Stability of functional equations in $(n,\beta)$-normed spaces

Abstract

In this paper, we first introduce the notions of $(n,\beta)$-normed space and non-Archimedean $(n,\beta)$-normed space, then we study the Hyers-Ulam stability of the Cauchy functional equation and the Jensen functional equation in non-Archimedean $(n,\beta)$-normed spaces and that of the pexiderized Cauchy functional equation in $(n,\beta)$-normed spaces.

1 Introduction

The stability problem of functional equations originated from a question of Ulam [1] in 1940 concerning the stability of group homomorphisms. Let $(G_{1},\cdot)$ be a group and let $(G_{2},\ast)$ be a metric group with the metric $d(\cdot,\cdot)$. Given $\varepsilon> 0$, does there exist $\delta>0$ such that if a mapping $h:G_{1}\rightarrow G_{2}$ satisfies the inequality $d(h(x\cdot y),h(x)\ast h(y))< \delta$ for all $x,y \in G_{1}$, then a homomorphism $H: G_{1} \rightarrow G_{2}$ exists with $d(h(x),H(x))< \varepsilon$ for all $x\in G_{1}$?

The case of approximately additive functions was solved by Hyers [2] under the assumption that $G_{1}$ and $G_{2}$ are Banach spaces. In 1978, Rassias [3] proved a generalization of the Hyers theorem for additive mappings. The result of Rassias has provided a lot of influence during the past 36 years in the development of a generalization of the Hyers-Ulam stability concept. This new concept is known as the Hyers-Ulam-Rassias stability of functional equation.

The stability problems of several functional equations have been extensively investigated by a number of authors, and there are many interesting results concerning this problem. A large list of references can be found in [4–11].

In [12, 13], Gähler introduced the theory of 2-norms and n-norms on a linear space. A systematic development of n-normed linear spaces is due to Kim and Cho [14], Malceski [15], Misiak [16] and Gunawan and Mashadi [17].

Recently, Park [18] investigated the approximate additive mappings, approximate Jensen mappings and approximate quadratic mappings in 2-Banach spaces. This is the first result for the stability problem of functional equations in 2-Banach spaces. In 2012, Xu and Rassias [19] examined the Hyers-Ulam stability of a general mixed additive and cubic functional equation in n-Banach spaces. In 2013, Xu [20] investigated approximate multi-Jensen, multi-Euler-Lagrange additive and quadratic mappings in n-Banach spaces.

In this paper, we first introduce the notions of $(n,\beta)$-normed space and non-Archimedean $(n,\beta)$-normed space, then we study the Hyers-Ulam stability of the Cauchy functional equation and the Jensen functional equation in non-Archimedean $(n,\beta)$-normed spaces in Section 2. Finally, in Section 3, we investigate the Hyers-Ulam stability of the pexiderized Cauchy functional equation in $(n,\beta)$-normed spaces.

Now, we give some concepts concerning the $(n,\beta)$-normed space.

Definition 1.1

Let X be a linear space over ℝ with $\dim X \geq n$, $n\in\mathbb{N}$ and $0<\beta\leq1$, let $\|\cdot,\ldots,\cdot\|_{\beta}:X^{n}\rightarrow\mathbb{R}$ be a function satisfying the following properties:

• (a) $\|x_{1},\ldots,x_{n}\|_{\beta}=0$ if and only if $x_{1},\ldots,x_{n}$ are linearly dependent;

• (b) $\|x_{1},\ldots,x_{n}\|_{\beta}$ is invariant under permutations of $x_{1},\ldots, x_{n}$;

• (c) $\|\alpha x_{1},\ldots, x_{n}\|_{\beta}= |\alpha|^{\beta}\|x_{1},\ldots, x_{n}\|_{\beta}$;

• (d) $\|x_{1},\ldots,x_{n-1},y+z\|_{\beta}\leq \|x_{1},\ldots,x_{n-1},y\|_{\beta}+\|x_{1},\ldots,x_{n-1},z\|_{\beta}$

for all $x_{1},\ldots,x_{n}\in X$ and $\alpha\in\mathbb{R}$.

Then the function $\|\cdot,\ldots,\cdot\|_{\beta}$ is called an $(n,\beta)$-norm on X and the pair $(X,\|\cdot,\ldots,\cdot\|_{\beta})$ is called a linear $(n,\beta)$-normed space or an $(n,\beta)$-normed space.

We remark that the concept of a linear $(n,\beta)$-normed space is a generalization of a linear n-normed space ($\beta=1$) and of a β-normed space ($n=1$). Now we present two examples about n-normed space.

Example 1.2

[19]

For $x_{1},\ldots, x_{n}\in\mathbb{R}^{n}$, the Euclidean n-norm $\|x_{1},\ldots,x_{n}\|_{E}$ is defined by

$$ \|x_{1},\ldots,x_{n}\|_{E}=\bigl\vert \operatorname{det}(x_{ij})\bigr\vert =\operatorname{abs} \left ( \left \vert \begin{array}{@{}c@{\quad}c@{\quad}c@{}} x_{11} & \cdots& x_{1n} \\ \vdots& \ddots& \vdots \\ x_{n1} & \cdots& x_{nn} \end{array} \right \vert \right ), $$

(1.1)

where $x_{i}=(x_{i1},x_{i2},\ldots,x_{in})\in\mathbb{R}^{n}$ for each $i=1,2,\ldots,n$.

Example 1.3

[19]

The standard n-norm on X, a real inner product space of dimension $\dim X\geq n$, is as follows:

$$ \|x_{1},x_{2},\ldots,x_{n}\|_{S}= \left \vert \begin{array}{@{}c@{\quad}c@{\quad}c@{}} \langle x_{1},x_{1}\rangle & \cdots& \langle x_{1},x_{n}\rangle \\ \vdots& \ddots& \vdots \\ \langle x_{n},x_{1}\rangle & \cdots& \langle x_{n},x_{n}\rangle \end{array} \right \vert ^{1/2}, $$

(1.2)

where $\langle \cdot,\cdot\rangle$ denotes the inner product on X. If $X=\mathbb{R}^{n}$, then this n-norm is exactly the same as the Euclidean n-norm $\|x_{1},\ldots,x_{n}\|_{E}$ mentioned earlier. For $n=1$, this n-norm is the usual norm $\|x_{1}\|=\langle x_{1},x_{1}\rangle^{1/2}$.

Lemma 1.4

Let $(X,\|\cdot,\ldots,\cdot\|_{\beta})$ be a linear $(n,\beta)$-normed space, $n\geq2$, $0<\beta\leq1$. If $x_{1}\in X$ and $\|x_{1},y_{1},\ldots,y_{n-1}\|_{\beta} = 0$ for all $y_{1},\ldots,y_{n-1}\in X$, then $x_{1} = 0$.

Proof

Since $\dim X\geq n$, we can take $y_{1},\ldots,y_{n}$ from X such that they are linearly independent. It follows from the assumption that $\| x_{1},y_{2},\ldots,y_{n}\|_{\beta} = 0$, then by the definition of linear $(n,\beta)$-normed space we have that $x_{1}, y_{2},\ldots,y_{n}$ are linearly dependent. Thus there exist $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\in\mathbb {R}$ with $(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})\neq(0,\ldots,0)$ such that

$$\alpha_{1}x_{1}+\alpha_{2}y_{2}+ \cdots+\alpha_{n}y_{n}=0. $$

Then we have $\alpha_{1}\neq0$. (If $\alpha_{1}=0$, since $y_{2},\ldots,y_{n}$ are linearly independent, then we have $\alpha_{2}=0,\ldots,\alpha_{n}=0$; this is a contradiction.) So we have

$$ x_{1}=-\frac{\alpha_{2}}{\alpha_{1}}y_{2}-\cdots-\frac{\alpha _{n}}{\alpha_{1}}y_{n}. $$

(1.3)

Hence $x_{1}\in \operatorname{span}\{y_{2},y_{3},\ldots, y_{n}\}$. Similarly, let $A_{i}=\{ y_{1},y_{2},\ldots, y_{n}\}\setminus\{y_{i}\}$, we can obtain that $x_{1}\in \operatorname{span} A_{i}$, $i=1,2,\ldots, n$. In the n-dimensional space $\operatorname{span} \{y_{1},y_{2},\ldots, y_{n}\}$, it is easy to get that $\bigcap_{i=1}^{n} \operatorname{span} A_{i}={0}$, from which it follows that $x_{1}=0$. □

Remark 1.5

Let $(X,\|\cdot,\ldots,\cdot\|_{\beta})$ be a linear $(n,\beta)$-normed space, $0<\beta\leq1$. One can show that conditions (b) and (d) in Definition 1.1 imply that

$$\bigl\vert \Vert x,z_{1},\ldots,z_{n-1}\Vert _{\beta}-\Vert y,z_{1},\ldots,z_{n-1}\Vert _{\beta}\bigr\vert \leq \Vert x-y,z_{1},\ldots,z_{n-1} \Vert _{\beta} $$

for all $x,y\in X$ and $z_{1},\ldots,z_{n-1}\in Y$.

Definition 1.6

A sequence $\{x_{m}\}$ in a linear $(n,\beta)$-normed space X is called a convergent sequence if there is $x\in X$ such that

$$\lim_{m\rightarrow\infty}\| x_{m}-x ,y_{1}, \ldots,y_{n-1} \|_{\beta} =0 $$

for all $y_{1},\ldots,y_{n-1} \in X$. In this case, we call that $\{x_{m}\}$ converges to x or that x is the limit of ${\{x_{m}\}}$, write ${x_{m}}\rightarrow x$ as $m \rightarrow\infty$ or $\lim_{m\rightarrow\infty} x_{m} = x$.

Definition 1.7

A sequence $\{x_{m}\}$ in a linear $(n,\beta)$-normed space X is called a Cauchy sequence if

$$\lim_{m,k\rightarrow\infty}\| x_{k}-x_{m}, z_{1},\ldots,z_{n-1} \| _{\beta} =0 $$

for all $z_{1},\ldots,z_{n-1}\in X$.

We can easily get the following lemma by Remark 1.5.

Lemma 1.8

For a convergent sequence $\{x_{m}\}$ in a linear $(n,\beta)$-normed space X,

$$\lim_{m\rightarrow\infty} \| x_{m} ,z_{1}, \ldots,z_{n-1} \|_{\beta}=\Bigl\| \lim_{m\rightarrow\infty} x_{m} ,z_{1},\ldots,z_{n-1} \Bigr\| _{\beta} $$

for all $z_{1},\ldots,z_{n-1}\in X$.

Definition 1.9

A linear $(n,\beta)$-normed space in which every Cauchy sequence is convergent is called a complete $(n,\beta)$-normed space.

In 1897, Hensel [21] introduced a normed space which does not have the Archimedean property. It turns out that non-Archimedean spaces have many nice applications (see [22–24]).

Definition 1.10

A field K equipped with a function (valuation) $| \cdot|$ from K into $[0,\infty)$ is called a non-Archimedean field if the function $| \cdot |:K\rightarrow[0,\infty)$ satisfies the following conditions:

(1):

$|r| =0$ if and only if $r=0$;

(2):

$|rs|=|r||s|$;

(3):

$|r+s|\leq \max \{|r|,|s|\}$ for all $r,s \in\mathbb{K}$;

(4):

there exists a member $a_{0}\in K$ such that $|a_{0}|\neq0,1$.

Definition 1.11

[25]

Let X be a vector space over a scalar field K with a non-Archimedean nontrivial valuation $|\cdot|$. A function $\|\cdot\|:X\rightarrow\mathbb{R}$ is a non-Archimedean norm (valuation) if it satisfies the following conditions:

(1′):

$\|x\| =0$ if and only if $x=0$;

(2′):

$\|rx\| =|r|\|x\|$;

(3′):

$\|x+y\|\leq \max \{\|x\|,\|y\|\}$ for all $x,y\in X$ and $r\in K$.

The pair $(X,\|\cdot\|)$ is called a non-Archimedean space if $\|\cdot\|$ is a non-Archimedean norm on X.

Definition 1.12

Let X be a real vector space with $\dim X \geq n$ over a scalar field K with a non-Archimedean nontrivial valuation $|\cdot|$, where n is a positive integer and β is a constant with $0<\beta\leq 1$. A real-valued function $\|\cdot,\ldots,\cdot\|_{\beta}:X^{n} \rightarrow\mathbb{R}$ is called an $(n,\beta)$-norm on X if the following conditions hold:

(N1′):

$\|x_{1},\ldots,x_{n}\|_{\beta}=0$ if and only if $x_{1},\ldots , x_{n}$ are linearly dependent;

(N2′):

$\|x_{1},\ldots,x_{n}\|_{\beta}$ is invariant under permutations of $x_{1},\ldots, x_{n}$;

(N3′):

$\|\alpha x_{1},x_{2},\ldots,x_{n}\|_{\beta}= |\alpha|^{\beta}\|x_{1},x_{2},\ldots,x_{n}\|_{\beta}$;

(N4′):

$\|x_{0}+x_{1},x_{2},\ldots,x_{n}\|_{\beta} \leq \max\{\|x_{0},x_{2},\ldots,x_{n}\|_{\beta},\|x_{1},x_{2},\ldots,x_{n}\| _{\beta}\}$

for all $\alpha\in K$ and $x_{0},x_{1},\ldots,x_{n}\in X$.

Then $(X,\|\cdot,\ldots,\cdot\|_{\beta})$ is called a non-Archimedean $(n,\beta)$-normed space.

It follows from the preceding definition that the non-Archimedean $(n,\beta)$-normed space is a non-Archimedean n-normed space if $\beta=1$, and a non-Archimedean β-normed space if $n=1$, respectively.

Remark 1.13

A sequence $\{x_{m}\}$ in a non-Archimedean $(n,\beta)$-normed space X is a Cauchy sequence if and only if $\{x_{m+1}-x_{m}\}$ converges to zero.

Proof

It follows from (N4′) that

$$\begin{aligned}& \|x_{m}-x_{k},y_{1},\ldots,y_{n-1} \|_{\beta} \\& \quad \leq \max \bigl\{ \Vert x_{j+1}-x_{j},y_{1}, \ldots,y_{n-1}\Vert _{\beta}:k\leq j\leq m-1\bigr\} \quad (m>k) \end{aligned}$$

for all $y_{1},\ldots,y_{n-1}\in X$. So a sequence $\{x_{m}\}$ is a Cauchy sequence in X if and only if $\{x_{m+1}-x_{m}\}$ converges to zero. □

Throughout this paper, let ℕ denote the set of positive integers and $j, k, m, n\in\mathbb{N}$, and let $n\ge2$ be fixed.

2 Cauchy functional equations

In this section, we assume that $|2|\neq1$. Under this condition we investigate the Hyers-Ulam stability of the Cauchy functional equation in which the target space Y is a complete non-Archimedean $(n,\beta)$-normed space. When the domain space X is a non-Archimedean β-normed space, we can formulate our result as follows.

Theorem 2.1

Suppose that X is a non-Archimedean $\beta_{1}$-normed space and that Y is a complete non-Archimedean $(n,\beta)$-normed space, where $n\ge2$, $0<\beta,\beta_{1} \leq1$. Let $\theta\in[0,\infty)$, $p, q\in(0,\infty)$ with $(p+q)\beta_{1}>\beta$, and let $\psi:\underbrace{Y\times Y\times\cdots\times Y}_{n-1} \rightarrow[0,\infty)$ be a function. Suppose that a mapping $f:X\rightarrow Y$ satisfies the inequality

$$ \bigl\Vert f(x+y)-f(x)-f(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\theta\|x\|_{\beta _{1}}^{p}\|y \|_{\beta_{1}}^{q}\psi(z_{1},\ldots,z_{n-1}) $$

(2.1)

for all $x,y\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Then there exists a unique additive mapping $A:X\rightarrow Y$ such that

$$ \bigl\Vert f(x)-A(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\theta\bigl\vert 2^{-\beta}\bigr\vert \|x\| _{\beta_{1}}^{p+q}\psi(z_{1},\ldots,z_{n-1}) $$

(2.2)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$.

Proof

Putting $y=x$ in (2.1) and dividing both sides by $|2^{\beta}|$, we get

$$ \biggl\Vert \frac{f(2x)}{2}-f(x),z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta}\leq \theta\bigl\vert 2^{-\beta}\bigr\vert \|x\|_{\beta_{1}}^{p+q}\psi(z_{1},\ldots,z_{n-1}) $$

(a)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Replacing x by $2^{m}x$ in (a) and dividing both sides by $|2^{m\beta}|$, we get

$$\begin{aligned}& \biggl\Vert \frac{f(2^{m+1}x)}{2^{m+1}}-\frac {f(2^{m}x)}{2^{m}},z_{1}, \ldots,z_{n-1}\biggr\Vert _{\beta} \\& \quad \leq \theta\biggl\vert \frac{1}{{2^{m\beta}}}\biggr\vert \biggl\vert \frac{1}{{2^{\beta }}}\biggr\vert \bigl\vert 2^{m(p+q)\beta_{1}}\bigr\vert \|x \|_{\beta_{1}}^{p+q}\psi(z_{1},\ldots ,z_{n-1}) \\& \quad = \theta\bigl\vert {2^{-\beta}}\bigr\vert \bigl\vert 2^{(p+q)\beta_{1}-\beta}\bigr\vert ^{m}\|x\|_{\beta_{1}}^{p+q} \psi(z_{1},\ldots,z_{n-1}) \end{aligned}$$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Since $(p+q)\beta _{1}>\beta$ and $|2|\neq1$, we have

$$\lim_{m\rightarrow\infty}\bigl\Vert 2^{-m-1}f\bigl(2^{m+1}x \bigr)-2^{-m}f\bigl(2^{m}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}=0 $$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Considering Remark 1.13, we get that $\{{2^{-m}f(2^{m}x)}\}$ is a Cauchy sequence in Y for all $x\in X$. Since Y is a complete space, we can define the mapping $A:X\rightarrow Y$ by

$$ A(x)=\lim_{m\rightarrow\infty}2^{-m}f\bigl(2^{m}x\bigr) $$

(b)

for all $x\in X$.

Next, we show that A is additive. It follows from (2.1), (b) and Lemma 1.8 that

$$\begin{aligned}& \bigl\Vert A(x+y)-A(x)-A(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad = \lim_{m\rightarrow\infty}\bigl\vert {2^{-m\beta}}\bigr\vert \bigl\Vert f\bigl(2^{m}x+2^{m}y\bigr)-f\bigl(2^{m}x \bigr)-f\bigl(2^{m}y\bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \lim_{m\rightarrow\infty}\theta\bigl\vert {2^{-m\beta}} \bigr\vert \bigl\Vert 2^{m}x\bigr\Vert _{\beta _{1}}^{p} \bigl\Vert 2^{m}y\bigr\Vert _{\beta_{1}}^{q} \psi(z_{1},\ldots,z_{n-1}) \\& \quad = \lim_{m\rightarrow\infty}\theta\bigl\vert 2^{(p+q)\beta_{1}-\beta}\bigr\vert ^{m}\Vert x\Vert _{\beta_{1}}^{p}\Vert y \Vert _{\beta_{1}}^{q}\psi(z_{1},\ldots,z_{n-1}) \end{aligned}$$

for all $x,y\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Since $(p+q)\beta _{1}>\beta$ and $|2|\neq1$, we get

$$\bigl\Vert A(x+y)-A(x)-A(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}=0 $$

for all $x,y\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. By Lemma 1.4, we get

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

for all $x,y\in X$. So the mapping A is additive.

Replacing x by 2x in (a) and dividing both sides by $|2^{\beta }|$, we get

$$ \biggl\Vert \frac{f(2^{2}x)}{{2^{2}}}-\frac{f(2x)}{2},z_{1},\ldots ,z_{n-1}\biggr\Vert _{\beta}\leq\theta\bigl\vert {2^{-2\beta}}\bigr\vert \|2x\|_{\beta _{1}}^{p+q} \psi(z_{1},\ldots,z_{n-1}). $$

(c)

Thus by (a) and (c), we get

$$\begin{aligned}& \biggl\Vert f(x)-\frac{f(2^{2}x)}{{2^{2}}},z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta} \\& \quad \leq \max \biggl\{ \biggl\Vert \frac{f(2x)}{2}-f(x),z_{1}, \ldots,z_{n-1}\biggr\Vert _{\beta},\biggl\Vert \frac{f(2^{2}x)}{{2^{2}}}-\frac{f(2x)}{2}, z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta} \biggr\} \\& \quad \leq \max\bigl\{ \theta\bigl\vert {2^{-\beta}}\bigr\vert \|x \|_{\beta_{1}}^{p+q}\psi (z_{1},\ldots,z_{n-1}), \theta\bigl\vert {2^{-2\beta}}\bigr\vert \|2x\|_{\beta_{1}}^{p+q} \psi(z_{1},\ldots ,z_{n-1}) \bigr\} \end{aligned}$$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Since $(p+q)\beta _{1}>\beta$ and $|2|\neq1$, we get

$$\bigl\Vert f(x)-{2^{-2}}f(2x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta}\leq\bigl\vert {2^{-\beta }}\bigr\vert \theta \|x\|_{\beta_{1}}^{p+q}\psi(z_{1},\ldots,z_{n-1}) $$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$.

By induction on m, we can conclude that

$$ \bigl\Vert f(x)-{2^{-m}}f\bigl(2^{m}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}\leq\bigl\vert {2^{-\beta }}\bigr\vert \theta\|x\|_{\beta_{1}}^{p+q} \psi(z_{1},\ldots,z_{n-1}) $$

(d)

for all $m\in\mathbb{N}$, $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Replacing x with 2x in (d) and dividing both sides by $|2^{\beta }|$, we get

$$ \bigl\Vert 2^{-1}f(2x)-2^{-m-1}f\bigl(2^{m+1}x \bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \leq \bigl\vert {2^{-2\beta}}\bigr\vert \theta\|2x\|_{\beta_{1}}^{p+q} \psi(z_{1},\ldots ,z_{n-1}) $$

(e)

for all $x\in X$, $z_{1},\ldots,z_{n-1}\in Y$ and $m\in\mathbb{N}$. It follows from (a) and (e) that

$$\bigl\Vert f(x)-2^{-m-1}f\bigl(2^{m+1}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}\leq\bigl\vert {2^{-\beta }}\bigr\vert \theta\|x\|_{\beta_{1}}^{p+q} \psi(z_{1},\ldots,z_{n-1}) $$

for all $x\in X$, $z_{1},\ldots,z_{n-1}\in Y$ and $m\in\mathbb{N}$. This completes the proof of (d).

Taking the limit as $m\rightarrow\infty$ in (d), we can obtain (2.2).

Finally, we need to prove the uniqueness of A. Let $A'$ be another additive mapping satisfying (2.2),

$$\begin{aligned}& \bigl\Vert A(x)-A'(x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta} \\& \quad = \bigl\vert 2^{-m\beta}\bigr\vert \bigl\Vert A \bigl(2^{m}x\bigr)-A'\bigl(2^{m}x \bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \bigl\vert 2^{-m\beta}\bigr\vert \max\bigl\{ \bigl\Vert A \bigl(2^{m}x\bigr)-f\bigl(2^{m}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta},\bigl\Vert f \bigl(2^{m}x\bigr)-A'\bigl(2^{m}x \bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\bigr\} \\& \quad \leq \bigl\vert 2^{-m\beta}\bigr\vert \bigl\vert {2^{-\beta}}\bigr\vert \theta\bigl\Vert 2^{m}x\bigr\Vert _{\beta_{1}}^{p+q}\psi (z_{1},\ldots,z_{n-1}) \\& \quad = \theta\bigl\vert 2^{(p+q)\beta_{1}-\beta}\bigr\vert ^{m}\bigl\vert {2^{-\beta}}\bigr\vert \Vert x\Vert _{\beta _{1}}^{p+q} \psi(z_{1},\ldots,z_{n-1}) \end{aligned}$$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Taking the limit as $m\rightarrow\infty$, we get

$$\bigl\Vert A(x)-A'(x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta}=0 $$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. By Lemma 1.4, we get $A(x)=A'(x)$ for all $x\in X$. So A is the unique additive mapping satisfying (2.2). □

When the domain space X is a vector space, we get the following theorems with a generalized control function.

Theorem 2.2

Let X be a vector space and Y be a complete non-Archimedean $(n,\beta)$-normed space, where $n\ge2$ and $0<\beta\leq1$. Let $\varphi:X^{2}\rightarrow[0,\infty)$ be a function such that

$$ \lim_{m\rightarrow\infty}\biggl\vert \frac{1}{2^{m\beta}}\biggr\vert \varphi \bigl(2^{m}x,2^{m}y\bigr)=0 $$

(2.3)

for all $x,y\in X$, and let $\psi:\underbrace{Y\times Y\times\cdots\times Y}_{n-1} \rightarrow[0,\infty)$ be a function. The limit

$$ \lim_{m\rightarrow\infty}\max\bigl\{ \bigl\vert 2^{-j\beta}\bigr\vert \varphi \bigl(2^{j-1}x,2^{j-1}x\bigr):1\leq j\leq m\bigr\} $$

(2.4)

exists for all $x\in X$, and it is denoted by $\widetilde{\varphi}(x)$. Suppose that a mapping $f:X\rightarrow Y$ satisfies the inequality

$$ \bigl\Vert f(x+y)-f(x)-f(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\varphi(x,y)\psi (z_{1},\ldots,z_{n-1}) $$

(2.5)

for all $x,y\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Then there exists an additive mapping $A:X\rightarrow Y$ such that

$$ \bigl\Vert f(x)-A(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\widetilde{\varphi}(x)\psi (z_{1}, \ldots,z_{n-1}) $$

(2.6)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Moreover, if

$$ \lim_{k\rightarrow\infty}\lim_{m\rightarrow\infty}\max\bigl\{ \bigl\vert 2^{-j\beta }\bigr\vert \varphi\bigl(2^{j-1}x,2^{j-1}x \bigr):1+k\leq j\leq m+k\bigr\} =0 $$

(2.7)

for all $x\in X$, then A is a unique additive mapping satisfying (2.6).

Proof

Putting $y=x$ in (2.5) and dividing both sides by $|2^{\beta}|$, we get

$$ \biggl\Vert \frac{f(2x)}{2}-f(x),z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta}\leq \bigl\vert {2^{-\beta}}\bigr\vert \varphi(x,x)\psi(z_{1},\ldots,z_{n-1}) $$

(f)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Replacing x by $2^{j}x$ in (f) and dividing both sides by $|2^{j\beta}|$, we get

$$ \biggl\Vert \frac{f(2^{j+1}x)}{2^{j+1}}-\frac{f(2^{j}x)}{2^{j}},z_{1},\ldots ,z_{n-1}\biggr\Vert _{\beta} \leq \bigl\vert {2^{-j\beta}}\bigr\vert \bigl\vert {2^{-\beta}}\bigr\vert \varphi \bigl(2^{j}x,2^{j}x\bigr) \psi(z_{1},\ldots ,z_{n-1}) $$

for all $x\in X$, $z_{1},\ldots,z_{n-1}\in Y$ and $j\in\mathbb{N}$. Taking the limit as $j\rightarrow\infty$ and considering (2.3), we get

$$\lim_{j\rightarrow\infty}\biggl\Vert \frac{f(2^{j+1}x)}{2^{j+1}}- \frac {f(2^{j}x)}{2^{j}},z_{1},\ldots,z_{n-1}\biggr\Vert _{\beta}=0 $$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Considering Remark 1.13, we know that $\{2^{-m}f(2^{m}x)\}$ is a Cauchy sequence. Since Y is a complete space, we can define the mapping $A:X\rightarrow Y$ by

$$A(x)=\lim_{m\rightarrow\infty}2^{-m}f\bigl(2^{m}x\bigr) $$

for all $x\in X$.

Next, we prove that A is additive:

$$\begin{aligned}& \bigl\Vert A(x+y)-A(x)-A(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \bigl\vert {2^{-m\beta}}\bigr\vert \bigl\Vert A \bigl(2^{m}x+2^{m}y\bigr)-A\bigl(2^{m}x\bigr)-A \bigl(2^{m}y\bigr),z_{1},\ldots ,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \bigl\vert {2^{-m\beta}}\bigr\vert \varphi \bigl(2^{m}x,2^{m}x\bigr) \psi(z_{1}, \ldots,z_{n-1}) \end{aligned}$$

for all $x,y\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Taking the limit as $m\rightarrow\infty$ and considering (2.3), we get

$$\bigl\Vert A(x+y)-A(x)-A(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}=0 $$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. By Lemma 1.4, we know that A is additive.

Replacing x by 2x in (f) and dividing both sides by $|2^{\beta}|$, we get

$$\biggl\Vert \frac{f(2^{2}x)}{2^{2}}-\frac{f(2x)}{2},z_{1},\ldots ,z_{n-1}\biggr\Vert _{\beta}\leq\bigl\vert {2^{-2\beta}} \bigr\vert \varphi(2x,2x)\psi (z_{1},\ldots,z_{n-1}) $$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Considering (f), we get

$$ \biggl\Vert f(x)-\frac{f(2^{2}x)}{2^{2}},z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta} \leq \max\bigl\{ \bigl\vert {2^{-\beta}}\bigr\vert \varphi(x,x),\bigl\vert {2^{-2\beta}}\bigr\vert \varphi(2x,2x) \bigr\} \psi(z_{1},\ldots,z_{n-1}) $$

for all $x\in X$, $z_{1},\ldots,z_{n-1}\in Y$.

By induction on m, we get

$$ \biggl\Vert f(x)-\frac{f(2^{m}x)}{2^{m}},z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta }\leq \max \biggl\{ \frac{\varphi(2^{k-1}x,2^{k-1}x)}{|2^{k\beta}|}:1\leq k \leq m \biggr\} \psi(z_{1},\ldots,z_{n-1}) $$

(g)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Replacing x by 2x in (g) and dividing both sides by $|2^{\beta}|$, we get

$$ \biggl\Vert \frac{f(2x)}{2}-\frac{f(2^{m+1}x)}{2^{m+1}},z_{1},\ldots ,z_{n-1}\biggr\Vert _{\beta}\leq \max \biggl\{ \frac{\varphi(2^{k}x,2^{k}x)}{|2^{(k+1)\beta}|}:1\leq k \leq m \biggr\} \psi(z_{1}, \ldots,z_{n-1}) $$

for all $x\in X$, $z_{1},\ldots,z_{n-1}\in Y$ and $m\in \mathbb{N}$, which together with (f) implies

$$\begin{aligned}& \biggl\Vert f(x)-\frac{f(2^{m+1}x)}{2^{m+1}},z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta} \\& \quad \leq \max \biggl\{ \frac{\varphi(x,x)}{|2^{\beta}|},\frac{\varphi (2^{k}x,2^{k}x)}{|2^{(k+1)\beta}|}:1\leq k \leq m \biggr\} \psi (z_{1},\ldots,z_{n-1}) \\& \quad = \max\bigl\{ \bigl\vert 2^{-(k+1)\beta}\bigr\vert \varphi \bigl(2^{k}x,2^{k}x\bigr):0\leq k\leq m\bigr\} \psi (z_{1},\ldots,z_{n-1}) \\& \quad = \max\bigl\{ \bigl\vert 2^{-k\beta}\bigr\vert \varphi \bigl(2^{k-1}x,2^{k-1}x\bigr):1\leq k\leq m+1\bigr\} \psi(z_{1},\ldots,z_{n-1}) \end{aligned}$$

for all $x\in X$, $z_{1},\ldots,z_{n-1}\in Y$ and $m\in \mathbb{N}$. This completes the proof of (g).

Taking the limit as $m\rightarrow\infty$ in (g), we can obtain (2.6).

Now we need to prove the uniqueness of A. Let $A'$ be another additive mapping satisfying (2.6). Since

$$\begin{aligned}& \lim_{k\rightarrow\infty}\bigl\vert 2^{-k\beta}\bigr\vert \widetilde{\varphi}\bigl(2^{k}x\bigr) \\& \quad = \lim_{k\rightarrow\infty}\bigl\vert 2^{-k\beta}\bigr\vert \lim_{m\rightarrow\infty} \max\bigl\{ \bigl\vert 2^{-j\beta}\bigr\vert \varphi\bigl(2^{j+k-1}x,2^{j+k-1}x\bigr):1\leq j \leq m\bigr\} \\& \quad = \lim_{k\rightarrow\infty}\lim_{m\rightarrow\infty} \max\bigl\{ \bigl\vert 2^{-j\beta}\bigr\vert \varphi\bigl(2^{j-1}x,2^{j-1}x \bigr):1+k \leq j \leq m+k\bigr\} \end{aligned}$$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$, it follows from (2.7) that

$$\begin{aligned}& \bigl\Vert A(x)-A'(x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta} \\& \quad = \lim_{k\rightarrow\infty}\bigl\vert 2^{-k\beta}\bigr\vert \bigl\Vert A\bigl(2^{k}x\bigr)-A'\bigl(2^{k}x \bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \lim_{k\rightarrow\infty}\bigl\vert 2^{-k\beta}\bigr\vert \max\bigl\{ \bigl\Vert A\bigl(2^{k}x\bigr)-f\bigl(2^{k}x \bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}, \\& \qquad \bigl\Vert f\bigl(2^{k}x\bigr)-A'\bigl(2^{k}x \bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\bigr\} \\& \quad \leq \lim_{k\rightarrow\infty}\bigl\vert 2^{-k\beta}\bigr\vert \widetilde{\varphi }\bigl(2^{k}x\bigr)\psi(z_{1}, \ldots,z_{n-1}) \\& \quad = 0 \end{aligned}$$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Considering Lemma 1.4, we prove that A is unique. □

Next, we study the Hyers-Ulam stability of Jensen functional equation in a non-Archimedean $(n,\beta)$-normed space.

Theorem 2.3

Let X be a vector space and Y be a complete non-Archimedean $(n,\beta)$-normed space, where $n\ge2$ and $0<\beta\leq1$. Let $\varphi:X^{2}\rightarrow[0,\infty)$ be a function such that

$$ \lim_{m\rightarrow\infty}\bigl\vert 2^{m\beta}\bigr\vert \varphi \biggl(\frac {x}{2^{m}},\frac{y}{2^{m}} \biggr)=0 $$

(2.8)

for all $x,y\in X$, and let $\psi:\underbrace{Y\times Y\times\cdots\times Y}_{n-1} \rightarrow[0,\infty)$ be a function. The limit

$$ \lim_{m\rightarrow\infty}\max \biggl\{ \bigl\vert 2^{j\beta}\bigr\vert \varphi \biggl(\frac {x}{2^{j}},0 \biggr):0\leq j\leq m-1 \biggr\} $$

(2.9)

exists for all $x\in X$, which is denoted by $\widetilde{\varphi}(x)$. Suppose that a mapping $f:X\rightarrow Y$ and $f(0)=0$ satisfies the inequality

$$ \biggl\Vert 2f \biggl(\frac{x+y}{2} \biggr)-f(x)-f(y),z_{1}, \ldots,z_{n-1} \biggr\Vert _{\beta}\leq\varphi(x,y) \psi(z_{1},\ldots,z_{n-1}) $$

(2.10)

for all $x,y\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Then there exists an additive mapping $A:X\rightarrow Y$ such that

$$ \bigl\Vert f(x)-A(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\widetilde{\varphi}(x)\psi (z_{1}, \ldots,z_{n-1}) $$

(2.11)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Moreover, if

$$ \lim_{k\rightarrow\infty}\lim_{m\rightarrow\infty}\max \biggl\{ \bigl\vert 2^{j\beta }\bigr\vert \varphi \biggl(\frac{x}{2^{j}},0 \biggr):k \leq j\leq m+k-1 \biggr\} =0 $$

(2.12)

for all $x\in X$, then A is a unique additive mapping satisfying (2.11).

Proof

Putting $y=0$ in (2.10), we get

$$ \biggl\Vert 2f \biggl(\frac{x}{2} \biggr)-f(x),z_{1}, \ldots,z_{n-1}\biggr\Vert _{\beta}\leq\varphi(x,0) \psi(z_{1},\ldots,z_{n-1}) $$

(a1)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Replacing x by $\frac{x}{2^{m}}$ in (a1) and multiplying both sides by $|2^{m\beta}|$, we get

$$\biggl\Vert 2^{m+1}f \biggl(\frac{x}{2^{m+1}} \biggr)-2^{m}f \biggl(\frac {x}{2^{m}} \biggr),z_{1},\ldots,z_{n-1}\biggr\Vert _{\beta}\leq\bigl\vert 2^{m\beta }\bigr\vert \varphi \biggl(\frac{x}{2^{m}},0 \biggr) \psi(z_{1},\ldots,z_{n-1}) $$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Taking the limit as $m\rightarrow\infty$ and considering (2.8), we get

$$\lim_{m\rightarrow\infty}\biggl\Vert 2^{m+1}f \biggl( \frac{x}{2^{m+1}} \biggr)-2^{m}f \biggl(\frac{x}{2^{m}} \biggr),z_{1},\ldots,z_{n-1}\biggr\Vert _{\beta}=0 $$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Considering Remark 1.13, we know that $\{2^{m}f (\frac{x}{2^{m}} )\}$ is a Cauchy sequence. Since Y is a complete space, we can define the mapping $A:X\rightarrow Y$ by

$$ A(x)=\lim_{m\rightarrow\infty}2^{m}f \biggl(\frac{x}{2^{m}} \biggr) $$

(b1)

for all $x\in X$.

By induction on m, we get

$$\begin{aligned}& \biggl\Vert 2^{m}f \biggl(\frac{x}{2^{m}} \biggr)-f(x),z_{1}, \ldots ,z_{n-1}\biggr\Vert _{\beta} \\& \quad \leq \max \biggl\{ \bigl\vert 2^{k\beta}\bigr\vert \varphi \biggl(\frac{x}{2^{k}},0 \biggr):0\leq k \leq m-1 \biggr\} \psi(z_{1}, \ldots,z_{n-1}) \end{aligned}$$

(c1)

for all $x\in X$, $z_{1},\ldots,z_{n-1}\in Y$ and $m\in\mathbb{N}$. Replacing x by $\frac{x}{2}$ in (c1) and multiplying both sides by $|2^{\beta}|$, we get

$$\begin{aligned}& \biggl\Vert 2^{m+1}f \biggl(\frac{x}{2^{m+1}} \biggr)-2f \biggl( \frac {x}{2} \biggr),z_{1},\ldots,z_{n-1}\biggr\Vert _{\beta} \\& \quad \leq \max \biggl\{ \bigl|2^{(k+1)\beta}\bigr|\varphi \biggl(\frac{x}{2^{k+1}},0 \biggr) :0\leq k \leq m-1 \biggr\} \psi(z_{1},\ldots,z_{n-1}) \end{aligned}$$

for all $x\in X$, $z_{1},\ldots,z_{n-1}\in Y$ and $m\in \mathbb{N}$. Considering the above inequality and (a1), we have

$$\begin{aligned}& \biggl\Vert 2^{m+1}f \biggl(\frac{x}{2^{m+1}} \biggr)-f(x),z_{1}, \ldots ,z_{n-1}\biggr\Vert _{\beta} \\& \quad \leq \max \biggl\{ \varphi(x,0),\bigl\vert 2^{(k+1)\beta }\bigr\vert \varphi \biggl(\frac{x}{2^{k+1}},0 \biggr):0\leq k \leq m-1 \biggr\} \psi (z_{1},\ldots,z_{n-1}) \\& \quad = \max \biggl\{ \bigl\vert 2^{k\beta}\bigr\vert \varphi \biggl( \frac{x}{2^{k}},0 \biggr):0\leq k \leq m \biggr\} \psi(z_{1}, \ldots,z_{n-1}) \end{aligned}$$

for all $x\in X$, $z_{1},\ldots,z_{n-1}\in Y$ and $m\in \mathbb{N}$. This completes the proof of (c1).

Taking the limit as $m\rightarrow\infty$ in (c1), we can obtain (2.11).

Next, we prove that A is additive. Considering (2.8), (2.10) and (b1), we have

$$\begin{aligned}& \biggl\Vert 2A \biggl(\frac{x+y}{2} \biggr)-A(x)-A(y),z_{1}, \ldots,z_{n-1}\biggr\Vert _{\beta} \\& \quad = \lim_{m\rightarrow\infty}\bigl\vert {2^{m\beta}}\bigr\vert \biggl\Vert 2f \biggl(\frac {x+y}{2^{m+1}} \biggr)-f \biggl(\frac{x}{2^{m}} \biggr)-f \biggl(\frac {y}{2^{m}} \biggr),z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta} \\& \quad \leq \lim_{m\rightarrow\infty}\bigl\vert {2^{m\beta}}\bigr\vert \varphi \biggl(\frac {x}{2^{m}},\frac{y}{2^{m}} \biggr) \psi(z_{1},\ldots,z_{n-1}) \\& \quad = 0 \end{aligned}$$

for all $x,y\in X$ and $z_{1}, \ldots,z_{n-1}\in Y$. Considering Lemma 1.4, we have $2A (\frac{x+y}{2} )-A(x)-A(y)=0$ for all $x,y\in X$. Since $f(0)=0$, $A(0)=0$, we know that A is additive.

Now we need to prove the uniqueness of A. Let $A'$ be another additive mapping satisfying (2.11). Since

$$\begin{aligned}& \lim_{k\rightarrow\infty}\bigl\vert 2^{k\beta}\bigr\vert \widetilde{\varphi} \biggl(\frac {x}{2^{k}} \biggr) \\& \quad = \lim_{k\rightarrow\infty}\bigl\vert 2^{k\beta}\bigr\vert \lim_{m\rightarrow\infty} \max \biggl\{ \bigl\vert 2^{(j+k)\beta}\bigr\vert \varphi \biggl(\frac{x}{2^{j+k}},0 \biggr):0\leq j \leq m-1 \biggr\} \\& \quad = \lim_{k\rightarrow\infty}\lim_{m\rightarrow\infty} \max \biggl\{ \bigl\vert 2^{j\beta}\bigr\vert \varphi \biggl(\frac{x}{2^{j}},0 \biggr):k \leq j \leq m+k-1 \biggr\} \end{aligned}$$

for all $x\in X$, it follows from (2.12) that

$$\begin{aligned}& \bigl\Vert A(x)-A'(x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta} \\& \quad = \lim_{k\rightarrow\infty}\bigl\vert 2^{k\beta}\bigr\vert \biggl\Vert A \biggl(\frac {x}{2^{k}} \biggr)-A' \biggl( \frac{x}{2^{k}} \biggr),z_{1},\ldots ,z_{n-1}\biggr\Vert _{\beta} \\& \quad \leq \lim_{k\rightarrow\infty}\bigl\vert 2^{k\beta}\bigr\vert \max\biggl\{ \biggl\Vert A \biggl(\frac{x}{2^{k}} \biggr)-f \biggl( \frac{x}{2^{k}} \biggr),z_{1},\ldots ,z_{n-1}\biggr\Vert _{\beta}, \\& \qquad \biggl\Vert f \biggl(\frac {x}{2^{k}} \biggr)-A' \biggl( \frac{x}{2^{k}} \biggr),z_{1},\ldots ,z_{n-1}\biggr\Vert _{\beta}\biggr\} \\& \quad \leq \lim_{k\rightarrow\infty}\bigl\vert 2^{k\beta}\bigr\vert \widetilde{\varphi} \biggl(\frac{x}{2^{k}} \biggr)\psi(z_{1}, \ldots,z_{n-1}) \\& \quad = 0 \end{aligned}$$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Considering Lemma 1.4, we prove that A is unique. □

3 Pexiderized Cauchy functional equations

In this section, we investigate the Hyers-Ulam stability of the pexiderized Cauchy functional equation in $(n,\beta)$-normed spaces.

Theorem 3.1

Let X be a vector space and Y be a complete $(n,\beta)$-normed space with $0<\beta\leq1$. Let $\varphi:X^{2}\rightarrow[0,\infty)$ be a function satisfying

$$ \Phi(x)=\sum_{i=1}^{\infty}2^{-i\beta} \bigl(\varphi\bigl(2^{i-1}x,0\bigr)+\varphi \bigl(0,2^{i-1}x \bigr)+\varphi\bigl(2^{i-1}x,2^{i-1}x\bigr)\bigr)< \infty $$

(3.1)

and

$$ \lim_{m\rightarrow\infty}2^{-m\beta}\varphi\bigl(2^{m}x,2^{m}y \bigr)=0 $$

(3.2)

for all $x,y\in X$. $\psi:\underbrace{Y\times Y\times\cdots\times Y}_{n-1} \rightarrow[0,\infty)$ is a function. If mappings $f,g,h:X\rightarrow Y$ satisfy the inequality

$$ \bigl\Vert f(x+y)-g(x)-h(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\varphi(x,y)\psi (z_{1},\ldots,z_{n-1}) $$

(3.3)

for all $x,y\in X$ and $z_{1},\ldots,z_{n-1}\in Y$, then there exists a unique additive mapping $A:X\rightarrow Y$ satisfying

$$\begin{aligned}& \bigl\Vert f(x)-A(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \Phi(x)\psi(z_{1},\ldots,z_{n-1})+\bigl\Vert h(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+\bigl\Vert g(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}, \end{aligned}$$

(3.4)

$$\begin{aligned}& \bigl\Vert g(x)-A(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \Phi(x)\psi (z_{1},\ldots,z_{n-1})+\bigl\Vert g(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+2\bigl\Vert h(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \qquad {}+\varphi(x,0)\psi(z_{1},\ldots,z_{n-1}), \end{aligned}$$

(3.5)

$$\begin{aligned}& \bigl\Vert h(x)-A(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \Phi(x)\psi(z_{1},\ldots,z_{n-1})+\bigl\Vert h(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+2\bigl\Vert g(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \qquad {}+\varphi(0,x)\psi (z_{1},\ldots,z_{n-1}) \end{aligned}$$

(3.6)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$.

Proof

Putting $y=x$ in inequality (3.3), we get

$$ \bigl\Vert f(2x)-g(x)-h(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\varphi(x,x)\psi (z_{1},\ldots,z_{n-1}) $$

(3.7)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Putting $y=0$ in inequality (3.3), we get

$$ \bigl\Vert f(x)-g(x)-h(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\varphi(x,0)\psi (z_{1},\ldots,z_{n-1}) $$

(3.8)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. It then follows from (3.8) that

$$ \bigl\Vert f(x)-g(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \leq\varphi(x,0)\psi(z_{1},\ldots,z_{n-1})+ \bigl\Vert h(0),z_{1},\ldots ,z_{n-1}\bigr\Vert _{\beta} $$

(3.9)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Putting $x=0$ in inequality (3.3), we get

$$\bigl\Vert f(y)-g(0)-h(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\varphi(0,y)\psi (z_{1},\ldots,z_{n-1}) $$

for all $y\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Thus, we obtain

$$ \bigl\Vert f(x)-h(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \leq\varphi(0,x)\psi(z_{1},\ldots,z_{n-1})+ \bigl\Vert g(0),z_{1},\ldots ,z_{n-1}\bigr\Vert _{\beta} $$

(3.10)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$.

Let us define

$$\begin{aligned}& u(x,z_{1},\ldots,z_{n-1}) \\& \quad = \bigl\Vert g(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+ \bigl\Vert h(0),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta}+\varphi(x,x)\psi(z_{1},\ldots,z_{n-1}) \\& \qquad {}+ \varphi(x,0)\psi(z_{1},\ldots,z_{n-1})+ \varphi(0,x)\psi(z_{1},\ldots,z_{n-1}). \end{aligned}$$

Using (3.7), (3.9) and (3.10), we have

$$\begin{aligned}& \bigl\Vert f(2x)-2f(x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta} \\& \quad \leq \bigl\Vert f(2x)-g(x)-h(x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta}+ \bigl\Vert g(x)-f(x),z_{1},\ldots ,z_{n-1}\bigr\Vert _{\beta} \\& \qquad {}+\bigl\Vert h(x)-f(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \bigl\Vert g(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+\bigl\Vert h(0),z_{1},\ldots ,z_{n-1} \bigr\Vert _{\beta}+\varphi(x,0)\psi(z_{1}, \ldots,z_{n-1}) \\& \qquad {}+\varphi (0,x)\psi(z_{1},\ldots,z_{n-1})+ \varphi(x,x)\psi(z_{1},\ldots,z_{n-1}) \\& \quad = u(x,z_{1},\ldots,z_{n-1}) \end{aligned}$$

(3.11)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Replacing x with 2x in (3.11), we get

$$ \bigl\Vert f\bigl(2^{2}x\bigr)-2f(2x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}\leq u(2x,z_{1}, \ldots ,z_{n-1}) $$

(3.12)

for all $x\in X$ and $z_{1}, \ldots,z_{n-1}\in Y$. It then follows from (3.11) and (3.12) that

$$\begin{aligned}& \bigl\Vert f\bigl(2^{2}x\bigr)-2^{2}f(x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \bigl\Vert f\bigl(2^{2}x\bigr)-2f(2x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}+2^{\beta}\bigl\Vert f(2x)-2f(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq u(2x,z_{1},\ldots,z_{n-1})+2^{\beta}u(x,z_{1}, \ldots,z_{n-1}) \end{aligned}$$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$.

Applying an induction argument on m, we will prove that

$$ \bigl\Vert f\bigl(2^{m}x\bigr)-2^{m}f(x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}\leq\sum _{i=1}^{m}2^{(i-1)\beta}u\bigl(2^{m-i}x,z_{1}, \ldots,z_{n-1}\bigr) $$

(3.13)

for all $x\in X$, $z_{1},\ldots,z_{n-1}\in Y$ and $m\in N$. In view of (3.11), inequality (3.13) is true for $m=1$. Assume that (3.13) is true for some $m>1$. Substituting 2x for x in (3.13), we obtain

$$\bigl\Vert f\bigl(2^{m+1}x\bigr)-2^{m}f(2x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}\leq\sum _{i=1}^{m}2^{(i-1)\beta}u\bigl(2^{m+1-i}x,z_{1}, \ldots,z_{n-1}\bigr) $$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Hence, it follows from (3.11) that

$$\begin{aligned}& \bigl\Vert f\bigl(2^{m+1}x\bigr)-2^{m+1}f(x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \bigl\Vert f\bigl(2^{m+1}x\bigr)-2^{m}f(x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}+2^{n\beta}\bigl\Vert f(2x)-2f(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \sum_{i=1}^{m}2^{(i-1)\beta}u \bigl(2^{m+1-i}x,z_{1},\ldots ,z_{n-1} \bigr)+2^{m\beta}u(x,z_{1},\ldots,z_{n-1}) \\& \quad = \sum_{i=1}^{m+1}2^{(i-1)\beta}u \bigl(2^{m+1-i}x,z_{1},\ldots,z_{n-1}\bigr) \end{aligned}$$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$, which proves inequality (3.13). By (3.13), we have

$$ \bigl\Vert 2^{-m}f\bigl(2^{m}x\bigr)-f(x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}\leq\sum _{i=1}^{m}2^{(i-1-m)\beta}u\bigl(2^{m-i}x,z_{1}, \ldots,z_{n-1}\bigr) $$

(3.14)

for all $x\in X$, $z_{1},\ldots,z_{n-1}\in Y$ and $m\in\mathbb{N}$. Moreover, if $m,k\in\mathbb{N}$ with $m< k$, then it follows from (3.11) that

$$\begin{aligned}& \bigl\Vert 2^{-k}f\bigl(2^{k}x\bigr)-2^{-m}f \bigl(2^{m}x\bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \sum_{i=m}^{k-1}\bigl\Vert 2^{-i}f\bigl(2^{i}x\bigr)-2^{-(i+1)}f \bigl(2^{i+1}x\bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \sum_{i=m}^{k-1}2^{-(i+1)\beta} \bigl\Vert 2f\bigl(2^{i}x\bigr)-f\bigl(2^{i+1}x \bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad = \sum_{i=m}^{k-1}2^{-(i+1)\beta} u \bigl(2^{i}x,z_{1},\ldots,z_{n-1}\bigr) \\& \quad = \sum_{i=m}^{k-1}2^{-(i+1)\beta} \bigl[\varphi\bigl(2^{i}x,0\bigr)\psi(z_{1},\ldots ,z_{n-1})+\varphi\bigl(0,2^{i}x\bigr)\psi(z_{1}, \ldots,z_{n-1}) \\& \qquad {}+\varphi \bigl(2^{i}x,2^{i}x\bigr) \psi(z_{1},\ldots,z_{n-1}) +\bigl\Vert h(0),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}+\bigl\Vert g(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\bigr] \\& \quad \leq \sum_{i=m}^{k-1}2^{-(i+1)\beta} \bigl[\varphi\bigl(2^{i}x,0\bigr)+\varphi \bigl(0,2^{i}x \bigr)+\varphi\bigl(2^{i}x,2^{i}x\bigr)\bigr] \psi(z_{1},\ldots,z_{n-1}) \\& \qquad {}+2^{-m}\bigl(\bigl\Vert h(0),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}+\bigl\Vert g(0),z_{1},\ldots ,z_{n-1}\bigr\Vert _{\beta}\bigr) \end{aligned}$$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Taking the limit as $m, k\rightarrow\infty$ and considering (3.1), we get

$$\lim_{m, k\rightarrow\infty} \bigl\Vert 2^{-k}f\bigl(2^{k}x \bigr)-2^{-m}f\bigl(2^{m}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}=0 $$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. According to Definition 1.7, we know that $\{{2^{-m}}f(2^{m}x)\}$ is a Cauchy sequence for every $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Since Y is a complete $(n,\beta)$-normed space, we can define a function $A:X\rightarrow Y$ by

$$A(x)=\lim_{m\rightarrow\infty} 2^{-m}f\bigl(2^{m}x\bigr). $$

Replacing x, y by $2^{m}x$, $2^{m}y$ in (3.3) and dividing both sides by $2^{m\beta}$, we get

$$\begin{aligned}& 2^{-m\beta}\bigl\Vert f\bigl(2^{m}x+2^{m}y\bigr)-g \bigl(2^{m}x\bigr)-h\bigl(2^{m}y\bigr),z_{1},\ldots ,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq 2^{-m\beta}\varphi\bigl(2^{m}x,2^{m}y\bigr) \psi(z_{1},\ldots,z_{n-1}) \end{aligned}$$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. It follows from (3.9) that

$$\begin{aligned} \begin{aligned}[b] &\bigl\Vert 2^{-m}f\bigl(2^{m}x \bigr)-2^{-m}g\bigl(2^{m}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta} \\ &\quad \leq 2^{-m\beta}\bigl[\bigl\Vert h(0),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}+\varphi\bigl(2^{m}x,0 \bigr)\psi (z_{1},\ldots,z_{n-1})\bigr] \end{aligned} \end{aligned}$$

(3.15)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Considering (3.1), we get

$$\begin{aligned}& 2^{-m\beta}\varphi\bigl(2^{m}x,0\bigr)\psi(z_{1}, \ldots,z_{n-1}) \\& \quad \leq2^{\beta }\sum_{i=m}^{\infty}2^{-(i+1)\beta} \bigl[\varphi\bigl(2^{i}x,0\bigr)\psi(z_{1},\ldots ,z_{n-1})+\varphi\bigl(0,2^{i}x\bigr)\psi(z_{1}, \ldots,z_{n-1}) \\& \qquad {}+\varphi \bigl(2^{i}x,2^{i}x\bigr) \psi(z_{1},\ldots,z_{n-1})\bigr] \\& \quad \rightarrow 0 \quad \mbox{as }m\rightarrow \infty. \end{aligned}$$

It follows from (3.15) that

$$ \lim_{m\rightarrow\infty} 2^{-m}g\bigl(2^{m}x\bigr)=\lim _{m\rightarrow\infty} 2^{-m}f\bigl(2^{m}x \bigr)=A(x) $$

(3.16)

for all $x\in X$. Also, by (3.10), we have

$$\begin{aligned}& \bigl\Vert 2^{-m}h\bigl(2^{m}x \bigr)-2^{-m}f\bigl(2^{m}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq 2^{-m\beta}\bigl[\bigl\Vert g(0),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}+\varphi \bigl(0,2^{m}x \bigr)\psi(z_{1},\ldots,z_{n-1})\bigr] \end{aligned}$$

(3.17)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Similarly, it follows from (3.17) that

$$ \lim_{m\rightarrow\infty} 2^{-m}h\bigl(2^{m}x\bigr)=\lim _{m\rightarrow\infty} 2^{-m}f\bigl(2^{m}x \bigr)=A(x) $$

(3.18)

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Thus, by (3.2), (3.16), (3.18) and Lemma 1.8, we get

$$\begin{aligned}& \bigl\Vert A(x+y)-A(x)-A(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad = \lim_{m\rightarrow\infty}\bigl\Vert 2^{-m}f \bigl(2^{m}x+2^{m}y\bigr)-2^{-m}g \bigl(2^{m}x\bigr)-2^{-m}h\bigl(2^{m}y \bigr),z_{1},\ldots ,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \lim_{m\rightarrow\infty}2^{-m\beta}\varphi \bigl(2^{m}x,2^{m}y\bigr)\psi (z_{1}, \ldots,z_{n-1}) \\& \quad = 0 \end{aligned}$$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$. Hence $A(x+y)-A(x)-A(y)=0$.

Taking the limit as $m\rightarrow\infty$ in (3.14), we get

$$\begin{aligned}& \bigl\Vert A(x)-f(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \lim_{m\rightarrow \infty}\sum_{i=1}^{m}2^{(i-1-m)\beta}u \bigl(2^{m-i}x,z_{1},\ldots,z_{n-1}\bigr) \\& \quad = \lim_{m\rightarrow\infty}\bigl(1-2^{-m\beta}\bigr) \bigl(\bigl\Vert g(0),z_{1},\ldots ,z_{n-1} \bigr\Vert _{\beta}+ \bigl\Vert h(0),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta}\bigr) \\& \qquad {}+\lim_{m\rightarrow\infty}\sum_{i=1}^{m}2^{(i-m-1)\beta} \bigl(\varphi \bigl(2^{m-i}x,0\bigr)\psi(z_{1}, \ldots,z_{n-1})+\varphi\bigl(0,2^{m-i}x\bigr)\psi (z_{1},\ldots,z_{n-1}) \\& \qquad {} +\varphi\bigl(2^{m-i}x,2^{m-i}x\bigr) \psi(z_{1},\ldots,z_{n-1})\bigr) \\& \quad = \bigl\Vert h(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+\bigl\Vert g(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+\Phi(x)\psi(z_{1},\ldots,z_{n-1}) \end{aligned}$$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$, which proves (3.4).

It remains to prove the uniqueness of A. Assume that $A':X\rightarrow Y$ is another additive mapping which satisfies (3.4). Then we have

$$\begin{aligned}& \bigl\Vert A(x)-A'(x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta} \\& \quad \leq 2^{-m\beta}\bigl\Vert A\bigl(2^{m}x\bigr)-f \bigl(2^{m}x\bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+2^{-m\beta}\bigl\Vert f\bigl(2^{m}x \bigr)-A'\bigl(2^{m}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq 2^{-m \beta+1}\bigl(\bigl\Vert g(0),z_{1}, \ldots,z_{n-1} \bigr\Vert _{\beta}+\bigl\Vert h(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+ \Phi \bigl(2^{m}x\bigr)\psi(z_{1},\ldots ,z_{n-1})\bigr) \\& \quad = 2^{-m\beta+1}\bigl(\bigl\Vert g(0),z_{1}, \ldots,z_{n-1} \bigr\Vert _{\beta}+\bigl\Vert h(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\bigr) \\& \qquad {}+2\sum_{i=m+1}^{\infty}2^{-i\beta} \bigl(\varphi\bigl(2^{i-1}x,0\bigr)+\varphi \bigl(0,2^{i-1}x \bigr)+\varphi\bigl(2^{i-1}x,2^{i-1}x\bigr)\bigr) \psi(z_{1},\ldots,z_{n-1}) \\& \quad \rightarrow 0 \quad \mbox{as }m \rightarrow \infty \end{aligned}$$

for all $x\in X$ and $z_{1},\ldots,z_{n-1}\in Y$, which together with Lemma 1.4 implies that $A(x)=A'(x)$ for all $x\in X$. Using (3.4) and (3.9), we can get (3.5), and also using (3.4) and (3.10), we can get (3.6). □