Stability and Instability of an Apollonius-Type Functional Equation

Abstract

For the inner product space, we have Appolonius’ identity. From this identity, Park and Th. M. Rassias induced and investigated the quadratic functional equation of the Apollonius type. And Park and Th. M. Rassias first introduced an Apollonius-type additive functional equation. In this work, we investigate an Apollonius-type additive functional equation in 2-normed spaces. We first investigate the stability of an Apollonius-type additive functional equation in 2-Banach spaces by using Hyers’ direct method. Then, we consider the instability of an Apollonius-type additive functional equation in 2-Banach spaces.

1. Introduction

The concept of stability of a functional equation occurs when one replaces a functional equation with an inequality that acts as a perturbation of the equation. The first stability problem of the functional equation was raised by Ulam [1] in 1940. Since then, this problem has attracted the attention of many researchers. The affirmative answer to this question was given in the next year by Hyers [2] in 1941. In 1950, Aoki [3] generalized Hyers’ theorem for additive mappings. Hyers’ result was generalized by Th. M. Rassias [4] for linear mappings by an unbounded Cauchy difference. In 1994, a further generalization of Th. M. Rassias’ theorem was obtained by Găvruta [5] (see also [6]).

After then, the stability problem of several functional equations has been extensively investigated by some authors, and there are many interesting results concerning the Ulam stability problems in [7,8,9,10,11,12,13,14,15,16,17,18]. In 2012, Chung and Park [13] investigated the generalized Hyers–Ulam stability of the functional equations

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

$$2f\left({\displaystyle \frac{x+y}{2}}\right)=f\left(x\right)+f\left(y\right)$$

and

$$f(x+y)+f(x-y)=2f\left(x\right)+2f\left(y\right)$$

in 2-Banach spaces. In 2013, B.M. Patel and A.B. Patel [7] investigated the Hyers-Ulam stability of the quadratic functional equation in 2-Banach spaces. And, in 2014, B.M. Patel [19] investigated the Hyers-Ulam stability of the quartic and additive functional equation in 2-Banach spaces. In 2018, Al-Ali and Elkettani [20] introduced a new type of radical cubic functional equation related to Jensen mapping of the form

$$f\left(\sqrt[3]{x^3+y^3}\right)+f\left(\sqrt[3]{x^3-y^3}\right)=2f\left(x\right).$$

They studied general solution and stability for the considered functional equation in 2-Banach spaces. In 2021, Cieplinski [21] proved the Ulam stability of general functional equations with multiple variables in 2-Banach spaces by applying the fixed-point method.

Recently, Arumugam and Najati [8] proved various types of Hyers–Ulam stability and hyperstability of a Jensen-type quadratic functional equation of the form

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

in 2-Banach spaces by using Hyers’ direct method.

Park and Rassias [22] investigated the quadratic functional equation,

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

(1)

For the inner product space, we have Apollonius’s identity as the form of the following:

$${\parallel z-x\parallel }^2+{\parallel z-y\parallel }^2={\displaystyle \frac{1}{2}}{\parallel x-y\parallel }^2+2\parallel z-{\displaystyle \frac{x+y}{2}}{\parallel }^2.$$

For this reason, the functional Equation (1) is called a quadratic functional equation of Apollonius type. In [23], Najati introduced and investigated a quadratic functional equation of n-Apollonius type.

For an Apollonius-type additive functional equation,

$$\begin{array}{c}\hfill f(x+y)+2f(z-x)+2f(z-y)=4f\left(z-{\displaystyle \frac{x+y}{4}}\right),\end{array}$$

(2)

which was first introduced in Park and Th. M. Rassias [24]. And by Kim and J.M. Rassias [25], the stability was investigated in Modular Spaces and Fuzzy Banach Spaces.

As far as we know, there is no result on the stability of (2) in 2-Banach spaces. Thus, we investigate the stability of Apollonius-type additive functional Equation (2) in 2-Banach spaces. Moreover, we discuss the instability of (2) in Section 4.

First, in Section 2, we introduce applicable definitions and lemmas concerning 2-normed spaces. In Section 3 and Section 4, we investigate the stability and instability of Apollonius-type additive functional equation. Finally, in Section 5, we give the conclusion of this paper.

2. Preliminaries

In this section, we provide some basic notations, definitions, theorems and lemmas, which will be very applicable to prove main results. In 1964, Gähler [26] introduced the concept of linear 2-normed spaces.

Gähler [26] stated: Let A be a linear space over $\mathbb{R}$ with $dimA>1$ and let $\parallel ·,·\parallel :A\times A\to \mathbb{R}$ be a function satisfying the following properties:

1.

$\parallel x,y\parallel =0$ if and only if x and y are linearly dependent,

2.

$\parallel x,y\parallel =\parallel y,x\parallel$,

3.

$\parallel \lambda x,y\parallel =\left|\lambda \right|\parallel x,y\parallel$,

4.

$\parallel x,y+z\parallel \le \parallel x,y\parallel +\parallel x,z\parallel$

for all $x,y,z\in A$ and $\lambda \in \mathbb{R}.$ Then the function $\parallel ·,·\parallel$ is called a 2-norm on A and the pair $(A,∥·,·∥)$ is called a linear 2-normed space.

For an example of 2-normed spaces, we can consider the Euclidean space ${\mathbb{R}}^3$ with $\Vert \mathbf{u},\mathbf{v}\Vert$ as $\Vert \mathbf{u}\times \mathbf{v}\Vert$. So, 2-normed spaces are obtained by an abstraction of the notion of area while usual normed spaces are obtained by an abstraction of the notion of length. Unfortunately, stability theory in two-norm space is not yet very developed. However, we think that this is a promising young branch in mathematics.

Basic properties for the linear 2-normed spaces can be found in [27,28].

Lemma 1 

([27]). Let $(A,∥·,·∥)$ be a 2-normed space. If $∥x,y∥=0,$ for all $y\in A,$ then $x=0$.

Definition 1 

([28]). A sequence $\left\{x_n\right\}$ in a linear 2-normed space A is called a Cauchy sequence if there are two points $y,z\in A$ such that y and z are linearly independent,

$$\underset{l,m\to \infty }{lim}∥x_l-x_m,y∥=0$$

and

$$\underset{l,m\to \infty }{lim}∥x_l-x_m,z∥=0.$$

Definition 2 

([28]). A sequence $\left\{x_n\right\}$ in a linear 2-normed space A is called a convergent sequence if there is an $x\in A$ such that

$$\underset{n\to \infty }{lim}∥x_n-x,y∥=0$$

for all $y\in A.$ If $\left\{x_n\right\}$ converges to x, we write $\underset{n\to \infty }{lim}{x}_n=x$.

Lemma 2 

([27]). For a convergent sequence $x_n$ in a linear 2-normed space A,

$$\underset{n\to \infty }{lim}∥x_n,y∥=\parallel \underset{n\to \infty }{lim}{x}_n,y\parallel$$

for all $y\in A.$

Definition 3 

([28]). A linear 2-normed space in which every Cauchy sequence is convergent is called a 2-Banach space.

Definition 4 

([18]). $(A,\parallel ·\parallel ,\parallel ·,·\parallel )$ is called a normed 2-Banach space if $(A,\parallel ·\parallel )$ is a normed space and $(A,\parallel ·,·\parallel )$ is a 2-Banach space.

From now on, let A be a normed 2-Banach space.

Lemma 3 

([29]). Let $f:A\to A$ be a mapping satisfying (2). Then f is an additive mapping.

For a function $f:A\to A$, we define a mapping $D_f:A\times A\times A\to A$ by

$$\begin{array}{cc}\hfill D_f(x,y,z)=& f(x+y)+2f(z-x)+2f(z-y)-4f\left(z-{\displaystyle \frac{x+y}{4}}\right)\hfill \end{array}$$

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

3. Stability of an Apollonius-Type Additive Functional Equation

In this section, we prove the Hyers–Ulam stability of an Apollonius-type additive functional equation.

Theorem 1. 

Let $ϵ\ge 0$ and $r,s,t\in (0,1)$. Assume that $f:(A,\parallel ·\parallel )\to (A,\parallel ·\parallel ,\parallel ·,·\parallel )$ is a function satisfying the inequality

$$∥D_f(x,y,z),w∥\le ϵ({∥x∥}^r+{∥y∥}^s+{∥z∥}^t)∥w∥$$

(3)

for all $x,y,z,w\in A$. Then there exists a unique additive mapping $\mathcal{M}:(A,\parallel ·\parallel )\to (A,\parallel ·\parallel ,\parallel ·,·\parallel )$ that satisfies the functional Equation (2) and

$$∥f\left(x\right)-\mathcal{M}\left(x\right),w∥\le \left({\displaystyle \frac{{∥x∥}^r}{4-2^{r+1}}}+{\displaystyle \frac{{∥x∥}^s}{4-2^{s+1}}}+{\displaystyle \frac{{∥x∥}^t}{4-2^{t+1}}}\right)ϵ∥w∥$$

for all $x,w\in A$.

Proof. 

First, by letting $x=y=z=0$ in the inequality (3), we have $f\left(0\right)$ = 0. Also, by setting $z=x,y=-x$ in (3), we have

$$\begin{array}{c}\hfill ∥f\left(2x\right)-2f\left(x\right),w∥\le {\displaystyle \frac{ϵ}{2}}\left({∥x∥}^r+{∥x∥}^s+{∥x∥}^t\right)\parallel w\parallel \end{array}$$

(4)

for all $x,w\in A$. Now, by dividing the above Equation (4) by 2, we obtain

$$\begin{array}{c}\hfill ∥f\left(x\right)-{\displaystyle \frac{1}{2}}f\left(2x\right),w∥\le {\displaystyle \frac{ϵ}{4}}\left({∥x∥}^r+{∥x∥}^s+{∥x∥}^t\right)\parallel w\parallel \end{array}$$

(5)

for all $x,w\in A$. Next, by replacing x by $2x$ and again diving by 2, in the inequality (5), we obtain

$$\begin{array}{c}\hfill ∥{\displaystyle \frac{1}{2}}f\left(2x\right)-{\displaystyle \frac{1}{4}}f\left(4x\right),w∥\le {\displaystyle \frac{1}{2}}.{\displaystyle \frac{ϵ}{4}}\left(2^r{∥x∥}^r+2^s{∥x∥}^s+2^t{∥x∥}^t\right)\parallel w\parallel .\end{array}$$

(6)

So, due to (5) and (6), we obtain

$$\begin{array}{cc}\hfill \parallel f\left(x\right)& -{\displaystyle \frac{1}{2^2}}f\left(2^{2}x\right),w\parallel =∥f\left(x\right)-{\displaystyle \frac{1}{2}}f\left(2x\right),w∥+∥{\displaystyle \frac{1}{2}}f\left(2x\right)-{\displaystyle \frac{1}{4}}f\left(4x\right),w∥\hfill \\ \hfill & \le {\displaystyle \frac{ϵ}{4}}{(\parallel x\parallel }^r+{\parallel x\parallel }^s+{\parallel x\parallel }^t)∥w∥+{\displaystyle \frac{1}{2}}.{\displaystyle \frac{ϵ}{4}}(2^r{∥x∥}^r+2^s{∥x∥}^s+2^t{∥x∥}^t)∥w∥\hfill \end{array}$$

for all $x,w\in A$. Therefore, $∥f\left(x\right)-{\displaystyle \frac{1}{2^2}}f\left(2^{2}x\right),w∥$ is bounded by

$${\displaystyle \frac{ϵ}{4}}\left[{(\parallel x\parallel }^r+{\parallel x\parallel }^s+{\parallel x\parallel }^t)+{\displaystyle \frac{1}{2}}(2^r{\parallel x\parallel }^r+2^s{\parallel x\parallel }^s+2^t{\parallel x\parallel }^t)\right]\parallel w\parallel$$

for all $x,w\in A$. And by using induction method on n, we obtain that

$$\begin{array}{cc}\hfill ∥f\left(x\right)-{\displaystyle \frac{1}{2^n}}f(2^{n}x,w)∥& \le {\displaystyle \frac{ϵ}{2}}\parallel w\parallel \sum _{k=0}^{n-1}{\displaystyle \frac{1}{2^k}}\left(2^{rk}{\parallel x\parallel }^r+2^{sk}{\parallel x\parallel }^s+2^{tk}{\parallel x\parallel }^t\right)\hfill \\ \hfill & \le {\displaystyle \frac{ϵ}{4}}\parallel w\parallel \left({\displaystyle \frac{2}{2-2^r}}{\parallel x\parallel }^r+{\displaystyle \frac{2}{2-2^s}}{\parallel x\parallel }^s+{\displaystyle \frac{2}{2-2^t}}{\parallel x\parallel }^t\right)\hfill \\ \hfill & \le \left({\displaystyle \frac{{\parallel x\parallel }^r}{4-2^{r+1}}}+{\displaystyle \frac{{\parallel x\parallel }^s}{4-2^{s+1}}}+{\displaystyle \frac{{\parallel x\parallel }^t}{4-2^{t+1}}}\right)ϵ\parallel w\parallel \hfill \end{array}$$

(7)

for all $x,w\in A$. Moreover, for $m,n\in \mathbb{N}$ and for all $x\in A$, we have

$$\begin{array}{cc}\hfill ∥{\displaystyle \frac{1}{2^m}}f\left(2^{m}x\right)-{\displaystyle \frac{1}{2^n}}f\left(2^{n}x\right),w∥& =∥{\displaystyle \frac{1}{2^{m-n+n}}}f\left(2^{m-n+n}x\right)-{\displaystyle \frac{1}{2^n}}f\left(2^{n}x\right),w∥\hfill \\ \hfill & ={\displaystyle \frac{1}{2^n}}∥{\displaystyle \frac{1}{2^{m-n}}}f(2^{m-n}.2^{n}x)-f\left(2^{n}x\right),w∥\hfill \end{array}$$

is bounded by

$${\displaystyle \frac{1}{2^n}}·{\displaystyle \frac{ϵ}{4}}\sum _{k=0}^{m-n-1}{\displaystyle \frac{1}{2^k}}\left(2^{rk}\parallel 2^n{x\parallel }^r+2^{sk}\parallel 2^n{x\parallel }^s+2^{tk}{\parallel 2^{n}x\parallel }^t\right)\parallel w\parallel ,$$

for all $x,w\in A$. So, $∥{\displaystyle \frac{1}{2^m}}f\left(2^{m}x\right)-{\displaystyle \frac{1}{2^n}}f\left(2^{n}x\right),w∥$ goes to zero as $m,n$ approach to infinity, for all $x,w\in A$. Therefore, $\left\{{\displaystyle \frac{1}{2^n}}f\left(2^{n}x\right)\right\}$ is a 2-Cauchy sequence in A, for all $x\in A$. Now, we define an additive function $\mathcal{M}:A\to A$ by

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

for all $x\in A$. Then, by using (7), we can obtain

$$\begin{array}{c}\hfill ∥f\left(x\right)-\mathcal{M}\left(x\right),w∥\le \left({\displaystyle \frac{{∥x∥}^r}{4-2^{r+1}}}+{\displaystyle \frac{{∥x∥}^s}{4-2^{s+1}}}+{\displaystyle \frac{{∥x∥}^t}{4-2^{t+1}}}\right)ϵ∥w∥\end{array}$$

(8)

for all $x,w\in A$. Next, we shall to prove that the function $\mathcal{M}$ satisfies the functional Equation (2). Now, for all $x,y,z,w\in A$, one can have

$$\begin{array}{cc}\hfill ∥D_{\mathcal{M}}(x,y,z),w∥& =\underset{n\to \infty }{lim}{\displaystyle \frac{1}{2^n}}∥D_f(2^{n}x,2^{n}y,2^{n}z),w∥\hfill \\ \hfill & =\underset{n\to \infty }{lim}{\displaystyle \frac{ϵ}{2^n}}\left({∥2^{n}x∥}^r+{∥2^{n}y∥}^s+{∥2^{n}z∥}^t\right)\parallel w\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}ϵ\left(2^{(r-1)n}{\parallel x\parallel }^r+2^{(s-1)n}{\parallel y\parallel }^s+2^{(t-1)n}{\parallel z\parallel }^t\right)\parallel w\parallel =0.\hfill \end{array}$$

Thus, $∥D_{\mathcal{M}}(x,y,z),w∥=0$ for all $x,y,z,w\in A$, and it implies that $D_{\mathcal{M}}(x,y,z)=0$ for all $x,y,z\in A$. Hence, by Lemma 3, $\mathcal{M}$ is additive.

Next, we prove the uniqueness of the function $\mathcal{M}$. Let ${\mathcal{M}}^{\prime }$ be another additive function that satisfies the inequality (8). Since $\mathcal{M}$ and ${\mathcal{M}}^{\prime }$ are additive functions, we obtain

$$\mathcal{M}\left(2^{n}x\right)=2^n\mathcal{M}\left(x\right)\mathrm{and}{\mathcal{M}}^{\prime }\left(2^{n}x\right)=2^n{\mathcal{M}}^{\prime }\left(x\right)$$

for all $x\in A$. Hence, for all $x\in A$, we have

$$\begin{array}{cc}\hfill ∥\mathcal{M}\left(x\right)-{\mathcal{M}}^{\prime }\left(x\right),w∥& ={\displaystyle \frac{1}{2^n}}∥\mathcal{M}\left(2^{n}x\right)-{\mathcal{M}}^{\prime }\left(2^{n}x\right),w∥\hfill \\ \hfill & ={\displaystyle \frac{1}{2^n}}\left(∥\mathcal{M}\left(2^{n}x\right)-f\left(2^{n}x\right),w∥+∥f\left(2^{n}x\right)-{\mathcal{M}}^{\prime }\left(2^{n}x\right),w∥\right)\hfill \\ \hfill & ={\displaystyle \frac{2}{2^n}}\left({\displaystyle \frac{{∥2^{n}x∥}^r}{4-2^{r+1}}}+{\displaystyle \frac{{∥2^{n}x∥}^s}{4-2^{s+1}}}+{\displaystyle \frac{{∥2^{n}x∥}^t}{4-2^{t+1}}}\right)ϵ∥w∥\hfill \\ \hfill & =2ϵ∥w∥\left({\displaystyle \frac{2^{(r-1)n}{∥x∥}^r}{4-2^{r+1}}}+{\displaystyle \frac{2^{(s-1)n}{∥x∥}^s}{4-2^{s+1}}}+{\displaystyle \frac{2^{(t-1)n}{∥x∥}^t}{4-2^{t+1}}}\right)=0,\hfill \end{array}$$

as $n\to \infty$, where $w\in A$. Therefore, $∥\mathcal{M}\left(x\right)-{\mathcal{M}}^{\prime }\left(x\right),w∥=0$ for all $w\in A$, and $\mathcal{M}\left(x\right)={\mathcal{M}}^{\prime }\left(x\right)$ for all $x\in A$. □

Theorem 2. 

Let $ϵ\ge 0$ and $r,s,t\in (1,\infty )$. Assume that $f:(A,∥.∥)\to (A,∥{.}_{,}.∥)$ satisfies the following inequality

$$\begin{array}{c}\hfill ∥D_f(x,y,z),w∥\le ϵ\left({∥x∥}^r+{∥y∥}^s+{∥z∥}^t\right)∥w∥\end{array}$$

(9)

for all $x,y,z,w\in A.$ Then there exists a unique additive mapping $\mathcal{M}:(A,∥.∥)\to (A,∥{.}_{,}.∥)$ that satisfies the functional Equation (2) and

$$∥f\left(x\right)-\mathcal{M}\left(x\right),w∥\le ϵ\left({\displaystyle \frac{{∥x∥}^r}{2^{r+1}-4}}+{\displaystyle \frac{{∥x∥}^s}{2^{s+1}-4}}+{\displaystyle \frac{{∥x∥}^t}{2^{t+1}-4}}\right)∥w∥,$$

for all $x,w\in A$.

Proof. 

By the inequality (4) of Theorem 1, we have

$$∥f\left(2x\right)-2f\left(x\right),w∥\le {\displaystyle \frac{ϵ}{2}}\left({∥x∥}^r+{∥x∥}^s+{∥x∥}^t\right)∥w∥$$

for all $x,w\in A$. Now, replacing x by ${\displaystyle \frac{x}{2}}$ in the above inequality, we obtain

$$∥f\left(x\right)-2f\left({\displaystyle \frac{x}{2}}\right),w∥\le {\displaystyle \frac{ϵ}{2}}\left(2^{-r}{∥x∥}^r+2^{-s}{∥x∥}^s+2^{-t}{∥x∥}^t\right)∥w∥$$

for all $x,w\in A$. Again, replacing x by ${\displaystyle \frac{x}{2}}$ in the last inequality, we have

$$∥f\left({\displaystyle \frac{x}{2}}\right)-2f\left({\displaystyle \frac{x}{4}}\right),w∥\le {\displaystyle \frac{ϵ}{2}}\left(2^{-2r}{∥x∥}^r+2^{-2s}{∥x∥}^s+2^{-2t}{∥x∥}^t\right)∥w∥$$

for all $x,w\in A$. Combining the above two inequalities, we have

$$\begin{array}{cc}\hfill ∥f\left(x\right)-2^{2}f\left({\displaystyle \frac{x}{2^2}}\right),w∥& =∥f\left(x\right)-2f\left({\displaystyle \frac{x}{2}}\right),w∥+2∥f\left({\displaystyle \frac{x}{2}}\right)-2f\left({\displaystyle \frac{x}{4}}\right),w∥\hfill \\ \hfill & ={\displaystyle \frac{ϵ}{2}}[\left(2^{-r}{\parallel x\parallel }^r+2^{-s}{\parallel x\parallel }^s+2^{-t}{\parallel x\parallel }^t\right)\hfill \\ \hfill & +2\left(2^{-2r}{\parallel x\parallel }^r+2^{-2s}{\parallel x\parallel }^s+2^{-2t}{\parallel x\parallel }^t\right)]\parallel w\parallel \hfill \end{array}$$

for all $x,w\in A$. Now, we apply the induction method on n, to obtain

$$\begin{array}{cc}\hfill & ∥f\left(x\right)-2^{n}f\left({\displaystyle \frac{x}{2^n}}\right),w∥\hfill \\ \hfill & \le {\displaystyle \frac{ϵ}{2}}∥w∥\sum _{k=0}^{n-1}{2}^k\left(2^{-rk}{∥x∥}^r+2^{-sk}{∥x∥}^s+2^{-tk}{∥x∥}^t\right)\hfill \\ \hfill & ={\displaystyle \frac{ϵ}{2}}∥w∥\sum _{k=0}^{n-1}\left(2^{k(1-r)-r}{∥x∥}^r+2^{k(1-s)-s}{∥x∥}^s+2^{k(1-t)-t}{∥x∥}^t\right)\hfill \end{array}$$

bounded by

$$ϵ∥w∥\left({\displaystyle \frac{1-2^{(1-r)n}}{2^{r+1}-4}}{\parallel x\parallel }^r+{\displaystyle \frac{1-2^{(1-s)n}}{2^{s+1}-4}}{∥x∥}^s+{\displaystyle \frac{1-2^{(1-t)n}}{2^{t+1}-4}}{∥x∥}^t\right)$$

(10)

for all $x,w\in A$.

Next, for $m,n\in \mathbb{N}$ and for $x\in A$, we have

$$\begin{array}{cc}\hfill & ∥2^{m}f\left({\displaystyle \frac{x}{2^m}}\right)-2^{n}f\left({\displaystyle \frac{x}{2^n}}\right),w∥\hfill \\ \hfill & =2^n∥2^{m-n}f\left({\displaystyle \frac{x}{2^{m-n}.2^{n}x}}\right)-f\left({\displaystyle \frac{x}{2^n}}\right),w∥\hfill \\ \hfill & \le 2^n{\displaystyle \frac{ϵ}{2}}∥w∥\sum _{k=0}^{m-n-1}\left(2^{k(1-r)-r}{∥{\displaystyle \frac{x}{2^n}}∥}^r+2^{k(1-s)-s}{∥{\displaystyle \frac{x}{2^n}}∥}^s+2^{k(1-t)-t}{∥{\displaystyle \frac{x}{2^n}}∥}^t\right)\hfill \\ \hfill & =\left(2^{n(1-r)-r}\left[{\displaystyle \frac{1-2^{(1-r)(m-n)}}{1-2^{1-r}}}\right]{∥x∥}^r+2^{n(1-s)-s}\left[{\displaystyle \frac{1-2^{(1-s)(m-n)}}{1-2^{1-s}}}\right]{∥x∥}^s\right.\hfill \\ \hfill & \left.+2^{n(1-t)-t}\left[{\displaystyle \frac{1-2^{(1-t)(m-n)}}{1-2^{1-t}}}\right]{∥x∥}^t\right){\displaystyle \frac{ϵ}{2}}∥w∥\hfill \\ \hfill & ⟶0\mathrm{as}n\to \infty \hfill \end{array}$$

for all $w\in A$. Therefore, $\left\{2^{n}f\left({\displaystyle \frac{x}{2^n}}\right)\right\}$ is a 2-Cauchy sequence in A, for all $x\in A$. Since A is a 2-Banach space, the sequence $\left\{2^{n}f\left({\displaystyle \frac{x}{2^n}}\right)\right\}$ is 2-converges, for all $x\in A$. Now, we define a mapping $\mathcal{M}:A\to A$ by

$$\mathcal{M}\left(x\right):=\underset{n\to \infty }{lim}{2}^{n}f\left({\displaystyle \frac{x}{2^n}}\right),$$

for all $x\in A$. Now, with the help of (10), we obtain

$$∥f\left(x\right)-\mathcal{M}\left(x\right),w∥\le ϵ\left({\displaystyle \frac{{∥x∥}^r}{2^{r+1}-4}}+{\displaystyle \frac{{∥x∥}^s}{2^{s+1}-4}}+{\displaystyle \frac{{∥x∥}^r}{2^{r+1}-4}}\right)∥w∥,$$

for all $x,w\in A$. The further part of the proof is similar to the proof of Theorem 1. □

4. Instability of an Apollonius-Type Additive Functional Equation

In this section, we propose an example that main theorem in previous section does not hold in some normed space.

Remark 1. 

Gajda [30] showed that for $ϵ>0$ one can find a function $f:\mathbb{R}\to \mathbb{R}$ such that

$$\left|f\right(x+y)-f(x)-f(y\left)\right|\le ϵ\left(\right|x|+|y\left|\right),$$

(11)

for all $x,y\in \mathbb{R}$, but, at the same time, there is no constant $\delta >0$ and no additive function $T:\mathbb{R}\to \mathbb{R}$ satisfying the condition

$$\left|f\right(x)-T(x\left)\right|\le \delta \left|x\right|,for all x\in \mathbb{R}.$$

(12)

In next theorem, we show that Gajda’s function $f:\mathbb{R}\to \mathbb{R}$ is a counterexample for the Hyers–Ulam stability of an Apollonius-type additive functional equation.

Theorem 3. 

For $ϵ>0$, one can find a function $f:\mathbb{R}\to \mathbb{R}$ such that

$$\left|f(x+y)+2f(z-x)+2f(z-y)-4f\left(z-{\displaystyle \frac{x+y}{4}}\right)\right|\le 12ϵ\left(\right|x|+\left|y\right|+\left|z\right|)$$

for all $x,y,z\in \mathbb{R}$, but there is no additive function $A:\mathbb{R}\to \mathbb{R}$ and constant $C>0$ such that

$$\left|f\right(x)-A(x\left)\right|\le Cϵ\left|x\right|$$

for all $x\in \mathbb{R}$.

Proof. 

To prove this, we use the method of proof by contradiction. Now, let us assume that there exists an additive mapping $A:\mathbb{R}\to \mathbb{R}$ and $C>0$ such that

$$\left|f\right(x)-A(x\left)\right|\le Cϵ\left|x\right|.$$

Consider the function $f:\mathbb{R}\to \mathbb{R}$ as the Gajda’s function in Remark 1.

Since

$$\left|f\right(x+y)-f(x)-f(y\left)\right|\le ϵ\left(\right|x|+|y\left|\right)$$

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

$$f\left(0\right)=0,\left|f\right(x)+f(-x\left)\right|\le 2ϵ\left|x\right|,\mathrm{and}\left|f\right(2x)-2f(x\left)\right|\le 2ϵ\left|x\right|$$

for all $x\in \mathbb{R}$. So, we obtain

$$\begin{array}{cc}\hfill |f& (x+y)+2f(z-x)+2f(z-y)-4f\left(z-{\displaystyle \frac{x+y}{4}}\right)|\hfill \\ \hfill =& \left|f\right(x+y)-f(x)-f(y)+2f(z-x)-2f(z)-2f(-x)+2f(z-y)-2f(z)-2f(-y)\hfill \\ \hfill & -4\left[f\left(z-{\displaystyle \frac{x+y}{4}}\right)-f\left(z\right)-f\left(-{\displaystyle \frac{x+y}{4}}\right)\right]\hfill \\ \hfill & +\left(f\right(x)+f(-x\left)\right)+\left(f\right(y)+f(-y\left)\right)+\left(f\right(-x)+f(-y)-f(-x-y\left)\right)\hfill \\ \hfill & +(f(-x-y)-2f\left(-{\displaystyle \frac{x+y}{2}}\right)+2\left[f\left(-{\displaystyle \frac{x+y}{2}}\right)-2f\left(-{\displaystyle \frac{x+y}{4}}\right)\right]|\hfill \\ \hfill \le & \left|f\right(x+y)-f(x)-f(y\left)\right|+2\left|f\right(z-x)-f(z)-f(-x\left)\right|+2\left|f\right(z-y)-f(z)-f(-y\left)\right|\hfill \\ \hfill & +4|f\left(z-{\displaystyle \frac{x+y}{4}}\right)-f\left(z\right)-f\left(-{\displaystyle \frac{x+y}{4}}\right)|\hfill \\ \hfill & +\left|f\right(x)+f(-x\left)\right)+\left(f\right(y)+f(-y\left)\right|+\left|f\right(-x)+f(-y)-f(-x-y\left)\right|\hfill \\ \hfill & +|f(-x-y)-2f\left(-{\displaystyle \frac{x+y}{2}}\right)|+2|f\left(-{\displaystyle \frac{x+y}{2}}\right)-2f\left(-{\displaystyle \frac{x+y}{4}}\right)|\hfill \\ \hfill \le & 12ϵ\left(\right|x|+|y|+|z\left|\right)\hfill \end{array}$$

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

Then, by assumption, there exists an additive mapping $A:\mathbb{R}\to \mathbb{R}$ and $C>0$ such that

$$\left|f\right(x)-A(x\left)\right|\le Cϵ\left|x\right|.$$

Hence, if we set $\delta =Cϵ$ then it contradicts the result of Remark 1. □

Theorem 4. 

For $ϵ>0$, one can find a function $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ such that

$$\parallel F(\mathbf{x}+\mathbf{y})+2F(\mathbf{z}-\mathbf{x})+2F(\mathbf{z}-\mathbf{y})-4F\left(\mathbf{z}-{\displaystyle \frac{\mathbf{x}+\mathbf{y}}{4}}\right)\parallel \le 12ϵ(\parallel \mathbf{x}\parallel +\parallel \mathbf{y}\parallel +\parallel \mathbf{z}\parallel )$$

(13)

for all $\mathbf{x},\mathbf{y},\mathbf{z}\in {\mathbb{R}}^2$, but, in same time, there is no additive function $A:{\mathbb{R}}^2\to {\mathbb{R}}^2$ and constant $C>0$ such that

$$\parallel F\left(\mathbf{x}\right)-A\left(\mathbf{x}\right)\parallel \le Cϵ\parallel \mathbf{x}\parallel$$

for all $\mathbf{x}\in {\mathbb{R}}^2$.

Proof. 

We use the method of proof by contradiction. Now, let us assume that there exists an additive mapping $A:{\mathbb{R}}^2\to {\mathbb{R}}^2$ and $C>0$ such that

$$\parallel F\left(\mathbf{x}\right)-A\left(\mathbf{x}\right)\parallel \le Cϵ\parallel \mathbf{x}\parallel .$$

Consider the function $f:\mathbb{R}\to \mathbb{R}$ as Gajda’s function in Remark 1. We set the function $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ by

$$F\left(\mathbf{x}\right)\equiv (f\left(x_1\right),x_2)$$

for all $\mathbf{x}=(x_1,x_2)\in {\mathbb{R}}^2$. Then, by Theorem 3 we obtain

$$\begin{array}{cc}\hfill & \parallel F(\mathbf{x}+\mathbf{y})+2F(\mathbf{z}-\mathbf{x})+2F(\mathbf{z}-\mathbf{y})-4F\left(\mathbf{z}-{\displaystyle \frac{\mathbf{x}+\mathbf{y}}{4}}\right)\parallel \hfill \\ \hfill & =\parallel (f(x_1+y_1)+2f(z_1-x_1)+2f(z_1-y_1)-4f\left(z_1-{\displaystyle \frac{x_1+y_1}{4}}\right),0)\parallel \hfill \\ \hfill & =|f(x_1+y_1)+2f(z_1-x_1)+2f(z_1-y_1)-4f\left(z_1-{\displaystyle \frac{x_1+y_1}{4}}\right)|\hfill \\ \hfill & \le 12ϵ\left(\right|x_1|+|y_1|+|z_1\left|\right)\le 12ϵ(\parallel \mathbf{x}\parallel +\parallel \mathbf{y}\parallel +\parallel \mathbf{z}\parallel ).\hfill \end{array}$$

So, $F\left(\mathbf{x}\right)$ satisfies the inequality (13) and by assumption, we have an additive mapping $A:{\mathbb{R}}^2\to {\mathbb{R}}^2$ and a constant $C>0$ such that

$$\parallel F\left(\mathbf{x}\right)-A\left(\mathbf{x}\right)\parallel \le Cϵ\parallel \mathbf{x}\parallel .$$

Since $A:{\mathbb{R}}^2\to {\mathbb{R}}^2$ is an additive mapping, for $i=1,2$, $A_i\left(x_1\right)={\pi }_{i}A(x_1,0)$ for $\mathbf{x}=(x_1,x_2)$ also is an additive mapping, where ${\pi }_i$ is the projection mapping. And we obtain

$$\begin{array}{cc}\hfill |f\left(x_1\right)-A_1\left(x_1\right)|& \le \parallel (f\left(x_1\right)-A_1\left(x_1\right),-A_2\left(x_1\right)\parallel \hfill \\ \hfill & =\parallel F\left(\mathbf{x}\right)-A\left(\mathbf{x}\right)\parallel \mathrm{for} \mathrm{all} \mathbf{x}=(x_1,0)\in {\mathbb{R}}^2\hfill \\ \hfill & \le Cϵ\parallel \mathbf{x}\parallel =Cϵ|x_1|\mathrm{for} \mathrm{all} x_1\in \mathbb{R},\hfill \end{array}$$

which contradicts the result of Theorem 3. □

Remark 2. 

For Theorem 4, one can generalize the result to ${\mathbb{R}}^n$ by defining the mapping

$$F\left(\mathbf{x}\right)\equiv (f\left(x_1\right),x_2,···,x_n)for all \mathbf{x}=(x_1,x_2,···,x_n)\in {\mathbb{R}}^n.$$

The above function is considered as a counterexample for additive functional equation in two-dimensional normed spaces.

Now, we consider ${\mathbb{R}}^2$ as a normed space with an usual inner product and a 2-normed space as

$$\parallel \mathbf{u},\mathbf{v}\parallel \equiv \parallel \mathbf{u}\parallel \parallel \mathbf{v}\parallel cos\theta ,$$

where $\theta$ is the angle between vectors $\mathbf{u}$ and $\mathbf{v}$. In other words, we define a 2-norm by the area of the parallelogram defined by vectors $\mathbf{u}$ and $\mathbf{v}$.

Theorem 5. 

For $ϵ>0$, one can find a function $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ such that

$$\begin{array}{cc}\hfill \parallel F(\mathbf{x}+\mathbf{y})+2F(\mathbf{z}-\mathbf{x})+2F(\mathbf{z}-\mathbf{y})& - 4F\left(\mathbf{z}-{\displaystyle \frac{\mathbf{x}+\mathbf{y}}{4}}\right),\mathbf{w}\parallel \hfill \\ & \le 12ϵ(\parallel \mathbf{x}\parallel +\parallel \mathbf{y}\parallel +\parallel \mathbf{z}\parallel )\parallel \mathbf{w}\parallel ,\hfill \end{array}$$

(14)

for all $\mathbf{x},\mathbf{y},\mathbf{z},\mathbf{w}\in {\mathbb{R}}^2$, but, in the same time, there is no additive function $A:{\mathbb{R}}^2\to {\mathbb{R}}^2$ and constant $C>0$ such that

$$\parallel F\left(\mathbf{x}\right)-A\left(\mathbf{x}\right),\mathbf{w}\parallel \le {Cϵ\parallel \mathbf{x}\parallel }^3\parallel \mathbf{w}\parallel for all \mathbf{x},\mathbf{w}\in {\mathbb{R}}^2.$$

(15)

Proof. 

As with the previous two Theorems, we use the method of proof by contradiction. Now, let us assume that there exists an additive mapping $A:{\mathbb{R}}^2\to {\mathbb{R}}^2$ and $C>0$ such that

$$\parallel F\left(\mathbf{x}\right)-A\left(\mathbf{x}\right),\mathbf{w}\parallel \le {Cϵ\parallel \mathbf{x}\parallel }^3\parallel \mathbf{w}\parallel .$$

We consider $\mathbf{w}$ as the vector perpendicular to

$$F(\mathbf{x}+\mathbf{y})+2F(\mathbf{z}-\mathbf{x})+2F(\mathbf{z}-\mathbf{y})-4F\left(\mathbf{z}-{\displaystyle \frac{\mathbf{x}+\mathbf{y}}{4}}\right).$$

Then, by (14) we have

$$\parallel F(\mathbf{x}+\mathbf{y})+2F(\mathbf{z}-\mathbf{x})+2F(\mathbf{z}-\mathbf{y})-4F\left(\mathbf{z}-{\displaystyle \frac{\mathbf{x}+\mathbf{y}}{4}}\right)\parallel \le 12ϵ(\parallel \mathbf{x}\parallel +\parallel \mathbf{y}\parallel +\parallel \mathbf{z}\parallel )$$

for all $\mathbf{x},\mathbf{y},\mathbf{z}\in {\mathbb{R}}^2$. Also, we can consider $\mathbf{w}$ as the vector perpendicular to $F\left(\mathbf{x}\right)-A\left(\mathbf{x}\right)$. Then, by (15) we have

$$\parallel F\left(\mathbf{x}\right)-A\left(\mathbf{x}\right)\parallel \le {Cϵ\parallel \mathbf{x}\parallel }^3$$

for all $\mathbf{x}\in {\mathbb{R}}^2$. Hence, it contradicts Theorem 4 and we complete the proof. □

5. Conclusions

In Theorem 1, we proved the Hyers–Ulam stability of the Apollonius-type additive functional Equation (2) for $0<r,s,t<1$. In Theorem 2, we proved the Hyers–Ulam stability for $r,s,t>1$. However, if $r=s=t=1$ then we do not have the stability property. For a counterexample of stability, it has been shown in Theorem 5.

For further research, in 2-Banach spaces, we can consider Hyers–Ulam–Găvruţa stability of the Apollonius-type additive functional Equation (2).