## 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

$f\left(x+y\right)+f\left(x-y\right)=2f\left(x\right)+2f\left(y\right)$
(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 [811].

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 [1316]. 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, 1923].

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 $N:X×\mathbb{R}\to \left[0,1\right]$ 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) $N\left(x,t\right)=0$ for all $t\le 0$;

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

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

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

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

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

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

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

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

is a fuzzy norm on X.

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

1. (1)

A sequence $\left\{{x}_{n}\right\}$ 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 $N\left({x}_{n}-x,t\right)>1-ϵ$ for all $n\ge {n}_{0}$. In this case, x is called the limit of the sequence $\left\{{x}_{n}\right\}$, which is denoted by $x={lim}_{n\to \mathrm{\infty }}{x}_{n}$.

2. (2)

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

3. (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 $\left\{{x}_{n}\right\}$ in X converging to a point ${x}_{0}\in X$, the sequence $\left\{f\left({x}_{n}\right)\right\}$ converges to $f\left({x}_{0}\right)$. 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 $N:\mathbb{R}×\mathbb{R}\to \left[0,1\right]$ be a fuzzy norm on ℝ defined by

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

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

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

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

$\mathrm{\Phi }\left(x,y\right):=\sum _{n=\frac{1-\ell }{2}}^{\mathrm{\infty }}\frac{1}{{2}^{\ell n}}\varphi \left({2}^{\frac{\ell n}{2}}x,{2}^{\frac{\ell n}{2}}y\right)+\varphi \left({2}^{\frac{\ell n+1}{2}}x,0\right)<\mathrm{\infty }$
(2.1)

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

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

$N\left(f\left(\sqrt{{x}^{2}+{y}^{2}}\right)-f\left(x\right)-f\left(y\right),\delta \varphi \left(x,y\right)\right)\ge \alpha$
(2.3)

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

$N\left(f\left(x\right)-Q\left(x\right),\frac{\delta }{2}\mathrm{\Phi }\left(x,x\right)\right)\ge \alpha$
(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 $t>0$,

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

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

$N\left(f\left(\sqrt{{x}^{2}+{y}^{2}}\right)-f\left(\frac{x+y}{\sqrt{2}}\right)-f\left(\frac{x-y}{\sqrt{2}}\right),t\varphi \left(\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}}\right)\right)\ge 1-ϵ$
(2.7)

for all $x,y\in \mathbb{R}$ and $t\ge {t}_{0}$. It follows from (2.6), (2.7), and (N4) that

$N\left(f\left(x\right)+f\left(y\right)-f\left(\frac{x+y}{\sqrt{2}}\right)-f\left(\frac{x-y}{\sqrt{2}}\right),t\left(\varphi \left(x,y\right)+\varphi \left(\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}}\right)\right)\right)\ge 1-ϵ$
(2.8)

for all $x,y\in \mathbb{R}$ and $t\ge {t}_{0}$. Letting $y=x$ in (2.8), we have

$N\left(2f\left(x\right)-f\left({2}^{\frac{1}{2}}x\right),t\stackrel{ˆ}{\varphi }\left(x,x\right)\right)\ge 1-ϵ$
(2.9)

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

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

$N\left(\frac{f\left({2}^{\frac{n}{2}}x\right)}{{2}^{n}}-\frac{f\left({2}^{\frac{n+p}{2}}x\right)}{{2}^{n+p}},\frac{{t}_{0}}{{2}^{n+p}}\sum _{k=0}^{p-1}{2}^{p-k-1}\stackrel{ˆ}{\varphi }\left({2}^{\frac{n+k}{2}}x,{2}^{\frac{n+k}{2}}x\right)\right)\ge 1-ϵ$
(2.11)

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

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

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

$\frac{{t}_{0}}{2}\sum _{k=n}^{n+p-1}\frac{1}{{2}^{k}}\stackrel{ˆ}{\varphi }\left({2}^{\frac{k}{2}}x,{2}^{\frac{k}{2}}x\right)<\delta$

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

$\begin{array}{c}N\left(\frac{f\left({2}^{\frac{n}{2}}x\right)}{{2}^{n}}-\frac{f\left({2}^{\frac{n+p}{2}}x\right)}{{2}^{n+p}},\delta \right)\hfill \\ \phantom{\rule{1em}{0ex}}\ge N\left(\frac{f\left({2}^{\frac{n}{2}}x\right)}{{2}^{n}}-\frac{f\left({2}^{\frac{n+p}{2}}x\right)}{{2}^{n+p}},\frac{{t}_{0}}{{2}^{n+p}}\sum _{k=0}^{p-1}{2}^{p-k-1}\stackrel{ˆ}{\varphi }\left({2}^{\frac{n+k}{2}}x,{2}^{\frac{n+k}{2}}x\right)\right)\ge 1-ϵ\hfill \end{array}$
(2.12)

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

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

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

${t}_{0}\stackrel{ˆ}{\varphi }\left({2}^{\frac{n}{2}}x,{2}^{\frac{n}{2}}y\right)<\frac{{2}^{n}t}{4}$

for all $n\ge {n}_{1}$. Then, by (N4), we have

$\begin{array}{c}N\left(Q\left(\sqrt{{x}^{2}+{y}^{2}}\right)-Q\left(x\right)-Q\left(y\right),t\right)\hfill \\ \phantom{\rule{1em}{0ex}}\ge min\left\{N\left(Q\left(x+y\right)-\frac{1}{{2}^{n}}f\left(\sqrt{{2}^{n}{x}^{2}+{2}^{n}{y}^{2}}\right),\frac{t}{4}\right),N\left(Q\left(x\right)-\frac{1}{{2}^{n}}f\left({2}^{\frac{n}{2}}x\right),\frac{t}{4}\right),\hfill \\ \phantom{\rule{2em}{0ex}}N\left(Q\left(y\right)-\frac{1}{{2}^{n}}f\left({2}^{\frac{n}{2}}y\right),\frac{t}{4}\right),N\left(f\left(\sqrt{{2}^{n}{x}^{2}+{2}^{n}{y}^{2}}\right)-f\left({2}^{\frac{n}{2}}x\right)-f\left({2}^{\frac{n}{2}}y\right),\frac{{2}^{n}t}{4}\right)\right\}\hfill \end{array}$
(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

$N\left(f\left(\sqrt{{2}^{n}{x}^{2}+{2}^{n}{y}^{2}}\right)-f\left({2}^{\frac{n}{2}}x\right)-f\left({2}^{\frac{n}{2}}y\right),{t}_{0}\stackrel{ˆ}{\varphi }\left({2}^{\frac{n}{2}},{2}^{\frac{n}{2}}y\right)\right)\ge 1-ϵ,$

we have

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

for all $x,y\in \mathbb{R}$, $t>0$ and $0<ϵ<1$. It follows from (N2) that $Q\left(\sqrt{{x}^{2}+{y}^{2}}\right)=Q\left(x\right)+Q\left(y\right)$ 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

${\psi }_{n}\left(x,y\right)=\sum _{k=0}^{n-1}\frac{1}{{2}^{k+1}}\stackrel{ˆ}{\varphi }\left({2}^{\frac{k}{2}}x,{2}^{\frac{k}{2}}y\right)$

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

$N\left({2}^{n}f\left(x\right)-f\left({2}^{\frac{n}{2}}x\right),\delta \sum _{k=0}^{n-1}{2}^{n-k-1}\stackrel{ˆ}{\varphi }\left({2}^{\frac{k}{2}}x,{2}^{\frac{k}{2}}x\right)\right)\ge \alpha$
(2.14)

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

$\begin{array}{c}N\left(f\left(x\right)-Q\left(x\right),\delta {\psi }_{n}\left(x,x\right)+t\right)\hfill \\ \phantom{\rule{1em}{0ex}}\ge min\left\{N\left(f\left(x\right)-\frac{f\left({2}^{\frac{n}{2}}x\right)}{{2}^{n}},\delta {\psi }_{n}\left(x,x\right)\right),N\left(\frac{f\left({2}^{\frac{n}{2}}x\right)}{{2}^{n}}-Q\left(x\right),t\right)\right\}.\hfill \end{array}$
(2.15)

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

$N\left(f\left(x\right)-Q\left(x\right),\delta {\psi }_{n}\left(x,x\right)+t\right)\ge \alpha$
(2.16)

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

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

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

$N\left(f\left(x\right)-Q\left(x\right),\frac{t}{2}\mathrm{\Phi }\left(x,x\right)\right)\ge 1-ϵ,\phantom{\rule{2em}{0ex}}N\left(f\left(x\right)-T\left(x\right),\frac{t}{2}\mathrm{\Phi }\left(x,x\right)\right)\ge 1-ϵ$

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

${t}_{0}\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{2}^{k}}\stackrel{ˆ}{\varphi }\left({2}^{\frac{k}{2}}x,{2}^{\frac{k}{2}}y\right)<\frac{c}{2}$

for all $x,y\in \mathbb{R}$ and $n\ge {n}_{0}$. It follows from

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

that

$\begin{array}{c}N\left(Q\left(x\right)-T\left(x\right),c\right)\hfill \\ \phantom{\rule{1em}{0ex}}\ge min\left\{N\left(\frac{f\left({2}^{\frac{n}{2}}x\right)}{{2}^{n}}-Q\left(x\right),\frac{c}{2}\right),N\left(T\left(x\right)-\frac{f\left({2}^{\frac{n}{2}}x\right)}{{2}^{n}},\frac{c}{2}\right)\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=min\left\{N\left(f\left({2}^{\frac{n}{2}}x\right)-Q\left({2}^{\frac{n}{2}}x\right),{2}^{n-1}c\right),N\left(T\left({2}^{\frac{n}{2}}x\right)-f\left({2}^{\frac{n}{2}}x\right),{2}^{n-1}c\right)\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\ge min\left\{N\left(f\left({2}^{\frac{n}{2}}x\right)-Q\left({2}^{\frac{n}{2}}x\right),{2}^{n}{t}_{0}\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{2}^{k}}\stackrel{ˆ}{\varphi }\left({2}^{\frac{k}{2}}x,{2}^{\frac{k}{2}}x\right)\right),\hfill \\ \phantom{\rule{2em}{0ex}}N\left(T\left({2}^{\frac{n}{2}}x\right)-f\left({2}^{\frac{n}{2}}x\right),{2}^{n}{t}_{0}\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{2}^{k}}\stackrel{ˆ}{\varphi }\left({2}^{\frac{k}{2}}x,{2}^{\frac{k}{2}}x\right)\right)\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\ge min\left\{N\left(f\left({2}^{\frac{n}{2}}x\right)-Q\left({2}^{\frac{n}{2}}x\right),{t}_{0}\mathrm{\Phi }\left({2}^{\frac{n}{2}}x,{2}^{\frac{n}{2}}x\right)\right),\hfill \\ \phantom{\rule{2em}{0ex}}N\left(T\left({2}^{\frac{n}{2}}x\right)-f\left({2}^{\frac{n}{2}}x\right),{t}_{0}\mathrm{\Phi }\left({2}^{\frac{n}{2}}x,{2}^{\frac{n}{2}}x\right)\right)\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\ge 1-ϵ\hfill \end{array}$

for all $x,y\in \mathbb{R}$ and $c>0$. Thus we have $N\left(Q\left(x\right)-T\left(x\right),c\right)=1$ for all $c>0$ and so $Q\left(x\right)=T\left(x\right)$ 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 $Q\left(x\right)={lim}_{n\to \mathrm{\infty }}{2}^{n}f\left({2}^{-\frac{n}{2}}x\right)$. This completes the proof. □

Corollary 2.2 Let $\left(\mathcal{Y},N\right)$ 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 $f\left(0\right)=0$ such that, for all $t>0$,

$\underset{t\to \mathrm{\infty }}{lim}N\left(f\left(\sqrt{{x}^{2}+{y}^{2}}\right)-f\left(x\right)-f\left(y\right),t\theta \left(|x{|}^{p}+|y{|}^{p}\right)\right)=1$
(2.17)

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

$\underset{t\to \mathrm{\infty }}{lim}N\left(f\left(x\right)-Q\left(x\right),\frac{2\left(2+{2}^{\frac{p}{2}}\right)}{2-{2}^{\frac{p}{2}}}\theta |x{|}^{p}t\right)=1$
(2.18)

uniformly on ℝ.

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

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

1. (a)

$\psi \left(ts\right)=\psi \left(t\right)\psi \left(s\right)$;

2. (b)

$\psi \left(\sqrt{2}\right)<2$.

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

$\underset{t\to \mathrm{\infty }}{lim}N\left(f\left(\sqrt{{x}^{2}+{y}^{2}}\right)-f\left(x\right)-f\left(y\right),t\theta \left(\psi \left(|x|\right)+\psi \left(|y|\right)\right)\right)=1$
(2.19)

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

$\underset{t\to \mathrm{\infty }}{lim}N\left(f\left(x\right)-Q\left(x\right),\frac{2\left(2+\psi \left(\sqrt{2}\right)\right)}{2-\psi \left(\sqrt{2}\right)}\theta \psi \left(|x|\right)t\right)=1$
(2.20)

uniformly on ℝ.

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

### 2.2 The fixed point method

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

1. (1)

$d\left(x,y\right)=0$ if and only if $x=y$;

2. (2)

$d\left(x,y\right)=d\left(y,x\right)$;

3. (3)

$d\left(x,z\right)\le d\left(x,y\right)+d\left(y,z\right)$ 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 $\left(\mathrm{\Omega },d\right)$ 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 $d\left({T}^{n}x,{T}^{n+1}x\right)=\mathrm{\infty }$ for all $n\ge 0$ or there exists a positive integer ${n}_{0}$ such that

1. (1)

$d\left({T}^{n}x,{T}^{n+1}x\right)<\mathrm{\infty }$ for all $n\ge {n}_{0}$;

2. (2)

the sequence $\left\{{T}^{n}x\right\}$ is convergent to a fixed point ${y}^{\ast }$ of T;

3. (3)

${y}^{\ast }$ is the unique fixed point of T in the set $\mathrm{\Lambda }=\left\{y\in \mathrm{\Omega }:d\left({T}^{{n}_{0}}x,y\right)<\mathrm{\infty }\right\}$;

4. (4)

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

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

$\varphi \left({2}^{\frac{1}{2}}x,{2}^{\frac{1}{2}}y\right)\le 2L\varphi \left(x,y\right)$
(2.21)

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

$N\left(f\left(\sqrt{{x}^{2}+{y}^{2}}\right)-f\left(x\right)-f\left(y\right),t\right)\ge \frac{t}{t+\varphi \left(x,y\right)}$
(2.22)

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

$N\left(f\left(x\right)-Q\left(x\right),t\right)\ge \frac{\left(1-L\right)t}{\left(1-L\right)t+\stackrel{ˆ}{\varphi }\left(x,x\right)}$
(2.23)

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

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

$N\left(f\left(\sqrt{{x}^{2}+{y}^{2}}\right)-f\left(\frac{x+y}{\sqrt{2}}\right)-f\left(\frac{x-y}{\sqrt{2}}\right),t\right)\ge \frac{t}{t+\varphi \left(\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}}\right)}$
(2.24)

for all $x,y\in \mathbb{R}$ and $t\ge {t}_{0}$. It follows from (2.22), (2.24), and (N4) that

$\begin{array}{c}N\left(f\left(x\right)+f\left(y\right)-f\left(\frac{x+y}{\sqrt{2}}\right)-f\left(\frac{x-y}{\sqrt{2}}\right),2t\right)\hfill \\ \phantom{\rule{1em}{0ex}}\ge min\left\{N\left(f\left(x\right)+f\left(y\right)-f\left(\sqrt{{x}^{2}+{y}^{2}}\right),t\right),\hfill \\ \phantom{\rule{2em}{0ex}}N\left(f\left(\sqrt{{x}^{2}+{y}^{2}}\right)-f\left(\frac{x+y}{\sqrt{2}}\right)-f\left(\frac{x-y}{\sqrt{2}}\right),t\right)\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\ge min\left\{\frac{t}{t+\varphi \left(x,y\right)},\frac{t}{t+\varphi \left(\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}}\right)}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\ge \frac{t}{t+\varphi \left(x,y\right)+\varphi \left(\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}}\right)}\hfill \end{array}$
(2.25)

for all $x,y\in \mathbb{R}$ and $t\ge {t}_{0}$. Letting $y=x$ in (2.25), we have

$N\left(f\left(x\right)-\frac{1}{2}f\left({2}^{\frac{1}{2}}x\right),t\right)\ge \frac{t}{t+\stackrel{ˆ}{\varphi }\left(x,x\right)}$
(2.26)

for all $x\in \mathbb{R}$ and $t\ge {t}_{0}$, where $\stackrel{ˆ}{\varphi }\left(x,y\right)=\varphi \left(x,y\right)+\varphi \left({2}^{\frac{1}{2}}x,0\right)$.

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

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

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

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

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

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

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

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

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

$Q\left(x\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{2}^{n}}f\left({2}^{\frac{n}{2}}x\right)$
(2.27)

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

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

which gives

$N\left(f\left(x\right)-Q\left(x\right),t\right)\ge \frac{\left(1-L\right)t}{\left(1-L\right)t+\stackrel{ˆ}{\varphi }\left(x,x\right)}$

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

$\begin{array}{c}N\left(Q\left(\sqrt{{x}^{2}+{y}^{2}}\right)-Q\left(x\right)-Q\left(y\right),t\right)\hfill \\ \phantom{\rule{1em}{0ex}}\ge \underset{n\to \mathrm{\infty }}{lim}N\left(f\left(\sqrt{{2}^{n}{x}^{2}+{2}^{n}{y}^{2}}\right)-f\left({2}^{\frac{n}{2}}x\right)-f\left({2}^{\frac{n}{2}}y\right),{2}^{n}t\right)\hfill \\ \phantom{\rule{1em}{0ex}}\ge \underset{n\to \mathrm{\infty }}{lim}\frac{t}{t+{L}^{n}\stackrel{ˆ}{\varphi }\left(x,y\right)}=1\hfill \end{array}$
(2.28)

for all $x,y\in \mathbb{R}$ and $t>0$. It follows from (N2) and $N\left(Q\left(\sqrt{{x}^{2}+{y}^{2}}\right)-Q\left(x\right)-Q\left(y\right),t\right)\ge 1$ that $Q\left(\sqrt{{x}^{2}+{y}^{2}}\right)=Q\left(x\right)+Q\left(y\right)$ 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 $\left(\mathcal{Y},N\right)$ be a fuzzy Banach space and $\varphi :{\mathbb{R}}^{2}\to \left[0,\mathrm{\infty }\right)$ be a mapping such that there exists $L<1$ with

$\varphi \left(\frac{x}{\sqrt{2}},\frac{y}{\sqrt{2}}\right)\le \frac{L}{2}\varphi \left(x,y\right)$
(2.29)

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

$N\left(f\left(x\right)-Q\left(x\right),t\right)\ge \frac{\left(1-L\right)t}{\left(1-L\right)t+L\stackrel{ˆ}{\varphi }\left(x,x\right)}$
(2.30)

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

Proof It follows from (2.26) that

$N\left(f\left(x\right)-2f\left(\frac{x}{\sqrt{2}}\right),Lt\right)\ge \frac{t}{t+\stackrel{ˆ}{\varphi }\left(x,x\right)}$
(2.31)

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

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

$x\in \mathbb{R}$. So, we have $d\left(Tg,Th\right)\le Ld\left(g,h\right)$ 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 $\left\{g\in \mathrm{\Omega }:d\left(f,g\right)<\mathrm{\infty }\right\}$ which is a unique fixed point of T such that

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

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

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

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 $\left(\mathcal{Y},N\right)$ 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 $f\left(0\right)=0$ such that, for all $t>0$,

$N\left(f\left(\sqrt{{x}^{2}+{y}^{2}}\right)-f\left(x\right)-f\left(y\right),t\right)\ge \frac{t}{t+\theta \left(|x{|}^{p}+|y{|}^{p}\right)}$
(2.32)

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

$N\left(f\left(x\right)-Q\left(x\right),t\right)\ge \left\{\begin{array}{cc}\frac{\left(2-{2}^{\frac{p}{2}}\right)t}{\left(2-{2}^{\frac{p}{2}}\right)t+2\left(2+{2}^{\frac{p}{2}}\right)\theta |x{|}^{p}},\hfill & p<2,\hfill \\ \frac{\left(2-{2}^{\frac{p}{2}}\right)t}{\left(2-{2}^{\frac{p}{2}}\right)t+{2}^{\frac{p}{2}}\left(2+{2}^{\frac{p}{2}}\right)\theta |x{|}^{p}},\hfill & p>2,\hfill \end{array}$
(2.33)

uniformly on ℝ.

Proof Taking $\varphi \left(x,y\right)=\theta \left(|x{|}^{p}+|y{|}^{p}\right)$ 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 $f\left(\sqrt{{x}^{2}+{y}^{2}}\right)=f\left(x\right)+f\left(y\right)$ is not stable for $p=2$ [11].