On Asymptotic Behavior of a 2-Linear Functional Equation

Abstract

In this paper, we deal with a 2-linear functional equation. The Hyers-Ulam stability of this functional equation is shown on some restricted unbounded domains, and the obtained results are applied to get several hyperstability consequences. Moreover, some asymptotic behaviors of 2-linear functions are investigated. We also study the Hyers-Ulam stability and superstability of the 2-linear functional equation in 2-Banach spaces.

1. Introduction

Let X be a linear space on the field 𝔽 and Y be a linear space on the field 𝕂. A function $f:X\times X\to Y$ is called 2-linear if it satisfies

$$\begin{array}{cc}\hfill f(a_1{x}_1+a_2{x}_2,y)& =A_{1}f(x_1,y)+A_{2}f(x_2,y)\hfill \\ \hfill f(x,b_1{y}_1+b_2{y}_2)& =B_{1}f(x,y_1)+B_{2}f(x,y_2),x,x_1,x_2,y,y_1,y_2\in X\hfill \end{array}$$

with some $a_1,a_2,b_1,b_2\in 𝔽$ and $A_1,A_2,B_1,B_2\in 𝕂$. If $a_1=a_2=b_1=b_2=1$ and $A_1=A_2=B_1=B_2=1$ in the above definition, we obtain a 2-additive (2-Cauchy) function. n-linear (n-additive) functions are defined similarly (see [1]).

Let $a_1,a_2,b_1,b_2\in 𝔽$ and $A_{11},A_{12},A_{21},A_{22}\in 𝕂$ be given scalars. In this paper, we deal with the following general functional equation:

$$\begin{array}{cc}\hfill f(a_1{x}_1+a_2{x}_2,b_1{y}_1+b_2{y}_2)& =A_{11}f(x_1,y_1)+A_{12}f(x_1,y_2)\hfill \\ \hfill & +A_{21}f(x_2,y_1)+A_{22}f(x_2,y_2),x_1,x_2,y_1,y_2\in X.\hfill \end{array}$$

(1)

It is clear that the bi-Jensen functional equation:

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

is a special case of (1) (see [2]). Another particular case of (1) is the functional equation:

$$f(x+y,z+w)=f(x,z)+f(x,w)+f(y,z)+f(y,w),$$

(2)

which characterizes 2-additive mappings (see [3]).

Example 1.

The function $f:ℝ\times ℝ\to ℝ$ defined by $f(x,y)=axy+bx+cy+d$, where $a,b,c,d$ are real constants, is bi-Jensen.

Example 2.

The function $f:ℝ\times ℝ\to ℝ$ defined by $f(x,y)=axy$, where a is a real constant, fulfills (2).

The stability problem of homomorphisms between groups was introduced by Ulam [4] in 1940. A year later, Hyers [5] presented his solution for Banach spaces. We recall that a functional equation $(†)$ is said to be Hyers–Ulam stable in a class of functions $\mathfrak{F}$ provided each function from $\mathfrak{F}$ fulfilling approximately in $(†)$ is near to its actual solution.

In recent years, various functional equations have been introduced by many researchers and their stability has been studied. For more information on the concept of Hyers–Ulam stability and its applications, we refer the reader to [6,7,8,9,10,11,12,13,14,15,16,17,18,19,20].

In [1], the Hyers–Ulam stability of functional Equation (1) was shown in Banach spaces and m-Banach spaces. However, for the study of an asymptotic behavior of Equation (1), we must prove its stability on an unbounded set.

In this paper, we prove the Hyers–Ulam stability of (1) on some restricted unbounded domains. Then, we apply the obtained results to obtain some asymptotic behaviors of functions fulfilling (1). We also study the Hyers–Ulam stability and superstability of the functional Equation (1) in 2-Banach spaces. Our results improve Theorem 7 of [1] and its consequences.

2. Stability

We now prove the Ulam stability of functional Equation (1) in Banach spaces on some restricted domains. For convenience, we set:

$$\begin{array}{cc}\hfill Df(x_1,x_2,y_1,y_2):=& f(a_1{x}_1+a_2{x}_2,b_1{y}_1+b_2{y}_2)-A_{11}f(x_1,y_1)-A_{12}f(x_1,y_2)\hfill \\ \hfill & -A_{21}f(x_2,y_1)-A_{22}f(x_2,y_2),\hfill \end{array}$$

for a given function $f:X\times X\to \mathcal{Y}$.

Theorem 1.

Assume that $\mathcal{Y}$ is a Banach space, $\epsilon ⩾0$ and $f:X\times X\to \mathcal{Y}$ is a function satisfying

$$\begin{array}{c}\hfill \parallel Df(x_1,x_2,y_1,y_2)\parallel ⩽\epsilon ,\parallel x_1\parallel +\parallel x_2\parallel +\parallel y_1\parallel +\parallel y_2\parallel ⩾d\end{array}$$

(3)

for some $d>0$. Suppose that $min\left\{\right|A_{11}+A_{12}+A_{21}+A_{22}|,|a_1+a_2|,|b_1+b_2\left|\right\}>1$. Then, there is a unique function $\mathcal{T}:X\times X\to \mathcal{Y}$ fulfilling Equation (1) and

$$\parallel f(x,y)-\mathcal{T}(x,y)\parallel ⩽\frac{M\epsilon }{\left|A\right|-1},x,y\in X,$$

(4)

where $A:=A_{11}+A_{12}+A_{21}+A_{22}$ and $M>0$ is a constant and is dependent on $A_{11},A_{12},A_{21},A_{22}$.

Proof. 

Put $a:=a_1+a_2$ and $b:=b_1+b_2$. Letting $x_1=x_2=x$ and $y_1=y_2=y$ in (3), we obtain

$$\parallel f(ax,by)-Af(x,y)\parallel ⩽\epsilon ,\parallel x\parallel +\parallel y\parallel ⩾d.$$

Then,

$$∥\frac{f(a^{n+1}x,b^{n+1}y)}{A^{n+1}}-\frac{f(a^{m}x,b^{m}y)}{A^m}∥⩽\sum _{k=m}^n\frac{\epsilon }{{\left|A\right|}^{k+1}},\parallel x\parallel +\parallel y\parallel ⩾d,n>m⩾0.$$

(5)

Thus, for each $x,y\in X$ with $\parallel x\parallel +\parallel y\parallel ⩾d$, the sequence ${\left\{\frac{f(a^{n}x,b^{n}y)}{A^n}\right\}}_n$ is Cauchy. It is easy to infer that the sequence ${\left\{\frac{f(a^{n}x,b^{n}y)}{A^n}\right\}}_n$ is Cauchy for each $x,y\in X$. From the fact that $\mathcal{Y}$ is a Banach space, we conclude that this sequence is convergent. Define

$$\mathcal{T}:X\times X\to \mathcal{Y},\mathcal{T}(x,y):=\underset{n\to \infty }{lim}\frac{f(a^{n}x,b^{n}y)}{A^n}.$$

From the definition of $\mathcal{T}$ and (3), we obtain

$$D\mathcal{T}(x_1,x_2,y_1,y_2)=0,(x_1,x_2,y_1,y_2)\ne \mathbf{0}.$$

Since $\mathcal{T}(0,0)={lim}_{n\to \infty }\frac{f(0,0)}{A^n}=0$, we conclude

$$D\mathcal{T}(x_1,x_2,y_1,y_2)=0,x_1,x_2,y_1,y_2\in X.$$

Hence, $\mathcal{T}$ fulfills Equation (1).

Putting $m=0$ and letting $n\to +\infty$ in (5), we obtain

$$\parallel f(x,y)-\mathcal{T}(x,y)\parallel ⩽\frac{\epsilon }{\left|A\right|-1},\parallel x\parallel +\parallel y\parallel ⩾d.$$

(6)

Now, we extend (6) to the whole $X\times X$. We may assume without loss of generality that $A_{11}\ne 0$. Then, two cases arise according to whether $|a_2|+|b_2|>0$ or $a_2=b_2=0$. First take the case $|a_2|+|b_2|>0$. Let $(x_1,y_1)\in X\times X$. If $b_2\ne 0$, we choose $(x_2,y_2)\in X\times X$ such that

$$\parallel b_2{y}_2\parallel ⩾(1+|b_2\left|\right)d+\parallel b_1{y}_1\parallel \mathrm{and}\parallel x_2\parallel ⩾d+\parallel x_1\parallel .$$

It is clear that $min\{\parallel b_1{y}_1+b_2{y}_2\parallel ,\parallel y_2\parallel ,\parallel x_2\parallel \}⩾d$. For the case $a_2\ne 0$, we can choose $(x_2,y_2)\in X\times X$ such that

$$\parallel a_2{x}_2\parallel ⩾(1+|a_2\left|\right)d+\parallel a_1{x}_1\parallel \mathrm{and}\parallel y_2\parallel ⩾d+\parallel y_1\parallel .$$

In this case, $min\{\parallel a_1{x}_1+a_2{x}_2\parallel ,\parallel x_2\parallel ,\parallel y_2\parallel \}⩾d$. Then, (6) yields that

$$\begin{array}{cc}\hfill & \parallel \mathcal{T}(a_1{x}_1+a_2{x}_2,b_1{y}_1+b_2{y}_2)-f(a_1{x}_1+a_2{x}_2,b_1{y}_1+b_2{y}_2)\parallel ⩽\frac{\epsilon }{\left|A\right|-1};\hfill \\ \hfill & \parallel A_{12}f(x_1,y_2)-A_{12}\mathcal{T}(x_1,y_2)\parallel ⩽\frac{|A_{12}|\epsilon }{\left|A\right|-1};\hfill \\ \hfill & \parallel A_{21}f(x_2,y_1)-A_{21}\mathcal{T}(x_2,y_1)\parallel ⩽\frac{|A_{21}|\epsilon }{\left|A\right|-1};\hfill \\ \hfill & \parallel A_{22}f(x_2,y_2)-A_{22}\mathcal{T}(x_2,y_2)\parallel ⩽\frac{|A_{22}|\epsilon }{\left|A\right|-1}.\hfill \end{array}$$

Adding these inequalities and (3) (by using the triangle inequality), we obtain

$$\begin{array}{cc}\hfill & \parallel \mathcal{T}(a_1{x}_1+a_2{x}_2,b_1{y}_1+b_2{y}_2)-A_{12}\mathcal{T}(x_1,y_2)-A_{21}\mathcal{T}(x_2,y_1)\hfill \\ \hfill & -A_{22}\mathcal{T}(x_2,y_2)-A_{11}f(x_1,y_1)\parallel ⩽\frac{\left|A\right|+|A_{12}|+|A_{21}|+|A_{22}|}{\left|A\right|-1}\epsilon \hfill \end{array}$$

Since $\mathcal{T}$ satisfies (1), we obtain

$$\parallel \mathcal{T}(x_1,y_1)-f(x_1,y_1)\parallel ⩽\frac{\left|A\right|+|A_{12}|+|A_{21}|+|A_{22}|}{|A_{11}\left|\right(\left|A\right|-1)}\epsilon ,x_1,y_1\in X.$$

(7)

Now, consider the case $a_2=b_2=0$. Then, $a_1,b_1\ne 0$. We distinguish two cases according to whether $|A_{12}|+|A_{21}|+|A_{22}|>0$ or $A_{12}=A_{21}=A_{22}=0$. First, suppose $|A_{12}|+|A_{21}|+|A_{22}|>0$. We may assume $A_{12}\ne 0$. Using the argument above, we conclude

$$\parallel \mathcal{T}(x_1,y_1)-f(x_1,y_1)\parallel ⩽\frac{\left|A\right|+|A_{11}|+|A_{21}|+|A_{22}|}{|A_{12}\left|\right(\left|A\right|-1)}\epsilon ,x_1,y_1\in X.$$

(8)

Now, assume $A_{12}=A_{21}=A_{22}=0$. Then, $A_{11}=A$, and in this case, (3) changes as follows:

$$\parallel f(a_1{x}_1,b_1{y}_1)-Af(x_1,y_1)\parallel ⩽\epsilon ,x_1,y_1\in X.$$

Therefore,

$$\parallel f(x_1,y_1)-\mathcal{T}(x_1,y_1)\parallel ⩽\frac{\epsilon }{\left|A\right|-1},x_1,y_1\in X.$$

(9)

By (7)–(9), we obtain (4). The uniqueness of $\mathcal{T}$ follows easily from (4). □

One can apply a similar argument as in the proof of Theorem 1 and prove the following theorem.

Theorem 2.

Assume that $\mathcal{Y}$ is a Banach space, $\epsilon ⩾0,d>0$, and $f:X\times X\to \mathcal{Y}$ is a function satisfying (3) for all $\parallel x_1\parallel +\parallel x_2\parallel +\parallel y_1\parallel +\parallel y_2\parallel ⩾d$. Suppose that $A_{ij}\ne 0,|a_j|+|b_i|>0$ for some $i,j\in \{1,2\}$ and

$$min\left\{\right|A_{11}+A_{12}+A_{21}+A_{22}|,|a_1+a_2|,|b_1+b_2\left|\right\}>1.$$

Then, there is a unique mapping $\mathcal{T}:X\times X\to \mathcal{Y}$ fulfilling Equation (1) and

$$\parallel \mathcal{T}(x,y)-f(x,y)\parallel ⩽\frac{\left|A\right|-|A_{ij}|+|A_{11}|+|A_{12}|+|A_{21}|+|A_{22}|}{|A_{ij}\left|\right(\left|A\right|-1)}\epsilon ,x,y\in X,$$

where $A:=A_{11}+A_{12}+A_{21}+A_{22}$.

Remark 1.

Since $\parallel x_1\parallel +\parallel x_2\parallel +\parallel y_1\parallel +\parallel y_2\parallel ⩾4\delta$ yields $max\{\parallel x_1\parallel ,\parallel x_2\parallel ,\parallel y_1\parallel ,\parallel y_2\parallel \}⩾\delta$ ($\delta >0$), Theorems 1 and 2 are still valid if we use $max\{\parallel x_1\parallel ,\parallel x_2\parallel ,\parallel y_1\parallel ,\parallel y_2\parallel \}⩾d$ instead of the condition $\parallel x_1\parallel +\parallel x_2\parallel +\parallel y_1\parallel +\parallel y_2\parallel ⩾d$.

Corollary 1.

Suppose z is a fixed point of Y and $min\left\{\right|A|,|a_1+a_2|,|b_1+b_2\left|\right\}>1$. For a function $f:X\times X\to Y$, the following conditions are equivalent:

$\left(i\right)$

${\displaystyle \underset{\parallel x_1\parallel +\parallel x_2\parallel +\parallel y_1\parallel +\parallel y_2\parallel \to \infty }{lim}Df(x_1,x_2,y_1,y_2)=z}$;

$\left(ii\right)$

${\displaystyle \underset{max\{\parallel x_1\parallel ,\parallel x_2\parallel ,\parallel y_1\parallel ,\parallel y_2\parallel \}\to \infty }{lim}Df(x_1,x_2,y_1,y_2)=z}$;

$\left(iii\right)$

$Df(x_1,x_2,y_1,y_2)=z$, $x_1,x_2,y_1,y_2\in X$.

Proof. 

It is clear that $\left(i\right)$ and $\left(ii\right)$ are equivalent. To prove $\left(i\right)\Rightarrow \left(iii\right)$, let f satisfy $\left(i\right)$. Define

$$g:X\times X\to Y,g(x,y)=f(x,y)+\frac{1}{A-1}z.$$

Then,

$${\displaystyle \underset{\parallel x_1\parallel +\parallel x_2\parallel +\parallel y_1\parallel +\parallel y_2\parallel \to \infty }{lim}Dg(x_1,x_2,y_1,y_2)=0.}$$

Let $\epsilon >0$ be an arbitrary real number. Then, there exists $d_{\epsilon }>0$ such that

$$\parallel Dg(x_1,x_2,y_1,y_2)\parallel ⩽\epsilon ,\parallel x_1\parallel +\parallel x_2\parallel +\parallel y_1\parallel +\parallel y_2\parallel ⩾d_{\epsilon }.$$

Let $\mathcal{Y}$ be the completion of Y. In view of Theorem 1, there exists unique function ${\mathcal{T}}_{\epsilon }:X\times X\to \mathcal{Y}$ fulfilling Equation (1) and

$$\parallel f(x,y)-{\mathcal{T}}_{\epsilon }(x,y)\parallel ⩽\frac{M\epsilon }{\left|A\right|-1},x,y\in X$$

where $A:=A_{11}+A_{12}+A_{21}+A_{22}$ and $M>0$ is a constant dependent on $A_{11},A_{12},A_{21},A_{22}$. Then,

$$\begin{array}{cc}\hfill \parallel Dg(x_1,x_2,y_1,y_2)\parallel & ⩽\parallel g(a_1{x}_1+a_2{x}_2,b_1{y}_1+b_2{y}_2)-{\mathcal{T}}_{\epsilon }(a_1{x}_1+a_2{x}_2,b_1{y}_1+b_2{y}_2)\parallel \hfill \\ \hfill & +\parallel A_{11}g(x_1,y_1)-{\mathcal{T}}_{\epsilon }(x_1,y_1)\parallel +\parallel A_{12}g(x_1,y_2)-{\mathcal{T}}_{\epsilon }(x_1,y_2)\parallel \hfill \\ \hfill & +\parallel A_{21}g(x_2,y_1)-{\mathcal{T}}_{\epsilon }(x_2,y_1)\parallel +\parallel A_{22}g(x_2,y_2)-{\mathcal{T}}_{\epsilon }(x_2,y_2)\parallel \hfill \\ \hfill & ⩽\frac{5M\epsilon }{\left|A\right|-1},x_1,x_2,y_1,y_2\in X.\hfill \end{array}$$

Since $\epsilon$ is arbitrary, we obtain $Dg(x_1,x_2,y_1,y_2)=0$ for all $x_1,x_2,y_1,y_2\in X$. Then,

$$Df(x_1,x_2,y_1,y_2)=z,x_1,x_2,y_1,y_2\in X.$$

This implies $\left(iii\right)$. The implication $\left(iii\right)\Rightarrow \left(ii\right)$ is obvious. Hence, the proof is complete. □

Example 3.

Let $f:X\times X\to Y$. Then,

$$f(x+y,z+w)=f(x,z)+f(x,w)+f(y,z)+f(y,w)$$

if and only if

$$\underset{\parallel x\parallel +\parallel y\parallel +\parallel z\parallel +\parallel w\parallel \to \infty }{lim}\parallel f(x+y,z+w)-f(x,z)-f(x,w)-f(y,z)-f(y,w)\parallel =0.$$

Theorem 3.

Assume that $\mathcal{Y}$ is a Banach space, $\epsilon ⩾0$, and $f:X\times X\to \mathcal{Y}$ is a function satisfying

$$\begin{array}{c}\hfill \parallel Df(x_1,x_2,y_1,y_2)\parallel ⩽\epsilon ,min\{\parallel x_1\parallel ,\parallel x_2\parallel ,\parallel y_1\parallel ,\parallel y_2\parallel \}⩾d\end{array}$$

(10)

for some $d>0$. Suppose that $min\left\{\right|A_{11}+A_{12}+A_{21}+A_{22}|,|a_1+a_2|,|b_1+b_2\left|\right\}>1$, and $a_1,a_2,b_1,b_2\ne 0$. Then, there is a unique function $\mathcal{T}:X\times X\to \mathcal{Y}$ fulfilling Equation (1) for all $x_1,x_2,y_1,y_2\in X\backslash \left\{0\right\}$, and

$$\parallel f(x,y)-\mathcal{T}(x,y)\parallel ⩽\frac{\left|A\right|+|A_{11}|+|A_{12}|+|A_{21}|+|A_{22}|-1}{\left|A\right|-1}\epsilon ,$$

(11)

for all $x,y\in X\backslash \left\{0\right\}$, where $A:=A_{11}+A_{12}+A_{21}+A_{22}$.

Proof. 

Letting $x_1=x_2=x$ and $y_1=y_2=y$ in (10), we obtain

$$\parallel f(ax,by)-Af(x,y)\parallel ⩽\epsilon ,min\{\parallel x\parallel ,\parallel y\parallel \}⩾d$$

where $a:=a_1+a_2$ and $b:=b_1+b_2$. Then, for integers $n>m⩾0$, we obtain

$$∥\frac{f(a^{n+1}x,b^{n+1}y)}{A^{n+1}}-\frac{f(a^{m}x,b^{m}y)}{A^m}∥⩽\sum _{k=m}^n\frac{\epsilon }{{\left|A\right|}^{k+1}},min\{\parallel x\parallel ,\parallel y\parallel \}⩾d.$$

(12)

Thus, for each $x,y\in X$ with $min\{\parallel x\parallel ,\parallel y\parallel \}⩾d$, the sequence ${\left\{\frac{f(a^{n}x,b^{n}y)}{A^n}\right\}}_n$ is Cauchy. It is easy to infer that the sequence ${\left\{\frac{f(a^{n}x,b^{n}y)}{A^n}\right\}}_n$ is Cauchy for $x=y=0$ and for each $x,y\in X$ with $min\{\parallel x\parallel ,\parallel y\parallel \}>0$, and then, it is convergent since $\mathcal{Y}$ is Banach. We now show that ${\left\{\frac{f(a^{n}x,0)}{A^n}\right\}}_n$ and ${\left\{\frac{f(0,b^{n}y)}{A^n}\right\}}_n$ are convergent. Let $x\in X$ be an arbitrary element, $e\in X\backslash \{-x,0\}$ be a fixed element, and m be an integer. We take

$$x_1=\frac{(m+1)x+me}{a_1},x_2=-\frac{m(x+e)}{a_2},y_1=\frac{me}{b_1},y_2=-\frac{me}{b_2}.$$

If m is large enough, then

$$min\{\parallel x_1\parallel ,\parallel x_2\parallel ,\parallel y_1\parallel ,\parallel y_2\parallel \}⩾d\mathrm{and}(a_1{x}_1+a_2{x}_2,b_1{y}_1+b_2{y}_2)=(x,0).$$

Since the sequences ${\left\{\frac{f(a^n{x}_i,b^n{y}_j)}{A^n}\right\}}_n$ (for $i,j=1,2$) are convergent, (10) implies that ${\left\{\frac{f(a^{n}x,0)}{A^n}\right\}}_n$ is convergent. The convergence of the sequence ${\left\{\frac{f(0,b^{n}y)}{A^n}\right\}}_n$ is similarly proven. Define

$$\mathcal{T}:X\times X\to \mathcal{Y},\mathcal{T}(x,y):=\underset{n\to \infty }{lim}\frac{f(a^{n}x,b^{n}y)}{A^n}.$$

From the definition of $\mathcal{T}$ and (10), we obtain

$$D\mathcal{T}(x_1,x_2,y_1,y_2)=0,x_1,x_2,y_1,y_2\ne 0.$$

Therefore, $\mathcal{T}$ fulfills Equation (1) for all $x_1,x_2,y_1,y_2\ne 0$.

Putting $m=0$ and letting $n\to +\infty$ in (12), we obtain

$$\parallel f(x,y)-\mathcal{T}(x,y)\parallel ⩽\frac{\epsilon }{\left|A\right|-1},min\{\parallel x\parallel ,\parallel y\parallel \}⩾d.$$

(13)

Now, we extend (13) to the whole $X\times X$. Let $(x,y)\in X\times X$ be an arbitrary element, $e\in X\backslash \{-x,-y,0\}$ be a fixed element, and n be an integer. We take

$$x_1=\frac{(n+1)x+ne}{a_1},x_2=-\frac{nx+ne}{a_2},y_1=\frac{(n+1)y+ne}{b_i},y_2=-\frac{ny+ne}{b_2}.$$

If n is large enough, then

$$min\{\parallel x_1\parallel ,\parallel x_2\parallel ,\parallel y_1\parallel ,\parallel y_2\parallel \}⩾d\mathrm{and}(a_1{x}_1+a_2{x}_2,b_1{y}_1+b_2{y}_2)=(x,y).$$

By (13), we have

$$\begin{array}{cc}\hfill \parallel A_{11}f(x_1,y_1)-A_{11}\mathcal{T}(x_1,y_1)\parallel & ⩽\frac{|A_{11}|}{\left|A\right|-1}\epsilon ;\hfill \\ \hfill \parallel A_{12}f(x_1,y_2)-A_{12}\mathcal{T}(x_1,y_2)\parallel & ⩽\frac{|A_{12}|}{\left|A\right|-1}\epsilon ;\hfill \\ \hfill \parallel A_{21}f(x_2,y_1)-A_{21}\mathcal{T}(x_2,y_1)\parallel & ⩽\frac{|A_{21}|}{\left|A\right|-1}\epsilon ;\hfill \\ \hfill \parallel A_{22}f(x_2,y_2)-A_{22}\mathcal{T}(x_2,y_2)\parallel & ⩽\frac{|A_{22}|}{\left|A\right|-1}\epsilon .\hfill \end{array}$$

Adding these inequalities and (10) (by using the triangle inequality), we obtain

$$\begin{array}{cc}\hfill & \parallel f(a_1{x}_1+a_2{x}_2,b_1{y}_1+b_2{y}_2)-A_{12}\mathcal{T}(x_1,y_2)-A_{21}\mathcal{T}(x_2,y_1)\hfill \\ \hfill & -A_{22}\mathcal{T}(x_2,y_2)-A_{11}f(x_1,y_1)\parallel ⩽\frac{\left|A\right|+|A_{11}|+|A_{12}|+|A_{21}|+|A_{22}|-1}{\left|A\right|-1}\epsilon \hfill \end{array}$$

Since $\mathcal{T}$ satisfies (1) and $(a_1{x}_1+a_2{x}_2,b_1{y}_1+b_2{y}_2)=(x,y)$, we obtain

$$\parallel f(x,y)-\mathcal{T}(x,y)\parallel ⩽\frac{\left|A\right|+|A_{11}|+|A_{12}|+|A_{21}|+|A_{22}|-1}{\left|A\right|-1}\epsilon ,x,y\in X.$$

The uniqueness of $\mathcal{T}$ follows easily from (11). □

Remark 2.

Since $min\{\parallel x_1\parallel ,\parallel x_2\parallel ,\parallel y_1\parallel ,\parallel y_2\parallel \}⩾d$ implies $\parallel x_1\parallel +\parallel x_2\parallel +\parallel y_1\parallel +\parallel y_2\parallel ⩾d$, Theorem 3 is still valid if we use $\parallel x_1\parallel +\parallel x_2\parallel +\parallel y_1\parallel +\parallel y_2\parallel ⩾d$ instead of the condition $min\{\parallel x_1\parallel ,\parallel x_2\parallel ,\parallel y_1\parallel ,\parallel y_2\parallel \}⩾d$. It should be noted that in this case, Theorem 3 is somewhat different from Theorems 1 and 2, and $\mathcal{T}$ fulfilling (1) for all $x_1,x_2,y_1,y_2\in X$.

In the following corollaries, we suppose that

$$min\left\{\right|A_{11}+A_{12}+A_{21}+A_{22}|,|a_1+a_2|,|b_1+b_2\left|\right\}>1\mathrm{and}{a}_1,a_2,b_1,b_2\ne 0.$$

Corollary 2.

Suppose z is a fixed point of Y. For a function $f:X\times X\to Y$, the following conditions are equivalent:

$\left(i\right)$

${\displaystyle \underset{min\{\parallel x_1\parallel ,\parallel x_2\parallel ,\parallel y_1\parallel ,\parallel y_2\parallel \}\to \infty }{lim}Df(x_1,x_2,y_1,y_2)=z}$;

$\left(ii\right)$

$Df(x_1,x_2,y_1,y_2)=z$, $x_1,x_2,y_1,y_2\in X\backslash \left\{0\right\}$.

Corollary 3.

Let $\epsilon ⩾0$ and $p,q,r,s<0$. Suppose z is a fixed point of Y. If a function $f:X\times X\to Y$ satisfies

$$\parallel Df(x_1,x_2,y_1,y_2)-z\parallel ⩽\epsilon (\parallel x_1{\parallel }^p+\parallel x_2{\parallel }^q+\parallel y_1{\parallel }^r+\parallel y_2{\parallel }^s),x_1,x_2,y_1,y_2\in X\backslash \left\{0\right\}$$

then $Df(x_1,x_2,y_1,y_2)=z$ for all $x_1,x_2,y_1,y_2\in X\backslash \left\{0\right\}$.

Corollary 4.

Let $\epsilon ⩾0$ and $p,q,r,s$ be real numbers with $p+q+r+s<0$. Suppose z is a fixed point of Y. If a function $f:X\times X\to Y$ satisfies

$$\parallel Df(x_1,x_2,y_1,y_2)-z\parallel ⩽\epsilon \parallel x_1{\parallel }^p\parallel x_2{\parallel }^q\parallel y_1{\parallel }^r{\parallel y_2\parallel }^s,x_1,x_2,y_1,y_2\in X\backslash \left\{0\right\}$$

then $Df(x_1,x_2,y_1,y_2)=z$ for all $x_1,x_2,y_1,y_2\in X\backslash \left\{0\right\}$.

With a slight modification in the proof of Theorem 3, the following theorem is proven, and we leave the proof to the reader.

Theorem 4.

Assume that $\mathcal{Y}$ is a Banach space, $\epsilon ⩾0$, and $f:X\times X\to \mathcal{Y}$ is a function satisfying

$$\begin{array}{c}\hfill \parallel Df(x_1,x_2,y_1,y_2)\parallel ⩽\epsilon ,min\{\parallel x_1+x_2\parallel ,\parallel y_1+y_2\parallel \}⩾d\end{array}$$

(14)

for some $d>0$. Suppose that $min\left\{\right|A_{11}+A_{12}+A_{21}+A_{22}|,|a_1+a_2|,|b_1+b_2\left|\right\}>1$, and $a_1,a_2,b_1,b_2\ne 0$ with $a_1\ne a_2,b_1\ne b_2$. Then, there is a unique function $\mathcal{T}:X\times X\to \mathcal{Y}$ fulfilling Equation (1) for all $x_1,x_2,y_1,y_2\in X\backslash \left\{0\right\}$, and

$$\parallel f(x,y)-\mathcal{T}(x,y)\parallel ⩽\frac{\left|A\right|+|A_{11}|+|A_{12}|+|A_{21}|+|A_{22}|-1}{\left|A\right|-1}\epsilon ,$$

(15)

for all $x,y\in X\backslash \left\{0\right\}$, where $A:=A_{11}+A_{12}+A_{21}+A_{22}$.

In the following corollaries, we suppose that $a_1,a_2,b_1,b_2\ne 0$ with $a_1\ne a_2,b_1\ne b_2$ and

$$min\left\{\right|A_{11}+A_{12}+A_{21}+A_{22}|,|a_1+a_2|,|b_1+b_2\left|\right\}>1.$$

Corollary 5.

Suppose z is a fixed point of Y. For a function $f:X\times X\to Y$, the following conditions are equivalent:

$\left(i\right)$

${\displaystyle \underset{min\{\parallel x_1+x_2\parallel ,\parallel y_1+y_2\parallel \}\to \infty }{lim}Df(x_1,x_2,y_1,y_2)=z}$;

$\left(ii\right)$

$Df(x_1,x_2,y_1,y_2)=z$, $x_1,x_2,y_1,y_2\in X\backslash \left\{0\right\}$.

Corollary 6.

Let $\epsilon ⩾0$ and $r,s<0$. Suppose z is a fixed point of Y. If a function $f:X\times X\to Y$ satisfies

$$\parallel Df(x_1,x_2,y_1,y_2)-z\parallel ⩽\epsilon (\parallel x_1+x_2{\parallel }^r+\parallel y_1+y_2{\parallel }^s),x_1,x_2,y_1,y_2\in X\backslash \left\{0\right\}$$

then $Df(x_1,x_2,y_1,y_2)=z$ for all $x_1,x_2,y_1,y_2\in X\backslash \left\{0\right\}$.

Corollary 7.

Let $\epsilon ⩾0$ and $r,s$ be real numbers with $r+s<0$. Suppose z is a fixed point of Y. If a function $f:X\times X\to Y$ satisfies

$$\parallel Df(x_1,x_2,y_1,y_2)-z\parallel ⩽\epsilon \parallel x_1+x_2{\parallel }^r{\parallel y_1+y_2\parallel }^s,x_1,x_2,y_1,y_2\in X\backslash \left\{0\right\}$$

then $Df(x_1,x_2,y_1,y_2)=z$ for all $x_1,x_2,y_1,y_2\in X\backslash \left\{0\right\}$.

3. Superstability and Stability in 2-Banach Spaces

The concept of 2-normed spaces was introduced by Gähler [21]. First, we recall (see for instance [22]) some basic definitions and facts concerning 2-normed spaces.

Definition 1.

Let X be a real linear space of dimension greater than one and $\parallel .,.\parallel :X\times X\to ℝ$ a function satisfying the following conditions:

1. 

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

$\parallel x,y+z\parallel ⩽\parallel x,y\parallel +\parallel x,z\parallel$.

for all $x,y,z\in X$ and $\lambda \in ℝ$. In this case, $(X,\parallel .,.\parallel )$ is called a linear 2-normed space.

By $\left(1\right),\left(3\right)$ and $\left(4\right)$ we infer that

$$0=\parallel 0,y\parallel =\parallel x-x,y\parallel ⩽2\parallel x,y\parallel ,x,y\in X.$$

Hence, $\parallel .,.\parallel$ is non-negative.

A sequence ${\left\{x_n\right\}}_n$ of elements of a 2-normed space X is called a Cauchy sequence if there exist linearly independent $y,z\in X$ with

$$\underset{m,n\to \infty }{lim}\parallel x_m-x_n,y\parallel =0=\underset{m,n\to \infty }{lim}\parallel x_m-x_n,z\parallel .$$

A sequence ${\left\{x_n\right\}}_n$ of a 2-normed space X is said to be convergent if there exists $x\in X$ such that ${lim}_{n\to \infty }\parallel x_n-x,y\parallel =0$ for all $y\in X$. In this case, x is called the limit of ${\left\{x_n\right\}}_n$ and denoted by $lim{x}_n=x$. It is easy to see that in a 2-normed space, a sequence has at most one limit, and every convergent sequence is Cauchy.

A 2-normed space X is called a 2-Banach space if every Cauchy sequence in X is convergent.

Lemma 1.

Let $(X,\parallel .,.\parallel )$ be a linear 2-normed space and $y,z\in X$ be linearly independent. If $\parallel x,y\parallel =\parallel x,z\parallel =0$, then $x=0$. In particular, if $\parallel x,w\parallel =0$ for all $w\in X$, then $x=0$.

Proof. 

Since $\parallel x,y\parallel =\parallel x,z\parallel =0$, there exist $\lambda ,\mu \in ℝ$ such that $x=\lambda y$ and $x=\mu z$. Then, $\lambda y-\mu z=0$, and we conclude that $\lambda =\mu =0$. Hence, $x=0$. □

Let $(X,\parallel .,.\parallel )$ be a linear 2-normed space and $y,z\in X$ be linearly independent. It is easy to verify that the function $\parallel .\parallel :X\to ℝ$ given by $\parallel x\parallel =\parallel x,z\parallel +\parallel y,z\parallel$ is a norm on X.

The following theorem improves Theorem 7 of [1] and its consequences.

Theorem 5.

Let $\epsilon ⩾0$, $\mathcal{X}$ be a linear space and $(Y,\parallel .,.\parallel )$ a 2-normed space. If a function $f:\mathcal{X}\times \mathcal{X}\to Y$ satisfies

$$\parallel Df(x_1,x_2,y_1,y_2),z\parallel ⩽\epsilon ,x_1,x_2,y_1,y_2\in \mathcal{X},z\in Y,$$

(16)

then $Df(x_1,x_2,y_1,y_2)=0$ for all $x_1,x_2,y_1,y_2\in \mathcal{X}$.

Proof. 

Replacing z by $nz$ in (16) and dividing the resultant inequality by n, we obtain

$$\parallel Df(x_1,x_2,y_1,y_2),z\parallel ⩽\frac{\epsilon }{n},x_1,x_2,y_1,y_2\in \mathcal{X},z\in Y,n\in N.$$

Allowing n to tend to infinity, we obtain $\parallel Df(x_1,x_2,y_1,y_2),z\parallel =0$ for all $x_1,x_2,y_1,y_2\in \mathcal{X}$ and $z\in Y$. Hence, by Lemma 1, $Df(x_1,x_2,y_1,y_2)=0$ for all $x_1,x_2,y_1,y_2\in \mathcal{X}$. □

Theorem 6.

Assume that $\epsilon ,\theta ⩾0$, $\mathcal{X}$ is a normed space, and Y is a 2-Banach space. Let $g:\mathcal{X}\to Y$ be a surjective function and

$$\left|A_{11}+A_{12}+A_{21}+A_{22}\right|>1.$$

(17)

If $f:\mathcal{X}\times \mathcal{X}\to Y$ is a function satisfying

$$\begin{array}{c}\parallel Df(x_1,x_2,y_1,y_2),g\left(z\right)\parallel ⩽\epsilon +\theta \parallel z\parallel \hfill \end{array}$$

(18)

for $x_1,x_2,y_1,y_2,z\in \mathcal{X}$, then there is a unique mapping $\mathcal{F}:\mathcal{X}\times \mathcal{X}\to Y$ fulfilling Equation (1) and

$$\parallel \mathcal{F}(x,y)-f(x,y),g\left(z\right)\parallel ⩽\frac{\epsilon +\theta \parallel z\parallel }{\left|A\right|-1},x,y,z\in \mathcal{X}.$$

Proof. 

Put $a:=a_1+a_2,b:=b_1+b_2$ and $A:=A_{11}+A_{12}+A_{21}+A_{22}$. Letting $x_1=x_2=x$ and $y_1=y_2=y$ in (18), we obtain

$$\parallel f(ax,by)-Af(x,y),g\left(z\right)\parallel ⩽\epsilon +\theta \parallel z\parallel ,x,y,z\in \mathcal{X}.$$

Then,

$$∥\frac{f(a^{n+1}x,b^{n+1}y)}{A^{n+1}}-\frac{f(a^{m}x,b^{m}y)}{A^m},g\left(z\right)∥⩽\sum _{k=m}^n\frac{\epsilon +\theta \parallel z\parallel }{{\left|A\right|}^{k+1}}$$

(19)

for all $x,y,z\in \mathcal{X}$ and $n>m⩾0$.

Thus, the sequence ${\left\{\frac{f(a^{n}x,b^{n}y)}{A^n}\right\}}_n$ is Cauchy. Since $\mathcal{Y}$ is a 2-Banach space, we conclude that this sequence is convergent. Define

$$\mathcal{F}:\mathcal{X}\times \mathcal{X}\to \mathcal{Y},\mathcal{F}(x,y):=\underset{n\to \infty }{lim}\frac{f(a^{n}x,b^{n}y)}{A^n}.$$

Putting now $m=0$ and letting $n\to \infty$ in (19), we see that

$$\parallel \mathcal{F}(x,y)-f(x,y),g\left(z\right)\parallel ⩽\frac{\epsilon +\theta \parallel z\parallel }{\left|A\right|-1},x,y,z\in \mathcal{X}.$$

In view of (17), we obtain

$$∥\frac{Df(a^n{x}_1,a^n{x}_2,b^n{y}_1,b^n{y}_2)}{{\left|A\right|}^n},g\left(z\right)∥⩽\frac{\epsilon +\theta \parallel z\parallel }{{\left|A\right|}^n},x_1,x_2,y_1,y_2,z\in \mathcal{X},n\in \mathbb{N}.$$

Letting now $n\to \infty$ and applying the definition of $\mathcal{F}$, we deduce that

$$\parallel D\mathcal{F}(x_1,x_2,y_1,y_2),g\left(z\right)\parallel =0,x_1,x_2,y_1,y_2,z\in \mathcal{X}.$$

Since g is surjective, we infer $D\mathcal{F}(x_1,x_2,y_1,y_2)=0$ for all $x_1,x_2,y_1,y_2,z\in \mathcal{X}$ by Lemma 1. The uniqueness of $\mathcal{F}$ is obvious. □

4. Conclusions

In this work, we studied the following 2-linear functional equation:

$$\begin{array}{cc}\hfill f(a_1{x}_1+a_2{x}_2,b_1{y}_1+b_2{y}_2)& =A_{11}f(x_1,y_1)+A_{12}f(x_1,y_2)\hfill \\ \hfill & +A_{21}f(x_2,y_1)+A_{22}f(x_2,y_2),x_1,x_2,y_1,y_2\in X,\hfill \end{array}$$

(20)

where $f:X\times X\to Y$ is the unknown function. We established a new strategy to the study Hyers–Ulam stability of the functional Equation (20) on some restricted unbounded domains. As a consequence, we applied the obtained results to investigate several asymptotic behaviors of functions fulfilling (20). We also investigated the Hyers–Ulam stability and superstability of the functional Equation (20) in 2-Banach spaces.