Stability Estimates for an Arithmetic Functional Equation with Brzdȩk Fixed Point Approaches

Abstract

We introduce an arithmetic functional equation $f(x^2+y^2)=f\left(x^2\right)+f\left(y^2\right)$ and then investigate stability estimates of the functional equation by using the Brzdȩk fixed point theorem on a non-Archimedean fuzzy metric space and a non-Archimedean fuzzy normed space. To apply the Brzdȩk fixed point theorem, the proof uses the linear relationship between two variables, x and y.

1. Introduction

In 1940, Ulam [1] proposed the stability problem of a group homomorphism. In other words, the question would be generalized as “Under what conditions a mathematical object satisfying a certain property approximately must be close to an object satisfying the property exactly?”. In 1941, Hyers [2] gave the first, affirmative, and partial solution to Ulam’s question with an additive function (Cauchy function) in Banach spaces. The Hyers stability result was first generalized in the stability involving p-powers of norm by Aoki [3]. In 1978, Rassias [4] provided a generalization of Hyers’ theorem that allows the Cauchy difference to become unbounded. For the last few decades, stability problems of various functional equations have been extensively investigated and generalized by many mathematicians (see [5,6,7,8,9]).

Baker [10] introduced for the first time the Ulam’s type stability by using the fixed point method and later applied it in numerous papers (see [11,12,13,14,15,16,17]). In fact, fixed point theory is a powerful resource for the research, study and applications of nonlinear functional analysis, optimization theory, and variational inequalities (see [18,19,20,21,22]). Many authors introduced new types of fixed point theorems in different directions. Moreover, Brzdȩk and Ciepliński [23] introduced the existence theorem of the fixed point for nonlinear operators in metric spaces:

Theorem 1

([23]). Let X be a non-empty set, $(Y,d)$ be a complete metric space and $\Lambda :Y^X\to Y^X$ be a non-decreasing operator satisfying the hypothesis

$$\underset{n\to \infty }{lim}\Lambda {\delta }_n=0 for every sequence {\left\{{\delta }_n\right\}}_{n\in \mathbb{N}} with\underset{n\to \infty }{lim}{\delta }_n=0.$$

Suppose that $\mathcal{T}:Y^X\to Y^X$ is an operator satisfying the inequality

$$d(\mathcal{T}\epsilon \left(x\right),\mathcal{T}\mu \left(x\right))\le \Lambda (▵(\epsilon ,\mu ))\left(x\right),\epsilon ,\mu \in Y^X,x\in X$$

where $▵:{\left(Y^X\right)}^2\to {\mathbb{R}}_{+}^X$ is a mapping, which is defined by

$$▵(\epsilon ,\mu )\left(x\right):=d(\epsilon \left(x\right),\mu \left(x\right)),\epsilon ,\mu \in Y^X,x\in X.$$

If there exist functions $\epsilon :X\to {\mathbb{R}}_{+}$ and $\varphi :X\to Y$ such that

$$d\left(\mathcal{T}\varphi \right(x),\varphi (x\left)\right)\le \epsilon \left(x\right)$$

and

$${\epsilon }^{*}\left(x\right):=\sum _{n\in {\mathbb{N}}_0}\left({\Lambda }^n\epsilon \right)\left(x\right)<\infty$$

for all $x\in X,$ then the limit

$$\underset{n\to \infty }{lim}\left({\mathcal{T}}^n\varphi \right)\left(x\right)$$

exists for each $x\in X.$ Moreover, the function $\varphi \in Y^X$ defined by

$$\psi \left(x\right):=\underset{n\to \infty }{lim}\left({\mathcal{T}}^n\varphi \right)\left(x\right)$$

is a fixed point of $\mathcal{T}$ with

$$d(\varphi \left(x\right),\psi \left(x\right))\le {\epsilon }^{*}\left(x\right)$$

for all $x\in X.$

Brzdȩk and Ciepliński [23] used this result to prove the stability problem of functional equations in non-Archimedean metric spaces and obtained the fixed point results in arbitrary metric spaces. In particular, the Brzdȩk’s fixed point method was also obtained from Theorem 1 (see [24]).

Theorem 2

([24]). Let X be a non-empty set, $(Y,d)$ be a complete metric space and $f_1,f_2:X\to X$ be the given mappings. Suppose that $\mathcal{T}:Y^X\to Y^X$ and $\Lambda :{\mathbb{R}}_{+}^X\to {\mathbb{R}}_{+}^X$ are two operators satisfying the following conditions:

$$d(\mathcal{T}\xi \left(x\right),\mathcal{T}\mu \left(x\right))\le d(\xi \left(f_1\left(x\right)\right),\mu \left(f_1\left(x\right)\right))+d(\xi \left(f_2\left(x\right)\right),\mu \left(f_2\left(x\right)\right))$$

(1)

and

$$\Lambda \delta \left(x\right):=\delta \left(f_1\left(x\right)\right)+\delta \left(f_2\left(x\right)\right)$$

(2)

for all $\xi ,\mu \in Y^X,$$\delta \in {\mathbb{R}}_{+}^X$ and $x\in X.$ If there exist $\epsilon :X\to {\mathbb{R}}_{+}$ and $\varphi :X\to Y$ such that

$$d(\mathcal{T}\varphi \left(x\right),\varphi \left(x\right))\le \epsilon \left(x\right) and {\epsilon }^{*}\left(x\right):=\sum _{n=0}^{\infty }\left({\Lambda }^n\epsilon \right)\left(x\right)<\infty$$

(3)

for all $x\in X,$ then the limit ${lim}_{n\to \infty }\left({\mathcal{T}}^n\varphi \right)\left(x\right)$ exists for each $x\in X.$ Moreover, the function $\psi \left(x\right):={lim}_{n\to \infty }\left({\mathcal{T}}^n\varphi \right)\left(x\right)$ is a fixed point of $\mathcal{T}$ with

$$d(\varphi \left(x\right),\psi \left(x\right))\le {\epsilon }^{*}\left(x\right)$$

for all $x\in X.$

The theory of fuzzy sets was introduced by Zadeh [25] in 1965. George and Veeramani [26] in 1994, introduced a fuzzy metric space by considering points in the crisp set and a fuzzy distance between them. Mirmostafaee and Moslehian [27] introduced a non-Archimedean fuzzy norm on a linear space over a non-Archimedean field. Many mathematicians considered the fuzzy normed spaces in different branches of pure and applied mathematics. In particular, Moslehian and Rassias [28] studied the stability problem of functional equations in non-Archimedean spaces. Moreover, Aiemsomboon and Sintunavarat [29,30] studied the stability problem due to Brzdȩk’s fixed point theorem.

The purpose of this paper is to introduce an arithmetic functional equation(see [31]) of the following form:

$$f(x^2+y^2)=f\left(x^2\right)+f\left(y^2\right)$$

(4)

and also to investigate the stability problem by using the Brzdȩk’s fixed point theorem on a non-Archimedean fuzzy normed space. In fact, for each real number $c,$ a function $f\left(x\right)=cx$ satisfies the functional Equation (4). Chung [31] characterized the Equation (4) for all positive integers x and $y.$ In terms of the stability problem, it is also important whether the equation still remains true in the range of real numbers.

In this paper, $\mathbb{N},{\mathbb{N}}_0,\mathbb{R}$ and ${\mathbb{R}}_{+}$ denote the set of positive integers, the set of non-negative integers, the set of real numbers and the set of non-negative real numbers, respectively.

2. Stability of Arithmetic Functional Equations

In this section, we will investigate the stability problem for the arithmetic functional Equation (4) by using the Brzdȩk fixed point method; see Theorem 3. Before proceeding, we will first reproduce the following definitions due to Mirmostafaee and Moslehian [27] and George and Veeramani [26].

Definition 1.

A binary operation $\ast :[0,1]\times [0,1]\to [0,1]$ is a continuous t-norm if it satisfies the following conditions:

1.

∗ is associative and commutative;

2.

∗ is continuous;

3.

$a\ast 1=a$ for all $a\in [0,1]$;

4.

$a\ast b\le c\ast d$ whenever $a\le c$ and $b\le d,$ and $a,b,c,d\in [0,1].$

Obviously, $a\ast b=a\times b$ and $a\ast b=min\{a,b\}$ are common examples of continuous t-norms.

Definition 2.

Let X be an arbitrary set. A fuzzy subset M of $M\times M\times [0,\infty ]$ is called a fuzzy metric on X if it satisfies the following conditions for all $x,y\in X$ and $t\in \mathbb{R}.$

1.

If $t\le 0,$ then $M(x,y,t)=0$;

2.

For all $t>0,$$M(x,y,t)=1$ if and only if $x=y$;

3.

For all $t>0,$$M(x,y,t)=M(y,x,t)$;

4.

For all $s,t\in \mathbb{R},$$M(x,z,s+t)\ge M(x,y,s)\ast M(y,z,t)$;

5.

$M(x,y,·)$ is a non-decreasing function on $\mathbb{R}$ and ${lim}_{t\to \infty }M(x,y,t)=1.$

The pair $(X,M,\ast )$ is called a fuzzy metric space. If we replace $\left(4\right)$ by

6.

$M(x,z,max\{s,t\left\}\right)\ge M(x,y,s)\ast M(y,z,t),$

then we call the triple $(X,M,\ast )$ a non-Archimedean fuzzy metric space.

Definition 3.

Let V be a real linear space. A function $N:X\times [0,\infty ]\to [0,1]$ is said to be a fuzzy norm on X if, for all $x,y\in X$ and all $t,s>0,$ if it satisfies the following conditions:

1.

$N(x,0)=0$;

2.

$x=0$ if and only if $N(x,t)=1$ for all $t>0$;

3.

$N(kx,t)=N(x,\frac{t}{\left|k\right|})$ for all $k\in \mathbb{R},k\ne 0$;

4.

$N(x+y,s+t)\ge N(x,t)\ast N(y,s)$;

5.

$N(x,·)$ is a non decreasing function on $\mathbb{R}$ and ${lim}_{t\to \infty }N(x,t)=1.$

The pair $(X,N,\ast )$ is called a fuzzy normed space. If we replace $\left(4\right)$ by

6.

$N(x+y,max\{s,t\left\}\right)\ge N(x,s)\ast N(y,t),$

then we call the triple $(X,N,\ast )$ a non-Archimedean fuzzy normed space.

Throughout this paper, we assume that

$$a\ast b=min\{a,b\}$$

for the continuous t-norm.

Example 1.

Let $X=[0,\infty )$ be a metric space with the usual metric d and the usual norm $\left|\right|·\left|\right|.$ Let $a\ast b\le a\times b$ for all $a,b\in [0,1].$ For each $t\in (0,\infty ),$ define $M:X\times X\times [0,\infty ]\to [0,1]$ by

$$M(x,y,t)=e^{-\frac{d(x,y)}{t}}$$

for all $x,y\in X$ and $t>0.$ Then, $(X,M,\ast )$ is a non-Archimedean fuzzy metric space.

Additionally, for each $t\in (0,\infty ),$ define $N:X\times [0,\infty ]\to [0,1]$ by

$$N(x,t)=e^{-\frac{\left|\right|x\left|\right|}{t}}$$

for all $x\in X$ and $t>0.$ Then, $(X,N,\ast )$ is a non-Archimedean fuzzy normed space.

There are some more examples such as: $M(x,y,t)={(1-e^{-t})}^{d(x,y)},M(x,y,t)={\left(\frac{t}{t+1}\right)}^{d(x,y)}.$ It is easy to see that each $(X,M,\ast )$ is a non-Archimedean fuzzy metric space.

Now, we will investigate stability estimates for an arithmetic functional equation by using Brzdȩk’s on a non-Archimedean fuzzy metric space and non-Archimedean fuzzy normed space.

Theorem 3.

Let $({\mathbb{R}}_{+},M,\ast )$ be a non-Archimedean fuzzy metric space, which is invariant (i.e., $M(x+z,y+z,t)=M(x,y,t)$ for $x,y,z\in {\mathbb{R}}_{+}$ and $t\in \mathbb{R}$), and $(\mathbb{R},N,\ast )$ be a non-Archimedean fuzzy normed space. Let $h:{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ be a function such that

$$L_0:=\{m\in \mathbb{N}|s\left(m^2\right)+s(1+m^2)<1\}\ne ⌀,$$

(5)

where

$$s\left(m\right):=inf\{t\in {\mathbb{R}}_{+}|h\left(mx\right)\le th\left(x\right)forallx\in {\mathbb{R}}_{+}\}.$$

Assume that

$$N\left(h\right(nx)+h(my),s)\ge N\left(s\right(n\left)h\right(x)+s(m\left)h\right(y),s),$$

(6)

for all $x,y\in {\mathbb{R}}_{+},$$n,m\in \mathbb{N}$ and $s>0.$ Suppose a function $f:{\mathbb{R}}_{+}\to \mathbb{R}$ satisfies the inequality

$$M(f(x^2+y^2),f\left(x^2\right)+f\left(y^2\right),t)\ge N(h\left(x\right)+h\left(y\right),t)$$

(7)

for all $x,y\in {\mathbb{R}}_{+}$ and $t>0.$ Then, there exists a unique solution $T:{\mathbb{R}}_{+}\to \mathbb{R}$ to the Equation (4) such that

$$M(f\left(x\right),T\left(x\right),t)\ge N(s_{0}h\left(x\right),t)$$

(8)

for $x\in {\mathbb{R}}_{+}$ and $t>0,$ where

$$s_0:=inf\left\{\frac{1+s\left(m\right)}{1-s\left(m^2\right)-s(1+m^2)}\right|m\in L_0\}.$$

(9)

Proof. 

Let $m\in L_0.$ By letting $y=mx$ in the inequality (7), the inequality (6) implies the following inequality:

$$M\left(f\left((1+m^2)x^2\right),f\left(x^2\right)+f\left(m^2{x}^2\right),t\right)\ge N(c_m\left(x\right),t)$$

(10)

for $t>0,$ where $c_m\left(x\right):=(1+s\left(m\right))h\left(x\right)$ for $x\in {\mathbb{R}}_{+}.$ To apply to the Brzdȩk fixed point method, we need to define two operators as in Theorem 2:

1.

${\mathcal{T}}_m:{\mathbb{R}}^{{\mathbb{R}}_{+}}\to {\mathbb{R}}^{{\mathbb{R}}_{+}}$ by

$${\mathcal{T}}_m\xi \left(x\right):=\xi \left((1+m^2)x\right)-\xi \left(m^{2}x\right)$$

(11)

2.

${\Lambda }_m:{\mathbb{R}}_{+}^{{\mathbb{R}}_{+}}\to {\mathbb{R}}_{+}^{{\mathbb{R}}_{+}}$ by

$${\Lambda }_m\mu \left(x\right):=\mu \left((1+m^2)x\right)+\mu \left(m^{2}x\right)$$

(12)

for all $x\in {\mathbb{R}}_{+}$ and $\xi \in {\mathbb{R}}^{{\mathbb{R}}_{+}},\mu \in {\mathbb{R}}_{+}^{{\mathbb{R}}_{+}}.$ Additionally, we have

$$\begin{array}{ccc}& & M({\mathcal{T}}_m\xi \left(x\right),{\mathcal{T}}_m\mu \left(x\right),t)\hfill \\ & =& M(\xi \left((1+m^2)x\right)-\xi \left(m^{2}x\right),\mu \left((1+m^2)x\right)-\mu \left(m^{2}x\right),t)\hfill \\ & \ge & min\{M(\xi \left((1+m^2)x\right)-\xi \left(m^{2}x\right),\mu \left((1+m^2)x\right)-\xi \left(m^{2}x\right),t),\hfill \\ & & M(\mu \left((1+m^2)x\right)-\xi \left(m^{2}x\right),\mu \left((1+m^2)x\right)-\mu \left(m^{2}x\right),t)\}\hfill \\ & =& min\{M(\xi \left((1+m^2)x\right),\mu \left((1+m^2)x\right),t),M(\xi \left(m^{2}x\right),\mu \left(m^{2}x\right),t)\}\hfill \end{array}$$

for $x\in {\mathbb{R}}_{+}$ and $t>0.$ If we may let $f_1\left(x\right)=(1+m^2)x$ and $f_2\left(x\right)=m^{2}x,$ then for each $m\in L_0,$ two operators ${\mathcal{T}}_m$ and ${\Lambda }_m$ satisfy the inequalities (1) and (2) in Theorem 2. To check the condition (3) in Theorem 2, we note that the inequality (10) implies

$$\begin{array}{ccc}\hfill M({\mathcal{T}}_{m}f\left(x^2\right),f\left(x^2\right),t)& =& M(f\left((1+m^2)x^2\right)-f\left(m^2{x}^2\right),f\left(x^2\right),t)\hfill \\ & =& M(f\left((1+m^2)x^2\right),f\left(x^2\right)+f\left(m^2{x}^2\right),t)\hfill \\ & \ge & N\left(\right(1+s\left(m\right)\left)h\right(x),t)=N\left(\right(1+s\left(m\right)\left)h\right(x),t)\hfill \end{array}$$

for $x\in {\mathbb{R}}_{+}$ and $t>0.$ Also, we note that

$$\begin{array}{ccc}\hfill {\Lambda }_m{c}_m\left(x\right)& =& (1+s\left(m\right))(h\left((1+m^2)x\right)+h\left(m^{2}x\right))\hfill \\ & \le & (1+s\left(m\right))(s(1+m^2)+s\left(m^2\right))h\left(x\right)\hfill \end{array}$$

for all $x\in {\mathbb{R}}_{+}.$ Using the mathematical induction, for each $k\in \mathbb{N},$ we obtain

$${\Lambda }_m^k{c}_m\left(x\right)=(1+s\left(m\right)){[s(1+m^2)+s\left(m^2\right)]}^{k}h\left(x\right)$$

for all $x\in {\mathbb{R}}_{+}$ and $m\in L_0.$ For each $m\in L_0$ and $x\in {\mathbb{R}}_{+},$ we have the following condition:

$$c_m^{\ast }\left(x\right):=\sum _{j=0}^{\infty }{\Lambda }_m^j{c}_m\left(x\right)=\frac{1+s\left(m\right)}{1-s(1+m^2)-s\left(m^2\right)}h\left(x\right)$$

where ${\Lambda }_m^0{c}_m\left(x\right)=c_m\left(x\right).$

Hence, the Brzdȩk fixed point theorem implies that

$$T_m\left(x\right):=\underset{k\to \infty }{lim}{\mathcal{T}}_m^{k}f\left(x\right)$$

exists for each $m\in L_0$ and $x\in {\mathbb{R}}_{+},$ and we have

$$M(f\left(x\right),T_m\left(x\right),t)\ge N(c_m^{\ast }\left(x\right),t)$$

for all $m\in L_0,$$x\in {\mathbb{R}}_{+}$ and $t>0$ (see Theorem 2). Now, we will prove that $T_m$ satisfies the Equation (4) for each $m\in L_0.$ Hence, we may conclude that the solution of the Equation (4) is uniquely determined. First, we will inductively check that for $k\in {\mathbb{N}}_0$

$$\begin{array}{ccc}& & M({\mathcal{T}}_m^{k}f(x^2+y^2),{\mathcal{T}}_m^{k}f\left(x^2\right)+{\mathcal{T}}_m^{k}f\left(y^2\right),t)\hfill \\ & & \ge N\left({(s(1+m^2)+s\left(m^2\right))}^k(h\left(x\right)+h\left(y\right)),t\right)\hfill \end{array}$$

for $x\in {\mathbb{R}}_{+}$ and $t>0.$ If $k=0,$ we note that ${\mathcal{T}}_m^{0}f\left(x^2\right)=f\left(x^2\right)$ and hence the base case follows from the inequality (10). Then,

$$\begin{array}{cccc}& M({\mathcal{T}}_m^{k+1}f(x^2+y^2),{\mathcal{T}}_m^{k+1}f\left(x^2\right)+{\mathcal{T}}_m^{k+1}f\left(y^2\right),t)\hfill & & \\ & =M({\mathcal{T}}_m^{k}f\left((1+m^2)(x^2+y^2)\right)-{\mathcal{T}}_m^{k}f\left(m^2(x^2+y^2)\right),\hfill & & \\ & {\mathcal{T}}_m^{k}f\left((1+m^2)x^2\right)-{\mathcal{T}}_m^{k}f\left(m^2{x}^2\right)+{\mathcal{T}}_m^{k}f\left((1+m^2)y^2\right)-{\mathcal{T}}_m^{k}f\left(m^2{y}^2\right),t)\hfill & & \\ & \ge min\{M({\mathcal{T}}_m^{k}f\left((1+m^2)(x^2+y^2)\right),{\mathcal{T}}_m^{k}f\left((1+m^2)x^2\right)+{\mathcal{T}}_m^{k}f\left((1+m^2)y^2\right),t),\hfill & & \\ & M({\mathcal{T}}_m^{k}f\left(m^2(x^2+y^2)\right),{\mathcal{T}}_m^{k}f\left(m^2{x}^2\right)+{\mathcal{T}}_m^{k}f\left(m^2{y}^2\right),t)\}\hfill & & \\ & \ge min\{N(s{(1+m^2)}^k(h\left((1+m^2)x\right)+h\left((1+m^2)y\right)),t),\hfill & & \\ & N(s{\left(m^2\right)}^k(h\left(m^{2}x\right)+h\left(m^{2}y\right)),t)\}\hfill & & \\ & \ge min\left\{N\left(h\left(x\right)+h\left(y\right),\frac{t}{s{(1+m^2)}^{k+1}}\right),N\left(h\left(x\right)+h\left(y\right),\frac{t}{s{\left(m^2\right)}^{k+1}}\right)\right\}\hfill & \end{array}$$

We note that

$$0<s\left(m^2\right)<s(1+m^2)+s\left(m^2\right)<1and0<s(1+m^2)<s(1+m^2)+s\left(m^2\right)<1.$$

Since $N(x,·)$ is a non-decreasing, we have

$$\begin{array}{ccc}& & M({\mathcal{T}}_m^{k+1}f(x^2+y^2),{\mathcal{T}}_m^{k+1}f\left(x^2\right)+{\mathcal{T}}_m^{k+1}f\left(y^2\right),t)\hfill \\ & & \ge N\left(h\left(x\right)+h\left(y\right),\frac{1}{{(s(1+m^2)+s\left(m^2\right))}^{k+1}}t\right)\hfill \end{array}$$

for $t>0.$ As $k\to \infty ,$ we have

$$N\left(h\left(x\right)+h\left(y\right),\frac{1}{{(s(1+m^2)+s\left(m^2\right))}^{k+1}}t\right)\to 1,$$

where $0<s(1+m^2)+s\left(m^2\right)<1.$

Hence, we obtain

$$T_m(x^2+y^2)=T_m\left(x^2\right)+T_m\left(y^2\right)$$

(13)

for all $x,y\in {\mathbb{R}}_{+}$ and $m\in L_0.$ That is, for each $m\in L_0,$$T_m$ is a solution of the Equation (4). Now, assume that a mapping $T:{\mathbb{R}}_{+}\to \mathbb{R}$ satisfies the Equation (4) such that

$$M\left(f\right(x),T(x),t)\ge N\left(Lh\right(x),t)$$

for $L>0$ constant. Let

$$T(x^2+y^2)=T\left(x^2\right)+T\left(y^2\right)$$

(14)

for all $x,y\in {\mathbb{R}}_{+}.$ Then, we will show that $T=T_m$ for each $m\in L_0.$ By letting $y=mx$ in the inequality (14), we have

$$T\left(x^2\right)=T\left((1+m^2)x^2\right)-T\left(m^2{x}^2\right)$$

for $x\in {\mathbb{R}}_{+}$ and $t>0.$ Let $m_0\in L_0.$ Then,

$$\begin{array}{ccc}& & M(T\left(x^2\right),T_{m_0}\left(x^2\right),t)\ge min\{M(T\left(x^2\right),f\left(x^2\right),t),M(f\left(x^2\right),T_{m_0}\left(x^2\right),t)\}\hfill \\ & & \ge min\left\{N(Lh\left(x\right),t),N\left(\frac{1+s\left(m_0\right)}{1-s(1+m_0^2)-s\left(m_0^2\right)}h\left(x\right),t\right)\right\}.\hfill \end{array}$$

By letting $S_0=((1+s\left(m_0\right))+(1-s(1+m_0^2)-s\left(m_0^2\right))L),$ we have

$$\frac{1+s\left(m_0\right)}{1-s(1+m_0^2)-s\left(m_0^2\right)}+L=S_0·\sum _{k=0}^{\infty }{(s(1+m_0^2)+s\left(m_0^2\right))}^k$$

Since $N(s,·)$ is non-decreasing, we note

$$\begin{array}{ccc}& & min\left\{N(Lh\left(x\right),t),N\left(\frac{1+s\left(m_0\right)}{1-s(1+m_0^2)-s\left(m_0^2\right)}h\left(x\right),t\right)\right\}\hfill \\ & & \ge N\left(h\left(x\right),\frac{1}{\frac{1+s\left(m_0\right)}{1-s(1+m_0^2)-s\left(m_0^2\right)}+L}t\right)\hfill \\ & & =N\left(S_0·\sum _{k=0}^{\infty }{(s(1+m_0^2)+s\left(m_0^2\right))}^{k}h\left(x\right),t\right)\hfill \end{array}$$

Hence, we obtain

$$M(T\left(x^2\right),T_{m_0}\left(x^2\right),t)\ge N\left(S_0·\sum _{k=0}^{\infty }{(s(1+m_0^2)+s\left(m_0^2\right))}^{k}h\left(x\right),t\right)$$

(15)

for $x\in {\mathbb{R}}_{+}$ and $t>0.$ For $l\in {\mathbb{N}}_0,$ assume that

$$M(T\left(x^2\right),T_{m_0}\left(x^2\right),t)\ge N\left(S_0·\sum _{k=l}^{\infty }{(s(1+m_0^2)+s\left(m_0^2\right))}^{k}h\left(x\right),t\right)$$

for $x\in {\mathbb{R}}_{+}$ and $t>0.$ We will check it by using mathematical induction on $l.$ If $l=0,$ it follows from the inequality (15). Then,

$$\begin{array}{cccc}& M(T\left(x^2\right),T_{m_0}\left(x^2\right),t)\hfill & & \\ & =M(T\left((1+m_0^2)x^2\right)-T\left(m_0^2{x}^2\right),T_{m_0}\left((1+m_0^2)x^2\right)-T_{m_0}\left(m_0^2{x}^2\right),t)\hfill & & \\ & \ge min\{M(T\left((1+m_0^2)x^2\right),T_{m_0}\left((1+m_0^2)x^2\right),t),M(T\left(m_0^2{x}^2\right),T_{m_0}\left(m_0^2{x}^2\right),t)\}\hfill & & \\ & \ge min\{N\left(s(1+m_0^2)S_{0}h\left(x\right)·\sum _{k=l}^{\infty }{(s(1+m_0^2)+s\left(m_0^2\right))}^k,t\right),\hfill & & \\ & N\left(s\left(m_0^2\right)S_{0}h\left(x\right)·\sum _{k=l}^{\infty }{(s(1+m_0^2)+s\left(m_0^2\right))}^k,t\right)\}\hfill & & \\ & \ge N(S_{0}h\left(x\right)·\sum _{k=l+1}^{\infty }{(s(1+m_0^2)+s\left(m_0^2\right))}^k,t)\hfill & \end{array}$$

for $x\in {\mathbb{R}}_{+}$ and $t>0.$ Hence, it holds whenever $l\in {\mathbb{N}}_0.$

As $l\to \infty ,$ we have

$$N\left(S_{0}h\left(x\right)·\sum _{k=l+1}^{\infty }{(s(1+m_0^2)+s\left(m_0^2\right))}^k,t\right)\to 1$$

Hence, we have $T=T_{m_0},$ for $m_0\in L_0.$ Thus, for each $m_0\in L_0,$

$$T_m=T_{m_0}$$

as desired. □

Example 2.

Let $p<0$ be a real number and a function $h:{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ is defined by

$$h\left(x\right)=A{\left(x\right)}^p,x\in {\mathbb{R}}_{+}$$

where the map $A:{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ is additive. Then, it is easily seen that the set $L_0$ is not empty as in Theorem 3. Hence, the Equation (5) is valid.

Corollary 1.

Let $h:{\mathbb{R}}_{+}\to (0,\infty )$ be a mapping such that

$$\underset{n\to \infty }{lim}{\inf \sup }_{x\in {\mathbb{R}}_{+}}\frac{h\left((1+n^2)x\right)+h\left(n^{2}x\right)+h\left(nx\right)}{h\left(x\right)}=0$$

(16)

Suppose $f:{\mathbb{R}}_{+}\to \mathbb{R}$ satisfies

$$M(f(x^2+y^2),f\left(x^2\right)+f\left(y^2\right),t)\ge N(h\left(x\right)+h\left(y\right),t)$$

(17)

for all $x,y\in {\mathbb{R}}_{+}$ and $t>0.$ Then, there exists a unique arithmetic functional equation $T:{\mathbb{R}}_{+}\to \mathbb{R}$ such that

$$M\left(f\right(x),T(x),t)\ge N\left(h\right(x),t)$$

(18)

for all $x\in {\mathbb{R}}_{+}$ and $t>0.$

Proof. 

For each $n\in \mathbb{N},$ let

$$a_n={\sup }_{x\in {\mathbb{R}}_{+}}\frac{h\left((1+n^2)x\right)+h\left(n^{2}x\right)+h\left(nx\right)}{h\left(x\right)}.$$

By the definition $s\left(n\right)$ as in Theorem 3, we will see that

$$s(1+n^2)={\sup }_{x\in {\mathbb{R}}_{+}}\frac{h\left((1+n^2)x\right)}{h\left(x\right)}\le a_n,$$

$$s\left(n^2\right)={\sup }_{x\in {\mathbb{R}}_{+}}\frac{h\left(n^{2}x\right)}{h\left(x\right)}\le a_n$$

and

$$s\left(n\right)={\sup }_{x\in {\mathbb{R}}_{+}}\frac{h\left(nx\right)}{h\left(x\right)}\le a_n$$

These inequalities imply that

$$s(1+n^2)+s\left(n^2\right)+s\left(n\right)\le 3a_n$$

(19)

for all $x\in {\mathbb{R}}_{+}.$ By our assumption, the sequence $\left\{a_n\right\}$ has a subsequence $\left\{a_{n_k}\right\}$ such that ${lim}_{k\to \infty }{a}_{n_k}=0,$ that is,

$$\underset{k\to \infty }{lim}{\sup }_{x\in {\mathbb{R}}_{+}}\frac{h\left((1+{n_k}^2)x\right)+h\left({n_k}^{2}x\right)+h\left(n_{k}x\right)}{h\left(x\right)}=0.$$

(20)

The inequalities (19) and (20) imply that

$$\underset{k\to \infty }{lim}s(1+{n_k}^2)+s\left({n_k}^2\right)+s\left(n_k\right)=0,$$

that is, ${lim}_{k\to \infty }s(1+{n_k}^2)=0,$${lim}_{k\to \infty }s\left({n_k}^2\right)=0$ and ${lim}_{k\to \infty }s\left(n_k\right)=0.$ Thus, we have

$$\underset{k\to \infty }{lim}\frac{1+s\left(n_k\right)}{1-s(1+{n_k}^2)-s\left({n_k}^2\right)}=1.$$

On letting $s_0=1$ as in Theorem 3, the inequality (18) follows from the inequality (8). □

Remark 1.

From the main result of the stability estimates in the Brzdȩk fixed point method, the method requires the non-Archimedean fuzzy metric has the invariant property, i.e., $M(x+z,y+z,t)=M(x,y,t)$ for all $x,y,z\in {\mathbb{R}}_{+}$ and $t\in \mathbb{R}.$ In fact, this property is not required in the different stability methods. Additionally, the use of $y=mx$ in the Brzdȩk fixed point method should be remarked. This linear relationship between two variables x and y makes it possible to prove the result and hence obtains the very nice stability approach. Some fixed point approaches required strictly contractive mapping and scaling processes. One of the main purposes of this paper is whether the Brzdȩk fixed point method can be applied in various spaces such as a non-Archimedean fuzzy normed space. We would like to propose open problems: (1) Can the Brzdȩk fixed point method be applied in various fuzzy normed spaces? (2) Can the Brzdȩk fixed point method be applied without the linear relationship between two variables x and y?