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.
Anzeige
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.
Anzeige
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: for all \(x_{1},\ldots,x_{n}\in X \) and \(\alpha\in\mathbb{R} \).
(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)
\(\|x_{1},\ldots,x_{n-1},y+z\|_{\beta}\leq \|x_{1},\ldots,x_{n-1},y\|_{\beta}+\|x_{1},\ldots,x_{n-1},z\|_{\beta}\)
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
where \(x_{i}=(x_{i1},x_{i2},\ldots,x_{in})\in\mathbb{R}^{n}\) for each \(i=1,2,\ldots,n\).
$$ \|x_{1},\ldots,x_{n}\|_{E}=\bigl\vert \operatorname{det}(x_{ij})\bigr\vert =\operatorname{abs} \left ( \left \vert \begin{array}{@{}c@{\quad}c@{\quad}c@{}} x_{11} & \cdots& x_{1n} \\ \vdots& \ddots& \vdots \\ x_{n1} & \cdots& x_{nn} \end{array} \right \vert \right ), $$
(1.1)
Example 1.3
[19]
The standard n-norm on X, a real inner product space of dimension \(\dim X\geq n\), is as follows:
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 \(\|x_{1}\|=\langle x_{1},x_{1}\rangle^{1/2}\).
$$ \|x_{1},x_{2},\ldots,x_{n}\|_{S}= \left \vert \begin{array}{@{}c@{\quad}c@{\quad}c@{}} \langle x_{1},x_{1}\rangle & \cdots& \langle x_{1},x_{n}\rangle \\ \vdots& \ddots& \vdots \\ \langle x_{n},x_{1}\rangle & \cdots& \langle x_{n},x_{n}\rangle \end{array} \right \vert ^{1/2}, $$
(1.2)
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
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
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 \(x_{1}=0\). □
$$\alpha_{1}x_{1}+\alpha_{2}y_{2}+ \cdots+\alpha_{n}y_{n}=0. $$
$$ x_{1}=-\frac{\alpha_{2}}{\alpha_{1}}y_{2}-\cdots-\frac{\alpha _{n}}{\alpha_{1}}y_{n}. $$
(1.3)
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
for all \(x,y\in X \) and \(z_{1},\ldots,z_{n-1}\in Y\).
$$\bigl\vert \Vert x,z_{1},\ldots,z_{n-1}\Vert _{\beta}-\Vert y,z_{1},\ldots,z_{n-1}\Vert _{\beta}\bigr\vert \leq \Vert x-y,z_{1},\ldots,z_{n-1} \Vert _{\beta} $$
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
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 \(\lim_{m\rightarrow\infty} x_{m} = x\).
$$\lim_{m\rightarrow\infty}\| x_{m}-x ,y_{1}, \ldots,y_{n-1} \|_{\beta} =0 $$
Definition 1.7
A sequence \(\{x_{m}\}\) in a linear \((n,\beta)\)-normed space X is called a Cauchy sequence if
for all \(z_{1},\ldots,z_{n-1}\in X\).
$$\lim_{m,k\rightarrow\infty}\| x_{k}-x_{m}, z_{1},\ldots,z_{n-1} \| _{\beta} =0 $$
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,
for all
\(z_{1},\ldots,z_{n-1}\in X\).
$$\lim_{m\rightarrow\infty} \| x_{m} ,z_{1}, \ldots,z_{n-1} \|_{\beta}=\Bigl\| \lim_{m\rightarrow\infty} x_{m} ,z_{1},\ldots,z_{n-1} \Bigr\| _{\beta} $$
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:
for all \(\alpha\in K\) and \(x_{0},x_{1},\ldots,x_{n}\in X\).
(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′)
\(\|x_{0}+x_{1},x_{2},\ldots,x_{n}\|_{\beta} \leq \max\{\|x_{0},x_{2},\ldots,x_{n}\|_{\beta},\|x_{1},x_{2},\ldots,x_{n}\| _{\beta}\}\)
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
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. □
$$\begin{aligned}& \|x_{m}-x_{k},y_{1},\ldots,y_{n-1} \|_{\beta} \\& \quad \leq \max \bigl\{ \Vert x_{j+1}-x_{j},y_{1}, \ldots,y_{n-1}\Vert _{\beta}:k\leq j\leq m-1\bigr\} \quad (m>k) \end{aligned}$$
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
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
for all
\(x\in X \)
and
\(z_{1},\ldots,z_{n-1}\in Y\).
$$ \bigl\Vert f(x+y)-f(x)-f(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\theta\|x\|_{\beta _{1}}^{p}\|y \|_{\beta_{1}}^{q}\psi(z_{1},\ldots,z_{n-1}) $$
(2.1)
$$ \bigl\Vert f(x)-A(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\theta\bigl\vert 2^{-\beta}\bigr\vert \|x\| _{\beta_{1}}^{p+q}\psi(z_{1},\ldots,z_{n-1}) $$
(2.2)
Proof
Putting \(y=x\) in (2.1) and dividing both sides by \(|2^{\beta}|\), we get
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
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
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
for all \(x\in X\).
$$ \biggl\Vert \frac{f(2x)}{2}-f(x),z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta}\leq \theta\bigl\vert 2^{-\beta}\bigr\vert \|x\|_{\beta_{1}}^{p+q}\psi(z_{1},\ldots,z_{n-1}) $$
(a)
$$\begin{aligned}& \biggl\Vert \frac{f(2^{m+1}x)}{2^{m+1}}-\frac {f(2^{m}x)}{2^{m}},z_{1}, \ldots,z_{n-1}\biggr\Vert _{\beta} \\& \quad \leq \theta\biggl\vert \frac{1}{{2^{m\beta}}}\biggr\vert \biggl\vert \frac{1}{{2^{\beta }}}\biggr\vert \bigl\vert 2^{m(p+q)\beta_{1}}\bigr\vert \|x \|_{\beta_{1}}^{p+q}\psi(z_{1},\ldots ,z_{n-1}) \\& \quad = \theta\bigl\vert {2^{-\beta}}\bigr\vert \bigl\vert 2^{(p+q)\beta_{1}-\beta}\bigr\vert ^{m}\|x\|_{\beta_{1}}^{p+q} \psi(z_{1},\ldots,z_{n-1}) \end{aligned}$$
$$\lim_{m\rightarrow\infty}\bigl\Vert 2^{-m-1}f\bigl(2^{m+1}x \bigr)-2^{-m}f\bigl(2^{m}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}=0 $$
$$ A(x)=\lim_{m\rightarrow\infty}2^{-m}f\bigl(2^{m}x\bigr) $$
(b)
Next, we show that A is additive. It follows from (2.1), (b) and Lemma 1.8 that
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
for all \(x,y\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\). By Lemma 1.4, we get
for all \(x,y\in X\). So the mapping A is additive.
$$\begin{aligned}& \bigl\Vert A(x+y)-A(x)-A(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad = \lim_{m\rightarrow\infty}\bigl\vert {2^{-m\beta}}\bigr\vert \bigl\Vert f\bigl(2^{m}x+2^{m}y\bigr)-f\bigl(2^{m}x \bigr)-f\bigl(2^{m}y\bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \lim_{m\rightarrow\infty}\theta\bigl\vert {2^{-m\beta}} \bigr\vert \bigl\Vert 2^{m}x\bigr\Vert _{\beta _{1}}^{p} \bigl\Vert 2^{m}y\bigr\Vert _{\beta_{1}}^{q} \psi(z_{1},\ldots,z_{n-1}) \\& \quad = \lim_{m\rightarrow\infty}\theta\bigl\vert 2^{(p+q)\beta_{1}-\beta}\bigr\vert ^{m}\Vert x\Vert _{\beta_{1}}^{p}\Vert y \Vert _{\beta_{1}}^{q}\psi(z_{1},\ldots,z_{n-1}) \end{aligned}$$
$$\bigl\Vert A(x+y)-A(x)-A(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}=0 $$
$$A(x+y)-A(x)-A(y)=0 $$
Replacing x by 2x in (a) and dividing both sides by \(|2^{\beta }|\), we get
Thus by (a) and (c), we get
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
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\).
$$ \biggl\Vert \frac{f(2^{2}x)}{{2^{2}}}-\frac{f(2x)}{2},z_{1},\ldots ,z_{n-1}\biggr\Vert _{\beta}\leq\theta\bigl\vert {2^{-2\beta}}\bigr\vert \|2x\|_{\beta _{1}}^{p+q} \psi(z_{1},\ldots,z_{n-1}). $$
(c)
$$\begin{aligned}& \biggl\Vert f(x)-\frac{f(2^{2}x)}{{2^{2}}},z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta} \\& \quad \leq \max \biggl\{ \biggl\Vert \frac{f(2x)}{2}-f(x),z_{1}, \ldots,z_{n-1}\biggr\Vert _{\beta},\biggl\Vert \frac{f(2^{2}x)}{{2^{2}}}-\frac{f(2x)}{2}, z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta} \biggr\} \\& \quad \leq \max\bigl\{ \theta\bigl\vert {2^{-\beta}}\bigr\vert \|x \|_{\beta_{1}}^{p+q}\psi (z_{1},\ldots,z_{n-1}), \theta\bigl\vert {2^{-2\beta}}\bigr\vert \|2x\|_{\beta_{1}}^{p+q} \psi(z_{1},\ldots ,z_{n-1}) \bigr\} \end{aligned}$$
$$\bigl\Vert f(x)-{2^{-2}}f(2x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta}\leq\bigl\vert {2^{-\beta }}\bigr\vert \theta \|x\|_{\beta_{1}}^{p+q}\psi(z_{1},\ldots,z_{n-1}) $$
By induction on m, we can conclude that
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
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
for all \(x\in X\), \(z_{1},\ldots,z_{n-1}\in Y\) and \(m\in\mathbb{N}\). This completes the proof of (d).
$$ \bigl\Vert f(x)-{2^{-m}}f\bigl(2^{m}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}\leq\bigl\vert {2^{-\beta }}\bigr\vert \theta\|x\|_{\beta_{1}}^{p+q} \psi(z_{1},\ldots,z_{n-1}) $$
(d)
$$ \bigl\Vert 2^{-1}f(2x)-2^{-m-1}f\bigl(2^{m+1}x \bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \leq \bigl\vert {2^{-2\beta}}\bigr\vert \theta\|2x\|_{\beta_{1}}^{p+q} \psi(z_{1},\ldots ,z_{n-1}) $$
(e)
$$\bigl\Vert f(x)-2^{-m-1}f\bigl(2^{m+1}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}\leq\bigl\vert {2^{-\beta }}\bigr\vert \theta\|x\|_{\beta_{1}}^{p+q} \psi(z_{1},\ldots,z_{n-1}) $$
Finally, we need to prove the uniqueness of A. Let \(A'\) be another additive mapping satisfying (2.2),
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\). Taking the limit as \(m\rightarrow\infty\), we get
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). □
$$\begin{aligned}& \bigl\Vert A(x)-A'(x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta} \\& \quad = \bigl\vert 2^{-m\beta}\bigr\vert \bigl\Vert A \bigl(2^{m}x\bigr)-A'\bigl(2^{m}x \bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \bigl\vert 2^{-m\beta}\bigr\vert \max\bigl\{ \bigl\Vert A \bigl(2^{m}x\bigr)-f\bigl(2^{m}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta},\bigl\Vert f \bigl(2^{m}x\bigr)-A'\bigl(2^{m}x \bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\bigr\} \\& \quad \leq \bigl\vert 2^{-m\beta}\bigr\vert \bigl\vert {2^{-\beta}}\bigr\vert \theta\bigl\Vert 2^{m}x\bigr\Vert _{\beta_{1}}^{p+q}\psi (z_{1},\ldots,z_{n-1}) \\& \quad = \theta\bigl\vert 2^{(p+q)\beta_{1}-\beta}\bigr\vert ^{m}\bigl\vert {2^{-\beta}}\bigr\vert \Vert x\Vert _{\beta _{1}}^{p+q} \psi(z_{1},\ldots,z_{n-1}) \end{aligned}$$
$$\bigl\Vert A(x)-A'(x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta}=0 $$
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
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
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
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
for all
\(x\in X \)
and
\(z_{1},\ldots,z_{n-1}\in Y\). Moreover, if
for all
\(x\in X\), then
A
is a unique additive mapping satisfying (2.6).
$$ \lim_{m\rightarrow\infty}\biggl\vert \frac{1}{2^{m\beta}}\biggr\vert \varphi \bigl(2^{m}x,2^{m}y\bigr)=0 $$
(2.3)
$$ \lim_{m\rightarrow\infty}\max\bigl\{ \bigl\vert 2^{-j\beta}\bigr\vert \varphi \bigl(2^{j-1}x,2^{j-1}x\bigr):1\leq j\leq m\bigr\} $$
(2.4)
$$ \bigl\Vert f(x+y)-f(x)-f(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\varphi(x,y)\psi (z_{1},\ldots,z_{n-1}) $$
(2.5)
$$ \bigl\Vert f(x)-A(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\widetilde{\varphi}(x)\psi (z_{1}, \ldots,z_{n-1}) $$
(2.6)
$$ \lim_{k\rightarrow\infty}\lim_{m\rightarrow\infty}\max\bigl\{ \bigl\vert 2^{-j\beta }\bigr\vert \varphi\bigl(2^{j-1}x,2^{j-1}x \bigr):1+k\leq j\leq m+k\bigr\} =0 $$
(2.7)
Proof
Putting \(y=x\) in (2.5) and dividing both sides by \(|2^{\beta}|\), we get
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
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
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
for all \(x\in X\).
$$ \biggl\Vert \frac{f(2x)}{2}-f(x),z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta}\leq \bigl\vert {2^{-\beta}}\bigr\vert \varphi(x,x)\psi(z_{1},\ldots,z_{n-1}) $$
(f)
$$ \biggl\Vert \frac{f(2^{j+1}x)}{2^{j+1}}-\frac{f(2^{j}x)}{2^{j}},z_{1},\ldots ,z_{n-1}\biggr\Vert _{\beta} \leq \bigl\vert {2^{-j\beta}}\bigr\vert \bigl\vert {2^{-\beta}}\bigr\vert \varphi \bigl(2^{j}x,2^{j}x\bigr) \psi(z_{1},\ldots ,z_{n-1}) $$
$$\lim_{j\rightarrow\infty}\biggl\Vert \frac{f(2^{j+1}x)}{2^{j+1}}- \frac {f(2^{j}x)}{2^{j}},z_{1},\ldots,z_{n-1}\biggr\Vert _{\beta}=0 $$
$$A(x)=\lim_{m\rightarrow\infty}2^{-m}f\bigl(2^{m}x\bigr) $$
Next, we prove that A is additive:
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
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\). By Lemma 1.4, we know that A is additive.
$$\begin{aligned}& \bigl\Vert A(x+y)-A(x)-A(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \bigl\vert {2^{-m\beta}}\bigr\vert \bigl\Vert A \bigl(2^{m}x+2^{m}y\bigr)-A\bigl(2^{m}x\bigr)-A \bigl(2^{m}y\bigr),z_{1},\ldots ,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \bigl\vert {2^{-m\beta}}\bigr\vert \varphi \bigl(2^{m}x,2^{m}x\bigr) \psi(z_{1}, \ldots,z_{n-1}) \end{aligned}$$
$$\bigl\Vert A(x+y)-A(x)-A(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}=0 $$
Replacing x by 2x in (f) and dividing both sides by \(|2^{\beta}|\), we get
for all \(x\in X \) and \(z_{1},\ldots,z_{n-1}\in Y\). Considering (f), we get
for all \(x\in X \), \(z_{1},\ldots,z_{n-1}\in Y\).
$$\biggl\Vert \frac{f(2^{2}x)}{2^{2}}-\frac{f(2x)}{2},z_{1},\ldots ,z_{n-1}\biggr\Vert _{\beta}\leq\bigl\vert {2^{-2\beta}} \bigr\vert \varphi(2x,2x)\psi (z_{1},\ldots,z_{n-1}) $$
$$ \biggl\Vert f(x)-\frac{f(2^{2}x)}{2^{2}},z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta} \leq \max\bigl\{ \bigl\vert {2^{-\beta}}\bigr\vert \varphi(x,x),\bigl\vert {2^{-2\beta}}\bigr\vert \varphi(2x,2x) \bigr\} \psi(z_{1},\ldots,z_{n-1}) $$
By induction on m, we get
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
for all \(x\in X \), \(z_{1},\ldots,z_{n-1}\in Y\) and \(m\in \mathbb{N}\), which together with (f) implies
for all \(x\in X \), \(z_{1},\ldots,z_{n-1}\in Y\) and \(m\in \mathbb{N}\). This completes the proof of (g).
$$ \biggl\Vert f(x)-\frac{f(2^{m}x)}{2^{m}},z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta }\leq \max \biggl\{ \frac{\varphi(2^{k-1}x,2^{k-1}x)}{|2^{k\beta}|}:1\leq k \leq m \biggr\} \psi(z_{1},\ldots,z_{n-1}) $$
(g)
$$ \biggl\Vert \frac{f(2x)}{2}-\frac{f(2^{m+1}x)}{2^{m+1}},z_{1},\ldots ,z_{n-1}\biggr\Vert _{\beta}\leq \max \biggl\{ \frac{\varphi(2^{k}x,2^{k}x)}{|2^{(k+1)\beta}|}:1\leq k \leq m \biggr\} \psi(z_{1}, \ldots,z_{n-1}) $$
$$\begin{aligned}& \biggl\Vert f(x)-\frac{f(2^{m+1}x)}{2^{m+1}},z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta} \\& \quad \leq \max \biggl\{ \frac{\varphi(x,x)}{|2^{\beta}|},\frac{\varphi (2^{k}x,2^{k}x)}{|2^{(k+1)\beta}|}:1\leq k \leq m \biggr\} \psi (z_{1},\ldots,z_{n-1}) \\& \quad = \max\bigl\{ \bigl\vert 2^{-(k+1)\beta}\bigr\vert \varphi \bigl(2^{k}x,2^{k}x\bigr):0\leq k\leq m\bigr\} \psi (z_{1},\ldots,z_{n-1}) \\& \quad = \max\bigl\{ \bigl\vert 2^{-k\beta}\bigr\vert \varphi \bigl(2^{k-1}x,2^{k-1}x\bigr):1\leq k\leq m+1\bigr\} \psi(z_{1},\ldots,z_{n-1}) \end{aligned}$$
Now we need to prove the uniqueness of A. Let \(A'\) be another additive mapping satisfying (2.6). Since
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\), it follows from (2.7) that
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\). Considering Lemma 1.4, we prove that A is unique. □
$$\begin{aligned}& \lim_{k\rightarrow\infty}\bigl\vert 2^{-k\beta}\bigr\vert \widetilde{\varphi}\bigl(2^{k}x\bigr) \\& \quad = \lim_{k\rightarrow\infty}\bigl\vert 2^{-k\beta}\bigr\vert \lim_{m\rightarrow\infty} \max\bigl\{ \bigl\vert 2^{-j\beta}\bigr\vert \varphi\bigl(2^{j+k-1}x,2^{j+k-1}x\bigr):1\leq j \leq m\bigr\} \\& \quad = \lim_{k\rightarrow\infty}\lim_{m\rightarrow\infty} \max\bigl\{ \bigl\vert 2^{-j\beta}\bigr\vert \varphi\bigl(2^{j-1}x,2^{j-1}x \bigr):1+k \leq j \leq m+k\bigr\} \end{aligned}$$
$$\begin{aligned}& \bigl\Vert A(x)-A'(x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta} \\& \quad = \lim_{k\rightarrow\infty}\bigl\vert 2^{-k\beta}\bigr\vert \bigl\Vert A\bigl(2^{k}x\bigr)-A'\bigl(2^{k}x \bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \lim_{k\rightarrow\infty}\bigl\vert 2^{-k\beta}\bigr\vert \max\bigl\{ \bigl\Vert A\bigl(2^{k}x\bigr)-f\bigl(2^{k}x \bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}, \\& \qquad \bigl\Vert f\bigl(2^{k}x\bigr)-A'\bigl(2^{k}x \bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\bigr\} \\& \quad \leq \lim_{k\rightarrow\infty}\bigl\vert 2^{-k\beta}\bigr\vert \widetilde{\varphi }\bigl(2^{k}x\bigr)\psi(z_{1}, \ldots,z_{n-1}) \\& \quad = 0 \end{aligned}$$
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
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
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
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
for all
\(x\in X \)
and
\(z_{1},\ldots,z_{n-1}\in Y\). Moreover, if
for all
\(x\in X\), then
A
is a unique additive mapping satisfying (2.11).
$$ \lim_{m\rightarrow\infty}\bigl\vert 2^{m\beta}\bigr\vert \varphi \biggl(\frac {x}{2^{m}},\frac{y}{2^{m}} \biggr)=0 $$
(2.8)
$$ \lim_{m\rightarrow\infty}\max \biggl\{ \bigl\vert 2^{j\beta}\bigr\vert \varphi \biggl(\frac {x}{2^{j}},0 \biggr):0\leq j\leq m-1 \biggr\} $$
(2.9)
$$ \biggl\Vert 2f \biggl(\frac{x+y}{2} \biggr)-f(x)-f(y),z_{1}, \ldots,z_{n-1} \biggr\Vert _{\beta}\leq\varphi(x,y) \psi(z_{1},\ldots,z_{n-1}) $$
(2.10)
$$ \bigl\Vert f(x)-A(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\widetilde{\varphi}(x)\psi (z_{1}, \ldots,z_{n-1}) $$
(2.11)
$$ \lim_{k\rightarrow\infty}\lim_{m\rightarrow\infty}\max \biggl\{ \bigl\vert 2^{j\beta }\bigr\vert \varphi \biggl(\frac{x}{2^{j}},0 \biggr):k \leq j\leq m+k-1 \biggr\} =0 $$
(2.12)
Proof
Putting \(y=0\) in (2.10), we get
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
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
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
for all \(x\in X\).
$$ \biggl\Vert 2f \biggl(\frac{x}{2} \biggr)-f(x),z_{1}, \ldots,z_{n-1}\biggr\Vert _{\beta}\leq\varphi(x,0) \psi(z_{1},\ldots,z_{n-1}) $$
(a1)
$$\biggl\Vert 2^{m+1}f \biggl(\frac{x}{2^{m+1}} \biggr)-2^{m}f \biggl(\frac {x}{2^{m}} \biggr),z_{1},\ldots,z_{n-1}\biggr\Vert _{\beta}\leq\bigl\vert 2^{m\beta }\bigr\vert \varphi \biggl(\frac{x}{2^{m}},0 \biggr) \psi(z_{1},\ldots,z_{n-1}) $$
$$\lim_{m\rightarrow\infty}\biggl\Vert 2^{m+1}f \biggl( \frac{x}{2^{m+1}} \biggr)-2^{m}f \biggl(\frac{x}{2^{m}} \biggr),z_{1},\ldots,z_{n-1}\biggr\Vert _{\beta}=0 $$
$$ A(x)=\lim_{m\rightarrow\infty}2^{m}f \biggl(\frac{x}{2^{m}} \biggr) $$
(b1)
By induction on m, we get
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
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
for all \(x\in X \), \(z_{1},\ldots,z_{n-1}\in Y\) and \(m\in \mathbb{N}\). This completes the proof of (c1).
$$\begin{aligned}& \biggl\Vert 2^{m}f \biggl(\frac{x}{2^{m}} \biggr)-f(x),z_{1}, \ldots ,z_{n-1}\biggr\Vert _{\beta} \\& \quad \leq \max \biggl\{ \bigl\vert 2^{k\beta}\bigr\vert \varphi \biggl(\frac{x}{2^{k}},0 \biggr):0\leq k \leq m-1 \biggr\} \psi(z_{1}, \ldots,z_{n-1}) \end{aligned}$$
(c1)
$$\begin{aligned}& \biggl\Vert 2^{m+1}f \biggl(\frac{x}{2^{m+1}} \biggr)-2f \biggl( \frac {x}{2} \biggr),z_{1},\ldots,z_{n-1}\biggr\Vert _{\beta} \\& \quad \leq \max \biggl\{ \bigl|2^{(k+1)\beta}\bigr|\varphi \biggl(\frac{x}{2^{k+1}},0 \biggr) :0\leq k \leq m-1 \biggr\} \psi(z_{1},\ldots,z_{n-1}) \end{aligned}$$
$$\begin{aligned}& \biggl\Vert 2^{m+1}f \biggl(\frac{x}{2^{m+1}} \biggr)-f(x),z_{1}, \ldots ,z_{n-1}\biggr\Vert _{\beta} \\& \quad \leq \max \biggl\{ \varphi(x,0),\bigl\vert 2^{(k+1)\beta }\bigr\vert \varphi \biggl(\frac{x}{2^{k+1}},0 \biggr):0\leq k \leq m-1 \biggr\} \psi (z_{1},\ldots,z_{n-1}) \\& \quad = \max \biggl\{ \bigl\vert 2^{k\beta}\bigr\vert \varphi \biggl( \frac{x}{2^{k}},0 \biggr):0\leq k \leq m \biggr\} \psi(z_{1}, \ldots,z_{n-1}) \end{aligned}$$
Next, we prove that A is additive. Considering (2.8), (2.10) and (b1), we have
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.
$$\begin{aligned}& \biggl\Vert 2A \biggl(\frac{x+y}{2} \biggr)-A(x)-A(y),z_{1}, \ldots,z_{n-1}\biggr\Vert _{\beta} \\& \quad = \lim_{m\rightarrow\infty}\bigl\vert {2^{m\beta}}\bigr\vert \biggl\Vert 2f \biggl(\frac {x+y}{2^{m+1}} \biggr)-f \biggl(\frac{x}{2^{m}} \biggr)-f \biggl(\frac {y}{2^{m}} \biggr),z_{1},\ldots,z_{n-1} \biggr\Vert _{\beta} \\& \quad \leq \lim_{m\rightarrow\infty}\bigl\vert {2^{m\beta}}\bigr\vert \varphi \biggl(\frac {x}{2^{m}},\frac{y}{2^{m}} \biggr) \psi(z_{1},\ldots,z_{n-1}) \\& \quad = 0 \end{aligned}$$
Now we need to prove the uniqueness of A. Let \(A'\) be another additive mapping satisfying (2.11). Since
for all \(x\in X\), it follows from (2.12) that
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\). Considering Lemma 1.4, we prove that A is unique. □
$$\begin{aligned}& \lim_{k\rightarrow\infty}\bigl\vert 2^{k\beta}\bigr\vert \widetilde{\varphi} \biggl(\frac {x}{2^{k}} \biggr) \\& \quad = \lim_{k\rightarrow\infty}\bigl\vert 2^{k\beta}\bigr\vert \lim_{m\rightarrow\infty} \max \biggl\{ \bigl\vert 2^{(j+k)\beta}\bigr\vert \varphi \biggl(\frac{x}{2^{j+k}},0 \biggr):0\leq j \leq m-1 \biggr\} \\& \quad = \lim_{k\rightarrow\infty}\lim_{m\rightarrow\infty} \max \biggl\{ \bigl\vert 2^{j\beta}\bigr\vert \varphi \biggl(\frac{x}{2^{j}},0 \biggr):k \leq j \leq m+k-1 \biggr\} \end{aligned}$$
$$\begin{aligned}& \bigl\Vert A(x)-A'(x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta} \\& \quad = \lim_{k\rightarrow\infty}\bigl\vert 2^{k\beta}\bigr\vert \biggl\Vert A \biggl(\frac {x}{2^{k}} \biggr)-A' \biggl( \frac{x}{2^{k}} \biggr),z_{1},\ldots ,z_{n-1}\biggr\Vert _{\beta} \\& \quad \leq \lim_{k\rightarrow\infty}\bigl\vert 2^{k\beta}\bigr\vert \max\biggl\{ \biggl\Vert A \biggl(\frac{x}{2^{k}} \biggr)-f \biggl( \frac{x}{2^{k}} \biggr),z_{1},\ldots ,z_{n-1}\biggr\Vert _{\beta}, \\& \qquad \biggl\Vert f \biggl(\frac {x}{2^{k}} \biggr)-A' \biggl( \frac{x}{2^{k}} \biggr),z_{1},\ldots ,z_{n-1}\biggr\Vert _{\beta}\biggr\} \\& \quad \leq \lim_{k\rightarrow\infty}\bigl\vert 2^{k\beta}\bigr\vert \widetilde{\varphi} \biggl(\frac{x}{2^{k}} \biggr)\psi(z_{1}, \ldots,z_{n-1}) \\& \quad = 0 \end{aligned}$$
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
and
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
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
for all
\(x\in X\)
and
\(z_{1},\ldots,z_{n-1}\in Y\).
$$ \Phi(x)=\sum_{i=1}^{\infty}2^{-i\beta} \bigl(\varphi\bigl(2^{i-1}x,0\bigr)+\varphi \bigl(0,2^{i-1}x \bigr)+\varphi\bigl(2^{i-1}x,2^{i-1}x\bigr)\bigr)< \infty $$
(3.1)
$$ \lim_{m\rightarrow\infty}2^{-m\beta}\varphi\bigl(2^{m}x,2^{m}y \bigr)=0 $$
(3.2)
$$ \bigl\Vert f(x+y)-g(x)-h(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\varphi(x,y)\psi (z_{1},\ldots,z_{n-1}) $$
(3.3)
$$\begin{aligned}& \bigl\Vert f(x)-A(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \Phi(x)\psi(z_{1},\ldots,z_{n-1})+\bigl\Vert h(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+\bigl\Vert g(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}, \end{aligned}$$
(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)
Proof
Putting \(y=x\) in inequality (3.3), we get
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\). Putting \(y=0\) in inequality (3.3), we get
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\). It then follows from (3.8) that
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\). Putting \(x=0\) in inequality (3.3), we get
for all \(y\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\). Thus, we obtain
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\).
$$ \bigl\Vert f(2x)-g(x)-h(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\varphi(x,x)\psi (z_{1},\ldots,z_{n-1}) $$
(3.7)
$$ \bigl\Vert f(x)-g(x)-h(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\varphi(x,0)\psi (z_{1},\ldots,z_{n-1}) $$
(3.8)
$$ \bigl\Vert f(x)-g(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \leq\varphi(x,0)\psi(z_{1},\ldots,z_{n-1})+ \bigl\Vert h(0),z_{1},\ldots ,z_{n-1}\bigr\Vert _{\beta} $$
(3.9)
$$\bigl\Vert f(y)-g(0)-h(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\leq\varphi(0,y)\psi (z_{1},\ldots,z_{n-1}) $$
$$ \bigl\Vert f(x)-h(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \leq\varphi(0,x)\psi(z_{1},\ldots,z_{n-1})+ \bigl\Vert g(0),z_{1},\ldots ,z_{n-1}\bigr\Vert _{\beta} $$
(3.10)
Let us define
Using (3.7), (3.9) and (3.10), we have
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\). Replacing x with 2x in (3.11), we get
for all \(x\in X\) and \(z_{1}, \ldots,z_{n-1}\in Y\). It then follows from (3.11) and (3.12) that
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\).
$$\begin{aligned}& u(x,z_{1},\ldots,z_{n-1}) \\& \quad = \bigl\Vert g(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+ \bigl\Vert h(0),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta}+\varphi(x,x)\psi(z_{1},\ldots,z_{n-1}) \\& \qquad {}+ \varphi(x,0)\psi(z_{1},\ldots,z_{n-1})+ \varphi(0,x)\psi(z_{1},\ldots,z_{n-1}). \end{aligned}$$
$$\begin{aligned}& \bigl\Vert f(2x)-2f(x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta} \\& \quad \leq \bigl\Vert f(2x)-g(x)-h(x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta}+ \bigl\Vert g(x)-f(x),z_{1},\ldots ,z_{n-1}\bigr\Vert _{\beta} \\& \qquad {}+\bigl\Vert h(x)-f(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \bigl\Vert g(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+\bigl\Vert h(0),z_{1},\ldots ,z_{n-1} \bigr\Vert _{\beta}+\varphi(x,0)\psi(z_{1}, \ldots,z_{n-1}) \\& \qquad {}+\varphi (0,x)\psi(z_{1},\ldots,z_{n-1})+ \varphi(x,x)\psi(z_{1},\ldots,z_{n-1}) \\& \quad = u(x,z_{1},\ldots,z_{n-1}) \end{aligned}$$
(3.11)
$$ \bigl\Vert f\bigl(2^{2}x\bigr)-2f(2x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}\leq u(2x,z_{1}, \ldots ,z_{n-1}) $$
(3.12)
$$\begin{aligned}& \bigl\Vert f\bigl(2^{2}x\bigr)-2^{2}f(x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \bigl\Vert f\bigl(2^{2}x\bigr)-2f(2x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}+2^{\beta}\bigl\Vert f(2x)-2f(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq u(2x,z_{1},\ldots,z_{n-1})+2^{\beta}u(x,z_{1}, \ldots,z_{n-1}) \end{aligned}$$
Applying an induction argument on m, we will prove that
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
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\). Hence, it follows from (3.11) that
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\), which proves inequality (3.13). By (3.13), we have
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
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
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
Replacing x, y by \(2^{m}x\), \(2^{m}y\) in (3.3) and dividing both sides by \(2^{m\beta}\), we get
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y\). It follows from (3.9) that
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y \). Considering (3.1), we get
It follows from (3.15) that
for all \(x\in X\). Also, by (3.10), we have
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y \). Similarly, it follows from (3.17) that
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
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y \). Hence \(A(x+y)-A(x)-A(y)=0\).
$$ \bigl\Vert f\bigl(2^{m}x\bigr)-2^{m}f(x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}\leq\sum _{i=1}^{m}2^{(i-1)\beta}u\bigl(2^{m-i}x,z_{1}, \ldots,z_{n-1}\bigr) $$
(3.13)
$$\bigl\Vert f\bigl(2^{m+1}x\bigr)-2^{m}f(2x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}\leq\sum _{i=1}^{m}2^{(i-1)\beta}u\bigl(2^{m+1-i}x,z_{1}, \ldots,z_{n-1}\bigr) $$
$$\begin{aligned}& \bigl\Vert f\bigl(2^{m+1}x\bigr)-2^{m+1}f(x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \bigl\Vert f\bigl(2^{m+1}x\bigr)-2^{m}f(x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}+2^{n\beta}\bigl\Vert f(2x)-2f(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \sum_{i=1}^{m}2^{(i-1)\beta}u \bigl(2^{m+1-i}x,z_{1},\ldots ,z_{n-1} \bigr)+2^{m\beta}u(x,z_{1},\ldots,z_{n-1}) \\& \quad = \sum_{i=1}^{m+1}2^{(i-1)\beta}u \bigl(2^{m+1-i}x,z_{1},\ldots,z_{n-1}\bigr) \end{aligned}$$
$$ \bigl\Vert 2^{-m}f\bigl(2^{m}x\bigr)-f(x),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}\leq\sum _{i=1}^{m}2^{(i-1-m)\beta}u\bigl(2^{m-i}x,z_{1}, \ldots,z_{n-1}\bigr) $$
(3.14)
$$\begin{aligned}& \bigl\Vert 2^{-k}f\bigl(2^{k}x\bigr)-2^{-m}f \bigl(2^{m}x\bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \sum_{i=m}^{k-1}\bigl\Vert 2^{-i}f\bigl(2^{i}x\bigr)-2^{-(i+1)}f \bigl(2^{i+1}x\bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \sum_{i=m}^{k-1}2^{-(i+1)\beta} \bigl\Vert 2f\bigl(2^{i}x\bigr)-f\bigl(2^{i+1}x \bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad = \sum_{i=m}^{k-1}2^{-(i+1)\beta} u \bigl(2^{i}x,z_{1},\ldots,z_{n-1}\bigr) \\& \quad = \sum_{i=m}^{k-1}2^{-(i+1)\beta} \bigl[\varphi\bigl(2^{i}x,0\bigr)\psi(z_{1},\ldots ,z_{n-1})+\varphi\bigl(0,2^{i}x\bigr)\psi(z_{1}, \ldots,z_{n-1}) \\& \qquad {}+\varphi \bigl(2^{i}x,2^{i}x\bigr) \psi(z_{1},\ldots,z_{n-1}) +\bigl\Vert h(0),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}+\bigl\Vert g(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\bigr] \\& \quad \leq \sum_{i=m}^{k-1}2^{-(i+1)\beta} \bigl[\varphi\bigl(2^{i}x,0\bigr)+\varphi \bigl(0,2^{i}x \bigr)+\varphi\bigl(2^{i}x,2^{i}x\bigr)\bigr] \psi(z_{1},\ldots,z_{n-1}) \\& \qquad {}+2^{-m}\bigl(\bigl\Vert h(0),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}+\bigl\Vert g(0),z_{1},\ldots ,z_{n-1}\bigr\Vert _{\beta}\bigr) \end{aligned}$$
$$\lim_{m, k\rightarrow\infty} \bigl\Vert 2^{-k}f\bigl(2^{k}x \bigr)-2^{-m}f\bigl(2^{m}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}=0 $$
$$A(x)=\lim_{m\rightarrow\infty} 2^{-m}f\bigl(2^{m}x\bigr). $$
$$\begin{aligned}& 2^{-m\beta}\bigl\Vert f\bigl(2^{m}x+2^{m}y\bigr)-g \bigl(2^{m}x\bigr)-h\bigl(2^{m}y\bigr),z_{1},\ldots ,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq 2^{-m\beta}\varphi\bigl(2^{m}x,2^{m}y\bigr) \psi(z_{1},\ldots,z_{n-1}) \end{aligned}$$
$$\begin{aligned} \begin{aligned}[b] &\bigl\Vert 2^{-m}f\bigl(2^{m}x \bigr)-2^{-m}g\bigl(2^{m}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta} \\ &\quad \leq 2^{-m\beta}\bigl[\bigl\Vert h(0),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}+\varphi\bigl(2^{m}x,0 \bigr)\psi (z_{1},\ldots,z_{n-1})\bigr] \end{aligned} \end{aligned}$$
(3.15)
$$\begin{aligned}& 2^{-m\beta}\varphi\bigl(2^{m}x,0\bigr)\psi(z_{1}, \ldots,z_{n-1}) \\& \quad \leq2^{\beta }\sum_{i=m}^{\infty}2^{-(i+1)\beta} \bigl[\varphi\bigl(2^{i}x,0\bigr)\psi(z_{1},\ldots ,z_{n-1})+\varphi\bigl(0,2^{i}x\bigr)\psi(z_{1}, \ldots,z_{n-1}) \\& \qquad {}+\varphi \bigl(2^{i}x,2^{i}x\bigr) \psi(z_{1},\ldots,z_{n-1})\bigr] \\& \quad \rightarrow 0 \quad \mbox{as }m\rightarrow \infty. \end{aligned}$$
$$ \lim_{m\rightarrow\infty} 2^{-m}g\bigl(2^{m}x\bigr)=\lim _{m\rightarrow\infty} 2^{-m}f\bigl(2^{m}x \bigr)=A(x) $$
(3.16)
$$\begin{aligned}& \bigl\Vert 2^{-m}h\bigl(2^{m}x \bigr)-2^{-m}f\bigl(2^{m}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq 2^{-m\beta}\bigl[\bigl\Vert g(0),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta}+\varphi \bigl(0,2^{m}x \bigr)\psi(z_{1},\ldots,z_{n-1})\bigr] \end{aligned}$$
(3.17)
$$ \lim_{m\rightarrow\infty} 2^{-m}h\bigl(2^{m}x\bigr)=\lim _{m\rightarrow\infty} 2^{-m}f\bigl(2^{m}x \bigr)=A(x) $$
(3.18)
$$\begin{aligned}& \bigl\Vert A(x+y)-A(x)-A(y),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad = \lim_{m\rightarrow\infty}\bigl\Vert 2^{-m}f \bigl(2^{m}x+2^{m}y\bigr)-2^{-m}g \bigl(2^{m}x\bigr)-2^{-m}h\bigl(2^{m}y \bigr),z_{1},\ldots ,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \lim_{m\rightarrow\infty}2^{-m\beta}\varphi \bigl(2^{m}x,2^{m}y\bigr)\psi (z_{1}, \ldots,z_{n-1}) \\& \quad = 0 \end{aligned}$$
Taking the limit as \(m\rightarrow\infty\) in (3.14), we get
for all \(x\in X\) and \(z_{1},\ldots,z_{n-1}\in Y \), which proves (3.4).
$$\begin{aligned}& \bigl\Vert A(x)-f(x),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq \lim_{m\rightarrow \infty}\sum_{i=1}^{m}2^{(i-1-m)\beta}u \bigl(2^{m-i}x,z_{1},\ldots,z_{n-1}\bigr) \\& \quad = \lim_{m\rightarrow\infty}\bigl(1-2^{-m\beta}\bigr) \bigl(\bigl\Vert g(0),z_{1},\ldots ,z_{n-1} \bigr\Vert _{\beta}+ \bigl\Vert h(0),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta}\bigr) \\& \qquad {}+\lim_{m\rightarrow\infty}\sum_{i=1}^{m}2^{(i-m-1)\beta} \bigl(\varphi \bigl(2^{m-i}x,0\bigr)\psi(z_{1}, \ldots,z_{n-1})+\varphi\bigl(0,2^{m-i}x\bigr)\psi (z_{1},\ldots,z_{n-1}) \\& \qquad {} +\varphi\bigl(2^{m-i}x,2^{m-i}x\bigr) \psi(z_{1},\ldots,z_{n-1})\bigr) \\& \quad = \bigl\Vert h(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+\bigl\Vert g(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+\Phi(x)\psi(z_{1},\ldots,z_{n-1}) \end{aligned}$$
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
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). □
$$\begin{aligned}& \bigl\Vert A(x)-A'(x),z_{1},\ldots,z_{n-1} \bigr\Vert _{\beta} \\& \quad \leq 2^{-m\beta}\bigl\Vert A\bigl(2^{m}x\bigr)-f \bigl(2^{m}x\bigr),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+2^{-m\beta}\bigl\Vert f\bigl(2^{m}x \bigr)-A'\bigl(2^{m}x\bigr),z_{1}, \ldots,z_{n-1}\bigr\Vert _{\beta} \\& \quad \leq 2^{-m \beta+1}\bigl(\bigl\Vert g(0),z_{1}, \ldots,z_{n-1} \bigr\Vert _{\beta}+\bigl\Vert h(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}+ \Phi \bigl(2^{m}x\bigr)\psi(z_{1},\ldots ,z_{n-1})\bigr) \\& \quad = 2^{-m\beta+1}\bigl(\bigl\Vert g(0),z_{1}, \ldots,z_{n-1} \bigr\Vert _{\beta}+\bigl\Vert h(0),z_{1},\ldots,z_{n-1}\bigr\Vert _{\beta}\bigr) \\& \qquad {}+2\sum_{i=m+1}^{\infty}2^{-i\beta} \bigl(\varphi\bigl(2^{i-1}x,0\bigr)+\varphi \bigl(0,2^{i-1}x \bigr)+\varphi\bigl(2^{i-1}x,2^{i-1}x\bigr)\bigr) \psi(z_{1},\ldots,z_{n-1}) \\& \quad \rightarrow 0 \quad \mbox{as }m \rightarrow \infty \end{aligned}$$
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.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors conceived of the study, participated in its design and coordination, drafted the manuscript, participated in the sequence alignment, and read and approved the final manuscript.