Stability of the Apollonius Type Additive Functional Equation in Modular Spaces and Fuzzy Banach Spaces

Abstract

In this work, we investigate the generalized Hyers-Ulam stability of the Apollonius type additive functional equation in modular spaces with or without ${\Delta }_2$-conditions. We study the same problem in fuzzy Banach spaces and $\beta$-homogeneous Banach spaces. We show the hyperstability of the functional equation associated with the Jordan triple product in fuzzy Banach algebras. The obtained results can be applied to differential and integral equations with kernels of non-power types.

1. Introduction and Preliminaries

The research on modulars and modular spaces was begun by Nakano [1] as generalizations of normed spaces. Since the 1950s, many prominent mathematicians like Luxemburg, Mazur, Musielak, and Orlicz [2,3,4,5] developed it extensively. Modulars and modular spaces have broad branches of applications, e.g., interpolation theory and Orlicz spaces.

We start by considering some basic relevant notions.

Definition 1.

([1]) Let X be a vector space over a field $\mathbb{K}$ ($\mathbb{R}$ or $\mathbb{C}$). A generalized function $\rho :X\to [0,\infty ]$ is called a modular if for any $\alpha ,\beta \in \mathbb{K}$ and $x,y\in X$,

(1) 

$\rho \left(x\right)=0$ if and only if $x=0$,

(2) 

$\rho \left(\alpha x\right)=\rho \left(x\right)$ for every α with $\left|\alpha \right|=1$,

(3) 

$\rho (\alpha x+\beta y)\le \rho \left(x\right)+\rho \left(y\right)$ if $\alpha +\beta =1$ and $\alpha ,\beta \ge 0$.

If the condition (3) is replaced with

(4) 

$\rho (\alpha x+\beta y)\le {\alpha }^s\rho \left(x\right)+{\beta }^s\rho \left(y\right)$ if ${\alpha }^s+{\beta }^s=1$ and $\alpha ,\beta \ge 0$ with an $s\in (0,1]$,

then ρ is called an s-convex modular. We call 1-convex modulars as convex modulars.

A modular $\rho$ on X generates a linear subspace $X_{\rho }$ of X naturally defined by

$$X_{\rho }=\{x\in X|\underset{\lambda \to 0}{lim}\rho \left(\lambda x\right)=0\}.$$

$X_{\rho }$ is called a modular space.

Definition 2.

Let $X_{\rho }$ be a modular space and $\left\{x_n\right\}$ be a sequence in $X_{\rho }$.

(1) 

$\left\{x_n\right\}$ is ρ-convergent to a point $x\in X_{\rho }$ if $\rho (x_n-x)\to 0$ as $n\to \infty$. The point x is called the ρ-limit of the sequence $\left\{x_n\right\}$.

(2) 

$\left\{x_n\right\}$ is called a ρ-Cauchy sequence if $\rho (x_n-x_m)\to 0$ as $n,m\to \infty$.

(3) 

$X_{\rho }$ is called ρ-complete if every ρ-Cauchy sequence in $X_{\rho }$ is ρ-convergent.

Remark 1.

If ρ is a convex modular and $0\le \lambda \le 1$, we have $\rho \left(\lambda x\right)\le \lambda \rho \left(x\right)$ for all $x\in X_{\rho }$. If ρ is a convex modular, and ${\lambda }_i\ge 0,i=1,2,\dots n$ and ${\lambda }_1+{\lambda }_2+\dots +{\lambda }_n\le 1$, then $\rho ({\lambda }_1{x}_1+{\lambda }_2{x}_2+\dots +{\lambda }_n{x}_n)\le {\lambda }_1\rho \left(x_1\right)+{\lambda }_2\rho \left(x_2\right)+\dots +{\lambda }_n\rho \left(x_n\right)$. If $\left\{x_n\right\}$ is ρ-convergent to x, then $\left\{\alpha x_n\right\}$ is ρ-convergent to $\alpha x$, where $0\le \alpha \le 1$. But the ρ-convergence of a sequence $\left\{x_n\right\}$ to x does not imply that $\left\{c{x}_n\right\}$ is ρ-convergent to $cx$ for scalars c with $\left|c\right|>1$.

There are two notions that play important roles when we study modulars. A modular $\rho$ is said to have the Fatou property if $\rho \left(x\right)\le {lim\; inf}_{n\to \infty }\rho \left(x_n\right)$ for every sequence $\left\{x_n\right\}$ that is $\rho$-convergent to x. $\rho$ is said to satisfy the ${\Delta }_2$-condition if there exists a constant $\tau \ge 0$ such that $\rho \left(2x\right)\le \tau \rho \left(x\right)$ for all $x\in X_{\rho }$.

Example 1.

For a measure space $(\mathsf{\Omega },\mathsf{\Sigma },\mu )$, let $L^0\left(\mu \right)$ be the collection of all measurable functions on Ω. Let

$$L^{\varphi }\left(\mu \right)=\{f\in L^0\left(\mu \right)|{\int }_{\mathsf{\Omega }}\varphi \left(\right|\lambda f\left(x\right)\left|\right)d\mu \left(x\right)\to 0\mathrm{as}\lambda \to 0\},$$

where $\varphi :[0,\infty )\to \mathbb{R}$ is assumed to be a continuous, positive, convex and nondecreasing function increasing to infinity with $\varphi \left(0\right)=0$. We can take, e.g., $\varphi \left(t\right)=e^{t^2}-1$. $L^{\varphi }\left(\mu \right)$ is called an Orlicz space. Define for $f\in L^{\varphi }\left(\mu \right)$,

$${\rho }_{\varphi }\left(f\right)={\int }_{\mathsf{\Omega }}\varphi \left(\right|f\left|\right)d\mu .$$

Then ${\rho }_{\varphi }$ is a complete modular.

The question of stability of a functional equation concerns the existence of an exact solution near to the function satisfying the equation approximately. In 1940, Ulam [6] raised the first stability problem. He proposed a question whether there exists an exact homomorphism near an approximate homomorphism. Hyers [7] gave an answer in Banach spaces. Since then, many authors have investigated the stability problems. We refer to [8,9,10,11,12,13] for more information on the stability of functional equations.

The equality

$${\parallel z-x\parallel }^2{+\parallel z-y\parallel }^2=\frac{1}{2}{\parallel x-y\parallel }^2+2{∥z-\frac{x+y}{2}∥}^2,$$

which is called the Apollonius identity, holds in inner product spaces. It motivated the following quadratic functional equation,

$$Q(z-x)+Q(z-y)=\frac{1}{2}Q(x-y)+2Q\left(z-\frac{x+y}{2}\right).$$

(1)

So, Equation (1) is called the quadratic functional equation of Apollonius type. Jun and Kim [14] initiated the investigation of this functional equation.

Park and Rassias [15] introduced the following functional equation:

$$f(z-x)+f(z-y)=-\frac{1}{2}f(x+y)+2f\left(z-\frac{x+y}{4}\right),$$

(2)

called the Apollonius type additive functional equation and investigated homomorphisms in $C^{\ast }$-ternary algebras and $J{B}^{\ast }$-triples associated with (2). In addition, in [16], the authors studied Jordan mappings in $C^{\ast }$-ternary algebras and $J{B}^{\ast }$-triples associated with (2).

When studying the stability of functional equations, many authors work in normed spaces. However, there exist a number of topological spaces that are not normable. The concept of modulars is wider than that of norms, so modulars have less properties than norms have, but they make more sense in many particular situations. Working in a modular space, many authors often assume that the modular satisfies the Fatou property or ${\Delta }_2$-condition or both (see, e.g., [17,18]).

Recently, many authors investigated the stability of various functional equations on modular spaces. We refer the readers to [19,20,21,22,23,24,25,26].

This paper consists of six sections. In Section 2, we show the stability of the following functional equation without any condition on the modular;

$$2f(z-x)+2f(z-y)=-f(x+y)+4f\left(z-\frac{x+y}{4}\right).$$

(3)

In Section 3, we prove the stability of Apollonius type additive functional Equation (2) in modular spaces under the condition that the modular fulfills the ${\Delta }_2$-condition but not necessarily Fatou property.

In Section 4, we obtain a similar result for $\beta$-homogeneous Banach spaces.

In Section 5, we show the fuzzy stability of the functional Equation (2) in fuzzy Banach spaces by using a fixed point method.

In Section 6, we show the hyperstability of the functional Equation (2) associated with the Jordan triple product in fuzzy Banach algebras.

2. Stability of (3) in Modular Spaces Without ${\Delta }_2$-Conditions

This section is concerned with the stability of the functional Equation (3). Dividing the functional equation by 2, this equation is reduced to the Apollonius type additive functional equation (2). Note that in the following theorem, the convex modular $\rho$ is not assumed to satisfy any other condition.

Lemma 1.

[15] Let V and X be linear spaces and $f:V\to X$ be a mapping such that

$$f(z-x)+f(z-y)+\frac{1}{2}f(x+y)-2f\left(z-\frac{x+y}{4}\right)=0$$

for all $x,y,z\in V$. Then f is additive.

Theorem 1.

Let V be a linear space, ρ be a convex modular, and $X_{\rho }$ be a ρ-complete modular space. Let $\phi :V^3\to [0,\infty )$ be a function with

$$\widehat{\phi }(x,y,z):=\sum _{k=1}^{\infty }\frac{1}{4^k}\phi (4^{k-1}x,4^{k-1}y,4^{k-1}z)<\infty$$

(4)

for all $x,y,z\in V$. Assume that $f:V\to X_{\rho }$ is a mapping satisfying $f\left(0\right)=0$ and

$$\rho \left(2f(z-x)+2f(z-y)+f(x+y)-4f\left(z-\frac{x+y}{4}\right)\right)\le \phi (x,y,z)$$

(5)

for all $x,y,z\in V$. Then there exists a unique additive mapping $T:V\to X_{\rho }$ such that

$$\rho \left(f\left(x\right)-T\left(x\right)\right)\le \widehat{\phi }(2x,2x,2x),x\in V.$$

(6)

The mapping T is defined by $T\left(x\right)=\rho -\mathit{limit}\frac{f\left(4^{n}x\right)}{4^n},x\in V$.

Proof. 

Replacing $(x,y,z)$ with $(2x,2x,2x)$ in (5) and letting $\psi \left(x\right)=\phi (2x,2x,2x)$, we have

$$\rho \left(f\left(4x\right)-4f\left(x\right)\right)\le \phi (2x,2x,2x)=\psi \left(x\right),x\in V,$$

(7)

hence,

$$\rho \left(\frac{1}{4}f\left(4x\right)-f\left(x\right)\right)\le \frac{1}{4}\psi \left(x\right).$$

(8)

Then by induction, we write

$$\rho \left(\frac{f\left(4^{k}x\right)}{4^k}-f\left(x\right)\right)\le \sum _{j=1}^k\frac{1}{4^j}\psi \left(4^{j-1}x\right)$$

(9)

for all $x\in V$ and all positive integer k. Indeed, the case $k=1$ follows from (8). Assume that (9) holds for $k\in \mathbb{N}$. Then we have the following inequality:

$$\begin{array}{cc}\hfill \rho \left(\frac{f\left(4^{k+1}x\right)}{4^{k+1}}-f\left(x\right)\right)& =\rho \left(\frac{1}{4}\left(\frac{f(4^k·4x)}{4^k}-f\left(4x\right)\right)+\frac{1}{4}(f\left(4x\right)-4f\left(x\right))\right)\hfill \\ \hfill & \le \frac{1}{4}\rho \left(\frac{f(4^k·4x)}{4^k}-f\left(4x\right)\right)+\frac{1}{4}\rho \left(f\left(4x\right)-4f\left(x\right)\right)\hfill \\ \hfill & \le \frac{1}{4}\sum _{j=1}^k\frac{1}{4^j}\psi \left(4^{j}x\right)+\frac{1}{4}\psi \left(x\right)\hfill \\ \hfill & =\sum _{j=1}^k\frac{1}{4^{j+1}}\psi \left(4^{j}x\right)+\frac{1}{4}\psi \left(x\right)\hfill \\ \hfill & =\sum _{j=1}^{k+1}\frac{1}{4^j}\psi \left(4^{j-1}x\right).\hfill \end{array}$$

Hence, (9) holds for every $k\in \mathbb{N}$.

Let m and n be nonnegative integers with $n>m$. By (9), we have

$$\begin{array}{cc}\hfill \rho \left(\frac{f\left(4^{n}x\right)}{4^n}-\frac{f\left(4^{m}x\right)}{4^m}\right)& =\rho \left(\frac{1}{4^m}\left(\frac{f(4^{n-m}·4^{m}x)}{4^{n-m}}-f\left(4^{m}x\right)\right)\right)\hfill \\ \hfill & \le \frac{1}{4^m}\sum _{j=1}^{n-m}\frac{\psi (4^{j-1}·4^{m}x)}{4^j}\hfill \\ \hfill & =\sum _{j=1}^{n-m}\frac{\psi \left(4^{m+j-1}x\right)}{4^{m+j}}\hfill \\ \hfill & =\sum _{k=m+1}^n\frac{\psi \left(4^{k-1}x\right)}{4^k}.\hfill \end{array}$$

(10)

Then (4) and (10) yield that $\left\{\frac{f\left(4^{n}x\right)}{4^n}\right\}$ is a $\rho$-Cauchy sequence in $X_{\rho }$. The $\rho$-completeness of $X_{\rho }$ guarantees its $\rho$-convergence. Hence, there exists a mapping $T:V\to X_{\rho }$ defined by

$$T\left(x\right)=\rho -\mathrm{limit}\frac{f\left(4^{n}x\right)}{4^n},x\in V.$$

(11)

We see that

$$\begin{array}{cc}\hfill \rho \left(\frac{T\left(4x\right)-4T\left(x\right)}{4^3}\right)& =\rho \left(\frac{1}{4^3}\left(T\left(4x\right)-\frac{f\left(4^{n+1}x\right)}{4^n}\right)+\frac{1}{4}\left(\frac{1}{4}·\frac{f\left(4^{n+1}x\right)}{4^{n+1}}-\frac{1}{4}T\left(x\right)\right)\right)\hfill \\ \hfill & \le \frac{1}{4^3}\rho \left(T\left(4x\right)-\frac{f\left(4^{n+1}x\right)}{4^n}\right)+\frac{1}{16}\rho \left(\frac{f\left(4^{n+1}x\right)}{4^{n+1}}-T\left(x\right)\right)\hfill \end{array}$$

(12)

for all $x\in V$. Then by (11), the right hand side of (12) tends to 0 as $n\to \infty$. Therefore, it follows that

$$T\left(4x\right)=4T\left(x\right),x\in V.$$

(13)

Next, we calculate $\rho \left(T\right(x)-f(x\left)\right)$. Note that for every $n\in \mathbb{N}$, by (13) we write

$$\begin{array}{cc}\hfill & \rho \left(T\left(x\right)-f\left(x\right)\right)\hfill \\ \hfill & =\rho \left(\sum _{k=1}^n\frac{f\left(4^{k}x\right)-4f\left(4^{k-1}x\right)}{4^k}+\left(T\left(x\right)-\frac{f\left(4^{n}x\right)}{4^n}\right)\right)\hfill \\ \hfill & =\rho \left(\sum _{k=1}^n\frac{f\left(4^{k}x\right)-4f\left(4^{k-1}x\right)}{4^k}+\frac{1}{4}\left(T\left(4x\right)-\frac{f(4^{n-1}·4x)}{4^{n-1}}\right)\right).\hfill \end{array}$$

(14)

Since ${\sum }_{k=1}^n\frac{1}{4^k}+\frac{1}{4}<1$, it follows from (7) and (14) that

$$\begin{array}{cc}\hfill & \rho \left(T\left(x\right)-f\left(x\right)\right)\hfill \\ \hfill & \le \sum _{k=1}^n\frac{1}{4^k}\rho \left(f\left(4^{k}x\right)-4f\left(4^{k-1}x\right)\right)+\frac{1}{4}\rho \left(T\left(4x\right)-\frac{f\left(4^{n-1}·4x\right)}{4^{n-1}}\right)\hfill \\ \hfill & \le \sum _{k=1}^n\frac{1}{4^k}\psi \left(4^{k-1}x\right)+\frac{1}{4}\rho \left(T\left(4x\right)-\frac{f\left(4^{n-1}·4x\right)}{4^{n-1}}\right)\hfill \\ \hfill & =\sum _{k=1}^n\frac{1}{4^k}\phi \left(4^{k-1}·2x,4^{k-1}·2x,4^{k-1}·2x\right)+\frac{1}{4}\rho \left(T\left(4x\right)-\frac{f\left(4^{n-1}·4x\right)}{4^{n-1}}\right),x\in V.\hfill \end{array}$$

(15)

Letting $n\to \infty$ in (15), we obtain

$$\rho (T\left(x\right)-f\left(x\right))\le \widehat{\phi }(2x,2x,2x),x\in V.$$

Therefore, we arrive at (6).

Now, we prove that T is additive. We note that

$$\begin{array}{cc}\hfill & \rho \left(\frac{2f\left(4^j(z-x)\right)}{4^j}+\frac{2f\left(4^j(z-y)\right)}{4^j}+\frac{f\left(4^j(x+y)\right)}{4^j}-\frac{4f\left(4^j(z-\frac{x+y}{4})\right)}{4^j}\right)\hfill \\ \hfill & \le \frac{1}{4^j}\rho \left(2f\left(4^j(z-x)\right)+2f\left(4^j(z-y)\right)+f\left(4^j(x+y)\right)-4f\left(4^j\left(z-\frac{x+y}{4}\right)\right)\right)\hfill \\ \hfill & \le \frac{1}{4^j}\phi (4^{j}x,4^{j}y,4^{j}z)\to 0\mathrm{as}j\to \infty \hfill \end{array}$$

(16)

for all $x,y,z\in V$.

Now, by (16) we have the following inequality

$$\begin{array}{cc}\hfill & \rho \left(\frac{1}{24}\left(2T(z-x)+2T(z-y)+T(x+y)-4T\left(z-\frac{x+y}{4}\right)\right)\right)\hfill \\ \hfill & \le \frac{1}{12}\left[\rho \left(T(z-x)-\frac{f\left(4^j(z-x)\right)}{4^j}\right)+\rho \left(T(z-y)-\frac{f\left(4^j(z-y)\right)}{4^j}\right)\right.\hfill \\ \hfill & +\left.\frac{1}{2}\rho \left(T(x+y)-\frac{f\left(4^j(x+y)\right)}{4^j}\right)+2\rho \left(T\left(z-\frac{x+y}{4}\right)-\frac{f\left(4^j(z-\frac{x+y}{4})\right)}{4^j}\right)\right]\hfill \\ \hfill & +\frac{1}{24}\rho \left(\frac{2f\left(4^j(z-x)\right)}{4^j}+\frac{2f\left(4^j(z-y)\right)}{4^j}+\frac{f\left(4^j(x+y)\right)}{4^j}-\frac{4f\left(4^j(z-\frac{x+y}{4})\right)}{4^j}\right)\hfill \\ \hfill & \to 0\mathrm{as}j\to \infty .\hfill \end{array}$$

Hence, we get

$$T(z-x)+T(z-y)+\frac{1}{2}T(x+y)-2T\left(z-\frac{x+y}{4}\right)=0$$

for all $x,y,z\in V$. Then by Lemma 1, it follows that T is an additive mapping.

Finally, to show the uniqueness of T, assume that $T_1$ and $T_2$ are additive mappings satisfying (6). Then we write

$$\begin{array}{cc}\hfill \rho \left(\frac{T_1\left(x\right)-T_2\left(x\right)}{2}\right)& =\rho \left(\frac{1}{2}\left(\frac{T_1\left(4^{k}x\right)}{4^k}-\frac{f\left(4^{k}x\right)}{4^k}\right)+\frac{1}{2}\left(\frac{f\left(4^{k}x\right)}{4^k}-\frac{T_2\left(4^{k}x\right)}{4^k}\right)\right)\hfill \\ \hfill & \le \frac{1}{2}\rho \left(\frac{T_1\left(4^{k}x\right)}{4^k}-\frac{f\left(4^{k}x\right)}{4^k}\right)+\frac{1}{2}\rho \left(\frac{f\left(4^{k}x\right)}{4^k}-\frac{T_2\left(4^{k}x\right)}{4^k}\right)\hfill \\ \hfill & \le \frac{1}{2}·\frac{1}{4^k}\left\{\rho \left(T_1\left(4^{k}x\right)-f\left(4^{k}x\right)\right)+\rho \left(T_2\left(4^{k}x\right)-f\left(4^{k}x\right)\right)\right\}\hfill \\ \hfill & \le \frac{1}{4^k}\widehat{\phi }(4^k·2x,4^k·2x,4^k·2x)\hfill \\ \hfill & =\sum _{l=k+1}^{\infty }\frac{1}{4^l}\phi \left(4^{l-1}·2x,4^{l-1}·2x,4^{l-1}·2x\right)\hfill \\ \hfill & \to 0\mathrm{as}k\to \infty .\hfill \end{array}$$

This implies that $T_1=T_2$. □

Now, we have the classical Ulam stability of (3) by putting $\phi \equiv \epsilon >0$.

Corollary 1.

Let V be a linear space, ρ be a convex modular and $X_{\rho }$ be a ρ-complete modular space. Assume $f:V\to X_{\rho }$ is a mapping such that $f\left(0\right)=0$ and

$$\rho \left(2f(z-x)+2f(z-y)+f(x+y)-4f\left(z-\frac{x+y}{4}\right)\right)\le \epsilon$$

for all $x,y,z\in V$. Then there exists a unique additive mapping $T:V\to X_{\rho }$ such that

$$\rho \left(f\left(x\right)-T\left(x\right)\right)\le \frac{\epsilon }{3},x\in V.$$

Corollary 2.

Let V be a normed linear space, ρ be a convex modular and $X_{\rho }$ be a ρ-complete modular space. Let $\theta >0$ and $0<p<1$ be real numbers. Assume that $f:V\to X_{\rho }$ is a mapping satisfying

$$\rho \left(2f(z-x)+2f(z-y)+f(x+y)-4f\left(z-\frac{x+y}{4}\right)\right)\le {\theta (\parallel x\parallel }^p+{\parallel y\parallel }^p+{\parallel z\parallel }^p)$$

for all $x,y,z\in V$. Then there exists a unique additive mapping $T:V\to X_{\rho }$ such that

$$\rho \left(f\left(x\right)-T\left(x\right)\right)\le \frac{3·2^p\theta }{4-4^p}{\parallel x\parallel }^p,x\in V.$$

3. Stability of (2) in Modular Spaces with ${\Delta }_2$-Conditions

We show the generalized Hyers-Ulam stability of Apollonius type additive functional equation from linear spaces to modular spaces.

Theorem 2.

Let V be a linear space, ρ be a convex modular satisfying the ${\Delta }_2$-condition with τ and $X_{\rho }$ be a ρ-complete modular space. Let $\phi :V^3\to [0,\infty )$ be a function with

$$\widehat{\phi }(x,y,z):=\sum _{k=1}^{\infty }\frac{1}{4^k}\phi (4^{k-1}x,4^{k-1}y,4^{k-1}z)<\infty$$

for all $x,y,z\in V$. Assume that $f:V\to X_{\rho }$ is a mapping satisfying $f\left(0\right)=0$ and

$$\rho \left(f(z-x)+f(z-y)+\frac{1}{2}f(x+y)-2f\left(z-\frac{x+y}{4}\right)\right)\le \phi (x,y,z)$$

(17)

for all $x,y,z\in V$. Then there exists a unique additive mapping $T:V\to X_{\rho }$ such that

$$\rho \left(f\left(x\right)-T\left(x\right)\right)\le \tau \widehat{\phi }(2x,2x,2x),x\in V.$$

Proof. 

Since $\rho$ satisfies the ${\Delta }_2$-condition with $\tau$, (17) implies

$$\rho \left(2f(z-x)+2f(z-y)+f(x+y)-4f\left(z-\frac{x+y}{4}\right)\right)\le \tau ·\phi (x,y,z)$$

for all $x,y,z\in V$. Then the conclusion is a direct consequence of Theorem 1. □

Putting $\phi \equiv \epsilon >0$ in Theorem 2, we have the following result on classical Ulam stability of the Apollonius type additive functional equation.

Corollary 3.

Let V be a linear space, ρ be a convex modular satisfying the ${\Delta }_2$-condition with τ and $X_{\rho }$ be a ρ-complete modular space. Assume $f:V\to X_{\rho }$ is a mapping satisfying $f\left(0\right)=0$ and

$$\rho \left(f(z-x)+f(z-y)+\frac{1}{2}f(x+y)-2f\left(z-\frac{x+y}{4}\right)\right)\le \epsilon$$

for all $x,y,z\in V$. Then there exists a unique additive mapping $T:V\to X_{\rho }$ such that

$$\rho \left(f\left(x\right)-T\left(x\right)\right)\le \frac{\tau ·\epsilon }{3},x\in V.$$

4. Stability of (2) in $\beta$-homogeneous Spaces

Definition 3.

Let X be a linear space over $\mathbb{C}$. An F-norm is a function $\parallel ·\parallel :X\to [0,\infty ]$ such that

(1) 

$\parallel x\parallel =0$ if and only if $x=0$,

(2) 

$\parallel \lambda x\parallel =\parallel x\parallel$ for every $x\in X$ and every λ with $\left|\lambda \right|=1$,

(3) 

$\parallel x+y\parallel \le \parallel x\parallel +\parallel y\parallel$ for all $x,y\in X$,

(4) 

$\parallel {\lambda }_{n}x\parallel \to 0$ provided ${\lambda }_n\to 0$,

(5) 

$\parallel \lambda x_n\parallel \to 0$ provided $x_n\to 0$.

$(X,d)$ is a metric space by letting $d(x,y)=\parallel x-y\parallel$. It is called an F-space if d is complete.

If, in addition, $\parallel tx\parallel ={\left|t\right|}^{\beta }\parallel x\parallel$ for all $x\in X$ and $t\in \mathbb{C}$, then $\parallel ·\parallel$ is called β-homogeneous $(\beta >0)$. A $\beta$-homogeneous F-space is called a β-homogeneous complex Banach space.

Remark 2.

For an s-convex modular ρ, if we define

$${\parallel x\parallel }_{\rho }=inf\left\{{\alpha }^s>0|\rho \left(\frac{x}{\alpha }\right)\le 1\right\},x\in X_{\rho }$$

then ${\parallel ·\parallel }_{\rho }$ is an F-norm on $X_{\rho }$ such that ${\parallel \lambda x\parallel }_{\rho }={\left|\lambda \right|}^s{\parallel x\parallel }_{\rho }$. Hence, ${\parallel ·\parallel }_{\rho }$ is s-homogeneous. For $s=1$, this norm is called the Luxemburg norm.

Considering Remark 2, we prove the generalized Hyers-Ulam stability of (2) from linear spaces to $\beta$-homogeneous Banach spaces.

Theorem 3.

Let V be a linear space, X be a β-homogeneous complex Banach space $(0<\beta \le 1)$, and $\phi :V^3\to [0,\infty )$ be a function with

$$\widehat{\phi }(x,y,z):=\frac{1}{2^{\beta }}\sum _{j=1}^{\infty }\frac{1}{4^{(j-1)\beta }}\phi (4^{j-1}x,4^{j-1}y,4^{j-1}z)<\infty$$

(18)

for all $x,y,z\in V$. Assume that $f:V\to X$ is a mapping satisfying $f\left(0\right)=0$ and

$$∥f(z-x)+f(z-y)+\frac{1}{2}f(x+y)-2f\left(z-\frac{x+y}{4}\right)∥\le \phi (x,y,z)$$

(19)

for all $x,y,z\in V$. Then there exists a unique additive mapping $T:V\to X$ such that

$$\parallel f\left(x\right)-T\left(x\right)\parallel \le \widehat{\phi }(2x,2x,2x),x\in V.$$

(20)

Proof. 

Replacing $(x,y,z)$ with $(2x,2x,2x)$ in (19), we get

$$∥\frac{1}{2}f\left(4x\right)-2f\left(x\right)∥\le \phi (2x,2x,2x),x\in V.$$

(21)

By induction on $k\in \mathbb{N}$, using (21) and putting $\psi \left(x\right)=\phi (2x,2x,2x)$, it is easy to see that

$$∥\frac{f\left(4^{k}x\right)}{4^k}-f\left(x\right)∥\le \frac{1}{2^{\beta }}\sum _{j=1}^k\frac{1}{4^{(j-1)\beta }}\psi \left(4^{j-1}x\right),x\in V$$

(22)

for all $k\in \mathbb{N}$. Let m and n be nonnegative integers with $n>m$. Then, by (22), we have

$$\begin{array}{cc}\hfill ∥\frac{f\left(4^{n}x\right)}{4^n}-\frac{f\left(4^{m}x\right)}{4^m}∥& =∥\frac{1}{4^m}\left(\frac{f\left(4^{n}x\right)}{4^{n-m}}-f\left(4^{m}x\right)\right)∥\hfill \\ \hfill & \le \frac{1}{4^{m\beta }}·\frac{1}{2^{\beta }}\sum _{j=1}^{n-m}\frac{1}{4^{(j-1)\beta }}\psi \left(4^{j+m-1}x\right)\hfill \\ \hfill & \le \frac{1}{2^{\beta }}\sum _{j=1}^{n-m}\frac{1}{4^{(j+m-1)\beta }}\psi \left(4^{j+m-1}x\right)\hfill \\ \hfill & =\frac{1}{2^{\beta }}\sum _{l=m+1}^n\frac{1}{4^{(l-1)\beta }}\psi \left(4^{l-1}x\right),x\in V.\hfill \end{array}$$

(23)

Since the last expression of (23) goes to 0 by (18), it follows that, for every $x\in V$, the sequence $\left\{\frac{f\left(4^{n}x\right)}{4^n}\right\}$ is a Cauchy sequence in X. Since X is complete, we know that the sequence is convergent. Hence, there exists a mapping $T:V\to X$ defined by

$$T\left(x\right)=\underset{n\to \infty }{lim}\frac{f\left(4^{n}x\right)}{4^n},x\in V.$$

Letting $m=0$ and passing the limit $n\to \infty$ in (23), we obtain the estimate (20).

In order to show that T is additive, we write

$$\begin{array}{cc}\hfill & ∥T(z-x)+T(z-y)+\frac{1}{2}T(x+y)-2T\left(z-\frac{x+y}{4}\right)∥\hfill \\ \hfill & \le ∥T(z-x)-\frac{f\left(4^j(z-x)\right)}{4^j}∥+∥T(z-y)-\frac{f\left(4^j(z-y)\right)}{4^j}∥\hfill \\ \hfill & +\frac{1}{2^{\beta }}∥T(x+y)-\frac{f\left(4^j(x+y)\right)}{4^j}∥+2^{\beta }∥T\left(z-\frac{x+y}{4}\right)-\frac{f\left(4^j(z-\frac{x+y}{4})\right)}{4^j}∥\hfill \\ \hfill & +∥\frac{f\left(4^j(z-x)\right)}{4^j}+\frac{f\left(4^j(z-y)\right)}{4^j}+\frac{f\left(4^j(x+y)\right)}{2·4^j}-\frac{2f\left(4^j(z-\frac{x+y}{4})\right)}{4^j}∥\hfill \\ \hfill & \to 0\mathrm{as}j\to \infty .\hfill \end{array}$$

Hence, we get

$$T(z-x)+T(z-y)+\frac{1}{2}T(x+y)-2T\left(z-\frac{x+y}{4}\right)=0$$

for all $x,y,z\in V$. Then by Lemma 1, it follows that T is an additive mapping.

Next, assume that $S:V\to X$ is another additive mapping satisfying (20). Then we have

$$\begin{array}{cc}\hfill & ∥T\left(x\right)-S\left(x\right)∥\hfill \\ \hfill & \le ∥\frac{T\left(4^{k}x\right)-f\left(4^{k}x\right)}{4^k}∥+∥\frac{S\left(4^{k}x\right)-f\left(4^{k}x\right)}{4^k}∥\hfill \\ \hfill & \le \frac{2}{4^{k\beta }}\frac{1}{2^{\beta }}\sum _{j=1}^{\infty }\frac{1}{4^{(j-1)\beta }}\phi \left(2·4^{k+j-1}x,2·4^{k+j-1}x,2·4^{k+j-1}x\right)\hfill \\ \hfill & =2^{1-\beta }\sum _{l=k+1}^{\infty }\frac{1}{4^{(l-1)\beta }}\phi \left(2·4^{l-1}x,2·4^{l-1}x,2·4^{l-1}x\right)\hfill \\ \hfill & \to 0\mathrm{as}k\to \infty \hfill \end{array}$$

for all $x\in V$, from which it follows that $T=S$. □

Letting $\phi \equiv \epsilon >0$ in Theorem 3, we obtain a result on classical Ulam stability of the Apollonius type additive functional equation.

Corollary 4.

Let V be a linear space, and X be a β-homogeneous complex Banach space with $0<\beta \le 1$. If $f:V\to X$ is a mapping satisfying $f\left(0\right)=0$ and

$$∥f(z-x)+f(z-y)+\frac{1}{2}f(x+y)-2f\left(z-\frac{x+y}{4}\right)∥\le \epsilon$$

for all $x,y,z\in V$, then there exists a unique additive mapping $T:V\to X$ such that

$$\parallel f\left(x\right)-T\left(x\right)\parallel \le \frac{2^{\beta }ϵ}{4^{\beta }-1},x\in V.$$

5. Fuzzy Stability of (2) in Fuzzy Banach Spaces

In this section we consider the stability of the Apollonius type additive functional Equation (2). The following theorem is a fundamental result in fixed point theory.

Theorem 4.

[27,28] Let $(X,d)$ be a complete generalized metric space and let $J:X\to X$ be a strictly contractive mapping with Lipschitz constant $L<1$. Then for each given element $x\in X$, either

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

for all nonnegative integers n or there exists a positive integer $n_0$ such that

1. 

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

2. 

the sequence $\left\{J^{n}x\right\}$ converges to a fixed point $y^{\ast }$ of J;

3. 

$y^{\ast }$ is the unique fixed point of J in the set $Y=\{y\in X|d(J^{n_0}x,y)<\infty \}$;

4. 

$d(y,y^{\ast })\le \frac{1}{1-L}d(y,Jy)$ for all $y\in Y$.

We use the definition of fuzzy normed spaces given in [29,30,31].

Definition 4.

[29,30,31] Let X be a real vector 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 all $s,t\in \mathbb{R}$,

(1) 

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

(2) 

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

(3) 

$N(cx,t)=N(x,\frac{t}{\left|c\right|})$ if $c\ne 0$;

(4) 

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

(5) 

$N(x,·)$ is a non-decreasing function of $\mathbb{R}$ and ${lim}_{t\to \infty }N(x,t)=1$;

(6) 

for $x\ne 0$, $N(x,·)$ is continuous on $\mathbb{R}$.

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

Example 2.

Let $(X,\parallel ·\parallel )$ be a normed linear space. Then

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

is a fuzzy norm on X.

Definition 5.

[29,30,31] Let $(X,N)$ be a fuzzy normed vector space. A sequence $\left\{x_n\right\}$ in X is said to be convergent to $x\in X$ if ${lim}_{n\to \infty }N(x_n-x,t)=1$ for all $t>0$ and we denote it by N-${lim}_{n\to \infty }{x}_n=x$.

Definition 6.

[29,30,31] Let $(X,N)$ be a fuzzy normed vector space. A sequence $\left\{x_n\right\}$ in X is called a Cauchy sequence if ${lim}_{n,m\to \infty }N(x_n-x_m,t)=1$ for all $t>0$.

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.

Definition 7.

[32] Let X be an algebra and $(X,N)$ a fuzzy Banach space. Then the space $(X,N)$ is said to be a fuzzy Banach algebra if

$$N(xy,ab)\ge N(x,a)\circ N(y,b),x,y\in X,a,b\in \mathbb{R},$$

(24)

where ∘ is a continuous t-norm.

Example 3.

Let X be an algebra and $(X,N)$ be a fuzzy Banach space, where $N(x,t)$ is given as in Example 2. Then

$$N(xy,ab)\ge N(x,a)·N(y,b),x,y\in X,a,b\in \mathbb{R}.$$

Hence, $(X,N)$ is a fuzzy Banach algebra with $\circ =·$, the product, (see [32]).

Let $(X,N)$ be a fuzzy Banach algebra and $x_n\to x$ and $y_n\to y$ be convergent sequences in $(X,N)$. It is easy to show that N-$lim{x}_n{y}_n=xy$ (see [32]).

Now, we show the generalized Hyers-Ulam stability of (2) in fuzzy normed vector spaces via a fixed point method.

Theorem 5.

Let X be a real vector space, $(Y,N)$ a fuzzy Banach space. Let $\phi :X^3\to [0,\infty )$ be a function such that $\phi (0,0,0)=0$ and there exists an $0<L<1$ satisfying

$$\phi (4x,4y,4z)\le 4L\phi (x,y,z)$$

(25)

for all $x,y,z\in X$. Let $f:X\to Y$ be a mapping that satisfies

$$N\left(f(z-x)+f(z-y)+\frac{1}{2}f(x+y)-2f\left(z-\frac{x+y}{4}\right),t\right)\ge \frac{t}{t+\phi (x,y,z)}$$

(26)

for all $x,y,z\in X,t>0$. Then there exists a unique additive mapping $F:X\to Y$ such that

$$N\left(f\left(x\right)-F\left(x\right),t\right)\ge \frac{4(1-L)t}{4(1-L)t+2\phi (2x,2x,2x)},x\in X,t>0.$$

(27)

The mapping F is defined by $F\left(x\right)=N-{lim}_{n\to \infty }\frac{1}{4^n}f\left(4^{n}x\right),x\in X$.

Proof. 

Replacing $(x,y,z)$ with $(0,0,0)$ in (26), we get

$$N\left(\frac{1}{2}f\left(0\right),t\right)\ge \frac{t}{t+\phi (0,0,0)}=1,t>0,$$

so that $f\left(0\right)=0$.

Replacing $(x,y,z)$ with $(2x,2x,2x)$, we see

$$N\left(\frac{1}{2}f\left(4x\right)-2f\left(x\right),t\right)\ge \frac{t}{t+\phi (2x,2x,2x)}.$$

Then, putting

$$M(x,t):=\frac{t}{t+2\phi (2x,2x,2x)},x\in X,t>0,$$

we have

$$N\left(f\left(4x\right)-4f\left(x\right),2t\right)=N\left(\frac{1}{2}f\left(4x\right)-2f\left(x\right),t\right)\ge \frac{t}{t+\phi (2x,2x,2x)},$$

hence,

$$N(f\left(4x\right)-4f\left(x\right),t)\ge \frac{t}{t+2\phi (2x,2x,2x)}=M(x,t),x\in X,t>0.$$

(28)

By the definition of $M(x,t)$ and (25), we note that for every $n\in \mathbb{N}$,

$$M\left(4^{n}x,4^n{L}^{n}t)\right)\ge M(x,t),x\in X,t>0.$$

Consider the set

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

and introduce the generalized metric on S by

$$d(p,q)=inf\left\{\mu \in {\mathbb{R}}^{+}|N\left(p\left(x\right)-q\left(x\right),\mu t\right)\ge M(x,t),x\in X,t>0\right\}.$$

Then $(S,d)$ is a complete generalized metric space (see [33], Lemma 2.1). Now, we consider the map $J:S\to S$ given by

$$Jp\left(x\right)=\frac{1}{4}p\left(4x\right),x\in X.$$

Then by a standard argument, we know that J is a contractive mapping.

Since $N\left(f\left(4x\right)-4f\left(x\right),t\right)\ge M(x,t)$ by (28), we have

$$N\left(\frac{1}{4}f\left(4x\right)-f\left(x\right),\frac{1}{4}t\right)\ge M(x,t),x\in X,t>0.$$

Hence, it follows that

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

Then by the fixed point alternative, i.e., Theorem 4, there exists a mapping $F:X\to Y$ such that

• F is a fixed point of J, i.e.,

  $$F\left(4x\right)=4F\left(x\right),x\in X.$$
• $d\left(J^{n}f,F\right)\to 0$, i.e.,

  $$N-\underset{n\to \infty }{lim}\frac{1}{4^n}f\left(4^{n}x\right)=F\left(x\right),x\in X.$$
• $d(f,F)\le \frac{1}{1-L}d(f,Jf)$, i.e.,

  $$d(f,F)\le \frac{1}{4(1-L)}.$$
  This implies

  $$N\left(f\left(x\right)-F\left(x\right),t\right)\ge M\left(x,4(1-L)t\right),x\in X,t>0.$$

  (29)

Note that (29) shows the inequality (27).

Now, we show that F is an additive mapping.

Replacing $(x,y,z)$ with $(4^{n}x,4^{n}y,4^{n}z)$ in (26), we get

$$\begin{array}{cc}\hfill & N\left(\frac{1}{4^n}f\left(4^n(z-x)\right)+\frac{1}{4^n}f\left(4^n(z-y)\right)+\frac{1}{2·4^n}f\left(4^n(x+y)\right)-2·\frac{1}{4^n}f\left(4^{n}z-\frac{1}{4}\left(4^n(x+y)\right)\right),t\right)\hfill \\ \hfill & \ge \frac{4^{n}t}{4^{n}t+\phi (4^{n}x,4^{n}y,4^{n}z)}\hfill \\ \hfill & \ge \frac{t}{t+L^n\phi (x,y,z)}.\hfill \end{array}$$

Hence, it follows that

$$F(z-x)+F(z-y)+\frac{1}{2}F(x+y)-2F\left(z-\frac{x+y}{4}\right)=0$$

for all $x,y,z\in X$.

Then by Lemma 1, it follows that F is additive.

Finally, assume that $F_1$ and $F_2$ are two additive mappings that satisfy (27). Then

$$N\left(f\left(x\right)-F_i\left(x\right),t\right)\ge M(x,4(1-L)t),x\in X,t>0,i=1,2.$$

Then, by (29), we have

$$\begin{array}{cc}\hfill N\left(F_2\left(x\right)-F_1\left(x\right),2t\right)& =N\left(\frac{1}{4^n}(F_2\left(4^{n}x\right)-F_1\left(4^{n}x\right)),2t\right)\hfill \\ \hfill & \ge min\left\{N\left(\frac{1}{4^n}\left(f\left(4^{n}x\right)-F_1\left(4^{n}x\right)\right),t\right),N\left(\frac{1}{4^n}\left(f\left(4^{n}x\right)-F_2\left(4^{n}x\right)\right),t\right)\right\}\hfill \\ \hfill & \ge M\left(4^{n}x,(1-L)4^{n+1}t\right)\ge \frac{(1-L)4^{n+1}t}{(1-L)4^{n+1}t+2·{\left(4L\right)}^n\phi (2x,2x,2x)}\hfill \\ \hfill & \to 1\mathrm{as}n\to \infty .\hfill \end{array}$$

This yields that $F_1=F_2$, as desired. □

Corollary 5.

Let X be a real normed linear space, and $(Y,N)$ be a fuzzy Banach space. Let $\theta >0$ and $0<r<1$ be real numbers. Let $f:X\to Y$ be a mapping satisfying

$$N\left(f(z-x)+f(z-y)+\frac{1}{2}f(x+y)-2f\left(z-\frac{x+y}{4}\right),t\right)\ge \frac{t}{t+\theta (\parallel x{\parallel }^r+\parallel y{\parallel }^r+\parallel z{\parallel }^r)}$$

for $x,y,z\in X$ and $t>0$. Then there exists a unique additive mapping $F:X\to Y$ such that

$$N\left(f\left(x\right)-F\left(x\right),t\right)\ge \frac{\left(1-4^{r-1}\right)t}{\left(1-4^{r-1}\right)t+3·2^{r-1}\theta {\parallel x\parallel }^r}$$

for $x\in X,t>0$.

Proof. 

Taking $\phi (x,y,z)=\theta (\parallel x{\parallel }^r+\parallel y{\parallel }^r+\parallel z{\parallel }^r)$ from Theorem 5, we can choose $L=4^{r-1}$ to get the result. □

6. Hyperstability of (2) in Fuzzy Banach Algebras

In this section, we consider the Apollonius type additive functional equation on fuzzy Banach algebras. Let X be a real algebra. An additive mapping $D:X\to X$ is called a derivation if

$$D\left(xy\right)=D\left(x\right)y+xD\left(y\right),x,y\in X.$$

An additive mapping $D:X\to X$ is called a Jordan derivation if

$$D\left(x^2\right)=D\left(x\right)x+xD\left(x\right),x\in X.$$

In addition, an additive mapping $D:X\to X$ is called a Jordan triple derivation in the sense from [34] if

$$D\left(xyx\right)=D\left(x\right)yx+xD\left(y\right)x+xyD\left(x\right),x,y\in X.$$

It is well-known that every derivation is a Jordan derivation and every Jordan derivation is a Jordan triple derivation. However, the converse implications do not hold in general.

Theorem 6.

Let $(X,N)$ be a fuzzy Banach algebra. Let $\phi :X^3\to [0,\infty )$ be a function such that $\phi (0,0,0)=0$ and there exists an $0<L<1$ satisfying

$$\phi (4x,4y,4z)\le 4L\phi (x,y,z)$$

(30)

for all $x,y,z\in X$. Assume $f:X\to X$ is a mapping that satisfies

$$\begin{array}{cc}\hfill & \left(a\right)N\left(f(z-x)+f(z-y)+\frac{1}{2}f(x+y)-2f\left(z-\frac{x+y}{4}\right),t\right)\ge \frac{t}{t+\phi (x,y,z)},\hfill \\ \hfill & \left(b\right)N\left(f\left(xyx\right)-f\left(x\right)yx-xf\left(y\right)x-xyf\left(x\right),t\right)\ge \frac{t}{t+\phi (x,y,0)}\hfill \end{array}$$

(31)

for all $x,y,z\in X,t>0$. Then there exists a unique Jordan triple derivation $F:X\to X$ such that

$$N\left(f\left(x\right)-F\left(x\right),t\right)\ge \frac{4(1-L)t}{4(1-L)t+2\phi (2x,2x,2x)},x\in X,t>0.$$

The mapping F is defined by $F\left(x\right)=N-{lim}_{n\to \infty }\frac{1}{4^n}f\left(4^{n}x\right),x\in X$.

Proof. 

By Theorem 5, the mapping F is additive. Replacing $(x,y)$ with $(4^{n}x,4^{n}y)$ in (31), we have by (30)

$$\begin{array}{cc}\hfill & N\left(\frac{1}{4^{3n}}f\left(4^{3n}xyx\right)-\frac{1}{4^{3n}}{4}^{2n}f\left(4^{n}x\right)yx-\frac{1}{4^{3n}}{4}^{2n}xf\left(4^{n}y\right)x-\frac{1}{4^{3n}}{4}^{2n}xyf\left(4^{n}x\right),t\right)\hfill \\ \hfill & =N\left(f\left(4^{3n}xyx\right)-4^{2n}f\left(4^{n}x\right)yx-4^{2n}xf\left(4^{n}y\right)x-4^{2n}xyf\left(4^{n}x\right),4^{3n}t\right)\hfill \\ \hfill & \ge \frac{4^{3n}t}{4^{3n}t+\phi (4^{n}x,4^{n}y,0)}\hfill \\ \hfill & \ge \frac{4^{3n}t}{4^{3n}t+{\left(4L\right)}^n\phi (x,y,0)}\hfill \\ \hfill & =\frac{t}{t+{\left(\frac{L}{16}\right)}^n\phi (x,y,0)},\hfill \end{array}$$

from which we infer that

$$F\left(xyx\right)=F\left(x\right)yx+xF\left(y\right)x+xyF\left(x\right),x,y\in X.$$

(32)

Therefore, F is a Jordan triple derivation. □

An algebra A is called semiprime if whenever $aAa=\left\{0\right\}$ for $a\in A$, then $a=0$. $B\left(X\right)$ for Banach spaces X and all $C^{\ast }$-algebras are examples of semiprime algebras. A ring R is said to be 2-torsion free if $2r=0$ implies $r=0$ for $r\in R$.

In the following theorem, we show that the mapping f in Theorem 6 is a derivation if the algebra is semiprime.

Theorem 7.

Let $(X,N)$ be a unital 2-torsion free semiprime fuzzy Banach algebra. Let $\phi :X^3\to [0,\infty )$ be a function such that $\phi (0,0,0)=0$ and there exists an $0<L<1$ satisfying, for all $x,y,z\in X$,

$$\begin{array}{cc}\hfill & \left(a\right)\phi (4x,4y,4z)\le 4L\phi (x,y,z),\hfill \\ \hfill & \left(b\right)\left\{\frac{1}{4^n}\phi \left(x,\frac{y}{4^n},0\right)|n\in \mathbb{N}\right\}\mathit{is}\mathit{bounded}.\hfill \end{array}$$

(33)

Assume $f:X\to X$ is a mapping such that, for all $x,y,z\in X$ and $t>0$,

$$\begin{array}{cc}\hfill & \left(c\right)N\left(f(z-x)+f(z-y)+\frac{1}{2}f(x+y)-2f\left(z-\frac{x+y}{4}\right),t\right)\ge \frac{t}{t+\phi (x,y,z)},\hfill \\ \hfill & \left(d\right)N\left(f\left(xyx\right)-f\left(x\right)yx-xf\left(y\right)x-xyf\left(x\right),t\right)\ge \frac{t}{t+\phi (x,y,0)}.\hfill \end{array}$$

(34)

Then f is an additive derivation.

Proof. 

Recall that the mapping F defined by $F\left(x\right)=N-{lim}_{n\to \infty }\frac{1}{4^n}f\left(4^{n}x\right),x\in X$ from Theorem 6 is an additive Jordan triple derivation. Replacing $(x,y)$ with $(4^{n}x,y)$ in (34), we have by (33)

$$\begin{array}{cc}\hfill & N\left(\frac{1}{4^{2n}}f\left(4^{2n}xyx\right)-\frac{1}{4^{2n}}{4}^{n}f\left(4^{n}x\right)yx-\frac{1}{4^{2n}}{4}^{2n}xf\left(y\right)x-\frac{1}{4^{2n}}{4}^{n}xyf\left(4^{n}x\right),t\right)\hfill \\ \hfill & =N\left(f\left(4^{2n}xyx\right)-4^{n}f\left(4^{n}x\right)yx-4^{2n}xf\left(y\right)x-4^{n}xyf\left(4^{n}x\right),4^{2n}t\right)\hfill \\ \hfill & \ge \frac{4^{2n}t}{4^{2n}t+\phi (4^{n}x,y,0)}\hfill \\ \hfill & \ge \frac{4^{2n}t}{4^{2n}t+{\left(4L\right)}^n\phi (x,\frac{y}{4^n},0)}\hfill \\ \hfill & =\frac{t}{t+{\left(\frac{L}{4}\right)}^n\phi (x,\frac{y}{4^n},0)},\hfill \end{array}$$

from which we get

$$F\left(xyx\right)=F\left(x\right)yx+xf\left(y\right)x+xyF\left(x\right)$$

(35)

for all $x,y\in X$. Comparing (35) with (32), it follows that

$$xF\left(y\right)x=xf\left(y\right)x$$

for all $x,y\in X$. Letting $x=1$, we conclude that $F=f$. Therefore, f is a Jordan triple derivation. By [34], (Theorem 4.3), every Jordan triple derivation on a 2-torsion free semiprime ring is a derivation. Hence, we conclude that f is an additive derivation. □

Now, we have an application of Theorem 7 to simple $C^{\ast }$-algebras.

Corollary 6.

Let X be a unital simple $C^{\ast }$-algebra, $\theta >0$ and $0<r<1$ be real numbers. Assume that $f:X\to X$ is a mapping such that

$$∥f(z-x)+f(z-y)+\frac{1}{2}f(x+y)-2f\left(z-\frac{x+y}{4}\right)∥\le {\theta (\parallel x\parallel }^r+{\parallel y\parallel }^r+{\parallel z\parallel }^r),$$

$$∥f\left(xyx\right)-f\left(x\right)yx-xf\left(y\right)x-xyf\left(x\right)∥\le {\theta (\parallel x\parallel }^r+{\parallel y\parallel }^r)$$

for all $x,y,z\in X$. Then there exists an element $a\in X$ such that

$$f\left(x\right)=ax-xa,x\in X.$$

Proof. 

Letting $L=4^{r-1}$, $N(x,t)$ be as in Example 2 and $\phi (x,y,z)=\theta (\parallel x{\parallel }^r+\parallel y{\parallel }^r+\parallel z{\parallel }^r)$, we have that f is a derivation by Theorem 7. It is well-known that every derivation on simple $C^{\ast }$-algebras is inner. Hence, we get the result. □

7. Conclusions

Using the direct method, we have proved the Hyers-Ulam stability of Apollonius type additive functional equation from linear spaces to modular spaces with or without the ${\Delta }_2$-conditions. Since spaces with s-convex modulars are s-homogeneous Banach spaces, we also have investigated the same problem for $\beta$-homogeneous Banach spaces. The obtained results can be applied to normed spaces as well. We also have shown the fuzzy stability of the functional equation in fuzzy Banach spaces by using a fixed point method. Finally, we have shown the hyperstability of the functional Equation (2) associated with the Jordan triple product in fuzzy Banach algebras. Removing the ${\Delta }_2$-condition in Theorem 2 will be a challenging problem.