# Non-Archimedean and random HUR-approximation of a Cauchy-Jensen additive mapping

## Abstract

In this paper, using the fixed point and direct methods, we prove the Hyers-Ulam-Rassias approximation (briefly, HUR-approximation) of a Cauchy-Jensen additive (briefly, CJA) functional equation in various normed spaces.

MSC:39B52, 39B82, 47H10, 46S10.

## 1 Introduction

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 must be close to an exact solution of the equation? If the problem accepts a solution, we say that the equation is stable. The first stability problem concerning group homomorphisms was raised by Ulam  in 1940. In 1941, Hyers  gave a positive answer to the above question for additive groups under the assumption that the groups are Banach spaces. Aoki  proved a generalization of Hyers’ theorem for additive mappings and Rassias  proved a generalization of Hyers’ theorem for linear mappings.

Theorem 1.1 (ThM Rassias)

Let $f:E\to {E}^{\prime }$ be a mapping from a normed vector space E into a Banach space ${E}^{\prime }$ subject to the inequality $\parallel f\left(x+y\right)-f\left(x\right)-f\left(y\right)\parallel \le \epsilon \left({\parallel x\parallel }^{p}+{\parallel y\parallel }^{p}\right)$, for all $x,y\in E$, where ε and p are constants with $\epsilon >0$ and $0\le p<1$. Then the limit $L\left(x\right)={lim}_{n\to \mathrm{\infty }}\frac{f\left({2}^{n}x\right)}{{2}^{n}}$ exists, for all $x\in E$, and $L:E\to {E}^{\prime }$ is the unique additive mapping which satisfies

$\parallel f\left(x\right)-L\left(x\right)\parallel \le \frac{2\epsilon }{2-{2}^{p}}{\parallel x\parallel }^{p},$

for all $x\in E$. Also, if for each $x\in E$ the function $f\left(tx\right)$ is continuous in $t\in \mathbb{R}$, then L is linear.

This new concept is known as a the Hyers-Ulam stability or the Hyers-Ulam-Rassias stability of functional equations. Furthermore, in 1994, a generalization of Rassias’ theorem was obtained by Gǎvruta  by replacing the bound $\epsilon \left({\parallel x\parallel }^{p}+{\parallel y\parallel }^{p}\right)$ by a general control function $\phi \left(x,y\right)$.

In 1983, a generalized Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof  for mappings $f:X\to Y$, where X is a normed space and Y is a Banach space. In 1984, Cholewa  noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group and, in 2002, Czerwik  proved the generalized Hyers-Ulam stability of the quadratic functional equation. The readers are referred to  and references therein for detailed information on stability of functional equations.

In 1897, Hensel  has introduced a normed space which does not have the Archimedean property. It turned out that non-Archimedean spaces have many nice applications (see ).

Definition 1.1 By a non-Archimedean field we mean a field equipped with a function (valuation) $|\cdot |:\mathbb{K}\to \left[0,\mathrm{\infty }\right)$ such that, for all $r,s\in \mathbb{K}$, the following conditions hold: (a) $|r|=0$ if and only if $r=0$; (b) $|rs|=|r||s|$; (c) $|r+s|\le max\left\{|r|,|s|\right\}$.

Clearly, by (b), $|1|=|-1|=1$ and so, by induction, it follows from (c) that $|n|\le 1$, for all $n\ge 1$.

Definition 1.2 Let X be a vector space over a scalar field with a non-Archimedean non-trivial valuation $|\cdot |$.

1. (1)

A function $\parallel \cdot \parallel :X\to \mathbb{R}$ is a non-Archimedean norm (valuation) if it satisfies the following conditions: (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 (ultra-metric) holds, that is, $\parallel x+y\parallel \le max\left\{\parallel x\parallel ,\parallel y\parallel \right\}$, for all $x,y\in X$.

2. (2)

The space $\left(X,\parallel \cdot \parallel \right)$ is called a non-Archimedean normed space (briefly, NAN-space).

Note that $\parallel {x}_{n}-{x}_{m}\parallel \le max\left\{\parallel {x}_{j+1}-{x}_{j}\parallel :m\le j\le n-1\right\}$, for all $m,n\in \mathbb{N}$ with $n>m$.

Definition 1.3 Let $\left(X,\parallel \cdot \parallel \right)$ be a non-Archimedean normed space.

1. (a)

A sequence $\left\{{x}_{n}\right\}$ is a Cauchy sequence in X if $\left\{{x}_{n+1}-{x}_{n}\right\}$ converges to zero in X.

2. (b)

The non-Archimedean normed space $\left(X,\parallel \cdot \parallel \right)$ is said to be complete if every Cauchy sequence in X is convergent.

The most important examples of non-Archimedean spaces are p-adic numbers. A key property of p-adic numbers is that they do not satisfy the Archimedean axiom: for all $x,y>0$, there exists a positive integer n such that $x.

Example 1.1 Fix a prime number p. For any nonzero rational number x, there exists a unique positive integer ${n}_{x}$ such that $x=\frac{a}{b}{p}^{{n}_{x}}$, where a and b are positive integers not divisible by p. Then ${|x|}_{p}:={p}^{-{n}_{x}}$ defines a non-Archimedean norm on . The completion of with respect to the metric $d\left(x,y\right)={|x-y|}_{p}$ 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 $|{a}_{k}|\le p-1$. 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 ${\mathbb{Q}}_{p}$ is a locally compact field.

In Section 3, we adopt the usual terminology, notions and conventions of the theory of random normed spaces as in . Throughout this paper, let + denote the set of all probability distribution functions $F:\mathbb{R}\cup \left[-\mathrm{\infty },+\mathrm{\infty }\right]\to \left[0,1\right]$ such that F is left-continuous and nondecreasing on and $F\left(0\right)=0$, $F\left(+\mathrm{\infty }\right)=1$. It is clear that the set ${D}^{+}=\left\{F\in {△}^{+}:{l}^{-}F\left(-\mathrm{\infty }\right)=1\right\}$, where ${l}^{-}f\left(x\right)={lim}_{t\to {x}^{-}}f\left(t\right)$, is a subset of +. The set + is partially ordered by the usual point-wise ordering of functions, that is, $F\le G$ if and only if $F\left(t\right)\le G\left(t\right)$, for all $t\in \mathbb{R}$. For any $a\ge 0$, the element ${H}_{a}\left(t\right)$ of ${D}^{+}$ is defined by

We can easily show that the maximal element in + is the distribution function ${H}_{0}\left(t\right)$.

Definition 1.4 A function $T:{\left[0,1\right]}^{2}\to \left[0,1\right]$ is a continuous triangular norm (briefly, a t-norm) if T satisfies the following conditions: (a) T is commutative and associative; (b) T is continuous; (c) $T\left(x,1\right)=x$, for all $x\in \left[0,1\right]$; (d) $T\left(x,y\right)\le T\left(z,w\right)$ whenever $x\le z$ and $y\le w$, for all $x,y,z,w\in \left[0,1\right]$.

Three typical examples of continuous t-norms are as follows: ${T}_{P}\left(x,y\right)=xy$, ${T}_{\mathrm{max}}\left(x,y\right)=max\left\{a+b-1,0\right\}$, ${T}_{M}\left(x,y\right)=min\left(a,b\right)$. Recall that, if T is a t-norm and $\left\{{x}_{n}\right\}$ is a sequence in $\left[0,1\right]$, then ${T}_{i=1}^{n}{x}_{i}$ is defined recursively by ${T}_{i=1}^{1}{x}_{1}={x}_{1}$ and ${T}_{i=1}^{n}{x}_{i}=T\left({T}_{i=1}^{n-1}{x}_{i},{x}_{n}\right)$, for all $n\ge 2$. ${T}_{i=n}^{\mathrm{\infty }}{x}_{i}$ is defined by ${T}_{i=1}^{\mathrm{\infty }}{x}_{n+i}$.

Definition 1.5 A random normed space (briefly, RN-space) is a triple $\left(X,\mu ,T\right)$, where X is a vector space, T is a continuous t-norm and $\mu :X\to {D}^{+}$ is a mapping such that the following conditions hold:

1. (a)

${\mu }_{x}\left(t\right)={H}_{0}\left(t\right)$, for all $t>0$ if and only if $x=0$;

2. (b)

${\mu }_{\alpha x}\left(t\right)={\mu }_{x}\left(\frac{t}{|\alpha |}\right)$, for all $\alpha \in \mathbb{R}$ with $\alpha \ne 0$, $x\in X$ and $t\ge 0$;

3. (c)

${\mu }_{x+y}\left(t+s\right)\ge T\left({\mu }_{x}\left(t\right),{\mu }_{y}\left(s\right)\right)$, for all $x,y\in X$ and $t,s\ge 0$.

Every normed space $\left(X,\parallel \cdot \parallel \right)$ defines a random normed space $\left(X,\mu ,{T}_{M}\right)$, where ${\mu }_{u}\left(t\right)=\frac{t}{t+\parallel u\parallel }$, for all $t>0$ and ${T}_{M}$ is the minimum t-norm. This space X is called the induced random normed space.

If the t-norm T is such that ${sup}_{0, then every RN-space $\left(X,\mu ,T\right)$ is a metrizable linear topological space with the topology τ (called the μ-topology or the $\left(\epsilon ,\delta \right)$-topology, where $\epsilon >0$ and $\lambda \in \left(0,1\right)$) induced by the base $\left\{U\left(\epsilon ,\lambda \right)\right\}$ of neighborhoods of θ, where

$U\left(\epsilon ,\lambda \right)=\left\{x\in X:{\mu }_{x}\left(\epsilon \right)>1-\lambda \right\}.$

Definition 1.6 Let $\left(X,\mu ,T\right)$ be an RN-space.

1. (a)

A sequence $\left\{{x}_{n}\right\}$ in X is said to be convergent to a point $x\in X$ (write ${x}_{n}\to x$ as $n\to \mathrm{\infty }$) if ${lim}_{n\to \mathrm{\infty }}{\mu }_{{x}_{n}-x}\left(t\right)=1$, for all $t>0$.

2. (b)

A sequence $\left\{{x}_{n}\right\}$ in X is called a Cauchy sequence in X if ${lim}_{n\to \mathrm{\infty }}{\mu }_{{x}_{n}-{x}_{m}}\left(t\right)=1$, for all $t>0$.

3. (c)

The RN-space $\left(X,\mu ,T\right)$ is said to be complete if every Cauchy sequence in X is convergent.

Theorem 1.2 If $\left(X,\mu ,T\right)$ is RN-space and $\left\{{x}_{n}\right\}$ is a sequence such that ${x}_{n}\to x$, then ${lim}_{n\to \mathrm{\infty }}{\mu }_{{x}_{n}}\left(t\right)={\mu }_{x}\left(t\right)$.

Definition 1.7 Let X be a set. A function $d:X×X\to \left[0,\mathrm{\infty }\right]$ is called a generalized metric on X if d satisfies the following conditions:

1. (a)

$d\left(x,y\right)=0$ if and only if $x=y$, for all $x,y\in X$;

2. (b)

$d\left(x,y\right)=d\left(y,x\right)$, for all $x,y\in X$;

3. (c)

$d\left(x,z\right)\le d\left(x,y\right)+d\left(y,z\right)$, for all $x,y,z\in X$.

Theorem 1.3 ([37, 38])

Let $\left(X,d\right)$ 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 $d\left({J}^{n}x,{J}^{n+1}x\right)=\mathrm{\infty }$, for all nonnegative integers n, or there exists a positive integer ${n}_{0}$ such that

1. (a)

$d\left({J}^{n}x,{J}^{n+1}x\right)<\mathrm{\infty }$, for all ${n}_{0}\ge {n}_{0}$;

2. (b)

the sequence $\left\{{J}^{n}x\right\}$ converges to a fixed point ${y}^{\ast }$ of J;

3. (c)

${y}^{\ast }$ is the unique fixed point of J in the set $Y=\left\{y\in X:d\left({J}^{{n}_{0}}x,y\right)<\mathrm{\infty }\right\}$;

4. (d)

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

In this paper, using the fixed point and direct methods, we prove the HUR-approximation of the following CJA functional equation:

$2f\left(\frac{x+y+z}{2}\right)=f\left(x\right)+f\left(y\right)+f\left(z\right)$
(1.1)

in various normed spaces.

## 2 NAN-stability

In this section, we deal with the stability problem for the Cauchy-Jensen additive functional equation (1.1) in non-Archimedean normed spaces.

Theorem 2.1 Let X be a non-Archimedean normed space and Y is a complete non-Archimedean space. Let $\phi :{X}^{3}\to \left[0,\mathrm{\infty }\right)$ be a function such that there exists an $\alpha <1$ with

$\phi \left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right)\le \frac{\alpha \phi \left(x,y,z\right)}{|2|},$
(2.1)

for all $x,y,z\in X$. Let $f:X\to Y$ be a mapping satisfying

${\parallel 2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right)\parallel }_{Y}\le \phi \left(x,y,z\right),$
(2.2)

for all $x,y,z\in X$. Then there exists a unique additive mapping $\mathrm{\Im }:X\to Y$ such that

${\parallel f\left(x\right)-\mathrm{\Im }\left(x\right)\parallel }_{Y}\le \alpha \phi \left(x,2x,x\right){\left(|2|-|2|\alpha \right)}^{-1},$
(2.3)

for all $x\in X$.

Proof Putting $y=2x$ and $z=x$ in (2.1), we get ${\parallel f\left(2x\right)-2f\left(x\right)\parallel }_{Y}\le \phi \left(x,2x,x\right)$, for all $x\in X$. So

${\parallel f\left(x\right)-2f\left(\frac{x}{2}\right)\parallel }_{Y}\le {|2|}^{-1}\alpha \phi \left(x,2x,x\right),$
(2.4)

for all $x\in X$. Consider the set $S:=\left\{h:X\to Y\right\}$ and introduce the generalized metric on S:

$d\left(g,h\right)=\underset{\mu \in \left(0,+\mathrm{\infty }\right)}{inf}{\parallel g\left(x\right)-h\left(x\right)\parallel }_{Y}\le \mu \phi \left(x,2x,x\right),$

for all $x\in X$, where, as usual, $inf\varphi =+\mathrm{\infty }$. It is easy to show that $\left(S,d\right)$ is complete (see ). Now we consider the linear mapping $J:S\to S$ such that $Jg\left(x\right):=2g\left(\frac{x}{2}\right)$, for all $x\in X$. Let $g,h\in S$ be given such that $d\left(g,h\right)=\epsilon$. Then ${\parallel g\left(x\right)-h\left(x\right)\parallel }_{Y}\le \epsilon \phi \left(x,2x,x\right)$, for all $x\in X$. Hence

${\parallel Jg\left(x\right)-Jh\left(x\right)\parallel }_{Y}={\parallel 2g\left(\frac{x}{2}\right)-2h\left(\frac{x}{2}\right)\parallel }_{Y}\le \alpha \cdot \epsilon \phi \left(x,2x,x\right),$

for all $x\in X$. So $d\left(g,h\right)=\epsilon$ implies that $d\left(Jg,Jh\right)\le \alpha \epsilon$. This means that $d\left(Jg,Jh\right)\le \alpha d\left(g,h\right)$, for all $g,h\in S$. It follows from (2.4) that $d\left(f,Jf\right)\le {|2|}^{-1}\alpha$. By Theorem 1.3, there exists a mapping $\mathrm{\Im }:X\to Y$ satisfying the following:

1. (1)

is a fixed point of J, i.e.,

$\mathrm{\Im }\left(x\right)=2\mathrm{\Im }\left(\frac{x}{2}\right),$
(2.5)

for all $x\in X$. The mapping is a unique fixed point of J in the set $M=\left\{g\in S:d\left(h,g\right)<\mathrm{\infty }\right\}$. This implies that is a unique mapping satisfying (2.5) such that there exists a $\mu \in \left(0,\mathrm{\infty }\right)$ satisfying ${\parallel f\left(x\right)-\mathrm{\Im }\left(x\right)\parallel }_{Y}\le \mu \phi \left(x,2x,x\right)$, for all $x\in X$;

1. (2)

$d\left({J}^{n}f,\mathrm{\Im }\right)\to 0$ as $n\to \mathrm{\infty }$. This implies the equality

$\underset{n\to \mathrm{\infty }}{lim}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)=\mathrm{\Im }\left(x\right),$
(2.6)

for all $x\in X$;

1. (3)

$d\left(f,\mathrm{\Im }\right)\le \frac{d\left(f,Jf\right)}{1-\alpha }$, which implies the inequality $d\left(f,\mathrm{\Im }\right)\le \alpha {\left(|2|-|2|\alpha \right)}^{-1}$. This implies that the inequalities (2.3) holds. It follows from (2.1) and (2.2) that

$\begin{array}{r}{\parallel 2\mathrm{\Im }\left(\frac{x+y+z}{2}\right)-\mathrm{\Im }\left(x\right)-\mathrm{\Im }\left(y\right)-\mathrm{\Im }\left(z\right)\parallel }_{Y}\\ \phantom{\rule{1em}{0ex}}=\underset{n\to \mathrm{\infty }}{lim}\frac{{\parallel 2f\left(\frac{x+y+z}{{2}^{n+1}}\right)-f\left(\frac{x}{{2}^{n}}\right)-f\left(\frac{y}{{2}^{n}}\right)-f\left(\frac{z}{{2}^{n}}\right)\parallel }_{Y}}{{|2|}^{-n}}\\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{lim}{\alpha }^{n}\phi \left(x,y,z\right)=0,\end{array}$

for all $x,y,z\in X$. So $2\mathrm{\Im }\left(\frac{x+y+z}{2}\right)=\mathrm{\Im }\left(x\right)+\mathrm{\Im }\left(y\right)+\mathrm{\Im }\left(z\right)$, for all $x,y,z\in X$. Hence $\mathrm{\Im }:X\to Y$ is an CJA mapping and we get the desired results. □

Corollary 2.1 Let θ be a positive real number and r is a real number with $0. Let $f:X\to Y$ be a mapping satisfying

${\parallel 2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right)\parallel }_{Y}\le \theta \left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}+{\parallel z\parallel }^{r}\right),$
(2.7)

for all $x,y,z\in X$. Then there exists a unique CJA mapping $\mathrm{\Im }:X\to Y$ such that

${\parallel f\left(x\right)-\mathrm{\Im }\left(x\right)\parallel }_{Y}\le |2|\theta \left(2+{|2|}^{r}\right){\left({|2|}^{r+1}-{|2|}^{2}\right)}^{-1}{\parallel x\parallel }^{r},$

for all $x\in X$.

Proof The proof follows from Theorem 2.1 by taking $\phi \left(x,y,z\right)=\left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}+{\parallel z\parallel }^{r}\right)$, for all $x,y,z\in X$. Then we can choose $\alpha ={|2|}^{1-r}$ and we get the desired result. □

Theorem 2.2 Let X be a non-Archimedean normed space and Y is a complete non-Archimedean space. Let $\phi :{X}^{3}\to \left[0,\mathrm{\infty }\right)$ be a function such that there exists an $\alpha <1$ with $\phi \left(x,y,z\right)\le |2|\alpha \phi \left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right)$, for all $x,y,z\in X$. Let $f:X\to Y$ be a mapping satisfying (2.2). Then there exists a unique CJA mapping $\mathrm{\Im }:X\to Y$ such that

${\parallel f\left(x\right)-\mathrm{\Im }\left(x\right)\parallel }_{Y}\le \phi \left(x,2x,x\right){\left(|2|-|2|\alpha \right)}^{-1},$
(2.8)

for all $x\in X$.

Proof Let $\left(S,d\right)$ be the generalized metric space defined in the proof of Theorem 2.1. Now we consider the linear mapping $J:S\to S$ such that $Jg\left(x\right):=\frac{g\left(2x\right)}{2}$, for all $x\in X$. Let $g,h\in S$ be given such that $d\left(g,h\right)=\epsilon$. Then ${\parallel g\left(x\right)-h\left(x\right)\parallel }_{Y}\le \epsilon \phi \left(x,2x,x\right)$, for all $x\in X$. Hence

${\parallel Jg\left(x\right)-Jh\left(x\right)\parallel }_{Y}={\parallel \frac{g\left(2x\right)}{2}-\frac{h\left(2x\right)}{2}\parallel }_{Y}\le \frac{|2|\alpha \cdot \epsilon \phi \left(x,2x,x\right)}{|2|},$

for all $x\in X$. So $d\left(g,h\right)=\epsilon$ implies that $d\left(Jg,Jh\right)\le \alpha \epsilon$. This means that $d\left(Jg,Jh\right)\le \alpha d\left(g,h\right)$, for all $g,h\in S$. It follows from (2.4) that $d\left(f,Jf\right)\le {|2|}^{-1}$. By Theorem 1.3, there exists a mapping $\mathrm{\Im }:X\to Y$ satisfying the following:

1. (1)

is a fixed point of J, i.e.,

$\frac{\mathrm{\Im }\left(2x\right)}{2}=\mathrm{\Im }\left(x\right),$
(2.9)

for all $x\in X$. The mapping is a unique fixed point of J in the set $M=\left\{g\in S:d\left(h,g\right)<\mathrm{\infty }\right\}$. This implies that is a unique mapping satisfying (2.9) such that there exists a $\mu \in \left(0,\mathrm{\infty }\right)$ satisfying ${\parallel f\left(x\right)-\mathrm{\Im }\left(x\right)\parallel }_{Y}\le \mu \phi \left(x,2x,x\right)$, for all $x\in X$;

1. (2)

$d\left({J}^{n}f,\mathrm{\Im }\right)\to 0$ as $n\to \mathrm{\infty }$. This implies the equality ${lim}_{n\to \mathrm{\infty }}\frac{f\left({2}^{n}x\right)}{{2}^{n}}=\mathrm{\Im }\left(x\right)$, for all $x\in X$;

2. (3)

$d\left(f,\mathrm{\Im }\right)\le \frac{d\left(f,Jf\right)}{1-\alpha }$, which implies the inequality $d\left(f,\mathrm{\Im }\right)\le {\left(|2|-|2|\alpha \right)}^{-1}$. This implies that the inequalities (2.8) holds. The rest of the proof is similar to the proof of Theorem 2.1. □

Corollary 2.2 Let θ be a positive real number and r is a real number with $r>1$. Let $f:X\to Y$ be a mapping satisfying (2.7). Then there exists a unique CJA mapping $\mathrm{\Im }:X\to Y$ such that

${\parallel f\left(x\right)-\mathrm{\Im }\left(x\right)\parallel }_{Y}\le \theta \left(2+{|2|}^{r}\right){\left(|2|-{|2|}^{r}\right)}^{-1}{\parallel x\parallel }^{r},$

for all $x\in X$.

Proof The proof follows from Theorem 2.2 by taking $\phi \left(x,y,z\right)=\left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}+{\parallel z\parallel }^{r}\right)$, for all $x,y,z\in X$. Then we can choose $\alpha ={|2|}^{r-1}$ and we get the desired result. □

Theorem 2.3 Let G be an additive semigroup and X is a non-Archimedean Banach space. Assume that $\lambda :{G}^{3}\to \left[0,+\mathrm{\infty }\right)$ be a function such that

$\underset{n\to \mathrm{\infty }}{lim}{|2|}^{n}\lambda \left(\frac{x}{{2}^{n}},\frac{y}{{2}^{n}},\frac{z}{{2}^{n}}\right)=0,$
(2.10)

for all $x,y,z\in G$. Suppose that, for any $x\in G$, the limit

$\text{£}\left(x\right)=\underset{n\to \mathrm{\infty }}{lim}\underset{0\le k
(2.11)

exists and $f:G\to X$ be a mapping satisfying

${\parallel 2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right)\parallel }_{X}\le \lambda \left(x,y,z\right).$
(2.12)

Then the limit $\mathrm{\Im }\left(x\right):={lim}_{n\to \mathrm{\infty }}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)$ exists, for all $x\in G$, and defines an CJA mapping $\mathrm{\Im }:G\to X$ such that

$\parallel f\left(x\right)-\mathrm{\Im }\left(x\right)\parallel \le \text{£}\left(x\right).$
(2.13)

Moreover, if ${lim}_{j\to \mathrm{\infty }}{lim}_{n\to \mathrm{\infty }}{max}_{j\le k then is the unique CJA mapping satisfying (2.13).

Proof Putting $y=2x$ and $z=x$ in (2.12), we get

${\parallel f\left(2x\right)-2f\left(x\right)\parallel }_{Y}\le \lambda \left(x,2x,x\right),$
(2.14)

for all $x\in G$. Replacing x by $\frac{x}{{2}^{n+1}}$ in (2.14), we obtain

$\parallel {2}^{n+1}f\left(\frac{x}{{2}^{n+1}}\right)-{2}^{n}f\left(\frac{x}{{2}^{n}}\right)\parallel \le |2{|}^{n}\lambda \left(\frac{x}{{2}^{n+1}},\frac{x}{{2}^{n}},\frac{x}{{2}^{n+1}}\right).$
(2.15)

Thus, it follows from (2.10) and (2.15) that the sequence ${\left\{{2}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}}_{n\ge 1}$ is a Cauchy sequence. Since X is complete, it follows that ${\left\{{2}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}}_{n\ge 1}$ is convergent. Set $\mathrm{\Im }\left(x\right):={lim}_{n\to \mathrm{\infty }}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)$. By induction on n, one can show that

$\parallel {2}^{n}f\left(\frac{x}{{2}^{n}}\right)-f\left(x\right)\parallel \le \underset{0\le k
(2.16)

for all $n\ge 1$ and $x\in G$. By taking $n\to \mathrm{\infty }$ in (2.16) and using (2.11), one obtains (2.13). By (2.10) and (2.12), we get

$\begin{array}{r}\parallel 2\mathrm{\Im }\left(\frac{x+y+z}{2}\right)-\mathrm{\Im }\left(x\right)-\mathrm{\Im }\left(y\right)-\mathrm{\Im }\left(z\right)\parallel \\ \phantom{\rule{1em}{0ex}}=\underset{n\to \mathrm{\infty }}{lim}{|2|}^{n}\parallel 2f\left(\frac{x+y+z}{{2}^{n+1}}\right)-f\left(\frac{x}{{2}^{n}}\right)-f\left(\frac{y}{{2}^{n}}\right)-f\left(\frac{z}{{2}^{n}}\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{lim}{|2|}^{n}\lambda \left(\frac{x}{{2}^{n}},\frac{y}{{2}^{n}},\frac{z}{{2}^{n}}\right)=0,\end{array}$

for all $x,y,z\in X$. So

$\mathrm{\Im }\left(\frac{x+y+z}{2}\right)=\mathrm{\Im }\left(x\right)+\mathrm{\Im }\left(y\right)+\mathrm{\Im }\left(z\right).$
(2.17)

Letting $x=y=z=0$ in (2.17), we get $\mathrm{\Im }\left(0\right)=0$. Letting $z=x+y$ in (2.17), we get $\mathrm{\Im }\left(x+y\right)=\mathrm{\Im }\left(x\right)+\mathrm{\Im }\left(y\right)$, for all $x,y\in X$. Hence the mapping $\mathrm{\Im }:X\to Y$ is Cauchy additive.

To prove the uniqueness property of , let be another mapping satisfying (2.13). Then we have

$\begin{array}{rcl}{\parallel \mathrm{\Im }\left(x\right)-\mathrm{\Re }\left(x\right)\parallel }_{X}& =& \underset{n\to \mathrm{\infty }}{lim}|2{|}^{n}{\parallel \mathrm{\Im }\left(\frac{x}{{2}^{n}}\right)-\mathrm{\Re }\left(\frac{x}{{2}^{n}}\right)\parallel }_{X}\\ \le & \underset{k\to \mathrm{\infty }}{lim}|2{|}^{n}max\left\{{\parallel \mathrm{\Im }\left(\frac{x}{{2}^{n}}\right)-f\left(\frac{x}{{2}^{n}}\right)\parallel }_{X},{\parallel f\left(\frac{x}{{2}^{n}}\right)-\mathrm{\Re }\left(\frac{x}{{2}^{n}}\right)\parallel }_{X}\right\}\\ \le & \underset{j\to \mathrm{\infty }}{lim}\underset{n\to \mathrm{\infty }}{lim}\underset{j\le k

for all $x\in G$. Therefore, $\mathrm{\Im }=\mathrm{\Re }$. This completes the proof. □

Corollary 2.3 Let $\xi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ be a function satisfying $\xi \left(\frac{t}{|2|}\right)\le \xi \left(\frac{1}{|2|}\right)\xi \left(t\right)$, $\xi \left(\frac{1}{|2|}\right)<\frac{1}{|2|}$, for all $t\ge 0$. Assume that $\kappa >0$ and $f:G\to X$ be a mapping such that

${\parallel 2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right)\parallel }_{Y}\le \kappa \left(\xi \left(|x|\right)+\xi \left(|y|\right)+\xi \left(|z|\right)\right),$

for all $x,y,z\in G$. Then there exists a unique CJA mapping $\mathrm{\Im }:G\to X$ such that

$\parallel f\left(x\right)-\mathrm{\Im }\left(x\right)\parallel \le {|2|}^{-1}\left(2+|2|\right)\xi \left(|x|\right).$

Proof If we define $\lambda :{G}^{3}\to \left[0,\mathrm{\infty }\right)$ by $\lambda \left(x,y,z\right):=\kappa \left(\xi \left(|x|\right)+\xi \left(|y|\right)+\xi \left(|z|\right)\right)$, then we have ${lim}_{n\to \mathrm{\infty }}|2{|}^{n}\lambda \left(\frac{x}{{2}^{n}},\frac{y}{{2}^{n}},\frac{z}{{2}^{n}}\right)=0$, for all $x,y,z\in G$. On the other hand, it follows that $\text{£}\left(x\right)={|2|}^{-1}\left(2+|2|\right)\xi \left(|x|\right)$ exists, for all $x\in G$. Also, we have

$\underset{j\to \mathrm{\infty }}{lim}\underset{n\to \mathrm{\infty }}{lim}\underset{j\le k

Thus, applying Theorem 2.3, we have the conclusion. This completes the proof. □

Theorem 2.4 Let G be an additive semigroup and X is a non-Archimedean Banach space. Assume that $\lambda :{G}^{3}\to \left[0,+\mathrm{\infty }\right)$ be a function such that ${lim}_{n\to \mathrm{\infty }}\frac{\lambda \left({2}^{n}x,{2}^{n}y,{2}^{n}z\right)}{{2}^{n}}=0$, for all $x,y,z\in G$. Suppose that, for any $x\in G$, the limit

$\text{£}\left(x\right)=\underset{n\to \mathrm{\infty }}{lim}\underset{0\le k
(2.18)

exists and $f:G\to X$ be a mapping satisfying (2.12). Then the limit $\mathrm{\Im }\left(x\right):={lim}_{n\to \mathrm{\infty }}\frac{f\left({2}^{n}x\right)}{{2}^{n}}$ exists, for all $x\in G$, and

$\parallel f\left(x\right)-\mathrm{\Im }\left(x\right)\parallel \le \frac{\text{£}\left(x\right)}{|2|},$
(2.19)

for all $x\in G$. Moreover, if ${lim}_{j\to \mathrm{\infty }}{lim}_{n\to \mathrm{\infty }}{max}_{j\le k, then is the unique CJA mapping satisfying (2.19).

Proof It follows from (2.14) that

${\parallel f\left(x\right)-\frac{f\left(2x\right)}{2}\parallel }_{X}\le \frac{\lambda \left(x,2x,x\right)}{|2|},$
(2.20)

for all $x\in G$. Replacing x by ${2}^{n}x$ in (2.20), we obtain

${\parallel \frac{f\left({2}^{n}x\right)}{{2}^{n}}-\frac{f\left({2}^{n+1}x\right)}{{2}^{n+1}}\parallel }_{X}\le \frac{\lambda \left({2}^{n}x,{2}^{n+1}x,{2}^{n}x\right)}{{|2|}^{n+1}}.$
(2.21)

Thus it follows from (2.21) that the sequence ${\left\{\frac{f\left({2}^{n}x\right)}{{2}^{n}}\right\}}_{n\ge 1}$ is convergent. Set $\mathrm{\Im }\left(x\right):={lim}_{n\to \mathrm{\infty }}\frac{f\left({2}^{n}x\right)}{{2}^{n}}$. On the other hand, it follows from (2.21) that

$\begin{array}{rcl}\parallel \frac{f\left({2}^{p}x\right)}{{2}^{p}}-\frac{f\left({2}^{q}x\right)}{{2}^{q}}\parallel & =& \parallel \sum _{k=p}^{q-1}\frac{f\left({2}^{k+1}x\right)}{{2}^{k+1}}-\frac{f\left({2}^{k}x\right)}{{2}^{k}}\parallel \le \underset{p\le k

for all $x\in G$ and $p,q\ge 0$ with $q>p\ge 0$. Letting $p=0$, taking $q\to \mathrm{\infty }$ in the last inequality and using (2.18), we obtain (2.19).

The rest of the proof is similar to the proof of Theorem 2.3. This completes the proof. □

Corollary 2.4 Let $\xi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ be a function satisfying $\xi \left(|2|t\right)\le \xi \left(|2|\right)\xi \left(t\right)$, $\xi \left(|2|\right)<|2|$, for all $t\ge 0$. Let $\kappa >0$ and $f:G\to X$ be a mapping satisfying

$\parallel 2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right)\parallel \le \kappa \left(\xi \left(|x|\right)\cdot \xi \left(|y|\right)\cdot \xi \left(|z|\right)\right),$

for all $x,y,z\in G$. Then there exists a unique CJA mapping $\mathrm{\Im }:G\to X$ such that

$\parallel f\left(x\right)-\mathrm{\Im }\left(x\right)\parallel \le \kappa \xi {\left(|x|\right)}^{3}.$

Proof If we define $\lambda :{G}^{3}\to \left[0,\mathrm{\infty }\right)$ by $\lambda \left(x,y,z\right):=\kappa \left(\xi \left(|x|\right)\cdot \xi \left(|y|\right)\cdot \xi \left(|z|\right)\right)$ and apply Theorem 2.4, then we get the conclusion. □

## 3 RNS-stability

In this section, using the fixed point and direct methods, we prove the HUR-approximation of the functional equation (1.1) in random normed spaces.

Theorem 3.1 Let X be a real linear space, $\left(Z,{\mu }^{\prime },min\right)$ be an RN-space and $\phi :{X}^{3}\to Z$ be a function such that there exists $0<\alpha <\frac{1}{2}$ such that

${\mu }_{\phi \left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right)}^{\prime }\left(t\right)\ge {\mu }_{\phi \left(x,y,z\right)}^{\prime }\left(\frac{t}{\alpha }\right),$
(3.1)

for all $x,y,z\in X$ and $t>0$ and ${lim}_{n\to \mathrm{\infty }}{\mu }_{\phi \left(\frac{x}{{2}^{n}},\frac{y}{{2}^{n}},\frac{z}{{2}^{n}}\right)}^{\prime }\left(\frac{t}{{2}^{n}}\right)=1$, for all $x,y,z\in X$ and $t>0$. Let $\left(Y,\mu ,min\right)$ be a complete RN-space. If $f:X\to Y$ be a mapping such that

${\mu }_{2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right)}\left(t\right)\ge {\mu }_{\phi \left(x,y,z\right)}^{\prime }\left(t\right),$
(3.2)

for all $x,y,z\in X$ and $t>0$. Then the limit $\mathrm{\Im }\left(x\right)={lim}_{n\to \mathrm{\infty }}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)$ exists, for all $x\in X$, and defines a unique CJA mapping $\mathrm{\Im }:X\to Y$ such that

${\mu }_{f\left(x\right)-A\left(x\right)}\left(t\right)\ge {\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(\frac{\left(1-2\alpha \right)t}{\alpha }\right),$
(3.3)

for all $x\in X$ and $t>0$.

Proof Putting $y=2x$ and $z=x$ in (3.2), we see that

${\mu }_{f\left(2x\right)-2f\left(x\right)}\left(t\right)\ge {\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(t\right).$
(3.4)

Replacing x by $\frac{x}{2}$ in (3.4), we obtain

${\mu }_{2f\left(\frac{x}{2}\right)-f\left(x\right)}\left(t\right)\ge {\mu }_{\phi \left(\frac{x}{2},x,\frac{x}{2}\right)}^{\prime }\left(t\right)\ge {\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(\frac{t}{\alpha }\right),$
(3.5)

for all $x\in X$. Replacing x by $\frac{x}{{2}^{n}}$ in (3.5) and using (3.1), we obtain

${\mu }_{{2}^{n+1}f\left(\frac{x}{{2}^{n+1}}\right)-{2}^{n}f\left(\frac{x}{{2}^{n}}\right)}\left(t\right)\ge {\mu }_{\phi \left(\frac{x}{{2}^{n+1}},\frac{x}{{2}^{n}},\frac{x}{{2}^{n+1}}\right)}^{\prime }\left(\frac{t}{{2}^{n}}\right)\ge {\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(\frac{t}{{2}^{n}{\alpha }^{n+1}}\right)$

and so

$\begin{array}{rcl}{\mu }_{{2}^{n}f\left(\frac{x}{{2}^{n}}\right)-f\left(x\right)}\left(\sum _{k=0}^{n-1}{2}^{k}{\alpha }^{k+1}t\right)& =& {\mu }_{{\sum }_{k=0}^{n-1}{2}^{k+1}f\left(\frac{x}{{2}^{k+1}}\right)-{2}^{k}f\left(\frac{x}{{2}^{k}}\right)}\left(\sum _{k=0}^{n-1}{2}^{k}{\alpha }^{k+1}t\right)\\ \ge & {T}_{k=0}^{n-1}\left({\mu }_{{2}^{k+1}f\left(\frac{x}{{2}^{k+1}}\right)-{2}^{k}f\left(\frac{x}{{2}^{k}}\right)}\left({2}^{k}{\alpha }^{k+1}t\right)\right)\\ \ge & {T}_{k=0}^{n-1}\left({\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(t\right)\right)={\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(t\right).\end{array}$

This implies that

${\mu }_{{2}^{n}f\left(\frac{x}{{2}^{n}}\right)-f\left(x\right)}\left(t\right)\ge {\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(\frac{t}{{\sum }_{k=0}^{n-1}{2}^{k}{\alpha }^{k+1}}\right).$
(3.6)

Replacing x by $\frac{x}{{2}^{p}}$ in (3.6), we obtain

${\mu }_{{2}^{n+p}f\left(\frac{x}{{2}^{n+p}}\right)-{2}^{p}f\left(\frac{x}{{2}^{p}}\right)}\left(t\right)\ge {\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(\frac{t}{{\sum }_{k=p}^{n+p-1}{2}^{k}{\alpha }^{k+1}}\right).$
(3.7)

Since ${lim}_{p,n\to \mathrm{\infty }}{\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(\frac{t}{{\sum }_{k=p}^{n+p-1}{2}^{k}{\alpha }^{k+1}}\right)=1$, it follows that ${\left\{{2}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}}_{n=1}^{\mathrm{\infty }}$ is a Cauchy sequence in a complete RN-space $\left(Y,\mu ,min\right)$ and so there exists a point $\mathrm{\Im }\left(x\right)\in Y$ such that ${lim}_{n\to \mathrm{\infty }}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)=\mathrm{\Im }\left(x\right)$. Fix $x\in X$ and put $p=0$ in (3.7) and so, for any $\epsilon >0$,

${\mu }_{\mathrm{\Im }\left(x\right)-f\left(x\right)}\left(t+\epsilon \right)\ge T\left({\mu }_{\mathrm{\Im }\left(x\right)-{2}^{n}f\left(\frac{x}{{2}^{n}}\right)}\left(\epsilon \right),{\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(\frac{t}{{\sum }_{k=0}^{n-1}{2}^{k}{\alpha }^{k+1}}\right)\right).$
(3.8)

Taking $n\to \mathrm{\infty }$ in (3.8), we get ${\mu }_{\mathrm{\Im }\left(x\right)-f\left(x\right)}\left(t+\epsilon \right)\ge {\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(\frac{\left(1-2\alpha \right)t}{\alpha }\right)$. Since ε is arbitrary, by taking $\epsilon \to 0$ in the previous inequality, we get

${\mu }_{\mathrm{\Im }\left(x\right)-f\left(x\right)}\left(t\right)\ge {\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(\frac{\left(1-2\alpha \right)t}{\alpha }\right).$

Replacing x, y and z by $\frac{x}{{2}^{n}}$, $\frac{y}{{2}^{n}}$ and $\frac{z}{{2}^{n}}$ in (3.2), respectively, we get

${\mu }_{{2}^{n+1}f\left(\frac{x+y+z}{{2}^{n+1}}\right)-{2}^{n}f\left(\frac{x}{{2}^{n}}\right)-{2}^{n}f\left(\frac{y}{{2}^{n}}\right)-{2}^{n}f\left(\frac{z}{{2}^{n}}\right)}\left(t\right)\ge {\mu }_{\phi \left(\frac{x}{{2}^{n}},\frac{y}{{2}^{n}},\frac{z}{{2}^{n}}\right)}^{\prime }\left(\frac{t}{{2}^{n}}\right),$

for all $x,y,z\in X$ and $t>0$. Since ${lim}_{n\to \mathrm{\infty }}{\mu }_{\phi \left(\frac{x}{{2}^{n}},\frac{y}{{2}^{n}},\frac{z}{{2}^{n}}\right)}^{\prime }\left(\frac{t}{{2}^{n}}\right)=1$, we conclude that satisfies (1.1). On the other hand

$2\mathrm{\Im }\left(\frac{x}{2}\right)-\mathrm{\Im }\left(x\right)=\underset{n\to \mathrm{\infty }}{lim}{2}^{n+1}f\left(\frac{x}{{2}^{n+1}}\right)-\underset{n\to \mathrm{\infty }}{lim}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)=0.$

This implies that $\mathrm{\Im }:X\to Y$ is an CJA mapping. To prove the uniqueness of the CJA mapping , assume that there exists another CJA mapping $\mathrm{\Re }:X\to Y$ which satisfies (3.3). Then we have

$\begin{array}{rcl}{\mu }_{\mathrm{\Im }\left(x\right)-\mathrm{\Re }\left(x\right)}\left(t\right)& =& \underset{n\to \mathrm{\infty }}{lim}{\mu }_{{2}^{n}\mathrm{\Im }\left(\frac{x}{{2}^{n}}\right)-{2}^{n}\mathrm{\Re }\left(\frac{x}{{2}^{n}}\right)}\left(t\right)\\ \ge & \underset{n\to \mathrm{\infty }}{lim}min\left\{{\mu }_{{2}^{n}\mathrm{\Im }\left(\frac{x}{{2}^{n}}\right)-{2}^{n}f\left(\frac{x}{{2}^{n}}\right)}\left(\frac{t}{2}\right),{\mu }_{{2}^{n}f\left(\frac{x}{{2}^{n}}\right)-{2}^{n}\mathrm{\Re }\left(\frac{x}{{2}^{n}}\right)}\left(\frac{t}{2}\right)\right\}\\ \ge & \underset{n\to \mathrm{\infty }}{lim}{\mu }_{\phi \left(\frac{x}{{2}^{n}},\frac{2x}{{2}^{n}},\frac{x}{{2}^{n}}\right)}^{\prime }\left(\frac{\left(1-2\alpha \right)t}{{2}^{n}}\right)\ge \underset{n\to \mathrm{\infty }}{lim}{\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(\frac{\left(1-2\alpha \right)t}{{2}^{n}{\alpha }^{n}}\right).\end{array}$

Since ${lim}_{n\to \mathrm{\infty }}{\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(\frac{\left(1-2\alpha \right)t}{{2}^{n}{\alpha }^{n}}\right)=1$. Therefore, we have ${\mu }_{\mathrm{\Im }\left(x\right)-\mathrm{\Re }\left(x\right)}\left(t\right)=1$, for all $t>0$, and so $\mathrm{\Im }\left(x\right)=\mathrm{\Re }\left(x\right)$. This completes the proof. □

Corollary 3.1 Let X be a real normed linear space, $\left(Z,{\mu }^{\prime },min\right)$ be an RN-space and $\left(Y,\mu ,min\right)$ be a complete RN-space. Let r be a positive real number with $r>1$, ${z}_{0}\in Z$ and $f:X\to Y$ be a mapping satisfying

${\mu }_{2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right)}\left(t\right)\ge {\mu }_{\left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}+{\parallel z\parallel }^{r}\right){z}_{0}}^{\prime }\left(t\right),$
(3.9)

for all $x,y\in X$ and $t>0$. Then the limit $\mathrm{\Im }\left(x\right)={lim}_{n\to \mathrm{\infty }}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)$ exists, for all $x\in X$, and defines a unique CJA mapping $\mathrm{\Im }:X\to Y$ such that

${\mu }_{f\left(x\right)-\mathrm{\Im }\left(x\right)}\left(t\right)\ge {\mu }_{{\parallel x\parallel }^{p}{z}_{0}}^{\prime }\left(\frac{\left({2}^{r}-2\right)t}{{2}^{r}+2}\right),$

for all $x\in X$ and $t>0$.

Proof Let $\alpha ={2}^{-r}$ and $\phi :{X}^{3}\to Z$ be a mapping defined by $\phi \left(x,y,z\right)=\left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}+{\parallel z\parallel }^{r}\right){z}_{0}$. Then, from Theorem 3.1, the conclusion follows. □

Theorem 3.2 Let X be a real linear space, $\left(Z,{\mu }^{\prime },min\right)$ be an RN-space and $\phi :{X}^{3}\to Z$ be a function such that there exists $0<\alpha <2$ such that ${\mu }_{\phi \left(2x,2y,2z\right)}^{\prime }\left(t\right)\ge {\mu }_{\alpha \phi \left(x,y,z\right)}^{\prime }\left(t\right)$, for all $x\in X$ and $t>0$, and

$\underset{n\to \mathrm{\infty }}{lim}{\mu }_{\phi \left({2}^{n}x,{2}^{n}y,{2}^{n}z\right)}^{\prime }\left({2}^{n}t\right)=1,$

for all $x,y,z\in X$ and $t>0$. Let $\left(Y,\mu ,min\right)$ be a complete RN-space. If $f:X\to Y$ be a mapping satisfying (3.2). Then the limit $\mathrm{\Im }\left(x\right)={lim}_{n\to \mathrm{\infty }}\frac{f\left({2}^{n}x\right)}{{2}^{n}}$ exists, for all $x\in X$, and defines a unique CJA mapping $\mathrm{\Im }:X\to Y$ such that

${\mu }_{f\left(x\right)-\mathrm{\Im }\left(x\right)}\left(t\right)\ge {\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(\left(2-\alpha \right)t\right),$
(3.10)

for all $x\in X$ and $t>0$.

Proof It follows from (3.4) that

${\mu }_{\frac{f\left(2x\right)}{2}-f\left(x\right)}\left(t\right)\ge {\mu }_{\phi \left(x,2x,x\right)}^{\prime }\left(2t\right).$
(3.11)

Replacing x by ${2}^{n}x$ in (3.11), we obtain

${\mu }_{\frac{f\left({2}^{n+1}x\right)}{{2}^{n+1}}-\frac{f\left({2}^{n}x\right)}{{2}^{n}}}\left(t\right)\ge {\mu }_{\phi \left({2}^{n}x,{2}^{n+1}x,{2}^{n}x\right)}^{\prime }\left({2}^{n+1}t\right)\ge {\mu }_{\phi \left(x,2x,x\right)}\left(\frac{{2}^{n+1}t}{{\alpha }^{n}}\right).$

The rest of the proof is similar to the proof of Theorem 3.1. □

Corollary 3.2 Let X be a real normed linear space, $\left(Z,{\mu }^{\prime },min\right)$ be an RN-space and $\left(Y,\mu ,min\right)$ be a complete RN-space. Let r be a positive real number with $0, ${z}_{0}\in Z$ and $f:X\to Y$ be a mapping satisfying (3.9). Then the limit $\mathrm{\Im }\left(x\right)={lim}_{n\to \mathrm{\infty }}\frac{f\left({2}^{n}x\right)}{{2}^{n}}$ exists, for all $x\in X$, and defines a unique CJA mapping $\mathrm{\Im }:X\to Y$ such that

${\mu }_{f\left(x\right)-\mathrm{\Im }\left(x\right)}\left(t\right)\ge {\mu }_{{\parallel x\parallel }^{p}{z}_{0}}^{\prime }\left(\frac{\left(2-{2}^{r}\right)t}{{2}^{r}+2}\right),$

for all $x\in X$ and $t>0$.

Proof Let $\alpha ={2}^{r}$ and $\phi :{X}^{3}\to Z$ be a mapping defined by $\phi \left(x,y,z\right)=\left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}+{\parallel z\parallel }^{r}\right){z}_{0}$. Then, from Theorem 3.2, the conclusion follows. □

Theorem 3.3 Let X be a linear space, $\left(Y,\mu ,{T}_{M}\right)$ be a complete RN-space and Φ be a mapping from ${X}^{3}$ to ${D}^{+}$ ($\mathrm{\Phi }\left(x,y,z\right)$ is denoted by ${\mathrm{\Phi }}_{x,y.z}$) such that there exists $0<\alpha <\frac{1}{2}$ such that

${\mathrm{\Phi }}_{2x,2y,2z}\left(t\right)\le {\mathrm{\Phi }}_{x,y,z}\left(\alpha t\right),$
(3.12)

for all $x,y,z\in X$ and $t>0$. Let $f:X\to Y$ be a mapping satisfying

${\mu }_{2f\left(\frac{x+y+z}{2}\right)f\left(x\right)-f\left(y\right)-f\left(z\right)}\left(t\right)\ge {\mathrm{\Phi }}_{x,y,z}\left(t\right),$
(3.13)

for all $x,y,z\in X$ and $t>0$. Then, for all $x\in X$, $\mathrm{\Im }\left(x\right):={lim}_{n\to \mathrm{\infty }}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)$ exists and $\mathrm{\Im }:X\to Y$ is a unique CJA mapping such that

${\mu }_{f\left(x\right)-\mathrm{\Im }\left(x\right)}\left(t\right)\ge {\mathrm{\Phi }}_{x,2x,x}\left(\frac{\left(1-2\alpha \right)t}{\alpha }\right),$
(3.14)

for all $x\in X$ and $t>0$.

Proof Putting $y=2x$ and $z=x$ in (3.13), we have

${\mu }_{2f\left(\frac{x}{2}\right)-f\left(x\right)}\left(t\right)\ge {\mathrm{\Phi }}_{x,2x,x}\left(\frac{t}{\alpha }\right),$
(3.15)

for all $x\in X$ and $t>0$. Consider the set $S:=\left\{g:X\to Y\right\}$ and the generalized metric d in S defined by

$d\left(f,g\right)=\underset{u\in \left(0,\mathrm{\infty }\right)}{inf}\left\{{\mu }_{g\left(x\right)-h\left(x\right)}\left(ut\right)\ge {\mathrm{\Phi }}_{x,2x,x}\left(t\right),\mathrm{\forall }x\in X,t>0\right\},$
(3.16)

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

$Jh\left(x\right):=2h\left(\frac{x}{2}\right),$
(3.17)

for all $x\in X$. First, we prove that J is a strictly contractive mapping with the Lipschitz constant 2α. In fact, let $g,h\in S$ be such that $d\left(g,h\right)<\epsilon$. Then we have ${\mu }_{g\left(x\right)-h\left(x\right)}\left(\epsilon t\right)\ge {\mathrm{\Phi }}_{x,2x,x}\left(t\right)$, for all $x\in X$ and $t>0$, and so

$\begin{array}{rcl}{\mu }_{Jg\left(x\right)-Jh\left(x\right)}\left(2\alpha \epsilon t\right)& =& {\mu }_{2g\left(\frac{x}{2}\right)-2h\left(\frac{x}{2}\right)}\left(2\alpha \epsilon t\right)={\mu }_{g\left(\frac{x}{2}\right)-h\left(\frac{x}{2}\right)}\left(\alpha \epsilon t\right)\\ \ge & {\mathrm{\Phi }}_{\frac{x}{2},x,\frac{x}{2}}\left(\alpha t\right)\\ \ge & {\mathrm{\Phi }}_{x,2x,x}\left(t\right),\end{array}$

for all $x\in X$ and $t>0$. Thus $d\left(g,h\right)<\epsilon$ implies that $d\left(Jg,Jh\right)<2\alpha \epsilon$. This means that $d\left(Jg,Jh\right)\le 2\alpha d\left(g,h\right)$, for all $g,h\in S$. It follows from (3.15) that $d\left(f,Jf\right)\le \alpha$. By Theorem 1.3, there exists a mapping $A:X\to Y$ satisfying the following:

1. (1)

is a fixed point of J, that is,

$\mathrm{\Im }\left(\frac{x}{2}\right)=\frac{1}{2}\mathrm{\Im }\left(x\right),$
(3.18)

for all $x\in X$. The mapping is a unique fixed point of J in the set $\mathrm{\Omega }=\left\{h\in S:d\left(g,h\right)<\mathrm{\infty }\right\}$. This implies that is a unique mapping satisfying (3.18) such that there exists $u\in \left(0,\mathrm{\infty }\right)$ satisfying ${\mu }_{f\left(x\right)-\mathrm{\Im }\left(x\right)}\left(ut\right)\ge {\mathrm{\Phi }}_{x,2x,x}\left(t\right)$, for all $x\in X$ and $t>0$.

1. (2)

$d\left({J}^{n}f,\mathrm{\Im }\right)\to 0$ as $n\to \mathrm{\infty }$. This implies the equality ${lim}_{n\to \mathrm{\infty }}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)=\mathrm{\Im }\left(x\right)$, for all $x\in X$.

2. (3)

$d\left(f,\mathrm{\Im }\right)\le \frac{d\left(f,Jf\right)}{1-2\alpha }$ with $f\in \mathrm{\Omega }$, which implies the inequality $d\left(f,\mathrm{\Im }\right)\le \frac{\alpha }{1-2\alpha }$ and so

${\mu }_{f\left(x\right)-\mathrm{\Im }\left(x\right)}\left(t\right)\ge {\mathrm{\Phi }}_{x,2x,x}\left(\frac{\left(1-2\alpha \right)t}{\alpha }\right),$

for all $x\in X$ and $t>0$. This implies that the inequality (3.14) holds. On the other hand

${\mu }_{{2}^{n+1}f\left(\frac{x+y+z}{{2}^{n+1}}\right)-{2}^{n}f\left(\frac{x}{{2}^{n}}\right)-{2}^{n}f\left(\frac{y}{{2}^{n}}\right)-{2}^{n}f\left(\frac{z}{{2}^{n}}\right)}\left(t\right)\ge {\mathrm{\Phi }}_{\frac{x}{{2}^{n}},\frac{y}{{2}^{n}},\frac{z}{{2}^{n}}}\left(\frac{t}{{2}^{n}}\right),$

for all $x,y,z\in X$, $t>0$ and $n\ge 1$. By (3.12), we know that ${\mathrm{\Phi }}_{\frac{x}{{2}^{n}},\frac{y}{{2}^{n}},\frac{z}{{2}^{n}}}\left(\frac{t}{{2}^{n}}\right)\ge {\mathrm{\Phi }}_{x,y,z}\left(\frac{t}{{\left(2\alpha \right)}^{n}}\right)$. Since ${lim}_{n\to \mathrm{\infty }}{\mathrm{\Phi }}_{x,y,z}\left(\frac{t}{{\left(2\alpha \right)}^{n}}\right)=1$, for all $x,y,z\in X$ and $t>0$, we have ${\mu }_{2\mathrm{\Im }\left(\frac{x+y+z}{2}\right)-\mathrm{\Im }\left(x\right)-\mathrm{\Im }\left(y\right)-\mathrm{\Im }\left(z\right)}\left(t\right)=1$, for all $x,y,z\in X$ and $t>0$. Thus the mapping $\mathrm{\Im }:X\to Y$ satisfying (1.1). Furthermore

$\begin{array}{rcl}\mathrm{\Im }\left(2x\right)-2\mathrm{\Im }\left(x\right)& =& \underset{n\to \mathrm{\infty }}{lim}{2}^{n}f\left(\frac{x}{{2}^{n-1}}\right)-2\underset{n\to \mathrm{\infty }}{lim}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)\\ =& 2\left[\underset{n\to \mathrm{\infty }}{lim}{2}^{n-1}f\left(\frac{x}{{2}^{n-1}}\right)-\underset{n\to \mathrm{\infty }}{lim}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)\right]\\ =& 0.\end{array}$

This completes the proof. □

Corollary 3.3 Let X be a real normed space, $\theta \ge 0$ and r be a real number with $r>1$. Let $f:X\to Y$ be a mapping satisfying

${\mu }_{2f\left(\frac{x+y+z}{2}\right)-f\left(x\right)-f\left(y\right)-f\left(z\right)}\left(t\right)\ge \frac{t}{t+\theta \left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}+{\parallel z\parallel }^{r}\right)},$
(3.19)

for all $x,y,z\in X$ and $t>0$. Then $\mathrm{\Im }\left(x\right)={lim}_{n\to \mathrm{\infty }}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)$ exists, for all $x\in X$, and $\mathrm{\Im }:X\to Y$ is a unique CJA mapping such that

${\mu }_{f\left(x\right)-\mathrm{\Im }\left(x\right)}\left(t\right)\ge \frac{\left({2}^{r}-2\right)t}{\left({2}^{r}-2\right)t+\left({2}^{r}+2\right)\theta {\parallel x\parallel }^{r}},$

for all $x\in X$ and $t>0$.

Proof The proof follows from Theorem 3.3 if we take ${\mathrm{\Phi }}_{x,y,z}\left(t\right)=\frac{t}{t+\theta \left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}+{\parallel z\parallel }^{r}\right)}$, for all $x,y,z\in X$ and $t>0$. In fact, if we choose $\alpha ={2}^{-r}$, then we get the desired result. □

Theorem 3.4 Let X be a linear space, $\left(Y,\mu ,{T}_{M}\right)$ be a complete RN-space and Φ be a mapping from ${X}^{3}$ to ${D}^{+}$ ($\mathrm{\Phi }\left(x,y,z\right)$ is denoted by ${\mathrm{\Phi }}_{x,y,z}$) such that for some $0<\alpha <2$, ${\mathrm{\Phi }}_{\frac{x}{2},\frac{y}{2},\frac{z}{2}}\left(t\right)\le {\mathrm{\Phi }}_{x,y,z}\left(\alpha t\right)$, for all $x,y,z\in X$ and $t>0$. Let $f:X\to Y$ be a mapping satisfying (3.13). Then the limit $\mathrm{\Im }\left(x\right):={lim}_{n\to \mathrm{\infty }}\frac{f\left({2}^{n}x\right)}{{2}^{n}}$ exists, for all $x\in X$, and $\mathrm{\Im }:X\to Y$ is a unique CJA mapping such that

${\mu }_{f\left(x\right)-\mathrm{\Im }\left(x\right)}\left(t\right)\ge {\mathrm{\Phi }}_{x,2x,x}\left(\left(2-\alpha \right)t\right),$
(3.20)

for all $x\in X$ and $t>0$.

Proof Putting $y=2x$ and $z=x$ in (3.13), we have

${\mu }_{\frac{f\left(2x\right)}{2}-f\left(x\right)}\left(t\right)\ge {\mathrm{\Phi }}_{x,2x,x}\left(2t\right),$
(3.21)

for all $x\in X$ and $t>0$. Let $\left(S,d\right)$ be the generalized metric space defined in the proof of Theorem 3.1. Now, we consider a linear mapping $J:\left(S,d\right)\to \left(S,d\right)$ such that $Jh\left(x\right):=\frac{1}{2}h\left(2x\right)$, for all $x\in X$. It follows from (3.21) that $d\left(f,Jf\right)\le \frac{1}{2}$. By Theorem 1.3, there exists a mapping $\mathrm{\Im }:X\to Y$ satisfying the following:

1. (1)

is a fixed point of J, that is,

$\mathrm{\Im }\left(2x\right)=2\mathrm{\Im }\left(x\right),$
(3.22)

for all $x\in X$. The mapping is a unique fixed point of J in the set $\mathrm{\Omega }=\left\{h\in S:d\left(g,h\right)<\mathrm{\infty }\right\}$. This implies that is a unique mapping satisfying (3.22) such that there exists $u\in \left(0,\mathrm{\infty }\right)$ satisfying ${\mu }_{f\left(x\right)-\mathrm{\Im }\left(x\right)}\left(ut\right)\ge {\mathrm{\Phi }}_{x,2x,x}\left(t\right)$, for all $x\in X$ and $t>0$.

1. (2)

$d\left({J}^{n}f,\mathrm{\Im }\right)\to 0$ as $n\to \mathrm{\infty }$. This implies the equality

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

for all $x\in X$.

1. (3)

$d\left(f,\mathrm{\Im }\right)\le \frac{d\left(f,Jf\right)}{1-\frac{\alpha }{2}}$ with $f\in \mathrm{\Omega }$, which implies the inequality ${\mu }_{f\left(x\right)-\mathrm{\Im }\left(x\right)}\left(\frac{t}{2-\alpha }\right)\ge {\mathrm{\Phi }}_{x,2x,x}\left(t\right)$, for all $x\in X$ and $t>0$. This implies that the inequality (3.20) holds. The rest of the proof is similar to the proof of Theorem 3.3. □

Corollary 3.4 Let X be a real normed space, $\theta \ge 0$ and r be a real number with $0. Let $f:X\to Y$ be a mapping satisfying (3.19). Then the limit $\mathrm{\Im }\left(x\right)={lim}_{n\to \mathrm{\infty }}\frac{f\left({2}^{n}x\right)}{{2}^{n}}$ exists, for all $x\in X$, and $\mathrm{\Im }:X\to Y$ is a unique CJA mapping such that

${\mu }_{f\left(x\right)-\mathrm{\Im }\left(x\right)}\left(t\right)\ge \frac{\left(2-{2}^{r}\right)t}{\left(2-{2}^{r}\right)t+\left({2}^{r}+2\right)\theta {\parallel x\parallel }^{r}},$

for all $x\in X$ and $t>0$.

Proof The proof follows from Theorem 3.4 if we take ${\mathrm{\Phi }}_{x,y,z}\left(t\right)=\frac{t}{t+\theta \left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}+{\parallel z\parallel }^{r}\right)}$, for all $x,y,z\in X$ and $t>0$. In fact, if we choose $\alpha ={2}^{r}$, then we get the desired result. □

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