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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Variables

Stefan Czerwik Silesian University of Technology, Poland


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


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


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


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


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 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

/.• 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


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) 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(*;)+ #( [—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


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 [—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


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 = 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


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 = \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),


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


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

iêR,

(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])-


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


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


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


/(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


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.


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


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


which means that Yôcsa(xys)

$â«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

:

= 5â 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.


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)ô)

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


By a character of a topological abelian group G we mean a function ip: G —» C such that s{x) ^ G ,

(8.9)

s=l m

a r (x) = 2Jasf3srsl3s,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


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]).

•


59

Theorem 8.3 Let (• 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


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)


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


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 ô)) + 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>)


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«(ô) = 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


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


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^+---+Â)- (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


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

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

1

= Afc 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).


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


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.


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)

= Âhl

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


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)


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


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)]


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


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,


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


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 i r f c (8 +k *) E(-!) s - n (s +n x) / K M + ( i - 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


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


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


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)}

• 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 )|| 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


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 • 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


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 ,

\ llîll /

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)\ 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)

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


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))\ 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


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


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


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 • 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


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].


Part II Ulam-Hyers-Rassias Stability of Functional Equations


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)]• 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


for some e > 0. Then there exists exactly one additive mapping A: El -Ê2: 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)\\ 0 and some 0 < p < 1, then there exists a unique additive mapping A: Ei —> E2 such that

ll/W-^)ll 2 (X,\\ • ||) a Banach space, cp: G x G —> [0,oo) a mapping such that oo

2

-

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

for k>3.

(13.13)

136


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
\\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)\\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


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
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)\\ 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 • 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


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 ) = 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

• oo, we get „ / T. 4 - ?/ \

mf(y) = f2,x as claimed.

+y

for

x,y

eG,

166


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)\ 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,


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)\\f{x)\\

+\a\v\\f{anx)

- anvf(x)\\

< ||/(a n+1 a;) - avf(anx)\\ < h(a, anx) + ... + 169

+

\a\nvh{a,x),

170


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)\\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

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