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 [1] in 1940. In 1941, Hyers [2] gave a positive answer to the above question for additive groups under the assumption that the groups are Banach spaces. Aoki [3] proved a generalization of Hyers’ theorem for additive mappings and Rassias [4] 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 $\parallel f(x+y)-f(x)-f(y)\parallel \le \epsilon ({\parallel x\parallel }^p+{\parallel y\parallel }^p)$, for all $x,y\in E$, where ε and p are constants with $\epsilon >0$ and $0\le p<1$. Then the limit $L(x)={lim}_{n\to \mathrm{\infty }}\frac{f(2^{n}x)}{2^n}$ exists, for all $x\in E$, and $L:E\to E^{\prime }$ is the unique additive mapping which satisfies

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

for all $x\in E$. Also, if for each $x\in E$ the function $f(tx)$ 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 [5] by replacing the bound $\epsilon ({\parallel x\parallel }^p+{\parallel y\parallel }^p)$ by a general control function $\phi (x,y)$.

In 1983, a generalized Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof [6] for mappings $f:X\to Y$, where X is a normed space and Y is a Banach space. In 1984, Cholewa [7] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group and, in 2002, Czerwik [8] proved the generalized Hyers-Ulam stability of the quadratic functional equation. The readers are referred to [9–29] and references therein for detailed information on stability of functional equations.

In 1897, Hensel [30] has introduced a normed space which does not have the Archimedean property. It turned out that non-Archimedean spaces have many nice applications (see [31–35]).

Definition 1.1 By a non-Archimedean field we mean a field equipped with a function (valuation) $|\cdot |:\mathbb{K}\to [0,\mathrm{\infty })$ 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) $|r+s|\le max\{|r|,|s|\}$.

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) 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, $\parallel x+y\parallel \le max\{\parallel x\parallel ,\parallel y\parallel \}$, for all $x,y\in X$.

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

Note that $\parallel x_n-x_m\parallel \le max\{\parallel x_{j+1}-x_j\parallel :m\le j\le n-1\}$, for all $m,n\in \mathbb{N}$ with $n>m$.

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

• (a) A sequence $\{x_n\}$ is a Cauchy sequence in X if $\{x_{n+1}-x_n\}$ converges to zero in X.

• (b) The non-Archimedean normed space $(X,\parallel \cdot \parallel )$ 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 $x<ny$.

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 ℚ. The completion of 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 $|a_k|\le p-1$. 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 [36]. Throughout this paper, let △⁺ denote the set of all probability distribution functions $F:\mathbb{R}\cup [-\mathrm{\infty },+\mathrm{\infty }]\to [0,1]$ such that F is left-continuous and nondecreasing on ℝ and $F(0)=0$, $F(+\mathrm{\infty })=1$. It is clear that the set $D^{+}=\{F\in {△}^{+}:l^{-}F(-\mathrm{\infty })=1\}$, where $l^{-}f(x)={lim}_{t\to x^{-}}f(t)$, 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 $F(t)\le G(t)$, for all $t\in \mathbb{R}$. For any $a\ge 0$, the element $H_a(t)$ of $D^{+}$ is defined by

$$H_a(t)=\{\begin{array}{cc}0,\hfill & \text{if }t\le a,\hfill \\ 1,\hfill & \text{if }t>a.\hfill \end{array}$$

We can easily show that the maximal element in △⁺ is the distribution function $H_0(t)$.

Definition 1.4 A function $T:{[0,1]}^2\to [0,1]$ 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) $T(x,1)=x$, for all $x\in [0,1]$; (d) $T(x,y)\le T(z,w)$ whenever $x\le z$ and $y\le w$, for all $x,y,z,w\in [0,1]$.

Three typical examples of continuous t-norms are as follows: $T_P(x,y)=xy$, $T_{\max }(x,y)=max\{a+b-1,0\}$, $T_M(x,y)=min(a,b)$. Recall that, if T is a t-norm and $\{x_n\}$ is a sequence in $[0,1]$, then $T_{i=1}^n{x}_i$ is defined recursively by $T_{i=1}^1{x}_1=x_1$ and $T_{i=1}^n{x}_i=T(T_{i=1}^{n-1}{x}_i,x_n)$, 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 $(X,\mu ,T)$, 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:

• (a) ${\mu }_x(t)=H_0(t)$, for all $t>0$ if and only if $x=0$;

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

• (c) ${\mu }_{x+y}(t+s)\ge T({\mu }_x(t),{\mu }_y(s))$, for all $x,y\in X$ and $t,s\ge 0$.

Every normed space $(X,\parallel \cdot \parallel )$ defines a random normed space $(X,\mu ,T_M)$, where ${\mu }_u(t)=\frac{t}{t+\parallel u\parallel }$, 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 ${sup}_{0<a<1}T(a,a)=1$, then every RN-space $(X,\mu ,T)$ is a metrizable linear topological space with the topology τ (called the μ-topology or the $(\epsilon ,\delta )$-topology, where $\epsilon >0$ and $\lambda \in (0,1)$) induced by the base $\{U(\epsilon ,\lambda )\}$ of neighborhoods of θ, where

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

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

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

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

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

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

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

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

• (b) $d(x,y)=d(y,x)$, for all $x,y\in X$;

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

Theorem 1.3 ([37, 38])

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) $d(y,y^{\ast })\le \frac{d(y,Jy)}{1-L}$, 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:

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

(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 $\phi :X^3\to [0,\mathrm{\infty })$ be a function such that there exists an $\alpha <1$ with

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

(2.1)

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

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

(2.2)

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

$${\parallel f(x)-\mathrm{\Im }(x)\parallel }_Y\le \alpha \phi (x,2x,x){(|2|-|2|\alpha )}^{-1},$$

(2.3)

for all $x\in X$.

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

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

(2.4)

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

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

for all $x\in X$, where, as usual, $inf\varphi =+\mathrm{\infty }$. It is easy to show that $(S,d)$ is complete (see [39]). Now we consider the linear mapping $J:S\to S$ such that $Jg(x):=2g(\frac{x}{2})$, for all $x\in X$. Let $g,h\in S$ be given such that $d(g,h)=\epsilon$. Then ${\parallel g(x)-h(x)\parallel }_Y\le \epsilon \phi (x,2x,x)$, for all $x\in X$. Hence

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

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

• (1) ℑ is a fixed point of J, i.e.,

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

  (2.5)

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

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

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

  (2.6)

for all $x\in X$;

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

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

for all $x,y,z\in X$. So $2\mathrm{\Im }(\frac{x+y+z}{2})=\mathrm{\Im }(x)+\mathrm{\Im }(y)+\mathrm{\Im }(z)$, 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 $0<r<1$. Let $f:X\to Y$ be a mapping satisfying

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

(2.7)

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

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

for all $x\in X$.

Proof The proof follows from Theorem 2.1 by taking $\phi (x,y,z)=({\parallel x\parallel }^r+{\parallel y\parallel }^r+{\parallel z\parallel }^r)$, 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 $\phi :X^3\to [0,\mathrm{\infty })$ be a function such that there exists an $\alpha <1$ with $\phi (x,y,z)\le |2|\alpha \phi (\frac{x}{2},\frac{y}{2},\frac{z}{2})$, 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

$${\parallel f(x)-\mathrm{\Im }(x)\parallel }_Y\le \phi (x,2x,x){(|2|-|2|\alpha )}^{-1},$$

(2.8)

for all $x\in X$.

Proof Let $(S,d)$ 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 $Jg(x):=\frac{g(2x)}{2}$, for all $x\in X$. Let $g,h\in S$ be given such that $d(g,h)=\epsilon$. Then ${\parallel g(x)-h(x)\parallel }_Y\le \epsilon \phi (x,2x,x)$, for all $x\in X$. Hence

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

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

• (1) ℑ is a fixed point of J, i.e.,

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

  (2.9)

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

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

• (3) $d(f,\mathrm{\Im })\le \frac{d(f,Jf)}{1-\alpha }$, which implies the inequality $d(f,\mathrm{\Im })\le {(|2|-|2|\alpha )}^{-1}$. 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

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

for all $x\in X$.

Proof The proof follows from Theorem 2.2 by taking $\phi (x,y,z)=({\parallel x\parallel }^r+{\parallel y\parallel }^r+{\parallel z\parallel }^r)$, 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 $\lambda :G^3\to [0,+\mathrm{\infty })$ be a function such that

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

(2.10)

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

$$\text{£}(x)=\underset{n\to \mathrm{\infty }}{lim}\underset{0\le k<n}{max}|2{|}^k\lambda (\frac{x}{2^{k+1}},\frac{x}{2^k},\frac{x}{2^{k+1}})$$

(2.11)

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

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

(2.12)

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

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

(2.13)

Moreover, if ${lim}_{j\to \mathrm{\infty }}{lim}_{n\to \mathrm{\infty }}{max}_{j\le k<n+j}|2{|}^k\lambda (\frac{x}{2^{k+1}},\frac{x}{2^k},\frac{x}{2^{k+1}})=0$ then ℑ is the unique CJA mapping satisfying (2.13).

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

$${\parallel f(2x)-2f(x)\parallel }_Y\le \lambda (x,2x,x),$$

(2.14)

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

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

(2.15)

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

$$\parallel 2^{n}f\left(\frac{x}{2^n}\right)-f(x)\parallel \le \underset{0\le k<n}{max}|2{|}^k\lambda (\frac{x}{2^{k+1}},\frac{x}{2^k},\frac{x}{2^{k+1}}),$$

(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

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

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

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

(2.17)

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

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

$$\begin{array}{rcl}{\parallel \mathrm{\Im }(x)-\mathrm{\Re }(x)\parallel }_X& =& \underset{n\to \mathrm{\infty }}{lim}|2{|}^n{\parallel \mathrm{\Im }\left(\frac{x}{2^n}\right)-\mathrm{\Re }\left(\frac{x}{2^n}\right)\parallel }_X\\ \le & \underset{k\to \mathrm{\infty }}{lim}|2{|}^{n}max\{{\parallel \mathrm{\Im }\left(\frac{x}{2^n}\right)-f\left(\frac{x}{2^n}\right)\parallel }_X,{\parallel f\left(\frac{x}{2^n}\right)-\mathrm{\Re }\left(\frac{x}{2^n}\right)\parallel }_X\}\\ \le & \underset{j\to \mathrm{\infty }}{lim}\underset{n\to \mathrm{\infty }}{lim}\underset{j\le k<n+j}{max}|2{|}^k\lambda (\frac{x}{2^{k+1}},\frac{x}{2^k},\frac{x}{2^{k+1}})=0,\end{array}$$

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

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

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

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

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

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

$$\underset{j\to \mathrm{\infty }}{lim}\underset{n\to \mathrm{\infty }}{lim}\underset{j\le k<n+j}{max}|2{|}^k\lambda (\frac{x}{2^{k+1}},\frac{x}{2^k},\frac{x}{2^{k+1}})=\underset{j\to \mathrm{\infty }}{lim}|2{|}^j\lambda (\frac{x}{2^{j+1}},\frac{x}{2^j},\frac{x}{2^{j+1}})=0.$$

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 $\lambda :G^3\to [0,+\mathrm{\infty })$ be a function such that ${lim}_{n\to \mathrm{\infty }}\frac{\lambda (2^{n}x,2^{n}y,2^{n}z)}{2^n}=0$, for all $x,y,z\in G$. Suppose that, for any $x\in G$, the limit

$$\text{£}(x)=\underset{n\to \mathrm{\infty }}{lim}\underset{0\le k<n}{max}\frac{\lambda (2^{k}x,2^{k+1}x,2^{k}x)}{{|2|}^k}$$

(2.18)

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

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

(2.19)

for all $x\in G$. Moreover, if ${lim}_{j\to \mathrm{\infty }}{lim}_{n\to \mathrm{\infty }}{max}_{j\le k<n+j}\frac{\lambda (2^{k}x,2^{k+1}x,2^{k}x)}{{|2|}^k}=0$, then ℑ is the unique CJA mapping satisfying (2.19).

Proof It follows from (2.14) that

$${\parallel f(x)-\frac{f(2x)}{2}\parallel }_X\le \frac{\lambda (x,2x,x)}{|2|},$$

(2.20)

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

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

(2.21)

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

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

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

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

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

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

Proof If we define $\lambda :G^3\to [0,\mathrm{\infty })$ by $\lambda (x,y,z):=\kappa (\xi (|x|)\cdot \xi (|y|)\cdot \xi (|z|))$ 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, $(Z,{\mu }^{\prime },min)$ be an RN-space and $\phi :X^3\to Z$ be a function such that there exists $0<\alpha <\frac{1}{2}$ such that

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

(3.1)

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

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

(3.2)

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

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

(3.3)

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

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

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

(3.4)

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

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

(3.5)

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

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

and so

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

This implies that

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

(3.6)

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

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

(3.7)

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

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

(3.8)

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

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

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

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

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

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

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

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

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

Corollary 3.1 Let X be a real normed linear space, $(Z,{\mu }^{\prime },min)$ be an RN-space and $(Y,\mu ,min)$ 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

$${\mu }_{2f(\frac{x+y+z}{2})-f(x)-f(y)-f(z)}(t)\ge {\mu }_{({\parallel x\parallel }^r+{\parallel y\parallel }^r+{\parallel z\parallel }^r)z_0}^{\prime }(t),$$

(3.9)

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

$${\mu }_{f(x)-\mathrm{\Im }(x)}(t)\ge {\mu }_{{\parallel x\parallel }^p{z}_0}^{\prime }\left(\frac{(2^r-2)t}{2^r+2}\right),$$

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

Proof Let $\alpha =2^{-r}$ and $\phi :X^3\to Z$ be a mapping defined by $\phi (x,y,z)=({\parallel x\parallel }^r+{\parallel y\parallel }^r+{\parallel z\parallel }^r)z_0$. Then, from Theorem 3.1, the conclusion follows. □

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

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

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

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

(3.10)

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

Proof It follows from (3.4) that

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

(3.11)

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

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

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, $(Z,{\mu }^{\prime },min)$ be an RN-space and $(Y,\mu ,min)$ be a complete RN-space. Let r be a positive real number with $0<r<1$, $z_0\in Z$ and $f:X\to Y$ be a mapping satisfying (3.9). Then the limit $\mathrm{\Im }(x)={lim}_{n\to \mathrm{\infty }}\frac{f(2^{n}x)}{2^n}$ exists, for all $x\in X$, and defines a unique CJA mapping $\mathrm{\Im }:X\to Y$ such that

$${\mu }_{f(x)-\mathrm{\Im }(x)}(t)\ge {\mu }_{{\parallel x\parallel }^p{z}_0}^{\prime }\left(\frac{(2-2^r)t}{2^r+2}\right),$$

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

Proof Let $\alpha =2^r$ and $\phi :X^3\to Z$ be a mapping defined by $\phi (x,y,z)=({\parallel x\parallel }^r+{\parallel y\parallel }^r+{\parallel z\parallel }^r)z_0$. Then, from Theorem 3.2, the conclusion follows. □

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

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

(3.12)

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

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

(3.13)

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

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

(3.14)

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

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

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

(3.15)

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

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

(3.16)

where $inf\mathrm{\varnothing }=+\mathrm{\infty }$. It is easy to show that $(S,d)$ is complete (see [39], Lemma 2.1). Now, we consider a linear mapping $J:(S,d)\to (S,d)$ such that

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

(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 $d(g,h)<\epsilon$. Then we have ${\mu }_{g(x)-h(x)}(\epsilon t)\ge {\mathrm{\Phi }}_{x,2x,x}(t)$, for all $x\in X$ and $t>0$, and so

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

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

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

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

  (3.18)

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

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

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

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

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

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

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

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

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

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

(3.19)

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

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

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

Proof The proof follows from Theorem 3.3 if we take ${\mathrm{\Phi }}_{x,y,z}(t)=\frac{t}{t+\theta ({\parallel x\parallel }^r+{\parallel y\parallel }^r+{\parallel z\parallel }^r)}$, 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, $(Y,\mu ,T_M)$ be a complete RN-space and Φ be a mapping from $X^3$ to $D^{+}$ ($\mathrm{\Phi }(x,y,z)$ is denoted by ${\mathrm{\Phi }}_{x,y,z}$) such that for some $0<\alpha <2$, ${\mathrm{\Phi }}_{\frac{x}{2},\frac{y}{2},\frac{z}{2}}(t)\le {\mathrm{\Phi }}_{x,y,z}(\alpha t)$, for all $x,y,z\in X$ and $t>0$. Let $f:X\to Y$ be a mapping satisfying (3.13). Then the limit $\mathrm{\Im }(x):={lim}_{n\to \mathrm{\infty }}\frac{f(2^{n}x)}{2^n}$ exists, for all $x\in X$, and $\mathrm{\Im }:X\to Y$ is a unique CJA mapping such that

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

(3.20)

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

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

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

(3.21)

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

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

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

  (3.22)

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

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

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

for all $x\in X$.

• (3) $d(f,\mathrm{\Im })\le \frac{d(f,Jf)}{1-\frac{\alpha }{2}}$ with $f\in \mathrm{\Omega }$, which implies the inequality ${\mu }_{f(x)-\mathrm{\Im }(x)}(\frac{t}{2-\alpha })\ge {\mathrm{\Phi }}_{x,2x,x}(t)$, 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 $0<r<1$. Let $f:X\to Y$ be a mapping satisfying (3.19). Then the limit $\mathrm{\Im }(x)={lim}_{n\to \mathrm{\infty }}\frac{f(2^{n}x)}{2^n}$ exists, for all $x\in X$, and $\mathrm{\Im }:X\to Y$ is a unique CJA mapping such that

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

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

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