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# Fuzzy Hyers-Ulam stability of an additive functional equation

Journal of Inequalities and Applications20112011:140

https://doi.org/10.1186/1029-242X-2011-140

• Received: 10 October 2011
• Accepted: 19 December 2011
• Published:

## Abstract

In this paper, using the fixed point and direct methods, we prove the Hyers-Ulam stability of the following additive functional equation

$2f\left(\frac{x+y+z}{2}\right)=f\left(x\right)+f\left(y\right)+f\left(z\right)$
(0.1)

in fuzzy normed spaces.

Mathematics Subject Classification (2010): 39B22; 39B52; 39B82; 46S10; 47S10; 46S40.

## Keywords

• Hyers-Ulam stability
• additive functional equation
• fuzzy normed space

## 1. Introduction

A classical question in the theory of functional equations is the following: When is it true that a function which approximately satisfies a functional equation must be close to an exact solution of the equation? If the problem accepts a solution, we say that the equation is stable. The first stability problem concerning group homomorphisms was raised by Ulam  in 1940. In the next year, Hyers  gave a positive answer to the above question for additive groups under the assumption that the groups are Banach spaces. In 1978, Rassias  proved a generalization of the Hyers' theorem for additive mappings.

Theorem 1.1. (Th.M. Rassias) Let f : XY be a mapping from a normed vector space X into a Banach space Y subject to the inequality
$\parallel f\left(x+y\right)-f\left(x\right)-f\left(y\right)\parallel \phantom{\rule{2.77695pt}{0ex}}\le \epsilon \left(\parallel x{\parallel }^{p}+\parallel y{\parallel }^{p}\right)$
for all x, y X, where ε and p are constants with ε > 0 and 0 ≤ p < 1. Then the limit
$L\left(x\right)=\underset{n\to \infty }{lim}\frac{f\left({2}^{n}x\right)}{{2}^{n}}$
exists for all x X and L : XY is the unique additive mapping which satisfies
$\parallel f\left(x\right)+L\left(x\right)\parallel \le \frac{2\epsilon }{2-{2}^{p}}\parallel x{\parallel }^{p}$

for all x X. Also, if for each x X, the function f(tx) is continuous in t , then L is -linear.

Furthermore, in 1994, a generalization of Rassias' theorem was obtained by Gǎvruta  by replacing the bound ε(||x|| p + ||y|| p ) by a general control function φ(x, y).

In 1983, a Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof  for mappings f : XY, where X is a normed space and Y is a Banach space. In 1984, Cholewa  noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group and, in 2002, Czerwik  proved the Hyers-Ulam stability of the quadratic functional equation. The reader is referred to () and references therein for detailed information on stability of functional equations.

Katsaras  defined a fuzzy norm on a vector space to construct a fuzzy vector topological structure on the space. Some mathematicians have defined fuzzy norms on a vector space from various points of view (see [22, 23]). In particular, Bag and Samanta , following Cheng and Mordeson , gave an idea of fuzzy norm in such a manner that the corresponding fuzzy metric is of Karmosil and Michalek type . They established a decomposition theorem of a fuzzy norm into a family of crisp norms and investigated some properties of fuzzy normed spaces .

Definition 1.2. Let X be a real vector space. A function N : X × → [0, 1] is called a fuzzy norm on X if for all x, y X and all s, t ,

(N 1) N(x, t) = 0 for t ≤ 0;

(N 2) x = 0 if and only if N(x, t) = 1 for all t > 0;

(N 3) $N\left(cx,\phantom{\rule{2.77695pt}{0ex}}t\right)=N\left(x,\phantom{\rule{2.77695pt}{0ex}}\frac{t}{\mid c\mid }\right)$if c ≠ 0;

(N 4) N(x + y, c + t) ≥ min{N(x, s), N(y, t)};

(N 5) N(x,.) is a non-decreasing function of and limt→∞N(x, t) = 1;

(N 6) for x ≠ 0, N(x,.) is continuous on .

The pair (X, N) is called a fuzzy normed vector space.

Example 1.3. Let (X, ||.||) be a normed linear space and α, β > 0. Then
$N\left(x,\phantom{\rule{2.77695pt}{0ex}}t\right)=\left\{\begin{array}{cc}\hfill \frac{\alpha t}{\alpha t+\beta \parallel x\parallel }\hfill & \hfill t>0,x\in X\hfill \\ \hfill 0\hfill & \hfill t\le 0,x\in X\hfill \end{array}\right\$

is a fuzzy norm on X.

Definition 1.4. Let (X, N) be a fuzzy normed vector space. A sequence {x n } in X is said to be convergent or converge if there exists an x X such that limt→∞N(x n - x, t) = 1 for all t > 0. In this case, x is called the limit of the sequence {x n } in X and we denote it by N - limt→∞x n = x.

Definition 1.5. Let (X, N) be a fuzzy normed vector space. A sequence {x n } in X is called Cauchy if for each ε > 0 and each t > 0 there exists an n0 such that for all nn0and all p > 0, we have N(xn+p- x n , t) > 1 - ε.

It is well known that every convergent sequence in a fuzzy normed vector space is Cauchy. If each Cauchy sequence is convergent, then the fuzzy norm is said to be complete and the fuzzy normed vector space is called a fuzzy Banach space.

We say that a mapping f : XY between fuzzy normed vector spaces X and Y is continuous at a point x X if for each sequence {x n } converging to x0 X, then the sequence {f(x n )} converges to f(x0). If f : XY is continuous at each x X, then f : XY is said to be continuous on X.

Definition 1.6. Let X be a set. A function d : X × X → [0, ∞] is called a generalized metric on X if d satisfies the following conditions:

(a) d(x, y) = 0 if and only if x = y for all x, y X;

(b) d(x, y) = d(y, x) for all x, y X;

(c) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z X.

Theorem 1.7. ([28, 29]) Let (X, d) be a complete generalized metric space and J : XX be a strictly contractive mapping with Lipschitz constant L < 1. Then, for all x X, either d(J n x, Jn+1x) = ∞ for all nonnegative integers n or there exists a positive integer n0such that

(a) d(J n x, Jn+1x) < ∞ for all n0n0;

(b) the sequence {J n x} converges to a fixed point y* of J;

(c) y* is the unique fixed point of J in the set$Y=\left\{y\in X:d\left({J}^{{n}_{0}}x,\phantom{\rule{2.77695pt}{0ex}}y\right)<\infty \right\}$;

(d)$d\left(y,\phantom{\rule{2.77695pt}{0ex}}{y}^{*}\right)\le \frac{d\left(y,Jy\right)}{1-L}$for all y Y.

## 2. Fuzzy stability of the functional Eq. (0.1)

Throughout this section, using the fixed point and direct methods, we prove the Hyers-Ulam stability of functional Eq. (0.1) in fuzzy normed spaces.

### 2.1. Fixed point alternative approach

Throughout this subsection, using the fixed point alternative approach, we prove the Hyers-Ulam stability of functional Eq. (0.1) in fuzzy Banach spaces.

In this subsection, assume that X is a vector space and that (Y, N) is a fuzzy Banach space.

Theorem 2.1. Let φ : X3 → [0, ∞) be a function such that there exists an L < 1 with
$\phi \left(x,\phantom{\rule{2.77695pt}{0ex}}y,\phantom{\rule{2.77695pt}{0ex}}z\right)\phantom{\rule{2.77695pt}{0ex}}\le \frac{L\phi \left(2x,2y,2z\right)}{2}$
for all x, y, z X. Let f : XY be a mapping satisfying
$N\left(2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right),\phantom{\rule{2.77695pt}{0ex}}t\right)\ge \frac{t}{t+\phi \left(x,y,z\right)}$
(2.1)
for all x, y, z X and all t > 0. Then the limit
$A\left(x\right)\phantom{\rule{2.77695pt}{0ex}}:=N-\underset{n\to \infty }{lim}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)$
exists for each x X and defines a unique additive mapping A : XY such that
$N\left(f\left(x\right)-A\left(x\right),\phantom{\rule{2.77695pt}{0ex}}t\right)\ge \frac{\left(2-2L\right)t}{\left(2-2L\right)t+L\phi \left(x,2x,x\right)}.$
(2.2)
Proof. Putting y = 2x and z = x in (2.1) and replacing x by $\frac{x}{2}$, we have
$N\left(2f\left(\frac{x}{2}\right)-f\left(x\right),\phantom{\rule{2.77695pt}{0ex}}t\right)\ge \frac{t}{t+\phi \left(\frac{x}{2},x,\frac{x}{2}\right)}$
(2.3)
for all x X and t > 0. Consider the set
$S\phantom{\rule{2.77695pt}{0ex}}:=\left\{g:X\to Y\right\}$
and the generalized metric d in S defined by
$d\left(f,\phantom{\rule{2.77695pt}{0ex}}g\right)=inf\left\{\mu \in {ℝ}^{+}:N\left(g\left(x\right)-h\left(x\right),\phantom{\rule{2.77695pt}{0ex}}\mu t\right)\ge \frac{t}{t+\phi \left(x,2x,x\right)},\phantom{\rule{2.77695pt}{0ex}}\forall x\in X,\phantom{\rule{2.77695pt}{0ex}}t>0\right\},$
where inf = +∞. It is easy to show that (S, d) is complete (see [30, Lemma 2.1]). Now, we consider a linear mapping J : SS such that
$Jg\left(x\right):=2g\left(\frac{x}{2}\right)$
for all x X. Let g, h S be such that d(g, h) = ε. Then
$N\left(g\left(x\right)-h\left(x\right),\phantom{\rule{2.77695pt}{0ex}}\epsilon t\right)\ge \frac{t}{t+\phi \left(x,2x,x\right)}$
for all x X and t > 0. Hence,
$\begin{array}{cc}\hfill N\left(Jg\left(x\right)-Jh\left(x\right),\phantom{\rule{2.77695pt}{0ex}}L\epsilon t\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}N\left(2g\left(\frac{x}{2}\right)-2h\left(\frac{x}{2}\right),\phantom{\rule{2.77695pt}{0ex}}L\epsilon t\right)\hfill \\ =\phantom{\rule{1em}{0ex}}N\left(g\left(\frac{x}{2}\right)-h\left(\frac{x}{2}\right),\phantom{\rule{2.77695pt}{0ex}}\frac{L\epsilon t}{2}\right)\hfill \\ \ge \phantom{\rule{1em}{0ex}}\frac{\frac{Lt}{2}}{\frac{Lt}{2}+\phi \left(\frac{x}{2},x,\frac{x}{2}\right)}\hfill \\ \ge \phantom{\rule{1em}{0ex}}\frac{\frac{Lt}{2}}{\frac{Lt}{2}+\frac{L\phi \left(x,2x,x\right)}{2}}\hfill \\ =\phantom{\rule{1em}{0ex}}\frac{t}{t+\phi \left(x,2x,x\right)}\hfill \end{array}$
for all x X and t > 0. Thus, d(g, h) = ε implies that d(Jg, Jh) ≤ . This means that
$d\left(Jg,\phantom{\rule{2.77695pt}{0ex}}Jh\right)\le Ld\left(g,\phantom{\rule{2.77695pt}{0ex}}h\right)$
for all g, h S. It follows from (2.3) that
$\begin{array}{cc}\hfill N\left(f\left(x\right)-2f\left(\frac{x}{2}\right),\phantom{\rule{2.77695pt}{0ex}}t\right)\phantom{\rule{1em}{0ex}}\ge \phantom{\rule{1em}{0ex}}\frac{t}{t+\phi \left(\frac{x}{2},x,\frac{x}{2}\right)}\phantom{\rule{1em}{0ex}}& \ge \phantom{\rule{1em}{0ex}}\frac{t}{t+\frac{L\phi \left(x,2x,x\right)}{2}}\hfill \\ =\phantom{\rule{1em}{0ex}}\frac{\frac{2t}{L}}{\frac{2t}{L}+\phi \left(x2x,x\right)}.\hfill \end{array}$
(2.4)
Therefore,
$N\left(f\left(x\right)-2f\left(\frac{x}{2}\right),\phantom{\rule{2.77695pt}{0ex}}\frac{Lt}{2}\right)\ge \frac{t}{t+\phi \left(x,2x,x\right)}.$
(2.5)
This means that
$d\left(f,\phantom{\rule{2.77695pt}{0ex}}Jf\right)\le \frac{L}{2}.$
By Theorem 1.7, there exists a mapping A : XY satisfying the following:
1. (1)
A is a fixed point of J, that is,
$A\left(\frac{x}{2}\right)=\frac{A\left(x\right)}{2}$
(2.6)

for all x X. The mapping A is a unique fixed point of J in the set
$\mathrm{\Omega }=\left\{h\in S:d\left(g,\phantom{\rule{2.77695pt}{0ex}}h\right)<\infty \right\}.$
This implies that A is a unique mapping satisfying (2.6) such that there exists μ (0, ∞) satisfying
$N\left(f\left(x\right)-A\left(x\right),\phantom{\rule{2.77695pt}{0ex}}\mu t\right)\ge \frac{t}{t+\phi \left(x,2x,x\right)}$
for all x X and t > 0.
1. (2)
d(J n f, A) → 0 as n → ∞. This implies the equality
$N-\underset{n\to \infty }{lim}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)=A\left(x\right)$

for all x X.
1. (3)
$d\left(f,\phantom{\rule{2.77695pt}{0ex}}A\right)\le \frac{d\left(f,Jf\right)}{1-L}$ with f Ω, which implies the inequality
$d\left(f,\phantom{\rule{2.77695pt}{0ex}}A\right)\le \frac{L}{2-2L}.$

This implies that the inequality (2.2) holds. Furthermore, since
$\begin{array}{c}N\left(2A\left(\frac{x+y+z}{2}\right)-A\left(x\right)-A\left(y\right)-A\left(z\right),\phantom{\rule{2.77695pt}{0ex}}t\right)\\ \phantom{\rule{1em}{0ex}}\ge N-\underset{n\to \infty }{lim}\left({2}^{n+1}f\left(\frac{x+y+z}{{2}^{n+1}}\right)-{2}^{n}f\left(\frac{x}{{2}^{n}}\right)-{2}^{n}f\left(\frac{y}{{2}^{n}}\right)-{2}^{n}f\left(\frac{z}{{2}^{n}}\right),\phantom{\rule{2.77695pt}{0ex}}t\right)\\ \phantom{\rule{1em}{0ex}}\ge \underset{n\to \infty }{lim}\frac{\frac{t}{{2}^{n}}}{\frac{t}{{2}^{n}}+\frac{{L}^{n}\phi \left(x,y,z\right)}{{2}^{n}}}\to 1\end{array}$

for all x, y, z X, t > 0. So $N\left(A\left(\frac{x+y+z}{2}\right)-A\left(x\right)-A\left(y\right)-A\left(z\right),t\right)=1$ for all x, y, z X and all t > 0. Thus the mapping A : XY is additive, as desired.    □

Corollary 2.2. Let θ ≥ 0 and let p be a real number with p > 1. Let X be a normed vector space with norm ||.||. Let f : XY be a mapping satisfying
$N\left(2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right),\phantom{\rule{2.77695pt}{0ex}}t\right)\ge \frac{t}{t+\theta \left(\parallel x{\parallel }^{p}+\parallel y{\parallel }^{p}+\parallel z{\parallel }^{p}\right)}$
for all x, y, z X and all t > 0. Then the limit
$A\left(x\right)\phantom{\rule{2.77695pt}{0ex}}:=N-\underset{n\to \infty }{lim}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)$
exists for each x X and defines a unique additive mapping A : XY such that
$N\left(f\left(x\right)-A\left(x\right),\phantom{\rule{2.77695pt}{0ex}}t\right)\ge \frac{\left({2}^{p}-1\right)t}{\left({2}^{p}-1\right)t+\left({2}^{r-1}+1\right)\theta \parallel x{\parallel }^{p}}$

for all x X.

Proof. The proof follows from Theorem 2.1 by taking φ(x, y, z): = θ(||x|| p + ||y|| p + ||z|| p ) for all x, y, z X. Then we can choose L = 2-pand we get the desired result.    □

Theorem 2.3. Let φ : X3 → [0, ∞) be a function such that there exists an L < 1 with
$\phi \left(2x,\phantom{\rule{2.77695pt}{0ex}}2y,\phantom{\rule{2.77695pt}{0ex}}2z\right)\phantom{\rule{2.77695pt}{0ex}}\le 2L\phi \left(x,\phantom{\rule{2.77695pt}{0ex}}y,\phantom{\rule{2.77695pt}{0ex}}z\right)$
for all x, y, z X. Let f : XY be a mapping satisfying (2.1). Then
$A\left(x\right):=N-\underset{n\to \infty }{lim}\frac{f\left({2}^{n}x\right)}{{2}^{n}}$
exists for each x X and defines a unique additive mapping A : XY such that
$N\left(f\left(x\right)-A\left(x\right),\phantom{\rule{2.77695pt}{0ex}}t\right)\ge \frac{\left(2-2L\right)t}{\left(2-2L\right)t+\phi \left(x,2x,x\right)}$
(2.7)

for all x X and all t > 0.

Proof. Let (S, d) be the generalized metric space defined as in the proof of Theorem 2.1.

Consider the linear mapping J : SS such that
$Jg\left(x\right):=\frac{g\left(2x\right)}{2}$
for all x X. Let g, h S be such that d(g, h) = ε. Then
$N\left(g\left(x\right)-h\left(x\right),\phantom{\rule{2.77695pt}{0ex}}\epsilon t\right)\phantom{\rule{2.77695pt}{0ex}}\ge \frac{t}{t+\phi \left(x,2x,x\right)}$
for all x X and t > 0. Hence,
$\begin{array}{cc}\hfill N\left(Jg\left(x\right)-Jh\left(x\right),\phantom{\rule{2.77695pt}{0ex}}L\epsilon t\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}N\left(\frac{g\left(2x\right)}{2}-\frac{h\left(2x\right)}{2},\phantom{\rule{2.77695pt}{0ex}}L\epsilon t\right)\hfill \\ =\phantom{\rule{1em}{0ex}}N\left(g\left(2x\right)-h\left(2x\right),\phantom{\rule{2.77695pt}{0ex}}2L\epsilon t\right)\hfill \\ \ge \phantom{\rule{1em}{0ex}}\frac{2Lt}{2Lt+\phi \left(2x,\phantom{\rule{2.77695pt}{0ex}},\phantom{\rule{2.77695pt}{0ex}}4x,\phantom{\rule{2.77695pt}{0ex}}2x\right)}\hfill \\ \ge \phantom{\rule{1em}{0ex}}\frac{2Lt}{2Lt+2L\phi \left(x,\phantom{\rule{2.77695pt}{0ex}}2x,\phantom{\rule{2.77695pt}{0ex}}x\right)}\hfill \\ =\phantom{\rule{1em}{0ex}}\frac{t}{t+\phi \left(x,\phantom{\rule{2.77695pt}{0ex}}2x,\phantom{\rule{2.77695pt}{0ex}}x\right)}\hfill \end{array}$
for all x X and t > 0. Thus, d(g, h) = ε implies that d(Jg, Jh) ≤ . This means that
$d\left(Jg,\phantom{\rule{2.77695pt}{0ex}}Jh\right)\le Ld\left(g,\phantom{\rule{2.77695pt}{0ex}}h\right)$
for all g, h S. It follows from (2.3) that
$N\left(\frac{f\left(2x\right)}{2}-f\left(x\right),\phantom{\rule{2.77695pt}{0ex}}\frac{t}{2}\right)\ge \frac{t}{t+\phi \left(x,2x,x\right)}.$
Therefore,
$d\left(f,\phantom{\rule{2.77695pt}{0ex}}Jf\right)\le \frac{1}{2}.$
By Theorem 1.7, there exists a mapping A : XY satisfying the following:
1. (1)
A is a fixed point of J, that is,
$2A\left(x\right)=A\left(2x\right)$
(2.8)

for all x X. The mapping A is a unique fixed point of J in the set
$\mathrm{\Omega }=\left\{h\in S:d\left(g,\phantom{\rule{2.77695pt}{0ex}}h\right)<\infty \right\}.$
This implies that A is a unique mapping satisfying (2.8) such that there exists μ (0, ∞) satisfying
$N\left(f\left(x\right)-A\left(x\right),\phantom{\rule{2.77695pt}{0ex}}\mu t\right)\ge \frac{t}{t+\phi \left(x,2x,x\right)}$
for all x X and t > 0.
1. (2)
d(J n f, A) → 0 as n → ∞. This implies the equality
$N-\underset{n\to \infty }{lim}\frac{f\left({2}^{n}x\right)}{{2}^{n}}$

for all x X.
1. (3)
$d\left(f,\phantom{\rule{2.77695pt}{0ex}}A\right)\le \frac{d\left(f,Jf\right)}{1-L}$ with f Ω which implies the inequality
$d\left(f,\phantom{\rule{2.77695pt}{0ex}}A\right)\le \frac{1}{2-2L}.$

This implies that the inequality (2.7) holds.

The rest of the proof is similar to that of the proof of Theorem 2.1.    □

Corollary 2.4. Let θ ≥ 0 and let p be a real number with$0. Let X be a normed vector space with norm || . ||. Let f : XY be a mapping satisfying
$N\left(2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right),\phantom{\rule{2.77695pt}{0ex}}t\right)\ge \frac{t}{t+\theta \left(\parallel x{\parallel }^{p}.\parallel y{\parallel }^{p}.\parallel z{\parallel }^{p}\right)}$
for all x, y, z X and all t > 0. Then
$A\left(x\right):=N-\underset{n\to \infty }{lim}\frac{f\left({2}^{n}x\right)}{{2}^{n}}$
exists for each x X and defines a unique additive mapping A : XY such that
$N\left(f\left(x\right)-A\left(x\right),\phantom{\rule{2.77695pt}{0ex}}t\right)\ge \frac{\left({2}^{3p}-1\right)t}{\left({2}^{3p}-1\right)t+{2}^{3p-1}\theta \parallel x{\parallel }^{3p}}.$

for all x X.

Proof. The proof follows from Theorem 2.3 by taking φ(x, y, z): = θ(||x|| p · ||y|| p · ||z|| p ) for all x, y, z X. Then we can choose L = 2-3pand we get the desired result.    □

2.2. Direct method. In this subsection, using direct method, we prove the Hyers-Ulam stability of the functional Eq. (0.1) in fuzzy Banach spaces.

Throughout this subsection, we assume that X is a linear space, (Y, N) is a fuzzy Banach space and (Z, N') is a fuzzy normed spaces. Moreover, we assume that N(x,.) is a left continuous function on .

Theorem 2.5. Assume that a mapping f : XY satisfies the inequality
$\begin{array}{c}N\left(2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right),t\right)\\ \phantom{\rule{1em}{0ex}}\ge {N}^{\prime }\left(\phi \left(x,y,z\right),t\right)\end{array}$
(2.9)
for all x, y, z X, t > 0 and φ : X3Z is a mapping for which there is a constant r satisfying$0<\phantom{\rule{2.77695pt}{0ex}}\mid r\mid \phantom{\rule{2.77695pt}{0ex}}<\frac{1}{2}$and
${N}^{\prime }\left(\phi \left(x,y,z\right),t\right)\ge {N}^{\prime }\left(\phi \left(2x,2y,2z\right),\frac{t}{\mid r\mid }\right)$
(2.10)
for all x, y, z X and all t > 0. Then there exist a unique additive mapping A : XY satisfying (0.1) and the inequality
$N\left(f\left(x\right)-A\left(x\right),t\right)\ge {N}^{\prime }\left(\phi \left(x,2x,x\right),\frac{\left(1-2\mid r\mid \right)t}{\mid r\mid }\right)$
(2.11)

for all x X and all t > 0.

Proof. It follows from (2.10) that
${N}^{\prime }\left(\phi \left(\frac{x}{{2}^{j}},\frac{y}{{2}^{j}},\frac{z}{{2}^{j}}\right),t\right)\ge {N}^{\prime }\left(\phi \left(x,y,z\right),\frac{t}{\mid r{\mid }^{j}}\right).$
(2.12)
So
${N}^{\prime }\left(\phi \left(\frac{x}{{2}^{j}},\frac{y}{{2}^{j}},\frac{z}{{2}^{j}}\right),\mid r{\mid }^{j}t\right)\ge {N}^{\prime }\left(\phi \left(x,y,z\right),t\right)$
for all x, y, z X and all t > 0. Substituting y = 2x and z = x in (2.9), we obtain
$N\left(f\left(2x\right)-2f\left(x\right),t\right)\ge {N}^{\prime }\left(\phi \left(x,2x,x\right),t\right)$
(2.13)
So
$N\left(f\left(x\right)-2f\left(\frac{x}{2}\right),t\right)\ge {N}^{\prime }\left(\phi \left(\frac{x}{2},x,\frac{x}{2}\right),t\right)$
(2.14)
for all x X and all t > 0. Replacing x by $\frac{x}{{2}^{j}}$ in (2.14), we have
$\begin{array}{cc}\hfill N\left({2}^{j+1}f\left(\frac{x}{{2}^{j+1}}\right)-{2}^{j}f\left(\frac{x}{{2}^{j}}\right){,2}^{j}t\right)& \ge {N}^{\prime }\left(\phi \left(\frac{x}{{2}^{j+1}},\frac{x}{{2}^{j}},\frac{x}{{2}^{j+1}}\right),t\right)\hfill \\ \ge {N}^{\prime }\left(\phi \left(x,2x,x\right),\frac{t}{\mid r{\mid }^{j+1}}\right)\hfill \end{array}$
(2.15)
for all x X, all t > 0 and any integer j ≥ 0. So
$\begin{array}{c}N\left(f\left(x\right)-{2}^{n}f\left(\frac{x}{{2}^{n}}\right),\sum _{j=0}^{n-1}{2}^{j}\mid r{\mid }^{j+1}t\right)\\ =N\left(\sum _{j=0}^{n-1}\left[{2}^{j+1}f\left(\frac{x}{{2}^{j+1}}\right)-{2}^{j}f\left(\frac{x}{{2}^{j}}\right)\right],\sum _{j=0}^{n-1}{2}^{j}\mid r{\mid }^{j+1}t\right)\\ \phantom{\rule{1em}{0ex}}\ge \underset{0\le j\le n-1}{min}\left\{N\left({2}^{j+1}f\left(\frac{x}{{2}^{j+1}}\right)-{2}^{j}f\left(\frac{x}{{2}^{j}}\right){,2}^{j}\mid r{\mid }^{j+1}t\right)\right\}\\ \phantom{\rule{1em}{0ex}}\ge {N}^{\prime }\left(\phi \left(x,2x,x\right),t\right).\end{array}$
(2.16)
Replacing x by $\frac{x}{{2}^{p}}$ in the above inequality, we find that
$\begin{array}{cc}\hfill N\left({2}^{n+p}f\left(\frac{x}{{2}^{n+p}}\right)-{2}^{p}f\left(\frac{x}{{2}^{p}}\right),\sum _{j=0}^{n-1}{2}^{j}\mid r{\mid }^{j+1}t\right)& \ge {N}^{\prime }\left(\phi \left(\frac{x}{{2}^{p}},\frac{2x}{{2}^{p}},\frac{x}{{2}^{p}}\right),t\right)\hfill \\ \ge {N}^{\prime }\left(\phi \left(x,2x,x\right),\frac{t}{\mid r\mid p}\right)\hfill \end{array}$
for all x X, t > 0 and all integers n ≥ 0, p ≥ 0. So
$N\left({2}^{n+p}f\left(\frac{x}{{2}^{n+p}}\right)-{2}^{p}f\left(\frac{x}{{2}^{p}}\right),\sum _{j=0}^{n-1}{2}^{j+p}\mid r{\mid }^{j+p+1}t\right)\ge {N}^{\prime }\left(\phi \left(x,2x,x\right),t\right)$
for all x X, t > 0 and all integers n > 0, p ≥ 0. Hence, one obtains
$N\left({2}^{n+p}f\left(\frac{x}{{2}^{n+p}}\right)-{2}^{p}f\left(\frac{x}{{2}^{p}}\right),t\right)\ge {N}^{\prime }\left(\phi \left(x,2x,x\right),\frac{t}{{\sum }_{j=0}^{n-1}{2}^{j+p}\mid r{\mid }^{j+p+1}}\right)$
(2.17)
for all x X, t > 0 and all integers n > 0, p ≥ 0. Since the series ${\sum }_{j=0}^{\infty }{2}^{j}\mid r{\mid }^{j}$ is convergent, by taking the limit p → ∞ in the last inequality, we know that a sequence $\left\{{2}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}$ is a Cauchy sequence in the fuzzy Banach space (Y, N) and so it converges in Y. Therefore, a mapping A: XY defined by
$A\left(x\right):=N-\underset{n\to \infty }{lim}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)$
is well defined for all x X. It means that
$\underset{n\to \infty }{lim}N\left(A\left(x\right)-{2}^{n}f\left(\frac{x}{{2}^{n}}\right),t\right)=1$
(2.18)
for all x X and all t > 0. In addition, it follows from (2.17) that
$N\left({2}^{n}f\left(\frac{x}{{2}^{n}}\right)-f\left(x\right),t\right)\ge {N}^{\prime }\left(\phi \left(x,2x,x\right),\frac{t}{{\sum }_{j=0}^{n-1}{2}^{j}\mid r{\mid }^{j+1}}\right)$
for all x X and all t > 0. So
$\begin{array}{cc}\hfill N\left(f\left(x\right)-A\left(x\right),t\right)& \ge min\left\{N\left(f\left(x\right)-{2}^{n}f\left(\frac{x}{{2}^{n}}\right),\left(1-\epsilon \right)t\right),N\left(A\left(x\right)-{2}^{n}f\left(\frac{x}{{2}^{n}}\right),\epsilon t\right)\right\}\hfill \\ \ge {N}^{\prime }\left(\phi \left(x,2x,x\right),\frac{t}{{\sum }_{j=0}^{n-1}{2}^{j}\mid r{\mid }^{j+1}}\right)\hfill \\ \ge {N}^{\prime }\left(\phi \left(x,2x,x\right),\frac{\left(1-2\mid r\mid \right)\epsilon t}{\mid r\mid }\right)\hfill \end{array}$
for sufficiently large n and for all x X, t > 0 and N with 0 < N < 1. Since N is arbitrary and N' is left continuous, we obtain
$N\left(f\left(x\right)-A\left(x\right),t\right)\ge {N}^{\prime }\left(\phi \left(x,2x,x\right),\frac{\left(1-2\mid r\mid \right)t}{\mid r\mid }\right)$
for all x X and t > 0. It follows from (2.9) that
$\begin{array}{cc}\hfill N& \left({2}^{n+1}f\left(\frac{x+y+z}{{2}^{n+1}}\right)-{2}^{n}f\left(\frac{x}{{2}^{n}}\right)-{2}^{n}f\left(\frac{y}{{2}^{n}}\right)-{2}^{n}f\left(\frac{z}{{2}^{n}}\right),t\right)\hfill \\ \ge {N}^{\prime }\left(\phi \left(\frac{x}{{2}^{n}},\frac{y}{{2}^{n}},\frac{z}{{2}^{n}}\right),\frac{t}{{2}^{n}}\right)\hfill \\ \ge {N}^{\prime }\left(\phi \left(x,y,z\right),\frac{t}{{2}^{n}\mid r{\mid }^{n}}\right)\hfill \end{array}$
for all x, y, z X, t > 0 and all n . Since
$\underset{n\to \infty }{lim}{N}^{\prime }\left(\phi \left(x,y,z\right),\frac{t}{{2}^{n}\mid r{\mid }^{n}}\right)=1$
and so
$N\left({2}^{n+1}f\left(\frac{x+y+z}{{2}^{n+1}}\right)-{2}^{n}f\left(\frac{x}{{2}^{n}}\right)-{2}^{n}f\left(\frac{y}{{2}^{n}}\right)-{2}^{n}f\left(\frac{z}{{2}^{n}}\right),t\right)\to 1$
for all x, y, z X and all t > 0. Therefore, we obtain in view of (2.18)
which implies
$2A\left(\frac{x+y+z}{2}\right)=A\left(x\right)+A\left(y\right)+A\left(z\right)$

for all x, y, z X. Thus, A: XY is a mapping satisfying the Eq. (0.1) and the inequality (2.11).

To prove the uniqueness, assume that there is another mapping L : XY which satisfies the inequality (2.11). Since $L\left(x\right)={2}^{n}L\left(\frac{x}{{2}^{n}}\right)$ for all x X, we have

for all t > 0. Therefore, A(x) = L(x) for all x X, this completes the proof.    □

Corollary 2.6. Let X be a normed spaces and (, N') a fuzzy Banach space. Assume that there exist real numbers θ ≥ 0 and 0 < p < 1 such that a mapping f : XY satisfies the following inequality
$N\left(2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right),t\right)\ge {N}^{\prime }\left(\theta \left(\parallel x{\parallel }^{p}+\parallel y{\parallel }^{p}+\parallel z{\parallel }^{p}\right),t\right)$
for all x, y, z X and t > 0. Then there is a unique additive mapping A : XY satisfying (0.1) and the inequality
$N\left(f\left(x\right)-A\left(x\right),t\right)\ge {N}^{\prime }\left(\theta \parallel x{\parallel }^{p},\frac{2t}{{2}^{r}+2}\right)$

Proof. Let φ(x, y, z): = θ(||x|| p + ||y|| p + ||z|| p ) and $\mid r\mid \phantom{\rule{2.77695pt}{0ex}}=\frac{1}{4}$. Applying Theorem 2.5, we get the desired result.    □

Theorem 2.7. Assume that a mapping f : XY satisfies (2.9) and φ : X2Z is a mapping for which there is a constant r satisfying 0 < |r| < 2 and
${N}^{\prime }\left(\phi \left(2x,2y,2z\right),\mid r\mid t\right)\ge {N}^{\prime }\left(\phi \left(x,y,z\right),t\right)$
(2.19)
for all x, y, z X and all t > 0. Then there is a unique additive mapping A : XY satisfying (0.1) and the following inequality
$N\left(f\left(x\right)-A\left(x\right),t\right)\ge {N}^{\prime }\left(\phi \left(x,2x,x\right),\left(2-\mid r\mid \right)t\right).$
(2.20)

for all x X and all t > 0.

Proof. It follows from (2.13) that
$N\left(\frac{f\left(2x\right)}{2}-f\left(x\right),\frac{t}{2}\right)\ge {N}^{\prime }\left(\phi \left(x,2x,x\right),t\right)$
(2.21)
for all x X and all t > 0. Replacing x by 2 n x in (2.21), we obtain
$N\left(\frac{f\left({2}^{n+1}x\right)}{{2}^{n+1}}-\frac{f\left({2}^{n}x\right)}{{2}^{n}},\frac{t}{{2}^{n+1}}\right)\ge {N}^{\prime }\left(\phi \left({2}^{n}x,{2}^{n+1}x,{2}^{n}x\right),t\right)\ge {N}^{\prime }\left(\phi \left(x,2x,x\right),\frac{t}{\mid r{\mid }^{n}}\right).$
So
$N\left(\frac{f\left({2}^{n+1}x\right)}{{2}^{n+1}}-\frac{f\left({2}^{n}x\right)}{{2}^{n}},\frac{\mid r{\mid }^{n}t}{{2}^{n+1}}\right)\ge {N}^{\prime }\left(\phi \left(x,2x,x\right),t\right)$
(2.22)
for all x X and all t > 0. Proceeding as in the proof of Theorem 2.5, we obtain that
$N\left(f\left(x\right)-\frac{f\left({2}^{n}x\right)}{{2}^{n}},\sum _{j=0}^{n-1}\frac{\mid r{\mid }^{j}t}{{2}^{j+1}}\right)\ge {N}^{\prime }\left(\phi \left(x,2x,x\right),t\right)$
for all x X, all t > 0 and all integers n > 0. So
$N\left(f\left(x\right)-\frac{f\left({2}^{n}x\right)}{{2}^{n}},t\right)\ge {N}^{\prime }\left(\phi \left(x,2x,x\right),\frac{t}{{\sum }_{j=0}^{n-1}\frac{\mid r{\mid }^{j}}{{2}^{j+1}}}\right)\ge {N}^{\prime }\left(\phi \left(x,2x,x\right),\left(2-\mid r\mid \right)t\right).$

The rest of the proof is similar to the proof of Theorem 2.5.    □

Corollary 2.8. Let X be a normed spaces and (, N') a fuzzy Banach space. Assume that there exist real numbers θ ≥ 0 and$0such that a mapping f : XY satisfies the following inequality
$N\left(2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right),t\right)\ge {N}^{\prime }\left(\theta \left(\parallel x{\parallel }^{p}\cdot \parallel y{\parallel }^{p}\cdot \parallel z{\parallel }^{p}\right),t\right)$
for all x, y, z X and t > 0. Then there is a unique additive mapping A : XY satisfying (0.1) and the inequality
$N\left(f\left(x\right)-A\left(x\right),t\right)\ge {N}^{\prime }\left(\theta \parallel x{\parallel }^{p},\frac{t}{{2}^{r}+2}\right)$

Proof. Let $\phi \left(x,y,z\right):=\theta \left(\parallel x{\parallel }^{{p}_{1}}\cdot \parallel y{\parallel }^{{p}_{2}}\cdot \parallel z{\parallel }^{{p}_{3}}\right)$ and |r| = 1. Applying Theorem 2.7, we get the desired result.    □

## Authors’ Affiliations

(1)
Department of Mathematics, College of Sciences, Yasouj University, 75914-353 Yasouj, Iran
(2)
Department of Mathematics, Firoozabad Branch, Islamic Azad University, Firoozabad, Iran
(3)
Department of Mathematics, Research Institute for Natural Sciences, Hanyang University, Seoul, 133-791, Korea
(4)
Department of Mathematics, Daejin University, Kyeonggi, 487-711, Korea

## References 