Non-Archimedean Hyers-Ulam-Rassias stability of m-variable functional equation

Abstract

The main goal of this paper is to study the Hyers-Ulam-Rassias stability of the following Euler-Lagrange type additive functional equation:

$$\sum _{j=1}^{m}f(-r_j{x}_j+\sum _{1\le i\le m,i\ne j}{r}_i{x}_i)+2\sum _{i=1}^m{r}_{i}f(x_i)=mf\left(\sum _{i=1}^m{r}_i{x}_i\right),$$

where $r_1,\dots ,r_m\in \mathbb{R}$, ${\sum }_{i=k}^m{r}_k\ne 0$, and $r_i,r_j\ne 0$ for some $1\le i<j\le m$, in non-Archimedean Banach spaces.

MSC:39B22, 39B52, 46S10.

1 Introduction and preliminaries

A valuation is a function $|\cdot |$ from a field $\mathbb{K}$ into $[0,\mathrm{\infty })$ such that 0 is the unique element having the 0 valuation, $|rs|=|r||s|$ and the triangle inequality is replaced by $|r+s|\le max\{|r|,|s|\}$.

The field $\mathbb{K}$ is called a valued field if $\mathbb{K}$ carries a valuation. The usual absolute values of $\mathbb{R}$ and $\mathbb{C}$ are the examples of valuations.

Let us consider the valuation which satisfies a stronger condition than the triangle inequality. If the triangle inequality is replaced by $|r+s|\le max\{|r|,|s|\}$ for all $r,s\in \mathbb{K}$, then the function $|\cdot |$ is called a non-Archimedean valuation and the field is called a non-Archimedean field. Clearly, $|1|=|-1|=1$ and $|n|\le 1$ for all integers $n\ge 1$. A trivial example of a non-Archimedean valuation is the function $|\cdot |$ taking everything except for 0 into 1 and $|0|=0$.

Definition 1.1 Let X be a vector space over a field $\mathbb{K}$ with a non-Archimedean valuation $|\cdot |$. A function $\parallel \cdot \parallel :X\to [0,\mathrm{\infty })$ is called a non-Archimedean norm if the following conditions hold:

• (a) $\parallel x\parallel =0$ if and only if $x=0$ for all $x\in X$;

• (b) $\parallel rx\parallel =|r|\parallel x\parallel$ for all $r\in \mathbb{K}$ and $x\in X$;

• (c) the strong triangle inequality holds:

  $$\parallel x+y\parallel \le max\{\parallel x\parallel ,\parallel y\parallel \}$$

for all $x,y\in X$.

Then $(X,\parallel \cdot \parallel )$ is called a non-Archimedean normed space.

Definition 1.2 Let $\{x_n\}$ be a sequence in a non-Archimedean normed space X.

• (a) A sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ in a non-Archimedean space is a Cauchy sequence iff the sequence ${\{x_{n+1}-x_n\}}_{n=1}^{\mathrm{\infty }}$ converges to zero;

• (b) The sequence $\{x_n\}$ is said to be convergent if, for any $\epsilon >0$, there is a positive integer N and $x\in X$ such that $\parallel x_n-x\parallel \le \epsilon$, for all $n\ge N$. Then the point $x\in X$ is called the limit of the sequence $\{x_n\}$, which is denoted by ${lim}_{n\to \mathrm{\infty }}{x}_n=x$;

• (c) If every Cauchy sequence in X converges, then the non-Archimedean normed space X is called a non-Archimedean Banach space.

Example 1.1 Fix a prime number p. For any nonzero rational number x, there exists a unique integer $n_x\in \mathbb{Z}$ such that $x=\frac{a}{b}{p}^{n_x}$, where a and b are integers not divisible by p. Then ${|x|}_p:=p^{-n_x}$ defines a non-Archimedean norm on $\mathbb{Q}$. The completion of $\mathbb{Q}$ with respect to the metric $d(x,y)={|x-y|}_p$ is denoted by ${\mathbb{Q}}_p$ which is called the p-adic number field. In fact, ${\mathbb{Q}}_p$ is the set of all formal series $x={\sum }_{k\ge n_x}^{\mathrm{\infty }}{a}_k{p}^k$ where $|a_k|\le p-1$ are integers. The addition and multiplication between any two elements of ${\mathbb{Q}}_p$ are defined naturally. The norm ${|{\sum }_{k\ge n_x}^{\mathrm{\infty }}{a}_k{p}^k|}_p=p^{-n_x}$ is a non-Archimedean norm on ${\mathbb{Q}}_p$ and it makes ${\mathbb{Q}}_p$ a locally compact field.

Theorem 1.1 Let$(X,d)$be a complete generalized metric space and$J:X\to X$be a strictly contractive mapping with Lipschitz constant$L<1$. Then, for all$x\in X$, either$d(J^{n}x,J^{n+1}x)=\mathrm{\infty }$for all nonnegative integers n or there exists a positive integer$n_0$such that:

• (a) $d(J^{n}x,J^{n+1}x)<\mathrm{\infty }$for all$n_0\ge n_0$;

• (b) the sequence$\{J^{n}x\}$converges to a fixed point$y^{\ast }$of J;

• (c) $y^{\ast }$is the unique fixed point of J in the set$Y=\{y\in X:d(J^{n_0}x,y)<\mathrm{\infty }\}$;

• (d) $d(y,y^{\ast })\le \frac{1}{1-L}d(y,Jy)$for all$y\in Y$.

In this paper, we prove the generalized Hyers-Ulam stability of the following functional equation:

$$\sum _{j=1}^{m}f(-r_j{x}_j+\sum _{1\le i\le m,i\ne j}{r}_i{x}_i)+2\sum _{i=1}^m{r}_{i}f(x_i)=mf\left(\sum _{i=1}^m{r}_i{x}_i\right),$$

(1.1)

where $r_1,\dots ,r_m\in \mathbb{R}$, ${\sum }_{k=1}^m{r}_k\ne 0$, and $r_i,r_j\ne 0$ for some $1\le i<j\le m$, in non-Archimedean Banach spaces. A classical question in the theory of functional equations is the following: ‘When is it true that a function which approximately satisfies a functional equation D must be close to an exact solution of D?’.

If the problem accepts a solution, we say that the equation D is stable. The first stability problem concerning group homomorphisms was raised by Ulam [34] in 1940.

In the next year D. H. Hyres [17], gave a positive answer to the above question for additive groups under the assumption that the groups are Banach spaces. In 1978, Th. M. Rassias [24] proved a generalization of Hyres’ theorem for additive mappings.

The result of Th. M. Rassias has influenced the development of what is now called the Hyers-Ulam-Rassias stability theory for functional equations. In 1994, a generalization of Rassias’ theorem was obtained by Gǎvruta [15] by replacing the bound $ϵ({\parallel x\parallel }^p+{\parallel y\parallel }^p)$ by a general control function $\phi (x,y)$.

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 (see [1–33]).

2 Non-Archimedean stability of the functional equation (1.1): a fixed point approach

In this section, using a fixed point alternative approach, we prove the generalized Hyers-Ulam stability of the functional equation (1.1) in non-Archimedean normed spaces. Throughout this section, let X be a non-Archimedean normed space and Y be a non-Archimedean Banach space. Also $|2|\ne 1$.

Lemma 2.1 Let$\mathcal{X}$and$\mathcal{Y}$be linear spaces and let$r_1,\dots ,r_n$be real numbers with${\sum }_{k=1}^n{r}_k\ne 0$and$r_i,r_j\ne 0$for some$1\le i<j\le n$. Assume that a mapping$f:\mathcal{X}\to \mathcal{Y}$satisfies the functional equation (1.1) for all$x_1,\dots ,x_n\in \mathcal{X}$. Then the mapping f is Cauchy additive. Moreover, $f(r_{k}x)=r_{k}f(x)$for all$x\in \mathcal{X}$and all$1\le k\le n$.

Proof Since ${\sum }_{k=1}^n{r}_k\ne 0$, putting $x_1=\cdots =x_n=0$ in (1.1), we get $f(0)=0$. Without loss of generality, we may assume that $r_1,r_2\ne 0$. Letting $x_3=\cdots =x_n=0$ in (1.1), we get

$$f(-r_1{x}_1+r_2{x}_2)+f(r_1{x}_1-r_2{x}_2)+2r_{1}f(x_1)+2r_{2}f(x_2)=2f(r_1{x}_1+r_2{x}_2)$$

(2.1)

for all $x_1,x_2\in \mathcal{X}$. Letting $x_2=0$ in (2.1), we get

$$2r_{1}f(x_1)=f(r_1{x}_1)-f(-r_1{x}_1)$$

(2.2)

for all $x_1\in \mathcal{X}$. Similarly, by putting $x_1=0$ in (2.1), we get

$$2r_{2}f(x_2)=f(r_2{x}_2)-f(-r_2{x}_2)$$

(2.3)

for all $x_1\in \mathcal{X}$. It follows from (2.1), (2.2) and (2.3) that

(2.4)

for all $x_1,x_2\in \mathcal{X}$. Replacing $x_1$ and $x_2$ by $\frac{x}{r_1}$ and $\frac{y}{r_2}$ in (2.4), we get

$$f(-x+y)+f(x-y)+f(x)+f(y)-f(-x)-f(-y)=2f(x+y)$$

(2.5)

for all $x,y\in \mathcal{X}$. Letting $y=-x$ in (2.5), we get that $f(-2x)+f(2x)=0$ for all $x\in \mathcal{X}$. So the mapping L is odd. Therefore, it follows from (2.5) that the mapping f is additive. Moreover, let $x\in \mathcal{X}$ and $1\le k\le n$. Setting $x_k=x$ and $x_l=0$ for all $1\le l\le n$, $l\ne k$, in (1.1) and using the oddness of f, we get that $f(r_{k}x)=r_{k}f(x)$. □

Using the same method as in the proof of Lemma 2.1, we have an alternative result of Lemma 2.1 when ${\sum }_{k=1}^n{r}_k=0$.

Lemma 2.2 Let$\mathcal{X}$and$\mathcal{Y}$be linear spaces and let$r_1,\dots ,r_n$be real numbers with$r_i,r_j\ne 0$for some$1\le i<j\le n$. Assume that a mapping$f:\mathcal{X}\to \mathcal{Y}$with$f(0)=0$satisfying the functional equation (1.1) for all$x_1,\dots ,x_n\in \mathcal{X}$. Then the mapping f is Cauchy additive. Moreover, $f(r_{k}x)=r_{k}f(x)$for all$x\in \mathcal{X}$and all$1\le k\le n$.

Remark 2.1 Throughout this paper, $r_1,\dots ,r_m$ will be real numbers such that $r_i,r_j\ne 0$ for fixed $1\le i<j\le m$ and ${\phi }_{i,j}(x,y):=\phi (0,\dots ,0,\underset{i\mathrm{th}}{\underbrace{x}},0,\dots ,0,\underset{j\mathrm{th}}{\underbrace{y}},0,\dots ,0)$ for all $x,y\in X$ and all $1\le i<j\le m$.

Theorem 2.1 Let $\phi :X^m\to [0,\mathrm{\infty })$ be a function such that there exists an $L<1$ with

$$\phi (\frac{x_1}{2},\dots ,\frac{x_m}{2})\le \frac{L\phi (x_1,\dots ,x_m)}{|2|}$$

(2.6)

for all$x_1,\dots ,x_m\in X$. Let$f:X\to Y$be a mapping with$f(0)=0$satisfying the following inequality:

(2.7)

for all$x_1,\dots ,x_m\in X$. Then there is a unique Euler-Lagrange type additive mapping$\mathit{EL}:X\to Y$such that

$$\begin{array}{rcl}\parallel f(x)-\mathit{EL}(x)\parallel & \le & \frac{L}{|2|-|2|L}max\{max\{{\phi }_{i,j}(\frac{x}{2r_i},-\frac{x}{2r_j}),{\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,-\frac{x}{2r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{r_i},\frac{x}{r_j}),{\phi }_{i,j}(\frac{x}{r_i},0),{\phi }_{i,j}(0,\frac{x}{r_j})\}\}\end{array}$$

(2.8)

for all$x\in X$.

Proof For each $1\le k\le m$ with $k\ne i,j$, let $x_k=0$ in (2.7). Then we get the following inequality:

(2.9)

for all $x_i,x_j\in X$. Letting $x_i=0$ in (2.9), we get

$$\parallel f(-r_j{x}_j)-f(r_j{x}_j)+2r_{j}f(x_j)\parallel \le {\phi }_{i,j}(0,x_j)$$

(2.10)

for all $x_j\in X$. Similarly, letting $x_j=0$ in (2.9), we get

$$\parallel f(-r_i{x}_i)-f(r_i{x}_i)+2r_{i}f(x_i)\parallel \le {\phi }_{i,j}(x_i,0)$$

(2.11)

for all $x_i\in X$. It follows from (2.9), (2.10) and (2.11) that for all $x_i,x_j\in X$

(2.12)

Replacing $x_i$ and $x_j$ by $\frac{x}{r_i}$ and $\frac{y}{r_j}$ in (2.12), we get that

(2.13)

for all $x,y\in X$. Putting $y=x$ in (2.13), we get

$$\parallel f(x)-f(-x)-f(2x)\parallel \le \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{r_i},\frac{x}{r_j}),{\phi }_{i,j}(\frac{x}{r_i},0),{\phi }_{i,j}(0,\frac{x}{r_j})\}$$

(2.14)

for all $x\in X$. Replacing x and y by $\frac{x}{2}$ and $-\frac{x}{2}$ in (2.13) respectively, we get

$$\parallel f(x)+f(-x)\parallel \le max\{{\phi }_{i,j}(\frac{x}{2r_i},-\frac{x}{2r_j}),{\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,-\frac{x}{2r_j})\}$$

(2.15)

for all $x\in X$. It follows from (2.14) and (2.15) that

(2.16)

for all $x\in X$. Replacing x by $\frac{x}{2}$ in (2.16), we obtain

(2.17)

Consider the set $S:=\{g:X\to Y;g(0)=0\}$ and the generalized metric d in S defined by

$$\begin{array}{rcl}d(f,g)& =& \underset{\mu \in {\mathbb{R}}^{+}}{inf}\{\parallel g(x)-h(x)\parallel \le \mu max\{max\{{\phi }_{i,j}(\frac{x}{2r_i},-\frac{x}{2r_j}),\\ {\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,-\frac{x}{2r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{r_i},\frac{x}{r_j}),{\phi }_{i,j}(\frac{x}{r_i},0),{\phi }_{i,j}(0,\frac{x}{r_j})\}\},\mathrm{\forall }x\in X\},\end{array}$$

where $inf\mathrm{\varnothing }=+\mathrm{\infty }$. It is easy to show that $(S,d)$ is complete (see [18], Lemma 2.1). Now, we consider a linear mapping $J:S\to S$ such that

$$Jh(x):=2h\left(\frac{x}{2}\right)$$

for all $x\in X$. Let $g,h\in S$ be such that $d(g,h)=ϵ$. Then

$$\begin{array}{rcl}\parallel g(x)-h(x)\parallel & \le & ϵmax\{max\{{\phi }_{i,j}(\frac{x}{2r_i},-\frac{x}{2r_j}),{\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,-\frac{x}{2r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{r_i},\frac{x}{r_j}),{\phi }_{i,j}(\frac{x}{r_i},0),{\phi }_{i,j}(0,\frac{x}{r_j})\}\}\end{array}$$

for all $x\in X$, and so

$$\begin{array}{rcl}\parallel Jg(x)-Jh(x)\parallel & =& \parallel 2g\left(\frac{x}{2}\right)-2h\left(\frac{x}{2}\right)\parallel \\ \le & |2|ϵmax\{max\{{\phi }_{i,j}(\frac{x}{4r_i},-\frac{x}{4r_j}),{\phi }_{i,j}(\frac{x}{4r_i},0),{\phi }_{i,j}(0,-\frac{x}{4r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{2r_i},\frac{x}{2r_j}),{\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,\frac{x}{2r_j})\}\}\\ \le & |2|\frac{Lϵ}{|2|}max\{max\{{\phi }_{i,j}(\frac{x}{2r_i},-\frac{x}{2r_j}),{\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,-\frac{x}{2r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{r_i},\frac{x}{r_j}),{\phi }_{i,j}(\frac{x}{r_i},0),{\phi }_{i,j}(0,\frac{x}{r_j})\}\}\end{array}$$

for all $x\in X$. Thus $d(g,h)=ϵ$ implies that $d(Jg,Jh)\le Lϵ$. This means that

$$d(Jg,Jh)\le Ld(g,h)$$

for all $g,h\in S$. It follows from (2.17) that

$$d(f,Jf)\le \frac{L}{|2|}.$$

By Theorem 1.1, there exists a mapping $\mathit{EL}:X\to Y$ satisfying the following:

• (1) EL is a fixed point of J, that is,

  $$\mathit{EL}\left(\frac{x}{2}\right)=\frac{1}{2}\mathit{EL}(x)$$

  (2.18)

for all $x\in X$. The mapping EL is a unique fixed point of J in the set

$$\mathrm{\Omega }=\{h\in S:d(g,h)<\mathrm{\infty }\}.$$

This implies that EL is a unique mapping satisfying (2.18) such that there exists $\mu \in (0,\mathrm{\infty })$ satisfying

$$\begin{array}{rcl}\parallel f(x)-\mathit{EL}(x)\parallel & \le & \mu max\{max\{{\phi }_{i,j}(\frac{x}{2r_i},-\frac{x}{2r_j}),{\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,-\frac{x}{2r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{r_i},\frac{x}{r_j}),{\phi }_{i,j}(\frac{x}{r_i},0),{\phi }_{i,j}(0,\frac{x}{r_j})\}\}\end{array}$$

(2.19)

for all $x\in X$.

• (2) $d(J^{n}f,\mathit{EL})\to 0$ as $n\to \mathrm{\infty }$. This implies the equality

  $$\underset{n\to \mathrm{\infty }}{lim}{2}^{n}f\left(\frac{x}{2^n}\right)=\mathit{EL}(x)$$

for all $x\in X$.

• (3) $d(f,\mathit{EL})\le \frac{d(f,Jf)}{1-L}$ with $f\in \mathrm{\Omega }$, which implies the inequality

  $$d(f,\mathit{EL})\le \frac{L}{|2|-|2|L}.$$

This implies that the inequality (2.8) holds.

By (2.6) and (2.7), we obtain

for all $x_1,\dots ,x_m\in X$ and $n\in \mathbb{N}$. So EL satisfies (1.1). Thus, the mapping $\mathit{EL}:X\to Y$ is Euler-Lagrange type additive, as desired. □

Corollary 2.1 Let$\theta \ge 0$and r be a real number with$0<r<1$. Let$f:X\to Y$be a mapping with$f(0)=0$satisfying the inequality

(2.20)

for all$x_1,\dots ,x\in X$. Then, the limit$\mathit{EL}(x)={lim}_{n\to \mathrm{\infty }}{2}^{n}f(\frac{x}{2^n})$exists for all$x\in X$and$\mathit{EL}:X\to Y$is a unique Euler-Lagrange additive mapping such that

$$\begin{array}{rcl}\parallel f(x)-\mathit{EL}(x)\parallel & \le & \frac{|2|}{{|2|}^{r+1}-{|2|}^2}max\{max\{\frac{{|2|}^r\theta {\parallel x\parallel }^r({|r_i|}^r+{|r_j|}^r)}{{|4|}^r{|r_i{r}_j|}^r},\frac{\theta {\parallel x\parallel }^r}{{|2|}^r{|r_i|}^r},\frac{\theta {\parallel x\parallel }^r}{{|2|}^r{|r_j|}^r}\},\\ \frac{1}{|2|}max\{\frac{\theta {\parallel x\parallel }^r({|r_i|}^r+{|r_j|}^r)}{{|r_i{r}_j|}^r},\frac{\theta {\parallel x\parallel }^r}{{|r_i|}^r},\frac{\theta {\parallel x\parallel }^r}{{|r_j|}^r}\}\}\\ \le & \frac{\theta {\parallel x\parallel }^r({|r_i|}^r+{|r_j|}^r)}{{|r_i{r}_j|}^r({|2|}^{r+1}-{|2|}^2)}\end{array}$$

for all$x\in X$.

Proof The proof follows from Theorem 2.1 by taking $\phi (x_1,\dots ,x_m)=\theta ({\sum }_{i=1}^m{\parallel x_i\parallel }^r)$ for all $x_1,\dots ,x_m\in X$. In fact, if we choose $L={|2|}^{1-r}$, then we get the desired result. □

Theorem 2.2 Let $\phi :X^m\to [0,\mathrm{\infty })$ be a function such that there exists an $L<1$ with

$$\phi (x_1,\dots ,x_m)\le |2|L\phi (\frac{x_1}{2},\dots ,\frac{x_m}{2})$$

(2.21)

for all$x_1,\dots ,x_m\in X$. Let$f:X\to Y$be a mapping with$f(0)=0$satisfying the inequality (2.7). Then, there is a unique Euler-Lagrange additive mapping$\mathit{EL}:X\to Y$such that

$$\begin{array}{rcl}\parallel f(x)-\mathit{EL}(x)\parallel & \le & \frac{1}{|2|-|2|L}max\{max\{{\phi }_{i,j}(\frac{x}{2r_i},-\frac{x}{2r_j}),{\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,-\frac{x}{2r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{r_i},\frac{x}{r_j}),{\phi }_{i,j}(\frac{x}{r_i},0),{\phi }_{i,j}(0,\frac{x}{r_j})\}\}.\end{array}$$

(2.22)

Proof By (2.16), we have

$$\begin{array}{rcl}\parallel \frac{f(2x)}{2}-f(x)\parallel & \le & \frac{1}{|2|}max\{max\{{\phi }_{i,j}(\frac{x}{2r_i},-\frac{x}{2r_j}),{\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,-\frac{x}{2r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{r_i},\frac{x}{r_j}),{\phi }_{i,j}(\frac{x}{r_i},0),{\phi }_{i,j}(0,\frac{x}{r_j})\}\}\end{array}$$

(2.23)

for all $x\in X$ . Let $(S,d)$ be the generalized metric space defined in the proof of Theorem 2.1. Now, we consider a linear mapping $J:S\to S$ such that

$$Jh(x):=\frac{1}{2}h(2x)$$

for all $x\in X$. Let $g,h\in S$ be such that $d(g,h)=ϵ$. Then

$$\begin{array}{rcl}\parallel g(x)-h(x)\parallel & \le & ϵmax\{max\{{\phi }_{i,j}(\frac{x}{2r_i},-\frac{x}{2r_j}),{\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,-\frac{x}{2r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{r_i},\frac{x}{r_j}),{\phi }_{i,j}(\frac{x}{r_i},0),{\phi }_{i,j}(0,\frac{x}{r_j})\}\}\end{array}$$

for all $x\in X$, and so

$$\begin{array}{rcl}\parallel Jg(x)-Jh(x)\parallel & =& \parallel \frac{g(2x)}{2}-\frac{h(2x)}{2}\parallel \\ \le & \frac{1}{|2|}ϵmax\{max\{{\phi }_{i,j}(\frac{x}{r_i},-\frac{x}{r_j}),{\phi }_{i,j}(\frac{x}{r_i},0),{\phi }_{i,j}(0,-\frac{x}{r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{2x}{r_i},\frac{2x}{r_j}),{\phi }_{i,j}(\frac{2x}{r_i},0),{\phi }_{i,j}(0,\frac{2x}{r_j})\}\}\\ \le & |2|L\frac{ϵ}{|2|}max\{max\{{\phi }_{i,j}(\frac{x}{2r_i},-\frac{x}{2r_j}),{\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,-\frac{x}{2r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{r_i},\frac{x}{r_j}),{\phi }_{i,j}(\frac{x}{r_i},0),{\phi }_{i,j}(0,\frac{x}{r_j})\}\}\end{array}$$

for all $x\in X$ . Thus $d(g,h)=ϵ$ implies that $d(Jg,Jh)\le Lϵ$. This means that

$$d(Jg,Jh)\le Ld(g,h)$$

for all $g,h\in S$. It follows from (2.23) that

$$d(f,Jf)\le \frac{1}{|2|}.$$

By Theorem 1.1, there exists a mapping $\mathit{EL}:X\to Y$ satisfying the following:

• (1) EL is a fixed point of J, that is,

  $$\mathit{EL}(2x)=2\mathit{EL}(x)$$

  (2.24)

for all $x\in X$. The mapping EL is a unique fixed point of J in the set

$$\mathrm{\Omega }=\{h\in S:d(g,h)<\mathrm{\infty }\}.$$

This implies that EL is a unique mapping satisfying (2.24) such that there exists $\mu \in (0,\mathrm{\infty })$ satisfying

$$\begin{array}{rcl}\parallel g(x)-h(x)\parallel & \le & \mu max\{max\{{\phi }_{i,j}(\frac{x}{2r_i},-\frac{x}{2r_j}),{\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,-\frac{x}{2r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{r_i},\frac{x}{r_j}),{\phi }_{i,j}(\frac{x}{r_i},0),{\phi }_{i,j}(0,\frac{x}{r_j})\}\}\end{array}$$

for all $x\in X$.

• (2) $d(J^{n}f,\mathit{EL})\to 0$ as $n\to \mathrm{\infty }$. This implies the equality

  $$\underset{n\to \mathrm{\infty }}{lim}\frac{f(2^{n}x)}{2^n}=\mathit{EL}(x)$$

for all $x\in X$.

• (3) $d(f,\mathit{EL})\le \frac{d(f,Jf)}{1-L}$ with $f\in \mathrm{\Omega }$, which implies the inequality

  $$d(f,\mathit{EL})\le \frac{1}{|2|-|2|L}.$$

This implies that the inequality (2.22) holds. The rest of the proof is similar to the proof of Theorem 2.1. □

Corollary 2.2 Let$\theta \ge 0$and r be a real number with$r>1$. Let$f:X\to Y$be a mapping with$f(0)=0$satisfying (2.20). Then, the limit$\mathit{EL}(x)={lim}_{n\to \mathrm{\infty }}\frac{f(2^{n}x)}{2^n}$exists for all$x\in X$and$\mathit{EL}:X\to Y$is a unique cubic mapping such that

$$\begin{array}{rcl}\parallel f(x)-\mathit{EL}(x)\parallel & \le & \frac{1}{|2|-{|2|}^r}max\{max\{\frac{{|2|}^r\theta {\parallel x\parallel }^r({|r_i|}^r+{|r_j|}^r)}{{|4|}^r{|r_i{r}_j|}^r},\frac{\theta {\parallel x\parallel }^r}{{|2|}^r{|r_i|}^r},\frac{\theta {\parallel x\parallel }^r}{{|2|}^r{|r_j|}^r}\},\\ \frac{1}{|2|}max\{\frac{\theta {\parallel x\parallel }^r({|r_i|}^r+{|r_j|}^r)}{{|r_i{r}_j|}^r},\frac{\theta {\parallel x\parallel }^r}{{|r_i|}^r},\frac{\theta {\parallel x\parallel }^r}{{|r_j|}^r}\}\}\le \frac{\theta {\parallel x\parallel }^r({|r_i|}^r+{|r_j|}^r)}{{|r_i{r}_j|}^r({|2|}^{r+1}-{|2|}^{r+2})}\end{array}$$

for all$x\in X$ .

Proof The proof follows from Theorem 2.2 by taking $\phi (x_1,\dots ,x_m)=\theta ({\sum }_{i=1}^m{\parallel x_i\parallel }^r)$ for all $x_1,\dots ,x_m\in X$. In fact, if we choose $L={|2|}^{r-1}$, then we get the desired result. □

3 Non-Archimedean stability of the functional equation (1.1): a direct method

In this section, using a direct method, we prove the generalized Hyers-Ulam stability of the cubic functional equation (1.1) in non-Archimedean normed spaces. Throughout this section, we assume that G is an additive semigroup and X is a non-Archimedean Banach space.

Theorem 3.1 Let $\phi :G^m\to [0,+\mathrm{\infty })$ be a function such that

$$\underset{n\to \mathrm{\infty }}{lim}{|2|}^n\phi (\frac{x_1}{2^n},\dots ,\frac{x_m}{2^n})=0$$

(3.1)

for all $x_1,\dots ,x_m\in G$ and let for each $x\in G$ the limit

$$\begin{array}{rcl}\mathrm{\Theta }(x)& =& \underset{n\to \mathrm{\infty }}{lim}max\{{|2|}^{k}max\{max\{{\phi }_{i,j}(\frac{x}{2^{k+2}{r}_i},-\frac{x}{2^{k+2}{r}_j}),{\phi }_{i,j}(\frac{x}{2^{k+2}{r}_i},0),\\ {\phi }_{i,j}(0,-\frac{x}{2^{k+2}{r}_j})\},\frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{2^{k+1}{r}_i},\frac{x}{2^{k+1}{r}_j}),{\phi }_{i,j}(\frac{x}{2^{k+1}{r}_i},0),\\ {\phi }_{i,j}(0,\frac{x}{2^{k+1}{r}_j})\left\}\right\}|0\le k<n\}\end{array}$$

(3.2)

exist. Suppose that$f:G\to X$is a mapping with$f(0)=0$satisfying the following inequality:

(3.3)

for all$x_1,\dots ,x_m\in X$. Then, the limit$\mathit{EL}(x):={lim}_{n\to \mathrm{\infty }}{2}^{n}f(\frac{x}{2^n})$exists for all$x\in G$and defines an Euler-Lagrange type additive mapping$\mathit{EL}:G\to X$such that

$$\parallel f(x)-\mathit{EL}(x)\parallel \le \mathrm{\Theta }(x).$$

(3.4)

Moreover, if

then EL is the unique mapping satisfying (3.4).

Proof By (2.17), we know

$$\begin{array}{rcl}\parallel f(x)-2f\left(\frac{x}{2}\right)\parallel & \le & max\{max\{{\phi }_{i,j}(\frac{x}{4r_i},-\frac{x}{4r_j}),{\phi }_{i,j}(\frac{x}{4r_i},0),{\phi }_{i,j}(0,-\frac{x}{4r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{2r_i},\frac{x}{2r_j}),{\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,\frac{x}{2r_j})\}\}\end{array}$$

(3.5)

for all $x\in G$. Replacing x by $\frac{x}{2^n}$ in (3.5), we obtain

(3.6)

It follows from (3.1) and (3.6) that the sequence ${\{2^{n}f(\frac{x}{2^n})\}}_{n\ge 1}$ is a Cauchy sequence. Since X is complete, so ${\{2^{n}f(\frac{x}{2^n})\}}_{n\ge 1}$ is convergent. Set

$$\mathit{EL}(x):=\underset{n\to \mathrm{\infty }}{lim}{2}^{n}f\left(\frac{x}{2^n}\right).$$

Using induction on n, one can show that

(3.7)

for all $n\in \mathbb{N}$ and all $x\in G$. By taking n to approach infinity in (3.7) and using (3.2), one obtains (3.4). By (3.1) and (3.3), we get

for all $x_1,\dots ,x_m\in X$. Therefore the function $\mathit{EL}:G\to X$ satisfies (1.1).

To prove the uniqueness property of EL, let $A:G\to X$ be another function satisfying (3.4). Then

for all $x\in G$. Therefore $A=\mathit{EL}$, and the proof is complete. □

Corollary 3.1 Let $\xi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ be a function satisfying

$$\xi \left(\frac{t}{|2|}\right)\le \xi \left(\frac{1}{|2|}\right)\xi (t)(t\ge 0)\xi \left(\frac{1}{|2|}\right)<{|2|}^{-1}.$$

(3.8)

Let$\kappa >0$and$f:G\to X$be a mapping with$f(0)=0$satisfying the following inequality:

(3.9)

for all$x_1,\dots ,x_m\in G$. Then there exists a unique Euler-Lagrange type additive mapping$\mathit{EL}:G\to X$such that

$$\parallel f(x)-\mathit{EL}(x)\parallel \le \frac{\kappa }{|4|}\{\xi \left(\right|\frac{x}{r_i}\left|\right)+\xi \left(\right|\frac{x}{r_j}\left|\right)\}.$$

(3.10)

Proof Defining $\zeta :G^m\to [0,\mathrm{\infty })$ by $\phi (x_1,\dots ,x_m):=\kappa ({\sum }_{k=1}^m\xi (|x_k|))$, then we have

$$\underset{n\to \mathrm{\infty }}{lim}{|2|}^n\phi (\frac{x_1}{2^n},\dots ,\frac{x_m}{2^n})\le \underset{n\to \mathrm{\infty }}{lim}{(|2|\xi \left(\frac{1}{|2|}\right))}^n\phi (x_1,\dots ,x_m)=0$$

for all $x_1,\dots ,x_m\in G$. On the other hand,

$$\begin{array}{rcl}\mathrm{\Theta }(x)& =& \underset{n\to \mathrm{\infty }}{lim}max\{{|2|}^{k}max\{max\{{\phi }_{i,j}(\frac{x}{2^{k+2}{r}_i},-\frac{x}{2^{k+2}{r}_j}),{\phi }_{i,j}(\frac{x}{2^{k+2}{r}_i},0),\\ {\phi }_{i,j}(0,-\frac{x}{2^{k+2}{r}_j})\},\frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{2^{k+1}{r}_i},\frac{x}{2^{n+1}{r}_j}),{\phi }_{i,j}(\frac{x}{2^{k+1}{r}_i},0),\\ {\phi }_{i,j}(0,\frac{x}{2^{k+1}{r}_j})\left\}\right\}|0\le k<n\}\\ =& max\{max\{{\phi }_{i,j}(\frac{x}{4r_i},-\frac{x}{4r_j}),{\phi }_{i,j}(\frac{x}{4r_i},0),{\phi }_{i,j}(0,-\frac{x}{4r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{2r_i},\frac{x}{2r_j}),{\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,\frac{x}{2r_j})\}\}\\ =& \frac{\kappa }{|4|}\{\xi \left(\right|\frac{x}{r_i}\left|\right)+\xi \left(\right|\frac{x}{r_j}\left|\right)\}\end{array}$$

for all $x\in G$, exists. Also

Applying Theorem 3.1, we get the desired result. □

Theorem 3.2 Let $\phi :G^m\to [0,+\mathrm{\infty })$ be a function such that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\phi (2^n{x}_1,\dots ,2^n{x}_m)}{{|2|}^n}=0$$

(3.11)

for all $x_1,\dots ,x_m\in G$ and let for each $x\in G$ the limit

$$\begin{array}{rcl}\mathrm{\Theta }(x)& =& \underset{n\to \mathrm{\infty }}{lim}max\{\frac{1}{{|2|}^k}max\{max\{{\phi }_{i,j}(\frac{2^{k-1}x}{r_i},-\frac{2^{k-1}x}{r_j}),\\ {\phi }_{i,j}(\frac{2^{k-1}x}{r_i},0),{\phi }_{i,j}(0,-\frac{2^{k-1}x}{r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{2^{k}x}{r_i},\frac{2^{k}x}{r_j}),{\phi }_{i,j}(\frac{2^{k}x}{r_i},0),{\phi }_{i,j}(0,\frac{2^{k}x}{r_j})\}\left\}\right|0\le k<n\}\end{array}$$

(3.12)

exist. Suppose that$f:G\to X$is a mapping with$f(0)=0$satisfying (3.3). Then, the limit$\mathit{EL}(x):={lim}_{n\to \mathrm{\infty }}\frac{f(2^{n}x)}{2^n}$exists for all$x\in G$and defines an Euler-Lagrange type additive mapping$\mathit{EL}:G\to X$, such that

$$\parallel f(x)-\mathit{EL}(x)\parallel \le \frac{1}{|2|}\mathrm{\Theta }(x).$$

(3.13)

Moreover, if

then EL is the unique Euler-Lagrange type additive mapping satisfying (3.13).

Proof It follows from (2.16) that

(3.14)

for all $x\in G$. Replacing x by $2^{n}x$ in (3.14), we obtain

(3.15)

It follows from (3.11) and (3.15) that the sequence ${\{\frac{f(2^{n}x)}{2^n}\}}_{n\ge 1}$ is convergent. Set

$$\mathit{EL}(x):=\underset{n\to \mathrm{\infty }}{lim}\frac{f(2^{n}x)}{2^n}.$$

On the other hand, it follows from (3.15) that

for all $x\in G$ and all nonnegative integers p, q with $q>p\ge 0$. Letting $p=0$ and passing the limit $q\to \mathrm{\infty }$ in the last inequality and using (3.12), we obtain (3.13). The rest of the proof is similar to the proof of Theorem 3.1. □

Corollary 3.2 Let $\xi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ be a function satisfying

$$\xi (|2t|)\le \xi (|2|)\xi (t)(t\ge 0)\xi (|2|)<|2|.$$

(3.16)

Let$\kappa >0$and$f:G\to X$be a mapping with$f(0)=0$satisfying the following inequality (3.9). Then there exists a unique Euler-Lagrange type additive mapping$\mathit{EL}:G\to X$such that

$$\begin{array}{rcl}\parallel f(x)-\mathit{EL}(x)\parallel & \le & \frac{\kappa }{|2|}max\{\xi \left(\right|\frac{x}{2r_i}\left|\right)+\xi \left(\right|\frac{x}{2r_j}\left|\right),\frac{1}{|2|}\xi \left(\right|\frac{x}{r_i}\left|\right)+\xi \left(\right|\frac{x}{r_j}\left|\right)\}\\ =& \frac{\kappa }{|2|}[\xi \left(\right|\frac{x}{2r_i}\left|\right)+\xi \left(\right|\frac{x}{2r_j}\left|\right)].\end{array}$$

(3.17)

Proof Defining $\zeta :G^m\to [0,\mathrm{\infty })$ by $\phi (x_1,\dots ,x_m):=\kappa ({\sum }_{k=1}^m\xi (|x_k|))$, then, we have

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\phi (2^n{x}_1,\dots ,2^n{x}_m)}{{|2|}^n}\le \underset{n\to \mathrm{\infty }}{lim}{\left(\frac{\xi (|2|)}{|2|}\right)}^n\phi (x_1,\dots ,x_m)=0$$

for all $x_1,\dots ,x_m\in G$. On the other hand,

$$\begin{array}{rcl}\mathrm{\Theta }(x)& =& \underset{n\to \mathrm{\infty }}{lim}max\{\frac{1}{{|2|}^k}max\{max\{{\phi }_{i,j}(\frac{2^{k-1}x}{r_i},-\frac{2^{k-1}x}{r_j}),\\ {\phi }_{i,j}(\frac{2^{k-1}x}{r_i},0),{\phi }_{i,j}(0,-\frac{2^{k-1}x}{r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{2^{k}x}{r_i},\frac{2^{k}x}{r_j}),{\phi }_{i,j}(\frac{2^{k}x}{r_i},0),{\phi }_{i,j}(0,\frac{2^{k}x}{r_j})\}\left\}\right|0\le k<n\}\\ =& max\{max\{{\phi }_{i,j}(\frac{x}{2r_i},-\frac{x}{2r_j}),{\phi }_{i,j}(\frac{x}{2r_i},0),{\phi }_{i,j}(0,-\frac{x}{2r_j})\},\\ \frac{1}{|2|}max\{{\phi }_{i,j}(\frac{x}{r_i},\frac{x}{r_j}),{\phi }_{i,j}(\frac{x}{r_i},0),{\phi }_{i,j}(0,\frac{x}{r_j})\}\}\end{array}$$

for all $x\in G$, exists. Also

Applying Theorem 3.14, we get the desired result. □

Remark 3.1 We remark that if $\xi (|2|)=0$, then $\xi =0$ identically, and so f is itself additive. Thus, for the nontrivial ξ, we observe that $\xi (|2|)\ne 0$ and

$$1\le \xi (|1|)\le \xi (|2|)\xi \left(\frac{1}{|2|}\right)\le |2|\xi \left(\frac{1}{|2|}\right)$$

implies that $\frac{1}{|2|}\le \xi (\frac{1}{|2|})$.