic different from 3. If a function f : X → Y satisfies equation (13.9.1), then it satisfies also equation (13.8.5). Proof. Suppose that f does not satisfy (13.8.5). By Lemma 13.9.1 there exists an x0 ∈ X such that f (2x0 ) = 2f (x0 ). By Lemma 13.9.2 d0 = f (x0 ) fulfils (13.9.8) and (13.9.9). Thus y0 = 2d0 ∈ Y is an element of order 3, and hence the characteristic of Y must be 3, contrary to the assumption.  Note that if the characteristic of Y is 3, then every constant function f : X → Y , f = 0, satisfies equation (13.9.1), but not (13.8.5). In fact, then for every d ∈ Y we have 3d = 0, whence 2d = −d, and if f (x) = d = const in X, then f (x) + f (y) = d + d = 2d = −d, and

2

2 f (x) + f (y) = (−d)2 = d2 = f (x + y) .

On the other hand (d = 0), f (x + y) = d = −d = f (x) + f (y), since otherwise we would have 2d = 0, which together with 3d = 0 implies d = 0. Since (R; +, ·) is a field, and hence an integral domain, and the characteristic of R is zero, we get from Theorem 13.9.1 Theorem 13.9.2. If a function f : RN → R satisfies equation (13.9.1), then it is additive. Equation (13.9.1) for functions f : X → Y , where (X, +) is a semigroup, and (Y ; +, ·) is a commutative ring, as well as equation (13.9.2) for functions f : X → Y , where (X, +) is a semigroup, and (Y, +) is a group, are dealt with in Kuczma [186].

13.10 The general linear equation The general linear equation7 f (ax + by + c) = Af (x) + Bf (y) + C,

abAB = 0,

(13.10.1)

has been studied by J. Acz´el [2], Z. Dar´ oczy [54], [55], and L. Losonczi [209]; cf. also Acz´el [5]. The Cauchy equation (5.2.1) is the particular case a = b = A = B = 1, c = 0, C = 0, of (13.10.1). The Jensen equation (13.2.1) corresponds to a = b = A = 1 B = , c = 0, C = 0. The question arises what are the solutions f : RN → R of 2 (13.10.1) for other values of a, b, c, A, B, C. First we determine the constant solutions of (13.10.1). If a function f (x) = β = const satisfies (13.10.1), then β = (A + B)β + C, i.e., (A + B − 1)β + C = 0.

(13.10.2)

In equation (13.10.1) it is understood that a, b, A, B, C ∈ R and c ∈ RN are fixed constants. So the expression “f : RN → R satisfies (13.10.1)” means that (13.10.1) holds for all x, y ∈ RN with fixed a, b, A, B, C ∈ R and c ∈ RN such that none of a, b, A, B is zero. 7

13.10. The general linear equation

383

If A + B = 1, then equation (13.10.2) has the unique solution β = −C(A + B − 1)−1 , and the function C , x ∈ RN , (13.10.3) f (x) = − A+B−1 is the only constant solution of (13.10.1). If A + B = 1 and C = 0, then every β ∈ R satisfies (13.10.2), and consequently every constant function f : RN → R is a solution of (13.10.1). Finally, if A + B = 1 and C = 0, then (13.10.2) has no solutions, whence also (13.10.1) has no constant solutions. We summarize this discussion in the following Theorem 13.10.1. If A + B = 1, then function (13.10.3) is the only constant solution of (13.10.1). If A + B = 1 and C = 0, then every constant function f : RN → R satisfies (13.10.1). If A + B = 1 and C = 0, then equation (13.10.1) has no constant solution. Now we turn to non-constant solutions of (13.10.1). We start with the following Lemma 13.10.1. If a function f : RN → R satisfies equation (13.10.1), then there exist an additive function g : RN → R and a constant β ∈ R such that f (x) = g(x) + β,

x ∈ RN .

(13.10.4)

Proof. Take arbitrary u, v ∈ RN . Equation (13.10.1) with x = u/a, y = (v − c)/b goes into   u v−c + Bf + C. (13.10.5) f (u + v) = Af a b Setting in (13.10.1) x = u/a, y = −c/b, we obtain u  c f (u) = Af + Bf − + C, a b and setting in (13.10.1) x = 0, y = (v − c)/b, we get   v−c f (v) = Af (0) + Bf + C. b Finally, with x = 0, y = −c/b, equation (13.10.1) becomes  c f (0) = Af (0) + Bf − + C. b

(13.10.6)

(13.10.7)

(13.10.8)

Relations (13.10.5), (13.10.6), (13.10.7), (13.10.8) yield f (u + v) − f (u) − f (v) + f (0) = 0. Define the function g : R

N

(13.10.9)

→ R by g(x) = f (x) − f (0),

x ∈ RN .

(13.10.10)

Then we have by (13.10.9) and (13.10.10) g(u + v) = g(u) + g(v), i.e., g is additive, and (13.10.4) follows with β = f (0).



384

Chapter 13. Related Equations

But now not every function of form (13.10.4) satisfies equation (13.10.1). Theorem 13.10.2. If a non-constant function f : RN → R satisfies equation (13.10.1), and f is measurable, or bounded above, or below, on a set T ∈ A, then necessarily a = A, b = B, and there exists a constant α ∈ RN \ {0} such that f (x) = αx + β,

x ∈ RN ,

(13.10.11)

where, in the case where A + B = 1, β = (αc − C)(A + B − 1)−1 . If A + B = 1, then α additionally fulfils the relation αc = C, and then β ∈ R may be arbitrary. In particular, f is continuous. Proof. By Lemma 13.10.1 f has form (13.10.4), where g : RN → R is additive. The conditions on f enforce the analogous conditions on g, therefore g must be continuous (cf., in particular, Theorem 9.4.2), and it follows from Theorem 5.5.2 that there exists an α ∈ RN such that g(x) = αx, and so f has form (13.10.11). We must have α = 0, since otherwise f would be constant. Inserting (13.10.11) into (13.10.1) yields aαx + bαy + αc + β = Aαx + Aβ + Bαy + Bβ + C for all x, y ∈ RN , which is possible only if a = A and b = B. Hence we get (A + B − 1)β = αc − C. Thus, if A + B = 1, β must have the form asserted. If A + B = 1, then α must fulfil the condition αc = C, and there is no constraint on β.  The argument presented shows also that every function f : RN → R of the form described actually satisfies equation (13.10.1). In order to determine non-measurable solutions f : RN → R of (13.10.1) we will need the following Lemma 13.10.2. Let f : RN → R be a non-constant solution of (13.10.1), and let r ∈ Q(x, y) be a rational function in x, y, with the coefficients from Q. If one of the expressions r(a, b), r(A, B) has sense, then the other has sense, too, and function (13.10.10) satisfies   g r(a, b)x = r(A, B)g(x) (13.10.12) for every x in RN . Proof. First we prove that for every x ∈ RN and k ∈ N ∪ {0}     g ak x = Ak g(x), g bk x = B k g(x).

(13.10.13)

For k = 0 (13.10.13) is trivial. Assume it valid for a k ∈ N∪{0}. Replacing in (13.10.1) x by ak x, and setting y = −c/b, we obtain  c     f ak+1 x = Af ak x + Bf − + C. b Setting in (13.10.1) x = 0, y = −c/b, we get  c f (0) = Af (0) + Bf − + C. b

13.10. The general linear equation

385

Subtracting the two expressions, and making use of the induction hypothesis (13.10.13), we obtain    

  g ak+1 x = f ak+1 x − f (0) = A f (ak x) − f (0) = Ag ak x = Ak+1 g(x). Induction shows that the first formula in (13.10.13) is generally valid. The second formula in (13.10.13) is established similarly. Now let p ∈ Q[x, y] be an arbitrary polynomial in x, y with rational coefficients: p(x, y) =

n k

αki xi y k−i ,

αki ∈ Q,

k = 1, . . . , n;

i = 1, . . . , k.

k=0 i=0

It follows from Lemma 13.10.1 that g is additive. In fact, if f has form (13.10.4), then f (0) = β, and g(x) = f (x) − β = f (x) − f (0) coincides with function (13.10.10). Hence, by Theorem 5.2.1, g(αki u) = αki g(u) for every u ∈ RN and k = 1, . . . , n; i = 1, . . . , k, whence by (13.10.13), for every x ∈ RN , n k      g p(a, b)x = g αki ai bk−i x k=0 i=0

=

n k 

 αki Ai B k−i g(x) = p(A, B)g(x).

(13.10.14)

k=0 i=0

Since f is non-constant, there exists a u ∈ RN such that g(u) = 0. Suppose that p(a, b) = 0. Setting in (13.10.14) x =  u/p(a, b) we obtain g(u) = p(A, B)g(x), whence p(A, B) = 0. If p(A, B) = 0, then g p(a, b)u = p(A, B)g(u) = 0, whence p(a, b) = 0, since g(0) = 0. Now take an arbitrary r ∈ Q(x, y): r(x, y) =

p(x, y) , q(x, y)

where p, q ∈ Q[x, y]. If one of q(a, b), q(A, B) is different from zero, so is the other, as has been just shown, and consequently if one of r(a, b), r(A, B) has sense, then the other has sense, too. Now for arbitrary x ∈ RN , we have by (13.10.14), if r(a, b) and r(A, B) have sense,       p(A, B)g(x) = g p(a, b)x = g q(a, b r(a, b)x) = q(A, B)g r(a, b)x , 

whence we obtain (13.10.12).

Now consider the extensions Q(a, b) and Q(A, B) of Q by a, b and A, B, respectively (cf. 4.7). Lemma 13.10.3. Let ϕ : Q(a, b) → Q(A, B) be an isomorphism such that ϕ(a) = A,

ϕ(b) = B.

(13.10.15)

Then for every x ∈ Q we have ϕ(x) = x, and for every rational function r ∈ Q(x, y) such that r(a, b), r(A, B) have sense   ϕ r(a, b) = r(A, B). (13.10.16)

386

Chapter 13. Related Equations

Proof. Since ϕ is an isomorphism, ϕ is one-to-one. Now, ϕ(0) = ϕ(0) + ϕ(0), whence ϕ(0) = 0, and so ϕ(x) = 0 for x = 0. We have ϕ(1) = ϕ(1)ϕ(1), whence ϕ(1) = 1. By induction ϕ(k) = k for k ∈ N. Further ϕ(1) = ϕ(−1)ϕ(−1), whence ϕ(−1) = −1, since ϕ is one-to-one and 1 = ϕ(1). So ϕ(−k) = ϕ(−1)ϕ(k) = −ϕ(k) = −k for k ∈ N, and since ϕ(0)  =0, we have ϕ(k) = k for k ∈ Z. Now, for arbitrary k ∈ Z and m ∈ N k = ϕ(k) = k, whence ϕ(x) = x for x ∈ Q. we have mϕ m     Since ϕ is an isomorphism, we have by (13.10.15) ϕ r(a, b) = r ϕ(a), ϕ(b) = r(A, B).    RN may be regarded as the linear space RN ; Q(a, b); +; · over Q(a, b), and as such has a base H in virtue of Corollary 4.2.1. If c = 0, then {c} is linearly independent over Q(a, b), and by Theorem 4.2.1 we may require that c ∈ H. Theorem 13.10.3. Suppose that there exists an isomorphism ϕ : Q(a, b) → Q(A, B) fulfilling (13.10.15). Let H ⊂ RN be a base of RN over Q(a, b) such that c ∈ H whenever c = 0. Put H0 = H ∪ {0}. Further assume that if A + B = 1 and c = 0, then C = 0. For every function f0 : H0 → R such that f0 (c) = (A + B)f0 (0) + C

(13.10.17)

there exists a unique function f : RN → R satisfying equation (13.10.1) and such that f (x) = f0 (x) for x ∈ H0 . Proof. Every x ∈ RN has a representation x = α1 h1 + · · · + αn hn ,

(13.10.18)

where αi ∈ Q(a, b), hi ∈ H, i = 1, . . . , n, and representation (13.10.18) is unique up to terms with coefficients zero. For such an x put f (x) =

n



ϕ(αi ) f0 (hi ) − f0 (0) + f0 (0).

(13.10.19)

i=1

Thus the function f : RN → R is unambiguously defined in the whole RN ; note that, since ϕ(0) = 0 (Lemma 13.10.3), adding to (13.10.18) terms with coefficients zero does not affect (13.10.19). Let c = 0 so that c ∈ H. Take arbitrary x, y ∈ RN , and let x have a representation (13.10.18). y has a representation y = β1 h1 + · · · + βn hn .

(13.10.20)

We may assume that the same hi occur in (13.10.18) and (13.10.20), adding, if necessary, suitable terms with coefficients zero. For the same reason we may assume that one of h1 , . . . , hn is c, and thus c can be written as c = γ1 h1 + · · · + γn hn ,

(13.10.21)

13.10. The general linear equation

387

where all γi are zero, except one, which is one. Now, by (13.10.18)–(13.10.21),  n 

f (ax + by + c) = f (αi a + βi b + γi )hi i=1

=

n



ϕ(αi a + βi b + γi ) f0 (hi ) − f0 (0) + f0 (0)

i=1

=

n





ϕ(αi )A + ϕ(βi )B + ϕ(γi ) f0 (hi ) − f0 (0) + f0 (0)

i=1

=A

n

n





ϕ(αi ) f0 (hi ) − f0 (0) + B ϕ(βi ) f0 (hi ) − f0 (0)

i=1

+

n

i=1



ϕ(γi ) f0 (hi ) − f0 (0) + f0 (0)

i=1





= A f (x) − f0 (0) + B f (y) − f0 (0) + f (c)

= Af (x) + Bf (y) + C − (A + B)f0 (0) − f (c) + C = Af (x) + Bf (y) + C, because the expression in brackets is zero in virtue of (13.10.17) and of the fact that, since c ∈ H, we have by (13.10.19) f (c) = ϕ(1) f0 (c) − f0 (0) + f0 (0) = f0 (c) − f0 (0) + f0 (0) = f0 (c). If c = 0, the argument is the same, only we do not assume that c is one of h1 , . . . , hn , and in representation (13.10.21) all γi are zeros. Thus f satisfies equation (13.10.1). For x ∈ H we have by (13.10.19)

f (x) = ϕ(1) f0 (x) − f0 (0) + f0 (0) = f0 (x). Similarly, for x = 0 we have x = 0h with an arbitrary h ∈ H, whence by (13.10.19)

f (0) = ϕ(0) f0 (h) − f0 (0) + f0 (0) = f0 (0). Thus f (x) = f0 (x) for x ∈ H0 . In order to prove uniqueness, suppose that a function f : RN → R satisfies equation (13.10.1) and f | H0 = f0 . By Lemma 13.10.1 f has form (13.10.4), where g : RN → R is additive, and β ∈ R is a constant. Since g(0) = 0, we have β = f (0), and thus g fulfils (13.10.10). Now take an arbitrary x ∈ RN , and let x have representation (13.10.18). Since α1 , . . . , αn ∈ Q(a, b), it follows by Corollary 4.8.1 that for every i = 1, . . . , n there exists a rational function ri ∈ Q(x, y) such that αi = ri (a, b). By Lemmas 13.10.2 and 13.10.3 n n n

  g(x) = g(αi hi ) = g ri (a, b)hi = ri (A, B)g(hi ) i=1

i=1

i=1

n n

  ϕ ri (a, b) g(hi ) = ϕ(αi )g(hi ), = i=1

i=1

388

Chapter 13. Related Equations

whence f (x) = g(x) + f (0) =

n

ϕ(αi )g(hi ) + f (0) =

i=1

n



ϕ(αi ) f (hi ) − f (0) + f (0),

i=1

i.e., since hi ∈ H0 , 0 ∈ H0 , and f | H0 = f0 , f (x) =

n



ϕ(αi ) f0 (hi ) − f0 (0) + f0 (0).

i=1

Thus f must be given by formula (13.10.19), which proves the uniqueness of f .



Theorem 13.10.4. Suppose that equation (13.10.1) has a non-constant solution f : RN → R . Then there exists an isomorphism ϕ : Q(a, b) → Q(A, B) fulfilling (13.10.15). Moreover, if A + B = 1 and c = 0, then also C = 0. Proof. Define the function ϕ : Q(a, b) → Q(A, B) by (13.10.16). We shall show that this definition is correct. Suppose that r1 (a, b) = r2 (a, b) for some r1 , r2 ∈ Q(x, y). Function (13.10.10) is not constant, since otherwise f would be constant. So there exists a u ∈ RN such that g(u) = 0. We have by Lemma 13.10.2     r1 (A, B)g(u) = g r1 (a, b)u = g r2 (a, b)u = r2 (A, B)g(u), whence also r1 (A, B) =  r2 (A,B). Also, if α = r(a, b) ∈ Q(a, b), then r(A, B) has sense, and so ϕ(α) = ϕ r(a, b) is by (13.10.16) well defined. It follows by Corollary 4.8.1 that ϕ is defined on the whole Q(a, b), and maps it onto Q(A, B). Also, the rational rx (x, y) = x/1, and ry (x, y) = y/1 belong  functions   to Q(x, y), whence ϕ(a) = ϕ rx (a, b) = rx (A, B) = A, and ϕ(b) = ϕ ry (a, b) = ry (A, B) = B. Thus ϕ satisfies (13.10.15). Now we show that ϕ is a homomorphism. Let α, β ∈ Q(a, b). Then α = r1 (a, b), β = r2 (a, b) for certain r1 , r2 ∈ Q(x, y). Then α+β = r1 (a, b)+r2 (a, b) = (r1 +r2 )(a, b) and αβ = r1 (a, b)r2 (a, b) = (r1 r2 )(a, b). Of course, r1 + r2 , r1 r2 ∈ Q(x, y). Hence   ϕ(α + β) = ϕ (r1 + r2 )(a, b) = (r1 + r2 )(A, B) = r1 (A, B) + r2 (A, B) = ϕ(α) + ϕ(β),   ϕ(αβ) = ϕ (r1 r2 )(a, b) = (r1 r2 )(A, B) = r1 (A, B)r2 (A, B) = ϕ(α)ϕ(β). Suppose that for a certain α = r(a, b) ∈ Q(a, b) we have ϕ(α) = r(A, B) = 0. Let u ∈ RN be such that g(u) = 0. Suppose that α = 0. Then by Lemma 13.10.2   u 0 = r(A, B)g = g(u) = 0. r(a, b) Consequently ϕ(α) = 0 implies α = 0, i.e., if ϕ(α) = ϕ(β) for some α, β ∈ Q(a, b), then ϕ(α − β) = ϕ(α) − ϕ(β) = 0, and α = β. Thus ϕ is one-to-one, i.e., it is a monomorphism. We have pointed out above that ϕ is an epimorphism, and so ϕ is an isomorphism.

13.10. The general linear equation

389

Setting in equation (13.10.1) x = y = 0, we obtain f (c) = (A + B)f (0) + C. Hence, if A + B = 1 and c = 0, we obtain C = 0.



Theorems 13.10.3 and 13.10.4 imply the following Theorem 13.10.5. Equation (13.10.1) has a non-constant solution f : RN → R if and only if there exists8 an isomorphism ϕ : Q(a, b) → Q(A, B) fulfilling (13.10.15) and C = 0 whenever A + B = 1 and c = 0.   Proof. Since R = Q(a, b) for any a, b ∈ R, we have dim R; Q(a, b); +; · > 1, and hence, in Theorem 13.10.3, card H > 1. Thus f0 fulfilling (13.10.17) can be chosen non-constant, and then also f is non-constant. The theorem results now from Theorems 13.10.3 and 13.10.4. 

Exercises 1. Find solutions f : (0, 1] → R of equation (13.1.3). 2. Determine all functions f : RN → R satisfying the condition: For every x, y, u, v ∈ RN , if x + y = u + v, then f (x) + f (y) = f (u) + f (v). ˙ be defined for x, y ∈ I by 3. Let I = [0, 1], and let the operation +  x+y if x + y  1, xy = x + y − 1 if x + y > 1. Find all monotonic solutions f : I → I of the equation f (x  y) = f (x)  f (y).

(∗)

[Hint: Note that if f : I → I satisfies (∗), and g(x) = 1 − f (x) for x ∈ I, then also g satisfies (∗).] 4. Find solutions f : RN → R of the equation

f (xy) f (xy) − f (x) − f (y) = 0.

2 5. Let f : RN → R and g : RN → R be additive functions. Prove that if f (x) =

2 g(x) for x ∈ RN , then either f = g, or f = −g in RN . (Grz¸a´slewicz [123].)

8 A condition for the existence of such an isomorphism is found in Theorem 4.12.2. (Note that by Lemma 13.10.3 we have ϕ | Q = identity.)

Chapter 14

Derivations and Automorphisms 14.1 Derivations In this chapter we will deal with functions satisfying the Cauchy equation (5.2.1) and also, simultaneously, another equations of a similar type. Let (Q; +, ·) be a commutative ring, and let (P ; +, ·) be a subring of (Q; +, ·). A function f : P → Q is called a derivation 1 (or a darivation of P ) iff it satisfies both the equations f (x + y) = f (x) + f (y), f (xy) = xf (y) + yf (x)

(14.1.1) (14.1.2)

for all x, y ∈ P . Before we give some examples, let us note the following simple fact. Lemma 14.1.1. Let P ⊂ Q be commutative rings, let f1 ,f2 : P → Q be derivations (of P ), and let a,b ∈ Q be arbitrary constants. Then f = af1 + bf2 also is a derivation of P . 

Proof. Obvious.

Example 14.1.1. Let (F ; +, ·) be a field, and let P = Q = F [x] be the ring of polynomials with coefficients from F . Let the function f : F [x] → F [x] be defined as f (p) = p , where the derivative p is defined in 4.7. We have clearly f (p + q) = (p + q) = p + q  = f (p) + f (q) , f (pq) = (pq) = pq  + qp = pf (q) + qf (p) . Consequently f is a derivation. Example 14.1.2. Let (F ; +, ·) be a field, and suppose that we are given a derivation f : F → F . We define a function f0 : F [x] → F [x] as follows. If p ∈ F [x], p(x) = n  ak xk , then f0 (p) is the polynomial (in sequel denoted by pf ): k=0

f0 (p) = pf (x) =

n

f (ak )xk .

k=0 1

The present exposition of the theory of derivations is based on Zariski-Samuel [325] (cf. also Horinouchi-Kannappan [144] concerning a generalization).

392

Chapter 14. Derivations and Automorphisms n 

If q ∈ F [x], q(x) =

bk xk is another polynomial (we can write p and q with the

k=0

same n, adding to that of the smaller degree terms with coefficients zero), then by (14.1.1) f0 (p + q) =

n

f (ak + bk )xk =

k=0

We have pq =

2n  k=0



n

f (ak )xk +

k=0

n

f (bk )xk = f0 (p) + f0 (q).

k=0

 ai bk−i xk (where aj = bj = 0 for j > n), whence by (14.1.1)

k 

i=0

and (14.1.2) f0 (pq) =

2n k 2n  k   

f ai bk−i xk = f (ai bk−i ) xk k=0

=

k=0

i=0

 ai f (bk−i ) + bk−i f (ai ) xk

k=0

=

i=0

2n  k 

i=0

2n  k

k=0

2n  k  

ai f (bk−i ) xk + f (ai )bk−i xk

i=1

k=0

i=0

= pf0 (q) + qf0 (p). 

 Note that by (14.1.1) f (0) = 0 so that f (aj ) = f (bj ) for j > n. Consequently f0 is a derivation. The derivations described in the above two examples have rather a fundamental importance for we have the following

Lemma 14.1.2. Let (K; +, ·) be a field, and let (F ; +, ·) be a subfield of (K; +, ·), and let f : F → K be a derivation. Then we have, for every a ∈ F and every polynomial p ∈ F [x],   f p(a) = pf (a) + f (a)p (a). Proof. For k ∈ N we have

  f ak = kak−1 f (a).

(14.1.3)

(14.1.3) is evidently true for k = 1. Assuming it valid for a k ∈ N, we have by (14.1.2)       f ak+1 = f aak = af ak + ak f (a) = akak−1 f (a) + ak f (a) = (k + 1)ak f (a), i.e., we obtain (14.1.3) for k + 1. Thus (14.1.3) is generally true. Moreover, setting in (14.1.2) x = y = 1 we obtain f (1) = 2f (1), whence f (1) = 0.

(14.1.4)

14.1. Derivations

393

Now let p(x) =

n 

ak xk . Then by (14.1.3) and (14.1.4)

k=0 n n n n           ak ak = f ak ak = ak f ak + ak f ak f p(a) = f k=0

= pf (a) +

k=0 n

k=0

k=1

ak kak−1 f (a) = pf (a) + f (a)p (a),

k=1



which proves the lemma.

In R the function f = 0 evidently is a (trivial) derivation (of R). It is difficult to find another example. We have, in particular, Lemma 14.1.3. If f : R → R is a derivation, then f (x) = 0 for every x ∈ Q. Proof. By (14.1.1) f is an additive function. Hence we have by (14.1.4) and Theorem 5.2.1 f (x) = xf (1) = 0 for every x ∈ Q.  This result can be strengthened as follows. Lemma 14.1.4. If f : R → R is a derivation, then f (x) = 0 for every x ∈ alg cl Q. Proof. Let a ∈ alg cl Q, and let p ∈ Q [x] be its minimal polynomial. Write p(x) = n  ak xk , ak ∈ Q, k = 0, . . . , n. By Lemma 14.1.3 f (ak ) = 0 for k = 0, . . . , n, whence k=0

pf (x) =

n

  f ak xk = 0. k=0

    So by Lemma 14.1.2 f p(a) = f (a)p (a). Now, p(a) = 0, whence f p(a) = f (0) = 0. On the other hand p (a) = 0, because degree p (a) < degree p(a), and p is the minimal polynomial of a. Hence f (a) = 0.  From Lemma 14.1.3 we get also Theorem 14.1.1. If f : R → R is a derivation, and f is measurable, or bounded above, or below, on a set T ∈ A, then f = 0. Proof. Since, by (14.1.1), f is additive, the conditions on f imply its continuity. Since Q is dense in R, we obtain from Lemma 14.1.3 that f = 0 identically in R.  It is not difficult to solve equation (14.1.2) alone for f : R → R. Taking x, y = 0 and dividing (14.1.2) by xy, we obtain f (xy) f (x) f (y) = + , xy x y which means that the function f0 (x) = f (x)/x satisfies equation (13.1.2) in (−∞, 0)∪ (0, ∞). By 13.1.2 there exists an additive function g : R → R such that  Theorem  f0 (x) = g log |x| , whence   f (x) = xg log |x| . (14.1.5)

394

Chapter 14. Derivations and Automorphisms

Formula (14.1.5), established so far only for x = 0, remains true also for x = 0 if we agree that 0g(log 0) = 0. But, in general, function (14.1.5) will not be additive, i.e., it will not satisfy (14.1.1). Is it possible to choose g in such a manner that f given by (14.1.5) should be additive? In the light of our experience up to now it seems doubtful whether there may exist non-trivial derivations f : R → R. In the next section we will answer this question.

14.2 Extensions of derivations In the present section we study the possibility of extending a derivation from its domain of definition onto a larger algebraic structure. Lemma 14.2.1. Let (P ; +, ·) be an integral domain, let (F ; +, ·) be its field of fractions, and let (K; +, ·) be a field, P ⊂ F ⊂ K. If f : P → K is a derivation, then there exists a unique derivation g : F → K such that g | P = f . Proof. Every x ∈ F can be written as x = u/v, where u, v ∈ P , v = 0. For such an x we put  u  vf (u) − uf (v) = g(x) = g . (14.2.1) v v2 We must check that definition (14.1.1) is unambiguous, i.e., if u/v = z/w, v, w = 0, then g(u/v) = g(z/w). Now, u/v = z/w means uw = vz,

(14.2.2)

whence by (14.1.2) uf (w) + wf (u) = vf (z) + zf (v), i.e., vf (z) − uf (w) = wf (u) − zf (v). Multiplying this by uw we get v 2 wf (z) − uvwf (w) = vw2 f (u) − zvwf (v), or, by (14.2.2) v 2 wf (z) − v 2 zf (w) = vw2 f (u) − uw2 f (v). Dividing this by v 2 w2 we obtain wf (z) − zf (w) vf (u) − uf (v) = , 2 w v2 i.e., g(z/w) = g(u/v). Thus expression (14.2.1) does not depend on the representation of x as a fraction u/v.

14.2. Extensions of derivations

395

Now take arbitrary x, y ∈ F , x = u/v, y = z/w, u, v, z, w ∈ P , v, w = 0. Then by (14.1.1) and (14.1.2)   u z uw + zv vwf (uw + zv) − (uw + zv)f (uw) g(x + y) = g + =g = v w vw v 2 w2     vw uf (w) + wf (u) + zf (v) + vf (z) − (uw + zv) vf (w) + wf (v) = v 2 w2 vw2 f (u) + v 2 wf (z) − uw2 f (v) − zv 2 f (w) = v 2 w2 z u vf (u) − uf (v) wf (z) − zf (w) +g = + =g 2 2 v w v w = g(x) + g(y) , and

= = = =

 uz 

vwf (uz) − uzf (vw) = 2 2 vw  v w   vw uf (z) + zf (u) − uz vf (w) + wf (v) v2 w2    uv wf (z) − zf (w) + zw vf (u) − uf (v) v 2 w2 u wf (z) − zf (w) z vf (u) − uf (v) + 2 v w w v2 u  z  z u g + g = xg(y) + yg(x). v w w v

g(xy) = g

Consequently g is a derivation of F . If x ∈ P , then x = x/1 and by (14.1.4) g(x) = g

x 1

=

f (x) − xf (1) = f (x), 1

i.e., g | P = f . Now let g : F → K be an arbitrary derivation such that g | P = f . Take an arbitrary x ∈ F , x = u/v, u, v ∈ P , v = 0. We have f (u) = g(u) = g(vx) = vg(x) + xg(v) = vg(x) + xf (v), whence g(x) =

f (u) − xf (v) vf (u) − uf (v) = . v v2

This means that g has form (14.2.1), which proves the uniqueness of the extension.  Lemma 14.2.2. Let (K; +, ·) be a field, let (F ; +, ·) be a subfield of (K; +, ·), and let f : F → K be a derivation. Further, let a, u ∈ K. There exists a derivation g : F (a) → K such that g | F = f and g(a) = u if and only if rf (a) + ur (a) = 0 for every r ∈ F [x] such that r(a) = 0. If it exists, the extension g is unique.

(14.2.3)

396

Chapter 14. Derivations and Automorphisms

Proof. First we define a derivation f0 : F [a] → K such that f0 | F = f and f0 (a) = u. If x ∈ F [a], then there exists a polynomial p ∈ F [t] such that p(a) = x. For such an x we put   (14.2.4) f0 (x) = f0 p(a) = pf (a) + up (a). We must check that definition (14.2.4) is unambiguous, i.e., if p(a) = q(a), q ∈ F [t], then pf (a) + up (a) = q f (a) + uq  (a). (14.2.5) Put r = p − q. Then r(a) = 0, whence it follows that (14.2.3) is fulfilled. But rf = pf − q f and r = p − q  , so (14.2.3) means (14.2.5). It is easily seen that f0 is a derivation, and if x ∈ F so that x = p(a) for p(t) = x = const, then by (14.2.4)   f0 (x) = f0 p(a) = pf (a) + up (a) = f (x), since pf (t) = f (x) for every t ∈ F , and p = 0. Thus f0 | F = f . Similarly, a = p(a) for p(t) = t. By (14.2.4)   f0 (a) = f0 p(a) = pf (a) + up (a) = u , because pf (t) = f (1)t = 0 by (14.1.4), and p (t) = 1. Thus f0 (a) = u. The extension f0 is unique, for if f0 : F [a] → K is an arbitrary derivation such that f0 | F = f and f0 (a) = u, then, by Lemma 14.1.2, for arbitrary x ∈ F [a], x = p(a), p ∈ F [t],   f0 (x) = f0 p(a) = pf0 (a) + f0 (a)p (a) = pf (a) + up (a). So f0 has to have form (14.2.4). Before proceeding further, let us make sure that F [a] is an integral domain. F [a] clearly is commutative, and the constant polynomial p(t) = 1 fulfils p(a) = 1, so 1 ∈ F [a]. Now let w, v ∈ F [a] be such that wv = 0. Since F [a] ⊂ K, we have w, v ∈ K. But a field has no divisors of zero, so wv = 0 implies that either w = 0, or v = 0. Consequently F [a] has no divisors of zero, and so it is an integral domain. By Lemma 4.8.1 F (a) is the field of fractions of F [a]. By Lemma 14.2.1 f0 can be uniquely extended onto F (a) to a derivation g : F (a) → K such that g | F [a] = f0 . Hence g | F = g | F [a] | F = f0 | F = f , and, since a ∈ F [a], g(a) = f0 (a) = u. In the opposite direction, if g is the required extension, then, in particular, g satisfies equation (14.1.1) for x, y ∈ F (a), whence g(0) = 0. Hence, by Lemma 14.1.2, if r ∈ F [t] is such that r(a) = 0, then   rf (a) + ur (a) = rg (a) + g(a)r (a) = g r(a) = g(0) = 0. Consequently condition (14.2.3) is fulfilled.



Lemma 14.2.3. Let (K; +, ·) be a field, let (F ; +, ·) be a subfield of (K; +, ·) and let S ⊂ K algebraically independent over F . Let f : F → K be a derivation, and let u : S → K be an arbitrary function. Then there exists a unique derivation g : F (S) → K such that g | F = f and g | S = u.

14.2. Extensions of derivations

397

Proof. Let R be the collection of all couples (Sα , gα ) such that Sα ⊂ S, gα : F (Sα ) → K is a derivation, and gα | F = f , gα | Sα = u. The couple (∅, f ) ∈ R so that R = ∅. We order R as follows: (Sα , gα )  (Sβ , gβ ) iff Sα ⊂ Sβ and gβ | F (Sα ) = gα . Thus (R, ) in an  ordered set, and if L ⊂ R is a chain, then the couple (S0 , g0 ) such that S0 = Sα , g0 | F (Sα ) = gα for (Sα , gα ) ∈ L, is an upper bound (Sα ,gα )∈L

of L in R. By Theorem 1.8.1 there exists a maximal element (Smax , gmax ) in R. Thus, in particular, Smax ⊂ S. Suppose that there exists an a ∈ S \ Smax . Since S is algebraically independent over F and a ∈ S\Smax , we have (cf. Exercise 4.19) r(a) = 0 for every r ∈ F (Smax )[x]. So the condition in Lemma 14.2.2 is trivially fulfilled. By Lemma 14.2.2 the derivation gmax can be extended onto F (Smax )(a) to a derivation g ∗ : F (Smax )(a) → K such that g ∗ | F (Smax ) = gmax , g ∗ (a) = u(a). Hence g ∗ satisfies g ∗ | F = g ∗ | F (S | F = f , g ∗ | Smax = gmax | Smax = u, whence  max ) | F = gmax ∗ ∗ g | Smax ∪ {a} = u. Writing S = Smax ∪ {a} we obtain hence that (S ∗ , g ∗ ) ∈ R and (Smax , gmax ) < (S ∗ , g ∗ ), which contradicts the maximality of (Smax , gmax ). Consequently Smax = S, and g = gmax is the required extension. It remains to prove the uniqueness. Let x ∈ F (S). By Lemma 4.8.2 there exists a finite set S1 = {a1 , . . . , an } ⊂ S such that x ∈ F (S1 ). But F (S1 ) = F (a1 )(a2 ) · · · (an ). Since, by Lemma 14.2.2, the extension of a derivation onto a simple extension of its field of definition (with the prescribed value at the element by which we extend the basic field) is unique, g is uniquely determined on F (a1 ) . . . (ak ) for every k = 1, . . . , n, and so, in particular, g(x) is uniquely determined. Thus g is uniquely determined at every point x ∈ F (S), and so g is unique.  Lemma 14.2.4. Let (K; +, ·) be a field of characteristic zero, let (K1 ; +, ·) be a subfield of (K; +, ·), and let (F ; +, ·) be a subfield of (K1 ; +, ·) such that K1 ⊂ alg cl F . Let f : F → K be a derivation. Then there exists a unique derivation g : K1 → K such that g | F = f . Proof. Let R be the collection of all couples (Kα , gα ) such that (Kα ; +, ·) is a subfield of (K1 ; +, ·) , F ⊂ Kα , gα : Fα → K is a derivation and gα | F = f . (F, f ) ∈ R, so R = ∅. We order R similarly as in the proof of Lemma 14.2.3: (Kα , gα )  (Kβ , gβ ) iff (Kα ; +, ·) is a subfield of (Kβ ; +, ·) and gβ | Kα = gα . Then (R, ) is an ordered set, and as previously we verify that every chain in R has an upper bound (K0 , g0 ) ∈ R. By Theorem 1.8.1 there exists in R a maximal element (Kmax , gmax ). In particular, (Kmax ; +, ·) is a subfield of (K1 ; +, ·). Suppose that there exists an a ∈ K1 \ Kmax . Thus a is algebraic over F , and let p ∈ F [x] be its minimal polynomial. We have p = 0, since the characteristic of K is zero (whence also the characteristic of F is zero), whence p (a) = 0, since degree p < degree p, and p is the minimal polynomial of a. Put u = −pgmax (a)/p (a), and let r ∈ Kmax [x] be a polynomial such that r(a) = 0. There exist polynomials q, s ∈ Kmax [x] such that r = qp + s and degree s < degree p. Hence 0 = r(a) = q(a)p(a) + s(a) = s(a), whence it follows that s = 0, and r = qp.

398

Chapter 14. Derivations and Automorphisms

Now rgmax (a) + ur (a) = q(a)pgmax (a) + p(a)q gmax (a) + up (a)q(a) + up(a)q  (a)   = q(a) pgmax (a) + up (a) , since p(a) = 0. By the choice of u we have pgmax (a) + up (a) = 0, whence also rgmax (a) + ur (a) = 0. By Lemma 14.2.2 there exists a derivation g ∗ : Kmax (a) → K such that g ∗ | Kmax = gmax . Write K ∗ = Kmax (a). Since Kmax ⊂ K1 and a ∈ K1 , the field (K ∗ ; +, ·) is a subfield of (K1 ; +, ·), and, of course, F ⊂ Kmax ⊂ K ∗ . Moreover, g ∗ | F = gmax | F = f . Thus (K ∗ , g ∗ ) ∈ R, and (Kmax , gmax ) < (K ∗ , g ∗ ), which contradicts the maximality of (Kmax , gmax ). Hence Kmax = K1 , and g = gmax is the desired extension. To prove uniqueness, suppose that g : K1 → K is a derivation such that g | F = f . Take an a ∈ K1 . Let p ∈ F [x] be the minimal polynomial of a. As we have seen above, we have p (a) = 0. By Lemma 14.1.2   pf (a) + g(a)p (a) = pg (a) + g(a)p (a) = g p(a) = g(0) = 0. Hence g(a) = −pf (a)/p (a) is uniquely determined. Thus all the values of g on K1 are uniquely determined, whence g is unique.  Theorem 14.2.1. Let (K; +, ·) be a field of characteristic zero, let (F ; +, ·) be a subfield of (K; +, ·), let S be an algebraic base of K over F , if it exists, and let S = ∅ otherwise. Let f : F → K be a derivation. Then, for every function u : S → K, there exists a unique derivation g : K → K such that g | F = f and g | S = u. Proof. If K = alg cl F , this results from Lemma 14.2.4. (Then S = ∅, and there is no u involved). If K = alg cl F , then, by Theorem 4.10.1, there exists an algebraic base S of K over F so that alg cl F (S) = K. S is algebraically independent over F , so by Lemma 14.2.3 there exists a unique derivation g0 : F (S) → K such that g0 | F = f and g0 | S = u. By Lemma 14.2.4 the derivation g0 can be uniquely extended onto alg cl F (S) = K to a derivation g : K → K such that g | F (S) = g0 , whence g | F = g | F (S) | F = g0 | F = f , and g | S = g | F (S) | S = g0 | S = u.  Theorem 14.2.2. There exist non-trivial derivations of R. Proof. We have R = alg cl Q, so there exists, by Theorem 4.10.1, an algebraic base of R over Q. Let u : S → R be an arbitrary function, u = 0. The characteristic of R is zero. Now take in Theorem 14.2.1 F = Q, K = R. The trivial derivation f0 : Q → R, f0 = 0, can, by Theorem 14.2.1, be uniquely extended onto R to a derivation f : R → R such that f | S = u, whence f = 0.  Incidentally, in this way we have obtained a description of all the derivations of R. Every such derivation can be arbitrary prescribed on an algebraic base S of R over Q, and then it is already uniquely determined. Unless identically equal to zero, every such derivation is discontinuous (and hence non-measurable, unbounded above and below on every set T ∈ A) in virtue of Theorem 14.1.1 .

14.3. Relations between additive functions

399

14.3 Relations between additive functions Let f : R → R be a derivation, and let x ∈ R \ {0}. We have by Lemma 14.1.3     1 1 1 = xf + f (x), 0 = f (1) = f x x x x whence f (x) = −x2 f

  1 . x

(14.3.1)

We will show that relation (14.3.1) characterizes derivations among additive functions2 f : R → R. Theorem 14.3.1. Let f : R → R be an additive function satisfying condition (14.3.1) for all real x = 0. Then f is a derivation. Proof. Take an arbitrary x ∈ R, x = 0, 1, −1. Then we have by (14.3.1) and (14.1.1)    2    1 x −1 1 1 =f x− =f f (x) + 2 f (x) = f (x) − f x x x x 2    2 2    2 x −1 1 x 1 x −1 = − − f f =− x x2 − 1 x x − 1 x2 − 1 2    2 2    2 x −1 1 1 x −1 + f f =− x x−1 x x2 − 1 2  2 2  2  2  2   1 1 x −1 x −1 f (x − 1) − f x2 − 1 = x x−1 x x2 − 1 2   x+1 1  = f (x − 1) − 2 f x2 − 1 . x x Setting in (14.3.1) x = 1 or −1, we obtain f (1) = f (−1) = 0.

(14.3.2)

Hence f (u − 1) = f (u) − f (1) = f (u) for every u ∈ R, and we get 1 f (x) + 2 f (x) = x



x+1 x

2 f (x) −

1  2 f x , x2

  2 x2 f (x) + f (x) = (x + 1) f (x) − f x2 , i.e.,

  f x2 = 2xf (x).

(14.3.3)

2 Kurepa [201]; cf. also Jurkat [159], Kurepa [202], Nishiyama-Horinouchi [243], Kannappan-Kurepa [163], Grz¸a´slewicz [122], [123]. Also the remaining results in this section are due to S. Kurepa [201], [202].

400

Chapter 14. Derivations and Automorphisms

By (14.1.1) f (0) = 0. Thus, in view of (14.3.2), relation (14.3.3), so far obtained for x = 0, 1, −1, is valid for all x ∈ R. We have by (14.3.3) for every x, y ∈ R   f (x + y)2 = 2 (x + y) f (x + y), i.e.,

    f x2 + 2f (xy) + f y 2 = 2xf (x) + 2xf (y) + 2yf (x) + 2yf (y),

and by (14.3.3) 2xf (x) + 2f (xy) + 2yf (y) = 2xf (x) + 2xf (y) + 2yf (x) + 2yf (y). Relation (14.3.4) yields (14.1.2), i.e., f is a derivation.

(14.3.4) 

Lemma 14.3.1. Let f : R → R and g : R → R be additive functions, f = 0, and let P : R → R be a continuous function. If   1 g(x) = P (x)f (14.3.5) x for all x ∈ R, x = 0, then

P (x) = P (1)x2

(14.3.6)

for all x ∈ R. Proof. Since f (0) = 0, but f = 0, there exists an x0 ∈ R \ {0} such that f (x0 ) = 0. Put u = x−1 0 , and take an arbitrary α ∈ Q, α = 0. We have by (14.3.5) and Theorem 5.2.1   1 1 αg(u) = g (αu) = P (αu) f = P (αu) f (x0 ) . αu α On the other hand, also by (14.3.5) αg(u) = αP (u) f (x0 ) . Hence

 αP (u) −

and, since f (x0 ) = 0,

 1 P (αu) f (x0 ) = 0, α

P (αu) = α2 P (u)

for all α ∈ Q \ {0}. Setting αu = y we obtain hence P (y) = y 2

P (u) u2

(14.3.7)

for all y ∈ Qu \ {0}. Since P is continuous and the set Qu \ {0} is dense in R, it follows that (14.3.7) is valid for all y ∈ R. Setting y = 1, we obtain from (14.3.7) P (1) = P (u) /u2 , and so (14.3.7) implies (14.3.6). 

14.3. Relations between additive functions

401

Theorem 14.3.2. Let f : R → R and g : R → R be additive functions satisfying the relation   1 2 g(x) = x f (14.3.8) x for all x ∈ R, x = 0. Then the function f + g is continuous, and the functions F, G : R → R : F (x) = f (x) − xf (1), G(x) = g(x) − xg(1) (14.3.9) are derivations. Proof. We have by (14.3.8) g (1) = f (1), whence by (14.3.8) and (14.3.9), for x = 0,     1 1 x2 F = x2 f − xf (1) = g(x) − xg(1) = G (x) x x so that G (x) = x2 F

  1 x

(14.3.10)

for every x = 0. Now, F (1) = G (1) = 0 by (14.3.9), whence, since F and G clearly are additive, F (x + 1) = F (x) + F (1) = F (x) and G (x + 1) = G (x) + G (1) = G (x) for every x ∈ R. Hence and by (14.3.10) we have for every x = −1     x 1 2 2 G (x) = G (x + 1) = (x + 1) F = (x + 1) F 1 − x+1 x+1  2     x x x+1 2 2 = − (x + 1) G = − (x + 1) F x+1 x+1 x     1 1 = −x2 G = −F (x) , = −x2 G 1 + x x i.e., G (x) = −F (x) .

(14.3.11)

For x = −1 we have G (−1) = −G (1) = 0 = F (1) = −F (−1), so (14.3.11) holds for x = −1, too, and thus (14.3.11) is valid for all x ∈ R. (14.3.11) yields g(x) − xg(1) = −f (x) + xf (1), whence   f (x) + g(x) = x f (1) + g(1) , and consequently f + g is a continuous function. Further, we get by (14.3.10) and (14.3.11)   1 2 F (x) = −x F x for x = 0. By Theorem 14.3.1 F is a derivation, and by Lemma 14.1.1 G = −F also is a derivation. 

402

Chapter 14. Derivations and Automorphisms

Theorem 14.3.3. Let f : R → R be an additive function, and let P : R → R be a continuous function. If   1 f (x) = P (x) f (14.3.12) x for all x ∈ R, x = 0, then if P (1) = −1, f is a derivation, and if P (1) = −1, then f is continuous. Proof. By Lemma 14.3.1 P (x) = cx2 ,

x ∈ R,

(14.3.13)

where c = P (1). If c = 0, then f = 0 by (14.3.12), and f is continuous. If c = 0, write g(x) = c−1 f (x), x ∈ R. Thus g : R → R is an additive  function,  and (14.3.12) with (14.3.13) imply (14.3.8). By Theorem 14.3.2 f + g = 1 + c−1 f is continuous, and hence f is continuous if c = −1. If c = −1, then P (x) = −x2 , and by Theorem 14.3.1 f is a derivation. 

14.4 Automorphisms of R Let (F ; +, ·) and (K; +, ·) be fields. As introduced in 14.2 a function f : F → K is called a homomorphism iff the equations f (x + y) = f (x) + f (y)

(14.4.1)

f (xy) = f (x)f (y)

(14.4.2)

and are fulfilled for every x, y ∈ F . If K = F , then f fulfilling (14.4.1) and (14.4.2) is called an endomorphism. If K = F and f is one-to-one and onto, then f is called an automorphism of F . In this and the next section we will be concerned with the automorphisms of R and C. In R the situation is very simple (Darboux [53]). Theorem 14.4.1. The only function f : R → R satisfying both equations (14.4.1) and (14.1.2) are f1 = 0 and f2 = identity. Proof. Let a function f : R → R satisfy (14.4.1) and (14.4.2). Then f is additive, and we have by (14.4.2) for every x > 0 √  2 f (x) = f x  0. Thus f is bounded below on (0, ∞). By Lemma 9.3.1 f is continuous, and hence (Theorem 5.5.2) f (x) = cx with a certain c ∈ R. Inserting this into (14.4.2) we obtain, cxy = c2 xy for every x, y ∈ R, whence (on setting x = y = 1) c = c2 , and either c = 0, or c = 1. Thus f is either zero, or the identity.  The function f (x) = x clearly is an automorphism (of every field). We call it a trivial automorphism. On the other hand, the function f = 0 is neither invertible, nor onto, and thus it is not an automorphism. So we obtain from Theorem 14.4.1 the following

14.5. Automorphisms of C

403

Theorem 14.4.2. The only automorphism of R is the trivial automorphism f (x) = x. By Theorem 14.4.1 the only endomorphisms of R are f = 0 and f (x) = x (trivial endomorphisms).

14.5 Automorphisms of C As we shall presently see, the situation in C is completely different (Kestelman [170]). Lemma 14.5.1. The only continuous endomorphisms f : C → C are f = 0, f (x) = x and f (x) = x 3 . Proof. By Theorem 5.6.1, if f : C → C is a continuous solution of equation (14.4.1), there exist constants a, b ∈ C such that f (x) = ax + bx.

(14.5.1)

Inserting (14.5.1) into (14.4.2) we obtain axy + bx y = a2 xy + b2 x y + ab (xy + xy)

(14.5.2)

for every x, y ∈ C. Setting in (14.5.2) x = y = 1 we get a + b = a2 + b2 + 2ab,

(14.5.3)

whereas setting in (14.5.2) x = y = i we obtain −a − b = −a2 − b2 + 2ab.

(14.5.4)

Adding (14.5.3) and (14.5.4) yields 4ab = 0, whence either a = 0, or b = 0. If a = b = 0, then we get by (14.5.1) f = 0. If a = 0, then b = 0, and (14.5.3) reduces to a = a2 , whence a = 1, and by (14.5.1) f (x) = x for x ∈ C. If b = 0, then a = 0, and (14.5.3) reduces to b = b2 , whence b = 1, and by (14.5.1) f (x) = x. It is readily seen that all the three functions actually satisfy both equations (14.4.1) and (14.4.2).  The endomorphisms furnished by Lemma 14.5.1 are referred to as trivial. Of these f = 0 is only an endomorphism, whereas f (x) = x and f (x) = x are automorphisms. So far the problem of the existence of non-trivial endomorphisms and automorphisms is unsettled. By Lemma 14.5.1, such endomorphisms and automorphisms, if they exist, must necessarily be discontinuous. In the sequel we use the terminology and notation from 4.12. Lemma 14.5.2. Let (F ; +, ·) be a subfield of (C; +, ·), and let a ∈ C be transcendental over F . Further, let f ∗ : F → F be an automorphism of F . Then there exists an automorphism f : F (a) → F (a) such that f | F = f ∗ . Proof. Since a is transcendental over F , a = 0 and a is f ∗ -conjugate with itself. By  Theorem 4.12.1 there exists an isomorphism f : F (a) → F (a) i.e., an automorphism  of F (a) such that f (a) = a and f | F = f ∗ . f is the desired extension.  3

The complex conjugate of x.

404

Chapter 14. Derivations and Automorphisms

Lemma 14.5.3. Let (F ; +, ·) be a subfield of (C; +, ·), and let a ∈ C be algebraic over F . Further, let f ∗ : F → C be a homomorphism, f = 0. Then there exists a homomorphism f : F (a) → C such that f | F = f ∗ .     Proof. It is easily checked that f ∗ (F ); +; · is a field subfield of (C; +, ·) . By ∗ 4.12.2 ·) onto  f is a∗ monomorphism, and hence an isomorphism of (F ; +, Lemma ∗ ∗ (F ); +; · . Since f satisfies (14.4.2), we have (on setting x = y = 1) f (1) =1 f  ∗ ∗ (1) = 0, whence f = 0 results on setting y = 1 in for otherwise we would have f  (14.4.2) . Let q ∈ F [x] be the minimal polynomial of a, and put Q = If ∗ (q). If we had Q = Q1 Q2 , Q1 , Q2 ∈ f ∗ (F )[x], then since by Lemma 4.12.1 If ∗ is an isomorphism, −1 −1 −1 so is also If−1 ∗ , and we would have q = If ∗ (Q) = If ∗ (Q1 )If ∗ (Q2 ) and q would be reducible, which is not the case. So also Q is irreducible. The coefficient of x in the highest power in Q is f ∗ (1) = 1. Consequently Q is the minimal polynomial of every of its roots. By the fundamental theorem of algebra Q has a root A ∈ C. Thus Q is the minimal polynomial of A. Consequently a and A are f ∗ -conjugate. By Theorem 4.12.1 there exists an isomorphism f : F (a) → f ∗ (F )(A) ⊂ C such that f (a) = A and f | F = f ∗ . f is a homomorphism of (F ; +, ·) into (C; +, ·) fulfilling f | F = f ∗ , and hence it is the desired extension.  Let (F0 ; +, ·) be a subfield of (C; +, ·), and let F be the collection of all couples (F, f ) such that F0 ⊂ F , (F ; +, ·) is a subfield of (C; +, ·) and f : F → C is a homomorphism. The order  is defined in F as usually: for (F1 , f1 ), (F2 , f2 ) ∈ F we write (F1 , f1 )  (F2 , f2 ) iff F1 ⊂ F2 and f2 | F1 = f1 . Lemma 14.5.4. Let F be as described above, and let L ⊂ F be a chain. Define F ∗ =  F , and let f ∗ : F ∗ → C be defined by the condition f ∗ | F = f for every (F,f )∈L

(F, f ) ∈ L. Then (F ∗ , f ∗ ) ∈ F, and (F, f )  (F ∗ , f ∗ ) for every (F, f ) ∈ L.

Proof. Take arbitrary x, y ∈ F ∗ . Since L is a chain, there exists an (F˜ , f˜) ∈ L such that x, y ∈ F˜ . By Lemma 4.7.1 x − y ∈ F˜ ⊂ F∗ and, if y = 0, xy −1 ∈ F˜ ⊂ F ∗ . Thus again by Lemma 4.7.1, (F ∗ ; +, ·) is a field a subfield of (C; +, ·) , and clearly F0 ⊂ F ∗ . Consequently we have also x + y, xy ∈ F˜ . Hence f ∗ (x + y) = f˜(x + y) = f˜(x) + f˜(y) = f ∗ (x) + f ∗ (y), f ∗ (xy) = f˜(xy) = f˜(x)f˜(y) = f ∗ (x)f ∗ (y), since f˜ is a homomorphism. Consequently f ∗ also is a homomorphism. Thus (F ∗, f ∗ ) ∈ F , and evidently (F, f )  (F ∗ , f ∗ ) for every (F, f ) ∈ L.  √ 2 u = 2 is algebraic over Q of degree 2, q(x) = x − 2 being its minimal polynomial. By Lemma 4.11.3 the system {1, u} forms a base of the linear space (Q(u); Q; +, ·). Consequently every x ∈ Q (u) can be uniquely written as x = α + βu,

α, β ∈ Q.

(14.5.5)

14.5. Automorphisms of C

405

Define the function f0 : Q(u) → Q(u) by f0 (x) = f0 (α + βu) = α − βu.

(14.5.6)

Thus, unless β = 0, f0 (x) = x = x and f0 (x) = 0, and so f0 is not a restriction to Q(u) of a trivial automorphism of C. √ Lemma 14.5.5. Let u = 2, and let the function f0 : Q(u) → Q(u) be defined by (14.5.6). Then f0 is an automorphism of Q(u). Proof. Take arbitrary x, y ∈ Q(u). Then x can be written in form (14.5.5), and similarly y = γ + δu, γ, δ ∈ Q. Hence (x + y) = (α + γ) + (β + δ)u, xy = (αγ + 2βδ) + (αδ + βγ)u, and f0 (x + y) = (α + γ) − (β + δ)u = (α − βu) − (γ − δu) = f0 (x) + f0 (y), f0 (xy) = (αγ + 2βδ) − (αδ + βγ)u = (α − βu)(γ − δu) = f0 (x)f0 (y). Consequently f0 is a homomorphism. Moreover, we have

f0 f0 (x) = f0 (α − βu) = α + βu = x. By Lemma 15.5.2 f0 is one-to-one and onto. Consequently f0 is an automorphism of Q(u).  Theorem 14.5.1. There exist non-trivial automorphisms of C. √ Proof. Let u = 2. Let R be the collection of all couples (F, f ) such that (F ; +, ·) is a subfield of (C; +, ·), Q(u) ⊂ F , f : F → F is an automorphism of F, and  f | Q(u) = f0 , where f0 is defined by (14.5.6). By Lemma 14.5.5 Q(u), f0 ∈ R , so R = ∅. Let L ⊂ R be an arbitrary chain, and put F ∗ = F , and let (F,f )∈L

f ∗ : F ∗ → F ∗ be defined by the condition that f ∗ | F = f for every (F, f ) ∈ L. We will show that (F ∗ , f ∗ ) ∈ R. By Lemma 14.5.4 (F ∗ ; +, ·) is a subfield of (C; +, ·), and f ∗ is a homomorphism. Clearly Q(u) ⊂ F ∗ and f ∗ = 0, since f ∗ | Q(u) = f0 . By Lemma 4.12.2 f ∗ is one-to-one. Take an arbitrary y ∈ F ∗ . Then there exists an (F, f ) ∈ L such that y ∈ F . Since f is an automorphism of F , there exists an x ∈ F ⊂ F ∗ such that f (x) = y. Hence also f ∗ (x) = f (x) = y. Thus F ∗ ⊂ f ∗ (F ∗ ). On the other hand, since for every (F, f ) ∈ L we have f (F ) = F ⊂ F ∗ , we have f ∗ (F ∗ ) ⊂ F ∗ . Consequently f ∗ maps F ∗ onto F ∗ , and so is an automorphism. Consequently (F ∗ , f ∗ ) ∈ R. By Lemma 14.5.4 (F ∗ , f ∗ ) is an upper  bound  of L. Thus in virtue of Theorem ˜ ˜ 1.8.1 there exists in R a maximal element F , f . Suppose that there exists an a ∈ C ˜ transcendental over F˜ . Then, by Lemma 14.5.2,f˜ can be extended   onto F (a)  to an ˜ ˜ ˜ ˜ ˜ automorphism f of F (a), and obviously F (a), f ∈ R and F , f < F (a), f , which   contradicts the maximality of F˜ , f˜ . Consequently every element x ∈ C is algebraic over F˜ . Let F be the collection of all pairs (F, f ) such that (F ; +, ·) is a subfield of (C; +, ·), F˜ ⊂ F , and f : F → C

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  is a homomorphism such that f | F˜ = f˜. We have F˜ , f˜ ∈ F, so F = ∅. If we order F in the same way as R above, then it follows from Lemma 14.5.4 that every chain in F has an upper bound. By Theorem 1.8.1 there exists in F a maximal element (Fmax , fmax ). Suppose that Fmax = C, and let a ∈ C \ Fmax . Then a is algebraic over F˜ , and hence also over Fmax . By Lemma 14.5.3 fmax can be extended onto Fmax (a)  to a homomorphism f : Fmax (a) → C. But then (Fmax , fmax ) < Fmax (a), f ∈ F, which contradicts the maximality of (Fmax , fmax ). Consequently Fmax = C. Thus fmax : C → C is a homomorphism, fmax | F˜ = f˜, and f˜ | Q(u) = f0 = 0, whence fmax = 0. By Lemma 4.12.2 fmax is one-to-one. We shall show that fmax (C) = C. Since C is algebraically closed and fmax is an isomorphism of C onto fmax (C), also fmax (C) is algebraically closed (cf. Exercise 4.18). Since F˜ ⊂ C, we ˜ so fmax (F˜ ) = f˜(F˜ ) = F˜ , since f˜ is an have fmax (F˜ ) ⊂ fmax (C). But fmax | F˜ = f, ˜ automorphism. Hence F = fmax (F ) ⊂ fmax (C). Now, every x ∈ C is algebraic over F˜ . This means that C = alg cl F˜ ⊂ alg cl fmax (C) ⊂ fmax (C) (cf. Exercise 4.17). Consequently fmax (C) = C. Thus fmax is onto C, and consequently it is an automorphism of C. We have fmax | Q(u) = fmax | F˜ | Q(u) = f˜ | Q(u) = f0 . If fmax were a trivial automorphism, then f0 would be a restriction of a trivial automorphism, which, as pointed out before Lemma 14.5.5, is not the case. Consequently fmax is a non-trivial automorphism of C. 

14.6 Non-trivial endomorphisms of C Now we will prove a number of theorems indicating at certain irregularities of discontinuous functions f : C → C satisfying (14.4.1) and (14.4.2) (non-trivial endomorphisms of C). If f is such a non-trivial endomorphism, then, in particular, f = 0, and hence by Lemma 4.12.2 f is one-to-one. Since, as results from (14.4.1), f (0) = 0, we have f (x) = 0 for every x ∈ C, x = 0. Theorem 14.6.1. If f : C → C is a non-trivial endomorphism, then f | R is discontinuous. Proof. Write f0 = f | R. For every x ∈ C we have x = Re x + i Im x, whence by (14.4.1) and (14.4.2) f (x) = f (Re x + i Im x) = f (Re x) + f (i)f (Im x) = f0 (Re x) + f (i)f0 (Im x) . If f0 were continuous, then also f would be continuous, and f would be a trivial endomorphism.  Lemma 14.6.1. If f : C → C satisfies (14.4.1), then we have for every α ∈ Q and x∈C f (αx) = αf (x). Proof. This results from Theorems 5.6.1 and 5.2.1.



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407

Theorem 14.6.2. Let f : C → C be a non-trivial endomorphism, and let A ⊂ R, mi (A) > 0, or A of the second category with the Baire property. Then f is unbounded on A, and f (A) is not contained in R. Proof. Suppose that f is bounded on A. Write f0 = f | R and g = |f0 |. It follows from Lemma 14.6.1 that g : R → R is a convex function, and our supposition implies that g is bounded on A. By Theorems 9.3.1 and 9.3.2 g is continuous. In particular lim |f0 (t)| = lim g(t) = g(0) = |f0 (0)| = f (0) = 0.

t→0 t∈R

t→0 t∈R

Hence lim f0 (t) = 0, and for every x ∈ R t→0 t∈R



lim f0 (x + t) = lim f0 (x) + f0 (t) = f0 (x),

t→0 t∈R

t→0 t∈R

i.e., f0 is continuous, which contradicts Theorem 14.6.1. Suppose that f (A) ⊂ R. Let x ∈ A + A. Then x = u + v, u, v ∈ A, and f (x) = f (u) + f (v) ∈ R. Thus f (A + A) ⊂ R. By Theorems 3.7.1 and 2.9.1 the set A + A contains an interval I. Take an x0 ∈ int I. If x ∈ I − x0 , then x = y − x0 , where y ∈ I ⊂ A + A, and f (x) = f (y) − f (x0 ) ∈ R. Thus f (I − x0 ) ⊂ R. But I − x0 is a neighbourhood of 0. Consequently for every x ∈ R there exists an n ∈ N such that x/n ∈ I − x0 . Hence f (x/n) ∈ R, and by Lemma 14.6.1 f (x) = nf (x/n) ∈ R. Consequently f (R) ⊂ R, and f0 is an endomorphism of R. By Theorem 14.4.1 f0 is continuous, which is incompatible with Theorem 14.6.1.  Theorem 14.6.3. Let f : C → C be a non-trivial endomorphism. Then cl f (R) = C. Proof. By Theorem 14.6.2 there exists a u ∈ R such that f (u) ∈ / R. The set Z = {z ∈ C | z = α + βf (u), α, β ∈ Q}

(14.6.1)

is dense in C. For every z ∈ Z we have by (14.4.1), (14.4.2) and Lemma 14.6.1 z = α + βf (u) = f (α + βu) ∈ f (R) . Consequently Z ⊂ f (R), and f (R) is dense in C.



Lemma 14.6.2. Let f : C → C be a non-trivial endomorphism, and let U = {x ∈ C | |x − x0 | < r} ⊂ C be an open disc in C centered at an x0 ∈ C and with a radius r > 0. Then the set A = R ∩ f −1 (U ) ⊂ R fulfils mi (A) = 0, me (A) > 0. Also, A is of the second category, but does not contain any subset of the second category with the Baire property. Proof. Let B = {x ∈ R | f (x) ∈ R}. Thus for every u ∈ R \ B we have f (u) ∈ / R and set (14.6.1) is dense in C. It will remain dense when we subject β to the additional condition β = 0. Consequently for every u ∈ R \ B there exist α, β ∈ Q, β = 0,

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such that f (α + βu) = α + βf (u) ∈ U , whence α + βu ∈ f −1 (U ). Since evidently 1 α + βu ∈ R, we have α + βu ∈ A, whence u ∈ (A − α). Consequently β R\B ⊂

 1 (A − α). β

α,β∈Q β =0

If we had m(A) = 0 or A were of the first category, then the same would also be true about R \ B, whence B would have positive measure or would be of the second category with the Baire property. By the definition of B we have f (B) ⊂ R, which is incompatible with Theorem 14.6.2. Consequently me (A) > 0 and A is of the second category. On the other hand, for x ∈ A we have f (x) ∈ U , which means that f is bounded on A. By Theorem 14.6.2 mi (A) = 0 and A cannot contain any subset of the second category and with the Baire property.  Lemma 14.6.3. Let A ⊂ R be a measurable set. Then m(A) > 0 if and only if  −1  m A \ {0} > 0. Similarly, let A ⊂ R have the Baire property. Then A is of  −1 is of the second category with the Baire the second category if and only if A \ {0} property. Proof. Let ϕ : (−∞, 0) ∪ (0, ∞) → R be given by ϕ(x) = x−1 , and let ϕ1 : (−∞, 0) → (−∞, 0) and ϕ2 : (0, ∞) → (0, ∞) be defined as ϕ1 = ϕ | (−∞, 0), ϕ2 = ϕ | (0, ∞). The function ϕ2 is continuous and convex, and ϕ1 is continuous and concave. Moreover, ϕ1 and ϕ2 are involutions (cf. 12.5), whence ϕ1 = ϕ−1 and ϕ2 = ϕ−1 1 2 . By Theorem 7.4.6 ϕ1 and ϕ2 are absolutely continuous, whence also ϕ−1 and ϕ−1 are 1 2 absolutely continuous. Also, evidently, ϕ1 and ϕ2 are homeomorphisms. For any set A ⊂ R write A+ = A ∩ (0, ∞), and A− = A ∩ (−∞, 0). Then, the following statement are equivalent: (i) A is measurable and m(A) > 0; (ii) A+ and A− are measurable and m (A+ ) > 0 or m(A− ) > 0;     (iii) ϕ2 (A+ ) and ϕ1 (A− ) are measurable, and m ϕ2 (A+ ) > 0 or m ϕ1 (A− ) > 0;      (iv) ϕ A \ {0} is measurable, and m ϕ A \ {0} > 0. Similarly, since ϕ1 and ϕ2 are homeomorphisms, the following statements are equivalent: (i’) A is of the second category and with the Baire property; (ii’) A+ and A− have the Baire property, and A+ or A− is of the second category; (iii’) ϕ2 (A+ ) and ϕ1 (A− ) have the Baire property, and ϕ2 (A+ ) or ϕ1 (A− ) is of the second category;   (iv’) ϕ A \ {0} is of the second category and with the Baire property. 

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Lemma 14.6.4. Let f : C → C be a non-trivial endomorphism, and let U ⊂ C be an open disc. Put A = R ∩ f −1 (U ). Then mi (R \ A) = 0 and R \ A does not contain a subset of the second category and with the Baire property. Proof. At first we consider the case, where U = {x ∈ C | |x| < r} is centered at the origin. Let B ⊂ R be an arbitrary measurable set with m(B) > 0 or a second category set with the Baire property. Suppose that |f (x)|  r for x ∈ B. Take a  −1   y ∈ B \ {0} . Then y −1 ∈ B \ {0} ⊂ B, and 1 = f (1) = |f (1)| = |f yy −1 | =  −1 . By |f (y)|f (y −1 )| ≥ r|f (y)|, whence |f (y)|  1/r. Thus f is bounded on B \ {0}  −1  −1 > 0 or B \ {0} is of the second category and with Lemma 14.6.3 m (B \ {0}) the Baire property, and we get a contradiction with Theorem 14.6.2. Consequently there must exist an x ∈ B ⊂ R such that |f (x)| < r, i.e., f (x) ∈ U . Thus x ∈ R∩f −1 (U ) = A, i.e., A∩B = ∅. So we have A∩B = ∅ for every measurable set B ⊂ R with m(B) > 0 and for every set B ⊂ R of the second category with the Baire property. By Theorem 3.3.1 mi (R \ A) = 0. It is also evident that R \ A cannot contain a subset of the second category and with the Baire property, for otherwise, if B ⊂ R \ A were such a set, then we would have A ∩ B = ∅, contrary to what has just been established. Now let U = V + f (d), where d∈ R and V ⊂ C is an open disc centered at the origin. Take an arbitrary x ∈ R ∩ f −1 (V ) + d . Then there exists a t ∈ f −1 (V ) such that x = t + d. Thus t = x − d ∈ R, since x ∈ R and d ∈ R. We have also f (x) = f (t + d) = f (t) + f(d) ∈ V + f (d) = U . Thus x ∈ f −1 (U ), and since −1 x ∈ R, x ∈ A. Hence  R ∩ f (V ) + d−1⊂ A.  By what we have already shown −1 m R \ R ∩ f (V ) = 0 and R \ R ∩ f (V ) does not contain any subset of the second category and with the Baire property. Further,    

R \ R ∩ f −1 (V ) + d = R \ R ∩ f −1 (V ) + d. (14.6.2) By Corollary 3.3.2 the set on the right-hand side of (14.6.2) has inner measure zero, and since translation is a homeomorphism, this set does not contain any subset of the second category and with the Baire property. Consequently also R \ A has the analogical properties. Finally, let U ⊂ C be an arbitrary open disc. By Theorem 14.6.3 there exists a d ∈ R such that f (d) ∈ U . Further, there exists an open disc W centered at f (d) and contained in U . Put V = W −f (d). Then V is an open disc centered at the origin, and  W = V + f (d). By what we have already proved the set R \ R ∩ f −1 (W ) has inner measure zero and does not contain any subset of the second category and with the Baire property. But we have W ⊂ U , whence f −1 (W ) ⊂ f−1 (U ) and R ∩ f −1 (W ) ⊂ R ∩ f −1 (U ) = A. Consequently R \ A ⊂ R \ R ∩ f −1 (W ) , and it follows that R \ A has inner measure zero and does not contain any subset of the second category and with the Baire property.  Corollary 14.6.1. Let f : C → C be a non-trivial endomorphism, and let U ⊂ C be an open disc. Then the set A = R ∩ f −1 (U ) is a saturated non-measurable and has property (∗) from 3.3.

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Proof. This results from Lemmas 14.6.2 and 14.6.4.



Theorem 14.6.4. Let f : C → C be a non-trivial endomorphism, and let S ⊂ C be a set such that int S = ∅ and int(C \ S) = ∅. Then the set A = R ∩ f −1 (S) is saturated non-measurable and has property (∗) from 3.3. Proof. There exist open discs U ⊂ S and V ⊂ C \ S. We have S ⊂ C \ V and −1 C\S ⊂ C\U , whence f −1 (S) ⊂ f −1 (C\V ) ⊂ C\f −1 C\f (S) ⊂ C\f −1 (U ).  (V ) and  −1 −1 −1 Hence A = R ∩ f (S) ⊂ R ∩ C \ f (V ) = R \ R ∩ f (V ) . By Lemma 14.6.4 A has inner measure zero and does not contain any of the and  subset  second  category  −1 −1 with the Baire property. Similarly, R \ A = R \ R ∩ f (S) = R ∩ C \ f (S) ⊂    R ∩ C \ f −1 (U ) = R \ R ∩ f −1 (U ) . By Lemma 14.6.4 also R \ A has inner measure zero and does not contain any subset of the second category and with the Baire property.  Theorem 14.6.5. Let f : C → C be a non-trivial endomorphism, let l ⊂ C be a straight line, and let S ⊂ C be a set such that int S = ∅ and int(C \ S) = ∅. Then the set l ∩ f −1 (S) is saturated non-measurable and has property (∗) from 3.3. Proof. Let x0 , x1 ∈ l, x0 = x1 . The assertion of the theorem means that the set A = {t ∈ R | x0 + t(x1 − x0 ) ∈ f −1 (S)} is saturated non-measurable and has property (∗) from 3.3 in R. By Lemma 4.12.2 f (x0 ) = f (x1 ). Put   S − f (x0 ) . B = R ∩ f −1 f (x1 ) − f (x0 )





If t ∈ B, then f (t) ∈ S −f (x0 ) / f (x1 )−f (x0 ) and f (x0 )+f (t) f (x1 )−f (x0 ) ∈ S. But by (14.4.1) and (14.4.2)

  (14.6.3) f (x0 ) + f (t) f (x1 ) − f (x0 ) = f x0 + t(x1 − x0 ) . Thus x0 + t(x1 − x0 ) ∈ f −1 (S), i.e., t ∈ A. Hence B ⊂ A. On the other hand, if t ∈ A, then t ∈ R and x0 + t(x1 − x0 ) ∈ f −1 (S). This means that, by (14.6.3), f (x0 ) + f (t) f (x1 ) − f (x0 ) ∈ S, and t ∈ B, whence A ⊂ B. Consequently A = B.



Now, the set S1 = S −f (x0 ) / f (x1 )−f (x0 ) fulfils int S1 = ∅ and int(C\S1 ) = ∅. The theorem results now from Theorem 14.6.4.  Theorem 14.6.6. Let f : C → C be a non-trivial endomorphism, and let A ⊂ R be a set of positive inner measure or of the second category and with the Baire property. Then cl f (A) = C. Proof. Let U ⊂ C be an arbitrary open disc. By Theorem 14.6.4 the set B = R ∩ f −1 (U ) is saturated non-measurable and has property (∗) from 3.3. Consequently (Theorem 3.3.1) B intersects every measurable set in R of positive measure.

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411

If mi (A) > 0, there exists a measurable set C ⊂ A with m (C) > 0. Therefore B ∩ C = ∅, whence also B ∩ A = ∅. (14.6.4) If A is of the second category and with the Baire property, then (14.6.4) results from Corollary 3.3.1. Let t ∈ B ∩ A. Then f (t) ∈ U and f (t) ∈ f (A), whence f (t) ∈ U ∩ f (A). Consequently U ∩f (A) = ∅. Consequently f (A) intersects every open disc contained in C, i.e., f (A) is dense in C.  In the sequel m2 denotes the two-dimensional Lebesgue measure on the complex plane C. Also, the expression “the set A is saturated non-measurable in C” means that m2i (A) = m2i (C \ A) = 0. Similarly, the expression “A has property (∗) in C” means that neither A, nor C \ A, contains a subset which is of the second category and has the Baire property in the topology of C. Theorem 14.6.7. Let f : C → C be a non-trivial endomorphism, and let S ⊂ C be a set such that m2i (S) > 0 or S is of the second category and with the Baire property (in the topology of C). Then cl f (S) = C. Proof. If m2i (S) > 0, then there exists a measurable (with respect to m2 ) set C ⊂ S with m2 (C) > 0. By the Fubini theorem there exists a straight line l ⊂ C such that the one-dimensional measure m(l ∩ C) > 0. Hence mi (l ∩ C) > 0. If S is of the second category and with the Baire property (in the topology of C), then S = (G ∪ P ) \ R, where G is open (and non-empty), and P, R are of the first category (in the topology of C). It follows from Theorem 2.1.7 that there exists a straight line l ⊂ C such that the sets l ∩ P and l ∩ R are of the first

category (in the topology of l), whereas l ∩ G = ∅. Thus l ∩ S = (l ∩ G) ∪ (l ∩ P ) \ (l ∩ R) is of the second category and with the Baire property. Suppose that f (S) is not dense in C. Then there exists an open disc U ⊂ C −1 such that f (S) ∩ U = ∅. \ U fulfils Hence f (S)

⊂ C \ U and S ⊂ f (C \ U ). But C−1 int(C \ U ) = ∅ and int C \ (C \ U ) = ∅. By Theorem 14.6.5 the set l ∩ f (C \ U ) is saturated non-measurable and has property (∗) (in the topology of l). Since l ∩ S ⊂ l ∩ f −1 (C \ U ), we obtain hence that l ∩ S has inner (one-dimensional) measure zero and does not contain a subset which is of the second category and with the Baire property, a contradiction.  Theorem 14.6.8. Let f : C → C be a non-trivial endomorphism, and let S ⊂ C be a set such that int S = ∅ and int(C \ S) = ∅. Then the set f −1 (S) is saturated non-measurable in C and has property (∗) in C.   a second Proof. If we had m2i f −1 (S) > 0 or f −1 (S) contained   category  subset S1 with the Baire property, then f f −1 (S) resp. f f −1 (S1 ) ⊂ f f −1 (S) would be dense in C according to Theorem 14.6.7, and since f f −1 (S) = S, the set  S would  be dense in C, which is not the case, since int(C\ S) = ∅. Consequently m2i f −1 (S) = 0 and f −1 (S) does not contain any subset of the second category and with the Baire property.

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Because of the symmetry  of the assumptions we may interchange S and C \ S, and we get m2i f −1 (C \ S) = 0 and f −1 (C \ S) does not contain any subset of the second category and with the Baire property. But f −1 (C \ S) = C \ f −1 (S), whence also inner (two-dimensional) measure of C \ f −1 (S) is zero and C \ f −1 (S) does not contain any subset of the second category and with the Baire property.    Theorem 14.6.9. Let f : C → C be a non-trivial endomorphism. Then f (R); +, ·   is a proper subfield of (C; +, ·), card f (R) = c, and either m2 f (R) = 0, or f (R) is saturated non-measurable in C. Similarly, f (R) either is of the first category, or has property (∗) in C. Proof. Take arbitrary x, y ∈ f (R). Then there exist u, v ∈ R such that f (u) = x, f (v) = y. We have u − v ∈ R, whence x − y = f (u) − f (v) = f (u − v) ∈ f (R). Similarly, if y = 0, then by Lemma 4.12.2 also v = 0, since f (0) = 0. Thus u/v ∈ R, and we have u = v(u/v), whence by (14.4.2) f (u) = f (v)f (u/v)and f (u/v)  = f (u)/f (v). Consequently x/y = f (u/v) ∈ f (R). By Lemma 4.7.1 f (R); +, · is a field, and thus a subfield of (C, +, ·). We have card R = c, whence also card f (R) = c, since by Lemma 4.12.2 f is one-to-one. Lemma 4.12.2 implies also that f (R) = f (C), whence it follows that necessarily f (R) = C, since f (R) ⊂ f (C) ⊂ C, and thus  f (R); +, · is a proper subfield of (C; +, ·).   It follows easily from the fact that f (R); +, · is a field, that f (R) + f (R) = f (R) − f (R) = f (R).

(14.6.5)

    Assume that we do not have m2 f (R) = 0. Then m2e f (R) > 0. By Theorem  14.6.3 2 2 f (R) is dense in C.  By Theorem 3.6.1 (C is to be regarded as R ) mi C \ f (R) = 0. Suppose that m2i f (R) > 0. By (14.6.5) and Theorem 3.7.1 int f (R) = ∅. Take an x0 ∈ int f (R) ⊂ f (R). The set V = int f (R) − x0 is open, contains the origin, and by (14.6.5) V ⊂ f (R) − f (R) = f (R).

(14.6.6)

Thus V is a neighbourhood of zero. For every x ∈ C there exists an n ∈ N such that x/n ∈ V . In other words, there exists a y ∈ V such that x = ny. But by (14.6.6) and using repeatedly (14.6.5) we obtain ny ∈ f (R). Hence C ⊂ f(R), i.e., f (R) = C, contrary to what we have already established. Thus m2i f (R) = 0, and f (R) is saturated non-measurable in C. If we assume that f (R) is of the second category, then from (14.6.5), Theorem 14.6.3, and Theorem 3.6.2 we infer that C \ f (R) does not contain any subset of the second category and with the Baire property. If the set f (R) contained a subset of the second category and with the Baire property, then by (14.6.5) and Theorem 2.9.1 we would get int f (R) = ∅, which, as we have just seen, leads to a contradiction. Consequently also f (R) does not contain any subset of the second category and with the Baire property, and f (R) has property (∗) in C. 

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Exercises 1. Show that there exists a discontinuous additive function f : R → R such that ∞ ∞   the series an f (xn ) converges if the series an xn converges (Baker-Segal n=0

n=0

[17]; cf. also Kemperman [166]). 2. Let f1 : R → R and f2 : R → R be derivations. Show that the composition f1 ◦ f2 is a derivation if and only if either f1 = 0 or f2 = 0. 3. Let f1 : R → R and f2 : R → R be derivations. Show that the product f1 f2 is a derivation if and only if either f1 = 0 or f2 = 0. 4. Let R be a ring. Show that if f1 , f2 : R → R are derivations, then so is also [f1 , f2 ] = f2 ◦ f1 − f1 ◦ f2 .

The derivation [f1 , f2 ] is called the bracket of f1 and f2 . 5. Let F be a field and K = F (x1 , . . . , xn ) be the field of rational functions (in n variables) over F . Prove that, for each i (1 ≤ i ≤ n), there exists a unique derivation Di : K → K such that Di | F = 0,

Di (xi ) = 1,

Di (xj ) = 0 for j = i, 1 ≤ j ≤ n.



The derivation Di is called the partial derivation (with respect to xi ) of K. ˆ be a field such that K ⊂ K. ˆ 6. Let F and K be as in Exercise 14.5, and let K ˆ such that f | F = 0 is a Prove that the family D0 of all derivations f : K → K ˆ and the partial derivations D1 , . . . , Dn form its base. linear space over K n

 Hint: Show that for every f ∈ D0 we have f = f (xi )Di . i=1

7. Show that there exists no Hamel basis H of R such that ab ∈ H, whenever a, b ∈ H. Hint: Let H ⊂ R be a Hamel basis such that ab ∈ H whenever a, b ∈ H, and put f0 (h) = 1 for h ∈ H. Let f : R → R be the additive extension of f0 . Show that f satisfies equation (14.4.2).

Chapter 15

Convex Functions of Higher Orders 15.1 The difference operator Let D ⊂ RN be a convex set, let f : D → R be an arbitrary function, and let h ∈ RN be arbitrary. The difference operator Δh with the span h is defined by the equality Δh f (x) = f (x + h) − f (x).

(15.1.1)

Thus Δh f is a real-valued function defined for x ∈ D such that x + h ∈ D.  

N function, then in Δh f g(x) =    generalthe expressions   If g : R →R is another  f g(x + h) − f g(x) and (Δh f ) g(x) = f g(x) + h − f g(x) do not coincide. In other words the order of effecting the operations of taking the difference and of substitution is essential. Only if g is a translation: g(x) = x + a, with a constant a ∈ RN , then  

Δh f g(x) = (Δh f )g(x),   and in the sequel this common value will be denoted simply by Δh f g(x) = Δh f (x + a). The iterates Δph of Δh , p = 0, 1, 2, . . ., are defined by the recurrence Δ0h f = f ,

Δp+1 = Δh (Δph f ) , h

p = 0, 1, 2 . . .

(15.1.2)

In particular , we have Δ1h f = Δh f . More generally, the superposition of several difference operators will be denoted shortly Δh1 ...hp f = Δh1 Δh2 . . . Δhp f,

p ∈ N.

(15.1.3)

Of course, if h1 = · · · = hp = h, the expression (15.1.3) reduces to Δh . . . h f = Δph f, 5 67 8

p ∈ N.

(15.1.4)

p times

Expression (15.1.3) is a function defined for all x ∈ D such that x + ε1 h1 + · · · + εp hp ∈ D for every choice of εi = 0 or 1, i = 1, . . . , p. Generally, in the sequel of this

416

Chapter 15. Convex Functions of Higher Orders

section we will write expressions containing difference operators without mentioning the domain of definition of the functions represented by these expressions, being understood that the identities derived are valid for x in the domain of definition of f such that all the occurring expressions are meaningful.   Lemma 15.1.1. The operator Δh and hence also all operators (15.1.3) is linear, i.e., for arbitrary functions f1 , f2 and for arbitrary constants a, b ∈ R, we have Δh (af1 + bf2 ) = aΔh f1 + bΔh f2 . 

Proof. This results immediately from (15.1.1). Lemma 15.1.2. For arbitrary h1 , h2 ∈ RN the operators Δh1 and Δh2 commute: Δh1 Δh2 f = Δh2 Δh1 f. Proof. We have by (15.1.1)   Δh1 Δh2 f = Δh1 f (x + h2 ) − f (x) = f (x + h1 + h2 ) − f (x + h1 ) − f (x + h2 ) + f (x) = Δh2 Δh1 f (x),



which was to be shown.

Corollary 15.1.1. Operator (15.1.3) is symmetric under the permutation of h1 , . . . , hp . Lemma 15.1.3. For arbitrary h1 , h2 ∈ RN we have Δh1 +h2 f − Δh1 f − Δh2 f = Δh1 h2 f. Proof. We have by (15.1.1) Δh1 +h2 f (x) − Δh1 f (x) − Δh2 f (x) = f (x + h1 + h2 ) − f (x) − f (x + h1 ) + f (x) − f (x + h2 ) + f (x) = f (x + h1 + h2 ) − f (x + h1 ) − f (x + h2 ) + f (x) = Δh1 h2 f (x) 

(cf. the proof of Lemma 15.1.2). The following two lemmas are immediate consequences of (15.1.1).

Lemma 15.1.4. Let D ⊂ RN be a convex set, and let H ⊂ R2N be the set of those (x, h) ∈ D × RN for which x + h ∈ D. If the function f : D → R is continuous in D, then the function g : H → R given by g(x, h) = Δh f (x) is continuous in H. Lemma 15.1.5. If {fn }n∈N is a pointwise convergent sequence of functions, then lim Δh fn = Δh lim fn .

n→∞

n→∞

The next lemma, of a slightly similar character, refers to the one-dimensional case N = 1.

15.1. The difference operator

417

Lemma 15.1.6. Let J ⊂ R be an interval, and let f : J → R be a continuous function. Then, for arbitrary x ∈ J and α, β, h ∈ R such that x + h, x + α, x + β, x + α + h, x + β + h ∈ J we have x+β 

Δh

x+β 

f (t)dt = x+α

Δh f (t)dt. x+α

Proof. We have x+β 

Δh

x+β+h 

f (t)dt −

f (t)dt = x+α

x+β 

x+β 



=

f (t + h) − f (t) dt =

x+α

x+β 

f (t + h)dt −

f (t)dt = x+α

x+α+h

x+β 

x+α

f (t)dt x+α

x+β 

Δh f (t)dt , x+α



which is the desired formula. Theorem 15.1.1. We have for p ∈ N Δh1 ...hp f (x) =

1

(−1)p−(ε1 +···+εp ) f (x + ε1 h1 + · · · + εp hp ).

(15.1.5)

ε1 ,...,εp =0

Proof. For p = 1 formula (15.1.5) becomes Δh1 f (x) =

1

(−1)1−ε1 f (x + ε1 h1 ) = −f (x) + f (x + h1 ) = f (x + h1 ) − f (x),

ε1 =0

which is consistent with (15.1.1). Now assume (15.1.5) true for a p ∈ N. Then we have by Corollary 15.1.1, induction hypothesis, and Lemma 15.1.1,

Δh1 ...hp hp+1 f (x) = Δhp+1 Δh1 ...hp f (x) 1

= Δhp+1

(−1)p−(ε1 +···+εp ) f (x + ε1 h1 + · · · + εp hp )

ε1 ,...,εp =0

=

1

(−1)p−(ε1 +···+εp ) Δhp+1 f (x + ε1 h1 + · · · + εp hp )

ε1 ,...,εp =0

=

1

(−1)p−(ε1 +···+εp )

ε1 ,...,εp =0

=

1

1

(−1)1−εp+1 f (x + ε1 h1 + · · · + εp hp + εp+1 hp+1 )

εp+1 =0

(−1)p+1−(ε1 +···+εp+1 ) f (x + ε1 h1 + · · · + εp+1 hp+1 ),

ε1 ,...,εp+1 =0

and so we obtain formula (15.1.5) for p + 1. Induction completes the proof.



418

Chapter 15. Convex Functions of Higher Orders

Corollary 15.1.2. We have for p ∈ N Δph f (x) =

p

(−1)p−k

k=0

  p f (x + kh). k

(15.1.6)

Proof. If exactly k among ε1 , . . . , εp in (15.1.5) are ones, the expression under the sum sign h1 = · · · = hp = h) reduces to (−1)p−k f (x + kh). But there are  (with  p exactly choices of k ones from among ε1 , . . . , εp , where k may run from 0 to p. k Thus (15.1.6) results from (15.1.4) and (15.1.5).  Corollary 15.1.3. We have for p ∈ N Δp−h f (x) = (−1)p Δph f (x − ph). Proof. We have by (15.1.6) Δp−h f (x)

  p f (x − kh) = (−1) k k=0   p

 p  f x − ph + (p − k)h (−1)p−k = k k=0   p

p j f (x − ph + jh) (−1) = p−j j=0   p

p p−j p = (−1) f (x − ph + jh) (−1) j j=0 p

p−k

= (−1)p Δph f (x − ph),     p p due to the equalities = and (−1)2p−j = (−1)j , j = 0, . . . , p. j p−j



Theorem 15.1.2.1 Let f : RN → R be an arbitrary function, and let h1 , . . . , hp ∈ RN be arbitrary. For any ε1 , . . . , εp ∈ R put hε1 ...εp = −

p

εj hj /j,

hε1 ...εp =

j=1

p

ε j hj .

j=1

Then we have for every x ∈ RN Δh1 ...hp f (x) =

1

(−1)ε1 +···+εp Δphε

ε1 ,...,εp =0 1

A private communication of J. H. B. Kemperman.

1 ···εp

f (x + hε1 ···εp ).

(15.1.7)

15.1. The difference operator

419

Proof. We will prove the formula (−1)p Δy1 2y2 ...pyp f (x) =

1

(−1)p−(ε1 +···+εp ) ΔpYε

1 ...εp

ε1 ,··· ,εp =0

where

p

Yε1 ...εp = −

Yε1 ...εp =

εj yj ,

j=1

p

f (x + Yε1 ...εp ), (15.1.8)

jεj yj ,

j=1

valid for all x ∈ RN and all y1 , . . . , yp ∈ RN . (15.1.7) results from (15.1.8) on multiplying by (−1)p and putting hj = jyj , j = 1, . . . , p. We have by Corollary 15.1.2 1

(−1)p−(ε1 +···+εp ) ΔpYε

1 ...εp

ε1 ,...,εp =0

=

=

1

f (x + Yε1 ...εp )

(−1)p−(ε1 +···+εp )

p

ε1 ,...,εp =0

k=0

1

p

(−1)p−(ε1 +···+εp )

ε1 ,...,εp =0

(−1)p−k

    p f x + Yε1 ...εp + kYε1 ...εp k

(−1)p−k

  p

 p  f x+ (j − k)εj yj k j=1

k=0

  p 1

  p (−1)p−k (−1)p−(ε1 +···+εp ) f x + (j − k)εj yj = k ε ,...,ε =0 j=1 k=0 1 p   p

p = Δ(1−k) y1 ...(p−k) yk f (x), (−1)p−k k p

k=0

in virtue of Theorem 15.1.1. For every k = 1, . . . , p one of the increments (1−k)y1 , . . . , (p − k)yk is zero, whence by Corollary 15.1.1

Δ(1−k)y1 ...(p−k)yk = Δ(1−k)y1 ...−yk−1 yk+1 ...(p−k)yk Δ0 f (x) = 0, since by (15.1.1) Δ0 f (x) = 0. Consequently in the last sum only the term corresponding to k = 0 remains, which yields (15.1.8).  Theorem 15.1.3. Let D ⊂ RN be a convex set, and let f : D → R be an arbitrary function. Then for every α1 , . . . , αp ∈ Q ∩ (0, ∞) there exists an α ∈ Q ∩ (0, ∞) and m, k0 , . . . , km ∈ N such that for every x ∈ D and h ∈ RN such that x + (α1 + · · · + αp )h ∈ D we have x + (m + p)αh ∈ D and Δα1 h...αp h f (x) =

m

i=0

ki Δpαh f (x + iαh).

(15.1.9)

420

Chapter 15. Convex Functions of Higher Orders

Proof. By (15.1.1) Δh f (x + ih) = f (x + (i + 1)h) − f (x + ih).

(15.1.10)

Summing up (15.1.10) over i from 0 to an n − 1 ∈ N ∪ {0} we obtain (supposing that x, x + nh ∈ D) Δnh f (x) = f (x + nh) − f (x) =

n−1

n ∈ N.

Δh f (x + ih),

(15.1.11)

i=0

Fix arbitrarily x ∈ D, h ∈ RN , and take arbitrary α1 , . . . , αp ∈ Q ∩ (0, ∞) such that x + (α1 + · · · + αp )h ∈ D. There exist numbers M, m1 . . . , mp ∈ N such that αi = mi /M , i = 1, . . . , p. Put α = 1/M so that α ∈ Q ∩ (0, ∞), and Mq = m1 + · · · + mq ∈ N, q = 1, . . . , p. We have αi = mi α, i = 1, . . . , p, and x + Mp αh = x + (m1 + · · · + mp )αh = x + (α1 + · · · + αp )h ∈ D, whence by the convexity of D we have x + Mq αh ∈ D for q = 1, . . . , p. We will prove by induction on q that for every q = 1, . . . , p there exist numbers k0q , . . . , kMq −q,q ∈ N such that Mq −q

Δα1 h...αq h f (x) =

kiq Δqαh f (x + iαh).

(15.1.12)

i=0

For q = 1 we have by (15.1.11) Δα1 h f (x) = Δm1 αh f (x) =

m 1 −1

Δαh f (x + iαh) =

i=0

M 1 −1

Δ1αh f (x + iαh),

i=0

so (15.1.12) holds with k01 = · · · = kM1 −1,1 = 1. Assume (15.1.12) true for a q ∈ {1, . . . , p − 1}. We have by Corollary 15.1.1, induction hypothesis, and Lemma 15.1.1, Mq −q



Δα1 h...αq hαq+1 h f (x) = Δαq+1 h Δα1 h...αq h f (x) = Δαq+1 h kiq Δqαh f (x + iαh) i=0 Mq −q

=

kiq Δαq+1 h Δqαh f (x + iαh),

i=0

and again by Corollary 15.1.1, (15.1.11), and Lemma 15.1.1, Mq −q

Δα1 h...αq+1 h f (x) =

kiq Δqαh Δmq+1 αh f (x + iαh)

i=0 Mq −q

=

i=0

mq+1 −1

kiq Δqαh

j=0

Δαh f (x + iαh + jαh)

15.2. Divided differences

421 Mq −q mq+1 −1

=



i=0

j=0

kiq Δq+1 αh f (x + (i + j)αh)

Mq −q mq+1 +i−1

=



i=0

l=i

kiq Δq+1 αh f (x + lαh)

Mq +mq+1 −q−1

=

min(l,Mq −q)

l=0

i=max(0,l−mq+1 +1)

kiq Δq+1 αh f (x + lαh)

Mq+1 −q−1

=

kl,q+1 Δq+1 αh f (x + lαh),

l=0 min(l,M  q −q)

where kl,q+1 =

kiq . Thus we have obtained (15.1.12) for q + 1.

i=max(0,l−mq+1 +1)

Induction shows that (15.1.12) is valid for q = 1, . . . , p. For q = p we obtain hence (15.1.9) with m = Mp − p and ki = kip , i = 0, . . . , m. Moreover, we have x + (m + p)αh = x + Mp αh ∈ D. 

15.2 Divided differences N In the sequel of this chapter RN + denotes the set of those x ∈ R whose first non-zero coordinate is positive:

RN + =

N  ! " x = (ξ1 , . . . , ξN ) ∈ RN | ξ1 = · · · = ξi−1 = 0, ξi > 0 . i=1

N N we write x > y iff x − y ∈ RN Then RN = RN + ∪ (−R+ ) ∪ {0}. For x, y ∈ R + and N x < y iff y > x. Then for any x, y ∈ R we have either x = y, or x < y, or x > y. If x > y and α ∈ R is positive, then αx > αy, and if α < 0, then αx < αy. We write

⎧ ⎪ ⎨1 sgn x = 0 ⎪ ⎩ −1

if x > 0, if x = 0, if x < 0.

Let D ⊂ RN be a convex set, and let f : D → R be a function. Let x1 , . . . , xp ∈ D be distinct collinear points. Put h=

xp − x1 sgn(xp − x1 ) |xp − x1 |

(15.2.1)

so that always h > 0. Since x1 , . . . , xp are collinear, they may be represented in the form xi = x1 + λi h, i = 1, . . . , p. (15.2.2)

422

Chapter 15. Convex Functions of Higher Orders

The divided difference [x1 , . . . , xp ; f ] of f at points x1 , . . . , xp is defined by recurrence (N¨ orlund [244], Popoviciu [259], Ger [101]): [x1 ; f ] = f (x1 ), [x1 , . . . , xp ; f ] =

[x2 , . . . , xp ; f ] − [x1 , . . . , xp−1 ; f ] , λp − λ1

p  2.

(15.2.3)

It is seen from (15.2.3) that the divided difference [x1 , . . . , xp ; f ] depends on the differences of λ’s rather than on λ’s themselves. Therefore if we represent x1 , . . . , xp with the aid of another collinear point x0 as xi = x0 + λi h, where h is given by (15.2.1), and i = 1, . . . , p, and form the divided difference (15.2.3) with the aid of λi instead of λi , the result will be the same. This can also be shown by induction from (15.2.3). Also, if we build h according to (15.2.1) with other points xi = xp lying on the same straight line, we obtain the same result. For p = 2 we have always λ1 = 0, λ2 = |x2 − x1 | sgn(x2 − x1 ), whence by (15.2.3) [x1 , x2 ; f ] =

Δx2 −x1 f (x1 ) f (x2 ) − f (x1 ) = . |x2 − x1 | sgn(x2 − x1 ) |x2 − x1 | sgn(x2 − x1 )

(15.2.4)

For higher values of p the situation is more complicated. Let, for collinear x1 , . . . , xp ∈ D, U (x1 , . . . , xp ; f ) denote the determinant # # #1, λ1 , λ21 , . . . , λp−2 , f (x1 )## 1 # # # # # #1, λ2 , λ2 , . . . , λp−2 , f (x2 )# 2 2 # # #, U (x1 , . . . , xp ; f ) = ## # #. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .# # # # # # # #1, λp , λ2 , . . . , λp−2 , f (xp )# p

(15.2.5)

p

 where λ1 , . . . , λp are defined by (15.2.2) and (15.2.1). Let V (x1 ,...,xp ) = U x1 ,...,xp ; |x − x1 |p−1 be the Van der Monde determinant of λ1 , . . . , λp . We have, as is well known from the elementary theory of determinants, # # #1, λ1 , λ21 , . . . , λp−2 # , λp−1 1 1 # # # # # # #1, λ2 , λ2 , . . . , λp−2 , λp−1 # p 2 2 2 # # #= V (x1 , . . . , xp ) = ## (λi − λj ), # #. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .# i,j=1 # # i>j # # # # #1, λp , λ2 , . . . , λp−2 , λp−1 # p p p

(15.2.6)

and so V (x1 , . . . , xp ) = 0 if and only if the points x1 , . . . , xp are distinct; and if, moreover, x1 < · · · < xp , then V (x1 , . . . , xp ) > 0.

15.2. Divided differences

423

We have the following Theorem 15.2.1. Let D ⊂ RN be a convex set, and let f : D → R be an arbitrary function. Then, for every p  2, p ∈ N, and for every p distinct collinear points x1 , . . . , xp ∈ D, we have [x1 , . . . , xp ; f ] =

U (x1 , . . . , xp ; f ) . V (x1 , . . . , xp )

(15.2.7)

Proof. The proof runs by induction. For p = 2 we have by (15.2.5) and (15.2.6) # # #1 f (x )# 1 # # U (x1 , x2 ; f ) = # # = f (x2 ) − f (x1 ), #1 f (x2 )# # # #1 λ # 1# # V (x1 , x2 ) = # # = λ2 − λ1 . #1 λ2 # By (15.2.2) and (15.2.1) λ1 = 0, λ2 = |x2 − x1 | sgn(x2 − x1 ), and thus (15.2.7) holds in virtue of (15.2.4). Suppose (15.2.7) true for a p ∈ N, p  2. We have by (15.2.3) [x1 , . . . , xp+1 ; f ] =

[x2 , . . . , xp+1 ; f ] − [x1 , . . . , xp ; f ] , λp+1 − λ1

i.e., by the induction hypothesis,   U (x2 , . . . , xp+1 ; f ) U (x1 , . . . , xp ; f ) 1 − . [x1 , . . . , xp+1 ; f ] = λp+1 − λ1 V (x2 , . . . , xp+1 ) V (x1 , . . . , xp ) Developing the determinants U according to the last column we have2 U (x2 , . . . , xp+1 ; f ) =

p+1

(−1)p+k+1 f (xk )V (x2 , . . . , xk−1 , xk+1 , . . . , xp+1 )

k=2

and U (x1 , . . . , xp ; f ) =

p

(−1)p+k f (xk )V (x1 , . . . , xk−1 , xk+1 , . . . , xp ).

k=1

On the other hand, by (15.2.6), V (x1 , . . . , xp+1 ) = V (x2 , . . . , xp+1 )

p+1 -

(λi − λ1 )

i=2

= (λp+1 − λ1 )V (x2 , . . . , xp+1 )

p -

(λi − λ1 ),

i=2 2 The remarks after formula (15.2.3) apply also to the determinants U and V . Therefore during the whole this proof we may calculate the occurring differences and determinants with the same λ’s determined from (15.2.1) and (15.2.2) for i = 1, . . . , p + 1.

424

Chapter 15. Convex Functions of Higher Orders p V (x1 , . . . , xp+1 ) = V (x1 , . . . , xp ) (λp+1 − λi ) i=1

= (λp+1 − λ1 )V (x1 , . . . , xp )

p -

(λp+1 − λi ).

i=2

Hence we get 1 (15.2.8) V (x1 , . . . , xp+1 ) p+1 p  × (−1)p+k+1 f (xk )V (x2 , . . . , xk−1 , xk+1 , . . . , xp+1 ) (λi − λ1 )

[x1 , . . . , xp+1 ; f ] =

i=2

k=2

+

p

(−1)p+k+1 f (xk )V (x1 , . . . , xk−1 , xk+1 , . . . , xp )

p -

 (λp+1 − λi ) .

i=2

k=1

Next again by (15.2.6), for k = 2, . . . , p, V (x1 , . . . , xk−1 ,xk+1 , . . . , xp+1 )(λp+1 − λk )

(15.2.9)

= (λp+1 − λ1 )V (x1 , . . . , xk−1 , xk+1 , . . . , xp )

p -

(λp+1 − λi ),

i=2

V (x1 , . . . , xk−1 ,xk+1 , . . . , xp+1 )(λk − λ1 )

(15.2.10)

= (λp+1 − λ1 )V (x2 , . . . , xk−1 , xk+1 , . . . , xp+1 )

p -

(λi − λ1 ).

i=2

From (15.2.8) , (15.2.9) and (15.2.10) we get [x1 ,...,xp+1 ;f ] =

p  1 f (xp+1 )V (x2 ,...,xp ) (λi − λ1 ) V (x1 ,...,xp+1 ) i=2

+ (−1)p f (x1 )V (x2 ,...,xp )

p -

(λp+1 − λi )

i=2 p  λ −λ

k 1 (−1)p+k+1 f (xk ) V (x1 ,...,xk−1 ,xk+1 ,...,xp+1 ) λp+1 − λ1 k=2  λp+1 − λk V (x1 ,...,xk−1 ,xk+1 ,...,xp+1 ) + λp+1 − λ1

+

1 (−1)p+k+1 f (xk )V (x1 ,...,xk−1 ,xk+1 ,...,xp+1 ) V (x1 ,...,xp+1 ) p+1

=

k=1

=

U (x1 ,...,xp+1 ;f ) . V (x1 ,...,xp+1 )

Thus we obtain (15.2.7) for p + 1. Induction completes the proof.



15.2. Divided differences

425

From Theorem 15.2.1 we get the following lemmas. Lemma 15.2.1. Let D ⊂ RN be a convex set, and let fn : D → R, n ∈ N, be pointwise convergent sequence of functions. Then, for arbitrary p distinct collinear points x1 , . . . , xp ∈ D,   lim [x1 , . . . , xp ; fn ] = x1 , . . . , xp ; lim fn . n→∞

n→∞

Lemma 15.2.2. Let D ⊂ R be a convex set, and let f : D → R be a continuous function. Then the divided difference [x1 , . . . , xp ; f ] is a continuous function of x1 , . . . , xp ∈ D, under the condition that these points vary in such a way that they remain collinear and distinct. N

Lemma 15.2.3. Let D ⊂ RN be a convex set, and let f : D → R be an arbitrary function. Then, for every distinct collinear points x1 , . . . , xp ∈ D, the divided difference [x1 , . . . , xp ; f ] is symmetric with respect to permutations of x1 , . . . , xp . Proof. If we permute x1 , . . . , xp in (15.2.7), then U (x1 , . . . , xp ; f ) and V (x1 , . . . , xp ) either remain unchanged (if the permutation is even), or both change the sign (if the permutation is odd). Consequently [x1 , . . . , xp ; f ] remains unchanged in both cases.  Let D ⊂ RN be a convex set, let f : D → R be a function, and let x1 , . . . , xp ∈ D be distinct and collinear. For x ∈ RN of the form x = x1 + λh, where h is given by (15.2.1), the function P is defined by the formula # # #1, λ1 , . . . , λp−2 , f (x1 ) ## 1 # # # # # #. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .# # # # = 0. # (15.2.11) # # #1, λp−1 , . . . , λp−2 , f (xp−1 )# p−1 # # # # # # #1, λ , . . . , λp−2 , P (x) # It is seen that P is polynomial in λ of degree at most p − 2 fulfilling the conditions P (xi ) = f (xi ) for i = 1, . . . , p − 1. (The Lagrange interpolation polynomial). As is well known, and also easily seen, such a polynomial is unique. In the sequel P defined by (15.2.11) will be denoted by P (x1 , . . . , xp−1 ; f | x). Lemma 15.2.4. Let D ⊂ RN be a convex set, and let f : D → R be an arbitrary function. Let p  2, and let x1 , . . . , xp ∈ D be distinct and collinear. Then for every x = x1 + λh lying on the straight line determined by x1 and xp (x = x1 , . . . , xp−1 ) we have f (x) − P (x1 , . . . , xp−1 ; f | x) =

V (x1 , . . . , xp−1 , x) [x1 , . . . , xp−1 , x; f ] . V (x1 , . . . , xp−1 )

(15.2.12)

Proof. We have, developing the determinant U according to the last row, U (x1 , . . . , xp−1 , x; f ) = V (x1 , . . . , xp−1 )f (x) + R,

(15.2.13)

426

Chapter 15. Convex Functions of Higher Orders

where the rest R depends only on λ and on the first p − 1 rows of the determinant (15.2.5). Similarly, by (15.2.11), 0 = V (x1 , . . . , xp−1 )P (x1 , . . . , xp−1 ; f | x) + R,

(15.2.14)

where R is the same expression as in (15.2.13). From (15.2.13) and (15.2.14) we obtain f (x) − P (x1 , . . . , xp−1 ; f | x) =

U (x1 , . . . , xp−1 , x; f ) , V (x1 , . . . , xp−1 ) 

whence (15.2.12) follows in virtue of Theorem 15.2.1.

The next lemma concerns the case, where the points x1 , . . . , xp are equidistributed on the segment x1 xp . Lemma 15.2.5. Let D ⊂ RN be a convex set, and let f : D → R be an arbitrary function. Let x1 ∈ RN , d ∈ RN + , and let the points x2 , . . . , xp be given by xi = x1 + (i − 1)d, Then [x1 , . . . , xp ; f ] =

i = 2, . . . , p. Δp−1 f (x1 ) d

(p − 1)! |d|p−1

.

(15.2.15)

(15.2.16)

Proof. The proof runs by induction on p. For p = 2 we have by (15.2.4) [x1 , x2 ; f ] =

Δd f (x1 ) Δx2 −x1 f (x1 ) = , |x2 − x1 | sgn(x2 − x1 ) |d|

since by (15.2.15) x2 − x1 = d > 0. Thus (15.2.16) holds for p = 2. Now we assume (15.2.16) true for a p − 1 ∈ N, p − 1  2. By (15.2.1) we have λ1 = 0, λp h = xp − x1 , whereas by (15.2.15) xp − x1 = (p − 1)d. Hence λp h = (p − 1)d, and since |h| = 1, we have |λp | = (p − 1)|d|. Finally h = (p − 1)/λp d, and since h > 0, d > 0, p − 1 > 0, we obtain λp > 0 and λp = (p − 1)|d|. Thus we obtain from (15.2.3) by the induction hypothesis [x1 , . . . , xp ; f ] = = = = =

[x2 , . . . , xp ; f ] − [x1 , . . . , xp−1 ; f ] λp − λ1 $ % p−2 Δ f (x ) f (x ) Δp−2 1 2 1 d d − (p − 1)|d| (p − 2)!|d|p−2 (p − 2)!|d|p−2   1 p−2 p−2 Δ f (x ) − Δ f (x ) 2 1 d d (p − 1)!|d|p−1   1 p−2 p−2 Δ f (x + d) − Δ f (x ) 1 1 d d (p − 1)!|d|p−1 1 Δp−1 f (x1 ). (p − 1)!|d|p−1 d

So (15.2.16) holds true also for p. Induction ends the proof.



15.2. Divided differences

427

Now let x1 , . . . , xp , with (15.2.2) and (15.2.1), be arbitrary p distinct collinear points in D, p  2. We prove the formula (λp − λ1 ) [x1 , . . . , xi−1 , xi+1 , . . . , xp ; f ] = (λi − λ1 ) [x1 , . . . , xp−1 ; f ] + (λp − λi ) [x2 , . . . , xp ; f ] (15.2.17) for i = 1, . . . , p. For i = 1 or p (15.2.17) is trivial, so we need only consider the case, where 1 < i < p. By (15.2.3) and Lemma 15.2.3 [x1 , . . . , xp ; f ] = [x1 , . . . , xi−1 , xi+1 , . . . , xp , xi ; f ] =

[x2 , . . . , xi−1 , xi+1 , . . . , xp , xi ; f ] − [x1 , . . . , xi−1 , xi+1 , . . . , xp ; f ] . λi − λ1

But, according to Lemma 15.2.3, [x2 , . . . , xi−1 , xi+1 , . . . , xp , xi ; f ] = [x2 , . . . , xp ; f ]. Hence and by (15.2.3) we obtain [x2 , . . . , xp ; f ] − [x1 , . . . , xp−1 ; f ] [x2 , . . . , xp ; f ] − [x1 , . . . , xi−1 , xi+1 , . . . , xp ; f ] = . λp − λ1 λi − λ1 After simple calculations we obtain hence (15.2.17). Lemma 15.2.6. Let D ⊂ RN be a convex set, and let f : D → R be an arbitrary function. Let x1 , . . . , xp ∈ D be p distinct collinear points, and let {x1 , . . . , xm } be a subsequence of {x1 , . . . , xp }. Then there exist constants A0 , . . . , Ap−m ∈ R such that p−m  Ai = 1 and i=0

[x1 , . . . , xm ; f ]

=

p−m

Ai [xi+1 , . . . , xi+m ; f ] .

(15.2.18)

i=0

If, moreover, x1 < · · · < xp , then Ai  0 for i = 0, . . . , p − m. Proof. Write k = p − m. Introduce λ1 , . . . , λp by means of (15.2.2) and (15.2.1). Note that x1 < · · · < xp means λ1 < · · · < λp . For k = 0, the lemma is trivial. If k = 1, the lemma results from (15.2.17). Observe that the coefficients (λi − λ1 )/(λp − λ1 ) and (λp − λi )/(λp − λ1 ) are nonnegative whenever λ1 < · · · < λp . Now assume the lemma true for a k ∈ N. Take p distinct collinear points x1 , . . . , xp ∈ D, and let {x1 , . . . , xm!} be a subsequence of {x1 , . . . , xp }, where p − m = " k + 1. Take another subsequence x1 , . . . , xm+1 of {x1 , . . . , xp } containing all the points x1 , . . . , xm and another point xj ∈ / {x1 , . . . , xm }. There exist constants B0 , B1 , where B0 + B1 = 1, such that

(15.2.19) [x1 , . . . , xm ; f ] = B0 [x1 , . . . , xm ; f ] + B1 x2 , . . . , xm+1 ; f (the lemma for k = 1). If x1 < · · · < xp , then B0 and B1 are non-negative. " ! Now, {x1 , . . . , xm } is a subsequence of {x1 , . . . , xp−1 }, and x2 , . . . , xm+1 is a subsequence of {x2 , . . . , xp }, and in both cases we have (p − 1) − m = p − m − 1 = k.

428

Chapter 15. Convex Functions of Higher Orders

By the induction hypothesis there exist constants A0 , . . . , Ak and A1 , . . . , Ak+1 (nonnegative, whenever x1 < · · · < xp ) such that k

Ai =

i=0

k+1

Ai = 1

(15.2.20)

i=1

and k

[x1 , . . . , xm ; f ] =

Ai [xi+1 , . . . , xi+m ; f ] ,

i=0



k+1



x2 , . . . , xm+1 ; f =

Ai [xi+1 , . . . , xi+m ; f ] .

i=1

Hence and by (15.2.19) [x1 , . . . , xm ; f ] =

k

B0 Ai [xi+1 , . . . , xi+m ; f ] +

i=0

=

k+1

k+1

B1 Ai [xi+1 , . . . , xi+m ; f ]

i=1

Ai [xi+1 , . . . , xi+m ; f ] ,

i=0

where A0 = B0 A0 ,

Ak+1 = B1 Ak+1 ,

Ai = B0 Ai + B1 Ai

for i = 1, . . . , k.

Hence and by (15.2.20) k+1

Ai = B0

i=0

k

Ai + B1

i=0

k+1

Ai = B0 + B1 = 1.

i=1

B0 , B1 , A0 , . . . , Ak , A1 , . . . , Ak+1

If x1 < · · · < xp , then are non-negative, whence also A1 , . . . , Ak+1 are non-negative. Thus we have obtained (15.2.18) for p−m = k +1. Induction ends the proof.  Lemma 15.2.7. Let D ⊂ RN be a convex set, and let f : D → R be an arbitrary function. Let x1 , . . . , xp ∈ D and y1 , . . . , yp ∈ D be two systems of p distinct points such that x1 , . . . , xp , y1 , . . . , yp are collinear. Suppose that for h defined by (15.2.1) and for a certain x0 collinear with points xi , yi , we have xi = x0 + λi h,

yi = x0 + μi h,

i = 1, . . . , p.

If xi = yj for i = j, then [x1 , . . . , xp ; f ] − [y1 , . . . , yp ; f ] =

p

(λi − μi ) [x1 , . . . , xi , yi , . . . , yp ; f ] ,

(15.2.21)

i=1

where (λi − μi ) [x1 , . . . , xi , yi , . . . , yp ; f ] = 0

whenever xi = yi ,

i = 1, . . . , p. (15.2.22)

15.3. Convex functions of higher order

429

Proof. As observed immediately after formula (15.2.3), in the definition of the divided differences we may use the linear coordinates λ1 , . . . , λp , μ1 , . . . , μp related to an arbitrary point x0 collinear with x1 , . . . , xp , y1 , . . . , yp . We have by (15.2.3) and Lemma 15.2.3, for arbitrary i such that xi = yi , (λi − μi ) [x1 , . . . , xi , yi , . . . , yp ; f ] = (λi − μi ) [yi , x1 , . . . , xi−1 , yi+1 , . . . , yp , xi ; f ] = [x1 , . . . , xi−1 , yi+1 , . . . , yp , xi ; f ] − [yi , x1 , . . . , xi−1 , yi+1 , . . . , yp ; f ] = [x1 , . . . , xi−1 , xi , yi+1 , . . . , yp ; f ] − [x1 , . . . , xi−1 , yi , yi+1 , . . . , yp ; f ] , and, due to convention (15.2.22), this formula remains valid also if xi = yi . Summing up over i = 1, . . . , p we obtain (15.2.21). 

15.3 Convex functions of higher order Let D ⊂ RN be an open and convex set, and let f : D → R be a function. The function f is called convex of order p, or shortly p-convex (p ∈ N) iff Δp+1 h f (x)  0

(15.3.1)

for every x ∈ D and every h ∈ RN + such that x + (p + 1)h ∈ D. In virtue of Corollary 15.1.3 if p is odd and f : D → R is p-convex, then Δp+1 h f (x)  0 for every x ∈ D and every h ∈ RN such that x + (p + 1)h ∈ D. Lemma 15.3.1. Let D ⊂ RN be a convex and open set, let f1 , f2 : D → R be p-convex functions, and let a, b ∈ R be non-negative constants. Then the function f = af1 +bf2 also is p-convex. Proof. This results from Lemma 15.1.1.



If Δp+1 h f (x)  0

(15.3.2)

such that x + (p + 1)h ∈ D, then f is called for every x ∈ D and every h ∈ concave of order p, or shortly p-concave (p ∈ N). A function f is p-concave if and only if −f is p-convex (cf. Lemma 15.1.1), so in the sequel we will deal only with p-convex functions. p-concave functions will not be considered separately. If a function f : RN → R is at the same time p-convex and p-concave, i.e., if we have Δp+1 (15.3.3) h f (x) = 0 RN +

for every x ∈ RN and (cf. Corollary 15.1.3) every h ∈ RN , then f is called polynomial function of order p (p ∈ N). Conversely, every function f : RN → R fulfilling (15.3.3) for all x, h ∈ RN is at the same time p-convex and p-concave. Polynomial functions will be considered more in detail in the last section of this chapter. The above notions are due to T. Popoviciu [259], who also proved a number of basic properties of p-convex functions. Concerning a generalization cf. Moldovan [228], [229], [230] and Kemperman [167].

430

Chapter 15. Convex Functions of Higher Orders

Lemma 15.3.2. Let f1 , f2 : RN → R be polynomial functions of order p, and let a, b ∈ R be arbitrary constants. Then the function f = af1 + bf2 also is a polynomial function of order p. 

Proof. This results from Lemma 15.1.1.

Let D ⊂ RN be a convex and open set, and let f : D → R be a convex function of order 1. Then, by Corollary 15.1.2 f (x + 2h) − 2f (x + h) + f (x)  0,

(15.3.4)

and, as pointed out above, since p = 1 is odd, (15.3.4) holds for every x ∈ D and 1 h ∈ RN such that x + 2h ∈ D. Take arbitrary u, v ∈ D, and put x = u, h = (v − u). 2 1 Then x + h = (u + v), x + 2h = v ∈ D, and we get by (15.3.4) 2   u+v + f (u)  0, f (v) − 2f 2 or

 f

u+v 2

 

f (u) + f (v) . 2

Thus f fulfils in D the Jensen inequality (5.3.1). In other words, convex functions of order 1 are ordinary convex functions as defined in 5.3. It is evident that every convex function f : D → R satisfies inequality (15.3.4) for every x ∈ D and h ∈ RN such that x + 2h ∈ D, i.e., is convex of order 1 . The notion of p-convex functions is a generalization of the notion of ordinary convex functions. Similarly, if f : RN → R is a polynomial function of order 1, then f satisfies the Jensen equation (13.2.1). Thus the notion of polynomial functions generalizes the notion of functions fulfilling the Jensen equation, i.e., those functions f : RN → R which are of the form (cf. Theorem 13.2.1) f (x) = g(x) + a, where g : RN → R is additive, and a ∈ R is a constant. Theorem 15.3.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a p-convex function. Further, let x1 , . . . , xp+2 ∈ D be p + 2 distinct collinear points. Let h ∈ RN + and λ1 , . . . , λp+2 be such that xi = x1 + λi h,

i = 1, . . . , p + 2 .

If λ1 , . . . , λp+2 are rational, then [x1 , . . . , xp+2 ; f ]  0.

(15.3.5)

15.3. Convex functions of higher order

431

Proof. According to Lemma 15.2.3 we may assume that x1 < · · · < xp+2 . Let3 λi = ki /m, ki , m ∈ N, i = 1, . . . , p + 2, and put yi = x1 +

i−1 h, m

i = 1, . . . , q = kp+2 + 1 .

Then xi = yki +1 , i = 1, . . . , p + 2, so that {x1 ,...,xp+2 } is a subsequence of {y1 ,...,yq }. By Lemma 15.2.6 [x1 , . . . , xp+2 ; f ] =

q−p−2

Ai [yi+1 , . . . , yi+p+2 ; f ] ,

i=0

where Ai  0, i = 0, . . . , q − p − 2. Put d=

yi+p+2 − yi h = . m p+2

We have by Lemma 15.2.5, since yi+j = yi+1 + (j − 1)d, j = 1, . . . , p + 2 (1  i  q − p − 2), [yi+1 , . . . , yi+p+2 ; f ] = Hence [x1 , . . . , xp+2 ; f ] =

Δp+1 f (yi+1 ) d , (p + 1)!|d|p+1

i = 1, . . . , q − p + 2 .

q−p−2

1 Ai Δp+1 f (yi+1 )  0 , d p+1 (p + 1)!|d| i=0



i.e., we have (15.3.5). Similarly we have

Theorem 15.3.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a pconvex function. Let λ1 , . . . , λp+1 ∈ Q be positive. Then for every x ∈ D and h ∈ RN + such that x + (λ1 + · · · + λp+1 )h ∈ D, we have Δλ1 h . . . Δλp+1 h f (x)  0.

(15.3.6)

Proof. By Theorem 15.1.3 there exist α ∈ Q ∩ (0, ∞) and m, k1 , . . . , km ∈ N such that Δλ1 h . . . Δλp+1 h f (x) =

m

ki Δp+1 αh f (x + iαh)  0

i=1



so that (15.3.6) holds. Theorem 15.3.3. If f : R

N

→ R is a polynomial function of order p, then Δh1 ...hp+1 f (x) = 0

for every x, h1 , . . . , hp+1 ∈ RN . Proof. This results from Theorem 15.1.2.



3 x < ··· < x 1 p+2 implies 0 = λ1 < · · · < λp+2 and 0 = k1 < · · · < kp+2 , whence ki  i − 1 for i = 1, . . . , p + 2.

432

Chapter 15. Convex Functions of Higher Orders

15.4 Local boundedness of p-convex functions First we note that it is purposeless to consider boundedness above of p-convex functions for p > 1. A p-convex function f : D → R can be bounded above in D, and nevertheless unbounded and discontinuous in D. E. g., let g : RN → R be a discon

2 tinuous additive function, and put f (x) = − g(x) for x ∈ RN . Then f : RN → R is bounded above (by zero) in RN , but unbounded below at every point of RN , and is discontinuous at every point of RN . We also have by (15.1.6), for every x, h ∈ RN ,

2

2

2

2 Δ3h f (x) = − g(x + 3h) + 3 g(x + 2h) − 3 g(x + h) + g(x)

2

2

2

2 = − g(x) + 3g(h) + 3 g(x) + 2g(h) − 3 g(x) + g(h) + g(x)

2

2

2 = − g(x) − 6g(x)g(h) − 9 g(h) + 3 g(x) + 12g(x)g(h)

2

2

2

2 + 12 g(h) − 3 g(x) − 6g(x)g(h) − 3 g(h) + g(x) = 0 , whence also p−2 3 p−2 Δp+1 h f (x) = Δh Δh f (x) = Δh 0 = 0

for p > 2. Thus f is a polynomial function of every order p  2, and hence f is p-convex for every p  2. Theorem 15.4.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a p-convex function. If f is locally bounded at a point x0 ∈ D, then f is locally bounded at every point x ∈ D. Proof. Suppose that f is locally bounded at a point x0 ∈ D, and take an arbitrary z ∈ D, z = x0 . Since D is open, we can find a point y ∈ D, collinear with x0 and z, and such that z lies between y and x0 . There exists an open ball K = K(x0 , r), centered at x0 and with the radius r > 0, such that f is bounded on K, i.e., for a certain M > 0 we have |f (x)|  M

for x ∈ K .

(15.4.1)

/ cl K. We may choose r so small that r < |x0 − z| so that z ∈ Let C be the part of D formed by the rays issuing from y and passing through   1 1 the points of the ball K1 = K x0 , r centered at x0 and with the radius r. 2 2 There exists an open ball B centered at z and contained in C, B ∩ K = ∅. We will show that f is bounded on B. Take an arbitrary x ∈ B, and assume that y < x. (If y > x, the proof is similar). Let l be the ray from y through x. The ray l intersects √ K1 , and hence also intersects K along a segment S, whose length is greater than 3 r. Choose on S points x1 < · · · < xp (then we have also x < x1 ) in such a manner that all the ratios 1√ |xi − y|/|x − y| are rational, and |xi+1 − xi |  3 r, which is possible, because the 2p √ length of S is greater that 3 r, and the points u ∈ S such that |u − y|/|x − y| ∈ Q from a dense subset of S. Let h = (x − y)/|x − y|, and x = y + λ0 h ,

xi = y + λi h,

i = 1, . . . , p .

15.4. Local boundedness of p-convex functions

433

Further put λ−1 = 0 so that y = y + λ−1 h. The conditions on xi imply that |λi+1 − 1√ λi |  3 r for i = 1, . . . , p, 0 = λ−1 < λ0 < λ1 < · · · < λp , and all the ratios 2p αi = λi /λ0 , i = −1, 0, 1, . . . , p are rational. Put h0 = λ0 h ∈ RN + . Then y = y + α−1 h0 ,

x = y + α0 h0 ,

xi = y + αi h0 ,

i = 1, . . . , p .

We have by Theorems 15.2.1 and 15.3.1 U (y, x, x1 , . . . , xp ; f ) = [y, x, x1 , . . . , xp ; f ]  0 . V (y, x, x1 , . . . , xp ) Also V (y, x, x1 , . . . , xp ) > 0, since y < x < x1 < · · · < xp . Consequently U (y, x, x1 , . . . , xp ; f )  0 .

(15.4.2)

Developing the determinant U in (15.4.2) according to the last column, we have (−1)p+1 V (x, x1 , . . . , xp ) f (y) + (−1)p V (y, x1 , . . . , xp ) f (x) +

p

(−1)p+i V (y, x, x1 , . . . , xi−1 , xi+1 , . . . , xp )f (xi )  0,

i=1

whence, if p is odd, V (x, x1 , . . . , xp ) f (y) V (y, x1 , . . . , xp ) p

V (y, x, x1 , . . . , xi−1 , xi+1 , . . . , xp ) f (xi ). (−1)i−1 + V (y, x1 , . . . , xp ) i=1

f (x) 

(15.4.3)

If p is even, we obtain the opposite inequality. The Van der Monde determinants V (y, x1 , . . . , xp ), V (x, x1 , . . . , xp ), and V (y, x, x1 , . . . , xi−1 , xi+1 , . . . , xp ) are bounded above and below by constants, which may be made independent of x, since all the differences λi −λj , i, j = −1, 0, 1, . . . , p, i > j, may be estimated by r and the distances between y and B and between y and K, and the radius of B. Moreover, f (xi ) fulfil (15.4.1), i = 1, . . . , p. Consequently we obtain from (15.4.3) that f is bounded above (bounded below, if p is even) on B. The boundedness of f on B in the opposite direction may be established similarly. One must choose points x1 , · · · , xp+1 ∈ S such that x1 < · · · < xp+1 , all the 1√ distances |xi − x| are rational, i = 1, . . . , p + 1, and |xi+1 − xi | > 3 r, i = 1, . . . , p. 2p Then it is proved quite similarly as above that U (x, x1 , . . . , xp+1 ; f )  0, and further the argument is the same.  Theorem 15.4.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a pconvex function. If f is locally bounded at a point x0 ∈ D, then f is continuous at x0 .

434

Chapter 15. Convex Functions of Higher Orders

Proof. It is enough to show that lim f (x) = f (x0 ). lim f (x) = x→x

x→x0 x<x0

(15.4.4)

0

x>x0

We will show that x→x lim f (x) = f (x0 ). The other equality in (15.4.4) can be established 0

x<x0

similarly. Let f be bounded on a ball K = K(x0 , r) centered at x0 and with the radius r. 1 Take an arbitrary x ∈ K such that x < x0 , and |x0 − x| < r. Let l be the straight 2 line passing through x and x0 , and write S = l ∩ K. We choose on S two sequences of points: I. x1 < · · · < xp such that xp < x. Put h = (x0 − x1 )/|x0 − x1 | > 0, and write x = x1 + λp+1 h,

x0 = x1 + λp+2 h,

xi = x1 + λi h.

We choose x1 , . . . , xp ∈ S in such a manner that all the ratios λi /λp+2 , i = 1, . . . , p+2, are rational, and the distances |λi+1 − λi |, i = 1, . . . , p, are bounded below by a constant which may be made independent of x. Clearly λp+1 → λp+2 when x → x0 . II. x1 < · · · < xp−1 < y such that xp−1 < x < x0 < y. Write x = x1 +λp h,

x0 = x1 +λp+1 h,

y = x1 +λp+2 h,

xi = x1 +λi h,

i = 1, . . . , p−1.

We choose x1 , . . . , xp−1 , y ∈ S in such a manner that all the ratios λi /λp+2 , i = 1, . . . , p + 2, are rational, and |λi+1 − λi |, i = 1, . . . , p − 1, p + 1, are bounded below by a constant which may be made independent of x. Now x → x0 implies λp → λp+1 . Similarly as in the proof of Theorem 15.4.1 we have U (x1 , . . . , xp , x, x0 ; f )  0 and U (x1 , . . . , xp−1 , x, x0 , y; f )  0, whence  V (x1 , . . . , xp , x) f (x0 ) − f (x)  f (x0 ) 1 − V (x1 , . . . , xp , x0 ) p

V (x1 , . . . , xi−1 , xi+1 , . . . , xp , x, x0 ) f (xi ), (15.4.5) (−1)p+i+1 + V (x1 , . . . , xp , x0 ) i=1 

and

$

V (x1 , . . . , xp−1 , x, y) f (x0 ) − f (x)  f (x0 ) 1 − V (x1 , . . . , xp−1 , x0 , y) +

p−1

(−1)p+i

i=1

+

%

V (x1 , . . . , xi−1 , xi+1 , . . . , xp−1 , x, x0 , y) V (x1 , . . . , xp−1 , x0 , y)

V (x1 , . . . , xp−1 , x, x0 ) f (y). V (x1 , . . . , xp−1 , x0 , y)

(15.4.6)

15.5. Operation H

435

Every determinant V (x1 , . . . , xi−1 , xi+1 , . . . , xp , x, x0 ) contains the factor (λp+2 − λp+1 ) cf. (15.2.6) , and hence tends to zero when x → x0 , whereas f (xi ) remain bounded. The determinant V (x1 , . . . , xp , x0 ) can be estimated from below by a constant independent of x. The ratio V (x1 , . . . , xp , x)/V (x1 , . . . , xp , x0 ) tends to 1 when x → x0 . So the right-hand side of (15.4.5) tends to zero when x → x0 . Similarly, the right-hand side of (15.4.6) tends to zero when x → x0 . Hence f (x) approaches f (x0 ) when x approaches x0 .  Theorem 15.4.3. Let D ⊂ RN be a convex and open set, and let f : D → R be a p-convex function. If f is bounded on a set T ⊂ D such that int T = ∅, then f is continuous in D. Proof. Results from Theorems 15.4.1 and 15.4.2.



15.5 Operation H Fix a p ∈ N. For every set T ⊂ RN we define the set H(T ) as follows (Ger [98], [101], [104]): ! H(T ) = x ∈ RN | there exists an h ∈ RN (15.5.1) " such that x + ih ∈ T , i = ±1, . . . , ±(p + 1) . Choosing, in particular, h = 0, we get from (15.5.1) T ⊂ H(T ). Lemma 15.5.1. If D ⊂ RN is a convex set, and T ⊂ D, then H(T ) ⊂ D. Proof. Suppose that x ∈ H(T ). Then there exists an h ∈ RN such that x + h ∈ T

1  and x − h ∈ T . Hence also x = (x + h) + (x − h) ∈ D. 2 The main property of the operation H is contained in the following (Ger [98], [101],[104]) Theorem 15.5.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a p-convex function. If f is bounded on a set T ⊂ D, then f is bounded also on H(T ). Proof. Let |f (x)|  M

for x ∈ T.

(15.5.2)

Take an x ∈ H(T ). By (15.5.1) there exists an h ∈ RN such that x + ih ∈ T for i = ±1, . . . , ±(p + 1). Replacing, if necessary, h by −h, we  may assume that h > 0. If h = 0, then x ∈ T , and we get |f (x)|  M by (15.5.2) . We must distinguish two cases. 1. p is even. We have Δp+1 h f (x)  0, or, by Corollary 15.1.2, p+1

p+1−k

(−1)

k=0

  p+1 f (x + kh)  0. k

(15.5.3)

436

Chapter 15. Convex Functions of Higher Orders

Hence, by (15.5.2), since p is even, f (x) 

p+1

(−1)p+1−k

k=1

   p+1 

p+1 p+1 f (x + kh)  M = (2p+1 − 1)M. (15.5.4) k k k=1

Similarly, we have Δp+1 h f (x − h)  0, whence p+1

p+1−k

(−1)

k=0

i.e.,

p

  p+1 f (x − h + kh)  0, k

(15.5.5)

 (−1)p−k

k=−1

 p+1 f (x + kh)  0, k+1

and (p + 1)f (x)  (−1)p f (x − h) +

p

 (−1)p−k−1

k=1

 p+1 f (x + kh) k+1

(15.5.6)

 p 

p+1 M = −(2p+1 − p − 1)M.  −M − k+1 k=1

We clearly have

2p+1 − p − 1 2p+1   2p+1 − 1, p+1 p+1

whence by (15.5.6) f (x)  −(2p+1 − 1)M. This together with (15.5.4) yields |f (x)|  (2p+1 − 1)M.

(15.5.7)

2. p is odd. Again we have (15.5.3), whence f (x) 

p+1

(−1)p−k

k=1

   p+1 

p+1 p+1 f (x + kh)  − M = −(2p+1 − 1)M, k k k=1

whereas by (15.5.5) (p + 1)f (x)  (−1)p+1 f (x − h) +

p

 (−1)p−k

k=1

 p+1 f (x + kh) k+1

 (2p+1 − p − 1)M  (2p+1 − 1)M, and consequently we obtain again (15.5.7). Since 2p+1 − 1 > 1, if h = 0, then we obtain (15.5.7) from (15.5.2). Thus (15.5.7) holds for all x ∈ H(T ), i.e., f is bounded on H(T). 

15.5. Operation H

437

The following two lemmas (Ger [98], [101], [104]) show an important feature of the operation H. Lemma 15.5.2. Let T ⊂ RN be such that mi (T ) > 0. Then int H(T ) = ∅. Proof. There exists a measurable set T0 ⊂ T such that m(T0 ) > 0. Let x0 be a density point of T0 . Then 0 is a density point of T0 − x0 as well as of λ(T0 − x0 ) for all λ ∈ R, λ = 0. There exists a cube K centered at 0 such that, for every i ∈ I = {−(p + 1), −p, . . . , −1, +1, . . . , p, (p + 1)},   1 1 m K \ (T0 − x0 ) < N +1 m(K) . (15.5.8) i 2 (p + 1) Let U =

1 K so that U + U ⊂ K. We will show that 2  1 (T0 − x0 − x) = ∅ i

(15.5.9)

i∈I

for every x ∈ U . If x ∈ U , then x/i ∈ U for i ∈ I,and henceU + (x/i) ∈ U + U ⊂ K, i.e., U ⊂ K − (x/i) for i ∈ I, and hence U ⊂ K − (x/i) . On the other hand, i∈I

 K∩

K=

   1 1 (T0 − x0 ) ∪ K \ (T0 − x0 ) , i i

and U⊂

      1 1 1 K− x = K ∩ (T0 − x0 ) − x i i i i∈I i∈I    1 1 ∪ K \ (T0 − x0 ) − x i i         1 1 1 1 K ∩ (T0 − x0 ) − x ∪ K \ (T0 − x0 ) − x ⊂ i i i i i∈I i∈I     1 1 (15.5.10) = K ∩ (T0 − x0 ) − x ∪ Q, i i i∈I

where by (15.5.8) m(Q) < m(K)/2N , whereas m(U ) = m(K)/2N . Hence     1 1 K ∩ (T0 − x0 ) − x = ∅. i i

(15.5.11)

i∈I

But        1 1 1 1 1 (T0 − x0 ) − x = (T0 − x0 − x) K ∩ (T0 − x0 ) − x ⊂ i i i i i

i∈I

i∈I

i∈I

438

Chapter 15. Convex Functions of Higher Orders

Thus we have proved (15.5.9). Now take an arbitrary x ∈ U . By (15.5.9) there exists an 1 1 h ∈ (T0 − x0 − x) ⊂ (T − x0 − x) i i for every i ∈ I, which means that x + ih ∈ T − x0 for i ∈ I. Thus for every x ∈ U + x0 we have x + ih ∈ T for i ∈ I. Consequently x ∈ H(T ) and U + x0 ⊂ H(T ). Hence int H(T ) = ∅.  Lemma 15.5.3. Let T ⊂ RN be of the second category and with the Baire property. Then int H(T ) = ∅. Proof. The proof is similar to that of Lemma 15.5.2. Since T has the Baire property, we have T = (G ∪ P ) \ R, where G is a non-empty open set, and the sets P, R are of the first category. Put T0 = G \ R, and let x0 be an arbitrary point of T0 . Let K be an open ball centered 1 (G − x0 ). (Note that 0 belongs to this intersection.) at zero and contained in i∈I i Then, for every i ∈ I, we have K\

1 1 1 1 (T0 − x0 ) ⊂ (G − x0 ) \ (T0 − x0 ) = (G \ T0 ) − x0 . i i i i

But G \ T0 = G \ (G \ R) ⊂ R, and so G \ T0 is of the first category, whence also 1 K \ (T0 − x0 ) is of the first category for i ∈ I. This replaces relation (15.5.8). Define i 1 U = K. Again we have (15.5.10) for every x ∈ U , where now Q is a set of the first 2 category, and (15.5.11) follows, which implies (15.5.9). Further the proof runs exactly like that of Lemma 15.5.2.  From Lemmas 15.5.2 and 15.5.3 and from Theorems 15.5.1 and 15.4.3 we obtain the following theorem of Ciesielski (Ciesielski [46]; cf. also Kurepa [201]). Theorem 15.5.2. Let D ⊂ RN be a convex and open set, and let f : D → R be a p-convex function. If f is bounded on a set T ⊂ D such that mi (T ) > 0, or T is of the second category and with the Baire property, then f is continuous in D. The iterates H n of H are defined as usually:   H 1 (T ) = H(T ), H n+1 (T ) = H H n (T ) ,

n = 1, 2, 3, . . .

The following theorem, which may be regarded as a generalization of Theorem 15.5.2, is due to R. Ger [98], [101], [104]. Theorem 15.5.3. Let D ⊂ RN be a convex and open set, and let f : D → R be a p-convex function.  If f is bounded on a set T ⊂ D such that for a certain n ∈ N we have mi H n (T ) > 0 or H n (T ) is of the second category and has the Baire property, then f is continuous in D. Proof. This is an immediate consequence of Theorems 15.5.1 and 15.5.2.



15.6. Continuous p-convex functions

439

From Theorem 15.5.2 we obtain also the following (cf. also Gajda [95]). Theorem 15.5.4. Let D ⊂ RN be a convex and open set, and let f : D → R be a p-convex function. If f is measurable, then it is continuous in D. Proof. Put En = {x ∈ D | |f (x)|  n} ,

n ∈ N.

(15.5.12)

The sets En are measurable, since f is measurable, and  En = D. n∈N

Therefore for a certain n0 ∈ N the set En0 must have positive measure: m(En0 ) > 0. According to (15.5.12) the function f is bounded on En0 . By Theorem 15.5.2 f is continuous in D.  We conclude this section with an example (Ger [98]) showing that there exist sets T , of measure zero and nowhere dense, such that int H(T ) = ∅. Thus if f is a p-convex function bounded on such a set, its continuity can be inferred from Theorem 15.5.3, but not from Theorem 15.5.2. Example 15.5.1. Let N = 1. Put q = 2p + 4, and let T be the set of those x ∈ [0, 1] which may be written as ∞

ci x= , (15.5.13) i q i=1 where ci ∈ {0, . . . , p, p + 2, . . . , 2p + 3}, i ∈ N. The set T has measure zero (Theorem 3.6.4) and is nowhere dense. Take an arbitrary x ∈ [0, 1], and suppose that x has a representation (15.5.13), where ci ∈ {0, . . . , 2p + 3}, i ∈ N. Define the number h as h=

∞

di i=1



where di =

0 1

qi

,

if ci = p + 1, if ci = p + 1,

i ∈ N.

Then x + kh ∈ T for k = ±1, . . . , ±(p + 1), which means that x ∈ H(T ). Thus4 [0, 1] ⊂ H(T ), whence int H(T ) = ∅.

15.6 Continuous p-convex functions Let D ⊂ RN be a convex and open set, and let f : D → R be a p-convex function. If f is continuous in D, then we obtain by Theorem 15.3.1 and Lemma 15.2.2 that [x1 , . . . , xp+2 ; f ]  0 for every system of distinct collinear points x1 , . . . , xp+2 ∈ D. 4

By Lemma 15.5.1 H(T ) ⊂ [0, 1] so that actually H(T ) = [0, 1].

(15.6.1)

440

Chapter 15. Convex Functions of Higher Orders

The converse is also true: Theorem 15.6.1. Let D ⊂ RN be a convex and open set, and let f : D → R be a function. The function f is p-convex and continuous in D if and only if (15.6.1) holds for every system of distinct collinear points x1 , . . . , xp+2 ∈ D. Proof. As pointed out above, the necessity results from Theorem 15.3.1 and Lemma 15.2.2. Now assume that (15.6.1) holds for every system of distinct collinear points x1 , . . . , xp+2 ∈ D. Take an arbitrary x ∈ D and h ∈ RN + such that x + (p + 1)h ∈ D. Put xi = x + (i − 1)h, i = 1, . . . , p + 2. By Lemma 15.2.5 and by (15.6.1) p Δp+1 h f (x) = p!|h| [x1 , . . . , xp+2 ; f ]  0, N whence Δp+1 h f (x)  0 for every x ∈ D, and h ∈ R+ such that x + (p + 1)h ∈ D. Thus f is p-convex. Consequently in order to prove that f is continuous, it is sufficient to prove that f is bounded on a subset of D with a non-empty interior (cf. Theorem 15.4.3). We will prove by induction on k that f is bounded on every closed k-dimensional rectangular parallelopiped P contained in D, k = 1, . . . , N . For k = N we have int P = ∅, and the theorem follows. For k = 1, P is a closed segment contained in D. Since D is open, there exists a closed segment S ⊂ D such that P ⊂ int S. Let S = ab, P = cd, and let, e.g., a < c < d < b. (The other possibility is b < d < c < a, which may be reduced to the former by re-naming the points a, b, c, d). Fix p + 1 distinct points x1 , . . . , xp+1 on the segment ac such that a  x1 < · · · < xp+1 < c. We have by (15.6.1) for arbitrary x∈P [x1 , . . . , xp+1 , x; f ]  0

whence also, according to Theorem 15.2.1, U (x1 , . . . , xp+1 , x; f )  0,

(15.6.2)

since x1 < · · · < xp+1 < x so that V (x1 , . . . , xp+1 , x) > 0. Developing the determinant in (15.6.2) according to the last column we obtain f (x) 

p+1

i=1

(−1)p+1+i

V (x1 , . . . , xi−1 , xi+1 , . . . xp+1 , x) f (xi ). V (x1 , . . . , xp+1 )

(15.6.3)

The determinant V (x1 , . . . , xi−1 , xi+1 , . . . xp+1 , x) (for every i = 1, . . . , p + 1) is a continuous function of x, and thus remains bounded when x ranges over the compact set P . The remaining expressions on the right-hand side of (15.6.3) are independent of x (the points x1 , . . . , xp+1 are fixed). Consequently the whole expression on the right-hand side of (15.6.3) remains bounded when x ranges over P , and consequently f is bounded below on P .

15.6. Continuous p-convex functions

441

Similarly, we have for arbitrary x ∈ P [x1 , . . . , xp , x, b; f ]  0, whence also U (x1 , . . . , xp , x, b; f )  0, and f (x) 

p

i=1

(−1)p+i

V (x1 , . . . , xi−1 , xi+1 , . . . xp , x, b) V (x1 , . . . , xp , x) f (xi ) + f (b). V (x1 , . . . , xp , b) V (x1 , . . . , xp , b)

Consequently f is also bounded above on P , and hence f is bounded on P . Now suppose that our statement is true for a certain k < N , and let P ⊂ D be a closed (k + 1)-dimensional rectangular parallelopiped. Let P = S × P0 , where S is a closed segment and P0 is a closed k-dimensional rectangular parallelopiped. We may find a closed segment S0 such that S ⊂ int S0 and S0 × P0 ⊂ D. Let S0 = ab, S = cd. Again we may assume that a < c < d < b. Choose p + 1 distinct points x1 , . . . , xp+1 on the segment ac such that a  x1 < · · · < xp+1 < c. Let Pb = {b} × P0 ,

Pi = {xi } × P0 ,

i = 1, . . . , p + 1.

P1 , . . . , Pp+1 , Pb are closed k-dimensional rectangular parallelopipeds, consequently, by the induction hypothesis, the function f is bounded in the set E = Pb ∪

p+1 

Pi .

i=1

For every x ∈ P let l(x) be the straight line passing through x and parallel to S. Write xb = l(x) ∩ Pb , xi = l(x) ∩ Pi , i = 1, . . . , p + 1. Thus, for every x ∈ P , x1 , . . . , xp+1 , x, xb are distinct collinear points fulfilling x1 < · · · < xp+1 < x < xb , and (15.6.4) x1 , . . . , xp+1 , xb ∈ E. The Van der Monde determinants V (x1 , . . . , xp+1 ), V (x1 , . . . , xi−1 , xi+1 , . . . xp+1 , x), V (x1 , . . ., xi−1 , xi+1 , . . . xp , x, xb ), V (x1 , . . . xp , xb ) and V (x1 , . . . xp , x) depend on x1 , . . . , xp+1 , xb only through their distances, which remain constant when x ranges over P , so as a function of x the above mentioned determinants are continuous, and hence bounded on P . Also, by (15.6.4), the values f (x1 ), . . . , f (xp+1 ), f (xb ) remain bounded when x ranges over P . From the inequalities 



x1 , . . . , xp+1 , x; f  0, x1 , . . . , xp , x, xb ; f  0, we infer that f is bounded on P quite similarly as in the case k = 1. Induction ends the proof. 

442

Chapter 15. Convex Functions of Higher Orders

15.7 Continuous p-convex functions. Case N = 1 In this and in the next section, in order to avoid burdensome details, we restrict ourselves to the one-dimensional case N = 1. Moreover, many results are not longer valid in the general, N -dimensional case. At first let us note the following consequence of Theorem 15.3.2 and Lemma 15.1.4: Theorem 15.7.1. Let D ⊂ R be an open interval, and let f : D → R be a continuous p-convex function. Then, for arbitrary x ∈ D and arbitrary positive h1 , . . . , hp+1 such that x + h1 + · · · + hp+1 ∈ D, we have Δh1 ...hp+1 f (x)  0.

(15.7.1)

But now condition (15.7.1) does not imply that f is continuous. E.g., let f : R → R be a discontinuous additive function. Then we have by (15.1.5) for arbitrary x, h1 , h2 ∈ R Δh1 h2 f (x) = f (x + h1 + h2 ) − f (x + h1 ) − f (x + h2 ) + f (x) = 0, but f is not continuous. On the other hand, condition (15.7.1) clearly implies that f is p-convex. (15.3.1) results from (15.7.1) with h1 = · · · = hp+1 = h. Hence we get the following Theorem 15.7.2. Let D ⊂ R be an open interval, and let f : D → R be a continuous p-convex function, p > 1. Let h > 0 be such that Dh = D ∩ (D − h) = ∅, and define the function g : Dh → R by g(x) = Δh f (x). Then g is (p − 1)-convex. Proof. The set Dh evidently is open, and by Theorems 5.1.1 and 5.1.2 Dh is convex. By Theorem 15.7.1 f satisfies (15.7.1). Take hp+1 = h. Then (15.7.1) yields Δh1 ...hp g(x)  0 for every x ∈ Dh and every positive h1 , . . . , hp such that x + h1 + · · · + hp ∈ Dh . This implies that g is (p − 1)-convex.  Concerning the case p = 1, cf. Theorem 7.3.4. According to the remarks after formula (15.2.3), in the one-dimensional case in the definition of the divided differences [x1 , . . . , xp ; f ] we may take λi = xi . Similarly, due to the uniqueness of P (x1 , . . . , xp+1 ; f ), in the definition (15.2.11) we may take λi = xi . Now, making use of Lemma 15.2.4 we can give a geometrical interpretation of the p-convexity for continuous functions. Let the points x1 , . . . , xp+1 , x ∈ D satisfy x1 < · · · < xp+1 < x. Then V (x1 , . . . , xp+1 , x) > 0 and V (x1 , . . . , xp+1 ) > 0, and consequently, by (15.2.12), f (x) − P (x1 , . . . , xp+1 ; f | x)  0 (cf. Theorem 15.6.1). Thus if we take arbitrary p + 1 points x1 , . . . , xp+1 ∈ D, x1 < · · · < xp+1  , and  draw the La grange interpolation polynomial P through the points x1 ,f (x1 ) ,..., xp+1 ,f (xp+1 ) , then to the right of xp+1 the graph of f lies above (or on) that of P . When the point x moves leftward, the graph of f will lie in the intervals (xp , xp+1 ), . . . , (x1 , x2 ) alter-

15.7. Continuous p-convex functions. Case N = 1

443

natively below and above (or on) that of P , since with every crossing of a point xi the determinant V (x1 , . . . , xp+1 , x) changes the sign. Finally, to the left of x, the graph of f lies below resp. above (or on) that of P , depending on whether p is even or odd. In the case p = 1, when f is continuous and convex in theusual sense, if we draw   a straight line L (a polynomial of degree 1) through the points u, f (u) and v, f (v) , u, v ∈ D, then the graph of f lies below (or on) L between u and v, and above (or on) L outside of the segment uv, which is the usual geometrical interpretation of convexity for continuous functions (cf. 7.1). Below we prove some lemmas concerning the boundedness of the divided differences of a function. Lemma 15.7.1. Let D ⊂ R be a finite interval, and let f : D → R be a function. If, for a p ∈ N, p > 1, the divided difference [x1 , . . . , xp ; f ] remains bounded for all x1 , . . . , xp ∈ D, x1 , . . . , xp distinct, then also the divided difference [x1 , . . . , xp−1 ; f ] remains bounded for all x1 , . . . , xp−1 ∈ D, x1 , . . . , xp−1 distinct. Proof. There exist an M > 0 such that | [t1 , . . . , tp ; f ] |  M

(15.7.2)

for all t1 ,...,tp ∈ D, t1 ,...,tp distinct. Fix arbitrarily p − 1 distinct points y1 ,...,yp−1 ∈ D, and take arbitrary p − 1 distinct points x1 , . . . , xp−1 ∈ D. By Lemma 15.2.3 we may assume that the point x1 , . . . , xp−1 have been numbered in such a manner that if some of the points x1 , . . . , xp−1 coincide with some of the points y1 , . . . , yp−1 , then they have the same indices as the corresponding y’s, i.e., if xj = yi , then j = i. Put (15.7.3) A = | [y1 , . . . , yp−1 ; f ] | . A is a fix finite number. By Lemma 15.2.7 [x1 , . . . , xp−1 ; f ] − [y1 , . . . , yp−1 ; f ] =

p−1

(xi − yi ) [x1 , . . . , xi , yi , . . . , yp−1 ; f ] ,

i=1

whence by (15.7.2) and (15.7.3) |[x1 , . . . , xp−1 ; f ]|  A + M

p−1

|xi − yi |  A + (p − 1) M |D|,

i=1

where |D| is the length of D.



Now fix m > p + 1 points x1 , . . . , xm ∈ D such that x1 < · · · < xm , and put di = [xi , . . . , xi+p ; f ] ,

i = 1, . . . , m − p.

(15.7.4)

Lemma 15.7.2. Let D ⊂ R be an open interval, and let f : D → R be a continuous pconvex function. Then, for every x1 , . . . , xm ∈ D, x1 < · · · < xm , m > p+ 1, sequence (15.7.4) is increasing.

444

Chapter 15. Convex Functions of Higher Orders

Proof. We have by (15.2.3) di+1 − di [xi+1 , . . . , xi+p+1 ; f ] − [xi , . . . , xi+p ; f ] = = [xi , . . . , xi+p+1 ; f ]  0 xi+p+1 − xi xi+p+1 − xi 



cf. Theorem 15.6.1 . Hence di  di+1 , since xi < xi+p+1 , i = 1, . . . , m − p.



Lemma 15.7.3. Let D ⊂ R be an open interval, and let f : D → R be a continuous p-convex function. Then, for every compact interval I ⊂ D, the divided difference [x1 , . . . , xp+1 ; f ] remains bounded for all x1 , . . . , xp+1 ∈ I, x1 , . . . , xp+1 distinct. Proof. Let D = (a, b), I = [c, d], a < c < d < b. Fix arbitrarily p + 1 distinct points y1 , . . . yp+1 ∈ (a, c) and p + 1 distinct points z1 , . . . , zp+1 ∈ (d, b) such that y1 < · · · < yp+1 < z1 < · · · < zp+1 . Take arbitrary p + 1 distinct points x1 , . . . , xp+1 ∈ I. According to Lemma 15.2.3 we may assume that x1 , . . . , xp+1 have been numbered in such a manner that x1 < · · · < xp+1 . Then we have y1 < · · · < yp+1 < x1 < · · · < xp+1 < z1 < · · · < zp+1 . By Lemma 15.7.2 (m = 3p + 3) we have [y1 , . . . , yp+1 ; f ]  [x1 , . . . , xp+1 ; f ]  [z1 , . . . , zp+1 ; f ] , and the lemma follows.



From Lemmas 15.7.1 and 15.7.3 we obtain Corollary 15.7.1. Let D ⊂ R be an open interval, and let f : D → R be a continuous p-convex function. Then the divided differences of f of all orders  p + 1 are bounded on every compact interval I ⊂ D.

15.8 Differentiability of p-convex functions Also in this section we restrict ourselves to the one-dimensional case N = 1. Theorem 15.8.1. Let D ⊂ R be an open interval, and let f : D → R be a continuous p-convex function, p > 1. Then f is of class C 1 in D, and the derivative f  : D → R is a (p − 1)-convex function. Proof. Take an arbitrary x ∈ D and an η > 0 such that I = [x − η, x + η] ⊂ D. By Corollary 15.7.1 the divided difference [t1 , t2 , t3 ; f ] is bounded in I, i.e., there exists a constant M > 0 such that # # # [t1 , t2 , t3 ; f ] #  M for every distinct points t1 , t2 , t3 ∈ I. In other words, by (15.2.3) # # # [t2 , t3 ; f ] − [t1 , t2 ; f ] # # #M # # t 3 − t1 for every distinct points t1 , t2 , t3 ∈ I. Hence # # # [t2 , t3 ; f ] − [t1 , t2 ; f ] #  M |t3 − t1 |

(15.8.1)

15.8. Differentiability of p-convex functions

445

for every distinct points t1 , t2 , t3 ∈ I. Now take arbitrary distinct points x1 , x2 , x1 , x2 ∈ I. We have by (15.8.1) # # # #   #[x1 , x2 ; f ] − [x1 , x2 ; f ]#  #[x2 , x1 ; f ] − [x1 , x2 ; f ]#+ # # + #[x1 , x2 ; f ] − [x2 , x1 ; f ]# # # # # (15.8.2)  M #x1 − x1 # + M #x2 − x2 #  4M η Inequality (15.8.2) shows that there exists the limit lim [x1 , x2 ; f ] = x ,x lim→x

x1 ,x2 →x x1 =x2

1

2

x1 =x2

f (x2 ) − f (x1 ) = f  (x). x2 − x1

Consequently f is differentiable at every point x ∈ D. Moreover, it follows from (15.8.2), when x1 , x2 → x ∈ I, x1 = x2 , and x1 , x2 → y ∈ I, x1 = x2 , that # # #  #f (y) − f  (x)|  2M #y − x#, which proves the continuity of f  in D. Consequently f is of class C 1 in D. Since the derivative f  exists in D, we have for every x ∈ D lim

h→0

Δh f (x) = f  (x). h

Take arbitrary x ∈ D and arbitrary positive h1 , . . . , hp ∈ D such that x+h1 +· · ·+hp ∈ D. For sufficiently small h > 0 we have x + h1 + · · · + hp + h ∈ D, whence by Theorem 15.7.1 1 Δh ...h h f (x)  0 . h 1 p Hence also (cf. Lemma 15.1.5) Δh1 ...hp f  (x) = Δh1 ...hp lim

h→0

Δh f (x) 1 = lim Δh1 ...hp h f (x)  0. h→0 h h

Consequently f  is (p − 1)-convex.



Theorem 15.8.2. Let D ⊂ R be an open interval, and let f : D → R be a function of class C 1 such that f  is (p − 1)-convex (p > 1). Then f is p-convex. Proof. We have for arbitrary x ∈ D and h > 0 such that x + h ∈ D x+h 

f  (t)dt,

Δh f (x) = f (x + h) − f (x) = x

whence by Lemma 15.1.6 x+h 

Δh1 ...hp h f (x) = Δh1 ...hp Δh f (x) = Δh1 ...hp



x+h 

Δh1 ...hp f  (t)dt  0,

f (t)dt = x

x

for arbitrary positive h1 , . . . , hp , h such that x + h1 + · · · + hp + h ∈ D. Consequently f is p-convex. 

446

Chapter 15. Convex Functions of Higher Orders

Using Theorem 15.8.1 several times, we arrive at the following Theorem 15.8.3. Let D ⊂ R be an open interval, and let f : D → R be a continuous p-convex function (p > 1). Then f is of class C p−1 in D, and for every r ∈ N, 1  r  p − 1, the derivative f (r) is (p − r)-convex. From Theorems 15.8.2 and 15.8.3 we obtain Theorem 15.8.4. Let D ⊂ R be an open interval, and let f : D → R be a continuous function. Then f is p-convex (p > 1) if and only if f is of class C p−1 in D and the derivative f (p−1) is convex. Hence and from Theorems 7.5.1 and 7.5.2 we obtain Theorem 15.8.5. Let D ⊂ R be an open interval, and let f : D → R be a function of class C p in D. Then f is p-convex if and only if the derivative f (p) is increasing in D. Theorem 15.8.6. Let D ⊂ R be an open interval, and let f : D → R be a function of class C p+1 in D. Then f is p-convex if and only if the derivative f (p+1) is nonnegative in D.

15.9 Polynomial functions The main objective of this section is to prove the basic theorem about the representation of polynomial functions with the aid of multiadditive functions (cf. 13.4). The results are due to S. Mazur-W. Orlicz [222] (cf. also McKiernan [223], Hossz´ u [146], Djokovi´c [71], Sz´ekelyhidi [310]). Given a function F : RpN → R, by the diagonalization of F we understand the function f : RN → R arising from F by putting all the variables (from RN ) equal: x ∈ RN .

f (x) = F (x, . . . , x), 5 67 8 p times We start with some preliminary lemmas.

Lemma 15.9.1. Let F : RkN → R be a symmetric k-additive function, and let f : RN → R be the diagonalization of F . Then, for every x, h ∈ RN , k−1

k  F (x, . . . , x, h, . . . , h). (15.9.1) Δh f (x) = 5 67 8 5 67 8 i i=0

i

k−i

Proof. It is enough to prove that k  

k F (x, . . . , x, h, . . . , h), F (x + h, . . . , x + h) = 5 67 8 5 67 8 5 67 8 i k

i=0

i

(15.9.2)

k−i

as (15.9.1) is an immediate consequence of (15.9.2). The proof will be by induction on k. For k = 1 relation (15.9.2) becomes F (x + h) = F (x) + F (h),

15.9. Polynomial functions

447

which is true in virtue of the additivity (i.e., 1-additivity) of F . Now assume (15.9.2) true for a k ∈ N, and consider a symmetric (k+1)-additive function F : R(k+1)N → R. We have for every fixed u ∈ RN , according to the induction hypothesis, k  

k F (x, . . . , x, h, . . . , h, u), F (x + h, . . . , x + h, u) = 5 67 8 5 67 8 67 8 5 i i=0 i

k

k−i

whence, on setting u = x + h and using the (k + 1)-additivity and symmetry of F , F (x + h,...,x + h) = 67 8 5 k+1

=

k  

k F (x,...,x, h,...,h,x + h) 5 67 8 5 67 8 i i=0

i=0

=

i

k−i

i

k−i

i+1

k−i

k 

 k  

k k F (x,...,x, h,...,h,x) + F (x,...,x, h,...,h) 5 67 8 5 67 8 5 67 8 5 67 8 i i

k 

i=0

i=0

i

k+1−i

 k  

k k F (x,...,x, h,...,h) + F (x,...,x, h,...,h). 5 67 8 5 67 8 5 67 8 5 67 8 i i i=0

i

k+1−i

Replacing in the first sum i+ 1 by j, and  in the  second  i by j, and then making use k k k+1 of the well-known identity + = we obtain j−1 j j  k 

k F (x + h, . . . , x + h) = F (x, . . . , x) + F (x, . . . , x, h, . . . , h) 5 5 67 8 5 67 8 5 67 8 67 8 j−1 j=1 k+1

j

k+1

+

k 

j=1

k+1−j

 k F (x, . . . , x, h, . . . , h) + F (h, . . . , h) 5 67 8 5 67 8 5 67 8 j j

k+1−j

k+1

   k 

k k = F (x, . . . , x) + + F (x, . . . , x, h, . . . , h) + F (h, . . . , h) 5 67 8 5 67 8 5 67 8 5 67 8 j−1 j j=1 j

k+1

=

k+1

 j=0

 k+1 F (x, . . . , x, h, . . . , h). 5 67 8 5 67 8 j j

k+1−j

k+1

k+1−j

Thus we have obtained (15.9.2) for k + 1. Induction completes the proof.



Lemma 15.9.2. Let F : RkN → R be a symmetric k-additive function, and let f : RN → R be the diagonalization of F . For every n ∈ N, n  k, and for every x, h1 . . . , hn ∈ RN we have  k! F (h1 , . . . , hk ) if n = k, (15.9.3) Δh1 ...hn f (x) = 0 if n > k.

448

Chapter 15. Convex Functions of Higher Orders

Proof. The proof is by induction on k. If k = 1, then F = f is an additive function, and Δh f (x) = f (x + h) − f (x) = f (x) + f (h) − f (x) = F (h) for every x, h ∈ RN . Hence also, for n > 1, Δh1 ...hn f (x) = Δh1 ...hn−1 Δhn f (x) = Δh1 ...hn−1 f (hn ) = 0, since f (hn ), as a function of x, is constant. Now assume that (15.9.3) is true for 1, . . . , k ∈ N, and consider a (k + 1)-additive function F : R(k+1)N → R, and its diagonalization f : RN → R. We have by Lemmas 15.9.1 and 15.1.1, for arbitrary x, h1 . . . , hk+1 ∈ RN , Δh1 ...hk+1 f (x) = Δh1 ...hk Δhk+1 f (x)  k 

k+1 Δh1 ...hk F (x, . . . , x, hk+1 , . . . , hk+1 ). = 5 67 8 5 67 8 i i=0 i

k+1−i

For every i, 1  i  k, the function F (x, . . . , x, hk+1 , . . . , hk+1 ) is i-additive, whereas 5 67 8 5 67 8 i

k+1−i

for i = 0 the expression F (hk+1 , . . . , hk+1 ) (as a function of x) is constant. By the induction hypothesis we have (0  i  k)  k! F (h1 , . . . , hk+1 ) if i = k , Δh1 ...hk F (x, . . . , x, hk+1 , . . . , hk+1 ) = 5 67 8 5 67 8 0 if i < k . i

k+1−i

Hence Δh1 ...hk+1 f (x) = (k + 1) k! F (h1 , . . . , hk+1 ) = (k + 1)! F (h1 , . . . , hk+1 ). For n > k + 1 we have hence, since F (h1 . . . , hk+1 ) (as a function of x) is constant, for arbitrary x, h1 , . . . , hn ∈ RN , according to Corollary 15.1.1 Δh1 ...hn f (x) = Δhk+2 ...hn Δh1 ...hk+1 f (x) = (k + 1)! Δhk+2 ...hn F (h1 , . . . , hk+1 ) = 0 . Consequently (15.9.3) is true also for k + 1, and induction completes the proof.



→ R be a symmetric k-additive function, and let Corollary 15.9.1. Let F : R f : RN → R be the diagonalization of F . If f = 0 in RN , then F = 0 in RkN . kN

Proof. Since f = 0, we have also for arbitrary x, x1 , . . . , xk ∈ RN Δx1 ...xk f (x) = 0, whence by Lemma 15.9.2 F (x1 , . . . , xk ) = 0.



In the sequel by a 0-additive function we understand a constant. A diagonalization of such a function is also the same constant. Observe that Lemma 15.9.2 remains valid for k = 0.

15.9. Polynomial functions

449

Lemma 15.9.3. Let Fk : RkN → R, k = 0, . . . , p, be symmetric k-additive functions, and let f0 , . . . , fp be the diagonalizations of F0 , . . . , Fp , respectively. Let the function f : RN → R be given by f (x) =

p

fk (x),

x ∈ RN .

(15.9.4)

k=0

If f is bounded in RN , then Fk = 0 for k = 1, . . . , p. Proof. By Theorem 13.4.1 we have for every x ∈ RN , λ ∈ Q, and for k = 0, . . . , p fk (λx) = λk fk (x). Hence f (λx) =

p

λk fk (x).

(15.9.5)

k=0

For every x ∈ RN expression (15.9.5) is a polynomial in λ, which, according to the assumptions of the lemma, remains bounded for all rational values of λ. But this is possible only if the polynomial is constant, i.e., if fk (x) = 0

for k = 1, . . . , p .

(15.9.6)

(15.9.6) holds for every x ∈ RN . In virtue of Corollary 15.9.1 this implies that Fk = 0

for k = 1, . . . , p , 

which was to be shown.

Theorem 15.9.1. Let Fk : RkN → R, k = 0, . . . , p, be symmetric k-additive functions, and let f0 , . . . , fp be the diagonalizations of F0 , . . . , Fp , respectively. If the function f : RN → R is given by (15.9.4), then f is a polynomial function of order p. Proof. By (15.1.4) and Lemma 15.9.2 we have for every x, h ∈ RN Δp+1 h fk (x) = 0,

k = 0, . . . , p ,

whence by Lemma 15.1.1 Δp+1 h f (x) = 0, which means that f is a polynomial function of order p.  The converse is expressed by the following Theorem 15.9.2. Let f : RN → R be a polynomial function of order p. Then there exist unique k-additive functions Fk : RkN → R, k = 0, . . . , p, such that (15.9.4) holds, where f0 , . . . , fp are the diagonalizations of F0 , . . . , Fp , respectively. Proof. 5 First we prove the existence of Fk , k = 0, . . . , p. The proof will be by induction on p. If p = 1, then f satisfies Δ2h f (x) = 0 for every x, h ∈ RN , which means, 5

Albert-Baker [11]

450

Chapter 15. Convex Functions of Higher Orders

as was pointed out in 15.3, that f satisfies the Jensen equation (13.2.1). By Theorem 13.2.1 f (x) = f1 (x) + f0 , where f1 : RN → R is an additive function, and f0 ∈ R is a constant. Consequently f admits representation (15.9.4). Now assume that for every polynomial function g : RN → R of an order p−1 ∈ N there exist symmetric k-additive functions Fk : RkN → R, k = 0, . . . , p − 1, such that g(x) =

p−1

fk (x),

(15.9.7)

k=0

where f0 , . . . , fp−1 are the diagonalizations of F0 , . . . , Fp−1 , respectively. Let f : RN → R be a polynomial function of order p, and define the function Fp : RpN → R by 1 Fp (x1 , . . . , xp ) = Δx1 ...xp f (0), x1 , . . . , xp ∈ RN . (15.9.8) p! By Corollary 15.1.1 Fp is symmetric, and we have by Corollary 15.1.1, Lemma 15.1.1, Lemma 15.1.3 and Theorem 15.3.3 for every i, 1  i  p, and for every x1 , . . . , xp , y i ∈ RN Fp (x1 ,...,xi−1 ,xi + yi ,xi+1 ,...,xp ) − Fp (x1 ,...,xp ) − Fp (x1 ,...,xi−1 ,yi ,xi+1 ,...,xp )

1 1 = Δx1 ...xi−1 ,xi+1 ...xp Δxi +yi f (0) − Δxi f (0) − Δyi f (0) = Δx1 ...xp yi f (0) = 0, p! p! which means that Fp is p-additive. Also, for arbitrary x, x1 , . . . , xp ∈ RN we have by Lemma 15.1.1 and Theorem 15.3.3

Δx1 ...xp f (x) − Δx1 ...xp f (0) = Δx1 ...xp f (x) − f (0) = Δx1 ...xp Δx f (0) = Δx1 ...xp x f (0) = 0, whence by (15.9.8) Δx1 ...xp f (x) = p! Fp (x1 , . . . , xp ).

(15.9.9)

Now let fp : R → R be the diagonalization of Fp , and put g = f − fp . We have by Lemmas 15.1.1 and 15.9.2 and by (15.9.9), for arbitrary x, h1 , . . . , hp ∈ RN , N

Δh1 ...hp g(x) = Δh1 ...hp f (x) − Δh1 ...hp fp (x) = Δh1 ...hp f (x) − p! Fp (h1 , . . . , hp ) = 0. This implies that g is a polynomial function of order p − 1. By the induction hypothesis there exist symmetric k-additive functions Fk : RkN → R, k = 0, . . . , p − 1, such that (15.9.7) holds, where f0 , . . . , fp−1 are the diagonalizations of F0 , . . . , Fp−1 , respectively. Hence we obtain (15.9.4), which ends the induction. To prove the uniqueness suppose that f has also a representation f (x) =

p

k=0

gk (x),

(15.9.10)

15.9. Polynomial functions

451

where each gk is the diagonalization of a symmetric k-additive function Gk : RkN → R, k = 0, . . . , p. Every function Fk − Gk is symmetric and k-additive (cf. Lemma 13.4.1), and fk − gk is the diagonalization of Fk − Gk , k = 0, . . . , p. By (15.9.4) and (15.9.10) p

  fk (x) − gk (x) = 0, (15.9.11) k=0

whence by Lemma 15.9.3 Fk = Gk for k = 1, . . . , p, and (15.9.11) reduces to f0 − g0 = 0. Hence F0 = f0 = g0 = G0 so that Fk = Gk for k = 0, . . . , p, which completes the proof of the uniqueness.  Corollary 15.9.2. Let f : RN → R be a polynomial function. If f is bounded in RN , then f = const in RN . Proof. This results from Theorem 15.9.2 and Lemma 15.9.3.



Since the general construction of symmetric multiadditive functions is known (cf. Theorem 13.4.2 and Corollary 13.4.1), Theorems 15.9.1 and 15.9.2 yield implicitely the general construction of polynomial functions. Let us note also that R. Ger [100] proved that for every p ∈ N there exists a set Hp ⊂ RN such that every function g : Hp → R can be uniquely extended onto RN to a polynomial function of order p. Every polynomial function of order p is, in particular, p-convex, therefore every condition implying the continuity of a p-convex function (cf. 15.5) implies also the continuity of a polynomial function of order p (cf. also Gajda [94] concerning analogues of Theorems 9.3.7 and 9.8.1). For continuous polynomial functions we have the following Theorem 15.9.3. A function f : RN → R is a continuous polynomial function of order p if and only if [x1 , . . . , xp+2 ; f ] = 0 (15.9.12) for every system of p + 2 distinct collinear points x1 , . . . , xp+2 ∈ RN . Proof. A function f : RN → R is a polynomial function of order p if and only if it is p-convex and p-concave, or, equivalently, if and only if the functions f and −f are p-convex. If f is moreover continuous, then, by Theorem 15.6.1, this is equivalent to the condition that [x1 , . . . , xp+2 ; f ]  0 and

[x1 , . . . , xp+2 ; −f ]  0

(15.9.13)

for every system of p + 2 distinct collinear points x1 , . . . , xp+2 ∈ RN . But, as is easily seen from Theorem 15.2.1, we have [x1 , . . . , xp+2 ; −f ] = − [x1 , . . . , xp+2 ; f ] , and so condition (15.9.13) is equivalent to (15.9.12). Conversely, if f : RN → R satisfies (15.9.12), then, by Theorem 15.6.1, it is continuous and p-convex, and also −f is p-convex, whence f is p-concave. Consequently f is a continuous polynomial function of order p. 

452

Chapter 15. Convex Functions of Higher Orders

Actually we can say much more. First we prove the following Lemma 15.9.4. If a function f : RN → R, f (x) = f (ξ1 , . . . , ξN ) is a polynomial separately in each variable ξi , i = 1, . . . , N , then f is a polynomial jointly in all variables. Proof. The proof runs by induction on N . If N = 1, the lemma is trivial. Assume it true for an N − 1 ∈ N, and let f : RN → R be a function which is a polynomial separately in each variable. Let p be the maximal degree of f with respect to particular variables. Thus we have f (ξ1 , . . . , ξN ) =

p

i AN i (ξ1 , . . . , ξN −1 )ξN = · · · =

i=0

p

A1i (ξ2 , . . . , ξN )ξ1i .

(15.9.14)

i=0

In particular, p

i=0

i AN i (ξ1 , . . . , ξN −1 )ξN =

p

−1 i AN (ξ1 , . . . , ξN −2 , ξN )ξN −1 . i

(15.9.15)

i=0

Setting in (15.9.15) ξ1 = ξ¯1 , . . . , ξN −2 = ξ¯N −2 and ξN = 0, we obtain that ¯ ¯ AN 0 (ξ1 , . . . , ξN −2 , ξN −1 ) is a polynomial in ξN −1 . Setting next in (15.9.15) ξ1 = ¯ ξ1 , . . . , ξN −2 = ξ¯N −2 , and ξN in turn 1, . . . , p, we obtain a system of linear equa¯ ¯ tions for AN i (ξ1 , . . . , ξN −2 , ξN −1 ), i = 1, . . . , p, whose determinant is # # #1 . . . 1 # # # #2 . . . 2p # # # # .. .. # = 0 , #. . ## # #p . . . pp # and the right-hand sides are polynomials in ξN −1 . Consequently also the solutions are polynomials in ξN −1 . Using other equations in (15.9.14) we obtain that all AN i , i = 0, . . . , p, are polynomials in each variable separately. By the induction hypothesis AN i , i = 0, . . . , p, are polynomials. Thus we see from (15.9.14) that also f is a polynomial.  Theorem 15.9.4. A function f : RN → R is a continuous polynomial function of order p if and only if f is a polynomial of degree at most p. Proof. Consider a monomial f : RN → R of a degree k ∈ N: αN f (x) = f (ξ1 , . . . , ξN ) = aξ1α1 · · · ξN ,

α1 + · · · + αN = k .

Put α0 = 0, and define the function F : RkN → R as F (x1 , . . . , xk ) = F (ξ11 , . . . , ξ1N , . . . , ξk1 , . . . , ξkN ) =

+αj N αj−1 a ξim j , k! +1 j=1 m=α j−1

15.9. Polynomial functions

453

where the sum extends over all permutations (i1 , . . . , ik ) of numbers (1, . . . , k), and r 2 we assume the convention that = 1 for every r ∈ N ∪ {0}. It is easily seen that r+1

the function F is symmetric and k-additive, and f is the diagonalization of F . In virtue of Theorem 15.9.1 f is a polynomial function of every order p  k. A polynomial of degree at most p is a sum of monomials of degrees  p, and thus of polynomial functions of order p. In virtue of Lemma 15.3.2 such a polynomial itself is a polynomial function of order p. Evidently every polynomial is continuous. Conversely, suppose that a continuous function f : RN → R, f (x) = f (ξ1 , . . . , ξN ), is a polynomial function of order p. Let l ⊂ RN be an arbitrary straight line, and take arbitrary distinct points x1 , . . . , xp+1 ∈ l. By Lemma 15.2.4 and Theorem 15.9.3 we have for every x ∈ l f (x) = P (x1 , . . . , xp+1 ; f | x). (15.9.16) (By the continuity of f and P (15.9.16) holds also if x equals one of x1 , . . . , xp+1 ). Consequently on every straight line f is a polynomial of degree at most p in the linear coordinate of x. In particular, f is a polynomial separately in each variable ξ1 , . . . , ξN . By Lemma 15.9.4 f is a polynomial. It remains to estimate the degree of f . Let the degree of f be q, and let f (x) =

q

gk (x),

(15.9.17)

k=0

where, for k = 0, . . . , q, gk is the homogeneous part of f of degree k. Since the degree of f is q, we have gq = 0, and consequently there exists an x0 ∈ RN , x0 = 0, such that gq (x0 ) = 0. Put h = (x0 /|x0 |) sgn x0 and λ0 = |x0 | sgn x0 . We have gq (x0 ) = gq (λ0 h) = λq0 gq (h), whence gq (h) = 0. By (15.9.17) f (λh) =

q

λk gk (h),

k=0

which is a polynomial in λ if degree q. On the other hand, by (15.9.16), on the straight line ! " l = x ∈ RN | x = λh, λ ∈ R f is a polynomial of a degree not exceeding p. Consequently q  p.



Thus, as we see, the polynomial functions generalize ordinary polynomials, and reduce to the latter under mild regularity assumptions. Consequently relation (15.3.3) has often been used to define polynomials in abstract spaces, and functions fulfilling this and related conditions have been thoroughly investigated. In this connection cf., e.g., Fr´echet [92], Mazur-Orlicz [222], Highberg [141], Hille-Phillips [142], Hyers [152].

454

Chapter 15. Convex Functions of Higher Orders

Exercises 1. Let D ⊂ R be an open interval, and let f : D → R be a continuous p-convex function, p > 1. Show that for every r, 1  r  p − 1, there exist at most r + 1 points c1 , . . . , cr+1 ∈ D, c1 < · · · < cr+1 , such that, if D = (a, b), c0 = a, cr+2 = b, then the function f is either (p − r)-convex or (p − r)-concave in each interval (ci , ci+1 ), i = 0, . . . , r + 1. (The statement is true also for r = p, if under 0-convex [0-concave] function we understand increasing [decreasing] function.) 2. Let D ⊂ R be an open interval, and let fn : D → R, n ∈ N, be continuous p-convex functions. Show that if the sequence {fn }n∈N converges pointwise on a dense subset of D, then it converges uniformly on every compact subset of D, and its limit is a continuous p-convex function. 3. Let D ⊂ R be an open interval, and let f : D → R be a continuous p-convex function. Without using the results of 15.8 show that f satisfies a Lipschitz condition on every compact subinterval of D. 4. Let Fk : RkN → R be symmetric k-additive functions, k = 0, . . . , p, and let f0 , . . . , fp be the diagonalizations of F0 , . . . , Fp , respectively. Show that if the p  fk (x) is continuous in RN , then every function Fk , k = 0, . . . , p, is function k=0

continuous in RkN .

Chapter 16

Subadditive Functions 16.1 General properties The Jensen inequality (5.3.1) is not the natural counterpart of the Cauchy equation (5.2.1). The natural counterpart of the Cauchy equation would be the inequality f (x + y)  f (x) + f (y).

(16.1.1)

It seems so the more interesting and astounding that the convex functions and the additive functions share so many properties that such is not the case of additive functions and functions satisfying inequality (16.1.1). Functions1 f : RN → [−∞, +∞] satisfying (16.1.1) for all x, y ∈ RN are called subadditive. Functions satisfying the converse inequality f (x + y)  f (x) + f (y)

(16.1.2)

for all x, y ∈ RN are called superadditive. Every additive function is, in particular, subadditive and superadditive2 . Thus there exist non-measurable subadditive functions. Many less trivial examples of subadditive functions may be found in Rosenbaum [269]. Subadditive functions (and superadditive functions) may also be considered which are defined not necessarily on the whole RN , but on subsets C of RN such that x + y ∈ C whenever x, y ∈ C, i.e., C + C ⊂ C.

(16.1.3)

According to Lemma 4.5.2, if a non-empty set C ⊂ RN fulfils (16.1.3), then (C, +) is semigroup a subsemigroup of (RN , +) . An example of a set C ⊂ RN fulfilling 1 So we admit also infinite values of f . Hereby it is understood that the expression f (x) + f (y) on the right hand side of (16.1.1) is always meaningful. Therefore a subadditive function cannot have both +∞ and −∞ in its range. For if we had f (x) = +∞ and f (y) = −∞ for some x, y in the domain of f , then f (x) + f (y) = ∞ − ∞ would be meaningless. 2 The converse is also true: a function f : RN → R which is both subadditive and superadditive is additive.

456

Chapter 16. Subadditive Functions

(16.1.3) is a cone3 as defined in 4.1. If C ⊂ RN is a set fulfilling (16.1.3) and f : C → [−∞, ∞] is a function satisfying inequality (16.1.1) resp. (16.1.2) for all x, y ∈ C, then f also will be called subadditive resp. superadditive. Subadditive functions defined on sets C other than RN will only briefly be mentioned in the present work. A function f : RN → R which is subadditive and satisfies f (nx) = nf (x)

for n ∈ N, x ∈ RN ,

(16.1.4)

will be called sublinear (Berz [31]). Sublinear functions are considered in 16.4–16.5. Except in the last section we will be mainly concerned with subadditive functions f : C → R, i.e., with functions with only finite values. Subadditive functions admitting also infinite values will be referred to as infinitary (cf. Rosenbaum [269], Hille-Phillips [142]). Such functions will be treated in the last section. First let us note the following two obvious lemmas. Lemma 16.1.1. If C ⊂ RN is a set fulfilling (16.1.3), f1 , f2 : C → [−∞, ∞] are subadditive functions, and a, b ∈ [0, ∞) are non-negative constants, then the function f = af1 + bf2 , if well defined, is subadditive. Lemma 16.1.2. If C ⊂ RN is a set fulfilling (16.1.3), and f : C → [−∞, ∞] is a subadditive function, then the function −f is superadditive, and conversely, if f is superadditive, then −f is subadditive. Because of Lemma 16.1.2 we will not deal with superadditive functions. In the present section we prove some elementary general properties of subadditive functions (Hille-Phillips [142], Cooper [49]). Lemma 16.1.3. Let C ⊂ RN be a set fulfilling (16.1.3) such that 0 ∈ C, and let f : C → R be a subadditive function. Then f (0)  0. Proof. We have from (16.1.1) for x = y = 0 f (0)  f (0) + f (0), whence f (0)  0.



Lemma 16.1.4. Let C ⊂ RN be a cone, and let f : C → [−∞, ∞] be a subadditive function. Then f (nx)  nf (x), (16.1.5)   1 1 f x  f (x) (16.1.6) n n for all x ∈ C and n ∈ N. All theorems in the present chapter where f is supposed to be defined on a cone C ⊂ RN remain valid if C is replaced by C0 = C \ {0}.

3

16.1. General properties

457

Proof. For n = 1 (16.1.5) is trivial. Assume (16.1.5) to hold for an n ∈ N and for all x ∈ C. Then by (16.1.1) f ((n + 1)x) = f (nx + x)  f (nx) + f (x)  nf (x) + f (x) = (n + 1)f (x). It follows by induction that (16.1.6) holds for all n ∈ N and x ∈ C. Replacing in 1 (16.1.5) x by x we obtain (16.1.6).  n Lemma 16.1.5. Let f : RN → R be a subadditive function. Then f (−x)  −f (x) for x ∈ RN . Proof. We have by (16.1.1) and Lemma 16.1.3 for arbitrary x ∈ RN 0  f (0) = f (x − x)  f (x) + f (−x), whence f (−x)  −f (x).



Lemma 16.1.6. Let C ⊂ RN be a set fulfilling (16.1.3), and let fα : C → [−∞, +∞], α ∈ A, be a family of subadditive functions. Let, for x ∈ C, f (x) = supα fα (x). If f (x) > −∞ for all x ∈ C, or f (x) < ∞ for all x ∈ C, then f is subadditive. Proof. Fix arbitrary x, y ∈ C. By the subadditivity of fα and the definition of f we have fα (x + y)  fα (x) + fα (y)  f (x) + f (y) for all α ∈ A, whence

fα (x + y)  f (x) + f (y)

for all α ∈ A. That the supremum over α we obtain (16.1.1).



Lemma 16.1.7. Let C ⊂ RN be a set fulfilling (16.1.3), and let fn : C → [−∞, +∞], n ∈ N, be subadditve functions. Assume that for every x ∈ C there exists a (finite or infinite) limit f (x) = lim fn (x). If f (x) > −∞ for all x ∈ C, or f (x) < ∞ for all n→∞ x ∈ C, then f is subadditive. Proof. We have for all x, y ∈ C fn (x + y)  fn (x) + fn (y), whence (16.1.1) follows on letting n → ∞.

n ∈ N, 

Lemma 16.1.8. Let f : RN → R be a subadditive function. If f is even, then f (x)  0 for x ∈ RN . Proof. By Lemma 16.1.5 we have for every x ∈ RN f (x) = f (−x)  −f (x), i.e., 2f (x)  0, whence also f (x)  0.



458

Chapter 16. Subadditive Functions

Lemma 16.1.9. Let f : RN → R be a subadditive function. If f is odd, then f is additive. Proof. By (16.1.1) f (x) = f (x + y − y)  f (x + y) + f (−y) = f (x + y) − f (y) for arbitrary x, y ∈ RN , since, due to the fact that f is odd, we have f (−y) = −f (y). Hence f (x) + f (y)  f (x + y). 

This together with (16.1.1) yields (5.2.1). Lemma 16.1.10. Let f : R

N

→ R be a sublinear function. Then f (0) = 0

(16.1.7)

f (λx) = λf (x)

(16.1.8)

and for all x ∈ R

N

and λ ∈ Q ∩ [0, ∞).

Proof. Let λ ∈ Q ∩ (0, ∞), λ = p/q, p, q ∈ N. Then we have by (16.1.4), for arbitrary x ∈ RN , pf (x) = f (px) = f (qλx) = qf (λx), whence (16.1.8) follows. In particular, for λ = 2 and x = 0, we obtain f (0) = 2f (0), which implies (16.1.7). Thus (16.1.8) is valid also for λ = 0, which completes the proof.  Lemma 16.1.11. Let f : RN → R be a sublinear function. Then f is convex. Proof. We have by (16.1.1) and Lemma 16.1.10, for arbitrary x, y ∈ RN ,  

f (x) + f (y) 1 1 x+y = f (x + y)  f (x) + f (y) = , f 2 2 2 2 

which means that f is convex.

16.2 Boundedness. Continuity For a function f : RN → R the functions mf : RN → [−∞ ∞) and Mf : RN → (−∞, +∞] are defined by (6.3.1) and (6.3.2), respectively. We always have mf (x)  f (x)  Mf (x),

x ∈ RN ,

(16.2.1)

and f is continuous at a point x if and only if mf (x) = Mf (x) (cf. Exercise 6.8). In this section K(t, r) denotes the open ball centered at t and with the radius r. Lemma 16.2.1. Let f : RN → R be a subadditive function. Then the functions mf and Mf also are subadditive.

16.2. Boundedness. Continuity

459

Proof. We will prove the subadditivity of mf , the proof for Mf is similar. Take an r > 0, arbitrary x, y ∈ RN , and a, b ∈ R such that a > inf f,

b > inf f.

K(x,r)

K(y,r)

There exist u ∈ K(x, r) and v ∈ K(y, r) such that b > f (v)  inf f.

a > f (u)  inf f, K(x,r)

K(y,r)

Then we have u + v ∈ K(x + y, 2r), whence inf K(x+y,2r)

f  f (u + v)  f (u) + f (v) < a + b.

When a → inf f and b → inf f , we obtain hence K(x,r)

K(y,r)

inf K(x+y,2r)

Letting r → 0, we get

f  inf f + inf f. K(x,r)

K(y,r)

mf (x + y)  mf (x) + mf (y), 

i.e., mf is subadditive. Lemma 16.2.2. Let f : RN → R be subadditive. Then, for every x ∈ RN , 0  Mf (x) − mf (x)  Mf (0).

(16.2.2)

Proof. The first inequality in (16.2.2) results from (16.2.1). Now take an arbitrary r > 0, an arbitrary x ∈ RN , and arbitrary a, b ∈ R such that a > inf f,

b < sup f.

K(x,r)

K(x,r)

Then there exist u, v ∈ K(x, r) such that b < f (u)  sup f.

a > f (u)  inf f, K(x,r)

K(x,r)

Hence b − a < f (v) − f (u). On the other hand, f (v) = f (u+v−u)  f (u)+f (v−u), whence f (v)−f (u)  f (v−u). Thus b − a < f (v − u). But v − u ∈ K(0, 2r), whence f (v − u)  sup f and K(0,2r)

b − a < sup f. K(0,2r)

Letting a tend to inf f , and b tend to sup f , we obtain hence K(x,r)

K(x,r)

sup f − inf f  sup f. K(x,r)

K(x,r)

K(0,2r)

When r → 0, we get hence the second inequality in (16.2.2).



460

Chapter 16. Subadditive Functions

Theorem 16.2.1. Let f : RN → R be subadditive. If f is continuous at zero and f (0) = 0, then f is continuous in RN . Proof. The continuity of f at zero implies in view of (16.2.1) that mf (0) = Mf (0) = f (0) = 0. Condition (16.2.2) then yields Mf (x) = mf (x) for every x ∈ RN , i.e., f is continuous in RN .  Theorem 16.2.2. Let f : RN → R be subadditive. If f is locally bounded above at a point, then f is locally bounded at every point of RN . Proof. Let f be locally bounded above at a point t0 ∈ RN . Suppose that f is not locally bounded above at 0. Then there exists a sequence {tn }n∈N such that lim tn = 0 and lim f (tn ) = ∞. But then lim (t0 + tn ) = t0 , and by (16.1.1) n→∞

n→∞

n→∞

f (tn ) = f (t0 + tn − t0 )  f (t0 + tn ) + f (−t0 ), i.e., f (t0 + tn )  f (tn ) − f (−t0 ), whence lim f (t0 + tn ) = ∞. But this means that n→∞ f is locally unbounded above at t0 , contrary to the hypothesis. Consequently f is locally bounded above at 0 so that Mf (0) < ∞. By Lemma 16.2.2 Mf (x) − mf (x) < ∞, which implies that Mf (x) < ∞ and mf (x) > −∞, for every x ∈ RN . Consequently f is locally bounded (above and below) at every point x ∈ RN .  Theorem 16.2.2 is fully analogous to the situation for convex functions (cf. Theorem 6.2.3). But now the local boundedness of f does not imply its continuity, as may be seen from the following Example 16.2.1. Let QN ⊂ RN be the set of those x ∈ RN which have rational coordinates, and let the function f : RN → R be given by  0 if x ∈ QN , f (x) = b if x ∈ / QN , where b > 0. Then f is subadditive and bounded on RN , but discontinuous at every point of RN . The condition of local boundedness above in Theorem 16.2.2 may still be weakened. Theorem 16.2.3. Let f : RN → R be subadditive. If f is bounded above on a set T ⊂ RN such that mi (T ) > 0, or T is of the second category and with the Baire property, then f is locally bounded at every point of RN . Proof. If t ∈ T + T , then t = x + y, where x, y ∈ T . Hence f (t) = f (x + y)  f (x) + f (y)

(16.2.3)

so that f is bounded above on T + T . By Theorems 3.7.1 and 2.9.1 int(T + T ) = ∅. The theorem results now from Theorem 16.2.2. 

16.2. Boundedness. Continuity

461

The operation Sn is defined in 9.2. Theorem 16.2.4. Let f : RN → R be subadditive. If f is bounded above on a set T ⊂ RN such that for a certain n ∈ N we have mi Sn (T ) > 0 or Sn (T ) is of the second category and with the Baire property, then f is locally bounded at every point of RN . Proof. By a repeated use of (16.2.3) we find that f is bounded above on Sn (T ), and the theorem follows from Theorem 16.2.3.  If f : RN → R is a measurable subadditive function, then f is bounded above on a set T ⊂ RN of positive measure (cf. the proof of Theorem 9.4.1), and hence is locally bounded at every point of RN . But actually we can prove this result for functions whose domain of definition is not necessarily the whole RN . Lemma 16.2.3. Let C ⊂ RN be a cone such that int C = ∅, and let f : C → [−∞, ∞) be a measurable subadditive function. Then f is locally bounded above at every point of int C. Proof. Suppose that f is not locally bounded above at a point t0 ∈ int C. We can find an r > 0 such that K(t0 , 2r) ⊂ int C. Write D = K(0, r) ∩ int C.

(16.2.4)

C is a convex set, so we have by Theorem 5.1.6 clintC = clC, and 0 ∈ C ⊂ clC = clintC. Consequently D = ∅. On the other hand evidently D is open, whence m(D) > 0. There exists a sequence {tn }n∈N converging to t0 and such that f (tn )  2n,

n ∈ N.

(16.2.5)

For every n ∈ N define the set En as En = {t ∈ K(t0 , 2r) ∪ D | f (t)  n}. Since f is measurable, also En is measurable for every n ∈ N. For n sufficiently large, say n > n0 , we have tn ∈ K(t0 , r). Take an n > n0 , an x ∈ D\En , and put yn = tn −x. We have by (16.2.4) |yn − t0 | = |tn − x − t0 |  |tn − t0 | + |x| < 2r, and so yn ∈ K(t0 , 2r). Moreover, by (16.1.1) and (16.2.5) 2n  f (tn ) = f (tn − x + x)  f (tn − x) + f (x) = f (yn ) + f (x).

(16.2.6)

Since x ∈ / En , we have f (x) < n, whence f (yn ) > n. Consequently yn ∈ En , i.e., x = tn −yn ∈ tn −En . Thus, if x ∈ D, then for n > n0 either x ∈ En , or x ∈ tn −En , whence D ⊂ En ∪ (tn − En ),

n > n0 ,

462

Chapter 16. Subadditive Functions

and by Corollary 3.2.2 m(D)  m[En ∪ (tn − En )]  m(En ) + m(tn − En ) = 2m(En ), Hence

n > n0 .

1 m(D), n > n0 , 2 ⊂ En ⊂ K(t0 , 2r) ∪ K(0, r) so that m(En ) < ∞,

m(En ) 

and (cf. Exercise 3.4), since En+1 n ∈ N,  ∞  1 m En = lim m(En )  m(D) > 0. n→∞ 2 n=1 Thus

∞

En = ∅, and if an x0 ∈

n=1

∞

En , then x0 ∈ En for all n ∈ N, and f (x0 )  n

n=1

for all n ∈ N, which contradicts the fact that f < ∞ in C.



Theorem 16.2.5. Let C ⊂ R be a cone such that int C = ∅, and let f : C → [−∞, ∞) be a measurable subadditive function. Then f is locally bounded at every point of int C. N

Proof. By Lemma 16.2.3 f is locally bounded above at every point of int C. Now suppose that f is not locally bounded below at a point t0 ∈ int C. Then there exists a sequence {tn }n∈N such that lim tn = t0 and limn→∞ f (tn ) = −∞. Let U ⊂ int C n→∞ be a neighbourhood of t0 such that f is bounded above on U : f (x)  M

for x ∈ U.

There exists an n0 ∈ N such that 2t0 − tn ∈ U for n > n0 . Thus we have by (16.1.1) for n > n0 f (2t0 ) = f (tn + 2t0 − tn )  f (tn ) + f (2t0 − tn )  f (tn ) + M, whence −f (tn )  M − f (2t0 ), and

∞ = lim − f (tn )  M − f (2t0 ) < ∞. n→∞



The contradiction obtained completes the proof.

or bounded Theorem 16.2.6. Let f : RN → R be subadditive. If f is either measurable,   above on a set T ⊂ RN such that for a certain n ∈ N either mi Sn (T ) > 0, or Sn (T ) is of the second category and with the Baire property, then f is bounded on every bounded subset of RN . Proof. By Theorems 16.2.4 and 16.2.5 f is locally bounded at every point of RN . For every x ∈ RN let Kx be an open ball centered at x such that f is bounded on  Kx . Let A ⊂ RN be a bounded set. Then the set cl A is compact. Clearly cl A ⊂ Kx , so x∈cl A

there exists a finite sequence of points x1 , . . . , xn ∈ cl A such that A ⊂ cl A ⊂ The function f , being bounded on every Kxi , i = 1, . . . , n, is also bounded on and hence on A.

n 

i=1 n  i=1

K xi . K xi , 

16.2. Boundedness. Continuity

463

Theorem 16.2.7. Let f : RN → [−∞, ∞] be subadditive, and put λ = lim inf f (x). x→0

Then λ  0, or λ = −∞. If |λ| = ∞, then f is either non-measurable, or infinitary. Proof. There exists a sequence {xn }n∈N of points xn ∈ RN such that lim xn = 0

n→∞

and

lim f (xn ) = λ.

n→∞

By Lemma 16.1.4 λ  lim inf f (2xn )  lim inf 2f (xn ) = 2 lim inf f (xn ) = 2λ. n→∞

n→∞

n→∞

If λ is finite, this yields λ  0. If f is measurable and finite, f : RN → R, then, by Theorem 16.2.5, f is locally bounded at 0, whence λ = lim inf f (x) = ±∞. Hence if f is measurable and |λ| = ∞, then f must be infinitary.

x→0



For subadditive functions assuming negative values we have the following Theorem 16.2.8. Let f : RN → R be a measurable subadditive function, and suppose that there exists an x0 ∈ RN such that f (x0 ) < 0. Then f (tx0 ) < 0 for all t ∈ R sufficiently large, and f (tx0 )  0 for all t  0. Proof. By Theorem 16.2.6 there exists an M > 0 such that |f (tx0 )|  M

for t ∈ [1, 2].

(16.2.7)

Now take an arbitrary t > 2, and let n = [t] be the integral part of t so that n  t < n + 1. Hence t − (n − 1) ∈ [1, 2]. Thus by (16.1.1), (16.2.7), and by Lemma 16.1.4       f (tx0 ) = f (n − 1)x0 + (t − n + 1)x0  f (n − 1)x0 + f (t − n + 1)x0  (n − 1)f (x0 ) + M = (n + 1)f (x0 ) + M − 2f (x0 )  tf (x0 ) + M − 2f (x0 ). Since f (x0 ) < 0, f (tx0 ) becomes negative for large values of t > 0. By Lemma 16.1.5 f (−tx0 )  −f (tx0 ), whence f (tx0 ) > 0 for large negative values of t. Suppose that there exists a t0 < 0 such that f (t0 x0 )  0. By Lemma 16.1.4 f (nt0 x0 )  nf (t0 x0 )  0

for n ∈ N,

which contradicts the fact (already established) that f (tx0 ) > 0 for large negative t. Thus f (tx0 ) > 0 for t < 0. Together with Lemma 16.1.3 this yields f (tx0 )  0 for t  0.  This proof shows also that, for large values of t, the function f (tx0 ) is majorized by a linear function of t. A better information about  the behavior of f (tx0 ) for t → ∞  without any assumption about the sign of f (x0 ) is obtained from the following Theorem 16.2.9. Let f : RN → R be a measurable subadditive function. Then for every x ∈ RN there exists the limit F (x) = lim

t→∞

f (tx) . t

(16.2.8)

The function F is finite, continuous in RN , positively homogeneous, and subadditive4 . 4

And hence sublinear.

464

Chapter 16. Subadditive Functions

Proof. If x = 0, then F (0) = 0 exists and is finite. Now fix an x ∈ RN , x = 0, and put f (tx) β = inf . t>0 t For every t > 0 the expression f (tx)/t is finite, whence −∞  β < ∞. Take an arbitrary b > β and choose a t0 > 0 such that f (t0 x)/t0 < b. Take an arbitrary t > 3t0 and let m = [t/t0 ] be the integral part of t/t0 . Put n = m − 2. Clearly n ∈ N, and by Theorem 16.2.6 f is bounded on the segment 2t0 x, 3t0 x: |f (sx)|  M

for s ∈ [2t0 , 3t0 ].

In particular, t − nt0 ∈ [2t0 , 3t0 ]. Hence, by Lemma 16.1.4     f (tx) = f (nt0 + t − nt0 )x  f (nt0 x) + f (t − nt0 )x  nf (t0 x) + M, and β

nt0 f (t0 x) M nt0 M f (tx)  < b+ . + t t t0 t t t

(16.2.9)

When n → ∞, the expression nt0 /t approaches5 1, and we get from (16.2.9) β  lim inf t→∞

f (tx) f (tx)  lim sup  b. t t t→∞

Letting b → β we obtain hence in view of (16.2.8) F (x) = β < ∞.

(16.2.10)

Now put α = sup t<0

f (tx) . t

We have −∞ < α  ∞. But

  f t(−x) α = − inf = −F (−x) . t>0 t

So −F (−x) > −∞.

(16.2.11)

By Lemma 16.1.5 we have f (tx) + f (−tx)  0, whence   f (tx) f (−tx) F (x) + F (−x) = lim +  0. t→∞ t t Consequently F (x)  −F (−x) > −∞, by (16.2.11). Hence by (16.2.10) |F (x)| < ∞, i.e., F is finite. 5

This follows from the inequalities n + 2  t/t0 < n + 3, whence

nt0 n n <  . n+3 t n+2

16.3. Differentiability

465

Take arbitrary x, y ∈ RN . We have for all t > 0   f t(x + y) = f (tx + ty)  f (tx) + f (ty), whence

    f t(x + y) f (tx) f (ty)  lim + = F (x) + F (y). F (x + y) = lim t→∞ t→∞ t t t

Thus F is subadditive. Similarly, for arbitrary x ∈ RN and u > 0 we have F (ux) = lim

t→∞

f (tux) f (tux) = u lim = uF (x). t→∞ t tu

So F is positively homogeneous, and, in particular, sublinear. By Lemma 16.1.11 F is convex. Since f is measurable, for every fixed t > 0 the function f (tx)/t (as a function of x) is measurable. Consequently also function (16.2.8) is measurable. By Theorem 9.4.2 F is continuous in RN .  Theorem 16.2.9 (Rosenbaum [269], Hille-Phillips [142]) shows that a subadditive function cannot increase too fast as x → ∞. If f is defined only on a cone C ⊂ RN , the theorem, and the argument presented, remain valid, except that now we cannot prove that F (x) > −∞. In such a cone it can actually happen that F (x) = −∞. Let N = 1, C = [0, ∞), let g : C → R be an arbitrary increasing positive function, and let f (x) = −xg(x) for x ∈ C. Then f is subadditive, and by a suitable choice of g we can make f to tend to −∞, when x → ∞, as fast as we wish. In particular, if lim g(x) = ∞, then for our f function (16.2.8) is identically −∞. x→∞

16.3 Differentiability Theorem 16.2.1 shows that in the theory of subadditive functions the origin plays a distinguished role. In particular, if f (0) = 0, then we can infer from the continuity of f at zero about its continuity at any other point. With the differentiability the situation is similar (cf., in particular, Wetzel [322]). Theorem 16.3.1. Let f : RN → R be subadditive, and let g : RN → R be a function such that g(0) = 0 and g has the Stolz differential at 0. If f (x)  g(x),

x ∈ RN ,

(16.3.1)

then there exists a c ∈ RN such that f (x) = cx,

x ∈ RN .

(16.3.2)

In particular, f is of class C 1 in RN . Proof. Put c = ∇g(0). Then g(x) = cx + r(x),

x ∈ RN ,

(16.3.3)

466

Chapter 16. Subadditive Functions

where lim

x→0

r(x) =0 |x|

(16.3.4)

(cf. 7.7). Now take an arbitrary x ∈ RN , x = 0. We have for n ∈ N by Lemma 16.1.4 and by (16.3.1) and (16.3.3) f (x)/n  f (x/n)  g(x/n) = cx/n + r(x/n), whence f (x) − cx  nr(x/n).

(16.3.5)

By (16.3.4) lim nr(x/n) = |x| lim

n→∞

n→∞

r(x/n) = 0, |x/n|

and so we obtain from (16.3.5), on letting n → ∞, f (x)  cx.

(16.3.6)

cx = −c(−x)  −f (−x)  f (x).

(16.3.7)

Hence and by Lemma 16.1.5

From (16.3.6) and (16.3.7) we obtain (16.3.2).



As an immediate consequence of Theorem 16.3.1 we obtain the following Theorem 16.3.2. Let f : RN → R be subadditive. If f (0) = 0 and f has the Stolz differential at zero, then there exists a c ∈ RN such that (16.3.2) holds. In particular, f is additive and of class C 1 in RN . Proof. Results from Theorem 16.3.1 on taking g = f .



The following example shows that the assumption f (0) = 0 in Theorem 16.3.2 is essential. Example 16.3.1. Let N = 1, and let f : R → R be defined by  1 for x  0, f (x) = −x2 e for x > 0. If x, y  0, then also x + y  0 and f (x + y) = 1 < 1 + 1 = f (x) + f (y). If x  0, 2 y > 0, and x + y  0, then f (x + y) = 1 < 1 + e−y = f (x) + f (y). Similarly, if x > 0, y  0, and x + y  0, then f (x + y) < f (x) + f (y). If x > 0, y  0, and 2 2 x + y > 0, then f (x + y) = e−(x+y)  1 < e−x + 1 = f (x) + f (y). Similarly if x  0, y > 0, and x + y > 0, then f (x + y) < f (x) + f (y). Finally, if x > 0 and y > 0, then 2 2 2 2 also x + y > 0, and f (x + y) = e−(x+y) < e−x < e−x + e−y = f (x) + f (y). So f is subadditive, and f is differentiable6 at zero, but f has not form (16.3.2) for any c ∈ R. 6

If N = 1, then the differentiability of a function at a point, and the existence of the Stolz differential at this point, are equivalent.

16.3. Differentiability

467

In the sequel of this section we assume that N = 1. Theorem 16.3.3. Let f : R → R be a measurable subadditive function, and let A = inf

t<0

f (t) , t

B = sup t>0

f (t) . t

(16.3.8)

If A resp. B is finite, then f (h) resp. h→0− h

A = lim

f (h) . h→0+ h

B = lim

(16.3.9)

Formulas (16.3.9) remain valid for A and/or B infinite under the additional assumption that lim f (x) = 0, or lim inf f (x) > 0. Moreover, in every case, x→0

x→0

AB

(16.3.10)

Proof. We prove only the first equality in (16.3.9), the proof of the other one is similar. Evidently we have −∞  A < ∞. Assume that A is finite, or A = −∞ and lim f (x) = 0.

(16.3.11)

x→0−

Suppose that A is finite and (16.3.11) does not hold. By Theorem 16.2.7 lim sup f (x)  lim inf f (x)  lim inf f (x)  0. x→0−

x→0−

x→0

If we had lim sup f (x) = 0, then the above inequalities yield lim inf f (x) = 0, whence x→0−

x→0−

(16.3.11) follows contrary to the supposition. Thus we must have lim sup f (x) > 0, x→0−

which means that there exists a sequence {xn }n∈N such that xn < 0 for n ∈ N, lim xn = 0, and there exists an α > 0 such that f (xn ) > α for n ∈ N. Then n→∞

f (xn ) α < → −∞, xn xn which shows that A = −∞. Consequently, if A is finite, relation (16.3.11) must hold. Thus we have (16.3.11) in either case. Now take an arbitrary a > A. There exists a point x0 < 0 such that f (x0 ) < a, x0 i.e., f (x0 ) > ax0 . Choose an arbitrary h such that x0 < h < 0, and let n = [x0 /h] be the integral part of x0 /h so that n  x0 /h < n + 1, whence nh  x0 > (n + 1)h. Put δ = x0 − nh so that 0  δ > h. (16.3.12) Further, we have x0 = nh + δ. Hence by Lemma 16.1.4 ax0 < f (x0 ) = f (nh + δ)  f (nh) + f (δ)  nf (h) + f (δ),

468

Chapter 16. Subadditive Functions

whence, since x0 < 0, a>

nf (h) f (δ) nh f (h) f (δ) f (x0 ) +  + = . x0 x0 x0 x0 h x0

We have f (h)/h  A and nh/x0 > 0, whence a>

nh f (δ) nh f (h) f (δ) +  A+ . x0 h x0 x0 x0

(16.3.13)

Now we let h → 0−. Then, by (16.3.12), also δ → 0−, and by (16.3.11) f (δ) → 0. On the other hand, nh/x0 = 1 − δ/x0 → 1. Thus we get by (16.3.13) a  lim sup h→0−

f (h) f (h)  lim inf  A. h→0− h h

Letting a tend to A, we obtain the first equality in (16.3.9). If lim inf f (x) > 0 (which implies A = −∞), then there exists an α > 0 such x→0

that f (x) > α in a neighbourhood of zero. Then also f (h) = −∞ = A. h→0− h lim

The second equality in (16.3.9) is established similarly. By Lemma 16.1.5 −f (−x)  f (x), whence for x > 0 f (x) f (−x)  , −x x and A = inf

x<0

f (x) f (−x) f (−x) f (x) = inf  sup  sup = B. x>0 −x x −x x>0 x>0 x 

This proves relation (16.3.10).

Theorem 16.3.3 may be formulated that for a measurable subadditive function f : R → R we have lim

h→0+

f (h) f (h) = sup , h h>0 h

lim

h→0−

f (h) f (h) = inf h<0 h h

provided that these quantities are finite (or, if they are infinite, provided that some further conditions concerning the behaviour of f near 0 are fulfilled). On the other hand, we have shown in the proof of Theorem 16.2.9 that f (h) f (h) = inf , h→∞ h h>0 h lim

f (h) f (h) = sup , h→−∞ h h<0 h lim

without any further assumptions except the measurability and subadditivity of f . In the next theorem there occur the Dini derivatives (cf. 7.5).

16.3. Differentiability

469

Theorem 16.3.4. Let f : R → R be a measurable subadditive function. Then, with notation (16.3.8), we have for every x ∈ R D+ f (x)  B,

D− f (x)  B,

d+ f (x)  A,

d− f (x)  A.

(16.3.14)

In particular, if A = B, then f (x) = Ax

for

x ∈ R.

(16.3.15)

Proof. If B = ∞, then the first two inequalities in (16.3.14) are trivial. Similarly, if A = −∞, then the last two inequalities in (16.3.14) are trivial. Suppose that B < ∞, whence by (16.3.8) B is finite. We have for arbitrary x ∈ R and h > 0 f (x + h)  f (x) + f (h), whence

f (x + h) − f (x) f (h)  , h h

(16.3.16)

and by Theorem 16.3.3 D+ f (x) = lim sup h→0+

f (x + h) − f (x) f (h)  lim = B. h→0+ h h

Similarly, since f (x)  f (h + x − h)  f (h) + f (x − h), we get for h > 0

f (x) − f (x − h) f (h)  , h h

(16.3.17)

and by Theorem 16.3.3 f (x + h) − f (x) f (x) − f (x − h) f (h) = lim sup  lim = B. h→0+ h h h h→0− h→0+   Now let A > −∞ and so, by (16.3.8), A is finite . For h < 0 the inequalities in (16.3.16) and (16.3.17) are reversed, whence by Theorem 16.3.3 D− f (x) = lim sup

d+ f (x) = lim inf h→0+

f (x + h) − f (x) f (x) − f (x − h) f (h) = lim inf  lim = A, h→0− h→0− h h h

and d− f (x) = lim inf h→0−

f (x + h) − f (x) f (h)  lim = A. h→0− h h

If A = B, then both these quantities are finite, and f  (x) = A for all x ∈ R, whence f (x) = Ax + b, with a certain b ∈ R. If we had b = 0, then   f (h) b = lim A + A = lim h→0− h h→0− h would be infinite. Consequently b = 0, and we obtain formula (16.3.15).



470

Chapter 16. Subadditive Functions

Theorem 16.3.5. Let f : R → R be a measurable subadditive function. If, with notation (16.3.8), A and B are finite, then f fulfils in R a Lipschitz condition.   Proof. Take an L > max |A|, |B| . It follows from Theorem 16.3.4 that for every x ∈ R there exists an rx > 0 such that # # # f (x) − f (y) # # #
Since the interval [x, y] is compact, we can choose a finite cover [x, y] ⊂

n 

(ti − rti , ti + rti ).

i=1

We may assume that x, y occur in the sequence {t1 , . . . , tn }. Suppose that x = t1 . If we had n  (x − rx , x + rx ) ∩ (ti − rti , ti + rti ) = ∅, i=2

the interval [x, y] would be disconnected. So in the sequence {t2 , . . . , tn } there is tj = x1 such that (we write rx1 = r1 ) (x − rx , x + rx ) ∩ (x1 − r1 , x1 + r1 ) = ∅. Proceeding in this way, and writing x = x0 , we may find a sequence x = x0 < x1 < · · · < xn = y such that (with the notation rxi = ri , i = 0, . . . , n) (xi − ri , xi + ri ) ∩ (xi+1 − ri+1 , xi+1 + ri+1 ) = ∅

for i = 0, . . . , n − 1.

Choose points y0 , . . . , yn−1 so that yi ∈ (xi − ri , xi + ri ) ∩ (xi+1 − ri+1 , xi+1 + ri+1 ),

i = 0, . . . , n − 1,

and x = x0 < y0 < x1 < y1 < · · · < yn−1 < xn = y. Then we have by (16.3.18) |f (y) − f (x)| 

n−1

|f (xi+1 ) − f (yi )| +

i=0 n−1

L

i=0

|xi+1 − yi | + L

n−1

|f (yi ) − f (xi )|

i=0 n−1

|yi − xi | = L|y − x|.

i=0

This means that f fulfils in R the Lipschitz condition with the constant L. (For x = y the inequality is trivial, and if y < x, we reverse the roles of x and y). 

16.4. Sublinear functions

471

Theorem 16.3.6. Let f : R → R be a measurable subadditive function. If, with notation (16.3.8), A and B are finite, then f is absolutely continuous in R. 

Proof. Results from Theorem 16.3.5.

16.4 Sublinear functions These are functions f : RN → R fulfilling (16.1.1) and (16.1.4). As we will show, there is a close connection between sublinear functions and additive functions (Berz [31]). Lemma 16.4.1. Let f : RN → R be a sublinear function, and let x0 ∈ RN . There exists an additive function g : RN → R such that g(x0 ) = f (x0 )

(16.4.1)

and g(x)  f (x)

for x ∈ RN .

(16.4.2)

Proof. Let L = Qx0 . L is a linear subspace of RN (over Q). We define the function g0 : L → R by g0 (x) = g0 (λx0 ) = λf (x0 ), λ ∈ Q. (16.4.3) By Lemma 16.1.10, if λ  0, we have g0 (x) = g0 (λx0 ) = λf (x0 ) = f (λx0 ) = f (x). If λ < 0, then by Lemmas 16.1.5 and 16.1.10 g0 (x) = g0 (λx0 ) = λf (x0 ) = −|λ|f (x0 )  |λ|f (−x0 ) = f (−|λ|x0 ) = f (x). Thus g0 (x)  f (x)

for x ∈ L.

Now, by Lemma 16.1.11 the function f is convex, and so it fulfils inequality (10.1.1). The set D = RN evidently is Q-convex and Q-radial at every point. It follows from (16.4.3) that g0 is a homomorphism from L into R. By Theorem 10.1.1 there exists an additive function g : RN → R fulfilling inequality (16.4.2) and the condition g | L = g0 . Hence by (16.4.3) g(x0 ) = g0 (x0 ) = f (x0 ), 

i.e., g fulfils (16.4.1).

Theorem 16.4.1. Let f : RN → R be a sublinear function. Then, for every x ∈ RN , f (x) = sup {g(x) | g : RN → R is additive and g  f }.

(16.4.4)

Proof. Let h(x) = sup {g(x) | g : RN → R is additive and g  f }. Then, of course, h(x)  f (x)

for x ∈ RN .

(16.4.5)

On the other hand, by Lemma 16.4.1, for every x0 ∈ R there exists an additive function g : RN → R fulfilling (16.4.1) and (16.4.2). Hence N

h(x0 )  f (x0 )

for x0 ∈ RN .

Inequalities (16.4.5) and (16.4.6) yield (16.4.4).

(16.4.6) 

472

Chapter 16. Subadditive Functions

Let F be the family of all sublinear functions f : RN → R. (F , ) is an ordered space. Theorem 16.4.2. Additive functions g : RN → R are the minimal elements of (F , ). Proof. Let g : RN → R be additive, let f : RN → R be sublinear, and suppose that f  g. Hence we have for every x ∈ RN −f (−x)  −g(−x) = g(x), whence by Lemma 16.1.5 f (x)  −f (−x)  g(x), i.e., f  g. Thus f = g, which means that g is a minimal element of (F, ). On the other hand, if f is a minimal element of (F , ), then, in particular, f is sublinear, and by Lemma 16.4.1 there exists an additive function g : RN → R fulfilling (16.4.2). Since f is minimal, (16.4.2) implies that f = g, i.e., f is additive.  In the one-dimensional case N = 1 we have the following Theorem 16.4.3. Let f : R → R be a measurable sublinear function. Then there exist real constants a  b such that  ax for x  0, f (x) = (16.4.7) bx for x < 0. Conversely, every function f : R → R of form (16.4.7), where a  b, is measurable and sublinear. Proof. Since f is measurable, by Theorem 9.4.1 every additive function g : R → R fulfilling (16.4.2) is continuous, and hence (Theorem 5.5.2) of the form g(x) = cg x, x ∈ R, with a certain constant cg ∈ R. Hence, by Theorem 16.4.1, f (x) = sup {cg x | cg t  f (t) for t ∈ R}. Put a = sup {cg | cg t  f (t) for t ∈ R}, b = − sup {−cg | cg t  f (t) for t ∈ R}. Then for x  0 f (x) = sup {cg x | cg t  f (t) for t ∈ R} = x sup {cg | cg t  f (t) for t ∈ R} = ax, whereas for x < 0 f (x) = sup { −cg (−x) | cg t  f (t) for t ∈ R } = −x sup { −cg | cg t  f (t) for t ∈ R } = bx. This proves relation (16.4.7). Moreover, we have a = f (1), b = −f (−1), so the inequality a  b results from Lemma 16.1.5.

16.5. Norm

473

Conversely, let a function f : R → R be given by (16.4.7), where a  b. Then f (x) = max(ax, bx), and by Lemma 16.1.6 f is subadditive. Further, if x  0 and n ∈ N, then nx  0, and f (nx) = anx = nax = nf (x). If x < 0 and n ∈ N, then nx < 0, and f (nx) = bnx = nbx = nf (x). Consequently f satisfies (16.1.4). This means that f is sublinear. Clearly, a function of form (16.4.7) is continuous, and so the more measurable.  Theorems 16.4.1, 16.4.2, 16.4.3 are due to E. Berz [31].

16.5 Norm Subadditive functions have many interesting applications in other branches of mathematics (cf., e.g., Hille-Phillips [142]). In particular, if E is any linear space (over R), then every norm f =  ·  in E (if it exists) is a subadditive function (and even sublinear). So is also the function f = g, where g : RN → E fulfils the Cauchy equation (5.2.1). P. Fischer [82] has expressed the following conjecture: If f : (0, ∞) → R is a non-negative subadditive function such that f (λx) = λf (x)

for x ∈ (0, ∞),

λ ∈ Q ∩ (0, ∞),

(16.5.1)

then there exist a normed space E and a function g : R → E satisfying equation (5.2.1) such that f (x) = g(x) for x > 0. (16.5.2)   Whereas this conjecture is not true in general cf. Moszner [232] , we have the following (Berz [31]) Theorem 16.5.1. If f : RN → R is an even sublinear function, then there exist a normed space E (over R) and a function g : RN → E satisfying equation (5.2.1) such that f (x) = g(x)

for x ∈ RN .

(16.5.3)

Proof. Let {gα }α∈A be the family of all additive functions gα : R → R such that gα  f , indexed by indices α from a certain set A. By Theorem 10.1.2 we have |gα |  f for α ∈ A. So gα  |gα |  f for α ∈ A, whence by Theorem 16.4.1 N

f (x) = sup gα (x)  sup |gα (x)|  f (x) α∈A

α∈A

so that f (x) = sup |gα (x)|.

(16.5.4)

α∈A

Let E be the set of all families {yα }α∈A , where yα ∈ R for α ∈ A, such that supα∈A |yα | < ∞. With the operations defined in the usual manner: {yα }α∈A + {zα }α∈A = {yα + zα }α∈A , λ{yα }α∈A = {λyα }α∈A

for λ ∈ R,

474

Chapter 16. Subadditive Functions

E becomes a linear space over R. We introduce in E the norm: {yα }α∈A  = sup |yα |.

(16.5.5)

α∈A

With this norm E becomes a normed linear space. Let the function g : RN → E be defined by g(x) = {gα (x)}α∈A x ∈ RN . Then, since gα are additive g(x + y) = {gα (x + y)}α∈A = {gα (x) + gα (y)}α∈A = {gα (x)}α∈A + {gα (y)}α∈A = g(x) + g(y), i.e., g satisfies equation (5.2.1). By (16.5.4) and (16.5.5) we get (16.5.3).



Theorem 16.5.2. Let f : (0, ∞) → R be an arbitrary function. Then the following conditions are equivalent: (i) f can be extended onto R to an even sublinear function f ∗ : R → R; (ii) f fulfils condition (16.5.1) and (setting f (0) = 0)       f |x + y|  f |x| + f |y| for x, y ∈ R; (16.5.6) (iii) there exist a normed linear space E (over R) and a function g : R → E satisfying equation (5.2.1) such that (16.5.2) holds. Proof. Suppose that f can be extended onto R to an even sublinear function f ∗ : R → R. Then, by Lemma 16.1.10, f ∗ , and hence also f = f ∗ | (0, ∞), fulfils (16.5.1). ∗ Moreover, by Lemma 16.1.10, f ∗ (0) = 0, so if we put f (0)  = 0, then we have f = f | ∗ ∗ ∗ [0, ∞). Since f is even, we have f (t) = f |t| = f |t| for every t ∈ R. Hence     f |x + y| = f ∗ |x + y| = f ∗ (x + y)  f ∗ (x) + f ∗ (y)         = f ∗ |x| + f ∗ |y| = f |x| + f |y| for arbitrary x, y ∈ R, i.e., f satisfies (16.5.6).  Consequently condition (i) implies (ii). Conversely, suppose that f fulfils (ii) we set f (0) = 0 , and put   f ∗ (x) = f |x| for x ∈ R. (16.5.7) Formula (16.5.7) yields an extension of f onto R, and it is obvious that f ∗ is even. By (16.5.1) f ∗ fulfils (16.1.4). Moreover, we have by (16.5.6)       f ∗ (x + y) = f |x + y|  f |x| + f |y| = f ∗ (x) + f ∗ (y) for arbitrary x, y ∈ R. Consequently f ∗ is subadditive, and hence sublinear. Thus (i) is true, and hence conditions (i), (ii) are equivalent. If (iii) is satisfied, then the formula f ∗ (x) = g(x)

for x ∈ R

16.6. Infinitary subadditive functions

475

defines an even sublinear function f ∗ : R → R, and by (16.5.2) f ∗ | (0, ∞) = f . Consequently (iii) implies (i). The converse implication results from Theorem 16.5.1. Thus all the three conditions (i), (ii), (iii) are equivalent to each other.  Theorem 16.5.2 (Berz [31]; cf. also Moszner [232]) gives precise conditions when Fischer’s conjecture is true: a function f : (0, ∞) → R fulfilling the hypotheses of Fischer’s conjecture admits representation (16.5.2) if and only if it satisfies also condition (16.5.6).

16.6 Infinitary subadditive functions The following theorem shows that infinitary subadditive functions only exceptionally can assume the value −∞. Theorem 16.6.1. Let f : RN → [−∞, ∞] be subadditive. If there exists an x0 ∈ RN such that f (x0 ) = −∞, then f (x) = −∞ for all x ∈ RN . Proof. Since f (x0 ) = −∞, we have f = +∞ in RN , whence for arbitrary x ∈ RN f (x) = f (x − x0 + x0 )  f (x − x0 ) + f (x0 ) = f (x − x0 ) − ∞ = −∞. Hence f (x) = −∞.



Hence it follows that if f : RN → [−∞, ∞] satisfies equation (5.2.1) and assumes infinite values, then either f (x) = −∞ for all x ∈ RN , or f (x) = ∞ for all x ∈ RN (Halperin [131]). In fact, such a function is both subadditive and superadditive, and for superadditive functions Theorem 16.6.1 holds with −∞ replaced by +∞. If the domain of f is only a subset of RN , we must assume more. Theorem 16.6.2. Let C ⊂ RN be a set fulfilling (16.1.3) and let f : C → [−∞, ∞] be a subadditive function. If there exists a sequence {tn }n∈N of points tn ∈ C such that lim tn = 0 and f (tn ) = −∞ for n ∈ N, then f (x) = −∞ for all x ∈ int C.

n→∞

Proof. Let x ∈ int C. Since f assumes the value −∞, we have f = +∞ in C. Further, there exists an n ∈ N such that x − tn ∈ C. Hence f (x) = f (tn + x − tn )  f (tn ) + f (x − tn ) = −∞ + f (x − tn ) = −∞. Thus f (x) = −∞.



With the value +∞ the situation is more complicated. A subadditive function may be equal +∞ on a proper subset of RN . Example 16.6.1. Define the function f : RN → [−∞, ∞] by (cf. Example 16.2.1)  +∞ for x ∈ / QN , f (x) = b for x ∈ QN , where b > 0. Then f is subadditive and infinitary, but f (x) = ∞ on a dense subset of RN .

476

Chapter 16. Subadditive Functions

Theorem 16.6.3. Let f : RN → [−∞, ∞] be a subadditive function. If there exists an r > 0 such that |f (x)| < ∞ for |x| < r, then |f (x)| < ∞ for all x ∈ RN . Proof. Take an x ∈ RN . There exists an n ∈ N such that |x/n| = |x|/n < r. Hence f (x/n) < ∞, and by Lemma 16.1.4 f (x)  nf (x/n) < ∞.

(16.6.1)

If we had f (x) = −∞, then we would get by Theorem 16.6.1 f ≡ −∞, which is incompatible with the assumption that |f (t)| < ∞ for |t| < r. Consequently f (x) > −∞, which together with (16.6.1) implies that |f (x)| < ∞.  If f : RN → [−∞, ∞] is a subadditive function, we put Pf = {x ∈ RN | f (x) < ∞}. This definition allows that f (x) = −∞ for some x ∈ Pf . But in view of Theorem 16.6.1, unless f ≡ −∞, we have Pf = {x ∈ RN | |f (x)| < ∞}. The set Pf is characterized by the following property. Theorem 16.6.4. Let C ⊂ RN be an arbitrary set. There exists a subadditive function f : RN → [−∞, ∞] such that Pf = C if and only if C fulfils (16.1.3). Proof. If f : RN → [−∞, ∞] is subadditive, and if u ∈ Pf + Pf , then there exist x, y ∈ Pf such that u = x + y, whence f (u) = f (x + y)  f (x) + f (y) < ∞, i.e., u ∈ Pf . Hence Pf + Pf ⊂ Pf . On the other hand, if a set C ⊂ RN fulfils (16.1.3), then the function  +∞ for x ∈ / C, f (x) = 0 for x ∈ C 

is subadditive, and Pf = C.

be an open ball centered at a point x0 ∈ R . Then there ∞  exists an α > 0 such that tx0 ∈ kB for t > α. Lemma 16.6.1. Let B ⊂ R

N

N

k=1

Proof. Let r > 0 denote the radius of B, and let α ∈ N be such that αr > |x0 |. Take an arbitrary t > α, and let n = [t] be the integral part of t. Then n  α and |t − n| < 1, whence |tx0 − nx0 | = |t − n||x0 |  |x0 | < αr  nr. But this means that tx0 belongs to the open ball centered at nx0 and with the radius ∞  nr, i.e., tx0 ∈ nB ⊂ kB.  k=1

16.6. Infinitary subadditive functions

477

Theorem 16.6.5. Let f : RN → [−∞, ∞] be a measurable subadditive function. If there exists a density point x0 of Pf such that7 x0 ∈ Pf and sx0 ∈ Pf for a certain s < 0, then Pf = RN . Proof. Since f is measurable, Pf ∈ L. By Theorem 16.6.4 Pf + Pf ⊂ Pf . By Theorem 3.7.2 2x0 = x0 + x0 ∈ int(Pf + Pf ) ⊂ int Pf . Let B ⊂ RN be an open ball centered at 2x0 and contained in Pf . By Lemma 16.6.1 there exists an α > 0 ∞  such that tx0 ∈ kB for t > α. Choose an m ∈ N such that −ms > α. Then there k=1

exists an n ∈ N such that −msx0 ∈ nB. The set U = msx0 + nB is a neighbourhood of zero. If z ∈ U , then z = msx0 + nx, where x ∈ B. Hence, by Lemma 16.1.4 f (z) = f (msx0 + nx)  f (msx0 ) + f (nx)  mf (sx0 ) + nf (x) < ∞, i.e., z ∈ Pf . Consequently U ⊂ Pf and there exists an r > 0 such that x ∈ Pf whenever |x| < r. If f (x) = −∞ for an x ∈ RN , then by Theorem 16.6.1 f (x) = −∞ for every x ∈ RN , i.e., Pf = RN . If f > −∞ in RN , then |f (x)| < ∞ for |x| < r, and by Theorem 16.6.3 |f | < ∞ in RN . Consequently Pf = RN .  The same argument may be used to prove the following Theorem 16.6.6. Let f : RN → [−∞, ∞] be subadditive. If mi (Pf ) > 0, or Pf contains a set of the second category, and with the Baire property, and if there exists8 an x0 ∈ int Pf such that sx0 ∈ Pf for a certain s < 0, then Pf = RN . Theorem 16.6.7. Let f : RN → [−∞, ∞] be subadditive. If there exists an x0 ∈ RN such that f (x0 ) = ∞ and tx0 ∈ Pf for all t > 1, then f (tx0 ) = ∞ for all t < 0. Proof. Since f (x0 ) = ∞, we have f = −∞ in RN . Consequently |f (tx0 )| < ∞ for t > 1. If t < 0, then 1 − t > 1, and     ∞ = f (x0 ) = f tx0 + (1 − t)x0  f (tx0 ) + f (1 − t)x0 . #  # But since #f (1 − t)x0 # < ∞, this implies that f (tx0 ) = ∞.  Theorem 16.2.9 says that if a measurable subadditive function f : RN → R is finite-valued, then, for every x0 ∈ RN , β = inf

t>0

f (tx0 ) t

(16.6.2)

is finite (cf., in particular, formula (16.2.10)). This is no longer true if f is infinitary. But in such a case we have the following Theorem 16.6.8. Let f : RN → [−∞, ∞] be measurable and subadditive, and suppose that for a certain x0 ∈ RN we have β = −∞, where β is given by (16.6.2). If |f (tx0 )| < ∞ for all t > 0, then f (tx0 ) = ∞ for all t < 0. 7 8

The condition x0 ∈ Pf is not indispensable and can be omitted. From Theorems 3.7.1, 2.9.1 and 16.6.4 it follows that int Pf = ∅.

478

Chapter 16. Subadditive Functions

Proof. Suppose that there exists a t0 < 0 such that f (t0 x0 ) < ∞. For every t  0 there exist a t ∈ [−t0 , −2t0 ) ⊂ (0, ∞) and a k ∈ N such that t = t + kt0 . Hence by Lemma 16.1.4 f (tx0 ) = f (t x0 + kt0 x0 )  f (t x0 ) + f (kt0 x0 )  f (t x0 ) + kf (t0 x0 ) < ∞. On the other hand, in view of Theorem 16.6.1, f (tx0 ) = −∞, since f (τ x0 ) = −∞ for τ > 0. Thus |f (tx0 )| < ∞ for all t ∈ R. Consequently the function ϕ : R → R given by ϕ(t) = f (tx0 ), t ∈ R, is finite, measurable, and subadditive, whence also ϕ(t) t>0 t

β = inf



is finite (cf. the proof of Theorem 16.2.9), contrary to the supposition. We have also the following annex to Theorem 16.2.7. Theorem 16.6.9. Let f : RN → [−∞, ∞) be a measurable subadditive function. If λ = lim inf f (x) = −∞, x→0

then f = −∞ in RN . Proof. There exists a sequence {tn }n∈N such that lim tn = 0

n→∞

and

lim f (tn ) = −∞.

n→∞

Take an arbitrary x ∈ RN . By Lemma 16.2.3 there exists a neighbourhood U of x and a constant M > 0 such that f (t) < M

for t ∈ U.

(16.6.3)

for n > n0 .

(16.6.4)

Further, there exist an n0 ∈ N such that x − tn ∈ U

Now, we have by (16.6.3) and (16.6.4), for n > n0 , f (x) = f (tn + x − tn )  f (tn ) + f (x − tn ) < f (tn ) + M. Letting n → ∞, we obtain hence f (x) = −∞.



In the one-dimensional case (N = 1) we can get further information about the set Pf . Theorems 16.6.10–16.6.12 say that if f : R → (−∞, ∞] is subadditive and infinitary, then the set of those x ∈ R at which f (x) = ∞ is rather large.   In the sequel I1 denotes either (0, ∞) or (−∞, 0), and I2 = R \ {0} \ I1 is either (−∞, 0), or (0, ∞), respectively. Theorem 16.6.10. Let f : R → [−∞, ∞] be subadditive. If I2 ∩ Pf = ∅, then either Pf = R, or f (x) = ∞ on a dense subset of I1 .

16.6. Infinitary subadditive functions

479

Proof. Suppose that the set {x ∈ I1 | f (x) = ∞} = I1 \ Pf is not dense in I1 . Then int(I1 ∩ Pf ) = ∅, whence also int Pf = ∅, and thus mi (Pf ) > 0. Take an arbitrary x0 ∈ int(I1 ∩Pf ) ⊂ int Pf . By hypothesis there exists a y ∈ I2 ∩Pf . So s = y/x0 < 0, and y = sx0 . By Theorem 16.6.6 Pf = R.  Theorem 16.6.11. Let f : R → [−∞, ∞] be a measurable subadditive function. If a ∈ I1 and f (a) = ∞, then   1 m {x ∈ I1 | |x| < |a| and f (x) = ∞}  |a|. 2

(16.6.5)

Proof. Let   A = {x ∈ I1 | |x| < |a| and f (x) = ∞} = I1 ∩ (−|a|, |a|) \ Pf . Since f is measurable, also the set A is measurable.   Let x ∈ I1 ∩ − |a|,|a| , and suppose that x ∈ / A so that x ∈ Pf . Put y = a − x. Then y ∈ I1 ∩ − |a|, |a| and x + y = a, whence f (a) = f (x + y)  f (x) + f (y).

(16.6.6)

Since f (x) < ∞ and f (a) = ∞, (16.6.6) implies that f (y) = ∞. Consequently y ∈ A  and x = a −y ∈ a − A. So if x ∈ I ∩ − |a|, |a| , then either x ∈ A, or x ∈ a − A, 1   whence I1 ∩ − |a|, |a| = A ∪ (a − A), and    |a| = m I1 ∩ − |a|, |a|  m(A) + m(a − A) = 2m(A). Hence we obtain (16.6.5)



Theorem 16.6.12. Let f : R → [−∞, ∞] be a measurable subadditive function. Then either I1 ∩ Pf = ∅, or I2 ∩ Pf = ∅, or m(Pf ) = 0, or Pf = R. Proof. Suppose that I1 ∩Pf = ∅, I2 ∩Pf = ∅, and m(Pf ) > 0 (Pf is measurable since f is measurable). Then either m(I1 ∩ Pf ) > 0, or m(I2 ∩ Pf ) > 0. Let I denote that of sets I1 , I2 for which m(I ∩ Pf ) > 0, and let J denote the other. By the Lebesgue density theorem (cf. also 3.5) the set I ∩ Pf contains a density point x0 of Pf , whereas J ∩ Pf = ∅, and hence there exists a y ∈ J ∩ Pf . Then s = y/x0 < 0, and y = sx0 . By Theorem 16.6.5 Pf = R.  Theorem 16.6.12 admits an extension to the N -dimensional case (Rosenbaum [269]), which we state here without proof. Theorem 16.6.13. Let f : RN → [−∞, ∞] be a measurable subadditive function. Then either there exists an (N − 1)-dimensional hyperplane π through the origin such that Pf lies entirely on one side of π, or m(Pf ) = 0, or Pf = RN .

480

Chapter 16. Subadditive Functions

Exercises 1. Let for x = (x1 , . . . , xN ) ∈ RN such that xi  0, i = 1, . . . , N , the function f be defined by N 1/r

r f (x) = xi i=1

with a certain r > 1. Show that f is subadditive. 2. Suppose that {An }n∈N is a sequence of arbitrary sets An ⊂ RN , and {cn }n∈N is a sequence of arbitrary positive constants. Further, let a, b ∈ R be such that 0  a  2b and 0  b  2a. Let the functions fn : RN → R be defined by  a for x ∈ An , fn (x) = b for x ∈ / An , and put f (x) =

∞

cn fn (x).

n=1

Show that the function f : RN → [0, +∞] is subadditive. The condition b  2a may be dropped if all the sets An , n ∈ N, fulfil the condition (16.1.3). 3. Let f : (0, ∞) → R be a measurable subadditive function. Show that if f is convex, then f (x)/x is decreasing in (0, ∞) (Hille-Philips [142]). 4. Let C ⊂ RN be a set such that for every x ∈ C and t > 1 also tx ∈ C. A function f : C → R is called quasi-homogeneous iff f (tx)  tf (x) for all x ∈ C and t > 1.The function f is called weakly-quasi-homogeneous iff for every x ∈ C and for every n ∈ N there exists a tn > n such that f (tn x)  tn f (x).

(∗)

Let f : (0, ∞) → R be a measurable function. Show that (Rosenbaum [269]): (i) If f is concave and lim f (x)  0, then f is quasi-homogeneous; x→0+

(ii) f is quasi-homogeneous if and only if f (x)/x is decreasing in (0, ∞); (iii) If f is quasi-homogeneous, then f is subadditive; (iv) If f is subadditive, then f is weakly-quasi-homogeneous. 5. Let f : RN → R be a continuous convex function. Show that f is subadditive if and only if f is quasi-homogeneous (Rosenbaum [269]). Hint: To show that f is quasi-homogeneous, represent t > 1 as t = λn + (1 −

λ)(n + 1), where n = [t] is the integral part of t. 6. Let f : RN → R be a weakly-quasi-homogeneous continuous convex function. Show that f is subadditive (Rosenbaum [269]).   −1 Hint: Use the relation f (x + y) = lim f t−1 n (tn x) + (1 − tn )y , where tn are n→∞

chosen according to (∗).

16.6. Infinitary subadditive functions

481

7. Let f : RN → R be a subadditive function such that lim f (x) = 0. Show that f x→0

is continuous in RN . 8. Show that in Theorem 16.2.1 the condition f (0) = 0 is essential (Hille-Philips [142]). 1 1 N∪ Hint: Take in Exercise 16.2 N = 1, b = , a = 1, cn = 2−n , An = 2 n   1 Z \ {0} , n ∈ N. Then f is discontinuous at the points x ∈ Q \ {0}, −(N) = n

1 and continuous elsewhere. Actually we have lim f (y) = for every x ∈ R. y→x 2 9. Let f : R → R be a measurable subadditive function, and put 1 If (x) = lim sup h→0 2h

if (x) = lim inf h→0

1 2h

h f (x + u)du, −h

h f (x + u)du. −h

Show that the function If , if are subadditive, and 0  If (x) − if (x)  If (0) for every x ∈ R (Hille-Philips [142]). Hint: Integrate the inequality f (x + y + (α + β)u)  f (x + αu) + f (y + βu) with

α and β suitably chosen. 10. Let f : (0, ∞) → R be a measurable and subadditive function, and set M = &∞ M for x > 0 (Phillips exp f (t)dt. Show that if M < ∞, then f (x)  2 log x 0 [255]).

2 Hint: Put g = exp f and show that 4g(x)  g(x − y) + g(y) . Then use the x/2

1/2

1/2

& relation x g(x) g(x) =2 dy. 0

11. Let f : RN → R be an arbitrary function. Show that if the limit lim f (tx)/t t→∞

exists for every x ∈ RN and is a convex function of x, then f = ϕ+ ψ, where ϕ is subadditive and continuous, and lim Ψ(tx)/t = 0 for every x ∈ RN (Rosenbaum t→∞

[269]). 12. Let f : R → [−∞, ∞] be a measurable subadditive function. Show that if lim f (x) = ∞, then f (x) = ∞ almost everywhere in (−∞, 0) or almost evx→0

erywhere in (0, ∞) (Hille-Phillips [142]).

Chapter 17

Nearly Additive Functions and Nearly Convex Functions 17.1 Approximately additive functions Under nearly additive and nearly convex functions we understand two modifications of additive functions and of convex functions: approximtely additive functions and approximtely convex functions, where the defining equation (5.2.1) resp. inequality (7.1.1) is satisfied only with some degree of accuracy, and almost additive functions and almost convex functions, where (5.2.1) resp. (5.3.1) is postulated only almost everywhere in the domain in question. (An interesting combination of both these conditions was considered by R. Ger [114].) Similar considerations will be carried out also for some other related classes of functions. We start our considerations with approximtely additive functions. A function f : RN → R is called ε-additive iff the inequality |f (x + y) − f (x) − f (y)|  ε

(17.1.1)

holds for all x, y ∈ RN . A function f : RN → R is called approximately additive iff it is ε-additive with a certain ε > 0. The main result concerning approximtely additive functions is due to D. H. Hyers [150] (however, cf. P´ olya-Szeg˝o [257], Chapter 3, 3.1, Problem 99; cf. also AlbertBaker [11], R¨atz [264]; and also Baker [15], Baker-Lawrence-Zorzitto [16], Cholewa [42], [43], [44], Forti [88], Moszner [233], Tabor [313], and further sections of this chapter concerning related results). Theorem 17.1.1. Let f : RN → R be an ε-additive function. Then there exists a unique additive function g : RN → R such that |f (x) − g(x)|  ε for x ∈ RN . Proof. Setting in (17.1.1) y = x we obtain |f (2x) − 2f (x)|  ε,

(17.1.2)

484

whence

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

# # #1 # # f (2x) − f (x)#  1 ε. #2 # 2

By induction we show that for every n ∈ N we have # −n #   #2 f (2n x) − f (x)#  1 − 2−n ε.

(17.1.3)

(17.1.4)

For n = 1 (17.1.4) reduces to (17.1.3). Assuming (17.1.4) true for an n ∈ N, we have by (17.1.3) # # # # # # # # # −(n+1) # n+1 −n # 1 n n # f (2 x) − f (x)#  2 # f (2 · 2 x) − f (2 x)# + #2−n f (2n x) − f (x)# #2 2     −(n+1) ε + 1 − 2−n ε = 1 − 2−(n+1) ε, 2 and thus we obtain (17.1.4) for n + 1, which completes the induction. Write gn (x) = 2−n f (2n x), x ∈ RN , n ∈ N. It follows by (17.1.4) that for arbitrary n, m ∈ N and x ∈ RN we have # # |gn+m (x) − gn (x)| = 2−n #2−m f (2m · 2n x) − f (2n x)#  2−n (1 − 2−m )ε < 2−n ε, which means that for every x ∈ RN the sequence {gn (x)}n∈N is a Cauchy sequence, and consequently converges. Let g(x) be its limit. We have by (17.1.1) #  #  2−n #f 2n (x + y) − f (2n x) − f (2n y)#  2−n ε, i.e.,

|gn (x + y) − gn (x) − gn (y)|  2−n ε,

whence, on letting n → ∞, we obtain g(x + y) − g(x) − g(y) = 0, which means that g is additive. (17.1.2) results from (17.1.4) on letting n → ∞. Now suppose that (17.1.2) holds with an additive function g : RN → R. By Theorem 5.2.1 g(nx) = ng(x) for n ∈ N and x ∈ RN , whence by (17.1.2) |f (nx) − ng(x)|  ε. Dividing by n and passing to the limit as n → ∞ we obtain g(x) = lim

n→∞

This proves the uniqueness of g.

f (nx) . n

(17.1.5) 

Corollary 17.1.1. Let f : RN → R be an ε-additive function, and let g : RN → R be the additive function fulfilling (17.1.2). Then g is given by formula (17.1.5).

17.2. Approximately multiadditive functions

485

Let us observe that every function f : RN → R fulfilling inequlity (17.1.2), where g : RN → R is an additive function, is approximtely additive. In fact, we have for arbitrary x, y ∈ RN , since g is additive, f (x + y) − f (x) − f (y) = f (x + y) − f (x) − f (y) − g(x + y) + g(x) + g(y), whence |f (x + y) − f (x) − f (y)|  |f (x + y) − g(x + y)| + |f (x) − g(x)| + |f (y) − g(y)|  3ε. Consequently f is 3ε-additive. Theorem 17.1.2. Let f : RN → R be an ε-additive function, and let g : RN → R be the additive function fulfilling (17.1.2). If f is measurable, or bounded above, or below, on a set T ∈ A, then g is continuous. Proof. Inequlity (17.1.2) can be written as f (x) − ε  g(x)  f (x) + ε. If f is bounded above, or below, on a set T ⊂ RN , then so is also g. If T ∈ A, then g is continuous in virtue of Theorems 9.1.2 and 10.2.2 and of the definition of the class B. If f is measurable, then g is continuous in virtue of Theorem 9.4.1. 

17.2 Approximately multiadditive functions The result in this section are due to M. Albert and J. A. Baker [11]. Let p ∈ N be fixed. Theorem 17.2.1. If a function f : RpN → R satisfies for every x1 , . . . , xp , y1 , . . . , yp ∈ RN the system of inequalities |f (x1 , . . . , xi−1 , xi + yi , xi+1 , . . . , xp ) − f (x1 , . . . , xp ) − f (x1 , . . . , xi−1 , yi , xi+1 , . . . , xp )|  εi ,

i = 1, . . . , p, (17.2.1)

where ε1 , . . . , εp are positive numbers, then there exists a unique p-additive function g : RpN → R such that |f (x1 , . . . , xp ) − g(x1 , . . . , xp )|  ε = min(ε1 , . . . , εp )

(17.2.2)

for all x1 , . . . , xp ∈ RN . Moreover, if f is symmetric, then g also is symmetric. Proof. By a suitable renumeration of the variables we may achieve, without violating (17.2.1), that min(ε1 , . . . , εp ) = ε1 . (17.2.3) By Theorem 17.1.1 and Corollary 17.1.1, for every x1 , . . . , xp ∈ RN the limit g(x1 , . . . , xp ) = lim

n→∞

f (nx1 , . . . , xp ) n

486

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

exists, and with fixed x2 , . . . , xp is an additive function of x1 and |f (x1 , . . . , xp ) − g(x1 , . . . , xp )|  ε1 for every x1 , . . . , xp ∈ RN , which together with (17.2.3) yields (17.2.2). For every i = 2, . . . , p, and every x1 , . . . , xp , yi ∈ RN we have by (17.2.1) |f (nx1 , x2 , . . . , xi−1 , xi + yi , xi+1 , . . . , xp ) − f (nx1 , x2 , . . . , xp ) − · · · · · · − f (nx1 , x2 , . . . , xi−1 , yi , xi+1 , . . . , xp )|  εi , whence, on dividing by n and passing to the limit as n → ∞, we obtain that g is p-additive. After permutating variables so as to return to the original numeration, g remains p-additive, and (17.2.2) still holds. To prove the uniqueness, assume that (17.2.2) holds with two p-additive functions g1 and g2 , and put g = g1 − g2 . Then g : RpN → R also is p-additive (Lemma 13.4.1), and |g(x1 , . . . , xp )| = |g1 (x1 , . . . , xp ) − g2 (x1 , . . . , xp )|  |g1 (x1 , . . . , xp ) − f (x1 , . . . , xp )| + |f (x1 , . . . , xp ) − g2 (x1 , . . . , xp )|  2ε so that g is bounded in RN . By Lemma 13.4.3 g = 0, i.e., g1 = g2 . Now assume that f is symmetric, and let (i1 , . . . , ip ) be an arbitrary permutation of the numbers (1, . . . , p). Define the function h : RpN → R by h(x1 , . . . , xp ) = g(xi1 , . . . , xip ).

(17.2.4)

By Lemma 13.4.2 h is p-additive, and we have due to the symmetry of f |f (x1 , . . . , xp ) − h(x1 , . . . , xp )| = |f (xi1 , . . . , xip ) − g(xi1 , . . . , xip )|  ε, whence by the uniqueness part of the theorem h(x1 , . . . , xp ) = g(x1 , . . . , xp ). This together with (17.2.4) yields g(xi1 , . . . , xip ) = g(x1 , . . . , xp ), i.e., g is symmetric.



17.3 Functions with bounded differences Now we will derive analogous results for polynomial functions (Hyers [151]1 , [152], Albert-Baker [11]; cf. also Whitney [323], [324]). 1

Added in the 2nd edition by K. Baron.

17.3. Functions with bounded differences

487

Theorem 17.3.1. Let f : RN → R satisfy (with a certain ε > 0) # # #Δh1 ...hp+1 f (x)#  ε

(17.3.1)

for all x, h1 , . . . , hp+1 ∈ RN , where p ∈ N ∪ {0} is fixed. Then there exist symmetric k-additive functions Fk : RkN → R, k = 0, . . . , p unique except for F0 , such that |f (x) − g(x)|  ε for all x ∈ RN , where g(x) =

p

(17.3.2)

fk (x),

(17.3.3)

k=0

and f0 , . . . , fp are the diagonalizations of F0 , . . . , Fp , respectively. Proof. The proof (Albert-Baker [11]) is similar to that of Theorem 15.9.2, and the proof of the existence of g will be by induction on p. For p = 0 f satisfies |Δx f (0)|  ε for every x ∈ RN , i.e., |f (x) − f (0)|  ε. With F0 = f0 = f (0) we obtain hence (17.3.2) with (17.3.3). Now assume that the theorem is true for a p − 1 ∈ N ∪ {0}, and let f : RN → R be a function fulfilling (17.3.1) for all x, h1 , . . . , hp+1 ∈ RN . For every x ∈ RN define the function Fx : RpN → R by Fx (x1 , . . . , xp ) = Δx1 ,...,xp f (x).

(17.3.4)

We have by Corollary 15.1.3, Lemma 15.1.3, and by (17.3.1), for every fixed x ∈ RN and arbitrary i, 1  i  p, and arbitrary x1 , . . . , xp , yi ∈ RN , # #Fx (x1 , . . . , xi−1 , xi + yi , xi+1 , . . . , xp ) − Fx (x1 , . . . , xp ) # − Fx (x1 , . . . , xi−1 , yi , xi+1 , . . . , xp )# #

# = #Δx1 ...xi−1 xi+1 ...xp Δxi +yi f (x) − Δxi f (x) − Δyi f (x) # # # # # = #Δx1 ...xi−1 xi+1 ...,xp Δxi yi f (x)# = #Δx1 ...xp yi f (x)#  ε. By Theorem 17.2.1, for every x ∈ RN , there exists a unique p-additive function Gx : RpN → R such that |Fx (x1 , . . . , xp ) − Gx (x1 , . . . , xp )|  ε

(17.3.5)

for all x1 , . . . , xp ∈ R . Moreover, since by Corollary 15.1.3 function (17.3.4) is symmetric, also the function Gx is symmetric (Theorem 17.2.1). For arbitrary x, y ∈ RN and arbitrary x1 , . . . , xp ∈ RN we have by (17.3.4), (17.3.1), and Lemma 15.1.1 # # #

# #Fy (x1 , . . . , xp ) − Fx (x1 , . . . , xp )# = #Δx1 ...xp f (y) − f (x) # (17.3.6) #

# # # = Δx1 ...xp f (x + (y − x)) − f (x) # # = #Δx1 ...xp Δy−x f (x)# # # = #Δx ...x (y−x) f (x)#  ε, N

1

p

488

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

whereas by (17.3.5) |Fy (x1 , . . . , xp ) − Gy (x1 , . . . , xp )|  ε.

(17.3.7)

Relations (17.3.5), (17.3.6) and (17.3.7) yield |Gy − Gx |  |Gy − Fy | + |Fy − Fx | + |Fx − Gx |  3ε. Consequently the function Gy − Gx is p-additive (Lemma 13.4.1) and bounded in RpN . By Lemma 13.4.3 Gx = Gy . This shows that Gx is independent of x, and we may define a symmetric p-additive function Fp : RpN → R by Fp (x1 , . . . , xp ) =

1 Gx (x1 , . . . , xp ). p!

(17.3.8)

Let fp : RN → R be the diagonalization of Fp , and define gp : RN → R as gp = f − fp . We have by (17.3.4), (17.3.8), (17.3.5), and by Lemmas 15.1.1 and 15.9.2, for arbitrary h1 , . . . , hp ∈ RN , # # # # #Δh1 ...hp gp (x)# = #Δh1 ...hp f (x) − Δh1 ...hp fp (x)# # # = #Fx (h1 , . . . , hp ) − p!Fp (h1 , . . . , hp )# # # = #Fx (h1 , . . . , hp ) − Gx (h1 , . . . , hp )#  ε. By the induction hypothesis there exist symmetric k-additive functions Fk : RkN → R, k = 0, . . . , p − 1, such that # # p−1 # #

# # (17.3.9) fk (x)#  ε, #gp (x) − # # k=0

where f0 , . . . , fp−1 are the diagonalizations of F0 , . . . , Fp−1 , respectively. Define g : RN → R by (17.3.3). Then (17.3.9) goes into (17.3.2) in view of the definition of gp . To prove the uniqueness, suppose that besides F0 , . . . , Fp , there exist also symmetric k-additive functions F3k : RkN → R, k = 0, . . . , p, such that |f (x) − gˆ(x)|  ε in RN , where gˆ(x) =

p

(17.3.10)

fˆk (x),

k=0

and fˆ0 , . . . , fˆp are the diagonalizations of F30 , . . . , F3p , respectively. By (17.3.2) and (17.3.10) |g(x) − gˆ(x)|  2ε. (17.3.11) We have g(x) − gˆ(x) =

p



 fk (x) − fˆk (x) ,

k=0

(17.3.12)

17.3. Functions with bounded differences

489

and every function fk − fˆk is the diagonalization of the symmetric k-additive function Fk − F3k , k = 0, . . . , p. Relation (17.3.11) says that function (17.3.12) is bounded in  RN . By Lemma 15.9.3 Fk = F3k for k = 0, . . . , p. Theorem 17.3.2. Let f : RN → R satisfy (with a certain ε > 0) # # # p+1 # #Δh f (x)#  ε

(17.3.13)

for all x, h ∈ RN , where p ∈ N is fixed. Then there exists a polynomial function g : RN → R of order p, unique up to an additive constant, such that |f (x) − g(x)|  2p+1 ε

(17.3.14)

for all x ∈ R . N

Proof. In view of Theorem 15.1.2 relation (17.3.13) implies that # # #Δh1 ...hp+1 f (x)#  2p+1 ε for all x, h1 , . . . , hp+1 ∈ RN . By Theorem 17.3.1 there exist symmetric k-additive functions Fk : RkN → R, k = 0, . . . , p, such that (17.3.14) holds in RN , where g is given by (17.3.3), and f0 , . . . , fp are the diagonalizations of F0 , . . . , Fp , respectively. By Theorem 15.9.1 g is a polynomial function of order p. Now suppose that g1 , g2 : RN → R are polynomial functions of order p fulfilling |f (x) − g1 (x)|  2p+1 ε and |f (x) − g2 (x)|  2p+1 ε for all x ∈ RN . Then

|g1 (x) − g2 (x)|  2p+2 ε.

(17.3.15)

By Lemma 15.3.2 g1 −g2 is a polynomial function, and by (17.3.15) g1 −g2 is bounded in RN . In virtue of Corollary 15.9.2 g1 − g2 = const.  The estimation in (17.3.14) can be improved. Actually, it can be shown (Whitney [323]) that, under the conditions of Theorem 17.3.2, the function g fulfils the inequality   p+1 |f (x) − g(x)|  2ε/ max k k for all x ∈ RN . Theorem 17.3.3. Let f : RN → R satisfy (17.3.13) (with certain ε > 0 and p ∈ N), and suppose that f is measurable or bounded on a set T ⊂ RN , where mi (T ) > 0 or T is of the second category and with the Baire property. Then there exists a polynomial g : RN → R of degree at most p, unique up to an additive constant, such that (17.3.14) holds for all x ∈ RN . Proof. If f is measurable, then the argument in the proof of Theorem 15.5.4 shows that f is bounded on a set of positive measure, so we need only consider the case where f is bounded on a suitble set T ⊂ RN . By Theorem 17.3.2 there exists a polynomial function g : RN → R of order p, unique up to an additive constant and such that (17.3.14) holds in RN . Consequently g is also bounded on T , and by Theorem 15.5.2 g is continuous. By Theorem 15.9.4 g is a polynomial of degree at most p. 

490

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

17.4 Approximately convex functions For convex functions we do not have a full analogue of Theorem 17.1.1 (cf. also Cholewa [43]). The corresponding result is weaker, and the proof (cf. Hyers-Ulam [153]) is much more complicated. Let D ⊂ RN be a convex and open set. A function f : D → R is called ε-convex iff   f λx + (1 − λ)y  λf (x) + (1 − λ)f (y) + ε (17.4.1) for every x, y ∈ D and λ ∈ [0, 1]. A function f : D → R is called approximately convex iff it is ε-convex with a certain ε > 0. Lemma 17.4.1. Let D ⊂ RN be a convex and open set, let n ∈ N, let x0 , . . . , xn ∈ D, and let α0 , . . . , αn be non-negative numbers such that α0 + · · · + αn = 1. Let x = α0 x0 + · · · + αn xn . If f : D → R is an ε-convex function, then f (x) 

n

αi f (xi ) +

i=0

n2 + 3n ε. 2n + 2

(17.4.2)

Proof. First note that by Lemma 5.1.3 x ∈ D, so that f (x) is meaningful. The proof of (17.4.2) will be by induction on n. For n = 1 (17.4.2) reduces to (17.4.1). Assume (17.4.2) to hold for an n ∈ N. Take arbitrary points x0 , . . . , xn+1 ∈ D and non-negative real numbers α0 , . . . , αn+1 such that n+1

αi = 1, (17.4.3) i=0

and put x = α0 x0 + · · · + αn+1 xn+1 . We may assume that the points xi have been numbered in such a way that αn+1  αi for i = 0, . . . , n. If αn+1 = 1, then by (17.4.3) αi = 0 for i = 0, . . . , n, x = xn+1 , and the inequality to be proved f (x)  f (xn+1 ) +

(n + 1)2 + 3(n + 1) ε 2(n + 1) + 2

is trivial. If αn+1 < 1, then put λ = 1 − αn+1 > 0,

(17.4.4)

βi = αi /λ, i = 0, . . . , n, z = β0 x0 + · · · + βn xn . All β’s are non-negative, and by (17.4.3) and (17.4.4) n

1 1 αi = (1 − αn+1 ) = 1. λ i=0 λ n

βi =

i=0

By the induction hypothesis f (z) 

n

i=0

βi f (xi ) +

n2 + 3n ε. 2n + 2

(17.4.5)

17.4. Approximately convex functions

491

On the other hand λz + (1 − λ)xn+1 = α0 x0 + · · · + αn xn + αn+1 xn+1 = x, whence by (17.4.1) f (x)  λf (z) + (1 − λ)f (xn+1 ) + ε.

(17.4.6)

Since αn+1  αi for i = 0, . . . , n, we have by (17.4.3) 1=

n+1

αi  (n + 2)αn+1 ,

i=0

whence αn+1  1/(n + 2) and λ = 1 − αn+1  (n + 1)(n + 2). Relations (17.4.5) and (17.4.6) yield f (x) 

n

αi f (xi ) + αn+1 f (xn+1 ) + λ

i=0

whence f (x) 

n+1

 αi f (xi ) +

i=0

Since

n2 + 3n ε + ε, 2n + 2

 n + 1 n2 + 3n + 1 ε. n + 2 2n + 2

(17.4.7)

n + 1 n2 + 3n (n + 1)2 + 3(n + 1) +1= , n + 2 2n + 2 2(n + 1) + 2

relation (17.4.7) yields (17.4.2) for n + 1. Induction completes the proof.



Lemma 17.4.2. Let D ⊂ R be a convex and open set, and let f : D → R be an ε-convex function. Then f is locally bounded at every point of D. N

Proof. Let x ¯ ∈ D. By Remark 5.1.1 there exists an N -dimensionl simplex S ⊂ D such that x ¯ ∈ int S. Let x0 , . . . , xN be the vertices of S so that S = conv{x0 , . . . , xN }. We have x0 , . . . , xN ∈ D and for every x ∈ S there exist non-negative numbers α0 , . . . , αN such that α0 + · · · + αN = 1 and x = α0 x0 + · · · + αN xN . Let M = max{f (x0 ), . . . , f (xN )}. We have by Lemma 17.4.1 f (x) 

N

i=0

αi f (xi ) +

N 2 + 3N N 2 + 3N εM+ ε, 2N + 2 2N + 2

which means that f is bounded above on S, and hence also on int S. But int S is a neighbourhood of x ¯, and so f is locally bounded above at x ¯. Let B ⊂ D be an open ball centered at x ¯ and such that f is bounded above on B: f (u)  K

for u ∈ B.

(17.4.8)

492

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

Take an arbitrary x ∈ B. Then also 2¯ x − x = x¯ − (x − x ¯) ∈ B, and we have by (17.4.1) and (17.4.8) f (¯ x) 

1 1 1 1 f (x) + f (2¯ x − x) + ε  f (x) + K + ε, 2 2 2 2

whence f (x)  2f (¯ x) − K − 2ε, which means that f is bounded below on B. Consequently f is locally bounded below at x ¯.  Corollary 17.4.1. Let D ⊂ RN be a convex and open set, and let f : D → R be an ε-convex function. Then f is bounded on every compact C ⊂ D. Proof. For every x ∈ C let Bx be an open ball centered at x and contained in D such that f is bounded on Bx . The compact set C can be covered by a finite number of balls Bx , whence f is bounded on C.  The next lemma is known in the theory of convex sets as the Carath´eodory theorem (cf., e.g., Eggleston [74]). Lemma 17.4.3. Let A ⊂ RN be an arbitrary non-empty set. For every x ∈ conv A there exist points x0 , . . . , xr ∈ A such that x ∈ conv{x0 , . . . , xr } and r  N . Proof. Let x ∈ conv A. By Theorem 5.1.3 there exist an n ∈ N, points x0 , . . . , xn ∈ A, and non-negative numbers α0 , . . . , αn such that α0 + · · · + αn = 1 and x = α0 x0 + · · · + αn xn . If n  N , there is nothing more to prove, so suppose that n > N . The conditions β0 x0 + · · · + βn xn = 0, (17.4.9) β0 + · · · + βn = 1,

(17.4.10)

represent system of N + 1 homogeneous liner equations with n + 1 > N + 1 unknowns β0 , . . . , βn . Consequently this system has a non-trivial solution. In other words, there exist numbers β0 , . . . , βn , not all zeroes and such that (17.4.9) and (17.4.10) hold. Let T be a set of all those real numbers t which satisfy αi + tβi  0

for i = 0, . . . , n.

(17.4.11)

Clearly 0 ∈ T , and T = R, since at least one of β0 , . . . , βn is different from zero. Let t0 be a frontier point of T . T is closed, consequently t0 ∈ T , which means that t = t0 satisfies (17.4.11). Moreover, we have by (17.4.9) and (17.4.10) n

i=0

and

(αi + t0 βi ) =

n

i=0

αi + t0

n

i=0

βi =

n

αi = 1

i=0

n n n n



(αi + t0 βi )xi = αi xi + t0 βi xi = αi xi = x. i=0

i=0

i=0

i=0

17.4. Approximately convex functions

493

Further, at least for one i = 0, . . . , n, we have αi + t0 βi = 0. Consequently x can be represented as a linear combintion of n−1 points from A with non-negative coefficients summing up to 1. Proceeding so further, we may reduce the number n to N .  Corollary 17.4.2. Let A ⊂ RN be an arbitrary non-empty set. Then  conv A = x ∈ RN | x = α0 x0 + · · · + αN xN , xi ∈ A, αi  0 for i = 0, . . . , N,

N

αi = 1 .

(17.4.12)

i=0

Proof. Let B denote the set on the right-hand side of (17.4.12). The inclusion B ⊂ conv A follows from Lemma 5.1.3, whereas the inclusion conv A ⊂ B results from Lemma 17.4.3. (Note that if x is a combination of a smaller number of points, we may always add terms with arbitrary xi ∈ A and αi = 0, so as to obtain a linear combination of N + 1 points.)  Corollary 17.4.3. Let A ⊂ RN be an arbitrary non-empty compact set. Then the set conv A also is compact. Proof. Let N

! " Ξ = (α0 , . . . , αN ) ∈ RN +1 | α0  0, . . . , αN  0, αi = 1 . i=0

Ξ is a closed subset of the compact set [0, 1]N +1 , and hence it is compact. Now define a function F : Ξ × AN +1 → RN by F (α0 , . . . , αN , x0 , . . . , xN ) = α0 x0 + · · · + αN xN . The function F is continuous and the set Ξ × AN +1 is compact, since Ξ and A are compact, therefore also the set F (Ξ × AN +1 ) is compact. But by Corollary 17.4.2 F (Ξ × AN +1 ) = conv A, and so conv A is compact.



Lemma 17.4.4. Let S ⊂ RN be an arbitrary non-empty set, and let K = conv S. For every frontier point x of K such that x ∈ K there exist points x0 , . . . , xN −1 ∈ S and non-negative numbers α0 , . . . , αN −1 such that α0 + · · · + αN −1 = 1 and x = α0 x0 + · · · + αN −1 xN −1 .   Proof. By Theorem 5.1.7 there exists a support hyperplane H (N − 1)-dimensional of K at x. If we had H ∩ S = ∅, then S would lie entirely in one open half-space π of RN determined by H and, since π is a convex set, we would have K = conv S ⊂ π, whence H ∩ K = ∅. But this is impossible, since x ∈ H ∩ K.

494

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

Consequently H ∩ S = ∅. We will prove that conv(H ∩ S) = H ∩ K.

(17.4.13)

Since H and K are convex, the set H ∩ K is convex by Theorem 5.1.2, and since evidently H ∩S ⊂ H ∩K, we have conv(H ∩S) ⊂ H ∩K. To prove the converse inclusion, take an arbitrary y ∈ H ∩ K. By Theorem 5.1.3 there exist points y1 , . . . , yn ∈ S and positive numbers β1 , . . . , βn such that β1 + · · · + βn = 1 and β1 y1 + · · · + βn yn = y. Let the equation of H be At + B = 0, where A ∈ RN , B ∈ R, and A, B are such that the closed half-space determined by H and containing S is described by the condition At + B  0. Hence Ayi + B  0 for i = 1, . . . , n, and since Ay + B =

n

βi (Ayi + B),

i=1

the condition y ∈ H, i.e., Ay + B = 0, implies tht Ayi + B = 0 for i = 1, . . . , n. Consequently yi ∈ H, and so yi ∈ H ∩ S, i = 1, . . . , n, whence by Theorem 5.1.3 y ∈ conv(H ∩ S). This proves the inclusion H ∩ K ⊂ conv(H ∩ S), and concludes the proof of (17.4.13). Now, we may treat H as RN −1 . By Corollary 17.4.2, since x ∈ H ∩ K, there exist points x0 , . . . , xN −1 ∈ H ∩ S and non-negative numbers α0 , . . . , αN −1 such that α0 + · · · + αN −1 = 1 and α0 x0 + · · · + αN −1 xN −1 = x, which was to be proved.  The following theorems are due to D. H. Hyers and S. M. Ulam [153] (cf. also Cholewa [43]). Theorem 17.4.1. Let D ⊂ RN be a convex and open set, and let f : D → R be an ε-convex function. Then there exists a continuous convex function g : D → R such that  2  N + 3N g(x)  f (x)  g(x) + +1 ε (17.4.14) 2N + 2 for all x ∈ D. Proof. Let {Cn }n∈N be a sequence of convex and compact subsets of D such that Cn ⊂ Cn+1 , and

∞ 

n ∈ N,

int Cn = D

(17.4.15)

(17.4.16)

n=1

(cf. Exercise 17.5). Put fn = f | Cn , and Sn = cl Gr(fn ), Kn = conv Sn . By Corollary 17.4.1 f is bounded on Cn , so every function fn is bounded, whence Sn is a compact

17.4. Approximately convex functions

495

subset of RN +1 . By Corollary 17.4.3, also Kn is a compct subset of RN +1 , n ∈ N. For every n ∈ N define the function gn : Cn → R by gn (x) = inf{y ∈ R | (x, y) ∈ Kn }.   Since Kn is compact, we have x, gn (x) ∈ Kn for every x ∈ Cn and n ∈ N. Consequently gn (x) is finite and gn : Cn → R is a function.   Take  arbitrary n ∈ N, x, y ∈ Cn , and λ ∈ [0, 1]. We have x, gn (x) ∈ Kn ,  y, gn (y) ∈ Kn , and, since Kn is convex,       λx + (1 − λ)y, λgn (x) + (1 − λ)gn (y) = λ x, gn (x) + (1 − λ) y, gn (y) ∈ Kn . Consequently, by the definition of gn ,   gn λx + (1 − λ)y  λgn (x) + (1 − λ)gn (y). (17.4.17)  This means that gn | int Cn by (17.4.15) and (17.4.16) int Cn =  ∅ for n sufficiently large, and by Theorem 5.1.5 for such n the set int Cn is convex is a continuous and convex function (Theorem 7.1.1). Fix an n ∈ N. Three cases may occur for x ∈ Cn :   (i) x, gn (x) ∈ Gr(fn ),   (ii) x, gn (x) ∈ Sn \ Gr(fn ),   (iii) x, gn (x) ∈ Kn \ Sn . In case (i) we have gn (x) = fn (x) = f (x), and so, in particular,   2 N + 3N + 1 ε. (17.4.18) gn (x)  f (x)  gn (x) + ε  gn (x) + 2N + 2 In case (ii) there exists a sequence {(tm , zm )}m∈N of distinct points such that tm ∈ Cn ,

(tm , zm ) ∈ Gr(fn ),

m ∈ N,

and lim tm = x,

m→∞

lim zm = gn (x).

m→∞

(17.4.19)

Note that if points (tm , zm ), (tm , zm ) ∈ Gr(fn ) are distinct, then necessarily tm = tm , and consequently all points tm , m ∈ N, are distinct. Hence tm = x for all m ∈ N, except possibly one exceptional value of m. Removing this exceptional element from the sequence {(tm , zm )}m∈N , we do not spoil the remaining properties of this sequence, and we may assume that tm = x for m ∈ N. Let B ⊂ D be a closed ball centered at x and with a radius r > 0, and put ym = x − r

tm − x , |tm − x|

m ∈ N.

Thus ym ∈ B for m ∈ N, and x = λm tm + (1 − λm )ym ,

m ∈ N,

(17.4.20)

496

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

where λm =

|x − ym | , m ∈ N. |tm − ym |

Since B is compact, we may assume, replacing possibly the sequences {tm }m∈N and {ym }m∈N by suitable subsequences, that the sequence {ym }m∈N converges lim ym = y0 .

m→∞

Hence by (17.4.19) lim λm =

m→∞

|x − y0 | = 1. |x − y0 |

(17.4.21)

By (17.4.1) and (17.4.20) f (x)  λm f (tm )+(1−λm )f (ym )+ε = λm zm +(1−λm )f (ym )+ε,

m ∈ N. (17.4.22)

By Corollary 17.4.1 f is bounded on B so that the sequence {f (ym )}m∈N is bounded. Letting in (17.4.22) m → ∞, we obtain in view of (17.4.19) and (17.4.21) f (x)  gn (x) + ε,  and since x, fn (x) ∈ Gr(fn ) ⊂ Sn ⊂ Kn , we have

(17.4.23)



gn (x)  fn (x) = f (x).

(17.4.24)

Relations (17.4.23) and (17.4.24) yield (17.4.18), which means that (17.4.18) holds in case (ii), too. In case (iii), according to Lemma 17.4.4, there exist points (x0 , u0 ), . . . , (xN , uN ) ∈ Sn and non-negative numbers α0 , . . . , αN such that N

αi = 1,

i=0

N

αi xi = x,

i=0

N

αi ui = gn (x).

(17.4.25)

i=0

Suppose that for a certain j we have uj = gn (xj ). Since (xi , ui ) ∈ Sn ⊂ Kn for i = 0, . . . , N , we must have ui  gn (xi ) for i = 0, . . . , N , and uj > gn (xj ). Hence N

i=0

αi ui >

N

αi gn (xi )

i=0

except when αj = 0, when we may replace uj by gn (xj ) without violating (17.4.25). On the other hand, since gn satisfies (17.4.17), we have by (17.4.25) and Lemma 5.3.2 N

i=0

αi ui = gn (x) 

N

αi gn (xi ).

i=0

This contradiction shows that ui = gn (xi ) for i = 0, . . . , N .

17.4. Approximately convex functions

497

   Now,  since xi , gn (xi ) ∈ Sn for i = 0, . . . , N , we have by (17.4.18) case (i) and (ii) f (xi )  gn (xi ) + ε, i = 0, . . . , N, whence by (17.4.25) and Lemma 17.4.1 N

  N 2 + 3N N 2 + 3N ε ε αi gn (xi ) + ε + 2N + 2 2N + 2 i=0 i=0    2  2 N

N + 3N N + 3N = + 1 ε = gn (x) + + 1 ε. αi ui + 2N + 2 2N + 2 i=0

f (x) 

N

αi f (xi ) +

The inequality gn (x)  f (x) results from the definition of gn . Consequently for every x ∈ Cn we have in all cases (i), (ii), (iii)   2 N + 3N + 1 ε. (17.4.26) gn (x)  f (x)  gn (x) + 2N + 2 From (17.4.15) it follows that Gr(fn ) ⊂ Gr(fn+1 ), whence Sn ⊂ Sn+1 and Kn ⊂ Kn+1 for every n ∈ N. Hence, for every x ∈ Cn , gn+1 (x) = inf{y ∈ R | (x, y) ∈ Kn+1 }  inf{y ∈ R | (x, y) ∈ Kn } = gn (x). For a fixed n ∈ N the sequence {gm | int Cn }mn is decreasing and bounded below, since f is bounded below on Cn and by (17.4.26) all gm are uniformly bounded below on Cn . Consequently the sequence {gm | int Cn } converges, and by Theorem 7.10.1 its limit is a continuous and convex function. In virtue of (17.4.16) we may define the function g : D → R by g(x) = lim gm (x), m→∞

and for every n ∈ N the function g | int Cn is continuous and convex. Take arbitrary x, y ∈ D. By (17.4.15) and (17.4.16) there exists an n ∈ N such that x, y ∈ int Cn . By Theorem 7.1.1 we have for arbitrary λ ∈ [0, 1]   g λx + (1 − λ)y  λg(x) + (1 − λ)g(y). (17.4.27) Thus g satisfies (17.4.27) for all x, y ∈ D and λ ∈ [0, 1]. By Theorem 7.1.1 g is a continuous and convex function. Letting in (17.4.26) n → ∞, we obtain (17.4.14).  Theorem 17.4.2. Let D ⊂ RN be a convex and open set, and let f : D → R be an ε-convex function. Then there exists a continuous and convex function g0 : D → R such that   1 N 2 + 3N |f (x) − g0 (x)|  +1 ε (17.4.28) 2 2N + 2 for all x ∈ D.

498

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

Proof. By Theorem 17.4.1 there exists a continuous and convex function g : D → R fulfilling (17.4.14) for all x ∈ D. The function g0 (x) = g(x) +

1 2



 N 2 + 3N +1 ε 2N + 2 

satisfies the conditions of the theorem.

Estimations (17.4.14) and (17.4.28) are not sharp. It can be shown (Hyers-Ulam [153]) that the function g constructed in the proof of Theorem 17.4.1 actully satisfies g(x)  f (x)  g(x) +

N 2 + 3N ε, 2N + 2

and then the proof of Theorem 17.4.2 shows that a continuous and convex function g0 : D → R exists fulfilling the inequality |f (x) − g0 (x)| 

N 2 + 3N ε 4N + 4

for x ∈ D. The problem of extending the above results to functions f : D → R satisfying in D the inequlity   x+y f (x) + f (y) f  +ε 2 2 remains still open (cf. also Cholewa [43])2 .

17.5 Set ideals Let X be an arbitrary set (space). A non-empty family I ⊂ P(X) of subsets of X is called an ideal iff it satisfies the two conditions: (i) If A ∈ I and B ⊂ A, then B ∈ I; (ii) If A, B ∈ I, then A ∪ B ∈ I. If condition (ii) is replaced by the stronger condition ∞  An ∈ I, (ii ) If An ∈ I, n ∈ N, then n=1

then I is called a σ-ideal. The family I = {∅} consisting only the empty set and the family I = P(X) of all the subsets of X are trivial examples of σ-ideals. Due to condition (i) we must have I = P(X) whenever X ∈ I. We wish to exclude such a trivial case. If an ideal [σ-ideal] I besides (i), (ii) (i), (ii ) satisfies also the condition (iii) X ∈ / I, 2

Added in the 2nd edition by K. Baron: This has recently been solved by Z. Kominek and J. Mrowiec in Nonstability results in the theory of convex functions, C. R. Math. Acad. Sci., Soc. R. Can. 28 (2006), 17–23.

17.5. Set ideals

499

it is called proper. If we are given a proper ideal I ⊂ P(X) in X,then we say that a condition is satisfied I-almost everywhere in X written I-(a.e.) iff there exists a set A ∈ I such that the condition in question is satisfied for every x ∈ X \ A. Now suppose that X is endowed with an inner operation + such that (X, +) is a (not necessarily commutative) group (cf. 4.5). We say that an ideal [σ-ideal] I ⊂ P(X) is linearly invariant iff besides conditions (i), (ii), [(i), (ii’)] it satisfies also the condition (iv) For every x ∈ X and A ∈ I the set x − A belongs to I.   A proper linearly invariant ideal i.e., a family I ⊂ P(X) fulfilling (i)–(iv) will in the sequel be referred to as a p.l.i. ideal. Lemma 17.5.1. If (X, +) is a group, and I ⊂ P(X) is a linearly invariant ideal, then, for every x ∈ X and A ∈ I, we have −A ∈ I,

x + A ∈ I,

A + x ∈ I.

Proof. −A = 0 − A ∈ I by (iv). Hence also x + A = x − (−A) ∈ I. Finally A + x = −(−x − A) ∈ I by what has already been shown.  Below we give some most important examples of ideals. I. Let X be an arbitrary set, and let Iℵ0 be the family of all at most countable subsets of X. Then Iℵ0 is a σ-ideal. If (X, +) is a group, then Iℵ0 is linearly invariant. Iℵ0 is proper if and only if card X > ℵ0 . If X = RN , we write IℵN0 instead of Iℵ0 . IℵN0 is a p.l.i. σ-ideal. II. Let X be a metric space, and let Ib be the family of all bounded subsets of X. Then Ib is an ideal, but in general not a σ-ideal. If (X, +) is a metric group, and the metric is invariant under translations, then Ib is linearly invariant. If X is unbounded, then Ib is proper. If X = RN , we write IbN instead of Ib . IbN is a p.l.i. ideal. III. Let X be a topological space, and let If be the family of all subsets of X which are of the first category. Then If is a σ-ideal. If (X, +) is a topological group, then If is linearly invariant. If X is of the second category, then If is proper. If X = RN , we write IfN instead of If . IfN is a p.l.i. ideal. IV. Let X = RN , and let I0N be the family of all subsets of RN of N -dimensional measure zero. I0N is a p.l.i. σ-ideal. N be the family of all subsets of RN of finite N V. Let X = RN , and let Im dimensional outer measure: N A ∈ Im ⇔ A ⊂ RN and me (A) < ∞. N is a p.l.i. ideal, but not a σ-ideal. Im VI. Let (X, +) be a group (not necessarily commutative), let B ⊂ X be an arbitrary set, and let  n 

J (B) = A ⊂ X | A = xi + (B ∪ (−B)) + yi , xi , yi ∈ X, i = 1, . . . , n; n ∈ N . i=1

500

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

Put



I(B) =

P(A).

A∈J (B)

Then I(B) is the smallest linerly invariant ideal in X containing B. In fact, if I is the smallest linearly invariant ideal in X containing B, then I(B) ⊂ I by (i), (ii), and Lemma 17.5.1. On the other hand, if U, V ∈ I(B), then there exist n, m ∈ N and points xi , yi , i = 1, . . . , n + m, such that U⊂

n 

xi + (B ∪ (−B)) + yi ,

V ⊂

i=1

n+m 



xi + (B ∪ (−B)) + yi .

i=n+1

Hence U ∪V ⊂

n+m 



  xi + B ∪ (−B) + yi ∈ J (B),

i=1

  n whence U ∪ V ∈ I(B). Similarly, if Z ⊂ U ⊂ i=1 xi + B ∪ (−B) + yi ∈ J (B), then Z ∈ I(B). Consequently I(B) fulfils (i) and (ii), and thus it is an ideal. If x ∈ X and U ∈ I(B), then x−U ⊂ x−

n 

n



    xi + B ∪ (−B) + yi = (x − yi ) + B ∪ (−B) − xi ∈ J (B),

i=1

i=1

and consequently x − U ∈ I(B). Thus I(B) fulfils (iv), which means that I(B) is linearly invariant. We have also B ⊂ B ∪ (−B) = 0 + B ∪ (−B) + 0 ∈ J (B), whence B ∈ I(B). Consequently I(B) is a linearly invariant ideal containing B, and since I is the smallest such ideal, we have I ⊂ I(B). Consequently I = I(B). I(B) is called the linearly invariant ideal generated by B. In general I(B) need not be proper, nor a σ-ideal. Some conditions on a set B ⊂ RN which guarantee that I(B) is proper were given by M. Sablik [272]. Let us note (Sablik [272], GerKuczma [116]) that a p.l.i. ideal in RN may contain quite large sets, e.g., sets of infinite measure and of the second category3 . Let X be a set (space). For any set A ⊂ X × X and x ∈ X we write A[x] = {y ∈ X | (x, y) ∈ A} (cf. 2.1). Suppose that we are given an ideal I1 in X and an ideal I2 in X × X. We say that ideals I1 and I2 are conjugate iff for every set A ∈ I2 we have A[x] ∈ I1

I1 -(a.e.) in X,

i.e., iff there exists a set U ∈ I1 such that A[x] ∈ I1 for x ∈ X \ U. This is an abstract version of the Fubini theorem. 3

Such is, e.g., the ideal I(B), where B ⊂ R2 is the first quadrant.

(17.5.1)

17.5. Set ideals

501

We may consider the space R2N as the product RN × RN . It is immediately seen from the definition that the ideals IℵN0 and Iℵ2N as well as the ideals IbN and Ib2N 0 N 2N are conjugate. The ideals I0 and I0 are conjugate in virtue of the Fubini theorem, N whereas IfN and If2N are conjugate in virtue of Theorem 2.1.7. Concerning Im and 2N Im we have the following4 (de Bruijn [64]) Lemma 17.5.2. Let α, β be positive numbers, and let A ⊂ R2N be a set such that m2N e (A)  β. Then !   " x ∈ RN | m2N A[x]  β/α  α. (17.5.2) m2N e e Proof. By Lemma 3.1.2 there exists a measurable set B ⊂ R2N such that A ⊂ B and 5 m2N (B) = m2N e (A). By the Fubini theorem almost all sections B[x] are measurable  (in the sense of the N -dimensional measure), the function ϕ(x) = mN B[x] is defined almost everywhere in RN and is measurable, and    2N mN B[x] dx. m (B) = RN

Thus we have, since mN (B[x])  0, β

m2N e (A)

=m

2N

 (B) = 

  mN B[x] dx

RN



mN (B[x])dx {x∈RN |mN (B[x])β/α}

 whence mN

  " β N ! m x ∈ RN | mN B[x]  β/α , α

!   " x ∈ RN | mN B[x]  β/α  α.

(17.5.3)

Since A ⊂ B, we have A[x] ⊂ B[x] for every x ∈ RN , and       N N B[x] mN e A[x]  me B[x] = m for almost all x ∈ RN . Hence   " !   " ! N N B[x]  β/α x ∈ R N | mN e A[x]  β/α ⊂ x ∈ R | m up to a set of N -dimensional measure zero, whence !   " !   " x ∈ RN | m N  mN x ∈ RN | mN B[x]  β/α , mN e e A[x]  β/α and from (17.5.3) we obtain (17.5.2). 4



We add superscripts to the symbols of the Lebesgue measure in order to indicate the dimension. In the present proof the expressions “almost all”, “almost everywhere” are to be understood in the sense of the ideal I0N .

5

502

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

N 2N Corollary 17.5.1. The ideals Im and Im are conjugate. 2N Proof. Take an arbitrary set A ⊂ R2N , A ∈ Im . This means that β = m2N e (A) < ∞. We have ! " !   !   " N N N x ∈ RN | A[x] ∈ = x ∈ RN | m N / Im e A[x] = ∞} ⊂ x ∈ R | me A[x]  β ,

whence by Lemma 17.5.2 (α = 1) ! " !   " N mN x ∈ RN | A[x] ∈  mN x ∈ RN | mN  1, / Im e e e A[x]  β i.e.,

!

" N N N x ∈ RN | A[x] ∈ ∈ Im / Im , or A[x] ∈ Im   N 2N and Im are conjugate. This means cf. (17.5.1) that Im

N Im -(a.e.) in RN .



Let (X, +) be a group (not necessarily commutative), and suppose that we are given a p.l.i. ideal I in X. With the aid of I we may define two families of subsets of X × X (Ger [105]): ! " Π(I) = A ⊂ X × X | A ⊂ (U × X) ∪ (X × U ), U ∈ I , ! " Ω(I) = A ⊂ X × X | A[x] ∈ I I-(a.e.) in X . The proof of the following simple lemma is left to the reader. Lemma 17.5.3. Let (X, +) be a group, and let I ⊂ P(X) be a p.l.i. ideal [σ-ideal ] in X. Then Π(I) and Ω(I) are p.l.i. ideals [σ-ideals] in X × X, and they are both conjugate with I. If I2 ⊂ P(X × X) is a p.l.i. ideal in X × X such that I and I2 are conjugate, then I2 ⊂ Ω(I). We have also the following Lemma 17.5.4. Let (X, +) be a group, and let I ⊂ P(X) be a p.l.i. ideal in X. If U ∈ I, then M = {(x, y) ∈ X × X | x + y ∈ U } ∈ Ω(I). (17.5.4) Proof. For every x ∈ X we have M [x] = {y ∈ X | x + y ∈ U } = {y ∈ X | y ∈ −x + U } = −x + U ∈ I by Lemma 17.5.1. Consequently M ∈ Ω(I).



Let (X, +) be a group, and let I ⊂ P(X) be a p.l.i. ideal in X, and let (S, +) be a subsemigroup of (X, +) such that S−S =X

(17.5.5)

and6 S∈ / I.

(17.5.6)

Condition (17.5.5) does not imply (17.5.6). E.g., if X =  S = (0, ∞) × (0, ∞) is the first quadrant, and I = I(S), then S fulfils (17.5.5) but not (17.5.6). This example is due to M. Sablik;  cf. Ger [112].

6

R2 ,

17.5. Set ideals

503

The following three lemmas are due to R. Ger [112]. Lemma 17.5.5. Let (X, +) be a group, and let I ⊂ P(X) be a p.l.i. ideal in X, and let (S, +) be a subsemigroup of (X, +) fulfilling (17.5.5) and (17.5.6). Then, for every s, t ∈ X, we have (s + S) ∩ (t + S) ∈ / I. (17.5.7) Proof. Suppose that (17.5.7) does not hold   for some s, t ∈ X. For arbitrary u, v ∈ S we have since (S, +) is a subsemigroup u + S ⊂ S and v + S ⊂ S, whence (s + u + S) ∩ (t + v + S) ⊂ (s + S) ∩ (t + S) ∈ I and (s + u + S) ∩ (t + v + S) ∈ I

(17.5.8)

by (i). By (17.5.5) we can find u, v ∈ S such that −s + t = u − v, whence s + u = t + v, and by (17.5.8) t + v + S ∈ I. By Lemma 17.5.1 S = −(t + v) + (t + v + S) ∈ I, 

which contradicts (17.5.6). Corollary 17.5.2. Under conditions of Lemma 17.5.5 (s + S) ∩ (t + S) = ∅

(17.5.9)

for arbitrary s, t ∈ S. Corollary 17.5.2 may also be formulated as follows (cf. 4.5). Corollary 17.5.3. Under conditions of Lemma 17.5.5 the semigroup (S, +) is left reversible. Lemma 17.5.6. Under conditions of Lemma 17.5.5 we have (S \ U ) − (S \ V ) = X

(17.5.10)

for every U, V ∈ I. Proof. Take an x ∈ X and sets U, V ∈ I. By (17.5.5) x = s − t, where s, t ∈ S. By Lemma 17.5.5







− s + (S \ U ) ∩ − t + (S \ V ) = (−s + S) ∩ (−t + S) \ (−s + U ) ∪ (−t + V ) ∈ / I, since (−s + U ) ∪ (−t + V ) ∈ I by (ii) and Lemma 17.5.1. Thus, in particular,



− s + (S \ U ) ∩ − t + (S \ V ) = ∅.



Consequently there exists a y ∈ − s + (S \ U ) ∩ − t + (S \ V ) , which means that s + y ∈ S \ U , t + y ∈ S \ V , and x = s − t = (s + y) − (t + y) ∈ (S \ U ) − (S \ V ). This means that X ⊂ (S \ U ) − (S \ V ). Since the converse inclusion is trivial, we obtain hence (17.5.10). 

504

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

Lemma 17.5.7. Under conditions of Lemma 17.5.5, for every U ∈ I and every u, s , t ∈ S \ U , there exist s, t ∈ S \ U such that t ∈ u + S, and s − t = s − t . Proof. Taking in Lemma 17.5.6 V = −u + U we obtain

(S \ U ) − S \ (−u + U ) = X. This means that for every y ∈ X there exist z ∈ S \ U and x ∈ S \ (−u + U ) such that y = z − x, i.e., y + x = z ∈ S \ U . Take y = s − t + u and choose a suitable x. Put s = s − t + u + x and t = u + x. Then s = s − t + t, whence s − t = s − t , and

t = u + x ∈ u + S \ (−u + U ) = (u + S) \ U ⊂ S \ U, (17.5.11) since u + S ⊂ S, and, on the other hand, (17.5.11) yields t ∈ u + S. Further, s = s − t + u + x = y + x ∈ S \ U. 

Thus s and t have the required properties.

Now suppose that X is an arbitrary set (space), and let I ⊂ P(X) be an arbitrary proper σ-ideal in X. For every function f : X → [−∞, ∞] and an arbitrary set B ⊂ X, B ∈ / I, we put (Smajdor [285]) I- infess f (x) = sup inf f (x). B

(17.5.12)

A∈I B\A

For X ⊂ RN and I = I0N ∩ P(X) this notion becomes the usual essential infimum of a function. Lemma 17.5.8. Let X be an arbitrary set, let I ⊂ P(X) be a proper σ-ideal in X, and let f : X → [−∞, ∞] be an arbitrary function. Further, let B ⊂ X, B ∈ / I, be an arbitrary set. Then there exists a set A ∈ I such that I- infess f (x) = inf f (x). B

(17.5.13)

B\A

Proof. If I-infess f (x) = −∞, then by (17.5.12) we have inf f (x) = −∞ for every B

B\A

A ∈ I and (17.5.13) holds with every set A ∈ I. So let a = I-infess f (x) > −∞, and B

take a sequence {an }n∈N of real numbers an such that an < an+1 < a for n ∈ N, and lim an = a. By (17.5.12) for every n ∈ N there exists a set An ∈ I such that n→∞

inf f (x) > an .

B\An

Put A =

∞ 

(17.5.14)

An . By (ii’) A ∈ I, and by (17.5.12)

n=1

a  inf f (x). B\A

(17.5.15)

17.6. Almost additive functions

505

On the other hand, by (17.5.14), since An ⊂ A, whence B \ A ⊂ B \ An for n ∈ N, inf f (x)  inf f (x) > an ,

B\A

whence, on letting n → ∞,

B\An

inf f (x)  a.

(17.5.16)

B\A

(17.5.13) results from (17.5.15) and (17.5.16) in view of the definition of a.



17.6 Almost additive functions In the year 1960 P. Erd˝ os [75] raised the following problem: Suppose that a function f : R → R satisfies the relation f (x + y) = f (x) + f (y)

(17.6.1)

for almost all (x, y) ∈ R2 (in the sense of the planar Lebesgue measure). Does there exist an additive function g : R → R such that f (x) = g(x)

(17.6.2)

almost everywhere in R (in the sense of the linear Lebesgue measure)? A positive answer to this question was given by N. G. de Bruijn [64], and, independently, by W. B. Jurkat [159]. J. L. Denny [66] also proved an analogous result (cf. also Denny [67]). N. G. de Bruijn [64] has put this problem into a more general setting. Let (X, +) and (Y, +) be groups, and suppose that we are given two p.l.i. ideals I1 and I2 in X and in X × X, respectively. A function f : X → Y is called I2 almost additive iff relation (17.6.1) holds I2 -(a.e.) in X × X. De Bruijn’s result was that if the groups (X, +) and (Y, +) are commutative, and if the ideals I1 and I2 are conjugate, then every I2 - almost additive function is equal I1 -(a.e.) in X to a function7 g : X → Y satisfying (17.6.1) everywhere in X × X. The Erd˝ os’ problem is the particular case X = Y = R, I1 = I01 , I2 = I02 . We start with the following Lemma 17.6.1. Let (X, +) and (Y, +) be groups (not necessarily commutative), and let I ⊂ P(X) be a p.l.i. ideal in X. Let g1 , g2 : X → Y be homomorphisms such that g1 = g2 I-(a.e.) in X. Then g1 = g2 in X. Proof. Let S ∈ I be such tht g1 (x) = g2 (x) for x ∈ X \ S.

(17.6.3)

Take an arbitrary x ∈ X. By (17.5.ii) and by Lemma 17.5.1 we have S ∪(−S +x) ∈ I, and consequently we can find a y ∈ X \ S ∪(−S +x) . Hence by (17.6.3) g1 (y) = g2 (y) and g1 (x − y) = g2 (x − y), whence, since g1 and g2 are homomorphisms, g1 (x) = g1 (x − y) + g1 (y) = g2 (x − y) + g2 (y) = g2 (x), which means that g1 = g2 in X. 7

I.e., g is a homomorphism from (X, +) into (Y, +).



506

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

R. Ger [111] (cf. also Ger [112]) generalized de Bruijn’s result to the case of non-commutative groups. His theorem runs as follows. Theorem 17.6.1. Let (X, +) and (Y, +) be groups (not necessarily commutative), and suppose that we are given two conjugate p.l.i. ideals I1 and I2 in X and in X × X, respectively. If f : X → Y is an I2 -almost additive function, then there exists a unique homomorphism g : X → Y such that (17.6.2) holds I1 -(a.e.) in X. Proof. Suppose that (17.6.1) holds for (x, y) ∈ (X × X) \ M , where M ∈ I2 . Since I1 and I2 are conjugate, there exists a set U ⊂ X, U ∈ I1 , such that M [x] ∈ I1 for x ∈ X \ U.

(17.6.4) For every x ∈ X the set U ∪(−U +x) belongs to I1 (Lemma 17.5.1), and X \ U ∪(−U

+ x) = ∅. So there exists a w(x),

w(x) ∈ X \ U ∪ (−U + x) . (17.6.5) Relation ∈ / U and x − w(x) ∈ / U , whence by (17.6.4)

(17.6.5) means that w(x)

M w(x) ∈ I1 and M x − w(x) ∈ I1 , and consequently also



 (17.6.6) Ax = M w(x) ∪ − w(x) + M x − w(x) ∈ I1 . Now we prove that for every x ∈ X the expression f (x + y) − f (x)

(17.6.7)

(considered as a function of y) is constant I1 -(a.e.) in X; more exactly, we will prove that for y ∈ X \ A x expression (17.6.7) does not

depend on y. Take a y ∈ X \ Ax . Then y ∈ / M w(x) and w(x) + y ∈ / M x − w(x) , i.e.,     w(x), y ∈ / M and x − w(x), w(x) + y ∈ / M. Hence         f (x + y) = f x − w(x) + f w(x) + y = f x − w(x) + f w(x) + f (y), or

    f (x + y) − f (y) = f x − w(x) + f w(x) ,

y∈ / Ax ,

and the right-hand side of (17.6.8) does not depend on y. Now we define the function g : X → Y by     g(x) = f x − w(x) + f w(x) , x ∈ X.

(17.6.8)

(17.6.9)

We will show that g is homomorphism. Fix arbitrary u, v ∈ X. We may find x, s, t ∈ X such that

x∈ / Au+v ∪ U ∪ − (u + v) + U ,

s∈ / (−x + Av ) ∪ M [x] ∪ U ∪ − (v + x) + U ,

  t∈ / − (v + x + s) + Au ∪ M [v + x + s] ∪ M [s] ∪ − s + M [u + v + x] ,

17.6. Almost additive functions

507

because by (17.5.ii), by Lemma 17.5.1, and by (17.6.4) and (17.6.6) the right-hand sides of the above relations belong to I1 . Put y = x + s, z = v + x + s + t. Then y∈ / Av and z ∈ / Au , whereas x has been chosen so that x ∈ / Au+v . Thus by (17.6.8) and (17.6.9) g(u) = f (u+z)−f (z), g(v) = f (v+y)−f (y), g(u+v) = f (u+v+x)−f (x). (17.6.10) Further, s ∈ / M [x], i.e., (x, s) ∈ / M , and t ∈ / M [v + x + s], i.e., (v + x + s, t) ∈ / M . This implies that f (y) = f (x + s) = f (x) + f (s) and f (z) = f (v + x + s + t) = f (v + x + s) + f (t) = f (v + y) + f (t), whence −f (x) + f (y) = f (s),

−f (v + y) + f (z) = f (t).

(17.6.11)

Finally, t ∈ / M [s], i.e., (s, t) ∈ / M , and s + t ∈ / M [u + v + x], i.e., (u + v + x, s + t) ∈ / M. Hence f (s)+f (t) = f (s+t),

f (u+v+x)+f (s+t) = f (u+v+x+s+t) = f (u+z). (17.6.12)

Now, by (17.6.10), (17.6.11) and (17.6.12) we have g(u + v) − g(v) − g(u)





= f (u + v + x) − f (x) − f (v + y) − f (y) − f (u + z) − f (z)

= f (u + v + x) − f (x) + f (y) − f (v + y) + f (z) − f (u + z) = f (u + v + x) + f (s) + f (t) − f (u + z) = f (u + v + x) + f (s + t) − f (u + z) = 0, i.e., g(u + v) = g(u) + g(v). This holds for arbitrary u, v ∈ X, which means that g is a homomorphism. Now take an arbitrary x ∈ X\ U . By (17.6.4) and (17.6.6) Ax ∪ M [x] ∈ I1 , so  we may find a y ∈ X \ Ax ∪ M [x] . Since y ∈ / Ax , we have by (17.6.8) and (17.6.9) g(x) = f (x + y) − f (y),

(17.6.13)

whereas, since y ∈ / M [x], i.e., (x, y) ∈ / M , relation (17.6.1) holds. (17.6.13) and (17.6.1) yield (17.6.2). Thus (17.6.2) certainly holds for all x ∈ X \ U , i.e., I1 -(a.e.) in X. The uniqueness results from Lemma 17.6.1.  Conversely, we have the following Theorem 17.6.2. Let (X, +) and (Y, +) be groups (not necessarily commutative), and suppose that we are given a p.l.i. ideal I in X. If f : X → Y is a function, and g : X → Y is a homomorphism, and if (17.6.2) holds I-(a.e.) in X, then f is Ω(I)almost additive.

508

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

Proof. Let (17.6.2) hold for x ∈ X \ U , where U ∈ I. Let A = {(x, y) ∈ X × X | x ∈ U or y ∈ U or x + y ∈ U } = (U × X) ∪ (X × U ) ∪ M, where M is defined by (17.5.4). Clearly (U × X) ∪ (X × U ) ∈ Ω(I), whence A ∈ Ω(I) in virtue of Lemmas 17.5.3 and 17.5.4. For (x, y) ∈ (X × X) \ A we have f (x) = g(x), f (y) = g(y) and f (x + y) = g(x + y), whence f (x + y) = g(x + y) = g(x) + g(y) = f (x) + f (y), so (17.6.1) holds for (x, y) ∈ (X × X) \ A, i.e., Ω(I)-(a.e.) in X × X.



The result of Theorem 17.6.1 can further be strengthened in the case of some particular ideals I2 . Suppose that we are given a p.l.i. ideal I in X. If a function f : X → Y is Π(I)-almost additive, then there exists a set U ∈ I such that (17.6.1) holds for all (x, y) ∈ (X × X) \ (U × X) ∪ (X × U ) , or, in other words, for all x, y ∈ X \ U . Conversely, if there exists a set U ∈ I such that (17.6.1) holds for all x, y ∈ X \ U , then f is Π(I)-almost additive. Actually we can say more, as is seen from the following theorem of S. Hartman [137] (cf. also de Bruijn [64]): Theorem 17.6.3. Let (X, +) and (Y, +) be groups (not necessarily commutative), and suppose that we are given a p.l.i. ideal I in X. If a function f : X → Y fulfils (17.6.1) for all x, y ∈ X \ U , where U ∈ I, then f is a homomorphism. Proof. As pointed out above, f is Π(I)-almost additive. Π(I) is a p.l.i. ideal in X × X and I and Π(I) are conjugate (Lemma 17.5.3). By Theorem 17.6.1 there exists a homomorphism g : X → Y such that (17.6.2) holds I-(a.e.) in X, i.e., for x ∈ X \ S, where S ∈ I. Take an arbitrary x ∈ X. By Lemma 17.5.1 −(S  ∪ U ) ∈ I and −x + (S ∪ U ) ∈ I, whence also − (S ∪

U ) ∪ − x + (S ∪ U ) ∈ I. Thus there exists a y ∈ X \ − (S ∪ U ) ∪ − x + (S ∪ U ) . This means that −y ∈ S and x + y ∈ / S, whence f (−y) = g(−y), f (x + y) = g(x + y), and −y ∈ / U, x + y ∈ / U , whence f (x) = f (x + y − y) = f (x + y) + f (−y). Hence f (x) = g(x + y) + g(−y) = g(x + y − y) = g(x), since g is a homomorphism. Thus f = g in X, which means that f is a homomorphism.  Theorem 17.6.3 can be formulated as follows. Corollary 17.6.1. Let (X, +) and (Y, +) be groups (not necessarily commutative), and suppose that we are given a p.l.i. ideal I in X. If a function f : X → Y is Π(I)-almost additive, then f is a homomorphism. Corollary 17.6.1 shows that in Theorem 17.6.2 Ω(I) cannot be replaced by an arbitrary p.l.i. ideal I2 in X × X.

17.6. Almost additive functions

509

A similar conclusion can also be obtained from the following (de Bruijn [64]) 2N -almost additive function. Then f is Theorem 17.6.4. Let f : RN → R be an Im 2N I0 -almost additive. 2N Proof. Suppose that (17.6.1) holds for all (x, y) ∈ R2N \ M , where M ∈ Im , and 2N choose a β < ∞ such that me (M ) < β. Take an arbitrary α > 0. By Lemma 17.5.2 !   " mN x ∈ RN | mN  α. e e M [x]  β/α   " ! Let U = x ∈ RN | mN e M [x]  β/α . Then

mN e (U )  α, and

  mN e M [x]  β/α

for x ∈ RN \ U.

(17.6.14) (17.6.15)

Take an arbitrary x ∈ RN . Since   N N N mN e U ∪ (x − U )  me (U ) + me (x − U ) = 2me (U )  2α < ∞, there exists a w(x) such that w(x) ∈ / U and x − w(x) ∈ / U . Hence for the set Ax defined by (17.6.6) we have mN (17.6.16) e (Ax )  2β/α. The argument in the proof of Theorem 17.6.1 shows that (17.6.8) holds. Define the   N function g : RN → R by (17.6.9) X = RN . By (17.6.14) and (17.6.16) U ∈ Im and N N Ax ∈ Im for arbitrary x ∈ R , and so we may repeat the argument from the proof of Theorem 17.6.1 to show that g is a homomorphism. For every x ∈ RN \U we have by N N (17.6.15) M [x] ∈ Im , whence Ax ∪M [x] ∈ Im , and there exists a y ∈ RN \ Ax ∪M [x] N so that we have (17.6.13) and (17.6.1). Hence g(x) = f (x) for x ∈ RN \ U , i.e., Im N (a.e.) in R . By Lemma 17.6.1 the function g is independent of U , and hence of α. Letting in (17.6.14) α → 0, we obtain (17.6.2) I0N -(a.e.) in RN . Let U ⊂ RN be a set of measure zero such that f (x) = g(x) for x ∈ RN \ U. Let T : R2N → R2N be the linear transform T (x, y) = (x, y − x),

x, y ∈ RN .

Put ! " Z = (x, y) ∈ R2N | x + y ∈ U ,

! " Z0 = (x, y) ∈ R2N | y ∈ U = RN × U.   Thus Z0 is Lebesgue measurable and m2N (Z0 ) = mN RN mN (U ) = 0. By Corollary 2N 2N 3.2.1 the set Z = T (Z0 ) isLebesgue Since  =N m (Z0 ) = 0.  measurable and  m N (Z) 2N N N evidently m U ×R ∪ R ×U = 0, we have U ×R ∪ R ×U ∪Z ∈ I02N , and the argument in the proof of Theorem 17.6.2 shows that f is I02N -almost additive. 

510

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

In the case where X = RN , Y = R, a homomorphism from X to Y is an additive function. In such a case we can infer from a regularity of f to the continuity of g fulfilling (17.6.2) I-(a.e.), at least for some p.l.i. ideals I. The reader will easily formulate the corresponding results (cf. also Exercises 17.12 and 17.13). Finally let us observe that the questions considered in the present section belong to the area of conditional Cauchy equations, or Cauchy equations on restricted domains (cf. the end of 13.6).

17.7 Almost polynomial functions Let I be a proper ideal in R2N . We say that a function f : RN → R is I-almost polynomial of order p iff Δp+1 f (x) = 0 y

I-(a.e.) in R2N .

(17.7.1)

In virtue of Corollary 15.1.2 relation (17.7.1) can also be written as p+1

p+1−k

(−1)

k=0

  p+1 f (x + ky) = 0 k

I-(a.e.) in R2N ,

(17.7.2)

or, equivalently, in the form (−1)p+1 f (x) =

p+1

(−1)p−k

k=1

  p+1 f (x + ky) I-(a.e.) in R2N , k

whence, after dividing both the sides by (−1)p+1 and observing that (−1)−k−1 = (−1)k−1 , f (x) =

p+1

k=1

 (−1)k−1

 p+1 f (x + ky) I-(a.e.) in R2N . k

(17.7.3)

In the present section we are going to prove analogues of Theorems 17.6.1–17.6.4 for almost polynomial functions (Ger [99]). Theorem 17.7.1. Let I be a p.l.i. ideal in RN fulfilling the condition If A ∈ I and a ∈ R, then aA ∈ I.

(i)

If g : RN → R is a polynomial function of order p, and f : RN → R is a function such that f (x) = g(x) I-(a.e.) in RN , (17.7.4) then f is an Ω(I)-almost polynomial function of order p. Proof. Suppose that f (x) = g(x) for x ∈ RN \ S, where S ∈ I. For every fixed k ∈ N ∪ {0} the set ! " Ak = (x, y) ∈ R2N | x + ky ∈ S

(17.7.5)

17.7. Almost polynomial functions

511

belongs to Ω(I). In fact, if k = 0, then ! " A0 = (x, y) ∈ R2N | x ∈ S = S × RN , which is in Ω(I). If k = 0, then  " ! 1 Ak = (x, y) ∈ R2N | ky ∈ S − x = (x, y) ∈ R2N | y ∈ (S − x) , k 1 (S − x) ∈ I by (i) and by Lemma 17.5.1, for every x ∈ RN . This k p+1  Ak ∈ Ω(I). means that Ak ∈ Ω(I). Consequently also

and so Ak [x] =

k=0

If (x, y) ∈ R2N \

p+1 

Ak , then x + ky ∈ / S for k = 0, . . . , p + 1, and by (17.7.5)

k=0

f (x + ky) = g(x + ky) for k = 0, . . . , p + 1. Hence, since g is a polynomial function, p+1

p+1−k

(−1)

k=0

    p+1

p+1 p+1−k p + 1 f (x + ky) = g(x + ky) = 0. (−1) k k k=0

Thus Δp+1 f (x) = 0 for (x, y) ∈ R2N \ y

p+1 

Ak , i.e., Ω(I)-(a.e.) in R2N .



k=0

Lemma 17.7.1. Let I be a p.l.i. ideal in RN fulfilling condition (i). If g1 , g2 : RN → R are polynomial functions of order p such that g1 (x) = g2 (x)

I-(a.e.) in RN ,

then g1 = g2 in RN . Proof. Let g1 (x) = g2 (x) for x ∈ RN \ S,

(17.7.6)

p+1 

1 (S − x). Then x+ ky ∈ /S k=1 k for  k = 1, . . . , p + 1, andby (17.7.6) g1 (x + ky) = g2 (x + ky), k = 1, . . . , p + 1. Hence comp. formula (17.7.3) where S ∈ I. Take an arbitrary x ∈ RN and a y ∈ RN \

g1 (x) =

p+1

k=1

(−1)k−1

    p+1

p+1 p+1 g1 (x + ky) = g2 (x + ky) = g2 (x). (−1)k−1 k k

This shows that g1 = g2 in RN .

k=1



512

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

Theorem 17.7.2. Let I1 and I2 be conjugte p.l.i. ideals in RN and R2N , respectively, such that I1 fulfils condition (i). If f : RN → R is an I2 -almost polynomial function of order p, then there exists a unique polynomial function g : RN → R of order p such that (17.7.4) holds with I = I1 . Proof. Suppose that Δp+1 f (x) = 0 for (x, y) ∈ R2N \ M, y

(17.7.7)

where M ∈ I2 . Since I1 and I2 are conjugate, there exists a set U ⊂ RN , U ∈ I1 , such that M [x] ∈ I1 for x ∈ RN \ U. (17.7.8) By (i) and Lemma 17.5.1 we have for every x ∈ RN Ax =

p+1  k=1

1 (U − x) ∈ I1 . k

(17.7.9)

Thus there exists a function ϕ : RN → RN such that for x ∈ / U,

ϕ(x) = 0 ϕ(x) ∈ / Ax We put g(x) =

p+1

k−1

(−1)

k=1

Since 0 = (1 − 1)p+1 = (17.7.10), for x ∈ RN \ U

g(x) =

k−1

(−1)

k=1

for x ∈ U.

(17.7.11)

   p+1  f x + kϕ(x) . k

p+1  p+1 k k (−1) , k=0

p+1

(17.7.10)

we have

p+1 

(−1)k−1

(17.7.12)

p+1

k=1

k

= 1, whence by

  p+1 f (x) = f (x). k

Thus f (x) = g(x) for x ∈ RN \ U , i.e., I1 -(a.e.) in RN . It remains to show that g is a polynomial function of order p. To this aim we show first that, for every x ∈ RN , g(x) =

p+1

k=1

k−1

(−1)

  p+1 f (x + ky) for y ∈ / Ax . k

(17.7.13)

Take an arbitrary x ∈ RN . If x ∈ U , then ϕ(x) ∈ / Ax by (17.7.11). If x ∈ / U , then by (17.7.10) ϕ(x) = 0. If we had 0 ∈ Ax , then there would exist a k ∈ {1, . . . , p + 1}   1 such that 0 ∈ (U − x) cf. (17.7.9) , whence 0 ∈ U − x, and x ∈ U , contrary to k

17.7. Almost polynomial functions

513

the supposition. Thus ϕ(x) ∈ / Ax for every x ∈ RN . Take an arbitrary y ∈ / Ax . By (17.7.9) we have x + kϕ(x) ∈ / U, x + ky ∈ / U for k = 1, . . . , p + 1, (17.7.14)

whence by (17.7.8) M x + kϕ(x) ∈ I1 and M [x + ky] ∈ I1 for k = 1, . . . , p + 1. Thus there exists a z ∈ RN such that z∈ /

p+1  k=1

 



1  M [x + ky] − ϕ(x) ∪ M x + kϕ(x) − y . k

(17.7.15)



Relation (17.7.15) implies that ϕ(x) + kz ∈ / M [x + ky] and y + kz ∈ / M x + kϕ(x) ,     k = 1, . . . , p + 1, i.e., x + ky, ϕ(x) + kz ∈ / M and x + kϕ(x)y + kz ∈ / M for k = 1, . . . , p + 1. Hence, by (17.7.7) and (17.7.3),   p+1     k−1 p + 1 f x + iϕ(x) = f x + iϕ(x) + k(y + iz) , (−1) k

i = 1, . . . , p + 1,

  p+1     k−1 p + 1 f x + ky + i ϕ(x) + kz , (−1) f x + iϕ(x) = k

i = 1, . . . , p + 1.

k=1

or 

k=1

(17.7.16) Similarly, we have by (17.7.7) and (17.7.3) f (x+ky) =

p+1

(−1)i−1

i=1

    p+1  f x+ky+i ϕ(x)+kz , i

k = 1, . . . , p+1. (17.7.17)

We have by (17.7.12) and (17.7.16)    p+1  f x + iϕ(x) i i=1   p+1   p+1

   i−1 p + 1 k−1 p + 1 = f x + ky + i ϕ(x) + kz (−1) (−1) i k i=1 k=1     p+1 p+1

   k−1 p + 1 i−1 p + 1 = f x + ky + i ϕ(x) + kz , (−1) (−1) k i i=1

g(x) =

p+1

(−1)i−1

k=1

whence (17.7.13) results in view of (17.7.17). Now we take arbitrary u, v ∈ RN . Choose an x ∈ RN \ Au (cf. (17.7.9)), and   p+1  1  

N (Au+kv − x) ∪ M [u + kx] − v . (17.7.18) y ∈ R \ Au ∪ k k=1

514

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

Since x ∈ / Au , we have by (17.7.9) u + kx ∈ / U for k = 1, . . . , p + 1, and by (17.7.8) M [u + kx] ∈ I1 for k = 1, . . . , p + 1 so that the set in brackets on the right-hand side of (17.7.18) belongs to I1 and such a choice of y is possible. By (17.7.18) we have x + ky ∈ / Au+kv for k = 0, . . . , p + 1 (for k = 0 this follows from the choice of x), whence by (17.7.13)   p+1

  i−1 p + 1 f u + kv + i(x + ky) (17.7.19) g(u + kv) = (−1) i i=1   p+1

  i−1 p + 1 = f u + ix + k(v + iy) , k = 0, . . . , p + 1. (−1) i i=1 Similarly (17.7.18) implies that v + iy ∈ / M [u + ix], i.e., (u + ix, v + iy) ∈ / M for i = 1, . . . , p + 1. Hence by (17.7.7) and (17.7.2)   p+1

  p+1−k p + 1 f u + ix + k(v + iy) = 0, i = 1, . . . , p + 1. (17.7.20) (−1) k k=0

Now we have by (17.7.19), (17.7.20), and Corollary 15.1.2,   p+1

p+1 p+1−k p + 1 g(u + kv) (−1) Δv g(u) = k k=0     p+1 p+1

  p+1−k p + 1 i−1 p + 1 = f u + ix + k(v + iy) (−1) (−1) k i i=1 k=0     p+1 p+1

  i−1 p + 1 p+1−k p + 1 = f u + ix + k(v + iy) = 0. (−1) (−1) i k i=1 k=0

This means that g is a polynomial function of order p. The uniqueness of g results from Lemma 17.7.1.  Theorem 17.7.3. Let I be a p.l.i. ideal in RN fulfilling condition (i). If a function f : RN → R fulfils the condition f (x) = 0 Δp+1 y

(17.7.21)

for all x, y ∈ RN \ U (p ∈ N fixed ), where U ∈ I, then f is a polynomial function of order p. Proof. The function f is Π(I)-almost polynomial. It follows by Theorem 17.7.2 that there exists a polynomial function g : RN → R, of order p, such that (17.7.5)  holds with an S ∈ I. Relation (17.7.21) can equivalently be written as cf. (17.7.3)     p+1

1 p + 1 f (x + y) = f (x) + f (x + ky) , (17.7.22) (−1)k p+1 k k=2

and (17.7.22) holds for all x, y ∈ R \ U . N

17.8. Almost convex functions

515

Take an arbitrary x ∈ RN , and choose an x ¯ such that   p+1  1 (S − kx) . x ¯ ∈ RN \ (S ∪ U ) ∪ (x − U ) ∪ 1−k

(17.7.23)

k=2

Put y = x − x ¯. By (17.7.23) x ¯ ∈ S, x ¯ + ky = x ¯ + k(x − x ¯) = kx + (1 − k)¯ x∈ / S, k = 2, . . . , p + 1, whence by (17.7.5) f (¯ x) = g(¯ x),

f (¯ x + ky) = g(¯ x + ky),

k = 2, . . . , p + 1.

(17.7.24)

Further, again by (17.7.23), x ¯∈ / U and y = x − x ¯∈ / U , whence by (17.7.22) we obtain in view of (17.7.24)     p+1

1 k p+1 f (x) = f (¯ x + y) = f (¯ x) + f (¯ x + ky) (−1) p+1 k k=2     p+1

1 k p+1 g(¯ x) + g(¯ x + ky) = g(¯ x + y) = g(x), (−1) = p+1 k k=2

since g is a polynomial function of order p. Thus f = g in RN , i.e., f is a polynomial function of order p.  2N Theorem 17.7.4. Let f : RN → R be an Im -almost polynomial function of order p. 2N Then f is an I0 -almost polynomial function of order p.

Proof. Suppose that (17.7.21) holds for (x, y) ∈ R2N \ M , where m2N e (M ) < ∞. Choose a β such that m2N e (M ) < β < ∞, and take an arbitrary α > 0. Let   U = {x ∈ RN | mN e M [x]  β/α}. By Lemma 17.5.2 mN e (U )  α.

(17.7.25)

By exactly the same argument as in the proof of Theorem 17.7.2 we show that there exists a unique polynomial function g : RN → R, of order p, such that f (x) = g(x) for N x ∈ RN \U , i.e., Im -(a.e.) in RN . By Lemma 17.7.1 the function g is independent of α. Letting in (17.7.25) α → 0 we see that f (x) = g(x) I0N -(a.e.) in RN . An argument as at the end of the proof of Theorem 17.6.4 shows that f is an I02N -almost polynomial function of order p. 

17.8 Almost convex functions In order to extend the most important results of 17.6 to the case of convex functions we must make stronger assumptions about the occurring ideals. Suppose that I1 is an ideal in RN , I2 is an ideal in R2N , and let T : R2N → R2N be a linear transform T (x, y) =

1 (x + y, x − y), 2

x, y ∈ RN .

(17.8.1)

516

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

We assume that I1 and I2 fulfil the conditions: (i) If A ∈ I1 and a ∈ R, then aA ∈ I1 ; (ii) If A ∈ I2 , then T (A) ∈ I2 , where T is given by (17.8.1). Observe that the ideals I1 = I0N , I2 = I02N fulfil conditions (i), (ii) in virtue of Corollaries 3.2.1 and 3.2.2. Similarly, the ideals I1 = IfN , I2 = If2N fulfil conditions (i), (ii), since  the multiplication by  a number a = 0, and transform T , are homeomorphisms. For a = 0 (i) is trivial . Lemma 17.8.1. Let I1 be a p.l.i. σ-ideal in RN fulfilling condition (i). If D ⊂ RN is a non-empty open set, then D ∈ / I1 . Proof. Suppose that there exists a non-empty open set D ⊂ RN such that D ∈ I1 . Let u ∈ D. By Lemma 17.5.1 D − u ∈ I1 , and by (i) n(D − u) ∈ I1 for all n ∈ N, whence, since I1 is a σ-ideal, ∞ 

n(D − u) ∈ I1 .

(17.8.2)

n=1

Take an arbitrary x ∈ RN . Since D − u is a neighborhood of zero in RN , there exists an n ∈ N such that x/n ∈ D − u, whence x ∈ n(D − u). This shows that ∞ 

n(D − u) = RN ,

n=1

and so (17.8.2) implies that RN ∈ I1 , which contradicts the fact that I1 is proper.  Lemma 17.8.2. Let I1 and I2 be conjugate p.l.i. ideals in RN and R2N , respectively, and let A ∈ I2 . Then ! " I1 -(a.e.) in RN . Ux = h ∈ RN | (x, x + h) ∈ A ∈ I1 Proof. h ∈ Ux means (x, x + h) ∈ A, or x + h ∈ A[x], i.e., h ∈ A[x] − x. Thus Ux = A[x] − x,

x ∈ RN .

(17.8.3)

Since A ∈ I2 and I1 and I2 are conjugate, we have A[x] ∈ I1 I1 -(a.e.) in RN whence, by Lemma 17.5.1, also A[x] − x ∈ I1 I1 -(a.e.) in RN . Hence by (17.8.3) Ux ∈ I1 I1 -(a.e.) in RN .  Lemma 17.8.3. Let I1 and I2 be conjugate p.l.i. ideals in RN and R2N , respectively, and assume that I2 fulfils condition (ii). Let A ∈ I2 . Then ! " I1 -(a.e.) in RN . (17.8.4) Vx = h ∈ RN | (x + h, x − h) ∈ A ∈ I1 Proof. Since transform (17.8.1) is invertible, we have (u, v) ∈ A if and only if T (u, v) ∈ T (A), for arbitrary u, v ∈ RN . Thus h ∈ Vx means (x + h, x − h) ∈ A,which is equivalent to T (x+h, x−h) ∈ T (A), or, by (17.8.1), to (x, h) ∈ T (A), i.e., h ∈ T (A)[x]. Consequently Vx = T (A)[x], x ∈ RN . (17.8.5)

17.8. Almost convex functions

517

By (ii) T (A) ∈ I2 , whence T (A)[x] ∈ I1 I1 -(a.e.) in RN , since I1 and I2 are conjugate. (17.8.4) now results from (17.8.5).  Lemma 17.8.4. Let I1 be a proper σ-ideal in RN , let D ⊂ RN , D ∈ / I1 , and let f : D → [−∞, ∞) and g : D → [−∞, ∞) be functions such that I1 -(a.e.) in D.

(17.8.6)

I1 - infess f (x) = I1 - infess g(x).

(17.8.7)

f (x) = g(x) Then D

D

Proof. Let U ∈ I1 be such that for x ∈ D \ U.

f (x) = g(x)

(17.8.8)

By Lemma 17.5.8 there exists a set A ∈ I1 such that I1 - infess f (x) = inf f (x). D

(17.8.9)

D\A

We have A ∪ U ∈ I1 , whence by (17.8.9) I1 - infess f (x) = inf f (x)  D

D\A

inf D\(A∪U)

f (x)  I1 - infess f (x), D

whence inf

D\(A∪U )

f (x) = I1 - infess f (x).

(17.8.10)

D

By (17.8.8) f (x) = g(x) for x ∈ D \ (A ∪ U ) ⊂ D \ U , whence by (17.8.10) I1 - infess f (x) = D

inf

D\(A∪U )

f (x) =

inf

D\(A∪U )

g(x)  I1 - infess g(x). D

Since the roles of f and g are symmetric, we may interchange f and g in the above inequality, whence (17.8.7) results.  Lemma 17.8.5. Let D ⊂ RN be a convex and open set, and for every x ∈ D put8 ! " Dx = h ∈ RN | x − h ∈ D and x + h ∈ D = (x − D) ∩ (D − x).

(17.8.11)

Let I1 be a p.l.i. σ-ideal in RN , and let f : D → [−∞, ∞) and g : D → [−∞, ∞) be functions fulfilling (17.8.6). Then, for every x ∈ D, I1 - infess h∈Dx

8



1 1 f (x + h) + f (x − h) = I1 - infess g(x + h) + g(x − h) . h∈Dx 2 2

Thus Dx is a non-empty (since 0 ∈ Dx ) open set, whence by Lemma 17.8.1 Dx ∈ / I1 .

518

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

Proof. Let U ∈ I1 be such that (17.8.8) holds. We have f (x + h) = g(x + h) for x+h ∈ D\U , i.e., for h ∈ (D \U )−x Similarly, f (x−h) = g(x−h) = (D −x)\(U −x).

for h ∈ (x−D)\(x−U ). If h ∈ Dx \ (U −x)∪(x−U ) , then

h ∈ Dx = (D−x)∩(x−D)

and h ∈ / (U − x) ∩ (x − U ), whence h ∈ (D − x) \ (U − x) ∩ (x − D) \ (x − U ) , and f (x + h) = g(x + h), f (x − h) = g(x − h). Consequently

1

1 f (x + h) + f (x − h) = g(x + h) + g(x − h) 2 2



for h ∈ Dx \ (U − x) ∪ (x − U ) .

By Lemma 17.5.1 U − x ∈ I1 and x − U ∈ I1 , whence also (U − x) ∪ (x − U ) ∈ I1 . Thus the lemma follows from Lemma 17.8.4.  Lemma 17.8.6. Let D ⊂ RN be a convex and open set, let I1 be a p.l.i. σ-ideal in RN fulfilling condition (i), and let f : D → R be a convex function. Then f (x) = I1 - infess h∈Dx

1 f (x + h) + f (x − h) . 2

(17.8.12)

Proof. Fix an x ∈ D, and let A ∈ I1 be arbitrary. By (i) ∞ 

B=



2n A ∪ (−A) ∈ I1 .

n=0

For arbitrary h ∈ Dx \ B we have9 ±2−n h ∈ Dx \ A, n = 0, 1, 2 . . . . By Theorem 6.1.1 (cf. also Exercise 6.2) lim f (x ± 2−n h) = f (x), n→∞

whence

1 f (x + 2−n h) + f (x − 2−n h) = f (x), n→∞ 2 lim

and

1 f (x + h) + f (x − h)  f (x). h∈Dx \A 2 inf

(17.8.13)

On the other hand, we have by the convexity of f f (x) 

1 f (x + h) + f (x − h) 2

for every h ∈ Dx , whence f (x) 

1 f (x + h) + f (x − h) . h∈Dx \A 2 inf

(17.8.14)

9 By Theorems 5.1.1 and 5.1.2 D is a convex set (D = ∅, since 0 ∈ D ), and clearly D is x x x x symmetric so that if h ∈ Dx , then also −h ∈ Dx . Hence, for every h ∈ Dx and n ∈ N,

±2−n h = 2−n (±h) + (1 − 2−n )0 ∈ Dx .

17.8. Almost convex functions

519

By (17.8.13) and (17.8.14) f (x) =

1 f (x + h) + f (x − h) . h∈Dx \A 2 inf

This is valid for arbitrary A ∈ I1 . Taking the supremum over all A ∈ I1 , we obtain (17.8.12).  Corollary 17.8.1. Let D ⊂ RN be a convex and open set, let I1 be a p.l.i. σ-ideal in RN fulfilling condition (i), and let f : D → R and g : D → R be convex functions such that (17.8.6) holds. Then f = g in D. Proof. By Lemmas 17.8.6 and 17.8.5

1 f (x + h) + f (x − h) h∈Dx 2

1 = I1 - infess g(x + h) + g(x − h) = g(x) h∈Dx 2

f (x) = I1 - infess

for every x ∈ D.



Now let D ⊂ RN be a convex and open set, and let f : D → [−∞, ∞) be a function. Let I2 be a proper ideal in R2N . If   f (x) + f (y) x+y  I2 -(a.e.) in D × D, f 2 2 then f is called I2 -almost convex in D. Lemma 17.8.7. Let D ⊂ RN be a convex and open set, and suppose that we are given conjugate p.l.i. σ-ideals I1 and I2 in RN and R2N , respectively, such that I1 fulfils condition (i). If f : D → [−∞, ∞) is an I2 -almost convex function fulfilling (17.8.12), then   f (x) + f (y) x+y  (17.8.15) f 2 2 for every x, y ∈ D. In particular, if f is finite, then it is convex. Proof. By Lemma 17.5.8 for every x ∈ D there exists a set Ax ∈ I1 such that I1 - infess h∈Dx



1 1 f (x + h) + f (x − h) = inf f (x + h) + f (x − h) , 2 h∈Dx \Ax 2

whence by (17.8.12) f (x)  Put

1 f (x + h) + f (x − h) for h ∈ Dx \ Ax . 2

   f (x) + f (y) x+y > Z = (x, y) ∈ D × D | f 2 2

(17.8.16)

.

(17.8.17)

520

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

Since f is I2 -almost convex, Z ∈ I2 , and consequently, since I1 and I2 are conjugate, Z[x] ∈ I1 I1 -(i.e.) in D. Thus there exists a set U ∈ I1 such that Z[x] ∈ I1

for x ∈ D \ U.

(17.8.18)

For every x ∈ D we choose an arbitrary ax > f (x) and we put  # 1

# f (x + h) + f (x − h) < ax . Bx = h ∈ D x # 2 Suppose that for an x ∈ D we have Bx ∈ I1 . For h ∈ Dx \ Bx we have

1 f (x + h) + f (x − h)  ax , 2 whence by (17.8.12) I1 - infess h∈Dx



1 1 f (x + h) + f (x − h)  inf f (x + h) + f (x − h) 2 h∈Dx \Bx 2

1  ax > f (x) = I1 - infess f (x + h) + f (x − h) . h∈Dx 2

The contradiction obtained shows that Bx ∈ / I1 for x ∈ D, and thus, for every x ∈ D and every set V ∈ I1 , we have Bx \ V = ∅,

x ∈ D,

Now take arbitrary x, y ∈ D, and put z = by (17.8.19) there exists an

V ∈ I1 .

(17.8.19)

1 (x + y) ∈ D. By Lemma 17.5.1 and 2



h ∈ Bx \ (x − U ) ∪ (U − x) .

(17.8.20)

In particular, x − h ∈ / U and x + h ∈ / U , whence by (17.8.18) Z[x − h ] ∈ I1 and  Z[x + h ] ∈ I1 . Again by Lemma 17.5.1 and by (17.8.19) there exists an    

h ∈ By \ y − Z[x − h ] ∪ Z[x + h ] − y ∪ (2Az − h ) . (17.8.21) Hence y − h ∈ / Z[x − h ] and y + h ∈ / Z[x + h ], i.e., (x − h , y − h ) ∈ / Z and   (x + h , y + h ) ∈ / Z. Hence by (17.8.17) f (z −

f (x − h ) + f (y − h ) h + h ) , 2 2

(17.8.22)

f (x + h ) + f (y + h ) h + h ) , (17.8.23) 2 2 1 Further, it follows from (17.8.21) that (h + h ) ∈ / Az , whereas by (17.8.20) and 2   (17.8.21) h ∈ Bx ⊂ Dx , h ∈ By ⊂ Dy . This means that x − h , x + h ∈ D and f (z +

17.8. Almost convex functions

521

1 1 (x−h )+(y −h ) ∈ y −h , y+h ∈ D, whence, since D is convex, z − (h +h ) = 2 2

1 1 1 (x + h ) + (y + h ) ∈ D. Consequently (h + h ) ∈ Dz , D, and z + (h + h ) = 2 2 2 1 and thus (h + h ) ∈ Dz \ Az . Hence by (17.8.16) 2      h + h h + h 1 f z+ +f z− . (17.8.24) f (z)  2 2 2 Since h ∈ Bx and h ∈ By , we have

1 f (x + h ) + f (x − h ) < ax , 2

(17.8.25)

1 f (y + h ) + f (y − h ) < ay . 2

(17.8.26)

Relations (17.8.24), (17.8.22), (17.8.23), (17.8.25) and (17.8.26) yield f (z) 

1 1 f (x − h ) + f (y − h ) + f (x + h ) + f (y + h ) < (ax + ay ). 4 2

Letting ax → f (x), ay → f (y), we obtain hence (17.8.15) in view of the definition of z.  With the notation from Lemma 17.8.7, we put g(x) = I1 - infess h∈Dx

1 f (x + h) + f (x − h) , 2

Then g(x) =

inf

h∈Dx \Ax

x ∈ D.

(17.8.27)

1 f (x + h) + f (x − h) < ∞. 2

Thus g is a function g : D → [−∞, ∞). Lemma 17.8.8. Let D ⊂ RN be a convex and open set, and suppose that we are given conjugate p.l.i. σ-ideals I1 and I2 in RN and R2N , respectively, fulfilling conditions (i) and (ii). If f : D → R is an I2 -almost convex function, and g : D → [−∞, ∞) is defined by (17.8.27), then (17.8.6) holds. Proof. Let Z denote set (17.8.17) so that Z ∈ I2 , and let Ux = {h ∈ Dx | (x, x + h) ∈ Z},

Vx = {h ∈ Dx | (x + h, x − h) ∈ Z}.

By Lemmas 17.8.2 and 17.8.3 Ux ∈ I1 and Vx ∈ I1 I1 -(a.e) in D, i.e., there exists a set S ∈ I1 such that Ux ∈ I1

and

Vx ∈ I1

for x ∈ D \ S.

(17.8.28)

522

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

Fix an arbitrary set A ∈ I1 and put for every x ∈ D \ S Bx =

∞ 



2n A ∪ (−A) ∪ Ux ∪ (−Ux ) ∪ Vx .

n=0

By (i) and (17.8.28) we have Bx ∈ I1 (x ∈ D \ S), since I1 is a σ-ideal. For every h ∈ Dx \ Bx we have ±2−n h ∈ / A,

±2−n h ∈ / Ux ,

2−n h ∈ / Vx ,

n = 0, 1, 2, . . .

(17.8.29)

In particular, h ∈ / Vx , whence (x + h, x − h) ∈ / Z, and by (17.8.17) f (x) 

1 f (x + h) + f (x − h) . 2

Consequently, in view of (17.8.27),

1 f (x + h) + f (x − h) h∈Dx \Bx 2

1  I1 - infess f (x + h) + f (x − h) = g(x). h∈Dx 2

f (x) 

inf

(17.8.30)

On the other hand, we have by (17.8.29) for h ∈ Dx \ Bx (x, x ± 2−n h) ∈ / Z,

n = 0, 1, 2, . . . ,

whence by (17.8.17) f (x ± 2−(n+1) h) = f



x + x ± 2−n h 2

 

f (x) + f (x ± 2−n h) , 2

n = 0, 1, 2, . . .

Hence  

1 f x ± 2−(n+1) h − f (x)  f (x ± 2−n h) − f (x) , 2

n = 0, 1, 2, . . . ,

and 



lim sup f (x ± 2−n h) − f (x) = lim sup f x ± 2−(n+1) h − f (x) n→∞

n→∞

 

1  lim sup f x ± 2−n h − f (x) . 2 n→∞

Hence

  



lim sup f x ± 2−n h − f (x) = lim sup f x ± 2−n h − f (x)  0, n→∞

or

n→∞

  lim sup f x ± 2−n h  f (x). n→∞

17.8. Almost convex functions

Hence also lim sup n→∞

523

  

1  f x + 2−n h + f x − 2−n h  f (x). 2

(17.8.31)

By (17.8.29) ±2−n h ∈ / A, n = 0, 1, 2, . . . , and so we obtain from (17.8.31)

1 f (x + h) + f (x − h)  f (x), h∈Dx \A 2 inf

  and, since A has been arbitrary set from I1 , cf. (17.8.27) ,

1 f (x + h) + f (x − h) = h∈Dx 2

1 f (x + h) + f (x − h)  f (x). = sup inf A∈I1 h∈Dx \A 2

g(x) = I1 - infess

(17.8.32)

Inequalities (17.8.30) and (17.8.32) yield the equality f (x) = g(x), valid for x ∈ D \ S, i.e., I1 -(a.e.) in D.  Lemma 17.8.9. Let D ⊂ RN be a convex and open set, and suppose that we are given a p.l.i. ideal I1 in RN fulfilling condition (i). If f : [−∞, ∞) is an Ω(I1 )-almost convex function, and g : D → [−∞, ∞) is a function fulfilling (17.8.6), then g is Ω(I1 )-almost convex. Proof. Let for x ∈ D \ S,

f (x) = g(x)

(17.8.33)

where S ∈ I1 , and put A = {(x, y) ∈ D × D | x ∈ S, or y ∈ S, or x + y ∈ 2S} = (S × D) ∪ (D × S) ∪ M,   where M is defined by (17.5.4) with U = 2S ∈ I1 cf. (i) . Clearly (S ×D)∪(D ×S) ∈ Ω(I1 ), whereas M ∈ Ω(I1 ) in virtue of Lemma 17.5.4. Hence A ∈ Ω(I1 ). Let Z be the set defined by (17.8.17). Since f is Ω(I1 )-almost convex, Z ∈ Ω(I1 ), whence also A ∪ Z ∈ Ω(I1 ). For (x, y) ∈ (D × D) \ (A ∪ Z) we have     x+y x+y =g , f (x) = g(x), f (y) = g(y), f 2 2 and

 f

Hence

 g

x+y 2



 =f

x+y 2 x+y 2

 

f (x) + f (y) . 2



f (x) + f (y) g(x) + g(y) = 2 2



holds for (x, y) ∈ (D × D) \ (A ∪ Z), i.e., Ω(I1 )-(a.e.) in D × D.



524

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

The following two theorems (Kuczma [178])are analogues of Theorems 17.6.2 and 17.6.1. Theorem 17.8.1. Let D ⊂ RN be a convex and open set, and let I1 be a p.l.i. ideal in RN fulfilling condition (i). If g : D → R is a convex function, and f : D → [−∞, ∞) is a function such that (17.8.6) holds, then f is Ω(I1 )-almost convex. Proof. We always have ∅ ∈ Ω(I1 ), so g is, in particular, Ω(I1 )-almost convex. The theorem results now from Lemma 17.8.9 (the roles of f and g are interchanged).  Theorem 17.8.2. Let D ⊂ RN be a convex and open set, and suppose that we are given conjugate p.l.i. σ-ideals I1 and I2 in RN and R2N , respectively, fulfilling conditions (i) and (ii). If f : D → R is an I2 -almost convex function, then there exists a unique convex function g : D → R such that f (x) = g(x) I1 -(a.e.) in D. Proof. Define g : D → [−∞, ∞) by (17.8.27). By Lemma 17.8.8 relation (17.8.6) holds. By Lemma 17.8.5 and (17.8.27)

1 f (x + h) + f (x − h) 2

1 = I1 - infess g(x + h) + g(x − h) . h∈Dx 2

g(x) = I1 - infess h∈Dx

(17.8.34)

By Lemma 17.5.3 we have I2 ⊂ Ω(I1 ), whence it follows that f is Ω(I1 )-almost convex. By (17.8.6) and Lemma 17.8.9 the function g is Ω(I1 )-almost convex. Using Lemmas 17.8.7 and 17.5.3 we infer from (17.8.34) that   x+y g(x) + g(y) g  (17.8.35) 2 2 for all x, y ∈ D. In virtue of (17.8.6) there exists a set S ∈ I1 such that (17.8.33) holds. Take an arbitrary x ∈ D. We may find a y ∈ D \ S ∪ (2S −x) . Then and, since  y ∈D \ S  x+y x+y x+y D is convex, ∈ D \ S. Hence g(y) = f (y), g =f , and by 2 2 2 (17.8.35)     x+y x+y − g(y) = 2f − f (y) > −∞. g(x)  2g 2 2   Thus g is finite, and hence cf. (17.8.35) it is a convex function g : D → R. The uniqueness of g results from Corollary 17.8.1. 

17.9 Almost subadditive functions The results concerning almost subadditive functions (Ger [110]) are at least satisfactory among those in this chapter. Let I2 be a p.l.i. ideal in R2N fulfilling the following condition. Let T : R2N → R2N be the linear transform: T (x, y) = (x + y, y),

x, y ∈ RN .

(17.9.1)

17.9. Almost subadditive functions

525

We assume that: (i) For every A ∈ I2 also T (A) ∈ I2 , where T is given by (17.9.1). However, let us observe that most important ideals in R2N (listed in 17.5) fulfil condition (i).  Let C ⊂ RN be a set fulfilling condition (16.1.3). This means that (C, +) is a subsemigroup of (RN , +) . A function f : C → R is called I2 -almost subadditive in C iff the inequality f (x + y)  f (x) + f (y) (17.9.2) holds I2 -(a.e.) in C × C. For every x ∈ C write Cx = C ∩ (x − C) = { h ∈ C | x − h ∈ C }.

(17.9.3)

We will need the following Lemma 17.9.1. Let C ∈ RN be a set fulfilling (16.1.3). For every x, y ∈ C and h ∈ Cx , we have Cy ⊂ Cx+y − h. (17.9.4) Proof. Take arbitrary x, y ∈ C, h ∈ Cx , and t ∈ Cy . By (17.9.3) t, y − t, h, x − h ∈ C, whence also t + h ∈ C and y − t + x − h ∈ C, since C fulfils (16.1.3). Hence t ∈ C − h and t ∈ (x + y − C) − h, i.e.,

t ∈ C ∩ (x + y − C) − h = Cx+y − h. 

Hence (17.9.4) follows.

In the sequel we assume that I1 is a p.l.i. σ-ideal in RN fulfilling the condition Cx ∈ / I1

for every x ∈ C,

(17.9.5)

where Cx is defined by (17.9.3). Condition (17.9.5) implies, since Cx ⊂ C, that also C∈ / I1 .

(17.9.6)

Lemma 17.9.2. Let C ⊂ RN be a set fulfilling (16.1.3), and suppose that we are given conjugate p.l.i. σ-ideal I1 in RN fulfilling condition (17.9.5) and p.l.i. ideal I2 in R2N . Let f : C → R be an I2 -almost subadditive function. Then the function ϕ : C → R defined by10

x ∈ C, (17.9.7) ϕ(x) = −I1 - infess f (h) − f (x + h) , h∈C

is subadditive, and fulfils the condition ϕ(x)  f (x) 10

I1 -(a.e.) in C.

(17.9.8)



The expression −I1 - infess − F (x) is also denoted by I1 - supess F (x) = inf sup F (x), and is B

B

A∈I1 B\A

/ I1 . referred to as the essential supremum (with respect to I1 ) of F on the set B ∈

526

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

Proof. By Lemma 17.5.8 for every x ∈ C there exists a set Ax ∈ I1 such that

ϕ(x) = sup f (x + h) − f (h) . (17.9.9) h∈C\Ax

Take arbitrary x, y ∈ C. Then we have by (17.9.9) ϕ(x)  f (x + h) − f (h)

for h ∈ C \ Ax ,

(17.9.10)

ϕ(y)  f (y + k) − f (k)

for k ∈ C \ Ay .

(17.9.11)

Let B = Ax ∪ (Ay − x) so that B ∈ I1 . By (17.9.6) C \ B = ∅, and thus there exists a t ∈ C \ B. For such a t we have t ∈ / Ax and x + t ∈ / Ay , whence, taking in (17.9.10) and (17.9.11) h = 1 and k = x + t, we get ϕ(x)  f (x + t) − f (t),

ϕ(y)  f (y + x + t) − f (x + t),

and ϕ(x) + ϕ(y)  f (x + y + t) − f (t). Hence ϕ(x) + ϕ(y)  sup



f (x + y + t) − f (t)

t∈C\B



= − inf f (t) − f (x + y + t) t∈C\B 

  inf − inf f (t) − f (x + y + t) B∈I1 t∈C\B

= − sup inf f (t) − f (x + y + t) t∈C\B B∈I1

= −I1 - infess f (t) − f (x + y + t) = ϕ(x + y). t∈C

Thus ϕ is subadditive. Since f is I2 -almost subadditive, there exists a set M ∈ I2 such that (17.9.2) holds for all (x, y) ∈ (C × C) \ M . Since I1 and I2 are conjugate, there exists a set U ⊂ RN , U ∈ I1 , such that M [x] = {y ∈ RN | (x, y) ∈ M } ∈ I1

for x ∈ RN \ U.

Take an x ∈ C \ U ⊂ RN \ U . Then for h ∈ C \ M [x] we have f (x + h)  f (x) + f (h), and

ϕ(x) = I1 - infess f (h) − f (x + h) h∈C

= − inf sup f (x + h) − f (h) A∈I1 h∈C\A



sup h∈C\M[x]



sup h∈C\M[x]

Thus ϕ satisfies (17.9.8).



f (x + h) − f (h)



f (x) + f (h) − f (h) = f (x).

17.9. Almost subadditive functions

527

It remains

to show that ϕ is finite. Take an arbitrary x ∈ C and an h ∈ Cx \ U ∪ (x − U ) , where Cx is given by (17.9.3). Such an h exists in virtue of (17.9.5). Then h ∈ C, x − h ∈ C, h ∈ / U, x − h ∈ / U . As we have just proved, this implies that ϕ(x)  f (x) and ϕ(x − h)  f (x − h). Since ϕ is subadditive, we have ϕ(x)  ϕ(x − h) + ϕ(h)  f (x − h) + f (h) < ∞. On the other hand, by (17.9.9) we have for arbitrary h ∈ C \ Ax ϕ(x)  f (x + h) − f (h) > −∞. 

Thus ϕ is finite.

Lemma 17.9.3. Let C ⊂ R be a set fulfilling (16.1.3), and suppose that we are given conjugate p.l.i. σ-ideal I1 in RN fulfilling condition (17.9.5) and p.l.i. ideal I2 in R2N fulfilling condition (i). Let f : C → R be an I2 -almost subadditive function. Then the function Φ : C → R defined by

Φ(x) = I1 - infess f (x − h) − f (h) , x ∈ C, (17.9.12) N

h∈Cx

is subadditive and fulfils the condition f (x)  Φ(x)

I1 -(a.e.) in C.

(17.9.13)

Proof. By Lemma 17.5.8 for every x ∈ C there exists a set Ax ∈ I1 such that

f (x − h) + f (h) . (17.9.14) Φ(x) = inf h∈Cx \Ax

Relation (17.9.14) implies that Φ(x) < ∞

for x ∈ C.

(17.9.15)

Let M ∈ I2 and U ∈ I1 have the same meaning as in the proof of Lemma 17.9.2.  Take arbitrary x, y ∈ C and a > Φ(x), b > Φ(y) cf. (17.9.15) . Since U ∪ (x − / I1 cf. (17.9.5) , we have by (17.9.12) U ) ∈ I1 and Cx ∈

a > Φ(x)  inf f (x − h) + f (h) , h∈Cx \[U ∪(x−U)]



and so there exists an h ∈ Cx \ U ∪ (x − U ) such that f (x − h) + f (h) < a.

(17.9.16)

By Lemma 17.9.1 we have (17.9.4), whence also  

Cy \ M [h] ∪ y − M [x − h] ∪ (Ax+y − h) ⊂  

⊂ (Cx+y − h) \ M [h] ∪ y − M [x − h] ∪ (Ax+y − h)

 

= (Cx+y \ Ax+y ) − h \ M [h] ∪ y − M [x − h] .

528

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

Hence, since h ∈ / U and x − h ∈ / U, b > Φ(y)  



inf

k∈Cy \[M[h]∪(y−M[x−h])∪(Ax+y −h)

inf

f (y − k) + f (k)

f (y − k) + f (k) ,

k∈[(Cx+y \Ax+y )−h]\[M[h]∪(y−M[x−h])]

 

and so there exists a k ∈ (Cx+y \ Ax+y ) − h \ M [h] ∪ y − M [x − h] such that f (y − k) + f (k) < b.

(17.9.17)

We have by (17.9.14) Φ(x + y) =

inf

t∈Cx+y \Ax+y



f (x + y − t) + f (t) ,

whence, since h + k ∈ Cx+y \ Ax+y , we get   Φ(x + y)  f (x + y) − (h + k) + f (h + k).

(17.9.18)

Since k ∈ / M [h], i.e., (h, k) ∈ / M , we have f (h + k)  f (h) + f (k),

(17.9.19)

and since k ∈ / y − M [x − h], i.e., (x − h, y − k) ∈ / M , we have   f (x + y) − (h + k) = f (x − h + y − k)  f (x − h) + f (y − k).

(17.9.20)

Relations (17.9.18), (17.9.20), (17.9.19), (17.9.16) and (17.9.17) imply Φ(x + y)  f (x − h) + f (y − k) + f (h) + f (k) < a + b, whence, on letting a → Φ(x) and b → Φ(y), we obtain Φ(x + y)  Φ(x) + Φ(y). Consequently Φ is subadditive. By (i) we have T (M ) ∈ I2 , where T is given by (17.9.1), whence T (M )[x] ∈ I1 I1 -(a.e.) in RN . Consequently there exists a set V ∈ I1 such that T (M )[x] ∈ I1

for x ∈ / V.

(17.9.21)

−1 But h ∈ T (M )[x] means (x, h) ∈ T (M ), or, equivalently, T   (x, h) ∈ M . According −1 to (17.9.1) T (x, h) = (x − h, h). Therefore cf. (17.9.21)

{h ∈ RN | (x − h, h) ∈ M } ∈ I1

for x ∈ / V.

Now take arbitrary x ∈ C \ V and a > Φ(x). Then by (17.9.22) W = {h ∈ Cx | (x − h, h) ∈ M } ⊂ {h ∈ RN | (x − h, h) ∈ M } ∈ I1 ,

(17.9.22)

17.9. Almost subadditive functions

529

whence by (17.9.12) a > Φ(x) 

inf



h∈Cx \W

f (x − h) + f (h) .

So we can find an h ∈ Cx \ W such that f (x − h) + f (h) < a.

(17.9.23)

But since h ∈ / W , we have (x − h, h) ∈ / M , i.e., f (x − h) + f (h)  f (x − h + h) = f (x).

(17.9.24)

Relations (17.9.23) and (17.9.24) yield f (x) < a, whence on letting a → Φ(x) we obtain f (x)  Φ(x). Hence (17.9.13) results. It remains to show that Φ is finite. Take an arbitrary x ∈ C. Since V ∈ I1 , we have V − x ∈ I1 , and by (17.9.6) C \ (V − x) ∈ / I1 . Therefore there exists a y ∈ C \ (V − x). This means that x + y ∈ / V , whereas by (16.1.3) x + y ∈ C. Thus x + y ∈ C \ V , which implies, as we have just shown, that Φ(x + y)  f (x + y) > −∞. Since Φ is subadditive, we have −∞ < Φ(x + y)  Φ(x) + Φ(y), which shows that Φ(x) > −∞. This together with (17.9.15) implies that Φ is finite.  In order to prove a weak analogue of Theorem 17.6.1 we must make still stronger assumptions. For any function f : RN → R and any ε > 0 put Af (ε) = {x ∈ RN | f (x) < ε and f (−x) < ε}.

(17.9.25)

Theorem 17.9.1. Let I1 and I2 be conjugate p.l.i. σ-ideal in RN and p.l.i. in R2N fulfilling condition (i), respectively. Let f : RN → R be an I2 -almost subadditive function fulfilling the condition Af (ε) ∈ / I1

for every ε > 0,

(17.9.26)

where Af (ε) is given by (17.9.25). Then there exists a subadditive function g : RN → R such that f (x) = g(x) I1 -(a.e) in RN . (17.9.27)

530

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

Proof. In this case C = RN , whence by (17.9.3)   Cx = C ∩ (x − C) = RN ∩ x − RN = RN ∩ RN = RN ∈ / I1 , since I1 is proper. Hence condition (17.9.5) is satisfied, and we can use Lemmas 17.9.2 and 17.9.3. Let ϕ and Φ be defined by (17.9.7) and (17.9.12), respectively, with C = RN . Our theorem will be proved when we show that I1 -(a.e.) in RN .

ϕ(x) = Φ(x)

Then we may take as g any of ϕ and Φ. Then g is subadditive and satisfies (17.9.27) in virtue of Lemmas 17.9.2 and 17.9.3. Clearly we have by (17.9.8) and (17.9.13) ϕ(x)  Φ(x)

I1 -(a.e.) in RN .

Thus S = {x ∈ RN | ϕ(x) > Φ(x)} ∈ I1 . We will show that ϕ(x) = Φ(x) for x ∈ / S. For an indirect proof suppose that there exists an x ∈ / S such that ϕ(x) = Φ(x). By the definition of S we must have ϕ(x) < Φ(x), and we can find an ε > 0 such that ϕ(x) + 2ε < Φ(x). Put

(17.9.28)

! " Z0 = h ∈ RN | f (x + h) − f (h) + 2ε > f (x − h) + f (h) ,

/ I1 . Let A1 and A2 be sets from I1 such that and suppose that Z0 ∈



I1 - infess f (h) − f (x + h) = inf f (h) − f (x + h) , h∈RN h∈RN \A1



I1 - infess f (x − h) + f (h) = inf f (x − h) + f (h) , h∈RN

h∈RN \A2

/ I1 . For (cf. Lemma 17.5.8). Then we have A1 ∪ A2 ∈ I1 , whence Z0 \ (A1 ∪ A2 ) ∈ h ∈ Z0 \ (A1 ∪ A2 ) we have

f (x + h) − f (h)  sup f (x + h) − f (h) h∈Z0 \A1



sup



h∈RN \A1

f (x + h) − f (h)



f (h) − f (x + h) 1

= −I1 - infess f (h) − f (x + h) = ϕ(x), =−

inf

h∈RN \A

h∈RN

whence by (17.9.28) f (x + h) − f (h) + 2ε  ϕ(x) + 2ε < Φ(x).

17.9. Almost subadditive functions

531

Further, for such h,

f (x − h) + f (h) h∈Z0 \A2

 inf f (x − h) + f (h) N h∈R \A2

= I1 - infess f (x − h) + f (h) = Φ(x).

f (x − h) + f (h) 

inf



h∈RN

Thus, for h ∈ Z0 \ (A1 ∪ A2 ), f (x + h) − f (h) + 2ε < f (x − h) + f (h), which contradicts the definition on Z0 . Consequently Z0 belongs to I1 , and so also the set Z = Z0 ∪ (−Z0 ). / I1 , and hence there exists an h ∈ Af (ε) \ Z. For such an h By (17.9.26) Af (ε) \ Z ∈ we have ε − f (h) > 0, ε − f (−h) > 0, f (x + h) − f (h) + 2ε  f (x − h) + f (h), and f (x − h) − f (−h) + 2ε  f (x + h) + f (−h), whence f (x + h) < f (x + h) − f (h) + ε  f (x − h) + f (h) − ε < f (x − h) < f (x − h) − f (−h) + ε  f (x + h) + f (−h) − ε < f (x + h), which is impossible. Consequently we must have ϕ(x) = Φ(x).



Note that there is no uniqueness statement in Theorem 17.9.1. Condition (17.9.26) is certainly fulfilled if the limit lim f (x) exists and is nonx→0

positive (cf. Exercise 17.7), but it can also be fulfilled for very irregular f (cf. Ger [110]). If the conditions of Theorem 17.9.1 are not fulfilled, then the estimation ϕ(x)  f (x)  Φ(x)

I1 -(a.e.) in C

resulting from Lemmas 17.9.2 and 17.9.3, can be very rough (cf. Exercise 17.15).

Exercises 1. Let M ∈ N be fixed, and let E = {t ∈ R | t = n/M, n = 0, . . . , M }. Let f : RN → R be a function such that f (x + y) − f (x) − f (y) ∈ E

for every x, y ∈ RN .

(∗)

Show that f can be represented as f = g +f0, where g : RN → R is additive, and f0 : RN → R fulfils (∗) and the condition |f0 (x)|  1 for x ∈ RN . (Forti [89].) 2. Let f, g : RN → R be functions such that for a certain p ∈ N and ε > 0 |Δph f (x) − g(h)|  ε

532

Chapter 17. Nearly Additive Functions and Nearly Convex Functions

holds for all x, h ∈ RN . Show that there exist a polynomial function γ : RN → R of order p and a symmetric p-additive function F : RpN → R such that   |f (x) − γ(x)|  2p+2 ε, |g(x) − ϕ(x)|  1 + 2p+2 ε for all x ∈ RN , where ϕ : RN → R is the diagonalization of F (Albert-Baker [11]). 3. Let f : RN → R be a function such that for a certain p ∈ N and ε > 0 |Δph f (x) − p!f (h)|  ε holds for all x, h ∈ RN . Show that there exists a unique symmetric p-additive function F : RpN → R such that   for all x ∈ RN , |f (x) − ϕ(x)|  1 + 2p+2 ε where ϕ : RN → R is the diagonalization of F (Albert-Baker [11]). 4. Let D ⊂ RN be a convex and open set, and put, for every ε > 0, !   " Dε = x ∈ RN | d x, RN \ D  ε ,

5.

6.

7. 8. 9. 10.

11.

where d is the Euclidean distance in RN . Show that if Dε = ∅, then Dε is convex. Let D ⊂ RN be an open and convex set. Show that there exists a sequence {Cn }n∈N of convex and compact sets such that Cn ⊂ Cn+1 for n ∈ N, and ∞  int Cn = D. n=1 Hint: Let Bn , n ∈ N, be the closed ball

centered at the origin and with the radius n. Consider the sets Bn ∩ D1/n . Show that if A ⊂ RN is a closed, but not compact set, then the set conv A need not be closed.

Hint: Take A = Gr(f ), where f : RN → R is given by f (x) = e−|x|, x ∈ RN −1 . Let I be a p.l.i. σ-ideal in RN . Show that I contains no non-empty open sets. Let (X, +) be a group, and let I be a p.l.i. ideal in X. Show that if the sets P, Q ⊂ X are such that Q ∈ / I and X \ P ∈ I, then P + Q = X. (Ger [106]). Let (X, +) be a group, and let (K, +) be a subgroup of (X, +). Let I be a p.l.i. ideal in X. Show that if X \ K ∈ I, then K = X. (Ger [106].) Let (X, +) be a group, let (K, +) be a subgroup of (X, +), and let K  = X \ K. Let I be a p.l.i. ideal in X. Show that if S, T ∈ I, then (K \ S) + (K  \ T ) = K  . (Ger [106].) Let (X, +) and (Y, +) be abelian groups, and for every set A ⊂ X × X write AT = {(x, y) ∈ X × X | (y, x) ∈ A}. Let I1 and I2 be conjugate p.l.i. ideals in X and in X × X, respectively, such that AT ∈ I2 for every set A ∈ I2 . Let f, g, h : X → Y be functions such that f (x + y) = g(x) + h(y)

I2 -(a.e.) in X × X.

17.9. Almost subadditive functions

533

Show that there exists a unique homomorphism F : X → Y and constants a, b ∈ Y such that

12.

13.

14.

15.

16.

f (x) = F (x) + a + b g(x) = F (x) + a

I1 -(a.e.) in X, I1 -(a.e.) in X,

h(x) = F (x) + b

I1 -(a.e.) in X.

(Ger [105]). Hint: Let M0 = {(x, y) ∈ X × X | f (x + y) = g(x) + h(y)}, and let M = M0 ∪ (M0 )T . There exists a set U ∈ I1 such that M [x] ∈ I1 for x ∈ X \ U . Take an x0 ∈ X \U . Then g(x0 )+h(x) = g(x)+h(x0 ) = f (x0 +x) for x ∈ / M [x0 ]. With a = −h(x0 ), b = −g(x0 ), this yields g(x) = f (x0 + x) + a , h(x) = f (x0 + x) + b I1 -(a.e.) in X. Hence f (x + y) = f (x0 + x) + f (x0 + y) + a + b Ω(I1 )-(a.e.)

in X × X. The function Φ(x) = f (2x0 + x) + a + b is Ω(I1 )-almost additive. Let D ⊂ RN be a convex and open set, and let f : D → R be a measurable I02N almost convex function. Show that there exists a continuous convex function g : D → R such that f = g I0N -(a.e.) in D. Let D ⊂ RN be a convex and open set, and let f : D → R be an If2N -almost convex function which is bounded above on a set T ⊂ D, where T is of the second category and has the Baire property. Show that there exists a continuous convex function g : D → R such that f = g I0N -(a.e.) in D. Let C ⊂ RN be a set fulfilling (16.1.3), and let f : C → R be a subadditive function. Let I be a p.l.i. ideal in RN , and let f : C → R be a function such that f = g I-(a.e.) in C. Show that f is Ω(I)-almost subadditive. The function f : R → R given by f (x) = 3 + sin x is subadditive, and hence, in particular, I02N -almost subadditive. Show that for this particular f the functions ϕ and Φ given by (17.9.7)# and #(17.9.12), respectively, (I1 = I0N ), are ϕ(x) = x# x # 2| sin | and Φ(x) = 6 − 2 #sin #. (Ger [110].) 2 2 With the notation and assumptions of Lemmas 17.9.2 and 17.9.3, show that the function ϕ˜ : C → R given by ϕ(x) ˜ = ϕ(x)+I1 - infess f (h)−ϕ(h) is subadditive h∈C ! and satisfies ϕ  ϕ˜  f I1 -(a.e.) in C. Similarly, if c = min I1 - infess Φ(h) − h∈C



" 9 : C → R given f (h) , inf Φ(s) + Φ(t) − Φ(s + t) , then the function Φ (s,t)∈C×C

9 = Φ − c is subadditive and satisfies f  Φ 9 < Φ I1 -(a.e.) in C. (Ger [110].) by Φ

Chapter 18

Extensions of Homomorphisms 18.1 Commutative divisible groups Let (X, +) be a group, and (S, +) a subsemigroup (i.e., a semigroup such that S ⊂ G and the operation + is the same as in X; cf. 4.5). Let (Y, +) ba another group, and let g : S → Y be a homomorphism1 . The problem with which we shall deal in 18.1–18.4 is the following. Does there exist a homomorphism f : X → Y such that f | S = g? The main result in this section (cf. Dhombres-Ger [70], Balcerzyk [18]) reads as follows. Theorem 18.1.1. Let (X, +) and (Y, +) be commutative groups such that Y is divisible, let (S, +) be a subsemigroup of (X, +), and let g : S → Y be a homomorphism. Then there exists a homomorphism f : X → Y such that f | S = g. Proof. Let R be the family of all couples (R, h) such that (R, +) is a subsemigroup of (X, +), and h : R → Y is a homomorphism: h(x + y) = h(x) + h(y)

(18.1.1)

for all x, y ∈ R. An order can be defined in R in the usual way: (R1 , h1 ) ≺ (R2 , h2 ) iff (R1 , +) is a subsemigroup of (R2 , +) and h2 | R1 = h1 . It is easy to see that every chain L ⊂ R has  an upper bound in R: this upper bound is the couple (R0 , h0 ) such that R0 = R and h0 | R = h on every R such that (R, h) ∈ L. By the Lemma (R, h)∈L

of Kuratowski-Zorn (Theorem 1.8.1) there exists in R a maximal element (T, f ) such that (S, g) ≺ (T, f ). The proof will be complete if we show that T = X. So suppose that T = X (of course, necessarily T ⊂ X), and let x ∈ X \ T . Consider the set R = {a ∈ X | a = b + nx, b ∈ T, n ∈ N0 = N ∪ {0}}.      Let a, a ∈ R so that a = b + nx, a =  b + n x, b, b ∈ T , n, n ∈ N0 . Then a + a =    b + b + (n + n )x, where b + b ∈ T since (T, +) is a semigroup and n + n ∈ N0 . Consequently a + a ∈ R, which shows that (R, +) is a semigroup, and obviously a 1

This means that g satisfies g(x + y) = g(x) + g(y) for all x, y ∈ S.

536

Chapter 18. Extensions of Homomorphisms

subsemigroup of (X, +). Now we define an element y ∈ Y . If (Nx) ∩ (T − T ) = ∅, then y may be an arbitrary element of Y . If there is an n ∈ N such that nx ∈ T − T , i.e., nx = t1 − t2 with t1 , t2 ∈ T , then y=

f (t1 ) − f (t2 ) , n

(18.1.2)

which is meaningful, since Y is divisible. To show that definition (18.1.2) is correct, we must show that whenever nx = t1 − t2 , mx = s1 − s2 , with n, m ∈ N, t1 , t2 , s1 , s2 ∈ T , then f (t1 ) − f (t2 ) f (s1 ) − f (s2 ) = . (18.1.3) n m But then mt1 − mt2 = mnx = ns1 − ns2 , whence mt1 + ns2 = ns1 + mt2 .

(18.1.4)

It follows by induction from (4.5.15) that f (pu) = pf (u) for every p ∈ N and u ∈ T . Hence and by (4.5.15) we get from (18.1.4) mf (t1 ) + nf (s2 ) = nf (s1 ) + mf (t2 ), and (18.1.3) follows. Consequently definition (18.1.2) is correct. Now we define a map h : R → Y . If a ∈ R, a = b + nx with b ∈ T , n ∈ N0 , we put h(a) = f (b) + ny. (18.1.5) Again we must show that definition (18.1.5) is correct. This amounts to showing that whenever b + nx = b + n x (18.1.6) with b, b ∈ T , n, n ∈ N0 , then f (b) + ny = f (b ) + n y.

(18.1.7)

If b = b , n = n , this is obvious. If one of b, n is not equal to the corresponding element b , n , then the other must be unequal, too, as results from (18.1.6). So now we may assume that n = n , say n > n . (If n > n, we interchange the roles of n and n ). From (18.1.6) we get (n − n )x = b − b ∈ T − T.

(18.1.8)

n − n ∈ N, so (18.1.8) means that (Nx) ∩ (T − T ) = ∅. But then we have in virtue of (18.1.2) f (b ) − f (b) = y, (18.1.9) n − n and (18.1.7) results. Thus definition (18.1.5) is correct.

18.2. The simplest case of S generating X

537

Let a, a ∈ R, say, a = b + nx, a = b + n x, b, b ∈ T , n, n ∈ N0 . Then a + a = b + b + (n + n )x (note that X is commutative), whence by (18.1.5) h(a + a ) = f (b + b ) + (n + n )x = f (b) + f (b ) + nx + n x = f (b) + nx + f (b ) + n x = h(a) + h(a ), since Y is commutative and f is a homomorphism. Consequently h is a homomorphism from R into Y . Finally note that T ⊂ R, for every b ∈ T may be written as b = b + 0x ∈ R. Consequently (T, +) is a subsemigroup of (R, +). For b ∈ T (b = b + 0x) we have by (18.1.5) h(b) = f (b) + 0y = f (b). This means that h | T = f . Consequently (R, h) ∈ R and (T, f ) ≺ (R, h), whereas clearly (T, f ) = (R, h). But this contradicts the maximality of (T, f ). Thus we must have T = X, i.e., f is a homomorphism from X into Y . Since (S, g) ≺ (T, f ), we have f | S = g. This completes the proof.  Note that the theorem fails to hold if (Y, +) is not divisible. E.g., take X = Y = Z, S = 2Z (the set of even integers). Those sets endowed with the operation + of the usual addition are commutative groups, but (Y, +) = (Z, +) is not divisible. Let g : 2Z → Z be the homomorphism defined by g(2k) = k for k ∈ Z. There does not exist a homomorphism f : Z → Z such that f | 2Z = g, for otherwise we would have f (1) + f (1) = f (2) = g(2) = 1, i.e., 2f (1) = 1, which has no solution in Z. It can be shown (Dhombres-Ger [70]) that if S does not generate X, then the extension f of g furnished by Theorem 18.1.1 is not unique. On the other hand, if S generates X, then f is unique, as follows easily from Corollary 18.2.1 below.

18.2 The simplest case of S generating X In this section we consider the case where S generates X in such a way that 

X = S − S.

(18.2.1)

It follows from (18.2.1) that S generates X, for then any group containing S must contain also S − S = X. The results are due to Acz´el-Baker-Djokovi´c-KannappanRad´o [8]. We start with a lemma. Lemma 18.2.1. Let (X, +) and (Y, +) be groups, and let (S, +) be a subsemigroup of (X, +) such that (18.2.1) holds. Let g : S → Y be a homomorphism: g(x + y) = g(x) + g(y)

(18.2.2)

for all x, y ∈ S. Then, if x, y, z, x , y  , z  ∈ S are such that

then

x − y + z = x − y  + z  ,

(18.2.3)

g(x) − g(y) + g(z) = g(x ) − g(y  ) + g(z  ).

(18.2.4)

538

Chapter 18. Extensions of Homomorphisms

Proof. By (18.2.1) an arbitrary u ∈ X may be represented as u = t − s, with t, s ∈ S. Hence u + s = t ∈ S. So (18.2.1) implies that (i) for every u ∈ X there exists an s ∈ S such that u + s ∈ S. So, in particular, taking u = −y + z, we may find an s ∈ S such that −y + z + s ∈ S.

(18.2.5)

Similarly, we can find an s ∈ S such that −y  + z  + s + s ∈ S.

(18.2.6)

With s = s + s ∈ S relations (18.2.5) and (18.2.6) yield −y + z + s ∈ S,

−y  + z  + s ∈ S.

(18.2.7)

On the other hand, relation (18.2.3) implies x − y + z + s = x − y  + z  + s.

(18.2.8)

  Now, if u ∈ S and −u + v ∈ S which implies that v = u + (−u + v) ∈ S , then g(−u + v) = −g(u) + g(v),

(18.2.9)

since taking in (18.2.2) x = u, y = −u + v, we obtain g(v) = g(u) + g(−u + v), which implies (18.2.9). It follows by (18.2.8) and (18.2.2) that g(x) + g(−y + z + s) = g(x ) + g(−y  + z  + s).   Applying now (18.2.9) with u = y ∈ S and v = z + s by (18.2.7) −u +v ∈ S , and also with u = y  ∈ S, v = z  + s again −u + v ∈ S in virtue of (18.2.7) , we obtain hece g(x) − g(y) + g(z + s) = g(x ) − g(y  ) + g(z  + s). Hence, again by (18.2.2) g(x) − g(y) + g(z) + g(s) = g(x ) − g(y  ) + g(z  ) + g(s). Cancelling in (18.2.10) g(s) yields (18.2.4).

(18.2.10) 

Theorem 18.2.1. Let (X, +) and (Y, +) be groups, and let (S, +) be a subsemigroup of (X, +) such that (18.2.1) holds. Let g : S → Y be a homomorphism. Then there exists a unique homomorphism f : X → Y such that f | S = g. Proof. Every x ∈ X can be represented as x = t − s with t, s ∈ S. For such an x define f (x) as f (x) = g(t) − g(s). (18.2.11)

18.2. The simplest case of S generating X

539

To prove that this definition is correct we must show that whenever t − s = t − s , then

t, s, t , s ∈ S,

g(t) − g(s) = g(t ) − g(s ).

But this follows from Lemma 18.2.1 on choosing x = t, y = s + s, z = s, x = t , y  = s + s , z  = s . Thus formula (18.2.11) unambiguously defines a map f : X → Y . We will show that f is a homomorphism. Take arbitrary x, y ∈ X, x = t − s, y = t − s , t, s, t , s ∈ S. Then x + y = t − s + t − s is an element of X, and so x + y = t − s with t , s ∈ S. Hence we get t − s + t = t − s + s , and, by Lemma 18.2.1, g(t) − g(s) + g(t ) = g(t ) − g(s ) + g(s ), or

g(t ) − g(s ) = g(t) − g(s) + g(t ) − g(s ).

(18.2.12)

According to definition (18.2.11) f (x) = g(t) − g(s), and

f (y) = g(t ) − g(s ),

f (x + y) = g(t ) − g(s ).

(18.2.13) (18.2.14)

Relations (18.2.13), (18.2.14) and (18.2.12) imply that f (x + y) = f (x) + f (y),

(18.2.15)

which means that f is a homomorphism. Let x ∈ S. Then x = (x + x) − x, whence by (18.2.11) and (18.2.2) f (x) = g(x + x) − g(x) = g(x) + g(x) − g(x) = g(x). Consequently f | S = g. In order to prove the uniqueness, let h : X → Y be an arbitrary homomorphism such that h | S = g. Take an arbitrary x ∈ X. Then x = t − s with t, s ∈ S, whence x + s = t. Relation (18.1.1) implies that h(t) = h(x) + h(s), whence h(x) = h(t) − h(s) = g(t) − g(s) = f (x), since h | S = g and f satisfies (18.2.11). This shows that h coincides with f on the whole of X, i.e., f is unique. 

540

Chapter 18. Extensions of Homomorphisms

Theorems 18.2.1 and 4.5.1 yield the following Corollary 18.2.1. Let (X, +) be a commutative group, let (Y, +) be a group, and let (S, +) be a subsemigroup of (X, +) generating X. Let g : S → Y be a homomorphism. Then there exists a unique homomorphism f : X → Y such that f | S = g. Similarly, we obtain by Lemma 4.5.6 Corollary 18.2.2. Let (X, +) and (Y, +) be groups, and let (S, +) be a subsemigroup of (X, +) such that for every x ∈ X, x = 0, either x ∈ S, or −x ∈ S (or both). Let g : S → Y be a homomorphism. Then there exists a unique homomorphism f : X → Y such that f | S = g. Corollary 18.2.2 generalizes a result of Acz´el-Erd˝ os [9] (cf. also 13.5). Theorem 18.2.1 is also related to Theorem 4.4.1.

18.3 A generalization Now we are going to consider the case, where X=

n

(−1)i S

(18.3.1)

i=0

(Grz¸a´slewicz-Sikorski [125], Martin [221]). Here n is a fixed positive integer, n  2. It is easy to see that also in case (18.3.1) S generates X. For n = 1 (18.3.1) reduces to (18.2.1). But now (n  2) it is no longer true that every homomorphism g : S → Y can be extended to a homomorphism f : X → Y . Some additional assumptions are necessary. We will include also case (18.2.1) into our considerations, thus allowing n to be 1. In this way Theorem 18.2.1 will become a particular case of Theorem 18.3.1 below. But again first we must prove a lemma. Lemma 18.3.1. Let (X, +) and (Y, +) be groups, and let (S, +) be a subsemigroup of (X, +) such that relation (18.3.1) holds for a certain fixed n ∈ N. Let g : S → Y be a homomorphism. Assume that for every x0 , . . . , xn−1 , y0 , . . . , yn−1 ∈ S, whenever n−1

n−1

i=0

i=0

(−1)i xi =

then also2

n−1

i

(−1) g(xi ) =

i=0

(−1)i yi ,

n−1

(−1)i g(yi ).

(18.3.2)

(18.3.3)

i=0

Then, for arbitrary k ∈ N0 = N ∪ {0} and for arbitrary x0 , . . . , xn , y0 , . . . , yn+k ∈ S, whenever n n+k

(−1)i xi = (−1)i yi , (18.3.4) i=0 2

i=0

For n = 1 this condition becomes: x = y implies g(x) = g(y), so it is trivially fulfilled.

18.3. A generalization

541

then also

n

(−1)i g(xi ) =

i=0

n+k

(−1)i g(yi ).

(18.3.5)

i=0

Proof. We proceed by induction on k. Let k = 0. Take arbitrary points x0 , . . . , xn , y0 , . . . , yn ∈ S such that n n

(−1)i xi = (−1)i yi . (18.3.6) i=0

It follows that x0 +

n

i=0

(−1)i xi − (−1)n yn =

i=1

n−1

(−1)i yi .

(18.3.7)

i=0

The expression following x0 on the left-hand side of (18.3.7) is an element of X, so by (18.3.1) there exist z0 , . . . , zn ∈ S such that n

(−1)i xi − (−1)n yn =

i=1

n

(−1)i zi .

(18.3.8)

i=0

Again, (18.3.8) may be written as n

(−1) xi − (−1) yn − (−1) zn = x1 + z0 + i

n

n

i=2

n−1

(−1)i zi .

(18.3.9)

i=1

From (18.3.7) and (18.3.8) we obtain n

x0 +

n−1

(−1)i zi =

(−1)i yi ,

i=0

i=0

whence x0 + z0 +

n−1

n−2

i=1

i=0

(−1)i zi =

(−1)i yi + (−1)n−1 yn−1 − (−1)n zn .

(18.3.10)

Now we must consider separately the cases where n is even or odd. First assume that n is even. Then −(−1)n yn − (−1)n zn = −yn − zn = −(zn + yn ), whereas (−1)n−1 yn−1 − (−1)n zn = −yn−1 − zn = −(zn + yn−1 ). Thus we obtain from (18.3.9) and (18.3.10) n

(−1)i xi − (zn + yn ) = (x1 + z0 ) +

i=2

(x0 + z0 ) +

n−1

(−1)i zi ,

i=1 n−1

n−2

i=1

i=0

(−1)i zi =

(−1)i yi − (zn + yn−1 ).

542

Chapter 18. Extensions of Homomorphisms

By assumption (18.3.3) we get hence n

(−1) g(xi ) − g(zn + yn ) = g(x1 + z0 ) + i

i=2

n−1

(−1)i g(zi ),

i=1

g(x0 + z0 ) +

n−1

n−2

i=1

i=0

(−1)i g(zi ) =

(−1)i g(yi ) − g(zn + yn−1 ),

or, since g is a homomorphism, n

(−1)i g(xi ) − g(yn ) =

i=1

n

(−1)i g(zi ),

g(x0 ) +

i=0

n

(−1)i g(zi ) =

i=0 n 

Hence we obtain, eliminating

n−1

(−1)i g(yi ).

i=0

(−1)i g(zi ),

i=0 n

(−1)i g(xi ) =

i=0

n

(−1)i g(yi ),

(18.3.11)

i=0

i.e., (18.3.5) for k = 0. Now let n be odd. Then −(−1)n yn − (−1)n zn = yn + zn , whereas (−1)n−1 yn−1 − n (−1) zn = yn−1 + zn and we obtain from (18.3.9) and (18.3.10) n

(−1)i xi + (yn + zn ) = (x1 + z0 ) +

i=2

n−1

(−1)i zi ,

i=1

(x0 + z0 ) +

n−1

n−2

i=1

i=0

(−1)i zi =

(−1)i yi + (yn−1 + zn ),

and by assumption (18.3.3) n

(−1)i g(xi ) + g(yn + zn ) = g(x1 + z0 ) +

i=2

n−1

(−1)i g(zi ),

i=1

g(x0 + z0 ) +

n−1

n−2

i=1

i=0

(−1)i g(zi ) =

(−1)i g(yi ) + g(yn−1 + zn ),

or, since g is a homomorphism, n

(−1)i g(xi ) + g(yn ) =

i=1

whence, eliminating

n

i=0

n 

(−1)i g(zi ),

g(x0 ) +

n

i=0

(−1)i g(zi ) =

n−1

(−1)i g(yi ),

i=0

(−1)i g(zi ), we obtain (18.3.11), i.e., (18.3.5) for k = 0.

i=0

18.3. A generalization

543

Now assume that the lemma is true for a k ∈ N0 , and let x0 , . . . , xn , y0 , . . . , yn+k+1 be arbitrary points from S such that n

(−1)i xi =

i=0

n+k+1

(−1)i yi .

(18.3.12)

i=0

n+k+1 

(−1)i yi is an element of X, so, according to (18.3.1), there exist points z0 , . . . ,

i=1

zn ∈ S such that

n+k+1

(−1)i yi =

i=1

n

(−1)i zi .

(18.3.13)

i=0

Now take an arbitrary u ∈ S. Again we must distinguish two cases, according as n + k is even or odd. Suppose that n + k is even. We have u + yn+k+1 ∈ S (since S is a semigroup) and u ∈ S, and so (18.3.13) may be written as n+k

(−1)i yi − (u + yn+k+1 ) + u = (y1 + z0 ) +

i=2

n

(−1)i zi .

i=1

By the induction hypothesis it follows n+k

(−1)i g(yi ) − g(u + yn+k+1 ) + g(u) = g(y1 + z0 ) +

i=2

n

(−1)i g(zi ),

i=1

or, since g is a homomorphism, n+k+1

(−1)i g(yi ) =

i=1

n

(−1)i g(zi ).

(18.3.14)

i=0

If n + k is odd, we have yn+k+1 + u ∈ S and u ∈ S, and so (18.3.13) may be written as n+k

(−1)i yi + (yn+k+1 + u) − u = (y1 + z0 ) +

i=2

n

(−1)i zi ,

i=1

and by the induction hypothesis n+k

(−1)i g(yi ) + g(yn+k+1 + u) − g(u) = g(y1 + z0 ) +

i=2

n

(−1)i g(zi ),

i=1

and, since g is a homomorphism, we arrive again at formula (18.3.14). Thus (18.3.14) is valid in both cases. We get by (18.3.12) and (18.3.13) n

i=0

(−1)i xi = (y0 + z0 ) +

n

i=1

(−1)i zi .

544

Chapter 18. Extensions of Homomorphisms

Applying to this our lemma for k = 0 (which has already been proved), we infer hence that n n

(−1)i g(xi ) = g(y0 + z0 ) + (−1)i g(zi ), i=0

i=1

or, since g is a homomorphism, n

i

(−1) g(xi ) = g(y0 ) +

i=0

Eliminating

n 

n

(−1)i g(zi ).

(18.3.15)

i=0

(−1)i g(zi ) from (18.3.14) and (18.3.15), we obtain

i=0 n

n+k+1

(−1)i g(xi ) =

i=0

(−1)i g(yi ),

i=0



i.e., (18.3.5) for k + 1. Induction ends the proof.

Theorem 18.3.1. Let (X, +) and (Y, +) be groups, and let (S, +) be a subsemigroup of (X, +) such that relation (18.3.1) holds for a certain n ∈ N. Let g : S → Y be a homomorphism. There exists a homomorphism f : X → Y such that f | S = g if and only if for every x0 , . . . , xn−1 , y0 , . . . , yn−1 ∈ S relation (18.3.2) implies (18.3.3). When this is the case, the homomorphism f is unique. Proof. First assume that condition (18.3.2) implies (18.3.3). It follows by (18.3.1) that for every x ∈ X there exist x0 , . . . , xn ∈ S such that x=

n

(−1)i xi .

(18.3.16)

i=0

For such an x define f (x) as f (x) =

n

(−1)i g(xi ).

(18.3.17)

i=0

It follows by Lemma 18.3.1 (k = 0) that this definition is correct. So f is a mapping f : X → Y . In order to show that f is a homomorphism take arbitrary x, y ∈ X. There exist x0 , . . . , xn , y0 , . . . , yn , z0 , . . . , zn ∈ S such that x=

n

(−1)i xi ,

y=

i=0

n

(−1)i yi ,

z=

i=0

n

(−1)i zi .

(18.3.18)

i=0

According to definition (18.3.17) f (x) =

n

i=0

(−1)i g(xi ),

f (y) =

n

i=0

(−1)i g(yi ),

f (x + y) =

n

(−1)i g(zi ).

i=0

(18.3.19)

18.3. A generalization

545

Clearly we have by (18.3.18) n

(−1)i zi =

i=0

n

(−1)i xi +

n

i=0

(−1)i yi .

(18.3.20)

i=0

If n is odd, this implies in virtue of Lemma 18.3.1 (k = n) n

n

(−1)i g(zi ) =

i=0

(−1)i g(xi ) +

i=0

n

(−1)i g(yi ),

i=0

i.e., by (18.3.19), we obtain (18.2.15). If n is even, we write (18.3.20) in the form n

(−1)i xi =

i=0

n

(−1)i zi +

i=0

n

(−1)i+1 yn−i ,

i=0

and, using again Lemma 18.3.1 and relation (18.3.19), we arrive at f (x) = f (x + y) − f (y), which yields (18.2.15). Thus (18.2.15) holds in both cases, which shows that f is a homomorphism. Now take an arbitrary x ∈ S. If n is even, we have x=

n

(−1)i xi ,

i=0

whence by (18.3.17) f (x) =

n

(−1)i g(x) = g(x).

i=0

If n is odd, we have x = (x + x) +

n

(−1)i x,

i=1

and by (18.3.17) f (x) = g(x + x) +

n

(−1)i g(x) = g(x) + g(x) − g(x) = g(x).

i=1

Thus in either case f (x) = g(x), which means that f | S = g. Now assume that a homomorphism f : X → Y such that f | S = g exists. It follows from (18.2.15) by induction that for every m ∈ N and every u1 , . . . , um ∈ X we have f (u1 + · · · + um ) = f (u1 ) + · · · + f (um ).   Also, (18.2.15) implies easily that f (−x) = −f (x) so that f (−1)i x = (−1)i f (x) for every i ∈ N0 and x ∈ X. Now, relation (18.3.2) implies  n−1  n−1 n−1 n−1



i i i (−1) f (xi ) = f (−1) xi = f (−1) yi = (−1)i f (yi ), i=0

i=0

i=0

i=0

546

Chapter 18. Extensions of Homomorphisms

whence, taking into account the fact that f | S = g and that x0 , . . . , xn−1 , y0 , . . . , yn−1 ∈ S, we obtain (18.3.3). Let h : X → Y be an arbitrary homomorphism such that h | S = g. Take an arbitrary x ∈ X. Then x has a representation (18.3.16) with x0 , . . . , xn ∈ S. By the same argument as above we show that then h(x) =

n n

(−1)i h(xi ) = (−1)i g(xi ), i=0

i=0

since h | S = g. So in view of (18.3.17) h = f , which proofs the uniqueness of f and completes the proof of the theorem. 

18.4 Further extension theorems The results in this section (Grz¸a´slewicz [121]) are closely related to the problems treated in 18.1–18.3. Here we will assume that (S, +) is a subgroup of (X, +). Theorem 18.4.1. Let (X, +) and (Y, +) be groups, and let (S, +) be a subgroup of (X, +). Let g : S → Y be a homomorphism. There exists a homomorphism f : X → Y such that f | S = g and f (X) = g(S) if and only if there exists a normal subgroup (X0 , +) of (X, +) such that Ker g = S ∩ X0 , (18.4.1) X = S + X0 . Here Ker g = g

−1

(18.4.2)

(0) = {x ∈ S | g(x) = 0} denotes the kernel of g.

Proof. Suppose that a normal subgroup (X0 , +) of (X, +) fulfilling conditions (18.4.1) and (18.4.2) exists. Let elements of X/X0 be denoted by x, x ∈ X. By Lemma 4.5.7 (Ker g, +) is a normal subgroup of (S, +). Let elements of S/ Ker g be denoted by [x], x ∈ S. In virtue of Theorem 4.5.3 (X/X0 , +) and (S/ Ker g, +) are groups, where the operations + are defined according to (4.5.14). By (18.4.1), (18.4.2) and Theorem 4.6.3 S/ Ker g is a selective subpartition of X/X0. Therefore for every x ∈ X/X0 there exists a unique class [u] ∈ S/ Ker g such that [u] ⊂ x. With these notations we define a function f : X → Y as f (x) = g(u).

(18.4.3)

From what has just been told and from Corollary 4.5.4 it follows that this definition is unambiguous. It is also immediately seen from (18.4.3) that f (X) ⊂ g(S). Conversely, for every t ∈ g(S) there exists a u ∈ S such that t = g(u). Since S/ Ker g is a subpartition of X/X0 , there exists an x ∈ X such that [u] ⊂ x. But according to (18.4.3) we have then f (x) = g(u) = t. So t ∈ f (X), and g(S) ⊂ f (X). Consequently f (X) = g(S). Let x ∈ S. Then, of course x ∈ [x] ∩ x, and consequently [x] ∩ x = ∅. By Lemma 4.6.1 we then have [x] ⊂ x. But according to definition (18.4.3) this implies that f (x) = g(x). Thus f | S = g. Take arbitrary x, y ∈ X, and let u, v ∈ S be such that [u] ⊂ x, [v] ⊂ y. This means, in particular, that u ∈ x, v ∈ y, and so u = x, v = y, whence it

18.4. Further extension theorems

547

follows that x + y = x + y = u + v = u + v. Thus u + v ∈ x + y, and since evidently u+v ∈ [u+v], we get [u+v]∩x+y = ∅. By Lemma 4.6.1 [u+v] ⊂ x+y. Hence f (x + y) = g(u + v) = g(u) + g(v) = f (x) + f (y). This shows that f is a homomorphism. Conversely, suppose that there exists a homomorphism f : X → Y with the required properties. Put X0 = Ker f = {x ∈ X | f (x) = 0}. By Lemma 4.5.7 (X0 , +) is a normal subgroup of (X, +). x ∈ Ker g means x ∈ S and g(x) = 0, whence, since f | S = g, also f (x) = 0, and x ∈ Ker f = X0 . Thus Ker g ⊂ X0 . By definition Ker g ⊂ S. So Ker g ⊂ S ∩ X0 . (18.4.4) Now take an arbitrary x ∈ S ∩ X0 . Then x ∈ X0 = Ker f and so f (x) = 0. But since also x ∈ S, we have f (x) = g(x), whence g(x) = 0, and x ∈ Ker g. Thus S ∩ X0 ⊂ Ker g, which together with (18.4.4) proves relation (18.4.1). Since both, S and X0 , are contained in X, we have S + X0 ⊂ X + X ⊂ X. On the other hand, take an arbitrary x ∈ X. Since f (X) = g(S), there exists a u ∈ S such that g(u) = f (x), whence, since f | S = g, we obtain f (u) = f (x). Since f is a homomorphism, we have   f (x) = f u + (−u + x) = f (u) + f (−u + x) so that f (−u + x) = 0 and −u + x ∈ Ker f = X0 . Hence x = u + (−u + x) ∈ S + X0 and X ⊂ S + X0 , and ultimately we get relation (18.4.2).  Theorem 18.4.2. Let (X, +) and (Y0 , +) be groups, and let (S, +) be a subgroup of (X, +). Let g : S → Y0 be a homomorphism from S onto Y0 . Then there exists a group (Y, +) and a homomorphism f : X → Y such that (Y0 , +) is a subgroup of (Y, +) and f | S = g if and only if there exists a normal subgroup (X0 , +) of (X, +) such that relation (18.4.1) holds. Proof. Suppose that there exists a normal subgroup of (X0 , +) of (X, +) such that relation (18.4.1) holds. We will employ the notation used in the proof of Theorem 18.4.1. It follows from (18.4.1) that Ker g ⊂ X0 , whence, by Theorem 4.6.1, S/ Ker g is a subpartition of X/X0. Hence for every [u] ∈ S/ Ker g there exists a (unique) x ∈ X/X0 such that [u] ⊂ x. Therefore card S/ Ker g  card X/X0

(18.4.5)

On the other hand, by Lemma 4.5.8, to every [u] ∈ S/ Ker g there exists exactly one t ∈ Y0 such that g([u]) = {t}. Conversely, since the map g : S →  Y0 is onto, for every t ∈ Y0 there exists a unique [u] ∈ S/ Ker g such that g [u] = {t}. Thus g induces a one-to-one mapping between the elements of S/ Ker g and these of Y0 . Let this mapping be α : S/ Ker g → Y0 , which is thus defined for [u] ∈ S/ Ker g:     (18.4.6) α [u] = t ∈ Y0 iff g [u] = {t}.

548

Chapter 18. Extensions of Homomorphisms

As a consequence we obtain that card(S/ Ker g) = card Y0 .

(18.4.7)

By (18.4.5) and (18.4.7) we obtain card Y0  card(X/X0 ). Let Y be an arbitrary set such that Y0 ⊂ Y and card Y = card(X/X0 ). So we may define a new mapping ϕ : X/X0 → Y (one-to-one and onto) in such a way that, whenever for some u ∈ S and x ∈ X we have [u] ⊂ x, then     ϕ x = α [u] . (18.4.8)  Note that by Theorem 4.6.2 S/ Ker g is a semiselective subpartition of X/X0 , and so for a given x ∈ X/X0 there may existat most one [u] ∈ S/ Ker g such that [u] ⊂ x. Thus condition (18.4.8) is meaningful. Now we define an operation ⊕ in Y . Take arbitrary t, s ∈ Y . There exist unique classes x, y ∈ X/X0 such that     t = ϕ x , s = ϕ y . (18.4.9) We put

  t ⊕ s = ϕ x + y .

(18.4.10)

It is easy to check that the operation ⊕ thus defined (Y, ⊕) is a group . Now  suppose  that t, s ∈ Y . Then there exist classes [u], [v] ∈ S/ Ker g such that t = α [u] , s = 0   α [v] . Since S/ Ker g is a subpartition of X/X0 , there exist classes x, y ∈ X/X0 such that [u] ⊂ x, [v] ⊂ y, (18.4.11)         whence, by (18.4.8),  ϕ x = α [u]  = t, ϕ y = α [v] = s. According to (18.4.6) this means that g [u] = {t}, g [v] = {s}. In particular, 3

g(u) = t,

g(v) = s.

Hence g(u + v) = g(u) + g(v) = t + s (here the operation + on the right-hand side denotes the operation originally existing in Y0 ), whence, again by (18.4.6),   α [u + v] = t + s. (18.4.12) Further, (18.4.11) implies that u ∈ x and v ∈ y, whence u = x and v = y. Thus u+v ∈ u+v = u+v = x+y = x+y. Since evidently u+v ∈ [u+v], we     The neutral element of (Y, ⊕) is ϕ 0 , and, to a given t = ϕ x the inverse element is given by   ϕ −x . 3

18.4. Further extension theorems

549

get hence [u+v]∩x+y   = ∅,  whence  it follows by Lemma4.6.1 that  [u+v] ⊂ x+y. By (18.4.8) ϕ x + y = α [u + v] , and by (18.4.10) α [u + v] = t ⊕ s. Now we obtain by (18.4.12) t ⊕ s = t + s, which shows that on Y0 both the operations, ⊕ and +, coincide. Thus in the sequel, without a fear of ambiguity, we may use the same symbol + to denote both these operations. The argument above shows that (Y0 , +) is a subgroup of (Y, +). Now we define a mapping f : X → Y as follows:   f (x) = ϕ x . (18.4.13) For arbitrary x, y ∈ X we have according to (18.4.13), (18.4.10) and (18.4.9)       f (x + y) = ϕ x + y = ϕ x + ϕ y = f (x) + f (y). So f is a homomorphism. If x ∈ S, then x ∈ [x] ∩ x, whence by Lemma 4.6.1 [x] ⊂ x, and by (18.4.8) and (18.4.13)   f (x) = α [x] . (18.4.14)   But in virtue of (18.4.6) α [x] = g(x), whence by (18.4.14) f (x) = g(x). Thus f | S = g. The proof of necessity of condition (18.4.1) is identical as the corresponding part of the proof of Theorem 18.4.1.  Theorem 18.4.3. Let (X, +) and (Y, +) be groups, and let (S, +) be a subgroup of (X, +) such that with S  = X \ S the following conditions are fulfilled: (i) For every x, y ∈ S  we have x + y ∈ S. (ii) There exists a u ∈ S  such that4 2u = 0 and for every x ∈ X we have x + u = u + x. Let g : S → Y be a homomorphism. Then there exists a homomorphism f : X → Y such that f | S = g. Proof. Put for x ∈ X

 f (x) =

g(x), if x ∈ S, g(x + u) if x ∈ S  ,

(18.4.15)

It is obvious from definition (18.4.15) that f | S = g. In order to prove that f is a homomorphism we must distinguish four cases. Take arbitrary x, y ∈ S. 1. x ∈ S, y ∈ S. Then also x + y ∈ S, and by (18.4.15) f (x) = g(x), 4

f (y) = g(y),

f (x + y) = g(x + y),

So u is an element of order 2. Similarly, it follows from condition (i) that (S, +) is a subgroup of (X, +) of index 2 (cf. Lemma 4.5.4; cf. also Exercise 4.7).

550

Chapter 18. Extensions of Homomorphisms

and f (x + y) = g(x + y) = g(x) + g(y) = f (x) + f (y), since g is a homomorphism. 2. x ∈ S, y ∈ S  . Then x + y ∈ S  in virtue of Lemma 4.5.3, whence f (x) = g(x),

f (y) = g(y + u),

f (x + y) = g(x + y + u),

and f (x + y) = g(x + y + u) = g(x) + g(y + u) = f (x) + f (y). 3. x ∈ S  , y ∈ S. Then again x + y ∈ S  (Lemma 4.5.3), whence f (x) = g(x + u),

f (y) = g(y),

f (x + y) = g(x + y + u).

and by (ii) f (x + y) = g(x + y + u) = g(x + u + y) = g(x + u) + g(y) = f (x) + f (y). 4. x ∈ S  , y ∈ S  . Then by (i) x + y ∈ S and f (x) = g(x + u),

f (y) = g(y + u),

f (x + y) = g(x + y).

By (ii) f (x+y) = g(x+y) = g(x+y+2u) = g(x+u+y+u) = g(x+u)+g(y+u) = f (x)+f (y). Consequently (18.2.15) holds in all cases, which means that f is a homomorphism.  K. Dankiewicz and Z. Moszner [52] showed that Theorem 18.4.3 can be derived directly from Theorem 18.4.1. They also proved a number of related results. We terminate this section with the following simple theorem (cf., e.g., Kuczma [183]). Theorem 18.4.4. Let (X, +) be a commutative group, and let (S, +) be a subgroup of (X, +) such that, for a certain n ∈ N and every x ∈ X, we have nx ∈ S. Let (Y, +) be a commutative group in which the division by n is performable, and let g : S → Y be a homomorphism. Then there exists a unique homomorphism f : X → Y such that f | S = g. This homomorphism is given by f (x) =

g(nx) , n

x ∈ X.

(18.4.16)

Proof. Let g : S → Y be a homomorphism, and let f be given by (18.4.16). The definition is meaningful, since nx ∈ S for every x ∈ X and the division by n is performable in Y . Take arbitrary x, y ∈ X. Since X is commutative, we have n(x + y) = nx + ny, and since Y is commutative,   g(nx) g(ny) g(nx) g(ny) n + =n +n = g(nx) + g(ny), n y n n

18.5. Cauchy equation on a cylinder

whence

551

g(nx) + g(ny) g(nx) g(ny) = + . n n n

Consequently   g n(x + y) g(nx + ny) g(nx) + g(ny) g(nx) g(ny) = = = + = f (x) + f (y). f (x + y) = n n n n n Thus f is a homomorphism. Now, we have by induction from (18.2.2) g(mx) = mg(x),

m ∈ N, x ∈ S,

whence, in particular, g(nx) = ng(x) for x ∈ S. Hence, if x ∈ S, we have by (18.4.16) f (x) =

g(nx) ng(x) = = g(x). n n

This shows that f | S = g. Finally, if h : X → Y is an arbitrary homomorphism such that h | S = g, then, similarly as above, we have h(nx) = nh(x) for x ∈ X. As nx ∈ S and h | S = g, this yields g(nx) = nh(x), x ∈ X. (18.4.17) Relations (18.4.17) and (18.4.16) yield h(x) =

g(nx) = f (x), n

x ∈ X.

This shows that h coincides with f on the whole of X, which proves the uniqueness of f , and also formula (18.4.16). 

18.5 Cauchy equation on a cylinder Let (X, +) and (Y, +) be arbitrary groups, and let Z ⊂ Y be an arbitrary non-empty set. Suppose that a function f : X → Y satisfies the equation f (x + y) = f (x) + f (y)

(18.5.1)

for all x ∈ X, y ∈ Z. Then (18.5.1) is a restricted Cauchy equation, the condition restricting the validity of the equation being (x, y) ∈ X × Z. A set of the form X × Z, ∅ = Z ⊂ X, is called a cylinder, therefore the present problem is referred to as the Cauchy equation on a cylinder. Let f : X → Y be an arbitrary mapping. The set Nf = {y ∈ X | f (x + y) = f (x) + f (y) for all x ∈ X}

(18.5.2)

is called the Cauchy nucleus of f (cf. Grz¸a´slewicz-Pow¸azka-Tabor[124]). The Cauchy equation (18.5.1) on a cylinder X × Z simply means that Z ⊂ Nf .

552

Chapter 18. Extensions of Homomorphisms

The following lemma (Kuczma [183], Dhombres-Ger [70]) is very useful. Lemma 18.5.1. Let (X, +) and (Y, +) be groups, and let f : X → Y be an arbitrary mapping. Then either Nf = ∅, or (Nf , +) is a subgroup of (X, +). Proof. Suppose that Nf = ∅. Then there exists a y ∈ Nf . With this y (18.5.1) holds for all x ∈ X. Taking x = 0, we get f (y) = f (0) + f (y), whence f (0) = 0. Now again, with y an arbitrary element of Nf and x = −y, we get from (18.5.1) 0 = f (0) = f (−y) + f (y), whence f (−y) = −f (y) for every y ∈ Nf .

(18.5.3)

We have for arbitrary x ∈ X and y ∈ Nf   f (x) = f (x − y) + y = f (x − y) + f (y), whence or by (18.5.3)

f (x − y) = f (x) − f (y),

(18.5.4)

  f x + (−y) = f (x) + f (−y).

(18.5.5)

Relation (18.5.5) holds for every x ∈ X, which means that −y ∈ Nf whenever y ∈ Nf . Hence we have, for arbitrary x ∈ X and u, v ∈ Nf ,     f x + (u − v) = f (x + u) + (−v) = f (x + u) + f (−v), and by (18.5.3)   f x + (u − v) = f (x + u) − f (v) = f (x) + f (u) − f (v), since u ∈ Nf . Relation (18.5.4) implies that f (u − v) = f (u) − f (v), whence   f x + (u − v) = f (x) + f (u − v). This holds for all x ∈ X so that u − v ∈ Nf . Lemma 4.5.1 completes the proof.



If Nf = ∅, then relation (18.5.1) holds, in particular, for all x, y ∈ Nf . This means that f | Nf is a homomorphism. Therefore, instead of (18.5.1), we will investigate the equation f (x + y) = f (x) + g(y) (18.5.6) valid for x ∈ X, y ∈ S, where (S, +) is a subgroup of (X, +), and g : S → Y is a homomorphism (Tabor [312], Dhombres-Ger [70]). Theorem 18.5.1. Let (X, +) and (Y, +) be groups, let (S, +) be a subgroup of (X, +), and let g : S → Y be a homomorphism. If f : X → Y is a function such that (18.5.6) holds for every x ∈ X, y ∈ S, then there exist a lifting ξ : X/S → X and a mapping h : X/S → Y such that     f (x) = h [x] + g − ξ([x]) + x , x ∈ X. (18.5.7)

18.5. Cauchy equation on a cylinder

553

Conversely, every function f of form (18.5.7), where h : X/S → Y is an arbitrary function and ξ : X/S → X is an arbitrary lifting, satisfies equation (18.5.6) for all x ∈ X, y ∈ S. Proof. Let ξ : X/S → X be a quite arbitrary lifting 5 . Define a function h : X/S → Y by      h [x] = f ξ [x] , [x] ∈ X/S. (18.5.8)  

According to definition, a lifting is characterized by the property that ξ [x] = [x]. In other words, the elements ξ [x] and x generate the same equivalence  class (left coset), i.e., they are equivalent   with  respect   to S. This means that −ξ [x] + x ∈ S. Hence, writing x as x = ξ [x] + − ξ [x] + x , we get by (18.5.6) and (18.5.8)              f (x) = f ξ [x] + g − ξ [x] + x = h [x] + g − ξ [x] + x , which proves relation (18.5.7). Conversely, suppose that the function f has form (18.5.7), where h : X/S → Y is an arbitrary function, and ξ : X/S → X is an arbitrary lifting. Take arbitrary x ∈ X, y ∈ S. Then we have −x + (x + y) = y ∈ S, which means that x and x + y are equivalent with respect to S. Thus [x] = [x + y]. Hence we have by (18.5.7), since g : S → Y is a homomorphism,       f (x + y) = h [x + y] + g − ξ [x + y] + x + y             = h [x] + g − ξ [x] + x + y = h [x] + g − ξ [x] + x + g(y) = f (x) + g(y), i.e., f satisfies equation (18.5.6).



Hence we derive the following result (Tabor [312], Dhombres-Ger [70]): Theorem 18.5.2. Let (X, +) and (Y, +) be groups, and let (S, +) be a subgroup of (X, +). If f : X → Y is a function such that (18.5.1) holds for every x ∈ X, y ∈ S, then there exist a homomorphism g : S → Y , a lifting ξ : X/S → X, and a mapping h : X/S → Y such that (18.5.7) holds and   h(S) = g ξ[S] . (18.5.9) Conversely, every function f of form (18.5.7), where g : S → Y is a homomorphism, ξ : X/S → X is a lifting, and h : X/S → Y is a mapping fulfilling condition (18.5.9), satisfies (18.5.1) for every x ∈ X, y ∈ S. Proof. Suppose that a function f : X → Y satisfies equation (18.5.1) for x ∈ X, y ∈ S. As we have pointed out after the proof of Lemma 18.5.1, the function g = f | S is a homomorphism, and it is readily seen that f satisfies equation (18.5.6) for x ∈ X, y ∈ S. In virtue of Theorem 18.5.1 there exist a lifting ξ : X/S → X and a mapping 5

As is clear from the final remarks in 4.5, the existence of such a lifting results from the Axiom of Choice.

554

Chapter 18. Extensions of Homomorphisms

h : X/S → Y such that relation (18.5.7) holds. It remains to show that also condition (18.5.9) is fulfilled.   Since g = f | S is a homomorphism, we have by (18.2.2) f (0) = g(0) = 0. Hence, setting in (18.5.7) x = 0, and taking into account the fact that [0] = S, we get   0 = h(S) + g − ξ(S) .     Of course, ξ(S) ∈ S, and, g being homomorphism, g − ξ(S) = −g ξ(S) . Hence (18.5.9) results. Conversely, suppose that f is given by (18.5.7), where g : S → Y is a homomorphism, ξ : X/S → X is a lifting, and h : X/S → Y is a mapping fulfilling condition (18.5.9). By Theorem 18.5.1 f satisfies   equation (18.5.6) with g occurring in (18.5.7). If x ∈ S, then [x] = S, and also ξ [x] ∈ S. Hence       g − ξ([x]) + x = −g ξ([x]) + g(x) = −g ξ(S) + g(x), and we get by (18.5.7) and (18.5.9) f (x) = g(x) for x ∈ S. Thus (18.5.1) results from (18.5.6).



Theorem 18.5.2 yields the general solution of equation (18.5.1) on the cylinder X ×Z. Let (S, +) be the subgroup of (X, +) generated by Z. Then clearly Z ⊂ S ⊂ Nf (Lemma 18.5.1) so that every solution of (18.5.1) on the cylinder X × Z must also be a solution of (18.5.1) on the cylinder X × S, and hence must be of form (18.5.7). Conversely, every solution of (18.5.1) on X × S is also a solution of (18.5.1) on X × Z. From Theorems 18.5.1 and 18.5.2 we deduce two corollaries (Tabor [312], Kuczma [183]). Corollary 18.5.1. Let (X, +) and (Y, +) be groups, and let (S, +) be a subgroup of (X, +). Let g0 : X → Y be a homomorphism, and let g = g0 | S. If f : X → Y is a function such that (18.5.6) holds for every x ∈ X, y ∈ S, then there exists a mapping h : X/S → Y such that   f (x) = h [x] + g0 (x). (18.5.10) Conversely, every function of form (18.5.10), where h : X/S → Y is an arbitrary function, satisfies equation (18.5.6) for all x ∈ X, y ∈ S. Proof. Suppose that f : X → Y satisfies equation (18.5.6) for all x ∈ X, y ∈ S. By Theorem 18.5.1 there exist a mapping h0 : X/S → Y and a lifting ξ : X/S → X such that             f (x) = h0 [x] + g − ξ [x] + x = h0 [x] + g0 − ξ [x] + x     = h0 [x] + g0 − ξ([x]) + g0 (x).   With h(u) = h0 (u) + g0 − ξ(u) , u ∈ X/S, we obtain hence (18.5.10).

18.5. Cauchy equation on a cylinder

555

Conversely, suppose that f is given by (18.5.10), where h : X/S → Y is an arbitrary function. Then, for x ∈ X, y ∈ S, we have −(x + y) + x = −y − x + x = −y ∈ S, which means that x + y and x are equivalent with respect to S, whence [x + y] = [x], and     f (x + y) = h [x + y] + g0 (x + y) = h [x] + g0 (x) + g0 (y)   = h [x] + g0 (x) + g(y) = f (x) + g(y), 

and so (18.5.6) holds.

Corollary 18.5.2. Let (X, +) and (Y, +) be groups, and let (S, +) be a subgroup of (X, +). Let f : X → Y be a function such that (18.5.1) holds for all x ∈ X, y ∈ S. Suppose that there exists6 a homomorphism g0 : X → Y such that g0 | S = f | S. Then there exists a mapping h : X/S → Y such that (18.5.10) holds. Conversely, every function of form (18.5.10), where h : X/S → Y is an arbitrary function and g0 : X → Y is a homomorphism such that g0 | S = f | S, satisfies equation (18.5.1) for all x ∈ X, y ∈ S. Proof. Let f : X → Y satisfy (18.5.1) for all x ∈ X, y ∈ S. Put g = f | S. Then g0 is a homomorphism from X to into Y, g0 | S = g, and f satisfies (18.5.6) for x ∈ X, y ∈ S. By Corollary 18.5.1 there exists a mapping h : X/S → Y such that (18.5.10) holds. Conversely, if f has form (18.5.10), then, by Corollary 18.5.1, f satisfies (18.5.6) for x ∈ X, y ∈ S. But for y ∈ S we have g(y) = f (y), so (18.5.6) reduces to (18.5.1).  One can also consider the problems symmetric to those dealt with the present section: the Cauchy equation (18.5.1) on a cylinder Z × X, and the equation f (x + y) = g(x) + f (y)

(18.5.11)

for x ∈ S, y ∈ X. Such problems can be reduced to those treated in the present section as follows. Introduce in X and Y new operations ⊕ defined in the following way: x ⊕ y = y + x for x, y ∈ X, t⊕s=s+t

for t, s ∈ Y.

With these operations (X, ⊕) and (Y, ⊕) are groups, and, whenever (S, +) is a subgroup of (X, +), (S, ⊕) is a subgroup of (X, ⊕). Then equation (18.5.1) on the cylinder Z × X is equivalent to the equation f (x ⊕ y) = f (x) ⊕ f (y) for x ∈ X, y ∈ Z, whereas equation (18.5.11) (x ∈ S, y ∈ X) is equivalent to f (x ⊕ y) = f (x) ⊕ g(y) for x ∈ X, y ∈ S. 6

Some conditions for the existence of such a homomorphism are found in 18.1 and 18.4. Note that f | S in a homomorphism from S into Y .

556

Chapter 18. Extensions of Homomorphisms

18.6 Cauchy nucleus Let (X, +) and (Y, +) be arbitrary groups, and let f : X → Y be an arbitrary function. According to Lemma 18.5.1 the Cauchy nucleus Nf of f , if non-empty, is a subgroup of X. A natural question arises as to whether every subgroup of X is the Cauchy nucleus of a certain mapping f : X → Y ? An answer to this question (Grz¸a´slewicz-Pow¸azka-Tabor [124]) is the subject matter of this section. Theorem 18.6.1. Let (X, +) and (Y, +) be arbitrary groups, and let (S, +) be a subgroup of (X, +). If the group Y contains an element t = 0 which is not of order two, then there exists a function f : X → Y such that Nf = S. Proof. The assumption on Y means that 2t = 0. Put  0, for x ∈ S f (x) = t, for x ∈ X \ S.

(18.6.1)

Take arbitrary x ∈ X, y ∈ S. Then either x ∈ S, or x ∈ X \ S. In the former case we have x + y ∈ S, and f (x) = f (y) = f (x + y) = 0 so that (18.5.1) holds. If x ∈ X \ S, then also x + y ∈ X \ S (Lemma 4.5.3), and f (x) = f (x + y) = t, whereas f (y) = 0, and again (18.5.1) holds. Thus (18.5.1) holds for every x ∈ X, y ∈ S, which shows that S ⊂ Nf . (18.6.2) Now take an arbitrary x ∈ X \ S. Then, by Lemma 4.5.3, also −x ∈ X \ S, and, since obviously 0 ∈ S, f (−x + x) = f (0) = 0 = 2t = f (−x) + f (x) so that x ∈ / Nf . In other words,

Nf ⊂ S,

which, together with (18.6.2), yields the desired equality Nf = S.

(18.6.3) 

Theorem 18.6.2. Let (X, +) and (Y, +) be arbitrary groups such that card Y > 1, and let (S, +) be a subgroup of (X, +), which is not of index 2. Then there exists a function f : X → Y such that Nf = S. Proof. Since card Y > 1, then there exists a t ∈ Y such that t = 0.

(18.6.4)

We define the function f : X → Y by (18.6.1). The same argument as in the proof of Theorem 18.6.1 shows that relation (18.5.1) holds for all x ∈ X, y ∈ S, whence we obtain (18.6.2). If S = X, then (18.6.2) implies Nf = S. If S = X, then take an arbitrary y ∈ X \ S. Since S is not of index 2, there exists an x ∈ X \ S such that x + y ∈ X \ S (cf. Exercise 4.7). Then by (18.6.1) and (18.6.4) f (x + y) = t = t + t = f (x) + f (y) whence y ∈ / Nf . Hence we get (18.6.3), which together with (18.6.2) yields Nf = S. 

18.6. Cauchy nucleus

557

If card Y = 1, then the only element of Y is 0, whence f = 0 is the only function f : X → Y . This function fulfills (18.5.1) for all x, y ∈ X so that Nf = X, and no proper subgroup of X can be the Cauchy nucleus of any f : X → Y . If card Y > 1, then the only case not covered by Theorems 18.6.1 and 18.6.2 is when every element of Y has order 2, and S is a subgroup of X of index 2. Then not every subgroup of X is the Cauchy nucleus of an f : X → Y (cf. Theorem 18.6.3 and Example 18.6.1 below). Lemma 18.6.1. Let (X, +) be a group. If every element x = 0 of X has order 2, then (X, +) is commutative. Proof. The assumption about X amounts to the fact 2x = 0, or x = −x for every7 x ∈ X .

(18.6.5)

Take arbitrary x, y ∈ X. By (18.6.5) x + y = −(x + y) = −y − x = y + x, which means that (X, +) is commutative.  Let (X, +) and (Y, +) be groups. A homomorphism f : X → Y is called an isomorphism iff it is one-to-one and onto. If an isomorphism f : X → Y exists, then (X, +) and (Y, +) are said to be isomorphic. Theorem 18.6.3. Let (X, +) and (Y, +) be groups such that every element t = 0 of Y has order 2, and let (S, +) be a subgroup of (X, +) whose index is 2. Then there exists a function f : X → Y such that Nf = S if and only if there exists a normal subgroup (T, +) of (S, +) such that the group (S/T, +) is isomorphic to a subgroup of (Y, +), and either there exist x ∈ S, u ∈ X \ S such that x−u−x+u∈ / T,

(18.6.6)

or there exist x ∈ X \ S, and u ∈ X \ S such that −u + x − u − x ∈ / T.

(18.6.7)

Proof. Take arbitrary x, y ∈ X \ S. Then also −x, −y ∈ X \ S (cf. Lemma 4.5.3), whence −x+ y = (−x)− (−y) ∈ S, since the index of s is two. Thus every x, y ∈ X \ S are equivalent with respect to S, which means that the set X/S consists of two classes only: S and X \ S. Suppose that there exists a function f : X → Y such that Nf = S. By Theorem 18.5.2 there exist a lifting ξ : X/S → X, a homomorphism g : S → Y , and a mapping h : X/S → Y fulfilling condition (18.5.9) and such that (18.5.7) holds. For x ∈ S we have [x] = S and           f (x) = h [x] + g − ξ [x] + x = h(S) + g − ξ(S) + x = h(S) + g − ξ(S) + g(x),     whence, since g is a homomorphism, and so g − ξ(S) = −g ξ(S) , we obtain by (18.5.9) f (x) = g(x). So in view of the preceding remarks, f can be written as  g(x) for x ∈ S, f (x) = (18.6.8) h(X \ S) + g(−u + x) for x ∈ X \ S, 7

For x = 0 formula (18.6.5) is evident.

558

Chapter 18. Extensions of Homomorphisms

where u = ξ(X \ S) ∈ X \ S. Take a y ∈ X \ S. In virtue of the condition Nf = S there exists an x ∈ X such that f (x + y) = f (x) + f (y).

(18.6.9)

If x ∈ S, then x + y ∈ X \ S, since otherwise we would have −x ∈ S and y = (−x) + (x + y) ∈ S, and so we get by (18.6.8) and (18.6.9) h(X \ S) + g(−u + x + y) = g(x) + h(X \ S) + g(−u + y)

(18.6.10)

= h(X \ S) + g(x) + g(−u + y), since according to Lemma 18.6.1 (Y, +) is commutative. Thus we get by (18.6.10) g(−u + x + y) = g(x) + g(−u + y) = g(x − u + y).

(18.6.11)

If x ∈ X \ S, then x + y ∈ S (Corollary 4.5.1) and we get by (18.6.8) and (18.6.9) and Lemma 18.6.1 g(x + y) = h(X \ S) + g(−u + x) + h(X \ S) + g(−u + y) = 2h(X \ S) + g(−u + x) + g(−u + y).

(18.6.12)

But h(X \ S) ∈ Y is an element of order 2. So we get from (18.6.12) g(x + y) = g(−u + x) + g(−u + y) = g(−u + x − u + y),

(18.6.13)

g being a homomorphism. Put T = Ker g. By Lemma 4.5.7 (T, +) is a normal subgroup of (S, +). By  Lemma 4.5.8 the function g is constant on every class from S/T in the sequel, these  classes will be denoted by x, x ∈ S , so the formula   γ x = g(x) unambiguously defines a mapping γ : S/T → Y . If we put Y0 = γ(S/T ) = g(S), this mapping clearly is one-to-one (cf. Lemma 4.5.8)   and onto Y0 . Take  arbitrary u, v ∈ Y0 . Then there exist x, y ∈ S such that u = γ x = g(x), v = γ y = g(y), and   u − v = g(x) − g(y) = g(x − y) = γ x − y . So u − v ∈ Y0 , and we infer from Lemma 4.5.1 that (Y0 , +) is a subgroup of (Y, +). Finally, we have for arbitrary x, y ∈ S/T         γ x + y = γ x + y = g(x + y) = g(x) + g(y) = γ x + γ y , which means that γ : S/T → Y0 is a homomorphism, and being one-to-one and onto, is an isomorphism. Thus the groups (S/T, +) and (Y0 , +) are isomorphic. Now, condition (18.6.11) means that   0 = g(x − u + y) − g(−u + x + y) = g (x − u + y) − (−u + x + y) = g(x − u + y − y − x + u) = g(x − u − x + u), i.e., x − u − x + u ∈ / Ker g = T . So we get condition (18.6.6).

18.6. Cauchy nucleus

559

Condition (18.6.13) means that   0 = g(−u + x − u + y) − g(x + y) = g (−u + x − u + y) − (x + y) = g(−u + x − u + y − y − x) = g(−u + x − u − x), i.e., −u + x − u − x ∈ / Ker g = T . So we get condition (18.6.7). Conversely, suppose that there exists a normal subgroup (T, +) of (S, +) such that (S/T, +) is isomorphic to a subgroup (Y0 , +) of (Y, +) and that there exist either x ∈ S, u ∈ X \S fulfilling (18.6.6), or, x, u ∈ X \S fulfilling (18.6.7). Let ϕ : S/T → Y0 be the isomorphism, and let π : S → S/T be the canonical homomorphism (cf. Lemma 4.5.9). Put g = ϕ ◦ π. It is easily We  −1 seen  that g : S → Y0 ⊂ Y is a homomorphism. −1 have Ker g = g −1 (0) = π −1 ϕ (0) . But since ϕ is one-to-one, we have ϕ (0) =   −1 −1 0 = π (T ) = T . {0}, whence Ker g = π Let condition (18.6.6) hold for certain x ∈ S, u ∈ X \ S. Take arbitrary y0 ∈ Y and v ∈ S. The formulas  ξ(S) = v, h(S) = g(v) (18.6.14) ξ(X \ S) = u, h(X \ S) = y0 , define a lifting ξ : X/S → X and a mapping h : X/S → Y fulfilling condition (18.5.9), and formula (18.5.7) defines a function f : X → Y satisfying, according to Theorem 18.5.2, equation (18.5.1) on the cylinder X × S. Thus, for this particular f , inclusion (18.6.2) holds. By the same argument, as in the first part of the proof, the function f can be written in form (18.6.8). Take an arbitrary y ∈ X \ S. Let x and u denote points occurring in condition (18.6.6). Then x + y ∈ X \ S, and (x − u + y), (−u + x + y) ∈ S. Moreover, by (18.6.6), (x − u + y) − (−u + x + y) = x − u − x + u ∈ / T = Ker g so that   g(x − u + y) − g(−u + x + y) = g (x − u + y) − (−u + x + y) = 0, and consequently we get, in turn (18.6.11), (18.6.10) and (18.6.9). This means that y ∈ / Nf , whence relation (18.6.3) follows, which together with (18.6.2) implies that Nf = S. Now suppose that condition (18.6.7) holds for certain x, u ∈ X\S. Take arbitrary y0 ∈ Y, v ∈ S, and define the lifting ξ : X/S → X and the mapping h : X/S → Y fulfilling (18.5.9) by formulas (18.6.14). Then formula (18.5.7), which again can be written in form (18.6.8), defines a function f : X → Y satisfying the Cauchy equation (18.5.1) on the cylinder X × S, whence inclusion (18.6.2) follows. Take an arbitrary y ∈ X \ S. Let x and u denote points occurring in condition (18.6.7). Then x + y ∈ S, −u + x − u + y ∈ S, and, by (18.6.7), (−u + x − u + y) − (x + y) = −u + x − u − x ∈ / T = Ker g.

560

Chapter 18. Extensions of Homomorphisms

Hence   g(−u + x − u + y) − g(x + y) = g (−u + x − u + y) − (x + y) = 0, and consequently we get, in turn, (18.6.13) , (18.6.12) and (18.6.9). This means that y ∈ / Nf , whence relation (18.6.3) follows, which together with (18.6.2) implies that Nf = S.  Example 18.6.1. Let (X, +) and (Y, +) be arbitrary groups in which every element has8 order 2, and let (S, +) be a subgroup of (X, +), whose index is two. By Lemma 18.6.1 all these groups are commutative. So we have x − u − x + u = 0 ∈ T and −u + x − u − x = 0 ∈ T for every subgroup (T, +) of (S, +) and for every x, u ∈ X. By Theorem 18.6.3 there does not exist a function f : X → Y such that Nf = S. Example 18.6.2. Let X = Z, and let S = 2Z be the set of all even integers. With the usual addition of numbers (X, +) is a commutative group, and (S, +) is its subgroup, whose index is two. Let T = 4Z be the set of all integers divisible by 4. (T, +) is a subgroup of (S, +), and all these groups being commutative, (T, +) is a normal subgroup of (S, +). Put Y = S/T . Then (Y, +) is group isomorphic to (S/T, +); the identity function yields an isomorphism of (S/T, +) onto (Y, +). Take u = 1 ∈ X \ S. Then −u + x − u − x = −2 ∈ / T for every x ∈ X. It follows by Theorem 18.6.3 that there exists a function f : X → Y such that Nf = S.

18.7 Theorem of Ger Let (X, +) and (Y, +) be arbitrary groups, and let (S, +) be a subsemigroup of (X, +) such that X = S − S. (18.7.1) Further, suppose that we are given two conjugate p.l.i. ideals I1 and I2 in X and in X × X, respectively (cf. 17.5), and S∈ / I1 .

(18.7.2)

Finally, suppose that we are given a function f : S → Y such that f (x + y) = f (x) + f (y) I2 -(a.e.) in S × S.

(18.7.3)

M = {(x, y) ∈ S × S | f (x + y) = f (x) + f (x)}.

(18.7.4)

Let By (18.7.3) we have M ∈ I2 . Since I1 and I2 are conjugate, there exists a set U ∈ I1 such that / U. (18.7.5) M [x] = {y ∈ X | (x, y) ∈ M } ∈ I1 for x ∈ There exist quite large groups in which every element has order 2. Let n ∈ N, let X be the set of all n × n diagonal matrices with only +1 and −1 on the main diagonal. If · denotes the multiplication of matrices, then (X, ·) is a group, every element of which has order 2, and card X = 2n . Similarly, there exist groups (X, +) every element of which has order 2 and such that the set X is infinite.

8

18.7. Theorem of Ger

561

Now, under the conditions specified above, and with the above notation, we have the following Lemma 18.7.1. For every x, y, u, v ∈ S \ U the equality x − y = u − v implies f (x) − f (y) = f (u) − f (v). Proof. Take x, y, u, v ∈ S \ U such that x − y = u − v. By Lemma 17.5.5 (y + S) ∩ (v +

S) ∈ / I1 , whence also (−v+y+S)∩S = −v+ (y+S)∩(v+S) ∈ / I . On the 1   other hand, since x, u ∈ / U , we have M [x], M [u] ∈ I1 , whence also − v + y + M [x] ∪ M [u] ∈ I1 , and     − v + y + S \ M [x] ∩ S \ M [u]

   = (−v + y + S) ∩ S \ − v + y + M [x] ∪ M [u] ∈ / I1 .   Similarly, M [v] ∪ − v + y + M [y] ∈ I1 , whence 

   

A = − v + y + (S \ M [x]) ∩ S \ M [u] \ M [v] ∪ − v + y + M [y] ∈ / I1 , and, in particular, A = ∅. Thus there exists an s ∈ A. For such an s, with the notation z = −y + v + s, (18.7.6) we have s ∈ S, s ∈ / M [u], s ∈ / M [v], z ∈ S, z ∈ / M [x], z ∈ / M [y], whence (u, s) ∈ / M, (v, s) ∈ / M, (x, z) ∈ / M, (y, z) ∈ / M.

(18.7.7)

On the other hand, we have x − y + v + s = u − v + v + s = u + s, i.e., x + z = u + s. Hence by (18.7.4) and (18.7.7) f (x) + f (z) = f (x + z) = f (u + s) = f (u) + f (s).

(18.7.8)

Relation (18.7.6) yields y + z = v + s, whence by (18.7.4) and (18.7.7) f (y) + f (z) = f (y + z) = f (v + s) = f (v) + f (s). Now, (18.7.8) and (18.7.9) yield that f (x) − f (y) = f (u) − f (v).

(18.7.9) 

The following result is due to R. Ger [112]: Theorem 18.7.1. Let (X, +) and (Y, +) be arbitrary groups, and let (S, +) be a subsemigroup of (X, +). Further, let I1 and I2 be conjugate p.l.i. ideals in X and in X × X, respectively, and suppose that conditions (18.7.1) and (18.7.2) hold. If f : S → Y is a function fulfilling (18.7.3), then there exists a unique homomorphism g : X → Y such that g | S = f I1 -(a.e.) in S. (18.7.10)

562

Chapter 18. Extensions of Homomorphisms

Proof. Let the set M ∈ I2 be defined by (18.7.4) and U ∈ I1 by (18.7.5). Take a z ∈ X. By Lemma 17.5.6 there exist x, y ∈ S \ U such that z = x − y. Define g(z) as g(z) = f (x) − f (y).

(18.7.11)

In virtue of Lemma 18.7.1, definition (18.7.11) is independent of the choice of x, y ∈ S \ U fulfilling z = x − y. Consequently formula (18.7.11) unambiguously defines a function g : X → Y . Now we show that g is a homomorphism. Take arbitrary x, y ∈ X. There exist s , t , u, v, p, q ∈ S \ U such that x = s  − t ,

y = u − v,

x + y = p − q.

(18.7.12)

By Lemma 17.5.7 there exist s, t ∈ S \ U such that t∈u+S

(18.7.13)

s − t = s − t = x.

(18.7.14)

and By (18.7.12) and (18.7.14) p − q = x + y = s − t + u − v, whence s − t + u = p − q + v.

(18.7.15)

By Lemma 17.5.5 (t + S) ∩ (y + q + S) ∈ / I1 , whence

(−y + t + S) ∩ (q + S) = −y + (t + S) ∩ (y + q + S) ∈ / I1 , and, since by (18.7.12) −y + t = v − u + t, we have (v − u + t + S) ∩ (q + S) ∈ / I1 , whence also

A = (−u + t + S) ∩ (−v + q + S) = −v + (v − u + t + S) ∩ (q + S) ∈ / I1 . (18.7.16) On the other hand, since p, q, s, t, u, v ∈ / U , we have according to (18.7.5) M [p], M [q], M [s], M [t], M [u], M [v] ∈ I1 , and       B = − v + q + M [p] ∪ − v + q + M [q] ∪ − u + t + M [s] (18.7.17)   ∪ − u + t + M [t] ∪ M [u] ∪ M [v] ∈ I1 ,   and cf. (18.7.16) and (18.7.17) A \ B ∈ / I1 . In particular, A \ B = ∅, which means that there exists a w ∈ A \ B. Thus, since by (18.7.13) −u + t ∈ S, w ∈ −u + t + S ⊂ S, and with z1 = −t + u + w, z2 = −q + v + w, we have     z1 ∈ S \ M [s] ∪ M [t] , z2 ∈ S \ M [p] ∪ M [q] .

(18.7.18)

18.7. Theorem of Ger

563

Moreover, w ∈ / M [u] ∪ M [v], whence (u, w) ∈ /M

and (v, w) ∈ / M,

(18.7.19)

whereas by (18.7.18) (s, z1 ) ∈ / M, (t, z1 ) ∈ / M, (p, z2 ) ∈ / M, (q, z2 ) ∈ / M.

(18.7.20)

Relation (18.7.15) yields s + z1 = p + z2 , whence by (18.7.20) and (18.7.4) f (s) + f (z1 ) = f (s + z1 ) = f (p + z2 ) = f (p) + f (z2 ).

(18.7.21)

On the other hand, by the definition of z1 and z2 we have t + z1 = u + w and q + z2 = v + w, whence by (18.7.19), (18.7.20) and (18.7.4) f (t) + f (z1 ) = f (t + z1 ) = f (u + w) = f (u) + f (w), and f (q) + f (z2 ) = f (q + z2 ) = f (v + w) = f (v) + f (w), whence we get by (18.7.21) f (s) − f (t) + f (u) + f (w) = f (p) − f (q) + f (v) + f (w) and f (p) − f (q) = f (s) − f (t) + f (u) + f (v).

(18.7.22)

According to (18.7.12), (18.7.14) and definition (18.7.11), relation (18.7.22) means g(x + y) = g(x) + g(y), i.e., g is a homomorphism. Now we prove relation (18.7.10). Take an arbitrary x ∈ S \ U . In particular, x ∈ X, whence by (18.7.1) there exist s, t ∈ S such that x = s − t. We have U ∈ I1 , and, by (18.7.5), M [x] ∈ I1 , whence also U ∪ M [x] ∈ I1 , and    (−s + U ) ∪ − t + U ∪ M [x] ∈ I1 , and in virtue of Lemma 17.5.5 

     − s + [S \ U ] ∩ − t + S \ U ∪ M [x]  

= (−s + S) ∩ (−t + S) \ (−s + U ) ∪ − t + U ∪ M [x] ∈ / I1 . (18.7.23)

In particular, set (18.7.23) is non-empty, and thus it contains an element y. Such a y fulfils   s + y ∈ S \ U, t + y ∈ S \ U ∪ M [x] , (18.7.24) and clearly (s + y) − (t + y) = s + y − y − t = s − t = x.

(18.7.25)

564

Chapter 18. Extensions of Homomorphisms

By definition (18.7.11) g(x) = f (s + y) − f (t + y).

(18.7.26)

On the other hand, we get by (18.7.25) x + (t + y) = s + y, whereas by (18.7.24) we have (x, t + y) ∈ / M , whence by (18.7.4) f (x) + f (t + y) = f (s + y), i.e., f (x) = f (s + y) − f (t + y).

(18.7.27)

Relations (18.7.26) and (18.7.27) yield g(x) = f (x),

(18.7.28)

and (18.7.28) holds for every x ∈ X \ U , i.e., I1 -(a.e.) in S. This proves relation (18.7.10). It remains to show the uniqueness of g. Suppose that homomorphisms g1 , g2 : X → Y both fulfil (18.7.10). Thus there exist sets V1 , V2 ∈ I1 such that g1 (x) = f (x) for x ∈ S \ V1 and g2 (x) = f (x) for x ∈ S \ V2 . We have V = V1 ∪ V2 ∈ I1 , and g1 (x) = g2 (x) = f (x)

for x ∈ S \ V.

(18.7.29)

Take arbitrary x ∈ X. By Lemma 17.5.6 there exist s, t ∈ S \ V such that x = s − t. Since g1 and g2 are homomorphisms, we have according to (18.7.29) g1 (x) = g1 (s − t) = g1 (s) − g1 (t) = g2 (s) − g2 (t) = g2 (s − t) = g2 (x). Consequently g1 = g1 in X, which proves the uniqueness of g.



Let us observe that if in Theorem 18.7.1 we take I1 = {∅} and I2 = {∅}, then we obtain Theorem 18.2.1, and if in Theorem 18.7.1 we take S = X, then we obtain Theorem 17.6.1. Consequently both, Theorem 18.2.1 and Theorem 17.6.1, are particular cases of Theorem 18.7.1. This indicates at the great generality (and hence also importance) of Ger’s Theorem (Theorem 18.7.1 above).

18.8 Inverse additive functions Let (S, +) be a semigroup, and let (F ; +, ·) be a field. A function f : S → F is said to be inverse additive iff 1 1 1 = + (18.8.1) f (x + y) f (x) f (y) holds for all x, y ∈ S such that f (x + y) = 0, f (x) = 0, f (y) = 0.

(18.8.2)

In order to bring equation (18.8.1) closer to Cauchy’s functional equation, we introduce the operation ˆ defined for functions f : S → F as follows  1/f (x) if f (x) = 0, fˆ(x) = x∈ S. 0 if f (x) = 0, Thus fˆ again is a function fˆ : S → F , and the following lemma is obvious.

18.8. Inverse additive functions

565

Lemma 18.8.1. Let (S, +) be a semigroup, let (F ; +, ·) be a field. The function f0 : S → F is inverse additive if and only if the function fˆ0 satisfies the Cauchy equation f (x + y) = f (x) + f (y)

(18.8.3)

for all x, y ∈ S such that f0 fulfils (18.8.2). The passage back from fˆ0 to f0 is well possible, because the operation ˆ is ˆ ˆ then f = gˆ. involutory: fˆ = f . In other words, if g = f, The problem of solving equation (18.8.1) under condition (18.8.2) was raised by J. Acz´el [6], and  was dealt with by R. Ger-M. Kuczma [116], and R. Ger [112]. Equation (18.8.1) or, equivalently, (18.8.3) with condition (18.8.2) cannot be solved unless we make some assumptions about the set Z = f −1 (0) = {x ∈ S | f (x) = 0}.

(18.8.4)

In fact, if the set (18.8.4) is very large, then condition (18.8.2) is very seldom (possibly never) fulfilled, and equation (18.8.1) furnishes very few, or even no, information about the behaviour of the function f . This is well seen from the following example (Ger-Kuczma [116]): Example 18.8.1. Let S = F = R (with the ordinary operations of addition and multiplication). We define by induction a sequence of intervals Jn = (an , bn ) ⊂ R. We put J1 = (a1 , b1 ) = (1, 2), and we define the sequences {an } and {bn } by recurrence as follows an+1 = 2bn , bn+1 = 2bn + 1, n ∈ N. We clearly have 1 = a1 < b1 < a2 < b2 < · · · < an < bn < an+1 < bn+1 , . . . , n ∈ N, and bn = an + 1, Put U=

∞ 

n ∈ N. Jn ,

n=1

and take arbitrary x, y ∈ U . There exist k, m ∈ N such that x ∈ Jk , y ∈ Jm , and let, e.g., k  m. We have a k < x < b k , am < y < b m , whence bm = am + 1  am + ak < x + y < bm + bk  2bm = am+1 . Consequently, bm < x + y < am+1 , which means that x + y does not belong to any interval Jn , and hence does not belong to U . Consequently the set U has the property x, y ∈ U implies x + y ∈ / U.

(18.8.5)

566

Chapter 18. Extensions of Homomorphisms

Define a function f :→ R as

 0 for x ∈ R \ U, f (x) = arbitrary for x ∈ U.

(18.8.6)

According to (18.8.5), condition (18.8.2) cannot be fulfilled for any x, y ∈ R, and thus function (18.8.6) is inverse additive. Other examples are found in Ger-Kuczma [116]. Instead of equation (18.8.1), we consider first equation (18.8.3) under condition (18.8.2). Of course, (18.8.3) with (18.8.2) is a conditional Cauchy equation (cf. the end of 13.6). Actually, now we do not need the operation of multiplication in the range of f , and so we may consider equation (18.8.3) with (18.8.2) for functions f : S → Y , where (Y, +) is an arbitrary group. Lemma 18.8.2. Let (S, +) be a semigroup, let (Y, +) be a group, and let Z ⊂ S be an arbitrary set. Further, let h : S → Y be a homomorphism. Then the function  0 for x ∈ Z, f (x) = (18.8.7) h(x) for x ∈ S \ Z, satisfies equation (18.8.3) with condition (18.8.2). Proof. Take arbitrary x, y ∈ S fulfilling (18.8.2). In view of (18.8.7) this means that x, y, x + y ∈ S \ Z, and consequently f (x) = h(x), f (y) = h(y), f (x + y) = h(x + y).  Now (18.8.3) results from the fact that h is a homomorphism. In order to prove a result in the converse direction we will need the following (Ger [112]). Lemma 18.8.3. Let (X, +) be a group, and let (S, +) be a subsemigroup of (X, +) such that (18.7.1) holds. Suppose that a set Z ⊂ S satisfies the condition for every k ∈ N and for every s, s1 , . . . , sk , t1 , . . . , tk ∈ S there

(18.8.8)

/ Z + si , s i ∈ / Z + ti + t, i = 1, . . . , k. exists a t ∈ S + s such that ti + t ∈ Then the linearly invariant ideal I(Z) generated9 by Z is proper. More exactly, S∈ / I(Z).

(18.8.9)

Proof. For an indirect proof suppose that S ∈ I(Z), i.e., there exist a k ∈ N, and x1 , . . . , xk , y1 , . . . , yk ∈ X such that S⊂

k 

  xi + Z ∪ (−Z) + yi .

i=1   According to (18.7.1) to every x ∈ Z there   exists an s ∈ S such that z + s ∈ S cf. the proof of Lemma 18.2.1, condition (i) . We define si ∈ S, i = 1, . . . , k, as 9

Cf. 17.5

18.8. Inverse additive functions

567

follows: s1 ∈ S is such that y1 + s1 ∈ S, and if we know s1 , . . . , sm , 1  m < k, then sm+1 ∈ S is such that (ym+1 + s1 + · · · + sm ) + sm+1 ∈ S. Thus i

yi +

sj ∈ S,

i = 1, . . . , k,

j=1

and with s = s1 + · · · + sk ∈ S, also ⎛ ⎞ i k

sj ⎠ + si ∈ S, si = y i + s = ⎝ y i + j=1

We have S +s⊂

i = 1, . . . , k.

j=i+1

k 

  xi + Z ∪ (−Z) + si , i=1

whence −s − S ⊂

k 

  − si + Z ∪ (−Z) − xi .

(18.8.10)

i=1

Repeating the above construction, we may find a t˜ ∈ S such that ti = −xi + t˜ ∈ S,

i = 1, . . . , k,

whence by (18.8.10) −s − S + t˜ ⊂

k 

  − si + Z ∪ (−Z) + ti ,

i=1

or −t˜ + S + s ⊂

k 

  − ti + Z ∪ (−Z) + si .

(18.8.11)

i=1

Now take an arbitrary u ∈ S. Then v = t˜ + u ∈ S, whence u = −t˜ + v ∈ −t˜ + S. Consequently S ⊂ −t˜ + S, and S + s ⊂ −t˜ + S + s, and by (18.8.11) S +s⊂

k 

  − ti + Z ∪ (−Z) + si .

i=1

Therefore for every t ∈ S + s there exists an i ∈ N, 1  i  k, such that t ∈ −ti + Z + si

or t ∈ −ti − Z + si ,

i.e., ti + t ∈ Z + si But it contradicts (18.8.8).

or si ∈ Z + ti + t. 

Corollary 18.8.1. Let (X, +) be a group, and let (S, +) be a subsemigroup of (X, +) such that (18.7.1) holds. Let Z ⊂ S be an arbitrary set. Then conditions (18.8.8) and (18.8.9) are equivalent.

568

Chapter 18. Extensions of Homomorphisms

Proof. That (18.8.8) implies (18.8.9) is the contents of Lemma 18.8.3. Conversely, suppose that (18.8.9) holds, and take arbitrary s, s1 , . . . , sk , t1 , . . . , tk ∈ S. The set k

   − ti + Z ∪ (−Z) + si belong to I(Z), and since by (18.8.9) S + s ∈ / I(Z), there i=1

exists a t, t ∈ (S + s) \

k 

  [−ti + Z ∪ (−Z) + si ].

i=1

This means that t ∈ S +s, and ti +t ∈ / Z +si , si ∈ / Z +ti +t, i = 1, . . . , k. Consequently condition (18.8.8) is fulfilled.  The importance of condition (18.8.8) lies in that it is expressed in terms of S only, and it is meaningful even if we do not know the group (X, +), and hence also the ideal I(Z). An example of verifying condition (18.8.8) may be found in Ger [112]. Theorem 18.8.1. Let (S, +) be a left reversible cancellative semigroup, and let (Y, +) be a group. Suppose that a function f : S → Y satisfies (18.8.3) for x, y ∈ S fulfilling (18.8.2). If the set (18.8.4) fulfils condition (18.8.8), then there exists a homomorphism h : S → Y such that (18.8.7) holds. Proof. By Theorem 4.5.2 there exists a group (X, +) such that (S, +) is a subsemigroup of (X, +), and condition (18.7.1) holds. By Lemma 18.8.3 condition (18.8.9) holds. Thus I1 = I(Z) is a p.l.i. ideal in X fulfilling (18.7.2). Put M = {(x, y) ∈ S × S | x ∈ Z or y ∈ Z or x + y ∈ Z} = (Z × S) ∪ (S × Z) ∪ {(x, y) ∈ S × S | x + y ∈ Z}. Clearly Z × S ∈ Ω(I1 ) and S × Z ∈ Ω(I1 ), because we have  S for x ∈ Z ∈ I1 , (Z × S)[x] = ∅ ∈ I1 otherwise, 

and (S × Z)[x] =

Z ∈ I1 ∅ ∈ I1

for x ∈ S, otherwise,

so that (Z × S)[x] ∈ I1 I1 -(a.e.) in X, and (S × Z)[x] ∈ I1 for all x ∈ X. By Lemma 17.5.4 {(x, y) ∈ S × S | x + y ∈ Z} ⊂ {(x, y) ∈ X × X | x + y ∈ Z} ∈ Ω(I1 ), whence {(x, y) ∈ S × S | x + y ∈ Z} ∈ Ω(I1 ) and M ∈ Ω(I1 ). Write I2 = Ω(I1 ). By Lemma 17.5.3 I1 and I2 are conjugate p.l.i. ideals. Now, if (x, y) ∈ (S × S)\ M , then x ∈ / Z, y ∈ / Z, and x+ y ∈ / Z, which means that condition (18.8.2) is fulfilled, and (18.8.3) holds. Consequently the function f satisfies equation (18.8.3) for all (x, y) ∈ (S × S) \ M , i.e., I2 -(a.e.) in S × S. By Theorem 18.7.1 there exists a unique homomorphism g : X → Y such that (18.7.10) holds. Write E = {x ∈ S | f (x) = g(x)}.

18.9. Concluding remarks

569

By (18.7.10) E ∈ I1 , whence also T = E ∪ Z ∈ I1 . By (18.7.1) and Lemma 17.5.6 we have X = (S \ T ) − (S \ T ). (18.8.12) Take an arbitrary x ∈ S \ Z ⊂ X. By (18.8.12) there exist u, v ∈ S \ T such that x = u − v. Thus u = x + v, and since u, v, x ∈ / Z, we have f (u) = 0, f (v) =  0, f (x) = 0, and by (18.8.3) f (u) = f (x + v) = f (x) + f (v), whence f (x) = f (u) − f (v).

(18.8.13)

Since also u, v ∈ / E, we have f (u) = g(u), f (v) = g(v), and by (18.8.13) f (x) = g(u) − g(v) = g(u − v) = g(x), since g is a homomorphism. Consequently f (x) = g(x)

for x ∈ S \ Z.

(18.8.14)

Now, if we put h = g | S, then h : S → Y is a homomorphism, and (18.8.7) is a consequence of (18.8.4) and (18.8.14).  Theorem 18.8.2. Let (S, +) be a left reversible and cancellative semigroup, and let (F ; +, ·) be a field. Let f : S → F be an inverse additive function. If the set (18.8.4) fulfils condition (18.8.8), then there exists a homomorphism10 h : S → F such that ⎧ ⎨0 for x ∈ Z, 1 f (x) = ⎩ for x ∈ S \ Z. h(x) Proof. This results immediately from Theorem 18.8.1 and Lemma 18.8.1.



18.9 Concluding remarks Even in so large a book it was impossible even to touch upon, leave alone to discuss more thoroughly, many topics connected with the Cauchy equation, or convex functions. So we did not considered extensions to more general spaces like, e.g., topological vector spaces, since otherwise we would have to include the whole functional analysis. Cauchy’s equation on restricted domains (conditional Cauchy equations) have only perfunctorily been mentioned. The interested reader will find further informations and references in Dhombres [68], Dhombres-Ger [69], [70], Kuczma [187], L. Paganoni-S. Paganoni Marzegalli [252]. Also of various generalizations of convex functions only p-convex functions have been discussed in the present book. Concerning may other generalizations consult the references quoted in 5.3. In connection with convex functions cf. also the monographs When we speak about a homomorphism h : S → F , we have in mind only additive structure of F , i.e., F is considered as a group (F, +).

10

570

Chapter 18. Extensions of Homomorphisms

Rockafellar [268] and Roberts-Varberg [267] as well as the article Beckenbach [22], and the numerous references given therein. Unfortunately, the lack of place detained us from discussing here extremely interesting in generalizations of additive functions and convex functions to stochastic processes. Cf. Nagy [235] and Nikodem [238]. Also set-valued additive functions and convex functions (Fifer [79], Godini [118], Henney [140], Nikodem [239], [240], [242], [241], R˚ adstr¨ om [261]) have not found place in this book. A generalization of additive functions to the case of relations is due to Sz´ az-Sz´az [308]. Another generalization of additive functions was dealt with by Aleksandrov [12] and Forti [86], [87]. There may exist further results pertinent to the material dealt with in this book, which has been consciously or unconsciously omitted. We suspect that if we wanted to include here absolutely everything related to additive functions and convex functions, this book would have never been finished, because new results would arise faster than we could proceed with writing. So we must have stopped at some moment. This moment has just come, and we trust that, imperfect as it is, this book will prove useful to many readers. If such is the case, our task will be accomplished.

Exercises 1. Let X be the set of all differentiable functions ϕ : R → R such that ϕ(R) = R, ϕ(0) = 0 and ϕ (x) > 0 for all x ∈ R. Let S = {ϕ ∈ X | ϕ (0) > 1}. With ◦ denoting the composition of functions, show  that (X, ◦) is a group and (S, ◦) is its subsemigroup. Let g : S → R \ {0} where R \ {0} is considered as a multiplicative group be given by g(ϕ) = ϕ (0). Show that f (ϕ) = ϕ (0), ϕ ∈ X, defines the unique homomorphism f : X → R \ {0} such that f | S = g. 2. Let X be the set of all non-singular n × n matrices, n odd, and let · denote the usual multiplication of matrices. Then (X, ·) is a group. Put S = {A ∈ X | det A > 0}. Then (S, ·) is a subgroup of (X, ·).  Let g : S → R \ {0} where R \ {0} is considered as a multiplicative group be given by g(A) = γ(det A), where γ : (0, ∞) → (0, ∞) is a function such that γ(uv) = γ(u)γ(v) for all u, v ∈ (0, ∞). Using Theorem 18.4.3 show that there exists a homomorphism f : X → R \ {0} such that f | S = g. Show that there exist exactly two such homomorphisms. 3. Let X = {(a, b, c, d) ∈ Q4 | ac = 0}, and define a binary operation · in X by (a, b, c, d) · (a , b , c , d ) = (aa , ab + b, cc , dc + d ). Then (X, ·) is a group. Put S = {(a,  of (X, ·). Show  b, c, d)∈ X | a, b, c, d ∈ Z}. Then (S, ·) is a subsemigroup     1 1 / S · S −1 and so X = S · S −1 , 1, 0, 1, ∈ / S −1 · S and that 1, , 1, 0 ∈ 2 2x y u v   , , , for all so X = S −1 · S , but (1, 0, u, v) · (z, 0, w, 0)−1 · (x, y, 1, 0) = z z w w  −1 x, y, z, u, v, w ∈ Z such that xzuw = 0 and so X = S · S · S (Benz [27]).

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Index of Symbols N, xiv Z, xiv Q, xiv R, xiv C, xiv , xiv P(A), 3 P (x), 6 A, 7 ω, 7 Γ(α), 9 α, 10 M (m), 11 Ω, 11 ℵ1 , 11 F (α, X), 13 lim xβ , 17 β<α

D(A), 20 A[x0 ], 21 B(X), 25 G(X), 25 F(X), 25 Aα , 25 Mα , 25 z, 30 K(x, r), 30 d(A), 30 A(X), 32 A, 35 Ad , 39 A• , 39 A + B, 42 A + b, 42 b + A, 42 C, 45 m, 47 L, 47 me , 48 mi , 48

D+ f (x), 65 |h|, 68 E(A), 76 EF (A), 76 E + (A), 78 dim L, 81 Ker f , 85 G/H, 93 Ker f , 94 R[x1 , . . . , xm ], 98 R[x], 98 R[a], 98 R[a1 , . . . , am ], 98 F (x1 , . . . , xm ), 100 F (A), 101 F (a), 101 F (a1 , . . . , am ), 101 algcl F , 103 (q, m), 107 Iϕ , 109 (a, b), 114 D, 117 conv A, 120 Q(A), 120 J(A), 120 Q(a, b), 121 l(a, b), 121 Hf , 137 gab , 147 Gab , 147 ab, 147 ϕx (r), 150 mf (x), 150 Mf (x), 150 I(x, h), 165 Δh f (x), 167  (x), 168 f+  (x), 168 f− D+ f (x), 172

588 D− f (x), 172 d+ f (x), 172 d− f (x), 172 fy (x), 178 ∇f (x), 179 ∂f (x), 179 Mn (F ; a; w), 208 a, 208 An , 208 Gn , 208 Hn , 208 Mnr [a; w], 209 α (H; F ; a; w; I), 209 A, 227 B, 227 C, 227 AN , 227 BN , 227 CN , 227 Sn (A), 230 R(T ), 230 U (T ), 230 V0 , 236 Gr(ϕ), 242 HA , 245 D(D), 247 AC , 251 BC , 251 F n , 252 a < b, 253 a  b, 253 (a, b), 253 [a, b], 253 Wp , 270 Wp [xq ], 270 Z(H), 288 Af , 305 pf , 391 Δh , 415 Δph , 415 Δh1 ,...,hp , 415 [x1 , . . . , xp ; f ], 422 H(T ), 435 Pf , 476 I-(a.e.), 499 Iℵ0 , 499 IℵN0 , 499 Ib , 499 IbN , 499

Index of Symbols If , 499 IfN , 499 I0N , 499 N Im , 499 J (B), 499 I(B), 500 Π(I), 502 Ω(I), 502 I − infess f (x), 504 B

I1 − supess F (x), 525 B

Nf , 551

Subject Index

almost everywhere I-∼, 499 automorphism, 108, 402 trivial ∼, 402 Baire property, 23 base, 78 algebraic ∼, 105 cone-∼, 78 basis Burstin ∼, 285 Hamel ∼, 82 Borel separable sets, 44 Borel set, 25 bracket of derivations, 413 Burstin basis, 285 cancellation law left ∼, 92 right ∼, 92 Cantor set, 45 Cauchy equation, 128, 343 conditional ∼, 374 on a cylinder, 551 on a restricted domain, 374 restricted ∼, 374 Cauchy nucleus, 551 chain, 5 closure algebraic ∼, 103 collection, 3 combination linear ∼, 75 cone, 78 cone-base, 78 coset left ∼, 93 right ∼, 93 cylinder, 551

derivation, 391 partial ∼, 413 derivative Dini ∼, 65, 172 diagonalization, 446 differential Stolz ∼, 179 Dini derivative, 65, 172 divided difference, 421 domain characteristic of an integral ∼, 99 integral ∼, 99 element algebraic ∼, 103 degree of an algebraic ∼, 104 greatest ∼, 6 last ∼, 6 least ∼, 6 maximal ∼, 5 minimal ∼, 5 minimal polynomial of an algebraic ∼, 104 of order k, 99 smallest ∼, 6 transcendental ∼, 103 elements ϕ-conjugate ∼, 110 algebraically dependent ∼, 101 algebraically independent ∼, 101 conjugate ∼, 111 linearly dependent ∼, 75 linearly independent ∼, 75 endomorphism, 108, 402 trivial ∼, 403 epimorphism, 108 equation alternative functional ∼, 380 Cauchy’s functional ∼, 128, 343

590 Hossz´ u ∼, 374 Jensen ∼, 351 Mikusi´ nski ∼, 376 Pexider ∼, 355 Erd˝ os set, 288 field, 98 algebraically closed ∼, 114 extension of a ∼, 101 homogeneity ∼, 137 of fractions, 100 simple extension of a ∼, 101 function J-convex ∼, 130 Ω-decreasing ∼, 335 Ω-increasing ∼, 335 Ω-monotonic ∼, 335 I-almost polynomial ∼, 510 I2 -almost additive ∼, 505 I2 -almost convex ∼, 519 I2 -almost subadditive ∼, 525 ω-decreasing ∼, 335 ω-increasing ∼, 335 ε-additive ∼, 483 ε-convex ∼, 490 p-additive ∼, 363 p-concave ∼, 429 p-convex ∼, 429 additive ∼, 128 additive and positively homogeneous ∼, 87 approximately additive ∼, 483 approximately convex ∼, 490 biadditive ∼, 363 concave ∼, 131 concave ∼ of order p, 429 continuous and convex ∼, 130, 211 convex ∼, 130 convex ∼ of order p, 429 decreasing ∼, 65 diagonalization of a ∼, 446 differentiable ∼ at a point, 178 Hamel ∼, 130 increasing ∼, 65 infimum of a ∼ at a point, 150 infinitary ∼, 456 inverse additive ∼, 564 Jensen convex ∼, 130 linear ∼, 177

Subject Index locally bounded ∼ at a point, 148 locally bounded above ∼ at a point, 148 locally bounded below ∼ at a point, 148 lower hull of a ∼, 150 microperiodic ∼, 332 polynomial ∼ of order p, 429 positively homogeneous ∼, 177 quasi-homogeneous ∼, 480 rational ∼, 100 satisfying the Baire condition, 256 strictly concave ∼, 131 strictly convex ∼, 130 subadditive ∼, 455 sublinear ∼, 456 superadditive ∼, 455 supremum of a ∼ at a point, 150 twice differentiable ∼, 191 twice differentiable ∼ at a point, 170 upper hull of a ∼, 150 weakly-quasi-homogeneous ∼, 480 with big graph, 317 with small graph, 317 gradient, 179 group, 89 abelian ∼, 89 commutative ∼, 89 divisible ∼, 90 division in a ∼, 90 subgroup of a ∼, 89 topological ∼, 42 Hadamard inequality, 215 Hamel basis, 82 Hamel function, 130 Hardy-Littlewood-P´ olya majorization principle, 211 homomorphism, 83, 94, 108, 402 canonical ∼, 95 kernel of a ∼, 85 Hossz´ u equation, 374 hull C-convex ∼, 120 J-convex ∼, 120 Q-convex ∼, 120 convex ∼, 120 hyperplane

Subject Index k-dimensional rational ∼, 121 support ∼, 125 ideal, 498 σ-∼, 498 linearly invariant ∼, 499 linearly invariant ∼ generated by a set, 500 p.l.i. ∼, 499 principal ∼ of a ring, 114 proper ∼, 499 ideals conjugate ∼, 500 inequalities for means, 208 Jensen’s ∼, 197 Popoviciu ∼, 211 Rado ∼, 211 inequality general ∼ of convexity, 223 Hadamard ∼, 215 Jensen’s ∼, 130 Lim’s ∼, 214 Mulholland’s ∼, 218 Petrovi´c ∼, 217 Steffensen ∼, 201 interval open ∼, 56, 163 isomorphism, 108 Jensen convex function, 130 Jensen equation, 351 Jensen’s inequality, 130, 197 kernel, 85 lifting, 95 Lim’s inequality, 214 line rational ∼, 121 support ∼, 126 Lusin set, 59 M-algebra, 37 Marczewski algebra, 37 mean harmonic ∼, 208 weighted quasiarithmetic ∼, 208 measure

591 inner ∼, 47 outer ∼, 47 Mikusi´ nski equation, 376 monomorphism, 108 Mulholland’s inequality, 218 nucleus Cauchy ∼, 551 number cardinality of an ordinal ∼, 10 cofinal ordinal ∼, 17 diadic ∼, 117 limit ∼, 7 ordinal ∼, 7 operation A-conservative ∼, 229 C-conservative ∼, 229 binary ∼, 89 inner ∼, 89 order, 5 linear ∼, 5 partial ∼, 5 partition, 95 subpartition of a ∼, 95 Petrovi´c inequality, 217 Pexider equations, 355 point density ∼, 61 of accumulation, 39 of condensation, 39 of outer density, 61 polynomial, 98 degree of a ∼, 99 derivative of a ∼, 99 irreducible ∼, 104 minimal ∼, 104 reducible ∼, 104 Popoviciu inequality, 211 projection canonical ∼, 95 property (∗) ∼, 72 Baire ∼, 23 Rado inequalities, 211 relation type, 3 ring, 98 commutative ∼, 98

592 segment initial ∼, 6 rational ∼, 121 semigroup, 89 cancellative ∼, 92 left reversible ∼, 92 subsemigroup of a ∼, 89 sequence limit of a transfinite ∼, 17 transfinite ∼, 12 set, 3 C-convex ∼, 117 J-convex ∼, 117 Q-convex ∼, 117 Q-radial ∼ at a point, 257 algebraically dependent ∼, 101 algebraically independent ∼, 101 analytic ∼, 31 Borel ∼, 25 Cantor ∼, 45 convex ∼, 117 countable ∼, xiv denumerable ∼, xiv Erd˝ os ∼, 288 level ∼, 327 linearly dependent ∼, 76 linearly independent ∼, 76 linearly ordered ∼, 5 Lusin ∼, 59 nowhere dense ∼, 19 of the first category, 19 of the first category at a point, 20 of the second category, 19 of the second category at a point, 20 perfect ∼, 39 residual ∼, 19 saturated non-measurable ∼, 56 upper bound of a ∼, 6 well-ordered ∼, 6 sets Borel separable ∼, 44 congruent by decomposition ∼, 332 similarly ordered ∼, 6 simplex, 122 dimension of a ∼, 122 vertex of a ∼, 122 space base of a linear ∼, 78 dimension of a linear ∼, 81

Subject Index linear ∼, 75 separable ∼, 21 Steffensen inequality, 201 Stolz differential, 179 subgradient, 179 subgroup, 89 generated by a set, 91 normal ∼, 91 of index 2, 90 subpartition, 95 selective ∼, 96 semiselective ∼, 96 subset cofinal ∼, 17 system regular ∼, 35 relation ∼, 4 systems isomorphic ∼, 4 well-order, 6

Index of Names Abian, A., 282, 285 Acz´el, J., xiii, 128, 138, 282, 304, 338, 343, 351, 355, 364, 367, 369, 382, 537, 540, 565 Albert, M., 449, 483, 485–487, 532 Aleksandrov, A. D., 570 Alexiewicz, A., 241 Anastassiadis, J., 335 Aumann, G., 131 Baker, J. A., 413, 449, 483, 485–487, 532, 537 Balcerzyk, S., 535 Banach, S., 241 Baron, K., xiii, xiv, 376 Beckenbach, E. F., 131, 197, 211, 226, 570 Bellman, R., 197, 211 Benz, W., 338, 570 Bereanu, B., 335 Bergman, G. M., 338 Bernstein, F., 131, 155 Berz, E., 259, 456, 471, 473, 475 Bing, R. N., 131 Blanuˇsa, D., 374 Blumberg, H., 131, 241 Boas, R. P. Jr., 201, 205 Bonnesen, T., 117, 241 Borel, E.11 , 63 Bourbaki, N., 131 Bruckner, A. M., 322 Burstin, C., 285, 331 Carath´eodory, C., 47 Cauchy, A. L., 128, 343 11 The name of Borel is not quoted if it occurs in the context Borel sets. Similarly, the name of Baire is not quoted here as it occurs only in the context Baire property.

Ceder, J. G., 322 Cholewa, P. W., 483, 490, 494, 498 Ciesielski, Z., 42, 205, 438 Clifford, A. H., 92 Cohen, P. J., 12, 16 Cooper, R., 456 Cs´ asz´ ar, A., 250 Czerwik, S., 131 Dankiewicz, K., 550 Darboux, G., 402 Dar´ oczy, Z., 369, 372, 374, 382 Davison, T. M. K., 374, 376 de Bruijn, N. G., 501, 505, 506, 508, 509 De´ ak, E., 131 Denny, J. L., 505 Dhombres, J., xiii, 374, 535, 537, 552, 553, 569 ˇ 446, 537 Djokovi´c, D. Z., Doetsch, G., 131, 155 Dubikajtis, L., 42, 376 Dubreil, P., 92 Eggleston, H. G., 117, 127, 492 Erd˝ os, P., 63, 282, 288, 295, 304, 328, 329, 331, 369, 505, 540 Fenchel, W., 117, 241 Feny˝ o, I., 374 Ferens, C., 42, 376 Fifer, Z., 570 Figiel, T., 241 Fischer, P., 241, 380, 473 Forti, G. L., 483, 531, 570 Fr´echet, M., 241, 453 Fuchs, L., 211 Gajda, Z., 439, 451 Gauss, C. F., 128

594 Ger, R., xiv, 42, 131, 138, 227, 232, 233, 235, 244, 290, 374, 376, 377, 380, 422, 435, 437–439, 451, 483, 500, 502, 503, 506, 510, 524, 531–533, 535, 537, 552, 553, 561, 565, 566, 568, 569 Glowacki, E., 374, 376 Golab, S., 369 Godini, G., 570 Green, J. W., 117 Grz¸a´slewicz, A., 96, 114, 389, 399, 540, 546, 551, 556 Guerraggio, A., 131 Gustin, W., 117 Gy˝ ory, K., 369 Hadamard, J., 131, 215 Hahn, H., 331 Halmos, P., 5, 47, 63, 71 Halperin, I., 56, 328, 329, 331, 333, 475 Halpern, J. D., 79 Hamel, G., 82, 129 Hardy, G. H., 197, 211 Hartman, S., 47, 508 Haupt, O., 131 Henney, D., 570 Highberg, I. E., 453 Hille, E., 453, 456, 465, 473, 480, 481 H¨ older, O., 131 Horinouchi, S., 391, 399 Hossz´ u, M., 374, 380, 446 Hukuhara, M., 250 Hurewicz, W., 34 Hyers, D. H., 453, 483, 486, 490, 494, 498 Jech, T. J., 16 Jensen, J. L. W. V., 131, 351 Jones, F. B., 252, 285, 299, 301, 317, 322 Jurkat, W. B., 399, 505 Kac, M., 241 Kannappan, Pl., 338, 376, 380, 391, 399, 537 Karamata, J., 211 Kemperman, J. H. B., 68, 232, 418, 429 Kestelman, H., 67 Kominek, B., 251, 254, 276 Kominek, Z., 42, 232, 237, 251, 253, 254, 276, 280 Krasnosel’ski˘ı, M. A., 131

Index of Names Kuczma, M., 42, 58, 63, 64, 67, 87, 131, 139, 227, 232, 241, 242, 256, 281, 282, 285, 288, 290, 294, 308, 323, 331, 334, 335, 339, 374, 376, 380, 382, 500, 524, 550, 552, 554, 565, 566, 569 Kuczma, M. E., 67, 259, 263, 264 Kulpa, W., 322 Kuratowski, K., 5, 15, 20, 21, 34, 256 Kurepa, S., 67, 232, 284, 399, 438 Lackovi´c, I. B., 214–216 Lajk´ o, K., 369, 374 L¨ auchli, H., 79 Lawrence, J., 483 Legendre, A. M., 128 Lim, V. K., 214 Littlewood, J. E., 197, 211 L  ojasiewicz, S., 47, 54, 62, 63, 170, 194, 195 Losonczi, L., 369, 372, 374, 382 Lusin, N., 34, 35, 44, 59, 333 Makai, I., 322, 323 Marcus, S., 232, 250, 282, 285, 319, 322, 328, 329, 331 Marczewski, E., 37, 319, 322 Martin, S. C., 139, 540 Mazur, S., 446, 453 McKiernan, M. A., 446 Mehdi, M. R., 232 Mikusi´ nski, J., 47, 376 Mitrinovi´c, D. S., 131, 197, 201, 211 Mohr, E., 131 Moldovan, E., 131, 429 Mo´scicki, J., 266, 267 Mostowski, A., 5, 34 Moszner, Z., 473, 475, 483, 550 Mulholland, H. P., 218 Musz´ely, Gy., 380 Nagy, B., 570 Natanson, I. P., 47 Na˘ımark, M. A., 34 Nikodem, K., 570 Nishiyama, A., 399 N¨ orlund, N. E., 422 Obreanu, F., 322 Opial, Z., 62

Index of Names Øre, O., 92 Orlicz, W., 42, 241, 446, 453 Ostrowski, A., 63, 232, 308, 310, 313 Oxtoby, J. C., 25 Paganoni Marzegalli, S., 569 Paganoni, L., 67, 131, 241, 569 Petrovi´c, M., 217 Pexider, H. W., 355 Phillips, R. S., 453, 456, 465, 473, 480, 481 Piccard, S., 42, 67 P´ olya, G., 197, 211, 483 Ponstein, J., 131 Popoviciu, T., 131, 211, 422, 429 Pow¸azka, Z., 114, 551, 556 Preston, G. B., 92 Rad´ o, F., 537 Rado, R., 211 R˚ adstr¨ om, H., 570 Rasiowa, H., 5 R¨ atz, J., 137, 483 Redlin, L., 374 Rees, D., 92 Riesz, M., 47 Roberts, A. W., xiv, 130, 131, 162, 197, 225, 570 Rockafellar, R. T., 130, 131, 162, 570 Rosenbaum, R. A., 63, 455, 456, 465, 479–481 Rosenthal, A., 331 Rothberger, F., 338 ˇ 131 Rubinˇste˘ın, G. S., Ruticki˘ı, Ya. B., 131 Ruziewicz, S., 327, 331 Sablik, M., 500, 502 Samuel, P., 391 Sander, W., 42, 67 Segal, S. L., 413 Sierpi´ nski, W., 34, 60, 241, 282, 284, 285, 295, 320, 331, 333 Sikorski, P., 540 Slodkowski, Z., 241 Sm´ıtal, J., 42, 63, 64, 241, 247, 266, 267, 271, 274, 294, 297, 298, 304, 322, 327–331, 339 Smajdor, A., 504

595 Snoha, L., 42 Solovay, R. M., 3 Souslin, M., 34 Stankovi´c, L. R., 214 Stankovi´c, M. R., 215 Steffensen, J. F., 201 Steinhaus, H., 67 Stolz, O., 131 ´ Swiatak, H., 374, 376, 380 ´ 570 Sz´ az, A., Sz´ az, G., 570 Sz´ekelyhidi, L., 369, 446 Szeg˝ o, G., 483 Szymiczek, K., 139 Tabor, J., 114, 483, 551–554, 556 Tarski, A., 81 Tomi´c, M., 212 Ulam, S. M., 490, 494 Varberg, D. E., xiv, 130, 131, 162, 197, 225, 570 Vasi´c, P. M., 216 Vertge˘ın, B. A., 131 Vincze, E., 380 Weiss, M., 322 Wetzel, J. E., 465 Whitney, H., 486, 489 Zariski, O., 391 Zermelo, E., 14 Zorn, M., 15 Zorzitto, F., 483 Zygmund, A., 198