On stability of functional equations related to quadratic mappings in fuzzy Banach spaces

Abstract

In this paper, we establish the generalized Hyers-Ulam stability problem of radical quadratic functional equations $f(\sqrt{x^2+y^2})=f(x)+f(y)$ in fuzzy Banach spaces via the direct and fixed point methods.

MSC:39B72, 39B82, 39B52, 47H09.

1 Introduction

The stability problem concerning the stability of group homomorphisms of functional equations was originally introduced by Ulam [1] in 1940. The famous Ulam stability problem was partially solved by Hyers [2] for a linear functional equation of Banach spaces. Hyers’ theorem was generalized by Aoki [3] for additive mappings and by Rassias [4] for linear mappings by considering an unbounded Cauchy difference. The paper of Rassias has had a lot of influence in the development of what we call the generalized Hyers-Ulam stability of functional equations. A generalization of Rassias’ theorem was obtained by Gǎvruta [5] by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias’ approach. Cădariu and Radu [6] applied the fixed point method to the investigation of the Jensen functional equation. They could present a short and simple proof (different from the direct method initiated by Hyers in 1941) for the generalized Hyers-Ulam stability of the Jensen functional and the quadratic functional equations.

The functional equation

$$f(x+y)+f(x-y)=2f(x)+2f(y)$$

(1.1)

is called a quadratic functional equation. Quadratic functional equations were used to characterize inner product spaces. In particular, every solution of the quadratic equation is said to be a quadratic mapping. The generalized Hyers-Ulam stability problem for the quadratic functional equation (1.1) was proved by Skof [7]. Recently, the stability problem of the radical quadratic functional equations in various spaces was proved in the papers [8–11].

In 1984, Katsaras [12] defined a fuzzy norm on a linear 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 [13–16]. Cheng and Mordeson [17] introduced a definition of fuzzy norm on a linear space in such a manner that the corresponding induced fuzzy metric is of the Kramosil and Michálek type [14]. In 2003, Bag and Samanta [18] modified the definition of Cheng and Mordeson by removing a regular condition. Also, they investigated a decomposition theorem of a fuzzy norm into a family to crisp norms and gave some properties of fuzzy norm. The fuzzy stability problems of several functional equations have been extensively investigated by a number of authors and there are many interesting results concerning these problems [17, 19–23].

In the sequel, we use the definitions and some basic facts concerning fuzzy Banach spaces given in Bag and Samanta [18].

Definition 1.1 Let X be a real linear space. A function $N:X\times \mathbb{R}\to [0,1]$ is called a fuzzy norm on X if, for all $x,y\in X$ and $s,t\in \mathbb{R}$, N satisfies the following conditions:

(N1) $N(x,t)=0$ for all $t\le 0$;

(N2) $x=0$ if and only if $N(x,t)=1$ for all $t>0$;

(N3) $N(cx,t)=N(x,t/|c|)$ for all $c\in \mathbb{R}$ with $c\ne 0$;

(N4) $N(x+y,s+t)\ge min\{N(x,s),N(y,t)\}$;

(N5) $N(x,\cdot )$ is a nondecreasing function of ℝ and ${lim}_{t\to \mathrm{\infty }}N(x,t)=1$;

(N6) for all $x\in X$ with $x\ne 0$, $N(x,\cdot )$ is continuous on ℝ.

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

Example 1.2 Let $(X,\parallel \cdot \parallel )$ be a normed linear space and let $\alpha ,\beta >0$. Then

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

is a fuzzy norm on X.

Definition 1.3 Let $(X,N)$ be a fuzzy normed linear space.

• (1) A sequence $\{x_n\}$ in X is said to be convergent to a point $x\in X$ if, for any $ϵ>0$ and $t>0$, there exists $n_0\in {\mathbb{Z}}^{+}$ such that $N(x_n-x,t)>1-ϵ$ for all $n\ge n_0$. In this case, x is called the limit of the sequence $\{x_n\}$, which is denoted by $x={lim}_{n\to \mathrm{\infty }}{x}_n$.

• (2) A sequence $\{x_n\}$ in X is called a Cauchy sequence if, for any $ϵ>0$ and $t>0$, there exists $n_0\in {\mathbb{Z}}^{+}$ such that $N(x_{n+p}-x_n,t)>1-ϵ$ for all $n\ge n_0$ and $p\in {\mathbb{Z}}^{+}$.

• (3) If every Cauchy sequence is convergent, then the fuzzy norm is said to be complete and the fuzzy normed linear space is called a fuzzy Banach space.

A mapping $f:X\to Y$ between fuzzy normed linear spaces X and Y is said to be continuous at a point $x_0\in X$ if, for any sequence $\{x_n\}$ in X converging to a point $x_0\in X$, the sequence $\{f(x_n)\}$ converges to $f(x_0)$. If $f:X\to Y$ is continuous at every point $x\in X$, then f is said to be continuous on X.

Example 1.4 Let $N:\mathbb{R}\times \mathbb{R}\to [0,1]$ be a fuzzy norm on ℝ defined by

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

Then $(\mathbb{R},N)$ is a fuzzy Banach space.

In this paper, we establish the generalized Hyers-Ulam stability problem of a radical quadratic functional equation $f(\sqrt{x^2+y^2})=f(x)+f(y)$ in fuzzy Banach spaces via the direct and fixed point methods.

2 Fuzzy stability of the radical quadratic functional equations

In this section, we study a fuzzy version of the generalized Hyers-Ulam stability of functional equation which approximate uniformly a radical quadratic mapping in fuzzy Banach spaces.

2.1 The direct method

Theorem 2.1 Let $\ell \in \{-1,1\}$ be fixed, $(\mathcal{Y},N)$ be a fuzzy Banach space and $\varphi :{\mathbb{R}}^2\to [0,\mathrm{\infty })$ be a mapping such that

$$\mathrm{\Phi }(x,y):=\sum _{n=\frac{1-\ell }{2}}^{\mathrm{\infty }}\frac{1}{2^{\ell n}}\varphi (2^{\frac{\ell n}{2}}x,2^{\frac{\ell n}{2}}y)+\varphi (2^{\frac{\ell n+1}{2}}x,0)<\mathrm{\infty }$$

(2.1)

for all $x,y\in \mathbb{R}$. Suppose that $f:\mathbb{R}\to \mathcal{Y}$ is a mapping with $f(0)=0$ such that, for all $t>0$,

$$\underset{t\to \mathrm{\infty }}{lim}N(f\left(\sqrt{x^2+y^2}\right)-f(x)-f(y),t\varphi (x,y))=1$$

(2.2)

uniformly on ${\mathbb{R}}^2$. Then there exists a unique quadratic mapping $Q:\mathbb{R}\to \mathcal{Y}$ such that, if there exist $\delta >0$ and $\alpha >0$ such that

$$N(f\left(\sqrt{x^2+y^2}\right)-f(x)-f(y),\delta \varphi (x,y))\ge \alpha$$

(2.3)

for all $x,y\in \mathbb{R}$, then

$$N(f(x)-Q(x),\frac{\delta }{2}\mathrm{\Phi }(x,x))\ge \alpha$$

(2.4)

for all $x\in \mathbb{R}$. Furthermore, the quadratic mapping $Q:\mathbb{R}\to \mathcal{Y}$ is a unique mapping such that, for all $t>0$,

$$N(f(x)-Q(x),t\mathrm{\Phi }(x,x))=1$$

(2.5)

uniformly on ℝ.

Proof Assume that $\ell =1$. For any $ϵ>0$, by (2.2), we can find some $t_0>0$ such that

$$N(f\left(\sqrt{x^2+y^2}\right)-f(x)-f(y),t\varphi (x,y))\ge 1-ϵ$$

(2.6)

for all $x,y\in \mathbb{R}$ and $t\ge t_0$. Replacing x and y by $\frac{x+y}{\sqrt{2}}$ and $\frac{x-y}{\sqrt{2}}$ in (2.6), respectively, we have

$$N(f\left(\sqrt{x^2+y^2}\right)-f\left(\frac{x+y}{\sqrt{2}}\right)-f\left(\frac{x-y}{\sqrt{2}}\right),t\varphi (\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}}))\ge 1-ϵ$$

(2.7)

for all $x,y\in \mathbb{R}$ and $t\ge t_0$. It follows from (2.6), (2.7), and (N4) that

$$N(f(x)+f(y)-f\left(\frac{x+y}{\sqrt{2}}\right)-f\left(\frac{x-y}{\sqrt{2}}\right),t(\varphi (x,y)+\varphi (\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}})))\ge 1-ϵ$$

(2.8)

for all $x,y\in \mathbb{R}$ and $t\ge t_0$. Letting $y=x$ in (2.8), we have

$$N(2f(x)-f\left(2^{\frac{1}{2}}x\right),t\hat{\varphi }(x,x))\ge 1-ϵ$$

(2.9)

for all $x\in \mathbb{R}$ and $t\ge t_0$, where $\hat{\varphi }(x,x)=\varphi (x,x)+\varphi (2^{\frac{1}{2}}x,0)$. By induction on n, we have

$$N(2^{n}f(x)-f\left(2^{\frac{n}{2}}x\right),t\sum _{k=0}^{n-1}{2}^{n-k-1}\hat{\varphi }(2^{\frac{k}{2}}x,2^{\frac{k}{2}}x))\ge 1-ϵ$$

(2.10)

for all $x\in \mathbb{R}$, $t\ge t_0$ and $n\in {\mathbb{Z}}^{+}$. Let $t=t_0$. Replacing n and x by p and $2^{\frac{n}{2}}x$ in (2.10), respectively, we have

$$N(\frac{f(2^{\frac{n}{2}}x)}{2^n}-\frac{f(2^{\frac{n+p}{2}}x)}{2^{n+p}},\frac{t_0}{2^{n+p}}\sum _{k=0}^{p-1}{2}^{p-k-1}\hat{\varphi }(2^{\frac{n+k}{2}}x,2^{\frac{n+k}{2}}x))\ge 1-ϵ$$

(2.11)

for all $n\ge 0$ and $p>0$. It follows from (2.1) and the equality

$$\sum _{k=0}^{p-1}\frac{1}{2^{n+k+1}}\hat{\varphi }(2^{\frac{n+k}{2}}x,2^{\frac{n+k}{2}}x)=\frac{1}{2}\sum _{k=n}^{n+p-1}\frac{1}{2^k}\hat{\varphi }(2^{\frac{k}{2}}x,2^{\frac{k}{2}}x)$$

that, for any $\delta >0$, there exists some $n_0\in {\mathbb{Z}}^{+}$ such that

$$\frac{t_0}{2}\sum _{k=n}^{n+p-1}\frac{1}{2^k}\hat{\varphi }(2^{\frac{k}{2}}x,2^{\frac{k}{2}}x)<\delta$$

for all $n\ge n_0$ and $p>0$. Now, it follows from (2.11) that

$$\begin{array}{c}N(\frac{f(2^{\frac{n}{2}}x)}{2^n}-\frac{f(2^{\frac{n+p}{2}}x)}{2^{n+p}},\delta )\hfill \\ \ge N(\frac{f(2^{\frac{n}{2}}x)}{2^n}-\frac{f(2^{\frac{n+p}{2}}x)}{2^{n+p}},\frac{t_0}{2^{n+p}}\sum _{k=0}^{p-1}{2}^{p-k-1}\hat{\varphi }(2^{\frac{n+k}{2}}x,2^{\frac{n+k}{2}}x))\ge 1-ϵ\hfill \end{array}$$

(2.12)

for all $n\ge n_0$ and $p>0$. Thus the sequence $\{\frac{f(2^{\frac{n}{2}}x)}{2^n}\}$ is a Cauchy sequence in a fuzzy Banach space and so it converges to some $Q(x)\in \mathcal{Y}$. We can define a mapping $Q:\mathbb{R}\to \mathcal{Y}$ by

$$Q(x)=\underset{n\to \mathrm{\infty }}{lim}\frac{f(2^{\frac{n}{2}}x)}{2^n},$$

that is, ${lim}_{n\to \mathrm{\infty }}N(\frac{f(2^{\frac{n}{2}}x)}{2^n}-Q(x),t)=1$ for all $x\in \mathbb{R}$ and $t>0$. Let $x,y\in \mathbb{R}$, $t>0$ and $0<ϵ<1$. Since ${lim}_{n\to \mathrm{\infty }}\frac{1}{2^n}\hat{\varphi }(2^{\frac{n}{2}}x,2^{\frac{n}{2}}y)=0$, there exists $n_1\in {\mathbb{Z}}^{+}$ with $n_1>n_0$ such that

$$t_0\hat{\varphi }(2^{\frac{n}{2}}x,2^{\frac{n}{2}}y)<\frac{2^{n}t}{4}$$

for all $n\ge n_1$. Then, by (N4), we have

$$\begin{array}{c}N(Q\left(\sqrt{x^2+y^2}\right)-Q(x)-Q(y),t)\hfill \\ \ge min\{N(Q(x+y)-\frac{1}{2^n}f\left(\sqrt{2^n{x}^2+2^n{y}^2}\right),\frac{t}{4}),N(Q(x)-\frac{1}{2^n}f\left(2^{\frac{n}{2}}x\right),\frac{t}{4}),\hfill \\ N(Q(y)-\frac{1}{2^n}f\left(2^{\frac{n}{2}}y\right),\frac{t}{4}),N(f\left(\sqrt{2^n{x}^2+2^n{y}^2}\right)-f\left(2^{\frac{n}{2}}x\right)-f\left(2^{\frac{n}{2}}y\right),\frac{2^{n}t}{4})\}\hfill \end{array}$$

(2.13)

for all $n\ge n_1$. Since the first three terms on the right-hand side of the above inequality tend to 1 as $n\to \mathrm{\infty }$ and

$$N(f\left(\sqrt{2^n{x}^2+2^n{y}^2}\right)-f\left(2^{\frac{n}{2}}x\right)-f\left(2^{\frac{n}{2}}y\right),t_0\hat{\varphi }(2^{\frac{n}{2}},2^{\frac{n}{2}}y))\ge 1-ϵ,$$

we have

$$N(Q\left(\sqrt{x^2+y^2}\right)-Q(x)-Q(y),t)\ge 1-ϵ$$

for all $x,y\in \mathbb{R}$, $t>0$ and $0<ϵ<1$. It follows from (N2) that $Q(\sqrt{x^2+y^2})=Q(x)+Q(y)$ for all $x,y\in \mathbb{R}$. This means that Q is a quadratic mapping [10].

Now, suppose that (2.3) holds for some $\delta >0$ and $\alpha >0$. Then assume that

$${\psi }_n(x,y)=\sum _{k=0}^{n-1}\frac{1}{2^{k+1}}\hat{\varphi }(2^{\frac{k}{2}}x,2^{\frac{k}{2}}y)$$

for all $x,y\in \mathbb{R}$. For all $x\in \mathbb{R}$, by a similar method to the beginning of the proof, we have

$$N(2^{n}f(x)-f\left(2^{\frac{n}{2}}x\right),\delta \sum _{k=0}^{n-1}{2}^{n-k-1}\hat{\varphi }(2^{\frac{k}{2}}x,2^{\frac{k}{2}}x))\ge \alpha$$

(2.14)

for all $n\in {\mathbb{Z}}^{+}$. Let $t>0$. Then we have

$$\begin{array}{c}N(f(x)-Q(x),\delta {\psi }_n(x,x)+t)\hfill \\ \ge min\{N(f(x)-\frac{f(2^{\frac{n}{2}}x)}{2^n},\delta {\psi }_n(x,x)),N(\frac{f(2^{\frac{n}{2}}x)}{2^n}-Q(x),t)\}.\hfill \end{array}$$

(2.15)

Combining (2.14) and (2.15) and using the fact ${lim}_{n\to \mathrm{\infty }}N(\frac{f(2^{\frac{n}{2}}x)}{2^n}-Q(x),t)=1$, we obtain

$$N(f(x)-Q(x),\delta {\psi }_n(x,x)+t)\ge \alpha$$

(2.16)

for large enough $n\in {\mathbb{Z}}^{+}$. It follows from the continuity of the function $N(f(x)-Q(x),\cdot )$ that

$$N(f(x)-Q(x),\frac{\delta }{2}\mathrm{\Phi }(x,x)+t)\ge \alpha .$$

Letting $t\to 0$, we conclude (2.5).

Next, assume that there exists another quadratic mapping T which satisfies (2.5). For any $ϵ>0$, by applying (2.5) for the mappings Q and T, we can find some $t_0>0$ such that

$$N(f(x)-Q(x),\frac{t}{2}\mathrm{\Phi }(x,x))\ge 1-ϵ,N(f(x)-T(x),\frac{t}{2}\mathrm{\Phi }(x,x))\ge 1-ϵ$$

for all $x\in \mathbb{R}$ and $t\ge t_0$. Fix $x\in \mathbb{R}$ and $c>0$. Then we find some $n_0\in {\mathbb{Z}}^{+}$ such that

$$t_0\sum _{k=n}^{\mathrm{\infty }}\frac{1}{2^k}\hat{\varphi }(2^{\frac{k}{2}}x,2^{\frac{k}{2}}y)<\frac{c}{2}$$

for all $x,y\in \mathbb{R}$ and $n\ge n_0$. It follows from

$$\begin{array}{rl}\sum _{k=n}^{\mathrm{\infty }}\frac{1}{2^k}\hat{\varphi }(2^{\frac{k}{2}}x,2^{\frac{k}{2}}y)& =\frac{1}{2^n}\sum _{k=n}^{\mathrm{\infty }}\frac{1}{2^{k-n}}\hat{\varphi }(2^{k-n}\left(2^{\frac{n}{2}}x\right),2^{k-n}\left(2^{\frac{n}{2}}y\right))\\ =\frac{1}{2^n}\sum _{m=0}^{\mathrm{\infty }}\frac{1}{2^m}\hat{\varphi }(2^m\left(2^{\frac{n}{2}}x\right),2^m\left(2^{\frac{n}{2}}y\right))\\ =\frac{1}{2^n}\mathrm{\Phi }(2^{\frac{n}{2}}x,2^{\frac{n}{2}}y)\end{array}$$

that

$$\begin{array}{c}N(Q(x)-T(x),c)\hfill \\ \ge min\{N(\frac{f(2^{\frac{n}{2}}x)}{2^n}-Q(x),\frac{c}{2}),N(T(x)-\frac{f(2^{\frac{n}{2}}x)}{2^n},\frac{c}{2})\}\hfill \\ =min\{N(f\left(2^{\frac{n}{2}}x\right)-Q\left(2^{\frac{n}{2}}x\right),2^{n-1}c),N(T\left(2^{\frac{n}{2}}x\right)-f\left(2^{\frac{n}{2}}x\right),2^{n-1}c)\}\hfill \\ \ge min\{N(f\left(2^{\frac{n}{2}}x\right)-Q\left(2^{\frac{n}{2}}x\right),2^n{t}_0\sum _{k=n}^{\mathrm{\infty }}\frac{1}{2^k}\hat{\varphi }(2^{\frac{k}{2}}x,2^{\frac{k}{2}}x)),\hfill \\ N(T\left(2^{\frac{n}{2}}x\right)-f\left(2^{\frac{n}{2}}x\right),2^n{t}_0\sum _{k=n}^{\mathrm{\infty }}\frac{1}{2^k}\hat{\varphi }(2^{\frac{k}{2}}x,2^{\frac{k}{2}}x))\}\hfill \\ \ge min\{N(f\left(2^{\frac{n}{2}}x\right)-Q\left(2^{\frac{n}{2}}x\right),t_0\mathrm{\Phi }(2^{\frac{n}{2}}x,2^{\frac{n}{2}}x)),\hfill \\ N(T\left(2^{\frac{n}{2}}x\right)-f\left(2^{\frac{n}{2}}x\right),t_0\mathrm{\Phi }(2^{\frac{n}{2}}x,2^{\frac{n}{2}}x))\}\hfill \\ \ge 1-ϵ\hfill \end{array}$$

for all $x,y\in \mathbb{R}$ and $c>0$. Thus we have $N(Q(x)-T(x),c)=1$ for all $c>0$ and so $Q(x)=T(x)$ for all $x\in \mathbb{R}$.

For the case $\ell =-1$, we can state the proof in the same method as in the first case. In the case, the mapping Q is defined by $Q(x)={lim}_{n\to \mathrm{\infty }}{2}^{n}f(2^{-\frac{n}{2}}x)$. This completes the proof. □

Corollary 2.2 Let $(\mathcal{Y},N)$ be a fuzzy Banach space, θ and $p\in \mathbb{R}$ with $p<2$ be positive real numbers. Suppose that $f:\mathbb{R}\to \mathcal{Y}$ is a mapping with $f(0)=0$ such that, for all $t>0$,

$$\underset{t\to \mathrm{\infty }}{lim}N(f\left(\sqrt{x^2+y^2}\right)-f(x)-f(y),t\theta (|x{|}^p+|y{|}^p))=1$$

(2.17)

uniformly on ℝ. Then the limit $Q(x)={lim}_{n\to \mathrm{\infty }}\frac{f(2^{\frac{n}{2}}x)}{2^n}$ exists for all $x\in X$ and there exists a unique quadratic mapping $Q:\mathbb{R}\to \mathcal{Y}$ such that

$$\underset{t\to \mathrm{\infty }}{lim}N(f(x)-Q(x),\frac{2(2+2^{\frac{p}{2}})}{2-2^{\frac{p}{2}}}\theta |x{|}^{p}t)=1$$

(2.18)

uniformly on ℝ.

Proof The proof follows from Theorem 2.1 by taking $\varphi (x,y)=\theta (|x{|}^p+|y{|}^p)$ for all $x,y\in \mathbb{R}$. □

Corollary 2.3 Let $(\mathcal{Y},N)$ be a fuzzy Banach space and $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ be a mapping such that, for all $s,t>0$,

• (a) $\psi (ts)=\psi (t)\psi (s)$;

• (b) $\psi (\sqrt{2})<2$.

Suppose that $f:\mathbb{R}\to \mathcal{Y}$ is a mapping with $f(0)=0$ such that, for all $t>0$,

$$\underset{t\to \mathrm{\infty }}{lim}N(f\left(\sqrt{x^2+y^2}\right)-f(x)-f(y),t\theta (\psi (|x|)+\psi (|y|)))=1$$

(2.19)

uniformly on ${\mathbb{R}}^2$, where $\theta >0$ is fixed. Then the limit $Q(x)={lim}_{n\to \mathrm{\infty }}\frac{f(2^{\frac{n}{2}}x)}{2^n}$ exists for all $x\in \mathbb{R}$ and defines a quadratic mapping $Q:\mathbb{R}\to \mathcal{Y}$ such that, for all $t>0$,

$$\underset{t\to \mathrm{\infty }}{lim}N(f(x)-Q(x),\frac{2(2+\psi (\sqrt{2}))}{2-\psi (\sqrt{2})}\theta \psi (|x|)t)=1$$

(2.20)

uniformly on ℝ.

Proof The proof follows from Theorem 2.1 by taking $\varphi (x,y)=\theta (\psi (|x|)+\psi (|y|))$ for all $x,y\in \mathbb{R}$. □

2.2 The fixed point method

Recall that a mapping $d:X^2\to [0,+\mathrm{\infty }]$ is called a generalized metric on a nonempty set X if

• (1) $d(x,y)=0$ if and only if $x=y$;

• (2) $d(x,y)=d(y,x)$;

• (3) $d(x,z)\le d(x,y)+d(y,z)$ for all $x,y,z\in X$.

A set X with the generalized metric d is called a generalized metric space.

In [24], Diaz and Margolis proved the following fixed point theorem, which plays an important role for the main results in this section.

Theorem 2.4 [24]

Suppose that $(\mathrm{\Omega },d)$ is a complete generalized metric space and $T:\mathrm{\Omega }\to \mathrm{\Omega }$ is a strictly contractive mapping with Lipshitz constant L. Then, for any $x\in \mathrm{\Omega }$, either $d(T^{n}x,T^{n+1}x)=\mathrm{\infty }$ for all $n\ge 0$ or there exists a positive integer $n_0$ such that

• (1) $d(T^{n}x,T^{n+1}x)<\mathrm{\infty }$ for all $n\ge n_0$;

• (2) the sequence $\{T^{n}x\}$ is convergent to a fixed point $y^{\ast }$ of T;

• (3) $y^{\ast }$ is the unique fixed point of T in the set $\mathrm{\Lambda }=\{y\in \mathrm{\Omega }:d(T^{n_0}x,y)<\mathrm{\infty }\}$;

• (4) $d(y,y^{\ast })\le \frac{1}{1-L}d(y,Ty)$ for all $y\in \mathrm{\Lambda }$.

Theorem 2.5 Let $(\mathcal{Y},N)$ be a fuzzy Banach space and $\varphi :{\mathbb{R}}^2\to [0,\mathrm{\infty })$ be a mapping such that there exists $L<1$ with

$$\varphi (2^{\frac{1}{2}}x,2^{\frac{1}{2}}y)\le 2L\varphi (x,y)$$

(2.21)

for all $x,y\in \mathbb{R}$. If $f:\mathbb{R}\to \mathcal{Y}$ is a mapping with $f(0)=0$ and

$$N(f\left(\sqrt{x^2+y^2}\right)-f(x)-f(y),t)\ge \frac{t}{t+\varphi (x,y)}$$

(2.22)

for all $x,y\in \mathbb{R}$ and $t>0$, then the limit $Q(x)={lim}_{n\to \mathrm{\infty }}\frac{1}{2^n}f(2^{\frac{n}{2}}x)$ exists for all $x\in \mathbb{R}$ and a unique quadratic mapping $Q:\mathbb{R}\to \mathcal{Y}$ satisfies the inequality

$$N(f(x)-Q(x),t)\ge \frac{(1-L)t}{(1-L)t+\hat{\varphi }(x,x)}$$

(2.23)

for all $x\in \mathbb{R}$, where $\hat{\varphi }(x,x)=\varphi (x,x)+\varphi (2^{\frac{1}{2}}x,0)$.

Proof Letting x and y by $\frac{x+y}{\sqrt{2}}$ and $\frac{x-y}{\sqrt{2}}$ in (2.22), respectively, we have

$$N(f\left(\sqrt{x^2+y^2}\right)-f\left(\frac{x+y}{\sqrt{2}}\right)-f\left(\frac{x-y}{\sqrt{2}}\right),t)\ge \frac{t}{t+\varphi (\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}})}$$

(2.24)

for all $x,y\in \mathbb{R}$ and $t\ge t_0$. It follows from (2.22), (2.24), and (N4) that

$$\begin{array}{c}N(f(x)+f(y)-f\left(\frac{x+y}{\sqrt{2}}\right)-f\left(\frac{x-y}{\sqrt{2}}\right),2t)\hfill \\ \ge min\{N(f(x)+f(y)-f\left(\sqrt{x^2+y^2}\right),t),\hfill \\ N(f\left(\sqrt{x^2+y^2}\right)-f\left(\frac{x+y}{\sqrt{2}}\right)-f\left(\frac{x-y}{\sqrt{2}}\right),t)\}\hfill \\ \ge min\{\frac{t}{t+\varphi (x,y)},\frac{t}{t+\varphi (\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}})}\}\hfill \\ \ge \frac{t}{t+\varphi (x,y)+\varphi (\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}})}\hfill \end{array}$$

(2.25)

for all $x,y\in \mathbb{R}$ and $t\ge t_0$. Letting $y=x$ in (2.25), we have

$$N(f(x)-\frac{1}{2}f\left(2^{\frac{1}{2}}x\right),t)\ge \frac{t}{t+\hat{\varphi }(x,x)}$$

(2.26)

for all $x\in \mathbb{R}$ and $t\ge t_0$, where $\hat{\varphi }(x,y)=\varphi (x,y)+\varphi (2^{\frac{1}{2}}x,0)$.

Let Ω be a set of all mapping from ℝ to and introduce a generalized metric on Ω as follows:

$$d(g,h)=inf\{\mu \in [0,\mathrm{\infty }):N(g(x)-h(x),\mu t)\ge \frac{t}{t+\hat{\varphi }(x,x)},\mathrm{\forall }x\in \mathbb{R},t>0\}.$$

It is easy to show that $(\mathrm{\Omega },d)$ is a generalized complete metric space [25]. We consider the mapping $T:\mathrm{\Omega }\to \mathrm{\Omega }$ defined by

$$Tg(x)=\frac{1}{2}g\left(2^{\frac{1}{2}}x\right)$$

for all $g\in \mathrm{\Omega }$ and $x\in \mathbb{R}$. Let $g,h\in \mathrm{\Omega }$ such that $d(g,h)\le \mu$. Then we have

$$N(Tg(x)-Th(x),t\mu L)=N(g\left(2^{\frac{1}{2}}x\right)-h\left(2^{\frac{1}{2}}x\right),2t\mu L)\ge \frac{t}{t+\hat{\varphi }(x,x)}$$

for all $x\in \mathbb{R}$, and so

$$d(Tg,Th)\le Ld(g,h)$$

for all $g,h\in \mathrm{\Omega }$. This means that T is a strictly contractive self-mapping of Ω with the Lipschitz constant L.

It follows from (2.26) that $d(f,Tf)\le 1<\mathrm{\infty }$. Now, it follows from Theorem 2.4 that the sequence $\{T^{n}f\}$ converges to a unique fixed point Q of T. So there exists a fixed point Q of T in Ω such that

$$Q(x)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2^n}f\left(2^{\frac{n}{2}}x\right)$$

(2.27)

for all $x\in \mathbb{R}$ since ${lim}_{n\to \mathrm{\infty }}d(T^n,Q)=0$. Again, using the fixed point method, since Q is the unique fixed point of T in ${\mathrm{\Omega }}^{\ast }=\{g\in \mathrm{\Omega }:d(f,g)<\mathrm{\infty }\}$, we have

$$d(f,Q)\le \frac{1}{1-L}d(f,Tf)\le \frac{1}{1-L},$$

which gives

$$N(f(x)-Q(x),t)\ge \frac{(1-L)t}{(1-L)t+\hat{\varphi }(x,x)}$$

for all $x\in \mathbb{R}$ and $t>0$. Further, we have

$$\begin{array}{c}N(Q\left(\sqrt{x^2+y^2}\right)-Q(x)-Q(y),t)\hfill \\ \ge \underset{n\to \mathrm{\infty }}{lim}N(f\left(\sqrt{2^n{x}^2+2^n{y}^2}\right)-f\left(2^{\frac{n}{2}}x\right)-f\left(2^{\frac{n}{2}}y\right),2^{n}t)\hfill \\ \ge \underset{n\to \mathrm{\infty }}{lim}\frac{t}{t+L^n\hat{\varphi }(x,y)}=1\hfill \end{array}$$

(2.28)

for all $x,y\in \mathbb{R}$ and $t>0$. It follows from (N2) and $N(Q(\sqrt{x^2+y^2})-Q(x)-Q(y),t)\ge 1$ that $Q(\sqrt{x^2+y^2})=Q(x)+Q(y)$ for all $x,y\in \mathbb{R}$. This means that Q is a quadratic mapping on ℝ. This completes the proof. □

Theorem 2.6 Let $(\mathcal{Y},N)$ be a fuzzy Banach space and $\varphi :{\mathbb{R}}^2\to [0,\mathrm{\infty })$ be a mapping such that there exists $L<1$ with

$$\varphi (\frac{x}{\sqrt{2}},\frac{y}{\sqrt{2}})\le \frac{L}{2}\varphi (x,y)$$

(2.29)

for all $x,y\in \mathbb{R}$. If $f:\mathbb{R}\to \mathcal{Y}$ is a mapping with $f(0)=0$ and (2.22), then the limit $Q(x)={lim}_{n\to \mathrm{\infty }}{2}^{n}f(\frac{x}{2^{\frac{n}{2}}})$ exists for all $x\in \mathbb{R}$ and there exists a unique quadratic mapping $Q:\mathbb{R}\to \mathcal{Y}$ satisfying the inequality

$$N(f(x)-Q(x),t)\ge \frac{(1-L)t}{(1-L)t+L\hat{\varphi }(x,x)}$$

(2.30)

for all $x\in \mathbb{R}$ and $t>0$, where $\hat{\varphi }(x,x)=\varphi (x,x)+\varphi (2^{\frac{1}{2}}x,0)$.

Proof It follows from (2.26) that

$$N(f(x)-2f\left(\frac{x}{\sqrt{2}}\right),Lt)\ge \frac{t}{t+\hat{\varphi }(x,x)}$$

(2.31)

for all $x\in \mathbb{R}$ and $t\ge t_0$, where $\hat{\varphi }(x,y)=\varphi (x,y)+\varphi (2^{\frac{1}{2}}x,0)$. Let Ω and d be as in the proof of Theorem 2.5. Then $(\mathrm{\Omega },d)$ becomes a generalized complete metric space and we consider the mapping $T:\mathrm{\Omega }\to \mathrm{\Omega }$ defined by

$$(Tg)(x)=2g\left(\frac{x}{\sqrt{2}}\right),$$

$x\in \mathbb{R}$. So, we have $d(Tg,Th)\le Ld(g,h)$ for all $g,h\in \mathrm{\Omega }$. It follows from Theorem 2.4 that there exists a unique mapping $Q:\mathbb{R}\to \mathcal{Y}$ in the set $\{g\in \mathrm{\Omega }:d(f,g)<\mathrm{\infty }\}$ which is a unique fixed point of T such that

$$Q(x)=\underset{n\to \mathrm{\infty }}{lim}{2}^{n}f\left(\frac{x}{2^{\frac{n}{2}}}\right)$$

for all $x\in \mathbb{R}$. Also, from (2.31) we have $d(f,Tf)\le L$. So, we can conclude that

$$d(f,Q)\le \frac{1}{1-L}d(f,Tf)\le \frac{L}{1-L},$$

which implies the inequality (2.30). The remaining assertion goes through in a similar way to the corresponding part of Theorem 2.4. This completes the proof. □

Corollary 2.7 Let $(\mathcal{Y},N)$ be a fuzzy Banach space and θ, $p\ne 2$ be positive real numbers. Suppose that $f:\mathbb{R}\to \mathcal{Y}$ is a mapping with $f(0)=0$ such that, for all $t>0$,

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

(2.32)

uniformly on ℝ. Then there exists a unique quadratic mapping $Q:\mathbb{R}\to \mathcal{Y}$ such that

$$N(f(x)-Q(x),t)\ge \{\begin{array}{cc}\frac{(2-2^{\frac{p}{2}})t}{(2-2^{\frac{p}{2}})t+2(2+2^{\frac{p}{2}})\theta |x{|}^p},\hfill & p<2,\hfill \\ \frac{(2-2^{\frac{p}{2}})t}{(2-2^{\frac{p}{2}})t+2^{\frac{p}{2}}(2+2^{\frac{p}{2}})\theta |x{|}^p},\hfill & p>2,\hfill \end{array}$$

(2.33)

uniformly on ℝ.

Proof Taking $\varphi (x,y)=\theta (|x{|}^p+|y{|}^p)$ for all $x,y\in \mathbb{R}$ and choosing $L=2^{\frac{p}{2}}$, we have the desired result. □

Remark 2.8 The radical quadratic functional equation $f(\sqrt{x^2+y^2})=f(x)+f(y)$ is not stable for $p=2$ [11].