# Functional Inequalities Associated with Additive Mappings

Authors: Jaiok Roh; Ick-Soon Chang
Journal: Abstract and Applied Analysis
 (2008)

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## Abstract

The functional inequality∥f(x)+2f(y)+2f(z)∥≤∥2f(x/2+y+z)∥+ϕ(x,y,z)(x,y,z∈G) is investigated, where G is a group divisible by 2,f:G→X and ϕ:G3→[0,∞) are mappings, and X is a Banach space. The main result of the paper states that the assumptions above together with (1) ϕ(2x,−x,0)=0=ϕ(0,x,−x)(x∈G) and (2) limn→∞(1/2n)ϕ(2n+1x,2ny,2nz)=0, or limn→∞2nϕ(x/2n−1,y/2n,z/2n)=0(x,y,z∈G), imply that f is additive. In addition, some stability theorems are proved.


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## Body

## 1. Introduction and Preliminaries

The concept of stability for a functional equation
arises when we replace the functional equation by an inequality which acts as a
perturbation of the equation. The study of stability problems had been
formulated by Ulam [1] during a talk in 1940: under what condition does there
exist a homomorphism near an approximate
homomorphism? In the following year 1941, Hyers [2]
had answered affirmatively the question of Ulam for
Banach spaces, which states that ifε≥0 and f:𝒳→𝒴 is a mapping with 𝒳 a normed space, 𝒴 a Banach space
such that(1.1)∥f(x+y)−f(x)−f(y)∥≤εfor allx,y∈𝒳, then there exists a unique additive mapping L:𝒳→𝒴 such that(1.2)∥f(x)−L(x)∥≤εfor allx∈𝒳. Then, Aoki [3] in
1950 and Rassias [4]
in 1978 proved the following generalization of Hyers' theorem [2] by
considering the case when the inequality (1.1) is unbounded.Proposition 1.1.
Letf:𝒳→𝒴 be a mapping from a normed space 𝒳 into a Banach space 𝒴 subject to the inequality (1.3)∥f(x+y)−f(x)−f(y)∥≤ε(∥x∥p+∥y∥p)for all x,y∈𝒳, where ε and p are constants with ε≥0 and p<1. Then, there exists a unique additive mapping L:𝒳→𝒴 such that(1.4)∥f(x)−L(x)∥≤2ε2−2p∥x∥pfor all x∈𝒳. If p<0, then inequality (1.3) holds for x,y≠0 and
(1.4) for x⁢≠0.Following the
techniques of the proof of the corollary of Hyers [2], we observed that Hyers
introduced (in 1941) Hyers continuity condition about the continuity of the
mapping f(tx) in t∈ℝ for each fixed x, and then he proved homogenouity of degree one
and, therefore, the famous linearity. This condition has been assumed further
till now through the complete Hyers direct method in order to prove linearity
for generalized Hyers-Ulam stability problem forms (cf., [5]).In 1991, Gajda [6] provided an affirmative answer to Rassias'
question whether his theorem can be extended for values of p greater than one. However, it was shown by
Gajda [6] as well as
by Rassias and Šemrl [7] that one cannot prove a
theorem similar to [4] when p=1. On the other hand, Rassias [8–10] generalized Hyers'
stability result by presenting a weaker condition controlled by (or involving)
a product of different powers of norms (from the right-hand side of assumed
conditions) as follows.Proposition 1.2.
Suppose that there exist constantsε≥0 and p,q∈ℝ such that r=p+q≠1, and f:𝒳→𝒴 is a mapping with 𝒳 a normed space, 𝒴 is a Banach space
such that the inequality (1.5)∥f(x+y)−f(x)−f(y)∥≤ε∥x∥p∥y∥qholds for all x,y∈𝒳. Then, there exists a unique additive mapping L:𝒳→𝒴 such that(1.6)∥f(x)−L(x)∥≤ε|2−2r|∥x∥rfor all x∈𝒳.Since then, more generalizations and applications of the generalized Hyers-Ulam stability
to a number of functional equations and mappings have been investigated in
[11–32].Recently, Roh and Shin [33] proved that if f:X→Y is a mapping from a normed space X into a Banach space Y satisfying the inequality(1.7)∥f(x)+2f(y)+2f(z)∥≤∥2f(x2+y+z)∥+ε∥x∥r∥y∥r∥z∥rfor all x,y,z∈X and some ε≥0, then it is additive. In addition, they
investigated the stability in Banach spaces.In this paper, we will consider a mapping on a group
instead of a normed space which satisfies the following
inequality:(1.8)∥f(x)+2f(y)+2f(z)∥≤∥2f(x2+y+z)∥+ϕ(x,y,z)for all x,y,z∈G, where G is a group, f:G→X and ϕ:G3→[0,∞) are mappings, and X is a Banach space.Inequalities of the type above and some corresponding
equations were examined by several authors [34–36].
In [35], it has been
proved that if G is a (not necessarily 2-divisible) group, X is an inner product space and the mapping f:G→X satisfies(1.9)∥f(x)+f(y)∥≤∥f(x+y)∥for all x,y∈G, then it is additive. By replacing x=0 in (1.8), we obtain(1.10)∥f(0)+2f(y)+2f(z)∥≤∥2f(0+y+z)∥+0for all y,z∈G, which implies the inequality (1.9). Therefore,
Theorem 2.1 in this paper is a special case of the result cited above for
mappings which maps into Hilbert spaces. However, Theorem 2.1 is also valid for
Banach spaces, while the result in [35] is not. (A counter example was constructed in [34]).


## 2. Main Results

Theorem 2.1.
Let(G,+) be a 2-divisible group and (X,∥⋅∥) a Banach space. Assume that a mapping ϕ:G3→[0,∞) satisfies the assumptions
(1)
ϕ(2x,−x,0)=0=ϕ(0,x,−x)(x∈G),(2)
lim⁡n→∞(1/2n)ϕ(2n+1x,2ny,2nz)=0 or lim⁡n→∞2nϕ(x/2n−1,y/2n,z/2n)=0(x,y,z∈G),
and that the
mapping f:G→X satisfies the inequality (1.8). Then, f is additive.Proof.
By lettingx=y=z=0 in (1.8), we get f(0)=0;
and by letting x=2x,⁢y=−x and z=0 in (1.8), we have(2.1)f(2x)=−2f(−x)for all x∈G. Also, by letting x=0 and z=−y in (1.8), we obtain(2.2)f(−y)=−f(y)for all y∈G.
Next, we are in the position to show thatf is additive. We will consider two different
cases for second assumption of ϕ.
Case 1.
Assumelim⁡n→∞2nϕ(x/2n−1,y/2n,z/2n)=0 for all x,y,z∈G. We get by (2.1) and (2.2)(2.3)f(2x)=2f(x),f(x)=2f(x2)=4f(x4)=⋯=2nf(x2n)for all positive integer n and all x∈G. Therefore, we can define f(x):=lim⁡n→∞2nf(x/2n) for all x∈G. Due to (1.8), (2.1), and (2.2), we
obtain(2.4)∥f(x)+f(y)−f(x+y)∥=lim⁡n→∞2n∥f(x2n)+f(y2n)+f(−x−y2n)∥≤lim⁡n→∞2n∥f(x2n)+f(y2n)+12f(−x−y2n−1)∥≤lim⁡n→∞2n−1ϕ(−x−y2n−1,x2n,y2n)=0for all x,y∈G. Thus, f(x+y)=f(x)+f(y).
Case 2.
Assumelim⁡n→∞(1/2n)ϕ(2n+1x,2ny,2nz)=0 for all x,y,z∈G. We get by (2.1) and (2.2)(2.5)f(2x)=2f(x),f(x)=12f(2x)=14f(4x)=⋯=12nf(2nx)for all positive integer n and all x∈G. Therefore, we can define f(x):=lim⁡n→∞(1/2n)f(2nx) for all x∈G. Due to (1.8), (2.1), and (2.2), we
obtain(2.6)∥f(x)+f(y)−f(x+y)∥=lim⁡n→∞12n∥f(2nx)+f(2ny)+f(2n(−x−y))∥≤lim⁡n→∞12n+1ϕ(2n+1(−x−y),2nx,2ny)=0for all x,y∈G. Thus, f(x+y)=f(x)+f(y).Next, we will study the generalized Hyers-Ulam
stability of functional inequality (1.8).Theorem 2.2.
Let(G,+) be a 2-divisible abelian group and (X,∥⋅∥) a Banach space. Assume that a mapping ϕ:G3→[0,∞) satisfies the assumptions
(1)
ρ(x)=∑j=0∞2j[ϕ(x/2j,−x/2j+1,0)+ϕ(0,x/2j+1,−x/2j+1)]<∞(x∈G),(2)
lim⁡n→∞2nϕ((−x−y)/2n−1,x/2n,y/2n)=0(x,y∈G),and the
inequality (1.8). Then, there exists a
unique additive mapping  L:G→X such that(2.7)∥L(x)−f(x)∥≤ρ(x)for all x∈G.Proof.
Lettingx=y=z=0 in (1.8), we get ∥f(0)∥≤(1/3)ϕ(0,0,0). By assumption, we should have ϕ(0,0,0)=0, since lim⁡n→∞2nϕ(0,0,0)=0. Hence, f(0)=0. So, by letting y=−x/2 and z=0 in (1.8), we have(2.8)∥f(x)+2f(−x2)∥≤ϕ(x,−x2,0)for all x∈G. Letting x=0,y=x,
and z=−x in (1.8), we obtain(2.9)∥f(x)+f(−x)∥≤12ϕ(0,x,−x)for all x∈G. Therefore, we have(2.10)∥2lf(x2l)−2mf(x2m)∥≤∑j=lm−1∥2jf(x2j)−2j+1f(x2j+1)∥≤∑j=lm−1[∥2jf(x2j)+2j+1f(−x2j+1)∥+∥2j+1f(−x2j+1)+2j+1f(x2j+1)∥]≤∑j=lm−1[2jϕ(x2j,−x2j+1,0)+2jϕ(0,x2j+1,−x2j+1)]for all nonnegative integers m and l with m>l and all x∈G. It means that the sequence {2nf(x/2n)} is a Cauchy sequence. Since X is complete, the sequence {2nf(x/2n)} converges. So we can define a mapping L:G→X by L(x):=lim⁡n→∞2nf(x/2n) for all x∈G. Moreover, by letting l=0 and passing m→∞, we get (2.7).
Now, we claim that the mappingL is additive. We note by (2.9)
that(2.11)∥L(x)+L(−x)∥=lim⁡n→∞2n∥f(x2n)+f(−x2n)∥≤lim⁡n→∞2n−1ϕ(0,x2n,−x2n)=0.So we have L(−x)=−L(x). By (1.8), (2.8), and (2.9), we
have(2.12)∥L(x)+L(y)−L(x+y)∥=lim⁡n→∞2n∥f(x2n)+f(y2n)+f(−x−y2n)∥≤lim⁡n→∞2n∥12f(−x−y2n−1)+f(x2n)+f(y2n)∥+lim⁡n→∞2n∥f(−x−y2n)+12f(x+y2n−1)∥+lim⁡n→∞2n−1∥f(−x−y2n−1)+f(x+y2n−1)∥≤lim⁡n→∞2n−1ϕ(−x−y2n−1,x2n,y2n)+lim⁡n→∞2n−1ϕ(x+y2n−1,−x−y2n,0)+lim⁡n→∞2n−2ϕ(0,−x−y2n−1,x+y2n−1)=0for all x,y∈G.
Now, to prove uniqueness of the mappingL, let us assume that T:G→X is an additive mapping satisfying (2.7). Then,
we obtain due to (2.7) that(2.13)∥L(x)−T(x)∥=2n∥L(x2n)−T(x2n)∥≤2n[∥L(x2n)−f(x2n)∥+∥T(x2n)−f(x2n)∥]≤2n+1ρ(x2n)⟶0for all x∈G, as n→∞.Theorem 2.3.
Let(G,+) be a 2-divisible abelian group and (X,∥⋅∥) a Banach space. Assume that a mapping ϕ:G3→[0,∞) satisfies the assumptions
(1)
∑j=0∞(1/2j)[ϕ(−2j+1x,2jx,0)+(1/2)ϕ(0,2j+1x,−2j+1x)]<∞(x∈G),(2)
lim⁡n→∞(1/2n)ϕ(2n+1(−x−y),2nx,2ny)=0(x,y∈G),
and the
inequality (1.8). Then, there exists a unique additive mapping L:G→X such that(2.14)∥L(x)−f(x)∥≤η(x)for all x∈G, where(2.15)η(x)=∑j=0∞12j+1[ϕ(−2j+1x,2jx,0)+12ϕ(0,2j+1x,−2j+1x)+116ϕ(0,0,0)].Proof.
By lettingx=y=z=0 in (1.8), we get(2.16)∥f(0)∥≤13ϕ(0,0,0).We also have, by letting y=−x/2 and z=0 in (1.8), that(2.17)∥12f(x)+f(−x2)∥≤12ϕ(x,−x2,0)+23ϕ(0,0,0)for all x∈G. Next, by letting x=0,y=x, and z=−x in (1.8), we obtain(2.18)∥f(x)+f(−x)∥≤12ϕ(0,x,−x)+12ϕ(0,0,0)for all x∈G. Therefore, we have(2.19)∥12lf(2lx)−12mf(2mx)∥≤∑j=lm−1∥12jf(2jx)−12j+1f(2j+1x)∥≤∑j=lm−1[∥12jf(2jx)+12j+1f(−2j+1x)∥+∥12j+1f(2j+1x)+12j+1f(−2j+1x)∥]≤∑j=lm−1[12j+1ϕ(−2j+1x,2jx,0)+12j+2ϕ(0,2j+1x,−2j+1x)+12j+1116ϕ(0,0,0)]for all nonnegative integers m and l with m>l and all x∈G. It follows that the sequence {(1/2n)f(2nx)} is a Cauchy and so it is convergent since X is complete. So one can define a mapping L:G→X by L(x):=lim⁡n→∞(1/2n)f(2nx) for all x∈G. By letting l=0 and taking the limit m→∞, we arrive at (2.14).
Now, we claim that the mappingL is additive. By (2.18), one
notes(2.20)∥L(x)+L(−x)∥=lim⁡n→∞12n∥f(2nx)+f(−2nx)∥≤lim⁡n→∞12n+1[ϕ(0,2nx,−2nx)+ϕ(0,0,0)]=0.So we have L(−x)=−L(x). By (1.8), (2.17), and (2.18), we
obtain(2.21)∥L(x)+L(y)−L(x+y)∥=lim⁡n→∞12n∥f(2nx)+f(2ny)+f(2n(−x−y))∥≤lim⁡n→∞12n∥12f(2n+1(−x−y))+f(2nx)+f(2ny)∥+lim⁡n→∞12n∥f(2n(−x−y))+12f(2n+1(x+y))∥+lim⁡n→∞12n+1∥f(2n+1(−x−y))+f(2n+1(x+y))∥≤lim⁡n→∞12n+1[ϕ(2n+1(−x−y),2nx,2ny)+23ϕ(0,0,0)]+lim⁡n→∞12n+1[ϕ(2n+1(x+y),2n(−x−y),0)+43ϕ(0,0,0)]+lim⁡n→∞12n+2[ϕ(0,2n+1(−x−y),2n+1(x+y))+ϕ(0,0,0)]=0for all x,y∈G.
Now, to show uniqueness of the mappingL, let us assume that T:G→X is another additive mapping satisfying (2.14).
Then, by (2.14), and assumptions of ϕ, we have(2.22)∥L(x)−T(x)∥=12n∥L(2nx)−T(2nx)∥≤12n[∥L(2nx)−f(2nx)∥+∥T(2nx)−f(2nx)∥]≤12n−1η(2nx)⟶0for all x∈G,
as n→∞.With the help of Theorems2.2 and 2.3, we obtain the
following corollaries.
Corollary 2.4.
Suppose thatf:E→X is a mapping from a normed space E into a Banach space X subject to the inequality(2.23)∥f(x)+2f(y)+2f(z)∥≤∥2f(x2+y+z)∥+ε(∥x∥p+∥y∥p+∥z∥p)for all x,y,z∈E, where ε and p are constants with ε≥0 and p>1. Then, there exists a unique additive mapping L:E→X such that(2.24)∥L(x)−f(x)∥≤2p+32p−2ε∥x∥pfor all x∈E.Corollary 2.5.
Suppose thatf:E→X is a mapping from a normed space E into a Banach space X subject to the inequality (2.23) for all x,y,z∈E, where ε and p are constants with ε≥0 and p<1. Then, there exists a unique additive mapping L:E→X such that(2.25)∥L(x)−f(x)∥≤2p+2+22−2pε∥x∥pfor all x∈E.Theorem 2.6.
Let(G,+) be a 2-divisible abelian group and (X,∥⋅∥) a Banach space. Assume that a mapping ϕ:G3→[0,∞) satisfies the assumptions
(1)
∑j=0∞2j[ϕ(x/2j,−x/2j+1,0)+ϕ(−x/2j,x/2j+1,0)]<∞(x∈G),(2)
lim⁡n→∞2nϕ(−x−y/2n−1,x/2n,y/2n)=0(x,y∈G),
and the
inequality (1.8). Then, there exists a
unique additive mapping L:G→X such that(2.26)∥L(x)−f(x)−f(−x)2∥≤γ(x)for all x∈G, where(2.27)γ(x)=∑j=0∞2j−1[ϕ(x2j,−x2j+1,0)+ϕ(−x2j,x2j+1,0)].Proof.
Lettingx=y=z=0 in (1.8), we get ∥f(0)∥≤(1/3)ϕ(0,0,0). By assumption, we should have ϕ(0,0,0)=0, since lim⁡n→∞2nϕ(0,0,0)=0. Hence, we obtain f(0)=0. By letting y=−x/2 and z=0 in (1.8), we have(2.28)∥f(x)+2f(−x2)∥≤ϕ(x,−x2,0)for all x∈G. Let g(x)=(f(x)−f(−x))/2. Then, we obtain(2.29)∥2g(x)−g(2x)∥≤∥f(x)+f(−2x)2∥+∥f(−x)+f(2x)2∥≤12[ϕ(−2x,x,0)+ϕ(2x,−x,0)]for all x∈G. Hence, for all nonnegative integers m and l with m>l and all x∈G,(2.30)∥2lg(x2l)−2mg(x2m)∥≤∑j=lm−1∥2jg(x2j)−2j+1g(x2j+1)∥≤∑j=lm−12j−1[ϕ(−x2j,x2j+1,0)+ϕ(x2j,−x2j+10)].So the sequence {2ng(x/2n)} is a Cauchy sequence. Due to the completeness
of X, this sequence is convergent. Let L:G→X be a mapping defined by L(x):=lim⁡n→∞2ng(x/2n) for all x∈G. Letting l=0 and sending m→∞, we get (2.26).
Next, we claim that the mappingL is additive. We first note that L(−x)=−L(x) because g(−x)=−g(x). So, by (1.8) and (2.28), we
obtain(2.31)∥L(x)+L(y)−L(x+y)∥=lim⁡n→∞2n∥g(x2n)+g(y2n)+g(−x−y2n)∥≤lim⁡n→∞2n[∥f(x/2n)2+f(y/2n)2+f(−x−y/2n−1)4∥⁢+∥f(−x/2n)2+f(−y/2n)2+f(x+y/2n−1)4∥]+lim⁡n→∞2n[∥f(−x−y/2n)2+f(x+y/2n−1)4∥+∥f(x+y/2n)2+f(−x−y/2n−1)4∥]≤lim⁡n→∞2n−2[ϕ(−x−y2n−1,x2n,y2n)+ϕ(x+y2n−1,−x2n,−y2n)]+lim⁡n→∞2n−2[ϕ(x+y2n−1,−x−y2n,0)+ϕ(−x−y2n−1,x+y2n,0)]=0for all x,y∈G. So we have L(x+y)=L(x)+L(y).
The proof of uniqueness forL is similar to the proof of Theorem 2.2.Theorem 2.7.
Let(G,+) be a 2-divisible abelian group and (X,∥⋅∥) a Banach space. Assume that a mapping ϕ:G3→[0,∞) satisfies the assumptions
(1)
∑j=0∞(1/2j)[ϕ(−2j+1x,2jx,0)+ϕ(2j+1x,−2jx,0)]<∞(x∈G),(2)
lim⁡n→∞(1/2n)ϕ(2n+1(−x−y),2nx,2ny)=0(x,y∈G),
and the
inequality (1.8). Then, there exists a
unique additive mapping L:G→X such that(2.32)∥L(x)−f(x)−f(−x)2∥≤δ(x)for all x∈G, where(2.33)δ(x)=∑j=0∞12j+2[ϕ(−2j+1x,2jx,0)+ϕ(2j+1x,−2jx,0)+83ϕ(0,0,0)].Proof.
Due to (2.17), we
have(2.34)∥12f(x)+f(−x2)∥≤12ϕ(x,−x2,0)+23ϕ(0,0,0)for all x∈G. So, we obtain(2.35)∥12lg(2lx)−12mg(2mx)∥≤∑j=lm−1∥12jg(2jx)−12j+1g(2j+1x)∥≤∑j=lm−112j+2[ϕ(−2j+1x,2jx,0)+ϕ(2j+1x,−2jx,0)+83ϕ(0,0,0)]for all nonnegative integers m and l with m>l and all x∈G. This means that the sequence {(1/2n)g(2nx)} is a Cauchy sequence. Since X is complete, the sequence {(1/2n)g(2nx)} converges. Thus, we may define a mapping L:G→X by L(x):=lim⁡n→∞(1/2n)g(2nx) for all x∈G. Letting l=0 and passing the limit m→∞, we get (2.32).
Now, we claim that the mappingL is additive. By (2.16), (1.8), and (2.34), we
have(2.36)∥L(x)+L(y)−L(x+y)∥=lim⁡n→∞12n∥g(2nx)+g(2ny)+g(2n(−x−y))∥≤lim⁡n→∞12n+2[ϕ(2n+1(−x−y),2nx,2ny)+ϕ(2n+1(x+y),−2nx,−2ny)]+lim⁡n→∞12n+2[ϕ(2n+1(x+y),2n(−x−y),0)+ϕ(2n+1(−x−y),2n(x+y),0)]+lim⁡n→∞12nϕ(0,0,0)=0for all x,y∈G.
The proof of uniqueness forL is similar to the proof of Theorem 2.3.



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*Source: 136592-2008-09-02.xml*