Fuzzy Hyers-Ulam stability of an additive functional equation

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.

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 [1] in 1940. In the next year, Hyers [2] gave a positive answer to the above question for additive groups under the assumption that the groups are Banach spaces. In 1978, Rassias [3] proved a generalization of the Hyers' theorem for additive mappings.

Theorem 1.1. (Th.M. Rassias) Let f : X → Y be a mapping from a normed vector space X into a Banach space Y subject to the inequality

$$\parallel f\left(x+y\right)-f\left(x\right)-f\left(y\right)\parallel \le \epsilon \left(\parallel x{\parallel }^p+\parallel y{\parallel }^p\right)$$

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 : X → Y 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 ∈ ℝ, then L is ℝ-linear.

Furthermore, in 1994, a generalization of Rassias' theorem was obtained by Gǎvruta [4] 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 [5] for mappings f : X → Y, where X is a normed space and Y is a Banach space. In 1984, Cholewa [6] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group and, in 2002, Czerwik [7] proved the Hyers-Ulam stability of the quadratic functional equation. The reader is referred to ([8–20]) and references therein for detailed information on stability of functional equations.

Katsaras [21] 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 [24], following Cheng and Mordeson [25], gave an idea of fuzzy norm in such a manner that the corresponding fuzzy metric is of Karmosil and Michalek type [26]. They established a decomposition theorem of a fuzzy norm into a family of crisp norms and investigated some properties of fuzzy normed spaces [27].

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

(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) $N\left(cx,t\right)=N\left(x,\frac{t}{\mid c\mid }\right)$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 ℝ and lim_t→∞N(x, t) = 1;

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

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

$$N\left(x,t\right)=\left\{\begin{array}{cc}\hfill \frac{\alpha t}{\alpha t+\beta \parallel x\parallel }\hfill & \hfill t>0,x\in X\hfill \\ \hfill 0\hfill & \hfill t\le 0,x\in X\hfill \end{array}\right.$$

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 lim_t→∞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 - lim_t→∞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 ∈ ℕ such that for all n ≥ n₀and all p > 0, we have N(x_n+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 : X → Y 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 : X → Y is continuous at each x ∈ X, then f : X → Y 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 : X → X be a strictly contractive mapping with Lipschitz constant L < 1. Then, for all x ∈ X, either d(Jⁿx, Jⁿ⁺¹x) = ∞ for all nonnegative integers n or there exists a positive integer n₀such that

(a) d(Jⁿx, Jⁿ⁺¹x) < ∞ for all n₀ ≥ n₀;

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

(c) y* is the unique fixed point of J in the set$Y=\left\{y\in X:d\left(J^{n_0}x,y\right)<\infty \right\}$;

(d)$d\left(y,y^{*}\right)\le \frac{d\left(y,Jy\right)}{1-L}$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

$$\phi \left(x,y,z\right)\le \frac{L\phi \left(2x,2y,2z\right)}{2}$$

for all x, y, z ∈ X. Let f : X → Y be a mapping satisfying

$$N\left(2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right),t\right)\ge \frac{t}{t+\phi \left(x,y,z\right)}$$

(2.1)

for all x, y, z ∈ X and all t > 0. Then the limit

$$A\left(x\right):=N-\underset{n\to \infty }{lim}{2}^{n}f\left(\frac{x}{2^n}\right)$$

exists for each x ∈ X and defines a unique additive mapping A : X → Y such that

$$N\left(f\left(x\right)-A\left(x\right),t\right)\ge \frac{\left(2-2L\right)t}{\left(2-2L\right)t+L\phi \left(x,2x,x\right)}.$$

(2.2)

Proof. Putting y = 2x and z = x in (2.1) and replacing x by $\frac{x}{2}$, we have

$$N\left(2f\left(\frac{x}{2}\right)-f\left(x\right),t\right)\ge \frac{t}{t+\phi \left(\frac{x}{2},x,\frac{x}{2}\right)}$$

(2.3)

for all x ∈ X and t > 0. Consider the set

$$S:=\left\{g:X\to Y\right\}$$

and the generalized metric d in S defined by

$$d\left(f,g\right)=inf\left\{\mu \in {ℝ}^{+}:N\left(g\left(x\right)-h\left(x\right),\mu t\right)\ge \frac{t}{t+\phi \left(x,2x,x\right)},\forall x\in X,t>0\right\},$$

where inf ∅ = +∞. It is easy to show that (S, d) is complete (see [30, Lemma 2.1]). Now, we consider a linear mapping J : S → S 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

$$N\left(g\left(x\right)-h\left(x\right),\epsilon t\right)\ge \frac{t}{t+\phi \left(x,2x,x\right)}$$

for all x ∈ X and t > 0. Hence,

$$\begin{array}{cc}\hfill N\left(Jg\left(x\right)-Jh\left(x\right),L\epsilon t\right)& =N\left(2g\left(\frac{x}{2}\right)-2h\left(\frac{x}{2}\right),L\epsilon t\right)\hfill \\ =N\left(g\left(\frac{x}{2}\right)-h\left(\frac{x}{2}\right),\frac{L\epsilon t}{2}\right)\hfill \\ \ge \frac{\frac{Lt}{2}}{\frac{Lt}{2}+\phi \left(\frac{x}{2},x,\frac{x}{2}\right)}\hfill \\ \ge \frac{\frac{Lt}{2}}{\frac{Lt}{2}+\frac{L\phi \left(x,2x,x\right)}{2}}\hfill \\ =\frac{t}{t+\phi \left(x,2x,x\right)}\hfill \end{array}$$

for all x ∈ X and t > 0. Thus, d(g, h) = ε implies that d(Jg, Jh) ≤ Lε. This means that

$$d\left(Jg,Jh\right)\le Ld\left(g,h\right)$$

for all g, h ∈ S. It follows from (2.3) that

$$\begin{array}{cc}\hfill N\left(f\left(x\right)-2f\left(\frac{x}{2}\right),t\right)\ge \frac{t}{t+\phi \left(\frac{x}{2},x,\frac{x}{2}\right)}& \ge \frac{t}{t+\frac{L\phi \left(x,2x,x\right)}{2}}\hfill \\ =\frac{\frac{2t}{L}}{\frac{2t}{L}+\phi \left(x2x,x\right)}.\hfill \end{array}$$

(2.4)

Therefore,

$$N\left(f\left(x\right)-2f\left(\frac{x}{2}\right),\frac{Lt}{2}\right)\ge \frac{t}{t+\phi \left(x,2x,x\right)}.$$

(2.5)

This means that

$$d\left(f,Jf\right)\le \frac{L}{2}.$$

By Theorem 1.7, there exists a mapping A : X → Y satisfying the following:

• (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

$$\mathrm{\Omega }=\left\{h\in S:d\left(g,h\right)<\infty \right\}.$$

This implies that A is a unique mapping satisfying (2.6) such that there exists μ ∈ (0, ∞) satisfying

$$N\left(f\left(x\right)-A\left(x\right),\mu t\right)\ge \frac{t}{t+\phi \left(x,2x,x\right)}$$

for all x ∈ X and t > 0.

• (2) d(Jⁿf, 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.

• (3) $d\left(f,A\right)\le \frac{d\left(f,Jf\right)}{1-L}$ with f ∈ Ω, which implies the inequality

  $$d\left(f,A\right)\le \frac{L}{2-2L}.$$

This implies that the inequality (2.2) holds. Furthermore, since

$$\begin{array}{c}N\left(2A\left(\frac{x+y+z}{2}\right)-A\left(x\right)-A\left(y\right)-A\left(z\right),t\right)\\ \ge N-\underset{n\to \infty }{lim}\left(2^{n+1}f\left(\frac{x+y+z}{2^{n+1}}\right)-2^{n}f\left(\frac{x}{2^n}\right)-2^{n}f\left(\frac{y}{2^n}\right)-2^{n}f\left(\frac{z}{2^n}\right),t\right)\\ \ge \underset{n\to \infty }{lim}\frac{\frac{t}{2^n}}{\frac{t}{2^n}+\frac{L^n\phi \left(x,y,z\right)}{2^n}}\to 1\end{array}$$

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 : X → Y 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 : X → Y be a mapping satisfying

$$N\left(2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right),t\right)\ge \frac{t}{t+\theta \left(\parallel x{\parallel }^p+\parallel y{\parallel }^p+\parallel z{\parallel }^p\right)}$$

for all x, y, z ∈ X and all t > 0. Then the limit

$$A\left(x\right):=N-\underset{n\to \infty }{lim}{2}^{n}f\left(\frac{x}{2^n}\right)$$

exists for each x ∈ X and defines a unique additive mapping A : X → Y such that

$$N\left(f\left(x\right)-A\left(x\right),t\right)\ge \frac{\left(2^p-1\right)t}{\left(2^p-1\right)t+\left(2^{r-1}+1\right)\theta \parallel x{\parallel }^p}$$

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

$$\phi \left(2x,2y,2z\right)\le 2L\phi \left(x,y,z\right)$$

for all x, y, z ∈ X. Let f : X → Y 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 : X → Y such that

$$N\left(f\left(x\right)-A\left(x\right),t\right)\ge \frac{\left(2-2L\right)t}{\left(2-2L\right)t+\phi \left(x,2x,x\right)}$$

(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 : S → S 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

$$N\left(g\left(x\right)-h\left(x\right),\epsilon t\right)\ge \frac{t}{t+\phi \left(x,2x,x\right)}$$

for all x ∈ X and t > 0. Hence,

$$\begin{array}{cc}\hfill N\left(Jg\left(x\right)-Jh\left(x\right),L\epsilon t\right)& =N\left(\frac{g\left(2x\right)}{2}-\frac{h\left(2x\right)}{2},L\epsilon t\right)\hfill \\ =N\left(g\left(2x\right)-h\left(2x\right),2L\epsilon t\right)\hfill \\ \ge \frac{2Lt}{2Lt+\phi \left(2x,,4x,2x\right)}\hfill \\ \ge \frac{2Lt}{2Lt+2L\phi \left(x,2x,x\right)}\hfill \\ =\frac{t}{t+\phi \left(x,2x,x\right)}\hfill \end{array}$$

for all x ∈ X and t > 0. Thus, d(g, h) = ε implies that d(Jg, Jh) ≤ Lε. This means that

$$d\left(Jg,Jh\right)\le Ld\left(g,h\right)$$

for all g, h ∈ S. It follows from (2.3) that

$$N\left(\frac{f\left(2x\right)}{2}-f\left(x\right),\frac{t}{2}\right)\ge \frac{t}{t+\phi \left(x,2x,x\right)}.$$

Therefore,

$$d\left(f,Jf\right)\le \frac{1}{2}.$$

By Theorem 1.7, there exists a mapping A : X → Y satisfying the following:

• (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

$$\mathrm{\Omega }=\left\{h\in S:d\left(g,h\right)<\infty \right\}.$$

This implies that A is a unique mapping satisfying (2.8) such that there exists μ ∈ (0, ∞) satisfying

$$N\left(f\left(x\right)-A\left(x\right),\mu t\right)\ge \frac{t}{t+\phi \left(x,2x,x\right)}$$

for all x ∈ X and t > 0.

• (2) d(Jⁿ 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.

• (3) $d\left(f,A\right)\le \frac{d\left(f,Jf\right)}{1-L}$ with f ∈ Ω which implies the inequality

  $$d\left(f,A\right)\le \frac{1}{2-2L}.$$

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$0<p<\frac{1}{3}$. Let X be a normed vector space with norm || . ||. Let f : X → Y be a mapping satisfying

$$N\left(2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right),t\right)\ge \frac{t}{t+\theta \left(\parallel x{\parallel }^p.\parallel y{\parallel }^p.\parallel z{\parallel }^p\right)}$$

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 : X → Y such that

$$N\left(f\left(x\right)-A\left(x\right),t\right)\ge \frac{\left(2^{3p}-1\right)t}{\left(2^{3p}-1\right)t+2^{3p-1}\theta \parallel x{\parallel }^{3p}}.$$

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 ℝ.

Theorem 2.5. Assume that a mapping f : X → Y satisfies the inequality

$$\begin{array}{c}N\left(2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right),t\right)\\ \ge N^{\prime }\left(\phi \left(x,y,z\right),t\right)\end{array}$$

(2.9)

for all x, y, z ∈ X, t > 0 and φ : X³ → Z is a mapping for which there is a constant r ∈ ℝ satisfying$0<\mid r\mid <\frac{1}{2}$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 : X → Y 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

$$\begin{array}{c}N\left(f\left(x\right)-2^{n}f\left(\frac{x}{2^n}\right),\sum _{j=0}^{n-1}{2}^j\mid r{\mid }^{j+1}t\right)\\ =N\left(\sum _{j=0}^{n-1}\left[2^{j+1}f\left(\frac{x}{2^{j+1}}\right)-2^{j}f\left(\frac{x}{2^j}\right)\right],\sum _{j=0}^{n-1}{2}^j\mid r{\mid }^{j+1}t\right)\\ \ge \underset{0\le j\le n-1}{min}\left\{N\left(2^{j+1}f\left(\frac{x}{2^{j+1}}\right)-2^{j}f\left(\frac{x}{2^j}\right){,2}^j\mid r{\mid }^{j+1}t\right)\right\}\\ \ge N^{\prime }\left(\phi \left(x,2x,x\right),t\right).\end{array}$$

(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: X → Y 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 ∈ 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)

$$\begin{array}{cc}\hfill N& \left(2A\left(\frac{x+y+z}{2}\right)-A\left(x\right)-A\left(y\right)-A\left(z\right),t\right)\hfill \\ \ge min\left\{N(A\left(\frac{x+y+z}{2}\right)-A\left(x\right)-A\left(y\right)-A\left(z\right)-2^{n+1}f\left(\frac{x+y+z}{2^{n+1}}\right)\right.\hfill \\ \left.-2^{n}f\left(\frac{x}{2^n}\right)-2^{n}f\left(\frac{y}{2^n}\right)-2^{n}f\left(\frac{z}{2^n}\right),\frac{t}{2}\right),\hfill \\ N\left.\left(2^{n+1}f\left(\frac{x+y+z}{2^{n+1}}\right)-2^{n}f\left(\frac{x}{2^n}\right)-2^{n}f\left(\frac{y}{2^n}\right)-2^{n}f\left(\frac{z}{2^n}\right),\frac{t}{2}\right)\right\}\hfill \\ =N\left(2^{n+1}f\left(\frac{x+y+z}{2^{n+1}}\right)-2^{n}f\left(\frac{x}{2^n}\right)-2^{n}f\left(\frac{y}{2^n}\right)-2^{n}f\left(\frac{z}{2^n}\right),\frac{t}{2}\right)\hfill \\ \ge N^{\prime }\left(\phi \left(x,y,z\right),\frac{t}{2^{n+1}\mid r{\mid }^n}\right)\to 1\mathsf{\text{as }}n\to \infty \hfill \end{array}$$

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: X → Y is a mapping satisfying the Eq. (0.1) and the inequality (2.11).

To prove the uniqueness, assume that there is another mapping L : X → Y 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

$$\begin{array}{cc}\hfill N& \left(A\left(x\right)-L\left(x\right),t\right)=\left(2^{n}A\left(\frac{x}{2^n}\right)-2^{n}L\left(\frac{x}{2^n}\right),t\right)\hfill \\ \ge min\left\{N\left(2^{n}A\left(\frac{x}{2^n}\right)-2^{n}f\left(\frac{x}{2^n}\right),\frac{t}{2}\right),N\left(2^{n}f\left(\frac{x}{2^n}\right)-2^{n}L\left(\frac{x}{2^n}\right),\frac{t}{2}\right)\right\}\hfill \\ \ge N^{\prime }\left(\phi \left(\frac{x}{2^n},\frac{2x}{2^n},\frac{x}{2^n}\right),\frac{\left(1-2\mid r\mid \right)t}{\mid r\mid 2^{n+1}}\right)\hfill \\ \ge N\left(\phi \left(x,2x,x\right),\frac{\left(1-2\mid r\mid \right)t}{\mid r{\mid }^{n+1}{2}^{n+1}}\right)\to 1\mathsf{\text{as }}n\to \infty \hfill \end{array}$$

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 (ℝ, N') a fuzzy Banach space. Assume that there exist real numbers θ ≥ 0 and 0 < p < 1 such that a mapping f : X → Y 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 : X → Y 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 $\mid r\mid =\frac{1}{4}$. Applying Theorem 2.5, we get the desired result.    □

Theorem 2.7. Assume that a mapping f : X → Y satisfies (2.9) and φ : X² → Z is a mapping for which there is a constant r ∈ 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 : X → Y 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 ⁿx 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 (ℝ, N') a fuzzy Banach space. Assume that there exist real numbers θ ≥ 0 and$0<p<\frac{1}{3}$such that a mapping f : X → Y 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 : X → Y 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.    □