Fuzzy Hyers-Ulam approximation of a mixed AQ mapping

Abstract

In this article, we adopt the fixed point and direct methods to prove the Hyers-Ulam stability for an additive-quadratic functional equation in fuzzy Banach spaces.

Mathematics Subject Classification 2010: 39B52, 46S40; 26E50.

1. Introduction and preliminaries

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 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] extended the theorem of Hyers by considering the unbounded Cauchy difference ∥f(x + y)- f(x) - f(y)∥ ≤ ε(∥x∥^p+ ∥y∥^p), (ε > 0, p ∈[0,1)). Furthermore, in 1994, a generalization of Rassias' theorem was obtained by Găvruta [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.

Katsaras [8] defined a fuzzy norm on a vector space to construct a fuzzy vector topological structure on the space. Some mathematicians have defined fuzzy norms of a vector space from various points of view (see [9–17]).

In particular, Bag and Samanta [18], following Cheng and Mordeson [19], gave an idea of fuzzy norm in such a manner that the corresponding fuzzy metric is of Karmosil and Michalek type [20]. They established a decomposition theorem of a fuzzy norm into a family of crisp norms and investigated some properties of fuzzy normed spaces [21].

In this article, we prove the Hyers-Ulam stability of the following additive-quadratic functional equation

$$\begin{array}{ll}\hfill r^{2}f\left(\frac{x+y+z}{r}\right)& +r^{2}f\left(\frac{x-y+z}{r}\right)+r^{2}f\left(\frac{x+y-z}{r}\right)+r^{2}f\left(\frac{-x+y+z}{r}\right)\\ =4f\left(x\right)+4f\left(y\right)+4f\left(z\right)\end{array}$$

(1.1)

where r is a positive real number, in fuzzy Banach spaces.

Lee and Jun [22] proved that an even (odd) mapping f : X → Y satisfies the functional Equation (1.1) if and only if the mapping f : X → Y is quadratic (additive). Moreover, they proved the Hyers-Ulam stability of the functional Equation (1.1) in non-Archimedean normed spaces. The stability problems of several functional equations have been extensively investigated by a number of authors, and there are many interesting results concerning this problem (see [23–33]).

Definition 1.1. [18] 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) $N\left(cx,t\right)=N\left(x,\frac{t}{\left|c\right|}\right)$ 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 1.2. Let (X, ∥⋅∥) be a normed linear space and α, β > 0. Then

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

is a fuzzy norm on X.

Definition 1.3. [18] 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 1.4. [18] 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 ≥ n₀ and 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.

Example 1.5. Let N : X × ℝ → [0,1] be a fuzzy norm on ℝ defined by

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

Then (ℝ, N) is a fuzzy Banach space. Let {x _n } be a Cauchy sequence in ℝ, δ > 0 and$\epsilon =\frac{\delta }{1+\delta }$. Then there exist m ∈ ℕ such that for all n ≥ m and all p > 0, we have

$$\frac{1}{1+\left|x_{n+p}-x_n\right|}\ge 1-\epsilon .$$

So |x_n+p-x _n | < δ for all n ≥ m and all p > 0. Therefore {x _n } is a Cauchy sequence in (ℝ, |·|). Let x _n → x₀ ∈ ℝ as n → ∞. Then lim_n→∞N (x _n - x₀, t) = 1 for all t > 0.

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 ([21]).

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:

• (1) d(x, y) = 0 if and only if x = y for all x, y ∈ X;

• (2) d(x, y) = d(y, x) for all x, y ∈ X;

• (3) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X.

Theorem 1.7. 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

$$d\left(J^{n}x,J^{n+1}x\right)=\infty$$

for all nonnegative integers n or there exists a positive integer n ₀ such that

• (1) d(J ⁿ x, J ^{n+ 1} x) < ∞ for all n ₀ ≥ n ₀;

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

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

• (4) $d\left(y,y^{*}\right)\le \frac{1}{1-L}d\left(y,Jy\right)$ for all y ∈ Y.

2. Fuzzy stability of the functional equation (1.1): fixed point method

In this section, using the fixed point method, we prove the Hyers-Ulam stability of the additive-quadratic functional equation (1.1) in fuzzy Banach spaces.

Remark 2.1. If f be an odd mapping satisfying (1.1) and r = 2, then an additive mapping f is nonzero in general. But if r is a rational number and r ≠ 2 in (1.1), then f ≡ 0. So, to avoid the trivial case, let r = 2 or r is an positive irrational real number.

Theorem 2.2. Let φ: X³ → [0, ∞) be a function such that there exists an α < 1 with

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

(2.1)

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

$$\begin{array}{c}N\left(r^{2}f\left(\frac{x+y+z}{r}\right)+r^{2}f\left(\frac{x-y+z}{r}\right)+r^{2}f\left(\frac{x+y-z}{r}\right)+r^{2}f\left(\frac{-x+y+z}{r}\right)\right.\\ \left.-4f\left(x\right)-4f\left(y\right)-4f\left(z\right),t\right)\ge \frac{1}{t+\phi \left(x,y,z\right)}\end{array}$$

(2.2)

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

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

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

$$N\left(f\left(x\right)-A\left(x\right),t\right)\ge \frac{\left(4-4\alpha \right)t}{\left(4-4\alpha \right)t+\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)}$$

(2.3)

Proof. Note that f (0) = 0 and f (-x) = -f(x) for all x ∈ X since f is an odd function. Putting y = z = 0 in (2.2) and replacing x by 2x, we get

$$N\left(2r^{2}f\left(\frac{2x}{r}\right)-4f\left(2x\right),t\right)\ge \frac{t}{t+\phi \left(2x,0,0\right)}$$

(2.4)

for all x ∈ X and all t > 0. Putting y = x and z = 0 in (2.2), we have

$$N\left(2r^{2}f\left(\frac{2x}{r}\right)-8f\left(x\right),t\right)\ge \frac{t}{t+\phi \left(x,x,0\right)}$$

(2.5)

for all x ∈ X and all t > 0. By (2.4), (2.5), (N 3) and (N 4), we get

$$\begin{array}{ll}\hfill N\left(\frac{f\left(2x\right)}{2}-f\left(x\right),\frac{t}{4}\right)& =N\left(2r^{2}f\left(\frac{2x}{r}\right)-8f\left(x\right)-2r^{2}f\left(\frac{2x}{r}\right)+4f\left(2x\right),2t\right)\\ \ge \text{min}\left\{N\left(2r^{2}f\left(\frac{2x}{r}\right)-4f\left(2x\right),t\right),N\left(2r^{2}f\left(\frac{2x}{r}\right)-8f\left(x\right),t\right)\right\}\\ \ge \frac{t}{t+\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)}.\end{array}$$

(2.6)

for all x ∈ X and all t > 0. Consider the set S := {h : X → Y} and introduce the generalized metric on S:

$$d\left(g,h\right)=\underset{\mu \in \left(0,+\infty \right)}{\text{inf}}\left\{N\left(g\left(x\right)-h\left(x\right),\mu t\right)\ge \frac{t}{t+\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)},\forall x\in X\right\},$$

where, as usual, inf ϕ = + ∞. It is easy to show that (S, d) is complete (see [11]). Now we consider the linear mapping J : (S, d) → (S, d) such that

$$Jg\left(x\right):=\frac{1}{2}g\left(2x\right)$$

for all x ∈ X.

Let g, h ∈ S be given such that d(g, h) = β. Then

$$N\left(g\left(x\right)-h\left(x\right),\beta t\right)\ge \frac{t}{t+\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)}$$

for all x ∈ X and all t > 0. Hence

$$\begin{array}{ll}\hfill N\left(Jg\left(x\right)-Jh\left(x\right),\alpha \beta t\right)=N\left(\frac{1}{2}g\left(2x\right)-\frac{1}{2}h\left(2x\right),\alpha \beta t\right)& =N\left(g\left(2x\right)-h\left(2x\right),2\alpha \beta t\right)\\ \ge \frac{2\alpha t}{2\alpha t+\phi \left(4x,0,0\right)+\phi \left(2x,2x,0\right)}\\ \ge \frac{2\alpha t}{2\alpha t+2\alpha \left(\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)\right)}\\ =\frac{t}{t+\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)}\end{array}$$

for all x ∈ X and all t > 0. So d (g, h) = β implies that d (Jg, Jh) ≤ αβ. This means that d (Jg, Jh) ≤ αd(g, h) for all g, h ∈ S. It follows from (2.6) that

$$d\left(f,Jf\right)\le \frac{1}{4}.$$

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

• (1) A is a fixed point of J, i.e.,

  $$2A\left(x\right)=A\left(2x\right)$$

  (2.7)

for all x ∈ X. The mapping A is a unique fixed point of J in the set M = {g ∈ S : d (h, g) < ∞}. This implies that A is a unique mapping satisfying (2.7) such that there exists a μ ∈ (0, ∞) satisfying

$$N\left(f\left(x\right)-A\left(x\right),\mu t\right)\ge \frac{t}{t+\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)}$$

for all x ∈ X;

• (2) d(J ⁿ f, A) → 0 as n → ∞. This implies the equality $\underset{n\to \infty }{\text{lim}}\frac{f\left(2^{n}x\right)}{2^n}=A\left(x\right)$ for all x ∈ X;

• (3) $d\left(f,A\right)\le \frac{1}{1-\alpha }d\left(f,Jf\right)$, which implies the inequality $d\left(f,A\right)\le \frac{1}{4-4\alpha }$. This implies that the inequalities (2.3) holds.

It follows from (2.1) and (2.2) that

$$\begin{array}{l}N\left(\frac{1}{2^n}\left[r^{2}f\left(\frac{2^n\left(x+y+z\right)}{r}\right)+r^{2}f\left(\frac{2^n\left(x-y+z\right)}{r}\right)+r^{2}f\left(\frac{2n\left(x+y-z\right)}{r}\right)\right.\right.\\ \left.\left.+r^{2}f\left(\frac{2^n\left(-x+y+z\right)}{r}\right)-4f\left(2^{n}x\right)-4f\left(2^{n}y\right)-4f\left(2^{n}z\right)\right],\frac{1}{2^n}\right)\\ \ge \frac{t}{t+\phi \left(2^{n}x,2^{n}y,2^{n}z\right)}\end{array}$$

for all x, y, z ∈ X, all t > 0 and all n ∈ ℕ. So

$$\begin{array}{l}N\left(\frac{1}{2^n}\left[r^{2}f\left(\frac{2^n\left(x+y+z\right)}{r}\right)+r^{2}f\left(\frac{2^n\left(x-y+z\right)}{r}\right)+r^{2}f\left(\frac{2^n\left(x+y-z\right)}{r}\right)\right.\right.\\ \left.\left.+r^{2}f\left(\frac{2^n\left(-x+y+z\right)}{r}\right)-4f\left(2^{n}x\right)-4f\left(2^{n}y\right)-4f\left(2^{n}z\right)\right],t\right)\\ \ge \frac{2^{n}t}{2^{n}t+{\left(2\alpha \right)}^n\phi \left(x,y,z\right)}\end{array}$$

for all x, y, z ∈ X, all t > 0 and all n ∈ ℕ. Since

$$\underset{n\to \infty }{\text{lim}}\frac{2^{n}t}{2^{n}t+{\left(2\alpha \right)}^n\phi \left(x,y,z\right)}=1$$

for all x, y, z ∈ X and all t > 0, we obtain that

$$\begin{array}{l}N\left(r^{2}A\left(\frac{x+y+z}{r}\right)+2^{r}A\left(\frac{x-y+z}{r}\right)\right.\\ \left.+r^{2}A\left(\frac{x+y-z}{r}\right)+r^{2}A\left(\frac{-x+y+z}{s}\right)-4A\left(x\right)-4A\left(y\right)-4A\left(z\right),t\right)=1\end{array}$$

for all x, y, z ∈ X and all t > 0. Hence the mapping A : X → Y is additive, as desired.

Corollary 2.3. Let θ ≥ 0 and r be a real positive number with r < 1. Let X be a normed vector space with norm ∥·∥. Let f : X → Y be an odd mapping satisfying

$$\begin{array}{l}N\left(r^{2}f\left(\frac{x+y+z}{r}\right)+r^{2}f\left(\frac{x-y+z}{r}\right)+r^{2}f\left(\frac{x+y-z}{r}\right)+r^{2}f\left(\frac{-x+y+z}{r}\right)\right.\\ \left.-4f\left(x\right)-4f\left(y\right)-4f\left(z\right),t\right)\ge \frac{t}{t+\theta \left({∥x∥}^r+{∥y∥}^r+{∥z∥}^r\right)}\end{array}$$

(2.8)

for all x, y, z ∈ X and all t > 0. Then the limit$A\left(x\right):=N-\underset{n\to \infty }{\text{lim}}\frac{f\left(2^{n}x\right)}{2^n}$exists for each x ∈ X and defines a unique additive mapping A: X → Y such that

$$N\left(f\left(x\right)-A\left(x\right),t\right)\ge \frac{4\left(1-2^{r-1}\right)t}{4\left(1-2^{r-1}\right)t+\left(2^r+2\right)\theta {∥x∥}^r}.$$

for all x ∈ X and all t > 0.

Proof. The proof follows from Theorem 2.2 by taking φ(x, y, z) := θ(∥x∥^r+ ∥y∥^r+ ∥z∥^r) for all x, y, z ∈ X. Then we can choose α = 2^r-1and we get the desired result.

Theorem 2.4. Let φ : X³ → [0, ∞) be a function such that there exists an α < 1 with

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

(2.9)

for all x, y, z ∈ X. Let f : X → Y be an odd mapping satisfying (2.2). Then the limit$A\left(x\right):=N-\underset{n\to \infty }{\text{lim}}{2}^{n}f\left(\frac{x}{2^n}\right)$exists for each x ∈ X and defines a unique additive mapping A : X → Y such that

$$N\left(f\left(x\right)-A\left(x\right),t\right)\ge \frac{\left(4-4\alpha \right)t}{\left(4-4\alpha \right)t+\alpha \left(\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)\right)}$$

(2.10)

Proof. Let (S, d) be the generalized metric space defined as in the proof of Theorem 2.2. Consider the linear mapping J : S → S such that

$$Jg\left(x\right):=2g\left(\frac{x}{2}\right).$$

Let g,h ∈ S be given such that d (g, h) = β. Then

$$N\left(g\left(x\right)-h\left(x\right),\beta t\right)\ge \frac{t}{t+\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)}$$

for all x ∈ X and all t > 0. Hence

$$\begin{array}{ll}\hfill N\left(Jg\left(x\right)-Jh\left(x\right),\alpha \beta t\right)=N\left(2g\left(\frac{x}{2}\right)-2h\left(\frac{x}{2}\right),\alpha \beta t\right)& =N\left(g\left(\frac{x}{2}\right)-g\left(\frac{x}{2}\right),\frac{\alpha \beta t}{2}\right)\\ \ge \frac{\frac{\alpha }{2}t}{\frac{\alpha }{2}t+\phi \left(x,0,0\right)+\phi \left(\frac{x}{2},\frac{x}{2},0\right)}\\ \ge \frac{t}{t+\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)}\end{array}$$

for all x ∈ X and all t > 0. So d (g, h) = β implies that d (Jg, Jh) ≤ αβ. This means that d(Jg, Jh) ≤ αd (g, h) for all g, h ∈ S. It follows from (2.6) that

$$N\left(2f\left(\frac{x}{2}\right)-f\left(x\right),\frac{t}{2}\right)\ge \frac{t}{t+\phi \left(x,0,0\right)+\phi \left(\frac{x}{2},\frac{x}{2},0\right)}\ge \frac{t}{t+\frac{\alpha }{2}\left(\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)\right)}$$

for all x ∈ X and t > 0. Therefore

$$N\left(2f\left(\frac{x}{2}\right)-f\left(x\right),\frac{\alpha t}{4}\right)\ge \frac{t}{t+\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)}.$$

(2.11)

So $d\left(f,Jf\right)\le \frac{\alpha }{4}$. By Theorem 1.7, there exists a mapping A : X → Y satisfying the following:

• (1) A is a fixed point of J, that is,

  $$A\left(\frac{x}{2}\right)=\frac{1}{2}A\left(x\right)$$

  (2.12)

for all x ∈ X. The mapping A is a unique fixed point of J in the set Ω = {h ∈ S : d(g, h) < ∞}. This implies that A is a unique mapping satisfying (2.12) such that there exists μ ∈ (0, ∞) satisfying

$$N\left(f\left(x\right)-A\left(x\right),\mu t\right)\ge \frac{t}{t+\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)}$$

for all x ∈ X and t > 0.

• (2) d(J ⁿ f, A) → 0 as n → ∞. This implies the equality $N-\underset{n\to \infty }{\text{lim}}{2}^{n}f\left(\frac{x}{2^n}\right)=A\left(x\right)$ 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 $d\left(f,A\right)\le \frac{\alpha }{4-4\alpha }$. This implies that the inequality (2.10) holds.

The rest of the proof is similar to that of the proof of Theorem 2.2.

Corollary 2.5. Let θ ≥ 0 and let r be a real number with r > 1. Let X be a normed vector space with norm ∥·∥. Let f : X → Y be an odd mapping satisfying (2.8). Then, the limit$A\left(x\right):=N-\underset{n\to \infty }{\text{lim}}{2}^{n}f\left(\frac{x}{2^n}\right)$exists for each x ∈ X and defines a unique additive mapping A : X → Y such that

$$N\left(f\left(x\right)-A\left(x\right),t\right)\ge \frac{4\left(2^r-2\right)t}{4\left(2^r-2\right)t+\left(2^{r+1}+4\right)\theta {∥x∥}^r}.$$

for all x ∈ X and all t > 0.

Proof. The proof follows from Theorem 2.4 by taking φ(x, y, z) := θ(∥x∥^r+ ∥y∥^r+ ∥z∥^r) for all x, y, z ∈ X. Then we can choose α = 2^1-rand we get the desired result.

Theorem 2.6. Let φ : X³ → [0, ∞) be a function such that there exists an α < 1 with

$$\phi \left(2x,2y,2z\right)\le 4\alpha \phi \left(x,y,z\right)$$

(2.13)

for all x, y, z ∈ X. Let f : X → Y be an even mapping with f(0) = 0 and satisfying (2.2). Then$Q\left(x\right):=N-\underset{n\to \infty }{\text{lim}}\frac{f\left(2^{n}x\right)}{4^n}$exists for each x ∈ X and defines a unique quadratic mapping Q : X → Y such that

$$N\left(f\left(x\right)-Q\left(x\right),t\right)\ge \frac{\left(8-8\alpha \right)t}{\left(8-8\alpha \right)t+3\left(\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)\right)}$$

(2.14)

Proof. Putting y = x and z = 0 in (2.2), we get

$$N\left(2r^{2}f\left(\frac{2x}{r}\right)-8f\left(x\right),t\right)\ge \frac{t}{t+\phi \left(x,x,0\right)}$$

(2.15)

for all x ∈ X and all t > 0. Putting y = z = 0 and replacing x by 2x in (2.2), we have

$$N\left(4r^{2}f\left(\frac{2x}{r}\right)-4f\left(2x\right),t\right)\ge \frac{t}{t+\phi \left(2x,0,0\right)}$$

(2.16)

for all x ∈ X and all t > 0. By (2.15), (2.16), (N 3) and (N 4), we get

$$\begin{array}{ll}\hfill N\left(\frac{f\left(2x\right)}{4}-f\left(x\right),\frac{t}{16}\right)& =N\left(2\left(2r^{2}f\left(\frac{2x}{r}\right)-8f\left(x\right)\right)-\left(4r^{2}f\left(\frac{2x}{r}\right)-4f\left(2x\right)\right),t\right)\\ \ge \text{min}\left\{N\left(2r^{2}f\left(\frac{2x}{r}\right)-8f\left(x\right),\frac{t}{4}\right),N\left(4r^{2}f\left(\frac{2x}{r}\right)-4f\left(2x\right),\frac{t}{2}\right)\right\}\end{array}$$

for all x ∈ X and all t > 0. So

$$\begin{array}{ll}\hfill N\left(\frac{f\left(2x\right)}{4}-f\left(x\right),\frac{3t}{8}\right)& \ge \text{min}\left\{N\left(2r^{2}f\left(\frac{2x}{r}\right)-8f\left(x\right),t\right),N\left(4r^{2}f\left(\frac{2x}{r}\right)-4f\left(x\right),t\right)\right\}\\ \ge \frac{t}{t+\phi \left(x,x,0\right)+\phi \left(2x,0,0\right)}.\end{array}$$

(2.17)

Consider the set S* := {h : X → Y; h(0) = 0} and introduce the generalized metric on S*:

$$d\left(g,h\right)=\underset{\mu \in \left(0,+\infty \right)}{\text{inf}}\left\{N\left(g\left(x\right)-h\left(x\right),\mu t\right)\ge \frac{t}{t+\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)},\forall x\in X\right\},$$

where, as usual, inf ϕ = + ∞. It is easy to show that (S*, d) is complete (see [11]). Now we consider the linear mapping J : (S*, d) → (S*, d) such that $Jg\left(x\right):=\frac{1}{4}g\left(2x\right)$ for all x ∈ X. Proceeding as in the proof of Theorem 2.2, we obtain that d(g, h) = β implies that d(Jg, Jh) ≤ αβ. This means that d(Jg, Jh) ≤ αd (g, h) for all g, h ∈ S. It follows from (2.17) that

$$d\left(f,Jf\right)\le \frac{3}{8}.$$

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

• (1) Q is a fixed point of J, i.e.,

  $$4Q\left(x\right)=Q\left(2x\right)$$

  (2.18)

for all x ∈ X. The mapping Q is a unique fixed point of J in the set M = {g ∈ S* : d(h, g) < ∞}. This implies that Q is a unique mapping satisfying (2.18) such that there exists a μ ∈ (0, ∞) satisfying $N\left(f\left(x\right)-Q\left(x\right),\mu t\right)\ge \frac{t}{t+\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)}$ for all x ∈ X;

• (2) d(J ⁿ f, Q) → 0 as n → ∞. This implies the equality $\underset{n\to \infty }{\text{lim}}\frac{f\left(2^{n}x\right)}{4^n}=Q\left(x\right)$ for all x ∈ X;

• (3) $d\left(f,Q\right)\le \frac{1}{1-\alpha }d\left(f,Jf\right)$, which implies the inequality $d\left(f,Q\right)\le \frac{3}{8-8\alpha }$. This implies that the inequalities (2.14) holds.

The rest of the proof is similar to the proof of Theorem 2.2.

Corollary 2.7. Let θ ≥ 0 and r be a real positive number with r < 1. Let X be a normed vector space with norm ∥·∥. Let f : X → Y be an even mapping satisfying (2.8) and f(0) = 0. Then the limit$Q\left(x\right):=N-\underset{n\to \infty }{\text{lim}}\frac{f\left(2^{n}x\right)}{4^n}$exists for each x ∈ X and defines a unique quadratic mapping Q : X → Y such that

$$N\left(f\left(x\right)-Q\left(x\right),t\right)\ge \frac{2\left(4-4^r\right)t}{2\left(4-4^r\right)t+3\left(2^r+2\right)\theta {∥x∥}^r}.$$

for all x ∈ X and all t > 0.

Proof. The proof follows from Theorem 2.6 by taking φ(x, y, z) := θ(∥x∥^r+ ∥y∥^r+ ∥z∥^r) for all x, y, z ∈ X. Then we can choose α = 4^r-1and we get the desired result.

Theorem 2.8. Let φ : X³ → [0, ∞) be a function such that there exists an α < 1 with

$$\phi \left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right)\le \frac{\alpha }{4}\phi \left(x,y,z\right)$$

for all x, y, z ∈ X. Let f : X → Y be an even mapping satisfying (2.2) and f(0) = 0. Then the limit$Q\left(x\right):=N-\underset{n\to \infty }{\text{lim}}{4}^{n}f\left(\frac{x}{2^n}\right)$exists for each x ∈ X and defines a unique quadratic mapping Q : X → Y such that

$$N\left(f\left(x\right)-Q\left(x\right),t\right)\ge \frac{\left(8-8\alpha \right)t}{\left(8-8\alpha \right)t+3\alpha \left(\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)\right)}$$

Proof. Let (S*, d) be the generalized metric space defined as in the proof of Theorem 2.6.

It follows from (2.17) that

$$N\left(f\left(x\right)-4f\left(\frac{x}{2}\right),\frac{3t}{2}\right)\ge \frac{t}{t+\phi \left(x,0,0\right)+\phi \left(\frac{x}{2},\frac{x}{2},0\right)}\ge \frac{t}{t+\frac{\alpha }{4}\left(\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)\right)}$$

for all x ∈ X and t > 0. So

$$N\left(f\left(x\right)-4f\left(\frac{x}{2}\right),\frac{3\alpha t}{8}\right)\ge \frac{t}{t+\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)}.$$

(2.19)

The rest of the proof is similar to the proof of Theorems 2.2 and 2.4.

Corollary 2.9. Let θ ≥ 0 and r be a real number with r > 1. Let X be a normed vector space with norm ∥·∥. Let f : X → Y be an even mapping satisfying f(0) = 0 and (2.8). Then$Q\left(x\right):=N-\underset{n\to \infty }{\text{lim}}{4}^{n}f\left(\frac{x}{2^n}\right)$exists for each x ∈ X and defines a unique quadratic mapping Q : X → Y such that

$$N\left(f\left(x\right)-Q\left(x\right),t\right)\ge \frac{2\left(4^{r+1}-16\right)t}{2\left(4^{r+1}-16\right)t+3\left(2^{r+2}+8\right)\theta ∥x∥r}.$$

for all x ∈ X and all t > 0.

Proof. The proof follows from Theorem 2.8 by taking φ(x, y, z) := θ(∥x∥^r+ ∥y∥^r+ ∥z∥^r) for all x, y, z ∈ X. Then we can choose α = 4^1-rand we get the desired result.

Let f : X → Y be a mapping satisfying f(0) = 0 and (1.1). Let $f_e\left(x\right):=\frac{f\left(x\right)+f\left(-x\right)}{2}$ and $f_o\left(x\right)=\frac{f\left(x\right)-f\left(-x\right)}{2}$. Then f _e is an even mapping satisfying (1.1) and f _o is an odd mapping satisfying (1.1) such that f(x) = f _e (x) + f _o (x). So we obtain the following.

Theorem 2.10. Let φ : X³ → [0, ∞) be a function such that there exists an α < 1 with

$$\phi \left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right)\le \frac{\alpha }{4}\phi \left(x,y,z\right)$$

for all x, y, z ∈ X. Let f : X → Y be a mapping satisfying (2.2) and f(0) = 0. Then there exist a unique additive mapping A: X → Y and a unique quadratic mapping Q : X → Y such that

$$\begin{array}{l}N\left(f\left(x\right)-A\left(x\right)-Q\left(x\right),2t\right)\\ =N\left(\frac{f\left(x\right)-f\left(-x\right)}{2}+\frac{f\left(x\right)+f\left(-x\right)}{2}-A\left(x\right)-Q\left(x\right),2t\right)\\ \ge \text{min}\left\{N\left(\frac{f\left(x\right)-f\left(-x\right)}{2}-A\left(x\right),t\right),N\left(\frac{f\left(x\right)+f\left(-x\right)}{2}-Q\left(x\right),t\right)\right\}\\ \ge \text{min}\left\{\frac{\left(8-8\alpha \right)t}{\left(8-8\alpha \right)t+\alpha \left(\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)\right)},\frac{\left(8-8\alpha \right)t}{\left(8-8\alpha \right)t+3\alpha \left(\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)\right)}\right\}\\ =\frac{\left(8-8\alpha \right)t}{\left(8-8\alpha \right)t+3\alpha \left(\phi \left(2x,0,0\right)+\phi \left(x,x,0\right)\right)}.\end{array}$$

Corollary 2.11. Let θ ≥ 0 and let r be a real number with r > 1. Let X be a normed vector space with norm ∥·∥. Let f : X → Y be a mapping with f(0) = 0 and satisfying (2.8). Then there exist a unique additive mapping A : X → Y and a unique quadratic mapping Q : X → Y such that

$$N\left(f\left(x\right)-A\left(x\right)-Q\left(x\right),t\right)\ge \frac{2\left(2^r-2\right)t}{2\left(2^r-2\right)t+3\left(2^r+2\right)\theta {∥x∥}^r}.$$

for all x ∈ X and all t > 0.

Proof. The proof follows from Theorem 2.10 by taking φ(x, y,z) := θ(∥x∥^r+ ∥y∥^r+ ∥z∥^r) for all x, y, z ∈ X. Then we can choose α = 2^1-rand we get the desired result.

3. Fuzzy stability of the functional equation (1.1): a direct method

In this section, using direct method, we prove the Hyers-Ulam stability of functional equation (1.1) in fuzzy Banach spaces. Throughout this section, 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 3.1. Assume that a mapping f : X → Y is an odd mapping satisfying the inequality

$$\begin{array}{l}N\left(r^{2}f\left(\frac{x+y+z}{r}\right)+r^{2}f\left(\frac{x-y+z}{r}\right)\right.+r^{2}f\left(\frac{x+y-z}{r}\right)+r^{2}f\left(\frac{-x+y+z}{r}\right)\\ \left.-4f\left(x\right)-4f\left(y\right)-4f\left(z\right),t\right)\ge N^{\prime }\left(\phi \left(x,y,z\right),t\right)\end{array}$$

(3.1)

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

$$N^{\prime }\left(\phi \left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right),t\right)\ge N^{\prime }\left(\phi \left(x,y,z\right),\frac{t}{\left|r\right|}\right)$$

(3.2)

for all x, y, z ∈ X and all t > 0. Then there exists a unique additive mapping A : X → Y satisfying (1.1) and the inequality

$$N\left(f\left(x\right)-A\left(x\right),t\right)\ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),2\left(1-2\left|r\right|\right)t\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{2\left(1-2\left|r\right|t\right)}{\left|r\right|}\right)\right\}$$

(3.3)

for all x ∈ X and all t > 0.

Proof. It follows from (3.2) that

$$N^{\prime }\left(\phi \left(\frac{x}{2^j},\frac{y}{2^j},\frac{z}{2^j}\right),t\right)\ge N^{\prime }\left(\phi \left(x,y,z\right),\frac{t}{{\left|r\right|}^j}\right)$$

for all x, y, z ∈ X and all t > 0. Putting y = z = 0 in (3.1) and replacing x by 2x, we get

$$N\left(2r^{2}f\left(\frac{2x}{r}\right)-4f\left(2x\right),t\right)\ge N^{\prime }\left(\phi \left(2x,0,0\right),t\right)$$

(3.4)

for all x ∈ X and all t > 0. Putting y = x and z = 0 in (3.1), we have

$$N\left(2r^{2}f\left(\frac{2x}{r}\right)-8f\left(x\right),t\right)\ge N^{\prime }\left(\phi \left(x,x,0\right),t\right)$$

(3.5)

for all x ∈ X and all t > 0. By (3.4), (3.5), (N 3) and (N 4), we get

$$\begin{array}{ll}\hfill N\left(\frac{f\left(2x\right)}{2}-f\left(x\right),\frac{t}{4}\right)& =N\left(2r^{2}f\left(\frac{2x}{r}\right)-8f\left(x\right)-2r^{2}f\left(\frac{2x}{r}\right)+4f\left(2x\right),2t\right)\\ \ge \text{min}\left\{N\left(2r^{2}f\left(\frac{2x}{r}\right)-4f\left(2x\right),t\right),N\left(2r^{2}f\left(\frac{2x}{r}\right)-8f\left(x\right),t\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(2x,0,0\right),t\right),N^{\prime }\left(\phi \left(x,x,0\right),t\right)\right\}\end{array}$$

(3.6)

for all x ∈ X and all t > 0. Replacing x by $\frac{x}{2}$ in (3.6), we have

$$N\left(f\left(x\right)-2f\left(\frac{x}{2}\right),\frac{t}{2}\right)\ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),t\right),N^{\prime }\left(\phi \left(\frac{x}{2},\frac{x}{2},0\right),t\right)\right\}.$$

(3.7)

Replacing x by $\frac{x}{2^j}$ in (3.7), we have

$$\begin{array}{l}N\left(2^{j+1}f\left(\frac{x}{2^{j+1}}\right)-2^{j}f\left(\frac{x}{2^j}\right),2^{j-1}t\right)\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(\frac{x}{2^j},0,0\right),t\right),N^{\prime }\left(\phi \left(\frac{x}{2^{j+1}},\frac{x}{2^{j+1}},0\right),t\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{t}{{\left|r\right|}^j}\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{t}{{\left|r\right|}^{j+1}}\right)\right\}\end{array}$$

(3.8)

for all x ∈ X, all t > 0 and any integer j ≥ 0. So

$$\begin{array}{ll}\hfill N\left(f\left(x\right)-2^{n}f\left(\frac{x}{2^n}\right),\sum _{j=0}^{n-1}{2}^{j-1}{\left|r\right|}^{j}t\right)& =N\left(\sum _{j=0}^{n-1}{2}^{j+1}f\left(\frac{x}{2^{j+1}}\right)-2^{j}f\left(\frac{x}{2^j}\right),\sum _{j=0}^{n-1}{2}^{j-1}{\left|r\right|}^{j}t\right)\\ \ge \underset{0\le j\le n-1}{\text{min}}\left\{N\left(2^{j+1}f\left(\frac{x}{2^{j+1}}\right)-2^{j}f\left(\frac{x}{2^j}\right),2^{j-1}{\left|r\right|}^{j}t\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),t\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{t}{\left|r\right|}\right)\right\}\end{array}$$

which yields

$$\begin{array}{l}N\left(2^{n+p}f\left(\frac{x}{2^{n+p}}\right)-2^{p}f\left(\frac{x}{2^p}\right),\sum _{j=0}^{n-1}{2}^{j+p-1}{\left|r\right|}^j\right)\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(\frac{x}{2^p},0,0\right),t\right),N^{\prime }\left(\phi \left(\frac{x}{2^p},\frac{x}{2^p},0\right),\frac{t}{\left|r\right|}\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{t}{{\left|r\right|}^p}\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{t}{{\left|r\right|}^{p+1}}\right)\right\}\end{array}$$

(3.9)

for all x ∈ X, t > 0 and any integers n > 0, p ≥ 0. So

$$N\left(2^{n+p}f\left(\frac{x}{2^{n+p}}\right)-2^{p}f\left(\frac{x}{2^p}\right),\sum _{j=0}^{n-1}{2}^{j+p-1}{\left|r\right|}^{j+p}t\right)\ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),t\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{t}{\left|r\right|}\right)\right\}$$

for all x ∈ X, t > 0 and any integers n > 0, p ≥ 0. Hence one obtains

$$\begin{array}{l}N\left(2^{n+p}f\left(\frac{x}{2^{n+p}}\right)-2^{p}f\left(\frac{x}{2^p}\right),t\right)\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{t}{{\sum }_{j=0}^{n-1}{2}^{j+p-1}{\left|r\right|}^{j+p}}\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{t}{{\sum }_{j=0}^{n-1}{2}^{j+p-1}{\left|r\right|}^{j+p+1}}\right)\right\}\end{array}$$

(3.10)

for all x ∈ X, t > 0 and any integers n > 0, p ≥ 0. Since the series ${\sum }_{j=0}^{+\infty }{2}^j{\left|r\right|}^j$ is a convergent series, we see by taking the limit p → ∞ in the last inequality that the 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 $A\left(x\right):=N-\underset{n\to \infty }{\text{lim}}{2}^{n}f\left(\frac{x}{2^n}\right)$ is well-defined for all x ∈ X. It means that

$$\underset{n\to \infty }{\text{lim}}N\left(A\left(x\right)-2^{n}f\left(\frac{x}{2^n}\right),t\right)=1$$

(3.11)

for all x ∈ X and all t > 0. In addition, it follows from (3.10) that

$$N\left(f\left(x\right)-2^{n}f\left(\frac{x}{2^n}\right),t\right)\ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{2t}{{\sum }_{j=0}^{n-1}{2}^j{\left|r\right|}^j}\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{2t}{{\sum }_{j=0}^{n-1}{2}^j{\left|r\right|}^{j+1}}\right)\right\}$$

for all x ∈ X and all t > 0. So

$$\begin{array}{ll}\hfill N\left(f\left(x\right)-A\left(x\right),t\right)& \ge \text{min}\left\{N\left(f\left(x\right)-2^{n}f\left(\frac{x}{2^n}\right),\left(1-\epsilon \right)t\right),N\left(A\left(x\right)-2^{n}f\left(\frac{x}{2^n}\right),\epsilon t\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{2\epsilon t}{{\sum }_{j=0}^{n-1}{2}^j{\left|r\right|}^j}\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{2\epsilon t}{{\sum }_{j=0}^{n-1}{2}^j{\left|r\right|}^{j+1}}\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0\right),2\left(1-2\right)\left|r\right|\epsilon t\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{2\left(1-2\left|r\right|\right)\epsilon t}{\left|r\right|}\right)\right\}\end{array}$$

for sufficiently large n and for all x ∈ X, t > 0 and ϵ with 0 < ϵ < 1. Since ϵ is arbitrary and N' is left continuous, we obtain

$$N\left(f\left(x\right)-A\left(x\right),t\right)\ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),2\left(1-2\left|r\right|\right)t\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{2\left(1-2\left|r\right|t\right)}{\left|r\right|}\right)\right\}$$

for all x ∈ X and t > 0. It follows from (3.1) that

$$\begin{array}{c}N\left(2^n{r}^{2}f\left(\frac{x+y+z}{r{.2}^n}\right)+2^n.r^{2}f\left(\frac{x-y+z}{r{.2}^n}\right)+2^n.r^{2}f\left(\frac{x+y-z}{r{.2}^n}\right)+2^n.r^{2}f\left(\frac{-x+y+z}{r{.2}^n}\right)\right.\\ \left.-2^{n+2}f\left(\frac{x}{2^n}\right)-2^{n+2}f\left(\frac{y}{2^n}\right)-2^{n+2}f\left(\frac{z}{2^n}\right),t\right)\\ \ge N^{\prime }\left(\phi \left(\frac{x}{2^n},\frac{y}{2^n},\frac{z}{2^n}\right),\frac{t}{2^n}\right)\ge N^{\prime }\left(\phi \left(x,y,z\right),\frac{t}{2^n{\left|r\right|}^n}\right)\to 1\mathsf{\text{as}}n\to +\infty \end{array}$$

for all x, y, z ∈ X, t > 0. So

$$\begin{array}{c}N\left(2^n{r}^{2}f\left(\frac{x+y+z}{r{.2}^n}\right)+2^n.r^{2}f\left(\frac{x-y+z}{r{.2}^n}\right)+2^n.r^{2}f\left(\frac{x+y-z}{r{.2}^n}\right)+2^n.r^{2}f\left(\frac{-x+y+z}{r{.2}^n}\right)\right.\\ \left.-2^{n+2}f\left(\frac{x}{2^n}\right)-2^{n+2}f\left(\frac{y}{2^n}\right)-2^{n+2}f\left(\frac{z}{2^n}\right),t\right)\to 1\end{array}$$

for all x, y, z ∈ X and all t > 0. Therefore, we obtain in view of (3.11)

$$\begin{array}{c}N\left(r^{2}A\left(\frac{x+y+z}{r}\right)+r^{2}A\left(\frac{x-y+z}{r}\right)+r^{2}A\left(\frac{x+y-z}{r}\right)+r^{2}A\left(\frac{-x+y+z}{r}\right)\right.\\ \left.-4A\left(x\right)-4A\left(y\right)-4A\left(z\right),t\right)\\ \ge \text{min}\left\{N\left(r^{2}A\left(\frac{x+y+z}{r}\right)+r^{2}A\left(\frac{x-y+z}{r}\right)+r^{2}A\left(\frac{x+y-z}{r}\right)+r^{2}A\left(\frac{-x+y+z}{r}\right)\right.\right.\\ -4A\left(x\right)-4A\left(y\right)-4A\left(z\right)-2^n{r}^{2}f\left(\frac{x+y+z}{r{.2}^n}\right)+2^n.r^{2}f\left(\frac{x-y+z}{r{.2}^n}\right)+2^n.r^{2}f\left(\frac{x+y-z}{r{.2}^n}\right)\\ \left.+2^n.r^{2}f\left(\frac{-x+y+z}{r{.2}^n}\right)-2^{n+2}f\left(\frac{x}{2^n}\right)-2^{n+2}f\left(\frac{y}{2^n}\right)-2^{n+2}f\left(\frac{z}{2^n}\right),\frac{t}{2}\right)\\ ,N\left(2^n{r}^{2}f\left(\frac{x+y+z}{r{.2}^n}\right)+2^n.r^{2}f\left(\frac{x-y+z}{r{.2}^n}\right)+2^n.r^{2}f\left(\frac{x+y-z}{r{.2}^n}\right)+2^n.r^{2}f\left(\frac{-x+y+z}{r{.2}^n}\right)\right.\\ \left.\left.-2^{n+2}f\left(\frac{x}{2^n}\right)-2^{n+2}f\left(\frac{y}{2^n}\right)-2^{n+2}f\left(\frac{z}{2^n}\right),\frac{t}{2}\right)\right\}\\ =N\left(2^n{r}^{2}f\left(\frac{x+y+z}{r{.2}^n}\right)\right.+2^n.r^{2}f\left(\frac{x-y+z}{r{.2}^n}\right)+2^n.r^{2}f\left(\frac{x+y-z}{r{.2}^n}\right)+2^n.r^{2}f\left(\frac{-x+y+z}{r{.2}^n}\right)\\ \left.-2^{n+2}f\left(\frac{x}{2^n}\right)-2^{n+2}f\left(\frac{y}{2^n}\right)-2^{n+2}f\left(\frac{z}{2^n}\right),\frac{t}{2}\right)\left(\mathsf{\text{for}}\mathsf{\text{sufficiently}}\mathsf{\text{large}}n\right)\\ \ge N^{\prime }\left(\phi \left(x,y,z\right),\frac{t}{2^{n+1}{\left|r\right|}^n}\right)\to 1\mathsf{\text{as}}n\to \infty \end{array}$$

which implies

$$r^{2}A\left(\frac{x+y+z}{r}\right)+r^{2}A\left(\frac{x-y+z}{r}\right)+r^{2}A\left(\frac{x+y-z}{r}\right)+r^{2}A\left(\frac{-x+y+z}{r}\right)=4A\left(x\right)+4A\left(y\right)+4A\left(z\right)$$

for all x, y, z ∈ X. Thus A : X → Y is a mapping satisfying the Equation (1.1) and the inequality (3.3). Thus the mapping A : X → Y is additive, as desired.

To prove the uniqueness, assume that there is another mapping L : X → Y which satisfies the inequality (3.3). Since L(2ⁿx) = 2ⁿL(x) for all x ∈ X, we have

$$\begin{array}{ll}\hfill N\left(A\left(x\right)-L\left(x\right),t\right)& =N\left(2^{n}A\left(\frac{x}{2^n}\right)-2^{n}L\left(\frac{x}{2^n}\right),t\right)\\ \ge \text{min}\left\{N\left(2^{n}A\left(\frac{x}{2^n}\right)-2^{n}f\left(\frac{x}{2^n}\right),\frac{t}{2}\right),N\left(2^{n}f\left(\frac{x}{2^n}\right)-2^{n}L\left(\frac{x}{2^n}\right),\frac{t}{2}\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(\frac{x}{2^{n+1}},0,0\right),\frac{\left(1-2\left|r\right|\right)t}{2^{n+1}}\right),N^{\prime }\left(\phi \left(\frac{x}{2^n},\frac{x}{2^n},0\right),\frac{\left(1-2\left|r\right|\right)t}{2^{n+1}\left|r\right|}\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{\left(1-2\left|r\right|\right)t}{{\left|r\right|}^n{2}^{n+1}}\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{\left(1-2\left|r\right|\right)t}{2^{n+1}{\left|r\right|}^{n+1}}\right)\right\}\to 1\mathsf{\text{as}}n\to \infty \end{array}$$

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

Corollary 3.2. Let X be a normed spaces and (ℝ, N') a fuzzy Banach space. Assume that there exist real numbers θ ≥ 0 and p > 1 such that an odd mapping f : X → Y satisfies the inequality

$$\begin{array}{c}N\left(r^{2}f\left(\frac{x+y+z}{r}\right)+r^{2}f\left(\frac{x-y+z}{r}\right)+r^{2}f\left(\frac{x+y-z}{r}\right)+r^{2}f\left(\frac{-x+y+z}{r}\right)\right.\\ \left.-4f\left(x\right)-4f\left(y\right)-4f\left(z\right),t\right)\ge N^{\prime }\left(\theta \left({∥x∥}^p+{∥y∥}^p+{∥z∥}^p\right),t\right)\end{array}$$

(3.12)

for all x, y, z ∈ X and t > 0. Then there is a unique additive mapping A : X → Y satisfying (1.1) and the inequality

$$N\left(f\left(x\right)-A\left(x\right),t\right)\ge \text{min}\left\{N^{\prime }\left(\theta {∥x∥}^p,\frac{\left(2^{p+1}-4\right)t}{2^p}\right),N^{\prime }\left(\theta {∥x∥}^p,\left(2^p-2\right)t\right)\right\}$$

Proof. Let φ(x, y, z) := θ (∥x∥^p+ ∥y∥^p+ ∥z∥^p) and |r| = 2^-p. Applying Theorem 3.1, we get the desired result.

Theorem 3.3. Assume that an odd mapping f : X → Y satisfies the inequality (3.1) and that φ:X³ → Z is a mapping for which there is a constant r ∈ ℝ satisfying 0 < |r| < 2 such that

$$N^{\prime }\left(\phi \left(x,y,z\right),\left|r\right|t\right)\ge N^{\prime }\left(\phi \left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right),t\right)$$

(3.13)

for all x, y, z ∈ X and all t > 0. Then there exists a unique additive mapping A : X → Y satisfying (1.1) and the inequality

$$N\left(f\left(x\right)-A\left(x\right),t\right)\ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{2\left(2-\left|r\right|t\right)}{\left|r\right|}\right),N^{\prime }\left(\phi \left(x,x,0\right),2\left(2-\left|r\right|\right)t\right)\right\}$$

(3.14)

for all x ∈ X and all t > 0.

Proof. It follows from (3.6) that

$$N\left(\frac{f\left(2x\right)}{2}-f\left(x\right),\frac{t}{4}\right)\ge \text{min}\left\{N^{\prime }\left(\phi \left(2x,0,0\right),t\right),N^{\prime }\left(\phi \left(x,x,0\right),t\right)\right\}$$

(3.15)

for all x ∈ X and all t > 0. Replacing x by 2ⁿx in (3.15), we obtain

$$\begin{array}{ll}\hfill N\left(\frac{f\left(2^{n+1}x\right)}{2^{n+1}}-\frac{f\left(2^{n}x\right)}{2^n},\frac{t}{2^{n+2}}\right)& \ge \text{min}\left\{N^{\prime }\left(\phi \left(2^{n+1}x,0,0\right),t\right),N^{\prime }\left(\phi \left(2^{n}x,2^{n}x,0\right),t\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{t}{{\left|r\right|}^{n+1}}\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{t}{{\left|r\right|}^n}\right)\right\}.\end{array}$$

(3.16)

So

$$N\left(\frac{f\left(2^{n+1}x\right)}{2^{n+1}}-\frac{f\left(2^{n}x\right)}{2^n},\frac{{\left|r\right|}^{n}t}{2^{n+2}}\right)\ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{t}{\left|r\right|}\right),N^{\prime }\left(\phi \left(x,x,0\right),t\right)\right\}$$

(3.17)

for all x ∈ X and all t > 0. Proceeding as in the proof of Theorem 3.1, we obtain that

$$N\left(f\left(x\right)-\frac{f\left(2^{n}x\right)}{2^n},\sum _{j=0}^{n-1}\frac{{\left|r\right|}^{j}t}{2^{j+2}}\right)\ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{t}{\left|r\right|}\right),N^{\prime }\left(\phi \left(x,x,0\right),t\right)\right\}$$

for all x ∈ X, all t > 0 and any integer n > 0. So

$$\begin{array}{ll}\hfill N\left(f\left(x\right)-\frac{f\left(2^{n}x\right)}{2^n},t\right)& \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{t}{\left|r\right|{\sum }_{j=0}^{n-1}\frac{{\left|r\right|}^j}{2^{j+2}}}\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{t}{{\sum }_{j=0}^{n-1}\frac{{\left|r\right|}^j}{2^{j+2}}}\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{2\left(2-\left|r\right|t\right)}{\left|r\right|}\right),N^{\prime }\left(\phi \left(x,x,0\right),2\left(2-\left|r\right|\right)t\right)\right\}.\end{array}$$

The rest of the proof is similar to the proof of Theorem 3.1.

Corollary 3.4. Let X be a normed spaces and that (ℝ, N') a fuzzy Banach space. Assume that there exist real numbers θ ≥ 0 and 0 < p < 1 such that an odd mapping f : X → Y satisfies (3.12). Then there is a unique additive mapping A : X → Y satisfying (1.1) and the inequality

$$N\left(f\left(x\right)-A\left(x\right),t\right)\ge \text{min}\left\{N^{\prime }\left(\theta {∥x∥}^p,\frac{2\left(2-2^p\right)}{2^p}\right),N^{\prime }\left(\theta {∥x∥}^p,\left(2-2^p\right)t\right)\right\}$$

Proof. Let φ(x, y, z) := θ(∥x∥^p+ ∥y∥^p+ ∥z∥^p) and |r| = 4^p. Applying Theorem 3.3, we get the desired results.

Theorem 3.5. Assume that a mapping f : X → Y is an even mapping satisfies (3.1) and f(0) = 0 and that φ : X³ → Z is a mapping for which there is a constant r ∈ ℝ satisfying 0 < |r| < 4 such that

$$N^{\prime }\left(\phi \left(x,y,z\right),\left|r\right|t\right)\ge N^{\prime }\left(\phi \left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right),t\right)$$

(3.18)

for all x, y, z ∈ X and all t > 0. Then there exists a unique quadratic mapping Q : X → Y satisfying (1.1) and the inequality

$$N\left(f\left(x\right)-Q\left(x\right),t\right)\ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{2\left(4-\left|r\right|\right)t}{\left|r\right|}\right),N^{\prime }\left(\phi \left(x,x,0\right),\left(4-\left|r\right|\right)t\right)\right\}$$

(3.19)

for all x ∈ X and all t > 0.

Proof. Putting y = x and z = 0 in (3.1), we get

$$N\left(2r^{2}f\left(\frac{2x}{r}\right)-8f\left(x\right),t\right)\ge N^{\prime }\left(\phi \left(x,x,0\right),t\right)$$

(3.20)

for all x ∈ X and all t > 0. Putting y = z = 0 and replacing x by 2 x in (3.1), we have

$$N\left(4r^{2}f\left(\frac{2x}{r}\right)-4f\left(2x\right),t\right)\ge N^{\prime }\left(\phi \left(2x,0,0\right),t\right)$$

(3.21)

for all x ∈ X and all t > 0. By (3.20), (3.21), (N 3) and (N 4), we get

$$\begin{array}{ll}\hfill N\left(\frac{f\left(2x\right)}{4}-f\left(x\right),\frac{t}{16}\right)& =N\left(2\left(2r^{2}f\left(\frac{2x}{r}\right)-8f\left(x\right)\right)-\left(4r^{2}f\left(\frac{2x}{r}\right)-4f\left(2x\right)\right),t\right)\\ \ge \text{min}\left\{N\left(2r^{2}f\left(\frac{2x}{r}\right)-8f\left(x\right),\frac{t}{4}\right),N\left(4r^{2}f\left(\frac{2x}{r}\right)-4f\left(2x\right),\frac{t}{2}\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,x,0\right),\frac{t}{4}\right),N^{\prime }\left(\phi \left(2x,0,0\right),\frac{t}{2}\right)\right\}\end{array}$$

(3.22)

for all x ∈ X and all t > 0. Replacing x by 2ⁿx in (3.22), we obtain

$$\begin{array}{ll}\hfill N\left(\frac{f\left(2^{n+1}\right)x}{4^{n+1}}-\frac{f\left(2^{n}x\right)}{4^n},\frac{t}{4^{n+2}}\right)& \ge \text{min}\left\{N^{\prime }\left(\phi \left(2^{n}x,2^{n}x,0\right),\frac{t}{4}\right),N^{\prime }\left(\phi \left(2^{n+1}x,0,0\right),\frac{t}{2}\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,x,0\right),\frac{t}{4{\left|r\right|}^n}\right),N^{\prime }\left(\phi \left(x,0,0\right),\frac{t}{2{\left|r\right|}^{n+1}}\right)\right\}\end{array}$$

for all x ∈ X and all t > 0. Proceeding as in the proof of Theorem 3.1, we obtain that

$$N\left(f\left(x\right)-\frac{f\left(2^{n}x\right)}{4^n},\sum _{j=0}^{n-1}\frac{{\left|r\right|}^{j}t}{4^{j+2}}\right)\ge \text{min}\left\{N^{\prime }\left(\phi \left(x,x,0\right),\frac{t}{4}\right),N^{\prime }\left(\phi \left(x,0,0\right),\frac{t}{2\left|r\right|}\right)\right\}$$

for all x ∈ X, all t > 0 and any integer n > 0. So

$$\begin{array}{ll}\hfill N\left(f\left(x\right)-\frac{f\left(2^{n}x\right)}{4^n},t\right)& \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,x,0\right),\frac{4t}{{\sum }_{j=0}^{n-1}\frac{{\left|r\right|}^j}{4^j}}\right),N^{\prime }\left(\phi \left(x,0,0\right),\frac{8t}{\left|r\right|{\sum }_{j=0}^{n-1}\frac{{\left|r\right|}^j}{4^j}}\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{2\left(4-\left|r\right|t\right)}{\left|r\right|}\right),N^{\prime }\left(\phi \left(x,x,0\right),\left(4-\left|r\right|\right)t\right)\right\}.\end{array}$$

The rest of the proof is similar to the proof of Theorem 3.5.

Corollary 3.6. 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 an even mapping f : X → Y satisfies (3.12). Then there is a unique quadratic mapping Q: X → Y satisfying (1.1) and the inequality

$$N\left(f\left(x\right)-Q\left(x\right),t\right)\ge \text{min}\left\{N^{\prime }\left(\theta {∥x∥}^p,\frac{2\left(4-4^p\right)t}{4^p}\right),N^{\prime }\left(\theta {∥x∥}^p,\frac{\left(4-4^p\right)t}{2}\right)\right\}$$

Proof. Let φ(x, y, z) := θ(∥x∥^p+ ∥y∥^p+ ∥z∥^p) and |r| = 4^p. Applying Theorem 3.1, we get the desired result.

Theorem 3.7. Assume that an even mapping f : X → Y satisfies the inequality (3.1) and f(0) = 0 and that φ : X³ → Z is a mapping for which there is a constant r ∈ ℝ satisfying$0<\left|r\right|<\frac{1}{4}$such that

$$N^{\prime }\left(\phi \left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right),t\right)\ge N^{\prime }\left(\phi \left(x,y,z\right),\frac{t}{\left|r\right|}\right)$$

(3.23)

for all x, y, z ∈ X and all t > 0. Then there exists a unique quadratic mapping Q : X → Y satisfying (1.1) and the inequality

$$N\left(f\left(x\right)-Q\left(x\right),t\right)\ge \text{min}\left\{N^{\prime }\left(\phi \left(x,x,0\right),\frac{\left(1-4\left|r\right|t\right)}{\left|r\right|}\right),N^{\prime }\left(\phi \left(x,0,0\right),\left(2-8\left|r\right|\right)t\right)\right\}$$

(3.24)

for all x ∈ X and all t > 0.

Proof. It follows from (3.22) that

$$N\left(f\left(x\right)-4f\left(\frac{x}{2}\right),\frac{t}{4}\right)\ge \text{min}\left\{N^{\prime }\left(\phi \left(\frac{x}{2},\frac{x}{2},0\right),\frac{t}{4}\right),N^{\prime }\left(\phi \left(x,0,0\right),\frac{t}{2}\right)\right\}$$

(3.25)

for all x ∈ X and all t > 0. Replacing x by $\frac{x}{2^j}$ in (3.25), we have

$$N\left(4^{j+1}f\left(\frac{x}{2^{j+1}}\right)-4^{j}f\left(\frac{x}{2^j}\right),4^{j}t\right)\ge \text{min}\left\{N^{\prime }\left(\phi \left(x,x,0\right),\frac{t}{{\left|r\right|}^{j+1}}\right),N^{\prime }\left(\phi \left(x,0,0\right),\frac{2t}{{\left|r\right|}^j}\right)\right\}$$

for all x ∈ X, all t > 0 and any integer j ≥ 0. By the same technique as in Theorem 3.1, we find that

$$\begin{array}{ll}\hfill N\left(f\left(x\right)-4^{n}f\left(\frac{x}{2^n}\right),t\right)& \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,x,0\right),\frac{t}{{\sum }_{j=0}^{n-1}{4}^j{\left|r\right|}^{j+1}}\right),N^{\prime }\left(\phi \left(x,0,0\right),\frac{2t}{{\sum }_{j=0}^{n-1}{4}^j{\left|r\right|}^j}\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left(\phi \left(x,x,0\right),\frac{\left(1-4\left|r\right|\right)t}{\left|r\right|}\right),N^{\prime }\left(\phi \left(x,0,0\right),\left(2-8\left|r\right|\right)t\right)\right\}\end{array}$$

The rest of the proof is similar to the proof of Theorem 3.1.

Corollary 3.8. Let X be a normed spaces and (ℝ, N') a fuzzy Banach space. Assume that there exists real numbers θ ≥ 0 and p > 1 such that an even mapping f : X → Y satisfies (3.12) and f(0) = 0. Then there is a unique quadratic mapping Q : X → Y satisfying (1.1) and the inequality

$$N\left(f\left(x\right)-Q\left(x\right),t\right)\ge \text{min}\left\{N^{\prime }\left(\theta {∥x∥}^p,\frac{\left(4^p-4t\right)}{2}\right),N^{\prime }\left(\theta {∥x∥}^p,\left(2-\frac{8}{4^p}\right)t\right)\right\}$$

Proof. Let φ(x, y, z) := θ (∥x∥^p+ ∥y∥^p+ ∥z∥^p) and |r| = 4^-p. Applying Theorem 3.7, we get the desired results.

Let f : X → Y be a mapping satisfying f(0) = 0 and (1.1). Let $f_e\left(x\right):=\frac{f\left(x\right)+f\left(-x\right)}{2}$ and $f_o\left(x\right)=\frac{f\left(x\right)-f\left(-x\right)}{2}$. Then f _e is an even mapping satisfying (1.1) and f _o is an odd mapping satisfying (1.1) such that f(x) = f _e (x) + f _o (x). So we obtain the following.

Theorem 3.9. Assume that a mapping f : X → Y is a mapping satisfying (3.1) and f(0) = 0 and that φ : X³ → Z is a mapping for which there is a constant r ∈ ℝ satisfying$0<\left|r\right|<\frac{1}{4}$such that

$$N^{\prime }\left(\phi \left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right),t\right)\ge N^{\prime }\left(\phi \left(x,y,z\right),\frac{t}{\left|r\right|}\right)$$

for all x, y, z ∈ X and all t > 0. Then there exist a unique additive mapping A : X → Y and a unique quadratic mapping Q : X → Y satisfying (1.1) and the inequality

$$\begin{array}{c}N\left(f\left(x\right)-A\left(x\right)-Q\left(x\right),2t\right)\\ \ge \text{min}\left\{\text{min}\left\{N^{\prime }\phi \left(x,0,0\right),2\left(1-2\left|r\right|t\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{2\left(1-2\right)\left|r\right|t}{\left|r\right|}\right)\right\}\right.,\\ \left.\text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{\left(1-4\left|r\right|\right)t}{\left|r\right|}\right),N^{\prime }\left(\phi \left(x,0,0\right),\left(2-8\left|r\right|\right)t\right)\right\}\right\}\end{array}$$

for all x ∈ X and all t > 0.

Proof. It follows from Theorems 3.1 and 3.7 that

$$\begin{array}{c}N\left(f\left(x\right)-A\left(x\right)-Q\left(x\right),2t\right)\\ =N\left(\frac{f\left(x\right)+f\left(-x\right)}{2}+\frac{f\left(x\right)-f\left(-x\right)}{2}-A\left(x\right)-Q\left(x\right),2t\right)\\ \ge \text{min}\left\{N\left(\frac{f\left(x\right)+f\left(-x\right)}{2}-Q\left(x\right),t\right),N\left(\frac{f\left(x\right)-f\left(-x\right)}{2}-A\left(x\right),t\right)\right\}\\ \ge \text{min}\left\{\text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),2\left(1-2\left|r\right|\right)t\right),N^{\prime }\left(\phi \left(x,x,0\right),\frac{2\left(1-2\left|r\right|\right)t}{\left|r\right|}\right)\right\},\right.\\ \left.\text{min}\left\{N^{\prime }\left(\phi \left(x,0,0\right),\frac{\left(1-4\left|r\right|\right)t}{\left|r\right|}\right),N^{\prime }\left(\phi \left(x,0,0\right),\left(2-8\left|r\right|\right)t\right)\right\}\right\}.\end{array}$$

Corollary 3.10. Let X be a normed spaces and that (ℝ, N') a fuzzy Banach space. Assume that there exists real number θ ≥ 0 and 0 < p < 1 such that a mapping f : X → Y satisfies (3.12) and f(0) = 0. Then there are unique quadratic and additive mappings Q : X → Y and A : X → Y (respectively) satisfying (1.1) and the inequality

$$N\left(f\left(x\right)--A\left(x\right)-Q\left(x\right),t\right)\ge \text{min}\left\{N^{\prime }\left(\theta {∥x∥}^p,\frac{\left(1-4^{p+1}\right)t}{4^p}\right),N^{\prime }\left(\theta {∥x∥}^p,\left(2-{2.4}^{p+1}\right)t\right)\right\}$$

Proof. Let φ(x, y, z) := θ (∥x∥^p+ ∥y∥^p+ ∥z∥^p) and |r| = 4^p. Applying Theorem 3.7, we get the desired results.