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

Journal of Inequalities and Applications20152015:112

https://doi.org/10.1186/s13660-015-0628-1

Received: 23 December 2014

Accepted: 11 March 2015

Published: 27 March 2015

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.

Keywords

  • \((n,\beta)\)-normed spaces
  • non-Archimedean \((n,\beta)\)-normed space
  • Cauchy functional equation
  • Jensen functional equation
  • pexiderized Cauchy functional equation

MSC

  • 39B82
  • 39B72

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 [411].

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 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:
  1. (a)

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

     
  2. (b)

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

     
  3. (c)

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

     
  4. (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 [2224]).

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 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). □

Declarations

Acknowledgements

XY was supported by the National Natural Science Foundation of China (grant No. 11371119) and all authors were supported by the Natural Science Foundation of Education Department of Hebei Province (grant No. Z2014031). The authors also express their thanks to the referees for their constructive suggestions in improving the quality of the paper.

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
College of Mathematics and Information Science, Hebei Normal University, Shijiazhuang, People’s Republic of China
(2)
Hebei Key Laboratory of Computational Mathematics and Applications, Shijiazhuang, People’s Republic of China

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