Asymptotic Stability of the Pexider–Cauchy Functional Equation in Non-Archimedean Spaces

Abstract

In this paper, we investigated the asymptotic stability behaviour of the Pexider–Cauchy functional equation in non-Archimedean spaces. We also showed that, under some conditions, if $\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel ⩽\epsilon$, then $f,g$ and h can be approximated by additive mapping in non-Archimedean normed spaces. Finally, we deal with a functional inequality and its asymptotic behaviour.

1. Introduction

The classical problem of the stability of homomorphisms was first posed by Ulam [1] in 1940 as follows: Given a group $(G_1,\ast )$, a metric group $(G_2,\diamond ,d)$ and $\epsilon >0$. Does $\delta >0$ exist such that if the function $f:G_1\to G_2$ satisfies the inequality:

$$d(f(x\ast y),f\left(x\right)\diamond f\left(y\right))<\delta ,x,y\in G_1,$$

then there exists a homomorphism $T:G_1\to G_2$ such that $d\left(f\right(x),T(x\left)\right)<\epsilon$ for all $x\in G_1$? In 1941, Donald H. Hyers [2] gave a partial affirmative answer to this question under the assumption that $f:X\to Y$ is a function between Banach spaces X and Y. Indeed, D. H. Hyers’ classical theorem states that, if X is a linear space and Y is a Banach space and, for some $\epsilon ⩾0$, a function $f:X\to Y$ is $\epsilon$-additive, meaning that:

$$\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel ⩽\epsilon ,x,y\in X$$

(1)

then there exists a unique additive function $A:X\to Y$ such that f is $\epsilon$-close to A, i.e., $\parallel f\left(x\right)-A\left(x\right)\parallel ⩽\epsilon$ for all $x\in X$. Ulam’s question can be expressed in a more general way as follows: when is it true that a function which approximately satisfies a functional equation $(†)$ must be close to an exact solution of $(†)$?

Aoki [3] and Th. M. Rassias [4] extended the Hyers’ theorem by considering an unbounded Cauchy difference. They tried to weaken the condition for the bound of the norm of the Cauchy difference in (1) as follows:

$$\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel ⩽\epsilon (\parallel x{\parallel }^p+\parallel y{\parallel }^p),x,y\in X$$

where $0⩽p<1$. In 1994, Găvruta [5] provided a generalization of Rassias’ theorem by replacing the bound $\epsilon (\parallel x{\parallel }^p+\parallel y{\parallel }^p)$ by a general control function $\phi (x,y)$. In recent decades, several stability problems for various functional equations and also for mappings with more general domains and ranges have been investigated by a number of mathematicians. We refer the interested reader the following books and surveys [6,7,8,9,10] and the references therein for more detailed information.

Let us recall (see, for instance, [11]) some basic definitions and facts concerning non-Archimedean normed spaces. By a non-Archimedean field, we mean a field $\mathbb{F}$ equipped with a function (valuation) $|.|$ from $\mathbb{F}$ into $[0,+\infty )$ such that:

$\left(i\right)$

$\left|r\right|=0$ if and only if $r=0$;

$\left(ii\right)$

$\left|rs\right|=\left|r\right|\left|s\right|,r,s\in \mathbb{F}$;

$\left(iii\right)$

$|r+s|⩽max\left\{\right|r|,|s\left|\right\},r,s\in \mathbb{F}$.

Clearly $\left|1\right|=|-1|=1$ and $\left|n\right|⩽1$ for all integers n. An obvious example is the trivial valuation that can be defined on any field $\mathbb{F}$ by setting:

$$\left|x\right|:=\left\{\begin{array}{cc}0,\hfill & \mathrm{if}x=0;\hfill \\ 1,\hfill & \mathrm{if}x\ne 0.\hfill \end{array}\right.$$

Example 1.

For a given prime number p, any non-zero rational number x can be uniquely written as $x=\frac{a}{b}{p}^{n_x}$, where $a,b,n_x\in \mathbb{Z}$ and $a,b$ are integers not divisible by p. Then, the function ${|.|}_p:\mathbb{Q}\to [0,+\infty )$ defined by

$$\left|x\right|:=\left\{\begin{array}{cc}0,\hfill & \mathrm{if}x=0;\hfill \\ p^{-n_x},\hfill & \mathrm{if}x\ne 0.\hfill \end{array}\right.$$

is a non-Archimedean (nontrivial) valuation on ℚ.

Let X be a vector space over a non-Archimedean field $\mathbb{F}$ with a non-trivial valuation $|.|$. A function $\parallel .\parallel :X\to [0,+\infty )$ is called a non-Archimedean norm (valuation) if it satisfies the following conditions:

$\left(i\right)$

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

$\left(ii\right)$

$\parallel rx\parallel =\left|r\right|\parallel x\parallel$ for all $r\in \mathbb{F}$ and all $x\in X$;

$\left(iii\right)$

the strong triangle inequality; namely,

$$\parallel x+y\parallel ⩽max\{\parallel x\parallel ,\parallel y\parallel \},x,y\in X.$$

Then, $(X,\parallel .\parallel )$ is called a non-Archimedean normed space. A sequence ${\left\{x_n\right\}}_{n=1}^{\infty }$ in X is called a Cauchy sequence if ${lim}_{m,n\to \infty }\parallel x_n-x_m\parallel =0$. A sequence ${\left\{x_n\right\}}_{n=1}^{\infty }$ in X is said to converge to $x\in X$ if ${lim}_{n\to \infty }\parallel x_n-x\parallel =0$. By the strong triangle inequality $\left(iii\right)$, we have

$$\parallel x_n-x_m\parallel ⩽max\left\{\parallel x_{j+1}-x_j\parallel :m⩽j⩽n-1\right\},n>m.$$

Then, the sequence ${\left\{x_n\right\}}_{n=1}^{\infty }$ in a non-Archimedean space is Cauchy if and only if ${\{x_{n+1}-x_n\}}_{n=1}^{\infty }$ converges to zero. By a Banach non-Archimedean space, we mean that every Cauchy sequence is convergent. It is necessary to remember that ${|.|}_p$ is a non-Archimedean norm on ℚ, and that the completion of ℚ with respect to ${|.|}_p$ is denoted by ${\mathbb{Q}}_p$ and is called the p-adic number field. Indeed, p-adic numbers do not satisfy the Archimedean axiom which states: for each $x,y>0$, there exists an integer n such that $nx>y$.

In [12], the authors showed that if $f:{\mathbb{Q}}_p\to \mathbb{R}$ is a continuous $\epsilon$-additive function, then there exists a unique additive function $A:X\to Y$ such that f is $\epsilon$-close to A. In [13,14], the authors investigated the stability of some functional equations in non-Archimedean normed spaces. In this paper, we investigated the asymptotic stability behaviour of the Pexider–Cauchy functional equation in non-Archimedean spaces. We will use the methods and ideas presented in [15,16,17].

2. Main Results

Throughout this section, we assume that X and Y are the non-Archimedean normed spaces over a non-Archimedean field $\mathbb{F}$ with a valuation $|.|$.

Theorem 1.

Let $\epsilon ⩾0$ and $f,g,h:X\to Y$ satisfy the following asymptotic stability property:

$$\underset{min\{\parallel x\parallel ,\parallel y\parallel \}\to \infty }{lim\; sup}\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel ⩽\epsilon .$$

(2)

If $\left|n\right|<1$ for some integer n, then:

$$\parallel f(x+y)-f\left(x\right)-f\left(y\right)+f\left(0\right)\parallel ⩽\epsilon ,x,y\in X.$$

(3)

Proof. 

Let $\eta >\epsilon$ be an arbitrary real number. Then, by (2) there exists $d>0$ such that:

$$\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel <\eta ,\parallel x\parallel ,\parallel y\parallel ⩾d.$$

(4)

Let $x,y\in X$ be arbitrary elements. Since $\left|n\right|<1$ for some integer n, we obtain:

$$\underset{m\to +\infty }{lim}∥\frac{1}{n^m}z∥=+\infty$$

for some $z\in X \backslash \left\{0\right\}$. Hence we can choose $u\in X$ such that $\parallel u\parallel ⩾d+\parallel x\parallel +\parallel y\parallel$. It is easy to see that:

$$min\left\{\parallel x-u\parallel ,\parallel y+u\parallel ,\parallel u\parallel \right\}⩾d.$$

Applying (4) and the strong triangle inequality, we obtain:

$$\begin{array}{cc}\hfill \parallel f(x+y)-f\left(x\right)-f\left(y\right)+f\left(0\right)\parallel ⩽max\{& \parallel g(x-u)+h\left(u\right)-f\left(x\right)\parallel ,\hfill \\ \hfill & \parallel g(-u)+h(y+u)-f\left(y\right)\parallel ,\hfill \\ \hfill & \parallel f(x+y)-g(x-u)-h(y+u)\parallel ,\hfill \\ \hfill & \parallel f\left(0\right)-g(-u)-h\left(u\right)\parallel \}<\eta \hfill \end{array}$$

for all $x,y\in X.$ Taking the limit as $\eta \to \epsilon$ in the last inequality, we obtain (3): □

Corollary 1.

Let $\epsilon ⩾0$ and $f:X\to Y$ satisfy the following asymptotic stability property:

$$\underset{min\{\parallel x\parallel ,\parallel y\parallel \}\to \infty }{lim\; sup}\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel ⩽\epsilon .$$

(5)

If $\left|n\right|<1$ for some integer n, then:

$$\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel ⩽\epsilon ,x,y\in X.$$

(6)

Proof. 

Let $\eta >\epsilon$ be an arbitrary real number. By (5), there exists: $d>0$ such that

$$\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel <\eta ,\parallel x\parallel ,\parallel y\parallel ⩾d.$$

(7)

By the proof of Theorem 1, we obtain:

$$\parallel f(x+y)-f\left(x\right)-f\left(y\right)+f\left(0\right)\parallel <\eta ,x,y\in X.$$

(8)

By using the strong triangle inequality, it follows from (7) and (8) that $\parallel f\left(0\right)\parallel <\eta$. Then:

$$\begin{array}{c}\hfill \parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel ⩽max\left\{\parallel f(x+y)-f\left(x\right)-f\left(y\right)+f\left(0\right)\parallel ,\parallel f\left(0\right)\parallel \right\}<\eta \end{array}$$

for all $x,y\in X$. Taking the limit as $\eta \to \epsilon$ in the last inequality, we obtain (6): □

Corollary 2.

Let $\left|n\right|<1$ for some integer n, and $f,g,h:X\to Y$ fulfil the following asymptotic property:

$$\underset{min\{\parallel x\parallel ,\parallel y\parallel \}\to \infty }{lim}\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel =0.$$

Then, f is affine. Moreover, if $f=g=h$, then f is additive.

The following example shows that with the conditions of Corollary 2, functions g and h may not be affine even if $g=h$.

Example 2.

Let ℚ be the field of rational numbers and let ${|.|}_p$ be the p-adic absolute value, where $p>2$ is a prime number. We define $f,g,h:\mathbb{Q}\to \mathbb{Q}$ by

$$f\left(x\right)=x\mathrm{and}g\left(x\right)=h\left(x\right)=x+\frac{1}{1+x^2},x\in \mathbb{Q}.$$

Let $n>1$ be an integer. For ${\left|x\right|}_p,{\left|y\right|}_p⩾p^n$, we have:

$$\begin{array}{cc}\hfill {|f(x+y)-g\left(x\right)-h\left(y\right)|}_p& ={\left|\frac{1}{1+x^2}+\frac{1}{1+y^2}\right|}_p\hfill \\ \hfill & ⩽max\left\{\frac{1}{|1+x^2{|}_p},\frac{1}{|1+y^2{|}_p}\right\}\hfill \\ \hfill & ⩽\frac{1}{p^{2n}-1}⟶0,\mathrm{as}n\to \infty .\hfill \end{array}$$

Then:

$$\underset{min\left\{\right|x{|}_p,\left|y{|}_p\right\}\to \infty }{lim}{|f(x+y)-g\left(x\right)-h\left(y\right)|}_p=0,$$

but g and h are not affine.

Example 3.

Let $p>2$ be a prime number. Suppose that a function $f:\mathbb{Q}\to \mathbb{Q}$ satisfies

$$f\left(1\right)=1\mathrm{and}\underset{min\left\{\right|x{|}_p,\left|y{|}_p\right\}\to \infty }{lim}{|f(x+y)-f\left(x\right)-f\left(y\right)|}_p=0.$$

Then f is the identity, i.e., $f\left(x\right)=x$ for all $x\in X$.

Theorem 2.

Let $\epsilon ⩾0$ and Y be a Banach non-Archimedean space. Suppose that $f,g,h:X\to Y$ are functions that satisfy:

$$\underset{min\{\parallel x\parallel ,\parallel y\parallel \}\to \infty }{lim\; sup}\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel ⩽\epsilon .$$

(9)

If $\left|n\right|<1$ for some integer n, then there exists a unique additive mapping $A:X\to Y$ such that:

$$\parallel f\left(x\right)-f\left(0\right)-A\left(x\right)\parallel ⩽\epsilon ,x\in X.$$

(10)

Proof. 

It follows by Theorem 1 that:

$$\parallel f(x+y)-f\left(x\right)-f\left(y\right)+f\left(0\right)\parallel ⩽\epsilon ,x,y\in X.$$

Defining $F:X\to Y$ by $F\left(x\right):=f\left(x\right)-f\left(0\right)$, we obtain:

$$\parallel F(x+y)-F\left(x\right)-F\left(y\right)\parallel ⩽\epsilon ,x,y\in X.$$

(11)

By ([13] [Theorem 2.1]), there is a unique additive mapping $A:X\to Y$ such that $\parallel F\left(x\right)-A\left(x\right)\parallel ⩽\epsilon$ for all $x\in X$ (see also [14,18]). This completes the proof. □

By Corollary 1 and using the argument in Theorem 2, we obtain the following result.

Corollary 3.

Let $\epsilon ⩾0$ and Y be a Banach non-Archimedean space. Let $f:X\to Y$ be functions satisfying:

$$\underset{min\{\parallel x\parallel ,\parallel y\parallel \}\to \infty }{lim\; sup}\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel ⩽\epsilon .$$

(12)

If $\left|n\right|<1$ for some integer n, then there exists a unique additive mapping $A:X\to Y$ such that:

$$\parallel f\left(x\right)-A\left(x\right)\parallel ⩽\epsilon ,x\in X.$$

(13)

Theorem 3.

Let $\epsilon ⩾0$ and $f,g,h:X\to Y$ satisfy the following asymptotic stability property:

$$\underset{\parallel x\parallel +\parallel y\parallel \to \infty }{lim\; sup}\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel ⩽\epsilon .$$

(14)

If $\left|n\right|<1$ for some integer n, then:

$$\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel ⩽\epsilon ,x,y\in X.$$

(15)

Proof. 

Let $\eta >\epsilon$ be an arbitrary real number. Then, by (14) there exists: $d>0$ such that

$$\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel <\eta ,\parallel x\parallel +\parallel y\parallel ⩾d.$$

(16)

Let $x,y\in X$ be arbitrary elements. Since $\left|n\right|<1$ for some integer n, we can choose $u\in X$ such that $\parallel 2u\parallel ⩾d+\parallel x\parallel +\parallel y\parallel$. It is easy to verify that:

$$\begin{array}{cc}\hfill & \parallel x-u\parallel +\parallel y+u\parallel ⩾d;\parallel 2u\parallel +\parallel x-u\parallel ⩾d;\parallel y\parallel +\parallel 2u\parallel ⩾d;\hfill \\ \hfill & \parallel y+u\parallel +\parallel u\parallel ⩾d;\parallel x\parallel +\parallel u\parallel ⩾d.\hfill \end{array}$$

(17)

Then, (16) and (17) yield:

$$\begin{array}{cc}\hfill \parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel ⩽max\{& \parallel f(x+y)-g(y+u)-h(x-u)\parallel ,\hfill \\ \hfill & \parallel f(x+u)-g\left(2u\right)-h(x-u)\parallel ,\hfill \\ \hfill & \parallel f(y+2u)-g\left(2u\right)-h\left(y\right)\parallel ,\hfill \\ \hfill & \parallel f(y+2u)-g(y+u)-h\left(u\right)\parallel ,\hfill \\ \hfill & \parallel f(x+u)-g\left(x\right)-h\left(u\right)\parallel \}<\eta \hfill \end{array}$$

for all $x,y\in X$. Taking the limit as $\eta \to \epsilon$ in the last inequality, we obtain (15). □

Corollary 4.

Let $\epsilon ⩾0$ and $f,g,h:X\to Y$ satisfy the following asymptotic property:

$$\underset{\parallel x\parallel +\parallel y\parallel \to \infty }{lim}\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel =0$$

(18)

If $\left|n\right|<1$ for some integer n, then $f,g$ and h are affine. Moreover, if $g\left(0\right)=h\left(0\right)=0$, then $f=g=h$ are additive.

Theorem 4.

Let $\epsilon ⩾0$ and $f,g,h:X\to Y$ be functions satisfying (14). If $\left|n\right|<1$ for some integer n, then there exists a unique additive mapping $A:X\to Y$ such that:

$$\parallel f\left(x\right)-f\left(0\right)-A\left(x\right)\parallel ⩽\epsilon ,\parallel g\left(x\right)-g\left(0\right)-A\left(x\right)\parallel ⩽\epsilon ,\parallel h\left(x\right)-h\left(0\right)-A\left(x\right)\parallel ⩽\epsilon$$

for all $x\in X$.

Proof. 

By Theorem 3, we have $\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel ⩽\epsilon$ for all $x,y\in X$. Then, the result easily follows from ([13] [Theorem 2.4]) (see also [18]). □

3. Results on the Hyperstability

Theorem 5.

Let $\phi :[0,+\infty )\to \mathbb{R}$ be a function such that ${lim}_{t\to +\infty }\phi \left(t\right)=+\infty$ and let $f,g,h:X\to Y$ satisfy:

$$\underset{\parallel x\parallel +\parallel y\parallel \to \infty }{lim\; sup}\phi (\parallel x-y\parallel )\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel <\infty .$$

(19)

If $\left|n\right|<1$ for some integer n, then $f(x+y)=g\left(x\right)+h\left(y\right)$ for all $x,y\in X$, and consequently $f,g$ and h are affine. Furthermore, $f-f\left(0\right)=g-g\left(0\right)=h-h\left(0\right)$.

Proof. 

According to (19), there exist constants $d>0$ and $K>0$ such that:

$$\phi (\parallel x-y\parallel )\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel <K,\parallel x\parallel +\parallel y\parallel ⩾d.$$

(20)

Let $\epsilon >0$ be arbitrary. We can choose $N>0$ such that $\phi \left(t\right)⩾\frac{K}{\epsilon }$ for all $t⩾N$. Then, (20) implies that:

$$\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel <\epsilon ,\parallel x\parallel +\parallel y\parallel ⩾d\mathrm{and}\parallel x-y\parallel ⩾N.$$

(21)

Let $x,y\in X$ with $\parallel x\parallel +\parallel y\parallel ⩾2d+2N$. Since $\left|n\right|<1$ for some integer n, we can choose $u\in X$ such that:

$$min\{\parallel 2u\parallel ,\parallel 3u\parallel \}⩾d+N+\parallel x\parallel +\parallel y\parallel +\parallel x-y\parallel +\parallel x-2y\parallel .$$

It is easy to verify that:

$$\begin{array}{cc}\hfill & min\{\parallel y\pm u\parallel ,\parallel x\pm u\parallel ,\parallel x-y+2u\parallel ,\parallel u\parallel \}⩾d,\hfill \\ \hfill & min\{\parallel y-x\pm 2u\parallel ,\parallel 3u-x\parallel ,\parallel 2u-y\parallel ,\parallel x-u\parallel ,\parallel x-2y+2u\parallel ,\parallel x-2y+3u\parallel \}⩾N.\hfill \end{array}$$

Since $\parallel x\parallel +\parallel y\parallel ⩾2d+2N$, two cases arise according to whether $\parallel x\parallel ⩾d+N$ or $\parallel y\parallel ⩾d+N$. First, take the case $\parallel x\parallel ⩾d+N$. Then, applying (21), we obtain:

$$\begin{array}{cc}\hfill \parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel ⩽max\{& \parallel f(x+y)-g(x+u)-h(y-u)\parallel ,\hfill \\ \hfill & \parallel f(x+2u)-g(x+u)-h\left(u\right)\parallel ,\hfill \\ \hfill & \parallel f(x+u)-g(x-y+2u)-h(y-u)\parallel ,\hfill \\ \hfill & \parallel f(x+u)-g\left(x\right)-h\left(u\right)\parallel ,\hfill \\ \hfill & \parallel f(x+2u)-g(x-y+2u)-h\left(y\right)\parallel \}<\epsilon .\hfill \end{array}$$

Now, consider the case $\parallel y\parallel ⩾d+N$. Then, applying (21) yields:

$$\begin{array}{cc}\hfill \parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel ⩽max\{& \parallel f(x+y)-g(y+u)-h(x-u)\parallel ,\hfill \\ \hfill & \parallel f(x+u)-g\left(2u\right)-h(x-u)\parallel ,\hfill \\ \hfill & \parallel f(y+2u)-g\left(2u\right)-h\left(y\right)\parallel ,\hfill \\ \hfill & \parallel f(y+2u)-g(y+u)-h\left(u\right)\parallel ,\hfill \\ \hfill & \parallel f(x+u)-g\left(x\right)-h\left(u\right)\parallel \}<\epsilon .\hfill \end{array}$$

Therefore, $\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel ⩽\epsilon$ for $\parallel x\parallel +\parallel y\parallel ⩾2d+2N$. Then, by Theorem 3, we have $\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel ⩽\epsilon$ for all $x,y\in X$. Since $\epsilon >0$ was arbitrary, this inequality implies that $f(x+y)=g\left(x\right)+h\left(y\right)$ for all $x,y\in X$. □

Remark 1.

Using the argument presented above, it can be concluded that Theorem 5 still holds if we have:

$$\underset{\parallel x\parallel +\parallel y\parallel \to \infty }{lim\; sup}\phi \left(max\{\parallel x\parallel ,\parallel y\parallel \}\right)\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel <\infty$$

instead of (20).

Theorem 6.

Let $\phi :[0,+\infty )\to \mathbb{R}$ be a function such that ${lim}_{t\to +\infty }\phi \left(t\right)=+\infty$ and let $f,g,h:X\to Y$ satisfy:

$$\underset{min\{\parallel x\parallel ,\parallel y\parallel \}\to \infty }{lim\; sup}\phi (\parallel x-y\parallel )\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel <\infty .$$

(22)

If $\left|n\right|<1$ for some non-zero integer n, then f is affine.

Proof. 

According to (22), there exist constants $d>0$ and $K>0$ such that:

$$\phi (\parallel x-y\parallel )\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel <K,\parallel x\parallel ,\parallel y\parallel ⩾d.$$

(23)

Let $\epsilon >0$ be arbitrary. We can choose $N>0$ such that $\phi \left(t\right)⩾\frac{K}{\epsilon }$ for all $t⩾N$. Then, (23) implies that

$$\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel <\epsilon ,\parallel x\parallel ,\parallel y\parallel ⩾d\mathrm{and}\parallel x-y\parallel ⩾N.$$

(24)

Let $x,y\in X$ be arbitrary elements. Since $0<\left|n\right|<1$ for some integer n, we can choose $u\in X$ such that $\parallel 2u\parallel ⩾d+N+\parallel x\parallel +\parallel y\parallel +\parallel x-y\parallel$. It is easy to see that:

$$\begin{array}{cc}\hfill & min\left\{\parallel x-u\parallel ,\parallel y+u\parallel ,\parallel x-2u\parallel ,\parallel y+2u\parallel ,\parallel x-y-2u\parallel ,\parallel 2u\parallel ,\parallel u\parallel \right\}⩾d,\hfill \\ \hfill & min\left\{\parallel x-2u\parallel ,\parallel y+2u\parallel ,\parallel x-y-2u\parallel ,\parallel 2u\parallel \right\}⩾N.\hfill \end{array}$$

Applying (24) and the strong triangle inequality, we obtain:

$$\begin{array}{cc}\hfill \parallel f(x+y)-f\left(x\right)-f\left(y\right)+f\left(0\right)\parallel ⩽max\{& \parallel g(x-u)+h\left(u\right)-f\left(x\right)\parallel ,\hfill \\ \hfill & \parallel g(-u)+h(y+u)-f\left(y\right)\parallel ,\hfill \\ \hfill & \parallel f(x+y)-g(x-u)-h(y+u)\parallel ,\hfill \\ \hfill & \parallel f\left(0\right)-g(-u)-h\left(u\right)\parallel \}<\epsilon \hfill \end{array}$$

for all $x,y\in X$. Since $\epsilon >0$ was arbitrary, this inequality implies that $f(x+y)=f\left(x\right)+f\left(y\right)-f\left(0\right)$ for all $x,y\in X$. Then, $f-f\left(0\right)$ is additive, and so f is affine. □

Remark 2.

Based on the argument presented above, it can be concluded that the assertion of Theorem 6 is still true by considering:

$$\underset{min\{\parallel x\parallel ,\parallel y\parallel \}\to \infty }{lim\; sup}\phi \left(max\{\parallel x\parallel ,\parallel y\parallel \}\right)\parallel f(x+y)-g\left(x\right)-h\left(y\right)\parallel <\infty$$

instead of (22).

4. Functional Inequalities

We assume that X and Y are non-Archimedean normed spaces over a non-Archimedean field $\mathbb{F}$ with a valuation $|.|$.

Lemma 1.

Let $0<\left|k\right|<1$ for some integer k, and $f:X\to Y$ with $f\left(0\right)=0$ satisfy:

$$\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel ⩽\parallel f(x-y)-f\left(x\right)+f\left(y\right)\parallel ,x,y\in X.$$

(25)

Then, f is rational homogeneous, i.e., $f\left(rx\right)=rf\left(x\right)$ for all $x\in X$ and $r\in Q$.

Proof. 

Letting $y=x$ in (25) and applying $f\left(0\right)=0$, we obtain $f\left(2x\right)=2f\left(x\right)$ for all $x\in X.$ Now, we will prove by induction that $f\left(nx\right)=nf\left(x\right)$ for all $x\in X$ and $n\in N$. Suppose this result holds for all $1⩽k<n$. Then, by letting $x=(n-1)y$ in (25), we obtain:

$$\parallel f\left(ny\right)-(n-1)f\left(y\right)-f\left(y\right)\parallel ⩽\parallel (n-2)f\left(y\right)-(n-1)f\left(y\right)+f\left(y\right)\parallel =0,y\in X.$$

Hence, $f\left(ny\right)=nf\left(y\right)$ for all $y\in X$. Putting $y=-k^{2}x$ in (25), we obtain:

$$\parallel f(-x)+f\left(x\right)\parallel ⩽{\left|k\right|}^2\parallel f\left(x\right)+f(-x)\parallel ,x\in X.$$

Since $0<\left|k\right|<1$, we infer that $f(-x)=-f\left(x\right)$ for all $x\in X.$ Therefore, $f\left(nx\right)=nf\left(x\right)$ for all $x\in X$ and $n\in Z$. Let $x\in X$ and $r=\frac{m}{n}$ be a rational number. Then, $nf\left(rx\right)=f\left(nrx\right)=f\left(mx\right)=mf\left(x\right)$ for all $x\in X$. This implies that $f\left(rx\right)=rf\left(x\right)$, and the proof is completed. □

The following example shows that the assumption $f\left(0\right)=0$ is necessary in Lemma 1.

Example 4.

Let ${\mathbb{Q}}_p$ be equipped with the p-adic norm ${|.|}_p$, and $f:{\mathbb{Q}}_p\to {\mathbb{Q}}_p$ be defined by $f\left(x\right)=x+3$ for all $x\in {\mathbb{Q}}_p$. Then, f satisfies (25), but f is not rational homogeneous.

Theorem 7.

Let $\left|2\right|<1,\epsilon ⩾0$ and Y be a Banach non-Archimedean space. Suppose that the function $f:X\to Y$ satisfies:

$$\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel ⩽\parallel f(x-y)-f\left(x\right)+f\left(y\right)\parallel +\epsilon ,\parallel x\parallel ,\parallel y\parallel ⩾d$$

(26)

for some $d>0$. Then, there exists a unique rational homogeneous mapping $A:X\to Y$ such that:

$$\parallel f\left(x\right)-A\left(x\right)\parallel ⩽\parallel f\left(0\right)\parallel +\epsilon ,\parallel x\parallel ⩾d.$$

(27)

Proof. 

Letting $y=x$ in (26), we obtain:

$$\parallel f\left(2x\right)-2f\left(x\right)\parallel ⩽\parallel f\left(0\right)\parallel +\epsilon ,\parallel x\parallel ⩾d.$$

Replacing x by $\frac{x}{2^{n+1}}$ in the above inequality and multiplying the resulting inequality by ${\left|2\right|}^n$ yields:

$$∥2^{n}f\left(\frac{x}{2^n}\right)-2^{n+1}f\left(\frac{x}{2^{n+1}}\right)∥⩽{\left|2\right|}^n(\parallel f\left(0\right)\parallel +\epsilon ),\parallel x\parallel ⩾d,n⩾0.$$

(28)

Then:

$$\begin{array}{cc}\hfill ∥2^{n}f\left(\frac{x}{2^n}\right)-2^{m}f\left(\frac{x}{2^m}\right)∥& =∥\sum _{k=m}^{n-1}\left[2^{k+1}f\left(\frac{x}{2^{k+1}}\right)-2^{k}f\left(\frac{x}{2^k}\right)\right]∥\hfill \\ \hfill & ⩽max\left\{∥2^{k+1}f\left(\frac{x}{2^{k+1}}\right)-2^{k}f\left(\frac{x}{2^k}\right)∥:m⩽k⩽n-1\right\}\hfill \\ \hfill & ⩽{\left|2\right|}^m(\parallel f\left(0\right)\parallel +\epsilon ),\hfill \end{array}$$

(29)

for all $\parallel x\parallel ⩾d$ and $n>m⩾0$. It follows from (28) that the sequence ${\left\{2^{n}f\left(\frac{x}{2^n}\right)\right\}}_{n=1}^{\infty }$ is a Cauchy sequence for all $x\in X$. Since Y is a Banach non-Archimedean space, the sequence ${\left\{2^{n}f\left(\frac{x}{2^n}\right)\right\}}_{n=1}^{\infty }$ is convergent. We define:

$$A:X\to Y,A\left(x\right):=\underset{n\to \infty }{lim}{2}^{n}f\left(\frac{x}{2^n}\right).$$

Since $\left|2\right|<1$, we obtain $A\left(0\right)=0$. By the definition of A, (26) yields:

$$\parallel A(x+y)-A\left(x\right)-A\left(y\right)\parallel ⩽\parallel A(x-y)-A\left(x\right)+A\left(y\right)\parallel ,x,y\in X.$$

By Lemma 1, A is rational homogeneous. Taking $m=0$ and letting $n\to \infty$ in (29), we obtain (27) for all $\parallel x\parallel ⩾d$. The uniqueness of A easily follows from (27). □

Lemma 2.

Let $0<\left|k\right|<1$ and $f:X\to Y$ with $f\left(0\right)=0$ satisfy:

$$∥f\left(\frac{x+y}{k}\right)-\frac{f\left(x\right)+f\left(y\right)}{k}∥⩽\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel ,x,y\in X.$$

(30)

Then, f is additive.

Proof. 

Letting $y=0$ in (30), we obtain: $f\left(\frac{x}{k}\right)=\frac{f\left(x\right)}{k}$ for all $x\in X$. Then, (30) yields:

$$\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel ⩽\left|k\right|\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel ,x,y\in X.$$

Then, $f(x+y)=f\left(x\right)+f\left(y\right)$ for all $x,y\in X$ because of $\left|k\right|<1$. □

Remark 3.

Let ${\mathbb{Q}}_2$ be equipped with the two-adic norm ${|.|}_2$. The function $f:{\mathbb{Q}}_2\to {\mathbb{Q}}_2$ given by $f\left(x\right)=2$ (for all $x\in {\mathbb{Q}}_2$) fulfils (30) with $k=2$, but f is not additive.

For convenience, let us define:

$$\begin{array}{cc}\hfill & {\mathcal{A}}_d:=\{(x,y)\in X\times Y:\parallel x\parallel ⩾d\},\hfill \\ \hfill & {\mathcal{B}}_d:=\{(x,y)\in X\times Y:\parallel y\parallel ⩾d\},\hfill \\ \hfill & {\mathcal{C}}_d:=\{(x,y)\in X\times Y:\parallel x\parallel +\parallel y\parallel ⩾d\}\hfill \end{array}$$

for some $d>0$.

Theorem 8.

Let $\left|k\right|<1$ for some non-zero integer, $\epsilon ⩾0$ and Y be a Banach non-Archimedean space. Suppose that $f:X\to Y$ satisfies:

$$∥f\left(\frac{x+y}{k}\right)-\frac{f\left(x\right)+f\left(y\right)}{k}∥⩽\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel +\epsilon$$

(31)

for all $(x,y)\in {\mathcal{A}}_d(respectively,(x,y)\in {\mathcal{B}}_{d}or(x,y)\in {\mathcal{C}}_d)$, where $d>0$ is a constant. Then, there exists a unique additive mapping $A:X\to Y$ such that:

$$\parallel f\left(x\right)-A\left(x\right)\parallel ⩽\frac{\left|k\right|}{1-\left|k\right|}max\left\{\parallel f\left(0\right)\parallel +2\epsilon ,\frac{\parallel f\left(0\right)\parallel +\epsilon }{\left|k\right|}\right\}.$$

Proof. 

Since the both sides of (31) are symmetric in x and y, and ${\mathcal{A}}_d\subseteq {\mathcal{C}}_d$, it is sufficient to consider (31) for all $(x,y)\in {\mathcal{A}}_d$. Letting $y=0$ in (31), we obtain:

$$∥f\left(\frac{x}{k}\right)-\frac{f\left(x\right)+f\left(0\right)}{k}∥⩽\parallel f\left(0\right)\parallel +\epsilon ,\parallel x\parallel ⩾d.$$

Then:

$$\begin{array}{cc}\hfill ∥k^{n+1}f\left(\frac{x}{k^{n+1}}\right)-k^{n}f\left(\frac{x}{k^n}\right)∥& ⩽{\left|k\right|}^n\left(\parallel f\left(0\right)\parallel +\epsilon \right),\parallel x\parallel ⩾d,n⩾0,\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill ∥k^{n+1}f\left(\frac{x}{k^{n+1}}\right)-k^{m}f\left(\frac{x}{k^m}\right)∥& ⩽{\left|k\right|}^m\left(\parallel f\left(0\right)\parallel +\epsilon \right),\parallel x\parallel ⩾d,n>m⩾0.\hfill \end{array}$$

(33)

Similar to the process of proving Theorem 7, it can be concluded that the sequence ${\left\{k^{n}f\left(\frac{x}{k^n}\right)\right\}}_{n=1}^{\infty }$ is convergent and the function $A:X\to Y$ can be defined as follows:

$$A\left(x\right):=\underset{n\to \infty }{lim}{k}^{n}f\left(\frac{x}{k^n}\right),x\in X.$$

By the definition of A and (31), we obtain $A\left(0\right)=0$ and:

$$∥A\left(\frac{x+y}{k}\right)-\frac{A\left(x\right)+A\left(y\right)}{k}∥⩽\parallel A(x+y)-A\left(x\right)-A\left(y\right)\parallel ,x,y\in X.$$

Hence, A is additive by Lemma 2. Taking $m=0$ and letting $n\to \infty$ in (33), we obtain:

$$\parallel f\left(x\right)-A\left(x\right)\parallel ⩽\parallel f\left(0\right)\parallel + \epsilon ,\parallel x\parallel ⩾d.$$

(34)

To achieve the inequality (8), suppose that x is a non-zero and arbitrary element of X. According to $0<\left|k\right|<1$, we can choose $y\in X$ such that $min\{\parallel y\parallel ,\parallel x+y\parallel \}⩾d$. Then, (34) yields:

$$\begin{array}{cc}\hfill ∥f\left(\frac{x+y}{k}\right)-A\left(\frac{x+y}{k}\right)∥& ⩽\parallel f\left(0\right)\parallel + \epsilon ,\hfill \\ \hfill ∥\frac{f\left(y\right)}{k}-\frac{A\left(y\right)}{k}∥& ⩽\frac{\parallel f\left(0\right)\parallel + \epsilon }{\left|k\right|},\hfill \\ \hfill \parallel f(x+y)-A(x+y)\parallel & ⩽\parallel f\left(0\right)\parallel + \epsilon .\hfill \end{array}$$

Using the strong triangle inequalities and the additivity of A, we obtain the following results from the above inequalities:

$$\begin{array}{cc}\hfill ∥f\left(\frac{x+y}{k}\right)-\frac{f\left(y\right)}{k}-A\left(\frac{x}{k}\right)∥& ⩽\frac{\parallel f\left(0\right)\parallel + \epsilon }{\left|k\right|},\hfill \\ \hfill \parallel f(x+y)-f\left(y\right)-A\left(x\right)\parallel & ⩽ \parallel f\left(0\right)\parallel + \epsilon .\hfill \end{array}$$

Then:

$$\begin{array}{cc}\hfill \parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel & ⩽max\{\parallel f(x+y)-f(y)-A(x)\parallel ,\parallel A(x)-f(x)\parallel \}\hfill \\ \hfill & ⩽max\{\parallel f(0)\parallel +\epsilon ,\parallel A(x)-f(x)\parallel \}.\hfill \end{array}$$

According to this inequality and (31), we obtain:

$$\begin{array}{cc}\hfill ∥\frac{f\left(x\right)}{k}-\frac{A\left(x\right)}{k}∥& ⩽max\left\{∥f\left(\frac{x+y}{k}\right)-\frac{f\left(x\right)+f\left(y\right)}{k}∥,∥\frac{A\left(x\right)}{k}+\frac{f\left(y\right)}{k}-f\left(\frac{x+y}{k}\right)∥\right\}\hfill \\ \hfill & ⩽max\left\{\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel + \epsilon ,\frac{\parallel f\left(0\right)\parallel + \epsilon }{\left|k\right|}\right\}\hfill \\ \hfill & ⩽max\left\{\parallel f\left(0\right)\parallel +2\epsilon ,\parallel A\left(x\right)-f\left(x\right)\parallel + \epsilon ,\frac{\parallel f\left(0\right)\parallel + \epsilon }{\left|k\right|}\right\}.\hfill \end{array}$$

Hence:

$$\parallel f\left(x\right)-A\left(x\right)\parallel ⩽\frac{\left|k\right|}{1-\left|k\right|}max\left\{\parallel f\left(0\right)\parallel +2\epsilon ,\frac{\parallel f\left(0\right)\parallel + \epsilon }{\left|k\right|}\right\}.$$

Obviously, the last inequality also holds true for $x=0$, and results in A being unique. □

Corollary 5.

Let $\left|k\right|<1$ be a Banach non-Archimedean space for some non-zero integer and Y. Suppose that $f:X\to Y$ with $f\left(0\right)=0$ satisfies one of the following asymptotic properties:

$$\begin{array}{cc}\hfill & \left(i\right)\underset{\parallel x\parallel +\parallel y\parallel \to \infty }{lim}\left|∥f\left(\frac{x+y}{k}\right)-\frac{f\left(x\right)+f\left(y\right)}{k}∥-\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel \right|=0;\hfill \\ \hfill & \left(ii\right)\underset{\parallel x\parallel \to \infty }{lim}\left|∥f\left(\frac{x+y}{k}\right)-\frac{f\left(x\right)+f\left(y\right)}{k}∥-\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel \right|=0;\hfill \\ \hfill & \left(iii\right)\underset{max\{\parallel x\parallel ,\parallel y\parallel \}\to \infty }{lim}\left|∥f\left(\frac{x+y}{k}\right)-\frac{f\left(x\right)+f\left(y\right)}{k}∥-\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel \right|=0.\hfill \end{array}$$

Then, f is additive.

Proof. 

Let $\epsilon >0$ be an arbitrary real number. By the assumption $\left(i\right)$, there exists $d>0$ such that (31) holds. By Theorem 8, there exists an additive mapping $A_{\epsilon }:X\to Y$ such that $\parallel f\left(x\right)-A_{\epsilon }\left(x\right)\parallel ⩽C\epsilon$ for all $x\in X$, where the constant $C>0$ is independent of $\epsilon$. Therefore, by the strong triangle inequality and the additivity of $A_{\epsilon }$, we obtain:

$$\begin{array}{cc}\hfill \parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel & ⩽max\{\parallel f(x+y)-A_{\epsilon }(x+y)\parallel ,\hfill \\ \hfill & \parallel A_{\epsilon }\left(x\right)-f\left(x\right)\parallel ,\parallel f\left(y\right)-A_{\epsilon }\left(y\right)\parallel \}\hfill \\ \hfill & ⩽C\epsilon ,x,y\in X.\hfill \end{array}$$

Since $\epsilon >0$ was arbitrary, we infer that f is additive. For the other cases, the proof will be the same. □

5. Conclusions

In this work, we studied the Pexider–Cauchy functional equation. We established a new strategy to study the Hyers–Ulam stability and hyperstability of the Pexider–Cauchy functional equation in the non-Archimedean normed spaces with restricted domains. We also showed that, subject to certain restrictions, every function which fulfils certain inequalities can be approximated by an additive function in non-Archimedean normed spaces. As a consequence, we used the results to investigate the asymptotic stability behaviour of the Cauchy and Pexider–Cauchy functional equations.