Stability of Bi-Additive Mappings and Bi-Jensen Mappings

Abstract

Symmetry is repetitive self-similarity. We proved the stability problem by replicating the well-known Cauchy equation and the well-known Jensen equation into two variables. In this paper, we proved the Hyers-Ulam stability of the bi-additive functional equation $f(x+y,z+w)=f(x,z)+f(y,w)$ and the bi-Jensen functional equation $4f\left(\frac{x+y}{2},\frac{z+w}{2}\right)=f(x,z)+f(x,w)+f(y,z)+f(y,w).$

1. Introduction

A functional equation is stable if there is a function that exactly satisfies the given equation in the vicinity of a function that approximately satisfies it. Any approximate solution can actually be an exact solution. In Cauchy’s equation $f(x+y)=f\left(x\right)+f\left(y\right)$ we can deal with a class of approximate solutions defined by the functional inequality introduced by Rassias.

$$\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel \le \epsilon (\parallel x{\parallel }^p+\parallel y{\parallel }^p).$$

It turns out that for $p\ne 1$ each solution of the above inequality can be approximated by an additive function A in such a way that the inequality

$$\parallel f\left(x\right)-A\left(x\right)\parallel \le {k\epsilon \parallel x\parallel }^p.$$

holds, with a suitable k, on the whole domain (for $p=0$ it coincides with the classical Hyers–Ulam result).

Let us say $\mathcal{X}$ and $\mathcal{Y}$ are vector spaces. The mapping $h:\mathcal{X}\to \mathcal{Y}$ is called an additional mapping (respectively, an affine mapping) if h satisfies the Cauchy functional equation $h(x+y)=h\left(x\right)+h\left(y\right)$ (respectively, the Jensen functional equation $2h\left(\frac{x+y}{2}\right)=h\left(x\right)+h\left(y\right)$). T. Aoki [1] and Th. M. Rassias [2,3] extended Hyers-Ulam stability taking into account the variables for the Cauchy equation. S.-M. Jung [4] got the result of the Jensen equation. It was also generalized as a functional case by P. Găvruta [5] and S.-M. Jung [6] and Y.-H. Lee and K.-W. Jun [7].

The following functional Equations (1) and (3) are functional equations those combine the existing well-known the Cauchy equation and the Jensen equation.

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

(1)

The authors [8] introduce the system of equations

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

(2)

and the bi-Jensen functional equation

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

(3)

We made the above functional equations with a symmetrical structure. Symmetry is repetitive self-similarity. The solution of (2) is coincide with the solution of (3). The solution of (1) is of the form $A_1\left(x\right)+A_2\left(y\right)$, where $A_1$ and $A_2$ are additive mappings. The solution of (2) is of the form $A_1\left(x\right)+A_2\left(y\right)+f(0,0)$, where $A_1$ and $A_2$ are additive mappings. The solution of (2) contains the solution of (1). The difference of the solutions (1) and (2) is merely a constant, that is, the solutions (1) and (2) are similar.

Jun, Jung, and Lee [9] obtained the stability on a bi-Jensen functional equation in Banach spaces. Additionally, the authors [10] proved the stability on a Cauchy-Jensen functional equation Banach spaces.

In this paper, we investigate the generalized Hyers-Ulam stability of (1) in Banach spaces and 2-Banach spaces. We proved the Hyers-Ulam stability of (2) and (3) in quasi-Banach spaces.

2. Solution and Stability of a Bi-Additive Functional Equation

In the following theorem, we find out the general solution of the bi-additive functional Equation (1).

Theorem 1.

A mapping $f:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ satisfies $\left(1\right)$ if and only if there exist two additive mappings $A_1,A_2:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ such that

$$f(x,y)=A_1\left(x\right)+A_2\left(y\right)$$

for all $x,y\in \mathcal{X}$.

Proof. 

We first assume that f is a solution of (1). Define $A_1,A_2:\mathcal{X}\to \mathcal{Y}$ by $A_1\left(x\right):=f(x,0)$ and $A_2\left(x\right):=f(0,x)$ for all $x\in \mathcal{X}$. One can easily verify that $A_1$, $A_2$ are additive. Letting $y=z=0$ in (1), we get

$$f(x,w)=f(x,0)+f(0,w)=A_1\left(x\right)+A_2\left(w\right)$$

for all $x,w\in \mathcal{X}$.

Conversely, we assume that there is two additive mappings $A_1,A_2:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$, such that $f(x,y)=A_1\left(x\right)+A_2\left(y\right)$ for all $x,y\in \mathcal{X}$. Since $A_1$, $A_2$ are additive, we gain

$$\begin{array}{ccc}\hfill f(x+y,z+w)& =& A_1(x+y)+A_2(z+w)\hfill \\ & =& A_1\left(x\right)+A_1\left(y\right)+A_2\left(z\right)+A_2\left(w\right)\hfill \\ & =& A_1\left(x\right)+A_2\left(z\right)+A_1\left(y\right)+A_2\left(w\right)\hfill \\ & =& f(x,z)+f(y,w)\hfill \end{array}$$

for all $x,y,z,w\in \mathcal{X}$. □

From now on, let $\mathcal{X}$ and $\mathcal{Y}$ be a normed linear space and a Banach space, respectively.

Theorem 2.

Let $0<p<1$, $\epsilon >0$, $\delta \ge 0$ and $f:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ be a mapping such that

$$\begin{array}{ccc}& & \parallel f(x+y,z+w)-f(x,z)-f(y,w)\parallel \le \epsilon +\delta (\parallel x{\parallel }^p+\parallel y{\parallel }^p+\parallel z{\parallel }^p+\parallel w{\parallel }^p)\hfill \end{array}$$

(4)

for all $x,y,z,w\in \mathcal{X}$. Then there is unique bi-additive mapping $F:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$, such that

$$\parallel f(x,y)-F(x,y)\parallel \le \epsilon +\frac{2\delta }{2-2^p}{(\parallel x\parallel }^p+{\parallel y\parallel }^p)$$

(5)

for all $x,y\in \mathcal{X}$. The mapping F is given by $F(x,y):={lim}_{j\to \infty }\frac{1}{2^j}f(2^{j}x,2^{j}y)$ for all $x,y\in \mathcal{X}$.

Proof. 

Putting $y=x$ and $w=z$ in (4), we have

$$\begin{array}{c}\hfill ∥f(x,z)-\frac{1}{2}f(2x,2z)∥\le \frac{\epsilon }{2}+{\delta (\parallel x\parallel }^p+{\parallel z\parallel }^p)\end{array}$$

for all $x,z\in \mathcal{X}$. Thus, we obtain

$$\begin{array}{c}\hfill ∥\frac{1}{2^j}f(2^{j}x,2^{j}z)-\frac{1}{2^{j+1}}f(2^{j+1}x,2^{j+1}z)∥\le \frac{\epsilon }{2^{j+1}}+2^{j(p-1)}{\delta (\parallel x\parallel }^p+{\parallel z\parallel }^p)\end{array}$$

for all $x,z\in \mathcal{X}$ and all j. Replacing z by y in the above inequality, we see that

$$\begin{array}{c}\hfill ∥\frac{1}{2^j}f(2^{j}x,2^{j}y)-\frac{1}{2^{j+1}}f(2^{j+1}x,2^{j+1}y)∥\le \frac{\epsilon }{2^{j+1}}+2^{j(p-1)}{\delta (\parallel x\parallel }^p+{\parallel y\parallel }^p)\end{array}$$

for all $x,y\in \mathcal{X}$ and all j. For given integers $l,m(0\le l<m)$, we get

$$\begin{array}{c}\hfill ∥\frac{1}{2^l}f(2^{l}x,2^{l}y)-\frac{1}{2^m}f(2^{m}x,2^{m}y)∥\le \sum _{j=l}^{m-1}\left[\frac{\epsilon }{2^{j+1}}+2^{j(p-1)}{\delta (\parallel x\parallel }^p+{\parallel y\parallel }^p)\right]\end{array}$$

(6)

for all $x,y\in \mathcal{X}$. By (6), the sequence $\left\{\frac{1}{2^j}f(2^{j}x,2^{j}y)\right\}$ is a Cauchy sequence for all $x,y\in \mathcal{X}$. Since $\mathcal{Y}$ is complete, the sequence $\left\{\frac{1}{2^j}f(2^{j}x,2^{j}y)\right\}$ converges for all $x,y\in \mathcal{X}$. Define $F:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ by

$$F(x,y):=\underset{j\to \infty }{lim}\frac{1}{2^j}f(2^{j}x,2^{j}y)$$

for all $x,y\in \mathcal{X}$. By (4), we have

$$\begin{array}{ccc}& & \frac{1}{2^j}\parallel f\left(2^j(x+y),2^j(z+w)\right)-f(2^{j}x,2^{j}z)-f(2^{j}y,2^{j}w)\parallel \hfill \\ & & \le \frac{\epsilon }{2^j}+2^{j(p-1)}{\delta (\parallel x\parallel }^p+{\parallel y\parallel }^p+{\parallel z\parallel }^p+{\parallel w\parallel }^p)\hfill \end{array}$$

for all $x,y,z,w\in \mathcal{X}$ and all $j\in \mathbb{N}$. Letting $j\to \infty$ in the above inequality, we see that F satisfies (1). Setting $l=0$ and taking $m\to \infty$ in (6), one can obtain the inequality (5). If $G:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ is another 2-variable additive mapping satisfying (5), we obtain

$$\begin{array}{ccc}& & \parallel F(x,y)-G(x,y)\parallel \hfill \\ & & =\frac{1}{2^n}\parallel F(2^{n}x,2^{n}y)-G(2^{n}x,2^{n}y)\parallel \hfill \\ & & \le \frac{1}{2^n}\parallel F(2^{n}x,2^{n}y)-f(2^{n}x,2^{n}y)\parallel +\frac{1}{2^n}\parallel f(2^{n}x,2^{n}y)-G(2^{n}x,2^{n}y)\parallel \hfill \\ & & \le \frac{1}{2^{n-1}}\left[\epsilon +\frac{2^{np+1}}{2-2^p}{\delta (\parallel x\parallel }^p+{\parallel y\parallel }^p)\right]\hfill \\ & & \to 0asn\to \infty \hfill \end{array}$$

for all $x,y\in \mathcal{X}$. Hence the mapping F is the unique bi-additive mapping, as desired. □

Corollary 1.

Let $f:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ be a mapping such that

$$\parallel f(x+y,z+w)-f(x,z)-f(y,w)\parallel \le \epsilon$$

for all $x,y,z,w\in \mathcal{X}$. Then, there exists a unique mapping $F:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ satisfying $\left(1\right)$, such that

$$\parallel f(x,y)-F(x,y)\parallel \le \frac{\epsilon }{2}$$

for all $x,y\in \mathcal{X}$.

Proof. 

If we insert $\delta =0$ in Theorem 2, we obtain $\epsilon$ as an estimate of the difference between the exact and the approximate solution of the considered equation. □

In the case $p>2$ in Theorem 2, one can also obtain the similar result.

We explain some definitions [11,12] on 2-Banach spaces.

Definition 1.

Let $\mathcal{X}$ be a vector space over $\mathbb{R}$ with dimension greater than 1 and $\parallel ·,·\parallel :{\mathcal{X}}^2\to \mathbb{R}$ be a function. Then we say $(\mathcal{X},\parallel ·,·\parallel )$ is a linear 2-normed space if

(a) $\parallel x,y\parallel =0$ if and only if x and y are linearly dependent;

(b) $\parallel x,y\parallel =\parallel y,x\parallel$;

(c) $\parallel \alpha x,y\parallel =\left|\alpha \right|\parallel x,y\parallel$;

(d) $\parallel x,y+z\parallel \le \parallel x,y\parallel +\parallel x,z\parallel$

for all $\alpha \in \mathbb{R}$ and $x,y,z\in \mathcal{X}$. In this case, the function $\parallel ·,·\parallel$ is called a 2-norm on $\mathcal{X}$.

Definition 2.

Let $\mathcal{X}$ be linear 2-normed space and $\left\{x_n\right\}$ a sequence in $\mathcal{X}$. The sequence $\left\{x_n\right\}$ is said to convergent in $\mathcal{X}$ if there is an $x\in \mathcal{X}$, such that

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

for all $y\in \mathcal{X}$. In this case, we say that a sequence $\left\{x_n\right\}$ converges to x, simply denoted by ${lim}_{n\to \infty }{x}_n=x$.

Definition 3.

Let $\mathcal{X}$ be linear 2-normed space and $\left\{x_n\right\}$ a sequence in $\mathcal{X}$ is called a Cauchy sequence if for any $\epsilon >0$, there exists $N\in \mathbb{N}$ such that for all $m,n\ge N$, $\parallel x_m-x_n,y\parallel <\epsilon$ for all $y\in \mathcal{X}$. For convenience, we will write ${lim}_{m,n\to \infty }\parallel x_n-x_m,y\parallel =0$ for a Cauchy sequence $\left\{x_n\right\}$. A 2-Banach space is defined to be a linear 2-normed space in which every Cauchy sequence is convergent.

In the following lemma, we get some primitive properties in linear 2-normed spaces that will be used to prove our stability results.

Lemma 1

([13]). Let $(\mathcal{X},\parallel ·,·\parallel )$ be a linear 2-normed space and $x\in \mathcal{X}$.

(a) If $\parallel x,y\parallel =0$ for all $y\in \mathcal{X}$, then $x=0$.

(b) $|\parallel x,z\parallel -\parallel y,z\parallel |\le \parallel x-y,z\parallel$ for all $x,y,z\in \mathcal{X}$.

(c) If a sequence $\left\{x_n\right\}$ is convergent in $\mathcal{X}$, then ${lim}_{n\to \infty }\parallel x_n,y\parallel =\parallel {lim}_{n\to \infty }{x}_n,y\parallel$ for all $y\in \mathcal{X}$.

In the rest of this section, let $\mathcal{X}$ be a normed space and $\mathcal{Y}$ a 2-Banach space.

Theorem 3.

Let $p\in (0,1)$, $\epsilon >0$, $\delta ,\eta \ge 0$ and let $f:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ be a surjective mapping such that

$$\begin{array}{cc}& \parallel f(x+y,z+w)-f(x,z)-f(y,w),f(u,v)\parallel \\ & \le \epsilon +{\delta (\parallel x\parallel }^p+{\parallel y\parallel }^p+{\parallel z\parallel }^p+{\parallel w\parallel }^p)+\eta (\parallel u\parallel +\parallel v\parallel )\end{array}$$

(7)

for all $x,y,z,w,u,v\in \mathcal{X}$. Then there exists a unique mapping $F:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ satisfying $\left(1\right)$, such that

$$\parallel f(x,y)-F(x,y),f(u,v)\parallel \le \frac{\epsilon }{2}+\frac{2\delta }{2-2^p}{(\parallel x\parallel }^p+{\parallel y\parallel }^p)+\frac{\eta }{2}(\parallel u\parallel +\parallel v\parallel )$$

(8)

for all $x,y,u,v\in \mathcal{X}$.

Proof. 

Letting $y=x$ and $w=z$ in (7), we have

$$\begin{array}{c}\hfill ∥f(x,z)-\frac{1}{2}f(2x,2z),f(u,v)∥\le \frac{\epsilon }{2}+{\delta (\parallel x\parallel }^p+{\parallel z\parallel }^p)+\frac{\eta }{2}(\parallel u\parallel +\parallel v\parallel )\end{array}$$

for all $x,z,u,v\in \mathcal{X}$. Thus, we obtain

$$\begin{array}{ccc}& & ∥\frac{1}{2^j}f(2^{j}x,2^{j}z)-\frac{1}{2^{j+1}}f(2^{j+1}x,2^{j+1}z),f(u,v)∥\hfill \\ & & \le \frac{\epsilon }{2^{j+1}}+2^{j(p-1)}{\delta (\parallel x\parallel }^p+{\parallel z\parallel }^p)+\frac{\eta }{2^{j+1}}(\parallel u\parallel +\parallel v\parallel )\hfill \end{array}$$

for all $x,z,u,v\in \mathcal{X}$ and all j. Replacing z by y in the above inequality, we see that

$$\begin{array}{ccc}& & ∥\frac{1}{2^j}f(2^{j}x,2^{j}y)-\frac{1}{2^{j+1}}f(2^{j+1}x,2^{j+1}y),f(u,v)∥\hfill \\ & & \le \frac{\epsilon }{2^{j+1}}+2^{j(p-1)}{\delta (\parallel x\parallel }^p+{\parallel y\parallel }^p)+\frac{\eta }{2^{j+1}}(\parallel u\parallel +\parallel v\parallel )\hfill \end{array}$$

for all $x,y,u,v\in \mathcal{X}$ and all j. For given integers $l,m(0\le l<m)$, we get

$$\begin{array}{ccc}& & ∥\frac{1}{2^l}f(2^{l}x,2^{l}y)-\frac{1}{2^m}f(2^{m}x,2^{m}y),f(u,v)∥\hfill \\ & & \le \sum _{j=l}^{m-1}\left[\frac{\epsilon }{2^{j+1}}+2^{j(p-1)}{\delta (\parallel x\parallel }^p+{\parallel z\parallel }^p)+\frac{\eta }{2^{j+1}}(\parallel u\parallel +\parallel v\parallel )\right]\hfill \end{array}$$

(9)

for all $x,y,u,v\in \mathcal{X}$. By (9), the sequence $\left\{\frac{1}{2^j}f(2^{j}x,2^{j}y)\right\}$ is a Cauchy sequence for all $x,y\in \mathcal{X}$. Since $\mathcal{Y}$ is complete, the sequence $\left\{\frac{1}{2^j}f(2^{j}x,2^{j}y)\right\}$ converges for all $x,y\in \mathcal{X}$. Define $F:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ by $F(x,y):={lim}_{j\to \infty }\frac{1}{2^j}f(2^{j}x,2^{j}y)$ for all $x,y\in \mathcal{X}$. By (7), we have

$$\begin{array}{ccc}& & \parallel \frac{1}{2^j}f\left(2^j(x+y),2^j(z+w)\right)-\frac{1}{2^j}f(2^{j}x,2^{j}z)-\frac{1}{2^j}f(2^{j}y,2^{j}w),f(u,v)\parallel \hfill \\ & & \le \frac{1}{2^j}\left[\epsilon +2^{jp}{\delta (\parallel x\parallel }^p+{\parallel y\parallel }^p+{\parallel z\parallel }^p+{\parallel w\parallel }^p)+\eta (\parallel u\parallel +\parallel v\parallel )\right]\hfill \end{array}$$

for all $x,y,z,w,u,v\in \mathcal{X}$ and all j. Letting $j\to \infty$, we see that F satisfies (1). Setting $l=0$ and taking $m\to \infty$ in (9), one can obtain the inequality (8). If $G:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ is another mapping satisfying (1) and (8), we obtain

$$\begin{array}{ccc}& & \parallel F(x,y)-G(x,y),f(u,v)\parallel =\frac{1}{2^n}\parallel F(2^{n}x,2^{n}y)-G(2^{n}x,2^{n}y),f(u,v)\parallel \hfill \\ & & \le \frac{1}{2^n}\parallel F(2^{n}x,2^{n}y)-f(2^{n}x,2^{n}y),f(u,v)\parallel +\frac{1}{2^n}\parallel f(2^{n}x,2^{n}y)-G(2^{n}x,2^{n}y),f(u,v)\parallel \hfill \\ & & \le \frac{2}{2^n}\left[\frac{\epsilon }{2}+\frac{2^{np+1}\delta }{2-2^p}{(\parallel x\parallel }^p+{\parallel y\parallel }^p)+\frac{\eta (\parallel u\parallel +\parallel v\parallel )}{2}\right]\hfill \\ & & \to 0asn\to \infty \hfill \end{array}$$

for all $x,y,u,v\in \mathcal{X}$. Hence the mapping F is the unique mapping satisfying (1), as desired. □

Corollary 2.

Let $f:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ be a mapping, such that

$$\parallel f(x+y,z+w)-f(x,z)-f(y,w),f(u,v)\parallel \le \epsilon$$

for all $x,y,z,w,u,v\in \mathcal{X}$. Then there exists a unique mapping $F:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ satisfying $\left(1\right)$ such that

$$\parallel f(x,y)-F(x,y),f(u,v)\parallel \le \frac{\epsilon }{2}$$

for all $x,y,u,v\in \mathcal{X}$.

Proof. 

Taking $\delta =\eta =0$ in Theorem 3, we have the desired result. □

In the case $p>2$ in Theorem 3, one can also obtain the similar result.

3. Solution and Stability of a Bi-Jensen Functional Equation

In [14,15], one can find the concept of quasi-Banach spaces.

Definition 4.

Let $\mathcal{X}$ be a real linear space. A quasi-norm is real-valued function on $\mathcal{X}$ satisfying the following:

(i) $\parallel x\parallel \ge 0$ for all $x\in \mathcal{X}$ and $\parallel x\parallel =0$ if, and only if, $x=0$.

(ii) $\parallel \lambda x\parallel =\left|\lambda \right|\parallel x\parallel$ for all $\lambda \in \mathbb{R}$ and all $x\in \mathcal{X}$.

(iii) There is a constant $K\ge 1$, such that $\parallel x+y\parallel \le K(\parallel x\parallel +\parallel y\parallel )$ for all $x,y\in \mathcal{X}$.

The pair $(\mathcal{X},\parallel ·\parallel )$ is called a quasi-normed space if $\parallel ·\parallel$ is a quasi-norm on $\mathcal{X}$. The smallest possible K is called the modulus of concavity of $\parallel ·\parallel$. A quasi-Banach space is a complete quasi-normed space. A quasi-norm $\parallel ·\parallel$ is called a p-norm ($0<p\le 1$) if

$${\parallel x+y\parallel }^p\le {\parallel x\parallel }^p+{\parallel y\parallel }^p$$

for all $x,y\in \mathcal{X}$. In this case, a quasi-Banach space is called a p-Banach space.

A quasi-norm gives rise to a linear topology on X, namely the least linear topology for which the unit ball $B=\{x\in \mathcal{X}:\parallel x\parallel \le 1\}$ is a neighborhood of zero. This topology is locally bounded, that is, it has a bounded neighborhood of zero. Actually, every locally bounded topology arises in this way.

From now on, assume that $\mathcal{X}$ is a quasi-normed space with quasi-norm ${\parallel ·\parallel }_{\mathcal{X}}$ and that $\mathcal{Y}$ is a p-Banach space with p-norm ${\parallel ·\parallel }_{\mathcal{Y}}$. Let K be the modulus of concavity of ${\parallel ·\parallel }_{\mathcal{Y}}$.

Let $\phi :\mathcal{X}\times \mathcal{X}\times \mathcal{X}\to [0,\infty )$ and $\psi :\mathcal{X}\times \mathcal{X}\times \mathcal{X}\to [0,\infty )$ be two functions such that

$$\underset{n\to \infty }{lim}\frac{1}{3^n}\phi (3^{n}x,3^{n}y,z)=0,\underset{n\to \infty }{lim}\frac{1}{3^n}\psi (3^{n}x,y,z)=0$$

(10)

and

$$\underset{n\to \infty }{lim}\frac{1}{3^n}\phi (x,y,3^{n}z)=0,\underset{n\to \infty }{lim}\frac{1}{3^n}\psi (x,3^{n}y,3^{n}z)=0$$

(11)

for all $x,y,z\in \mathcal{X}$, and

$$M(x,y,z):=\sum _{j=0}^{\infty }\frac{1}{3^{pj}}\phi {(3^{j}x,3^{j}y,z)}^p<\infty$$

(12)

and

$$N(z,x,y):=\sum _{j=0}^{\infty }\frac{1}{3^{pj}}\psi {(z,3^{j}x,3^{j}y)}^p<\infty$$

(13)

for all $x,z\in \mathcal{X}$ and all $y\in \{-x,-3x\}$.

We will use the following lemma in order to prove Theorem 4.

Lemma 2

([16]). Let $0\le p\le 1$ and let $x_1,x_2,\cdots ,x_n$ be non-negative real numbers. Then

$${\left(\sum _{j=1}^n{x}_j\right)}^p\le \sum _{j=1}^n{x_j}^p.$$

Theorem 4.

Let $0<p\le 1$ and suppose that a mapping $f:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ satisfies the inequalities

$$\begin{array}{cc}\hfill \parallel 2f\left(\frac{x+y}{2},z\right)-f(x,z)-f(y,z){\parallel }_Y& \le \phi (x,y,z),\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill \parallel 2f\left(x,\frac{y+z}{2}\right)-f(x,y)-f(x,z){\parallel }_Y& \le \psi (x,y,z)\hfill \end{array}$$

(15)

for all $x,y,z\in \mathcal{X}$. Then there exists a unique additive-Jensen mapping $J_1:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ satisfying

$$\parallel f(x,y)-f(0,y)-J_1{(x,y)\parallel }_{\mathcal{Y}}\le \frac{K}{3}{[M(x,-x,y)+M(-x,3x,y)]}^{\frac{1}{p}}$$

(16)

for all $x,y\in \mathcal{X}$. There exists a unique Jensen-additive mapping $J_2:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ satisfying

$$\parallel f(x,y)-f(x,0)-J_2{(x,y)\parallel }_{\mathcal{Y}}\le \frac{K}{3}{[N(x,y,-y)+N(x,-y,3y)]}^{\frac{1}{p}}$$

(17)

for all $x,y\in \mathcal{X}$.

Proof. 

Let $g(x,y):=f(x,y)-f(0,y)$ for all $x,y\in \mathcal{X}$. Then $g(0,y)=0$ for all $y\in \mathcal{X}$. Letting y by $-x$ in (14), we get

$${\parallel g(x,z)+g(-x,z)\parallel }_{\mathcal{Y}}\le \phi (x,-x,z)$$

for all $x,z\in \mathcal{X}$. Replacing x by $-x$ and y by $3x$ in (14), we have

$${\parallel 2g(x,z)-g(-x,z)-g(3x,z)\parallel }_{\mathcal{Y}}\le \phi (-x,3x,z)$$

for all $x,z\in \mathcal{X}$. By two above inequalities and replacing z by y, we get

$${∥3g(x,y)-g(3x,y)∥}_{\mathcal{Y}}\le K[\phi (x,-x,y)+\phi (-x,3x,y)]$$

for all $x,y\in \mathcal{X}$. Thus we have

$$\begin{array}{ccc}& & {∥\frac{1}{3^j}g(3^{j}x,y)-\frac{1}{3^{j+1}}g(3^{j+1}x,y)∥}_{\mathcal{Y}}\le \frac{K}{3^{j+1}}[\phi (3^{j}x,-3^{j}x,y)+\phi (-3^{j}x,3^{j+1}x,y)]\hfill \end{array}$$

for all $x,y\in \mathcal{X}$ and all j. For given integer $l,m(0\le l<m)$, by Lemma 2, we get

$$\begin{array}{c}\hfill {∥\frac{1}{3^l}g(3^{l}x,y)-\frac{1}{3^m}g(3^{m}x,y)∥}_{\mathcal{Y}}^p\le {\left(\frac{K}{3}\right)}^p\sum _{j=l}^{m-1}\frac{1}{3^{pj}}\left[\phi {(3^{j}x,-3^{j}x,y)}^p+\phi {(-3^{j}x,3^{j+1}x,y)}^p\right]\end{array}$$

(18)

for all $x,y\in \mathcal{X}$. By (12), the sequence $\left\{\frac{1}{3^j}g(3^{j}x,y)\right\}$ is a Cauchy sequence for all $x,y\in \mathcal{X}$. Since $\mathcal{Y}$ is complete, the sequence $\left\{\frac{1}{3^j}g(3^{j}x,y)\right\}$ converges for all $x,y\in \mathcal{X}$. Define $J_1:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ by

$$J_1(x,y):=\underset{j\to \infty }{lim}\frac{1}{3^j}g(3^{j}x,y)$$

for all $x,y\in \mathcal{X}$. Putting $l=0$ and taking $m\to \infty$ in (18), one can obtain the inequality (16). From the definition of $J_1$, we get

$$3^j{J}_1(x,y)=J_1(3^{j}x,y)\mathrm{and}{J}_1(0,y)=0$$

(19)

for all $x,y\in \mathcal{X}$ and all j. By (14), (16) and (19), we gain

$$\begin{array}{cc}& \parallel 2J_1(2x,y)-4J_1{(x,y)\parallel }_{}Y\hfill \\ & =\parallel 2J_1(2x,y)-J_1(3x,y)-J_1{(x,y)\parallel }_{}Y\hfill \\ & \le 3^{-j}{∥2J_1(3^j·2x,y)-J_1(3^j·3x,y)-J_1(3^{j}x,y)∥}_{}Y\hfill \\ & \le 3^{-j}\left[{∥2J_1(3^j·2x,y)-2f(3^j·2x,y)∥}_{\mathcal{Y}}+{∥J_1(3^j·3x,y)-f(3^j·3x,y)∥}_{\mathcal{Y}}\right]\hfill \\ & +3^{-j}{∥J_1(3^{j}x,y)-f(3^{j}x,y)∥}_{\mathcal{Y}}+3^{-j}{∥2f\left(\frac{3^j(3x+x)}{2},y\right)-f(3^j·3x,y)-f(3^{j}x,y)∥}_{}Y\hfill \\ & \le 2·3^{-j-1}K{[M(3^j·2x,3^j(-2x),y)+M(3^j·(-2x),3^{j+1}·2x,y)]}^{\frac{1}{p}}\hfill \\ & +3^{-j-1}K{[M(3^j·3x,3^j(-3x),y)+M(3^j·(-3x),3^{j+1}·3x,y)]}^{\frac{1}{p}}\hfill \\ & +3^{-j-1}K{[M(3^{j}x,3^j(-x),y)+M(3^j·(-x),3^{j+1}x,y)]}^{\frac{1}{p}}\hfill \\ & +3^{-j}\phi (3^{j}x,3^{j+1}x,y)\hfill \end{array}$$

for all $x,y\in \mathcal{X}$ and all j. From this and (19), we obtain

$$2J_1(x,y)=J_1(2x,y)$$

(20)

for all $x,y\in \mathcal{X}$. From (12) and (14),

$$\begin{array}{cc}& {∥2J_1\left(\frac{x+y}{2},z\right)-J_1(x,z)-J_1(y,z)∥}_{}Y\hfill \\ & =\underset{j\to \infty }{lim}{3}^{-j}{∥2J_1\left(\frac{3^{j}x+3^{j}y}{2},z\right)-J_1(3^{j}x,z)-J_1(3^{j}y,z)∥}_{}Y\hfill \\ & \le \underset{j\to \infty }{lim}{3}^{-j}\phi (3^{j}x,3^{j}y,z)=0\hfill \end{array}$$

for all $x,y,z\in \mathcal{X}$. From (20) and the above inequality,

$$J_1(x+y,z)=2J_1\left(\frac{x+y}{2},z\right)=J_1(x,z)+J_1(y,z)$$

for all $x,y,z\in \mathcal{X}$. Hence

$$J_1(x+y,z)=J_1(x,z)+J_1(y,z)$$

for all $x,y,z\in \mathcal{X}$. That is, $J_1$ is an additive mapping with respect to the first variable. By (15), we get

$$\begin{array}{c}\hfill {∥\frac{2}{3^j}g\left(3^{j}x,\frac{y+z}{2}\right)+\frac{1}{3^j}g(3^{j}x,y)-\frac{1}{3^j}g(3^{j}x,z)∥}_{\mathcal{Y}}\le \frac{1}{3^j}\psi (3^{j}x,y,z)\end{array}$$

for all $x,y,z\in \mathcal{X}$ and all j. Letting $j\to \infty$ in the above inequality and using (10), $J_1$ is a Jensen mapping with respect to the second variable. To prove the uniqueness of $J_1$, let $S_1$ be another additive-Jensen mapping satisfying (16). Then we obtain

$$\begin{array}{cc}& \parallel 2S_1(2x,y)-4S_1{(x,y)\parallel }_{\mathcal{Y}}^p\hfill \\ & =\parallel 2S_1(2x,y)-S_1(3x,y)-S_1{(x,y)\parallel }_{\mathcal{Y}}^p\hfill \\ & =3^{-jp}{\parallel 2S_1(2·3^{j}x,y)-S_1(3·3^{j}x,y)-S_1(3^{j}x,y)\parallel }_{\}}\}{Y}^p\hfill \\ & \le 3^{-jp}{\parallel 2S_1(2·3^{j}x,y)-2g(2·3^{j}x,y)\parallel }_{\mathcal{Y}}^p\hfill \\ & +3^{-jp}\parallel S_1(3·3^{j}x,y)-g(3·3^{j}x,y){\parallel }_Y^p+3^{-jp}{\parallel S_1(3^{j}x,y)-g(3^{j}x,y)\parallel }_{\mathcal{Y}}^p\hfill \\ & +3^{-jp}{∥2g\left(3^j·\frac{3x+x}{2},y\right)-g(3·3^{j}x,y)-g(3^{j}x,y)∥}_{\mathcal{Y}}^p\hfill \end{array}$$

for all $x,y\in \mathcal{X}$ and all j. It follows from (16), we have

$$\begin{array}{cc}& \parallel J_1(x,y)-S_1{(x,y)\parallel }_{\mathcal{Y}}^p\hfill \\ & ={∥\frac{1}{3^j}{J}_1(3^{j}x,y)-\frac{1}{3^j}{S}_1(3^{j}x,y)∥}_{\mathcal{Y}}^p\hfill \\ & \le {∥\frac{1}{3^j}{J}_1(3^{j}x,y)-\frac{1}{3^j}f(3^{j}x,y)+\frac{1}{3^j}f(0,y)∥}_{\mathcal{Y}}^p+{∥\frac{1}{3^j}f(3^{j}x,y)-\frac{1}{3^j}f(0,y)-\frac{1}{3^j}{S}_1(3^{j}x,y)∥}_{\mathcal{Y}}^p\hfill \\ & \le \frac{2K^p}{3^{p(j+1)}}[M(3^{j}x,-3^{j}x,y)+M(-3^{j}x,3^{j+1}x,y)]\hfill \end{array}$$

for all $x,y\in \mathcal{X}$ and all j. Taking $j\to \infty$ in the above inequality and using (12), we get $J_1=S_1$.

Define $J_2:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ by

$$J_2(x,y):=\underset{j\to \infty }{lim}\frac{1}{3^j}f(x,3^{j}y)$$

for all $x,y\in \mathcal{X}$. By the same method in the above arguments, $J_2$ is a unique Jensen-additive mapping satisfying (17). □

Corollary 3.

Let $0<p\le 1$ and $\epsilon ,\delta >0$ be fixed. Suppose that a mapping $f:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ satisfies the inequalities

$$\begin{array}{cc}\hfill \parallel 2f\left(\frac{x+y}{2},z\right)-f(x,z)-f(y,z){\parallel }_{\mathcal{Y}}& \le \epsilon ,\hfill \\ \hfill \parallel 2f\left(x,\frac{y+z}{2}\right)-f(x,y)-f(x,z){\parallel }_{\mathcal{Y}}& \le \delta \hfill \end{array}$$

for all $x,y,z\in \mathcal{X}$. Then there exists a unique additive-Jensen mapping $J_1:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ satisfying

$$\parallel f(x,y)-f(0,y)-J_1{(x,y)\parallel }_{\mathcal{Y}}\le K\epsilon {\left(\frac{2}{3^p-1}\right)}^{\frac{1}{p}}$$

for all $x,y\in \mathcal{X}$. There exists a unique Jensen-additive mapping $J_2:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ satisfying

$$\parallel f(x,y)-f(x,0)-J_2{(x,y)\parallel }_{\mathcal{Y}}\le K\delta {\left(\frac{2}{3^p-1}\right)}^{\frac{1}{p}}$$

for all $x,y\in \mathcal{X}$.

Proof. 

Let $\phi (x,y,z):=\epsilon$ and $\psi (x,y,z):=\delta$ for all $x,y,z\in \mathcal{X}$. By Theorem 4, we have an additive-Jensen mapping $J_1$ and a Jensen-additive mapping $J_2$, as desired. □

From now on, let $\chi :\mathcal{X}\times \mathcal{X}\times \mathcal{X}\times \mathcal{X}\to [0,\infty )$ be a function such that

$$\underset{n\to \infty }{lim}\frac{1}{4^n}\chi (2^{n}x,2^{n}y,2^{n}z,2^{n}w)=0$$

(21)

and

$$L(x,y,z,w):=\sum _{j=0}^{\infty }\frac{1}{4^{pj}}\chi {(2^{j}x,2^{j}y,2^{j}z,2^{j}w)}^p<\infty$$

(22)

for all $x,y,z,w\in \mathcal{X}$.

Theorem 5.

Let $0<p\le 1$ and suppose that a mapping $f:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ satisfies $f(x,0)=0$ and the inequality

$$\parallel 4f\left(\frac{x+y}{2},\frac{z+w}{2}\right)-f(x,z)-f(x,w)-f(y,z)-f(y,w){\parallel }_{\mathcal{Y}}\le \chi (x,y,z,w)$$

(23)

for all $x,y,z,w\in \mathcal{X}$. Then the limit $F(x,y):={lim}_{j\to \infty }\frac{1}{4^j}f(2^{j}x,2^{j}y)$ exists for all $x,y\in X$ and the mapping $F:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ is the unique bi-Jensen mapping satisfying

$${\parallel f(x,y)-f(0,y)-F(x,y)\parallel }_{\mathcal{Y}}\le \tilde{\chi }{(x,y)}^{\frac{1}{p}},$$

(24)

where

$$\tilde{\chi }(x,y)=\sum _{j=0}^{\infty }\frac{1}{4^{p(j+1)}}\left[\chi {(2^{j+1}x,0,2^{j+1}y,0)}^p+\chi {(0,0,2^{j+1}y,0)}^p\right]$$

for all $x,y\in \mathcal{X}$.

Proof. 

Replacing x by $2^{j+1}x$ and putting $y=0,z=2^{j+1}y,w=0$ in (23), we gain

$${∥\frac{1}{4^j}f(2^{j}x,2^{j}y)-\frac{1}{4^{j+1}}f(2^{j+1}x,2^{j+1}y)-\frac{1}{4^{j+1}}f(0,2^{j+1}y)∥}_{\mathcal{Y}}\le \frac{1}{4^{j+1}}\chi (2^{j+1}x,0,2^{j+1}y,0)$$

(25)

for all $x,y\in \mathcal{X}$ and all j. Letting $x=0$ in (25), we get

$${∥\frac{1}{4^j}f(0,2^{j}y)-\frac{2}{4^{j+1}}f(0,2^{j+1}y)∥}_{\mathcal{Y}}\le \frac{1}{4^{j+1}}\chi (0,0,2^{j+1}y,0)$$

(26)

for all $y\in \mathcal{X}$ and all j. By (25) and(26), we have

$$\begin{array}{cc}& \parallel \frac{1}{4^j}\left[f(2^{j}x,2^{j}y)-f(0,2^{j}y)\right]-\frac{1}{4^{j+1}}\left[f(2^{j+1}x,2^{j+1}y)-f(0,2^{j+1}y)\right]{\parallel }_{\mathcal{Y}}^p\hfill \\ & \le \frac{1}{4^{p(j+1)}}\left[\chi {(2^{j+1}x,0,2^{j+1}y,0)}^p+\chi {(0,0,2^{j+1}y,0)}^p\right]\hfill \end{array}$$

(27)

for all $x,y\in \mathcal{X}$ and all j. Thus we have

$$\begin{array}{cc}& \parallel \frac{1}{4^l}\left[f(2^{l}x,2^{l}y)-f(0,2^{l}y)\right]-\frac{1}{4^m}\left[f(2^{m}x,2^{m}y)-f(0,2^{m}y)\right]{\parallel }_{\mathcal{Y}}^p\hfill \\ & \le \sum _{j=l}^{m-1}\frac{1}{4^{p(j+1)}}\left[\chi {(2^{j+1}x,0,2^{j+1}y,0)}^p+\chi {(0,0,2^{j+1}y,0)}^p\right]\hfill \end{array}$$

(28)

for all integers $l,m(0\le l<m)$ and all $x,y\in \mathcal{X}$. By (22), the sequence $\left\{\frac{1}{4^j}[f(2^{j}x,2^{j}y)-f(0,2^{j}y)]\right\}$ is a Cauchy sequence for all $x,y\in \mathcal{X}$. Since $\mathcal{Y}$ is complete, the sequence $\left\{\frac{1}{4^j}[f(2^{j}x,2^{j}y)-f(0,2^{j}y)]\right\}$ converges for all $x,y\in \mathcal{X}$. So one can define the mapping $F:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ by

$$F(x,y):=\underset{n\to \infty }{lim}\frac{1}{4^n}[f(2^{n}x,2^{n}y)-f(0,2^{n}y)]$$

(29)

for all $x,y\in \mathcal{X}$. Letting $l=0$ and taking the limit $m\to \infty$ in (28), we get (24). Now, we show that F is a bi-Jensen mapping.

On the other hand it follows from (22), (23) and (29) that

$$\begin{array}{cc}& \parallel 4F\left(\frac{x+y}{2},\frac{z+w}{2}\right)-F(x,z)-F(x,w)-F(y,z)-F(y,w){\parallel }_{\mathcal{Y}}^p\hfill \\ & =\underset{n\to \infty }{lim}\frac{1}{4^{pn}}\parallel 4f\left(\frac{2^{n}x+2^{n}y}{2},\frac{2^{n}z+2^{n}w}{2}\right)-f(2^{n}x,2^{n}z)-f(2^{n}x,2^{n}w)-f(2^{n}y,2^{n}z)\hfill \\ & -f(2^{n}y,2^{n}w)-4f\left(0,\frac{2^{n}z+2^{n}w}{2}\right)+2f(0,2^{n}z)+2f(0,2^{n}w){\parallel }_{\mathcal{Y}}^p\hfill \\ & =\underset{n\to \infty }{lim}\frac{1}{4^{pn}}[\chi {(2^{n}x,2^{n}y,2^{n}z,2^{n}w)}^p+\chi {(0,0,2^{n}z,2^{n}w)}^p]=0\hfill \end{array}$$

for all $x,y,z,w\in \mathcal{X}$. Hence the mapping F satisfies (3).

To prove the uniqueness of F, let $G:\mathcal{X}\to \mathcal{Y}$ be another bi-Jensen mapping satisfying (24). It follows from (22) that

$$\underset{n\to \infty }{lim}\frac{1}{4^{pn}}L(2^{n}x,2^{n}y,2^{n}z,2^{n}w)=\underset{n\to \infty }{lim}\sum _{j=n}^{\infty }\frac{1}{4^{pj}}\chi {(2^{j}x,2^{j}y,2^{j}z,2^{j}w)}^p=0$$

for all $x,y,z,w\in \mathcal{X}$. Hence ${lim}_{n\to \infty }\frac{1}{4^{pn}}\tilde{\chi }(2^{n}x,2^{n}y)=0$ for all $x,y\in \mathcal{X}$. It follows from (21), (27) and (29) the above equality that

$$\begin{array}{cc}& {\parallel F(2x,2y)-4F(x,y)\parallel }_{}Y\hfill \\ & =\underset{n\to \infty }{lim}\parallel \frac{1}{4^n}f(2^{n+1}x,2^{n+1}y)-f(0,2^{n+1}y)-\frac{1}{4^{n-1}}f(2^{n}x,2^{n}y)+f(0,2^{n}y){\parallel }_{}Y\hfill \\ & =4\underset{n\to \infty }{lim}\parallel \frac{1}{4^n}[f(2^{n}x,2^{n}y)-f(0,2^{n}y)]-\frac{1}{4^{n+1}}[f(2^{n+1}x,2^{n+1}y)-f(0,2^{n+1}y)]{\parallel }_{}Y\hfill \\ & \le 4\underset{n\to \infty }{lim}\frac{1}{4^{p(n+1)}}\left[\chi {(2^{n+1}x,0,2^{n+1}y,0)}^p+\chi {(0,0,2^{n+1}y,0)}^p\right]=0\hfill \end{array}$$

for all $x,y\in \mathcal{X}$. So $F(2x,2y)=4F(x,y)$ for all $x,y\in \mathcal{X}$. Thus it follows from (24) and (29) that

$$\begin{array}{cc}\hfill {\parallel F(x,y)-G(x,y)\parallel }_{\mathcal{Y}}^p& =\underset{n\to \infty }{lim}\frac{1}{4^{pn}}{\parallel f(2^{n}x,2^{n}y)-f(0,2^{n}y)-G(2^{n}x,2^{n}y)\parallel }_{\mathcal{Y}}^p\hfill \\ & \le \underset{n\to \infty }{lim}\frac{1}{4^{pn}}\tilde{\chi }(2^{n}x,2^{n}y)=0\hfill \end{array}$$

for all $x,y\in \mathcal{X}$. So $F=G$. □

Corollary 4.

Let $0<p\le 1$ and $\epsilon >0$ be fixed. Suppose that a mapping $f:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ satisfies the inequalities

$$\begin{array}{c}\hfill \parallel 4f\left(\frac{x+y}{2},\frac{z+w}{2}\right)-f(x,z)-f(x,w)-f(y,z)-f(y,w){\parallel }_{\mathcal{Y}}\le \epsilon \end{array}$$

for all $x,y,z,w\in \mathcal{X}$. Then there exists a unique bi-Jensen mapping $F:\mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ satisfying

$${\parallel f(x,y)-f(0,y)-F(x,y)\parallel }_{\mathcal{Y}}\le \epsilon {\left(\frac{2}{4^p-1}\right)}^{\frac{1}{p}}$$

for all $x,y\in \mathcal{X}$.

Proof. 

Taking $\chi (x,y,z,w):=\epsilon$ for all $x,y,z,w\in \mathcal{X}$ in Theorem 5, we obtain $\tilde{\chi }(x,y)=\frac{2{\epsilon }^p}{4^p-1}$ for all $x,y\in \mathcal{X}$. Thus we obtain the estimate value $\tilde{\chi }{(x,y)}^{\frac{1}{p}}=\epsilon {\left(\frac{2}{4^p-1}\right)}^{\frac{1}{p}}$ for all $x,y\in \mathcal{X}$. □