# Hyperstability of Some Functional Equations on Restricted Domain

Authors: Anna Bahyrycz; Jolanta Olko
Journal: Journal of Function Spaces
 (2017)

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## Abstract

The paper concerns functions which approximately satisfy, not necessarily on the whole linear space, a generalization of linear functional equation. A Hyers-Ulam stability result is proved and next applied to give conditions implying the hyperstability of the equation. The results may be used as tools in stability studies on restricted domains for various functional equations. We use the main theorem to obtain a few hyperstability results of Fréchet equation on restricted domain for different control functions.


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## Body

## 1. Introduction

The famous Cauchy equation(1)fx+y=fx+fyhas a very long history, which started with A. M. Legendre in 1791 and C. F. Gauss in 1809, but the first important result about the solutions of (1) was proved by Cauchy (see [1]) and due to this outcome (1) is today named after him, as is mentioned in [2]. In the meantime, many authors have dealt with this equation, and next Fréchet in [3] studied an important generalization of Cauchy equation, characterizing the polynomials among the continuous mappings, namely, the equation(2)Δh1⋯hnnfx=0,where Δ denotes the difference operator defined as usual by (3)Δhfx=Δh1fx≔fx+h-fx,Δh1⋯hn+1n+1fx=Δhn+1Δh1⋯hnnfx,for x,h,h1,…,hn+1∈R,n∈N. From the result of Fréchet it follows that a continuous function f:R→R satisfies (2) for all x and hj if and only if f is a polynomial of degree smaller than or equal to n-1 (cf. [2, 4]). As a generalization we can consider the functional equation(4)∑i=1mAig∑j=1naijxj+A=0,where g:X→Y and X,Y are the linear spaces over a field F∈{R,C}, A,aij∈F, and Ai∈F∖{0},i∈{1,…,m},j∈{1,…,n}.The stability and hyperstability of the above functional equation and its particular cases were studied by many mathematicians (cf., e.g., [5–16]). Let us recall (see [9]) that the theory of Hyers-Ulam stability, in general, was motivated by a problem raised in 1940 by Ulam [17] and a partial answer to it was provided by Hyers [18]. The termhyperstability, in the meaning used here, was introduced presumably in [19] (cf. also [20]).Our basic assumptions are the following.X is a normed space over a field F∈{R,C}, E1,E2 are the nonempty subsets of X such that E2⊆E1⊆X, and D is a nonempty subset of F such that 1∈D,R+≔[0,+∞). Denote N≤k≔{1,…,k} and Nk≔{l∈N:l≥k}, for k∈N, and a sum of numbers over an empty set is defined to be zero. In the sequel, P(A)={B⊂A:A≠∅} for nonempty set A.We say that the equation(5)∑i=1mAig∑j=1naijxj+A=0,x1,…,xn∈E2,∑j=1na1jxj,…,∑j=1namjxj∈E1is θ-hyperstable in the class of functions g:E1→Y (with a control function θ:E2n→R+), if g:E1→Y satisfies the inequality(6)∑i=1mAig∑j=1naijxj+A≤θx1,…,xn,for all x1,…,xn∈E2, and ∑j=1na1jxj,…,∑j=1namjxj∈E1 fulfills (5).The above notion of hyperstability for functions defined on subsets of a linear spaceX (not necessarily on the whole space but on the restricted domain) is a counterpart of hyperstability considered in [8]. In this paper we obtain, applying the fixed point theorem, some results concerning stability and hyperstability for (4) on restricted domain. Our outcomes allow studying the behavior of functions approximately fulfilling functional equations of this linear type on different subsets of the linear space, and this also applies to domain restrictions for control functions. This way we give a tool for the further Hyers-Ulam stability research. Its application is presented in the last section where some new as well as generalizations of known results on hyperstability of Cauchy and Fréchet equation are obtained.


## 2. Auxiliary Result

First we establish a counterpart of stability result from [7] on restricted domain, and to do this we use the fixed point theorem. This result will be needed to show our main theorem which provides criteria for hyperstability of many functional equations of the linear type on a restricted domain.Lemma 1.
Assume thatY is a Banach space, A=0 or “A≠0 and ∑i=1mAi≠0,” and ω:D→R+. Suppose that there exist ∅≠I⊂N≤m and c1,…,cn∈D such that(7)βi≔∑j=1naijcj∈D,cjx∈E2,βix∈E2,i∈N≤m,j∈N≤n,x∈E2,βl=1,l∈I,AI≔∑l∈IAl≠0,γ≔∑i∉IAiAIωβi<1. Let θ:E2n→R+ satisfy the inequality(8)θβix1,…,βixn≤ωβiθx1,…,xn,i∈N≤m,x1,…,xn∈E2.If g:E1→Y fulfills the estimation(9)∑i=1mAig∑j=1naijxj+A≤θx1,…,xn,x1,…,xn∈E2,∑j=1na1jxj,…,∑j=1namjxj∈E1,then there exists a unique function G:E2→Y satisfying (5) such that(10)gx-Gx≤θc1x,…,cnxAI1-γ,x∈E2.Proof.
Take∅≠I⊂N≤m and c1,…,cn∈D such that assumptions (7) and (8) hold.
First consider the caseA=0. Taking x∈E2 and substituting xj=cjx, j∈N≤n, in (9) we have(11)gx-∑i∉I-AiAIgβix≤θc1x,…,cnxAI,x∈E2.Define(12)Tξx≔∑i∉I-AiAIξβix,ξ∈YE2,x∈E2,(13)Λδx≔∑i∉I-AiAIδβix,δ∈R+E2,x∈E2,(14)εx≔θc1x,…,cnxAI,x∈E2.Then (11) may be rewritten as follows:(15)gx-Tgx≤εx,x∈E2.By induction, for every x∈E2 and k∈N0(16)Λkεx≤εxγk.Since γ<1, we see that(17)ε∗x≔∑n=0∞Λnεx≤εx∑n=0∞γn=εx1-γ,x∈E2.Notice that all assumptions of Theorem 1 in [21] are satisfied for operators (12) and (13) and X=E2. According to this fixed point theorem, the function G:E2→Y given by G(x)=limn→∞(Tng)(x) for x∈E2 is a fixed point of T fulfilling (10).
It is sufficient to show thatG satisfies (5) (with A=0). First we prove that for every k∈N∪{0} and every x1,…,xn∈E2 such that ∑j=1na1jxj,…,∑j=1namjxj∈E1(18)∑i=1mAiTkg∑j=1naijxj≤γkθx1,…,xn.Obviously, the case k=0 is just (9). Next, fix k∈N∪{0}, x1,…,xn∈E2, and ∑j=1na1jxj,…,∑j=1namjxj∈E1 and assume that (18) holds. Then (19)∑i=1mAiTk+1g∑j=1naijxj=∑i=1mAiTTkg∑j=1naijxj=∑i=1mAi∑p∉I-ApAITkgβp∑j=1naijxj=∑p∉I-ApAI∑i=1mAiTkg∑j=1naijβpxj.Thus by induction hypothesis and (8) we have (20)∑i=1mAiTk+1g∑j=1naijxj≤∑p∉I-ApAIγkθβpx1,…,βpxn≤∑p∉I-ApAIγkωβpθx1,…,xn=γk+1θx1,…,xn,which finishes the proof of (18). Letting k→∞ in (18), we prove that G satisfies (5), which completes the proof in the first case.
In the case whereA≠0 and ∑i=1mAi≠0, the function h(x)≔g(x)+A/∑i=1mAi, x∈E1, satisfies(21)∑i=1mAih∑j=1naijxj≤θx1,…,xn,for x1,…,xn∈E2, and ∑j=1na1jxj,…,∑j=1namjxj∈E1. Consequently, we conclude that there exists a unique function H:E2→Y such that(22)hx-Hx≤θc1x,…,cnxAI1-γ,x∈E2.Obviously G(x)≔H(x)-A/∑i=1mAi and x∈E2 is the desired function.Denote(23)X0≔X∖0,F0≔F∖0.Remark 2.
It is easily seen that functions(i)
θ1(x1,…,xn)=C;(ii)
θ2(x1,…,xn)=C∑j=1ncjxjkj;(iii)
θ3x1,…,xn=Cmax{cjxjkj:j∈{1,…,n}};(iv)
θ4(x1,…,xn)=C⊓j=1nxjkj;(v)
θ5(x1,…,xn)=C⊓j=1nxjtj+S∑j=1ncjxjkj;with some C,S∈[0,+∞), C+S>0, cj∈F0, and kj,tj≠0 fulfill condition (8) for some function ω. For instance, in case (iv) we can use ω4(β)=|β|∑j=1nkj for β∈D, and in case (v) we may take (24)ω5β≔βmax⁡k1,…,kn,tforβ≥1βmin⁡k1,…,kn,tforβ<1,where t=∑j=1ntj. If there exists at least one j0∈N≤n such that kj0<0 or tj0<0 then 0∉D.


## 3. The Main Result

We can now state a counterpart (on restricted domain) of Theorem2.1 in [8], concerning θ-hyperstability of (5).Theorem 3.
LetY be a normed space over F, A=0 or “A≠0 and ∑i=1mAi≠0,” g:E1→Y, ω:D→R+, and θ:E2n→R+ satisfy (9). If there exist ∅≠I⊂N≤m and a sequence {(c1k,…,cnk)}k∈N of elements of Dn such that(25)βik≔∑j=1naijcjk∈D,cjkx∈E2,βikx∈E2,i∈N≤m,j∈N≤n,k∈N,x∈E2,(26)βlk=1,l∈I,AI≔∑l∈IAl≠0,limk→∞∑i∉IAiAIωβik<1,(27)θβikx1,…,βikxn≤ωβikθx1,…,xni∈N≤m,k∈N,x1,…,xn∈E2,(28)limk→∞θc1kx,…,cnkx=0,then g satisfies (5).Proof.
We can certainly assume thatY is a Banach space, for if not we replace it by its completion.
Take∅≠I⊂N≤m and the sequence {(c1k,…,cnk)}k∈N of the elements of Dn such that conditions (25), (26), and (27) hold. Hence we can find ε∈(0,1) and k0∈N such that(29)γk≔∑i∉IAiAIωβik<1-ε,k≥k0.Therefore, according to Lemma 1, for each k∈Nk0 there exists a unique function Gk:E2→Y satisfying equation (5) such that(30)gx-Gkx≤θc1kx,…,cnkxAI1-γk,x∈E2.In this way we obtain the sequence {Gk}k∈Nk0 of functions satisfying (5). Letting in (30) k→∞, we get, using (28), that g satisfies (5).


## 4. Applications

Using the above outcome we can get the results for the particular form of (5), as, for instance, for Fréchet functional equation, presented in this section.Corollary 4.
Assume thatY is a normed space over F, n∈N2, E is a nonempty subset of X0 such that there exists a positive integer l0 with(31)-x,lx∈E,x∈E,l∈Nl0.If g:E→Y fulfills the inequality(32)Δx2⋯xnn-1gx1≤C⊓j=1nxjkj,∑j∈Txj∈E,T∈PN≤n,with C>0 and ∑j=1nkj<0, then g is a solution of the equation(33)Δx2⋯xnn-1gx1=0,∑j∈Txj∈E,T∈PN≤n.Proof.
PutE≔E2=E1,D≔Z0, and ω(β)≔|β|∑j=1nkj for β∈Z0 and consider two cases: (a)
∑j=2nkj<0,(b)
∑j=2nkj≥0.In the case (a) we define the sequence {(c1k,…,cnk)}k∈N as follows: (34)cjk≔k+l0forj≠11forj=1.It follows easily that the assumptions of Theorem 3 are satisfied with m=2n and I={2n}.
In case (b) we can findj0∈{2,…,n} such that kj0≥0. Define (35)cjk≔k+l0forj≠j01-k-l0forj=j0.There exists i0∈{1,…,2n} such that ai01=ai0j0=1 and ai0j=0 for j∈{2,…,n}∖{1,j0}. We will verify that we can apply Theorem 3 for I={i0}. Namely, (36)∑j=1nai0jcjk=k+l0+1-k-l0=1,k+l0≤∑j=1naijcjk≤k+l0n-1fori≠i0.Therefore (37)limk→∞∑i≠i0∑j=1naijcjk∑j=1nkj=0,limk→∞C⊓j=1ncjkxkj=limk→∞k+l0∑j≠j0kjk+l0-1kj0x∑j=1nkj≤limk→∞k+l0∑j=1nkjx∑j=1nkj=0.Thus conditions (26) and (28) hold, which finishes the proof.Corollary 5.
Assume thatY is a normed space over F, n∈N2, and E is a nonempty subset of X0 such that there exists a positive integer l0 with (31).
Ifg:E→Y fulfills the inequality(38)Δx2⋯xnn-1gx1≤∑j=1nCjxjkjw,∑j∈Txj∈E,T∈PN≤n,with Cj>0 and wkj<0, for j∈{1,…,n}, then g is a solution of (33).Proof.
PuttingE≔E2=E1, D≔Z0, and ω(β)≔|β|-v|w| for β∈Z0, where v≔min⁡{|k1|,…,|kn|}, I≔{1}, and(39)cjk≔n-1k+l0+1forj=1-k-l0forj>1,we get that (40)β1k=1,k+l0+1≤βik≤n-1k+l0+1fori>1,and therefore limk→∞|βik|=+∞ for i≠1, and consequently condition (26) is satisfied. Also, condition (28) holds, because (41)limk→∞C1n-1k+l0+1xk1+∑j=2nCjk+l0xkjw=0,which with Theorem 3 completes the proof.Denote(42)Ff,x1,…,xn=Δx1⋯xnnf0+-1n+1f0,f∈YX,x1,…,xn∈X.Corollary 6.
Assume thatY is a normed space over F, n∈N2, and E is a nonempty subset of X0 such that there exists a positive integer l0 with(43)lx∈E,x∈E,l∈Nl0.If g:E→Y fulfills the inequality(44)Fg,x1,…,xn≤C⊓j=1nxjkj,∑j∈Txj∈E,T∈PN≤n,with C>0 and ∑j=1nkj<0, then g is a solution of the equation(45)Fg,x1,…,xn=0,∑j∈Txj∈E,T∈PN≤n.Proof.
PutE≔E2=E1, D≔Z0, and ω(β)≔|β|∑j=1nkj for β∈Z0. There exist j0∈{1,…,n} and i0∈{1,…,2n-1}, such that ∑j≠j0nkj<0, ai0j0=1, and ai0j=0 for j≠j0. Define the sequence {(c1k,…,cnk)}k∈N as follows: (46)cjk≔k+l0forj≠j01forj=j0.The assumptions of Theorem 3 are fulfilled with m=2n-1 and I={i0}, as is easy to check, which finishes the proof.In the same way as Corollary5 we can prove the following corollary.Corollary 7.
Assume thatY is a normed space over F, n∈N2, and E is a nonempty subset of X0 such that there exists a positive integer l0 with (31).
Ifg:E→Y fulfills the inequality(47)Fg,x1,…,xn≤∑j=1nCjxjkjw,∑j∈Txj∈E,T∈PN≤nwith Cj>0 and wkj<0 for j∈{1,…,n}, then g is a solution of (45).Remark 8.
Observe that putting in the above two corollariesn=2 we get hyperstability results for Cauchy equation. Corollary 7 generalizes the result from [9] (see Theorem 1.2), where n=2 and k1=k2<0 (in our considerations k1,k2<0 may be different). Corollaries 6 and 7 generalize the outcomes from [6].We end the paper with an example, which shows that the additional assumption onE, appearing in the above corollaries, is essential.Example 9.
LetX=Y=R, E=[-1,0)∪(0,1], and f:E→R be defined by f(x)=sin⁡x/2n, x∈E. Then f satisfies the estimations (48)Δx2⋯xnn-1fx1<1≤⊓j=1nxj-1,Ff,x1,…,xn<1≤⊓j=1nxj-1,Δx2⋯xnn-1fx1<1≤∑j=1nxj-1,Ff,x1,…,xn<1≤∑j=1nxj-1,for ∑j∈Txj∈E, and T∈P(N≤n) but f satisfies neither (33) nor (45).



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