Approximate Multivariable Bi-Jensen Mappings

Abstract

In this paper, we determine the stability of multivariable bi-Jensen functional equation $mnf\left({\displaystyle \frac{x_1+\cdots +x_m}{m}},{\displaystyle \frac{y_1+\cdots +y_n}{n}}\right)={\sum }_{i=1}^m{\sum }_{j=1}^{n}f(x_i,y_j)$ in Banach spaces. We establish a Hyers-Ulam-Rassias stability result using a direct method and a fixed-point scheme, respectively.

1. Introduction

Functional equations classically link algebraic structure with analytic regularity. The archetypal instance is the Cauchy equation $f(x+y)=f\left(x\right)+f\left(y\right)$, whose Hyers-Ulam stability problem asks the following question: if a function approximately satisfies a given functional equation, must it be close to an exact solution? This question was posed by S. M. Ulam [1] (origins of the stability problem posed in 1940) for group homomorphisms and first answered by D. H. Hyers [2] (1941) on Banach spaces; quantitative variants were developed by T. Aoki [3] (1950) and Th. M. Rassias [4] (1978), with further refinements by Găvruta [5] and many others. A central modern technique for proving such stability is the fixed-point method, which converts error bounds of a functional equation into the contraction of a suitable operator acting on a function space. In recent, H. Koh [6] investigate the Hyers–Ulam–Rassias stability of reciprocal functional equations in non-Archimedean fuzzy normed spaces by using both the direct method and the fixed-point alternative. Hyers–Ulam stability provides a rigorous framework for approximating exact solutions by functions satisfying functional equations up to controlled errors.

Beyond the classical Hyers-Ulam stability (and Hyers-Ulam-Rassias stability), there are other significant stability concepts applicable to functional equations.

Hyperstability: This refers to the phenomenon where a mapping satisfying a functional equation approximately (under certain conditions) turns out to be an exact solution. This is a “stronger” form of stability.

Superstability: This is often observed in exponential or multiplicative functional equations, where a bounded difference implies the function is either bounded or an exact solution.

Stability in Generalized Structures: Stability can also be investigated using criteria defined in Fuzzy Banach spaces, Probabilistic Normed spaces, or non-Archimedean spaces.

In this section, let $X,Y$ and Z be vector spaces over $\mathbb{R}$ or $\mathbb{C}$. We first recall the definition of the Cauchy-Jensen equation and related results.

Definition 1

([7]). A mapping $f:X\times Y\to Z$ is called a Cauchy-Jensen mapping if f satisfies the system of equations

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

In 2006, W.-G. Park and J.-H. Bae [7] obtained the general solution of the Cauchy-Jensen functional equation

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

(1)

and its stability. Subsequent papers have been published since 2007 by several authors [8,9,10].

In 2012, J.-H. Bae and W.-G. Park [11] introduced the following multivariable Cauchy-Jensen functional equation:

$$nf\left(x_1+\cdots +x_n,{\displaystyle \frac{y_1+\cdots +y_n}{n}}\right)=\sum _{1\le i,j\le n}f(x_i,y_j),$$

(2)

where n is an integer greater than 1, and we observed that the functional Equations (1) and (2) are equivalent. The functional Equation (1) is a special case of functional Equation (2). In 2024, J.-H. Bae and W.-G. Park [12] obtained the stability of the multivariable Cauchy-Jensen functional Equation (2) in Banach spaces, quasi-Banach spaces, and normed two-Banach spaces.

Next, we introduce the bi-Jensen equation and its generalizations.

Definition 2

([13]). A mapping $f:X\times Y\to Z$ is called a bi-Jensen mapping if f satisfies the following system of equations:

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

(3)

In 2006, J.-H. Bae and W.-G. Park [13] obtained the general solution of the bi-Jensen functional equation

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

(4)

and its stability. In 2023, J.-H. Bae and W.-G. Park [14] introduced the following multivariable bi-Jensen functional equation:

$$n^{2}f\left({\displaystyle \frac{x_1+\cdots +x_n}{n}},{\displaystyle \frac{y_1+\cdots +y_n}{n}}\right)=\sum _{i=1}^n\sum _{j=1}^{n}f(x_i,y_j),$$

(5)

where n is an integer greater than 1. We also observe that the functional Equations (4) and (5) have the same solutions. The functional Equation (4) is the special case of functional Equation (5).

In 2024, A. Bodaghi and A. Sahami [15] introduced an alternative presentation of the multi-Jensen equation.

For two integers $m,n$ greater than 1, we consider the bi-Jensen functional equation

$$mn f\left({\displaystyle \frac{x_1+\cdots +x_m}{m}},{\displaystyle \frac{y_1+\cdots +y_n}{n}}\right)=\sum _{i=1}^m\sum _{j=1}^{n}f(x_i,y_j).$$

(6)

In this paper, we investigate the Hyers-Ulam-Rassias stability of the multivariable bi-Jensen functional Equation (6) in Banach spaces. Our main results establish the existence of a unique bi-additive mapping approximating a given function satisfying (6) approximately. The proofs are obtained via a direct method and a fixed-point approach, respectively.

2. Main Results

In this section, $\mathcal{X},\mathcal{Y}$ and $\mathcal{Z}$ are normed spaces.

Lemma 1. 

Let $f:\mathcal{X}\times \mathcal{Y}\to \mathcal{Z}$ satisfy (6). And let $g_x:\mathcal{Y}\to \mathcal{Z}$ and $g_y^{\prime }:\mathcal{X}\to \mathcal{Z}$ be given by $g_x\left(y\right):=f(x,y)-f(x,0)$ and $g_y^{\prime }\left(x\right):=f(x,y)-f(0,y)$ for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$. Then, $g_x$ is additive for all $x\in \mathcal{X}$, and $g_y^{\prime }$ is additive for all $y\in \mathcal{Y}$.

Proof. 

Letting $x_1=x_2=\cdots =x_m=x$, $y_1=y$ and $y_2=\cdots =y_n=0$ in (6), we have

$$nf\left(x,{\displaystyle \frac{y}{n}}\right)=f(x,y)+(n-1)f(x,0)$$

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$. So, we have

$$n{g}_x\left({\displaystyle \frac{y}{n}}\right)=nf\left(x,{\displaystyle \frac{y}{n}}\right)-nf(x,0)=f(x,y)-f(x,0)=g_x\left(y\right)$$

(7)

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$. Putting $x_1=x_2=\cdots =x_m=x$ and $y_3=\cdots =y_n=0$ in (6), we have

$$nf\left(x,{\displaystyle \frac{y_1+y_2}{n}}\right)=f(x,y_1)+f(x,y_2)+(n-2)f(x,0)$$

for all $x\in \mathcal{X}$ and all $y_1,y_2\in \mathcal{Y}$. So, we have

$$\begin{array}{cc}\hfill n{g}_x\left({\displaystyle \frac{y_1+y_2}{n}}\right)& =nf\left(x,{\displaystyle \frac{y_1+y_2}{n}}\right)-nf(x,0)=f(x,y_1)+f(x,y_2)-2f(x,0)\hfill \\ & =g_x\left(y_1\right)+g_x\left(y_2\right)\hfill \end{array}$$

for all $x\in \mathcal{X}$ and all $y_1,y_2\in \mathcal{Y}$. Given Equation (7) and the above equation, we know that $g_x$ is additive for all $x\in \mathcal{X}$. Similarly, we also know that $g_y^{\prime }$ is additive for all $y\in \mathcal{Y}$. □

Theorem 1. 

A mapping $f:\mathcal{X}\times \mathcal{Y}\to \mathcal{Z}$ satisfies (3) if and only if it satisfies (6).

Proof. 

First, we assume that f satisfies (3). As shown in [13], there exists a bi-additive mapping, $B:\mathcal{X}\times \mathcal{Y}\to \mathcal{Z}$, and two additive mappings, $A:\mathcal{X}\to \mathcal{Z}$ and $A^{\prime }:\mathcal{Y}\to \mathcal{Z}$, such that $f(x,y)=B(x,y)+A\left(x\right)+A^{\prime }\left(y\right)+f(0,0)$ for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$. Thus, we have

$$\begin{array}{cc}\hfill mnf& ({\displaystyle \frac{x_1+\cdots +x_m}{m}},{\displaystyle \frac{y_1+\cdots +y_n}{n}})\hfill \\ & =mnB\left({\displaystyle \frac{1}{m}}\sum _{i=1}^m{x}_i,{\displaystyle \frac{1}{n}}\sum _{j=1}^n{y}_j\right)+mnA\left({\displaystyle \frac{1}{m}}\sum _{i=1}^m{x}_i\right)+mn{A}^{\prime }\left({\displaystyle \frac{1}{n}}\sum _{j=1}^n{y}_j\right)+mnf(0,0)\hfill \\ & =\sum _{i=1}^m\sum _{j=1}^{n}B(x_i,y_j)+n\sum _{i=1}^{m}A\left(x_i\right)+m\sum _{j=1}^n{A}^{\prime }\left(y_j\right)+mnf(0,0)\hfill \\ & =\sum _{i=1}^m\sum _{j=1}^{n}B(x_i,y_j)+\sum _{i=1}^m\sum _{j=1}^{n}A\left(x_i\right)+\sum _{i=1}^m\sum _{j=1}^n{A}^{\prime }\left(y_j\right)+\sum _{i=1}^m\sum _{j=1}^{n}f(0,0)\hfill \\ & =\sum _{i=1}^m\sum _{j=1}^{n}f(x_i,y_j)\hfill \end{array}$$

for all $x_1,\cdots ,x_m\in \mathcal{X}$ and all $y_1,\cdots ,y_n\in \mathcal{Y}$. That is, f satisfies (6).

Conversely, assume that f satisfies (6). Define $g_x:\mathcal{Y}\to \mathcal{Z}$ and $g_y^{\prime }:\mathcal{X}\to \mathcal{Z}$ by $g_x\left(y\right):=f(x,y)-f(x,0)$ and $g_y^{\prime }\left(x\right):=f(x,y)-f(0,y)$ for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$. As shown in Lemma 1, $g_x$ is additive for all $x\in \mathcal{X}$, and $g_y^{\prime }$ is additive for all $y\in \mathcal{Y}$. Thus, we get

$$2g_x\left({\displaystyle \frac{y+z}{2}}\right)=g_x\left(y\right)+g_x\left(z\right)$$

and

$$2f\left(x,{\displaystyle \frac{y+z}{2}}\right)=2g_x\left({\displaystyle \frac{y+z}{2}}\right)+2f(x,0)=f(x,y)+f(x,z)$$

for all $x\in \mathcal{X}$ and all $y,z\in \mathcal{Y}$. Similarly, we get

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

for all $x,y\in \mathcal{X}$ and all $z\in \mathcal{Y}$. That is, f satisfies (3). □

From now on, let $\mathcal{Z}$ be complete.

Theorem 2. 

Let $r>0$ with $mn>\max \{m^r,n^r\}$, and let $f:\mathcal{X}\times \mathcal{Y}\to \mathcal{Z}$ be a mapping satisfying $f(x,0)=0$ for all $x\in \mathcal{X}$ such that

$$∥mnf\left({\displaystyle \frac{x_1+\cdots +x_m}{m}},{\displaystyle \frac{y_1+\cdots +y_n}{n}}\right)-\sum _{i=1}^m\sum _{j=1}^{n}f(x_i,y_j)∥\le \epsilon +\delta \left(\sum _{i=1}^m\parallel x_i{\parallel }^r+\sum _{j=1}^n{\parallel y_j\parallel }^r\right)$$

(8)

for all $x_1,\cdots ,x_m\in \mathcal{X}$ and all $y_1,\cdots ,y_n\in \mathcal{Y}$. If $g(x,y):=f(x,y)-f(0,y)$ for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$, then there exists a unique bi-additive mapping $F:\mathcal{X}\times \mathcal{Y}\to \mathcal{Z}$ such that

$$\parallel g(x,y)-F(x,y)\parallel \le {\displaystyle \frac{2\epsilon }{mn-1}}+\delta \left({\displaystyle \frac{m^r}{mn-m^r}}{\parallel x\parallel }^r+{\displaystyle \frac{2n^r}{mn-n^r}}{\parallel y\parallel }^r\right)$$

(9)

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$.

Proof. 

Note that $g(0,y)=0$ for all $y\in \mathcal{Y}$. Taking $x_1=m^{k+1}x,x_2=\cdots =x_m=0$ and $y_1=n^{k+1}y,y_2=\cdots =y_n=0$ in (8), we have

$$\begin{array}{c}∥mnf\left({\displaystyle \frac{m^{k+1}x+0+\cdots +0}{m}},{\displaystyle \frac{n^{k+1}y+0+\cdots +0}{n}}\right)-\sum _{i=1}^m\left[f\left(x_i,n^{k+1}y\right)+(n-1)f\left(x_i,0\right)\right]∥\hfill \\ =∥mnf\left({\displaystyle \frac{m^{k+1}x}{m}},{\displaystyle \frac{n^{k+1}y}{n}}\right)-\sum _{i=1}^m\left[f\left(x_i,n^{k+1}y\right)+(n-1)0\right]∥\hfill \\ =∥mnf\left(m^{k}x,n^{k}y\right)-\sum _{i=1}^{m}f\left(x_i,n^{k+1}y\right)∥\hfill \\ =∥mnf\left(m^{k}x,n^{k}y\right)-f\left(m^{k+1}x,n^{k+1}y\right)-\sum _{i=2}^{m}f\left(0,n^{k+1}y\right)∥\hfill \\ =∥mnf\left(m^{k}x,n^{k}y\right)-f\left(m^{k+1}x,n^{k+1}y\right)-(m-1)f\left(0,n^{k+1}y\right)∥\hfill \\ \le \epsilon +\delta \left(m^{r(k+1)}{\parallel x\parallel }^r+n^{r(k+1)}{\parallel y\parallel }^r\right)\hfill \end{array}$$

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$ and all non-negative integers k. Letting $x=0$ in the above inequality, we have

$$m∥nf\left(0,n^{k}y\right)-f\left(0,n^{k+1}y\right)∥\le \epsilon +\delta n^{r(k+1)}{\parallel y\parallel }^r$$

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$ and all k. Given the above two inequalities, we obtain that

$$\begin{array}{c}∥mng\left(m^{k}x,n^{k}y\right)-g\left(m^{k+1}x,n^{k+1}y\right)∥\hfill \\ =∥mnf\left(m^{k}x,n^{k}y\right)-f\left(m^{k+1}x,n^{k+1}y\right)+f\left(0,n^{k+1}y\right)-mnf\left(0,n^{k}y\right)∥\hfill \\ \le ∥mnf\left(m^{k}x,n^{k}y\right)-f\left(m^{k+1}x,n^{k+1}y\right)-(m-1)f\left(0,n^{k+1}y\right)∥\hfill \\ +∥mf\left(0,n^{k+1}y\right)-mnf\left(0,n^{k}y\right)∥\hfill \\ \le \epsilon +\delta \left(m^{r(k+1)}{\parallel x\parallel }^r+n^{r(k+1)}{\parallel y\parallel }^r\right)+m∥f\left(0,n^{k+1}y\right)-nf\left(0,n^{k}y\right)∥\hfill \\ \le 2\epsilon +\delta \left(m^{r(k+1)}{\parallel x\parallel }^r+2n^{r(k+1)}{\parallel y\parallel }^r\right)\hfill \end{array}$$

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$ and all k. If we denote for simplicity

$$a_k={\displaystyle \frac{1}{{\left(mn\right)}^k}}g\left(m^{k}x,n^{k}y\right),$$

dividing ${\left(mn\right)}^{k+1}$ in the above inequality, we have

$$\begin{array}{cc}\hfill \parallel a_k-a_{k+1}\parallel =\parallel {\displaystyle \frac{1}{{\left(mn\right)}^k}}g\left(m^{k}x,n^{k}y\right)& - {\displaystyle \frac{1}{{\left(mn\right)}^{k+1}}}g\left(m^{k+1}x,n^{k+1}y\right)\parallel \hfill \\ & \le {\displaystyle \frac{1}{{\left(mn\right)}^{k+1}}}\left[2\epsilon +\delta \left(m^{r(k+1)}{\parallel x\parallel }^r+2n^{r(k+1)}{\parallel y\parallel }^r\right)\right]\hfill \end{array}$$

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$ and all k. Given the above inequality, we also have

$$\parallel a_l-a_p\parallel \le \sum _{k=l}^{p-1}{\displaystyle \frac{1}{{\left(mn\right)}^{k+1}}}\left[2\epsilon +\delta \left(m^{r(k+1)}{\parallel x\parallel }^r+2n^{r(k+1)}{\parallel y\parallel }^r\right)\right]$$

(10)

for all integers $l,p(0\le l<p)$ and all $(x,y)\in \mathcal{X}\times \mathcal{Y}$. Since $mn>\max \{m^r,n^r\}$, we have

$$\begin{array}{cc}& \sum _{k=0}^{\infty }{\displaystyle \frac{1}{{\left(mn\right)}^{k+1}}}\left[2\epsilon +\delta \left(m^{r(k+1)}{\parallel x\parallel }^r+2n^{r(k+1)}{\parallel y\parallel }^r\right)\right]\hfill \\ \hfill =& \sum _{k=0}^{\infty }{\displaystyle \frac{2\epsilon }{{\left(mn\right)}^{k+1}}}+\delta \left[{\parallel x\parallel }^r\sum _{k=0}^{\infty }{\left({\displaystyle \frac{m^r}{mn}}\right)}^{k+1}+2{\parallel y\parallel }^r\sum _{k=0}^{\infty }{\left({\displaystyle \frac{n^r}{mn}}\right)}^{k+1}\right]\hfill \\ \hfill =& {\displaystyle \frac{2\epsilon }{mn-1}}+\delta \left({\displaystyle \frac{m^r}{mn-m^r}}{\parallel x\parallel }^r+{\displaystyle \frac{2n^r}{mn-n^r}}{\parallel y\parallel }^r\right)<\infty .\hfill \end{array}$$

(11)

Thus we have

$$\sum _{k=l}^{p-1}{\displaystyle \frac{1}{{\left(mn\right)}^{k+1}}}\left[2\epsilon +\delta \left(m^{r(k+1)}{\parallel x\parallel }^r+2n^{r(k+1)}{\parallel y\parallel }^r\right)\right]\to 0$$

as $l,p\to \infty$, that is, $\parallel a_l-a_p\parallel \to 0$ as $l,p\to \infty$. Hence, the sequence $\left\{{\displaystyle \frac{1}{{\left(mn\right)}^k}}g\left(m^{k}x,n^{k}y\right)\right\}$ is a Cauchy sequence. Since $\mathcal{Z}$ is complete, we can define $F:\mathcal{X}\times \mathcal{Y}\to \mathcal{Z}$ by

$$F(x,y):=\underset{k\to \infty }{\lim }{\displaystyle \frac{1}{{\left(mn\right)}^k}}g\left(m^{k}x,n^{k}y\right)$$

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$. By (11), the inequality (9) holds.

Let $G:\mathcal{X}\times \mathcal{Y}\to \mathcal{Z}$ be another bi-additive mapping satisfying (9). Then,

$$\begin{array}{cc}\hfill \parallel F(x,y)& - G(x,y)\parallel =\underset{k\to \infty }{\lim }{\displaystyle \frac{1}{{\left(mn\right)}^k}}∥g\left(m^{k}x,n^{k}y\right)-G\left(m^{k}x,n^{k}y\right)∥\hfill \\ & \le \underset{k\to \infty }{\lim }{\displaystyle \frac{1}{{\left(mn\right)}^k}}\left[{\displaystyle \frac{2\epsilon }{mn-1}}+\delta \left({\displaystyle \frac{m^r}{mn-m^r}}{∥m^{k}x∥}^r+{\displaystyle \frac{2n^r}{mn-n^r}}{∥n^{k}y∥}^r\right)\right]=0\hfill \end{array}$$

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$. □

We obtain Theorem 2 in [14] as a corollary.

Corollary 1. 

Let $0<r<2$ and $f:\mathcal{X}\times \mathcal{Y}\to \mathcal{Z}$ be a mapping such that

$$∥n^{2}f\left({\displaystyle \frac{x_1+\cdots +x_n}{n}},{\displaystyle \frac{y_1+\cdots +y_n}{n}}\right)-\sum _{i=1}^n\sum _{j=1}^{n}f(x_i,y_j)∥\le \epsilon +\delta \sum _{i=1}^n\left(\parallel x_i{\parallel }^r+{\parallel y_i\parallel }^r\right)$$

for all $x_1,\cdots ,x_n\in \mathcal{X}$ and all $y_1,\cdots ,y_n\in \mathcal{Y}$. If $g(x,y):=f(x,y)-f(0,y)$ for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$; then, there exists a unique bi-additive mapping $F:\mathcal{X}\times \mathcal{Y}\to \mathcal{Z}$ such that

$$\parallel g(x,y)-F(x,y)\parallel \le {\displaystyle \frac{2\epsilon }{n^2-1}}+{\displaystyle \frac{\delta n^r}{n^2-n^r}}\left({\parallel x\parallel }^r+2{\parallel y\parallel }^r\right)$$

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$.

While Theorem 2 establishes the theoretical existence of a unique bi-additive approximation, it is instructive to provide a concrete realization of such a mapping. The following example demonstrates that the inner product in a Hilbert space serves as a canonical exact solution to the functional Equation (6). This validates that the functional equation is consistent with fundamental geometric structures in mathematics.

Example 1. 

Let $\mathcal{H}$ be a complex Hilbert space with the inner product $\langle ·,·\rangle$. For $\lambda \in \mathbb{C}$ and $a,b\in \mathcal{H}$, define a mapping $f:\mathcal{H}\times \mathcal{H}\to \mathbb{C}$ by

$$f(x,y)=\lambda \langle x+a,y+b\rangle .$$

We can easily verify that f satisfies the Equation (6) exactly:

$$\begin{array}{cc}\hfill & mnf\left({\displaystyle \frac{1}{m}}\sum _{i=1}^m{x}_i,{\displaystyle \frac{1}{n}}\sum _{j=1}^n{y}_j\right)=mn·\lambda 〈{\displaystyle \frac{1}{m}}\sum _{i=1}^m{x}_i+a,{\displaystyle \frac{1}{n}}\sum _{j=1}^n{y}_j+b〉\hfill \\ \hfill & =mn·\lambda ·\left({\displaystyle \frac{1}{mn}}\sum _{i=1}^m\sum _{j=1}^n\langle x_i,y_j\rangle +{\displaystyle \frac{1}{m}}\sum _{i=1}^m\langle x_i,b\rangle +{\displaystyle \frac{1}{n}}\sum _{j=1}^n\langle a,y_j\rangle +\langle a,b\rangle \right)\hfill \\ \hfill & =\lambda \sum _{i=1}^m\sum _{j=1}^n\left(\langle x_i,y_j\rangle +\langle x_i,b\rangle +\langle a,y_j\rangle +\langle a,b\rangle \right)=\sum _{i=1}^m\sum _{j=1}^{n}f(x_i,y_j)\hfill \end{array}$$

for all $x_1,\cdots ,x_m,y_1,\cdots ,y_n\in \mathcal{H}$. Note that the inner product is linear in the first argument and is conjugate linear in the second argument. If $a=b=0$, f is bi-additive. For the case $\lambda =1$ and $a=b=0$, we know that the inner product function is an exact solution of the Equation (6).

Example 2. 

Let $r=2$ and $m=n$. We show that the stability of the functional Equation (6) fails in this singular case. Let $\varphi :\mathbb{R}\to \mathbb{R}$ be a bounded function defined by

$$\varphi \left(t\right)=\left\{\begin{array}{cc}C{t}^2,\hfill & \left|t\right|<1,\hfill \\ C,\hfill & \left|t\right|\ge 1,\hfill \end{array}\right.$$

where C is a constant. Consider the function $f:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ defined by

$$f(x,y):=\sum _{k=0}^{\infty }{\displaystyle \frac{\varphi \left(n^{k}x\right)+\varphi \left(n^{k}y\right)}{n^{2k}}}$$

for all $x,y\in \mathbb{R}$. The function f satisfies the inequality (8) with $r=2$ (the error is bounded by a quadratic term). However, there exists no bi-additive mapping F that approximates f in the sense of Theorem 2. In the proof of Theorem 2, the uniqueness and existence of F rely on the convergence of the sequence $\left\{{\displaystyle \frac{1}{n^{2k}}}f\left(n^{k}x,n^{k}y\right)\right\}$. For $r=2$, the scaling factor of the error term $(n^r=n^2)$ is equal to the scaling factor of the operator $(mn=n^2)$. Consequently, the geometric series in the error estimation becomes

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{n^r}{mn}}\right)}^k=\sum _{k=0}^{\infty }{\left({\displaystyle \frac{n^2}{n^2}}\right)}^k=\sum _{k=0}^{\infty }1,$$

which diverges to infinity. This justifies the condition $mn>\max \{m^r,n^r\}$ assumed in Theorem 2.

Example 3 (Bounded Perturbation).

Let $\mathcal{X}=\mathcal{Y}=\mathcal{Z}=\mathbb{R}$. Consider the mapping $f:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ defined by

$$f(x,y)=xy+\epsilon \sin (x+y),$$

where $\epsilon >0$ is a constant. Note that $\sin (x+y)$ is a bounded non-linear function. We observe that the error term for the functional Equation (6) is bounded. Specifically, since $\left|\sin \right(·\left)\right|\le 1$, the defect of f in the equation is bounded by a constant dependent on $m,n,ϵ$. This corresponds to the stability condition (8) with $\delta =0$ (generalized Hyers-Ulam stability) or can be viewed as satisfying the condition with arbitrary $r>0$ for sufficiently large δ. According to Theorem 2, the unique bi-additive mapping F approximating f is given by the limit formula (11). Let us compute it directly:

$$\begin{array}{cc}\hfill F(x,y)& =\underset{k\to \infty }{\lim }{\displaystyle \frac{1}{{\left(mn\right)}^k}}g(m^{k}x,n^{k}y)\hfill \\ & =\underset{k\to \infty }{\lim }{\displaystyle \frac{1}{{\left(mn\right)}^k}}\left[\left(m^{k}x\right)\left(n^{k}y\right)+\epsilon \sin (m^{k}x+n^{k}y)-\epsilon \sin \left(n^{k}y\right)\right]\hfill \\ & =\underset{k\to \infty }{\lim }\left[{\displaystyle \frac{{\left(mn\right)}^{k}xy}{{\left(mn\right)}^k}}+{\displaystyle \frac{\epsilon \sin (m^{k}x+n^{k}y)}{{\left(mn\right)}^k}}-{\displaystyle \frac{\epsilon \sin \left(n^{k}y\right)}{{\left(mn\right)}^k}}\right]\hfill \\ & =xy.\hfill \end{array}$$

Since the sine terms are bounded, they vanish as ${\left(mn\right)}^k\to \infty$. Thus, our theorem correctly filters out the bounded “noise” $\epsilon \sin (x+y)$ and recovers the exact bi-additive core $xy$.

Example 4 (Unbounded Logarithmic Perturbation).

Let $\mathcal{X}=\mathcal{Y}=\mathcal{Z}=\mathbb{R}$. Consider the mapping $f:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ defined by

$$f(x,y)=xy+\epsilon \ln (1+|x|+|y\left|\right),$$

where $\epsilon >0$ is a constant. Unlike bounded perturbations, the error term $\ln (1+|x|+|y\left|\right)\to \infty$ as $\left|x\right|,\left|y\right|\to \infty$. Let $r\in (0,1]$ be a constant; then, the function $h\left(u\right)=u^r$ is concave and so sub-additive for $u\ge 0$. This implies that $\left(\right|x|+{\left|y\right|)}^r\le {\left|x\right|}^r+{\left|y\right|}^r$ for all $x,y\in \mathbb{R}$. For a given $x,y\in \mathbb{R}$, let $u=\left|x\right|+\left|y\right|$. Then, we have

$$\underset{u\to \infty }{\lim }{\displaystyle \frac{\ln (1+u)}{u^r}}=\underset{u\to \infty }{\lim }{\displaystyle \frac{\frac{1}{1+u}}{r{u}^{r-1}}}=\underset{u\to \infty }{\lim }{\displaystyle \frac{1}{r(u^r+u^{r-1})}}=0.$$

Because the limit is 0, there exists some $M>0$ such that ${\displaystyle \frac{\ln (1+u)}{u^r}}\le 1$ for all $u>M$. Thus, $\ln (1+|x|+|y\left|\right)\le {\left|x\right|}^r+{\left|y\right|}^r$ for sufficiently large inputs. Thus, f satisfies the stability inequality (8) with the power r. Since $r\le 1$, the conditions $m^r<mn$ and $n^r<mn$ hold for any integers $m,n\ge 2$. Applying Theorem 2, we compute the unique bi-additive approximation $F(x,y)$:

$$\begin{array}{ccc}\hfill F(x,y)& =& \underset{k\to \infty }{\lim }{\displaystyle \frac{1}{{\left(mn\right)}^k}}g(m^{k}x,n^{k}y)\hfill \\ & =& \underset{k\to \infty }{\lim }{\displaystyle \frac{1}{{\left(mn\right)}^k}}\left[\left(m^{k}x\right)\left(n^{k}y\right)+\epsilon \ln {\displaystyle \frac{1+m^k\left|x\right|+n^k\left|y\right|}{1+n^k\left|y\right|}}\right]\hfill \\ & =& xy+\epsilon \underset{k\to \infty }{\lim }{\displaystyle \frac{1}{{\left(mn\right)}^k}}\ln {\displaystyle \frac{1+m^k\left|x\right|+n^k\left|y\right|}{1+n^k\left|y\right|}}=xy.\hfill \end{array}$$

The limit of the logarithmic term is zero because the exponential growth of the denominator ${\left(mn\right)}^k$ dominates the linear growth inside the logarithm. This example demonstrates that our stability result is robust enough to handle unbounded perturbations that grow slower than the bi-additive scaling.

Definition 3.

Let $\mathsf{\Omega }$ be a set. A function $d:\mathsf{\Omega }\times \mathsf{\Omega }\to [0,\infty ]$ is called a generalized metric on $\mathsf{\Omega }$ if d satisfies

(1) 

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

(2) 

$d(x,y)=d(y,x)$ for all $x,y\in \mathsf{\Omega }$;

(3) 

$d(x,z)\le d(x,y)+d(y,z)$ for all $x,y,z\in \mathsf{\Omega }$.

In this case, $(\mathsf{\Omega },d)$ is called a generalized metric space.

Note that the only substantial difference between the generalized metric and the metric is that the range of the generalized metric includes infinity.

Theorem 3

(The alternative of fixed point [16]). Suppose that we are given a complete generalized metric space $(\mathsf{\Omega },d)$ and a strictly contractive mapping $T:\mathsf{\Omega }\to \mathsf{\Omega }$ with Lipschitz constant L. Then, for each given $x\in \mathsf{\Omega }$, either

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

• or
• there exists a positive integer $n_0$ such that

• $d\left(T^{n}x,T^{n+1}x\right)<\infty$ for all $n\ge n_0;$
• the sequence $\left(T^{n}x\right)$ is convergent to a fixed point $y^{*}$ of $T;$
• $y^{*}$ is the unique fixed point of $T$ in the set $\Delta =\{y\in \mathsf{\Omega }|d(T^{n_0}x,y)<\infty \};$
• $d(y,y^{*})\le {\displaystyle \frac{1}{1-L}}d(y,Ty)$ for all $y\in \Delta$.

Theorem 4. 

Let $r>0$ with $mn>\max \{m^r,n^r\}$, and let $\epsilon >0,\delta \ge 0$. Suppose that the mapping $f:\mathcal{X}\times \mathcal{Y}\to \mathcal{Z}$ satisfies the inequality (8). If $f(x,0)=0$ for all $x\in \mathcal{X}$; then, there exists a unique bi-additive mapping $F:\mathcal{X}\times \mathcal{Y}\to \mathcal{Z}$ such that

$$\parallel f(x,y)-f(0,y)-F(x,y)\parallel \le {\displaystyle \frac{L}{1-L}}\mathsf{\Phi }(x,y)$$

(12)

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$, where $L:={\displaystyle \frac{1}{mn}}\max \{m^r,n^r\}$ and the control function $\mathsf{\Phi }:\mathcal{X}\times \mathcal{Y}\to [0,\infty )$ is given by

$$\mathsf{\Phi }(x,y):=2\epsilon +\delta (\parallel x{\parallel }^r+2\parallel y{\parallel }^r)$$

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$.

Proof. 

Put $g(x,y):=f(x,y)-f(0,y)$ for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$. Since $f(x,0)=0$, we must have $g(x,0)=0$ for all $x\in \mathcal{X}$. Letting $x_1=mx,x_2=\cdots =x_m=0$ and $y_1=ny$, $y_2=\cdots =y_n=0$ in (8), we have

$$∥g(x,y)-{\displaystyle \frac{1}{mn}}g(mx,ny)∥\le {\displaystyle \frac{1}{mn}}{[2\epsilon +\delta (\parallel mx\parallel }^r+{2\parallel ny\parallel }^r\left)\right]={\displaystyle \frac{1}{mn}}\mathsf{\Phi }(mx,ny)\le L\mathsf{\Phi }(x,y)$$

(13)

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$. Let $\mathsf{\Omega }$ be the set of all the mappings $h:\mathcal{X}\times \mathcal{Y}\to \mathcal{Z}$ satisfying $h(x,0)=0$ for all $x\in \mathcal{X}$. Consider a generalized metric d on $\mathsf{\Omega }$ given by

$$d(h,h^{\prime })=d_{\mathsf{\Phi }}(h,h^{\prime }):=\inf S_{\mathsf{\Phi }}(h,h^{\prime }),$$

where

$$S_{\mathsf{\Phi }}(h,h^{\prime }):=\left\{K\in [0,\infty ]|\parallel h(x,y)-h^{\prime }(x,y)\parallel \le K\mathsf{\Phi }(x,y)\mathrm{for}\mathrm{all}(x,y)\in \mathcal{X}\times \mathcal{Y}\right\}$$

for all $h,h^{\prime }\in \mathsf{\Omega }$. By the same argument used in the proof of Lemma 2.2 in [17], the generalized metric space $(\mathsf{\Omega },d)$ is complete. Now, we define a mapping $T:\mathsf{\Omega }\to \mathsf{\Omega }$ by

$$\left(Th\right)(x,y):={\displaystyle \frac{1}{mn}}h(mx,ny)$$

for all $h\in \mathsf{\Omega }$ and all $(x,y)\in \mathcal{X}\times \mathcal{Y}$. Given the inequality (13), we obtain

$$d(g,Tg)\le L<\infty .$$

(14)

Observe that

$$\begin{array}{c}{K}_1\in S_{\mathsf{\Phi }}(h,h^{\prime })\mathrm{and}{K}_1<K_2\hfill \\ \Rightarrow \parallel h(x,y)-h^{\prime }(x,y)\parallel \le K_1\mathsf{\Phi }(x,y)\le K_2\mathsf{\Phi }(x,y)\mathrm{for}\mathrm{all}(x,y)\in \mathcal{X}\times \mathcal{Y}\hfill \\ \Rightarrow K_2\in S_{\mathsf{\Phi }}(h,h^{\prime })\hfill \end{array}$$

for all $h,h^{\prime }\in \mathsf{\Omega }$.

Let $h,h^{\prime }\in \mathsf{\Omega },K\in [0,\infty ]$ and $d(h,h^{\prime })<K$. Then, we gain $K\in S_{\mathsf{\Phi }}(h,h^{\prime })$. So, we get $\parallel h(x,y)-h^{\prime }(x,y)\parallel \le K\mathsf{\Phi }(x,y)$ for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$. Thus, we have

$$\begin{array}{cc}\hfill \parallel \left(Th\right)(x,y)& -\left(T{h}^{\prime }\right)(x,y)\parallel =∥{\displaystyle \frac{1}{mn}}h(mx,ny)-{\displaystyle \frac{1}{mn}}{h}^{\prime }(mx,ny)∥\le {\displaystyle \frac{K}{mn}}\mathsf{\Phi }(mx,ny)\hfill \\ & \le {\displaystyle \frac{K}{mn}}\left[2\epsilon +\delta \left(m^{2r}{\parallel x\parallel }^r+2n^{2r}{\parallel y\parallel }^r\right)\right]\le LK\mathsf{\Phi }(x,y)\hfill \end{array}$$

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$. Hence, $d(Th,T{h}^{\prime })\le LK$. Therefore, $d(Th,T{h}^{\prime })\le Ld(h,h^{\prime })$ for all $h,h^{\prime }\in \mathsf{\Omega }$; that is, T is a strictly contractive mapping of $\mathsf{\Omega }$ with Lipschitz constant L. Applying the alternative of the fixed point, we see that there exist a fixed point F of T in $\mathsf{\Omega }$ such that

$$F(x,y)=\underset{k\to \infty }{\lim }\left(T^{k}g\right)(x,y)=\underset{k\to \infty }{\lim }{\displaystyle \frac{1}{{\left(mn\right)}^k}}g\left(m^{k}x,n^{k}y\right)$$

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$.

Given the inequality (8), we have

$$∥mng\left({\displaystyle \frac{x_1+\cdots +x_m}{m}},{\displaystyle \frac{y_1+\cdots +y_n}{n}}\right)-\sum _{i=1}^m\sum _{j=1}^{n}g(x_i,y_j)∥\le \epsilon +\delta \left(\sum _{i=1}^m\parallel x_i{\parallel }^r+\sum _{j=1}^n{\parallel y_j\parallel }^r\right)$$

for all $x_1,\cdots ,x_m\in \mathcal{X}$ and all $y_1,\cdots ,y_n\in \mathcal{Y}$. Thus, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{{\left(mn\right)}^k}}\parallel mng& \left({\displaystyle \frac{m^k{x}_1+\cdots +m^k{x}_m}{m}},{\displaystyle \frac{n^k{y}_1+\cdots +n^k{y}_n}{n}}\right)-\sum _{i=1}^m\sum _{j=1}^{n}g\left(m^k{x}_i,n^k{y}_j\right)\parallel \hfill \\ & \le {\displaystyle \frac{\epsilon }{{\left(mn\right)}^k}}+\delta \left[{\left({\displaystyle \frac{m^r}{mn}}\right)}^k\sum _{i=1}^m\parallel x_i{\parallel }^r+{\left({\displaystyle \frac{n^r}{mn}}\right)}^k\sum _{j=1}^n{\parallel y_j\parallel }^r\right]\hfill \end{array}$$

for all $x_1,\cdots ,x_m\in \mathcal{X}$ and all $y_1,\cdots ,y_n\in \mathcal{Y}$. Hence, we can determine that

$$\begin{array}{c}\parallel mnF\left({\displaystyle \frac{x_1+\cdots +x_m}{m}},{\displaystyle \frac{y_1+\cdots +y_n}{n}}\right)-\sum _{i=1}^m\sum _{j=1}^{n}F(x_i,y_j)\parallel \hfill \\ \hspace{1em}\hspace{1em}\le \underset{k\to \infty }{\lim }\left({\displaystyle \frac{\epsilon }{{\left(mn\right)}^k}}+\delta \left[{\left({\displaystyle \frac{m^r}{mn}}\right)}^k\sum _{i=1}^m\parallel x_i{\parallel }^r+{\left({\displaystyle \frac{n^r}{mn}}\right)}^k\sum _{j=1}^n{\parallel y_j\parallel }^r\right]\right)=0\hfill \end{array}$$

for all $x_1,\cdots ,x_m\in \mathcal{X}$ and all $y_1,\cdots ,y_n\in \mathcal{Y}$.

Given the inequality (13), we have

$$\begin{array}{c}\parallel T^{k}g(x,y)-T^{k+1}g(x,y)\parallel ={\displaystyle \frac{1}{{\left(mn\right)}^k}}∥g\left(m^{k}x,n^{k}y\right)-{\displaystyle \frac{1}{mn}}g\left(m^{k+1}x,n^{k+1}y\right)∥\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em} \le {\displaystyle \frac{L}{{\left(mn\right)}^k}}\mathsf{\Phi }\left(m^{k}x,n^{k}y\right)\le \cdots \le {\displaystyle \frac{L}{{\left(mn\right)}^k}}{\left(mnL\right)}^k\mathsf{\Phi }(x,y)=L^{k+1}\mathsf{\Phi }(x,y)\hfill \end{array}$$

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$ and all $k\in \mathbb{N}$; that is, $d\left(T^{k}g,T^{k+1}g\right)\le L^{k+1}<\infty$ for all $k\in \mathbb{N}$. Given the alternative of the fixed point, there exists a natural number $k_0$ such that the mapping F is the unique fixed point of T in the set $\Delta =\left\{h\in \mathsf{\Omega }|d\left(T^{k_0}g,h\right)<\infty \right\}$. So, we have $d\left(T^{k_0}g,F\right)<\infty$. Since

$$d\left(g,T^{k_0}g\right)\le d(g,Tg)+d\left(Tg,T^{2}g\right)+\cdots +d\left(T^{k_0-1}g,T^{k_0}g\right)<\infty ,$$

we get $g\in \Delta$. Thus, we have $d(g,F)\le d\left(g,T^{k_0}g\right)+d\left(T^{k_0}g,F\right)<\infty$. Hence, there exists a non-negative real number K such that

$$\parallel g(x,y)-F(x,y)\parallel \le K\mathsf{\Phi }(x,y)$$

for all $(x,y)\in \mathcal{X}\times \mathcal{Y}$. Again, using the fixed-point alternative, we have

$$d(g,F)\le {\displaystyle \frac{1}{1-L}}d(g,Tg).$$

Given (14), we conclude that

$$d(g,F)\le {\displaystyle \frac{L}{1-L}},$$

which implies the inequality (12). □

3. Conclusions

In this paper, we investigated the Hyers-Ulam stability of the multivariable bi-Jensen functional equation in Banach spaces using two distinct approaches. First, using the direct method, we established stability results under specific conditions in which the approximation error is controlled by the sum of powers of the norms (Theorem 2). This approach explicitly demonstrates the existence of a unique bi-additive mapping approximating the given function when the power r satisfies the conditions $m^r<mn$ and $n^r<mn$. Second, we extended our investigation by applying the fixed-point method. Utilizing the fixed point alternative theorem, we proved the stability for a more general control function $\mathsf{\Phi }(x,y)$ (Theorem 4). This method allows for a generalized metric space setting, confirming that the stability holds as long as the control function satisfies a specific contraction property. Consequently, this study confirms that the multivariable bi-Jensen functional equation is stable in Banach spaces using both the classical direct constructive method and the modern fixed-point approach. These results provide a comprehensive mathematical foundation for approximating multivariable bi-Jensen mappings.

Furthermore, the stability criteria can vary significantly depending on the underlying geometric structure of the spaces involved. It would be a valuable direction for future research to extend these results to more generalized frameworks, such as 2-Banach spaces, quasi-Banach spaces, non-Archimedean spaces, or Sobolev spaces, where different norm properties and stability conditions apply. In such spaces, the appropriate control functions and stability inequalities would differ from the standard Banach space setting, requiring new analytical approaches.