Sitzungsber. Abt. II (2005) 214: 111–132

Stability of the Homogeneity and Completeness

By Wojciech Jablon´ski and Jens Schwaiger (Vorgelegt in der Sitzung der math.-nat. Klasse am 13. Oktober 2005 durch das w. M. Ludwig Reich)

Abstract It is known that the classical homogeneity equation f ðxÞ¼f ðxÞ under rather weak conditions is stable (cf. [4, 8]). Following J. SCHWAIGER [14] we will consider here the stability of the -homogeneity equation f ðxÞ¼ðÞ f ðxÞ. We will prove, without assumption that is a homomorphism, the stability of the -homogeneity equation under some conditions. One of the necessary conditions is that the target space of the f be se- quentially complete. We will also prove that the converse is true, that is we will show that stability of the -homogeneity implies the fact that if the target space is a normed space then it has to be a .

1. Introduction Since the time when ULAM ([17]) had posed his famous problem many authors have considered stability of different functional equations (cf. [4,8]).Amongothers,in1992CZERWIK ([1]) examined the stability of the v-homogeneity equation v FðxÞ¼ FðxÞ for 2 Uv; x 2 X; where a function F maps a real linear space X into a real Banach space v Y, v 2 Rnf0g is fixed and Uv :¼f 2 R: is well definedg.Ithas been proved there that if a function f : X ! Y satisfies the inequality v k f ðxÞ f ðxÞk gð; xÞ for 2 Uv; x 2 X; 112 W. Jablonnski and J. Schwaiger with a function g: Uv X !½0; 1Þ then under some assumptions on g there exists a v-homogeneous function F: X ! Y such that k f ðxÞFðxÞk hðxÞ for x 2 X; with some function h: X !½0; 1Þ depending on g. Independently, JOOZEF TABOR proved in [16] that every mapping f from a real linear space X into a normed space Y satisfying k1f ðxÞf ðxÞk " for 2 Rnf0g; x 2 X; where " 0 is given, is homogeneous. These results then were generalized successively in different directions. In [15] the inequality k f ðxÞvf ðxÞk gð; xÞ;2 Rnf0g; x 2 X; with a constant v 2 Rnf0g and a function g mapping RX into R has been investigated. KOMINEK and MATKOWSKI began to investigate in [11] the stability of the homogeneity on a restricted domain. They have considered the condition 1f ðxÞf ðxÞ2V;2 A; x 2 S; for the mapping f from a cone S X into a sequentially complete locally convex linear topological Hausdorff space Y over R and a A ð1; 1Þ. This result has been generalized in [9] and [14]. SCHWAIGER [14] has examined the condition f ðxÞðÞf ðxÞ2VðÞ; for 2 A; x 2 X; where – G is a semigroup with unit acting on the non-empty set X; – Y is a sequentially complete locally convex linear topological Hausdorff space Y over K 2fR; Cg; – A G generates G as a semigroup; – : G ! K is a function; – V: G !BðYÞ is a mapping from G into the family BðYÞ of all bounded of Y. It is proved there that if f ðXÞ is unbounded, then is a multiplicative function, i.e. satisfies the equation ðÞ¼ðÞðÞ for ; 2 G; Stability of the Homogeneity and Completeness 113 and there is a function F: X ! Y satisfying FðxÞ¼ðÞFðxÞ for 2 G; x 2 X; (we say then that F is -homogeneous) and such that the difference F f is suitably bounded on X.

2. Notations We begin here with notations we will need in the following. A semigroup ðG; Þ we will call a semigroup with zero if there exists an element 0 2 G such that 0 ¼ 0 ¼ 0 for every 2 G. A semigroup ðG; Þ we will call amonoid(cf. [6]) if there exists 1 2 G such that 1 ¼ 1 ¼ for 2 G.Furthermore,atripleðG; ; 0Þ we will call agroupwithzero if ðG; Þ is a with zero and every element of the set G :¼ Gnf0g is invertible, that is for each g 2 G there is g1 2 G with g g1 ¼ g1 g ¼ 1. A subset A G of a group with zero ðG; ; 0Þ we will call a subgroup of the group G if A ¼;6 and A is a subgroup of the group G. As it is easy to see a set A is a subgroup of the group with zero G if and only if A ¼;6 and 1 2 A for 2 A and 2 A. In the following lemma we list some properties of homomorphisms of groups with zeros. Since they are analogous to similar properties of multiplicative functions on the real line, we omit the proof of them. Lemma 1. Let ðG; ; 0Þ and ðH; ; 0Þ be groups with zeros and let : G ! H be a homomorphism, i.e. satisfies the equation ðÞ¼ðÞðÞ for ; 2 G: Then (i) ð0Þ2f0; 1g; (ii) if ð0Þ¼1 then ¼ 1; (iii) if 6¼ 1 then ð0Þ¼0; (iv) if ð0Þ¼0 for some 0 2 G then ¼ 0; (v) if 6¼ 0 then ð1Þ¼1.

Let ðG; Þ be a semigroup and let ; 6¼ A G.ByhAis we denote the subsemigroup of G generated by A, whereas if ðG; ; 0Þ is a group with zero then by hAi we denote the subgroup of G generated by A. As one() can see we have Yn hAis :¼ i: i 2 A;i 2f1;...;ng;n 2 N ; ()i¼1 Yn "i hAi:¼ i : "i 2f1;1g;i 2 Aif "i ¼1;i 2 A if "i ¼1;n 2 N : i¼1 114 W. Jablonnski and J. Schwaiger

In the case where ðG; ; 0Þ is Abelian, for a set A G, A 6¼;,we 1 have (see [12], Theorem 1, p. 89) hAi¼hAishA is . Then in virtue of Theorem 1 ([12], p. 471), we get Lemma 2. Let ðG; ; 0Þ and ðH; ; 0Þ be Abelian groups with zeros and let A G be a subsemigroup such that A ¼;6 . Assume that : A ! H is a homomorphism such that ðAÞH. Then there exists exactly ~ ~ one homomorphism : hAi!HsuchthatjA ¼ . Finally, we will need the notion of a G-space. We will use here a generalization of the standard notion of a G-space to the semigroup case (cf. [13, I, x 5, pp. 25–33]). Definition 1. Let ðG; Þ be a semigroup and let X be a nonempty set with a fixed element which we will call zero. Assume that we are given a semigroup action on X,thatiswehaveafunction: GX ! X which satisfies the following conditions:

ðg1g2Þx ¼ g1ðg2xÞ for g1; g2 2 G; x 2 X; 1x ¼ x for x 2 X; if 1 2 G: Let moreover g ¼ for g 2 G; 0x ¼ for x 2 X; if 0 2 G: Then the pair ðX; GÞ satisfying these conditions we will call aG- space.

3. The Homogeneity Equation In the following we assume that ðG; Þ and ðH; Þ are semigroups and that ðX; GÞ and ðY; HÞ are G- and H-spaces, respectively. Let ; 6¼ A G and assume that ; 6¼ U X is a set such that AU U (we have then hAisU U). We will consider here solutions : A ! H and F: U ! Y of the -homogeneity equation FðxÞ¼ðÞFðxÞ for 2 A; x 2 U: ð1Þ We will assume additionally that the mapping H 3 7! x 2 Y is injective for every x 2 Y :¼ Ynfg: As one can easily see later, this assumption is crucial only in the case when we will not assume that is a homomorphism. On the other hand, we will use the results of this section in the case when the target Stability of the Homogeneity and Completeness 115 space for the function F is a linear space and then this condition is trivially fulfilled. At first we consider the case when there exists a homomorphism ~ ~ : hAis ! H such that jA ¼ . We have then ~ Lemma 3. Assume that a homomorphism : hAis ! H and a function F: U ! Y satisfy the equation FðxÞ¼~ðÞFðxÞ for 2 A; x 2 U: ð2Þ Then FðxÞ¼~ðÞFðxÞ for 2hAi ; x 2 U: ð3Þ Q s n Proof.Let ¼ i¼1 i 2hAi, i 2 A for i ¼ 1; ...; n with some n 2 N. Then for arbitrary x 2 U, from (2) we get Yn Yn FðxÞ¼F i x ¼ ~ðiÞ FðxÞ i¼1 i¼1 Yn ¼ ~ i FðxÞ¼~ðÞFðxÞ: 7 i¼1 Now we will show that in some cases a function satisfy- ing jointly with F Eq. (1) can be extended to a homomorphism ~ such that the functions ~ and F will satisfy the ~-homogeneity equation (3). Theorem 1. Assume that the functions : A ! HandF: U ! Y satisfy Eq. (1). Then either F ¼ or there exists exactly one ~ ~ homomorphism : hAis ! H such that jA ¼ and the function F satisfies then the ~-homogeneity equation (3).

Proof. Assume that Fðx0Þ 6¼ for some x0 2 U. Fix 2hAis arbitrarily.Q Then there are n 2 N and i 2 A for i ¼ 1; ...; n such that n ¼ i¼1i. Put Yn ~ðÞ :¼ ðiÞ: i¼1 Q Q ~ n m We will show that is well defined. Let ¼ i¼1 i ¼ j¼1 j with some i;j 2 A for i ¼ 1; ...; n, j ¼ 1; ...; m, where n; m 2 N. From (1) we get then

ð1ÞðnÞFðx0Þ¼Fð1 nx0Þ

¼ Fð1 mx0Þ¼ð1ÞðmÞFðx0Þ: 116 W. Jablonnski and J. Schwaiger

The mapping H 3 7! Fðx0Þ2Y is injective, so Yn Ym ðiÞ¼ ðjÞ: i¼1 j¼1 From the definition of ~ weQ have ~j ¼ Q. Moreover ~ is a ho- n A m momorphism. Indeed, for ¼ i¼1i; ¼ j¼1j 2hAis with some i;j 2 A for i ¼ 1; ...; n, j ¼ 1; ...; m,wheren; m 2 N,wehave Yn Ym Yn Ym ~ðÞ¼~ i j ¼ ðiÞ ðjÞ¼~ðÞ~ðÞ: i¼1 j¼1 i¼1 j¼1

Finally we show that ~ is unique. Suppose that ~1; ~2: hAi ! H are ~ ~ s homomorphisms such that 1j ¼ 2j ¼ . Fix Q2hAi . Then there A A n s exist i 2 A for i ¼ 1; ...; n, n 2 N such that ¼ i. Thus i¼1 Yn Yn ~1ðÞ¼~1 i ¼ ~1ðiÞ i¼1 i¼1 Yn Yn Yn ~ ~ ~ ~ ¼ ðiÞ¼ 2ðiÞ¼2 i ¼ 2ðÞ: i¼1 i¼1 i¼1 Hence from (1) we obtain FðxÞ¼~ðÞFðxÞ for 2 A; x 2 U: Then Lemma 3 finishes the proof. & In the following we will assume that ðG; ; 0Þ and ðH; ; 0Þ are Abelian groups with zeros. Moreover let ðX; GÞ and ðY; HÞ be G-andH-spaces, respectively. Let A G be such that A ¼;6 and let U X, AU U. As in the semigroup case we begin with the case when there exists a ~ ~ homomorphism : hAi!H such that jA ¼ .Theninvirtueof Lemma 3, from (1) we get ~ FðxÞ¼ðÞFðxÞ for 2hAis; x 2 U: ð4Þ Assume that ~ ¼ 0. Then from (4) we obtain

FðxÞ¼0 for 2hAis; x 2 U; ð5Þ 1 and then one can easily get FjAU ¼ 0. Now let ~ 6¼ 0. Since ~ is a homomorphism so (cf. Lemma 1 (iv)) ~ðhAiÞ H.

1 In this case we have AU ¼hAisU. Stability of the Homogeneity and Completeness 117

Theorem 2. Assume that a nonzero homomorphism ~: hAi!H and a function F: U ! Y satisfy Eq. (4). Then there exists a unique function F: hAiU ! Y such that FjU ¼ F and FðxÞ¼~ðÞFðxÞ for 2hAi; x 2hAiU: ð6Þ Proof.Fixx 2hAiU.Thenx ¼ u 2hAiU with some 2hAi and u 2 U. Put FðxÞ :¼ ~ðÞFðuÞ:

We will show that F is well defined. Let 1u1 ¼ 2u2 2hAiU. Since 1 hAi¼hAis hA is so there are elements 1;2 2hAis and 1;2 2 1 1 1 hA is such that i ¼ ii for i ¼ 1; 2. Since 11 u1 ¼ 22 u2 so (G is an Abelian group) 1 2u1 ¼ 12u2. From (4) we get then ~ ~ ~ ~ ð1Þð 2ÞFðu1Þ¼Fð1 2u1Þ¼Fð12u2Þ¼ð1Þð2ÞFðu2Þ: Hence we obtain 1 1 ~ð1Þ~ð1Þ Fðu1Þ¼~ð2Þ~ð 2Þ Fðu2Þ: Then using the fact that ~ is a homomorphism we get ~ ~ ~ 1 ~ ~ 1 ð1ÞFðu1Þ¼ð1Þð1Þ Fðu1Þ¼ð2Þð 2Þ Fðu2Þ ~ ¼ ð2ÞFðu2Þ; and therefore the function F is well defined. From the definition of F we obtain that FjU ¼ F (cf. Lemma 1(v)). We will show now that then Eq. (6) is fulfilled. Fix 2hAi and x 2hAiU. Then x ¼ u with 2hAi, u 2 U and 2hAi.Next~ is a homomorphism, so FðxÞ¼FðuÞ¼~ðÞFðuÞ¼~ðÞ~ðÞFðuÞ¼~ðÞFðxÞ:

Finally we will prove that F is unique. Suppose that F1; F2: hAiU ! Y are such that F1jU ¼ F2jU ¼ F and

FiðxÞ¼~ðÞFiðxÞ for 2hAi; x 2hAiU; i ¼ 1; 2: Fix x 2hAiU arbitrarily. Then one can find elements 2hAi and u 2 U such that x ¼ u. Thus

F1ðxÞ¼F1ðuÞ¼~ðÞF1ðuÞ¼~ðÞFðuÞ

¼ ~ðÞF2ðuÞ¼F2ðuÞ¼F2ðxÞ; which finishes the proof. & Finally we obtain 118 W. Jablonnski and J. Schwaiger

Theorem 3. Let : A ! H and F: U ! Y satisfy (1). Assume addi- tionally that ðAÞH. Then either F ¼ or there is a unique homomorphism ~: hAi!H and a unique function F: hAiU ! Y ~ such that A ¼ , FjU ¼ F and FðxÞ¼~ðÞFðxÞ for 2hAi; x 2hAiU: ð7Þ

Proof. Assume that and F satisfy (1). Let x0 2 U be such that Fðx0Þ 6¼ . By Theorem 1 there exists a unique homomorphism 1: hAis ! H such that 1jA ¼ and

FðxÞ¼1ðÞFðxÞ for 2hAis; x 2 U: ð8Þ

It follows from the definition of the homomorphism 1 that 1ðhA i ÞH . Then from Lemma 2 we get that there exists a s ~ ~ unique homomorphism : hAi!H such that jhAi ¼ 1. Theorem 2 then finishes the proof. s & 4. Stability of Homogeneity From now on let Y stand for a locally convex and sequentially complete linear topological Hausdorff space over K 2fR; Cg. Let V Y beanonemptyset.ByaconvV we denote the absolutely convex hull of the set V,byV we denote the smallest balanced superset of V, whereas seq cl V will denote the sequential closure of V. By BðYÞ we denote the family of all bounded subsets of Y. We will need several properties of convex sets and bounded sets, namely

Lemma 4. (i) Let 2 K and V; V1; V2 2BðYÞ. Then V; cl V; conv V; V ; V1 þ V2 2BðYÞ: In particular, seq cl V 2BðYÞ. (ii) The set convðVÞ is the smallest absolutely convex set con- taining the set V, i.e. aconv V ¼ convðVÞ. Thus if V 2BðYÞ, then also aconv V 2BðYÞ. (iii) Let V Y be convex (absolutely convex). Then the set seq cl V is convex (absolutely convex). (iv) If V Y is absolutely convex then for every 2 K we have V ¼jjV. (v) Let V Y and assume that ; 2 K are such that jjjj. Then V jj aconv V. (vi) Assume that V 2BðYÞ, n 0, n ! 0 and xn 2 nV. Then xn ! 0. (For facts about locally convex spaces consult for example [10].) Stability of the Homogeneity and Completeness 119

4.1. The Semigroup Case Let ðG; Þ be a semigroup. By ZðGÞ we denote the center of the semigroup G. Let ðX; GÞ be a G-space. Assume that A G and U X are nonempty sets such that AU U. Let V: A !BðYÞ, and let : hAis !½0; 1Þ be a homomorphism into the multiplicative semigroup ½0; 1Þ and assume that a function K: U ! K satisfies the inequality

jKðxÞj ðÞjKðxÞj for all 2hAis; x 2 U: ð9Þ Definition 2. A function f : U ! Y we will call K-bounded if there exists a bounded set W 2BðYÞ such that f ðxÞ2KðxÞW for every x 2 U. If not, then f will be called K-unbounded. We have the following theorem. Theorem 4. Assume that the functions : A ! K and f : U ! Y satisfy the condition f ðxÞðÞf ðxÞ2KðxÞVðÞ for 2 A; x 2 U; ð10Þ where K satisfies (9). If

A1 :¼f 2 ZðGÞ\A: jðÞj> ðÞg 6¼;; then there exists a unique function F: U ! Y such that FðxÞ¼ðÞFðxÞ for 2 A; x 2 U; and

FðxÞf ðxÞ2KðxÞV0; x 2 U; where \ 1 V :¼ seq cl aconv VðÞ 2BðYÞ: 0 jðÞj ðÞ 2 A1

Proof. Fix 0 2 A1 (then ð0Þ=jð0Þj<1). Thus for m; n 2 N0 and x 2 U we have ðmþnÞ mþn m m ð0Þ f ð0 xÞð0Þ f ð0 xÞ Xn ðmþkÞ mþk1 mþk1 ¼ ð0Þ ½ f ð0 0 xÞð0Þ f ð0 xÞ k¼1 Xn ðmþkÞ mþk1 2 ð0Þ Kð0 xÞVð0Þ: k¼1 120 W. Jablonnski and J. Schwaiger

Hence from (9), in virtue of Lemma 4 we get Xn ðmþkÞ mþk1 ð0Þ Kð0 xÞVð0Þ k¼1 Xn ðmþkÞ mþk1 jð0Þj jKð0 xÞj aconv Vð0Þ k¼1 Xn ðmþkÞ mþk1 jð0Þj ð0 ÞjKðxÞj aconv Vð0Þ k¼1 ð Þ m 1 ð Þ n ¼ 0 KðxÞ 1 0 jð0Þj jð0Þj ð0Þ jð0Þj aconv Vð Þ 0 m ð0Þ 1 KðxÞ aconv Vð0Þ: jð0Þj jð0Þj ð0Þ Thus we obtain ð ÞðmþnÞf ðmþnxÞð Þmf ðmxÞ 0 0 0 0 m ð0Þ 1 2 KðxÞ aconv Vð0Þ: ð11Þ jð0Þj jð0Þj ð0Þ

Hence, because the set aconv Vð0Þ is bounded we have that for n n every x 2 U, ðð0Þ f ð0xÞ: n 2 NÞ is a Cauchy sequence. Then the function F0 : U ! Y, n n F ðxÞ :¼ lim ð0Þ f ð xÞ; 0 n!1 0 is well defined. We will show that F0 satisfies the equation

F0 ðxÞ¼ðÞF0 ðxÞ for 2 A; x 2 U: ð12Þ n Put in (10) 0x in the place of x. Then we get n n n n f ð0xÞðÞf ð0xÞ2Kð0xÞVðÞ ð0Þ KðxÞ aconv VðÞ: n Since ð0Þ 6¼ 0 and 0 2 ZðGÞ (then also 0 2 ZðGÞ), so we obtain n n n n n ð0Þ ð0Þ f ð0xÞðÞð0Þ f ð0xÞ2 KðxÞ aconvVðÞ; jð0Þj and when n tends to infinity and using the fact that the set aconv VðÞ is bounded, from Lemma 4(v) we obtain (12). Stability of the Homogeneity and Completeness 121

From (11), for m ¼ 0 we get

n n 1 ð0Þ f ð0xÞf ðxÞ2 KðxÞ aconv Vð0Þ; jð0Þj ð0Þ and hence 1 F0 ðxÞf ðxÞ2 KðxÞseqclaconvVð0Þ for x 2 U: jð0Þj ð0Þ ð13Þ

Put F :¼ F0 . We will show that F is unique. Indeed, suppose that F1; F2: U ! Y satisfy

FiðxÞ¼ðÞFiðxÞ for 2 A; x 2 U; i ¼ 1; 2; and

FiðxÞf ðxÞ2KðxÞVi; for x 2 U; i ¼ 1; 2 with some sets V1; V2 2BðYÞ. Then we have (ð0Þ 6¼ 0) that n n Fið0xÞ¼ð0Þ FiðxÞ for 2 A; x 2 U; n 2 N; i ¼ 1; 2: Hence for arbitrary x 2 U we get n n n n F1ðxÞF2ðxÞ¼ð0Þ F1ð0xÞð0Þ F2ð0xÞ n n n n ¼ ð0Þ F1ð0xÞð0Þ f ð0xÞ n n n n þ ð0Þ f ð0xÞð0Þ F2ð0xÞ n n n n 2 ð0Þ Kð0xÞV1 ð0Þ Kð0xÞV2 1 n n jKð0xÞjaconvV1 jð0Þ j 1 þ jKðnxÞjaconvV n 0 2 j ð 0Þ j n ð0Þ jKðxÞjðaconvV1 þ aconvV2Þ jð0Þj for every n 2 N. Thus, since aconv V1 þ aconv V2 is bounded (cf. Lemma 4(i), (ii)), in virtue of Lemma 4(vi) we obtain F1ðxÞ F2ðxÞ¼0 for x 2 U. Since 0 2 A1 was arbitrarily fixed, we derive from (13)

FðxÞf ðxÞ2KðxÞV0 for x 2 U; 122 W. Jablonnski and J. Schwaiger where \ 1 V :¼ seq cl aconv VðÞ 2BðYÞ: 7 0 jðÞj ðÞ 2 A1 The following simply example shows that the estimation obtained in Theorem 4 is the best one. Example 1. Let f : R ! R be given by 8 < x þ 2 for x 2h2i; f ðxÞ¼: x 2 for x 2h2i; 0 for x 2 Rnhf2; 2gi: If we consider suitable cases one can check that the assumptions of Theorem 4 are fulfilled with A ¼f2; 2g, U ¼ R, ¼ idA, K ¼ 1, ¼ 1, VðÞ¼½2; 2 for 2f2; 2g. Next one can verify that the function F: R ! R, x for x 2 hf2; 2gi; FðxÞ¼ 0 for x 2 Rnhf2; 2gi satisfies FðxÞ¼FðxÞ for 2 hf2; 2gi [ f0g; x 2 R; and moreover 8 < 2 for x 2h2i; FðxÞf ðxÞ¼: 2 for x 2h2i; 0 for x 2 Rnhf2; 2gi:

On the other hand A1 ¼f2; 2g,so \ 1 V ¼ seq cl aconv ½2; 2¼½2; 2: 0 jj1 2 A1 Since FðxÞf ðxÞ2f2; 0; 2g; the estimation obtained in Theorem 4 is the best one. As a corollary, from Theorem 4 and Theorem 1 we obtain the following result. Corollary 1. Assume that the functions : A ! K and f : U ! Y satisfy condition (10). Let A1 :¼f 2 ZðGÞ\A: jðÞj> ðÞg 6¼; and assume that f is K-unbounded. Stability of the Homogeneity and Completeness 123

Then there exists a unique homomorphism ~: hAi ! K and there ~ s exists a unique function F: U ! Y such that jA ¼ , ~ FðxÞ¼ðÞFðxÞ for 2hAis; x 2 U; ð14Þ and

FðxÞf ðxÞ2KðxÞV0 for x 2 U; where \ 1 V :¼ seq cl aconv VðÞ 2BðYÞ: 0 jðÞj ðÞ 2 A1 Proof. From Theorem 4 we derive the existence of a unique function F: U ! Y and the existence of the bounded set \ 1 V :¼ seq cl aconv VðÞ 2BðYÞ; 0 jðÞj ðÞ 2 A1 such that FðxÞ¼ðÞFðxÞ for 2 A; x 2 U; ð15Þ and

FðxÞf ðxÞ2KðxÞV0; x 2 U: ð16Þ

Suppose that F ¼ 0. Then from (16) we get f ðxÞ2KðxÞ aconv V0 for every x 2 U, which is in contradiction to the assumption that f is K- unbounded. Thus F 6¼ 0. It follows from Theorem 1 for Eq. (15) that ~ ~ there exists a unique homomorphism : hAis ! K, jA ¼ such that the function F is ~-homogeneous, i.e. that Eq. (14) holds. & Remark 1. Note that in Corollary 1 the assumption that f is K- unbounded is essential. Indeed, if there exists a set W 2BðYÞ such that f ðxÞ2KðxÞW for x 2 U then 0 f ðxÞ2KðxÞ aconv W for x 2 U:

Clearly the function F1: U ! Y, F1ðxÞ¼ is a -homogeneous one. On the other hand, in the proof of Theorem 4 we have shown the existence of a unique -homogeneous function F regardless of the estimation of the difference F f .ThusF ¼ F1 ¼ 0. It is rather difficult to prove any property of such a function in this case (let us recall that in the case when F ¼6 0, there exists a unique homomorphism ~ such that ~ ). jA1 ¼ 124 W. Jablonnski and J. Schwaiger

4.2. The Group Case From now on we will assume that ðG; ; 0Þ is a group with zero and that ðX; GÞ is a G-space. Then directly from Theorem 4 and Theorem 3 we deduce the following corollary (we will omit the obvious proof). Corollary 2. Assume that ðG; ; 0Þ is an Abelian group, A G, A 6¼;and let the functions : A ! K, ðAÞK and f : U ! Y satisfy (10). Let A1 :¼f 2 A: jðÞj> ðÞg 6¼;and assume that f is a K-unbounded function. Then there exists a unique homomorphism ~: hAi!K and a ~ unique function F: hAiU ! Y such that jA ¼ , FjU ¼ F, FðxÞ¼~ðÞFðxÞ for 2hAi; x 2hAiU; and

FðxÞf ðxÞ2KðxÞV0 for x 2 U; where \ 1 V :¼ seq cl aconv VðÞ 2BðYÞ: 0 jðÞj ðÞ 2 A1

In previous theorems we have assumed that the set A1 is nonempty. It appears that now, when ðG; ; 0Þ is a group, we can consider a weaker condition, that is we may take ¼6 in a place of >.Butwemustassume that A G and U X are sets such that A ¼;6 , hAiU U, : hAi! ½0; 1Þ is a nonzero homomorphism and the function K satisfies the inequality jKðxÞj ðÞjKðxÞj for 2hAi; x 2 U: ð17Þ We prove the following theorem, where we use the convention that intersections with an empty index set should be equal to Y. Theorem 5. Assume that : A ! K, ðAÞK and f : U ! Y satisfy the condition f ðxÞðÞf ðxÞ2KðxÞVðÞ for 2 A; x 2 U: ð18Þ Put A1 :¼f 2 ZðGÞ\A : jðÞj> ðÞg and A2 :¼f 2 ZðGÞ\ A: jðÞj< ðÞg. If A1 [ A2 6¼;then there exists a unique function F: U ! Y such that FðxÞ¼ðÞFðxÞ for 2 Ae; x 2 U ð19Þ Stability of the Homogeneity and Completeness 125 and

FðxÞf ðxÞ2KðxÞV0; x 2 U; e e where A ¼ A provided A1 6¼;and A ¼ A in the case when A1 ¼;, and \ 1 V :¼ seq cl aconv VðÞ 0 jðÞj ðÞ 2 A1 \ 1 \ seq cl aconv VðÞ 2BðYÞ: ðÞjðÞj 2 A2

Proof. Assume that A1 6¼;. Then by Theorem 4 there exists a unique function F1: U ! Y such that

F1ðxÞ¼ðÞF1ðxÞ for 2 A; x 2 U ð20Þ and

F1ðxÞf ðxÞ2KðxÞV1; x 2 U; ð21Þ where \ 1 V :¼ seq cl aconv VðÞ : 1 jðÞj ðÞ 2 A1 1 Now let A2 6¼;and fix 2 A and x 2 U. Put x in the place of x in (18). Then from Lemma 4(iv), (v) we get f ðxÞðÞf ð1xÞ2Kð1xÞVðÞjKð1xÞj aconv VðÞ ðÞ1jKðxÞj aconv VðÞ ¼ ðÞ1KðxÞ aconv VðÞ: Thus, since 2 A and x 2 U were arbitrarily fixed f ð1xÞðÞ1f ðxÞ2ððÞ ðÞÞ1KðxÞ aconv VðÞ; 2 A; x 2 U; and hence f ðxÞð1Þ1f ðxÞ2ðjð1Þj ð1ÞÞ1KðxÞ aconv Vð1Þ for 2ðAÞ1; x 2 U: Put ðÞ :¼ ð1Þ1 and VeðÞ :¼jð1Þj1 ðÞ aconv Vð1Þ 126 W. Jablonnski and J. Schwaiger for 2ðAÞ1. Then we get f ðxÞðÞf ðxÞ2KðxÞVeðÞ for 2ðAÞ1; x 2 U: ð22Þ Note that ðAÞ1U U.Wehave f 2 ZðGÞ\ðAÞ1: jðÞj> ðÞg ¼f 2 ZðGÞ\A: jð1Þj> ð1Þg1 ¼f 2 ZðGÞ\A: jðÞj1 > ðÞ1g1 1 1 ¼f 2 ZðGÞ\A : jðÞj< ðÞg ¼ A2 6¼;: If we apply Theorem 4 to the condition (22) then we obtain the existence of a unique function F2: U ! Y such that 1 F2ðxÞ¼ðÞF2ðxÞ for 2ðA Þ ; x 2 U and

F2ðxÞf ðxÞ2KðxÞV2; x 2 U; where \ 1 e V2 :¼ seq cl aconv VðÞ : 1 jðÞj ðÞ 2ðA2Þ Using our notation we get 1 1 1 F2ðxÞ¼ð Þ F2ðxÞ for 2ðA Þ ; x 2 U ð23Þ and

F2ðxÞf ðxÞ2KðxÞV2; x 2 U; ð24Þ where \ 1 1 1 V2 :¼ jð Þj ðÞ seqclaconvVðÞ 1 1 1 jð Þj ðÞ 2ðA2Þ \ 1 ¼ seqclaconvVðÞ : ðÞjðÞj 2 A2 Then, from (23) we obtain 1 F2ðxÞ¼ðÞF2ð xÞ for 2 A ; x 2 U and then, if we put x in the place of x we get F2ðxÞ¼ðÞF2ðxÞ for 2 A ; x 2 U: ð25Þ Stability of the Homogeneity and Completeness 127

We are going to show that F1 ¼ F2 (provided A1 6¼;and A2 6¼;). From (20) and (25) we have n n Fið xÞ¼ðÞ F1ðxÞ for 2 A ; x 2 U; n 2 Z; i ¼ 1;2: ð26Þ

Fix x 2 U and 2 A1, 2 A2 (then we have ðÞ;ðÞ ¼6 0). From (26), (21) and (24), for every n 2 N we obtain F ðxÞF ðxÞ¼ðÞnF ðnxÞðÞnF ðnxÞ 2 1 2 1 ðÞ n ¼ ½ðÞnF ðnxÞðÞnF ðnxÞ 2 1 ð Þ ðÞ n ¼ ½F ðnnxÞf ðnnxÞ ðÞ 2 þ f ðnnxÞF ðnnxÞ 1 jðÞj ðÞ n 2 KðxÞðaconvV þ aconvV Þ: ðÞ jðÞj 2 1 Since jðÞj= ðÞ, ðÞ=jðÞj<1, and with n tending to infinity, and also using the fact that the set aconv V2 þ aconv V1 is bounded (cf. Lemma 4(i), (ii)), we get from Lemma 4(vi) F1ðxÞ¼F2ðxÞ. Put F ¼ F1 ¼ F2. From (20) and (25) we then obtain (19). Moreover, using (21) and (24) we get

FðxÞf ðxÞ2KðxÞðV1 \ V2Þ:

Clearly V1 \ V2 2BðYÞ. The uniqueness of F follows from the fact that F1 and F2 were the unique functions satisfying (20), (21), and (24), (25), respectively, while the function F satisfies all these conditions. & From Theorems 5 and 3 one can derive the following corollary. Corollary 3. Let ðG; ; 0Þ be an Abelian group with zero and assume that the functions : A ! K, ðAÞH and f : U ! Y satisfy (18). Let A1 ¼f 2 A : jðÞj> ðÞg and A2 :¼f 2 A : jðÞj< ðÞg. If A1 [ A2 6¼;and f is a K-unbounded function then there exists a unique homomorphism ~: hAei!K and a unique function F: hAeiU ! Y ~ such that jA~ ¼ jA~, FjU ¼ F, FðxÞ¼~ðÞFðxÞ for 2hAei; x 2 U; and

FðxÞf ðxÞ2KðxÞ V0 for x 2 U; 128 W. Jablonnski and J. Schwaiger where Ae ¼ A provided A 6¼;, Ae ¼ A when A ¼;, and 1 1 \ 1 V :¼ seq cl aconv VðÞ 0 jðÞj ðÞ 2 A1 \ 1 \ seq cl aconv VðÞ 2BðYÞ: ðÞjðÞj 2 A2 Remark 2. Note that in Corollaries 1, 2, 3 the assumption that f is K- unbounded is a crucial only in the case when we do not know whether ~ ~ there exists a homomorphism such that jA ¼ . In the case when such a homomorphism exists those corollaries remain true without the assumption on the K-unboundedness of f . Corollary 4 (cf. [1], [15]). Let X be a real linear space and let Y be a Banach space. Fix p 0 and assume that a function f : X ! Ysatisfies k f ðxÞf ðxÞk "jjp for 2 R; x 2 X; with some " 0. Then f is a homogeneous function. Proof.Put ¼ id , ¼ 1, K ¼ 1, VðÞ¼fy 2 Y: kyk"jjpg, R A1 ¼f 2 R: jj>1g,andA2 ¼f 2 R : jj<1g.Thenfrom Corollary 3 and from Remark 2 we obtain that there exists a unique homogeneous function F: X ! Y such that

FðxÞf ðxÞ2V0 for x 2 X; where \ 1 V ¼ seq cl aconv VðÞ 0 jj1 2 A1 \ 1 \ seq cl aconv VðÞ : 1 jj 2 A2

Since for p ¼ 0 we have inf 2 A 1=ðjj1Þ¼0, whereas for p>0, p 1 inf 2 A2 jj =ð1 jjÞ ¼ 0; this implies V0 ¼f0g. Thus f ¼ F. & We have also the following Corollary 5. Let ðG; ; 0Þ be a group with zero and let ðX; GÞ be a G-space. Assume that Y is a Banach space and let : G !½0; 1Þ be given. Let a nonzero homomorphism : A ! K and a function f : X ! Y satisfy the condition k f ðxÞðÞ f ðxÞk ðÞ for 2 G; x 2 X: Put A1 :¼f 2 ZðGÞ : jðÞj>1g and A2 :¼f 2 ZðGÞ : jðÞj<1g. Stability of the Homogeneity and Completeness 129

If A1 [ A2 6¼;then there exists a unique -homogeneous function F: U ! Y such that kFðxÞf ðxÞk C for x 2 U; where 1 1 C :¼ min inf ; inf : 2 A1 jðÞj 1 2 A2 1 jðÞj

5. Completeness In [5] it was shown that a normed space Y has to be a Banach space provided that for some Abelian group A containing an element of infinite order and for all functions f : A ! Y such that the Cauchy difference f ðx þ yÞf ðxÞf ðyÞ is bounded, there is some additive function h such that f h is bounded. (For a survey on the original stability question, whether for some f as above there is some additive function h, such that f h is bounded, see, for example, [3].) The aim of this section is to show a similar result for the -homogeneity equation, namely we have Theorem 6. Let A be a group which is isomorphic to H A0 with some subgroup H of K (2fR; Cg) such that H contains some element z0 of modulus different from 1. Assume that there is an action : AX ! X of A on some set X and assume that the stabilizer Ax0 :¼f 2 A: x0 ¼ x0g is trivial for some x0 2 X. Furthermore let Y be a normed space. Assume furthermore that for all functions f : X ! Y and all ho- momorphisms ’: A ! K :¼ Rnf0g such that for some positive "; k f ðxÞ’ðÞ f ðxÞk "j’ðÞj þ for all 2 A and x 2 X there is some h: X ! Y being ’-homogeneous – which means that hðxÞ¼’ðÞhðxÞ for all 2 A and x 2 X – such that f h is bounded. Then Y is a Banach space, i.e., a complete normed space.

Proof. Let z0 2 H and (without loss of generality) jz0j¼: r>1. Let ðynÞn 2 N be a Cauchy sequence in Y. Since it is enough to show that this sequence contains a convergent subsequence we may – again without loss of generality – assume that ky y krn for all n; m 2 N: ð27Þ nþm n Then we define f : X ! Y in the following way. We put f Ax :¼ 0if X 3 x ¼6 x0.(Ax :¼ AðxÞ :¼fx j 2 Ag is the orbit of x with respect to 130 W. Jablonnski and J. Schwaiger

0 the action of A on X.) On Ax0 we define f ðx0Þ for ¼ðz;Þ2 A ¼ H A0 as zy if jzj1 and rn jzj0 k f ððz;0ÞxÞzfðxÞk jzj" þ ð28Þ for all ðz;0Þ2A and all x 2 X. This is obvious if x 2= Ax0 because in this case the left-hand side of 0 (28) vanishes. So let us assume that x ¼ðz;Þx0 2 Ax0 and take any 0 ðz1;1Þ2A. We have to consider several cases.

Case 1. Let jzj; jz1j1. Then there are nonnegative m and n nþ1 m mþ1 nþm n such that r jzj

Case 2.Ifjzj1 and jz1j<1 we have to divide the consideration into two subcases. 0 (a) If jzz1j<1wehavef ððz1;ÞxÞ¼0 and f ðxÞ¼zyn, implying 0 k f ððz1;ÞxÞz1 f ðxÞk ¼ jz1jjzjkynkM :¼ sup kykk: k 2 N m mþ1 (b) If jzz1j1wehaver jzz1j

Case 3. In the case jzj<1, jz1j1 we again have two subcases. 0 (a) jzz1j<1 implies the desired estimate since then f ððz1;1ÞxÞ¼ f ðxÞ¼0. m mþ1 n (b) If jzz1j1 let m 2 N0 be such that r jz1j

Case 4. The last case jzj; jz1j<1 is trivial since here as in the first subcase of the previous case all interesting terms vanish. If we put " :¼ r and :¼ maxfM; rg we see that the hypotheses of the theorem are satisfied. Thus there is some ’-homogeneous h: X ! Y and some constant N such that k f ðxÞhðxÞk N for all n n x 2 X.Puttingxn :¼ðz0; 1Þ x0 we have f ðxnÞ¼z0yn and hðxnÞ¼ n z0hðx0Þ.Thisimplies n k f ðxnÞhðxnÞk ¼ jz0j kyn hðx0Þk N: n n Dividing by r ¼jz0j and letting n tending to infinity then shows that yn tends to hðx0Þ2Y, the desired result. &

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Authors’ addresses: Wojciech Jablonnski, Department of Mathematics, University of Rzeszow, Rejtana 16 A, 35-310 Rzeszoow, Poland. E-Mail: [email protected]. Jens Schwaiger, Institute of Mathematics, Karl-Franzens-University Graz, Heinrich- straße 36, 8010 Graz, Austria. E-Mail: [email protected].