Additive functional inequalities in 2-Banach spaces

Abstract

We prove the Hyers-Ulam stability of the Cauchy functional inequality and theCauchy-Jensen functional inequality in 2-Banach spaces.

Moreover, we prove the superstability of the Cauchy functional inequality and theCauchy-Jensen functional inequality in 2-Banach spaces under someconditions.

MSC: 39B82, 39B52, 39B62, 46B99, 46A19.

1 Introduction and preliminaries

In 1940, Ulam [1] suggested the stability problem of functional equations concerning thestability of group homomorphisms as follows: Let$(\mathcal{G},\circ )$be a group and let$(\mathcal{H},\star ,d)$be a metric group with the metric$d(\cdot ,\cdot )$. Given$\epsilon >0$, does there exist a$\delta =\delta (\epsilon )>0$such that if a mapping$f:\mathcal{G}\to \mathcal{H}$satisfies the inequality

$$d(f(x\circ y),f(x)\star f(y))<\delta$$

for all $x,y\in \mathcal{G}$ , then a homomorphism $F:\mathcal{G}\to \mathcal{H}$ exists with

$$d(f(x),F(x))<\epsilon$$

for all $x\in \mathcal{G}$ ?

In 1941, Hyers [2] gave a first (partial) affirmative answer to the question of Ulam forBanach spaces. Thereafter, we call that type the Hyers-Ulam stability.

Hyers’ theorem was generalized by Aoki [3] for additive mappings and by Rassias [4] for linear mappings by considering an unbounded Cauchy difference. Ageneralization of the Rassias theorem was obtained by Gǎvruta [5] by replacing the unbounded Cauchy difference by a general controlfunction.

Gähler [6, 7] introduced the concept of linear 2-normed spaces.

Definition 1.1 Let be a real linear space with $dim\mathcal{X}>1$, and let $\parallel \cdot ,\cdot \parallel :\mathcal{X}\times \mathcal{X}\to {\mathbb{R}}_{\ge 0}$ be a function satisfying the following properties:

• (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) $\parallel \alpha x,y\parallel =|\alpha |\parallel x,y\parallel$,

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

for all $x,y,z\in \mathcal{X}$ and $\alpha \in \mathbb{R}$. Then the function $\parallel \cdot ,\cdot \parallel$ is called 2-norm on and the pair $(\mathcal{X},\parallel \cdot ,\cdot \parallel )$ is called a linear 2-normed space.Sometimes condition (d) is called the triangle inequality.

See [8] for examples and properties of linear 2-normed spaces.

White [9, 10] introduced the concept of 2-Banach spaces. In order to definecompleteness, the concepts of Cauchy sequences and convergence are required.

Definition 1.2 A sequence $\{x_n\}$ in a linear 2-normed space is called a Cauchy sequence if

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

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

Definition 1.3 A sequence $\{x_n\}$ in a linear 2-normed space is called a convergent sequenceif there is an $x\in \mathcal{X}$ such that

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

for all $y\in \mathcal{X}$. If $\{x_n\}$ converges to x, write$x_n\to x$ as $n\to \mathrm{\infty }$ and call x the limit of$\{x_n\}$. In this case, we also write ${lim}_{n\to \mathrm{\infty }}{x}_n=x$.

The triangle inequality implies the following lemma.

Lemma 1.4[11]

For a convergent sequence$\{x_n\}$in a linear 2-normed space,

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

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

Definition 1.5 A linear 2-normed space, in which every Cauchy sequence is aconvergent sequence, is called a 2-Banach space.

Eskandani and Gǎvruta [12] proved the Hyers-Ulam stability of a functional equation in 2-Banachspaces.

In [13], Gilányi showed that if f satisfies the functionalinequality

$$\parallel 2f(x)+2f(y)-f\left(x{y}^{-1}\right)\parallel \le \parallel f(xy)\parallel ,$$

(1.1)

then f satisfies the Jordan-von Neumann functional equation

$$2f(x)+2f(y)=f(xy)+f\left(x{y}^{-1}\right).$$

See also [14]. Gilányi [15] and Fechner [16] proved the Hyers-Ulam stability of functional inequality (1.1).

Park et al.[17] proved the Hyers-Ulam stability of the following functional inequalities:

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

(1.2)

$$\parallel f(x)+f(y)+2f(z)\parallel \le \parallel 2f(\frac{x+y}{2}+z)\parallel .$$

(1.3)

In this paper, we prove the Hyers-Ulam stability of Cauchy functional inequality(1.2) and Cauchy-Jensen functional inequality (1.3) in 2-Banach spaces.

Moreover, we prove the superstability of Cauchy functional inequality (1.2) andCauchy-Jensen functional inequality (1.3) in 2-Banach spaces under someconditions.

Throughout this paper, let be a normed linear space, and let be a 2-Banach space.

2 Hyers-Ulam stability of Cauchy functional inequality (1.2) in 2-Banachspaces

In this section, we prove the Hyers-Ulam stability of Cauchy functional inequality(1.2) in 2-Banach spaces.

Proposition 2.1 Let $f:\mathcal{X}\to \mathcal{Y}$ be a mapping satisfying

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

(2.1)

for all$x,y,z\in \mathcal{X}$and all$w\in \mathcal{Y}$. Then the mapping$f:\mathcal{X}\to \mathcal{Y}$is additive.

Proof Letting $x=y=z=0$ in (2.1), we get $3\parallel f(0),w\parallel \le \parallel f(0),w\parallel$ and so $\parallel f(0),w\parallel =0$ for all $w\in \mathcal{Y}$. Hence $f(0)=0$.

Letting $y=-x$ and $z=0$ in (2.1), we get $\parallel f(x)+f(-x),w\parallel \le \parallel f(0),w\parallel =0$ and so $\parallel f(x)+f(-x),w\parallel =0$ for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. Hence $f(x)+f(-x)=0$ for all $x\in \mathcal{X}$.

Letting $z=-x-y$ in (2.1), we get

$$\parallel f(x)+f(y)+f(-x-y),w\parallel \le \parallel f(0),w\parallel =0$$

and so

$$\parallel f(x)+f(y)+f(-x-y),w\parallel =0$$

for all $x,y\in \mathcal{X}$ and all $w\in \mathcal{Y}$. Hence

$$0=f(x)+f(y)+f(-x-y)=f(x)+f(y)-f(x+y)$$

for all $x,y\in \mathcal{X}$. So, $f:\mathcal{X}\to \mathcal{Y}$ is additive. □

Theorem 2.2 Let$\theta \in [0,\mathrm{\infty })$, $p,q,r\in (0,\mathrm{\infty })$with$p+q+r<1$, and let$f:\mathcal{X}\to \mathcal{Y}$be a mapping satisfying

$$\parallel f(x)+f(y)+f(z),w\parallel \le \parallel f(x+y+z),w\parallel +\theta {\parallel x\parallel }^p{\parallel y\parallel }^q{\parallel z\parallel }^r\parallel w\parallel$$

(2.2)

for all$x,y,z\in \mathcal{X}$and all$w\in \mathcal{Y}$. Then there is a unique additive mapping$A:\mathcal{X}\to \mathcal{Y}$such that

$$\parallel f(x)-A(x),w\parallel \le \frac{2^r\theta }{2-2^{p+q+r}}{\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

(2.3)

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

Proof Letting $x=y=z=0$ in (2.2), we get $3\parallel f(0),w\parallel \le \parallel f(0),w\parallel$ and so $\parallel f(0),w\parallel =0$ for all $w\in \mathcal{Y}$. Hence $f(0)=0$.

Letting $y=-x$ and $z=0$ in (2.2), we get $\parallel f(x)+f(-x),w\parallel \le \parallel f(0),w\parallel =0$ and so $\parallel f(x)+f(-x),w\parallel =0$ for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. Hence $f(x)+f(-x)=0$ for all $x\in \mathcal{X}$.

Putting $y=x$ and $z=-2x$ in (2.2), we get

$$\parallel f(2x)-2f(x),w\parallel \le \parallel f(0),w\parallel +2^r\theta {\parallel x\parallel }^{p+q+r}\parallel w\parallel =2^r\theta {\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

(2.4)

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. So, we get

$$\parallel f(x)-\frac{1}{2}f(2x),w\parallel \le \frac{2^r\theta }{2}{\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

(2.5)

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. Replacing x by $2^{j}x$ in (2.5) and dividing by $2^j$, we obtain

$$\parallel \frac{1}{2^j}f\left(2^{j}x\right)-\frac{1}{2^{j+1}}f\left(2^{j+1}x\right),w\parallel \le 2^{(p+q+r-1)j+r-1}\theta {\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

for all $x\in \mathcal{X}$, all $w\in \mathcal{Y}$ and all integers $j\ge 0$. For all integers l, m with$0\le l<m$, we get

$$\parallel \frac{1}{2^l}f\left(2^{l}x\right)-\frac{1}{2^m}f\left(2^{m}x\right),w\parallel \le \sum _{j=l}^{m-1}{2}^{(p+q+r-1)j+r-1}\theta {\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

(2.6)

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. So, we get

$$\underset{l\to \mathrm{\infty }}{lim}\parallel \frac{1}{2^l}f\left(2^{l}x\right)-\frac{1}{2^m}f\left(2^{m}x\right),w\parallel =0$$

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. Thus the sequence $\{\frac{1}{2^j}f(2^{j}x)\}$ is a Cauchy sequence in for each $x\in \mathcal{X}$. Since is a 2-Banach space, the sequence $\{\frac{1}{2^j}f(2^{j}x)\}$ converges for each $x\in \mathcal{X}$. So, one can define the mapping$A:\mathcal{X}\to \mathcal{Y}$ by

$$A(x):=\underset{j\to \mathrm{\infty }}{lim}\frac{1}{2^j}f\left(2^{j}x\right)$$

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

$$\underset{j\to \mathrm{\infty }}{lim}\parallel \frac{1}{2^j}f\left(2^{j}x\right)-A(x),w\parallel =0$$

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

By (2.2), we get

$$\begin{array}{r}\underset{j\to \mathrm{\infty }}{lim}\parallel \frac{1}{2^j}(f\left(2^{j}x\right)+f\left(2^{j}y\right)+f\left(2^{j}z\right)),w\parallel \\ \le \underset{j\to \mathrm{\infty }}{lim}(\frac{1}{2^j}\parallel f(2^{j}x+2^{j}y+2^{j}z),w\parallel +\frac{2^{(p+q+r)j}}{2^j}\theta {\parallel x\parallel }^p{\parallel y\parallel }^q{\parallel z\parallel }^r\parallel w\parallel )\\ \le \underset{j\to \mathrm{\infty }}{lim}\frac{1}{2^j}\parallel f(2^{j}x+2^{j}y+2^{j}z),w\parallel \end{array}$$

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

$$\parallel A(x)+A(y)+A(z),w\parallel \le \parallel A(x+y+z),w\parallel$$

for all $x,y,z\in \mathcal{X}$ and all $w\in \mathcal{Y}$. By Proposition 2.1, $A:\mathcal{X}\to \mathcal{Y}$ is additive.

By Lemma 1.4 and (2.6), we have

$$\parallel f(x)-A(x),w\parallel =\underset{m\to \mathrm{\infty }}{lim}\parallel f(x)-\frac{1}{2^m}f\left(2^{m}x\right),w\parallel \le \frac{2^r\theta }{2-2^{p+q+r}}{\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

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

Now, let $B:\mathcal{X}\to \mathcal{Y}$ be another additive mapping satisfying (2.3). Then wehave

$$\begin{array}{rl}\parallel A(x)-B(x),w\parallel & =\frac{1}{2^j}\parallel A\left(2^{j}x\right)-B\left(2^{j}x\right),w\parallel \\ \le \frac{1}{2^j}[\parallel A\left(2^{j}x\right)-f\left(2^{j}x\right),w\parallel +\parallel f\left(2^{j}x\right)-B\left(2^{j}x\right),w\parallel ]\\ \le \frac{2\cdot 2^r\theta }{2-2^{p+q+r}}{\parallel x\parallel }^{p+q+r}\parallel w\parallel \cdot \frac{2^{(p+q+r)j}}{2^j},\end{array}$$

which tends to zero as $j\to \mathrm{\infty }$ for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. By Definition 1.1, we can conclude that$A(x)=B(x)$ for all $x\in \mathcal{X}$. This proves the uniqueness ofA. □

Theorem 2.3 Let$\theta \in [0,\mathrm{\infty })$, $p,q,r\in (0,\mathrm{\infty })$with$p+q+r>1$, and let$f:\mathcal{X}\to \mathcal{Y}$be a mapping satisfying (2.2). Then there is a unique additivemapping$A:\mathcal{X}\to \mathcal{Y}$such that

$$\parallel f(x)-A(x),w\parallel \le \frac{2^r\theta }{2^{p+q+r}-2}{\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

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

Proof It follows from (2.4) that

$$\parallel f(x)-2f\left(\frac{x}{2}\right),w\parallel \le \frac{\theta }{2^{p+q}}{\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

(2.7)

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. Replacing x by $\frac{x}{2^j}$ in (2.7) and multiplying by $2^j$, we obtain

$$\parallel 2^{j}f\left(\frac{x}{2^j}\right)-2^{j+1}f\left(\frac{x}{2^{j+1}}\right),w\parallel \le \frac{2^j\theta }{2^{p+q}\cdot 2^{(p+q+r)j}}{\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$ and all integers $j\ge 0$. For all integers l, m with$0\le l<m$, we get

$$\parallel 2^{l}f\left(\frac{x}{2^l}\right)-2^{m}f\left(\frac{x}{2^m}\right),w\parallel \le \sum _{j=l}^{m-1}\frac{2^j\theta }{2^{p+q}\cdot 2^{(p+q+r)j}}{\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. So, we get

$$\underset{l\to \mathrm{\infty }}{lim}\parallel 2^{l}f\left(\frac{x}{2^l}\right)-2^{m}f\left(\frac{x}{2^m}\right),w\parallel =0$$

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. Thus the sequence $\{2^{j}f(\frac{x}{2^j})\}$ is a Cauchy sequence in . Since is a 2-Banach space, the sequence$\{2^{j}f(\frac{x}{2^j})\}$ converges. So, one can define the mapping$A:\mathcal{X}\to \mathcal{Y}$ by

$$A(x):=\underset{j\to \mathrm{\infty }}{lim}{2}^{j}f\left(\frac{x}{2^j}\right)$$

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

$$\underset{j\to \mathrm{\infty }}{lim}\parallel 2^{j}f\left(\frac{x}{2^j}\right)-A(x),w\parallel =0$$

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

The further part of the proof is similar to the proof of Theorem2.2. □

Now we prove the superstability of the Cauchy functional inequality in 2-Banachspaces.

Theorem 2.4 Let$\theta \in [0,\mathrm{\infty })$, $p,q,r,t\in (0,\mathrm{\infty })$with$t\ne 1$, and let$f:\mathcal{X}\to \mathcal{Y}$be a mapping satisfying

$$\parallel f(x)+f(y)+f(z),w\parallel \le \parallel f(x+y+z),w\parallel +\theta {\parallel x\parallel }^p{\parallel y\parallel }^q{\parallel z\parallel }^r{\parallel w\parallel }^t$$

(2.8)

for all$x,y,z\in \mathcal{X}$and all$w\in \mathcal{Y}$. Then$f:\mathcal{X}\to \mathcal{Y}$is an additive mapping.

Proof Replacing w by sw in (2.8) for$s\in \mathbb{R}\setminus \{0\}$, we get

$$\parallel f(x)+f(y)+f(z),sw\parallel \le \parallel f(x+y+z),sw\parallel +\theta {\parallel x\parallel }^p{\parallel y\parallel }^q{\parallel z\parallel }^r{\parallel sw\parallel }^t$$

and so

$$\parallel f(x)+f(y)+f(z),w\parallel \le \parallel f(x+y+z),w\parallel +\theta {\parallel x\parallel }^p{\parallel y\parallel }^q{\parallel z\parallel }^r{\parallel w\parallel }^t\frac{{|s|}^t}{|s|}$$

(2.9)

for all $x,y,z\in \mathcal{X}$, all $w\in \mathcal{Y}$ and all $s\in \mathbb{R}\setminus \{0\}$.

If $t>1$, then the right-hand side of (2.9) tends to$\parallel f(x+y+z),w\parallel$ as $s\to 0$.

If $t<1$, then the right-hand side of (2.9) tends to$\parallel f(x+y+z),w\parallel$ as $s\to +\mathrm{\infty }$.

Thus

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

for all $x,y,z\in \mathcal{X}$ and all $w\in \mathcal{Y}$. By Proposition 2.1, $f:\mathcal{X}\to \mathcal{Y}$ is additive. □

3 Hyers-Ulam stability of Cauchy-Jensen functional inequality (1.3) in 2-Banachspaces

In this section, we prove the Hyers-Ulam stability of Cauchy-Jensen functionalinequality (1.3) in 2-Banach spaces.

Proposition 3.1 Let $f:\mathcal{X}\to \mathcal{Y}$ be a mapping satisfying

$$\parallel f(x)+f(y)+2f(z),w\parallel \le \parallel 2f(\frac{x+y}{2}+z),w\parallel$$

(3.1)

for all$x,y,z\in \mathcal{X}$and all$w\in \mathcal{Y}$. Then the mapping$f:\mathcal{X}\to \mathcal{Y}$is additive.

Proof Letting $x=y=z=0$ in (3.1), we get $4\parallel f(0),w\parallel \le 2\parallel f(0),w\parallel$ and so $\parallel f(0),w\parallel =0$ for all $w\in \mathcal{Y}$. Hence $f(0)=0$.

Letting $y=-x$ and $z=0$ in (3.1), we get $\parallel f(x)+f(-x),w\parallel \le 2\parallel f(0),w\parallel =0$ and so $\parallel f(x)+f(-x),w\parallel =0$ for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. Hence $f(x)+f(-x)=0$ for all $x\in \mathcal{X}$.

Letting $z=-\frac{x+y}{2}$ in (3.1), we get

$$\parallel f(x)+f(y)+2f(-\frac{x+y}{2}),w\parallel \le 2\parallel f(0),w\parallel =0$$

and so

$$\parallel f(x)+f(y)+2f(-\frac{x+y}{2}),w\parallel =0$$

for all $x,y\in \mathcal{X}$ and all $w\in \mathcal{Y}$. Hence

$$0=f(x)+f(y)+2f(-\frac{x+y}{2})=f(x)+f(y)-2f\left(\frac{x+y}{2}\right)$$

for all $x,y\in \mathcal{X}$. Since $f(0)=0$, $f:\mathcal{X}\to \mathcal{Y}$ is additive. □

Theorem 3.2 Let$\theta \in [0,\mathrm{\infty })$, $p,q,r\in (0,\mathrm{\infty })$with$p+q+r<1$, and let$f:\mathcal{X}\to \mathcal{Y}$be a mapping satisfying

$$\parallel f(x)+f(y)+2f(z),w\parallel \le \parallel 2f(\frac{x+y}{2}+z),w\parallel +\theta {\parallel x\parallel }^p{\parallel y\parallel }^q{\parallel z\parallel }^r\parallel w\parallel$$

(3.2)

for all$x,y,z\in \mathcal{X}$and all$w\in \mathcal{Y}$. Then there is a unique additive mapping$A:\mathcal{X}\to \mathcal{Y}$such that

$$\parallel f(x)-A(x),w\parallel \le \frac{\theta }{2-2^{p+q+r}}{\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

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

Proof Letting $x=y=z=0$ in (3.2), we get $4\parallel f(0),w\parallel \le 2\parallel f(0),w\parallel$ and so $\parallel f(0),w\parallel =0$ for all $w\in \mathcal{Y}$. Hence $f(0)=0$.

Letting $y=-x$ and $z=0$ in (3.2), we get $\parallel f(x)+f(-x),w\parallel \le 2\parallel f(0),w\parallel =0$ and so $\parallel f(x)+f(-x),w\parallel =0$ for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. Hence $f(x)+f(-x)=0$ for all $x\in \mathcal{X}$.

Letting $y=x$ and $z=-x$ in (3.2), we get

$$\parallel 2f(x)-f(2x),w\parallel \le \parallel 2f(0),w\parallel +\theta {\parallel x\parallel }^{p+q+r}\parallel w\parallel =\theta {\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

(3.3)

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. Replacing x by $2^{j}x$ in (3.3) and dividing by $3^j$, we obtain

$$\parallel \frac{1}{2^j}f\left(2^{j}x\right)-\frac{1}{2^{j+1}}f\left(2^{j+1}x\right),w\parallel \le \frac{2^{(p+q+r)j}}{2\cdot 2^j}\theta {\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$ and all integers $j\ge 0$. For all integers l, m with$0\le l<m$, we get

$$\parallel \frac{1}{2^l}f\left(2^{l}x\right)-\frac{1}{2^m}f\left(2^{m}x\right),w\parallel \le \sum _{j=l}^{m-1}\frac{2^{(p+q+r)j}}{2\cdot 2^j}\theta {\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. So, we get

$$\underset{l\to \mathrm{\infty }}{lim}\parallel \frac{1}{2^l}f\left(2^{l}x\right)-\frac{1}{2^m}f\left(2^{m}x\right),w\parallel =0$$

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. Thus the sequence $\{\frac{1}{2^j}f(2^{j}x)\}$ is a Cauchy sequence in for each $x\in \mathcal{X}$. Since is a 2-Banach space, the sequence $\{\frac{1}{2^j}f(2^{j}x)\}$ converges for each $x\in \mathcal{X}$. So, one can define the mapping$A:\mathcal{X}\to \mathcal{Y}$ by

$$A(x):=\underset{j\to \mathrm{\infty }}{lim}\frac{1}{2^j}f\left(2^{j}x\right)=\underset{j\to \mathrm{\infty }}{lim}\frac{1}{2^j}f\left(2^{j}x\right)$$

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

$$\underset{j\to \mathrm{\infty }}{lim}\parallel \frac{1}{2^j}f\left(2^{j}x\right)-A(x),w\parallel =\underset{j\to \mathrm{\infty }}{lim}\parallel \frac{1}{2^j}f\left(2^{j}x\right)-A(x),w\parallel =0$$

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

The further part of the proof is similar to the proof of Theorem2.2. □

Theorem 3.3 Let$\theta \in [0,\mathrm{\infty })$, $p,q,r\in (0,\mathrm{\infty })$with$p+q+r>1$, and let$f:\mathcal{X}\to \mathcal{Y}$be a mapping satisfying (3.2). Then there is a unique additivemapping$A:\mathcal{X}\to \mathcal{Y}$such that

$$\parallel f(x)-A(x),w\parallel \le \frac{\theta }{2^{p+q+r}-2}{\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

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

Proof It follows from (3.3) that

$$\parallel f(x)-2f\left(\frac{x}{2}\right),w\parallel \le \frac{1}{2^{p+q+r}}\theta {\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

(3.4)

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. Replacing x by $\frac{x}{2^j}$ in (3.4) and multiplying by $2^j$, we obtain

$$\parallel 2^{j}f\left(\frac{x}{2^j}\right)-2^{j+1}f\left(\frac{x}{2^{j+1}}\right),w\parallel \le \frac{2^j}{2^{(p+q+r)(j+1)}}\theta {\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$ and all integers $j\ge 0$. For all integers l, m with$0\le l<m$, we get

$$\parallel 2^{l}f\left(\frac{x}{2^l}\right)-2^{m}f\left(\frac{x}{2^m}\right),w\parallel \le \sum _{j=l}^{m-1}\frac{2^j}{2^{(p+q+r)(j+1)}}\theta {\parallel x\parallel }^{p+q+r}\parallel w\parallel$$

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. So, we get

$$\underset{l\to \mathrm{\infty }}{lim}\parallel 2^{l}f\left(\frac{x}{2^l}\right)-2^{m}f\left(\frac{x}{2^m}\right),w\parallel =0$$

for all $x\in \mathcal{X}$ and all $w\in \mathcal{Y}$. Thus the sequence $\{2^{j}f(\frac{x}{2^j})\}$ is a Cauchy sequence in for each $x\in \mathcal{X}$. Since is a 2-Banach space, the sequence $\{2^{j}f(\frac{x}{2^j})\}$ converges for each $x\in \mathcal{X}$. So, one can define the mapping$A:\mathcal{X}\to \mathcal{Y}$ by

$$A(x):=\underset{j\to \mathrm{\infty }}{lim}{2}^{j}f\left(\frac{x}{2^j}\right)$$

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

$$\underset{j\to \mathrm{\infty }}{lim}\parallel 2^{j}f\left(\frac{x}{2^j}\right)-A(x),w\parallel =0$$

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

The further part of the proof is similar to the proof of Theorem2.2. □

Now we prove the superstability of the Jensen functional equation in 2-Banachspaces.

Theorem 3.4 Let$\theta \in [0,\mathrm{\infty })$, $p,q,r,t\in (0,\mathrm{\infty })$with$t\ne 1$, and let$f:\mathcal{X}\to \mathcal{Y}$be a mapping satisfying

$$\parallel f(x)+f(y)+2f(z),w\parallel \le \parallel 2f(\frac{x+y}{2}+z),w\parallel +\theta {\parallel x\parallel }^p{\parallel y\parallel }^q{\parallel z\parallel }^r{\parallel w\parallel }^t$$

(3.5)

for all$x,y,z\in \mathcal{X}$and all$w\in \mathcal{Y}$. Then$f:\mathcal{X}\to \mathcal{Y}$is an additive mapping.

Proof Replacing w by sw in (3.5) for$s\in \mathbb{R}\setminus \{0\}$, we get

$$\parallel f(x)+f(y)+2f(z),sw\parallel \le \parallel 2f(\frac{x+y}{2}+z),sw\parallel +\theta {\parallel x\parallel }^p{\parallel y\parallel }^q{\parallel z\parallel }^r{\parallel sw\parallel }^t$$

and so

$$\parallel f(x)+f(y)+2f(z),w\parallel \le \parallel 2f(\frac{x+y}{2}+z),w\parallel +\theta {\parallel x\parallel }^p{\parallel y\parallel }^q{\parallel z\parallel }^r{\parallel w\parallel }^t\frac{{|s|}^t}{|s|}$$

for all $x,y,z\in \mathcal{X}$, all $w\in \mathcal{Y}$ and all $s\in \mathbb{R}\setminus \{0\}$.

The rest of the proof is similar to the proof of Theorem 2.4. □