# The  Ulam Type Stability of a Generalized Additive Mapping and Concrete Examples

Authors: Hiroyoshi Oda; Makoto Tsukada; Takeshi Miura; Yuji Kobayashi; Sin-Ei Takahasi
Journal: International Journal of Mathematics and Mathematical Sciences
 (2013)

---

## Abstract

We give an Ulam type stability result for the following functional equation:f(αx−αx′+x0)=βf(x)−βf(x′)+y0(for allx,x′∈X)
under a suitable condition. We also give a concrete stability result for the case taking up δ∥x∥p∥x′∥q as a control function.


---

## Body

## 1. Introduction

In 1940, Ulam [1] proposed the following stability problem: “When is it true that a function which satisfies some functional equation approximately must be close to one satisfying the equation exactly?” Next year, Hyers [2] gave an answer to this problem for additive mappings between Banach spaces. Furthermore, Aoki [3] and Rassias [4] obtained independently generalized results of Hyers’ theorem which allow the Cauchy difference to be unbounded.LetX and Y be normed spaces over 𝕂, which denotes either the real field ℝ or the complex field ℂ. Throughout the paper, we fix scalars a,b,c,d∈𝕂∖{0} and vectors x0∈X and y0∈Y. We say that a mapping f of X into Y is (a,b,c,d;x0,y0)-additive if
(1)f(ax+bx′+x0)=cf(x)+df(x′)+y0
for all x,x′∈X. When x0=y0=0, we say it to be (a,b,c,d)-additive. Aczél [5] specified what this generalized Cauchy equation is. The Ulam type stability problem for such an f has been investigated in [6–8]. However, these results have been obtained in cases where either a+b≠0 or c+d≠0 (see Theorems A and B). In this paper, we will investigate the problem for (a,-a,c,-c;x0,y0)-additive mappings, that is, in the case a+b=c+d=0. In Section 2, we state the details of (a,-a,c,-c;x0,y0)-additive mappings (Theorem 3). In Section 3,  we give our main results about the stability for them (see Theorems 7–10). In the final section, we apply the results to some concrete examples, where we take up δ∥x∥p∥x′∥q as a control function ε(x,x′) (see Corollaries 11–14).


## 2.(a,-a,c,-c;x0,y0)-Additive Mappings

The following result asserts that any(a,-a,c,-c;x0,y0)-additive mapping is transformed into some (a,-a,c,-c)-additive mapping by a certain translation and that any (a,-a,c,-c)-additive mapping is an additive mapping in usual sense with some extra condition.Proposition 1.
Letf and g be two mappings of X into Y such that g(x)=f(x+x0)-y0 for all x∈X. Then the following three statements are equivalent: (i)
f  is(a,-a,c,-c;x0,y0)-additive,(ii)
g  is(a,-a,c,-c)-additive,(iii)
g  is additive andg(ax)=cg(x)   for allx∈X.Proof.
(i)⇔(ii) Since
(2)f(ax-ax′+x0)=g(ax-ax′)+y0=g(a(x-x0)-a(x′-x0))+y0,c(x)-cf(x′)+y0=c(f(x)-y0)-c(f(x′)-y0)+y0=cg(x-x0)-cg(x′-x0)+y0
for all x,x′∈X, it follows that f:(a,-a,c,-c;x0,y0)-additive⇔g(a(x-x0)-a(x′-x0))+y0=cg(x-x0)-cg(x′-x0)+y0(forallx,x′∈X)⇔g(ax-ax′)=cg(x)-cg(x′)(forallx,x′∈X)⇔g:(a,-a,c,-c)-additive.
(ii)⇒(iii) Suppose that
(3)g(ax-ax′)=cg(x)-cg(x′)
for all x,x′∈X. When x=x′, we have g(0)=0. Using this, g(ax)=cg(x) and also g(-ax)=-cg(x) for all x∈X. Therefore,
(4)g(x+x′)=g(axa-a(-x′a))=cg(xa)-cg(-x′a)=cg(xa)+(-c)g(-x′a)=g(x)+g(x′)
for all x,x′∈X.
(iii)⇒(ii) Because g(-x)=-g(x) for all x∈X (see also the following remark), it is trivial.Remark 2.
We denote byℚ the field of all rational numbers. It is well known that if g is additive, then g(qx)=qg(x) for every q∈ℚ and x∈X, that is, g is ℚ-linear. Hence, if g is additive and continuous, g must be ℝ-linear. On the other hand, when 𝕂=ℂ, we have a lot of continuous additive nonlinear mappings by considering the composition of linear transformations on ℝ2 and the ℝ-linear isometry (x,y)↦x+iy.The constanty0 is a trivial (a,-a,c,-c;x0,y0)-additive mapping of X into Y. The following theorem says that unless it is a unique (a,-a,c,-c;x0,y0)-additive mapping, discontinuous one always exists.Theorem 3.
(I) Ifa=c, then there exists a discontinuous (a,-a,c,-c;x0,y0)-additive mapping of X into Y.
(II) Ifa≠c, then the followings hold: (i)
If botha and c are transcendental numbers, then there exists a discontinuous (a,-a,c,-c;x0,y0)-additive mapping of X into Y.(ii)
If one ofa and c is transcendental and the other is algebraic, then the constant y0 is a unique (a,-a,c,-c;x0,y0)-additive mapping of X into Y.(iii)
If botha and c are algebraic with a common minimal polynomial, then there exists a discontinuous (a,-a,c,-c;x0,y0)-additive mapping of X into Y.(iv)
Ifa and c are algebraic with distinct minimal polynomials, then the constant y0 is a unique (a,-a,c,-c;x0,y0)-additive mapping of X into Y. Moreover, when 𝕂=ℝ, there is no nontrivial continuous (a,-a,c,-c;x0,y0)-additive mapping f of X into Y. On the other hand, when 𝕂=ℂ, if a is not real and a=c¯ (the complex conjugate of c), then the mapping x↦f(x+x0)-y0 must be conjugate linear for every continuous (a,-a,c,-c;x0,y0)-additive mapping f of X into Y.In order to show Theorem3, we need some lemmas for 𝕂-valued (a,-a,c,-c)-additive functions defined on 𝕂. For any x∈𝕂, we denote by ℚ(x) the subfield of 𝕂 generated by x over ℚ.By the following proofs of Lemma6 and Theorem 3, if there is a discontinuous (a,-a,c,-c;x0,y0)-additive mapping, then there are sufficiently many such mappings in the sense that there exists such a mapping which separates any ℚ(a)-linear independent points of X.Lemma 4.
Any(a,-a,c,-c)-additive φ of 𝕂 into itself satisfies φ(p(a)x)=p(c)φ(x) for all x∈𝕂 and p(X)∈ℚ[X].Proof.
Note thatφ is additive and φ(ax)=cφ(x) for each x∈𝕂 by Proposition 1. Let p(X)=a0+a1X+⋯+anXn with a0,a1,…,an∈ℚ. Since φ is ℚ-linear,
(5)φ(p(a)x)=a0φ(x)+a1φ(ax)+⋯+anφ(anx)=a0φ(x)+a1cφ(x)+⋯+ancnφ(x)=p(c)φ(x)
for all x∈𝕂.Lemma 5.
LetE and F be subfields of 𝕂 and ψ an isomorphism of E onto F. Then, ψ has an additive bijective extension φ to the full space 𝕂 such that
(6)φ(εx)=ψ(ε)φ(x)
for all ε∈E and x∈𝕂. Moreover, one has a discontinuous one whenever ℝ∖E≠∅.Proof.
Let{ej:j∈J} and {fj:j∈J′} be an E-linear base and an F-linear base of 𝕂, respectively. Because both of them have same cardinality, we take J=J′. Moreover, we may assume without loss of generality that ej0=fj0=1 for some j0∈J. For any x∈𝕂, there exist a finite I⊂J and {xi}i∈I⊆E such that
(7)x=∑i∈Ixiei,
and this decomposition is unique. Hence, we can define φ:𝕂→𝕂 by
(8)φ(x)=∑i∈Iψ(xi)fi.
This φ is a desired extension.
In order to get a discontinuous extension, we consider the base{ej:j∈J} and {fj:j∈J} such that fj1≠ej1∈ℝ∖E for some j1∈J. Because ℚ⊆E, take a rational sequence {εn} converging to ej1. Suppose that φ is continuous. Then,
(9)φ(εn)→φ(ej1)=fj1,φ(εn)=ψ(εn)=εn→ej1,
as n→∞. This is contradiction. Thus φ is discontinuous.Lemma 6.
Suppose thata≠c. (i)
If botha and c are transcendental numbers, then there exists a discontinuous (a,-a,c,-c)-additive function of 𝕂 into itself.(ii)
If one ofa and c is transcendental and the other is algebraic, then the constant 0 is a unique (a,-a,c,-c)-additive function of 𝕂 into itself.(iii)
If botha and c are algebraic with a common minimal polynomial, then there exists a discontinuous (a,-a,c,-c)-additive function of 𝕂 into itself.(iv)
Ifa and c are algebraic with distinct minimal polynomials, then the constant 0 is a unique (a,-a,c,-c)-additive function of 𝕂 into itself. Moreover, when 𝕂=ℝ, every nontrivial (a,-a,c,-c)-additive function of 𝕂 into itself is discontinuous. On the other hand, when 𝕂=ℂ, if a is not real and a=c¯, then any continuous (a,-a,c,-c)-additive function f of ℂ into itself must be of form f(x)=αx¯(x∈ℂ) for some α∈ℂ.Proof.
(i) Suppose that botha and c are transcendental. Then, ℚ(a) (resp., ℚ(c)) is isomorphic to the rational function field in indeterminate a (resp., c). So, the substitution a→c induces an isomorphism φac:ℚ(a)→ℚ(c) of fields. By Lemma 5, because ℝ∖ℚ(a)≠∅, φac has a discontinuous additive extension φ to 𝕂 such that φ(εx)=φac(ε)φ(x) for every ε∈ℚ(a) and x∈𝕂. Then, φ(ax)=φac(a)φ(x)=cφ(x) for all x∈𝕂, and, hence, φ is (a,-a,c,-c)-additive by Proposition 1.
(ii) Letφ be any (a,-a,c,-c)-additive function. If a is transcendental and c is algebraic with nonzero polynomial such that p(c)=0, then from Lemma 4, we have
(10)φ(x)=φ(p(a)p(a)-1x)=p(c)φ(p(a)-1x)=0
for all x∈𝕂. If c is transcendental and a is algebraic with nonzero polynomial such that q(a)=0, then from Lemma 4, we have
(11)φ(x)=1q(c)q(c)φ(x)=1q(c)φ(q(a)x)=1q(c)φ(0)=0
for all x∈𝕂.
(iii) Ifa is algebraic with minimal polynomial p(X)∈ℚ[X], then ℚ(a) consists of all polynomials f(a) in a of degree up to degp-1. So, if c is also algebraic with the same minimal polynomial p, then the substitution a→c induces an automorphism φac:ℚ(a)→ℚ(c)=ℚ(a). As same as (i), φac has a discontinuous (a,-a,c,-c)-additive extension.
(iv) Suppose thata and c are algebraic with distinct minimal polynomials p and q over the field ℚ, respectively. Let φ be any (a,-a,c,-c)-additive function. To show φ=0, we assume, on the contrary, that there is an x0∈𝕂 with φ(x0)≠0. Then, from Lemma 5, we have p(c)φ(x0)=φ(p(a)x0)=φ(0)=0, and hence p(c)=0. This contradicts the prerequisite for a and c. Hence, φ must be zero.
When𝕂=ℝ, since every continuous additive function is ℝ-linear and a≠c, there is no continuous (a,-a,c,-c)-additive function by Proposition 1. Now, we consider the case 𝕂=ℂ. Let φ be a nontrivial continuous (a,-a,c,-c)-additive function. Note that φ is ℝ-linear. If a is not real and a=c¯, we can easily see that φ(ix)=-iφ(x) for all x∈ℂ. Thus, φ is conjugate linear, and hence φ(x)=αx¯(x∈ℂ), where α=φ(1).Proof of Theorem3.
(I) We assume without loss of generality thatx0=y0=0 with the help of Proposition 1. Given an (a,-a,c,-c)-additive function φ, take a y1∈Y with ∥y1∥=1 and a nonzero functional h in X*, the dual space of X. Put
(12)f(x)=φ(h(x))y1(x∈X).
Then, we can easily see that f is an (a,-a,c,-c)-additive mapping of X into Y. Also, if φ is discontinuous, then so is f. In fact, if φ is discontinuous, we can find a sequence {an} in 𝕂 such that limn→∞an=0 and |φ(an)|≥1 (n=1,2,…). Choose an x1 in X with h(x1)=1 and put xn=anx1 for each n∈ℕ. Then, ∥xn∥=|an|∥x1∥→0 as n→∞ and ∥f(xn)∥=|φ(an)|≥1(n=1,2,…), so f is discontinuous, as required. Therefore Theorem 3(I), (II)-(i), and (II)-(iii) follow easily from Lemmas 4 and 6.
Given an(a,-a,c,-c)-additive mapping f of X into Y, take x∈X and h∈Y* arbitrarily, and put φ(t)=h(f(tx)) for each t∈𝕂. Then, φ:𝕂→𝕂 is (a,-a,c,-c)-additive. If φ=0 for each h∈Y* and x∈X, then f=0 by the Hahn-Banach theorem. Therefore, Theorem 3(II)-(ii) and (II)-(iv) follow easily from Lemma 6. The final assertion in (II) also follows from Lemma 6 and its proof.


## 3. A Stability of Generalized Additive Mappings

In this section, we consider a couple of cases which are left out in [8] about the Ulam type stability. We take a nonnegative function ε (say a control function) on X×X and also a certain nonnegative function δ on X which depends on ε. We say that a system of all (a,b,c,d;x0,y0)-additive mappings is strictly(ε,δ)-stable whenever the following statement is true:“If a mappingfofXintoYsatisfies(†)∥f(ax+bx′+x0)-cf(x)-df(x′)-y0∥≤ε(x,x′)for allx,x′∈X, then there exists a unique(a,b,c,d;x0,y0)-additive mappingf∞such that(‡)∥f(x)-f∞(x)∥≤δ(x)for allx∈X.”Throughout the remainder of this paper, we assume thatY is a Banach space. This is because all of our results depend on the following theorems whose proofs need the Banach fixed point theorem.Theorem A (see [8, Theorem 3.1]).
Leta+b≠0 and K≥0 with K|c+d|<1. One takes a control function ε which satisfies
(13)ε(x,x′)≤Kε((a+b)x+x0,(a+b)x′+x0)
for all x,x′∈X and puts
(14)δ(x)=Kε(x,x)1-K|c+d|
for each x∈X.
Then, the strict(ε,δ)-stability holds for the system of (a,b,c,d;x0,y0)-additive mappings.Theorem B (see [8, Theorem 3.2]).
Letc+d≠0 and K≥0 with K<|c+d|. One takes a control function ε which satisfies
(15)ε((a+b)x+x0,(a+b)x′+x0)≤Kε(x,x′)
for all x,x′∈X and puts
(16)δ(x)=ε(x,x)|c+d|-K
for each x∈X.
Then, the strict(ε,δ)-stability holds for the system of (a,b,c,d;x0,y0)-additive mappings.Both of these theorems do not say about(a,-a,c,-c;x0,y0)-additive mappings at all; however, we will get the following stability theorems for them. Theorem 7 is of the case a≠-1, Theorem 8 is of the case b≠-1, and Theorems 9 and  10  are for (-1,1,-1,1;x0,y0)-additive mappings. These cover all of the systems of (a,-a,c,-c;x0,y0)-additive mappings.Theorem 7.
Leta+1≠0 and K≥0 with K|1+c|<|c|. One takes a control function ε which satisfies
(17)ε(x,x′)≤Kε((a-1+1)x-a-1x0,(a-1+1)x′-a-1x0)
for all x,x′∈X, and puts
(18)δ(x)=Kε((a-1+1)x-a-1x0,x)|c|-K|1+c|
for each x∈X.
Then, the strict(ε,δ)-stability holds for the system of (a,-a,c,-c;x0,y0)-additive mappings.Proof.
Putu=ax-ax′+x0 and u′=x′ for each x,x′∈X. Then, (†) changes into
(19)∥f(a-1u+u′-a-1x0)-c-1f(u)-f(u′)+c-1y0∥≤ε1(u,u′),
where
(20)ε1(u,u′)=1|c|ε(a-1u+u′-a-1x0,u′)
for all u,u′∈X. By denoting a1=a-1, b1=1, c1=c-1, d1=1, u0=-a-1x0 and v0=-c-1y0 in (19), (†) changes up to the following estimate of f by the control function ε1:
(21)∥f(a1u+b1u′+u0)-c1f(u)-d1f(u′)-v0∥≤ε1(u,u′)
for all u,u′∈X.
Under these transformations,a1,b1,c1,d1, and ε1 are equipped with
(22)a1+b1≠0,K|c1+d1|<1,ε1(u,u′)≤Kε1((a1+b1)u+u0,(a1+b1)u′+u0)
for all u,u′∈X. The latter follows from the inequality in which ε must satisfy because by using (20), we get
(23)|c|ε1(u,u′)=ε(a-1u+u′+u0,u′)=ε(x,x′),ε((a-1+1)x-a-1x0,(a-1+1)x′-a-1x0)=ε((a-1+1)x-a-1x0,(a1+b1)u′+u0)=ε((a-1+1)(a-1u+u′+u0)-a-1x0,(a1+b1)u′+u0)=ε(a-1((a1+b1)u+u0)+((a1+b1)u′+u0)-a-1x0,(a1+b1)u′+u0)=|c|ε1((a1+b1)u+u0,(a1+b1)u′+u0)
for all u,u′∈X. Since (21) and (22) hold, it follows from Theorem A that there exists a unique (a1,b1,c1,d1;u0,v0)-additive mapping f∞ such that
(24)∥f(x)-f∞(x)∥≤Kε1(x,x)1-K|c1+d1|
for all x∈X. However, we can easily see the following two assertions: (i)
f∞ is (a1,b1,c1,d1;u0,v0)-additive if and only if f∞ is (a,-a,c,-c;x0,y0)-additive,(ii)
(24) is equivalent to (‡).
This completes the proof.Theorem 8.
Letc+1≠0 and K≥0 with K|c|<|c+1|. One takes a control function ε which satisfies
(25)ε((a-1+1)x-a-1x0,(a-1+1)x′-a-1x0)≤Kε(x,x′)
for all x,x′∈X and puts
(26)δ(x)=ε((a-1+1)x-a-1x0,x)|c+1|-K|c|
for each x∈X.
Then the strict(ε,δ)-stability holds for the system of (a,-a,c,-c;x0,y0)-additive mappings.Proof.
We consider the same transformations and the same estimate (21) of f by ε1 in the proof of Theorem 7. Under these transformations, we have
(27)c1+d1≠0,K<|c1+d1|.
Moreover, for every u,u′∈X we have
(28)ε1((a1+b1)u+u0,(a1+b1)u′+u0)≤Kε1(u,u′),
because
(29)|c|ε1(u,u′)=ε(a-1u+u′+u0,u′)=ε(x,x′),(30)|c|ε1((a1+b1)u+u0,(a1+b1)u′+u0)=ε((a-1+1)x-a-1x0,(a-1+1)x′-a-1x0).
Since (21), (27) and (28) hold, it follows from Theorem B that there exists a unique (a1,b1,c1,d1;u0,v0)-additive mapping f∞ such that
(31)∥f(x)-f∞(x)∥≤ε1(x,x)|c1+d1|-K
for all x∈X. This means (‡) and f∞ is (a,-a,c,-c;x0,y0)-additive.Theorem 9.
Let0≤K<1/2. One takes a control function ε which satisfies
(32)ε(x,x′)≤Kε(2x-x0,2x′-x0)
for all x,x′∈X and puts
(33)δ(x)=Kε(x,2x-x0)1-2K
for each x∈X.
Then the strict(ε,δ)-stability holds for the system of (-1,1,-1,1;x0,y0)-additive mappings.Proof.
Putu=x,u′=-x+x′+x0 for each x,x′∈X. Then, (†) changes into the following estimate of f by the control function ε1:
(34)∥f(u+u′-x0)-f(u)-f(u′)+y0∥≤ε1(u,u′),
where
(35)ε1(u,u′)=ε(u,u+u′-x0).
for all u,u′∈X. Put a1=b1=c1=d1=1.
Under these transformations,a1,b1,c1,d1, and ε1 are equipped with
(36)a1+b1=2≠0,K|c1+d1|=2K<1,ε1(u,u′)≤Kε1((a1+b1)u-x0,(a1+b1)u′-x0)
for all u,u′∈X. The latter follows from the following inequality:
(37)ε1(u,u′)=ε(u,u+u′-x0)≤Kε(2u-x0,2(u+u′-x0)-x0)=Kε(2u-x0,(2u-x0)+(2u′-x0)-x0)=Kε1(2u-x0,2u′-x0)
for all u,u′∈X. Since (34), (36) hold, it follows from Theorem A that there exists a unique (a1,b1,c1,d1;-x0,-y0)-additive mapping f∞ such that
(38)∥f(x)-f∞(x)∥≤Kε1(x,x)1-K|c1+d1|
for all x∈X. However we can easily see the following two assertions: (i)
f∞ is (a1,b1,c1,d1;-x0,-y0)-additive if and only if f∞ is (-1,1,-1,1;x0,y0)-additive;(ii)
(38) is equivalent to (‡).
This completes the proof.Theorem 10.
Let0≤K<2. One takes a control function ε which satisfies
(39)ε(2x-x0,2x′-x0)≤Kε(x,x′)
for all x,x′∈X and puts
(40)δ(x)=ε(x,2x-x0)2-K
for each x∈X.
Then the strict(ε,δ)-stability holds for the system of (-1,1,-1,1;x0,y0)-additive mappings.Proof.
We consider the same transformation and the same estimate (34) of f by ε1 in the proof of Theorem 9. Under these transformations, c1,d1 and ε1 are equipped with
(41)c1+d1=2≠0,K<2=|c1+d1|,(42)ε1(2u-x0,2u′-x0)=ε(2u-x0,(2u-x0)+(2u′-x0)-x0)=ε(2x-x0,2(u+u′-x0)-x0)=ε(2x-x0,2x′-x0)≤Kε(x,x′)=Kε(u,u+u′-x0)=Kε1(u,u′)
for all u,u′∈X. So it follows that
(43)ε1((a1+b1)u-x0,(a1+b1)u′-x0)≤Kε1(u,u′)
for all u,u′∈X. Since (34), (41) and (43) hold, it follows from Theorem B that there exists a unique (a1,b1,c1,d1;-x0,-y0)-additive mapping f∞ such that
(44)∥f(x)-f∞(x)∥≤ε1(x,x)|c1+d1|-K
for all x,x′∈X. This means (‡) and f∞ is (-1,1,-1,1;x0,y0)-additive.


## 4. Concrete Examples

Throughout this section, letx0=y0=0. We fix nonnegative constants p,q and δ and take the control function ε defined by ε(x,x′)=δ∥x∥p∥x′∥q for every x,x′∈X.Corollary 11.
Whena+1≠0 and |a|p+q|c+1|<|c||a+1|p+q, one puts
(45)δ′(x)=δ|a+1|p|a|q∥x∥p+q|c||a+1|p+q-|c+1||a|p+q
for each x∈X.
Then, the system of(a,-a,c,-c)-additive mappings is strictly (ε,δ′)-stable.Proof.
PutK=|a-1+1|-(p+q). Then K|c+1|<|c| and
(46)ε(x,x′)=Kε((a-1+1)x,(a-1+1)x′)
for all x,x′∈X. By Theorem 7, for a mapping f of X into Y satisfying (†) for a=-b and c=-d, there exists a unique (a,-a,c,-c)-additive mapping f∞ such that
(47)∥f(x)-f∞(x)∥≤Kδ|a-1+1|p∥x∥p+q|c|-K|c+1|
for all x,x′∈X. Because of K=|a-1+1|-(p+q), we have the corollary.Corollary 12.
Whenc+1≠0 and |a+1|p+q|c|<|c+1||a|p+q, one puts
(48)δ′(x)=δ|a+1|p|a|q∥x∥p+q|a|p+q|c+1|-|a+1|p+q|c|
for each x∈X.
Then the system of(a,-a,c,-c)-additive mappings is strictly (ε,δ′)-stable.Proof.
PutK=|a-1+1|p+q. Then K|c|<|c+1| and
(49)ε((a-1+1)x,(a-1+1)x′)=Kε(x,x′)
for all x,x′∈X. By Theorem 8, for a mapping f of X into Y satisfying (†) for a=-b and c=-d, there exists a unique (a,-a,c,-c)-additive mapping f∞ such that
(50)∥f(x)-f∞(x)∥≤δ|a-1+1|p∥x∥p+q|c+1|-K|c|
for all x∈X. Because of K=|a-1+1|p+q, we have the corollary.Corollary 13.
Whenp+q>1, one puts
(51)δ′(x)=2qδ∥x∥p+q2p+q-2
for each x∈X.
Then, the system of(-1,1,-1,1)-additive mappings is strictly (ε,δ′)-stable.Proof.
PutK=2-(p+q). Then,
(52)ε(x,x′)=Kε(2x,2x′)
for all x,x′∈X. Since p+q>1, we also have 0≤K<1/2. By Theorem 9, for a mapping f of X into Y satisfying (†) for a=c=-1 and b=d=1, there exists a unique (-1,1,-1,1)-additive mapping f∞ such that
(53)∥f(x)-f∞(x)∥≤2qδK∥x∥p+q1-2K
for all x∈X. Because of K=2-(p+q), we have the corollary.Corollary 14.
Whenp+q<1, one puts
(54)δ′(x)=2qδ∥x∥p+q2-2p+q
for each x∈X.
Then, the system of(-1,1,-1,1)-additive mappings is strictly (ε,δ′)-stable.Proof.
PutK=2p+q. Then,
(55)ε(2x,2x′)=Kε(x,x′)
for all x,x′∈X. Since p+q<1, we also have 0≤K<2. By Theorem 10, for a mapping f of X into Y satisfying (†) for a=c=-1 and b=d=1, there exists a unique (-1,1,-1,1)-additive mapping f∞ such that
(56)∥f(x)-f∞(x)∥≤2qδ∥x∥p+q2-K
for all x∈X. Because of K=2p+q, we have the corollary.Remark 15.
In Corollary14, taking p=q=0, we can easily observe that the corollary is just the stability result due to Hyers [2, Theorem 1].



---

*Source: 109754-2013-03-20.xml*