Functional Equations and Inequalities in Several Variables

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Stefan Czerwik

I~

r

Functional Equations and Inequalities in S e v e r a l

Variables

World Scientific

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Functional Equations and Inequalities in S e v e r a l

Variables

Stefan Czerwik Silesian University of Technology, Poland

Functional Equations and Inequalities in S e v e r a l

Variables

10World Scientific m

New Jersey • London • Singapore • Hong Kong

Published by World Scientific Publishing Co. Pte. Ltd. P O Box 128, Farrer Road, Singapore 912805 USA office: Suite IB, 1060 Main Street, River Edge, NJ 07661 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

FUNCTIONAL EQUATIONS AND INEQUALITIES IN SEVERAL VARIABLES Copyright © 2002 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

ISBN 981-02-4837-7

Printed in Singapore by World Scientific Printers (S) Pte Ltd

Preface Functional equations have substantially grown to become an important branch of mathematics, particularly during the last two decades, with its special methods, a number of interesting results and several applications. Many aspects of functional equations containing several variables may be found by the reader in the books of J. Aczel and J. Dhombres [1], M. Kuczma [123] and D. H. Hyers, G. Isac and Th. M. Rassias [97]. For more information about the history of functional equations the reader may consult Aczel [2] and Dhombres [43]. This book combines the classical theory and examples as well as recentmost results in the subject. The recent book consists of three parts. The first one is devoted to additive functions and convex functions defined on linear spaces endowed with so-called semilinear topologies. Basic results concerning important functional equations are also included in the first part. In the second part of this book we study the problem of stability of functional equations of several variables is considered. This problem has originally been posed by S. Ulam in 1940. In 1941 D. H. Hyers gave a significant partial solution to this problem in his paper [100]. In 1978, Th. M. Rassias [178] generalized the Hyers' result, a fact which rekindled interest in the field. Since then a number of articles have appeared in the literature. Such type of stability is now called the Ulam-Hyers-Rassias stability of functional equations. Finally, functional equations with set-valued functions are dealt v

Preface

VI

with in Part 3. A systematic exposition of functional equations with set-valued functions is still lacking in the mathematical litarature, as the result are scattered over several articles. An attempt is made here to fill this gap. As for the backgroud, we assume that the reader has some knowledge of the mathematical analysis, algebra, set theory and topology. The specialized notation we use is the same as in the relevant articles. Standard symbols Z + , Z, Q, R, C, • denote the sets of positive integers, integers, rationals, reals, complex numbers and the end of the proof, respectively. Many other symbols are introduced in the text of the book. The list of references at the end of the book is by no means complete. We have included only those articles and books that we referred to in the text of the book. It is a pleasant duty to express here our gratitude to Professor Themistocles M. Rassias, for fruitful suggestions and comments. We wish to express our thanks to World Scientific Publishing Co. for an efficient and agreeable cooperation.

Stefan Czerwik June, 2001

Contents Preface

v

P A R T I Functional Equations and Inequalities in Linear Spaces

1

1

Linear spaces and semilinear topology

3

2

Convex functions

9

3

Lower hull of a convex function

19

4

Theorems of Bernstein-Doetsch, Piccard and Mehdi 5 Some set classes of continuous and J-convex functions

29

6

Cauchy's exponential equation

35

7

D ' A l e m b e r t ' s equation on abelian groups

43

8

D ' A l e m b e r t ' s equation on topological groups

49

23

9 Polynomial functions and their extensions

65

10 Quadratic mappings

89 vii

viii

Contents

11 Quadratic equation on an interval

113

12 Functional equations for quadratic differences

121

PART II Ulam-Hyers-Rassias Stability of Functional Equations

127

13 Additive Cauchy equation

129

14 Multiplicative Cauchy equation

141

15 Jensen's functional equation

147

16 Pexider's functional equation

153

17 G a m m a functional equation 18 D ' A l e m b e r t ' s and Lobaczevski's functional equations

157

19 Stability of homogeneous mappings

169

20 Quadratic functional equation

185

161

21 Stability of functional equations in function spaces

203

22 Cauchy difference operator in Lp spaces

213

23 Pexider difference operator in LP spaces

221

24 Cauchy and Pexider operators in X\ spaces

229

25 Stability in t h e Lipschitz norms

235

26 Round-off stability of iterations

245

Contents 27 Quadratic difference operator in LP spaces

ix 251

PART III Functional Equations in Set-Valued Functions

261

28 Cauchy's set-valued functional equation

263

29 Jensen's functional equation

279

30 Pexider's functional equation

287

31 Quadratic set-valued functions

293

32 Subadditive set-valued functions

301

33 Superadditive set-valued functions and generalization of Banach-Steinhaus t h e o r e m

323

34 Hahn-Banach t y p e t h e o r e m and applications

331

35 S u b q u a d r a t i c set-valued functions

345

36 K-convex and K-concave set-valued functions

355

37 Iteration semigroups of set-valued functions

379

References

387

Index

407

Part I Functional Equations and Inequalities in Linear Spaces

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Chapter 1 Linear spaces and semilinear topology Basic definitions Let X be a real linear space and A be a subset of X. A point a € A is said to be algebraically interior to A iff for every x E X there exists an e > 0 such that Xx + aeA

for all

\e(-e,e).

(1.1)

Denote core A := {a E A: a is algebraically interior to

A}.

(1.2)

We say that A is algebraically open iff A = core A. Evidently, X is algebraically open as well as the union of any number of algebraically open sets and the intersection of a finite number of such sets is an algebraically open set. This means that the family T(X) of algebraically open subsets of a linear space X is a topology (see [114], [181], [213]), which is called the core-topoplogy. Directly from (1.2) we get AcB

implies

core A c core B, 3

(1.3)

4

Functional Equations and Inequalities in Several Variables

and core ,4 c A

(1.4)

core core A C core A

(1.5)

Therefore It may happen that the inclusion in (1.5) is strict. We will show the following example due to Z.Kominek and M.Kuczma [116]. Example 1.1 ([116]) Let X = R 2 . We denote by (a, p) the point with the coordinates a, P and 9 = (0,0). Assume that A= A1uA2uA3, where A1 = {{a,P)eR2: {a + l) 2 + p2 < 1}, 2 A2 = {(a, P)eR :(al) 2 + P2 < 1}, A3 = {(a,P)eR2:a = 0,pe(-l,l)}. Directly from then definition, we get core A = Ai\jA2U{9}^

core core A — Ax U A2.

For A, B C X, where X is a group (linear space), we use the notation A + B = {xeX: x = a + b, aeA, be B}, A- B = (x £ X: x = a-b, a e A, be B}, and if a e R, then OLA

— \x e X: x = aa, a e A}.

Semilinear topology In this section we are going to present some basic information about semilinear topologies, the way these were introduced by Z. Kominek and M. Kuczma [114], [116], [115], [117].

5

Linear spaces and semilinear topology

Assume that X is a linear space (over K) with a topology T. If the function ip: RxXxX^-Xis given by the formula ip(X,x,y) = Xx + y, XeR,

x,y e X,

(1.6)

and (p is continuous as a function of three variables, then T is called a linear topology and the space (X, T) is called a linear topological space. Such spaces are very important and well known (see e.g. [19]). The topology T will be called semilinear provided that the function (1.6) is continuous with respect to each variable separately. Note that every linear topology is semilinear. Directly from the definition of core-topology we have the following Lemma 1.1 Let X be a linear space and A C X an algebraically open set. Then for every a G K\{0} and x G X the sets aA and A + x are algebraically open. In the sequel we will consider real linear spaces. Lemma 1.2 If X is a linear space, then the core-topology T(X) is semilinear. Proof. From Lemma 1.1 one can easily guess that ip is continuous with respect to x and y separately. To prove continuity with respect to A fix x, y G X and A0 G K. Let A be an algebraically open set such that A0 + y G A, then A0 G A — y and A — y is algebraically open. Therefore, there exists an e > 0 such that XQX +

ax G A — y

for all

a G (—£, s),

i.e. Ax + y G A

for

A G (A0 — e, A0 + e),

which implies continuity of (p at Ao- Since A0 is arbitrarily fixed, this completes the proof. •

6

Functional Equations and Inequalities in Several Variables

If X — R, then the core-tolopogy T(X) is just the natural topology on the real line, so it is a linear topology. However if dimX > 2, the assertion is not any longer true. For some more details see [114]. Remark 1.1 In order to prove that a topology T in a linear space X is semilinear, it is sufficient to verify that for every fixed a G R\{0} and a G X the function defined by f(x) = ax + a,

x G X,

(1.7)

is continuous, and for every fixed x, y G X the function UJ : R —> X given by co{X) = Xx + y, A e l , (1.8) is continuous at the point zero. The continuity of u at any point Ao follows from the continuity of u at zero and from the following condition ip(\, x, y) = (p(\0, x, y) + ip(X - A0, x, 0),

A <E R, x, y e X,

where (p is given by (1.6). Lemma 1.3 If T is a semilinear topology in a linear space X, then for every A C X int^CcoreA. (1.9) Proof. Let y G A and x £ X. Since the function to is continuous at zero, then there exists an e > 0, such that (1.1) is satisfied, which means that y G core A and therefore (1.9) follows. • Lemma 1.4 Let T be a semilinear topology in a linear space X. Then T C T{X) (i.e. every open set is algebraically open). Proof. Assume that A is an open set in X. Therefore from (1.9) and (1.4) we obtain A = int A c core Ac

A

which means that A — core A and A is algebraically open.

•

Under some additional conditions in (1.9) one has the equality.

7

Linear spaces and semilinear topology

Lemma 1.5 Assume that X is a linear space with a semilinear topology. If A C X is a convex subset such that int A / 0 ; then int A = core A

(1.10)

Proof. In vew of Lemma 1.3 it is enough to show that corei4 C int A. Let U ^ 0 be an open set such that U C A. Take a G U and z G core A. Then there exists an e > 0 such that z + a(z — a) E A

for all

ae(—s,s).

For any S G (0, e) we take y = z + 5(z-a),

\ = 8/5 + 1.

Then we have z = \a+(l-\)y,

A 6(0,1).

Define the function ip: X —> X by ip(x) = Xx + (1 — A)y,

x G X.

It is easy to show that y> is a homeomorphism of X onto X and
8

Functional Equations and Inequalities in Several Variables

Notes 1.1 Consider a linear topological space X with a linear topology T. In view of Lemma 1.4 we have T C T(X). If dim X > 1, then the above inclusion is strict. Indeed, we have the following Proposition 1.1 ([114]) Let (X,T) be a linear topological space. 7/dim X > 1, then the core-topology T(X) is finer than T: T$T(X). 1.2 Note that there exists a semilinear topology which is neither linear nor core topology (for a suitable example see Z. Kominek [117]). 1.3 For more information about the core-topology one may see, for example, [181], [213].

Chapter 2 Convex functions Properties of convex functions In this section we will present some basic properties of a very important class of functions — the convex functions in linear spaces. The results are due to M. Kuczma and Z. Kominek. Let X be a linear space and D C X be a convex set. A function f:D—t [—00, oo) is said to be J-convex (convex in the Jensen's sense) iff it satisfies the following Jensen's inequality

/ (x-±i) < M±iM

p.D

for all x,y G D. A function / : D —> (—00,00] such that —/is J-convex is called J-concave. A function / : X —» R satisfying Cauchy's functional equation

f(x + y) = f(x) + f(y)

(2.2)

for all x,y £ X is called additive. If in addition / is homogeneous, i.e. f{ax) = a f{x) for all x G X and a G R, then / is called linear. 9

(2.3)

10

Functional Equations and Inequalities in Several Variables By induction one can easily prove that f(\x

+ (1 - \)y) < \f(x)

+ (1 - A)/(y)

(2.4)

holds for every diadic A G [0,1] and all x,y G D. If (2.4) is true for all x,y G D and every A G [0,1], then such a function is called a convex function. If / : D —> (—00,00] is such that —/ is convex, then / is said to be concave.For an extensive account of such functions one may see [168],[165]. We start with the following lemma that is due to Kominek, Kuczma [115]. Lemma 2.1 Let X be a linear space with a semilinear topology and A C X be an open and convex set. If f: A —> [—00,00) is a J-convex function, then f = —oo; or f: A —>• R is a finite function. For the case where A is an interval contained in R the reader is referred to [184]. Corollary 2.1 Let X,A,f be as in Lemma 2.1. Then the inequality (2.4) is satisfied for all x,y G A and every rational number A G [0,1]. For details see [123], [181]. If g: D -» [—00, 00) is a solution of Jensen's functional equation

9{^)=l[g(x)+g(y)}

(2.5)

for all x,y G D, then g is said to be an affine function. Let /': D —>• [—00, 00) be a given function and let t G D. An affine function gt: D —>• [—00,00) is said to be a support function of / at t iff 9t(t) = f(t), (2.6) 9t(x) < f{x),

xeD.

(2.7)

The following very general and useful result is due to Z. Kominek [117] (see also Rode [184], Konig [120]).

Convex functions

11

Theorem 2.1 Assume that X is a linear space with a semilinear topology and A C X an open and convex set. Let f:A—t [—00, 00) be a J-convex function. Then for every t G A there exists a support function gt: A —> [—00,00) of f at t. Moreover, if f is finite, then gt is finite. The following theorem gives the formula for a general solution of Jensen's equation (2.5) (see Kominek [117], Ger-Kominek [68]). Theorem 2.2 Let X be a linear space with a semilinear topology and A C X an open and convex set. If f: A —> K satisfies Jensen's functional equation (2.5), then there exist an additive function a: X —>• R and a constant a £ R such that f(x) = a(x) + a

for all

x G A.

(2.8)

Proof. Take x0 G A and set A0 = A — x0- Define / 0 : A0 —>• K by the formula fo{x) = f(x + x0) - f(x0),

x G A\.

Then / 0 is a solution of Jensen's functional equation (2.5) and moreover, /o(0) = 0, fa{\x) = \fo{x) for all x G A0. Therefore, by (2.5) we obtain for x, y G AQ satisfying x + y G A0 that

/.<* + y) = 2/„ ( ^ )

= 2AM+AW

= Mx)

+ My).

(2 . 9)

The set A0 is a convex neighbourhood of zero, thus we have: 1. for x G A0 and f G [0,1], £3; G A0; 2. for any x,y G Ao, there exists a natural number n such that -(x + n

y)eA0,

12

Functional Equations and Inequalities in Several Variables

which means that A0 is "full" in the sense of Dhombres (see [44], p. 410). In view of (2.9) and Theorem 4.3 from [44] there exists an additive function a: X —» K such that fo(x) = a(x) for x G A0. Consequently, for any x G A, we have /(x) = /o(z-:zo)+/(zo) = a(a;-a;o)+/(^o) =

a(x)+f(x0)-a(xQ).

Putting a = /(a;o) — a(x0), we get the desired conclusion.

•

Now we present two technical lemmas. Lemma 2.2 Let X be a linear space with a semilinear topology, let A C X be an open and convex set and let f: A —>• R be a J-convex function. If f is bounded above in a neighbourhood of XQ £ A, then it is bounded above in a neighbourhood of every x G A. Proof. Assume that U C A is an open set such that XQ G U and f(u) < M

for all

u€U,

(2.10)

where M G R. Let x G A be an arbitrary point. By Lemma 1.4 the set A is algebraically open, thus there exists an e > 0 such that x + a(x — x0) £ D

for

a£(—e,e).

Denote A = j ^ _ . If a G (0,e), then A G (0, JT^)- Hence there exists an a G (0,e) such that A is a diadic number. Put y = x + a(x — x0), then y G A and x = Ax0 + (1 - X)y. We can assume that e < 1. Let V — XU + (1 — A)y, then it follows easily that V is an open set and since A is convex, we obtain V C A. Therefore for every v G V we have D = Au + (1 — \)y for some u € U and consequently by (2.10) and J-convexity of / , we get

f(v)<XM+(l-\)f(y), i.e. / is bounded above on V. This completes the proof.

•

Convex functions

13

Lemma 2.3 Let X,A be as in Lemma 2.2. If f: A -» R is J-convex and bounded above in a neighbourhood of XQ G A, then it is also bounded below in a neighbourhood of XQ. Proof. Let (2.10) holds for some M eR and a neighbourhood U of xQ. The set V — 2z 0 — U is open, xQ G V and every t i g V has the form v = 2x0-u, ueU. (2.11) Now, for every v G V H A and u G C/ given by (2.11) we have

/(^)=/(!L^) 2f{xQ) - M. This means that / is bounded below on an open set VnA containing

xQ.

• The next lemma is obvious.

Lemma 2.4 Let X, A be as in Lemma 2.2 and assume that (2.10) holds with some M G R and an open set U C D. If, moreover, f(x0) = M for some x0 G U, then f(x) — M for all x G U.

Continuous convex functions Lemma 2.5 Let X, A be as in Lemma 2.2. If moreover, f-.A^-R is J-convex and lower semicontinuous function in A, then it is convex. Proof. Take x,y G A and A G (0,1) and put z = Xx + (1 — \)y. Since / is lower semicontinuous in A, then for every e > 0 there exists a neighbourhood V C A of z such that f(t)>f(z)-e

forteV.

(2.12)

14

Functional Equations and Inequalities in Several Variables

It is easy to see that the function LO0 : R —> X, denned by u0(\) = \x + {l-\)y

= \(x-y)

is continuous for every x,y G X. that (A - v, A + v) C (0,1) and p,x + (1 - n)y G V

for all

+ y,

AGR

(2.13)

So there exists an v > 0 such pe (\-u,\

+ u).

(2.14)

Consider a rational number /iG (A, A + ^). By (2.12) and (2.14) we have f{z) - e < /(fix + (1 - fJL)y) < fif{x) + (1 -

fi)f{y).

For /z —>• A over rational values in (A, X + u) we get / W - e < A / ( a : ) + (l-A)/(y). Now letting e —¥ 0+, we obtain f(Xx + (1 - Ay)) = /(*) < Xf(x) + (1 - A)/(y). The inequality (2.4) for A = 0 and A = 1 is obvious. This ends the proof of convexity of / on A. • Now we shall prove the following. Lemma 2.6 Let X be a linear space and let A C X be a convex and algebraically open subset. If f: A —>• R is a convex function, then for every a G K the sets U = {x G A: f(x)
and

V = {x e A: f(x) > a}

are algebraically open. Proof. Let x G X and y G U. Then y G A and since A is algebraically open, there exists a /3 > 0 such that y + AzG.4

for all

AG (-/?,/?).

Convex functions

15

Let (p(X) = f(y + Xx), for A G (-/3,/3). Then (p is a convex function of a real variable and therefore it is continuous (see [123]). Moreover, because y G U, ip(0) = f(y) < a, then we have f(y + Xx) = 0 such that Xx + y G U

for A G (—£, e)

e.g. y is algebraically interior to U. Since y G U and x G X are arbitrary points, this proves that U is an algebraically open set. In the similar manner we can prove that V is also algebraically open.

•

The next theorem gives the sufficient condition for J-convex functions to be continuous (see [117]). Theorem 2.3 Let X be a linear space with a semilinear Baire topology and let D C X be an open and convex set. Assume, moreover, that f: D —> M is a lower semicontinuous J-convex function. Then f is continuous function in D. Proof. Let a G R and Ua = {xe D: f(x) < a},

Va = {x G D: f(x) < a}.

From Lemma 2.5 if follows that / is convex and by Lemma 2.6 we get that Ua are algebraically open. From the definitions of Ua, Va and the convexity of / we guess that Ua, Va are convex sets. Also lower semicontinuity of / implies that Va are closed in D. Let a G E be fixed. Consider two cases. 1. Ua = 0. Then Ua is an open set.

16

Functional Equations and Inequalities in Several Variables 2. Ua ^ 0. Thus Va / 0 and is convex. Therefore by (1.3) and Lemma 1.6 we get Ua = core Ua C core Va = int Va, i.e. UaCintVa.

(2.15)

Now we shall prove that f(x) < a for every x € intVQ. For an indirect proof assume that for some x G int VQ, f(x) = a. By Lemma 2.4 we would have equality f(x) = a for all x € int Va and consequently by (2.15), the set Ua would be empty; this, however, is incompatible with our assumption that Ua / 0. Therefore we have mtVa C Ua, and together with (2.15), Ua = mtV^, which means that Ua is open. Since Ua is open for every a G E, then / is an upper semicontinuous function. Consequently, / being lower semicontinuous, is continuous and therefore the proof is complete.

Notes 2.1 If a function / : D —>• [—oo, oo) satisfies the inequality f[Xx + (1 - X)y] + /[(l - X)x + Ay] < f(x) + f(y)

(2.16)

for all x,y E 'D and A G [0,1], then it is called W-convex (convex in the sense of Wright, see [141], [181]). Clearly every W-convex function is J-convex . To see this set A = \ in (2.16). The class of all W-convex functions is intermediate between the class of convex and J-convex functions. Les us note that both inclusions are strict (see [181]).

Convex functions

17

2.2 Assume that a: X -> E is an additive function and g: 2) ->• [-co, oo) is a convex function. Then the function /:£)—>• [-co, oo) defined by the formula: /(*,):=a(*;)+ #(<;),

xGD,

(2.17)

is obviously W-convex. A function / : 2) -»• [-oo, oo) is called Woconvex iff it admits the decomposition (2.17) with a additive and g convex and continous (see [125]). 2.3 Convex functions have been introduced by J. L. W. V. Jensen [102], [103]. Similar functions were considered also by O. Hodler [93], J. Hadamard [81] and O. Stolz [203]. Basic properties of convex functions were proved by Jensen, F. BersteinG. Doetsch [15], H. Blumberg [16] and E. Mohr [137]. 2.4 Several monographs (e.g. Rockafellar [183], RobertsVarberg [181], Popoviciu [157], Moldovan [138]) have been devoted to convex functions. 2.5 There are several generalizations of convex functions (cf., e.g. Beckenbach [13], Kemperman [112], Ger [77], Czerwik [33], Gueraggio-Paganoni [80], Sander [188]).

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Chapter 3 Lower hull of a convex function The idea of the lower hull of a convex function has been presented for the first time by F. Berstein and G. Doetsch [15] (for a functions of a single real variable). Here we present the results contained in [114], [115], [117]. Let X be a topological space with topology T • For x G X the symbol Tx will denote the family of all open subsets of X containing x. Let A be an open subset of X. The lower hull mj of a function / : A —> [—oo, oo) is defined by the formula mf(x) = sup inf f(t), x G A. uerx teu

(3.1)

UQA

Hence mf. A —> [—00,00). Directly from the definition (3.1) it follows that mf{x) < f(x) for all x G A. (3.2) Also directly from the definition (3.1) we get the following result. Theorem 3.1 Let (X, T) be a topological space and let A C X be an open set. Then for every f:A—> [—oo,oo); the function mj defined by (3.1) is lower semicontinuous in A. 19

20

Functional Equations and Inequalities in Several Variables

Proof. Fix x e A. If mj{x) = - o o , then by the definition mj is lower semicontinuous. Let m,f(x) be finite. On account of (3.1) for every e > 0 there exists U ETX, U C A such that

mf/(i)

>mf{x)-£.

Now for every z G U, U G Tz and consequently by the definition of mf(z) and the last inequality we get m z

f( )

>

mi

" f(z) > mf(x) — e,

z G U.

Since e > 0 is arbitrary and U G Tx, this means that m; is lower semicontinuous at a;. • The next theorem provides a condition concerning the convexity and continuity of the function mf. Theorem 3.2 Let X be a linear space with a semilinear topology T, let A C X be an open and convex set and f: A —> [—oo, oo) be a J-convex function. Then mf given by the formula (3.1) is a convex and lower semicontinuous function in A. Moreover if (X, T) is a Baire space, then mj is continuous in A. Proof. To prove that m/ is J-convex in A, take x, y G A and a,(3 eR such that irif(x) < a

and

mf(y) < (3,

(3.3)

and put z = ^ . Let W C A be a neighbourhood of z. Since T is semilinear topology, the function ipi: X —> X defined by
t e X

is continuous and (fi(x) = z. Thus there exists a neighbourhood U C A of X such that yx{U) C W. From (3.1) and (3.3) there exists a point u G U such that f(u) < a,

(3.4)

Lower hull of a convex function

21

and clearly, s = X defined by

is continuous and ^{v) = ¥>i(w) — s G W. Thus there exists a neighbourhood V C A of y such that
f(v) < P.

(3.5)

The point w =
]<^,

and therefore, i n f / H < ^ .

(3.6)

Now, let a —>• rrif(x)+, (3 —>• mf(y)+ in (3.6), then we obtain inf / ( « ; ) < ^m/OO + m/fo)].

(3-7)

Finally, taking the supremum of (3.7) over all w G Tz, W C A, we have

which copmpletes the proof of the J-convexity of rrif in A. From Lemma 2.1, either m/ = — oo and therefore nif is convex and continuous, or m/ is a finite function and by Theorem 3.1 and Lemma 2.5 nif is convex and semicontinuous too. Moreover, if (X, T) is a Baire topological space, then in virtue of Theorem 2.3, nif is continuous in A. This completes the proof. • The last theorem of this section is the following.

22

Functional Equations and Inequalities in Several Variables

Theorem 3.3 Let X be a topological space and A C X an open set. The function f:A—> [—00,00) is lower semicontinuous at x E A iff f(x) = mf(x). (3.8) Proof. Suppose that (3.8) holds. In the case f(x) = rrif(x) = —00, / is clearly lower semicontinuous at x. If f(x) — mf(x) is finite, then from (3.1) for every e > 0 there exists a U G Tx, U C A such that f(t) > inf /(«) > mf{x) -e = f{x) - e for t G U, which means that / is lower semicontinuous at x. Now, let us assume that / is lower semicontinuous at x. If f(x) = —00, then (3.8) follows from the inequality (3.2). If f(x) is finite, then for every e > 0 there exists &U G Tx, U C A such that /(*) > f(x) - e for all

teU

and therefore, by (3.1), we get mf(x) >Mf(x)

>f(x)-e.

Hence, for e -> 0, we obtain rrif(x) > f(x), which together with (3.1) gives the equality (3.8). This concludes the proof of Theorem 3.3. •

Chapter 4 Theorems of Bernstein-Doetsch, Piccard and Mehdi The theorems of Bernstein-Doetsch, Piccard and Mehdi are very important in the theory of convex functions. In this section we present them under some very general assumptions. The proofs are based on the paper [116]. Let's start with the following. Theorem 4.1 Let X be a linear space with a semilinear topology, let A C X be an open and convex set and let f: A -> [—00, 00) be a J-convex function. If f is bounded above on a nonempty open subset of A, then it is convex and continuous in A. Proof. In the case f(x) = —00 for x G A, f is convex and continuous in D. In view of Lemma 2.1, we may assume that / : A —>• R is a finite function. On acount of Lemma 2.2 it suffices to prove that if / is bounded above in a neighbourhood of x G A, then it is continuous at x. Let x G A and V C A be a neighbourhood of x such that f(v) < s

for all v G V, 23

(4.1)

24

Functional Equations and Inequalities in Several Variables

where s is a real number. If follows that s - f{x) > 0.

(4.2)

Let e > 0 be an aritrary number and take a rational number A G (0,1) such that

The function (p(y) = Xx + (1 — X)y, y G X is a homeomorphism and hence the set W =
< Xf(x)+(1-X)f(y) = f(x) + (l-

< Xf(x) + (1-X)s =

X){8- f{x)) <

f(x)+e

i.e. fiw) < fix) + e

for all w G W.

(4.4)

Denote

U=

Wni2x-W),

then U is a neighbourhood of x and [/ C W so (4.4) is satisfied for all w eU. Moreover, U = 2x -U and therefore by (4.4) we get /(2x -u)

< fix) + e

for all u e U.

Also by the J-convexity of / we get

/ (U + 2Z~U) = M < \\f^) + /(2a; - u)], i.e. /(«) > 2fix) - fi2x - u)

for

ueU.

Now applying (4.4) we obtain /(u) > 2/(x) - /(re) - e = /(x) - e

for all

u G U.

(4.5)

Theorems of Bernstein-Doetsch, Piccard and Mehdi

25

Consequently, (4.4) and (4.5) imply the inequality |/(u)-/(u)|<e

for all

u G U,

which establistes the continuity of / at x. We guess from Lemma 2.5 that / is also a convex function. • Theorem 4.1 is a generalization of a very important result in the theory of convex functions that is due to Bernstein and Doetsch [15] (for other extentions see also [123], [51], [115]). Let us point out that even global boundedness below does not imply the continuity of a J-convex function. For example, the function exp is continuous and J-convex on R and let a: R™ —>• R be a discontinuous additive function. Then the function / : R™ —>• R given by f(x) — expa(x) is J-convex , discontinuous and f(x) > 0 for i £ K " . If a function is continuous at a point, then it is locally bounded (i.e. bounded on a neighbourhood of this point) at this point. Thus Theorem 4.1 and Lemma 2.1 imply the following. Theorem 4.2 Let X, A, f be as in Theorem Jf..l. Then either f is continuous in A, or f is discontinuous at every point of A. Let us now recall some definitions and denotations. A subset A of a topological space X is said to be locally of the second category at a point x € X iff for every neighbourhood V of x the set V D A is of second category. By D(A) we denote the set of all points x G X at which A is locally of the second category. A group X endowed with a topology such that the group operation is separately continuous with respect to either variable is called a semitopological group (see [94]). Let us note that, of course, a linear space endowed with a semilinear topology is a semitopological group. Now, we present without proof the following lemma (see [116]).

26

Functional Equations and Inequalities in Several Variables

Lemma 4.1 Let X be a semitopological group and let U,V C X be subsets of the first category. If, moreover, one of the sets G, H C X is open and the other is locally of the second category at every of its points, then we have (G\U) - (H\V) = G-H.

(4.6)

Applying this lemma we are in a position to prove the following Theorem 4.3 Let X be a semitopological group. If A C X is of the second category Baire set and B C X is a set of the second category, then int (A-B)^ 0. (4.7) Proof. Let A, B C X be as in the Theorem. Since A is a set with Baire property, this means that there exist a non-empty open set G and two sets P, Q of the first category such that A = (G\P) U Q.

(4.8)

H = Bf}D(B).

(4.9)

Denote Now we get (see also [127], [123]) D(B) = D(H U (B\D{B))) and since D(B\D(B))

= D(H) U

D(B\D(B)),

= 0, it follows that D(B) = D{H).

(4.10)

Conseqently, in view of the Banach first category theorem (see [11] and [127]), we obtain that H ^ 0. Thus (4.9) and (4.10)) imply the relation H C D(H), i.e. H is locally of the second category at every of its points. Applying Lemma 4.1 with U — P, V = 0, we obtain

(G\P) -H = G-H=

\J(G-h). heH

(4.11)

Theorems of Bernstein-Doetsch, Piccard and Mehdi

27

The set \JheH(G — h) is non-empty and open and by (4.8), (4.9), (4.11) we arrive at the conclusion

\J{G-h)cA-B, and therefore the relation (4.7) is proved.

•

Remark 4.1 If we replace in (4-7)the set B by —B, we get mt(A + B)^$.

(4.12)

Remark 4.2 If X = R and A = B C K is a second category set with Baire property, then from Theorem 4.3 we obtain the result proved by S. Piccard [154]. Thus its generalisations are referred to as a theorem of Piccard. Theorem 4.4 Let X be a linear space with a semilinear topology, let A C X be an open and convex set, and let f: A —> [—oo, oo) be a J-convex function. If, moreover, f is bounded above on a second category Baire set V C A, then it is convex and continuous in A. Proof. We have f(v) < s,

for

v e V,

(4.13)

for some s € R Set

u = \
/(„)=/(£+«) < M +M < . ,

(4.14)

for all u € U. So / is bounded above on U. Take in (4.12) A = B = V, then we obtain

int U^t

28

Functional Equations and Inequalities in Several Variables

This together with (4.14) means that / is bounded above on a non-empty open subset of A. The continuity and the convexity of F follow directly from Theorem 4.1. • Remark 4.3 If X is a linear topological space, then from Theorem 4-4 we get the result proved by M. R. Mehdi [135] (see also [134])-

Chapter 5 Some set classes of continuous and J-convex functions Let's look again into the Theorem 4.1. If X is a linear space with a semilinear topology T, then every J-convex function defined on an open and convex set A C X and bounded above on a non-empty open subset of A has to be continuous in A. This justifies the following definition. A(X,T)

:= {T C X: every J-convex real-valued function defined on an open and convex subset A D T of X which is bounded above on T, is continuous } .

Therefore any non-empty open subset of X belongs to the family A(X,T). We also introduce another class of subsets. I (X, T) := {T C X: every additive function / : X -> R which is bounded on T, is continuous } . The above set classes have been introduced by R. Ger and M. Kuczma [66] for X = E n , and M. E. Kuczma [126] has proved 29

30

Functional Equations and Inequalities in Several Variables

that A (R*1, T) = B (R", 7"). More results about the class B (E n , T) one can find in [44], [123]. Now we will show that A{X,T)=M(X,T)

(5.1)

holds true in the case where X is a linear space with a semilinear Baire topology [117]. The following lemma will be used further on. Lemma 5.1 Let X be a linear space with a semilinear Baire topology T, let A C X be an open and convex set, and let f.D —> [—00,00) be a J-convex function. Then f is lower semicontinuous at t G A iff there exists a continuous support function gt: A —>• [—00,00). Proof. Suppose that / is a lower semicontinuous function at t £ A. According to Theorem 3.3, f(t) = m,f(t). In the case when fit) — —00, then gt = — 00 is a required continuous support function. If f(t) = mf(t), then by Theorem 2.1 there exists a support function gt of mt at t, which is of course also a support function of / at the point t. By Theorem 3.2 m/ is a continuous function. Hence there exists a neighbourhood V of t, V C A such that mj(x) < m,j{t) + 1 for x E V, and consequently gt(x) < m/(z) < 771/(t) + 1

for x e V.

Therefore the continuity of gt results now from Theorem 4.1. To prove the second part of the Lemma, let gt: A —>• [—00,00) be a continuous function with properties (2.5), (2.6), (2.7). If 9t{t) = —00, then also f(t) = —00 and clearly / is lower semicontinuous at t. If gt(t) > —00, then for every e > 0 there exists a neighbourhood V of t, V C A such that 9t{x) > gt(t) -e

for x £V.

Some set classes of continuous and J-convex functions

31

Therefore, by (2.7) and (2.6) we get f{x)>9t(x)>9t{t)-e

= f(t)-e

for

x G V,

which means that / is lower semicontinuous at the point t. The proof is complete. • Now we prove the following (see [117]). Theorem 5.1 Let X be a linear space with a semilinear Baire topology T. Then A (X, T) = B (X, T). Proof. Clearly, A(X,T) C M(X,T). So we will prove that M(X,T) C A(X,T). Let T G B(X,T),let A C X be an open and convex set such that T C A and let / : A —> R be a J-convex function which is upper bounded on T. By Theorem 2.1 for every t € T there exists a support function gt: A —>• R of / at t. From Theorem 2.2 there exists an additive function at: X —> R and a constant a ( 6 i such that gt(x) = at(x) + at,

x G A.

(5.2)

Hence and in view of (2.7), at is bounded above on T, and consequently by the condition T G B(X, 7"), a^ is continuous. So by (5.2) also gt is continuous. Applying Lemma 5.1 we remark that / is lower semicontinuous in A. Consequently, by Theorem 2.3, / is continuous which proves that T G A(X,T) and thus the proof of the theorem follows. H Note that in the space X = W1 convexity of / implies its continuity. However, in infinite-dimensional spaces, it is not any longer true. Taking this into account it is interesting to investigating the following set classes. A(X) := {T C X: every J-convex real-valued function / : A —>• R defined on an algebraically open and convex subset A of X containing T which is bounded above on T, is convex } ,

32

Functional Equations and Inequalities in Several Variables

B(X) : = { T C I : every additive function / : A -> E which is bounded above on T, is linear } . To prove that A (X) = B {X), we will need the following lemma. Lemma 5.2 Let X be a linear space and let f: X —>• R ie an additive function. If, moreover, f is convex, then it is linear. Proof. Let / : X —>• R be an additive and convex function, i.e. / satisfies the conditions f{x + y) = f{x) + f{y),

x,yeX,

(5.3)

and f(Xx + {l-X)y)<Xf(x)

+ (l-X)f{y),

x,y e X,

A € [0,1]. (5.4)

Putting into (5.3) y = 0, or y — —x, one gets /(0) = 0

(5.5)

and f(-x)

= ~f(x),

xeX,

(5.6)

respectivety. Setting y = 0 into (5.4), we get in view of (5.5) f(Xx)<Xf{x),

xeX,

A €[0,1],

(5.7)

whence, changing x by — x and applying (5.6), one gets / (Ax) > A/(a;).

(5.8)

Now, (5.7) and (5.8) give the relation / (Xx) = Xf(x) iovxeX

and A G [0,1].

Every a G E may be presented as a = n + A, where n is an integer and AG [0,1].

Some set classes of continuous and J-convex functions

33

From the additivity of / we get easily for any integer n (see also [126]) f(nx) = nf(x), x G X. Therefore we have / (ax) = f(nx + Xx) = f (nx) + f (Xx) = nf(x) + Xf(x) = Xf(x), which means that / is a linear function. This completes the proof.

• At the end of this section we prove the following. Theorem 5.2 Let X be a linear space. Then A(X)=M(X). Proof. Let T C X, T G A (X) and let / : X ->• R be an additive function bounded above on T. Hence / is J-convex and since T G A (X), f is convex too. By Lemma 5.2 the function / is linear, which means that T e B (X) and this proves that A(X) C B(X). Conversly, assume that T e B(X). Let A D T be an algebraically open and convex subset of X, and let / : A —> R be a J-convex function which is bounded above on T. From Theorem 2.1 for every t G A there exists a support function gt: A —> K of / at t. On account of Theorem 2.2 there exists an additive function at: A —> R and constant at G R such that gt(x) = at(x) + at,

x,t e A.

Since gt(x) < f(x) and / is bounded, then every function at, t G A is also bounded above on T, and consequently in view of the condition T G B (X) it is linear. Therefore, for all x,y,t G A and A G [0,1] we have gt (Xx + (1-X)y)

= Xgt(x) + (1 - A) gt(y).

(5.9)

34

Functional Equations and Inequalities in Several Variables

For x, y G A,

AG [0,1] put z — Xx + (1 — A) y.

Then we get by (2.6), (5.9) and (2.7) / (Ax + (1 - A) y) = f(z) = gz(z) = gz (Ax + (1 - A) y) =Xgz(x) + (1 - A) gz(y) < Xf(x) + (1 - A) f(y), which means that / is a convex function and proves that T G A (X). Hence M(X) C A(X) and consequently A(X) =M(X). • For an extensive study of convexity theory with applications to nonlinear analysis the reader is referred to the recent book of D. H. Hyers, G. Isac and Th. M. Rassias [98].

Notes 5.1 The following complements Theorems 5.1 and 5.2. Corollary 5.1 Let X be a linear space endowed with the coretopology T{X). Then A(X) =

A(X,T(X)),

B(X)=M(X,T(X)). 5.2 In the case of the core-topology T{X) we may prove the following stronger version of the Theorem of Berstein-Doetsch. Proposition 5.1 ([119]). Let X be linear space and letT G X be such that coreT ^ 0. Then T belongs to the class A{X).

Chapter 6 Cauchy's exponential equation The following functional equations are referred to as Cauchy's equations (cf. Cauchy [23], Aczel [3], Kuczma [123], D. H. Hyers, G. Isac and Th. M. Rassias [97] , Th. M. Rassias [176]). f(x + y) = f(x) + f(y), f(x + y) = f(x)f(y), f(xy) = f(x + y),

(6.1) (6.2) (6.3)

f(xy) = f(x)f(y).

(6.4)

Cauchy's additive equation We first recollect some important facts concerning the Cauchy's additive equation (6.1) (see D. H. Hyers and Th. M. Rassias [95], as well as references [170], [164]). Theorem 6.1 Let X and Y be linear spaces. satisfies the equation (6.1), then f (Ax) = A/(x) for every x G X and A € Q. 35

If f: X —> Y (6.5)

36

Functional Equations and Inequalities in Several Variables

Proof. From (6.1) for x = y ~ 0 we get /(0) = 0. Hence setting y = —x in (6.1), we obtain

0 = f(x-x)

= f(x) + f(-x),

and therefore f(—x) = —f(x), which means that / is an odd function. Now let x\ = ... = xn = x, then f[xx + ... + xn) =f(xi) + f(x2 + ...+xn) =/0&i) + f(x2) + f(x3 + ... + xn) =

nf(x),

i.e. f(nx) = nf{x) for n € N. Consequently we see that / (Xx) = Xf(x) for all A e Z. Let A = —, where m G Z, n £ N. Hence ?7j.a; = n (Xx) and by what has already been proved, f(mx) = mf(x) = / (n (Ax)) = nf (Xx), which establishes the equality (6.5) for all A G .

•

Theorem 6.2 If f: R -> C satisfies the additive Cauchy's equation (6.1) and is continuous at the point x0 6 R, then f(x) = cx,

xeR

(6.6)

with c = f(l) e C. Proof. By the continuity at x0, we have lim f(y) = lim [f(y + x0) - f(x0)] = f(x0) - f(x0) = 0, 2/-»0

y->0

i. e. / is continuous at zero. Now, we get for x E R lim f(y) = lim [f(y - x) + f(x)} = lim f(u) + f(x) = which implies the continuity of / at any point x e M.

f(x),

Cauchy's exponential equation

37

Take x € R and consider a sequence {An} of rational numbers such that An —> x as n —> oo. Then we have f{x) = lim / (An) = lim [A„/(l)] = f(l)x

= ex,

i.e. we obtain the formula (6.6). Theorem 6.3 If a function f: R —> R satisfies (6.1) and is bounded above on an interval of positive length, or is Lebesgue measurable, then there exists a constant c € K such that f has a form (6.6). Proof. In the case when / is bounded, our statement follows directly from Theorems 4.1. If / is measurable, we can apply Sierpihski [191] theorem about the continuity of measurable convex function (since every additive function is a convex one). •

Cauchy's exponential equation We shall now determine the complex valued solutions / : R —>• C of Cauchy's exponential equation (6.2). Theorem 6.4 Let f: R —> C satisfy the equation (6.2). f(x) - 0 for all x e R or f{x) = exp \F{x) + i h(x)]

for all x 6 R,

Then (6.7)

where F: R —> R is a solution of (6.1) and h: R —> R is a solution of the equation h(x + y) = h(x) + h(y)

(mod 2?r)

for all

x, y e R.

(6.8)

Proof. Assume that / is a solution of the equation (6.2). Write f(x) =g(x)exp[ih(x)],

(6.9)

38

Functional Equations and Inequalities in Several Variables

where g,h: R ->• R and g{x) > 0 for x G R (6.1), we obtain

Putting (6.9) into

g(x + y) exp [i h{x + y)] = g(x)g{y) exp [i (h{x) + h(y))]. Taking the absolute value of both sides, we get g{x + y) = g{x)g{y) for all

i^eR,

(6.10)

and (if g ^ 0) exrj[ih(x + y)] = exp{i(h{x) + h(y))}

for all

x,y£R

(6.11)

which is equivalent to (since exp is periodic with period 27ri) h(x + y) = h(x) + h(y)

(mod 2TT)

for all

x, y G R.

(6.12)

First, let's consider the equation (6.10). Suppose that g(xo) = 0 for some x0 G R- Then, for arbitrary x £ l , g(x) = g(x -xQ + XQ) = g(x - xQ)g(x0) = 0, i.e. g — 0. Assume now that g ^ 0, then by what has been shown, g(x) ^ 0 for every x G R. Thus we have for every x € R, '1

1

-X + -X

2

2

^

9

fl T:^M

^2 7.

which implies that ^(a;) > 0 for every x G R. So we can define F(a:) :=\ogg(x), x G R. Then, by (6.10), F : R —> R is an additive function, and g(x) = eF^ for xeR. Therefore, substituting this formula into (6.9), we obtain the formula (6.7) and the proof is complete. • We return now to continuous solutions of the equation (6.2)(see [I])-

Cauchy's exponential equation

39

Theorem 6.5 Let f: R —>• C be a measurable solution of the aquation (6.2). Then f(x) = 0 for all x EM. or f(x) = eax,

xER,

(6.13)

where a € C is a constant. Proof. Suppose that / : R —> C, / ^ 0, is a measurable solution of (6.2). As before, write f(x) = g(x)exp[ih(x)],

x E R.

Since \f(x)\ = g(x), then g is a measurable function satisfying the equation (6.10). From Theorems 6.3 and 6.4 9(x) = e F ( l ) , where F: R —>• R is measurable additive function and therefore F(x) = cx, cER. Thus g(x) = ecx

for

xER,

where c is a real constant. Consider the function G: R —>• C defined by G(x) :=exp[ih(x)}, cx

Then since G{x) = e~ f(x),

xER.

G is measurable, and moreover,

\G(x)\ = l

for all xER

(6.14)

which implies that G is integrable on every finite interval. Taking into account the equation (6.11) and the definition of G, we get G(x + y) = G(x) G(y)

for all

x,yER.

Let us note that there exists a number s ER such that /

G(x)dx

= ^ 0 .

(6.15)

40

Functional Equations and Inequalities in Several Variables

To verify this, assume that for every a > 0 we have /

G(x)dx

= 0,

Jo

then by the well known Lebesque theorem, G(x) = 0 almost everywhere on R+, which contradicts (6.14). Therefore, integrating both sides of (6.15) with respect to y on the interval [0, s], we obtain G(x) = 7 f G{x + y) dy = \ [ G(u)du. (6.16) o Jo b Jx This means that G is a continuous function (as the integral of measurable function). Thus the inegral on the right hand side of (6.16) as a function of x is differentiate, which means that G is differentiate as well. Therefore taking derivative of both sides of (6.15) at y = 0, we get G'{x) = (7(0) G(x) for all x 6 E. This diffential equation has the solution G(x) = ae0x where /? = G'(0) and a is arbitrary complex constant. Substituing this expression into equation (6.15), we obtain

which gives a = 0 or a = 1. In view of (6.14) we get a = 1. Thus G(x) = e" x and because of(6.14) it follows that (3 = di, where d is a real number. Consequently, f{x) = g(x) exp [i h(x)} = g(x) G(x) = ecx edix = eax, where a = c + id e C , that is, / has the form (6.13).

•

By the very similar simple calculations one can get the following result (see also Aczel and Dhombres [1]).

Cauchy's exponential equation Theorem 6.6 If f:C^C for all x € C or

41

is a solution of (6.2), then f(x) = 0

f{x) = exp [/i(a;i) + f2{x2) + i M ^ i ) + £2(2:2))], where x = X\ + ix2, f\, f2: R —> R are solutions of the equation (6.1) and g\,g2: R —>• R are solutions of (6.8). Moreover, if f is measurable, then f(x) = 0 for

i e l

or

f(x) — exp [ax + bx],

where a, b are constant complex numbers (x means the conjugate number to x).

Notes 6.1 Every base of the space (Rn; Q; +; •), where Q denotes the set of all rational numbers (i.e. the base of linear space W1 over the field of rationals) is referred to as a Hamel basis (see Hamel [83]). 6.2 For many years the existence of discontinuous additive functions was an open problem. It was G. Hamel who first succeeded in proving that there exist discontinuous additive functions. The following result (Kuczma [123]) is a usefull tool which can be used to solve this problem. Proposition 6.1 Let H be an arbitrary Hamel basis of the linear space (Mn; Q; +; •). Then for every function g: H —>• R there exists exactly one additive function / : Rn —> R such that f/H — g (f restricted to H is equal to g). From this result we obtain the following Corollary 6.1 Let H be an arbitrary Hamel basis of the linear space (R n ; Q; +; •) and let g: H —>• R be an arbitrary function. Let f: R —> M. be the unique additive extension of g. The function f is continuous iff ^ ^ = const, for x G H.

42

Functional Equations and Inequalities in Several Variables

Proof. Assume that / is continuous, thus f(x) = ax for x 6 K, and in particular, for x € H, we have g{x) = f(x) = ax, whence ^- = a = const. Conversly, if g(x) = ax for all x £ H, then the function ax is an additive extension of g. From Proposition 6.1 such extension is only one, therefore f(x) = ax for x 6 R and consequently / is continuous. • 6.3 To construct a discontinuous additive function, it is enough to consider any function g: H —)• K such that ^ ^ =^ const, on H and its additive extension / , on account of Corollary 6.1, will be such a function.

Chapter 7 D'Alembert's equation on abelian groups The following functional equation is refered to as d'Alembert's equation

f(x + y) + f(x-y)

= 2f(x)f(y).

(7.1)

The theorem below provides the general solution ( the formula for every solution) of this important equation on abelian groups (see Baker [10], Aczel Dhombres [1], D.H.Hyers, G.Isac and Th.M.Rassias [30]). Theorem 7.1 Let G be an abelian group and let: G —> C satisfy the equation (7.1) for all x, y G G. Then there exists a function m: G —)• C satysfying the equation m(x + y) — m(x)m(y).

(7.2)

for all x, y G G, such that f(x) = ^[m(x)+m(-x)},

xeG.

(7.3)

Proof. It is clear that / = 0 is a solution of (7.1) and is of the form (7.3) with m = 0. Thus let' s assume that / ^ 0 is a solution of (7.1). Then setting y = 0 and next x = 0 in (7.1), we get 43

44

Functional Equations and Inequalities in Several Variables

/(0) = 1,

f(-x)

= f(x),

xeG.

(7.4)

Define g(x) :=f(x + s) -

f(x-s),

(7.5)

x £ G

(7.6)

where s G G is fixed, and consider m(x) = f(x) + ag(x),

(a is a complex constant). We shall determine s and a in order that the function m satisfies the equation (7.2). To this end, using (7.6), (7.5) and (7.1) by simple calculation, we obtain successively m(x)m{y) = f(x)f(y)

+ a[g(x)f{y) + f{x)g{y)] + a2g(x)g{y)

= f(x)f(y) + a[f(x + 8)f(y) - f(x - s)f(y) + f(x)f(y + s) - f(x)f{y - s)] + a2 [f(x + s)f(y + s) - f(x - s)f(y + s) - f(x + s)f(y -s) + f(x - s)f(y - s)]

= 2 [f(x + y) + f(x - y)l + \ [ f(x + y +s) + f(x - y + s ) - f(x + y - s) - f(x - y - s) + f(x + y + s) + f(x - y - s) - f(x + y - s) - f{x - y + s) ] a2 + — [f{x + y + 2s) + f{x-y)-f{x + y)-f{x-y-2s) - f{x + y)-f{x-y 1 2

+ 2s) + f(x + y-

2s) + f{x - y) }

+ a [/(2s)-l] f(x + y)

+

~ a2 [/(2s) - 1] f(x-y)

+ ag(x + y).

In order for m to satisfy the equation (7.2), the last expression should be equal to m(x + y) = f{x + y) + ag(x + y),

D'Alembert's equation on abelian groups

45

and , as we can see this is satisfied only if a2[f{2s)-l]

= \.

Hence, if /(2s) — 1 ^ 0, we take a = ±[2/(2s)-2]~i Applying properties(7.4), we also get g(—x) = —g{x) for x € G, whence by definition (7.6), m(-x)

= f(x)

-ag(x).

The last equation together with (7.6) implies the formula f(x) = -\m{x) +m(-x)],

xeG.

Now we shall show that in the case when /(2s) — 1 = 0 for all s (E G, / has the form (7.3). Indeed, assume that f(2x) = 1

for all x

eG.

Then from (7.1) and (7.4) we get for y = x, 2f2(x) = f(2x) + 1 = 2

for

x G G,

i.e. /2(x) = l

for all

xeG.

This is equivalet to the following statement: for any

x 6 G,

either

f(x) = 1 or

f(x) = —1.

(7.7)

Now, assume that f(x)f(y) = 1 for x, y G G, then by (7.1), f(x + y) + f(x-y) = 2 and from (7.7) f(x + y) = f(x - y) = 1, so that f(x + y) = f(x)f(y),

x,yeG.

(7.8)

46

Functional Equations and Inequalities in Several Variables

On the other hand if f(x)f(y) = — 1, then similarly by (7.1) and (7.7), f(x + y)+f(x-y) = - 2 , thus f(x + y) = f{x-y) = -1 and the equation (7.8) is satisfied. Moreover, by (7.4) we have

/CO = 5 [/(*) +/(-*)], so taking m(x) := /(:r),we have proved that / is of the form (7.3) with m satysfying (7.8). This competes the proof. • Corollary 7.1 Let the assumptions of Theorem 7.1 be satisfied with G = R. //, moreover, f is measurable, then f(x) = 0 for x e R, or f(x) = cosh(car) xER (7.9) where c is a complex constant. Proof. If / is not a trivial solution of (7.1), then / has a form (7.3). Therefore m is also a measurable function (see formula (7.6)) and from Theorem 6.5, m(x) = ecx

for

x ER,

which gives the formula (7.9). Corollary 7.2 Let f:R^Rbea equation (7.1). Then f(x) = 0 or

f(x) = cosh (ax)

• measurable solution of the or

f(x) — cos(bx)

(7.10)

for all x G R where a, b are real constants. Proof. Write c = a + ib, and from Corrolary 7.2 we have for all x eR f{x) = cosh(a + i b)x = cosh(aar) cosh(z bx) + sinh(ax) sinh(z bx) = cosh(ax) cos(bx) — i sinh(ax) sin(6:r). Since / has to be a real valued function, then the last equality implies a = 0 or 6 = 0, which gives the formulae (7.10).

•

D'Alembert's equation on abelian groups

47

Note Representations and approximations of functions in several variables in terms of finite sums of products of functions have been extensively treated in the book by Th. M. Rassias and J. Simsa [122], as well as in the papers of A.Prastaro and Th. M. Rassias [183] , [159] , [161] , [158] . Functions of certain special forms had first been studied by J. D. Alembert in 1747.

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Chapter 8 D'Alembert's equation on topological groups At first we consider the fuctional equation n

f(x-y)

= ^2as(x)as{y),

(8.1)

s=l

where f,as: G —> C, s = 1,2, . . . , n are defined on a locally compact abelian group G. In [151] has been proved that for any system of continuous functions f,a,i,...,an satisfying equation (8.1) there exist continuous characters 4>u ••> 4>n of the group G and positive number a\,... ,an such that / has the form n

f(x) = ^2as(f)s(x),

xeG.

The author did not find the form of functions a 1 ; ..,a„. This problem (among other problems) has been solved by Z. Gajda in [62]. Now we are going to present the main ideas and results of Gajda's paper [62]. The equation (8.1) may be rewritten as f(x-y)

= (a(x)\a(y)), 49

x,yeG

(8.2)

50

Functional Equations and Inequalities in Several Variables

with a: G —> C" given by a(x) := (ai(x),..,an(x)),

xeG,

where (-|-) stands for the usual inner product in C™. Instead of C \ we shall consider an arbitrary Hilbert space (H, (-|-))- In such general setting, solutions of (8.2) can be described in terms of unitary representations of the group G. In the sequel, we will need the following notions. By a unitary representation of a topological group G in a Hilbert space H we mean a homomorphism of G into the group of all unitary operators in H (see, e. g. [140]). A unitary representation U is said to be continuous iff for every vector ( e f f , the transformation G3x^

U(x)£ e H

is continuous. The symbol Lin A stands for the linear space spanned by a set A C H whereas cl denotes the closure operation in the norm topology of H. A vector £0 € H is called a cyclic vector of a representation [/iff clLin{C/(x)£ 0 : x e G} = H.

Solution of equations (8.1) and (8.2) Note that for the group G not necesserily abelian we apply the multiplicative notation. By e we denote the identity element of G. So the equation (8.2) can be written as f{y-lx)

= {a{x)\a{y)),x,ytG

which proves (8.2) for the case of an Abelian group G. The first result is the following (see [62]).

(8.3)

D'Alembert's equation on topological groups

51

Theorem 8.1 Let (G,-) be a topological group and let (H,(-\-)) be a Hilbert space. Assume that f: G —> C and a: G —> H are continous functions. Then f and a fulfil the equation (8.3) iff there exists a continous unitary representation U of the group G in the space Ho := cl Lina(G) with the cyclic vector £o := a(e) such that (8.4)

a(x) = U(x)£o, x G G, f(x) = (I/(aOG)|£o),

(8.5)

xeG.

Proof. Assume that there exist an unitary representation U of G and a vector £0 G H such that the formulae (8.4) and (8.5) hold true. Then we obtain for x,y G G, f{y~lx)

=([/(y-1x)6|6) = ( 1 7 ( ^ - ^ ( ^ 0 I 6 ) = (tf(aO£o I U(y)£0) = (a(x)\a{y))

i.e. / and a satisty the equation (8.3). Conversely, suppose that / and a satisty the equation (8.3). Then, for x,y,z G G, we get = f((xz)~lxy)

[a(xy) | a{xz))

= f{z~lx~lxy)

=

f{z~ly)

= (a(y) I «(*))> i.e.

(a(xy)\a(xz))

= (a(y)\a(z)))

x,y,ze

Take now n G N, complex numbers a\,...,an G. By (8.6) we obtain Y2oiia(xys)

=

G.

and x,yi,...,yn

'^2'^2a8ar(a(xyl)\a(xyr))

s=l

n

n

= ^2J2asar(a(ys)\a(yr)) s=l

r=l

=

^2asa(ys) s=l

(8.6) G

52

Functional Equations and Inequalities in Several Variables

which means that Y^ocsa(xys)

$^a«a(y*)

5= 1

s=l

(8.7)

Define D := Lina(G). Take x € G and let £ = 1^" =1 asa(?/a) for some a s G C, ys € G, s = 1 , . . . , n, then we define an operator V{x) in £> by n

V(x)t

:

= 5^a a a(a;y,). 5=1

Observe that V{x) is correctly defined. In fact, assume that £ has two expansions £ = J ^ a s a ( y s ) = J ^ A-a(*r)i s=l

r=l

than (8.7) implies n

m

^2asa(xys) s=l

-^Pra(xzr)

^ a s a ( y s ) - J ^ / 3 r a ( z r ) = 0.

r=l

r=l

4=1

Clearly, for every x G G, V(x) is a linear operator in D and in view of (8.7) it is an isometry of D into itself (cf. Th. M. Rassias and C. S. Sharma [71]). Also for every x G G, V(x) is surjective. Taking x,y G G and ( e D such that n

£=

^2asa(ys), s=l

we obtain V(xy)£ = ^2asa{xyys)

=

V(x)V(y)£,

s=l

whence, V{xy) =

V{x)V{y)ioix,y€G.

D'Alembert's equation on topological groups

53

Moreover, for every n

€ = ^2asa(xys) € D, s=l

the operator G 3 x -^ V(x)£ = Yl™=i asa{xVs) £ D is continuous in view of continuity of the function a. Take x e G, and let [/(a;) be the unique continuous extension of V(x) to the space H0 = clD. Then we have obtained the transformation G 3 x ->• £/(:r), which is a continuous unitary representation of the group G in the space HQ. Now we shall verify that the representation U has all the required properties. If Co := a(e), then £0 e D C H0 and U(x)£0 = V(x)£0 = a(xe) = a(x) for x e G, which yields (8.4). Put y=e into the equation (8.3), then we get f(x) = (a(x)\a(e)) = (U(x)^o)

for

x€G,

i.e. (8.5). Moreover, from (8.4) we obtain, clLin{U(x)£0:

x £ G} = clLina(G)

= H0,

which means that £o is a cyclic vector of the representation U and this completes the proof. • In the following we will state the result concerning the equation (8.1) in the case of an abelian group G. By M(m x n; C) we denote the set of all complex matrices with m rows and n columns. As usual the delta of Kronecker is defined by 0 for i ^ k, 5\k 1 for i = k.

54

Functional Equations and Inequalities in Several Variables

By a character of a topological abelian group G we mean a function ip: G —» C such that
eG,

and \cp(x)\ = 1

for

xeG.

Theorem 8.2 Let (G, +) be an abelian topological group. Then continuous functions f,as: G —> C, s = l , . . . , n satisfy the equation (8.1) iff there exist continuous characters tpi,..., ipm, m < n of the group G, complex numbers a±,..., am and a matrix [f3s,r] ^ M{m x n; C) such that n

^2 Psr]h~r = 5sk s,k = l,...,m,

(8.8)

r=l

f(x) = J2 \<x, I2 V>s{x) ^ G ,

(8.9)

s=l m

a r (x) = 2Jasf3sr
.

(8.10)

Proof. Assume that / and ar, r = 1 , . . . , n are given by (8.9) and (8.10) where
D'Alembert's equation on topological groups

55

Now we have by (8.1), (8.10), (8.8) and (8.9), and properties of characters, n

Y^as{x)as{y) s=l

n

m

= X ! £ ockPks
k,j=l n

= £(£«!) k,j=l m

ak(Xj
s=l m

= £tf<XkajMx)
j - \

This means that / and ar, r = 1 , . . . , n satisfy the equation (8.1). Conversely, suppose that / and ar, r = 1 , . . . , n, are continuous functions satisfying the equation (8.1). Consider the orthonormal basis ( e i , . . . , en) of the n-dimensional space C" and put n

a{x) := 22as(x)es

>x £ G •

(8-11)

s=\

Then the equation (8.1) becomes f(x -y)

= (a(x)\a(y)),

x,yeG.

On account of Theorem 8.1, there exists a continuous unitary representation U of the group G in the space H0 := cl Lin a(e) C Cn such that a(x) = U(x)£o, f(x) = (U(x)Zo\to), for x E G, where £0 : = a(0). The space H0 admits a decomposition into the direct sum of orthogonal subspaces H\,...,Hm invariant under U (see [90], Chap.I). To each subspace Hs, s = l , . . . , m there corresponds a

56

Functional Equations and Inequalities in Several Variables

continuous unitary representation Us of the group G in the space Hs such that U(x)Z = U^xfa + ••• + Um{x)U

(8.12)

for x € G and £ = £1 H h ( m , ^ e f f s , s = l , . . . , m . Moreover, one can find continuous characters ipi,...,(pm of the group G such that U8(x)£s = (pa(x)£a foTxeG,£aeHa,s

= l,...,m,

(8.13)

(see [90], Chap.I). Now, for each s — l , . . . , m , choose a vector us € Hs with ||u s || = 1. Thus ui,...,um form an orthonormal basis of H0. Assume that m

£0 = a(0) = ] P a, V U S

S

s=l

for some a s € C, s = 1 , . . . , m. In view of (8.12) and (8.13) we have m

U{x)£o - ^2
y

£2(Ps{x)asus\1*>2as

/(x) = ( t / ( x ) 6 | 6 ) = (

\s=l m m

s=l

= 5^5^¥'s(aj)a«afc(«a|wAi) = J ^ \a,\2(pa(x), s=l

fc=l

s-l

i.e. (8.9). Select the matrix [/?S)fc] e M (ra x n, C) such that us = ^2 A»*efc' s = 1 , . . . , m. fc=i

Since ( « i , . . . , itm) and ( e i , . . . , en) are orthonormal systems of vectors in Cn, then the condition (8.8) holds true.

D'Alembert's equation on topological groups

57

Finally, we obtain for x E G, m

m

a(x) =U(x)£Q = ^2ips(x)asus

= ^^s(x)as ^ft^e*

s=\ n

n

s=l

fc=l

m a

=5 Z E fc=l

s^s,kfs(x))ek.

5=1

Therefore, taking into account (8.11), we get m

ar(x) = '^/a>sl3s,r(ps(x),

r = l,...,n,

s=l

which implies the formula (8.10). This ends the proof.

•

For both old and news properties of orthogonality we refer to Th. M. Rassias [166].

Solution of D'Alembert's equation For D'Alembert equation on arbitrary group (not necessarily abelian) f(xy) + f{xy-1)

= 2f(x)f(y),

x,yeG,

where / : G —>• C, two results are very important. The first one, due to Kannappan [156] states that for any solution / satisfying the condition f(xyz) = f(xzy),

x,y,ze

G,

(8.14)

there exists a homomorphism m: G —> C such that fix) = -[m{x) + mix-1)],

xeG.

The second one, due to Dacic [41] is analogous, but the condition (8.14) is replaced by the weaken one f(xyz0)

= f(xz0y)

for x,y E G,

58

Functional Equations and Inequalities in Several Variables

where z0 e G is fixed. We shall consider D'Alembert's equation in the following general form f{xy) + f(xy-1)

=2(a{x)\a(y)),x,y£G.

(8.15)

Lemma 8.1 Let f: G —>• C, a: G —>• H satisfy the equation (8.15). Then for all x,y G G, (i) f(x) = (a(x)\a(e)), (ii) f(x) = f (x- 1 ), (Hi) f(x)

=f(x),

(iv) f(xy) =

f(yx).

Proof. If we set y=e to (8.15), we get the formula (i). Now set x = e into (8.15), then f(y) + f (iT 1 ) = 2(a(e)|a(y)) = 2(a(j/)|a(e)) = 2/ffi, whence, f(y) + f{y-1) Inserting y

_1

= Wy)

for

yeG.

(8.16)

istead of y to (8.16), we get

f(y) +

f{y-1)=mrI),

which implies that f(y) = / (y _1 ) and consequently also (ii). To get (iii) we can apply (ii) to (8.16). Moreover by (8.15), (ii) and(iii), we obtain f(yx) + fiyx-1) = f{xy) + fixy-1)

= 2 (a(y) | a(x)) = 2(a(x) \ a(y)) = = f(xy) + f{xy~l)

= f{xy) +

f{yx~l),

i.e. f(yx) + f{yx-1)

= f{xy) +

f(yx~l),

which implies the condition (iv). We may now formulate the following (see Gajda [62]).

•

D'Alembert's equation on topological groups

59

Theorem 8.3 Let (<7, •) be a topological group and let f: G -» C, a: G —>• H be continuous solutions of the equation (8.15) such that there exists a z0 £ G with a(z0) = 0

(8.17)

a(xyzo) — a(xzoy) for all x,y € G.

(8.18)

and Then there exists a continuous unitary representation U of G in the space Ho := clLina(G) with the cyclic vector £o := «(e) such that for x,y £ G, (U(xy)Z0 | £o) = (£%x)£o 16),

(8.19)

/(*) = ^(£/(*)£0 + tfCz-% I 6),

(8.20)

a(x) = i [t/(x)Co + tf ( * - % ] .

(8.21)

Conversely, iff:G-±C and a: G -^ H are given by (8.20) and (8.21) with some unitary representation U and a vector £o fulfilling (8.19), then f and a safisty the equation (8.15). Proof. Suppose that / : G —> C and a: G —> H are continuous solutions of the equation (8.15) and such that (8.17) and (8.18) hold true. Then, by (i) and (8.18), we get for x,y G G f(xyzo) = (a(xyz0) | o(e)) = (a(xz0y) | a(e)) =

f(xz0y),

whence f(xyz0)

= f(xz0y)

for

x,yeG.

/(arj/so) = / 0 ^ o y ) =

f{yxz0),

(8.22)

Therefore by (iv) we guess that

i.e. f{xyz0) = f(yxz0)

for

x,y€G.

(8.23)

60

Functional Equations and Inequalities in Several Variables

By (8.17) from the equation (8.15) it follows that f(xz%) + f(x) = 2(a(xz0) | a(z0)) =

0,xeG,

whence, f(xzl) = -f(x),xeG.

(8.24)

Moreover, by (8.24) and (8.22) we infer that f{xz^y~x)

= -f(xzQly-lzl)

= - f (xz^1 zQy~l zQ) =

-/(xy^zo).

Hence we obtain f(xzt1y-1)

= -f{xy-1z0)

for x, y e G.

(8.25)

Consequenty, by (8.15), (8.22) and (8.25), we deduce 2(a(xz0) | a(yz0)) = f(xz0yz0) = f(xz0(yz0))

+ fixy'1)

+

/{xzoz^y'1)

= f{xyz%) +

f{xy~l)

= -f{xy) + f{xy~l), which together with (8.15) imply fixy-1)

= (a(x) | a(y)) + {a{xz0) \ a(yz0)),x,y&G.

(8.26)

In the sequel, applying two times the equation (8.15) and (8.22) and (8.25) we arrive to 2(a(xz0) | a{y)) = f(xzQy) + /(zzolT 1 ) = f{xyz0) +

f(xy-xz0),

and 2{a(x) | a(yz0)) = f{xyz0) + fixz^y'1)

= f{xyz0) - /(xy _ 1 z 0 )-

These last two equalities yield f{xy~lZQ) = (a(xz0) | a(y)) - (a(x) a(yzo)),x, y € G.

(8.27)

D'Alembert's equation on topological groups

61

New, we define: g: G —> C and b: G -> H by g(x) := f(x) + if{xz0), b(x) := a(x) + ia(xzo). By (iv) and (8.23) we obtain

g(xy) = g(yx) for j . y e G .

(8.28)

Now we shall verify that g and b safisty the equation g(y~lx) = (b(x)

\b(y)),x,yeG.

Indeed, by (8.26), (8.27), (8.28) we infer that (b(x) | b(y)) = (a(x) + ia(xz0) \ a(y) + ia(yz0)) =(a(x) | a(y)) + (a(xz0) \ a(yz0) + + i [(axz0) | a(y)) - (a(x) \ a(yz0))] =f(xy~1) + ifixy^zo) = gixy'1) = g{y~lx). Therefore, from Theorem 8.1, there exists a continuous unitary representation U of G in the space H0 := clLinb(G) with the cyclic vector £o := b(e) = a(e) + ia(z0) = a(e) such that for x <E G b(x) = U{x)£0,

(8.29)

g(x) = (U(x)£0 | 6). We have now by the last formula and (8.28), (U(xy)£0 | 6 ) = (g(xy) I 6 ) = (g(xy) I 6 ) = {U(yx)£0 I &), i.e. (8.19). Further, using (8.15), we obtain 2|| o(x) - a ^ " 1 ) ||2 =2(a(z) | a(x)) - 2(a(x) | a ^ " 1 ) ) - 2(a(oT1) \ a(x)) + + 2(a(aT1) | a ^ - 1 ) ) =f(x2) + /(e) - /(e) - f{x2) - /(e) - /(x~ 2 )+

+ / ( ^ 2 ) + /(e) = 0,

62

Functional Equations and Inequalities in Several Variables

whence, a(x) =a{x-l),xeG.

(8.30)

Similarly by (8.15), (8.23) and (8.24), we get 2||a(a;,Zo) + a(x_12;o|| = 2(a(xz0) \ a(xz0)) + 2(a(xz0) | a(x _1 ^o)) + 2(a(x~1z0) | a(xz0)) + 2(a(x~1z0 \ a(x~lz0)) =f{xz0{xz0)) + /(e) + f{xzQ{x-lz))) + f(x2) + f(x-lz0(xz0)) + f(x-2) + f{x-lz,{x-lz,)) + /(e) 2 2 2 2 =f(x z 0) + /(e) + f{zl) + /(:r ) + f(z 0) + f(X-2)+ + f(x-2z20) + f(e) = [f(x2z20) + f(x2)} + [f{x-24)

+ fix'2)]

+ 2 [fiz20) + fie)} = 0.

This is equivalent to aix~lZo) = — aixzo)

for x G G.

Hence in view of (8.30) and the definition of b, we derive for x G G, - [b(x) + 6(x -1 )]

= - [a(x) + aix~1)} + -i [a(xzo) + afc" 1 ^)] — a(x).

Consequantly, we get by (8.29) a(x) = \ [b(x) + bix-1)] = \ [U(x)t0 + Uix-l)Zo] , which yields (8.21).Moreover we get apparently from the equality a(x) = ±[b(x) + b(x-1)] the condition Lin b{G) = Lin a{G) and hence H0 = clLin a(G). To end the proof of the first part observe that from (i) and (8.21) it follows that fix) = iaix) | o(e)) = \iUix)^0

+ tf(*-% I G>)

D'Alembert's equation on topological groups

63

for x € G, which means that (8.20) also holds true. By a simple direct calculation one can verify that the converse statement is also true. The details are therefore omitted here. This completes the proof. • Using very similar arguments as in the proof of Theorem 8.2, one can derive from Theorem 8.3 the following. Theorem 8.4 Let (G, +) be an abelian topological group and let f,as:G-^C, s = 1 , . . . , n, be continuous solutions of the equation n

fix + y) + f{x - y) = 2 ^ a s ( a ; K ( y j ,

x,yeG,

(8.31)

8= 1

such that there exists z0 € G with o«(^o) = 0 for

s — 1 , . . . , n.

Then there exist continuous characters (pi,...,r] G M(m x n; C) such that n

J2 &>rK~r = Ssk,s,k = l,...,m,

(8.32)

r=l m

/Or) = Yl I a° I' Re ^ 0 * 0 ' * e G,

(8.33)

m

a

r{x) — y~]as(3S!rRe (ps(x),x G G, r = 1 , . . . ,n.

(8.34)

s=l

Conversely, all functions f and as,s — 1 , . . . , n given by (8.33) and (8.34) with fisr fulfilling (8.32) are solutions of the equation (8.31).

64

Functional Equations and Inequalities in Several Variables

Notes 8.1 Other facts on a functional equations of D'Alembert's type may be found in O'Connor [150] and Szekelyhidi [207]. 8.2 If/: E—>• C is a continuous solution of the equation f{x + y) + f{x-y)

= 2f{x)f(y),

x, y € R

(8.35)

then, by Corollary 7.2, f(x) = cosh (ex),

a;6l

for some complex constant c. Taking a such that a 2 = c, we can write 00

71

2n

=

/(*) E T ^ P *eR-

( 8 - 36 )

The following result may be found in Aczel, Dhombres [1]. Proposition 8.1 Let A be a Banach algebra with an identity. A continuous function f: R —>• A satisfies (8.35) and /(0) = 1 iff there exist an a E A such that (8.36) holds true. 8.3 For / : (0,00) -» A, Baker [9] has proved the following Proposition 8.2 Let A be a Banach algebra and let / : (0,00) —>• .4 satisfy f(x + y) + f{x -y)

= 2f(x)f(y),

for x > y > 0.

Suppose that lim fix) := k. TTien k2 = k and there exist a,b € A such that ka = ak = a, and ,, .

(,

/or a// x > 0.

ax1

a2x4

bk = b, \

, / ,

kb = 0, x3a

xho?

\

Chapter 9 Polynomial functions and their extensions Polynomial operators were investigated by S. Mazur and W. Orlicz in [132] and afterwards by several other mathematicians (cf. McKiernan [133], Frechet [55] Szekelyhidi [205], G. V. Milovanovic, D. S. Mitrinovic and Th. M. Rassias [136]). The main objective of this section is to prove the basic theorem about the representation of polynomial functions with the aid of multiadditive functions. To deal with, let us recall some new tools, which will be used further on.

The diference operator Let X and Y be two linear spaces over the field Q of rational numbers, and let / : X —> Y be an arbitrary function. The difference operator Ah with the span h G X is defined by the equality Ahf(x)

:= f(x + h)65

f(x),x

G X.

(9.1)

66

Functional Equations and Inequalities in Several Variables

The iterates Ash of A^, s = 0 , 1 , 2 , . . . , are defined by the natural recurrence A°hf = f,A'h+1

= Ah(A'hf),

8 = 0,1,2,...

. (9.2)

The superposition of several difference operators will be written shortly Ahl...hJ = Ahl...Ahsf,seN. (9.3) If hi = • • • = hs — h, then we write A^=AsJ,seN.

(9.4)

s

Lemma 9.1 For arbitrary functions fi,f2: X —> Y and for arbitrary constants a, f3 G Q, we have Ah(ah

+ 0f2) = aAhfx + PAhf2.

Proof. This is an immediate consequence of the definition (9.1).

Lemma 9.2 For arbitrary hi,h2 commute: AhxAh2f

€ X the operators

Ahl,Ah2

= Ah2Ahlf.

Proof. By (9.1) we get AhlAh2f(x)

= Ahl [f(x + h2) - f(x)] =

= f(x + hx + h2) ~ f(x + hx) - f(x + h2) + f{x) =

Ah2AhJ{x).

• Corollary 9.1 Operator (9.3) is symmetric under the permutation ofhi,...,hn.

Polynomial functions and their extensions

67

Lemma 9.3 For arbitrary hi,h2 G X, Afn+hif - A h l / - Ah2f

= Ahlh2f.

Proof. Directly by (9.1) we have &hl+hJ(x) - Ahlf(x) AhJ(x) =f(x + hx + hi) - f{x) - f{x + h) + fix) - fix + h2) + fix) =fix + hx + h2) - fix + h) - fix + h2) + fix) = Ahlh2fix).

Theorem 9.1 Let s e N, then l

J2 (-ir (£l+ - +ea) /(^+£i^+---+^A)- (9.5)

Afcl..Aj/(x)=

ei,...,e,=0

Proof. The proof will be given by induction on s. For s = 1 formula (9.5) becomes l

Afcx/Or) = ^ ( - l ) 1 " 5 1 / ^ + d M = -fix)

+ / ( * + h),

e1=0

which coincides with (9.1). Assume that (9.5)is true for an s e N. Then by Corollary 9.1,the induction hypothesis and Lemma 9.1,

68

Functional Equations and Inequalities in Several Variables

we obtain &h1...h.h,+1f(x) =

Ahs+1[Ahl...hJ(x)]

1

= Afc<+1

Yl

(-ir ( £ i + '" + £ s ) /(^+^i+-+£A)

.ei,...,e s =0 1

=

(-1)"iei+'"+e')^ht+1f{x

E

+ elh1 + --- + eah8)

ei,...,e s =0 1

E (-D

s—(eiH

he s X

£i,...,e s =0 X

( - l ) 1 - ^ ^ ! / ^ + £lhi + •••+ eshs +

^

es+lhs+l)

es+i=0 1

= £l

]£

(-l)s+1-{El+-+e3+l)f(x

+ eih + --- +

eshs+£s+lhs+1).

£s + l = 0

Thus we have got the formula (9.5) for s + 1. Induction completes the proof. •

Corollary 9.2 Let s e N, then

n=0

W

Proof. Assume that exactly n among £ l 5 . . . ,es in (9.5) are equal to one, then the expression on the right hand side of (9.5) reduces to (—l)s~nf(x + nh). But there are (*) choices of n ones from Si,... ,ss, where n may run from 0 to s. Therefore (9.6) follows from (9.4) and (9.5). •

Polynomial functions and their extensions

69

Theorem 9.2 (see [123]). Let f: X —>• Y be a function and let hi,...,hsGX be arbitrary. For£i,...,es € {0,1} define

S

E

S

h

£rtlr

_ ^—\ r=l

T—1

Then for every x 6 X,

I

&h,..U(x) =

( - l ) £ l + - + ' " A ^ ...J(x + bEl...J. (9.7)

E £l,-,Ej=0

Proof. Following M. Kuczma [123], p. 369, we shall prove the formula

(-lYAyu2y2...syJ(x)

=

(-iy-^+-+^A^J(x+BSv..J,

J] ei,...,es=0

(9.8) where s

•"•ei—Es

=

~ / j ^nVnt n=l

s

&eiEs

=

/ y n=l

n

^-nVni

for all x, y i , . . . , ys G X. Having done this, we get the formula (9.7) just on multiplying both sides of (9.8) by (—l)s and substituting hn = nyn for n - 1,.. . ,s.

70

Functional Equations and Inequalities in Several Variables Now by Corollary 9.2 and Theorem 9.1, we obtain l

E

(-l)^ ( £ l + - + £ i ) A^ i ... £ s /(x + JBei...es)

ei,...,e s =0

= E

(-ir(ei+--+£s) E(-i) s - n (*)/(*+ B «~*.+^i--.)

ei,...,e s =0

n=0

^

'

= E (-i) s " (£l+ - +£s) E(- 1 ) s " n Q/( a; +E( r - ri )^) ei,...,£ s =0

n=0

^

'

r=l

= D - ^ f f ) E (-i)Mei+-+")/(^+E(r-n)^) n=0

^

'

n=0

^

'

ei,...,e,=0

r=l

Observe that for every n = 1 , . . . , s, one of the increments {l-n)yi,...,(s-n)ys is equal to zero, and therefore in view of Corollary 9.1 \l-n)yu..(s-n)yj(x)

= ^•(l-n)yi...-yn-1yn+1...(s-n)y,[^of(x)]

= 0,

since A( n _ n ) yn /(ar) = A 0 /(:c) = f{x) - f(x) = 0. Consequently in the last sum only the term for A; = 0 remains, which implies the formula (9.8). This completes the proof. • For certain fundamental properties of the difference operator the reader is referred to the book of D. H. Hyers, G. Isac and Th. M. Rastias [97] where one can find an extensive study of functions with bounded differences, as well as the paper of Th. M. Rassias [177] in which the concept of difference operator is used for a generalization to the case of higher order derivatives of the Ulam - Hyers theorem on the stability of differential expressions.

Polynomial functions and their extensions

71

Polynomial functions A function f-.X—tY

fulfilling the condition Ash+1f(x) = 0

for every x G X and every h e X is called a polynomial function of order s (s G N). Lemma 9.4 Let fi,f2: X ^ Y be polynomial functions of order s and let a, /? G Q be arbitrary constants. Then the function f = a / i + /3 / 2 is also a polynomial function of order s. Proof. Since the composition of linear operators is also a linear operator, from Lemma 9.1 we get AJi +1 (a/i(x) + 0f2(x)) = aAsh+1h(x)

+ PAsh+1f2(x) = 0

for all x G X and h £ X, which was to be shown.

•

Theorem 9.3 / / /': X —> Y is a polynomial function of order s, then Afcl...hj+1/(a;) = 0 for every x, hi,...,

hs+i G X.

Proof. In fact, the difference &h1-h.+1f(x) in view of Theorem 9.2, can be expressed by the formula (9.7) containing only diferences of order s + 1, which on account of our assumption, all are zero. This yields our statement. • We shall recall the definition of multiadditive functions and prove two auxiliary lemmas concerning such type of functions.

72

Functional Equations and Inequalities in Several Variables

Suppose that s G N. A function / : Xs -» Y is called sadditive if for every r, 1 < r < s, and for every Xi,..., xs, yr G X, J \X\, =

. . . , Xr—i,

£ r ~r ^ri 2 - r + l ) • • • j ^ s j

. / 1,^-1) • • • > %s) T / v ^ l ! - • • i ^ r - l j j / n -^r+lj • • • i-^sjt

i.e. / is additive with respect to each of its variables xr G X, r — 1 , . . . , s. A 2-additive function is called shortly biadditive. Given a function F: Xs —>• Y, by the diagonalization of F we understand the function / : Xs —> Y given by the formula f(x):=F(x,...,x),

xeX.

Then we have Lemma 9.5 Let F: Xn -^ Y be a symmetric n-additive funtion, and let f: Xn —>• Y be the diagonalization of F. Then, for every x,h G X,

AJ(x) = J2 Q f ^ J ^ . r

r

(9.9)

n—r

Proof. By the definition of A^, (9.9) is equivalent to

F(x + h,..., x + h) = ^2 ( U ) F(x> ••-,x,h,. r=0 ^

.^h).

(9.10)

' r

n—r

The proof of (9.10) will follow by induction on n. If n = 1, the relation (9.10) becomes F{x + h) = F(h) + F(x), which is true since F is additive. Further on, suppose that (9.10) is true for an n G N and let F: Xn+1 —> Y be a symmetric (n+1)additive function. Then for u G X from the induction hyphothesis, we obtain Fix + h,..., x + h, u) — y^ ( r=0

I Fix,...,

x, h,.... h, u).

Polynomial functions and their extensions

73

From this equality for u = x + h by the (n+l)-additivity and symmetry of F, we get F(x + h,...,

x + h) = 2_\ I

) F(x, ...,x,h,...,h,x

w

n+l

r

n—r

=E(")^(j1^3^---,/»^)+E('!)i;,(^ r=0 ^ ' =

]C (

^

r=0 ^

+ h)

1 X * ft, > • • ,

'

It)

n+l—r

K ( z , ••-,£,/*• ••,/*) + V ( ) F ( a ; , . . . , a ; , / i , . . . > ) . r+1

n+1—r

n—r

Finally, by replacing in the first sum r + 1 by k and in the second one r by k and by using the identity

n \ k-l)

n+ l k

in +

[k

we derive F(x + h,...,x

+ h)

n+i

fc=1

n+l

fc

n+l-fc

+ Y l u ) F ( x > • • • > *> ^ • • •' h )+ F ( h > • • •' h ) "_i

fc

n+l-fc

n+l

= F ( x , . . . , x) + *_1

*

n+l

n+l =

^2\ fc=0

kt.

)F(x,...,x:h,...,h) n+l-fc

n+l-fc

74

Functional Equations and Inequalities in Several Variables

Comparing the first and the last terms we see that we have got (9.3) for n + 1. Induction principle completes the proof. • Lemma 9.6 Let F: Xn —>• Y be a symmetric n-additive function and let f: Xn —>• Y be the diagonalization of F. Then, for every m > n and for every x, hi,..., hm € X, we have A , , . , m / W = ( " ! F ( A — *"> 10

f = "' if m > n.

(9.11)

Proof. We shall apply the induction method on n. For n = 1, since F = / is an additive function, we have Ahf(x)

= f{x + h) - f{x) = fix) + fih) - fix) = fih) = F(h).

For m > 1, Ahl...hmfix) = Ahl...hm_1(Afcm/(ar)) = Ahl__,hm_Jihm) = 0, because fihm), as a function of x, is constant. Suppose now that (9.11) holds true for 1 , . . . , n € N and take an (n+l)-additive symmetric function F: Xn+1 —>• Y. If / : X -> Y is its diagonalization, then by Lemma 9.5 and Lemma 9.1 we have for x,hi,...,hn+i e X, Ahl...fcn+1/(a;) = A Al ... An (A hB+1 /(x)) = y2 (

7=1 \

r

} ^...hnFix,...,

x, hn+i,...,

^7^^

J

r

The function Fjx,...

,x,hn+i,... r

^

hn+i).

n+l—r

,hn+i)

is r-additive for every

n+l—r

1 < r < n. For r = 0, the expression F(hh+i, •.., hn+i) as a function of x is constant. Therefore on account of the induction hypothesis we have A, u Fir rh h \_\nW{K---,K+i) ^•hx...hnt [x,... ,x, nn+i,..., nn+i) — < N / —v ' ^^ v 0 r

n+l-r

K

if r = n, if r < n.

Polynomial functions and their extensions

75

Thus Ahl...hn+J(x)

= (n+l)n\F(hu

..., hn+1) = (n+l)\F(hu

..., hn+1).

On the other hand, if m > n + 1, then taking into account that F(h\,..., hn+i) as a function of x is constant, we get because of Corollary 9.1that khx...hj{x) = AA„+a...hm[Afcl...fcn+1/(a;)] =&hn+2...hm[(k + l)\F(hu . . . , hn+1)] = 0. This means exactly that (9.11) holds true for n + 1 and therefore induction concludes the proof. • Corollary 9.3 Let F: Xn —>• Y be a symmetric n-additive function and let f: X —>• Y be the diagonalization of F. If f = 0, then F = 0. Proof. Take x, hi,...,

hn G X. Then clearly, AAl>...)hB/(x) = 0,

whence, by Lemma 9.6, F(hu...,hn)

= ^Ahl

hnf(x)=0.

lb.

Since hi,...,

hn G X are arbitrary elements, this means that F = 0.

• For convinience, any constant function will be called a 0additive function. Now we can prove the following (see e.g. Kuczma [123]). Theorem 9.4 Let An: Xn —> Y, n = 0 , 1 , . . . , s be symmetric nadditive functions and let An: X —> Y, n = 0 , 1 , . . . , s, be their diagonalizations, respectively. Then the function f: X —>• Y given by s

f(x)=J2An(x), 71=0

is a polynomial function of order s.

xeX,

(9.12)

76

Functional Equations and Inequalities in Several Variables

Proof. Take x,h € X, then by Lemma 9.6 Ash+1An{x)=0

for

n = 0,l,...,s.

Hence

A?1 f{x) = A*+1 (J2An(x)\ =J2A?lAnW \n=0

/

which completes the proof.

•

Let us note that for s = 1, the equation A\f(x) the form f(x + 2h)-2f(x

= °'

n=0

+ h) + f(x) = 0,

x,heX,

— 0 assumes (9.13)

or, equivalently, by substitution h = |(y — x), f ( ^ )

= \\J{x)

+ f(y)},

for

x,yEX,

(9.14)

which is the well known Jensen's functional equation. We shall need further on the following theorem concerning the formula giving the general solution of the Jensen's equation (9.14). Theorem 9.5 A function f:X^Yisa solution of the equation 0 (9.14) iff there exist a constant A G Y and an additive mapping A1: X ->• Y such that f(x) = A0 + A1 (x),

xe X.

(9.15)

Proof. Assume that / : X -t Y is a solution of the equation (9.14). Define the function g: X —> Y by the formula g{x) = f(x)-f{0),

xeX.

(9.16)

Then 51(0) = 0 and by (9.14) we get for x, y 6 X, 9 ( ^ )

= l2l9(x) + g(y)].

(9.17)

Polynomial functions and their extensions

77

Now, inserting y = 0 in (9.14), we obtain 9 ( | ) = \g{x) for every x G X. x,y€X,

(9.18)

Consequently, (9.18) and (9.17) imply for

( —2~ J =g{x) + g{y), X ~\~ V\

which means that g is an additive function. Therefore, in view of (9.16), f{x) = A° + A1{x), xeX. where A0 = /(0), A1(x) = g(x), x G X and A1 is an additve mapping. Conversely, if / has the form (9.15) with A1 additive, then we have for x, h G s~\, Al f(x) = A2hA° + AlA\x) = A\A\x) 1 =A\x + 2h) - 2A\x + h) + A (x) = 0 i.e. / is a solution of the equation A2J(x)

= 0,

xeX,

which, as we already know, is equivalent the Jensen's functional equation. This completes the proof. • Now we are in a position to prove a main result of this section (Kuczma[123], see also [5]). Theorem 9.6 Let f: X —> Y be a polynomial function of order s. Then there exist n-additive symmetric mappings An: Xn —» Y, n = 0 , . . . , s, such that (9.12) holds, where An, n = 0 , . . . , s, are the diagonalizations of An, n = 0 , . . . , s, respectively.

78

Functional Equations and Inequalities in Several Variables

Proof. We shall prove the existence of An, n = o , . . . , s by induction on s. For s = 0 it is obvious. Let s = 1, then / satisfies the equation A2J(x)=0, for x,heX, which is equivalent to the Jensen's equation (9.14). Hence by Theorem 9.5, there exist a constant A0 E Y and an additive mapping A1: X —> Y such that f(x) = A° + A\x), which means that Next, suppose of order s — 1 E An: Xn ->• Y, n =

xEX,

/ admits representation (9.12). that for every polynomial function g: X —¥ Y N there exist symmetric n-additive mappings 0 , . . . , s - 1 such that s-l

g(x) = YtAn(x),

xEX,

(9.19)

n=0

where An, n — 0 , . . . , s — 1 are the diagonalizations of An, n = o,..., s — 1, respectively. Assume that / : X —> Y is a polynomial function of order s. Define the function As: Xs —> Y, by As(xu ...,x,):=-

AXl...Xs /(0),

i„eX,

n = 1 , . . . , s. (9.20)

Evidently, by Corollary 9.1, As is symmetric. Moreover, by Corollary 9.1, Lemma 9.1, Lemma 9.3 and Theorem 9.3, we have for every n = 1 , . . . , s and x\,..., xs, yn E X,

— Ji.s\X\,..., xs) =^ AXl...Xn_lXn+1...Xs =-jA X l ... X s J / r i / ( 0 ) = 0.

AS\X\)..., [AXn+yJ(0)

xn—i, yn, xn+i,..., - AxJ(0)

-

xs)

AyJ(0)]

Polynomial functions and their extensions

79

This proves that As is s-additive. Let's note that for X, by Lemma 9.1 and Theorem 9.3 we have AXl...x. f{x) - A Xl ... x ./(0) = = AX1...SM [f(x) - /(0)] = AXl...Xs = AX1...X,X /(0) = 0,

Ax /(0) =

and consequently by the formula (9.20) AXl...Xj/(ar) = s\Aa(xi,

...,xs).

Hence and in view of Lemma 9.1 and Lemma 9.6 for g = f — As, where As is the diagonalization of As, we get A hl ... hj g(x) - A fcl ... h ,/(x) - AAl...A,As(a;) = Afcl...h, /(x) - s!A s (/i 1 ; ..., /i,) = 0 for hi,... ,hs,x G X. This means exactly that g is a polynomial function of order s — 1. Therefore the induction hypothesis implies that there exist symmetric n-additive mappings An: Xn —>• Y, n = 0, . . . , s - 1, such that (9.19) holds with An, n = 0 , . . . , s — 1 as diagonalizations of An, n = 0, ...,s — 1, respectively. Therefore s

71 = 0

which completes the induction and the proof of the theorem.

•

For some interesting study of the Jensen's functional equation in the complex domain, as a special case of the equation for polynomial functions, the reader is referred to H. Haruki and Th. M. Rassias [86].

Extensions of polynomial functions The results we are going to report on in this section are contained in R. Ger [74] (see also Szekelyhidi [205],

80

Functional Equations and Inequalities in Several Variables

G. V. Milovanovic, D. S. Mitrinovic and Th. M. Rassias [136] for basic concepts and an extensive bibliography). First we recall some necessary definitions and notations. Let X be a linear space over Q. Then a subset D of X is called Q-convex iff for any points x,y G D the rational segment {Xx + (1 - X)y G X: A G [0,1] n Q} is contained in D. The intersection of all Q-convex sets B such that A C B C X is denoted by convQ A. The set of all algebraically interior points to A (see p. 11) will be denoted by algintQA Let oev :— ^j-j-, v = 1 , . . . , n + 1. For a nonempty subset D of X we define Sn(D) := {x G X: there exists a u e D such that ai,y + (1 — av)x € D for v = 1,..., n}.

(9-21)

Lemma 9.7 Let X be a Q-linear space and let D C X be a nonempty Q-convex set. Then, for arbitary n G N, the set Sn(D) is Q-convex, i.e. for any u,v G Sn(D) and some s,t G D) such that avs + (1 — av)u G D and avt+ (1 — av)v G D for v = 1 , . . . , n, we have av{Xs + (1 - \)t) + (1 - av){Xu + (1 - X)v) G D for allv = 1 , . . . , n and a// A G [0,1] n Q. Moreover, D C Sn(D). Proof. Let u,v £ Sn(D). Since D is Q-convex, then Xs-\-(l-X)t G .D for A G [0,1] n Q and s,t £ D. Hence we have for v = 1 , . . . , n, a„(As + (1 - X)t) + (1 - a„)(Au + (1 - A)u) = =X(avs + (1 - a„)«) + (1 - A)(at,i + (1 - a„)u) G £>, which means that Xu + (1 - A)i; G ^ ( D ) for A G [0,1] n Q and proves the Q-convexity of Sn(D). For every x G D, we have also a„:r + (1 — av)x = x £ D

for

v = 1 , . . . , n,

Polynomial functions and their extensions

81

whence by the definition of Sn(D) we infer that D C Sn(D).

•

Before we prove next lemma, let as remind that every base of a linear space over Q is referred to as a Hamel basis (see [83]). Lemma 9.8 Let X be a Q-linear space and let H be a Hamel basis

ofX.

Then 0 G alg int Q conQ[H U ( - # ) ] •

Proof. Denote A := convQ[iJ U (-H)]. Then A is Q-convex and symmetric with respect to zero. Consider a straight line / passing through zero. Take a point x 6 I \ {0}. Assume that x - aihi H

ha„/i„,

afcGQ\{0},

hk <E H,

k=

l,...,n,

is the Hamel expansion of x with respect to H. If we denote a := |ori| + ... + \an\, then 1 |ai| \an\ -x = s g n aiAii + • • • H

L

sgn anhn G A.

By the convexity and symmetricity of A, the rational segment with endpoints — ^x and ^x is contained in / D A. Obviously .

1/ I N 1,1 N z a La. is the interior point of this segment line, which ends the proof.

•

Let us emphasize that the equation

AJ+VOO = o by Corollary 9.2 can be rewitten with the subsitution h = ^ j in the from s+1

,

, -, x

Sn-0( - 1 ) S + 1 ""(^ 1 ) /

(1 y

)y +

s+ l

jy

x

0,

(9.22)

s+1

for all x,y E X. The next result concerning the extension of polynomial functions, reads as follows (see [74]).

82

Functional Equations and Inequalities in Several Variables

Lemma 9.9 Let X, Y be Q-linear spaces and let D be a nonempty Q-convex subset of X. For every polynomial function f: D —> Y of order s (function fulfilling the equation (9.22) for all x, y € D), there exists exactly one polynomial function F: SS(D) —>• Y of order s (i.e. a function fulfilling the equation (9.22) for all x,y € SS(D)) such that F/Df. Proof. Take an x G SS(D), then there exists a y G D such that zn := any + (I — anx £ D)

for

n = l,...,s.

(9.23)

Define

fW-D-rf^j/w71=1

^

(9-24)

'

Note that zs+i = y G D. We shall show that the formula (9.24) it correct i.e. does not depend on the choice of a point y G D satisfying (9.23). To this end suppose that for some z G D we have also tn '•= otnz + (1 — an)x G D

for

n = 1 , . . . , s.

From the equation (9.22) we obtain the formula /(*) = E ( - l ) ' " f c ( S f c 1 ) / ( « * * + ( ! - « * ) * )

f^ all

s,teD.

(9.25) Hence for (t,s) = (z,zn) G DxD and (t,s) = (akz+(l—ak)y,tk) G D x D we get respectively

/(*.) = E(-l) 5 _ f c (S + X) fl<*kz + (1 - ak)zn] fc=i

\

n

/

and s-,

/(**) = YX-l)s-n 71=1

(" ^

+ l n

\ f[an{akz '

+ (1 - afc)v) + (1 -

an)tk}.

Polynomial functions and their extensions

83

Therefore s+l

£(-l)~('+'W) n=l

^

'

=D-')" (' **) D - D - * (' ^') / M + a - <*)*] = n=l

^

'

k=\

^

'

+ (1 - an)tk]

=D-ir*('t1)^)t—i

V

/

which proves the correctness of the definition (9.24). Clearly F\D = f directly in view of the definition (9.24) and the equality (9.25). Moreover the uniqueness of the extension is obvious. To end the proof, we shall verify that F is a polynomial function of order s. It is sufficient to show that F(x) = Y,(-l)S-k(

k

\F{aky + (l-ak)x),

x,y G SS(D). (9.26)

To this end, take x,y G SS(D) and let s,t E D be such that xn :— ans + (1 - an)x G D

and

yn := ant + (1 - an)y G D

for all n — l , . . . , s . In view of Lemma 9.7, the elements aky + (1 — ak)x G SS(D) for k = 1 , . . . , s, and an[akt + (1 - ak)s] + (1 - cx„)hfc2/ + (1 - «fc)z] € D for all n = 1 , . . . , s. By (9.25), for n = 1 , . . . , s, one has /(zn) = ] T ( - l K f c r

fc

J f[akyn + (1 - ak)xn]

(9.27)

84

Functional Equations and Inequalities in Several Variables

which implies, by (9.24),

^)=D-irB(s^1)/w 71 = 1

^

'

= E E ("Dt+" ( s I') (s +k x) /[<«>*.+(i - * ) * j . 7 1 = 1 Jfc = l

On the other hand, taking into account the definition of F and (9.27), we obtain the following equalities

E(- 1 )'"*( s tV(«*y + (1-a*)^) = fc=i

^

'

= i > i r f c (8 +k *) E(-!) s - n (s +n x) / K M + ( i - <**)*] fc = l

^

'

71=1

^

'

+ (1 - an)[aky + (I - ak)x]] s+1 s+1

=EB-W'* 1 )('* 1 ) /[0MW+(1 -'*' )lJ 7 1 = 1 fc = l r. —1 1

1

\

/

\

/

Consequently comparing left and right sides, of two last sequences of equalities, we get the formula (9.26), which completes the proof.

The main result of this section reads as follows. Theorem 9.7 Let X and Y be Q-linear spaces and let D be a nonempty Q-convex subset of X. 7/alg intQ.D ^ 0, then for every polynomial mapping f': D —»• Y of order s there exists exactly one polynomial mapping F: X —>• Y of order s such that F \o= f • Proof. We may assume without loss of generality that D is symmetric with respect to zero and zero is an algebraically interior point for D.

Polynomial functions and their extensions

85

By Qo we denote the Minkowski functional of the set D. Then (X, QD) is a semi-normed space over rationals. The ball B := {x £ X: QD{X) < 1} is contained in D. Let 5" denote the n-th iterate of the operation Ss defined by (9.21). Now we want to prove that 71

Bn:={xeX:

QD<1

n = l,2,...

+ -}CS?(B),

.

(9.28)

Let n = 1; take x € B\, then for Xj. := afcO + (1 — ak)x,k = 1 , . . . , s we have QDM

= (l-ak)gD(x)

< ( l - a f c ) ( l + - ) < (1 ^ r ) ( l + - ) - 1, s s +1 s which means that xk £ B for k = 1 , . . . , s. Hence we get x e SS{B) = S}(B) since y = 0eB (see the definition of 5,(5)). Now assume that (9.28) holds true for some n G N. To prove it for n + 1, it is enough to show Bn+1 C S s (5 n ) on account of monotonicity of the operation Ss. Take x £ We have again

(9.29) JB„ + I.

QD(xk) = (1 - <**)Ms) < - T - r t 1 + ~ )< !+ ~ s +1 s s i.e. Xk € B n for A; = 1 , . . . , s. Because 0 £ Bn, this shows that :r £ Ss(Bn) and completes the proof of (9.29). Therefore Bn+l c S,(S?(B)) = S?+1(B) and the inductive proof of (9.28) is completed. Clearly, we have OO

00

X=[JBnc\J 71=1

00

S?(B) C |J S?(D) C X. 71=1

71=1

Consequently we can apply successively step-by-steep (for n = 1,2,...) Lemma 9.9, to get our statement. • Directly from this theorem we get

86

Functional Equations and Inequalities in Several Variables

Corollary 9.4 Assume that the assumptions of Theorem 9.1 are satisfied. Then for every polynomial mapping f: D —» Y of order s there exist k-additive symmetric mappings Ak: Xk —*Y,k = 0 , . . . , s such that f(x) = A°(x) + ... +

As(x),xeD,

where Ak,k = 0 , . . . , s are the diagonalizations of Ak,k = 0 , . . . , s respectively. Proof. The proof follows from Theorem 9.7 and Theorem 9.6 about the representation of polynomial functions. • Let us note another extention theorem for polynomial mappings (the proof can be found in [74]). I

T h e o r e m 9.8 Let X, Y be • Y of order s there exists a polynomial mapping F: X —» Y of order s such that F \D— f • The extension is unique iff the affine hull of D (i. e. linear variety spanned by D) is equal to the space X. Corollary 9.5 Under the assumptions of Theorem 9.8, for every polynomial mapping f:D-+Y of order s there exist symmetric k k-additive mappings Ak: X —>• Y, k — 0 , . . . , s, such that f(x) = A°{x) + ... +

As(x),xeD,

where Ak,k — 0 , . . . ,s are the diagonalisations of Ak,k = 0 , . . . ,s, respectively. The mappings Ak, k = 0 , . . . , s, are determined uniquely iff the affine hull of D is equal to the space X.

Notes 9.1 In the special case X = W1, Y = R, we have

Polynomial functions and their extensions

87

Proposition 9.1 ([123]). If a function f: Rn ->• R, /(x) =

f(xu...,xn)

is a polynomial separately in each variable Xk, k = 1 , . . . , n then f is a polynomial jointly in all variables. Proposition 9.2 ([123]). A function f:Rn^Risa continuous polynomial funcion of order s iff f is a polynomial of degree at most s. Thus we see that polynomial functions generalize ordinary polynomials and reduce to the latter under weak regularity assumptions. 9.2 Polynomial functions have been thoroughly investigated by many authors (cf. Frechet [56], Highberg [91], Hille-Philips [92], Djokovic [46]).

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Chapter 10 Quadratic mappings In this section we are going to consider the quadratic functional equation and we will prove some fundamental results concerning this equation, which have many applications in several areas of mathematics. Let us recall that a mapping / : G —> F, where G, F are groups is called a quadratic mapping provided that it satisfies the functional equation f{x + y) + f(x-y)

= 2f(x) + 2f(y),

x,yeG.

(10.1)

A mapping S: G x G —>• F is said to be biadditive iff S{xi + x2, y) = S(x1, y) + S(x2, y) and S(x, 3/1 + y2) = S(x, y{) + S(x, y2) for a\\xi,x2,y,x,yi,y2 e G. The first important result is the following Theorem 10.1 (representation formula). Let G be an abelian group and F an abelian group with the division by two. The mapping f:G —¥ F is quadratic iff there exists exactly one symmetric biadditive mapping S: G x G —>• F such that f{x) = S{x,x), 89

xeG.

(10.2)

90

Functional Equations and Inequalities in Several Variables

Proof. Assume that f(x) = S(x, x) for x € G, then f(x + y) + f{x -y) = S(x + y,x + y) + S(x =2S(x,x) + 2S(y,y) = 2f(x) + 2f(y)

-y,x-y)

for all x,y E G.

I

Conversely, let / be a quadratic mapping. mapping S(x,y) := i [ / ( x + y)-

f(x -y)],x,yE

We define the

G.

(10.3)

Seting y = 0 in (10.1) we obtain that /(0) = 0 and hence by choosing x = 0 in (10.1) we deduce that f(—y) = f(y), i-e. / is even. Hence and by the definition (10.3) we guess that S is symmetric i.e. S(x,y) = S(y,x) for all x,y £ S. We shall prove that S is biadditive. Indeed, we have by (10.1), 4[S(x + z,y)-S(x,y)-S(z,y)} = = [f(x + z + y)-f{x + z-y)} + [f{x - y) + f{z - y)]- [fix + y) + f(z + y)} = [fix + y + z)-f(x + z~ y)}+ + \[f(x + z-2y)

+ f{x - z)\ - \[fix + z + 2y) + fix - z)\

= [fix + z + y)-fix

+ z~y)]

+ -fix

+z-

2y)-

- -fix

+ z + 2y) = [fix + z + y) - fix + z - y)] + fix + z - 2y)

-^[fix

+ z-2y)

+ fix + z + 2y)} = [fix + z + y)-

-fix + z - y)\ + fix + z - 2y) - fix + z) = [fix + z + y)-fix + z-y)} + 2f(x + z -

f(2y) y)-

- 2 fix + z)- 2fiy) = fix + z + y) + fix + z-y)-

2 fix + z) - 2/(y) = 0.

This is the required additivity of S witch respect to the first variable (the symmetricity yields the additivity in the second variable).

Quadratic mappings

91

To prove the uniqueness, assume that Si,S2 biadditive mappings such that f(x) = Sk(x,x),

x&G,k

are symmetric

= 1,2.

We have for x,y e G, Si{x+y,x+y)-Si(x,x)-Si(y,y)

=

S2{x+y,x+y)-S2(x,x)-S2(y,y)

from which one gets 2Si(x,y) = 2S2(x,y)

for all

x,yEG.

Thus S\ = S2, and the proof is completed.

•

If we wish to get solution of the equation (10.1) to be homogeneous, we have to make some stronger assumptions. Let us emphasize that, in general, a quadratic mapping is not necessarily homogeneous (of order two). We shall call a function / : X —>• R, where X is a real linear space, ray-bounded (see [1]) if for all x ^ 0 and y in X, there exists a set Ax of positive Lebesgue measure in R and a constant KXty such that \f(ax + y)\< KXtV

for all

a G Ax.

Note the following result Theorem 10.2 Let X be a real linear space. Then f: X —>• R. is a ray-bounded quadratic function iff there exists exactly one symmetric bilinear (i.e. linear with respect to each variable) function 5 : I x l 4 l such that f{x) = S(x,x)

for

x e X.

Proof. If / is of the form f{x) = S(x,x),x G X, where S is bilinear, then from Theorem 10.1 it follows that / is a quadratic function. Moreover, / is ray-bounded, since f(ax + y) = a2S(x, x) + 2aS(x, y) + S(x, y).

92

Functional Equations and Inequalities in Several Variables

Conversely, assume that / is a ray-bounded solution of the equation (10.1). From Theorem 10.1 we get that there exists exactly one symmetric biadditive function 5 : 1 x 1 -» R such that

f(x) = S(x,x)

for

xEX.

We shall show that S is bilinear, too. In fact, define F: R —)• R by F(a) = S(ax, y), where x ^ 0 and y are fixed elements of X. Taking into account that S is given by (10.3) and / is raybounded, we get for a f 4 ,

\F(a)\<±(\f(ax

+ y)\ +

\f(ax-y)\)

Hence by the Corollary 2.5 from [1], we get F(a) = aF(l), so S(ax,y) = aS(x,y), which means that S linear with respect to the first variable. Symmetry of S yields the linearity with respect to the second variable. This proves the theorem. • We say that a function / : X —> R is ray-B-bounded, if for any x ^ 0 there exists a second category Baire subset Bx of R such that for any y G X there exists a constant Kxy such that \F(ax + y)\ < KXiV

for all

a e Bx.

Then we have Theorem 10.3 Let X be a real linear space. Then f: X —>• R is a ray-B-bounded quadratic function iff there exists exactly one symmetric bilinear function S : I x I - f R such that f(x) - S(x, x)

for all x £ X.

Proof. The proof runs quite similary as for Theorem 10.2. The only difference is that we utilize Theorem 4.4 instead of Corollary

2.5 from [1].

•

Quadratic mappings

93

Now we turn to the case of complex valued quadratic functions. A function S: I x X - > C , where X is a complex linear space is said to be sesquilinear provided that S is biadditive and satisfies S(\x, y) = XS(x, y)

and

S(x, Xy) = XS(x, y)

for all x, y G X and A G C, where A means the conjugate to A. Theorem 10.4 Let X be a complex linear space and let f: X —> C be a quadratic function satisfying the condition f{ax) = \a\2f(x)

for all xeX,ae

C.

(10.4)

Then there exists exactly one sesquilinear function U: X xX —> C, such that f(x) = U(x,x) for all x G X. Proof, (following Aczel, Dhombres [1], see also Vrbova [215] and Kurepa [129]). Consider first the case when / is a real valued quadratic function satisfying (10.4). Evidently, the space X can be considered as a real linear space. Therefore Theorem 10.1 implies that there exists a symmetric biadditive function S : I x I - ) E such that f(x) = S(x,x), x G X. Hence and by (10.4) we get S(ax,ax) = \a\2S(x,x) for all x G X and a G C. Using biadditivity and symmetry of S and the last property of S1, we get S[a(x + y), a{x + y)] = \a\2S{x + y,x + y) =\a\2[S(x,x) + 2S(x,y) + S(y,y)} =S(ax, ax) + 2S(ax, ay) + S(ay, ay) = \a\2S(x, x) + 2S(ax, ay) + \a\2S(y, y), whence S(ax,ay)

=

\a\2S(x,y).

94

Functional Equations and Inequalities in Several Variables

We can rewrite this equation in the form S(ax,y)

= \a\2S{x,y/a),

x,y e X,

a € C\{0}.

We have by additivity of S, S(x, -y) = -S(x,y), (10.5) for a = i, we get S(ix,y) = Now, let us define U: X x X

(10.5)

therefore from (10.6)

-S(iy,x). ^Cby

U(x, y) := S(x, y) - iS(ix, y).

(10.7)

Evidently, U is a biadditive function. Moreover, we can easily see by (10.6) that U(y,x)=U(x,y)

for

x,y £ X,

and S(ix, x) = 0 for x G X, so U(x,x) = S(x,x) = f(x)

for

x € X.

We shall show now that U is homogeneous with respect to the first variable. Take arbitrary x, y G X and define g: C —> C by g(a) := U(ax,y) -

U{x,ay).

Then g is an additive function on C. Equation (10.5) implies that g(a) =S (cnx, y) — iS (aix, y) — [S (x, ay) — iS (ix, ay)]

5

=\a\2[s(x,^-is(ix,^

(i' y )- i 5 ( i i« y ).

\a\29 Thus g satisfies the functional equation g{a) = -\a\2g(-\

for

a € C\{0}

(10.8)

Quadratic mappings

95

In the sequel, we want to find all additive g satisfying the equation (10.8). Take any a^ G R. If |o;i| < 1, then there exists a2 G R such that a — a.\ + zct2 and \a\ = 1. Then, (10.8) implies (/(a) = — g (^) = — 5(0;), whence by additivity of 9, 9M

+ 9{ia2) =g{u\ +102) = - p ( a i - *a2) = -g(a1) + g(ia2).

Consequently g(a.\) = 0 for a\ G R and such that |«i| < 1. On the other hand, if \a\\ > 1, then g(ai) = -ct\g ( — J

=0

< 1. This shows that g{ax) = 0 for all ax G R. Take since a2 G R\{0} then by (10.8) we obtain g(ia2) = -|a 2 | 2 i? I — ) = a 2 5 ( — I • Put G(a2) :— g(ia2), then we have G(a2) = a22G (—)

,

a2 G R\{0}.

(10.9)

It can be proved (see e.g. [146]) that additive G satisfying the equation (10.9) has to be continuous. Hence G(a2) = a2G{\) = a2g(i)

for all

a2 G R,

and consequently, g(cti + ia2) = a2g(i)

for all

ot\, a2 G R.

Turning back to the definition of g, we get U(aix,y)

= U{x,a\y)

for

x,y £ X

and

a\ G R.

(10.10)

96

Functional Equations and Inequalities in Several Variables

Moreover, in view of g{ia2) = os2g(i), U(a2ix, y) - U(x, a2iy) = a2[U(ix, y) - U(x, iy)]

(10.11)

for all x, y G X and all a2 G K. We also have U(ix, y) =S(ix, y) - iS(-x, y) = S(ix, y) + iS(x, y) =i[S(x, y) - iS{ix, y)} = iU(x, y), i.e. U(ix,y) = iU(x,y).

(10.12)

U{x, iy) = TJii^x) = -iU{x, y).

(10.13)

Furthemore, by (10.12)

By (10.12) and (10.13), we obtain U(ix,y) - U(x,iy) = 2iU(x,y), and hence by (10.12), (10.13) and (10.10), U(a2ix,y)

- U(x,a2iy)

=

i[U(a2x, y) + U{x, a2y)] =2iU(a2x, y). But (10.11), (10.12) and (10.13) imply U(a2ix,y)

- U(x,a2iy)

=a2[U(ix,y) =2ia2U(x,y).

U(x,iy)\

Thus U(a2x,y)

=

a2U(x,y).

Finally, by (10.12) U[{ai + ia2)x, y] = U(aix, y) + U(ia2x, y) =aiU(x, y) + ia2U(x, y) = (ai + ia2)U(x, y)

Quadratic mappings

97

i.e. U is homogeneous with respect to the first variable. In addition, we have U(x, ay) = U(ay, x) = aU(x, y), which proves that U is a sesquilinear. Consider now the case when / is a complex valued function. Then f(x) = h{x)+if2(x), where flt f2: X -> R satisfy (10.1) and (10.4). By what we have already proved, there exist sesquilinear functions U\, U2: X x X ->• C such that fi(x) = Ui(x,x),

f2(x) = U2{x,x),

xeX.

Take U := Ui + iU2, then U is a sesquilinear and for all x G X U(x,x) = Ui(x,x) + iU2(x,x) = /i(x) + i/ 2 (x) = f(x). The uniqueness part follows directly from Theorem 10.1. Thus the proof is finished. • Remark 10.1 The equation (10.8) has additive solutions g: R —> R which are not continuous. This is the reason why the situation in the real case differs from that in the complex case. For X = R2 we have the following (see [17])

98

Functional Equations and Inequalities in Several Variables

Theorem 10.5 (Decomposition Theorem). Every quadratic function / : R2 —> R can be uniquely decomposed in the form f(xlt x2) = B{xux2)

+ Ql{xl) + Q2(x2),

(10.14)

where B: R2 —>• R is hiadditive and Qi,Q2:'R^'R functions.

are quadratic

Proof. Define B{xux2)

= -[f(xux2)

- f{xu

-x2)},

S{xux2)

=-[f(xi,x2)

+ f{xi,

-x2)],

Q1(x)=S(x,0)

,

Q2{x)=S(0,x).

Clearly, B and S are quadratic functions. It is easy to see that the functions 5(-,0) and 5(0, •) as functions of first and second variable, respectively, are also quadratic functions of one variable, hence Q\ and Q2 are quadratic, too. Since /(0, —x) = f(0,x) because / is quadratic, then S(xu 0) = f(xlt 0)

,

5(0, x2) = ^[/(0, x 2 )+/(0, -x2)] = /(0, x),

whence, by (10.1) 5(xi,a;2)

=

^[f{xi,x2)

+ f(xi,-x2)]=f{xu0)

+ f{0,x2)

=

S{x1,0) + S{0,x2) = Q1{xl) + Q2{x2).

Clearly, then we get f{xi,x2)

= B(xi,x2)

+ S(xi,x2)

i.e. the decomposition (10.14).

= B(xux2)

+ Qi(xi) +Q 2 (^2),

Quadratic mappings

99

To prove the biadditivity of B, we find B(x, y + z) =^[f{x, y + z)= 2 ^ ^ ' y + z)

+

f{x, -y - z)\ / ( x ' ~y + *)]

- gl/fo "^ + z) + Z^' "^ - z)] =\f{x,z) + f(0,y)]-[f(x,-y)

+ mz)]

= 2 1 / ^ ' *) - /(*» ~z)\ + j ^ (*> 1/) - /(*> -?/)]

+ /(0,y)-/(0,«) =5(rr, z) + B(x, y) + hj{x,

z) + f(x,

-z)]

- \[f{x, V) + f(x, -y)] + /(0, y) - /(0, z) =B(x, z) + B(x, y) + -A[f(2x, 0) + /(0,2z)} - ^[f(2x, 0) + /(0,2i,)] + /(0, y) - /(0, z) = £ ( * , z) + B(x, y) + ±[/(0, 2*) - /(0, 2?/)] + /(0, y) -/(0,z) =B(a; J y) + B(a;,z)> since /(0,2z) = 4/(0, z) and /(0,2j/) = 4/(0, y) (quadratic functions are rational-homogeneous of order two). A similar argument shows that B is additive in its first variable, too, so B is biadditive. To prove the uniqueness, assume that f(xi,x2)

= Bi(x1,x2)

+ Qi(xi) + Q2(x2)

for x±,x2 G R, where Bi is biadditive and Qi,Q2

are

quadratic.

100

Functional Equations and Inequalities in Several Variables

Then we have B(xi,x2)

+Qi(xi)

+ Q2(x2) = Bi(xux2)

+ Qi(xi) + Q2(x2). (10.15) n For xi = 2 x , x2 = 0 since Q2(0) = Q2(0) = 0 , we get B{2nx, 0) + Qx{2nx) = Bx(2nx, 0) + Qi{2nx), whence, by Qi{2nx) = 22nQl(x) , Qi{2nx) = 22nQ1(x) for all n e N 2"7l[JB(:r,0) - ^(rr.O)] = Qi(x) - Qx{x). Letting n —> oo , we obtain hence, Q\ = Q\. Repeating a similar argument, one can deduce that also Q2 = Q2. Therefore, in view of (10.15) we get that B = B\ and the proof of the uniqueness is completed. •

Continuous quadratic functionals We start with the following L e m m a 10.1 Let X and Y be linear spaces over K. If f: X —>Y is quadratic, then for every rational number r and every x € X, we have f{rx)=r2f{x). (10.16) Proof. Setting x = y = 0 in (10.1) , we get /(0) = 0. For x = 0 we obtain f(—y) = f(y) for y e X, i.e. / is an even function. Substituting x = y in (10.1) gives f(2x) — 22f(x). Now assume that n is fixed and for every s G N, s < n, f(sx) = s2f{x). Then by (10.1), f[(n + l)rr] + f[(n - l)x] = 2f(nx) + 2f(x), whence f[(n + l)x] + f[(n - l)x] = 2n2f(x) + 2f(x) - (n - l)2f(x)

=

Quadratic mappings

101 = (n +

l)2f(x).

Thus f{nx) — n2f(x) holds for every n G N and every i £ l . Now we have for m G N, /Or) = / ( m - ) = m 2 / (1) so

,

/ (-) = "V^

which implies that for every positive rational s, and i £ l , /(S*)=S2/(x). But since / is even, this implies (10.16), and concludes the proof.

Lemma 10.2 (see [146]). Let A C K be a set of positive inner Lebesgue measure or of the second category with the Baire property and let Y be a real normed space. If a quadratic function f: K —> Y is bounded on A, then it is bounded on a neighbourhood of zero. Proof. Suppose that ||/(x)|| < M for x G A. Define that set B(A) := {x G R: A n {A + x) n (A - x) ^ 0} . We shall show that / is bounded on B(A). Indeed, if x G B(A), then there is a point j / 6 l such that y,y — x,y + x G A. Hence, by (10.1), we have

\\f(y) + m

-[f(y + x) +

f(y-x)}

<2 (11/(2/ + *)ll + 11/(2/ - ^ ) l l ) < ^ Hence ||/(a0|| = \\f{x) + f(y) - f(y)\\ < \\f(x) + f(y)\ + \\f(y)\\ < 2M.

102

Functional Equations and Inequalities in Several Variables

i.e. / is bounded on B(A). Since A is a set of positive inner Lebesgue measure or is of the second categoty with the Baire property, then B(A) contains a neighbourhood of zero (for details see [149]), which completes the proof. • The main result of this section is Theorem 10.6 Let A C R be a set of positive inner Lebesgue measure or of the second category with the Baire property, and let Y be a real normed space. If a quadratic mapping f: R —>• Y is bounded on A, then f{x) = x2f(l)

for all

x G R.

(10.17)

Proof. Consider the function

F(x)~f(x)-x2f(l),x£R. Then F is quadratic and by Lemma 10.2, there exists a positive rational number r such that F is bounded on the interval [0, r]. Put a : = sup \\F{x)\\. x€[0,r]

We shall prove that a — 0. For the contrary suppose that a > 0. Then there exists a number x0 G [0, r] such that \\F(x0)\\ > \a. We have by Lemma 10.1, ||F(2xo)||=4||F(x 0 )||>3a. The number 2x0 can be written in the form 2x0 = r + y, where y G [0,r]. We have now by (10.1) F(2x0) = F(r + y) = 2F{r) + 2F(y) - F(r - y)

Quadratic mappings

103

and since F(r) = 0 and r — y G [0, r], then ||F(2x 0 )||<2||F(y)|| + | | F ( r - y ) | | < 3 a . Thus we get 3a < ||F(2i 0 )|| < 3 a , which contradicts o > 0. The contradiction implies that a = 0 and consequently, /(x) = x2f(l)

for every

a; G [0, r].

Consider any a; G R and choose a rational s ^ 0 such that sx G [0, r]. By Lemma 10.2, it follows that f(x) = -2f(sx)

= -2s2x2f(l)

= x2/(l),

which completes the proof.

•

As an immediate consequence of this theorem we get Corollary 10.1 // a quadratic mapping f: R —» Y is continuous at one point XQ G R, then it is of the form (10.17). In the sequel we present another sufficient condition for a quadratic function to be of the form (10.17). Theorem 10.7 Let g: R —> R be a measurable function and let / : R —> R be a quadratic function. If A C R is a Lebesgue measurable set of positive measure and f(x) < g(x)

for

xeA,

(10.18)

then f is of the form (10.17). Proof. Define the sets Ak := {xe A: k-l

< g{x) < k},

keZ.

(10.19)

104

Functional Equations and Inequalities in Several Variables

In view of the measurability of g, the sets Ak are Lebesgue measurable sets. Moreover, the sets Ak are pairwise disjoint and oo

U Ak = A, k——cx>

whence oo

Y^ rn(Ak) = m(A) > 0. fc=—00

Thus there exists a k0 € Z such that m(Ak0) > 0. On the other hand, by (10.18) and (10.19), 1/0*01 < \g(x)\ < max[|fc 0 -l|,|fc 0 |]

for

x e Ako,

i.e. / is bounded on a set of positive Lebesque measure. By Theorem 10.6, / has the form (10.17) and the proof is completed.

Corollary 10.2 // a quadratic function / : R —>• R is measurable, then it is of the form (10.17). Proof. Take in Theorem 10.7, A = R and g = f, to get the assertion of the Corollary. • Lemma 10.3 ([129]). Let X,Y be normed real spaces. If a quadratic mapping f: X —> Y is bounded on one ball, then it is bounded on every ball. Proof. Suppose that / is bounded on the ball B(xQ,e), i.e. sup ||/(a; + a;o)|| < M. ||x||< e

From (10.1) we get 2f(x) = f(x + x0) + f{x - x0) - 2/(x 0 ) =

(10.20)

Quadratic mappings

105

= f(x + xQ) + f(-x + xQ) - 2/(x 0 ), and since —x + x0 € B(x0,e) for ||x|| < e, then by (10.20), sup ||/(z)|| < M + | | / ( x + 0)|| = M1 \\x\\<e

whence, for s G [—1,1] and ||x|| < e, suV\\f{sx)\\<Mx. \s\ Y, given by the formula
teR,

thus f(tx) = t2f(x)

for all teR

and x e X.

(10.21)

Consider the ball B(0,r),r > 0, then sup ||/(x)|| = sup ||/ (-y) || = ( - ) ' sup ||/(„)|| < \\x\\
\\y\\<£

K£

'

Xe

'

||y||<£

which means that / is bounded on every ball.

V £

(-)2MU '

•

Lemma 10.4 ([129]). Let X,Y be two real normed spaces. If a quadratic mapping f: X —>• Y is bounded on a ball, then it is continuous at the point zero and conversely. Proof. By Lemma 10.3, / is bounded on the unit ball, so for some M>0, sup | | / ( x ) | | < M . (10.22) IWKi

106

Functional Equations and Inequalities in Several Variables

Suppose that / is not continuous at zero. Therefore there exist a number e > 0 and a sequence xn —> 0, xn ^ 0, such that ||/(z„)||>e, n e N . As we have already seen in the proof of Lemma 10.3, (10.22) implies (10.21) for t e R and xeX. Hence, for n e N ,

\ ll^ill /

llxn|l

11"°rxJ|

which contradicts (10.22). Thus / is continuous at zero. The converse statement follows directly from the definition of limit at a point. • Now we can prove the following T h e o r e m 10.8 ([129]). Let X be a real normed space and let f: X —> M. be a quadratic function. If f is continuous at one point x0 € X, then it is continuous everywhere and it is bounded on every ball. Conversely, if f is bounded on one ball, then it is continuous. Proof. Let / be continuous at 2o € X. From the equation (10.1) we get for xn E X,n eN, f(xn) = 7,[f(xo + xn) + f(x0 - xn)\ -

f{x0).

Thus if xn —>• 0 as n —> oo, we obtain f(xn) —>• 0 = /(0), so / is continuous at zero. We shall prove that / is continuous at every point. Suppose that this is not so, then there exist an e > 0 and a z € X and a sequence {xn} C X, xn —> 0 as n -» oo, such that for nGN, \f{z + xn)-f(z)\>e (10.23) From Lemma 10.4 and Lemma 10.3, / is bounded on every ball, so there exist a constans L > 0 such that \f(z + x)\
for all

||z|| < 1.

(10.24)

Quadratic mappings

107

Hence, for ||x n || < 1, \f(z +

xn)-f(z)\
i.e. the sequence {f(z + xn) — f(z)} of real numbers is bounded. Therefore, there exists a subsequence {xnk} of the sequence {xn} and a real number b such that \im[f(z + xnk)-f(z)]

= b.

k—too

Without loss of generality we may assume that just the whole sequence {xn} has this property, therefore lim[f(z + xn)-f(z)]

= b.

(10.25)

n—•oo

Now, set x + y instead of x into (10.1), then f(x + 2y) = 2f{x + y) + 2f(y) - f(x).

(10.26)

Putting x = z,y = xn in (10.26), we obtain f(z + 2xn) = 2f{z + xn) + 2f(xn) -

f(z),

whence by (10.25) we find lim f(z + 2xn) = 2b+ f(z).

(10.27)

Setting again x = z, y = 2xn in (10.26) we get f(z + 4xn) = 2f(z + xn) + 2f(2xn) -

f(z),

which together with (10.27) leads to \imf{z

+ 22xn) = 22b + f(z).

n—>oo

By the induction we can prove that lim f{z + 2kxn) = 2kb + f(z).

(10.28)

108

Functional Equations and Inequalities in Several Variables

Now let b > 0. For k sufficiently large 2*6 + f(z) > M, so for n large enough, from (10.28), we get / ( 2 + 2kxn) > M, which contradicts (10.24) (for n such that ||2fc3;n|| < 1). Moreover, if b < 0, then for k sufficiently large, 2kb + f{z)

<-M,

so for n large enough, from (10.28) one has f(z + 2kxn) < -M, which contradicts (10.24). Therefore the hypothesis that / is not continuous at some point z leads to the countradiction. This proves that / is continuous everywhere. Conversely, if / is bounded on one ball, then from Lemma 10.4, it is continuous at zero. Therefore, from what we have just proved, / is continuous everywhere. This completes the proof. •

Continuous quadratic functionals on Hilbert spaces We shall prove the following fundamental result for continuous quadratic functionals. Theorem 10.9 ([129]) Let X be a real Hilbert space. Then f: X —> R is a continuous quadratic function iff there exists exactly one linear bounded mapping A: X —> X such that f(x) = (x\A(x)), for all x G X,

(10.29)

Quadratic mappings

109

Proof. Assume that / : X —> R is continuous quadratic functional. Define M : I x I ^ l b y

M(x,y) = -[f(x + y) - f(x - y)]. Thus by Theorem 10.1 we get that M is asymmetric biadditive maping such that f(x) = M(x, x),

for x G X.

Since / is continuous, than M is continuous function with respect to x and y, and therefore as an additive function, M is also homogeneous with respect to x and y, respectively. Thus M is bilinear and continuous, i.e. M is a bilinear bunded real functional. Now fix arbitralily y G X. From the well known theorem about form of a linear bounded functional on Hilbert space, for every fixed y G X, there exists exactly one A(y) EX such that M(x, y) = (x\A{y)),

for all x G X.

(10.30)

We want to show that A: X —> X defined by the relation (10.30) is a linear bounded operator. Indeed, if ?/i,?/2 G X, then for all xeX, (x\A(yi + y2)) =M(x,

Vl

+ y2) = M(x,

Vl)

+ M{x, y2)

=WMvi)) + WMV*)) = WAfa) + A(y2)) whence A{yi + 1/2) = A{yi) + A{y2)

for yu y2 e X.

Moreover, if A e R and y G X, (x\A(Xy)) = M(x,Xy) = XM(x,y) = X{x\A(y)) = (x\XA(y)), thus A(Xy) = XA(y)

for A G R,

y G X.

110

Functional Equations and Inequalities in Several Variables

Since M is bilinear bounded operator, then there exists a constant K such that \M(x,y)\ < if||:r||||y||

for all x,y e X.

Hence for x = A(y), we have \M(A(y),y)\

=

\(A(y)\A(y))\
that is, ||A(y)|| 2
P(y)||
foryeX

This finishes the proof that A is a bounded linear operator. To prove the uniqueness, let us assume that there exist two bounded linear operators A, B: X —> X such that f(x) = (x\A(x)) = (x\B(x)),

for all x e X.

Take Mi(x, y) = (x\A(y)),M2(x,y)

= (x\B(y)),

for all x,y e X.

Then Mi(x,x)

= M2(x,x) = f(x).

iorxeX.

and Mi,M 2 are symmetric biadditive mappings. From Theorem 10.1 Mi = M 2 , and therefore, (x\A(y)) = (x\B(y)),

for

x,yeX.

Hence (x\A(y)-B(y))

= 0,

for all x, y e X.

Taking x = A(y) - B(y), we get \\A(y)-B(y)\\2

= 0,

i.e. i4(y) - fi(y) = 0 for all y e X. This concludes the proof of the theorem. •

Quadratic mappings

111

Notes 10.1 Applying functional equations for quadratic functions one can get the following characterization of inner product spaces. Proposition 10.1 ([1]). If every two-dimensional subspace of a real (complex) normed space (E, \\ • ||) is a real (complex) inner product space, then E itself is a real (complex) inner product space. Proposition 10.2 A normed space (E, || • ||) is an inner product space iff the function p(x) = \\x\\2 satisfies the functional equation 9{x)g{y)g{x + y)=g[xg(y)

+ yg{x)],

x,y £ E.

(10.31)

10.2 An interesting result concerning quadratic functionals has been proved by Volkman [214]. Proposition 10.3 Let E be a normed space (real or complex). Assume that g: E —» R is continuous, g(0) = 0 ; g(x) > 0 for x G JE'\0 and lim g{x) — oo. ||x||—>oo

If, moreover, g satisfies the equation (10.31), then g is a quadratic functional over E. 10.3 Another result in this direction is the following Proposition 10.4 (Rohmel [185]). Let E be a linear space over a commutative field F of characteristic different from 2. If g: E —>• F satisfies g(x + \y) + g(x - \y) = (1 + X2)[g(x) + g(y)},

(10.32)

for all x,y e E, X G F, then there exists a bilinear symmetric S: E2 -> F such that g(x) = S(x, x)

for x G E.

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Chapter 11 Quadratic equation on an interval In this chapter, we give the general solutions of the quadratic functional equation on an interval of real number. All the results we shall present here are due to K. Dlutek [47]. In the first result we shall give the formula for the general solution of the generalized quadratic functional equation on a group. We shall recall that F: G\ —> G2, G\, G2 — groups, is called a quadratic function iff it satisfies the equation F(x + y) + F(x - y) = 2F{x) + 2F(y)

for x,y e G^

(11.1)

Theorem 11.1 Let G\, G2 be groups with division by two. Let A,B,C,D: G\ —> G2 satisfy the equation A{x)+B{y)

= C{x + y) + D{x-y)

forx,yeGx.

(11.2)

Then there exist a quadratic function K: G\ —> G2, additive functions E, F: G\ —> G2 and constants Si, S2, S3, S4 € G2 such 113

114

Functional Equations and Inequalities in Several Variables

that A(x) =2K{x) B{x) =2K{x) C{x)= K{x) D{x) = K{x)

+ E{x) + E(x) + E(x) + F(x)

+ F(x) + Su - F(x) + S2, + S3, + S4,

for x G Gi and S\ + S2 = S3 + S4. Proof. Put A+{x) := ±[A{x)+A(-x)],

A~(x) := ±[A(x)-A{-x)],

xeG,

and similarly B+, B~, C+, C~, D+, D'. From (11.2) we get A+{x) + B+(y) = C+{x + y) + D+(x - y), A~(x) + B'(y) = C-(x + y) + D~(x - y). Denote Sx = A+{6), S2 = B+(9), S3 = C+(0), 5 4 = D+(9). (11.2) we obtain S1 + S2 = S3 + S4. Denote A0(x) = A+(x)-Su C0(x) = C+(x) - S3, for x € G\. The functions A0,B0,C0,

(11.3) (11.4) From

B0(x) = B+(x)-S2, D0(x) = D+(x) - S4.

D0 are even and

A0(6) = B0{9) = C0{9) = D0(9) = 9. Moreover, for x,y E Gi, AQ{x) + B0(y) = CQ(x + y) + D0(x - y).

(11.5)

Putting first y = 9 and then x — 9 and y = x into (11.5) we get A0(x) = C0 Or) + D 0 (x),

B0(x) = C 0 (x) + £> o (-a0,

Quadratic equation on an interval

115

whence A0 — B0. Thus, for all x,y e Gi, A0(x) + A0{y) = C0{x + y)+ D0(x - y).

(11.6)

Now inserting into (11.6) x = | , y = f and x = | , y — — | respectively, we obtain

^Q)+A,Q)=Co(«), ^ o ( | ) + A 0 ( - | ) = A)(«), which implies the equality CQ = D0. Hence A0(x) + A0(y) = C0(x + y) + C0(x - y).

(11.7)

For y = 9, we get A0(o;) + A)(0) = C 0 (x) + C0{x), i.e. ^ o W = 2C 0 (z), x e G i . Denote C0 = K. Then (11.7) yields 2K(x) + 2K(y) = K{x + y) + K(x - j/) which shows that K is a quadratic function. Moreover, A+(x) = 2K(x) + Si, C+(x) = K{x) + S3,

B+(x) = 2K(x) + S2, D+(x) = K(x) + S4.

Puty = 6 into (11.4), A-{x) = C-(x) + D-(x),

(11.8)

then take x = 6 and y = x in (11.4), B-{x) = C-{x)-D-{x).

(11.9)

Denote C~ = E and D~ = F. In order to finish the proof we shall show that E and F are additive. Indeed, inserting (11.8), (11.9) into (11.4) and taking y = x, one gets E(x) + F(x) + E(x) - F(x) = E(2x), whence E(2x) - 2E(x). Now by suitable substitution we get from (11.4),

116

Functional Equations and Inequalities in Several Variables

(a) A-(x) + B~(x)

=2E(x),

(b) A-(y) + B-(y) = 2E(x)t (c) A~(x) + B-(y) = E(x + y) + F(x - y), (d) A~(y) + B~{x) = E{x + y) + F{x - y). Multiply the equations (c), (d) by —1 and then add side by side the equations (a), (b), (-c), (-d), to get 2E(x) + 2E{y) - 2E{x + y) = 0, i.e. E(x + y) = E(x) + E(y). Thus E is additive. By similar manner, one may verify that F is additive, too. This completes the proof • The inverse theorem can be proved in a more general setting. Namely, T h e o r e m 11.2 Let G\ be a group and G2 an abelian group and let K,E,F: Gi —>• G2 be such that K is quadratic and E,F are additive. If, moreover Si, S2, S3, S4 G G2 are constants with Si + S2 = S3 + S4, then the functions A(x) B(x) C(x) D(x)

:= := := :=

2K(x) 2K(x) K(x) K{x)

+ E(x) + E{x) + E{x) + F(x)

+ F(x) + Sx, - F(x) + S2, + S3, + 54,

satisfy the equation (11.2). Proof. Starightforward verification.

•

As we announced, we are going to consider the quadratic functional equation on an interval of real line. Let I = [a, a + b] for some a, b G K, b > 0. Obviously, 2 7 = [2a,2a+ 2 6 ] , / - / = [-&,&]. We start by introducing lemma on extension of a solution of the quadratic equation on / to the solution on J = [a — | , a + b\.

Quadratic equation on an interval

117

Lemma 11.1 (left extension lemma) Let G be an abelian group with division by two. Let A: I —> G, B: 21 —> G, C: I — I—¥ G be such that A(x)+A(y)

= B(x + y) + C(x-y),

x,yel.

Then there exist unique functions A[:

b a— -,a + b -)• G,

C[:

B[: [2a -b,2a + 2b] ->• G,

2 '2

such that A'1(x)+A'1{y)

= B[(x + y) + C'l{x-y)

for x,y e a — -,a + b Z>

andA'.lr = A, B[\2I = B, C{| 7 _ 7 = C. Proof. The proof, rather tedious and long, may be found in [47]

Lemma 11.2 (right extension lemma) Under the assumptions of Lemma 11.1 there exist unique functions

A[

a,a+

-b -> G,

B[: [2a,2a + 36] ->• G

ZJ

3 2

z

3' -+G, 2

z

such that A'^x) + A\{y) = B'1(x + y) + C[(x - y) andA'1\I = A,B[\2I

= B,C[\I_I

= C.

for x, y G

a,a + -b Li

118

Functional Equations and Inequalities in Several Variables

Proof. See [47]

•

Now we are in a position to prove the extension theorem. Theorem 11.3 (extension theorem) Let G be an abelian group with division by two. Let A: I —>• G, B: 21 —> G, C: I — / —>• G be such that A(x) + A{y) = B(x + y) + C(x - y) Then there exist unique functions

AQ,BO,CO-

A0(x) + A0(y) = BQ(x + y) + C0(x - y) and A0\i = A, B0\2i = B,

C0\I-I

for

x,yel.

K —> G such that for x, y e R

— C.

Proof. We shall consider two cases. I), /-finite interval. We define the sequences of functions {Ai}> {Bn}, {Cn} by the folowing way Ai+i = An,

Bn+i = Bn,

Cn+i = Cn,

where Ai - A, B1 = B, Ci — C and A'n, B'n, C'n for add n are right extensions of An, Bn, Cn (Lemma 11.2) and for even n left extensions of An: Bn, Cn (Lemma 11.1). Clearly, the functions A)Or) := An(x), BQ(X) := Bn(x), CQ(X) := Cn(x),

n = min{fc G N: Ak(x) exists} n = min{A; E N: Bk(x) exists} n = min{A; G N: Ck(x) exists}

are defined on K and satisfy all the required conditions. II). /-infinite interval. In this case we may consider any finite subinterval of / and repeat the proof presented for the case I). This concludes the proof. • As an immediate consequence of the obove theorem we obtain the result about the form of solutions of the quadratic equation on an interval.

Quadratic equation on an interval

119

Theorem 11.4 Let G be an abelian group with division by two and A: J -> G, B:2I -±G,C: I - I-> G be functions. The following conditions are equivalent: (a) A(x) + A(y) = B(x + y) + C(x - y)

for x, y G I;

(b) there exist a quadratic function K: R —>• G, an additive function E: R —» G and constants Si, S2, S3, S4 such that 2Si = 02 + 03

A(x) = 2K(x) + E(x) + SU xel, B(x) = K(x) + E(x) + S2, xe 27, C(x)= K(x)+S3, xel-I. Proof. We may apply Theorem 11.3 and Theorem 11.1 with F = 0 since A = B. •

Notes 11.1 Following Aczel [4], we state the following results concerning the Cauchy equation on an interval. Proposition 11.1 Let J C I C M be intervals such that J + and let f: I —>• R satisfy the equation f(x + y) = f{x) + f(y)

forx,yeJ.

Jcl

(11.10)

Then there exist an additive function g: R —> R and a constant a € R such that

[X)

~

\g(x) + 2a

forxeJ+J.

120

Functional Equations and Inequalities in Several Variables

Proposition 11.2 Let J C I C R be intervals such that J+J C I and let f: I —>• R satisfy the equation (11.10). If J D3 J ^ 0, then there exists an additive function g: R —>• R such that f{x) = g{x)

forxeJU{J

+ J).

Proposition 11.3 Let J c R be an interval such that 0 £ c/7, and let I = J + J = 2J. If a function /:/—>• K satisfies equation (11.10), i/ien / can be uniquely extended onto R to an additive function. 11.2 For some more details about the Cauchy equation on the so called restricted domain see e.g. Daroczy-Losonczi [42], Lajko [130], Kuczma [124], Ger [70], Dhombres [44].

Chapter 12 Functional equations for quadratic differences Let X and Y be groups and let / : X —> Y be a function. We define Qf, the quadratic difference of / , by the formula

Qf(x,y) = f(x + y) + f(x-y)-2f(x)-2f(y),

x,y e X. (12.1)

Following an idea of S. Czerwik [35], we get Lemma 12.1 Let X and Y be abelian groups and let f: X -» Y be an even function. Then Qf, given by the formula (12.1), satisfies the following functional equation Qf(x + z,y) + Qf(z -x,y) = Qf(z, x + y) + Qf{z, x-y) for all x,y,z

+ 2Qf(x,z) = + 2Qf(x, y)

[

' '

G X.

Proof. Using the definition of Qf and calculating left and right hand sides of the equation (12.2), we get easily the desired conclusion. •

121

122

Functional Equations and Inequalities in Several Variables

Remark 12.1 Let X and Y be abelian groups and f: X —» Y be a function. Put F(x):=f(x)

+ f(-x),

xeX.

Then F is even and QF(x, y) = Qf(x, y) + Qf(-x,

-y).

Moreover, QF satisfies the equation (12.2), i.e. we get the equation for quadratic differences for any function (not necessarily even). Let X, Y be normed spaces. By Dn(X, Y) we denote the space of all functions / : X —> Y that are n-times differentiable on X. By <9fc/, k — 1,2, we denote, as usual, the n-th partial derivative of / : X x X —>• Y with respect to the k-th variable. Lemma 12.2 Let X, Y be normed spaces and let f': X —>• F be an even function. Then for all x, y £ X: Qf(x,y)=Qf(y,x),

(12.3)

Qf(~x,y)=Qf(y,x).

(12.4)

// moreover, Qf G D2(X x X, Y), then for all x, y G X, dl(Qf)(x,y) d l(Qf)(-x,y) 2

= di(Qf)(x,y), = d22(Qf)(x,y).

(12.5) (12.6)

Proof. The properties (12.3) and (12.4) one can get directly form definition of Qf and the fact that / is even. Properties (12.5) and (12.6) are the immadiate consequance of (12.3) and (12.4) respectively. • Proposition 12.1 Let X,Y be normed spaces and let f: X —>• Y be even. If Qf e D2(X x X, Y), then for all x,yeX, 2dUQf)(x,y) 2

= d l(Qf)(0,x

+

2dUQf)(0,x)

+ y) + d2(Qf)(Q,x-y),

1d2{Qf){x,y) + 2dl{Qf)^y) = d2(Qf)(0,x + y) + d2(Qf)(0,x-y).

{

•<)

.

. *>

[

Functional equations for quadratic differences

123

Proof. Differentiating two times both sides of the equality (12.3) with respect to z at the point z = 0, we obtain d2(Qf)(x,y) = 2d2(Qf)(0,x

+ d2(Qf)(-x,y) + 2d2(Qf)(0,x) + y) + 2d2(Qf)(0,x-y).

Hence in view of (12.6), we get (12.7). We may rewrite the equation (12.2) in the following form Qf{x, y + z) + Qf(x, z-y) + 2Qf(z, x) = Qf{z, x + y) + Qf(z,xy) + 2Qf(x, y). Consequenlty, differentiating two times both sides of the above equality with respect to z at the point z — 0. we obtain d22Qf(x, y) + d2Qf(x, -y) + 20?Q/(O, y) = d2Qf(0,x + y) + d2Qf(0,x-y). Hence, in view of (12.5) and (12.6) we obtain the equation (12.8)

Proposition 12.2 Let X,Y be normed spaces and let f: X —>• Y be an even function. If, moreover, f G D2(X,Y), then for all x,y EX 2d212(Qf)(x, y) = d2(Qf)(0,x

+ y)-

d2(Qf)(0,x-y).

(12.9)

Proof. Since / is even, one has D2f{-x)

= D2f(x)

for

xeX.

Thus, calculating all derivatives occuring in (11.9) we may verify that our statement holds true. • Note that the equation (12.7) may be rewritten in the following form 2F(x, y) + 2F(0, x) = F(0, x + y) + F(0, x - y).

(12.10)

124

Functional Equations and Inequalities in Several Variables

Proposition 12.3 Let X be a group and Y be an abelian group with division by two. A function F: X x X —>• Y is a solution of the equation (12.10) iff there exists an even function f: X —>• Y such that F(x, y) = Qf(x,y)

+ 2f(y),

x,yeX.

(12.11)

Proof. Clearly, if F is a function given by the formula (12.11) where / is an even function, then it satisfies the equation (12.10). Conversely, assume that F is a solution of the equation (12.10). Putting into (12.10) y = 0, one gets 2F(x, 0) = 0,

for all x G X.

Thus for x = 0, from (12.10) we get F(0, y) = F(0, -y),

for all x <E X.

(12.12)

Define / : X -> Y by F(0,y) = 2/(y), then on account of (12.12) it follows that / is an even function. Moreover, by (12.10) one gets 2F(x, y) + 4/(a;) = 2f(x + y) + 2f(x - y), whence F(x, y) = f(x + y) + f(x - y) - 2fix) = Qf(x, y) + 2f(y) for all x,y G X, i.e. the formula (12.11), which completes the proof. • Remark 12.2 Equation (12.2) may be written in the form F(x + z,y) + F(z-x,y) = F(z,x + y) + F(z,x-y)

+ 2F(x,z) + 2F{x,y).

[

'

j

IfX, Y are abelian groups and f: X —>• Y is an even function, then F(x,y) = Qf{x,y),

x,y e X

is a solution of the equation (12.13). The question is what is the general solution of this equation.

Functional equations for quadratic differences

125

Notes 12.1 Let us note that the Cauchy difference Cf of a function

Cf(x,y)

:= f{x + y)-

f(x) - f(y),

x,y E X,

satisfies the following functional equation Cf(u, v) + Cf{x, u + v)= Cf{x + u,v) + Cfix, y)

(12.14)

for all u, v G X. 12.2 For some more information see Czerwik-Diutek [28].

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Part II Ulam-Hyers-Rassias Stability of Functional Equations

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Chapter 13 Additive Cauchy equation Very often instead of a functional equation, we consider a functional inequality and one can ask the following question: when can one assert that the solutions of the inequality lie near to the solutions of the equation? The problem of this kind had been formulated by S. M. Ulam [212] during his talk before the Mathematics Club of the University of Wisconsis, in 1940. Let G± be a group and let G2 be a metric group with the metric d. Given e > 0, does there exist a 5 > 0 such that if / : Gi —> G2 satisfies the inequality d[f(xy),f{x)f(y)]<6

for all x,y e Gu

then there exists a homomorphism H: Gi —>• G2 with d [f(x), H(x)] < e

for all x G d ?

In 1941 D. H. Hyers [100] has given the following answer to Ulam's question. Theorem 13.1 Let j ' : E\ —>• E2 be a mapping between Banach spaces Ei,E2 such that \\f(x + y)-f(x)-f(y)\\<e

for all x,y e Elt 129

(13.1)

130

Functional Equations and Inequalities in Several Variables

for some e > 0. Then there exists exactly one additive mapping A: El -^E2: A(x + y) = A(x) + A(y)

for all x, y G Ex

such that \\A{x) - f(x)\\ < e

for all x G Eu

(13.2)

given by the formula A(x) = lim 2'nf (2nx), xeEi.

(13.3)

Moreover, if f(tx) is continuous in t for each fixed x G E\, then A is linear. The Ulam's question and the Hyers' theorem became a source of the stability theory in the Ulam-Hyers sense. The most important results and the wide list of references concerning this kind of stability are included in the survey papers [54], [69], [95], [176] as well as in the book by D.H.Hyers,G.Isac and Th.M.Rassias [74]. In 1978, Th. M. Rassias [178] considerably weakened the condition for the bound of the norm of the Cauchy difference f(x +

y)-f(x)-f(y)

and obtained a very interesting result. Namely, he has proved the following Theorem 13.2 Let Ei,E2 satisfies the inequality \\f(x + y)-f(x)-f(y)\\
be Banach spaces. If f: E\ —¥ E2 + \\y\\')1

x,y G Eu

(13.4)

for some Q > 0 and some 0 E2 such that

ll/W-^)ll<2^ll*

*e£i-

(13.5)

If, moreover, f(tx) is continuous in t for each fixed x G E\, then the mapping A is linear.

Additive Cauchy equation

131

This result provided a remarkable generalization of the Theorem proved by Hyers. What is more important here is that Th. M. Rassias Theorem stimulated several mathematicians working in functional equations to investigate this kind of stability for many important functional equations. Taking this fact into consideration, the Ulam-Hyers stability of such type is therefore called the Ulam-Hyers-Rassias stability (breafly UH-R stability). During the last decade several results for the Ulam-Hyers-Rassias stability of very many functional equations have been proved by several researchers. Our goal is to present some of the most important developments in this area. The first result is the following Theorem 13.3 ([£#]). Let (G,+) be an abelian group, k an integer, k > 2 (X,\\ • ||) a Banach space, cp: G x G —> [0,oo) a mapping such that oo

for all

x,yeG.

n=0

(13.6) Let f:G-*X

satisfy the inequality

\\f(x + y)-f(x)-f(y)\\<
for all x,y G G. (13.7)

There exists exactly one additive mapping A: G -> X such that it-i

\\f(x) -A(x)\\

xeG.

(13.8)

m=l

Proof. First of all we will prove the inequality k-i

\\f(kx)-kf(x)\\
xeG,

keK

n=l

Setting y = x in (13.7) we get | | / ( 2 z ) - 2/(rc)|| <
xeG,

(13.9)

132

Functional Equations and Inequalities in Several Variables

i. e. the relation (13.9) for k = 2. Suppose now (13.9) is true for some k > 1 and let us prove it for k + 1. From (13.7) for y = kx, we obtain

\\f((k + i)x) - f[x) - f{kx)\\ <
f(kx)\\

k-i

+ \\f(kx) - kf(x)\\ < nx) 71=1

k

= ^2
which means (by the induction principle) that (13.9) holds true for every k £ N (for k = 1 the inequality is trivial). From (13.9) dividing by k one gets fc-i

Wk^fikx)

- f(x)\\ < Y^k~l
(13.10)

n=l

We claim that k—1 m—\ m

m

\\k- f(k x)

k {s+1)

- f(x)\\ <J2Y1

~

v(ksx,

nksx),

(13.11)

n = l s=0

for all x e G, m € N. In view of (13.10), the equality (13.11) is true for m — 1 . Assume that (13.11) is true for some m > 1. Substituting kx by x into (13.11) implies k—1 m—1 m

m

||AT /(A; fca;) -/(fca;)|| < ^

^

n = l *=0

k-^s+1^(ks+1x,nks+1x).

Additive Cauchy equation

133

Dividing this inequality by k one has, fc-1 m - 1 m+1)

\\k^

m+1

f(k

x)

- fc-7(fcc)|| < Y

k~{s+2) ip{ks+l x,

Y

n=l

nks+1x).

s=Q

Now we obtain for x G G \\k-{m+1)f(km+lx)-f{x)\\

< \\k-(m+Vf(km+1x) - k-1/^

+ Wk^fikx) - f(x)\\

fc—1 m—1

< Y

fc—1 {s+2)

Y

k-

p(k

s+1

x,

s+1

nk

x) + Y fc~ V f o nx)

n=l s=0 fc—1 m

n=l

YYk~(S+1)V(kSx,nkSx)-

= n-l

s=0

By induction, (13.11) holds for all m E K Consider the sequence that was Th. M. Rassias[175] := k~mf (kmx),

fm(x)

xeG,

introduced

meK

by

(13.12)

This sequence we call a Rassias' sequence. We will show that the sequence {fm{x)} is a Cauchy sequence. Take m > r, then we have ||ATm/(£;m:r) - k~rf(krx)\\

= k~r\\kr-m f{km-rkTx)

- /(fc r x)||

fc—1 TO—r—1

r

Y

k-{s+1)
S= 0

fc—1 771—1

- Y, Y, k~^+^ip(kBx, nksx) ->• 0

as r -> oo.

n = l s=r

Thus {/mOs)} is a Cauchy sequence and since X is a Banach space there exists A: G —> X such that lim fm(x) = A(x),

x e G.

134

Functional Equations and Inequalities in Several Variables

The mapping A is an additive mapping. Indeed, in view of (13.7) we get for x,y 6 G \\k-mf(km{x+y))-k-mf(kmx)-k-mf{kmy)\\

< r>(n,k

m

y).

From (13.6) it follows that lim k-mip(kmx,

kmy) = 0

m—>oo

and therefore taking the limit of both sides of the last inequality we obtain A(x + y) - A(x) - A(y) = 0

for

x,yeG,

i. e. A is an additive mapping. To verity that (13.8) holds true, one can take limit a s m - ^ o o in (13.11). Then we get for x e G fc-l

oo

fc-1

71=1

71 = 1 S = 0

To prove the uniqueness of the mapping A assume that there exist two additive mappings As: G —*• X, s = 1,2 with the property k-l

\\f(x)-As{x)\\<^2^k(x,mx)

5 = 1,2,

x(EG.

m=l

We get for x e G IIAxCx) - A2(x)\\ = Wk^A^x) k-mA2(kmx)\\ < k-m (WA^x) - f(kmx)\\ + \\f(kmx) - A2(kmx)\\) fc-i

2k-mJ2^k{kmx,nkmx)

<

71 = 1

fc—1

OO

= 2 Y^ Yl 7i=l

s=m

k {s+1)(

~

P(ksx,

nksx).

Additive Cauchy equation

135

Hence lim ||Ai(x) — A 2 (x)|| = 0 for x G G and proves that Ai(x) = A2(x) for all x € G. This concludes the proof of the theorem. • Remark 13.1 Take 0, p < 1, then one has

0 <

oo

**(*, y) = Y, k-{n+1)9knr (\\x\\p + \\y\n n=0

= k-10(\\z\\> +

\\y\\')1^.

Therefore hf)

Mx,nx)

1

= ]—rp.-\\x\ni

+ nn,

and hence k

l

kQ

,

k-kP"

" A; .

n=l

/

\

kd A;-feP

*

l

ra=2

^,p)dkin,

it-i

where s(k,p) = 1 + 1k ^ n p . TTiis implies the result proved in [175] n-2

by Th. M. Rassias. Remark 13.2 In 1991 Professor Th. M. Rassias asked what is the best possible value of k in the estimation (13.8). We will show (see [63]) that for the function cp considered in Remark 13.1 this value is 2. Set

«M-jh-

Q{k p)

' = T=4'

k>2

-

We claim that Rip) < Q(k,p)

for k>3.

(13.13)

136

Functional Equations and Inequalities in Several Variables

We will apply mathematical induction on k. For k = 3 we get easily 2 . 3? — 2P — Ap

Q(3,P)-fl(ri=(2_2p),(3_3P)>0 in view of the Jensen inequality for the concave f: (0,oo)->R, / ( i ) = ^ p e [ 0 , l ) : ( ^ ^ )

P >

^ Y ^ '

^^€(0,oo)

function

(13.14)

applied for x\ — 2 , X2 — 4. By simple calculation we obtain Q{k + l,p)-

k + 1_{k

+

1)p-

Assume that (13.13) holds true for some k > 3. Then we get Q(k +

l,p)-R(p)>

R(p){k-kp) k + l-(k 2(k + l)p (2 -2P)[k +

+ kp + l 2 p + l) 2 - 2P p p p 2 - 2 •k l-(k + l)p]'

Applying again the inequality (13.14) with x\ = 2k, x^ = 2, we ramark that the last expression is possitive. This concludes the proof of (13.13) for every k > 2. Remark 13.3 For the function

Additive Cauchy equation

137

Theorem 13.4 Let X be a real Banach space and let Y be a real normed linear space. Suppose that ip: X —> R is a nonnegative subadditive functional on X and f: X —>• Y is a mapping such that 11/(3 + y) - f(X) _ f(y)\\ < ^X)

+ ply) _ ^

+

y)

(13.15)

ZioWs irae for all x,y E X. If, moreover, f and

+ y(x),

XEX.

Another result of this kind is the following. Theorem 13.5 ([190]). Let 9 > 0 and assume that a continuous mapping f: R —> R satisfies <0^|:zn|, \n=l

)

n=\

xi,...,xm

G R, (13.16)

n=l

for all m G N. Then there exists an additive mapping A: R —>• R such that \f(x)-A(x)\<0\x\, xeR. A far-reaching result for the modified Ulam-Hyers-Rassias stability of the additive Cauchy equation has been achieved by G. Isac and Th. M. Rassias in [101]. They introduced the following definition. Definition 13.1 A mapping f:E\-+ E2, where Ei,E2 are two normed spaces, is <£>-additive iff there exist 6 > 0 and a mapping ip: R+ —> R+ such that l i m ^ = 0 4->00

t

and \\f(x + y)-

f(x) - f(y)\\ < 9 M\\x\\) +
138

Functional Equations and Inequalities in Several Variables

By making use of this notion, G. Isac and Th. M. Rassias obtained in [101] the following result. Theorem 13.6 Let Ei,E2 be a real normed space and a real Banach space, respectively. Let f: E\ —± E2 be a mapping such that f(tx) is continuous in t e l for each fixed x G E±. If, moreover, f is ^-additive and (p{st)<(p(s)
s.teRf,

and cp(s) < s,

s > 1,

then there exists exactly one linear mapping A: Ei —>• E2 such that 9f) Wttx)-AW\\<j=^
xeEr.

For further information about results of this kind, one may see for example [49], [54], [52], [6], [195], [20], [53], [179], [104], [110], [111], [113], [211]..

Notes 13.1 The historical background and important results for the Ulam-Hyers-Rassias stability of various functional equations are surveyed in the expository paper of Soon-Mo Jung [105]. 13.2 In 1991 Z. Gajda has extended the Rassias' Theorem 13.2 for the case p > 1 by a slight modification of the formula (13.3) by taking A(x) = lim 2nf(2-nx). n-»oo

For details see [61]. 13.3 Th. M. Rassias and P. Semrl [173] proved the following generalization of Hyers stability result.

Additive Cauchy equation

139

Proposition 13.1 Let H: R^. —>• R+ be a positively homogenous mapping of degree p with p / 1. Given a normed space E\ and a Banach space E-i, assume that f: E\ —> Ei satisfies the inequality \\f(x + y)-

f(x) - f(y)\\ < H(\\x\\, \\y\\),

x,y € JE?i-

Then there exists a unique additive mapping A: E\ —)• E2 such that \\f(x) - A(x)\\ < H(l, 1)|2 - 2p\-1\\x\\p,

x e Ex.

Moreover, [ lim 2-nf(2nx) K

'

n

n

I lim 2 f(2~ x)

forp

1.

13.3 For the stability of a generalized Cauchy equation of the form f(ax + by) = Af{x) + Bf{y) + L(x, y) one can refer to the papers [21], [8], [108].

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Chapter 14 Multiplicative Cauchy equation Let (G, •) be a semigroup and E a normed algebra with multiplicative norm, i.e. \\a-b\\ = ||a|| • \\b\\ for a,b G E. A mapping / : G —> E is said to be multiplicative iff / satisfies the equation (multiplicative Cauchy equation) f(x-y)

= f(x)f(y),

x,yeG.

(14.1)

In 1980 J. Baker [10] proved the following result. Theorem 14.1 Let G be a semigroup and E a normed algebra with the multiplicative norm. Given 5 > 0, let f: G —¥ E satisfy \\f(x-y)-f(x)f(y)\\<8,

x,yeG.

(14.2)

Then either \\f(x)\\ < | (1 + (1 + 45)2 J or f is multiplicative. We say that in this case we have the super stability. J. Baker observed in the same paper that the assumption that the norm is multiplicative is essential to prove the result. He presented the following example.

141

142

Functional Equations and Inequalities in Several Variables

Example 14.1 Given 5 > 0, let e be such that \e — e2\ = 8. By M2 (C) we denote the space of all 2 x 2 complex matrices with the usual norm. Let f: R —> M2 (C) be given by

m

ex

0

0

e

xG

It can be easily verified that f is an unbounded function satisfying the equation \\f(x + y)-f(x)f(y)\\

= 8,

x,y£R.

But clearly, f is not a multiplicative function. In the case when the range space is M2 (C) one can expect only the Hyers-Ulam stability instead of the superstability phenomen for the multiplicative Cauchy equation. In fact, J. Lawrence in the paper [131] has proved the following theorem. Theorem 14.2 / / G is an abelian group and a mapping f: G —> M2 (C) satisfies the inequality (1\.2) for some S > 0 ; then there exists a multiplicative mapping m: G —> M2 (C) such that f — m is bounded. A similar result for Mn (C) was proved by D. Dicks [45]. Very interesting results concerning the superstability of the mapping / one can find in the paper [76] of R. Ger, [67] of R. Ger and P. Semrl. Now we will present the result of P. Gavruta [64], who has given an answer to a problem posed by Th. M. Rassias and J. Tabor [6] concerning mixed stability of mappings. Theorem 14.3 Let e,s > 0 and 5 = 2s + (22s + 8e)^ / 2. Let B be a normed algebra with multiplicative norm and X be a real normed space. If f': X —>• B satisfies the inequality l|/(^ + 2 / ) - / ( ^ ) / ( 2 / ) | | < ^ ( N s + ||2/||s)

(14.3)

Multiplicative Cauchy equation for all

143

x,y E X, then either \\f(x)\\ < 5\\x\\s

for all xeX

with

\\x\\ > 1,

or f(x + y) = f(x)f(y)

(14.4)

for all x,y € X. Proof. Assume that there exists an x0 € X, \\x0\\ > 1 such that ||/(x 0 )|| > 5||xo||*. Hence there exists an a > 0 such that \\f(xQ)\\>(S

+

a)\\x0\\s.

From (14.3) we get 11/(2^0) - / ( x 0 ) 2 | | < 2 e | | ( r r 0 | | s and therefore ||/(2x0)||>||/(x0)||2-||/(2x0)-/(xo)2|| > (5 + a)2\\x0\\2s - 2e\\x0\\s a)2-2e]\\xQ\\s.

> [(5 +

From the definition of 8 it follows that 52 = 2SS + 2e

and 5 > 2s,

and consequently | | / ( 2 x 0 ) | | > ( 5 + 2a)2s||a;o||s.

(14.5)

We claim that \\f(2nx0)\\>(5

+ 2na)\\2nx0\\s

for n G N.

Indeed, by (14.3) we obtain ||/(2"+1a;o)-/(2^o)2||<2e||2"x0|r,

(14.6)

144

Functional Equations and Inequalities in Several Variables

and applying the inequality (14.6) we deduce that ||/(2" +1 *o)|| > \\f(2nx0)\\2 - \\f(2n+lx0) - f(2nx0)2\\ > (6 + 2 n a) 2 ||2 n :ro|| 2s - 2e||2 n x 0 || s > [{5 + 2naf

- 2e] \\2nx0\\s.

Finally we get \\f(2n+1x0)\\>(5

+

2n+1a)2s\\2nx0\\s,

which on account of induction principle, proves the inequality (14.6). Denote xn — 2nx0, then ||a;„|| > 1 for all n £ N and in view of (14.6)

lim .J*"11' , = 0.

(14.7)

n-oo||/(2™X 0 )||

Take x,y £ X and z £ X such that f(z) ^ 0. Applying (14.3), we obtain \\f(z)f(x + y)-f(x + y + z)\\<e (\\z\\s + \\x + y\\s), ||/(a; + y + z) - f(x)f(y + z)\\ < e (\\x\\s + \\y + z\\s), whence \\f(z)f(x

+ y)-f(x)f(y

+ z)\\

<e{\\z\\' + \\x\\' + \\x + y\\' + \\y + z\\').

^ l

"}

From (14.3) we get also \\f(x)f(y

+ z)-

f(x)f(y)f(z)\\

< e \\f{x)\\ {\\y\\* + IM| S ),

which together with (14.8) leads to the inequality \\f(z)f(y

+ z)-

f(x)f(y)f(z)\\

< e u(x, y, z)

(14.9)

where u(x,y,z)

:= llzll' + llary + llar+yll' + lly+^l' + ll/Cx)!! (||y||s + \\z\\').

Multiplicative Cauchy equation

145

Using the multiplicativity of the norm in B, we get from (14.9)

y)-f(x)f(y)\\<£-^j^.

\\f(x +

Let z = xn, then from (14.7) it follows u(x,y,xn) n->oo

||/(X„)||

and consequently we obtain the equality

f(x + y) = f{x)f{y), which completes the proof of the theorem.

•

We end this section with the following general result of L. Szekelyhidi [206]. Theorem 14.4 Let G be a semigroup and V be a right invariant vector space of complex-valued mappings on G. If f,m: G —> C are mappings such that the mapping f(xy) -

f(x)m(y)

in variable x belongs to V for each y e G, then either f G V or m is multiplicative. Proof. Suppose that m is non-multiplicative, thus there exist y,z € G such that m(yz) — m(y)m(z) / 0. We have the relation for all x G G: f(xyz) - f(xy)m(z)

= [f(xyz) - f(x)m(yz)] - m(z) [f(xy) - f(x)m(y)] + f(x) [m{yz) - m{y)m{z)}.

146

Functional Equations and Inequalities in Several Variables

Therefore, f{x) = [ [f{xyz) - f(xy)m(z)}

- [f{xyz) -

+ m(z) [f(xy) - f(x)m(y)}}

f{x)m{yz)}

• [m(yz) -

m{y)m(z)}~1.

But the right-hand side in the above relation as a mapping of x belongs to V, so also f €V. •

Notes 14.1 R. Ger proposed that it is more natural to pose the problem of stability of multiplicative Cauchy equation, instead of (14.2), in the following way

f{xy)

<e,

mm

x,yeG,

(14.10)

for mappings / : G -» C \ {0}. In [76] he proved the following result: Proposition 14.1 Let (G,+) be an amenable semigroup. If a mapping f: G —> C \ {0} satisfies (14.10) for some 0 < e < 1, then there exists a multiplicative mapping m: G —> C \ {0} such that m(x)

m m m(x)

< z , l- £ 2-e <

xeG, X

eG.

Let us note that the estimation j ^ - does not tend to zero as e tends to zero. Recently, R. Ger and R. Semrl improved this result (see [67].

Chapter 15 Jensen's functional equation An interesting result for the stability of Jensen's functional equation was obtained by Z. Kominek [118]. His result reads as follows. Theorem 15.1 Let D be a subset o/R™ with non-empty interior and Y be a real Banach space. Assume that there exists an XQ in the interior of D such that D0 = D — XQ fulfills the condition | c D o . Let a mapping f: D —)• Y satisfy the inequality

\\2f(^p^-f(x)-f(y)\\<5, for some 5 > 0 and for all x,y € D. Then there exist a mapping F: Rn —> Y and a constant K > 0 such that

2F

1 =F{x)+F{y)

{H )

for all x,y e K" and \\f(x)-F(x)\\
xeD.

Now we shall present some results of S. M. Jung concerning the Ulam-Hyers-Rassias stability of Jensen's equation and its application to the study of asymptotic behavior of the additive mapping ( see [106]). 147

148

Functional Equations and Inequalities in Several Variables

T h e o r e m 15.2 Let p > 0, p ^ 1, 5 > 0, 6 > 0, be given. Let X and Y be a real normed space and a real Banach space, respectively. Assume that f:X^Y satisfies the inequality ||2/ ( ^ )

- f{x) - f(y)\\ <S + 9 (||*||* + \\y\\>)

(15.1)

for all x, y € X. Further, assume /(0) = 0 and S — 0 in (15.1) for the case of p > 1. Then there exists exactly one additive mapping F: X -+ F such that \\f(x) - F(x)\\ < 6+ ||/(0)|| + ^JL—

(forp
\\X\\P

(15.2)

or op-l

\\tt*)-F(x)\\
(f°rP>1)

( 15 - 3 )

Proof. First, we shall prove the inequality ||2-"/(2 B s) - f(x)\\ <(S+ ||/(0)||) j ^ 2 ~ k + fc=i fe=i

e

WXWP E 2 ~ { l ~ P ) k (15.4)

for all x E X. Set y = 0 in (15.1), then l | 2 / ( | ) - / ( x ) | | < *+11/(0)11 + 011*11',

(15.5)

and substituting 2x for x and dividing by 2 both sides of the last inequality, we get the inequality (15.4) for n = 1. Assume now that (15.4) holds true for some n € N. Substituting 2n+1x for x in (15.5) and dividing by 2 both sides of (15.5), we get, in view of (15.4): ||2-(»+ 1 )/(2 n + 1 a;)-/(a;)|| < 2-"||2- 1 /(2 n + 1 *) - / ( 2 V ) | | + ||2""/(2"x) n+l n+1 < (*+ll/(0)||) E 2-fc + 0||:r|p> J2 2-{1-p)kfc=i fc=i

f(x)\\

Jensen's functional equation

149

This proves the inequality (15.4).

Define Fn(x) := 2~nf(2nx),

xeX,

n G N.

Take n > m, then by (15.4) we get for x G X \\Fn(x) - Fm(x)\\ = \\2-nf(2nx) - 2~mf(2mx)\\ = 2-" l ||2-( n -" l )/(2 n - m • 2mx) - f(2mx)\\ < 2~m (5+ ||/(0)|| + p p ^ ^ P H 0, as m - ^ o o . This means that the sequence {Fn(x)} is a Cauchy sequence for all x G X (note that Y is a Banach space). Let F(x) := lim Fn(x),

x e X.

n—¥oo

Take x,y eX.

\\F(x +

From (15.1) it follows

y)-F(x)-F(y)\\

= lim 2-("+ 1 )||2/

(*±£±K1)

_ /(2-+1Z) - /(2"+ 1 y)||

< lim 2-("+1) [5 + 02( n+1 > (||x||p + ||y||p)l = 0, i.e. F is an additive function. From the inequality (15.4) we get directly the estimation (15.2). To prove the uniquennes in the case p < 1, assume that G: X —y Y is an additive mapping satisfying the inequality (15.2). Then one gets \\F{x) - G{x)\\ = 2-n\\F{2nx) - G(2nx)\\ < 2~n (\\F(2nx) - f(2nx)\\ + \\f{2nx) - G(2"x)||) < 2-" (26 + 2||/(0) || + 202"? (21~P - I ) - 1 \\x\\A ->> 0,

as n ->• oo.

Therefore F(x) = G(x) for all x G X. If p > 1 and <5 = 0, then instead of (15.4) one can similarly prove the inequality n-l B

||2 /(2-"x) - f(x)\\ < 9 \\x\\* X > ( p - 1 ) f c , k=Q

x GX

150

Functional Equations and Inequalities in Several Variables

and the rest of the proof goes through in the similar way.

•

Remark 15.1 Similarly one can prove the following statement. If f:X—tY satisfies the inequality (15.1) with 0 = 0 (or p = 0), then there exists exactly one additive mapping F: X —» Y such that (15.2) holds true with 0 = 0. Remark 15.2 For the case p <0, the method used in the proof of Theorem 15.2 can not be applied, thus this is still an open problem. Remark 15.3 If p = 1, Theorem 15.2 is not true as can be seen from the following example (see also [174])Example 15.1 Define _ \ x log2(x + 1) for x>0, [ x log2 |rrr — 11 for x < 0. Then it is not difficult to verify that

2/ ( ^ ) - f(x) - f(y)' <2(M + |y|) for all x, y G R and the expression \f(x)-A(x)\/x

for

x^O

is unbounded for each additive mapping A: R —> R. In the following we present a result on the stablity of the Jensen's functional equation on a restricted domain (see [106]). Theorem 15.3 Let d > 0 and 5 > 0 be given real numbers. Suppose that f: X —> Y (X,Y-as in Theorem (15.2)) satisfies the inequality \\V(^)-m-f(y)\\<S

(15.6)

for all x, y 6 X with \\x\\ + \\y\\ > d. Then there exists exactly one additive mapping F: X —>Y such that \\f(x)-F(x)\\<56+\\f(0)l

*eX

Jensen's functional equation

151

Proof. One can prove that Pf(^-)-f(x)-f(y)\\<5S

for all x,y e X.

Hence in view of Theorem 15.2 we get the conclusion.

•

The last theorem can be applied to prove a result on asymptotic behavior of the additive mappings. Theorem 15.4 Let X,Y be as in Theorem 15.2 and / : X —>• Y satisfy the condition /(0) = 0. Then f is additive iff l|2/(£y^)-/(x)-/(y)|H0

as \\x\\ + \\y\\ -+ oo.

(15.7)

Proof. Assume that the condition (15.7) is satisfied. Then there exists a sequence {Sn} monotonically decreasing to zero such that

Pf(^)-f(x)-f(y)\\<5n for all x, y € X with ||x|| + ||y|| > n. Therefore from Theorem 15.3 it follows that there exists exactly one additive mapping Fn: X -> Y such that \\f(x)-Fn(x)\\<55n, xeX.

(15.8)

But uniquennes of Fn implies that Fm = Fn+k,

for A; = 1,2, • • - .

Therefore, by letting n —> oo in (15.8), we derive that / is an additive function. The proof of the reverse part of the theorem is trivial. •

152

Functional Equations and Inequalities in Several Variables

Notes 15.1 For convex functions the problem is more complicated and we do not have a full analogue of Hyers theorem. Let D C Rn be a convex and open set. A function / : D —>• R is called e-convex iff /[Ax + (1 - X)y] < Xf(x) + (1 - X)f(y) + e for every x,y G D and A G [0,1]. Proposition 15.1 ([99], [24])- Let D e R n be a convex and open set and let f: D —> R be an e-convex function. Then there exists a continuous and convex function g: D —> R such that

for all x G D. For further information see Cholewa [24] and the book by D. H. Hyers, G. Isac and Th. M. Rassias [74].

Chapter 16 Pexider's functional equation First we establish some notations. Let (5, +) be a semigroup and Y a topological vector space over the field of all rational numbers. A set-valued function F: S —> 2Y is said to be subadditive iff

F(s + t)cF(s)

+ F(t),

t,seS,

where „+" on the right side means the algebraic sum of sets. Let A C Y be a bounded set and U C Y a neighbourhood of zero. Then we define diam^ A := inf {q G Q n (0, oo): A - A C qll} . By seqcl A we denote the sequential closure of A. The following Lemma is due to Gajda and Ger [57]. Lemma 16.1 Let (S,+) be an abelian semigroup and let Y be sequentially complete topological vector space over Q. Assume that F: S —>• 2 y \ { 0 } is a subadditive set-valued function such that F(s) is Q-convex for all s 6 S and sup {diam[/ F(s): s E S} for every Q-balanced neighbourhood U of zero. Then there exists an additive function <j>: S —>• Y such that 4>{s) € seqclF(s) for all s G S. Now we shall present the result concerning the stability of the Pexider equation

f(s + t)=g(s) + h(t) 153

(16.1)

154

Functional Equations and Inequalities in Several Variables

with three unknown functions /, g,h proved by K. Nikodem [148]. T h e o r e m 16.1 Let (S,+) be an abelian semigroup with zero and let Y be a sequentially complete topological vector space over Q. Suppose that V is a non-empty Q-convex symmetric and bounded subset ofY. If, moreover, f,g,h: S —*Y satisfy the condition f(s + t)-g{s)-h{t)eV

s,teS,

(16.2)

then there exist functions fi,gi,h\: S —>• Y fulfilling the equation (16.1) for all s,t € S and such that h(s) - f(s)E 3seqciy, hi(s) - h(s) e 4seqcl V,

gi(s)

- g{s) € 4seqcl V,

for all s G S. Proof. Set a := g(0), b :— /i(0), /o := f — a — b. Substituting first t = 0, then s = 0 into (16.2), we get fo{s) + a- g(s) = f(s) - b - g(s) e V, fo{t) + b - h(t) = f(t) - a - h{t) EV,

s e S, teS,

whence g{s) e Ms) + a - V,

seS,

h(t) e f0(t) + b-V.

teS,

Define the set-valued function F0: S —>• 2 y by F0(s):=f0(s)

+ 3V,

s € S.

Then we have (by the simmetricity of V) F0{s + t) = f0{s +1) + W = f{s +1) - a - b + 3V C g(s) + h(t) - a - b + W Cfo(s)-V + f0(t)-V + W CF0(s) + F0{t), which means that F0 is subadditive.

(16.3) (16.4)

Pexider's functional equation

155

Since the set 3V — 3V is bounded, we have also for any neighbourhood U of zero sup {diaimy FQ(s): s G S} = i n f { # e Q n ( 0 , o o ) : 3V - 3V C gU} < oo. Consequently, by Lemma 16.1, there exists an additive selection ip: S —> Y such that (p(s) e seqcl F 0 (s),

s € S.

Define / i := (p + a + b, gi := ip + a, hx := y? 4- 6. Then /i(s + *) = Si(s) + M*) ,

M6S.

Moreover, we have for all s e 5, /i(s) - /(s) = p(s) + a + b - f0(s) -a-be

seqcl F 0 (s) -

f0(s)

= 3 seqcl V , i.e. h(s) - f{s)e

3 seqcl V .

Now, in view of (16.3), we obtain 01 (*) - #( s ) = <^(s) +

a

- #( s ) e seqcl FQ(s) + a-

g(s)

= /o(s) + seqcl 3V + a - g(s) C 3 seqcl V + V C4 seqcl V . By the same way, applying (16.4), we get hi(s) - h(s) e 4 seqcl V , This concludes the proof of the theorem.

se S . •

Let's note the followin corollary, which is an immediate consequence of the above theorem.

156

Functional Equations and Inequalities in Several Variables

Corollary 16.1 Let (S, +) be an abelian semigroup with zero and (Y, || • ||) a Banach space. If f,g,h:S—*Y satisfy the inequality \\f(s

+

t)-g(s)-h(t)\\<e

(16.5)

for all s,t € S and some e > 0, then there exist functions fi,gi,h\: S ->Y satisfying the equation (16.1) and such that \\h{s) - f(s)\\ < 3e, IM*) - 0(5)|| < 4e, ||M«) - h(a)\\ < Ae (16.6) for all s G S. Remark 16.1 The functions f\, g\, h\, need not be unique. Take, for example f,g,h:S—>Y any solution of the equation (16.1). Then the condition (16.5) is trivially satisfied, with an arbitrary e > 0 . Let a,b EY and IMI<4e,

||6||<4e,

||a + 6 | | < 3 e ,

Put fi:= f + a + b , then fi,gi,hi: S —> Y the conditions (16.6).

gx := g + a , satisfy

hx:=h

+b,

the equation

(16.1)

and

Remark 16.2 A new functional equation of Pexider type related to the complex exponential function was studied by H. Haruki and Th. M. Rassias [85] .

Chapter 17 Gamma functional equation The functional equation f(x + l)=xf(x)

(17.1)

or (under suitable assumptions) the equation / ( * + !) xf{x)

= 1

is well-known as the gamma functional equation. We are going to investigate the Ulam-Hyer-Rassias stability of this equation. Denote E° := R \ {0, - 1 , - 2 , . . . } and let S,e > 0 be given real numbers. Define the following functions 00

/

1

\

OO

:

,

N.

<**><= n l - < ^ ) • ^> =n(i+(^4 for x > eT+s. The following result is due to S. M. Jung [107]. Theorem 17.1 Let / : R° -> R \ {0} satisfy the inequality /(s + 1) xf(x)

157

xW

x > e i+s.

(17.2)

158

Functional Equations and Inequalities in Several Variables

Then there exists exactly one mapping F: K° —> K satisfying the gamma functional equation (17.1) and such that a(x)<^!-
x>e^.

(17.3)

Proof. For i 6 l ° and n G N define f{x + n) x(x + 1) • ... • (x + n — 1)

Pn{x) rFor n > m, we have

Pn(x) _ f(x + m + l) Pm(x) (x + m)f(x + m)

'"

f(x + n) (x + n-l)f(x +

n-l)'

Take m so large that x + m > e1^ and ^7T7 > 0

!-7

(x + m)l+s {x G R° is fixed). Then it holds by (17.2),

or, equivalently,

c—rn s=m n-1

*

< Elo§

*

1+

'

'

e

(x + sy+sj'

Since the series occuring in the last inequality are convexgent , then m—> lim

V l o g ( l - 7 {x + ^TTTT I =0, s)1

s—m oo

lim £ m—>oo

log [ 1 + -

^--j )=0,

Gamma functional equation

159

and consequently the sequence {logPn(x)}™=1 is a Cauchy sequence for all x € R°. Therefore, let L(x) := lim logF n (x),

x£R°

and put F(x):=eL^,

xGR°.

Now we get easily (F(x) := lim P n (x)), n—Kx>

F(x + 1) := lim \ogPn(x + 1) = lim zP n +i(x) = xF(:r) for £ 6 M ° i.e. F satisfies the gamma functional equation. Observe that

^x

P(X) = HE+21. n{

'

xf(x)

••• (x

+ n-l)f(x

+ n

) J[

+ n-l)

fix)

h

If x > e i+*, then (a; + l)1+<5 for all s — 0,1,2,

n fi

Hence by (17.2) we obtain e

-

1 < ^ i < TT7I+

which implies the validity of (17.3). To prove the uniqueness, assume that there is another mapping G: R° -> R satisfying (17.1) and (17.3). Let xeR° and take n G N such that i

£ + n > £*+*. We have, in view of (17.1), F(x)

F(x + n) x(x + 1) • ... • (x + n — 1)'

160

Functional Equations and Inequalities in Several Variables

G(x)

G(x + n) x(x + l) •... • (x + n - 1)'

whence,

F(x)

F(x + n)

F(x + n)f(i; + n)

G{x)

G(x + n)

f(x + n)G(x + n)'

Now, applying the estimation (17.3), it follows g(x + n) F(x) 0(x + n) P(x + n) ~ G{x) ~ a(x + n) for all sufficiently large n. However a(x + n) —> 1 and

/3(x + n) -+ 1 as

n —>• oo,

which implies that F(x) = G(x) for x G R°. This concludes the proof of the theorem. •

Chapter 18 D'Alembert's and Lobaczevski's functional equations We will now consider the D'Alembert's functional equation (called also the cosine equation) f{x + y) + f{x -y)

= 2fix)fiy).

(18.1)

For this equation we have a different kind of stability, usually called superstability. The first result of this type was obtained by Baker [10]. In the following we will present the result proved by Gavruta [65] and use his method of proof, which is very simple(see also [30]). Theorem 18.1 Let 5 > 0 and G be an abelian group and f: G —> C be a function satisfying the inequality \f{x + y) + f{x-y)-2fix)fiy)\<6

for all

x,yeG.

Then either f is bounded or satisfies the D'Alembert's equation (18.1).

(18.2) functional

Proof. Suppose that / is not bounded. Then there exists a sequence {xn}n€iq of elements of G such that lim |/(rc„)| = oo. 161

162

Functional Equations and Inequalities in Several Variables

Take x,y eG. By (18.2) we obtain for n e N \V(Xn)f(x) ~ f(x + Xn) ~ f{x ~Xn)\< S, which implies f(x) = lim

1 (f(x + xn) + 2f(xn)

f(x-xn))

(18.3)

The formula (18.3) gives (18.4)

2f(x)f(y) = lim An, f{x + y) + f{x -y)=

lim Bn, n-foo

(18.5)

where (for n sufficiently large) An R n

1

2PM

[f(x + xn) + f{x - xn)][fiy + xn) + f{y - xn)],

f(xn) [f{x + y + xn) + fix + y- xn) 2P(xn) +fix - y + xn) + fix - y - xn)].

Now, applying (18.2), we obtain the sequence of inequalities \2fix + xn)fiy + xn )-f{x

+ y + 2xn) - fix - y) <6, \2fix-xn)fiy + xn )-f{x + y)-f(x-y2xn) <6, \2fix + xn)fiy-xn ) - fix + y)- fix-y + 2xn) <s, \2fix - xn)f{y - xn ) - f(x + y- 2xn) - f{x - y) <8, \2fix + y + xn)fixn ) - fix + y + 2xn) - fix + y) <5, \2fix +

y-xn)fixn ) - fix + y)- fix +

y-2xn) <6, \2fix-y + xn)fixn ) - fix - y) - fix - y + 2xn) <6, \2fix -yxn)fixn ) ~ f{x - y) - fix - y - 2xn) <S.

D'Alembert's and Lobaczevski's functional equations

163

Hence we can easily get 14 — R I < \f(Xn)

for

n e N.

The last inequality together with (18.4), (18.5) and the condition lim |/(rc„)| = oo imply that / satisfies the equation (18.1) which n->oo

ends the proof.

•

In the sequel we will deal with the Lobaczevski's functional equation

f

x+ y

(18.6)

= f(x)f(y).

The following theorem can be found in [65]. Recall that a group G is 2-divisible if for every x e G there exists a unique y G G such that y + y = 2y = x and we denote

y= l Theorem 18.2 Let 5 > 0 and G be an abelian 2-divisible group. Let f: G —> C be such that

f

x+y

- mm

< 8 for all x,y G G.

(18.7)

Then either

\m\ < g l/(0)| +

(|/(0)| 2 + 4 < ^

forall

x e G,

(18.8)

or f satisfies the Lobaczevski's functional equation (18.6). Proof. Suppose that the condition (18.8) is not satisfied.Then there exists i 0 e G such that

l/(*o)|> ^ [1/(0)1 + (l/(0)|2 + ^Y

(18.9)

164

Functional Equations and Inequalities in Several Variables

We shall find a sequence {xnjneN with the property lim | / ( O i = oo.

(18.10)

Set y = 0 and 2x instead of x in (18.7), then |/ 2 (x) - /(0)/(2rr)| < <5 for all x G G.

(18.11)

If we had /(0) = 0, then from (18.11) would follow that |/(a;)| < V8,

xeG.

But, due to (18.9), |/(ajo)| > \/6, a contradiction. /(O) ± 0. Thus from (18.11) we get for x G G

Therefore,

|/(0)||/(2^)| = \f(x) + (f(0)f(2x)~p(x))\ >\f(x)\*-\f(0)f(2x)-p(x)\ >|/(x)|2-5. Hence \f(2x)\>-r^-[\f(xQ)\2~5],

for all xeG.

(18.12)

Denote a=

2V

1/(0)1 + (|/(0)| 2 + 45)"},

P=\f(x0)\-a,

then \f(x0)\ = a + /3, and in view of (18.9), 0 > 0. In the next step we will show by induction the inequality f(2nx0) >a + 2np,

neN.

(18.13)

To prove it for n = 1, take x = x0 in (18.12), then

|/(2*o)| > ]7^)fK" + P ) 2 - $ } = * + |^jj(2a/3 + /?2) > a + 2/?,

D'Alembert's and Lobaczevski's functional equations since

a2 = \f(Q)\a + 5,

a > |/(0)|,

165

(3 > 0.

This means that (18.13) is true for n = 1. Assume (18.13) to hold for some n e N. Take x = 2nx0 in (18.12), then we get \mn+1xo)\>JjJaj\{(«

+

= a + ~—

2nP)2-5]

[2"+1a/3 + 22"/32] > a + 2"+1/?,

which finishes the inductive proof of (18.13). Take any x, y e G. From (18.7) it follows that

- f2

2 1

fMM

'

~^n

< 5 for

n 6N

and consequently, in virtue of (18.10), hence we obtain the formula f(x) = lim

12 ( *^ ' "^n

lf(Xn)

" V

,

2

ieG.

Applying this formula, we get for x, y € G /(x)/(y) = lim /

2 f ^ + 2/

1

- / Z + £ W \ , (V + Vn

./M f2 f(xn)

lim

I x + y + 2xr,

Again, from (18.7) we obtain for x,y £ G and n G N . x + a:ra \

/M

2

/ y + yn

/ ' V

2

J ~fJ

x + y + 2xn V

4

<

i/wr

whence, by what we already stated above, on letting n —>• oo, we get „ / T. 4 - ?/ \

mf(y) = f2,x as claimed.

+y

for

x,y

eG,

166

Functional Equations and Inequalities in Several Variables

Remark 18.1 It is possible to prove that if a function f satisfying the inequality (18.2) is bounded, then \f(x)\ < |[1 + \ / l + 25] for xeG. Remark 18.2 For the sine functional equation f(xy)f(xy-1)

= f2(x) - f(y),

(18.14)

P. W. Cholewa [25] has proved the following superstability result. Theorem 18.3 Let G be an abelian group uniquely divisible by 2. Then, every unbounded mapping f: G -» C satisfying the inequality \f(xy)f(xy-1)-f2(x)

+ f2(y)\<6,

x,yeG,

(18.15)

for some 5 > 0, is a solution of the sine equation (18.14)He observed also in [25] that the situation for the equation (18.14) is different from the case for the cosine equation (18.1). Indeed, the functions fn{x) = nsina; H—,

n GN

satisfy the inequality (18.15) with 8 = 3 for all n G N, but the inequality |/ n (a;)| < M fails to hold for some x and n, which means that there is no common bound for all solutions of the inequality (18.15). Remark 18.3 Another interesting result about the Ulam-HyersRassias stability of functional equations which is satisfied by the cosine and sine mappings has been presented by R. Ger in [75]. Theorem 18.4 Let G be a semigroup and let H,K: G x G -> K+ be functions such that H(-, x) and K(x, •) are bounded for each fixed x G G. If, moreover, f,g: G —> C satisfy the system of inequalities \f(xy) - f(x)g(y) - g(x)f(y)\

< H(x, y),

\g(xy) - g(x)g(y) + f(x)f(y)\

< K(x, y),

D'Alembert's and Lobaczevski's functional equations

167

for all x,y G G, then there exist functions s: G —> C nadc: G —> C satisfying the system of equations s(xy) = s(x)c(y) + c(x)s(y), c(xy) = c(x)c(y) -

s{x)s(y),

for all x,y G G, such that f — s and g — c are bounded. Remark 18.4 The equation f(xy) = yf(x) + xf(y)

(18.16)

is called the functional equation of derivation. P. Semrl [189] proved the following superstability result for this equation. Theorem 18.5 Let E be a Banach space, let A be an algebra of operators on E and let B(E) be the algebra of all bounded operators on E. Assume that ip: R+ —> R+ is a function with the property lim t—>oo

-Mt) t

= 0.

If, moreover, f: A —> B(E) satisfies the inequality \\f(xy) - yf(x) - xf(y)\\ < p(||a;||||y||), then f satisfies the equation (18.16).

x,y € A,

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Chapter 19 Stability of homogeneous mappings We shall use the following symbols: N, R, Ko, R+ to denote the sets of all natural, real, non-zero real or nonnegative real numbers respectively. Moreover, denote Uv:= {a e R: a"exists}

for

v G RQ.

We start with the following (see [37]). Lemma 19.1 Let E be a linear space and F a normed space (over R). Let f: E —>• F and h: R x E —»• R+ satisfy the inequality \\f(a,x)-avf(x)\\
(19.1)

for all (a, x) e Uv x E, where v € RQ is fixed. Then n-l

\\f(anx) - anvf(x)\\

< J2 Hvh(a,an-s-lx)

(19.2)

s-0

for all n G N and (a, x) e Uv x E. Proof. If n = 1, then (19.2) follows directly from (19.1). For n + 1 we have, by (19.2), \\f(an+1x)

- dn+l>f{x)\\

+\a\v\\f{anx)

- anvf(x)\\

< ||/(a n+1 a;) - avf(anx)\\ < h(a, anx) + ... + 169

+

\a\nvh{a,x),

170

Functional Equations and Inequalities in Several Variables

and hence by the induction principle, (19.2) holds true for all n € N.

• Now we shall prove a result about the stability of homogeneous mappings. We follow the idea of S. Czerwik [37]. Theorem 19.1 Assume that the assumptions of Lemma 19.1 are satisfied and let F be a Banach space. Suppose, moreover, that for some 0 ^ /3 G Uv the series oo

Y, \P\-nvW, Pnx)

(19.3)

n=l

converges pointwise for every x G E and lim M{\p\-nvh(a,Pnx)\

=0

(19.4)

for all (a, x) G Uv x E. Then there exists exactly one vhomogeneous mapping g: E —> F: g(ax) = avg(x)

for all

(a,x) G Uv x E

such that 00

nv

\W)-f{x)\\
for xeE.

(19.5)

71=1

Proof. Put for n G N gn(x):=p-nvf(/3nx), From (19.2) we have for n G N and

xeE.

(19.6)

xeE

n

\\9n(x) - f(x)\\
(19.7)

Stability of homogeneous mappings

171

Take m,n EN, n > m, then in view of (19.2), we obtain p^n-m^f(Pmx)\\

\\9n(x) - gm(x)\\ < \p\-™\\f{p»x) oo

< Y, \P\-nh{p,p-lx). s=m+l

Hence {gn(x)}neN is a Cauchy sequence for every x E E. Define, therefore, g(x) := lim gn(x), x E E. n—>oo

By (19.6), (19.1) and (19.4), we get for

aEUv,xEE,

g(ax) - avg{x) = lim {/T™ [/(«/?V) - avf(/3nx)}}

= 0,

i.e. g is a ^-homogeneous mapping. Moreover, directly from (19.7) we get the estimation (19.5). It remains to prove that there is only one ^-homogeneous mapping satisfying (19.5). For the contrary, suppose that there are two such mappings, say g\ and g
f(Pmx)\\}

oo

<2 £

l^-^h^J^x).

s=m+l

Consequently, since the series (19.3) converges, it follows that 9\ = 9i a n d completes the proof. H Corollary 19.1 Let the assumptions of Lemma 19.1 be satisfied with h(a,x) =5 + \a\ve for given nonnegative numbers 6 and e, and let F be a Banach space. Then there exists exactly one vhomogeneous mapping g: E —>• F such that \\g(x)-f(x)\\<e

for

xEE.

(19.8)

172

Functional Equations and Inequalities in Several Variables

Proof. First assume that v > 0. From Theorem 19.1 for 2 < (3 = m G N, there exists a w-homogeneous mapping gm(x) = lim m~nvf(mnx),

x G E,

such that \\gm{x)-f{x)\\<{6

+ mve)(mv-l)-\

x G E.

(19.9)

Now we shall verify that for every 2 < TO, r G N, we have 9m = 9r- Indeed, by (19.9), for n G N, x G JE1, ||
gr{2nx)\\

< 2~nv [(8 + mve)(mv - l ) " 1 + (8 + rve)(rv - l ) " 1 ] , and, since v > 1, if we let n —>• oo, we get g m = gr. Take ^(x) = #2(2), x € E, then from (19.9), letting TO —»• 00, we find that \\g(x)-f(x)\\<e, for xeE, and the proof is completed. In the case v < 0, we take /5 = ^ and follow a similar argument. I

Example 19.1 Take f(x) — sin a;, I G K . |sin(o!x) - a"sin:r| < 1 + \a\v

for

T/ten (a,x) € Uv x K,

6ui / is no£ a v-homogeneous function, which means that we have not a superstability in this case. Corollary 19.2 Let the assumption of Lemma 19.1. be satisfied with h(a,x) = 8+\a\ve, for some 8,e G R+, and let F be a Banach space. Then, if either 8 or £ is zero, f(ax) = avf(x),

for all

(a, x) G (Uv \ {0}) x E.

(19.10)

Stability of homogeneous mappings

173

Proof. Suppose that 5 = 0. Then \\f{ax) - avf(x)\\

< \a\ve

for

(a,x)

eUvxE.

Let y & E and a e Uv \ {0}. Inserting ^ instead of x, we get < \a\v£

><»>-«•>(=)

(19.11)

and for v > 0, we have /(y) = lim a—•()

["(2)

for

y € E.

Therefore, for (/3, x) G (J7„ \ {0}) x E, we obtain /(/3x) = lim a-5-0

lim

^'T

"i !)''(£

a->0

= 0V(*), i.e. (19.10) holds true. If v < 0, then from (19.11) we get

/(y)=lim

Wf{l)

|a|->oo L

,

for

y € E

\Qj/

and one can carry out a similar proof. On the other hand, assume that e = 0. the inequality \\a~vf{ax)

- /(:r)|| < \a\~v8

for

Then we have

(a,x) e (Uv \ {0}) x X.

Hence, when v > 0, f(x) = lim [a-"f(ax)],

x e E,

\a\—>oo

and when u < 0, /(a;) = lim[a _ , 7(aa0],

x e E.

174

Functional Equations and Inequalities in Several Variables

As before, it is easily shown that our statement holds in these cases as well. • Now we will consider the Pexider version of stability of homogeneous mappings following the paper of S. Czerwik [39]. First we will present the following Lemma 19.2 Let E be a real linear space and F a real normed space. Let f: E ->• F, g: E ->• F, • R and h: R x E -> R+ be given functions. Assume that \\f(ax)-
(19.12)

for all (a, x) e R x £ and
\\f(anx) -yn{a)f(x)\\

<^

IvW^Hfaar-x),

\\g{anx) - ^{a)g{x)\\<J2W{a)rlG{a,an-sx),

(19.13)

(19.14)

s=l

where H(a,x) := h(a,x) + G(a,x) := h(a,x) +

\(p(a)\h(l,x), h(l,ax).

Proof. From (19.12) we get for (a,x) eRx \\f(ax)r\f(a)f(x)\\ 1

<

E,

\\f(ax)-
< h(a,x) + \
H(a,x).

This means that (19.13) is satisfied for n = 1. Now we have for n + 1,

Stability of homogeneous mappings

U/K+V)

175

-
\
+ n

n

J2Ma)\SH(a>an~Sx)

< H(a,a x) + s-l n+1 s=l

Therefore, by the induction principle, the inequality (19.13) holds true for all natural numbers n. The inequality (19.14) one can verify by the same way. • The main result in this part reads as follows (see S. Czerwik [39]). Theorem 19.2 Let the assumptions of Lemma 19.2 he satisfied and let F be a Banach space. Suppose that there exists /? G R such that
X>(/3)r#(/?,/3V)

(19.15)

n=l

converges pointwise for all x G E, and lim inf \y{p)\-nH{a,

pnx) = 0,

(19.16)

n->oo

for all (a,x) G R x E. Then there exists exactly one
(a, x) G R x E,

(19.17)

176

Functional Equations and Inequalities in Several Variables

such that 00

\\A{x)-f{x)\\
(19.18)

n=l oo

\\A{x)-g{x)\\ ^imrGfaF-'x),

(19.19)

n=l

/or x £ E, Proof. Define the following generalized Hyers-Rassias sequence An{x) := (p-n(/3)f(f3nx),

xeE,

neK

(19.20)

By (19.13) we get for a; G X and n G N,

\\An(x)-f(x)\\ <J2W)\-{n-s)~lH(PJn-sx), i.e. n

HAn^)-/^)!!^^!^)!-^,^-1!).

(19.21)

We shall verify that {An(x)}ne^ is a Cauchy sequence for every x E E. Indeed, in view of (19.13) we get for all natural numbers m, n, with n > m and x € E that

KCE) - ^(aOH < W)ri|/(/5V) - /(^M/*)"-"!! n+l

< J2 \m\-sHWjs~ix), s=m+l

and since the series (19.15) converges, we have proved our statement. Denote A{x) := lim An(x) for x e E. n—>oo

Stability of homogeneous mappings The function A satisfies the expression

177

(19.17).

To see this,

consider

A(ax) - ip{a)A(x) = Urn {^(/?)[/(c*/?V) -
< \
whence, from (19.16), we get for (a, x) G R x £ , A(ax) -
x G E.

As before, we can prove that there exists the limit B(x) := lim Bn(x),

x G E,

n—>oo

and taking into account (19.14), we get the inequality n

WBnW-gWW^Y.lipWl-G&p-'x),

xeE.

(19.22)

s=l

Moreover, we have \\An(x) - Bn(x)\\ = \
< \
l

-W{P)\-nH{l,^x),

whence by (19.16), A(x) = B{x) for all x G E. Therefore, from (19.22), we get the inequality (19.19). To prove the uniqueness part, let's assume that there exist two (^-homogeneous functions As: E —>• F, s = 1, 2, such that (19.18) and (19.19) hold for A = As, s = 1,2. If A,, s = 1,2, are zero

178

Functional Equations and Inequalities in Several Variables

functions, then Ai = A2. Therefore let's assume that say A\ ^ 0, i.e. there exists x0 G E such that Ai(xQ) ^ 0. Then we have for a, 0 G R, AifaPxo) =
a j e l

(19.23)

In the sequel, from (19.18), we get for all n e N and all x e E,

WA&) - A2(x)\\ = M/^HIAiGS"*) - A2(/3»x)|| < Wm^iWMiFx)

~ / ( M i l + \\A2(Pnx) - f(Pnx)\\}

00

Since the series (19.15) converges, the last inequality implies that Ai = A2 and the proof of the theorem is completed. • Remark 19.1 It follows from the proof of the theorem that if A is not the zero function, then the function ip must be multiplicative, i.e. satisfies the condition (19.23). Corollary 19.3 Let E be a real linear space and F a real Banach space. Let f': E —>• F and g: E —»• F satisfy the inequality \\f(ax) - \a\vg{x)\\ < S + \a\ve

aER, (19.24) where v > 0, 5 > 0, e > 0 are given real numbers. Then there exists exactly one absolutly v-homogeneous function A: E —>• F (i.e. A(ax) = \a\vA(x) for x G E, a G R ) such that \\A{x)-f{x)\\<8 \\A{x)-g{x)\\<e, for x G E.

for

x GE

+ 2e,

and

(19.25) (19.26)

Stability of homogeneous mappings

179

Proof. Applying (19.13) and (19.14) for n = 1, we get \\f{ax)-\a\vf(x)\\<6+\a\'>{6

+ 2e),

\\g{ax) - \a\vg(x)\\ <25 + e + \a\ve, for a E R, x € E. Hence, by Theorem 19.2 , for every 2 < f3 = m G N, there exists absolutly u-homogeneous function Am{x) := lim m-nvf(mnx),

xEE

satisfying the inequalities PmOr) - f(x)\\ <{S + mv(6 + 2e)]{rrf - 1 ) ~ \ \\Am(x) - g(x)\\ <(25 + e + mve){rrf - l ) " 1 , for x G E. To finish the proof, one can use the same method as in the proof of Corollary 19.2 . • Corollary 19.4 Let the assumptions of Corollary 19.3 be satisfied. If, moreover, 5 = 0, then f is absolutly v-homogeneous and \\f(x)-g(x)\\<£

for

xGE.

If e = 0, then g is absolutely v-homogeneous and \\f(x)-g(x)\\<6

for

x £ E.

Proof. See the proof of Corollary 19.3 and 19.2 .

•

Now we shall aplly Corollary 19.1 to quadratic functions. Let us recall that a function / : E —> F is called a quadratic function iff for all x,y G E,

f(x + y) + f(x~y)

= 2f(x) + 2f(y).

180

Functional Equations and Inequalities in Several Variables

Corollary 19.5 Let E be a real linear space and F a real Banach space. Let f,g,h: E —»• F be given functions and e > 0 a given number. Then f = g + h, where g is a quadratic 2-homogeneous function and \\h(x)\\ < e for x £ E iff \\f(x + y) + f(x -y)-

2f(x) - 2f(y)\\ < 6e, (19.27)

||/(aar) - a2f(x)\\ <e + a2e

(19.28)

for all x,y e E and a 6 t Proof. First assume that f = g + h, where g is quadratic and 2homogeneous and h bounded by e. Then, for x,y e E and a € K, ||/(a; + y) + / ( a ; - y ) - 2 / ( x ) - 2 / ( y ) | | = \\h{x + y) + h(x -y)2h(x) - 2h{y)\\ < 6e, and \\f{ax) - c?f{x)\\ = \\h{ax) - a2h(x)\\ <s + a2e, i.e. (19.27) and (19.28) hold true. In the sequel assume that (19.27) and (19.28) satisfied. Then, from Corollary 19.1 there exists 2-homogeneous function g: E —> F with \\f(x)-g{x)\\<e for x e E. Put / - g = h, then / = g + h and \\h(x)\\ <£foix£E. We shall show that g is a quadratic function. Indeed, in view of (19.27), we have \\g[a(x + y)\ + g[a(x - y)] - 2g(ax) - 2g(ay) + h[a(x + y)]+ +h[a(x - y)] - 2h(ax) - 2h(ay)\\ < 6e, whence, for a ^ 0, we get \\g(x + y) + g(x -y)-

2g{x) - 2g{y) + a-2[h(a(x

+h(a(x - y)) - 2h(ax) - 2h{ay)]\\ < 6ea~2.

+ y))+

Stability of homogeneous mappings

181

Consequently, since h is bounded, letting a —> oo, we obtain for x,y E E \\g(x + y) + g(x-y)-2g(x)-2g(y)\\

= 0,

wchich means that g is also a quadratic function. This concludes the proof. • Corollary 19.6 Let f: R ->• R satisfy the inequality (19.28). Then f satisfies the inequality (19.27). Proof. On account of (19.28), by Corollary 19.1 it follows that there exists a 2-homogeneous function A: R —>• R such that \\A(x)-f(x)\\<e

for

xER.

Evidently, A(x) — x2A(l) for i £ l Therefore, we obtain for all x, y € R, \\f(x + y) + f(x -y)2f(x) 2f(y)-A(x + y)and since (19.27).

A(x -y) + 2A(x) + 2A(y)\\ < 6e

2

A(x) — x A(l)

for

x 6 R, we get the inequality •

J. Chudziak in [26] has proved the following theorem about the superstability of the homogeneous equation. Theorem 19.3 Let K be a real or complex field, let X be a normed space and Y a Banach space over K. Let f: X —> Y satisfy the inequality \\f{ax)-af(x)\\<e{\a\*+\\x\\<) for all a G K and x £ X such that the right hand side of this inequality is well defined, where p, q G R and e > 0 are arbitrarily fixed. If p < 1 or q < 0, then f{ax) = af{x) //, moreover,

for

a e K \ {0},

x <E X \ {0}.

182

Functional Equations and Inequalities in Several Variables

A) p < 1 and q > 0 or B) p > 0 and q < 0 , then f(ax) = af(x)

for

a EK

and

x E X.

If V > 1 anc ^ 9 = 0, then there exists an unique homogeneous mapping h: X —>• Y such that \\f(x)-h(x)\\<e

for

i e l

Jacek and Jozef Tabor [208] have generalized this problem to the case where Y is topological vector space. To formulate the problem let X and Y be normed spaces. Denote V :={xeX: \\x\\ <e}. Then the inequality \\f(ax)-af(x)\\

<e\a\

for

«Gl,

x £ X,

can be written as f(ax) - af{x) € aV

for

a e R,

x E X.

Hence it is clear that the following condition f(ax) — af(x) E g(a,x)V

for

a E R,

x E X,

where V C X and g is a mapping from R x X into R, generalized our problem. Let us recall that a subset V of a topological vector space over K is said to be bounded iff for each neighbourhood U of zero there exists an r E K \ {0} such that rV C U. The following result is due to J. J. Tabor [208].

Stability of homogeneous mappings

183

Theorem 19.4 Let X be a vector space over K, Y a topological vector space over K and let X\ and Y\ be subsets of X an Y, respectively, such that KX\ C X\ and KY\ C Y\. Let V C Y be a bounded set and g: K x X\ —> K a mapping with the property that there exists a sequence {an} of non-zero elements of K such that lim g(aan, (an)'1 x) = 0 for

a E K,

x E Xx.

(19.29)

If f: Xi —>Yi satisfies the condition f(ax) - af(x) G g(a,x)V

for

a G K,

x G Xu

(19.30)

then f{ax) = af(x)

for

a E K,

xE Xx.

(19.31)

Proof. From (19.30) we get a~lf{anx)

G f(x) + a~1g(an,x)V

for

x E X\

and

n EN.

Hence we obtain aa~lf(anx)

E af{x) +

aa~1g(an,x)V,

and consequently, by (19.30), aa~lf(anx)

- f(ax) E aa~1g(an,x)V

-

g{a,x)V.

Now replacing a and x by aan and (an)~lx, respectively, we obtain af(x) - f(ax) E ag(aan,a'1x)V

-

g(aan,a~1x)V

for a E K, x E Xi, n E N . But V is bounded, so in view of (19.29) we get (19.31) and the proof is completed. •

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Chapter 20 Quadratic functional equation The equation f(x + y) + f(x -y)

= 2f(x) + 2f(y)

(20.1)

is called the quadratic functional equation. We define any solution of (20.1) to be a quadratic function. The Hyers-Ulam stability theorem for the quadratic functional equation was proved by F. Skof [194] and in more general setting by P.W. Cholewa [24]. The Rassias type stability has been studied by S. Czerwik [38]. Here we shall present the ideas and main results obtained by S. Czerwik in [40]. First we shall introduce some definitions and notations. Let X be a commutative semigroup with zero and with the following law of cancellation: a + c = b + c implies

a = b for all

a, b, c G X.

Assume, moreover, that in X is defined a multiplication by nonnegative real numbers with the properties: a(a + b) = era + aj3, a(f3a) = (a/3)a,

aa + (3a = (a + (3)a,

la = a, 185

186

Functional Equations and Inequalities in Several Variables

for all a, b G X and a, f3 G 1+. Let (X, d) be a metric space such that d(x + y,x + z) — d(y, z) d(tx,ty) = td(x,y)

for all

for all

x,y,z

x,y £ X,

G X, £ G R+.

Denote \\x\\:=d(x,0),

xeX.

A commutative semigroup with zero satisfying all stated above conditions is called a quasi-normed space. Now we shall construct very important example of such a space. Consider a normed space Y. Given sets A, B C Y and a number s £ R , w e define A + B := {x G Y: x = a + b, sA := {x G Y: x — sa,

a G A,b G B},

a G A}.

By CC(y) we denote the space of all non-empty compact convex subset of Y. Set H(A,B)

: = i n f { s > 0 : A C B + sK, B C A + sK},

where K is the closed unit ball in Y and A, B C Y non-empty closed bounded sets. Then the function H is a metric called the Hausdorff metric induced by the metric of the space Y. The space CC{Y) with Hausdorff metric is an example of a quasi-mormed space (for more details see [162]). Moreover, if Y is a Banach space, then CC(Y) is a complete metric space (see [22]). We now proceed to show that the law of cancelation holds for some classes of convex sets. This lemma will be useful in the next parts of this book, as well (see also [146]). Lemma 20.1 Let A,B and U be subsets of a topological vector space Y satisfying the TQ separation axiom. Suppose that B is

Quadratic functional equation

187

closed and convex and U is bounded. If A + U C B + U, then A C B. If, moreover, A is closed and convex and A + U = B + U, then A = B. Proof. Take any element a € A, and assume that A + U C B + U. We shall show that a G B, too. Take U\ G U, then a + U\ G B + U, so there exist &i G B and u 2 G U such that a + U\ = b± + w2Similarly there exist 62 G B and u 3 G £/ such that a + u2 = fr2 + ^3. Repeating this procedure and adding the first n equations, we get n

n

n+1

na + 'Y] uk = y~] bk + y ^ ttfc, fc=l

A;=l

fc=2

which implies na + ui = 2-,6ifc + « n + 1 , fc=i

or a = - > 6fc + -^n+i Denote ± £ 6fc

•

x n , then

fc=i 1

Wi

n

n

Note that xn £ B for all n G N because of the convexity of B. Let W be a neighbourhood of zero in Y and let V be a balanced neighbourhood of zero in Y such that V + V C W. Thus there exists an s > 0 with s[7 C V-. Choose n G N with ^ < s. Then ^ [ / C V, which implies that 1

-Mi,

! -u„+i G K

n

n

and

x n eBn(a + F + F)cBn(o + iy).

188

Functional Equations and Inequalities in Several Variables

This proves that

B n (a + W) ^ 0 for every neighbourhood W of zero in Y. Consequently a belongs to the closure of B which (since B is closed) is equal to B. Thus a e B which completes the proof of the inclusion A C B. Now assume that A is closed and convex and A + U = B + U. Then from the first part, we obtain that B C A, which gives the equality A = B, and completes the proof. • The next lemma deals with the basic properties of Hausdorff distance. Lemma 20.2 Let Y be a real normed space. If A,B,X and m is a positive number, then

6 CC(Y)

H(A + X,B + X) = H(A, B), H(mA, mB) = mH(A, B). Proof. Let K be the closed unit ball in Y. Consider the following four inclusions : (a) AcB + sK , (b) BcA + sK , (c) A + X cB + X + sK , (d) B + XcA + X + sK. Denote hx = H(A,B) and h2 = H(A + X,B + X). Then according to the definition h\ is equal to the infimum of all positive s for which (a) and (b) hold, and similarly, h% equals the infimum of all positive s such that (c) and (d) hold. Since (a) and (b) imply (c) and (d), so h2 < h\ . On the other hand, by Lemma 20.1, (c) and (d) imply (a) and (b), whence we get hi < h2, which gives hi = h2 and proves that H(A, B) = H(A + X,B + X). To prove the second property, consider the following inclusions (e) mA C mB + sK , (f) mB c mA + sK , (g) ACB + ^K, (h) BcA + f-K.

Quadratic functional equation Put hz — H(mA,mB). then we have hi =H(A,B)=M{a>0:

189

Since (e) and (f) imply (g) and (h), AcB

+ aK

and

BcA

+ aK}

< inf ( — > 0: mA C mB + sK lm

and

mB C mA + sK} J

= — inf{s > 0: mA C mB + sif m

and

mB C mA + sK}

m i.e. m/ii h3, which concludes the lemma. • Let E be a group and h: ExE

—>• M+ a given function. Denote

H(x, y) := /i(z,y) + /i(or, 0) + % , 0) + /i(0,0), K(x, y) := 2/i(z, j/) + /i(z + y, 0) + h(x - y, 0), for x,y £ E. First we establish the following. L e m m a 20.3 Let X be a group and (Y,d) a quasi-normed space. Let F: X —» Y', G: X —>• Y satisfy the inequality d[F(x + y) + F(x-y),G(x)

+ G(y)]
for all x,y e X. (20.2)

TTien, /or all x, y € X, d[F(x + y) + F ( z -y)

+ 2F(0), 2F(x) + 2F(y)] < H{x, y), (20.3)

d[G(z + y) + G{x -y) + 2G{0),2G{x) + 2G(y)} < K(x, y). (20.4) Proof. Let x, y € X then we have, d[G{x + y) + G{x - y) + 2G(0), 2G(x) + 2G(y)] < < d[G(x + y) + G{x -y) + G(0), 2F(x + j/) + G{x - y)}+ +d[2F(x + y) + G(x -y) + G(0), 2F(x + y) + 2F(x - y)}+ +d[2F(x + y) + 2F{x - y), 2G(x) + 2G(y)} < < h(x + y, 0) + h(x -y,0)

+ 2h(x, y) = K(x, y),

190

Functional Equations and Inequalities in Several Variables

i.e. the inequality (20.4). The proof of (20.3) can be given by the same way. •

Lemma 20.4 Let X be a group and Y a quasi-normed space. If F,G: X —> Y satisfy the inequality (20.2), then, for all x € X, n,k G N , where k > 2, d[F{knx) + {k2n - l)F(0),k2nF(x)] fc-l

n-l

2{n l)

< k

<

~ Y^ ^{k

~ m)H(mksx,

ksx)k~2s,

(20.5)

m = l s=0

d[G(knx) + (k2n - l)G(0),k2nG(x)} k-l

2

n-l

1

< A; ^- ) J2 J^(k ~ m)K(™ksx, m=\

<

ksx)k~2s.

(20.6)

s=0

Proof. The proof follows by induction with respect to n. When n = 1, we have to show the inequality fc-i 2

2

d[F{kx) + (k - 1)F(0), k F(x)} <^(k-

m)H(mx, x),

(20.7)

m=l

for all x G X and all integers k > 2. Inserting x = y in (20.3), we get d[F(2x) + 3F(0),AF(x)] < H{x,x) for all x G X. Next, let's make the induction hypothesis that (20.7)

Quadratic functional equation

191

holds true for some k > 2 and all x G X. Then we get d[F{{k + l)x) + ((k + l) 2 - 1)F(0), {k + l)2F{x)] < < d[F{{k + l)x) + F((k - l)x) + {k2 + 2k)F(0), (k2 + 2k- 2)F(0) + 2F(kx) + 2F(x)] + d[2F(kx) + 2F(x)+ + {k2 + 2k- 2)F(0), F{(k - l)x) + (k + l)2F{x)] < < H{kx,x) + d[2F{kx) + (A;2 + 2k - 2)F{0),F((k - l)x)+ + (k2 + 2k- l)F(x)] < < H(kx,x) + 2H[(k - l)x,x] + ... + iH[(k + 1 - i)x,x}+ + d[(i + l)F[(k + 1 - i)x] + [k2 + 2k- i(i + 1)]F(0), iF[(k - i)x] + [(k + l) 2 - i(i + l)]F(x)] < H(kx,x) + ... + kH(x,x). This proves (20.7). Now we assume that (20.5) holds true. Then, for n + 1 we obtain, d[F{kn+1x) + (A;2("+1) - 1)F(0), k^n+1^F(x)} < d[F(kn+1x) + {k2 - 1)F{0), k2F{knx)}+ +k2d[F{knx) + {k2n - 1)F(0), k2nF{x)] jfe-i

- m)H(mknx, knx)+

< ^(k m=l

fc-1

n-1

2n

+k J2 ^2(k - m)H(mkSxi

k$x k

)

m=l s=0 k—l n

= k2n ] T Yl(k - m)H(mksx, ksx)k~2 m = l s=0

This means exactly that (20.5) is valid for all x G X and n, k G N(k > 2). In order to proof the inequality (20.6) one can proceed similarly. I Now we are ready to formulate the following (Czerwik [40]).

192

Functional Equations and Inequalities in Several Variables

T h e o r e m 20.1 LetX be an abeliangroup andY a complete quasinormed space. Suppose that the inequality (20.2) is satisfied. Let for some integer k > 2 and m = 1 , . . . , k — 1 the series oo

^2h(mksx,ksx)k-2s,

(20.8)

s=0 00

J2Hksx,0)k-2s

(20.9)

s=0

be convergent for all x G X. If, moreover, lim inf h(knx, kny)k-2n

= 0 for all x, y G X,

(20.10)

n—>oo

then there exists exactly one quadratic mapping A: X —> Y such that k—l

oo

2

d[A{x) + F(0), F{x)] < AT Y^ ^2(k ~ m)H(mksx, ksx)k~2s, (20.11) m=l

s=0

k—l

oo

d[2A(x) + G(0),G(x)j < k~2 ] T ^(k

- m)K{mksx,ksx)k'2s,

(20.12)

m = l s=0

/or a// i £ l . Proof. Write An(x) := k-2nF(knx)

for

xeX

and

n G N.

(20.13)

In the first step, we shall check, that {A„(x)} is a Cauchy sequence for every x G X. Indeed, in view of Lemma 20.4 and the homogeneity of the metric d, we get for n > r and x G X, d[An(x),Ar(x)]

k-2nd[F{knx),k2^n-^F(krx)]

=

< AT 2r ||F(0)|| + k-2nd[F{knx) k—l 2r

2

< k- \\F(Q)\\ + k~ Y^ m=l k—l

n—r—l k

J2( ~

+ [k2^n-^ - 1]F(0), m)H{mks+rx,

ks+r x)k~2{s+^

s=0 oo

< £r 2 r ||F(0)|| + AT2 ] T ^2(k - m)H(mksx, m=l

k^n-r^F{krx)]

s=r

ksx)k~2s.

Quadratic functional equation

193

Since the series (20.8) converges, the conclusion follows. Thus there exists the limit A(x) := lim An{x)

for all x E X.

(20.14)

By the same way we can prove that the sequence {k~2nG(knx)} a Cauchy sequence. We shall prove that 2A(x) = lim k~2nG{knx),

xeX.

is

(20.15)

In fact, we have by (20.2) and the definition of || • ||, d[2A(x), k~2nG{knx)] < d[2A{x),2+d[2 • k~2nF(knx), +d[k-2nG{knx)

k~2nG{knx) + Ar2nG(0)]

+ k~2nG(0),

<2d[A(x),An(x)]

k-2nF(knx)}

+k-2nG{knx)}

+ k-2nh(knx,0)

+ k-2n\\G{0)\\.

Hence, by (20.10) and (20.14), it follows that lim d[2A(x), k-2nG{knx)]

= 0 for all x E X,

i.e. we have proved (20.15). In the next step we shall verify that A is a quadratic mapping. In fact, d[An(x + y) + An{x - y), k~2nG{knx) + k-2nG{kny)} <

2n

n

<

n

k- h(k x,k y).

Now, letting n —>• oo, in view of (20.9) and (20.15), we obtain the equality A(x + y) + A(x -y)

= 2A{x) + 2A(y)

for all x, y 6 X. The estimations (20.11) and (20.12) follow directly from (20.5) and (20.6) respectively. In the final step we shall prove

194

Functional Equations and Inequalities in Several Variables

the uniqueness part. To this end let assume that there are two quadratic mappings Cs: X -» Y, s = 1, 2, such that k—1 oo 2

d[d(x) + F(0), F(x)] < k- cn J2 ^2(k ~ m)H(mksx, m=l

ksx)k~2s,

s=0

for all x G X, and i = 1,2, where a* > 0, i = 1, 2, are some real constants. We leave the reader to verify that for i = 1,2 Ci(fcn:r) = A ; 2 " ^ ) ,

a; G X,

n G N.

Therefore, we get for a; G X, d[Ci(z),C2(x)] = A;-2nd[C1(fcnx),C2(A;nx)] fc —1

00

2

< (a! + a2)k~ J ^ J](A; - m)H{mks+nx,

ks+nx)k~2{s+n)

m = l s=0 k—1 oo

= (oi + a2)k~2 J2 J2(k ~ m)HimkSx, m=l

ksx)k~2s.

s=n

Since the series (20.8) and (20.9) converge, the right side of the last inequality can be made as small as we wish by taking n sufficiently enough. Hence it follows that Ci(x) = C2(x) for all x € X. Thus the proof is completed. • Lemma 20.5 Let X be a group divisible by k and Y a quasinormed space. If F,G: X —>• F satisfy the inequality (20.2), then d[F{x) + {k2n - l)F{0),k2nF(k-nx)] k—l

<

n

k 2

- ~ ] C H ( * ~ m)H(mk~Sx,

k~sx)k2s,

(20.16)

771=1 S = l

d[G(x) + {k2n - 1)G(0), k2nG(k-nx)] fc—1

71

k 2

- ~ ^2 J2(k 771=1

<

- m)K(mk-sx,

S=l

for all x € X and n, k G N, where

k>2.

k~sx)k2s,

(20.17)

Quadratic functional equation

195

Proof. Inserting x = k nt into (20.5) and (20.6) respectively, we get immediately (20.16) and (20.17). • Theorem 20.2 Let X be an abelian group divisible by k e N, k > 2 and Y a complete quasi-normed space. Let F,G: X —> Y satisfy the inequality (20.2). Suppose that for m = 1,2,..., k — 1, the series oo

^2h(mk'sx,k-sx)k2s,

(20.18)

s=l oo

^h(k-sx,0)k2s

(20.19)

s=l

are convergent for all x 6 X. If, moreover, \imMh{k-nx,k-ny)k2n

= 0 for all x,yeX,

(20.20)

7l->00

and F(0) = 0, then there exists exactly one quadratic mapping B: X ->Y such that fc—1 oo 2

^2(k ~ m)H{mk-sx,k-sx)k2s,

d[B{x),F{x)] < AT ^ m=l

(20.21)

s=l

fc—1 oo 2

d[B{x), G(x)] < k~ Y, 5 ^

_

m)K{mk-"x, k-sx)k2s

(20.22)

m=l s=l

/or a// a; e X . Proof. It follows from the convergence of the series oo

5>(0,0)A* that /i(0,0) = 0 and consequently by (20.2) we get F(0) = G(0) = 0. Define: Bn{x) := k2nF(k-nx),

xeX,

n e N.

196

Functional Equations and Inequalities in Several Variables

Applying Lemma 20.5, we can prove that for every x E X the sequence {Bn(x)} is a Cauchy sequence (see also proof of Theorem 20.1). So there exists the limit B{x) := lim Bn(x),

x E X.

(20.23)

7J-+00

It is easy to verify that 2B(x) = lim k2nG(k~nx),

x E X.

(20.24)

+ k2nG(k-ny)]

<

We also have d[Bn(x + y) + Bn(x-y), <

k2nG(k~nx)

k2nh{k-nx,k-ny).

Hence with respect to (20.23) and (20.24) taking limits of both sides as n —> oo, we get B(x + y) + B(x - y) = 2B(x) + 2B(y) for all x,y E X, i.e. B is a quadratic mapping. From (20.16) and (20.17) we derive immediately the estimations (20.21) and (20.22). The uniqueness part can be deduced similarly as in the proof of Theorem 20.1. • For the case F(0) ^ 0 we have the following result Theorem 20 Let X be an abelian group divisible by k G N, m c u i e m 20.3 > 22 andY and Y „a Banach space. Let the mappings F,G: X —>• Y k > „„. satisfy the ineaualit.ii inequality \\F(x + y) + F{x-y)-G(x)-G(y)\\
for all x,y E X. (20.25) Assume, moreover, that the series (20.18) and (20.19) are convergent for all x E X and the condition (20.20) is satisfied.

Quadratic functional equation

197

Then there exists exactly one quadratic mapping g: X —» Y such that k—l

oo

2

\\g{x) + F(0) - F(x)\\ < k' £ £(Jfc - m)H{rnk-'x, ATsx)fc2s, m=l

(20.26)

s=l

k—l

oo

||2ff(a:) + G(0) - G(x)\\ < fc"2 ^ J2(k - m)K{mk~sx, k~sx)k2s,

(20.27)

m=l s=l

/or a// x € X. //, moreover, X is a linear topological space and F is measurable (i.e. F _ 1 (f/) is a .Bore/ se£ in X for every open set U in Y) or the mapping R 3 t —> F(tx) is continuous for every fixed x € X, then g(tx) = t2g{x), xeX, t GR (20.28) Proof. Write f(x) := F(z) - F(0),

^(x) := G(x) - G(0)

for

x e X.

Then from Lemma 20.3 and (20.25) we get ||/(x + y) + /(a: - y) - 2/(x) - 2/(y)|| < //(a;,y) for all x,y £ X. Therefore, by Theorem 20.1 , g{x) := lim k2nf{k~nx),

xeX

n—too

exists and is the quadratic function satisfying the conditions (20.26) and (20.27). Now let L be any continuous linear functional defined on X. We define • R by
x 6 X,

teR.

where x is fixed. It is easy to see that (p is a quadratic function and, moreover, as the pointwise limit of the sequence
te R

198

Functional Equations and Inequalities in Several Variables

is measurable and therefore (for details see [128]) has the form tp(t) = t2cp(l) for t £ R. Consequently for every t G R and every xeX L[g(tx)] =
\a

x G ( - o o , - l ] U [l,+oo),

where a is a positive real number. Define / : R —> R by the formula oo

f{x) := ^k~2nf(knx),

xeR,

k>2.

n=0

Evidently, / is continuous and bounded by ak2(k2 — 1) _ 1 . We shall show that, in addition, we have \f(x + y) + f(x - y) - 2f{x) - 2/(y)| < &ak\k2 - l ) " 1 ^ 2 + y2) (20.29) for all x, y € R. Indeed, for x — y = 0 or x, y € R such that x2 + y2 > k~2, it is quite obvious. So consider the case 0<x2+y2

Then there is an r € N such that k~2r~2 <x2 + y2< k~2r whence k2r~2x2
and

k2r-2y2
(20.30)

Quadratic functional equation

199

Therefore, we get kr~xxt kr~ly, kr~\x

+ y), kr-\x

-y)e

(-1,1),

and consequently we have also for every n = 0, l , . . . , r — 1 knx,kny,kn(x

+ y),kn(x-y)e

(-1,1).

Thus we get if[kn{x + y)] +
\f(x

+ y)

+

f(x-y)-2f(x)-2f(y)\

00

< J2 k-2nWn{x

+ y)] + ip[kn(x - y)] - 2ip(knx) - 2<^(Py)|

n=0 oo

< ^ 6 a £ T 2 " - Qak2~2r{k2 - l ) " 1 < Qak4(k2 - l)-\x2

+ y2),

i.e. (20.29) holds true. Let us suppose that there exists a quadratic function g: R —> R such that \f(x) - g(x)\ < $x2 for all i £ l Then (see [128]) g has to be of the form g(x) = ex2, x € R for some constant c. Hence we get \f(x)\<(p

+ \c\)x2,

xeR.

(20.31)

Let us take r € N so that ra > /3 + \c\. If x £ (0, A;1_r), then knx G (0,1) for n < r — 1. Hence we have r—1

oo 2n

n

f{x) = ^k- ip(k x) n=0

>^2k-2na(knx)2 n=0

which contradicts the condition (20.31).

= rax2 > (0 + \c\)x2

200

Functional Equations and Inequalities in Several Variables Under some stronger assumptions, we may get the following

Corollary 20.1 (see also [24]). Let X be an abelian group and E a Banach space. If f: X —> E satisfies the inequality ||/Cr + y) + f(x -y)-

2f(x) - f(y)\\ < e

(20.32)

for all x, y G X, where e > 0 is fixed , then there exists exactly one quadratic function g: X —>• E such that \\f{x)-9{x)\\<\e

(20.33)

for all x G X. Moreover, the function g is given by the formula g(x) = lim [4-nf(2nx)],

x G X.

(20.34)

Notes 20.1 Let (G, +) be a group. A functional / : G —>• M. is said to be subquadratic iff

f(x + y) + f(x-y)<2f(x)

+ 2f(y)

for all x,y G G. R. Ger [71] found the following condition in order for a map F controlled by / to be asympotically close to a quadratic mapping Q. Proposition 20.1 Let (G, +) be abelian group and let (Y, || • ||) be an n-dimensional real normed linear space. Let further f: G —> R be a nonnegative subquadratic functional on G and let F': G —>• Y be a mapping such that \\F(x + y) +

F(x-y)-2F(x)-2F(y)\\<

< 2f{x) + 2/(y) - f(x + y)-

f(x - y)

Quadratic functional equation

201

for all x,y G G. Then there exists a quadratic function D: G -^Y

such that

\\F(x)-Q(x)\\
= 0 for all x G G.

fc-»oo

20.2 The stability of the quadratic equation on A orthogonal vectors has been studied by H. Drljevic in [50]. Namely he proved the following Proposition 20.2 Let X be a complex Hilbert space with dimX > 3 , A : I - } I f l bounded self-adjoint operator such that dim AX ^ 1,2 and h a continuous functional on X. Assume that 9 > 0 and

pe[o,2). if \h(x + y) + h(x -y)-

2h(x) - 2 % ) | < 6 [\(Ax,x)\* + \{Ay,y)\S

whenever (Ax, y) = 0, then H{x) := lim

2-2nh(2nx)

n—too

is a continuous functional on X such that H{x + y) + H(x -y)

= 2H(x) + 2H(y)

whenever (Ax,y) = 0. Moreover, there exists e > 0 such that \h(x)-H(x)\<\(Ax,x)\Se

forxeX.

20.3 S. M. Jung in [109] proved a Hyers-Ulam stability theorem on another quadratic equation of the form f(x + y + z) + f{x) + f(y) + f(z) = f(x + y) + f(y + z) + f(z + x)

[

'

'

202

Functional Equations and Inequalities in Several Variables

20.4 The general solution of the quadratic functional equation (20.35) for / defined on a field of characteristic zero has been determined by PI. Kannappan [155]. In particular, if / : X —> R is a solution of the functional equation (20.35), then there exist an additive mapping A: X —> M. and a biadditive mapping B: X x X ^R such that / has the form

f(x) = A(x) + B(x,x),

xeX.

Chapter 21 Stability of functional equations in function spaces Consider a group X and a normed space Y. Denote by Cf the Cauchy difference for a function f-.X-tY, Cf(x, y) := f(x + y)-

f(x) - f(y),

x, y G X.

Then the stability problem can be stated as follows. Given e > 0, does there exist a 5 > 0 such that if / : X —> Y satisfies \\Cf\\i<5, then an additive function a: X -^Y ll/-a||i<£

exists with ?

Here || • ||i denotes the supremum norm in the space of bounded functions defined in X, X x X respectively. Of course, several different norms in function spaces, can be taken into consideration. In the following we are going to discuss such a problem (see [193]). At first we shall study the stability of the Cauchy, Jensen and Pexider equations in generalized LP norms.

203

204

Functional Equations and Inequalities in Several Variables

Stability in Lp norms First we shall recall some definitions and notations from [193]. We assume that (X, +, J2-> I1) IS a complete measurable group, i.e. that the following conditions are satisfied : (a) (X, +) is a group, (b) (X, Y^,, /-0 is a cr-finite measure space, // is not identically zero and is a complete measure, (c) the cr-algebra Y, a n d the measure // are invariant with respect to left translations, (d) yu x JJ, is the completion of the product measure, (e) the transformation S: X x X 3 (x, y) -> (x, x + y) is measurability preserving, i.e. S and S~x are measurable. Moreover, in this section, we assume that y.{X) = oo. An example of such measurable group is R" with the Lebesgue measure. Let us note that the measure fj, with conditions (a) - (e) is invariant under translations and under symmetry with respect to zero and S and S~l preserve the measure \i x \i. The symbols f — goxf — g will denote that / and g are equal almost everywhere with respect to the measure fi or JJL X /X, respectively. As usual, for a measurable space (Y, v) by L(Y, R) we denote the space of all integrable functions / : Y —> R. If / : Y —> R is non-negative, then we define the upper integral by J

fdv :=inf IJ

or fY fdv — oo if there is no integrable ip for / .

f{x)<
Stability of functional equations in function spaces

205

Let (E, +) be an abelian group and d be a metric such that d (x, y) = d (a + x, a + y)

for all

a,x,y G E.

We denote ||z|| : = d ( z , 0 ) ,

z€£.

Then it is easy to see that the following conditions are satisfied : (f) ||a;|| = 0 ^ x = 0,

x£E,

(g) \\x + y\\ < \\x\\ + \\y\\, (h) | | - x | | = \\x\\,

x,y G E,

xeE.

Given a measure space (Y, v) and p > 0, we define L p + (F,£) := {/: X^Y;3
L(Y,R), \\f(x)\\p < tp{x)}.

The following Lemma shows that L+(Y,E) addition.

is closed under

L e m m a 21.1 Let Y be a measure space, let p > 0, and let f,geL+(Y,E). Thenf + geL;(Y,E). Proof. From the definition of L+(Y, E), there exist (p,ip,£ L+(Y,E) such that \\f(x)\\><
and

\\f(x)\\>(x).

Hence we get ||/(x) + g(x)\\' < (\\f(x)\\ + \\g(x)\\y < (2max{||/(x)||, \\g{x)\\})> < 2P(\\f(x)\\p + \\g(x)\\r) < 2P((*)), which proves that / + g G L+(Y,E). For any f:X—>E

•

and x0 e Y we put

fx0{x):=f{x

+ x0),

xeX.

206

Functional Equations and Inequalities in Several Variables

Lemma 21.2 Let f e L+(X, E) and let p > 0. Then fx0 G L+(X, E)

for every

x0 e X.

Proof. Fix arbitrarily an x0 £ X. There exists a ip € L(X, R) such that \\f(x)\\*<
for

xeX,

i.e. x0(x)

for

^ £ -X".

From the condition that u. is invariant under translations, cpXo £• L(X, R), which completes the proof. • By f{-,y) we mean the function obtained from / by fixing the second variable at the point y. Lemma 21.3 Let p > 0 and let f e L+(X x X,E). exists a set A C X such that fi(A) — 0 and f(-,y)eL+(XtE)

for

Then there

yeX\A.

Proof. There exists a function ip e L(X x X,R) such that \\f{x,y)\\p<
for

x,yeX.

From the well known Fubini Teorem (cf. [82]) there exists a subset Ac X such that u.(A) = 0 and
for

y€*\i4,

which proves our Lemma. Lemma 21.4 Let A C X and fx(A) = 0. / / D := {(z, i / ) 6 l x I : a ; e A V i / e ^ V 3 ; + ! / G 4 £/ien (/x x //)(£>) = 0.

•

Stability of functional equations in function spaces

207

Proof. We have D = (A x X) [J (X x A) U {(x,y) eXxX:x = (A x X) U (X x A) U S~\X

+

yeA}

x A),

where S is the transformation defined in (e). In view of the facts /J,(A x X) = 0, /J,(X x A) = 0, we have /i(5' _1 (X x A)) = 0, which yields the assertion of the Lemma. Lemma 21.5 Let G be an abelian group, let 61,62 £ G and let Oi, 02: X —y G be additive functions such that ai(x) + 61 = a2(x) + 62,

x E X.

Then a\ = a2 and 61 = 62 . Proof. From the hypotheses, there exists a subset U C X such that //(£/) = 0 and a 1 (x) + 6 1 = a 2 ( z ) + 6 2

for

x€X\U.

(21.1)

Take x £ X. Because //([/ U ( - [ / + x)) = 0, then there exists a y e X\{UU (-U + x)). Therefore, y e X \ U and also x - y E X\U. Hence by (21.1) we obtain ax(x) - a2(x) = al(x-y)-

a2(x - y) + ax{y) - a2(y) = 2(62 - 61),

i.e. ai(ar) - a2(x) = 2(62 - 6X)

for

x GX

(21.2)

Consequently in view of (21.1), we get 2(6 2 - 6 1 ) = 62 - 6 1 ,

which implies that 61 = 62 . Finally, by (21.2) we obtain ai = a2 and the proof is concluded. • Now we are in a position to formulate the following

208

Functional Equations and Inequalities in Several Variables

T h e o r e m 21.1 Assume that f: X -^ E is such that Cf G L+(Xx X, E) for a certain p > 0. Then f(x + y)^f(x)

+ f(y)

for x,y e X.

Moreover, there exists an unique additive mapping a: X —> E such that f(x) = a(x) for x G X. Proof. On account of Lemma 21.3 there exists a subset A C X such that n(A) = 0 and Cf{-,y)eL+(X,E)

for yeX\A.

(21.3)

For the proof of the first part of the theorem, it is enough to verify that Cf(u,v) = 0 for u,v,u + v eX\A. (21.4) Now, for u,v,x G X, we obtain Cf(u, v) = f(u + v)- f(u) - f(v) = [f((x + u) + v)- f(x + u) - / ( « ) ] - [/(* + (« + v)) - f(x) - f(u + v)}

+ [f(x + u)-f(x)-f(u)} = Cf(x + u,v)- Cf(x, u + v) + Cf(x, u). This means that Cf(u,v)

= (Cf(;v))u(x)-Cf(;U

+ v)(x)+Cf(;U)(x),

(21.5)

for x,u,v G X. Take u, v G X \ A such that u + v G X \ A. In view of Lemma 21.2 , Lemma 21.1 and (21.3) we see that the right side of (21.5) (as a function of x) belongs to L+(X, E), which means that

/

Jx

Cf(u,v)\\p dfi{x) < o o .

Stability of functional equations in function spaces

209

In view of the fact that fJ.(X) = oo, hence it follows Cf(u, v) = 0 for

u,veX\A,

u+v e

X\A

and concludes the proof of (21.4). One can rewrite the condition (21.4) in the form Cf(u, v) = 0 for

(u, v) e (X x X) \ D,

where D is as in the Lemma 21.4 . Since by this lemma, (u. x n)(D) = 0, we have proved the first part of the theorem. Hence and by the Theorem 17.6.1 from [123], we obtain the second assertion. This concludes the proof of the theorem. • Let us note that Theorem 21.1 states that the boundedness of the Cauchy difference of / by an integrable function implies that mapping / is a solution of the Cauchy equation almost everywhere. We could say that in this case we have a phenomenon of almost superstability. Passing to the Jensen equation, assume that Y, Z are uniquely 2-divisible groups. For / : Y —>• Z we define

J f ( ^ y ) : = / ( ^ ) - ^ ) + /(y)] for x,yeY Theorem 21.2 Assume that a group X is abelian and groups X,E are uniquely 2-divisible. Let for p > 0 and f:X —> E, Jf e L£(X x X,E). Then there exist a unique additive mapping a: X —>• E and a unique constant b G E such that f(x)=a{x)+b

for

x e X.

(21.6)

Proof. In view of Lemma 21.3 there exists a subset A C X such that fi(A) = 0 and H(;y)eL+(X,E)

for all y E X \ A.

(21.7)

210

Functional Equations and Inequalities in Several Variables

Let us fix an x0 £ X \ A and set g(x):=f(x

+ x0)-f(x0)

for

x £ X.

(21.8)

We shall prove that g(u + v) =g(u) + g(y)

for

u,v,u + v G X \ (A - x0).

(21.9)

We obtain for u, v, x £ X, g(u + v) - g(u) - g(v) = f(u + v + x0) - f(u + x0) - f(v + x0) +f(x0) — 2Jf(x + u + x0,v + v0) - 2Jf(x + x0,u + v + xQ) —2Jf (x + u + xQ,xQ) + 23i(x + XQ,U + XQ). We have obtained the equation Cg(u,v) = 2Jf(x + u + x0,v + x0) — 2J£(x + x0,u + v + xQ) -2Jf (x + u + x0,x0) + 2Jf (x + x0,u + xQ) (21.10) for u,v,x E X. If u,v,u+v G X\(A — x0), then u+xo,v+Xo,u+v+Xo G X\A and, by assumption, x0 G X\A. Therefore from (21.7) and Lemma 21.2, the right side of (21.10), as the function of x ,is an element oiL+(X,E). Thus /

Jx

Cg(u, v) dfji{x) = Cg(u,v)fj,(X)

< oo,

and since [i(X) = oo, we get Cg(u, v) = 0, i.e. (21.9) holds true. Denote

Dg := {(u, v) £ X xX: ue A-x0Vv

£ A-x0\/u

+v e

A-x0}.

Lemma 21.4 implies that (^ x fJ,)(Dg) = 0. Now (21.9) can be rewritten as g(u + v) = g(u) + g(v)

for

u + v £ (X x X) \ Dg.

Stability of functional equations in function spaces

211

From Theorem 17.6.1 [123] applied for the function g it follows that there exists an additive mapping a: X —>• E such that g{x) = a(x)

for x G X.

Hence and by the definition (21.8) we obtain f(x) = g(x - x0) + f(x0) = a{x - x0) + f(x0) = a(x) - a(x0) + f{x0) for x e X, i.e. f(x)=a(x)

+ b,

where b = f(x0) — a(x0). The uniqueness of a and b is a simple consequence of Lemma 21.5 , which concludes the proof. • The next theorem will provide an answer to the question when does a function / : X —> E with its Jensen difference belonging to L+(X x X,E) is also almost Jensen. Theorem 21.3 Assume that a group X is abelian and the groups X, Y are uniquely 2-divisible. Let, moreover, the following condition be satisfied fi(A) = 0 implies fi(2A) = 0. Let f:X —> E be a mapping. equivalent :

The following conditions are

(i) Jf € L+(X x X, E) for a certain p > 0, (ii) f is of the form (21.6), (in) f (*-¥)"= l[f(x) + f(y)} for x,y EX . Proof. From Theorem 21.2 it follows that (i) implies (ii). Now assume that (ii) is satisfied, i.e. f(x)=a(x)+b

for xEX\U,

(21.11)

212

Functional Equations and Inequalities in Several Variables

where a: X —»• E is an additive function, b G E, and U
x

X)\D,

which means that (iii) holds true. The inverse implication (iii) —>• (i) is obvious, and the proof is completed. • Now we will consider the Pexider equation. For mappings / , g, h: X -» E the Pexider difference P(f, g,h): X x X ->• E is defined by P(f,9,h)(x,y):=

f(x + y)-g(x)-h(y)

for

x,y € X.

Then the following theorem can be proved (see [193]). Theorem 21.4 Let X,E be abelian groups. conditions are equivalent :

The following

(i) P(f, g, h) G L+(X x X, E) for a certain p > 0, (ii) There exist an unique additive mapping a: X —> E and unique constants a, (3 G E such that f{x) = a{x) + a + P, g(x) = a(x) + a, h{x) =a(x) + /?, for x G X, (iii) f(x + y) ^

g(x) + h(y) for x, y G X.

Chapter 22 Cauchy difference operator in Lp spaces As we have seen, under the assumption that the measure of the space is +00, the case of superstability occurs. Now we shall investigate the case when this measure is finite (n(X) < 00). As in the proceding section, let (X, ^ , /j,) be a complete measure space, where (i is not identically zero. Given A € ^ , we define Ay = {t e X: ty e A},

yA

= {t<E X: yte

A},

where X is a nonempty set with the binary operation „•" . We will assume:

(r) AyeJ2

and

(0 yA^T,

and

KAy) = V(A)

for

rivA) = v(A) ^

A e

£>

V e X>

A e E, y£ x.

For / : X ->• E, y 6 X, we define yf,fy: X ^ E by y/(z) := /{yx),

fy{x) := /(xy).

Let us recall that for a complete measure space X and 1 • £ : / i s measurable and ||/|| p < 00}, 213

214

Functional Equations and Inequalities in Several Variables

where

\\f\\P=(Jx\\f(t)\\Pd»(t)y, ||/||oo = esssup t e X ||/(t)||,

if l
The following lemma will be used further on (for a proof see Siudut [193]). Lemma 22.1 Let (X,^,/^) be a complete measure space with fj,(X) < oo satisfying (r) or (I) and let E be a Banach space and y G X. If, moreover, f G L1(X,E), then fy G L\X,E) yf

G L\X,

E)

and

/ /„(*) dfj,(t) = f f(t) dfi(t), (22.1) Jx Jx

and

f yf(t) dfi(t) = f f(t) dfi(t). (22.2) Jx Jx

Stability of the Cauchy equation We start by presenting the result of S. Siudut contained in [193], p.27. Theorem 22.1 Let X be a semigroup and let (X, ^ A O be a complete measure space satisfying at least one of the conditions (r), (I). Assume that the mapping S: X x X 3 (x, y) —>• (x,xy) G X x X is measurable. Let E be a Banach space. If, moreover, f: X -> E is such that Cf G LP{X xX, E) for a certain 1 E such that: (i) Cg=0,

(ii)

A = fj, x fj,,

f-geIS(X,E),

(in) | | / - ^ | | p < / * W ^ | | C ! f | | p >

Cauchy difference operator in LP spaces

215

(iv) Qi = 9 for every map g\: X —>• E satisfying the conditions Cgi = Q,f-gieLP{X,E) . Proof. Assume that the condition (r) is satisfied, 1 {X xX, E)
(?/) := f IfM Jx

t

~ f(x) - f(y)] dn{x)

[f(xy)-f(x)]dli(x)-f(y).

Therefore by the Fubini theorem is defined for //-almost all y G X, thus there exists M G J^, n{M) = 0 such that 0 is defined for y G X \ M, and is also measurable. Define g: X —¥ X by the formula , f f[f(xy)-f(x)}dfi(x) g(y) = J ^

,xeX\M, (22.3)

I o

,xeM.

Then g — f ~4>- Moreover, by Lemma 22.1 we have for y,z,yz X\M, g{y) + g(z) = [ \f(xy) - /(*)] d»(x) + [ [f(xz) - f(x)] d/x(x) JX

JX

= / [f((xy)z) - f(xz) + f(xz) - /(*)] d^x(x) Jx = / [ / f a M ) - f(x)] dn(x) = g{yz). Jx This means that g(y) + g{z) = g{yz) 2

for (y,z) e X \(M

x X U X x M [J S'^X

x M)).

G

216

Functional Equations and Inequalities in Several Variables

Since S is measurable, the set S~1(X x M) is measurable too. We have also S~l(X

x M) = {(y,z) EXxX-.ye

Mz}.

By (r), Mz <EJ2 and X(MZ) = X(M) = 0 for every z € X. Hence by the Fubini theorem A[5-X(X x M)] = 0. Clearly X(M x X) = \{X x M) = 0. Therefore A j M x I U l x M u S ^ I x M)] = 0, which completes the proof of (i). We have also (^(X) = 1)

\g(v)-f(y)\\ <

=

[ Cf(x,y)d(i(x) Jx

< [ Jx

\\Qf(x,y)\\dn(x)

(J WCffayW'dnix)

for [x almost all y G X. By the Fubini theorem we get hence / h{y)-f{y)\\vdn.{y) Jx

= [

< [ diM(y) f \\Qf(x,y)\\>dvL(x) Jx Jx

\\Cf(x,y)\\pd\(x,y).

JxxX Thus, the conditions (ii) and (iii) for fi(X) = 1 and 1 1. Consequently by the first part of the proof, for every 1 < s < oo there exists a map g: X —» E satisfying the conditions (z), (ii) and (iii) (g does not depend on s). Thus, we have f-geLs(X,E),

||/-
for

1 < s < oo.

Cauchy difference operator in IP spaces

217

Further, we obtain f-geL°°,

Km | | / - 0||, < lim \\Cf\\s s—>oo

s-¥oo

i.e. ||/-S||oo<||C/||oo

(for details see [187], Chapter 3). Therefore, we have proved (i), (ii), (Hi) for n(X) = 1. In the general case 0 < p(X) < 00, it is enough to consider the measure /j,(X)~lfi . Finally, to prove (iv), assume that for gx: X —> E we have (7^ = 0 a n d / - # i eIS(X,E).

Put h := g - gx, then If(X, E) c Ll(X, E)

h = (f-gi)-(f-g)e and (7/i=0, so Ch e L\X

x X,E).

By Lemma 22.1 ,

/ Ch(x,y) d/j,(x) = / [hy(x) - h(x)] dji(x) Jx Jx =

h(y)n(X)

-h{y)fi(X)

for /i-almost all y € X, whence

fi(X)\\h(y)\\ < [ \\Ch{x,y)\\drtx). Jx

Consequently, by the Fubini theorem fi(X) [ ||%)||d/x(j,)< f JX

\\Ch(x,y)\\dX(x,y)

= 0,

JxxX

which implies \\h\\i = 0, i.e. gx = g. If the case (/) occurs, the proof can be given similarly. • Remark 22.1 Under some additional assumption on the measure \i, one can prove that there exists a unique additive map A: X —>• E such that / - A e L p (X, E) and \\f-A\\p
218

Functional Equations and Inequalities in Several Variables

We say also that the pair (LP {X, E), LP {X x X,E)) has the double difference property, i.e. for every / : X —»• E such that Cf G LP(X x X, E) there exists an additive map A: X —> E such that f -AeLP(X,E).

Cauchy difference operator In this section we shall consider the Cauchy difference as a linear operator. Some characteristic properties of such operators shall be established. We start by presenting the result of S. Siudut [193]. T h e o r e m 22.2 Let X be a non-empty set with a binary operation 0 . Let (G, X^>AO be a measure space satisfying at least one of the conditions (r), (I). If 0 < fJ,(X) < oo, / € LP for some 1 f(x o y) is measurable, then (i)

CfeLP(XxX,E),

(a) ||/||p<M*)^l|tf/llp, (in) \\cf\\p<MX)Hf\\P, (iv) (7/ = 0 = > / ^ 0 . Proof. Similarly as in the proof of Theorem 22.1 , we may assume that n(X) = 1. Let us also assume that (r) is satisfied. By Lemma 22.1 we have 11/11? = f \\f(x)\\pd„(x)= Jx

[ Jx

\\fy(x)\\'dpix)

= [ d»{y) f 11/^)11'd/i(a;)
Jx

and consequently by the Fubini theorem the mapping m\: (x, y) —> f(xoy) is an element of LP{X xX, E) and ||mi|| p < ||/|| p . Clearly

Cauchy difference operator in U spaces

219

the mappings ra2: (x,y) -» f(x) and m 3 : (x,y) -» /(y) belong to jy(A" x X , £ ) , thus C"/ belongs to U(X x X , £ ) as well. This proves (i). Now we have II^/IIP

= ll m i -mv-

m3\\p < \\mi\\p + ||m 2 || p +

+||m 3 || p < 3II/H,, which implies (iii). Let us note that

Cf(;y)=fv(-)-f(-)-f(v)eV(X,E!) CL\X,E) and again by Lemma 22.1 ,

L Cf(x,y)d ,(x) = f

-f(y).

Hence we get <7 = 0, where g is the mapping defined by (22.3). Moreover, by (i) from Theorem 22.1 we obtain (ii). Finally, (ii) implies (it;) and concludes the proof of the theorem. • Furthermore, let us denote Dc := {/ G Lp(X, E):Cfe

If(X

x X, E)}.

Then, we have the following. T h e o r e m 22.3 Let the assumptions of Theorem 22.2 be satisfied. Then the linear operator C: Dc 3 f ->• Cf € LP[X x X,E) is continuous and invertible. The inverse operator defined for h G C(DC) is continuous and has the form — (//(X) _1 / h(x, •) dfx(x) if (r) occurs, Jx

c-'hu = { -(fi(X)'1

h(-,y)dfi(y)

if (I) occurs.

220

Functional Equations and Inequalities in Several Variables

Proof. The continuity of C follows directly from (iii) of Theorem 22.2 , as well as the existence and continuity of C"1 results from Theorem 22.2 , (ii) and well known theorem from functional analysis concerning continuity and invertibility of linear operators. Finally, we have f Cf(x, y) dn(x) = / [f(x o y) - f(x) - f(y)} df,(x) JX

JX

= Jx =

f[fy(x)-f(x)]dlx(x)-^(X)f(y) -»(X)f(y),

i.e.

f(y) = -n(X)-1

[ Cf(x,y)dn(x) Jx if (r) occurs. This gives the formula for C _ 1 if (r) occurs. In the case when (/) is satisfied, we obtain the formula for C _ 1 by the same manner. This completes the proof. •

Chapter 23 Pexider difference operator in Lp spaces Let E be a Banach space and let (X, +,S,/x) be a semigroup with zero and with a complete not identically zero measure // such that /j,(x) < oo. For any mappings f,g,h: X —>• E we define the Pexider operator as follows P{f, g, h)(x, y) := f(x + y)- g{x) - h(y),

x, y € X.

Then we have the following.

Lemma 23.1 For every f,g,h: X —> E Cf(x,y) = P(f,g,h)(x,y)-P(f,g,h)(0,y)-P(f,g,h)(x,0) -g(0)-h(0), (23.1)

Cg(x,y) = P(f,g,h)(x,y)-P(f,g,h)(x -P(f,g,h)(0,y)-g(0),

+ y,0) +

P(f,g,h)(y,0) (23.2)

Ch{x,y) = P(f,g,h)(x,y)-P(f,g,h)(0,x -P(f,g,h)(x,0)-h(0).

+ y) +

P(f,g,h)(0,x) (23.3)

221

222

Functional Equations and Inequalities in Several Variables

Proof. Take x,y e X. Clearly, we have Cf(x, y) = f(x + y)- f(x) - f(y) = f(x + y)- g(x) - h(y) +g(x) + MO) - f{x) + h(y) + g(0) - f(y) - g(0) - h(0) = P(f,g,h) (x, y)-P(f,g, h) (x, 0)-P(f,g, h) (0, y) -g(0)-h(0). The proofs for the other two properties can be given quite similarly.

• Recall that yA

= {teX:y

+ teA},

Ay = {t G X: y + t G A},

A G E.

If / : X —> E and y G X, then we denote fy(x):=f(y

+ x),

fy(x) = f(x + y).

For / : X x X ->• E we put f°{x):=f{0,x), f+{x,y):=f(0,x

/o(x):=/(ar,0), + y).

f+(x,y)

:= f(x+

y,0),

Stability of the Pexider equation In this section we are going to give a positive answer to the problem of Hyers-Ulam-Rassias stability of the Pexider equation f(x + y) = g(x) + h(y),

x,y e X.

All the results in this spirit that we are going to present in the sequel are substantially due to S. Czerwik and K. Dlutek [28]. By A we understand the completion of the product of the measure // times itself. Now comes

Pexider difference operator in LP spaces

223

Theorem 23.1 Let X be a semigroup and let (X, +, E,//) be a complete measure space satisfying at least one of the conditions (r),(l). Assume that the mapping S: X x X 3 (x,y) —> (x,x + y) £ X x X is measurable. Let E be a Banach space and let f,g,h: X —y E be such that P(f,9, h), P(f,g, h)+, P(f,g, h)+ € U{X

xX,E),

P(f,g,h)°,P(f,g,h)0eL*>(X,E) for a fixed 1 (X,E)

, then B = A,

b) ifD:X->E,CD

= 0,g-D£LP(X,E)

, then D = A,

c) ifH:X-*E,CH

= 0,h-DeLP(X,E),

then H = A.

Proof. From Lemma 23.1 it follows that Cf,Cg,Ch

£Lp(XxX,E).

Therefore, making use of Theorem 22.1, we realize that there exist A, Ai, A2: X -¥ E such that f-A,g-Auh-A2eLp(X,E),

(23.4)

Wf-AWp^fxix^wc/Wp, b-^HP<M*)^IIQ?IU

(23-5)

CA = CAl = CA2 = 0,

\\h-A2\\p
224

Functional Equations and Inequalities in Several Variables

Claim that A = A\ = A
Pexider linear operator We are ready to prove the following Theorem 23.2 Let X be a nonempty set with a binary operation +. Let (X, E, (j,) be a measure space satisfying one of the conditions (r),(l) and 0 < n{x) < oo. If f,g,h G LP(X,E) for a certain 1 • g(x + y), X23(x,y)->h(x

+ y)

are measurable in the product space, then (v) Operator: (/, g, h) —> P(f, g, h) is linear, (vj)

P(f,g,h)eI/(XxX,E),

(vjj) Ifh{Q)=g(0)

= 0, then

\\f\\P + \\9\\P + \\h\\P < 3/i(X)f (||P(/, g, h)\\p + \\P(f, g, h)+\\p +\\P(f, g, h)+\\p) + 3(\\P(f, g, h)% + \\P(f, g, h)0\\p),

Pexider difference operator in IP spaces (m) lfh(0)=g(0)

225

= 0, then

l*(X)T(\\P(f, g, h) ||p + ||P(/, g, h)+\\p + \\P(f, g, h)+\\p) +\\P(f,g, h)\ + \\P(f,g, h)Q\\p < 5(||/|| p + \\g\\P + \\h\\p). Proof. Assume that f,g,h,k,l,m,£ IP and s G K (the set of real or complex numbers). Then by the definition of the Pexider diference, we obtain P(f + k,g + l,h + m) = P(f, g, h) + P(k, I, m), P(sf,sg,sh) Thus P is a linear operator.

= sP(f,g,h).

Define lr: X x X ->• E, for r =1,2,3, by h(x, y) := f(x + y), l2{x, y) := g(x), l3{x, y) := % ) . By the Fubini-Tonelli theorem, we get ||Z2||J= /

\\l2(x,y)\\<>d\(x,y) =

JXxX

= [ d/i(y) / \\g(x)rdf,(x) = n{X)\\g(x)\\'p < oo, Jx Jx p and hence l2 € £ . By the same arguments we can verify that h € Lp and ||Z3||£ = //(a;)||/i||P. In view of Lemma 22.1 we obtain

ll/lg= f l l / C ^ W * ) = f \\f(x + y)\\pd»(x) Jx

Jx

= ^X)~1

f d/i{y) [ \\h(x,y)\\pdLi(x) Jx Jx

= fi(X)-1 [

\\h(x,y)\\'d\(xty)

= M*) _1 ll'ilB,

JxxX

i.e. ||/|| p = n(X)^\\h\\p. Therefore P(f,g,h) = h - l2 - l3 £ P L (X x X, E) and (vj) has been proved. Moreover, we have also proved that \\P(f,g,h)\\p<\\i^

+ \\i2\\p + \\h\\p = Kx)H\\f\\P + \\9\\v + \\HP)(23.7)

226

Functional Equations and Inequalities in Several Variables

In the following we shall prove (vjj). get the inequalities

Aplying Theorem 22.2 we

11/11, < M*)'H3flU WQWP < i*(x)*\\Cg\\P, \\h\\P <

»(x)r\\ch\\P.

Hence by Lemma 23.1, we obtain

ll/ll, + \\9\\P + WHP < n(X)H\\Cf\\p + \\Cg\\P + \\Ch\\p) < ti(X)H\\P(f,9,h) - P(f,g,h)° - P(f,g,h)Q\\p + IIW, 9, h) - P(f, g, h)+ - P(f, g, h)° + P(f, g, h)0\\p + IIW, g, h) - P(f, g, h)+ - P(f, g, h)0 + P(f, g, h)%) < 3fi(X)H\\P(f, g, h)\\p + \\P(f,g, h)+\\p + \\P(f, g, h)+\\p + S(\\P(f,g,h)X + \\P(f,g,h)0\\p). i.e.

(vjj). Similarly, one can get

\\P(f, g, h)\

< ll/ll, + 11%,

||P(/, g, h)0\\p < ll/ll, + ||%,

\\P{f,g,h)+\\P
+ \\gU

\\P(f,g,h)+\\p
zVxVxU:

P(f,g,h)+

e U>,

P(f, g, h)+ € U, P(f, g, h) e V, g(0) = h(0) = 0}. In DP (linear subspace of Lp x IP x V) we define the norm:

\\(f,g,h)\\ := ||/||p + |M|„+ llfcllp, (f,g,h) e DP,

Pexider difference operator in LP spaces

227

&ndmD>(XxX,E):

\\\F\\\ := M * ) ^ ( I I * 1 I P +

II^IIP

FGLP(X

+ ll*+U +

II^IIP

+ ll^ollp,

xX,E).

Now we are in a position to present the following Theorem 23.3 Let the assumptions of Theorem 23.2 be satisfied and let g(0) = /i(0) = 0. Then the linear operator P-.DP3

(f,g,h)

- • P{f,g,h)

G LP(X x X,E)

is continuous and continuously invertible. Moreover, the inverse operator (defined for all m G P(DP)) has the form: p-'mi-)

= [-//(X)- 1 f [m(x, •) - m0(x) - m°(-)]d^(a;); Jx - n(X)-1

/ [m(x, •) - m+(x, •) - m°(-) + mQ(-)]dfi(x); Jx - ii{X)~x I [m(x, •) — m+(x,-) — m0(x) + m°(x)]du.(x)]

Jx if (r) occurs;

P -i m (.) =

[-^X)-1

f [m(-,y) - mo(-) - m°(j/)]d/i(y); Jx

- /i(X)- 1 f [m{-, y) - m+{; y) - m°{y) + m0(y)}dfj,(y); Jx - n(X)-1

f [m(; y) - m+(; y) - m0(-) + ro°(-)]d/i(y)] Jx

if (I) occurs. Proof. The continuity of P follows from Theorem 23.2, (vjjj). The existence of the inverse operator P _ 1 and the continuity of

228

Functional Equations and Inequalities in Several Variables

P~l are consequences of Theorem 23.2, (vjj) and Theorem 3, p. 43 from [216]. To find the form of P _ 1 , let's assume the condition (r) (in the case (1) proof runs quite similarly). Since Cf, Cg,Ch € Ll(X,E), from Lemma 22.1 we get

f(y) = -niX)-1 [ Cf(x,y)d^(x), Jx

g(y) = -^(X)-1 [ Cg(x,y)d^(x), Jx

h(y) = -ii(X)-1

[ Ch(x,y)d»(x). Jx Hence by using Lemma 23.1 we obtain the expressions f(y)=

-^(X)-1 Jx -P(f,g,h)(0,y)}dti(x), -HI)'

f[P(f,g,h)(x,y)-P(f,g,h)(x,0)

1

g(y)=

f[P(f,g,h)(x,y)-P(f,g,h)(x + y,0) Jx + P(f, g, h)(y, 0) - P(f, g, h)(0, y)W(x),

h(y)=

-n(X)-1

f[P(f,g,h)(x,y)-P(f,g,h)(0,x Jx

+ y)

+ P(f, g, h)(0, x) - P(f, g, h)(x, 0)}dfx(x), which give the form of P - 1 . The proof is concluded.

•

Chapter 24 Cauchy and Pexider operators in X\ spaces Let X and Y be normed spaces. Given A > 0 consider a function / : X —> Y satisfying the condition ||/(x)|| < M(/)eAIMI,

*eX,

where M(f) is a constant depending on / . The space whose elements are all such functions, will be denoted by Xx. Define

for

feXx

ll/H :=sup{e-^H||/(x)||}. xex Then cleary the following holds.

(24.1)

Lemma 24.1 The space Xx with norm (24-1) is a linear normed space. Recall that C(f)(x,y)

:= f(x + y) - f(x) - f{y),

x,y € X

Thus C(f): X x X -> Y and if / e Xx, then ||C(/)(x,y)||<3M(/)e^l x H + l | ! ' l l ) 229

for

x,y E X.

230

Functional Equations and Inequalities in Several Variables

The space of all mappings g: X x X —> Y satisfying the condition

\\g(x,y)\\<M(gy^+^\

x,yeX,

where M(g) is a constant depending on g, is denoted by X%. We define | M | : = sup{e- A (NI + W|| 5 (x ) ? / )||}. (24.2) x,y£X

Lemma 24.2 The space X\ with norm (24-2) is a linear normed space. The proof is just straightforward.

Cauchy operator Following the methods from the paper [27] of S. Czerwik and K. Dlutek, we shall prove the following. Theorem 24.1 The Cauchy difference operator C: X\ —> X\ is a linear bounded operator and it satisfies the inequality | | C ( / ) | | < 3||/||

for all

/ G Xx.

(24.3)

Proof. In fact, we have the inequalities for / G X\ ||C(/)|| := sup {e-^\+^\\f(x

+ y) - f(x) -

f(y)\\}

x,y£X

< sup { c - * M + » ( | | / ( s + y)\\ + \\f(x)\\ + \\f(y)\\)} x,y£X

< sup {e-x^+^(\\f(x x,y£X

+ y)\\} + sup{e- A ll x ll||/(x)||} x(zX

A

+ sup{c- «»»||/(y)||} = 3||/||, yex

i-e- I|C(/)||<3||/||. Under some additional assumptions we have

•

Cauchy and Pexider operators in Xx spaces

231

Theorem 24.2 LetR+ C X,R± cY and \\x\\ = \x\ Then \\C\\ = 3.

forxeR±. (24.4)

Proof. Consider a decreasing sequence {xn} of non-negative real numbers such that xn —>• 0 as n —>• oo. Define for n G N, fi^^^Tl

rp

e 0

rp

, x = Zxn> , otherwise.

Cleary we have ||/n(a:)|| < eAx"eA||a:|1

for x G X

and hence /„ G A\ for all n G N. Moreover,

-A||x||

ll/n(*)|| =

x

jX

—

ZXji)

0

, otherwise.

which implies that ||/ n || = eXxn for n G N. Finally, we get ||C(/„)|| = sup {e-x^+^\\fn(x

+ y)- fn(x) -

fn(y)\\}

x,y€X > e-2Xx»\fn(2xn)

- 2fn(xn)\

= e -2^| e 2Ax„

+

2e2Ax„| =

3

Now suppose that ||C|| < 3.Then there exists an e > 0 such that | | C ( / ) | | < ( 3 - e ) | | / | | for all / G Xx. On the other hand, for fn G X\, we have 3<||C(/n)||<(3-e)||/n|| = (3-e)eAi;"^3-e which is imposible.

as n ^ oo, •

232

Functional Equations and Inequalities in Several Variables

Pexider operator We define Xl:={(f,g,h):f,g,heXx},\$f,g,h)\\:=ma4\\f\\,\\9\\,\\h\\}, (24.5) for f,g,he X\. It is obvious that Xl with norm (24.5) is a linear normed space. We can prove the following (see[27]). Theorem 24.3 Pexider operator P: X% —> X\ is a linear bounded operator and ||P(u)||<3||«|| Proof. Take f,g,he

for all

u e X\.

X\, then we have by the definition (24.2),

\\P(f, g, h)\\ = sup {e-A(INWMI)||/(x + y) - g(x) - h{y)\$ x,y£X

< sup {e- A ll^ll(||/(a; + 2/ )||} + sup{e-Alla;ll(||^(a:)||} x,y€X

xeX A

+ 8up{e- ll"»||%)||} = H/ll + \\g\\ + \\h\\

yex < 3ma X (|j/||,||^||,||^||) = 3||(/,^,/ i )||, i.e. \\P(f,g,h)\\<3\\(f,g,h)\\, which completes the proof.

•

Corollary 24.1 // R+ C X, K+ C Y and \\x\\ = \x\ for x e 1+, then \\P\\ = 3. Proof. Assume on the contrary that ||P|| < 3. Then for / = g = h, we get

\\P{f,g,h)\\

= \\C(f)\\ < | | P | | | | ( / , / , / ) | | = \\P\\\\f\\,

Cauchy and Pexider operators in X\ spaces

233

whence l|C(/)||<||P||||/||

for all

f€Xx.

By the hypothesis, ||P|| < 3 and therefore we infer that ||C|| < 3, which is imposible in view of Theorem 24.2. •

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Chapter 25 Stability in the Lipschitz norms Following ideas of Jacek Tabor [193] we are going to deal with the problem of stability in the Lipschitz norms. Let us now recollect some definitions. Definition 25.1 Let E be a vector space and let S(E) denotes the given family of subsets of E. We say that this family is linearly invariant iff (i) x e E, a G R, V G S(E) =* x + aV G S{E), (ii) U, V G S(E) => U + V G S(E). For example, the family of all convex subsets of E or the family of all closed balls are linearly invariant. Let G be a semigroup. By B(G, S(E)) we denote the set of all functions f:G->E such that imf C V for some V G S(E). By x af for a G G we denote the function defined by af(%) '•— fifl + )Definition 25.2 Assume that G is a semigroup, E a vector space and S(E) a linearly invariant family of subsets of E . We say that B(G,S(E)) admits the left invariant mean (briefly LIM) iff there exists a linear operator M: B(G, S(E)) —> E such that 235

236

Functional Equations and Inequalities in Several Variables

(Hi) imf C V for some V G S(E) => M(f) G V, (iv) f e B(G, S(E)), aeG^

M(J)

= M (/).

Definition 25.3 We say that d: G x G -> S(E) is invariant iff for all x,y,a G G,

translation

d(x + a,y + a) = d(a + x,a + y) = d(x, y). The function f': G —>• E is said to be d-Lipschitz f(x) - f(y)e

d(x,y)

for all

iff

x,yeG.

Remark 25.1 Let G he a semigroup with metric g and E be a normed space. For L G R+ we define d{x,y) :=

Lg(x,y)B(0,l),

where B(0,1) denotes the ball in E with centre at 0 and radius equal to 1. Then the function f: G —)• E is Lipschitz with the constant L iff it is d-Lipschitz. By Z(G) we denote the centre of a semigroup G. Note that for any abelian semigroup or semigroup with zero, Z(G) is a nonempty set. The following very general result (see[193]) can be proved. Theorem 25.1 Let G be a semigroup such that Z(G) ^ 0 and E be a vector space with a linearly invariant family of subsets S(E) such that B(G,S(E)) admits LIM. Let d: G x G ->• S(E) be a translation invariant function. Asumme that Cf(-,y): G —> E is d-Lipschitz for every y E G. Then there exists an additive mapping A: G —>• E such that f — A is d-Lipschitz. If, moreover, im(Cf) C V for a certain V G S(E), then

im{A

-f)cV.

Stability in the Lipschitz norms

237

Proof. Take a e Z(G) and let M: B{G,S(E)) invariant mean. We have for x,y £ G f(a+x + y)-f(a+y)

-4 £ be a left

= [Cf(a+x,y)-Cf(a,y)} + f(a+x)-f(a) £ d(a + x,a) + f(a + x) — f(a).

This means that the function ip defined by ip(y) = f(a + x + y) — f(a + y) belongs to B(G,S(E)) (by properties of d). Thus we may define A{x) := Mz[f{a + x + z)-f(a + z)] (the subscript z idicates that m is applied to function of variable z). Take x,y e G , then we obtain A(x) + A(y) = Mz[f(a + x + z)-f(a + z)} + Mz[f(a + y + z)-f(a + z)] = Mz[f(a + x + (y + z))- f(a + (y + z))] + Mz[f(a + y + z)-f(a + z)} = Mz[f(a + (x + y) + z)- f(a + z)] = A(x + y), i.e. A is additive. Since a G Z{G), f(a + x + z) = f(a + x + z) and consequently A(x) - f(x) = Mz[f(x + a + z)-f(a

+ z)\ -

f(x).

Let us emphasize that if / is constant, then Mz[f] = Therefore, we get A{x)-f{x)

= Mz[f(a+x+z)-f(a+z)-f(x)}

=

imf.

Mz[Cf(x,a+z)},

i.e. A{x)-f(x)

= Mz[Cf(x,a

+ z)],

xeG.

(25.1)

Hence (A-f)(x)-(A-f)(y)=Mx[Cf(x,a

+ z)-Cf(y,a

+ z)]. (25.2)

238

Functional Equations and Inequalities in Several Variables

By the assumption we have Cf(x,a

+ z) - Cf(y,a + z) G d(x,y)

for all

z€G,

which, in view of the properties of the mean, implies Mz[Cf(x,a

+ z) -Cf(y,a

+ z)] G d(x,y)

for all

x,yeG,

which means, taking into account (25.2) that A — f is d-Lipschitz. To end the proof, assume that im(Cf) C V for a certain V G S(E). Making use of (Hi) and (25.1), we obtain A(x)-f(x)

= Mz[Cf(x,a

+ z)}

£V

for all

x £ G.

m Denote by B(E) the family of balls in E. Corollary 25.1 Asumme that G is a group and E a normed space such that the B(G,B(E)) admints LIM. Let, moreover, f:G—>E and g: G —>• R+ be such that \\Cf(x,y)-Cf(u,y)\\
for

x,u,y<EG.

Then there exist an additive mapping A: G —» E such that \\(f(x)-A(x))-(f(u)-A(u))\\
for

x,ueG.

Proof. The assertion follows directly from Theorem 25.1 by taking

d(x,y):=g(x-y)B(0,l) and the remark that in our case Z(G) / 0.

•

For further information one may see the paper by R. Ger [23]. Concider a semigroup G with a metric g invariant under translation, i.e. satisfying the condition g(x + a, y + a) = g(a + x, a + y) = g(x, y)

for

x,y,aeG.

We say that a metric ^ o n G x G i s a product metric iff it is an invariant metric and the following condition holds ~g[(x, a),(y, a)] = ~g[(a,x), (a,y)} = g(x,y)

for

x,y,a€G.

Stability in the Lipschitz norms

239

Example 25.1 Consider a semigroup G with an invariant metric Q. Define ~Q[(XU x2), {yi, 2/2)] := Q(XU yi) + g(x2,

y2),

Q[(xi,x2), (yi, y2)] := max[g(xi, j/i), p(x2, y2)], ^ i , ^ 2 ) , ( y i , 2 / 2 ) ] := [ ^ i , 2 / i ) 2 + ^ 2 , y 2 ) 2 ] ^

Then £ is the product metric of the metric g. Definition 25.4 Any function to: R+ —Y R+ is said to be module of continuity of f: G —> E iff for every 5 G R+

d(*,y)<*^ll/(*)-/(v)ll <"(*)• Now we present an interesting application of Theorem 25.1. Corollary 25.2 Let G be a semigroup with an invariant metric g and Z(G) ^ 0, and let E be a normed space such that B(G, B(E)) admits LIM. Assume,moreover,that ui: R+ -» R+ is the module of continuity of Cf: G x G —» E with the product metric on G x G. Then there exists an additive function A: G —> E such that u is the module of continuity of the function f — A. If moreover, Cf G B(G x G,B(E)), then

11/ - A\\sup < \\Cf\\sup

(25.3)

where \\ • \\sup denotes the supremum norm in the space of bounded functions defined in G(G x G respectively). Proof. Take d(x,y) := u[g(x,y)]B (0,1) for x,y G G. Then d: G x G —>• B(E) is a translation invariant function on G x G. Moreover, we have \\Cf(x,y)

- Cf(xu

y)\\ < u[d[(x, y), (xu y)]] = u>[g(x,xi)],

which means that Cf(x,y)

- Cf(xx,y)

G u[g(x,Xl)]B(0,1)

=

d(x,Xl)

240

Functional Equations and Inequalities in Several Variables

for all x,xx,y G G. Thus Cf(-, y): G —> E is d-Lipschitz for every y G G. Therefore from Theorem 25.1 we obtain that there exists an additive function A: G —> E such that / — A is rf-Lipschitz. Write h = f — A, then we have h(x) — h(y) G d(x, y)

for

x,y G G

i.e. \\h(x) - h(y)|| < w[e(a;, y)] for

x,y e G,

which means that u is the module of continuity of the function

h = f-A. Now, by (25.1) we have A{x) - f(x) = Mz[Cf(x, a + z)} = u(x) G U = {w G G: u = Cf(x,a + z),z G G}. Thus 11/ - A\\mp = sup \\f(x) - A(x)\\ = sup ||«(a;)|| < sup \\Cf(x,z)\\

= \\Cf\\sup,

x,z£G

which proves the inequality (25.3).

•

Stability in the Lipschitz norms Let us recall the following facts. Definition 25.5 Suppose that G is a semigroup with a metric g and E a normed space. A function f': G —>• E is called Lipschitz iff there exists an L eR+ such that \\f(x)-f(y)\\
for

x, y G G.

Stability in the Lipschitz norms

241

The smallest such constant will be denoted by lip(f). The symbol Lip(G, E) will stand for the space of all bounded Lipschitz functions with norm Wf\\iAp-\\S\\suP + lip{f)

for

feIAp{G,E).

If G is a semigroup with zero, by Lip°(G,E) space of all Lipschitz functions with the norm \\f\\LiPo~

11/(0)11 + lip(f)

for

we denote the

feIAp°{G,E).

As a consequence of Corollary 25.2, one can obtain the following. Theorem 25.2 Let G be a semigroup with an invariant metric Q and let E be a normed space such that B(G,B(E)) admits LIM. Assume that we are given a product metric on G x G. (a) Assume additionally that Z(G) ^ 0. Let f: G —» E be such that Cf G Lip(GxG,E). Then there exists an additive mapping A: G —>• E such that

\\f-A\\Lip<\\Cf\\Lip. (b) Let G be a semigroup with zero, and f: G —>• E be such that Cf € Lip°(G x G,E). Then there exists an additive mapping A: G -» E such that

11/ - A\\Lipo < \\Cf\\Lipo. Proof. Consider the case (a). Since Cf is Lipschitz, then it has the module of continuity u(x) := lip(Cf)x. Therefore from Corollary 25.2 it follows that there exists an additive mapping A: G —>• E such that u is the module of continuity of the function / — A and, moreover,

11/ - A\\sup < \\Cf\\sup.

242

Functional Equations and Inequalities in Several Variables

Hence we get that / — A is Lipschitz with constant

lip(f-A)
\\Cf\\sup + lip{f - A)

<\\Cf\\suP + lip(Cf) = \\Cf\\Lip, i.e.

11/ - A\\Lip < \\Cf\\Lip, which proves part (a). Part (b) one can prove using very similar arguments. This concludes the proof. •

Notes 25.1 Recall that

3f{x,y):=f^±^j-\[f(x)

+ f{y)].

Regarding the results that follow concerning the Jensen difference of / one may see J. Tabor [193]. Proposition 25.1 Let G be a uniquely 2-divisible abelian semigroup with an translation invariant metric d and let E be a normed space such that B{G,E) admits LIM. Let f': G —> E be arbitrary. Assume that ui: R+ —> R is the module continuity of Jf: G xG —» E. Then there exists an additive function A: G —» E such that 2u> is the module of continuity of function f — A. If, moreover, Jf G B(G,E),

then there exists b G E such that

\\J ~ A — b\\sup < 2||Jt \\sup. The above result is a simple corollary from Theorem 25.1.

Stability in the Lipschitz norms

243

25.2 We also have the following stability result for the Jensen equation. Proposition 25.2 Let G be a uniquely 2-divisible abelian semigroup with an invariant metric d and let E be a normed space such that B(G, E) admits LIM. We assume that we are given a product metric on G x G. (a) Let f: G ->• E be such that Jf G Lip(G xG,E). Then there exists a Jensen function J: G —»• E such that

||/-./|b P <2||Jf|| Lip . (b) Assume additionaly that G is a semigroup with zero. Let f:G^Ebea function such that Jf G Lip°(G x G,E). Then there exists a Jensen function J: G —>• E such that 11/ - J\\Lip° < 2||Jf HiipO. 25.3 Some results about Lipschitz stability of the quadratic equation one can find in [48].

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Chapter 26 Round-off stability of iterations In this section we shall present some results concerning the Ostrowski's type stability of iteration procedures for fixed points of Banach operators in generalized metric spaces, so-called b-metric spaces. We shall follow the paper of S. Czerwik and K. Dlutek [29]. Consider a metric space (X,d) and a mapping T: X —> X. The idea of iteration procedure for finding a fixed point of T is very important and fruitfull in many areas of mathematics and applications. However, in computation, an approximate sequence is used in place of a sequence of successive approximations. This problem leads to an idea of stability of iteration procedures with respect to the mapping T. The first stability result of this kind has been proved by A. M. Ostrowski [152]. His result has been extended by many authors to various classes of operators (see e.g.[84], [180], [192]).

Preliminaries We shall introduce a generalization of ordinary metric space (see Czerwik [36]). 245

246

Functional Equations and Inequalities in Several Variables

Let X be a set and let s > 1 be a given real number. A function d: X x X —> R+ is said to be a 6-metric iff for all x,y,z G X, the following conditions are satisfied: (a) d(x,y) = 0iSx

= y,

(b) d(x,y) = d(y,x), (c) d(x,z) <s[d{x,y) + d{y,z)]. The pair (X, d) is called a b-metric space. The following result about the existence and uniquennes of fixed points of Banach operators in b-metric spaces can be found in [34] Theorem 26.1 Let (X,d) be a complete b-metric space and let T: X -> X satisfy d[Tx,Ty) < [d(x,y)], x,y e X

(26.1)

where (f>: K+ —>• K+ is increasing function such that lirnn_>oo n{t) = 0 for each fixed t > 0. Then T has exactly one fixed point u (i.e. Tu = u) and lim d[Tnx, u] = 0 for each

x€X

(26.2)

n—Kx>

We shall apply the following Lemma 26.1 Let {en} be a sequence of nonnegative real numbers. Then lim en = 0 iff lim Sn = 0, (26.3) n-»oo

n->oo

where n

Sn:=^an-rer

(26.4)

r=0

and 0 < a < 1.

(26.5)

Round-off stability of iterations

247

Proof. If lim^oo Sn = 0, then l i m ^ o o ^ = 0, since 0 < en < Sn. Therefore, suppose en —> 0 as n —v oo. Fix arbitrary e > 0. We may assume that a > 0. So there exist a natural number m such that for all n > m, en < e. For n — m + k, k eN and /3 — a~l, we have Sn = an[eQ + p£l + ... + Pnen] = an[(eQ + ... + rem)

+ Wm+1em+i + ••• + /3m+kem+k)}

< an(eQ + ... + Pmem) + ane(Pm+1 + ••• + Pm+k)
pmem)+e{l-a)-1.

There exists no > m such that for n > n 0 we have Sn < e + e(l — a ) - 1 . Since Sn > 0 and e > 0 is arbitrary number, this means that lim^oo Sn = 0. The proof is completed. •

Stability of iteration procedures Concider a 6-metric space X and an operator T: X —> X. For any x0 G X we define (Picard iterative procedure) xn+l := Txn,

n = 0,1,2,...

.

(26.6)

Assume that the sequence {xn} is convergent to a fixed point u of T (i.e. Tu = u). Let {yn} be an arbitrary sequence of elements of X. Put £n = d(yn+i,Tyn), n = 0,1,2,.... The iteration process (26.6) is said to be T-stable (cf.[84]) provided that lim en — 0 implies lim yn = u. n—>oo

n—KX>

Now we are able to prove the first result about the T-stability of Picard iteration process.

248

Functional Equations and Inequalities in Several Variables

Theorem 26.2 Let X be a complete b-metric space and let T: X —>• X satisfy the inequality d[Tx, Ty] < qd{x, y)

for all

x, y <E X,

(26.7)

with a:=sq<

1.

(26.8)

Let xo 6 X be an arbitrary point in X and let {xn} be a sequance of iterates ofT given by (26.6). Let {yn} be a sequance in X. Then lim xn = u = Tu,

(26.9) n

d(u, yn+l)

n+1

< sd(u, xn+i) + sa

2

an~rer.

d(x0, y0) + s ^

(26.10)

r=0

Moreover, lim yn = u

iff

lim en = 0.

(26.11)

Proof. Define <j>(t) := gt, t > 0, then in view of Theorem 26.1, T has exactly one fixed point u £ X and the sequence {xn} for any x0 G X has the limit u, which means that (26.9) holds true. Appling (26.7) and the triangle inequality (c) for 6-mertic, we get for nonnegative integer n, d(xn+i,yn+1) = d(Txn,Tyn+1) < sd(Txn,Tyn) + < sqd(xn,yn) + sen < sq[sqd(xn-i,yn-i) < a2d{xn^uyn^x) + s[aen-i + en}.

sd(Tyn,Tyn+i) + s£ n -i] + sen

Consequently, by induction principle, we obtain the inequality n

d{xn+i,yn+i)

n+1

d(x0,y0)

s^an-r£r-

+ r=0

Hence, d(u,yn+i) < sd(u,xn+1)

+

sd(xn+l,yn+1) n 2

< sd(u, xn+l) + san+id(x0, y0) + s ^ r=0

an^rer,

Round-off stability of iterations

249

which proves the inequality (26.10). To prove (26.11), assume first that yn —> u as n —)• oo. Then we have £n = d{yn+i,Tyn) < sd(yn+l,u) + sd(u,Tyn) < sd(yn+l, u) + s[sd(u, Tu) + sd(Tu, Tyn)] < sd(yn+1, u) + s2qd(u, yn). Since yn —> u as n —>• oo, hence we infer that en —> 0 as n —> oo. Conversely, let en —> 0 as n —>• oo. Since yn —> u as n —>• oo, and 0 < a = sg < 1, the first two terms of the right hand side of (26.10) vanish in the limit. Moreover, the third term of (26.10) has also limit zero in view of (26.3) of Lemma 26.1, which concludes the proof. •

Theorem 26.3 Let X T: X ->X satisfy d[Tx,Ty]
be a complete b-metric space and let

for

x,y e X

and

0 < g < 1.

Let m be a natural number such that p:=qms
Moreover, n n+1

d(u, yn+1) < sd(u, xn+1) + sp

2

d(x0, y0) + s £

pn~rer,

(26.12)

r=0

lim yn = u n—>oo

iff

lim en = 0. n—^oo

(26.13)

250

Functional Equations and Inequalities in Several Variables

Proof. Consider <j)(t) — qt , t > 0, then on account of Theorem 26.1, it follows that T has exactly one fixed point u 6 X. For x, y G X we also have d[Tmx,Tmy]

< qd[Tm-lx,Tm-1y] < ...
q2d[Tm-2x,Tm-2y]

<

i.e d[Fx, Fy] < qmd{x, y)

for all

x, y e X.

Hence again by Theorem 26.1 for (f>(x) = qmt , t > 0, we infer that F has exactly one fixed point w e X and w = lim xn. Now we n—KX>

2

shall show that u = w. Indeed, u = Tu — T u = . . . = Tmu — Fu, so u is a fixed point of F, and therefore since F has only one fixed point, then u = w. The statements (26.12) and (26.13) one can prove repeating the argument used in the proof of Theorem 26.2. • Let us also observe that under the assumptions of Theorem 26.3, lim en = 0 implies lim en = lim d{yn+l,Tyn)

= 0.

In fact, since Tu = u, we get e« = d(yn+i,Tyn) < sd(yn+i,Tu) + sd(Tu,Tyn) < sd(yn+i,u) + sqd(u, yn) —> 0 as n —> oo. Remark 26.1 The stabilitty of iterations for multi-valued operators in b-metric spaces has been studied in [31].

Chapter 27 Quadratic difference operator in Lp spaces In this section we are going to discuss the problem of stability of the quadratic functional equation in LP spaces. All results presented here are due to S. Czerwik and K. Dlutek [30]. First we shall prove the following Lemma 27.1 Let X be an abelian complete measurable group and let Ac X be such that /j,(A) = 0. If D := {(x,y)

e X x X: x E A

or ye A or x + or x — y £ A} ,

yeA (27.1)

then v(D) = 0. Proof. Clearly, we have D = (AxX)l>(X

xA)U

S~\X

xA)U

S[X x (-A)],

where S is defined in (e) p. 194. Since X is cr-firrite and n(A) = 0, then v(X x A) = v{A x X) — 0. Moreover, taking into account that S and S~l preserve the measure u, we get 251

252

Functional Equations and Inequalities in Several Variables

v[S{X x A)] = ulS'^X

x {-A))] = 0.

Consequently, v(D) = 0, as claimed.

•

Superstability of the quadratic functional equation Let us recall that a function / : X —> E is called quadratic iff it satisfies the quadratic functional equation 2f(x) + 2f(y) = f(x + y) + f(x - y),

(27.2)

for x, y G X. Emphasize that the next result is interesting for itself. Lemma 27.2 Let E be an abelian metric group without elements of order two. Assume that f:X —>• E is given by f(x) — g(x) + c , x G X, where g: X —> E is quadratic and c G E is fixed. If f G L+(X, E) for a certain p > 0, then g = 0 and c = 0. Proof. Since / belongs to L+ , there exists a ip G L\ (X, E) with \\g(x) + c\\P<
,

xeX.

(27.3)

To prove that g = 0 assume on the contrary that ||^(xo)|| ^ 0, for some XQ G X. Because X does not contain elements of order two, | | 2 < 7 ( x 0 ) | | = / ^ 0 . Set

D := ix G X: ip{x)1/p > ^ j . Clearly, D is measurable and since

we get fi(D) <

tp(x + x0)1/p + y{x - XQ)IIV > \\g(x + x0) + c\\ \\g(x - x0) + c\\ > \\g(x + x0) + g(x - x0) + 2c|| = \\2g(x) + 2g(x0) + 2c\\ > \\2g(xQ)\\ - \\2g(x) + 2c\\ > \\2g(x0)\\ - 2\\g(x) + c\\ > /3 - /3/2 = 0/2.

Quadratic difference operator in LP spaces

253

Hence ^

or

> ^

which means that x + XQ G D or x — x0 G D. Consequently, X = DU{D +

x0)U{D-x0).

Hence on account of the condition ^(D) < oo, we obtain that n(X) < co, which is a contradiction. This proves that g = 0, whence c as an integrable function on the space X with fJ,(X) — +00 has to be equal to zero. The proof is completed. • We define the quadratic difference Qf of a function / : X —> Eby Qf(x,y):=2f(x)

+ 2f(y)-f{x

+ y)-f{x-y)

,

x,y e X. (27.4)

The next result reads as follows. Theorem 27.1 Let E be an abelain group with unique division by two and X be an abelian complete measurable group. Let f: X —> E be such that Qf(x,y)=0. Then there exists a quadratic function g: X —>• E such that f{x)=g{x)

for

xeX.

(27.5)

Proof. By the assumption, there exists a set V C X x X such that v(V) = 0 and

2f(x) + 2f(y) = f(x + y) +

f(x-y)

for all (x,y) G X x X\V. Thus by the Fubini Theorem (cf. [82], Th. 36.A) there exist sets Uu U2 C X such that fi(Ui) = fi(U2) = 0 and (a) for every x G X\U\ there exists K[x] C X such that n(K[x\) = 0 and for all y G X\K[x] we have 2f(x) + 2f(y) = f{x + y) + f(x - y);

(27.6)

254

Functional Equations and Inequalities in Several Variables

(b) for every y G X\U2 there exists L[y] C X such that fjL{L[y]) = 0 and for all x G X\L[y] (27.6) holds. Put U := Ui U f/2. Then, clearly, u.(U) = 0. For any x G X, we define

£4 : = £ / U ( x - £ / ) u ( - x + t/). Obviously, n(Ux) = 0, whence X\UX ^ 0. Consequently, for every x G X there exists a w(x) G X\UX, i.e. w(x)<£U

,

x + w(x)(£U

,

x-w(x)£?7.

(27.7)

Now we define the function g: X —> E by the formula g(x) := 1/2 [/ (x + w(x)) + f (x - w(x))] - f (w{x)).

(27.8)

First we shall verify that g does not depend on w(x) € X\UX. Take any x G X, then x + w(x) ^ U and x — w(x) ^ [/, thus by ( a ) we get 2f(x + w(x)) + 2f{y) = f(x + y + w(x)) + f(x-y

+ w(x)) (27.9)

for y G X\K[x + w(x)], and 2/(x - w{x)) + 2f(y) = f{x + y -w(x))

+ f(x-y

-w(x))

(27.10)

for y G X\K[x - w(x)]. Analogously, in view of ( b ) (substituting y — w(x) and x as x + y or x — y), we obtain 2/(x + y) + 2 / M x ) ) = / ( x + ?/ + ™(x)) + / ( x + 2 /-™(x)) (27.11) for x + y G X\L[io(x)], i.e. y G X$L[u;(x)] — x), and 2f(x-y)

+ 2f{w(x)) = f(x-y

+ w(x)) + f(x-y-w(x))

(27.12)

for x - y G X\L[w(x)], Denote

i.e. j/ G X\(-L[w(x)]

+ x).

^io(x) := K[x +

w(x)]UK[x-w(x)]\J(L[w(x)]-x)U(-L[w(x)]+x),

Quadratic difference operator in IP spaces

255

then we have fj,[Aw^) = 0. Adding first the equations (27.9) and (27.10) side by side and then substracting the equations (27.11) and (27.12) from the obtained sum, we derive the equality 4g(x) + 4/(y) = 2f(x + y) + 2f(x - y)

(27.13)

for y e X\Aw{x). Consider any two elements Wi(x), W2(x) G X\UX. We can find y G X\(AWl(x) U AW2(X)). Consequently by (27.13) we obtain Agn{x) = 2f(x + y) + 2f(x - y) - 4/(y) i.e. since in E the division by two is uniquely performed,

gn{x) = l/2[f{x + y) + f{x-y)]-f{y)

,

n = l,2,

where gn , n = 1,2 are defined by (27.8) for w(x) — wn(x). Therefore, gi(x) =92(x)

= g(x),

as claimed. Now we shall show that / = g. Indeed if x G X\U, then we can find w(x) G X\(UX U K[x$ and hence on account of (a), we infer that 2f(x) + 2f(w(x)) = f(x + w(x)) + f(x - w{x)). Consequently, f{x) = l/2[f(x + w(x)) + f{x - mix))} - fiw(x))

= g(x),

i.e. fix) = gix) for x G X\U. To finish the proof, we have to verify that g is quadratic. Let us note that for every x G X , fJ,(Ux) = 0. Let x, y G X be arbitrarily fixed. Thus for b G X\Uy, on account of De Morgan's law, we have Z := iX\Ux) H HX\Ux+y)

- 6) n iiX\Ux.y)

+ 6)n

(27.14)

256

Functional Equations and Inequalities in Several Variables (X\L[b}) n ((X\L[y + b]) -x)n

(X\(x - L[y - b}) ± 0

Hence, for b G X\Uy, there exists an o G Z, which means that a G X\UX , a + be X\Ux+y, x + aeX\L[y

+ b], y-beX\U,

a-be x-a

X\Ux-y,

y + be

G X\L[y-b],

X\U, (27.15) b G X\U,

a G X\L[b}. Taking into account that the definition of g(x) does not depend on the choice of w{x) G X\UX, we get g(x) = l/2[f{x + a) + f(x - a)} -

g(y) = l/2[f(y + b) +

f(a),

f(y-b)]-f(b),

g{x + y) = l/2[f(x + y + a + b) + f(x + y - a - b)] - f(a + b), g(x -y) = l/2[f(x - y + a - 6) + f(x - y - a + b)} - f(a - b). From the above equalities and on account of (a), (b) and (27.15) we obtain 2g(x) + 2g(y) = f(x + a) + f(x - a) - 2/(o) + f{x + b) + f(x-b)-2f{b) = l/2[f{x + y + a + b) + f{x-y + a-b) + f(x + y-a-b) + f(x~y-a + b)]-f(a + b)- f(a - b) = g{x + y)+g(x-y). This concludes the proof.

•

Remark 27.1 A similar problem for additive and polynominal functions in a more general setting (for the so called proper linearly invariant conjugate ideals in X and X x X) has been studied by R. Ger in [73], [72]. In the next lemma we shall establish the functional equation for quadratic differences.

Quadratic difference operator in LP spaces

257

Lemma 27.3 Let G and H be abelian groups and let f: G —»• H be a function. Then 2Qf(u, v) = Qf(x + u,v) + Qf{x -u,v)-Qf{x,u-v) + 2Qf{x,u)

Qf{x, u + v) {

'

]

for all x, u, v € G. Proof. The proof follows immediately from the definition of the quadratic difference of / and comparison of the left and right sides of the equality (27.16). • Now we are in a position to prove one on the main results of this section. Theorem 27.2 Let X be an abelian complete measurable group and E be a metric abelian group without elements of order two. Let f: X ->• E be such that Qf e L p + (I x X, E), for some p > 0. //, moreover, (J,(X) = +oo, then Qf(x,y)

= 0.

(27.17)

Proof. Define the function KUjV: X —>• E by Ku,v(x)

:==

Qf(x + u,v) + Qf(x-u,v)-Qf(x,u

+

v)-Qf(x,u-v)

+2Qf(x,u). Observe that for u,v,x G X, Kuv(x) = (Qf(-,v)u)(x) + (Qf(,v)u)(x)-(Qf(.,u+v))(x)-(Qf(.,u-v))(x)+2(Qf(.,u))(x). In view of Lemma 21.3, there exists a subset A C X such that fi(A) = 0 and Qf(.,y)eLZ(X,E)

for y G X\A.

(27.18)

We shall prove that Qf(u,v) = 0 for u,v,u + v,u-veX\A.

(27.19)

258

Functional Equations and Inequalities in Several Variables

Take u, v e X\A with u + v,u - v e X\A. On account of (27.18), Lemma 21.2, and, ultimately, Lemma 21.1 we infer that Ku,v £ L+(X, E) for u, v, u + v, u - v e X\A. Moreover, in virtue of Lemma 27.3, KUjV(x) = 2Qf(u,v) for x € X, which means that KUtV as a function of x is constant. Since +

+

J \\Ku,v(x)\\^(x)

=J

\\2Qf{u,v)\\^(x) < +oo

and /J,(X) = +oo, thus 2Qf(u, v) = 0 for u, v, u + v, u — v G X\A. Therefore from our assumptions, this is equivalent to the condition Qf(u, v) = 0 for

(u, v) e {X x

X)\D,

where D is defined in Lemma 27.1. Bearing in mind that v{D) = 0, we conclude the proof of (27.17). • As a consequence of Theorem 27.2, we obtain Theorem 27.3 Let X be an abelian complete measurable group, /x(X) = +oo ; and E be a metric abelian group without elements of order two. Let f: X —>• E be a function. The following conditions are equivalent: (i) Qf € L+{X x X, E) for some p > 0; (ii) there exists a quadratic function g: X g(x) = f{x).

—>• E such that

Proof. The implication (i)=^(ii) follows from Theorem 27.2 and Theorem 27.1. The converse implication is obvious. •

Remark 27.2 The assumption fJ,(X) = +oo is essential. For this, consider any nonzero constant function.

Quadratic difference operator in LP spaces

259

Remark 27.3 In Theorem 27.2 we have proved under the hypothesis that the quadratic difference of f is bounded by an integrable function, that such f satisfies the quadratic equation almost everywhere. Following R. Ger, we say that in such "superstability" occurs.

phenomenon of

Remark 27.4 The problem of stability of the quadratic functional equation in LP spaces under the assumption that the measure of the space X is finite, has been considered by S. Czerwik and K .Dlutek [32].

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Part III Functional Equations in Set-Valued Functions

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Chapter 28 Cauchy's set-valued functional equation Let us consider a topological vector space X which satisfies the T0 separation axiom. By 2X we denote the space of all subsets of X with Hausdorff topology. In this topology the sets NV(A) := {B C X: B C A + U, A C B + U}, where U belongs to a base U of neighbourhoods of zero in X form a base of neighbourhoods of sets A C X. The symbol An —> A will stand for the limit of the sequence {v4n}n€N to the set A in the introduced Hausdorff topology in the space 2X. By CL(X) we denote the family of all closed and non-empty subsets of X, by C(X) - the family of all compact non-empty subsets of X, and by CC(X) - the family of all convex compact and non-empty subsets of X respectively. A set-valued function (briefly s.v.f.) A: [0,oo) —» 2X is called additive iff it satisfies the Cauchy functional equation A(x + y)=A(x)

+ A(y)

for all x,y G [0, oo). Our first result is the following (see K. Nikodem [144]) 263

(28.1)

264

Functional Equations and Inequalities in Several Variables

Theorem 28.1 Let H G 2X and let A{x) = xH for every x G [0, oo). Then the s.v.f A : [0, oo) -> 2X is additive iff H is convex. Proof. Suppose that the s.v.f. A is additive. Take u,v G H and A G (0,1). Then we have AM +

(1 - X)v eXH + (l- X)H = A(X) + A(l - A) = A(l) = H,

which shows the convexity of H. Conversely, assume that the set H is convex. Let x, y G [0, oo). We must prove (x+y)H = xH+yH. Clearly, (x+y)H C xH+yH. Let z G xH + yH, then z = xu + yv for some u,v G H. We may suppose that x + y > 0. Since if is convex, we get w = (x + 2/) -1 |xu + 2/u] =

u+ x +y

v G if, x+y

whence 2 = (x + y)w e (x + y)H, which means that xH + yH C (x + y)H for all x, y G [0, oo). This completes the proof. • Remark 28.1 Theorem 28.1 states that a necessary condition for an additive s.v.f. A: [0, oo) —>• 2X to be of the form A(x) = xA(l)

for all x G [0, oo)

(28.2)

is the convexity of all its values. We shall show later that under some additional assumptions this condition is also sufficient. Let us emphasize that, in general, this is not true. To see this, consider the s.v.f. A: [0, oo) —> 2R defined by A(x) = f(x) + [0,oo), i e [ 0 , o o ) , where / : K —> R is a discontinuous additive function. Then A is additive with convex values, but it is not of the form (28.2). For details see [78].

Cauchy's set-valued functional equation

265 A: [0,oo) ->• 2R2

Remark 28.2 Consider (see [144]) the s.v.f defined by A(x)

= xK

+ Z2,

XG[0,OO),

where K denotes the unit ball in K2. Thus A is additive and for x > (\/2) _ 1 its values are convex sets, but for x < (\/2) _ 1 are not convex. The following theorem gives a fundamental property of solutions of the equation (28.1). Theorem 28.2 Let X be a real linear space and let a s.v.f. A: [0,oo) -» 2 X \{0} be additive. If A(x) forx G (0,00) are convex sets, then A(qx) = qA(x)

for all q G Q fl (0,00)

and

x G (0,00). (28.3)

If, moreover, A(0) is bounded, then A(qx) = qA{x)

for all

geQn[0,oo)

and

xG[0,oo). (28.4)

Proof. For U C X,set Sn(U) := {x G X: x — ax + . . . + an , ak G U , k =

l,...,n}.

Then, if U is convex, Sn(U) = nil. Indeed, taking a point x G Sn(U), x — a\ + ... + an, by the convexity of U, we have X_ _

Oi + • • • + On

n

^

n

whence a; G nC/. The inclusion n[/ c Sn(U) is trivial. Now, take x G (0,00), since A(x) is convex, hence we get A(nx) = A(x) + ... 4(:c) = ^ (A(x)) = nA(x) for all n G N. Analogously,

A(x)=Sm\A(-)]=mA(-),

266

Functional Equations and Inequalities in Several Variables

or A (—) = —A(x) for all m G N. \m/ m Consequently, for every q = ^ G Q n (0, oo) and x G (0,oo), we obtain

A(qx) = nA (—J = <^4(:r). i.e. we have got (28.3). Moreover, for n G N, from (28.1) we obtain A{0) = A(0) + ...+ A(0) D nA(0), which implies, in view of the boundedness of ^4(0), that A(0) = {0} and therefore (28.4) follow. •

Set-valued functions in /(R) Let 1(E) denote the family of all nonempty intervals contained in R. In this section we shall state same characterizations of s.v.f. A: [0, oo) —> I(M) satisfying the Cauchy equation (28.1). We start with some preliminary lemmas. Lemma 28.1 Let a s.v.f. A: [0, oo) —»• I(R) be additive. If for some x0 G (0, oo) the set A(x0) is upper (lower) bounded, then for every x G [0, oo) the set A(x) is upper (lower) bounded respectively. Proof. Take an x G (0, oo) and choose a q G Q fl (0, oo) such that qx < XQ. Since A is additive and has convex values, then by Theorem 28.2, A(x0) = qA(x) + A{XQ - qx). From the fact that A(x0) is upper (lower) bounded, it follows from this equality that so is A(x). We have also A(x0) = A(xQ) + A(0), which implies that ^4(0) has to also be upper or lower bounded, respectively. This concludes the proof. •

Cauchy's set-valued functional equation

267

Remark 28.3 Let A: [0,oo) ->• 7(E) be given by A[X)

- \ R

, x > 0.

Then A is an additive s.v.f. such that A(0) is lower bounded but A(x) for x > 0 is not a bounded set. This shows that in the Lemma 28.1 the assumption that A(XQ) is bounded for some XQ > 0 is essential. Lemma 28.2 Let a s.v.f. A: [0, oo) —>• 7(E) be additive. If for some XQ G (0, oo) the set A(x0) is open at the right (left) side, then for every x G (0, oo) the set A(x) is open at the right (left) side, respectively. Proof. Take an x 0 G (0, oo) and choose a q G Qn (0, oo) such that qxo < x. Since A is additive with convex values, then in view of Theorem 28.2, A(x) = qA(x0) + A(x - qx0). Hence, because A(x0) is open from one side, it follows that A(x) has to be open from the same side independently on the form of the set A(x — qx0). • Remark 28.4 Consider the s.v.f. A: [0,oo) —>• 7(E) defined by K J

\ (0,oo)

, x > 0.

v

'

Then A is additive and for every x > 0 , A(x) is open at the left side but A(0) is not open at the left side, which shows that Lemma 28.2 is not true for x = 0. Remark 28.5 A similar lemma is true for one side closed intervals.

268

Functional Equations and Inequalities in Several Variables

The next result reads as follows Theorem 28.3 A s.v.f. A: (0, oo) -> I(R) is additive iff there exists an additive function f: R —> R and a se£ [/ G /(M) SMC/I £/ia£ ^(x) = f(x) + xU for all x G (0, oo).

(28.6)

Proof. Assume that A: (0,oo) —> I(R) is an additive s.v.f. such that all its values are bounded and closed intervals (for the remaining cases one can proceed similarly). Define f(x) :='miA(x),

g(x) := supA(:c),

x G (0,oo).

It is easy to see that / and g are additive functions, f < g and A(x) = [f(x),g(x)] for x G (0,oo). Consequently the function h := g — f is additive and bounded from below by zero on (0, oo), which implies that h is continuous. Thus there exists a constant k G [0,oo) such that h(x) = kx for x G (0, oo). Take U := [0,A;]. Then for x G (0, oo), A(x) = [f(x),g(x)] = [f(x)J(x) + kx] = f(x) + [0, kx] = f(x) + xU, i.e. A has a form (28.6). Conversely, if A is given by the formula (28.6), then by Theorem 28.1 we can easily verify that A satisfies the Cauchy equation (28.1). Remark 28.6 Note that the s.v.f. (28.5) is of the form (28.6) on (0,oo), but is not of such form on [0, oo). In the sequel we shall apply the following idea of semi-continuity of a s.v.f. . A s.v.f. A: (0,oo) —> 1(E) is said to be lower semicontinuous provided that the set A~1(D) :— {x G (0, oo): A(x) D D ^ 0} is open for every open set O c t Let us note that if A is an ordinary function, this definition states that A is a continuous function.

269

Cauchy's set-valued functional equation

Lemma 28.3 Assume that a s.v.f. A: (0,oo) —» /(E) is given by the formula A(x) = f(x) + xU for x G (o, oo), where f is an additive function and U is an upper or lower bounded set in 1(B). Then A is lower semi-continuous iff f is continuous. Proof. Assume that / is a continuous function and consider an open set D e l . Then we have

A-\D)

= {xe (0, oo): [f(x) + xU] n D ^ 0} = = \J{xe(0,^):f(x) teu

+ xteD}

=

[jg^(D) teu

where gt: E —>• R, t G U are given by gt(x) = f(x) +xt,

x G (0, oo).

Since gt, t G U are continuous, the set A~l(D) is open which proves lower semi-continuity of A. Assume now that A is a lower semi-continuous s.v.f. and let U be an upper bounded set (if U is lower bounded interval of E, the proof is similar). Take x0 G (0, oo) and set a := sup U, b := sup A(x0) = f(x0) + x0a. Let e > 0 and consider the set D :— (b — e,b + e). From the assumption that A is lower semi-continuous, it follows that A~1(D) is open. Since x0 G A^1(D), there exists an s G (0,:Eo) such that V = (x0- s, x0 + s) C A~l(D). Now we shall prove that / is lower bounded on V. Assume the contrary. Thus for every n G N, there exists an xn G V with f(xn) < —n. Therefore, for n0 > (x0 + s)\a\ + e — b and corresponding xno, we have

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Functional Equations and Inequalities in Several Variables

supA(xno)

= f(xno)-\-xnoa < b-e,

< -n0 + (xQ + s)\a\

whence we get A(xno) D D — 0 which contradicts the fact that xno £ A~1(D). So / is additive and lower bounded on V. Thus / is a continuous function. • Therefore we can formulate the following Theorem 28.4 If a s.v.f. A: (0,oo) —>• I(R) is additive and lower semi-continuous, then A(x) = xA(l)

for all x e (0,oo).

(28.7)

Proof. Because of Theorem 28.3, A(x) = f(x) + xU

for all x e (0, oo),

where / : R —> R is an additive function and U E /(R). Consider first the case when U is neither upper nor lower bounded. So U = R and hence A(x) = xA(l) for x G (0,1). On the other side, if U is upper or lower bounded, then in view of Lemma 28.3, / is continuous and consequently f(x) = xf(l). Finally, we have for all are (0,1), A(x) = xf(l) + xU = x[f{\) + U) = i.e. the formula (28.7).

xA(l), •

Additive set-valued functions with values in topological vector spaces We start by presenting three lemmas by K. Nikodem [143] on the behaviour of sequences of sets.

Cauchy's set-valued functional equation Lemma 28.4 If {An}ne^ subsets of X, then

271

is a decreasing sequence of compact

An —> J J An. neN

Proof. Denote A := f] An and take any neighbourhood U of neN

zero. We have to prove that An G NJJ(A) for sufficiently large n G N, i.e. we must show that A C An + U, which is obvious and An C A + U for some n e N since {An} is a decreasing sequence. Assume that this is not true thus An\(A + U) ^ 0 for all n G N. But the sets An\(A + U) are compact and form a decreasing family, which implies that

f][An\(A + U)}^$. n£N

On the other hand,

n [An\(A+u)i c n (An\A) c n ^ n \^=A\A=0,

neN

neN

neN

a contradiction. This proves the lemma.

•

Lemma 28.5 / / {A n } n e N is a increasing sequence of subsets of a compact set, then

An -> cl (J An. neN

Proof. Denote A := cl \J An and take any neighbourhood U of neN

zero. It is enough to prove (see the proof of Lemma 28.3) that A C An + U for some n G N (since {An} is increasing). Suppose on the contrary that for every n G N, A\(A n + U) ^ 0. Then we have also f] ^4\(^4n + U) ^ 0, which contradicts the definition of neN

A.

m

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Functional Equations and Inequalities in Several Variables

L e m m a 28.6 If An -» A and An ->• B, then clA = clB. Proof. Take a neighbourhood U of zero and consider such a V € U that V + V C U. By the assumptions,one has An C A + V and A C An+ V as well as An C B + V and £ C -A„ 4- V for n large enough. Therefore,

AcB

+ V + V C B + U and B c A + F + V c A + t/.

Since the inclusions hold for all U G U, then

Ac

f~]{B + U)=clB uev

and £ C f | (A + U) = clA, uev

whence in view of the inclusions (see the definition of closure) clA C clB and clB C clA, one gets clA = clB and the proof is completed.

•

Let us now recall some essential definitions which will be usefull later on. We say that a s.v.f. A: [0,oo) -> 2X is bounded on a set B C [0, oo) iff the set (J A(x) is bounded in X, i.e. for every xeB neighbourhood U of zero there exists a number s > 0 such that

s({jA(x))cU. \xeB

J

A s.v.f. A: [0,oo) —> 2 X is called upper semicontinuous (shortly u.s.c.) at a point x0 e [0, oo) iff for every [ / e U there exists a number 8 > 0 such that A(x) C A(x0) + U for every

x € (x0 - 5, x0 + 5) n [0, oo).

Cauchy's set-valued functional equation

273

A s.v.f. A: [0, oo) -» 2X is said to be continuous at a point x G [0, oo) iff it is continuous at this point with respect to Hausdorff topology in 2X. An interesting and fundamental property of additive mapping is expressed by the following lemma. Lemma 28.7 If an additive s.v.f. A: [0, oo) —> CC(X) is right continuous at a point XQ G [0, oo), then it is right continuous at every point x > XQ. Proof. Take f / e U . Since A is right continuous at x0, then there exists a 5 > 0 such that A(x) C A(x0) + U and A(x0) C A(x) + U for all x G [xQ, xQ + 5). Let d :— x — x0, then s — d G [x0,xo + S) for every s £ [x,x + 6). Since A is additive, we obtain A{s) = A{s -d) + A{d) C A(x0) + U + A{d) = = A(x0 + d) + U = A(x) + U, i.e. A(s) cA(x) + U for all s G [x, x + 6).

(28.8)

Moreover, A(x) = A(x0) + A(d) cA(s-d)

+ U + A(d) = A(s) + U,

i.e. A(x)cA(s) + U for all se[x,x + 5). (28.9) Consequently (28.8) and (28.9) imply that A is right continuous at x > x0. • The next result due to K. Nikodem [143] gives the characterization of continuous additive set-valued functions. Theorem 28.5 Let a s.v.f. A: [0, oo) ->• CC(X) be additive. Then the following conditions are equivalent:

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Functional Equations and Inequalities in Several Variables

(a) A is continuous on [0, oo); (b) A is right u.s.c. at some point XQ G [0, oo); (c) A is bounded on a subset of [0, oo) with a non-empty interior; (d) A is bounded on a subset of [0, oo) with a positive Lebesque measure; (f) A is is of the form A(x) — xA(l) for x G [0, oo). Proof. The first implication (a) —>• (6) follows immediately from the definitions of continuity and u.s.c. at a point. To prove the implication (b) —>• (c), we shall show that A is bounded on the interval [0,1]. To this end, fix an arbitrary neighbourhood f / e U and consider a balanced neighbourhood V of zero such that V + V + V C U. Since A is right u.s.c. at x0, then there exists a 5 > 0 such that A(x) C A(x0) + V

for all x G [x0,x0 + S).

Let us choose a number q € Q such that 0 < q < min[5,1] and qA(x0) C V. Because we have XQ + qx G [x0, XQ + 5) for x G [0,1], so qA(x0 + qx) C q[A(x0) + V) C V + V. Therefore, by the additivity of A, one gets q2A(x) = qA(qx) C qA(qx + x0) - qA{x0) CV + V + V CU for all x G [0,1], i.e. for x G [0,1], q2A(x) C U. Since U is an arbitriay neighbourhood of zero, this means that A is bounded on the interval [0,1]. The implication (c) —> (d) is trivial. To prove that (d) implies (e) let us suppose that A is bounded on a set D c [0, oo) with positive Lebesque measure. We shall

Cauchy's set-valued functional equation

275

show that A is bounded on the set (D — D) D [0, oo). To do this, let us fix an arbitrary neighbourhood U € U and consider a balanced neighbourhood V of zero with V + V C U. From our assumption that A is bounded on D, there exists an s > 0 such that

\jA(x)

CV.

.xeD Now, letx G (D—D)C\[0,oo), i.e. x — xx—x2, where xi,x2 G D. The additivity of A implies then A(x) C sA(x) + sA(x2) - sA{x2) = sA(xi) - sA(x2) CV + V CU which means that A is bounded on the set (D — D) Pi [0, oo) . Consequently, by the classical Steinhaus theorem (see e.g. [202]) that if D is a set with a positive Lebesgue measure, the set D — D contains a neighbourhood of zero, we infer that there exist a 8 > 0 such that [0,6) C (D-D) n [0, oo). In the sequel, we want to prove that A is right continuous at the point zero. Taking into account that A is bounded on [0,s), there exist an s > 0 such that 5A(x) C U

for all

x e [0,5)

,

and, since U is balanced, we may suppose that s is a rational number. Hence, for every a = sx E [0, s5) , we obtain A(o) = A(sx) = sA(x) cU = {6} + U

,

(28.10)

.

(28.11)

which implies also for all a G [0, s5) , {6}cA(a)-A{a)cA(a) + U

Since A(0) is bounded, then from the equality A(0) = A(0) + A(0) we guess easily that A{0) = {6}. Thus, (28.10) and (28.11) show that A{a) e Nv[A{0)] for all a G [0, s5)

276

Functional Equations and Inequalities in Several Variables

and proves the right continuity of A at zero. Consequently, by Lemma 28.7, A is right continuous at every point x G [0, oo) . To finish the proof, consider two cases: 9 G A(l) and 9 ^ ^4(1)Let 9 G A(l). Let us fix an a; G [0, oo) and choose a decreasing sequence {qn}ne^ of rational numbers converging to x. So A(qn) —>• A(x). Moreover, in view of Lemma 28.4 and Theorem 28.2, A(qn) = qnA(l) -> f | qnA(l) = xA{\) . Consequently, Lemma 28.6, yelds A{x) — xA(l). Now, consider the case 9 £ A(l) . Fix a point u G ^4(1) and define the s.v.f. B(x) := A(x) — xu , x G [0, oo) . Obviously, B is an additive and right continuous function on [0, oo) with 9 G A(l). So, from the proof just presented for the first case, we get B(x) =xB(l)

,

which yelds the required equality: A(x) — xA(l), x G [0,oo). Finally, we shall prove the implication: (e) —> (a). Let us take a balanced neighbourhood U of zero and fix a point x0 G [0,oo). Because A(l) is bounded, then there exists a 5 > 0 such that sA(l) c U for alls G (-6,5). Thus, for x G (x0-5,x0 + 5)n[0,oo), we have A(x)

= xA(l) = [x0 + {x-x0)]A(l)

Cx0A(l)

+

(x-x0)A(l)

C A(x0) + U, and A(x0)

= x0A(l) = [x + {x0 - x)]A(l) C xA(l) + [x0 C A(x) + U,

x)A(l)

which shows that A is continuous at XQ G [0, oo) , and concludes the proof. • An important consequence of the last theorem, is the following

Cauchy's set-valued functional equation

277

Theorem 28.6 Assume that A: [0,oo) —> CC(X) is an additive s.v.f. and f: [0, oo) —> X is its additive selection (i.e. f(x) G A{x) for x G [0, oo) j . Then A is continuous iff f is continuous. Proof. Assume that A is continuous. Then by Theorem 28.5 A is bounded on some interval I C [0, oo) . Hence / is also bounded on / and as an additive function has to be continuous. Conversly, assume that / is a continuous function. Define the s.v.f. B: [0,oo) ^CC(X) by B(x) := A(x) - f(x)

for

x G [0, oo).

Clearly, B is an additive s.v.f. and 9 G B(x) for all x e [0, oo) . Therefore B{x) C B(x) + B(l - x) = B(l)

for all

a; €[0,1].

Since B(l) is compact, so B(l) is bounded and consequently B is bounded on [0,1]. Thus Theorem 28.5 implies that B is continuous. From the equality A = f + B we infer that A is continuous, too. • We conclude this section with the following: Theorem 28.7 Assume that a s.v.f. A: [0,oo) ->• CC(X) is additive and let f: [0, oo) —> X is its additive selection. Then there exists a set H G CC(X) such that A(x) =xH + f(x)

for all

x G [0, oo).

(28.12)

Proof. Consider the s.v.f. B: [0, oo) ->• CC(X) defined by B(x) := A(x) - f(x) for x G [0, oo) . Evidently B is additive and 6 G B(x) for all x G [0, oo) . Hence the function g{x) = 9 for x G [0, oo) is an additive and continuous selection of B . Thus by Theorem 28.6, B is continuous and consequently, from Theorem 28.5, we infer that B[x) = xB(l)

for all

x G [0, oo).

Taking H := B(l) , we get hence for all x G [0, oo) , A{x) = B{x) + f{x) = xB(l) + f(x) =xH + f(x), i.e. the formula (28.12), which concludes the proof.

•

278

Functional Equations and Inequalities in Several Variables

Notes 28.1 An analogous result to Theorem 28.5 under the assumpion that X is a reflexive Banach space has been proved by D. Henney [89]. 28.2 During the last twenty years a qualitative theory of the Cauchy's functional equation in set-valued functions has substantially been developed. The reader is referred to, e.g. [88], [122], [153], [163].

Chapter 29 Jensen's functional equation We say that a set-valued function (briefly s.v.f.) F: [0, oo) —> 2 is a solution of the Jensen functional equation iff X

F(^pL)

= \[nx)

+ F(y)}

for all

x,ye[0,co).

(29.1)

We are going to present basic results concerning this equation which are due to K. Nikodem [143]. Let us start with the following Lemma 29.1 // a s.v.f. F: [0,oo) —> CL(X) satisfies Jensen's functional equation (29.1), then for every x G [0, oo) , F(x) is a convex set. Proof. Setting x = y in the equation (29.1), we get F(x) = \[F{x) + F(x)]. Fix arbitrary u, v G F(x), then u = \{a 4- b), v — i(ai + &i) , where a,b, a\, b\ G F(x). Now

U + V = { {a+ h) + {ai + h)} E [F{X) + F{X)] =

\ \

\\

\

1

thus \u + \v G F{x) . Similary, by induction we can prove that

^-u+(l-^-)veF(x) On

v

On

279

v

'

280

Functional Equations and Inequalities in Several Variables

for all k = 0,1, ••-,2", n = 1,2,... . Taking a G [0,1] and ®n,k = | r such that anik -> a, by the fact that F(x) is closed, one obtains k k —u + (1 )t> ->• au + (1 - a)v G Fix), v 2« 2" which means that .F(:r) is convex.

•

The next lemma is the following Lemma 29.2 Assume that {An: n G N} and {Bn: n G N} are decreasing sequences of closed subsets of X and A\ is a compact set. Then f | (An + Bn) = f | An + f | Bn. (29.2) Proof. Assume that z E f] (An + Bn) . Then for every n G N nGN

there exist an G ^4n and bn G 5 n such that z — an + bn. Hence Ann(zBn) y£ 0 for all n G N. Moreover A„ n (z - Bn) C An C Ai

for all

n€N,

so yl„ n (2; — 5 n ) are compact sets for all n G N. Clearly A n + i n {z - Bn+l) C Ann(z-

Bn)

for

n G N,

and therefore by the well known theorem, the set P| [>ln fl (2 — 7i£N

£„)] 7^ 0. Let x G H [^n

n

z

B

( ~ n)] , then x G A n for n G N

neN

and there exist c G 5 n for n G N such that a; = 2 — c. Hence 2: = x + c G P | y l n + p ) S n . The inverse inclusion is obvious. This ends the proof.

•

Lemma 29.3 If A is a bounded subset of X and sequence converging to an s G K , then snA ->• sA.

(S„)„ 6 N

is a real

Jensen's functional equation

281

Proof. Let U be a balanced neighbourhood of zero. Thus there exists an r > 0 such that rA C U . Take n G N with \s — s\ L — < 1 , r then we get snA = ( s y j - s + s ) ^ C (sn-s)A+sA

c (sn-s)-U+sA r

C U + sA ,

i.e.

snA c U + sA. Moreover, sA = (s-sn+sn)A

C (s-s„)A+s n A c (s-sn)-U+snA r

C C/+snA ,

i.e. sAcU

+ snA.

Therefore, for n sufficiently large, we have snA € Nu(sA), which means that snA —> sA , as claimed.

•

Now we are in a position to prove the main result of this section (see [143]). Theorem 29.1 A s.v.f. F: [0, oo) —> C(X) is a solution of the Jensen's functional equation (29.1) iff there exists an additive s.v.f. A: [0,oo) -)• CC(X) and a set K e CC(X) such that F{x) = A{x) + K

for all

x e [0, oo).

(29.3)

Proof. If F has the form (29.3), then by simple calculations one can verify that F satisfies the Jensen's equation (29.1). Let us assume that F is a solution of the equation (29.1). Thus, by Lemma 29.1, for every x 6 [0, oo), F(x) is a convex set.

282

Functional Equations and Inequalities in Several Variables

Fix a point a G F(0) and consider the s.v.f. G: [0, oo) —> CC(X) given by the formula G{x) := F(x) - a , for

xG [0, oo).

It is a simple exercise to show that G is a solution of the equation (29.1), and clearly, 0 G G(0). Define the sets

An(x):=±G(2nx),neK Then we have for n G N and x G [0, oo)

4"i(*) = ^lG^+lx)

c

Yn • \[G(r+1x)

+ G(0)]

= ±G(2nx) = An(x), which means that {An(x)}ne^ is a decreasing family. Define A x for A(x) := fl n( ) x G [0,oo). Obviously, A(x) G CC(X) for all a; G [0, oo). Moreover, in view of Lemma 29.2, we obtain, for x,y G [0, oo),

A(x + y) = f | A>(* + y) = f | ^ [ 2 " + l r r + 2 n + l y ]

= n^[ G ( 2n+i ^)+ G ( 2 " +i ^ oo

= f][An(x) + An(y)} n=2 oo

oo

= p\An(x) + f]An(y) = A(x) + A(y), n=1

n=2

i.e. the s.v.f. A: [0,oo) -» CC(X) is additive. Next, by the equation (29.1), we obtain G(x) =

1

-[G(2x)+G(0)}

Jensen's functional equation

283

and by induction G{x) = ±G(rx)

+ ^±G(0)

for

n e N.

Now applying Lemma 28.4, 2~nG(2nx) = A„(x) ->• A(x) , and in view of Lemma 28.5, 2- n (2 n - 1)G(0) -> cZ (j 2~ n (2 n - 1)G(0) = G(0) . Consequently, on account of Lemma 28.6, we get clG(x) = d[A(x) + G(0)] which is equivalent to G(x) = A(x) + G(0) . Taking K := G(0) + a, we obtain F(x) = G(x) + a = A(x) + K, which concludes the proof. •

Continuous solutions of Jensen's functional equation In this section we give a characterization of s.v. functions fulfilling the Jensen's equation under some regularity conditions. Theorem 29.2 Assume that a s.v.f. F: [0,oo) ->• CC(X) satisfies Jensen's functional equation (29.1). Then the following conditions are equivalent: (a) F is continuous on [0, oo); (b) F is right u.s.c. at some point x0 e [0, oo) ; (c) F is bounded on a subset of [0, oo) with a non-empty interior; (d) F is bounded on a subset of [0, oo) with positive Lebesgue measure; (e) F is of the form F(x) H,KeCC(X).

= xH + K, x G [0, oo), where

Proof. The implication (a) —> (b) is obvious. Assume now that F satisfies (b). Then, from Theorem 29.1, F has the form F(x) = A(x) + K, where A: [0,oo) -» CC{X) is additive and K € CC(X). Consider a neighbourhood U of zero and take a

284

Functional Equations and Inequalities in Several Variables

balanced neighbourhood V of zero such that V + V + V + VcU. Since F is u.s.c. at x0 G [0, oo), then there exists a 5 > 0 such that F(x) C F(x0) + V for all x € [XQ,XQ + 5). Choose an s e Q such that 0 < s < min(S, 1), sK C V and sF(x0) C V. Hence, for every t G [0,1] we have s2F(t) + sF(xQ) = sA(x0 + st) + sK + s2K = sF(x0 + st) + s2K, and consequently, s2F(t) C sF(x0 + st) - sF(x0) + s2K C s[F(x0) + V]-V

+

VcU,

which implies that F is bounded on [0,1], and proves the implication (b) —> (c). The implication (c) —> (d) is trivial. To prove that (d) —> (e), let us write F in the form F(x) = A(x) + K, where A: [0, oo) ->• CC(X) is additive and K G CC(X), and assume that F is bounded on y C [0, oo) of positive Lebesgue measure. Take a neighbourhood V of zero such that V + V C U. From (d) there exist s > 0 such that s ( J F(z) C F xeY

and

sK c V,

whence, for every x EY sA{x) C s[A(x)+K-K]

= s[F{x)-K]

C sF(x)~sK

C V+V C U,

i.e. A is bounded on Y, and by Theorem 28.5 it follows that A is of the form A{x) = xA{\) for x G [0,oo). Take H := A(l), then F(x) = xi7 + K for all x G [0, oo), which proves that (e) holds true. Finally, if F(x) = xH + K, then clearly F is continuous on [0, oo) and the implication (e) —> (a) is true. • Let us recall that X is a locally bounded topological vector space iff there exists a bounded neighbourhood of zero in X. In

Jensen's functional equation

285

this case we have another condition for solution of Jensen's equation to be continuous on [0, oo). A s.v.f. F: [0,oo) —>• C(X) is called measurable iff for every Borel set B C C(X) the inverse image F " 1 ^ ) := {x G [0,oo): F(x) G B} is Lebesgue measurable. We shall prove the following theorem. T h e o r e m 29.3 Let X be a locally bounded topological vector space and let a s.v.f. F: [0,00) —> C(X) be a measurable solution of Jensen's equation (29.1). Then there exist sets H,K G CC(X) such that F(x) = xH + K

for all

ie[0,oo).

Proof. Let U be a bounded neighbourhood of zero. Define the sets Bn := {A G C(X): AcnU} for n G N. We shall show that £ „ , n G N are open sets in the Hausdorff topology on the space C(X). Let K0 G Bn, for a fixed n G N, then KQ is compact and K0 C nil. Therefore by Theorem 1.10 in [?] there exists a neighbourhood V of zero such that K0 + V C nil. Consequently the family NV(K0) :={KeC(X):

K CK0 + V

and

K0 C K + V}

is a neighbourhood of KQ and Ny(K0) C Bn. This proves that Bn is open. From our assumptions, F is measurable and thus F^1(Bn) are Lebesgue measurable subsets of [0, 00). Moreover, (J Bn ~ C(X), whence

[0,oo) = F - 1 [ C ( X ) ] = | J ^ 1 ( ^ ) .

286

Functional Equations and Inequalities in Several Variables

Therefore, there exists an m G N such that the set Y := F-l(Bm)

= {xe

[0, oo): F(x) C mU}

is of positive Lebesgue measure. But the set mU is bounded (since U is bounded). Thus, for every neighbourhood D of zero there exists a constant a > 0 such that avail C D. Hence, we get a ( J F(x) C amU C D, xeY which means, according to the definition, that F is bounded on the set Y. Consequently, Theorem 29.2 implies our assertion. •

Chapter 30 Pexider's functional equation In this section we give a characterization of set-valued solutions of the Pexider functional equation F{x + y) = G{x) + H(y),

(30.1)

where F, G and H are unknown s.v. functions. Following K. Nikodem [147], we shall prove the following basic result. Theorem 30.1 Let X be an abelian semigroup with zero and Y a T0 topological vector space. LetF: X -> CC{Y), G: X -> CC{Y) and H: X —> CC(Y) be s.v. functions satistying the Pexider equation (30.1). Then there exist an additive s.v.f. A: X —> CC{Y) and sets M,K € CC(Y) such that F(x) = A(x) + K + M, G(x) = A(x) + K, H(x) = A(x) + M (30.2) for all x e X. Proof. We consider two cases. Suppose first that 9 G H(6). Take x £ l , then F(2x)

= =

G(x) + H(x) C G{x) + H{6) + H{x) + G{6) F(x) + F(x) = 2F(x), 287

288

Functional Equations and Inequalities in Several Variables

i.e., 2 _1 F(2a;) C F(x), and by induction, one can verify that 2-(n+i) j p(2 n+1 £) c 2-nF(2nx) for n G N, which means that the sequence {2~nF(2nx)}ne® is decreasing. We define A(x) := p | 2~nF(2nx)

for

x G X.

(30.3)

Clearly for any x G X, A(x) G CC(F). Now, using the equation (30.1) three times, we obtain for x G X, G(2x) + H{6) = F(2x) = G(x) + H(x) C G{x) + [H(x) + G(9)] = G(x) + F(x) = G(x) + G(x) + H{8) = 2G{x) + H(9). By Lemma 20.1, we get hence G(2x) C 2G(x), which implies that the sequence {2~nG(2nx)}n€^ is decreasing. From the equation (30.1), we obtain 2~nF{2nx) = 2-nG{2nx) + 2~ n #(0), whence,by Lemma 29.2, A{x) = p | 2~nF(2nx) = p | 2-nG(2nx) n€N

n6N

+ p|

2-nH(9).

n£N

Since #(0)is bounded, f) 2" n if(0) = {0}. Thus A(rr) = P | 2'nG(2nx)

for

i e l

By the same argument, one can show that the sequence {2~nH(2nx)}neN is decreasing for every x e X and A{x) = p | 2-nH{2nx)

for

z G X

Pexider's functional equation

289

Therefore,applying again Lemma 29.2, one gets A{x + y) = P | 2-nF{2nx + 2ny) = f] 2-n[G{2nx) + H(2ny)} neN

neN

n

n

= P| 2- G{2 x) + P | 2~nH(2nx) neN

n£N

= A(x) + A(y) for all x,y £ X. This shows that the s.v.f. A is additive. In the sequel, we shall show the equality F(nx) + (n - 1)H{0) = F{x) + (n - l)H{x)

(30.4)

for all x G X and all n G N. Clearly, for n = 1 it is trivial. To apply the induction, assume that (30.4) holds for some s G N. Then, by the equation (30.1), we get F[(s + l)x] + sH{9) = = = =

G(sx) + H(x) + (s - 1)H(6) + H{0) F(sx) + H{x) + (s- 1)H(9) F{x) + (s- l)H(x) + H{x) F{x) + sH(x),

which proves that (30.4) holds for n = s + 1. Consequently, by induction, (30.4) holds true for all natural numbers n. Thus applying (30.4) for 2 n , we obtain F{2nx) + (2" - l)H(d) = F(x) + (2" -

l)H(x),

which gives the equality 2" n F(2"a;) + (1 - 2n)H(6) = 2~nF{x) + (1 By Lemma 28.4, for x G X, 2~nF{2nx) -> p | 2~nF{2nx) = A(x). neN

2~n)H(x).

290

Functional Equations and Inequalities in Several Variables

It is a simple exercise to show that (1 - 2n)H(9) -> H(8), 2~nF{x) ->• {0}, (1 - 2~n)H{x) -> H{x). Therefore, applying Lemma 28.6, we get cl[A(x) + H{9)] = clH{x)

for

xGX

and since A(x), H(9), H(x) are closed, then H(x) = A(x) + H(9)

for all

x G X.

Similary one may get the equality G(x) = A(x) + G{9)

for all

x e X.

Put K := G{9) and M := H{9). Hence G(x) = A(x) + K, H(x) = A(x) + M, F(x) - G(x) + H(9) = A(x) +K + M, for all x G X. This completes the proof in the first case. Assume now that 9 $ G(6), 9 <£ H{9). Fix arbitrarily a G G{9) and b G H{9) and define the s.v. functions F\,G\,H\\ X -» C C (Y) by the formulas Fi(:r) := F(x) -a-b,

G^x) := G{x) - a, Hx{x) := if(a;) - 6.

Clearly F\,Gi,Hi satisty the equation (30.1) and moreover, 9 G G\{9) and 9 G H\{9). Therefore, from the proof already presented, the functions Fi,G\,Hi are of the form (30.2). Consequently the s.v. functions F,G,H are of the form (30.2), as well, which concludes the proof. • Remark 30.1 Let us observe that in the case when X is a group, the s.v.f. A occuring in the formulas (30.2) has to be, in fact, a single valued function. Indeed, since the set A{9) is bounded, then from the equality A{9) + A(9) = A{9) we infer, that A{9) = {0}. Thus for every x G X, A(x) + A(—x) = A(9) = {9}, which implies that A(x) is the set consisting of one point only.

Pexider's functional equation

291

Notes 30.1 For the case when Y is a locally convex Hausdorff space, one can prove the following (see also [163]). Proposition 30.1 Let Y be a locally convex Hausdorff space. S.v. functions F: [0,oo) -> CC(Y), G: [0,oo) -» CC(Y) and H: [0, oo) —> CC(Y) satisfy the Pexider equation (30.1) iff there exist an additive function f: [0, oo) —> Y and sets K,A,B G CC{Y) such that F(x) = f{x) +XK + A + B, G(x) = f{x) +xK + A, H(x) = f(x)+xK

[

+B

"}

for all x G [0, oo). 30.2 Consider the functions F(x)=G(x):=(0,x

H(x) = i^,X^ \{0,x)

+ l),

l£[0,w)nQ

'

XG[0,OO)\Q.

Then F,G,H satisfy the equation (30.1), but they are not of the form (30.2) or (30.5). Let us note that the values of F,G,H are not compact sets. This shows that Theorem 30.1 and Proposition 30.1 do not hold without the assumption that the values of F, G, H are compact. It is possible to show that the hypothesis that the values of F, G, H are convex, is also essential.

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Chapter 31 Quadratic set-valued functions Consider a real normed space X. Let us remind that a setvalued function F: R —>• 2X is called quadratic provided that F(x + y) + F{x - y) = 2F(x) + 2F{y)

(31.1)

for all i , j / e l It is a simple exercise to show that if a set A C X is convex, then the s.v.f. F given by the formula F(t) = t2A

for

t G R,

(31.2)

is quadratic. Let us note that, in general, the inverse statement is not true. The main object of this section is to give a sufficient conditions for a s.v.f. with convex and compact values to be of the form (31.2). Let us also note that in normed spaces the s.v.f. F: E —>• 2X is bounded on a set A C E iff there exists a ball B(9,r) := {x € X: \\x\\ < r} with F{x) C B(0,r) for all x E A. We shall use the following lemmas (see [146]). Lemma 31.1 Let a s.v.f. F: E —> CC{X) be quadratic. Then for every rational number s and every x £ R, we have F{sx) = s2F(x). 293

(31.3)

294

Functional Equations and Inequalities in Several Variables

Proof. Using (31.1) with x = y = 0, we get F(0) + F(0) = 2F(0) = 2F(0), which, by the convexity of F(0) yields 2F(0) = 4F(0). Since F(0)is bounded, thus F(0) = {9}. Now, using (31.1) with x = y, we obtain F(2x) = AF(x) for all x £ R. Assuming that F(kx) = k2F(x) for every k G N, k < n, we have for n + 1: F((n + l)x) + (n - l)2F(x) = 2n2F(x) + 2F(x). By the convexity of F(x), we get hence F((n + l)x) + (n - l)2F(x) = (2n2 + 2)F(x) = [(n + l) 2 + (n - l) 2 ]F(x) = (n + l) 2 F(a;) + (n -

l)2F(x).

which, by Lemma 20.1, yields F(n + l)x) =

(n+l)2F(x).

We have thus proved the equality F(nx) = n2F(x) for every n £ N and i G l . Moreover, for m e N, F(x) = F(m—) = m

m2F(—), m

which gives F(-) = m~2F(x), i.e. F(sx) = s2F(x) for all positive rational s and all x G R. To prove our statement, note that F is an even function. In fact, taking x = 0 in the equation (31.1), we get F(y) + F(-y)

= 2{d) + 2F(y) = F(y) + F(y),

whence, using Lemma 20.1 once more, we deduce F(—y) = F(y) for every y e R. This concludes the proof. •

Quadratic set-valued functions

295

Lemma 31.2 Let A C R be a subset of positive inner Lebesgue measure or of the second category with the Baire property. If a quadratic s.v.f. F: R —>• CC(X) is bounded on A, then it is bounded on a neighbourhood of zero. Proof. Assume that B(9,r) is a ball such that F(x) C B(9,r) for all x G A. Define the set U(A) := {x ER: An (A + x) n (A - x) ^ 0}. Take x G U(A), then there exists s G R withs, s — x, s + x G A. Thus, by the equation (31.1), we get F(s)+F(x)

= ±[F(a+x)+F(8-x)]

C

1

-[B(9,r)+B(e,r)] = B(9,r),

since B(9, r) is a convex set. For any y € F(s), we have y G B(9, r) and F(x)

C F(x) + [F(s)-y]c[F(x) + C B(6,r) + B{d,r) = B{d,2r),

F(s)]-y

i.e. F(x) C B(9,2r) for every x e U(A). Thus we have proved that F is bounded on the set U(A). By the properties of the set A, one can deduce that U(A) contains a neighbourhood of zero. • Now we formulate and prove the main result of this section (see K. Nikodem [146]). Theorem 31.1 Let A C R be a subset of positive inner Lebesgue measure or of the second category with the Baire property. Let a quadratic s.v.f. F: R —>• CC(X) be bounded on A. Then F(x) = x2F(l)

for all

x GR

(31.4)

296

Functional Equations and Inequalities in Several Variables

Proof. Since F is bounded on A, then by Lemma 31.2 it follows that there exists an interval [0, e] (we may assume that e is a rational number) such that F is bounded on [0, e\. Thus taking into account that F(l) is compact, there exists a ball B(Q, r) such that F(x)cB{9,r)

and

x2F{\) C B(9,r)

for all

xe[0,e}.

Consequently (H is the Hausdorff distance) , H[F(x),x2F(l)}

< 2r

for

x G [0,e].

Denote h := sup{H[F(x),x2F(l)]:x

G [0,e]}.

We shall prove that h = 0. For the contrary, assume that h > 0. So there exists Xo G (0, e] with H[F(XQ),XQF(1)] > \h. Observe that without loss of generality we may suppose that XQ G [§,£]• Indeed, if x0 G (0, | ) , then we can find a rational number v such that vx0 G [|,£] and hence, by Lemma 31.1 and Lemma 20.2, H[F(vx0),v2x2QF(l)]

= w 2 //[F(x 0 ),^F(l)] > v2^h > ^h.

The number 2x0 can be represented as 2x0 = e + s with s G [0, e]. Hence, in view of Lemma 20.2, and the triangle inequality for the Hausdorff metric H, we obtain H[F(2x0),(2x0)2F(l)] = = H[F(2x0) + F(e - s), (2x0)2F(l) + F(e - s)] < H[F(2xQ) + F(e - s), (2xQ)2F(l) + (e - s)2F(l)} + H[(2x0)2F(l) + (£- s)2F(l), (2x0)2F(l) + F(e - s)}. Obviously, we have the equalities F(2x0) + F{e - s) = F{e + s) + F{e - s) = 2F{e) + 2F{s) = 2e2F(l) + 2F{s),

Quadratic set-valued functions

297

and (e + s)2F(l) + (e - s) 2 F(l) = 2e2F(l) + 2s2F(l). Hence we obtain (using properties of the metric H) H[F{2x0)+F(e - s), {2x0)2F{l) + (e - s)2F(l)} = H[2e2F(l) + 2F(s), 2e2F{l) + 2s2F(l)} = H[2F{s),2s2F(l)} = 2H[F(s),s2F{l)} < 2h. Clearly, we have also H[(2xQ)2F(l) + (e- s)2F(l), (2x 0 ) 2 F(l) + F(e - s)} = H[(e-s)2F(l),F(e-s)]
H[4F(x0),4x20F(l)} 4H[F(x0),x2F(l)]>3h,

= =

a contradiction. Therefore, h = 0, and hence F(x) = x 2 F(l) for every x € [0, e\. To finish the proof, let us fix arbitrarily x € R and consider a rational number v / 0 such that vx G [0, e]. Thus on account of Lemma 31.1, and by what we have just proved, F(x) =

\F(VX) V

= \v2x2F(l)

=

x2F{l),

2

V

which concludes the proof.

•

The simple consequence of this theorem is the following Corollary 31.1 // a quadratic s.v.f. F: R ->• CC(X) is continuous at some point XQ € R, then it is of the form (31.4).

298

Functional Equations and Inequalities in Several Variables

Let us now introduce some definitions that are needed for further considerations. The Haurdorff distance generates the topology in the space C(X). In particular we have the family of Borel sets. We say that a s.v.f. F: R —>• C(X) is measurable iff for every Borel set B C C(X) the inverse-image F_1(B) is a Lebesgue measurable set. A s.v.f. F: R ->• C(X) has the Baire property iff for every open set A c C(X) the set F~1(A) has the Baire property. Moreover, we say that a s.v.f. U: R —> 2* majorizes a s.v.f. F : R -> 2 X on a set F C R if F(x) C tf(a:) for all x£V. Another very general sufficient condition for a quadratic s.v.f. to be of the form (31.4) reads as follows. Theorem 31.2 Assume that U: R —>• C{X) is a measurable s.v.f. (or a s.v.f. with the Baire property). Let A C R be a set with positive Lebesgue measure (or a second category with the Baire property). If a quadratic s.v.f. F: R —>• C(X) is majorized by U on A, then F has the form (31.4). Proof. Consider the sets An := {D E C{X): D C B{6, n)},

n e N.

In view of Theorem II from [22], An, n € N are open sets. Clearly, C(X) = {J~=1An, whence oo

l

R = U- [C(X)] = | J U-\An) n=l

oo

= [j{x e R: U(x) C B(0,n)}. n=l

From the assumption that U is measurable, we guess that the sets {x G R: U(x) C B(9,n)} for n e N are Lesbegue measurable sets. Assume first that A is a set with positive Lebesgue measure. Then it follows that there exists on m € N such that the set

A f l j i e K : U(x) cB(9,n)}

Quadratic set-valued functions

299

has a positive Lebesgue measure, which means that U and therefore also F is bounded on a set with positive Lebesgue measure. Consequently, by Theorem 31.1, F has to be of the form (31.4). If we assume that U is a s.v.f. with the Baire property and A is a set of second category with the Baire property, the proof runs quite similarly. • Directly from the last theorem we obtain Corollary 31.2 If a quadratic s.v.f. F: R —> C(X) is measurable or has the Baire property, then it is of the form (31.4)Remark 31.1 Note that there are s.v. functions which can not be expressed in the form (31.4.). To see this, consider the s.v.f. F(x) := x2 + Q,

x e R,

where Q denotes the set of all rational numbers. In particular, the formula (31.4) does not hold for x — 0.

Notes 31.1 The results presented here form a generalization of theorems proved by D. Henney [87]. 31.2 W. Smajdor [200] contributed further to the theory of solutions of the quadratic equation in set-valued functions. Namely, she has proved the following generalization of Theorem 31.1 Proposition 31.1 Let X be a locally convex space and let A C R be a set of positive inner Lebesgue measure or of the second category with the Baire property. Let a s.v.f. F: R —>• C(X) be quadratric and bounded on A. Then F(x) = x2F(l)

for all xeR.

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Chapter 32 Subadditive set-valued functions Let X and Y be two normed spaces and consider a function / : X —>• Y satisfying the inequality \\f(x + y)-f(x)-f(y)\\<e

(32.1)

for x,y € X and some e > 0. Then by Theorem 13.1 there exists (exactly one) additive function A: X —> Y such that ||A(x) - f(x)\\ < e for all x £ X. Let B(9,e) denote the closed ball in Y with center at 6 and radius e. Then the inequality (32.1) may be rewritten as f(x +

y)-f(x)-f(y)eB(9,e),

or equivalently fix + y) + B{0,e) c f(x) + B(0,e) + f(y) + B{9,e). Hence, if we define a s.v.f. F: X -> 2Y by F(x) := f(x) + B{6,e), then we have F(x + y)cF(x) + F(y) (32.2) for all x,y, e X and, moreover, A(x) G F(x) for a; G X. In this section we shall deal with s.v. functions satistying the condition (32.2). 301

302

Functional Equations and Inequalities in Several Variables

Let X and Y be two semigroups with operation „+". A s.v.f. F: X —> 2Y is called subadditive iff the inclusion (32.2) holds for all x,y, G X. Assume that Y is a topological vector space. We will denote by: 2Y — the family of all nonempty subsets of Y, C(Y) — the family of all compact members of 2Y, CL(Y) — the family of all closed members of 2Y, CC(Y) — the family of all convex members of C(Y), B(Y) — the family of all bounded members of 2Y. In the sequel (unless otherwise stated) we shall assume that Y is a topological vector space. The theory of subadditive s.v. functions was widely developed by many authors. Here we present some results of W. Smajdor [200]. Lemma 32.1 Let a s.v.f. F: [0, oo) —> 2Y be subadditive and let F(0) be convex and closed. Then 9 G F(0). Proof. Inserting x = y = 0 in (32.2), we get F(0) C F(0)+F(0) = 2F(Q). Assume that x G F(0), then x G 2F(0), whence § G F(0). Therefore 2~nx G F(0) for n G N. Since 2~nx -> 9 and F(0) is closed, so 9 G F(0). • Now we introduce some further definitions. Definition 32.1 (cf. [7]). A s.v.f. F: X -> 2Y, where X,Y are topological vector spaces, is called upper semi-continuous (abbreviated to u.s.c.) at x G X iff for every neighbourhood V of zero in Y there exists a neighbourhood U of zero in X such that F(x + h) C F(x) + V

for every

heU.

Moreover, F is called lower semi-continuous (abbreviated to l.s.c.) at x G X if for every neighbourhood V of zero in Y there exists a neighbourhood U of zero in X such that F(x) CF(x + h) + V

for every

h G U.

Subadditive set-valued functions

303

As usual, F is called u.s.c. (l.s.c.) in X iff it is u.s.c. (l.s.c.) at every point x G X. Definition 32.2 A s.v.f. F: X -> 2Y is called continuous at x € X iff it is u.s.c. and l.s.c. at x. It is said to be continuous iff it is continuous at every point x G X. In the next result we shall apply the following theorem, which will be proved in the 36 paragraph (see also [145]) (Theorem 36.7). Theorem 32.1 Let D be a convex, open, non-empty subset of a topological vector space. Let f: D —> C(Y) be a midpoint concave s.v.f., i.e., F ( ^ l ) C \ [F{x) + F(y)]

for x,y e D.

If A C D is a set with a non-empty interior and F is bounded on A, then F is continuous. Let X and Y be a vector spaces. A s.v.f. F: X —¥ 2Y is called Q+ -homogeneous iff F(sx) = sF(x)

for s G Q, s > 0, x G X.

Now we are in a position to prove Theorem 32.2 ([200]) Let F: [0,oo) ->• C(Y) be a subadditive and Q+ -homogeneous s.v.f. . If there exists an interval (a,b) C (0, oo) such that F is bounded on (a,b), then F is continuous in [0, oo) and xF(l)cF(x) for xG[0,oo). (32.3) Moreover, F(0) = {0}. Proof. Note first that since F is subadditive and Q + homogeneous, that it is midpoint concave. Hence by Theorem 32.1, F is continuous in (0, oo). Now we shall show that it is continuous

304

Functional Equations and Inequalities in Several Variables

at zero. To this end observe that F(0) = {9}. Take any balanced neighbourhood V of zero in Y. Since F(0) is the bounded set, then there exists an s > 0 such that sF(Q) C V, i.e. F(0) C W. Consider n € N with ns > 1. The homogenity of F implies F(0) = F ( - 0 ) = - F ( 0 ) C ( n s ) - V C V, which means that F(0) = cl{9} = {6}. We are now going to show that F is bounded on the interval [0,b). As before, let s > 0 be such that sF[(a,b)] C V and let [r,w C (a,b), where r, iu G Q+. Cleary, sF[[r, io]] C V. Take any xG , then since jx G [r, iw], S F(x)

= sF (T- • ™x) = s^F (^X) \w r / w \r /

c^-VGV. w

Hence s F ( ^-,r 1 C V. By the same argument, sF

10^ 10

CV

and, in general, y.n+1

sF

n

W

n=l

„n

Wn~l

CV

for

n G N.

Subadditive set-valued functions

305

Taking into account that F(0) = {9}, we obtain sF([0,b)) C V. Denote U := i[0,6), where n G N is such that JF([0,6)) C V. If x GU, then x = \y with y G [0, 6) and hence F(x) = F ( - j / ) = -F{y) CV = V + F(0), ft

lb

i.e. F{x) C F(0) + V

for

ie[/,

(32.4)

which means that F is u.s.c. at zero. Moreover, since V is balanced, by (32.4) and F(0) = {9}, we get -F(x) cV fov x eV, whence F(0) = {9} C F(z) - F(x) C F(x) + V

for

a; G V.

Thus F is also l.s.c. at zero and consequently F is continuous at zero as claimed. Now we shall prove the inclusion (32.3). Fix arbitrarily x0 G (0, oo) and take a sequence xn —>• XQ, xn G Q+ for n G N. Let V be a neighbourhood of zero in Y such that clW + clW C V. Since F is continuous at x0, there exists an m G N such that for n > m, xnF(l)

= F(xn) C F{x0) + WC F(x0) + clW.

Thus the closedness of F(x0) + c^VF implies a) 0 F(l) C F(x0) + clW C F(x0) + clW + clW C F(x0) + V and consequently, x0F(l) C f|[F(a;o) + V] C cZF(z0) = F(xo), v For x0 = 0, we also have x0F(l) This completes the proof.

x0 G (0, oo).

= {9} C F(x0) = F(0) = {9}. •

Lemma 32.2 LetF: [0, oo) —> 2Y be a subadditive s.v.f. . Assume that there exists a subset A C (0, oo) of positive inner Lesbegue measure (or of the second category with the Baire property) such that F is bounded on A. Then there exists an interval (a, b) C (0,oo) such that the set F((a,b)) is bounded.

306

Functional Equations and Inequalities in Several Variables

Proof. Let U, V be neighbourhoods of zero in Y such that U + U C V. Since F is bounded on A, there exists an s > 0 with sF(A) C U. By the subadditivity of F, F(A + A)CF(A)

+ F(A),

whence, sF(A + A)C sF{A) + sF(A) cU + UcV, which means that the set F(A + A) is bounded. In view of the theorem of H. Steinhaus [123], p.69, or by the Remark 4.1, there exists an interval (a, b) with (a, b) C A + A. Since F((a, 6)) C F(A + A), thus F((a, 6)) is bounded. • Combining Theorem 32.2 and Lemma 32.2, we get the following result (see [200]) about the form of a subadditive and Q+homogeneous s.v.f. . Theorem 32.3 Let F: [0,oo) -> C(Y) be a subadditive and Q+homogeneous s.v.f. . If F is bounded on the set A C (0,oo) of positive inner Lesbegue measure (or of the second category with the Baire property), then F(x)=xF(l)

for rce[0,oo).

(32.5)

Proof. Fix an arbitrary x G (0, oo) and define the s.v.f. G(p) := F(px) for p G [0, oo). It is a simple exercise to show that G is subadditive, Q+-homogeneous and bounded on the set -A of positive inner Lesbegue measure (or of the second category with the Baire property). In view of Theorem 32.2 and Lemma 32.2, we obtain pF(x) = pG(l) C G(p) = F(px)

for p > 0.

Take p = -, then -F(x)

C F(s)

for s > 0,

Subadditive set-valued functions

307

whence -F(x) C -F(s) for x,s e (0,oo). x s By the same manner (changing roles x and s) we obtain the inclusion -F(s) C -F(x) for x, s G (0,oo), s a: which yields the equality -F(x) x

= -F(s) s

for

x, s G (0, oo).

Thus ^-F(x) = B = const, for £ G (0,oo), and consequently F(x) — xB for x > 0, where 5 = F ( l ) . For :r = 0, we get in view of Theorem 32.2, F(0) = {9} = 0F(1), which completes the proof of (32.5). • Let us now note the following Lemma 32.3 Let (X, +) be a semigroup and Fn: X —> C(Y), n G N be subadditive s.v. functions satisfying the condition Fn+\ C Fn, n G N. Then the s.v.f. oo

F(x) := f | Fn(x),

xeX

n=l

is subadditive with non-empty compact values. Proof. Cleary F{x) G C(Y) for every x G X. Lemma 29.2, we obtain for x, y G X, oo

F(x + y)=

Moreover, by

oo

f | Fn(x + y) C f | [Fn(x) + F(y)} n=l oo

n=l oo

= f l Fn(x) + f | Fn(y) = F(x) + F(y), which means that F is subadditive. Having done this, we can prove

•

308

Functional Equations and Inequalities in Several Variables

Theorem 32.4 Let F0: [0, oo) —>• CC(Y) be a subadditive s.v.f. Then there exists a subadditive and Q+ -homogeneous s.v.f. F: [0, oo) —>• CC(Y) such that F C F0, given by the formulas oo

F{x) = f]Fn(x),

ze[0,oo),

n=0

where oo

Fn(x) := f | ( n + l)-*F„_i[(n + 1 ) H

n = 1, 2 , . . . .

(32.6)

s=0

Proof. Since F0 is subadditive, then for x = ?/ from the inclusion F0(x +

y)cF0(x)+F0(y),

we obtain ^F 0 (2x) C F0(x).

(32.7)

Consider the sequence of non-empty and compact sets {2-sFo(2ax)},

S

= 0,l,2,....

Directly from (32.7) one can guess that this sequence is decreasing and hence by Lemma 32.3 the s.v.f. oo

F^x) := f | 2-sF0{2sx),

x e [0, oo)

s=0

is subadditive. Moreover, for every x e [0, oo), Fi(x) G CC(Y). We have also ..

oo

-F1(2x) = f)2-'F0(Vx)

= F1{x).

s-l

i.e. -Fl(2x)

= Fl{x)

for i e [ 0 , o o ) .

(32.8)

Subadditive set-valued functions

309

Cleary Fi C F0, i.e. Fx{x) C FQ(x) for all x € [0,oo). From the inclusion Fi(x + y) C Fi(x) + Fi(y) for y = 2x, we get ^F1{3x)cF1(x), which implies that the sequence {3~sFi(3sx)} Lemma 32.3, the s.v.f.

is decreasing. By

oo

F2(x):=f]S-aF1(3'x),

*G[0,oo),

is subadditive. Moreover, F 2 : [0,oo) ->• CC(Y) and F2 C Fi. By (32.8) we have, OO

00

F2{2x) = p ) 3 - ^ ( 3 * • 2x) = 2 P 3-^(3*2;) = 2F2(x), s=0

s=0

i.e., F 2 (2z) = 2F 2 (z),

xG[0,oo).

Directly from the definition of F 2 , we get F2{3x) = 3F 2 (x),

ie[0,oo).

Repeating this procedure one can construct a seaquence of subbaditive s.v.functions Fn: [0, oo) —> CC(y) with properties Fn(kx) = kFn(x)

for fc = 1, 2 , . . . ,n + 1,

neN

and Fn+lcFn

for

nGN.

(32.9)

Define oo

F(x):=f]Fn(x),

ze[0,oo),

71 = 0

then by Lemma 32.3 and (32.9), F is subadditive and F(x) G CC(Y) for every x G [0, oo). Moreover, F C F 0 . Fix any k G N. Since F 0 D F x D F2 D . . . and Fn(kx) = kFn{x)

for

n = /c, A; + 1 , . . . ,

310

Functional Equations and Inequalities in Several Variables

then 00

00

F[kx) = f)Fn(kx)=f)Fn(kx) n=0

=

00

00

= kf)Fn(x)

n=k

=

kf)Fn(x) n=0

n=k

kF(x),

i.e. F(kx) = kF{x)

for

k e N.

This implies that F(qx) = qF(x)

for all q e Q+ and

xe[0,oo),

which means that F is Q+-homogeneous and concludes the proof.

• The next lemma provides some examples of s.v.functions in M. L e m m a 32.4 Let A,B be convex subset ofY.

Let

(3210)

™-fct "<* Then F is subadditive s.v.f. iff 6eA-B.

(32.11)

Proof. First assume that F given by (32.10) is subadditive. We have F(0) = {6}, whence 6 E F(0) C F(l) + F(-l)

=

A-B

Conversly, assume that (32.11) holds true. If x > 0 and y > 0, then we have, since x + y > 0, F(x + y) = (x + y)A = xA + yA = F(x) + F(y). In the case x < 0 and y < 0 we have similarly, F(x + y) = (x + y)B = xB + yB = F(x) + F(y).

Subadditive set-valued functions

311

Consider now the case: x > 0 and y < 0, x + y > 0. Then F(x) + F(y) =

xA

+ yB = (x + y)A +

(~y){A-B)

D (x + y)A = F(x + y). If x > 0, y < 0, and x + y < 0, then F(x) + F(y) = xA + yB = (x + y)B + x(A - B) D {x + y)B + {9} = F(x + y). For the remaining cases: x < 0,

y>0,

x + y>0

x < 0,

y > 0,

x + y < 0,

or one may argue similarly.

•

The next theorems say that a subadditive s.v.f. defined in E contains the largest s.v.f. of the form (32.10). Theorem 32.5 Let FQ: R ->• CC(Y) be a subadditive s.v.f. . Let U C R be a set with positive inner Lebesgue measure(or of the second category with the Baire property) and let FQ be bounded on U. Then there exist the largest sets A,B£ CC(Y) such that the s.v.f. (32.10) is subadditive and contained in F0. The proof based on Lemma 32.4 and on the Kuratowski-Zorn Lemma is omitted here. For detailes see W. Smajdor [200], Th. M. Rassias [113]. In the sequel we shall present some results about the existence of additive selections of a subadditive s.v.functions. Theorem 32.6 Assume that the hypothesis of Theorem 32.5 are satisfied. Then there exists a u £Y such that xu E F0(x) for all x eR.

(32.12)

312

Functional Equations and Inequalities in Several Variables

Proof. By Theorem 32.5 there exist sets A,B G CC{Y) such that the s.v.f. (32.10) is subadditive and contained in F0. From Lemma 32.4, 9 G A — B, which implies that we can choose a u E A• F is said to be A-additive iff f(x + y) - f(x) - f(y) G A (32.13) for all x, y G E. Note that for a normed space Y and A = B(6, e), where B(9, e) denotes the closed ball in Y, a function / : E —>• Y is A-additive iff ||/(x + y ) - / ( : r ) - / ( y ) | | < e for all x,y eR. Theorem 32.7 Let / : R —>• F be an A-additive function, where A G CC(Y), and let U C t be a set with positive inner Lebesgue measure (or of the second category with the Baire property) such that f(U) is bounded. Then there exists a unigue u EY such that xu - f{x) G A for all x G E.

(32.14)

Proof. Consider the s.v.f. F: E ->• CC(Y) defined by F(x) := f(x) + A,

xeR.

Since / is A-additive, then f(x + y)ef(x)

+ f(y) + A,

x,yeR,

which yields f(x + y) + Acf(x)

+ A + f{y) + A,

x,y<ER.

Thus F is a subadditive s.v.f. . Take an arbitrary neighbourhood V of zero in F and choose a balansed neighbourhood V1 of zero

Subadditive set-valued functions

313

in Y such that Vi + Vi C V. Since A is bounded, there exist constants si,S2 > 0 such that SiA C V\ and S2f(U) C V\. Hence, for s = min[si, s2] sf{U) + sAcV1 + VlcV. Clearly, F(U) = f(U) + A, which by the last inclusion implies sF(U) = s[f(U) + A] c V and means that the s.v.f. F is bounded on U. Consequently, in view of Theorem 32.6 there exists a u e Y such that xu € f(x) + A

for all

iGR,

whence we get the condition (32.14). To prove the uniquennes of u, take x = n in (32.14), then we have u - -fin) e-A, ne N. (32.15) n n Let V be an arbitrary balanced neighbourhood of zero in Y. Since A is bounded, then there exists an s > 0 with sA C V. Thus by (32.15) we get

u--f(n) n

e—VcV ns

for n sufficiently large (ns > 1), which means that ^/(n) —> u and proves that if u satisfies the condition (32.14), then the limit of the sequence {^f(n)} is uniquelly determined. This concludes the proof of the uniquennes of u satisfying the condition (32.14). •

Subadditive s.v. functions in topological vector spaces Following W. Smajdor [200] in this section we shall prove theorems on boundedness of subadditive s.v. functions defined in a

314

Functional Equations and Inequalities in Several Variables

topological vector spaces. A connection between boundedness and semi-continuity will be studied too. In particular, a generalization of the well known Banach-Steinhaus theorem, will be presented. We say that a s.v.f. F: X —>• 2 y is bounded if for every bounded subset A in X, F(A) is bounded subset in Y. We also say that Y is a locally convex space iff it possesses a base of convex neighbourhoods of zero. T h e o r e m 32.8 Let X be a topological vector space and Y a locally convex space. Let a s.v.f. F: X —» 2Y be subadditive. If F{6) is bounded and F is u.s.c. at zero, then F is bounded. Proof. Let V be a balanced and convex neighbourhood of zero in Y. Since F is u.s.c. at zero, there exists a balanced neighbourhood U of zero in X such that F(x) C F(9) + V for every x 6 U. Let us take a bounded set A in X. Then there exist positive integers s, r such that -AcU and - F(6) C V. s r Take any x € A, then x € sU and x = su for some u e U. Therefore, F(x)

= F{su) C F{u) + ... + F{u\ C F{9) + ... + F(9) + sV s

C rsV + sV = s(r + 1)V, i.e. F(A)Cs(r

+ l)V ,

which shows that F is bounded on A. Since A is arbitrary bounded set in X, this proves the boundedness of F. •

Subadditive set-valued functions

315

Definition 32.3 ([14]) A s.v.f. F: X -»• 2Y is called upper semi-continous in Berge's sense (shortly B.u.s.c.) at x0 G X iff for every open set U in Y such that F(XQ) C U there exists a neighbourhood V of zero in X such that F(x) C U for all

x e x0 + V. F is called lower semi-continous in Berge's sense (abbreviated to B.l.s.c.) at XQ G X iff for every open U in Y such that F(xo) fl U ^ 0, there exists a neighbourhood V of zero in X such that F(x) DU / 0 for all x G x0 + V. A s.v.f. F is said to be B.u.s.c. or B.l.s.c. in X iff it is B.u.s.c. or B.l.s.c. at every point x G X, respectively. Note that if F is B.u.s.c, then F is also u.s.c. but not conversly. Also a s.v.f. which is l.s.c. is B.l.s.c, too. If F is a s.v.f. with compact values the definitons of Berge's semi-continuity are equivalent to the definitions of semi-continuity. As before, let X and Y denote arbitrary topological vector spaces. Theorem 32.9 Assume that a s.v. f. F: X -> 2Y is B.u.s.c. at zero, F{6) is bounded and F(nx) C nF(x)

for x G X , n G N.

(32.16)

Then F is bounded. Proof. Take any open neighbourhood U of zero in Y. By the boundedness of F(9), there exists an s > 0 with F(9) C ^ U. Since F is B.u.s.c. at zero, there exists an open balanced neighbourhood V of zero in X such that F(x) C -U for all x G V, which implies that F(V) C - U. Consider any bounded set A C X. Thus we can find an n G N such that - A C V and consequently, V n

J

s

Hence by (32.16) we get for every x £ A,

F{x) = F(n-) CnF(-) C - U , \ n) \n) s

316

Functional Equations and Inequalities in Several Variables

i.e. F(A) C j U. Thus F(A) is bounded for every bounded set A C X, which proves the theorem. • The next theorem is the generalization of the well-known Banach-Steinhaus theorem on the uniform boundedness of norms of linear continous operators. T h e o r e m 32.10 Let a s.v.f. F: X ->• 2Y be subadditive, Q+homogeneous and B.l.s.c. in X. If, moreover, F(—x) = —F(x) or F(—x) = F(x) and the set B:={xeX:

F(x) G B(Y)}

is of the second category, then B = X and F is bounded. Proof. Let A be a neighbourhood of zero in Y. Choose a balanced neighbourhood V of zero in Y such that clV + clV C A. Define D:= {xeX:

F(x) C clV} .

Then since F is B.l.s.c, the set D is closed. Indeed, we have X\D

= {xeX:

F(x) (jLclV) = {x e X: F(x) n (Y\clV)

^ 0}.

Take x0 G X \ D, then since Y \ clV is open, by the B.l.s.c. of F, there evists a neighbourhood U of x0 such that F(x)(~)(Y\clV) ^ 0 for x G U, i.e. U C X \ D, which shows that X \ D is open and consequently D is closed. Take any x G B,then we can find an n G N such that i F{x) C y,whence F(-\=-F{X)
\n/

n

This yields that ^ G D or x G n.D and proves the inclusion oo

B e |J nD. n=l

Subadditive set-valued functions

317

Hence, by the assumption that B is of the secoud category, we infer int D ^ 0 and consequenty there exists a neighbourhood U of zero in X such that U C D — x0 for some XQ E D. Take any x E U, then a; + XQ E -D, and F(z) = F ( z + % - a;0) C C

F(x + x0) + F(-xQ) clV ± F{x0) C clV + clV C A.

Thus F(C7) C A, which implies that F(6) C f]A, where A runs the family of all neighbourhoods of zero in Y. Hence F(9) = {6} and the inclusion F(x) C A

for

x eU

shows that F is B.u.s.c. at zero. Moreover, Q+ -homogeneity of F implies the relation (32.16). From Theorem 32.9, it follows that F is bounded and F(x) € B(Y) for every x € X, because the set {x} is bounded. This proves the equality B = X and completes the proof. • Remark 32.1 Let X be a Banach space and Y be a normed space and let fs: X —>• Y, s e /, I-the index set, be a family of linear continuous operators. Define the s.v.f. F: X —> 2Y by the formula F{x):={fs(x):sel},

i £ l

(32.17)

One can verify that F is a subadditive, Q+ -homogeneous s.v.f. . Moreover, F(—x) = —F(x) for x € X and F is B.l.s.c. in X. Therefore, in view of Theorem 32.10, F is bounded which means, for normed spaces, that there exists the ball B(9,r) with center at zero and radius rinY such that F(x)cB(6,r)

for

|| x\\ < 1, x E X.

Hence, II fs{x)\\
for

x G l , || x|| < 1,

and

s E I.

318

Functional Equations and Inequalities in Several Variables

Consequently \\fs\\
for all

sEl.

(32.18)

The condition (32.18) means that the norms of linear operators fs, s G / are uniformly bounded. Thus the Banach-Steinhaus theorem is a simple consequence of Theoremm 32.10. Next theorem gives a sufficient condition for a subadditive s.v.f. to be semi-continuous at a point. Theorem 32.11 Let X be a metrizable topological vector space. Let a s.v.f. F: X —> 2Y be subadditive, Q+ -homogeneous and bounded and let F{0) be convex and closed. Then F is B.u.s.c. at zero. Proof. Assume that {Un: n G N} is a base of neighbourhoods of zero in X. Without loss of generality, we may suppose that this sequence is decreasing. For the contrary, let us assume that F is not B.u.s.c. at zero. By the definition 32.3 it is equivalent to the condition that there exists an open set G in Y such that F(9) C G and for every neighbourhood U of zero in X there exists &nx E U with F(x) <£. G. Assume that xn G Un, n G N is such that F(xn) <£ G. Cleary xn

->• 0.

Moreover, we can find a sequence {an} C Q+ such that an —>• oo and anxn —>• 0 (for details see e.g. [186]). Since the sequence {anxn} is bounded, so the boundedness of F implies that the set F({anxn: n G N}) = U^Li F(anxn) is bounded as well. Lemma 32.1 yields that 0 G F(0). Take a balanced neighbourhood V of zero in Y such that V C G. Thus there exists a constant s > 0 for which s( \J™=1 F(an xn)) C V, whence

F{xn) = a-lF{anxn)

C-VcVcG, san

(for n sufficiently large), which contradicts our assumption that

F(xn) <£G.

•

Subadditive set-valued functions

319

Assume now that Y is a normed space and consider A G B(Y). Let us denote m i l :=sup{||x||: x G A}. We shall end this section with the following two theorems. Theorem 32.12 Let X be a Banach space and Y a normed space. Let a s.v.f. F: X —>• B(Y) be subbadditive and B.l.s.c. . Then L := sup{|| F{x)\\: || x\\ < 1} < oo

(32.19)

and \\F{x)\\<

L(l + \\x\\)

for

xeX.

(32.20)

Proof. Let An:={xeX:

\\F(x)\\
n G N.

If x G X, then we can find an n G N such that || F(a;)|| < n, so x G An and therefore oo

X = | J An.

(32.21)

n=l

Since F is B.l.s.c. and An := {x G X : F(x) C c/5(0,n)} , then A n are closed sets for n G N. From (32.21) and assumption that X is a Banach space, it follows that there exists an m G N such that intAm ^ 0. Hence, for some x0 G X and r > 0, B(x0,r)

C Am.

Take x G X such that || x|| < r, then x + x0 G A m and F(x) = F(x+a;o-2;o) C F(x+x0)+F{-x0)

C c/5(^,m)+F(-xo).

Thus for every x satisfying the condition || x|| < r, | | F ( x ) | | < | | c / 5 ( ^ , m ) | | + | | F ( - x 0 ) | | < m + a,

(32.22)

320

Functional Equations and Inequalities in Several Variables

where a = || F(—x0)\\. Let n > r"1 and || x\\ < 1. Since

then

||F(x)||<||F(£)|| + ... + ||f(i)||=n||F(i)||. Therefore from (32.22) for || f || < J < r, ||F(a;)||
+ \\x\\)L. •

Theorem 32.13 If a s.v.f. F: X —> 2 y is subadditive, u.s.c. at zero and F(6) = {9}, then F is continuous in X. Proof. Let U be a neighbourhood of zero in Y. Since F is u.s.c. at zero, there exists a balanced neighbourhood V of zero in X such that for h G V F(h) C F{9) + U = {6} + U = U,

Subadditive set-valued functions

321

and since V is balanced, then —h€.V and F(-h)

C F(0) + U = U.

Let us fix an arbitrary x € X. In view of the subadditivity of F, we have F(x + h)c F{x) + F(h) C F(x) + U for all h e V, which shows that F is u.s.c. at x. Similarly, F{x) C F ( z + h) + F(-h)

cF(x

+ h) + U

for all h € V, i.e. F is l.s.c. at x. Thus F is continuous at every point x E X, which completes the proof. •

Notes 32.1 The problem of existence of an additive selection for a given subadditive multifunction was also examined by Z. Gajda and R. Ger [59]. We shall state here the following result due to Gajda [60]. Proposition 32.1 Suppose that G is a right or left amenable semigroup and (X, <) is a boundedly complete linear tattice. Let f,g: G —> X be mappings for which the following conditions

f(x + y)>f(x) + f(y), g(x + y)
x,y£G, x,y<=G,

f(x) < g(x),

x G S,

hold true. Moreover, assume that the set {g(x)-f(x):xeG} is bounded from obove. Then there exists exacly one additive mapping A: G —> X such that f(x) < A(x) < g(x)

for all

x e G.

322

Functional Equations and Inequalities in Several Variables

32.2 For the case of subadditive s.v. functions with closed and convex values in Banach spaces, one can have (see [60]). Proposition 32.2 Let G be a right or left amenable semigroup and let X be a Banach space. If F: G —Y CCL(X) is a subadditive s.v.f. satisfying the condition sup{diam F(x): x G G} < oo, then there exists exacly one additive selection A: G —> X of F'.

Chapter 33 Superadditive set-valued functions and generalization of Banach-Steinhaus theorem Let X and Y be semigroups. superadditive iff F(x) + F(y)cF(x

A s.v.f.

+ y),

F: X

x,y e X.

—> 2Y is (33.1)

First of all we shall give some examples of superadditive s.v.functions. Example 33.1 Assume that X is a semigroup, and let f,g: X —>• K. satisfy the inquality f(x) < g(x) for i £ l , Let F(x):=[f(x),g(x)},

x G X.

Then F is superadditive iff f(x + y)
+ f(y),

x,yeX

g(x + y)>g(x)

+ g(y),x,yeX

(33.2)

and

323

(33.3)

324

Functional Equations and Inequalities in Several Variables

Example 33.2 Let f:X F: X - > 2 R be defined by

—> M. be a given function,

F(x):=[f(x),oo),

and let

xeX.

Then one can verify that F is superadditive iff f satisfies the condition (33.2). A s.v.f. F: [ 0, oo) -)• 2Y is called increasing iff F(x{) C F(x2) for all 0 < x\ < x2 < oo. Now we shall give two sufficient conditions for F to be superadditive (see [201]). Lemma 33.1 Let Y be a vector space and let F: [0,oo) —> 2Y be such that F(x) is a convex subset of Y for every x € [0, oo). //, moreover, the s.v.f. detined by : G(x) = -F(x) for x > 0 is increasing in (0, oo) and F(0) = {6}, then F is superadditive in [0,oo). Proof. Since F has convex values, then we have F(x + y) = ^-F(x x +y

+ y) = ^—F(x x +y

+ y) + -^—F{x x+y

+ y).

By the monotonicity of - F (x), we get x

\ V ^/ N Fix) F(y) y yyj x + y F(x + y) + — x +— y F(x + y)D x—^ x + y- y T-,/

Thus we get, for x, y G (0, oo) F(x + y)D x ^ - + y ^ = F(x) + F(y). x y Moreover, taking into account the condition F(0) = {0} we guess that the inclusion (33.1) holds for all x, y e [0, oo) and F is superadditive in [0,oo). • We say that s.v.f. F: [0, oo) ->• 2 y is concave iff for every x,y £ [0, oo) and every A G [0,1], F[Xx

+ (1 - X)y] C XF(x) + (1 - A)F(y),

(33.4)

Superadditive set-valued

functions...

325

Lemma 33.2 Let Y be a vector space and let F: [0, oo) —>• 2Y be a concave s.v.f. with convex values. If, moreover, F(0) — {9}, then F is superadditive. Proof. Take x = 0, y = b,l — a = f, where 0 < a < b, since F is concave, then we get F(a) = F[aO+{l-

a)b] C aF(0) + (1 - a) F(b) = °rF(b)

F(a) C

-bF(b),

F(a) c F(b) a b for every a, b, 0 < a < b, which means that - ^ is increasing in (0, oo). Therefore, Lemma 33.1 implies that F is superadditive in

[0,oo).

•

We shall prove a generalization of the well-known BanachSteinhaus theorem on the uniform boundedness of norms of linear continuous operators. In the sequel we shall assume that X and Y are topological vector spaces satisfying the T0 separation axiom. A set K C X is called a convex cone iff A x, x + y e K for all x,y € K and A > 0. Theorem 33.1 ([201]). Let K be a convex cone in X and let (Fn: n E I) be a family of superadditive and l.s.c. set-valued funtions Fn: K —> 2Y. Let, moreover, the set B:=

IxeK:

( J Fn(x) G 5 ( F ) }

(33.5)

be of the second category in K. Then (a) for every neighbourhood V of zero in Y there exists a neighbourhood U of zero in X such that Fn(U n K) c V for n E I.

326

Functional Equations and Inequalities in Several Variables

Proof. Let V be a neighbourhood of zero in Y. One can find a balanced neighbourhood D of zero in Y such that clD + dD C V. Consider the sets F + (clD)

:={x€K:Fn{x)CclD}.

Since F is l.s.c. , then the sets F+ (clD) are closed sets in K (see also proof of Theorem 32.10). Define

E := D F + {dD). Thus E is closed set in K. From the definition of the set B, for every x G B there exists an m G N with

The superadditivity of Fn yields mF n (x) C Fn(rc) + ... + Fn(x) C whence m F n U J

C F n (z)

or

F „ M

Fn(mx), C ^n(i)

C D for

all n E I. This shows that ^ G E. Consqently x € mE which proves the inclusion oo

B c |J m£. m=l

Taking into account that B is of the second cat gory set, thus int E ^ 0 (otherwise £? would had to be of the first category). Hence for some x0 G E there exists a neighbourhood U of zero in X such that (U + x0) D K C £\ Take any x G £/nif. Then y = x+x0 G £ and Fn(x)+Fn(x0) C F n (x + rr0) = Fn(y), which implies Fn{x) C F n (y) - Fn(x0) C cZD - clD = clD + clD C V.

Superadditive set-valued

functions...

327

Hence

Fn[UnK]= (J

Fn(x)cV

xeunK for every n € I. Hence (a) results.

•

The following three theorems are due to W. Smajdor [201]. Theorem 33.2 Let K be a convex cone in X. Assume that for s.v.functions Fn: K —>• 2 y , n E I, the condition (a) is satisfied. If, moreover, Fn(mx) C mFn(x),

m e N , x e K, n G I,

then the s.v.f. F defined by F(x) := [jneI bounded.

(33.6)

Fn(x), x G K, is

Proof. Let V and U be such that the condition (a) is stisfied (we may assume that U is a balanced neighbourhood of zero in X). Consider a bounded set A C K. Thus we may find an s G N such that

-A C s

UnK.

If x € A, then x = sy for some y G U D K. Hence and by (33.6) Fn(x) = Fn{sy) C sFn(y) C sV,

n G /,

i.e. \JneIFn(x) C sV, which means that the set F(A) is bounded. Since A is an arbitrary bounded set contained in K, this concludes the proof of our theorem. I Let us emphasize that since the s.v. function F(x) = (J n € N F n (x) is bounded, so the set (33.5) is equal to K, B = K. Therefore, immadiately from Theorem 33.1 and Theorem 33.2, we obtain Theorem 33.3 Let K be a convex cone in X and let (Fn: n E I) be a family of superadditive, l.s.c and Q+ -homogeneous s.v. funtions Fn: K —v 2Y. If, moreover, the set (33.5) is of the second category in K, then the s.v.f. F defined by F(x) :— \JneI Fn(x), x GK is bounded and B — K.

328

Functional Equations and Inequalities in Several Variables

Remark 33.1 Assume that K = X, and the family (Fn: n E I) satisfies all the hypotheses of Theorem 33.3. Then for every n E / , Fn is an ordinary single-valued function. Indeed, we have for every ne I, Fn(9) + Fn(9) C Fn(6). If x G Fn{9), then mx G Fn(8) for m G N, and since Fn(8) is bounded, x = 9. Thus Fn(9) — {9} for n E I. Now, we have for x G X, Fn(x) + Fn(-x) C Fn{6) = {9},nel. Take an a € Fn(x), then a + b = 9 for some b G Fn(—x). Suppose that there exists an element c / a, c G Fn(x), so c+b = 9, whence c = a, a contradiction. Moreover, for single valued mappings the inclusion (33.1) is equivalent to the equality Fn(x) + Fn(y) = Fn(x + y), and since Fn are l.s.c. functions for n G / , it means now that Fn are continous mappings, which implies that Fn are linear continuous operators. Thus in this case Theorem 33.3 is just the BanachSteinhaus theorem of the uniform boundedness. Let us recall that for a normal space Y and A G B(Y), \\A\\ :=sup{||a;||: x G .4}. Theorem 33.4 Let X and Y be two real normed spaces and let K be a convex cone in X. If K is of the second category in K and (Fn: n G /) is a family of superadditive, l.s.c. and Q+ -homogeneous in K s.v. funtions Fn: K —> 2Y such that Une/ Fn(%) G B(Y)for x G K, then there exists a constant 0 < L < oo such that sup||F„(x)|| < L | | x | | , ne I

xeK.

(33.7)

Superadditive set-valued

functions...

329

Proof. From Theorem 33.3 it follows that there is a ball r > 0 such that Fn(x)cB(9,r)

for

||x|| < 1 and

B(8,r),

xeK,neI,

whence \\Fn(x)\\
for

x G K, \\x\\ < 1 and

n G I.

Take any x G K, x ^ 9 and choose an s G Q with —p^

<

S <

2||a;||

-rr-rr.

(33.8)

||x||

Taking into account that sx G K and || sx\\ < 1, we get for n G / , s||F n (x)|| = || F n (sa;)|| < r. Hence in view of (33.8), we obtain \\Fn{x)\\
< 2r\\x\\,

nel.

If x = 8, then from the equality Fn(9) = {0}, the condition (33.7) holds for x = 9, too. Putting L := 2r, we get the condition (33.7).

Notes 33.1 The famous Banach-Steinhaus theorem of the uniform boundedness has been extended by N. X. Tan [210], [209], to the class of upper semi-continuous convex s.v.functions. 33.2 The problem of the existence of additive selections for superadditive s.v.functions was studied by A. Smajdor [196] and for subadditive s.v.functions by P. Kranz [121]. For further results in the subject the reader is referred to Th. M. Rassias [168], [167].

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Chapter 34 Hahn-Banach type theorem and applications Let Y be a subgroup of a group X and let E be a linear space. Suppose that F: X —> 2Y is a subadditive set-valued function. The question arises whether every additive selection of the restriction of F to Y can be extended to an additive selection of F defined on X. In this section we shall present some results to this effect which are due to Z. Gajda, A. Smajdor and W. Smajdor [58].

Hahn-Banach theorem The following result is well-known as the Hahn-Banach Theorem on the extension of linear functionals. Theorem 34.1 Let Y be a subspace of a real linear space X. Suppose that p: X —>• E has the properties (i) p (x + y) 0 and x € X. 331

332

Functional Equations and Inequalities in Several Variables

If, moreover, f:X condition

—> R is a linear functional satisfying the f(x) K with g{x)
(x)

for all x e X.

(34.2)

We are going to rewrite this theorem in terms of s.v.f. with values in CC(R). Note that the elements of CC(M) are just the non-empty compact intervals in R. First we shall show that (i) and (ii) imply the inqualities -V (~x) < f(x)
(x)

-P (~x) < g(x)

for xGY,

(34.3)

for x e X.

(34.4)

Indeed, from (ii) for s = 0, we get p(9) — 0, and from (i) 0 = p (9) = p (x — x) < p (x) + p (—x) we obtain -p(-x) < p{x), xeX. (34.5) Since / is linear, then f(6) = 0 and f(—x) = —f{x) and consequently by (34.1) for x 6 Y,

-f(x)=f(-x)
for x € Y,

This together with (34.1) gives —p (—x) < f(x) < p (x)

for x E Y.

In a similar way one may verify the inequalify (34.4). Taking into account (34.5) we may define a s.v.f. F: X —• CC(R) by F(x):=[-p(-x),p(x)], xeX. (34.6)

Hahn-Banach type theorem and applications

333

Cleary F is subadditive, positively homogeneous and F(—x) =

-F(x) for xeX. Conversely, every s.v.f F: X —v CC(R) which is subadditive, positively homogeneous and odd has to be of the form (34.6) with a functional p: X -> R satisfying the conditions (i), (ii). Thus theorem 34.1 may be considered as a result on extending of additive selections of a s.v.f. F: X —>• CC(R). Namely Theorem 34.2 Let Y be a linear subspace of a real linear space X. Let F: X —>• CC(R) be a subadditive, positievly homogeneous and odd s.v.f. . If f: Y —>• R is a linear functional such that f{x) e F(x)

for all

xeY,

then f can be extended to a linear functional g: X —>• R such that g(x) € F(x) for all x e X.

Generalizations of the Hahn-Banach theorem Let X be an abelian group and let ^ be a family of non-empty subsets of a linear space E over Q. We say that g: X —>• E is an additive selection of a s.v.f. F: X —> $ iff g(x + y)= g{x) + g(y) for x,y e X, g(x) e F(x)

for x € X.

We say that a family 5 has the binary intersection property provided that every subfamily of $, for which any two members intersect, has a non-empty intersection. The first generalization of the Hahn-Banach theorem reads as follows (see [58]). Theorem 34.3 Let Y be a subgroup of an abelian group (X, +) and let E be a linear space over Q. Moreover, let $ be a family of

334

Functional Equations and Inequalities in Several Variables

non-empty subsets of E with the binary intersection property and satisfying the following conditions: Ae$, Ae$,

aE E

imply

A + a G$ ;

n G Z 0 = Z \ {0} imply

-A G £ .

(34.7) (34.8)

ft

Suppose that F: X —> $ is a subadditive s.v.f. with F(nx) C nF(x)

for all

x €X

and

n G Z0.

(34.9)

TTaen, if f: Y -^ E is an additive selection of the restriction of F to Y (denoted by F\Y ) , then f can be extended to an additive selection of F. Proof. Let A be the family of all additive maps (j): X^ —> E such that Y C X$ C X , X^ is a subgroup of X, (j)(x) G F(x), x e X$ and <j){x) = f(x) for x G Y. Define the partial order in the family A by <j><%l)

iff

X^cX^p

and

cf) = ^\Xij>.

Then (A, -<) is partially ordered space. For any chain £ C A define the map $^ such that

x*£--=

U ( ^

:

££} •

Then obviously $^ is an upper bound of £ in A. Moreover, ^£\x f° r every G £. Thus by the Kuratowski-Zorn Lemma, there exists a maximal element g G A. We shall show that Xg — X. For the contrary, assume that z0 G X \ Xg. Define W := {x + nzQ : x G Xg, n G Z } . Cleary, W is a subgroup of X and X 5 C W, Xg ^ W. Put U : = {nG Z 0 : nz 0 G X fl } . We shall distinguish two cases.

Hahn-Banach type theorem and applications

335

Case 1: [ 7 / 0 . Let m,n G U, then ran G U and ng(mz0) = g(mnz0) = mg(nzQ), whence g(mz0) _ m

g(nz0) n

Put v := ^ for some m eU, then t; G E and w does not depend on the choice of m G U. Let us define: h: W —>• JE7 by /i(rc + n^ 0 ) := p(aj)+nv

xGXfl,nGZ.

(34.10)

Suppose that x + n z 0 = y + mz0 for x, y G X s , and m, n G Z. Then x — y = (m — n)zo G X s and conseqently either m = n o r m — n E U. IfTO= n, then a; = y and ^(rc) + nu = #(?/) + mv. In the second case, g((m - n) z0) v = — —-, m —n which yields g(x) - g(y) = g{x - y) = g((m - n) z0) =

(TO

- n)v,

i.e. h(x + n ZQ) = g(x) + nv = g(y) + mv = h(y + mz0) and proves the correctness of the definition of h. Moreover, h is additive and h\xg = gTake any x G Xg, n G Z,TOG U, then by (34.9) h(x + nz0) = ^(x) + nv = g(x) -\

ng(mz0) Til

= — # [mix + nz0)] C — F(m(x + nz0)) m m C F(x + nz0). This means that h is an additive selection of F\wSince X 5 ^ W, this is impossible because of the maximality of g in A.

336

Functional Equations and Inequalities in Several Variables

Case 2: [7 = 0. So for every n G Z 0 , nz0 € X \ X 9 . Let 1,1/6 X 5 , and m, n € Zo, then since F is subadditive, by (34.9) we obtain mg(x) — ng(y) = g(mx — ny) G F(mx — ny) = F(mx + nmzo — nmz0 — ny) C F(ra(n: + n^o)) + F(-n(y + mz0)) C mF(x + nzo) — nF(y + mzo). Hence 6> G m [F(x + nzQ) - g(x)] - n [F(y + mz0) - g{y)\ i.e. 9 G - [F{x + nz 0 ) - g(x)] ft

[F(y + mz0) - g(y)} . Ilb

Therefore for every x,y G Xg and m,n € Zo, the intersection - [F(x + n^ 0 ) - g(x)] P i —[F(y + mz0)-g(y)] n ' ' m is non-empty. By the binary intersection property of the family F, we get (){-[F(x

+ nz0)-g(x)]:

x € Xg , n G Z 0 } ^ 0.

Suppose that w is an element of this intersection. Then g(x) + nv G F(x + nz0) for all x G Xg and n G Z. Similary as in the Case 1 we define h: W —> E1 by the formula (34.10). Then ft is an additive section of F\w which contradicts the maximality of g in A. This concludes the proof. • Consider now the family F consisting with Q-convex sets, i.e. every A e F satysfies the condition aA + (l-a)Ac

A

for all

a€Qn[0,l].

Then we get another version of Theorem 34.3

Hahn-Banach type theorem and applications

337

Theorem 34.4 Let Y be a subgroup of an abelian group X and E be a linear space over Q. Moreover, let $ be a family of non-empty Q-convex subsets of E with the binary intersection property and satisfying the conditions (34-7) and (34.8). If a s.v.f. F: X —>• # is subadditive and F{-x)

C -F(x)

for all

x G X,

(34.11)

then every additive selection of F\y can be extended to an additive selection of F. Proof. In view of Theorem 34.3, it is enough to show that the condition (34.9) holds true. Let x G X, n G Z, n > 0 be given. Then by the subadditivity of F and since F(x) is convex, we get

F(nx) c F(x) + ... + F(x) C nF(x), If n < 0 , then by (34.11)

F(nx) C -F(-nx)

C - ( - n ) F(x) = nF{x),

which proves the condition (34.9). Thus Theorem 34.3 yields our assertion. • Corollary 34.1 Assume that all the hypotheses of either Theorem 34-3 or Theorem 34-4 concerning the family $ and s.v.f. F: X —> $ are satisfied. If, moreover, 8 G F(6), then F has an additive selection. Proof. Let Y := {6} and f(9) := 6. Our assertion follows now immediately from either Theorem 34.3 or Theorem 34.4. • Remark 34.1 The family CC(R) has the binary intersection property (for details see [139]). Moreover, this family satisfies all the conditions in both Theorem 34-3 and Theorem 34-4- Thus every linear selection f: Y —> K of F\y has an extension to an additive

338

Functional Equations and Inequalities in Several Variables

selection g: X —>• K of F. We shall show that g is in fact a linear selection. Indeed, for every x € X, we define gx: R —> R by the formula 9x(s) '•= g(sx) for all seR. Then gx is additive and for s > 0, 9x(s) < supF(sx)

= s supF(x),

which implies that gx is upper bounded on a non-empty open interval, and in view of Theorem 28.5 it is continuous and consequently also homogeneous. Thus g is a linear selection of F. This shows us that Theorem 34.2 is a simple consequence of both Theorem 34-3 and Theorem 34-4

Applications to stability of functional equations In the next theorem we shall prove a result on extending additive maps which approximate a function which Cauchy's differences belong to a given linear space. For the results presented, see [58]. Theorem 34.5 Let Y be a subgroup of an abelian group X and L be a subspace of a linear space E over Q. Let <j): X —)• E satisfy the condition (j)(x + y)-

(j)(x) -
for all

x, y G X,

(34.12)

and let f: Y —> E be an additive map such that f(x) - (f)(x) e i

for

xeY

.

Then f can be extended to an additive map g: X —» E such that g(x) — (j)(x) e L

for

x e X .

Hahn-Banach type theorem and applications

339

Proof. Define: $:={x

+ L: xeE}

.

Then # has the binary intersection property: in fact, every subfamily # which any two members intersect, consists of a single set only. Moreover, every element of $ is Q-convex and $ satisfies (34.7) and (34.8). Put F(x) := (j){x) + L, xeX. Then clearly F: X -> $ is a subadditive s.v.f. and / : Y —>• E is an additive selection of F\yWe shall verify that F satisfies (34.11). Seting x — y — 9 into (34.12), we get (#) G L. Taking again y = —x, we obtain

4>{e) - 0(x) - 4>{-x) e L , whence (-x) + (x) e L

or

{-x) G -<j>(x) + L.

Consequently F(-x)

= {-x) +Lc

-(x) +L =

-F(x),

i.e. F(—x) C —F(x) for all x G X. Theorem 34.4 completes the proof. I

Corollary 34.2 Let X be an abelian group and let L be a linear subspace of a linear space E over Q. If : X —> E fulfils the condition (34-12), then there exists an additive function g: X —>• E such that g(x) — (f>(x) e L for all x e X. Proof. Put Y := {6} and apply Theorem 34.5.

•

340

Functional Equations and Inequalities in Several Variables

Remark 34.2 Corollary 34-2 has been proved in a different way by K. Baron [12]. Consider a normed space(£T, || • ||). We say that E has the binary intersection property iff the family of all closed balls in E has the binary intersection property. Theorem 34.6 Let Y be subgroup of an abelian group X and let (E, || • ||) be a normed space with the binary intersection property. Let r: X —> E+ be an even, subadditive function and let (f>: X —> E be an odd map satisfying the condition || <j) (x + y) - (x) - 0 (j/)|| < r(x) + r(y) - r(x + y)

(34.13)

for all x,y E X. If moreover, f: Y —¥ E is an additive function satisfying the inequality || f(x) - (x)\\ < r(x) for xeY, (34.14) then f can be extended to an additive function g: X —» E such that \\g{x)-(f)(x)\\

< r(x)

for

x € X.

Proof. Take any x E E and s e [0, oo). Let B(x,s) closed ball in E with centre x and radius s. Define

(34.15) denote the

$ := {B{x, s): x G E , s e [0, oo) }. By our assumptions, $ is a family of convex sets with the binary intersection property. Clearly, it satisfies the conditions (34.7) and (34.8). We have also the equality B(9,Sl)

+ B{6,s2) = B(9 ,Sl + s2)

(34.16)

for all s i , s 2 > 0. Since 4> satisfies (34.13), then by (34.15) we get the inclusion (j)(x + y)-

(x) - (y) G B[9, r{x) + r(y) - r(x + y)] ,

Hahn-Banach type theorem and applications

341

or <j>{x + y)-(f>(x) - <j>{y) + B[0, r(x + y)} CB[6, r(x) + r(y) - r(x + y)] + B[6, r(x + y)} = B[9, r(x) + r(y)) = B[6, r{x)} + B[9, r(y)}. This yields (x + y) + B(9,r(x

+ y)) C (x) + B(0,r(x))

+ <j>(y) + B(0,r{y))

which can be written as B[(t>(x + y),r(x + y)} C B[{x),r{x)] + B[(y),r{y)]. (34.17) Let us define the s.v.f. F: X —> 5 by F(x) := B[(x),r(x)] for

x e X.

Thus, on account of (34.17), F is subadditive. Moreover, by the assumptions on <> / and r,we have F(-x)

= B[(-x),r(-x)] = B[-(x),r(x)] = -B[(x),r(x)] = -F{x),

for all x £ X. In view of (34.14) we guess that / is a selection of F\yConsequently, by Theorem (34.4), / can be extended to an additive selection g of F which implies that g satisfies the condition (34.15). The proof is completed. • Corollary 34.3 Let the assumptions of Theorem 34-6 on X,E,r and (j) be satisfied. Then there exists an additive function g: X —>• E such that \\g(x) - <j>{x)|| < r(x) for x € X. Proof. Put Y := {9} and f(0) := 6 and apply Theorem 34.6.

•

Now we shall drop the assumption that 0 is an odd mapping. Namely, we have

.

342

Functional Equations and Inequalities in Several Variables

Theorem 34.7 Let the assumptions of Theorem 34-6 on X,E,r and (j> be satisfied except that (f> is an odd map. Then there exists an additive function g: X —> E such that \\g{x)-

(j){x)\\ <2r(x)

for

x e X.

(34.18)

Proof. Denote

Then (f>i is even and 2 is odd and = <j>i + (f)2. If follows f (34.13) that ||0(0)||
< -(r(x) + r(-x)

- r(6)) = r(x) -

Li

-r(6). Z

Thus we get for x 6 X, since || (6)\\ < r(0), ||<M*)|| < r(x) + \{\$6)\\ ~ r(6)] < r(x).

(34.19)

We shall show that 2 satisfies the condition (34.13). In fact, we have for x,y G X, U2{x + y) - cj>2{x) - <j>2(y)\\ =

= g l l ^ + y) ~ (-x -y)~

{-x) - (y) + (-y)\\

< \[\\4>{x + y)~ m - m \ \ + \\(-y)\$ < ~[r(x) + r(y) - r(x + y) + r(-x) + r(-y) - r{-x - y)] = r(x) + r(y) - r(x + y).

Hahn-Banach type theorem and applications

343

Consequently, from Corollary 34.3, there exists an additive function g: X -> E such that || g(x) - (j)2(x)\\ < r(x)

for

x € X.

Hence, in view of (34.19), for x G X \\g(x) - cf>(x)\\ = \\g(x) - M*) ~ Mx)\\ <\\g{x)-
which was to be shown.

•

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Chapter 35 Subquadratic set-valued functions It follows from Corollary 20.1, that if X is an abelian group and Y is a Banach space and / : X —> Y satisfies the inequality ||/(x + y) + f(x -y)-

2f(x) - 2f(y)\\ < e

(35.1)

for all x, y G X, where e > 0 is fixed, then there exists exactly one function g: X —> Y satisfying the quadratic functional equation g{x + y) + g{x - y) = 2g(x) + 2g{y)

(35.2)

for all x, y G X and such that \\f(x)-g(x)\\<2e

(35.3)

for all x £ X. The inequality (35.1) one may rewrite in the following form fix + y) + f{x -y)-

2 fix) - 2f(y) G B (6, e),

where B (9, e) is the closed ball with centre at zero and radius e . Therefore, fix + y) + B(6,2s) + fix -y) + B{0,2e) C 2[f(x) + B{6,2e)) + 2[fiy) + Bi9,2e)}. 345

346

Functional Equations and Inequalities in Several Variables

Denoting F(x) := f(x) + B (9, 2e), we obtain inclusion F{x + y) + F{x -y)c

2F(x) + 2F{y)

(35.4)

and the condition g(x) e F(x)

(35.5)

for all x,y E X. A s.v.f. F: X —> 2Y, where Y is a group is said to be subquadratic iff it satisfies the inclusion (35.4) for all x,y G X. The condition (35.5) says that the subquadratic s.v.f. F has a quadratic selection. For the results presented, see [200] . We shall start with the following Lemma 35.1 Let X be an abelian group with division by two let Y be an abelian semigroup. If F is a subquadratic s. v. f. such that F(2x) = 4F(x)

for x G X,

(35.6)

then F is quadratic, i. e. F(x + y) + F{x -y) = 2F(x) + 2F{y)

(35.7)

for all x,y G X. Proof. Put x + y = u,x — y = t, then x = ^(u + t),y = ^(u — t) and by (35.4) and (35.6) we get 2F{u) + 2F(t) C 2 [2F ( ^ ) + 2F (*=!•)] = AF {*f) + AF ( ^ )

= F(u +

t)+F{u-t),

i.e. 2F(u) + 2F(t) C F(u + t)+ F{u - t). This inclusion together with (35.4) yield the equality (35.7) and shows that F is quadratic. • This lemma implies

Subquadratic set-valued functions

347

Theorem 35.1 Let X be an abelian group with division by two and let Y be a topological vector space. Let a s.v.f. F: X —> CC(Y) be subquadratic. Then there exists a quadratic s.v.f. G: X —> CC(Y) such that G{x) C F(x) for all x G X. (35.8) Proof. From (35.4) for x = y = 9, we get 2F(9) c 4F(0), whence {6} + 2F{9) C2F{9) + 2F{9). Therefore from Lemma 20.1 we get {9} c 2F{9), i.e. 9 G F(9). Put x = y in (35.4), then F(2x) + F(9) c 4F(y). Hence F(2x) = F(2x) + {0} C F(2x) + F(9) C 4F(z) and consequently, \F{2x) C F(rr). By induction, we get for k G N i

lIF(2^1x) C l F ( 2 H

Define oo

1

n=0

Since the sequence {4~"F(2n:r)}^L0 is decreasing, G(x) ^ 0 and G{x) G CC(Y) terxeX. Clearly, G(a;) C F(x) for a; G X . Moreover, we have OO

00

G(2x) = p ] 4- n F(2 n+1 a;) = 4 f]

4-n~1F(2n+1x)

oo

= 4f)4-nF(2nx)

= 4G(x),

n=l

whence G(2x) = AG(x),

xeX.

348

Functional Equations and Inequalities in Several Variables

We shall show that the s.v.f. G is subquadratic. Indeed, by Lemma 29.2, we obtain oo

oo

n

n

+ y)) + f | 4TnF(2n{x - y))

G{x + y) + G(x - y) = f | 4- F(2 (x n=0

n=0

oo

= p | [4-nF(2n(x + y)) +

4-nF(2n(x-y))]

n=0 oo

C p i 4" n [2F(2nx) + 2F(2ny)] n=0 oo

oo

= 2 P) 4-"F(2V) + 2 p | 4-nF{2ny) n=0

= 2G(x) + 2G{y).

n=0

It remains to apply Lemma 35.1.

•

Corollary 35.1 Let X be an abelian group with division by two and let f,g: X —>• R satisfy the inequalities f{x + y) + f{x -y)>

2f(x) + 2/(2/),

gix + y) + gix - y) < 2gix) + 2g(y),

(35.9) (35.10)

forx,y G X. If moreover, fix) < gix) forx G X, then there exists a quadratic functional h: X —y R such that fix) < h(x) < gix)

for x G X.

(35.11)

Proof. Define the s.v.f. F: X ->• CC(R) by F(x) := [ / ( a r ) , ^ ) ] ,

i € l

Then F is subquadratic, thus by Theorem 35.1 there exists a quadratic s.v.f. G: X -> CC(R) satistying the inclusion G(:r) C F(x) for x G X. Put hix) := maxG(a;),

i £ l

Subquadratic set-valued functions

349

Clearly h is quadratic and satisfies the inequalities (35.11).

•

In the sequel we shall consider the problem of existence of quadratic selections of subquadratic set-valued functions. To this end let us prove the following lemma concerning the existence of minimal quadratic s.v.f. contained in a given subquadratic s.v.f. . Lemma 35.2 Let X be an abelian group with division by two and let Y be a topological vector space. If F: X —>• CC(Y) is a subquadratic s.v.f. ,then there exists a minimal quadratic s.v.f. F0: X —> CC(Y) contained in F. Proof. Consider the family a := {G: X ->• CC(Y): G is quadratic, G C F}. In view of Theorem 35.1 , a is non-empty. Let a0 C a be a chain. Define G0: X -> CC(Y) by G0 := p | G(X), xeX.

(35.12)

Gea0

One can prove that f | G(x + y)+ f | G{x-y)= Gea0

f\[G{x

Geao

+ y) +

G(x-y)].

Geao

For details see [200], Lemma 1.5. Hence G0(x + y) + G0(x - y) = f | [G(x + y) + G(x - y)} Gea0

= f | [2G(x) + 2G(y)} = 2 f | G(x) + 2 f | G(y) Geao

Geao

Geao

= 2Go(x) + 2G0(y). This shows that G0 is quadratic. From the definition (35.12) if follows that GQ C F, i.e. G0 G a and G0 C G for every

350

Functional Equations and Inequalities in Several Variables

G G «o- Therefore we may apply Kuratowski-Zorn Lemma to get the existence of a minimal element F0 G a, which completes the proof. I The above lemma leads to the following. Theorem 35.2 Let X be an ahelian group with division by two and let Y be a topological vector space. Let F: X —>• CC(Y) be a subquadratic s.v.f. . Then there exists a quadratic selection of F. Remark 35.1 Note that a subquadratic s.v.f. may have more than one quadratic selections. For example, let F(x) = [—x2,x2] for x G M., then every function fix) = ax2, where \a\ < l,x G K. is a quadratic selection of F. The next theorem gives the sufficient condition for F to have exactly one quadratic selection. Theorem 35.3 Let X be an ahelian group and let Y be a Banach space. Assume that a s.v.f. F: G —>• CL(Y) is subadditive and has convex values. If, moreover, a := sup{diamF(z): x G X} < oo, where diamF(x) = sup{||a — b\\: a,b G F(x)}, then there exists exactly one quadratic selection f': X -^Y Proof. Define the s.v.f. Fn: X -> CL(Y),n£ Fn(x) := 4-nF(2nx),

of F.

N0 by xeX.

As in the proof of Theorem 35.1 one can show the inclusion 4~1F{2x) C F(x). Hance we get Fn+l{x) = 4-(" +1 )F(2 n+1 x) C 4 • 4^n+1^F(2nx)

= Fn{x),

Subquadratic set-valued functions

351

i.e. the sequence Fn is decreasing. Now, from the assumption on diamF(:r), we get diamF n (a;) = sup{||a — 6||: a,b G Fn(x)} = sup{||4-"a - 4- n 6||: a, b e F{2nx)} = 4~ n sup{||a -b\\:a,be

F{2nx)} < A~na.

Therefore, lim^oo diamF n (x) = 0 for every x G X. By the Cantor intersection theorem, the set oo

oo

f l Fn(x) = f |

4~nF(2nx)

n=0

n—0

contains exactly one point, say f(x) for every x € X. Clearly, f(x) G F(x) for each x G X. Now we shall show that / : X ->• Y is quadratic. Since F is subquadratic, we obtain oo

oo

f{x + y) + fix - y) E f| 4~"F(2"(x + y)) + f| 4~nF(2n(i - y)) n=0

n=0

CO

OO

C fl 4~" [F(2"(z + y)) + F{2n{x - y))] C f| 4"" • 2 [F(2"x) + F(2nj,)]. n=0

n=0

Therefore, we get oo

f(x

+ y)

4~" [F(2nx) + F(2 n y)].

+ f(x-y)ef)2-

(35.13)

71 = 0

Moreover, we have for all n G No, 2/(x) + 2f(y) e 2 • A-nF(2nx)

+ 2 • 4TnFi2ny),

so that oo

2/(x) + 2/(i/) G f l 2 • 4" n [F(2nx) + F{2ny)\. n=0

(35.14)

352

Functional Equations and Inequalities in Several Variables

Also, oo

diam P | 2 • 4" n [F(2nx) + F(2ny)} 71=0

< diam [2 • 4-nF(2nx) < diam 2 • 4rnF(2nx)

+ 2 • 4TnF{2ny)] + diam 2 • 4rnF(2ny)

< 2 • 4-"diamF(2 n a;) + 2 • 4-"diamF(2 n y) < 4" n + 1 a. Consequently oo

diam f] 2 • 4~n [F{2nx) + F{2ny)] = 0.

(35.15)

n=0

By (35.13), (35.14) and (35.15), f(x + y) + f{x-y)

= 2f(x) + 2f(y)

for all x,y e X.

Assume that / is not unique. Thus there would be two quadratic selections, say / i , /2 of F and such that for some x0 6 X, ll/i(zo)-/2(*o)||=£>0. Clearly, since / i , / 2 are quadratic, fk(2nx)

= 4nfk(x)

for

xeX,

n e N0,

A; = 1,2.

Obviously, HA (x) - f2(x)\\

< diamF(x) < a,

xeX.

On the other hand, a > H/i(2na;o) - / 2 (2 n x 0 )|| = ||4B/i(ar0) - 4 B /i(s 0 )|| = 4B& which implies that ^ = 0 and proves the uniqueness part of our theorem. •

Subquadratic set-valued functions

353

Corollary 35.2 Let X be an abelian group and let Y be a Banach space. If U is a bounded subset of Y and f: X —>• Y is subject to the condition f(x + y) + f(x-y)-2f(x)-2f(y)e2U

for

x,y £ X, (35.16)

then there exists a quadratic function g: X —> Y such that g(x) - f(x) e cl conv U.

(35.17)

Proof. Denote A := cl conv U and define F(x):=f(x)+A

for

x e X.

Clearly for every x £ X, F(x) is a non-empty, convex, closed and bounded subset of Y. We shall show that F is also subquadratic. In fact, for x, y € X, we have by (35.16) F(x + y) + F(x -y) C2f{x)

= f(x + y) + A + f{x-y)

+ A

+ 2f(y) + 2U + 2A

C 2f{x) + 2/(y) + 2A + 2A = 2F(x) + 2F(y). Moreover, sup{diamF(x): x e X} = = sup { sup{||a - 6||: a, 6 G F{x)} : x G l ) = sup{||a — 6||: a, b 6 A} = diamA < oo. Consequently, in view of Theorem 35.3, there exists exactly one quadratic function g: X —> Y such that

g(x) ef(x) + A which is equivalent to (35.17).

•

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Chapter 36 K-convex and K-concave set-valued functions Let X and Y convex subset of Recall that a sK C K for alls

be real vector spaces and let D be a non-empty X. set K c Y is called a cone iff K + K c K and e (0,oo).

Definition 36.1 ([18]) A s.v.f. F: D -> 2 y is said to be Kconvex iff sF{x) + (1 - s)F(y) C F(sx + (1 - s)y) + K

(36.1)

for all s e [0,1] and all x,y e D. Similarly, a s.v.f. F: D ->• 2 y is said to be K-concave iff F(sx + (1 - s)j/) C sF(x) + (1 - a)F(y) + K

(36.2)

for all s € [0,1] and all x,y € D. A s.v.f. F : D —>• 2 Y is called convex (concave) iff it satisfies the condition (36.1) (36.2) with K = {6}. Convex s.v. functions play an important role in the theory of convex optimization problems and convex analysis as well as in economic theory (see e.g. [182]). First of all we shall present some examples of K-convex and K-concave s.v. functions (see [142]). 355

356

Functional Equations and Inequalities in Several Variables

Example 36.1 Let D be a non-empty convex subset of X and let K = [0,oo). The s.v. functions F: D —y 2Y defined by F(x) := [f(x), g(x)]> x e D where f: D -» R is a convex (concave) function and g: D —» R is an arbitrary function satisfying the condition f(x) < d(x) for x E D, are K-convex (K-concave), respectively. Example 36.2 Let f:D —y R be a convex and g: D —>• R a concave function. If f < g on D, then the s.v.f. F \ D —¥ 2R defined by F(x) := [f(x),g(x)] for x G D, is convex. Moreover, if 9 < f on D, then F defined by F(x) := [g(x), f(x)], x G D, is concave. Example 36.3 Suppose that fs: X —>• Y, s G T, are linear mappings. Then the s.v.f. F: X —> 2Y defined by F(x) := {fs(x): s G T } , x G X, is concave. If fs,s G T are K-convex (K-concave) functions, then the defined above s.v.f. F is K-convex (K-concave), respectively. Example 36.4 The s.v.f. F: R ->• 2Y defined by F(x) := xA, x G R, where A is a subset of Y, is concave. In general, such s.v. functions are not convex. However, if A is convex, then F: [0,oo) -^ 2V or F: (-00, 0] -> T> defined by F(x) := xA for x G [0,oo) or x € (—oo,0], respectively, is convex. Now we shall present generalizations of some well known properties of real-valued convex (concave) functions (see [127]). Theorem 36.1 Let a s.v.f. F: D ->• 2Y be K-convex (K-concave). Then siFfri) + . . . + snF(xn)

C F(slXl

+ ... + snxn) + K,

(36.3)

+ ... + snxn) C siF{xi) + ... + snF(xn) + K,

(36.4)

or Ffaxt

for all n G N, Xi,...,xn G D and sx,...,sn Si + ... + sn = l, respectively.

G [0,1] such that

K-convex and K-concave set-valued functions

357

Proof. The proof will be given for the case where F is K-convex (if F is K-concave, the proof is similar). Since F is K-convex, the inclusion holds for n — 1. Assume now that (36.3) holds for some n € N and take points x\,..., xn+\ G D and numbers s\,..., sn+i G [0,1] with si + . . . + s n +i = 1. Then we get SiF(xi) + ... + sn+iF(xn+l) = = (si + . . . + s n )[(s x + . . . + 5„) _1 5iF(a;i) + . . . + + (si + . . . + {Xn)} + Sn+1F(xn+1) C (Si + . . . + Sn) F({Si + ... + Sn^SiXi + (si + . . . + s„) _ 1 s n x n J + K +

+ ...+

sn+iF(xn+1)

C F(s1x1 + . . . + s n+ irr„ + i) + K, whence, by induction, the inclusion holds true for all n G N.

•

Theorem 36.2 A s.v.f. F: D —> 2Y is K-convex (K-concave) iff for each pair of points X\,x• 2Y defined by G{s) := F(sxx + (1 - s)x2), s E [0,1], is K-convex (K-concave). Proof. Suppose that F is K-convex andfixarbitrarily two points xi,x2 G D. Thus for all s G [0,1] and Si, s 2 G [0,1], sG(Sl) + (1 - s)G(s2) = = sF{s1x1 + (1 - si)x2) + (1 - s)F(s2xi + (1 - s2)x2) C F[s(sixi + (1 - si)x2) + (1 - s)(s2xi + (1 - s2)x2)] + K = F[(ss! + (1 - s)s2)x1 + (1 - ssi - (1 - ssi - (1 - s)s2)x2] + K = G[sSl + (1 - s)s2] + K, which shows that the s.v.f. G is K-convex.

358

Functional Equations and Inequalities in Several Variables

Conversely, fix arbitrarily xi, x2 G D and suppose that the s.v.f. G is K-convex. Thus for every s e [0,1], sF(Xl) + (1 - s)F(x2) = sG(l) + (1 - s)G(O) C G(s) + K = F(sXl + (1 - s)x2) + K, which means that F is K-convex. The proof for the case F is K-concave is quite analogous. •

Continuity properties for K-midconvex set-valued functions A s.v.f. F: D —y 2Y is said to be K-midconvex iff 1

F(Xl) + F(x2)

CF Xl + X2 +K

\\

l)

(36 5)

'

for xi,x2 G D. If F satisfies the condition (36.5) with K = {9}, then we say that F is midconvex (Jensen convex). It is well known that convex functions need not be continuous. Nevertheless, under relatively weak additional conditions such functions are continuous. In this section, we shall present some strong results of this kind (see [142]). In the sequel, we shall assume that X, Y are topological vector spaces, D is a non-empty open subset of X and K is a cone with zero. By W we shall denote an arbitrarily fixed base of balanced neighbourhoods of zero in Y. Following K. Nikodem [142], we shall prove the following useful lemmas. Lemma 36.1 Let a s.v.f. F: D —> 2Y be K-midconvex. sF{x) + (1 - s)F(y) C F(sx + (1 - s)y) + K

Then (36.6)

for all x,y € D and all diadic s 6 [0,1], i.e. s = n2~m with n,m e N, n< 2m.

K-convex and K-concave set-valued functions

359

Proof. By (36.5) we have 1_

C

1

F(Xl) + F(x2) + F(x3) + F(x 4 )

F

xr + x2\

+F(x3

+ xA

+ 5*

1 C i ? (2" 2 (a; 1 + x2 + x3 + xA)) + -K + -K = F(2~2(xi + x2 + x3 + x4) + K. By induction, we get F{xi) + ... + F(a:2m)] C F[2~m{xx + ... + x2m)} + K, for all x\,..., x2m £ D. Putting xi = . . . = xn = x, x n + i = .. %2™ — y, we get hence sF(x) + (1 - s)F(y) C 2~m [F(ari) + ... +

F ^ )

C F[2- m (a;i + . . . + x2m)\ + K = F[sx + (1 - s)y] + K, which was to be shown.

•

In the space 2Y we define the Hausdorff topology. In this topology the sets N,u(A) := { B e2Y:

AcB

+U

and

B cA + U

}•

where U £ W, form a base of neighbourhoods of a set A £ 2Y. We say that a s.v.f. F: X —> 2Y is K-lower semicontinuous (abbreviated to K-l.s.c.) at a point x0 £ X iff for every neighbourhood V of zero in Y there exists a neighbourhood U of zero in X such that F(x0) C F{x) + V + K for all x £

XQ + U.

(36.7)

360

Functional Equations and Inequalities in Several Variables

We say that a s.v.f. F: X —> 2Y is K-upper semicontinuous (K-u.s.c.) at a point xo E X iff for every neighbourhood V of zero in Y there exists a neighbourhood U of zero in X such that F(a;) C F(x0) + V + K

(36.8)

for all x € XQ + U. We say that F is K-continuous at x0 iff it is both K-l.s.c. and K-u.s.c. at xo. F is K-continuous on X iff it is K-continuous at every point x0 E X. Note that in the case where K = {6}, the K-continuity of F means its continuity with respect to the topology on 2 y . Having done this, we now are ready to prove the following Lemma 36.2 If a s.v.f. F: D -> B(Y) is K-midconvex and Kl.s.c. at a point of D, then it is K-l.s.c. at every point of D. Proof. Suppose that F is K-l.s.c. at XQ E D. Take a point x E X, x / xQ. Thus, in view of the fact that D is open, it follows that there exist a point x\ E D and a diadic number s E (0,1) such that x = sx\ + (1 — s)x0. Let G E W and choose a. V E W with V + V C G. Since F is K-l.s.c. at x0, there exists a neighbourhood U of zero in X such that F(x0) C F(x0 + u) + V + K

for u E U.

(36.9)

Take y0 E F(x0) and yi E F(xi) and let y := syi + (1 — s)y0- Then by (36.9) and Lemma 36.1, we get for u E U y E sFfa) cF[sxl

+ (1 - s)[F(x0 + u) + V + K]

+ {l-s){x0

+ u)] + V + K

(36.10)

= F(x + (1 - s)u) + V + K. But F(x) is bounded, thus there exists a diadic number v E (0,1) such that v[F(x)-y] C V. Therefore, applying (36.10) and Lemma

K-convex and K-concave set-valued functions

361

36.1, we obtain F(x) C (1 - v)F(x) + v[F(x) -y] + vy C (1 - v)F(x) + V + vF[x + (1 - s)u] + V + K C F[x + (1 - s)vu] + G + K. This means that F(x) C F ( x + u i ) + ( j + Ar for every ux G u ( l - s ) C / and by the definition proves that F is K-l.s.c. at x. Since x is an arbitrary point of X, this concludes the proof. •

Remark 36.1 Let a s.v.f. F: D —> B(Y) be K-midconvex and let x0 G X and y0 G Y. Then the s.v.f. G: D — XQ —> B(Y) given by G(x) := F(x + XQ) — yo is K-midconvex, as well. Moreover, if x0 G D and y0 G F(x0), then clearly 9 G G(9) and F is K-u.s.c. (K-l.s.c.) at XQ iff G is K-U.S.C. (K-l.s.c.) at 6. Theorem 36.3 / / a s.v.f. F: D —¥ B{Y) is K-midconvex and K-l.s.c. at a point of D, then it is K-u.s.c. at this point. Proof. In view of Remark 36.1, we may assume that 9 G D and F is K-l.s.c. at 9. To prove that F is K-u.s.c. at zero assume that A G W and choose a. V G W such that V+ V+ V C A. Since F is K-l.s.c. at zero, there exists a balanced neighbourhood U of zero in X such that U C D and F{6) C F{u) + V + K

for all uE U.

From the boundedness of F(9) it follows that there is a diadic number s G (5,1) such that (s" 1 - 1)F(9) C V. Let u G (1 — s_1)C7 and put u\ := (1 — s~l)~xu. Clearly, ui G U and su + (1 — s)ui ~ 9. Take a point a G F(9). Since

362

Functional Equations and Inequalities in Several Variables

F{9) C F(u\) + V + K, so there exists a point y £ F(ui) such that a G y + V + K. Therefore, by Lemma 36.1, we obtain sF(u) + (1 -s)y + (1 - s)a C sF(u) + (1 - s)F(Ul) + (l-s)(y + V + K) C F{9) + (1 - s)y + (1 - s) V + K, whence sF(«) C F(9) + (l-s)V

+ K+(l-

s)a

or F(u) C s-'FiO) + V + K + (1 - s " 1 ) ^ ) .

(36.11)

Moreover, we have s^Ftf)

C (s" 1 - 1)F(6) + F(6) CV + F{9),

and (1 - s-l)F{9) whence, it follows by (36.11) that

c V,

F(u) cV + F{9) + V + K + V C F(9) + A + K, for all u G (1 — s~l)U. Thus, according to the definition, F is K-u.s.c. at zero, which was to be shown. • Directly from Lemma 36.2 and Theorem 36.3 we obtain the following important result. Theorem 36.4 / / a s.v.f. F: D —> B(Y) is K-midconvex and K-l.s.c. at some point of D, then it is K-continuous on D. Another stronger version of the above theorem is contained in the following Theorem 36.5 Let s.v. functions F: D -» B(Y) B(Y) satisfy the inclusion G{x) C F(x) + K

for

x e D.

and G: D -» (36.12)

If F is K-midconvex and G is K-l.s.c. at some point of D, then F is K-continuous on D.

K-convex and K-concave set-valued functions

363

Proof. In view of Remark 36.1 , we may assume that 9 € D, 9 G G(9) and G is K-l.s.c at zero. Fix a neighbourhood A e W and choose a V G W with V + V C A . Since G is K-l.s.c. at zero, there exists a neighbourhood U C D of zero in X such that G{9) c G(u) + V + K

for u G U.

We have 0 G G{9), thus by (36.12) we obtain 9zF(u)

+V +K

for u G U

(36.13)

Since F(#) is bounded, we can find a diadic number s G (0,1) with sF(9) C V. Hence by Lemma 36.1 and (36.13), we have for u eU, F(9) c sF{9) + (1 - s)F{9) + sF{u) + V + K CV + F(su) + V + K C F(su) + A + K. Consequently F is K-l.s.c. at zero. To finish the proof, it remains to apply Theorem 36.4 . • Immediate consequence of the above theorem is the following Corollary 36.1 Let F: D -> B(Y) be a K-midconvex s.v.f. and f:D-^Ybea selection of F. If f is K-l.s.c. at some point of D, then F is K-continuous on D. In the sequel we shall apply the definitions of upper boundedness of sets with respect to a cone K, which are natural extensions of the idea of the upper boundedness for real-valued functions. Namely, a s.v.f. F: D —> 2Y is said to be K-upper bounded on a set U C D iff there exists a bounded set B c Y such that F(x) CB-K

for all x G U.

(36.14)

We say that F is weakly K-upper bounded on a set U C D iff there exists a bounded set B C Y such that F(x) C)(B-K)^®

for all x e U.

(36.15)

364

Functional Equations and Inequalities in Several Variables

Undoubtedly, if F is K-upper bounded on A, then it is weakly Kupper bounded on A. Moreover, for single-valued functions these definitions coincide. Now we are in a position to prove the following far-reaching generalization of the famous theorem of Bernstein-Doetsch. Theorem 36.6 Let F: D -> B(Y) be a K-midconvex s.v.f. and G:D-^2Y be a s.v.f. satisfying the condition G(x) c F(x) + K

for all x G D.

If G is weakly K-upper bounded on a subset of D with a nonempty interior, then F is K-continuous on D. Proof. Suppose that, according to the assumption, G is weakly K-upper bounded on a set x0+U C D, where U is a neighbourhood of zero in X and XQ G D. Thus there exists a bounded set B C Y such that G{x) n ( £ - # ) # 0

for

X£XQ

(36.16)

+ U.

We want to prove that F is K-l.s.c. at the point XQ. TO this aim take an arbitrary neighbourhood A G W and choose a V € W with V + V C A. Boundedness of the sets B and F(xQ) imply that there exists a diadic number s G (0,1) such that sBcV

and

sF(x0) C V.

(36.17)

In view of (36.16) we get sG(x) n (V - K) / 0 for

xex0

+ U,

which yields, by our assumption, 6 G sG{x) -{V-K)= for all x G x0 + U.

sG{x) -V + Kc

sF(x) + V + K (36.18)

K-convex and K-concave set-valued functions

365

Take an arbitrarily fixed u £ U. Then, on account of (36.17), (36.18) and Lemma 36.1, one gets F(x0) C sF(xQ) + (1 -

s)F{x0)

C V + (1 - s)F(x0) + sF{x0 + u) + V + K C F(x0 + su) + A + K. Thus, bearing in mind the definition, we see that F is K-l.s.c. at x0. It remains to apply Theorem 36.4 to prove that F is K-continuous on D. M As an important consequence of the above theorem, we obtain the following T h e o r e m 36.7 / / a s.v.f. F: D -» B(Y) is K-midconvex and weakly K-upper bounded on a subset of D with a non-empty interior, then it is K-continuous on D. Another result which may be deduced directly from Theorem 36.6 reads as follows. Corollary 36.2 Let F: D —»• B(Y) be a K-midconvex s.v.f. and let f: D —> Y be its selection. If f is K-upper bounded on a subset of D with a non-empty interior, then F is K-continuous on D. Now we are going to present a generalization of the classical result stating that convex functions defined on an open subset of a finite-dimensional space are continuous. T h e o r e m 36.8 Let D be an open and convex subset of E n . / / a s.v.f. F: D —> B(Y) is K-convex, then it is K-continuous on D. Proof. Without loss of generality (see Remark 36.1) we may assume that 6 £ D. Bearing in mind that D is an open set, one can find points xi,... ,xn+\ & D linearly independent and such that 6 G int S(xi,..., xn+i), where S(xi,..., xn+\) denotes the simplex with vertices xi,..., xn+i.

366

Functional Equations and Inequalities in Several Variables

Now take a neighbourhood A G W and choose a V G W in such a manner that V + ... + V C A. Since the values of F are v

v

'

bounded sets, thus there exists a number s G (0,1) with sF{0)cV

and

sF(xk)cV,

k = l,...,n

+ l.

(36.19)

Put U := S^sxi,.. . , s x n + i ) . Clearly 9 G int [/. Let u G C/ be arbitrarily fixed. Choose n points rEjtj,..., xjfcn G { x i , . . . ,xn+i} in such a way that u G 5(0, sxfcu . . . , sxkn). Hence we may write u = sQ9 + s^x^

+ . . . + snsxkn

where So,... ,sn G [0,1] and so + ... + sn = 1. On the other hand, u can be represented in the form u = (1 - s + s0s)9 + Sisxkl + . . . +

snsxkn.

Hence, by Theorem 36.1 , we obtain (1 - s + s0s)F(9) + slSF(xkl)

+ ... + snsF(xkn)

C F(u) + K,

which in view of (36.19), implies (1 - s + s0s)F(9) C F{u) - slSF{xkl)

- . . . - snsF{xkn)

+ K

C F{u) + V + ... + V + K. Finally, we obtain for u G U, F{9) c (1 - s + s0s)F(9) + s(l - 30)F(9) C F(u) + V + ... + V + K + V C F(u) + A + K, which shows that F is K-l.s.c. at zero. Theorem 36.4 completes the proof. •

K-convex and K-concave set-valued functions

367

Midconvex set-valued functions with values in R In this section we shall give a characterization of midconvex s.v. functions whose values are compact subsets of R. Namely, we shall prove that such functions have to be of the form F = a + G, where a is an additive function and G is a convex and continuous set-valued function. First we shall prove Theorem 36.9 Let D be a non-empty, open and convex subset of X. A s.v.f. F: D —>• C(R) is midconvex iff there exist an additive function a: X —> R and a convex continuous s.v.f. G: £>->C(R) such that F(x) = a{x) + G{x),

xtD.

(36.20)

Proof. Define the functions / i , f2: D —>• R by the formulae /i(x):=infF(a;)

,

f2{x) := supF{x),

x G D.

(36.21)

Since F satisfies the inclusion ^ 1 ) +% ) ] C F ^ )

(36.22)

for all Xi,x2 € D, then we obtain i mf[F(Xl) + F(x2)} = l[inf F(Xl) + MF(x2)\

>

Smfli-^__j, i.e.

^[/i(a:i) + M*2)] > h (^1\x1,x2

6A

368

Functional Equations and Inequalities in Several Variables

which means that fi is midconvex real-valued function. Similarly, ^sup[F(x1)

+ F(x2)] = ±[supF(Xl)+supF(x2)]

<

(x1+x2\

< supi ^___j, which means that f2 is midconcave. Clearly, h(x)
for

xeD.

By Theorem 2.1 there exists a Jensen function f:D->R such that f(x) < fi(x) for x e D. Moreover, by Theorem 2.2 / has the form f(x) = a(x) + a,

x € D,

where a: X —>• R is additive and a is a real constant. We have also f2(x) - a(x) > fi(x) - a(x) > f(x) - a(x) > a for alia; G D. Since the function g2 = f2 — a is midconcave thus in view of the last inequality, ~92(x) = a{x) — f(x) < —a for

x £ D,

and consequently, by Theorem 4.1 , the function — g2, is continuous and convex. Thus g2 is a continuous and concave function. Moreover, the function g\ = f\ — a is midconvex and gi < g2 on D. Thus again it follows from Theorem 4.1 that gi is continuous and convex on D. Define the s.v.f. G: D -> C(R) by the formula G(x) := [9i(x),92(x)], x £ D. Obviously, G is convex, continuous and G(x) = [fax), f2(x)] - a(x),

x£D.

Take any x £ D. Then by assumption we have

F(x)=F(^-^JD1-[F(x)

+ F(x)},

(36.23)

K-convex and K-concave set-valued functions

369

i.e. F(x) + F{x) C 2F(x) for a; e D. Clearly 2F(z) C F(x) + F(x), which implies the equality -[F{x) + F(x)] = F(x),

xeD.

Taking into account that for every x G D, F(x) is closed, this yields that the sets F{x),x € D are convex. Hence we obtain for x € D F(x) = [MF(x),supF(x)}

=

[h(x),f2(x)}.

Consequently, by (36.23), we obtain the formula F(x)=a(x)

+ G{x),

xeD.

The proof of the converse implication is straightforward and is therefore left to the reader. • In the case where D is equal to the whole space X, the s.v.f. G occuring in the formula (36.20) has to be constant. Namely, we have the following Theorem 36.10 Let X be a vector space. A s.v.f. F: X -> C(E) is midconvex iff there exist an additive function a: X —>• R and a compact interval G c R such that F(x) = a(x) +G,

xeX.

(36.24)

Proof. Suppose that F is midconvex and consider the functions fi,f2'-X->R defined by (36.21). Thus / i is midconvex and /2 midconcave and hence h :— f2 — f\ is midconcave and nonnegative on X. We shall show that it is a constant function. Contrary to what claim, suppose that there exist x, y 6 X such that h(x) < h(y). Choose a rational number s > 1 in such a way that s[h(x) - h(y)] + h(y) < 0. (36.25) Since h is midconvex, we obtain h{x) = h[s-l(sx + (1 - s)y) + (1 - s _1 )y] > > s'^-hisx + (1 - s)y) + (1 - s- 1 )/^?/),

370

Functional Equations and Inequalities in Several Variables

whence, by (36.24) we get h(sx + (1 - s)y) < sh(x) + (1 - s)h(y) < 0, a contradiction with the non-negativity of h. Consequently, h = L = const. Note that fi = f2-L has to be midconvex and midconcave simultaneously, so /i is a Jensen function. By Theorem 2.2 , /i has to be of the form fi(x) — a(x) + a,

i£l,

where a: X —> E is additive and a is a real constant. Thus f2(x) = a(x) + a + L, x e X. Denote G := [a, a + L], then for x e X, F(x) = [f^x), f2{x)] = \a(x) + a, a{x) + a + L] = a(x) + [a, a + L] = a(x) + G. By simple calculations one can verify the converse implication.

•

K-midconcave set-valued functions As in the previous section, we shall assume that X and Y are topological vector spaces , D is a non-empty convex and open subset of X and K is a cone with zero in Y. A s.v.f F: D —> 2Y is said to be K-midconcave iff ^{xi+x2)

C^[F(xl)

+ F(x2)] + K

(36.26)

for all x\, x2 G X. We say that F is midconcave provided (36.26) is satisfied with K = {9}. Unfortunately, although properties of K-midconcave s.v. functions are, in general, analogous to the corresponding properties of K-midconvex s.v. functions presented in the previous

K-convex and K-concave set-valued functions

371

section, however, they can not be obtained from those theorems and proves can not be repeated in this case. As before, by W will be denoted an arbitrary base of balanced neighbourhoods of zero in Y. Concerning the results presented here see Nikodem [142]. By repeating similar arguments as in the proof of Lemma 36.1 one may get the following Lemma 36.3 / / a s.v.f. F: D —>• 2Y is K-midconcave and has convex values, then F[sx + (1 - s)y] C sF(x) + (1 - s)F(y) + K

(36.27)

for all x, y G X and all diadic s G [0,1]. We are now going to prove the following general result. Theorem 36.11 If a s.v.f. F: D -» BC(Y) is K-midconcave and K-u.s.c. at some point of D, then it is K-u.s.c. at every point of D. Proof. Suppose that F is K-u.s.c. at a point x0 G D. Fix an arbitrary x € D. Take an arbitrary A EW and choose &V EW in such a way that V + V + V + VcA. By the assumption on F, there exists a neighbourhood U of XQ, U C D such that F(u) C F(x0) + V + K

for

ueU.

(36.28)

Since F(XQ) is a bounded set, we can find a diadic s G (0,1) with y := —^— (x - x0) + x G D

and

sF(x0) C V.

(36.29)

Put Ux := (1 - s)y + sU. Then Vx is open, Ui C D and x — (1 — s)y + sx0 G U\. Moreover, in view of the boundedness of the sets F(y), F(x), there exists a diadic number a G (0,1) such that aF(y) C V and aF(x) C V. (36.30)

372

Functional Equations and Inequalities in Several Variables

Define U2 := (1 - o)x + aU\. Clearly U2 C D and U2 is a neighbourhood of x. Therefore, from Lemma 36.3 , for u2 e U2 , u2 — (1 — a)x 4- au\ , we get F{u2) C (1 - a)F(x) + aF{Ul) + K.

(36.31)

For U\ e U\ we can find an u e U with Ui = (1 — s)y + su . Thus applying Lemma 36.3 and (36.28), (36.29) we get F(u{) C (1 - s)F(y) + sF(u) + K C (1 - s)F{y) + s[F(ar0) + V + /T] + X C (1 - s)F{y) + V + V + K. Hence, on account of (36.30), we obtain aF(Ul)cV

+ V + V + K.

(36.32)

Moreover, since F(x) is convex, thus (1 - a)F(x) + aF(x) C F(x), whence, by (36.30), we infer that (1 - a)F(x) C F{x) - aF(x) C F(x) + V.

(36.33)

Consequently, in view of (36.31), (36.32) and, ultimately, (36.33), we have for u2 G U2, F(u2) C F(x) + V + V + V + V + K + K C F(x) + A + K, which, in turn, simply means that F is K-u.s.c. at x.

•

L e m m a 36.4 If a s.v.f. F: D —>• BC(Y) is K-midconcave and K-u.s.c. at a point of D, then it is K-l.s.c. at this point.

K-convex and K-concave set-valued functions

373

Proof. We may assume (see Remark 36.1) that 9 G D and F is K-u.s.c. at zero. We claim that F is K-l.s.c. at zero. To this end take A G W and choose ^Oj V £ W in such a manner that 2A0 C A and V + V + V C AQ . Since F is K-u.s.c. at zero, there exists U eW,U C D such that F(x) C F(0) + V + K

for

a; €

tf.

(36.34)

By the boundedness of F(ff), there exists a number c > 0 such that cF(9) C V, whence, for n satisfying the condition c2n > 1, we obtain ^-F{9) C -\-V C V. 2" v ; c2n Thus we can find a diadic number s € (0, | ) such that sF(9) C V (e.g. s = 2- n _ 1 ). Denote U0 :=

-?-U. s- 1 Obviously, i7o is a neighbourhood of zero contained in D, and for i € [/0, we have X\ :=

s- 1 s

£ G C/ and

(1 — s)x + sxi = 9.

Therefore, by Lemma 36.3 and (36.34), we get F{9) = F[(l - s)x + sari] C (1 - s)F(x) + sF(xi) + K C (1 - s)F(x) + s[F(9) + V + K] + K C (1 - s)F{x) + V + V + K, i.e. F(9) C (1 - s)F(x) + V + V + K for x G U0. By the convexity of F(9), we have (1 - s)F(6) + sF(9) C F(6)

(36.35)

374

Functional Equations and Inequalities in Several Variables

whence (1 - s)F(0) C F(9) - sF(8) C F(6) + V. Consequently, by (36.35) it follows that (l-s)F(9)

C (l-s)F(x) + V + V + V + K C (l-s)F(x)+A0 + K.

(36.36)

Taking into account that s < \ implies (1 — s ) - 1 < 2, then (1 - s T % C 2A0 C A. Clearly (36.36) yields the inclusion F{6) C F(x) + (1 - s ) - M 0 + (1 - s)~lK, whence, F(0) C F(x) + A + K, as long as x remains in C/0 • Thus F is K-l.s.c. at zero, as claimed.

• Let us note, that Theorem 36.11 and Lemma 36.4 yield directly the following result. Theorem 36.12 If a s.v.f. F: D ->• BC(Y) is K-midconcave and K-u.s.c. at some point of D, then it is K-continuous on D. We can even prove a stronger result. Theorem 36.13 Let a s.v.f. F: D —>• BC(Y) and let G: D —> B(Y) be such that

be a K-midconcave

F(x) C G(x) + K for x e D. If G is K-u.s.c. at a point x0 € D, then F is K-continuous on D.

K-convex and K-concave set-valued functions

375

Proof. Take a neighbourhood A £W and choose a V G W such that V + V + V C A. Since G is K-u.s.c. at xo, there exists a neighbourhood U of zero in X such that Xo + U C D and G{x0 + u)c G{x0) + V + K

for ueU.

(36.37)

Choose a diadic number s G (0,1) with (note that G(x0) and F(x0) are bounded) sG{x0) C V and sF(x0) C V.

(36.38)

Therefore, on account of (36.37), (36.38) and Lemma 36.3 , the following inclusions hold true F(x0 + su) C sF(x0 + u) + (l- s)F(x0) + K C s[G{x0 + u) + K] + F(x0) - sF(x0) + K C s[G(x0) + V + K] + F{x0) + V + K C V + V + F{x0) + V + K C F(x0) + A + K, provided u £ U. This simple means that F is K-u.s.c. at XQ. Theorem 36.12 completes the proof. • Let us introduce the following definition. A s.v.f. F: D —> 2Y is said to be K-lower bounded on a set A C D iff there exists a bounded set B C Y such that F{x) CB + K

for all x G A.

Now we are in a position to prove the following generalization of the Bernstein-Doetsch theorem. Theorem 36.14 If a s.v.f. F: D —> BC(Y) is K-midconcave and K-lower bounded on some subset of D with a non-empty iv.i-rior, then it is K-continuous on D.

376

Functional Equations and Inequalities in Several Variables

Proof. By the assumption, there exists a neighbourhood U of zero in X such that F is K-lower bounded on a set x0 + U C D. Thus, by the definition, there exists a bounded set B C Y with F(x0 + u)cB

+K

for ueU.

(36.39)

Take an arbitrary A eW and choose &V £W with V + V C A. Since the sets F(xo) and B are bounded, then we can find a diadic number s G (0,1) for which we have the inclusions sF(x0) C V and s £ c V.

(36.40)

Put C/0 : = ^o + sU, then, obviously UQ is a neighbourhood of XQ and C/o C D. Apply Lemma 36.4 for x G UQ, X — x0 + su, to get F(x) = F[s(xQ + u) + (1 - s)x0] C sF{x0 + u) + (1 - s)F(zo) + A" C sF(x0 + u)+ F(x0) - sF(x0) + K. Thus, on account of (36.39) and (36.40), we obtain F(x) CV + F(x0) + V + KC F{x0) + A + K, provided x G UQ, i.e. F is K-u.s.c. at x0. Theorem 36.12 completes the proof. • As an immediate consequence of Theorem 36.5 and Theorem 36.13, we obtain Theorem 36.15 Let X,Y be topological vector spaces and K be an open cone in X. Suppose that F: S -> BC(Y) is a s.v.f and A: S —> BC(Y) is an additive s.v.f. . If, moreover, F(x) C A(x) for x € S and F is l.s.c. at a point of S, or if A(x) C F(x) for x G S and F is u.s.c. at a point of S, then A is continuous on S. Let us recall that a s.v.f. A: K —>• 2Y is said to be linear iff it is additive and positively homogeneous, i.e. A(sx) = sA(x) for x G S and s G (0, oo).

K-convex and K-concave set-valued functions

377

Theorem 36.16 Let X, Y be topological vector spaces and let S be a cone in X. If a s.v.f. A: S —>• CC(Y) is additive and continuous, then it is linear. Proof. Take x € S and r € (0, oo). Choose a sequence {rn}„eN of positive rational numbers converging to r. Since A is continuous, so A(rnx) -)• A(rx). On the other hand, A(rnx) = rnA(x) ->• rA(x). Thus, since the sets A(rx) and rA(x) are closed, by Lemma 28.6 , we get A(rx) = rA(x), which concludes the proof.

•

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Chapter 37 Iteration semigroups of set-valued functions Let X be a non-empty set. A family {F*:s>0}

(37.1)

of s.v. functions Fs: X -» 2X is said to be an iteration semigroup iff F ( o F s = Ft+S (37.2) in X, where {F*oF')(x)

:= [jiF'iy):

y e Fs(x)}

(37.3)

for every t, s > 0 and x G X. In this chapter we deal with iteration semigroups of Jensen setvalued functions. For further considerations we shall need the following two lemmas. Lemma 37.1 Suppose that X is a real vector space and Y is a topological vector space. Let S be a cone with zero in X. A s.v.f. F: S —t C(Y) satisfies the Jensen equation F

2(xi+:c2)

= \[F(Xl) 379

+ F(x2)]

(37.4)

380

Functional Equations and Inequalities in Several Variables

for all xi,x2 G S iff there exist an additive s.v.f. A: S —>• CC(Y) and a set B G CC(Y) such that F(x) = A(x) + B

(37.5)

for all x G S. Proof. To prove this theorem, one may repeat arguments used to prove Theorem 29.1 (see also [142], Theorem 5.6). • The next lemma is due to A. Smajdor [198]. Lemma 37.2 Let S be a cone with zero in a vector space X and let Y be a normed space. If a s.v.f. F: S —>• CC(Y) is a solution of the equation (37.4) satisfying the condition sup{diamF(x): x G S} < oo,

(37.6)

then there exist an additive function f: S —>• Y and a set B G CC(Y) such that F{x) = f(x) + B (37.7) for all x G S. Proof. By Lemma 37.1 there exists an additive s.v.f. A: S -> CC{Y) and a set B G CC{Y) such that F(x) = A{x) + B,

xeS.

Hence we obtain F(2nx) = A(2nx) +B = 2nA(x) + B, or 2~nF{2nx) = A(x) + 2~nB for all x G S and n G N. Fix an arbitrary x G S and take an element b G B. Let u, v G A(x), then we have for n G N, ||« - v|| = ||(u + 2- n 6) -(v + 2-nb)\\ < diam(2-"F(2 n :r)) = 2""diamF(2 n a;).

Iteration semigroups of set-valued functions

381

In view of (37.6) it follows that u = v and shows that A(x) for every x G S is a set containing only one element. Therefore for every x G S there exists exactly one f(x) G Y such that A(x) = {/(:r)}. Thus F(x) = f(x) + B

for

xeS,

which completes the proof.

•

We may formulate the following theorem (see [198]). Theorem 37.1 Let S be a closed cone in a normed space X. Let s.v. functions F ' : S —)• CC(S), t > 0 be such that F* for every t > 0 satisfy the equation (37.4) in S and let sup{diamF'(x): x G S} < oo

(37.8)

for every t > 0. Then the family {F*: t > 0} is an iteration semigroup iff there exist an iteration semigroup {/*: t > 0} of additive functions fl: S -> S and the family {Bu. t > 0} of sets Bl G CC{S) such that for t > 0, s > 0 Ft(x) = f\x)

+ Bt,

xG5,

(37.9)

and Bs+t = fs{Bt)

+ Bs.

(37.10)

Proof. Assume that the family {F*: t > 0} of s.v. functions F*: S -> CC{S), t > 0 satisfying the equation (37.4) and the condition (37.8), is an iteration semigroup. Thus, by Lemma 37.2, for every t > 0 there exist an additive function p-.S-^X l and a set B G CC{S) such that F*(x) = f\x)

+ Bf,

xe S.

For x = 6, we get F\e) = ft(8) + Bt = Bt,

(37.11)

382

Functional Equations and Inequalities in Several Variables

whence Bl e CC(S) for t > 0. Fix arbitrarily x e S, t > 0 and take a b e Bl. Then for n € N, by (37.11), we obtain f{x) + -b = -\f{nx) ft

+ b] e -F\nx)

ft

C S.

ft

Letting n —> oo, by the closedness of S, we get /*(:r) G S, which means that /* is a map of S into itself. Moreover, applying (37.2) and (37.11), we get fs+t(x)

+ Bs+t = / s [/*(z)] + /*(£') + Bs,

(37.12)

whence putting x = 9, Bs+t = fs(Bt)

+ Bs.

(37.13)

Therefore Lemma 37.1 applied to (37.12) yields f'+t(x)

= f\f(x)]

(37.14)

for x E S and s, t > 0. The equality (37.14) means that the family {f-s>0}

(37.15)

is an iteration semigroup of additive functions. Conversely, assume that a family (37.1) is given by the formula (37.9) where the family (37.15) is an iteration semigroup of additive functions / ' : S —> S satisfying the condition (37.10). Then by (37.10) and (37.14) if follows that (37.12) holds true, which means that ps+t _ ps 0 pt and shows that the family (37.1) is an iteration semigroup of Jensen s.v. functions.

Iteration semigroups of set-valued functions

383

Moreover, sup{diamF s (x): x € S} = - sup j{sup \\a -b\\: a,b G Fs(x)}: = sup {{sup \\f'(x) +h-

x G S\

fs(x) -b2\\:bub2£Bs}:xe

= sup{{sup||6i -&2||: bub2EBs}:

s}

x G s}

= sup{sup \\b\ — b21|: h,b2 G Bs} < 00, since for every s > 0, Bs is bounded. The proof is completed.

•

Now we shall consider so called increasing iteration semigroups of set-valued functions. Namely, an iteration semigroup {Fs: s > 0} of s.v. functions Fs: X —>• CC(X) is said to be increasing iff F\x)

C Fs(x)

(37.16)

whenever x G X and 0 < t < s. Let S be a cone in a normed linear space X. Consider an increasing iteration semigroup {Fs: s > 0} of s.v. functions Fs: S —>• CC(S) satisfying the equation (37.4). Moreover, assume that the condition (37.8) is fulfilled. By (37.16), (37.9) and Theorem 37.1 , we have f{x) + Bl = F\x)

C Fs(x) = fs(x) + Bs

for all x G S, 0 < t < s. Fix arbitrarily 0 < t < s and choose an a G Bl. Thus, for x G S, ft(x)-fs(x)eBs-a. The additivity of /*, / " implies /'(*) - fs(x) = - [ / ' ( n i ) - fs(nx)}

G ~[BS - a]

384

Functional Equations and Inequalities in Several Variables

for x e S and n e N. This jointly with the boundedness of the set Bs — a yields ft(x) = fs(x) for xeS. Therefore, there exists exactly one additive function f: S —> S such that Fs(x) = f(x)+Bs, xeS, s>0. (37.17) The equations (37.13) and (37.14) one may rewritte in the form Bs+t = f(Bt) + Bs,

t,s>0,

(37.18)

and Define FUF2,F3: F1(t)=Bt,

f(x) = f(x), xeS. [0,oo) ->• CC(Y) by F2(t) = f(Bt),

(37.19)

F3{t) = B<,

then the equation (37.18) implies that Fi,F2, F 3 satisfy the Pexider functional equation

Fl{t + s)=F2(t) + F3(s) for t, s > 0. On account of Theorem 30.1, there exist an additive s.v.f. A: [0,oo) ->• CC(Y) and sets K,M G CC(F) such that F 1 (t) = JBt = y l ( 0 + ^ + M, F 2 (t) = /( J B i )=A(t)+ J fi:,

(37.20)

F 3 ( i ) = J B t = A(0 + M, for all t > 0. Hence, in view of Lemma 20.1 , K = {#}, and therefore A(t) = f(Bt) = f[A(t)] + f(M). Thus putting in the above equation t = 0, we get / ( M ) = {9},

(37.21)

Iteration semigroups of set-valued functions

385

and consequently A(t) = f[A(t)],

t > 0.

(37.22)

Since Bl = A(t) + M, then B° = M C S. Moreover, A(t) = / ( # ) C S. Let us note that the monotonicity of {Fs: s > 0}, the additivity of A, (37.17) and, ultimately, (37.20) imply f(x) + A(t) + M = F\x)

c Ft+S{x) = f(x) + A(t) + A(s) + M,

whence, applying once more Lemma 20.1 , we obtain 9 e A(s)

for

s > 0.

This condition states that additive s.v.f. A admits a continuous selection. Thus, in view of Theorem 36.15 (putting F(x) := {9}) we guess that A is continuous. Therefore by Theorem 36.16 A(t) = tA(l) = L40,

A0:=A{1),

t > 0.

Finally, condition (37.22) yields A) = /(A))-

(37.23)

All these considerations we may summarize in the following theorem (see [198]). Theorem 37.2 Let S be a closed cone in a normed linear space X. The family {Fs: s > 0} is an increasing iteration semigroup of s.v. functions Fs: S —> CC(S), s > 0 satisfying the equation (37.4) and condition (37.8) iff there exist an additive function f': S —>• S and sets A0,M e CC{S) such that (37.19), (37.21), (37.23) and 9 € AQ and Fs{x) = f{x) + sA0 + M,

s>0,

xeS,

hold true. Observe that the part „if" can be proved without any difficulties and is therefore omitted. •

386

Functional Equations and Inequalities in Several Variables

Notes 37.1 For midpoint convex s.v. functions one has the following (see A. Smajdor [199]). Proposition 37.1 Let X be a locally convex space and let {Fs: s > 0} be a family of midpoint convex s.v. functions Fs: X —>• C(X) of the form F°(x) = fs(x) + Fs(9), where {fs: s > 0} is the family of additive functions from X into X. Then {Fs: s > 0} is an iteration semigroup iff {fs: s > 0} is an iteration semigroup and Ft+s{9) =

ft[Fs{9))+Ft(9).

Moreover, if every Fs is upper semicontinuous at a point of X, then the functions fs are linear and continuous. An analogous result is contained in [79]. 37.2 More results concerning the iterations of s.v. functions can be found in [197].

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[166] Th. M. Rassias (ed.), Inner Product Spaces and Applications, Pitman Research Notes in Mathematics Series (Addison Wesley Longman Ltd), No. 376, 1997. [167] Th. M. Rassias (ed.), New Approaches in Nonlinear Analysis, Hadronic Press Inc., Florida, 1999. [168] Th. M. Rassias (ed.), Topics in Mathematical Analysis, Word scientific Publ. Co. (in the series in Pure Mathematics), 1989. [169] Th. M. Rassias and C. S. Sharma, Properties of Isometrics, Journal of Natural Geometry 3(1993), 1-38. [170] Th. M. Rassias and J. Tabor (eds.), Stability of Mapping of Hyers-Ulam Type, Hadronic Press Inc., Florida, 1994. [171] Th. M. Rassias and J. Tabor, What is Left of Hyers-Ulam stability?, J. Natural Geometry 1(1992), 65-69. [172] Th. M. Rassias and J. Simsa, Finite Sums Decompositions in Mathematical analysis, John Wiley and Sond Lts. (WileyInterscience series in Pure and Applied Mathematics), 1995. [173] Th. M. Rassias and P. Semrl, On the Hyers-Ulam stability of linear mappings, J. Math. Anal. Appl. 173(1993), 325-338. [174] Th. M. Rassias and P. Semrl, On the behaviour of mappings which do not satisfy Hyers-Ulam stability, Proc. Amer. Math. Soc. 114(1992), 989-993. [175] Th. M. Rassias, On a modified Hyers-Ulam sequence, Journal of Math. Analysis and Appl. 158(1991), 106-113. [176] Th. M. Rassias, On the stability of functional equations and a problem of Ulam, Acta Appl. Math. 62(2000), 23-130. [177] Th. M. Rassias, On the stability of minimum Mathematica (to appear).

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[178 Th. M. Rassias, On the stability of the linear mapping in Banach spaces, Proc. Amer. Math. Soc. 72(1978), 297-300. [179 Th. M. Rassias, Stability and set-valued functions , in: Analysis and Topology (eds. C. Andreian Cazacu, O. Lehto and Th. M. Rassias),World Scientific Publ. Co., 1998, pp.585-614. [180 B. E. Rhoades, Fixed point theorems and stability results for fixed point iteration procedures, Indian J. Pure Appl. Math. 21(1990), 1-9. [181 A. W. Roberts and D. E. Varberg, Convex Functions, Academic Press, New York and London, 1973. [182 S. M. Robinson, Regularity and stability for convex multivalued functions, Math. Oper. Res. 1(1976), 130-143. [183 R. T. Rockafellar, Convex Analysis, Princeton University Press, Princeton, N. J. 1970. [184 G. Rodee, Eine abstrakte Version des Satzes von HahnBanach, Arch. Math. 31(1978), 474-481. [185 J. Rohmel, Eine charakterisierung quadratischer Formen durch eine Funktionalgleichung, Aequationes Math. 15(1977), 163-168. [186 W. Rudin, Functional Analysis, New York, 1973. [187 W. Rudin, Real and Complex Analysis, 2nd ed., McGrawHill, New York, 1974. [188 W. Sander, A remark on convex functions, Glasnik Mat. [189 P. Semrl, The functional equation of multiplicative derivation is superstable on standard operator algebras, Integral Equations Operator Theory 18(1994), 118-122.

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[190] P. Semrl, The stability of approximately additive functions in: Stability of mappings of Hyers-Ulam type, (ed. Th. M. Rassias and J. Tabor), Hadronic Press, Palm Harbor, Florida, 1994, 135-140. [191] W. Sierpiriski, Sur les fonctions convexes mesurables, Fund. Math. 1(1920), 125-128. [192] S. L. Singh and V. Chadha, Round-off stability of iterations for multivalued operators, C. R. Math. Rep. Acad. Sci. Canada XVII (1995), 187-192. [193] S. Siudut, J. Tabor, J. Tabor, Stability of the Cauchy Equation in Function Spaces, Ped. Univ. Rzeszow, Inst, of Mathem., May 1996 (preprint). [194] F. Skof, Local properties and approximations of operators, Rend. Sem. Mat. Fis. Milano 53(1983), 113-129. [195] F. Skof, SulVapprossimazione delle applicazioni localmente S-additive, Atti Accad. Sc. Torino 117(1983), 377-289. [196] A. Smajdor, Additive selections of superadditive set-valued functions Aequationes Math. 39(1990), 121-128. [197] A. Smajdor, Iterations of multi-valued functions, Prace Naukowe Uniwersytetu Slaskiego 759, Katowice, 1985. [198] A. Smajdor, Semigroups of Jensen set-valued functions, Results in Mathematics 26(1994), 385-389. [199] A. Smajdor, Semigroups of midpoint convex set-valued functions Publicationes Mathematicae 46(1995), 161-170. [200] W. Smajdor, Subadditive and Subquadratic Set-Valued Functions, Prace Naukowe Uniwersytetu Slaskiego w Katowicach, Katowice, 889(1987).

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[201] W. Smajdor, Superadditive set-valued Functions and Banach-Steinhaus Theorem, Radovi Matem., 3(1987), 203214. [202] H. Steinhaus, Sur les distances des points des ensembles de mesure positive, Fund. Math. 1(1920), 99-104. [203] O. Stolz, Grundziige der Differenzial- und Integralrechnung, Vol. I, Teubner, Leipzig, 1983. [204] L. Szekelyhidi, An abstract superstability theorem, Abh. Math. Sem. Univ. Hamburg 59(1989), 81-83. [205] L. Szekelyhidi, Local polynomials and functional equations, Publ. Math. 30(1983), 283-290. [206] L. Szekelyhidi, On a theorem of Baker, Lawrence and Zorzitto, Proc. Amer. Math. Soc. 84(1992), 95-96. [207] L. Szekelyhidi, Functional equations on Abelian groups, Acta Math. Hungar. 37(1981). [208] J. Tabor and J. Tabor, Homogenity is superstable (to appear). [209] N. X. Tan, Banach-Steinhaus principle for convex miltivalued mappings, Math. Nachr. 126(1986), 45-54. [210] N. X. Tan, Banach-Steinhaus theorem for multivalued mappings, Math. Nachr. 102(1981), 157-167. [211] T. Trif, Hyers-Ulam-Rassias stability of a Jensen type functional equation, Journal of Math. Anal, and Appl. 250(2000), 579-588. [212] S. M. Ulam, Problems in Modern Mathematics, Chapter VI, Science Editions, Wiley, New York, 1960. [213] F. A. Valentine, Convex Sets, McGraw-Hill Book Company, New York-San Francisco-Toronto-London, 1964.

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[214] P. Volkmann, Eine Charakterisierung der positiv definiten quadratischen Formen, Aequationes Math. 11(1974), 174182. [215] P. Vrbova, Quadratic junctionals and bilinear forms, Casopis. Pest. Mat. 98(1973), 159-161. [216] K. Yosida, Functional Analysis, 3rd ed., Springer-Verlag, Berlin-Heidelberg New York, 1971.

Index A-additive mappings, 312 Q-convex, 80, 154 X\-spaces, 229 Xf-spaces, 230 er-convex, 152 y-additive mappings, 137 (/^-homogeneous mappings, 175 Q+ -homogeneous, 303 6-metric, 246 ^-homogeneous mappings, 170

325 Bernstein-Doestch theorem, 25 best posible approximation, 135 biadditive mapping, 89 bilinear mapping, 91 binary intersection property, 186, 333 bounded set, 182

abelian group, 131 additive Cauchy equation, 129 additive functions, 9 additive selection, 155 affine function, 10 algebraically interior point, 3 algebraically open set, 3

additive equation, 35 difference, 130, 203 exponential equation, 37 inverse operator, 219 operator, 218 sequence, 133 characters, 54 complete measurable group, 204 concave functions, 9 cone, 355 control functions, 136 convex functions, 9 convex hull, 19 core-topology, 3

Cauchy

Baire property, 26 Baire space, 7 balanced neighbourhood, 313, 314 Banach algebra, 64 Banach spaces, 138 Banach-Steinhaus theorem on uniform boundedness, 407

Index

408 D'Alembert's equation, 57, 161 d-Lipschitz function, 236 decomposition theorem, 98 decreasing sequence of sets, 271 diagonalization, 74 difference operator, 65 divisible group, 194 double difference property, 218 equation functional Cauchy, 263 D'Alembert, 161 exponential, 37 Gamma, 157 homogeneous, 170 Jensen, 147 Lobaczevski, 163 multiplicative, 141 Pexider, 153 quadratic, 89 sine, 166 extension of linear functionals, 331 extention theorem, 86 fixed point, 246 Fubini theorem, 218 functional equation of derivation, 167 functional inequality Jensen, 9 Gamma functional equation, 157

generalized Cauchy equation, 139 generalized polynomial, 87 group 2-divisible, 163, 253 group characters, 49 Hahn-Banach theorem, 331 Hamel basis, 81 Hausdorff metric, 186 Hilbert space, 201 homogeneous functions, 9 stability, 174 homomorphism, 129 Hyers-Rassias sequence, 176 increasing iteration semigroup, 383 increasing set-valued function, 324 induction principle, 144 invariant means, 235 invariant metric, 236 inverse-image, 298 iteration semigroup, 379 Jensen difference, 211 Jensen equation, 10 Jensen inequality, 9 K-bounded, 355, 363 K-continuous, 360 Kuratowski-Zorn lemma, 334 law of cancellation, 185 Lebesgue measurable, 39 linear functions, 9 linearly invariant family, 235

409

Index Lipschitz function, 240 Lipschitz norms, 241 locally convex space, 314 lower hull, 19

ray-B-bounded, 92 ray-bounded, 91 representation theorem, 77 round-off stability, 245

measurable set-valued function, 298 Mehdi theorem, 28 metric group, 240 mixed stability, 142 module of continuity, 239 multiadditive functions, 71

second category Baire set, 101 semigroup, 214 semitopological group, 25 sequential closure, 153 sesquilinear function, 93 set locally of the second category, 25 set valued function additive, 263 bounded on a set, 272, 314 continuous, 273 iteration semigroup, 379 K-concave, 355 K-continuous, 360 K-lower bounded, 375 K-lower semicontinuous, 359 K-midconcave, 370 K-midconvex, 358 K-upper semicontinuous, 360 measurable, 285 stability of iteration procedures, 245 subadditive s.v.f., 302 subquadratic function, 346 superadditive s.v.f, 323 superstability, 141 support function, 10

norm multiplicative, 141 normed algebra, 141 normed space, 148 nth difference, 66 Pexider equation, 153 inverse operator, 227 operator, 221 Pexider difference, 212 Piccard theorem, 27 polynomial function, 71 positive inner Lebesgue measure, 311 quadratic difference, 253 quadratic function, 89 quadratic functional equation, 185 quadratic selection, 350 quasi-normed space, 186 Rassias type stability, 185 Rassias' sequence, 133

topological groups, 54

410 topology Baire, 7 Hausdorff, 359 linear, 5 semilinear, 5 triangle inequality, 248 Ulam-Hyers-Rassias stability, 131 unitary representation, 50 upper integral, 204 upper semicontinuous s.v.f., 272 vector space, 302 weakly K-upper bounded, 363 Wright convex function, 16

Stefan Czerwik Silesian University of Technology, Poland

his book outlines the modern t h e o r y of f u n c t i o n a l equations and inequalities in several variables. It consists of three parts. The first is devoted to additive and convex functions defined on linear spaces with semilinear topologies. In the second part, the problems of stability of functional equations in the sense of Ulam-Hyers-Rassias and in some function spaces are considered. In the last part, the functional equations in set-valued functions are dealt with — for the first time in the mathematical literature. The book contains many fresh results concerning those problems.

T

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