Generalized Hyers-Ulam Stability of the Pexider Functional Equation

Abstract

In this paper, we investigate the generalized Hyers-Ulam stability of the Pexider functional equation $f(x+y,z+w)=g(x,z)+h(y,w)$.

1. Introduction

Throughout this paper, let V and W be real vector spaces and Y be a real Banach space. In 1940, Ulam [1] raised the question about the stability of group homomorphisms. In 1941, for a given mapping $f:V\to Y$, Hyers [2] solved the stability problem for the Cauchy additive functional equation

$$f(x+y)-f\left(x\right)-f\left(y\right)=0,$$

for all $x,y\in V$. In 1978, Rassias [3] generalized Hyers’ result and Găvruta [4] made Rassias’ result more general. The concept of stability shown by Găvruta is called the ’generalized Hyers-Ulam stability’.

For given mappings $f,g,h:V\to Y$, the stability of the Pexider functional equation

$$f(x+y)-g\left(x\right)-h\left(y\right)=0\forall x,y\in V$$

was investigated by Lee and Jun [5] (see also [6,7,8,9]).

Now, for given mappings $f,g,h:V\times V\to Y$, we consider the Pexider functional equation

$$f(x+y,z+w)-g(x,z)-h(y,w)=0,$$

(1)

for all $x,y,z,w$ in the vector space V. One of typical examples of solutions of the functional Equation (1) are the mappings $f,g,h:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ given by $f(x,z)=ax+bz+c+d$, $g(x,z)=ax+bz+c$ and $h(x,z)=ax+bz+d$ with real constants $a,b,c,d$.

In this paper, we will investigate the generalized Hyers-Ulam stability of the Pexider functional Equation (1).

2. Main Results

For given mappings $f,g,h:V\times V\to Y$, we use the following abbreviations:

$$\begin{array}{ccc}\hfill D_{fgh}(x,y,z,w)& :=& f(x+y,z+w)-g(x,z)-h(y,w),\hfill \end{array}$$

for all $x,y,z,w\in V$. We need the following lemma to prove the main theorems.

Lemma 1.

If the mappings $f,g,h:V\times V\to W$ satisfy $\left(1\right)$ for all $x,y,z,w\in V$, then there exists an additive mapping $k:V\times V\to W$ such that $k(x,z)=f(x,z)-f(0,0)=g(x,z)-g(0,0)=h(x,z)-h(0,0)$ for all $x,z\in V$.

Proof. 

From the equalities $f(0,0)=g(0,0)+h(0,0)$, $f(x,z)=g(x,z)+h(0,0)$, and $f(x,z)=g(0,0)+h(x,z)$ for all $x,z\in V$, we obtain the equalities $f(x,z)-f(0,0)=g(x,z)-g(0,0)=h(x,z)-h(0,0)$ for all $x,z\in V$. Put $k(x,z)=f(x,z)-f(0,0)=g(x,z)-g(0,0)=h(x,z)-h(0,0)$ for all $x,z\in V$, then

$$k(x+y,z+w)=f(x+y,z+w)-f(0,0)=g(x,z)-g(0,0)+h(y,w)-h(0,0)=k(x,z)+k(y,w)$$

for all $x,y,z,w\in V$. □

Using the previous lemma, we obtain the following generalized Hyers-Ulam stability results for Equation (1).

Theorem 1.

Suppose that $f,g,h:V\times V\to Y$ are mappings for which there exists a function $\phi :V^2\to [0,\infty )$ such that

$${\displaystyle \sum _{i=0}^{\infty }}\frac{\phi (2^{i}x,2^{i}y,2^{i}z,2^{i}w)}{2^i}<\infty$$

(2)

and

$$\parallel f(x+y,z+w)-g(x,z)-h(y,w)\parallel \le \phi (x,y,z,w)$$

(3)

for all $x,y,z,w\in V$. Then, there exists a unique additive mapping $F:V\times V\to Y$ such that

$$\begin{array}{cc}\hfill \parallel f\left(x,z\right)-f(0,0)-F\left(x,z\right)\parallel \le & {\displaystyle \sum _{n=0}^{\infty }}\frac{\mu (2^{n}x,2^{n}z)}{2^{n+1}},\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill \parallel g\left(x,z\right)-g(0,0)-F\left(x,z\right)\parallel \le & {\displaystyle \sum _{n=0}^{\infty }}\frac{\nu (2^{n}x,2^{n}z)}{2^{n+1}},\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill \parallel h\left(x,z\right)-h(0,0)-F\left(x,z\right)\parallel \le & {\displaystyle \sum _{n=0}^{\infty }}\frac{\xi (2^{n}x,2^{n}z)}{2^{n+1}},\hfill \end{array}$$

(6)

for all $x,z\in V$, where the functions $\mu ,\nu ,\xi :V^2\to \mathbb{R}$ are defined by

$$\begin{array}{cc}\hfill \mu (x,z)=& \phi (x,x,z,z)+\phi (x,0,z,0)+\phi (0,x,0,z)+\phi (0,0,0,0),\hfill \\ \hfill \nu (x,z)=& \phi (2x,-x,2z,-z)+\phi (x,0,z,0)+\phi (x,-x,z,-z)+\phi (0,0,0,0),\hfill \\ \hfill \xi (x,z)=& \phi (-x,2x,-z,2z)+\phi (0,x,0,z)+\phi (-x,x,-z,z)+\phi (0,0,0,0).\hfill \end{array}$$

Proof. 

Let $f^{\prime }(x,z):=f(x,z)-f(0,0)$, $g^{\prime }(x,z):=g(x,z)-g(0,0)$, and $h^{\prime }(x,z):=h(x,z)-h(0,0)$. Since the equalities

$$\begin{array}{cc}\hfill f^{\prime }(2x,2z)-2f^{\prime }(x,z)=& D_{fgh}(x,x,z,z)-D_{fgh}(x,0,z,0)\hfill \\ & -D_{fgh}(0,x,0,z)-D_{fgh}(0,0,0,0),\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill g^{\prime }(2x,2z)-2g^{\prime }(x,z)=& D_{fgh}(2x,-x,2z,-z)-D_{fgh}(x,0,z,0)\hfill \\ & -D_{fgh}(x,-x,z,-z)+D_{fgh}(0,0,0,0),\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill h^{\prime }(2x,2z)-2h^{\prime }(x,z)=& D_{fgh}(-x,2x,-z,2z)-D_{fgh}(0,x,0,z)\hfill \\ & -D_{fgh}(-x,x,-z,z)+D_{fgh}(0,0,0,0),\hfill \end{array}$$

(9)

hold for all $x,z\in V$, the inequalities

$$\begin{array}{cc}\hfill \parallel f^{\prime }(x,z)-\frac{f^{\prime }(2x,2z)}{2}\parallel \le & \frac{1}{2}\mu (x,z),\hfill \\ \hfill \parallel g^{\prime }(x,z)-\frac{g^{\prime }(2x,2z)}{2}\parallel \le & \frac{1}{2}\nu (x,z),\hfill \\ \hfill \parallel h^{\prime }(x,z)-\frac{h^{\prime }(2x,2z)}{2}\parallel \le & \frac{1}{2}\xi (x,z),\hfill \end{array}$$

for all $x,z\in V$, follow from Inequality (3). Using the above inequalities, and the equalities

$$\begin{array}{cc}\hfill \frac{f^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n}-\frac{f^{\prime }\left(2^{n+m}x,2^{n+m}z\right)}{2^{n+m}}=& {\displaystyle \sum _{k=n}^{n+m-1}}\left(\frac{f^{\prime }\left(2^{k}x,2^{k}z\right)}{2^k}-\frac{f^{\prime }\left(2^{k+1}x,2^{k+1}z\right)}{2^{k+1}}\right),\hfill \\ \hfill \frac{g^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n}-\frac{g^{\prime }\left(2^{n+m}x,2^{n+m}z\right)}{2^{n+m}}=& {\displaystyle \sum _{k=n}^{n+m-1}}\left(\frac{g^{\prime }\left(2^{k}x,2^{k}z\right)}{2^k}-\frac{g^{\prime }\left(2^{k+1}x,2^{k+1}z\right)}{2^{k+1}}\right),\hfill \\ \hfill \frac{h^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n}-\frac{h^{\prime }\left(2^{n+m}x,2^{n+m}z\right)}{2^{n+m}}=& {\displaystyle \sum _{k=n}^{n+m-1}}\left(\frac{h^{\prime }\left(2^{k}x,2^{k}z\right)}{2^k}-\frac{h^{\prime }\left(2^{k+1}x,2^{k+1}z\right)}{2^{k+1}}\right),\hfill \end{array}$$

we get the following inequalities

$$\begin{array}{cc}\hfill \parallel \frac{f^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n}-\frac{f^{\prime }\left(2^{n+m}x,2^{n+m}z\right)}{2^{n+m}}\parallel \le & {\displaystyle \sum _{k=n}^{n+m-1}}\frac{\mu (2^{k}x,2^{k}z)}{2^{k+1}},\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill \parallel \frac{g^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n}-\frac{g^{\prime }\left(2^{n+m}x,2^{n+m}z\right)}{2^{n+m}}\parallel \le & {\displaystyle \sum _{k=n}^{n+m-1}}\frac{\nu (2^{k}x,2^{k}z)}{2^{k+1}},\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill \parallel \frac{h^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n}-\frac{h^{\prime }\left(2^{n+m}x,2^{n+m}z\right)}{2^{n+m}}\parallel \le & {\displaystyle \sum _{k=n}^{n+m-1}}\frac{\xi (2^{k}x,2^{k}z)}{2^{k+1}},\hfill \end{array}$$

(12)

for all $x,z\in V$ and all $n,m\in \mathbb{N}\cup \left\{0\right\}$. So, the sequences ${\left\{\frac{f^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n}\right\}}_{n\in \mathbb{N}}$, ${\left\{\frac{g^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n}\right\}}_{n\in \mathbb{N}}$, and ${\left\{\frac{h^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n}\right\}}_{n\in \mathbb{N}}$ are Cauchy sequences for all $x,z\in V\backslash \left\{0\right\}$. As Y is a real Banach space, we can define the mappings $F,G,H:V\times V\to Y$ by

$$\begin{array}{cc}\hfill F(x,z)=& {\displaystyle \underset{n\to \infty }{lim}}\frac{f^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n},\hfill \\ \hfill G(x,z)=& {\displaystyle \underset{n\to \infty }{lim}}\frac{g^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n},\hfill \\ \hfill H(x,z)=& {\displaystyle \underset{n\to \infty }{lim}}\frac{h^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n},\hfill \end{array}$$

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

$$\begin{array}{cc}\hfill {\displaystyle \underset{n\to \infty }{lim}}∥\frac{g^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n}-\frac{f^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n}∥=& {\displaystyle \underset{n\to \infty }{lim}}∥\frac{-D_{fgh}(2^{n}x,0,2^{n}z,0)+D_{fgh}(0,0,0,0)}{2^n}∥\hfill \\ \hfill \le & {\displaystyle \underset{n\to \infty }{lim}}\left(\frac{\phi (2^{n}x,0,2^{n}z,0)+\phi (0,0,0,0)}{2^n}\right)=0,\hfill \\ \hfill {\displaystyle \underset{n\to \infty }{lim}}∥\frac{h^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n}-\frac{f^{\prime }\left(2^{n}x,2^{n}z\right)}{2^n}∥=& {\displaystyle \underset{n\to \infty }{lim}}∥\frac{-D_{fgh}(0,2^{n}x,0,2^{n}z)+D_{fgh}(0,0,0,0)}{2^n}∥\hfill \\ \hfill \le & {\displaystyle \underset{n\to \infty }{lim}}\left(\frac{\phi (0,2^{n}x,0,2^{n}z)+\phi (0,0,0,0)}{2^n}\right)=0,\hfill \end{array}$$

for all $x,z\in V$, we have $F(x,z)=G(x,z)=H(x,z)$ for all $x,z\in V$. By putting $n=0$ and letting $m\to \infty$ in Inequalities (10)–(12), we obtain Inequalities (4)–(6) for all $x,z\in V$.

From Inequality (3), we get

$$∥\frac{D_{fgh}(2^{n}x,2^{n}y,2^{n}z,2^{n}w)}{2^n}∥\le \frac{\phi \left(2^{n}x,2^{n}y,2^{n}z,2^{n}w\right)}{2^n},$$

for all $x,y,z,w\in V$. Since the right-hand side of the above equality tends to zero as $n\to \infty$, we obtain that $F,G,$ and H satisfy the functional Equation (1). Thus, by Lemma 1, F is an additive mapping.

If $F^{\prime }:V\times V\to Y$ is another additive mapping satisfying (4)–(6), we obtain

$$\begin{array}{cc}\hfill \parallel F^{\prime }\left(x,z\right)-F\left(x,z\right)\parallel \le & \parallel \frac{F^{\prime }\left(2^{k}x,2^{k}z\right)}{2^k}-\frac{f\left(2^{k}x,2^{k}z\right)}{2^k}\parallel +\parallel \frac{f\left(2^{k}x,2^{k}z\right)}{2^k}-\frac{F\left(2^{k}x,2^{k}z\right)}{2^k}\parallel \hfill \\ \hfill \le & {\displaystyle \sum _{n=k}^{\infty }}\frac{\mu (2^{n}x,2^{n}z)}{2^n},\hfill \end{array}$$

for all $x,z\in V$ and all $k\in \mathbb{N}$. As ${\sum }_{n=k}^{\infty }\frac{\mu (2^{n}x,2^{n}z)}{2^n}\to 0$ as $k\to \infty$, we have $F^{\prime }(x,z)=F(x,z)$ for all $x,z\in V$. Hence, the mapping is the unique additive mapping, as desired. □

Corollary 1.

Suppose that $f:V\times V\to Y$ is a mapping for which there exists a function $\phi :V^2\to [0,\infty )$ satisfying inequality (2) and

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

(13)

for all $x,y,z,w\in V$. Then, there exists a unique additive mapping $F:V\times V\to Y$ such that

$$\begin{array}{cc}\hfill \parallel f\left(x,z\right)-f(0,0)-F\left(x,z\right)\parallel \le & min\left\{{\displaystyle \sum _{n=0}^{\infty }}\frac{\mu (2^{n}x,2^{n}z)}{2^{n+1}},{\displaystyle \sum _{n=0}^{\infty }}\frac{\nu (2^{n}x,2^{n}z)}{2^{n+1}},{\displaystyle \sum _{n=0}^{\infty }}\frac{\xi (2^{n}x,2^{n}z)}{2^{n+1}}\right\}\hfill \end{array}$$

for all $x,z\in V$.

Theorem 2.

Suppose that $f:V\times V\to Y$ is a mapping for which there exists a function $\phi :V^2\to [0,\infty )$, such that

$${\displaystyle \sum _{i=0}^{\infty }}{2}^i\phi \left(\frac{x}{2^i},\frac{y}{2^i},\frac{z}{2^i},\frac{w}{2^i}\right)<\infty$$

(14)

and $\left(3\right)$ for all $x,y,z,w\in V$. Then, there exists a unique additive mapping $F:V\times V\to Y$ such that

$$\begin{array}{cc}\hfill \parallel f\left(x,z\right)-f(0,0)-F\left(x,z\right)\parallel \le & {\displaystyle \sum _{n=0}^{\infty }}{2}^n\mu \left(\frac{x}{2^{n+1}},\frac{z}{2^{n+1}}\right),\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill \parallel g\left(x,z\right)-g(0,0)-F\left(x,z\right)\parallel \le & {\displaystyle \sum _{n=0}^{\infty }}{2}^n\nu \left(\frac{x}{2^{n+1}},\frac{z}{2^{n+1}}\right),\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill \parallel h\left(x,z\right)-h(0,0)-F\left(x,z\right)\parallel \le & {\displaystyle \sum _{n=0}^{\infty }}{2}^n\xi \left(\frac{x}{2^{n+1}},\frac{z}{2^{n+1}}\right),\hfill \end{array}$$

(17)

for all $x,z\in V$, where the functions $\mu ,\nu ,\xi :V^2\to \mathbb{R}$ are defined as in Theorem 1.

Proof. 

The inequalities

$$\begin{array}{l}\parallel f^{\prime }(x,z)-2f^{\prime }\left(\frac{x}{2},\frac{z}{2}\right)\parallel \le \mu \left(\frac{x}{2},\frac{x}{2}\right),\\ \parallel g^{\prime }(x,z)-2g^{\prime }\left(\frac{x}{2},\frac{z}{2}\right)\parallel \le \nu \left(\frac{x}{2},\frac{z}{2}\right),\\ \parallel h^{\prime }(x,z)-2h^{\prime }\left(\frac{x}{2},\frac{z}{2}\right)\parallel \le \xi \left(\frac{x}{2},\frac{z}{2}\right),\end{array}$$

for all $x,z\in V$ follow from equalities (7)–(9) for all $x,z\in V$ and inequality (3). Using the above inequalities, we easily get the following inequalities

$$\begin{array}{cc}\hfill \parallel 2^n{f}^{\prime }\left(\frac{x}{2^n},\frac{z}{2^n}\right)-2^{n+m}{f}^{\prime }\left(\frac{x}{2^{n+m}},\frac{z}{2^{n+m}}\right)\parallel \le & {\displaystyle \sum _{k=n}^{n+m-1}}{2}^k\mu \left(\frac{x}{2^{k+1}},\frac{z}{2^{k+1}}\right),\hfill \end{array}$$

(18)

$$\begin{array}{cc}\hfill \parallel 2^n{g}^{\prime }\left(\frac{x}{2^n},\frac{z}{2^n}\right)-2^{n+m}{g}^{\prime }\left(\frac{x}{2^{n+m}},\frac{z}{2^{n+m}}\right)\parallel \le & {\displaystyle \sum _{k=n}^{n+m-1}}{2}^k\nu \left(\frac{x}{2^{k+1}},\frac{z}{2^{k+1}}\right),\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill \parallel 2^n{h}^{\prime }\left(\frac{x}{2^n},\frac{z}{2^n}\right)-2^{n+m}{h}^{\prime }\left(\frac{x}{2^{n+m}},\frac{z}{2^{n+m}}\right)\parallel \le & {\displaystyle \sum _{k=n}^{n+m-1}}{2}^k\xi \left(\frac{x}{2^{k+1}},\frac{z}{2^{k+1}}\right),\hfill \end{array}$$

(20)

for all $x,z\in V$ and all $n,m\in \mathbb{N}\cup \left\{0\right\}$. Thus, the sequences ${\left\{2^n{f}^{\prime }\left(\frac{x}{2^n},\frac{z}{2^n}\right)\right\}}_{n\in \mathbb{N}}$, ${\left\{2^n{g}^{\prime }\left(\frac{x}{2^n},\frac{z}{2^n}\right)\right\}}_{n\in \mathbb{N}}$, and ${\left\{2^n{h}^{\prime }\left(\frac{x}{2^n},\frac{z}{2^n}\right)\right\}}_{n\in \mathbb{N}}$ are Cauchy sequences for all $x,z\in V$. Since Y is a real Banach space, we can define the mappings $F,G,H:V\times V\to Y$ by

$$\begin{array}{cc}\hfill F\left(x\right)=& {\displaystyle \underset{n\to \infty }{lim}}{2}^n{f}^{\prime }\left(\frac{x}{2^n},\frac{z}{2^n}\right),\hfill \\ \hfill G\left(x\right)=& {\displaystyle \underset{n\to \infty }{lim}}{2}^n{g}^{\prime }\left(\frac{x}{2^n},\frac{z}{2^n}\right),\hfill \\ \hfill H\left(x\right)=& {\displaystyle \underset{n\to \infty }{lim}}{2}^n{h}^{\prime }\left(\frac{x}{2^n},\frac{z}{2^n}\right)\hfill \end{array}$$

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

$$\begin{array}{cc}\hfill {\displaystyle \underset{n\to \infty }{lim}}∥2^n{g}^{\prime }\left(\frac{x}{2^n},\frac{z}{2^n}\right)-2^n{f}^{\prime }\left(\frac{x}{2^n},\frac{z}{2^n}\right)∥=& {\displaystyle \underset{n\to \infty }{lim}}{2}^n∥-D_{fgh}\left(\frac{x}{2^n},0,\frac{z}{2^n},0\right)+D_{fgh}(0,0,0,0)∥\hfill \\ \hfill \le & {\displaystyle \underset{n\to \infty }{lim}}\left(2^n\phi \left(\frac{x}{2^n},0,\frac{z}{2^n},0\right)+2^n\phi (0,0,0,0)\right)=0,\hfill \\ \hfill {\displaystyle \underset{n\to \infty }{lim}}∥2^n{h}^{\prime }\left(\frac{x}{2^n},\frac{z}{2^n}\right)-2^n{f}^{\prime }\left(\frac{x}{2^n},\frac{z}{2^n}\right)∥=& {\displaystyle \underset{n\to \infty }{lim}}{2}^n∥-D_{fgh}\left(0,\frac{x}{2^n},0,\frac{z}{2^n}\right)+D_{fgh}(0,0,0,0)∥\hfill \\ \hfill \le & {\displaystyle \underset{n\to \infty }{lim}}\left(2^n\phi \left(0,\frac{x}{2^n},0,\frac{z}{2^n}\right)+2^n\phi (0,0,0,0)\right)=0,\hfill \end{array}$$

for all $x,z\in V$, we have $F(x,z)=G(x,z)=H(x,z)$ for all $x,z\in V$. By putting $n=0$ and letting $m\to \infty$ in Inequalities (18)–(20), we obtain Inequalities (15)–(17) for all $x,z\in V$.

From Inequality (3), we get

$$∥2^n{D}_{fgh}\left(\frac{x}{2^n},\frac{z}{2^n},\frac{z}{2^n},\frac{w}{2^n}\right)∥\le 2^n\phi \left(\frac{x}{2^n},\frac{z}{2^n},\frac{z}{2^n},\frac{w}{2^n}\right)$$

for all $x,y,z,w\in V$. Since the right-hand side of the above equality tends to zero as $n\to \infty$, we obtain that $F,G,$ and H satisfy the functional Equation (1). Hence, by Lemma 1, F is an additive mapping.

If $F^{\prime }:V\times V\to Y$ is another additive mapping satisfying (15)–(17), we obtain $G(0,0)=0=F(0,0)$ and

$$\begin{array}{cc}\hfill \parallel F^{\prime }\left(x,z\right)-F\left(x,z\right)\parallel \le & ∥2^k{F}^{\prime }\left(\frac{x}{2^k},\frac{z}{2^k}\right)-2^{k}f\left(\frac{x}{2^k},\frac{z}{2^k}\right)∥+∥2^{k}f\left(\frac{x}{2^k},\frac{z}{2^k}\right)-2^{k}F\left(\frac{x}{2^k},\frac{z}{2^k}\right)∥\hfill \\ \hfill \le & {\displaystyle \sum _{n=k}^{\infty }}{2}^{n+1}\mu \left(\frac{x}{2^{n+1}},\frac{z}{2^{n+1}}\right)\hfill \end{array}$$

for all $x,z\in V$ and all $k\in \mathbb{N}$. Since ${\sum }_{n=k}^{\infty }{2}^{n+1}\mu \left(\frac{x}{2^{n+1}},\frac{z}{2^{n+1}}\right)\to 0$ as $k\to \infty$, we have $F^{\prime }(x,z)=F(x,z)$ for all $x,z\in V$. Hence, the mapping F is the unique additive mapping, as desired. □

Corollary 2.

Suppose that $f:V\times V\to Y$ is a mapping for which there exists a function $\phi :V^2\to [0,\infty )$ satisfying inequality (14) and f satisfies inequality (13) for all $x,y,z,w\in V$. Then, there exists a unique additive mapping $F:V\times V\to Y$, such that

$$\begin{array}{cc}\hfill \parallel f\left(x,z\right)& -f(0,0)-F\left(x,z\right)\parallel \hfill \\ \hfill \le & min\left\{{\displaystyle \sum _{n=0}^{\infty }}{2}^n\mu \left(\frac{x}{2^{n+1}},\frac{z}{2^{n+1}}\right),{\displaystyle \sum _{n=0}^{\infty }}{2}^n\nu \left(\frac{x}{2^{n+1}},\frac{z}{2^{n+1}}\right),{\displaystyle \sum _{n=0}^{\infty }}{2}^n\xi \left(\frac{x}{2^{n+1}},\frac{z}{2^{n+1}}\right)\right\},\hfill \end{array}$$

for all $x,z\in V$.

Aoki ([10]) and Gajda ([11]) proved the generalized Hyers-Ulam stability for an additive mapping in the cases where $0\le p<1$ and $1<p$, respectively. It was also proved by Gajda ([11]) and Rassias and Semrl ([12]) that the generalized Hyers-Ulam stability for an additive mapping does not holds for the case $p\ne 1$.

The above results for the cases $p>0$ and $p\ne 1$ can be applied to the next results, due to Theorems 1 and 2.

Corollary 3.

Let X be a normed space and $p,\theta$ be positive real numbers with $p\ne 1$. Suppose that $f,g,h:X^2\to Y$ are mappings, such that

$$\parallel f(x+y,z+w)-g(x,z)-h(y,w)\parallel \le \theta (\parallel x{\parallel }^p+\parallel y{\parallel }^p+\parallel z{\parallel }^p+\parallel w{\parallel }^p)$$

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

$$\begin{array}{cc}\hfill \parallel f\left(x,z\right)-f(0,0)-F\left(x,z\right)\parallel \le & \frac{4}{|2-2^p|}{\theta (\parallel x\parallel }^p+{\parallel z\parallel }^p),\hfill \\ \hfill \parallel g\left(x,z\right)-g(0,0)-F\left(x,z\right)\parallel \le & \frac{4+2^p}{|2-2^p|}{\theta (\parallel x\parallel }^p+{\parallel z\parallel }^p),\hfill \\ \hfill \parallel h\left(x,z\right)-h(0,0)-F\left(x,z\right)\parallel \le & \frac{4+2^p}{|2-2^p|}{\theta (\parallel x\parallel }^p+{\parallel z\parallel }^p),\hfill \end{array}$$

for all $x,z\in X$.

Corollary 4.

Let X be a normed space and $p,\theta$ be positive real numbers with $p\ne 1$. If $f:X^2\to Y$ is a mapping satisfying Inequality (13) for all $x,y,z,w\in X$. Then, there exists a unique additive mapping $F:X\times X\to Y$, such that

$$\parallel f\left(x,z\right)-f(0,0)-F\left(x,z\right)\parallel \le \frac{4}{|2-2^p|}{\theta (\parallel x\parallel }^p+{\parallel z\parallel }^p),$$

for all $x,z\in X$.

Lemma 2.

If the odd mappings $f,g,h:V\times V\to W$ satisfy the equality (1) for all $x,y,z,w\in V\backslash \left\{0\right\}$. Then, the mapping f satisfies the equality $D_{fff}(x,y,z,w)=0$, for all $x,y,z,w\in V$.

Proof. 

Choose any $u\in V\backslash \left\{0\right\}$. Notice that the equality

$$f(2x,2z)=2f(x,z)$$

for all $x,z\in V$ follows from the equality

$$\begin{array}{cc}\hfill f(2x,2z)-2f(x,z)=& D_{fgh}(\frac{3x}{2},\frac{x}{2},\frac{3z}{2},\frac{z}{2})+D_{fgh}(\frac{x}{2},\frac{-x}{2},\frac{z}{2},\frac{-z}{2})\hfill \\ & -D_{fgh}(\frac{3x}{2},\frac{-x}{2},\frac{3z}{2},\frac{-z}{2})-D_{fgh}(\frac{x}{2},\frac{x}{2},\frac{z}{2},\frac{z}{2}),\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill f(0,2z)-2f(0,z)=& D_{fgh}(\frac{u}{2},\frac{-u}{2},\frac{3z}{2},\frac{z}{2})+D_{fgh}(\frac{u}{2},\frac{-u}{2},\frac{z}{2},\frac{-z}{2})\hfill \\ & -D_{fgh}(\frac{u}{2},\frac{-u}{2},\frac{z}{2},\frac{z}{2})-D_{fgh}(\frac{u}{2},\frac{-u}{2},\frac{3z}{2},\frac{-z}{2}),\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill f(2x,0)-2f(x,0)=& D_{fgh}(\frac{3x}{2},\frac{x}{2},\frac{u}{2},\frac{-u}{2})+D_{fgh}(\frac{x}{2},\frac{-x}{2},\frac{u}{2},\frac{-u}{2})\hfill \\ & -D_{fgh}(\frac{3x}{2},\frac{-x}{2},\frac{u}{2},\frac{-u}{2})-D_{fgh}(\frac{x}{2},\frac{-x}{2},\frac{u}{2},\frac{u}{2}),\hfill \end{array}$$

(23)

for all $x,z,u\in V\backslash \left\{0\right\}$. Using the equalities $f(2x,2z)=2f(x,z)$,

$$\begin{array}{cc}\hfill 2f(\frac{x+y}{2},& \frac{z+w}{2})-f(x,z)-f(y,w)=D_{fgh}(\frac{x}{2},\frac{y}{2},\frac{z}{2},\frac{w}{2})+D_{fgh}(\frac{y}{2},\frac{x}{2},\frac{w}{2},\frac{z}{2})\hfill \\ & -D_{fgh}(\frac{x}{2},\frac{x}{2},\frac{z}{2},\frac{z}{2})-D_{fgh}(\frac{y}{2},\frac{y}{2},\frac{w}{2},\frac{w}{2}),\hfill \\ \hfill 2f(\frac{y}{2},& \frac{z+w}{2})-f(0,z)-f(y,w)=D_{fgh}(\frac{3y}{4},\frac{-y}{4},\frac{w}{2},\frac{z}{2})+D_{fgh}(\frac{y}{4},\frac{y}{4},\frac{z}{2},\frac{w}{2})\hfill \\ & -D_{fgh}(\frac{y}{4},\frac{-y}{4},\frac{z}{2},\frac{z}{2})-D_{fgh}(\frac{3y}{4},\frac{y}{4},\frac{w}{2},\frac{w}{2}),\hfill \\ \hfill 2f(0,& \frac{z+w}{2})-f(0,z)-f(0,w)=D_{fgh}(\frac{x}{2},\frac{-x}{2},\frac{z}{2},\frac{w}{2})+D_{fgh}(\frac{x}{2},\frac{-x}{2},\frac{w}{2},\frac{z}{2})\hfill \\ & -D_{fgh}(\frac{x}{2},\frac{-x}{2},\frac{z}{2},\frac{z}{2})-D_{fgh}(\frac{x}{2},\frac{-x}{2},\frac{w}{2},\frac{w}{2}),\hfill \\ \hfill 2f(\frac{y}{2},& \frac{z}{2})-f(0,z)-f(y,0)=D_{fgh}(\frac{3y}{4},\frac{-y}{4},\frac{z}{4},\frac{z}{4})+D_{fgh}(\frac{y}{4},\frac{y}{4},\frac{3z}{4},\frac{-z}{4})\hfill \\ & -D_{fgh}(\frac{3y}{4},\frac{y}{4},\frac{z}{4},\frac{-z}{4})-D_{fgh}(\frac{y}{4},\frac{-y}{4},\frac{3z}{4},\frac{z}{4}),\hfill \end{array}$$

for all $x,y,z,w\in V\backslash \left\{0\right\}$, we have the equalities

$$\begin{array}{cc}\hfill D_{fff}(x,y,z,w)=& f(x+y,z+w)-f(x,z)-f(y,w)=0,\hfill \\ \hfill D_{fff}(0,y,z,w)=& f(y,z+w)-f(0,z)-f(y,w)=0,\hfill \\ \hfill D_{fff}(0,0,z,w)=& f(0,z+w)-f(0,z)-f(0,w)=0,\hfill \\ \hfill D_{fff}(0,y,z,0)=& f(y,z)-f(0,z)-f(y,0)=0,\hfill \end{array}$$

for all $x,y,z,w\in V\backslash \left\{0\right\}$. By the same method, we obtain the equalities

$$\begin{array}{cc}\hfill D_{fff}(x,0,z,w)=& f(x,z+w)-f(x,z)-f(0,w)=0,\hfill \\ \hfill D_{fff}(x,y,0,w)=& f(x+y,w)-f(x,0)-f(y,w)=0,\hfill \\ \hfill D_{fff}(x,y,z,0)=& f(x+y,z)-f(x,z)-f(y,0)=0,\hfill \\ \hfill D_{fff}(x,0,0,w)=& f(x,w)-f(x,0)-f(0,w)=0,\hfill \\ \hfill D_{fff}(x,y,0,0)=& f(x+y,0)-f(x,0)-f(y,0)=0,\hfill \end{array}$$

for all $x,y,z,w\in V\backslash \left\{0\right\}$. As $f(0,0)=0$, the equalities $D_{fff}(x,0,z,0)=0$, $D_{fff}(0,y,0,w)=0$, $D_{fff}(0,0,0,0)=0$, $D_{fff}(x,0,0,0)=0$, $D_{fff}(0,y,0,0)=0$, $D_{fff}(0,0,z,0)=0$, $D_{fff}(0,0,0,w)=0$, $D_{fff}(0,0,0,0)=0$ hold for all $x,y,z,w\in V\backslash \left\{0\right\}$. Hence, we conclude that $D_{fff}(x,y,z,w)=0$ for all $x,y,z,w\in V$. □

Theorem 3.

Suppose that $f,g,h:V\times V\to Y$ are odd mappings, for which there exists a function $\phi :{(V\backslash \left\{0\right\})}^4\to [0,\infty )$ such that

$${\displaystyle \sum _{i=0}^{\infty }}\frac{\phi (2^{i}x,2^{i}y,2^{i}z,2^{i}w)}{2^i}<\infty$$

(24)

and

$$\parallel f(x+y,z+w)-g(x,z)-h(y,w)\parallel \le \phi (x,y,z,w),$$

(25)

for all $x,y,z,w\in V\backslash \left\{0\right\}$. Then, there exists a unique mapping $F:V\times V\to Y$, such that $D_{FFF}(x,y,z,w)=0$ for all $x,y,z,w\in V$ and

$$\begin{array}{cc}\hfill \parallel f\left(x,z\right)-F\left(x,z\right)\parallel \le & {\displaystyle \sum _{n=0}^{\infty }}\frac{\mu (2^{n}x,2^{n}z)}{2^{n+1}},\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill \parallel f\left(x,0\right)-F\left(x,0\right)\parallel \le & {\displaystyle \sum _{n=0}^{\infty }}\frac{\nu (2^{n}x,2^{n}u)}{2^{n+1}},\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill \parallel f\left(0,z\right)-F\left(0,z\right)\parallel \le & {\displaystyle \sum _{n=0}^{\infty }}\frac{\xi (2^{n}u,2^{n}z)}{2^{n+1}},\hfill \end{array}$$

(28)

for all $x,z,u\in V\backslash \left\{0\right\}$, where the functions $\mu ,\nu ,\xi :V^2\to \mathbb{R}$ are defined by

$$\begin{array}{cc}\hfill \mu (x,z)=& \phi (\frac{3x}{2},\frac{x}{2},\frac{3z}{2},\frac{z}{2})+\phi (\frac{x}{2},\frac{-x}{2},\frac{z}{2},\frac{-z}{2})\hfill \\ & +\phi (\frac{3x}{2},\frac{-x}{2},\frac{3z}{2},\frac{-z}{2})+\phi (\frac{x}{2},\frac{x}{2},\frac{z}{2},\frac{z}{2}),\hfill \\ \hfill \nu (x,u)=& \phi (\frac{3x}{2},\frac{x}{2},\frac{u}{2},\frac{-u}{2})+\phi (\frac{x}{2},\frac{-x}{2},\frac{u}{2},\frac{-u}{2})\hfill \\ & +\phi (\frac{3x}{2},\frac{-x}{2},\frac{u}{2},\frac{-u}{2})+\phi (\frac{x}{2},\frac{-x}{2},\frac{u}{2},\frac{u}{2}),\hfill \\ \hfill \xi (u,z)=& \phi (\frac{u}{2},\frac{-u}{2},\frac{3z}{2},\frac{z}{2})+\phi (\frac{u}{2},\frac{-u}{2},\frac{z}{2},\frac{-z}{2})\hfill \\ & +\phi (\frac{u}{2},\frac{-u}{2},\frac{z}{2},\frac{z}{2})+\phi (\frac{u}{2},\frac{-u}{2},\frac{3z}{2},\frac{-z}{2}),\hfill \end{array}$$

for all $x,z,u\in V\backslash \left\{0\right\}$.

Proof. 

Since $f,g,h:V\times V\to Y$ are odd mappings, the equalities $f(x,y)=-f(-x,-y)$, $g(x,y)=-g(-x,-y)$, $h(x,y)=-h(-x,-y)$, and $f(0,0)=g(0,0)=h(0,0)=0$ hold for all $x,z\in V$. Choose any $u\in \backslash \left\{0\right\}$. Since the equalities (21)–(23) hold for all $x,z\in V\backslash \left\{0\right\}$, the inequalities

$$\begin{array}{cc}\hfill \parallel f(x,z)-\frac{f(2x,2z)}{2}\parallel \le & \frac{1}{2}\mu (x,z),\hfill \\ \hfill \parallel f(x,0)-\frac{f(2x,0)}{2}\parallel \le & \frac{1}{2}\mu (x,u),\hfill \\ \hfill \parallel f(0,z)-\frac{f(0,2z)}{2}\parallel \le & \frac{1}{2}\xi (u,z),\hfill \end{array}$$

for all $x,z,u\in V\backslash \left\{0\right\}$ follow from the inequality (25). Using the above inequalities, we get the following inequalities

$$\begin{array}{cc}\hfill \parallel \frac{f\left(2^{n}x,2^{n}z\right)}{2^n}-\frac{f\left(2^{n+m}x,2^{n+m}z\right)}{2^{n+m}}\parallel \le & {\displaystyle \sum _{k=n}^{n+m-1}}\frac{\mu (2^{k}x,2^{k}z)}{2^{k+1}},\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill \parallel \frac{f\left(2^{n}x,0\right)}{2^n}-\frac{f\left(2^{n+m}x,0\right)}{2^{n+m}}\parallel \le & {\displaystyle \sum _{k=n}^{n+m-1}}\frac{\nu (2^{k}x,2^{k}u)}{2^{k+1}},\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill \parallel \frac{f\left(0,2^{n}z\right)}{2^n}-\frac{f\left(0,2^{n+m}z\right)}{2^{n+m}}\parallel \le & {\displaystyle \sum _{k=n}^{n+m-1}}\frac{\xi (2^{k}u,2^{k}z)}{2^{k+1}},\hfill \end{array}$$

(31)

for all $x,z,u\in V\backslash \left\{0\right\}$ and all nonnegative integers $n,m$. So, the sequence ${\left\{\frac{f\left(2^{n}x,2^{n}z\right)}{2^n}\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence for all $x,z\in V$. As Y is a real Banach space, we can define a mapping $F:V\times V\to Y$ by

$$\begin{array}{cc}\hfill F(x,z)=& {\displaystyle \underset{n\to \infty }{lim}}\frac{f\left(2^{n}x,2^{n}z\right)}{2^n}.\hfill \end{array}$$

By putting $n=0$ and letting $m\to \infty$ in the inequalities (29)–(31), we obtain the inequalities (26)–(28) for all $x,z\in V$.

Since the equalities

$$\begin{array}{cc}\hfill {\displaystyle \underset{n\to \infty }{lim}}\frac{2g(2^{n}x,2^{n}z)-f(2^{n+1}x,2^{n+1}z)}{2^{n+1}}=& {\displaystyle \underset{n\to \infty }{lim}}(-\frac{D_{fgh}(2^{n}x,2^{n}x,2^{n}z,2^{n}z)}{2^{n+1}}\hfill \\ & -\frac{D_{fgh}(2^{n}x,-2^{n}x,2^{n}z,-2^{n}z)}{2^{n+1}})=0\hfill \\ \hfill {\displaystyle \underset{n\to \infty }{lim}}\frac{2h(2^{n}x,2^{n}z)-f(2^{n+1}x,2^{n+1}z)}{2^{n+1}}=& {\displaystyle \underset{n\to \infty }{lim}}(-\frac{D_{fgh}(2^{n}x,2^{n}x,2^{n}z,2^{n}z)}{2^{n+1}}\hfill \\ & -\frac{D_{fgh}(-2^{n}x,2^{n}x,-2^{n}z,2^{n}z)}{2^{n+1}})=0,\hfill \end{array}$$

hold for all $x,z\in V\backslash \left\{0\right\}$, we know that the limits of $\frac{g\left(2^{n}x,2^{n}z\right)}{2^n}$ and $\frac{h\left(2^{n}x,2^{n}z\right)}{2^n}$ are given by

$$\begin{array}{c}{\displaystyle \underset{n\to \infty }{lim}}\frac{g\left(2^{n}x,2^{n}z\right)}{2^n}=F(x,z),\hfill \\ {\displaystyle \underset{n\to \infty }{lim}}\frac{h\left(2^{n}x,2^{n}z\right)}{2^n}=F(x,z),\hfill \end{array}$$

for all $x,z\in V\backslash \left\{0\right\}$. From inequality (25), we get

$$D_{FFF}(x,y,z,w)={\displaystyle \underset{n\to \infty }{lim}}∥\frac{D_{fgh}(2^{n}x,2^{n}y,2^{n}z,2^{n}w)}{2^n}∥\le {\displaystyle \underset{n\to \infty }{lim}}\frac{\phi \left(2^{n}x,2^{n}y,2^{n}z,2^{n}w\right)}{2^n},$$

for all $x,y,z,w\in V\backslash \left\{0\right\}$. Since the right-hand side in the above equality equals zero, we obtain that the equality $D_{FFF}(x,y,z,w)=0$ holds for all $x,y,z,w\in V\backslash \left\{0\right\}$. By Lemma 2, F satisfies the equality $D_{FFF}(x,y,z,w)=0$ for all $x,y,z,w\in V$. If $F^{\prime }:V\times V\to Y$ is another mapping satisfying inequalities (26)–(28) and the equality $D_{F^{\prime }{F}^{\prime }{F}^{\prime }}(x,y,z,w)=0$ for all $x,y,z,w\in V$, then we obtain the inequalities

$$\begin{array}{cc}\hfill \parallel F^{\prime }\left(x,z\right)-F\left(x,z\right)\parallel \le & \parallel \frac{F^{\prime }\left(2^{k}x,2^{k}z\right)}{2^k}-\frac{f\left(2^{k}x,2^{k}z\right)}{2^k}\parallel +\parallel \frac{f\left(2^{k}x,2^{k}z\right)}{2^k}-\frac{F\left(2^{k}x,2^{k}z\right)}{2^k}\parallel \hfill \\ \hfill \le & {\displaystyle \sum _{n=k}^{\infty }}\frac{\mu (2^{n}x,2^{n}z)}{2^n},\hfill \\ \hfill \parallel F^{\prime }\left(x,0\right)-F\left(x,0\right)\parallel \le & {\displaystyle \sum _{n=k}^{\infty }}\frac{\nu (2^{n}x,2^{n}u)}{2^n},\hfill \\ \hfill \parallel F^{\prime }\left(0,z\right)-F\left(0,z\right)\parallel \le & {\displaystyle \sum _{n=k}^{\infty }}\frac{\xi (2^{n}u,2^{n}z)}{2^n},\hfill \end{array}$$

for all $x,z,u\in V\backslash \left\{0\right\}$ and all $k\in \mathbb{N}$. As ${\sum }_{n=k}^{\infty }\frac{\mu (2^{n}x,2^{n}z)}{2^n}$, ${\sum }_{n=k}^{\infty }\frac{\nu (2^{n}x,2^{n}u)}{2^n}$, ${\sum }_{n=k}^{\infty }\frac{\xi (2^{n}u,2^{n}z)}{2^n}\to 0$ as $k\to \infty$, we have $F^{\prime }(x,z)=F(x,z)$ for all $x,z\in V$. Hence, the mapping F is the unique additive mapping, as desired. □

Lee (Theorem 5 in [13] and Corollary 1 in [14]) proved the generalized Hyers-Ulam stability for an additive mapping in the case $n=1$ and $p<0$.

The next corollary, for the case $p<0$, follows from Theorem 3.

Corollary 5.

Let X be a normed space and p be a negative real number. Suppose that $f,g,h:X^2\to Y$ are odd mappings, such that

$$\parallel f(x+y,z+w)-g(x,z)-h(y,w)\parallel \le \theta (\parallel x{\parallel }^p+\parallel y{\parallel }^p+\parallel z{\parallel }^p+\parallel w{\parallel }^p)$$

for all $x,y,z,w\in X\backslash \left\{0\right\}$. Then $f,g,h$ satisfy the equality $D_{fff}(x,y,z,w)=0$ for all $x,y,z,w\in X$ and the inequalities

$$\begin{array}{cc}\hfill \parallel g(x,z)-f(x,z)\parallel \le & 2\theta (\parallel x{\parallel }^p+\parallel z{\parallel }^p),\hfill \\ \hfill \parallel h(x,z)-f(x,z)\parallel \le & 2\theta (\parallel x{\parallel }^p+\parallel z{\parallel }^p),\hfill \end{array}$$

for all $x,z\in X\backslash \left\{0\right\}$.

Proof. 

By Theorem 3, there exists a mapping $F:X\times X\to Y$, such that $D_{FFF}(x,y,z,w)=0$ for all $x,y,z,w\in X$ and

$$\parallel f\left(x,z\right)-F\left(x,z\right)\parallel \le \frac{(2·3^p+6){\theta (\parallel x\parallel }^p+{\parallel z\parallel }^p)}{2^p(2-2^p)},$$

for all $x,z,u\in X\backslash \left\{0\right\}$. Therefore, we obtain the inequalities

$$\begin{array}{cc}\hfill \parallel g(x,z)-F(x,z)\parallel \le & \left(2+\frac{(3^p+3)}{2-2^p}\right){\theta (\parallel x\parallel }^p+{\parallel z\parallel }^p)\hfill \\ \hfill \parallel h(x,z)-F(x,z)\parallel \le & \left(2+\frac{(3^p+3)}{2-2^p}\right){\theta (\parallel x\parallel }^p+{\parallel z\parallel }^p),\hfill \end{array}$$

for all $x,z\in X\backslash \left\{0\right\}$, from the inequalities

$$\begin{array}{cc}\hfill \parallel g(x,z)-F(x,z)\parallel \le & \parallel g(x,z)-\frac{f(2x,2z)}{2}\parallel +\parallel \frac{f(2x,2z)}{2}-\frac{F(2x,2z)}{2}\parallel \hfill \\ \hfill \le & \parallel \frac{D_{fgh}(x,x,z,z)}{2}\parallel +\parallel \frac{D_{fgh}(x,-x,z,-z)}{2}\parallel +\frac{(3^p+3){\theta (\parallel x\parallel }^p+{\parallel z\parallel }^p)}{2-2^p},\hfill \\ \hfill \parallel h(x,z)-F(x,z)\parallel \le & \parallel h(x,z)-\frac{f(2x,2z)}{2}\parallel +\parallel \frac{f(2x,2z)}{2}-\frac{F(2x,2z)}{2}\parallel \hfill \\ \hfill \le & \parallel \frac{D_{fgh}(x,x,z,z)}{2}\parallel +\parallel \frac{D_{fgh}(-x,x,-z,z)}{2}\parallel +\frac{(3^p+3){\theta (\parallel x\parallel }^p+{\parallel z\parallel }^p)}{2-2^p},\hfill \end{array}$$

for all $x,z\in X\backslash \left\{0\right\}$. Hence, we have the inequalities

$$\begin{array}{cc}\hfill \parallel f& \left(x,z\right)-F\left(x,z\right)\parallel \hfill \\ \hfill =& \parallel D_{fgh}((k+1)x,-kx,(k+1)z,-kz)-D_{FFF}((k+1)x,-kx,(k+1)z,-kz)\hfill \\ & +g\left(\right(k+1)x,(k+1\left)z\right)-F\left(\right(k+1)x,(k+1\left)z\right)+h(-kx,-kz)-F(-kx,-kz)\parallel \hfill \\ \hfill \le & ({(k+1)}^p+k^p)\left(1+\left(2+\frac{(3^p+3)}{2-2^p}\right)\right){\theta (\parallel x\parallel }^p+{\parallel z\parallel }^p),\hfill \\ \hfill \parallel f& \left(x,0\right)-F\left(x,0\right)\parallel \hfill \\ \hfill =& \parallel D_{fgh}((k+1)x,-kx,kz,-kz)-D_{FFF}((k+1)x,-kx,kz,-kz)\hfill \\ & +g\left(\right(k+1)x,kz)-F\left(\right(k+1)x,kz)+h(-kx,-kz)-F(-kx,-kz)\parallel \hfill \\ \hfill \le & \left(1+\left(2+\frac{(3^p+3)}{2-2^p}\right)\right)\theta (({(k+1)}^p+k^p){\parallel x\parallel }^p+2k^p{\parallel z\parallel }^p),\hfill \\ \hfill \parallel f& \left(0,z\right)-F\left(0,z\right)\parallel \hfill \\ \hfill =& \parallel D_{fgh}(kx,-kx,(k+1)z,-kz)-D_{FFF}(kx,-kx,(k+1)z,-kz)\hfill \\ & +g(kx,(k+1\left)z\right)-F(kx,(k+1\left)z\right)+h(-kx,-kz)-F(-kx,-kz)\parallel \hfill \\ \hfill \le & \left(1+\left(2+\frac{(3^p+3)}{2-2^p}\right)\right)\theta (2k^p{\parallel x\parallel }^p+{(k+1)}^p+k^p{)\parallel z\parallel }^p),\hfill \end{array}$$

for all $x,z\in X\backslash \left\{0\right\}$ and $k\in N$. Since the right-hand side of the above equalities tends to zero as $k\to \infty$ when $p<0$, we know that $f(x,z)=F(x,z)$ for all $x,z\in X$. So, f satisfies the equality $D_{fff}(x,y,z,w)=0$ for all $x,y,z,w\in X$, which implies the equality $f(x,z)=\frac{f(2x,2z)}{2}$ for all $x,z\in X$, and the inequalities

$$\begin{array}{cc}\hfill \parallel g(x,z)-f(x,z)\parallel =& \parallel g(x,z)-\frac{f(2x,2z)}{2}\parallel \hfill \\ \hfill \le & \parallel \frac{D_{fgh}(x,x,z,z)}{2}\parallel +\parallel \frac{D_{fgh}(x,-x,z,-z)}{2}\parallel \hfill \\ \hfill \le & 2\theta (\parallel x{\parallel }^p+\parallel z{\parallel }^p),\hfill \\ \hfill \parallel h(x,z)-f(x,z)\parallel =& \parallel h(x,z)-\frac{f(2x,2z)}{2}\parallel \hfill \\ \hfill \le & \parallel \frac{D_{fgh}(x,x,z,z)}{2}\parallel +\parallel \frac{D_{fgh}(-x,x,-z,z)}{2}\parallel \hfill \\ \hfill \le & 2\theta (\parallel x{\parallel }^p+\parallel z{\parallel }^p),\hfill \end{array}$$

for all $x,z\in X\backslash \left\{0\right\}$. □