Hyers–Ulam–Rassias Stability of Reciprocal-Type Functional Equations: Comparative Study of Direct and Fixed Point Methods

Abstract

In this paper, we investigate the Hyers–Ulam–Rassias stability of reciprocal functional equations in non-Archimedean fuzzy normed spaces by using both the direct method and the fixed point alternative. In addition, we study a modified reciprocal type functional equation within the same framework using Brzdȩk’s fixed point method. A brief remark is provided on the incidental role of symmetry in the structure of such functional equations. Finally, a comparative analysis highlights the distinctive features, strengths, and limitations of each approach.

1. Introduction

The stability problem for functional equations, a classical topic in functional analysis, concerns conditions under which a function that approximately satisfies an equation is close to an exact solution. This notion was first posed by Ulam [1] in 1940 for group homomorphisms. Hyers [2] provided the first partial solution in 1941 for additive functions in Banach spaces, establishing what is now known as Hyers–Ulam stability. Subsequent extensions include Aoki’s treatment of p-power norms [3], Rassias’ allowance of unbounded Cauchy differences [4], and Gâvruta’s use of general control functions [5], leading to what is now referred to as Hyers–Ulam–Rassias stability.

Definition 1

(Hyers–Ulam–Rassias Stability). Let X and Y be normed vector spaces and let $f:X\to Y$ be a mapping. We say that the functional equation $\mathcal{F}(f,x,y,\cdots )=0$ is Hyers–Ulam–Rassias stable if the following holds: If there exists a control function $\varphi :X\times X\times \cdots \to [0,\infty )$ such that

$$\left|\right|\mathcal{F}(f,x,y,\cdots )\left|\right|\le \varphi (x,y,\cdots )for all x,y,\cdots \in X,$$

then there exists an exact solution $T:X\to Y$ of the functional equation such that

$$\left|\right|f\left(x\right)-T\left(x\right)\left|\right|\le \psi \left(x\right)for all x\in X,$$

where $\psi :X\to [0,\infty )$ is a function that depends on $\varphi .$

Over time, stability theory has developed through distinct methodological approaches. One classical approach is the direct method, first investigated by Hyers [2] and later extended by Rassias [4]. This approach is constructive in nature: starting from a function that approximately satisfies a given functional equation, it derives exact solutions by explicitly generating sequences that converge to the desired solution.

In contrast, the fixed point method, systematically advanced by Radu [6] and Cădariu and Radu [7], reformulates stability problems as fixed point problems in suitable function spaces. By applying general fixed point theorems—such as the Banach fixed point theorem or the fixed point alternative—one can guarantee the existence and uniqueness of exact solutions. Brzdȩk and Ciepliński [8] further refined this method in non-Archimedean spaces, thereby establishing a powerful framework for generalized settings. More recently, Koh [9] applied Brzdȩk’s fixed point method to obtain stability estimates for an arithmetic functional equation, demonstrating the versatility of this approach within the broader stability theory.

Beyond these methodological advances, stability theory has also expanded to generalized structures. Fuzzy sets, introduced by Zadeh [10], and non-Archimedean fuzzy norms, developed by Mirmostafaee and Moslehian [11] and George and Veeramani [12], provide natural settings for studying stability in contexts involving uncertainty or non-Archimedean valuations. In this paper, we adopt a slightly modified framework by defining fuzzy metrics and norms on the set of positive real numbers ${\mathbb{R}}_{+}$, thereby aligning the analytic setting with the structure of the functional equations under consideration.

Definition 2

(Non-Archimedean Fuzzy Metric). Let X be a non-empty set. A non-Archimedean fuzzy metric on X is a 3-tuple $(X,M,\ast ),$ where $M:X\times X\times {\mathbb{R}}_{+}\to [0,1]$ is a fuzzy metric and ∗ is a continuous t-norm, which is the minimum operator: $a\ast b=min(a,b)$ for all $a,b\in [0,1].$ The function M satisfies the following conditions for all $x,y,zinX$ and $s,t\in {\mathbb{R}}_{+}$:

 (1) 

$M(x,y,t)=1$ if and only if $x=y$;

 (2) 

$M(x,y,t)=M(y,x,t)$;

 (3) 

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

 (4) 

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

Definition 3

(Non-Archimedean Fuzzy Norm). Let X be a real linear space. A function $N:X\times {\mathbb{R}}_{+}\to [0,1]$ is said to be a non-Archimedean fuzzy norm on $X$ if for all $x,y\in X,$$t,s\in {\mathbb{R}}_{+},$ and $k\in \mathbb{R},k\ne 0,$ the following conditions hold:

 (1) 

$x=0$ if and only if $N(x,t)=1$ for all $t\in {\mathbb{R}}_{+}$;

 (2) 

$N(kx,t)=N(x,{\displaystyle \frac{t}{\left|k\right|}})$;

 (3) 

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

 (4) 

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

 (5) 

for $x\ne 0,$$N(x,·)$ is continuous on ${\mathbb{R}}_{+}.$

In this paper, we focus on the reciprocal functional equation

$$f(x+y)={\displaystyle \frac{f\left(x\right)f\left(y\right)}{f\left(x\right)+f\left(y\right)}},$$

(1)

where $f:X\to Y$ is defined on the set of nonzero real numbers. Reciprocal-type functional equations inherently exhibit symmetric structures, as the variables x and y play interchangeable roles in the functional relation. For each real number $c,$ the reciprocal function $f\left(x\right)={\displaystyle \frac{c}{x}}$ is a solution of the functional Equation (1) on the space of non-zero real numbers. Such equations arise naturally in various applications. For instance, in physics, the concept of the reduced mass $\mu$ of a two-body system with masses $m_1$ and $m_2$ is given by

$${\displaystyle \frac{1}{\mu }}={\displaystyle \frac{1}{m_1}}+{\displaystyle \frac{1}{m_2}},\mathrm{equivalently},\mu ={\displaystyle \frac{m_1{m}_2}{m_1+m_2}}.$$

A similar structure appears in electrical engineering when computing the equivalent resistance of resistors connected in parallel. The stability of this equation has been investigated using several approaches. For instance, Jung [13] successfully applied the fixed point alternative method to establish its stability. A significant body of work by Ravi and his collaborators has explored its stability using the direct method, extending the results to reciprocal difference and adjoint functional equations [14] and to a multi-variable setting [15,16].

The aim of this paper is to study the Hyers–Ulam–Rassias stability of Equation (1) in non-Archimedean fuzzy normed spaces using both the direct method and the fixed point alternative method. Furthermore, we will study the stability of a slightly modified reciprocal-type equation via Brzdȩk’s fixed point method. Our primary goal is to compare and analyze the distinctive features and effectiveness of these stability methods. To do so, we first need to clarify the definition of Hyers–Ulam–Rassias stability in our context.

Definition 4

(Hyers–Ulam–Rassias Stability). Let X be a linear vector space, and $(Y,N_Y)$ and $(Z,N_Z)$ be non-Archimedean fuzzy norm spaces. Let $f:X\to Y$ be a mapping. We say that the functional equation $\mathcal{F}(f,x,y,\cdots )=0$ is Hyers–Ulam–Rassias stable if the following holds: If there exists a control function $\varphi :X\times X\times \cdots \to Z$ such that

$$N_Y(\mathcal{F}(f,x,y,\cdots ),t)\ge N_Z(\varphi (x,y,\cdots ),t)for all x,y,\cdots \in Xand t\in {\mathbb{R}}_{+},$$

then there exists an exact solution $T:X\to Y$ of the functional equation such that

$$N_Y(f\left(x\right)-T\left(x\right),t)\ge N_Z(\psi \left(x\right),t)for all x\in Xand t\in {\mathbb{R}}_{+}$$

for some function $\psi :X\to Z$ that depends on $\varphi .$

Throughout this paper, let $N:\mathbb{R}\times {\mathbb{R}}_{+}\to [0,1]$ be a non-Archimedean fuzzy norm on the real vector space $\mathbb{R}$. We restrict the domain of this norm to ${\mathbb{R}}_{+}$ and denote the resulting structure by $({\mathbb{R}}_{+},N)$. This allows us to perform all proofs using the same norm N for both spaces.

Moreover, we denote ${\mathbb{R}}^{*}=\mathbb{R}\setminus \left\{0\right\}$. For a mapping $f:{\mathbb{R}}^{*}\to \mathbb{R}$, we define

$$Df(x,y)=f(x+y)-{\displaystyle \frac{f\left(x\right)f\left(y\right)}{f\left(x\right)+f\left(y\right)}}$$

(2)

for all $x,y\in {\mathbb{R}}^{*}.$

2. Direct Method

In this section, we will investigate the Hyers–Ulam–Rassias stability of reciprocal functional equations in non-Archimedean fuzzy normed spaces using the direct method.

Theorem 1.

Let $\varphi :{\mathbb{R}}^{*}\times {\mathbb{R}}^{*}\to {\mathbb{R}}_{+}$ be a function and let γ satisfy ${\displaystyle \frac{1}{2}}<\gamma <1$. Assume that

$$N\left({\displaystyle \frac{1}{2}}\varphi \left({\displaystyle \frac{x}{2}},{\displaystyle \frac{y}{2}}\right),t\right)\ge N\left(\gamma \varphi (x,y),t\right)$$

(3)

for all $x,y\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$. Suppose that a mapping $f:{\mathbb{R}}^{*}\to \mathbb{R}$ satisfies

$$N\left(Df(x,y),t\right)\ge N\left(\varphi (x,y),t\right)$$

(4)

for all $x,y\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$. Then the limit

$$R\left(x\right):=\underset{n\to \infty }{lim}{\displaystyle \frac{1}{2^n}}f\left({\displaystyle \frac{x}{2^n}}\right)$$

exists for all $x\in {\mathbb{R}}^{*}$, and the mapping $R:{\mathbb{R}}^{*}\to \mathbb{R}$ is the unique reciprocal mapping satisfying

$$N\left(f\left(x\right)-R\left(x\right),t\right)\ge N\left(\varphi (x,x),{\displaystyle \frac{1-\gamma }{2\gamma }}t\right)$$

(5)

for all $x\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$.

Proof. 

Letting $x={\displaystyle \frac{1}{2}}x$ and $y={\displaystyle \frac{1}{2}}x$ in inequality (4), it follows from inequality (3) that

$$N\left(f\left(x\right)-{\displaystyle \frac{1}{2}}f\left({\displaystyle \frac{1}{2}}x\right),t\right)\ge N\left(\varphi \left({\displaystyle \frac{1}{2}}x,{\displaystyle \frac{1}{2}}x\right),t\right)\ge N\left(\varphi (x,x),{\displaystyle \frac{1}{2\gamma }}t\right)$$

(6)

for all $x\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$. Replacing x by ${\displaystyle \frac{1}{2^n}}x$ in inequality (6), we obtain

$$N\left(f\left({\displaystyle \frac{1}{2^n}}x\right)-{\displaystyle \frac{1}{2}}f\left({\displaystyle \frac{1}{2^{n+1}}}x\right),t\right)\ge N\left(\varphi \left({\displaystyle \frac{1}{2^n}}x,{\displaystyle \frac{1}{2^n}}x\right),{\displaystyle \frac{1}{2\gamma }}t\right),$$

which can be rewritten as

$$N\left({\displaystyle \frac{1}{2^n}}f\left({\displaystyle \frac{1}{2^n}}x\right)-{\displaystyle \frac{1}{2^{n+1}}}f\left({\displaystyle \frac{1}{2^{n+1}}}x\right),{\displaystyle \frac{1}{2^n}}t\right)\ge N\left(\varphi (x,x),{\displaystyle \frac{1}{{\left(2\gamma \right)}^{n+1}}}t\right)$$

(7)

for all non-negative integers n, $x\in {\mathbb{R}}^{*}$, and $t\in {\mathbb{R}}_{+}$. Now, replacing t by ${\left(2\gamma \right)}^{n+1}t$ in inequality (7), we arrive at

$$N\left({\displaystyle \frac{1}{2^n}}f\left({\displaystyle \frac{1}{2^n}}x\right)-{\displaystyle \frac{1}{2^{n+1}}}f\left({\displaystyle \frac{1}{2^{n+1}}}x\right),2{\gamma }^{n+1}t\right)\ge N\left(\varphi (x,x),t\right)$$

(8)

for all $x\in {\mathbb{R}}^{*}$, $t\in {\mathbb{R}}_{+}$, and $n\ge 0$. Inequality (8) implies

$$\begin{array}{cc}\hfill & N\left(f\left(x\right)-{\displaystyle \frac{1}{2^n}}f\left({\displaystyle \frac{1}{2^n}}x\right),2\sum _{k=0}^{n-1}{\gamma }^{k+1}t\right)\hfill \\ \hfill & =N\left(\sum _{k=0}^{n-1}\left[{\displaystyle \frac{1}{2^k}}f\left({\displaystyle \frac{1}{2^k}}x\right)-{\displaystyle \frac{1}{2^{k+1}}}f\left({\displaystyle \frac{1}{2^{k+1}}}x\right)\right],2\sum _{k=0}^{n-1}{\gamma }^{k+1}t\right)\hfill \\ \hfill & \ge N\left(\sum _{k=0}^{n-1}\left[{\displaystyle \frac{1}{2^k}}f\left({\displaystyle \frac{1}{2^k}}x\right)-{\displaystyle \frac{1}{2^{k+1}}}f\left({\displaystyle \frac{1}{2^{k+1}}}x\right)\right],\underset{0\le k\le n-1}{max}\left\{2{\gamma }^{k+1}t\right\}\right)\hfill \\ \hfill & \ge \underset{0\le k\le n-1}{min}\left\{N\left({\displaystyle \frac{1}{2^k}}f\left({\displaystyle \frac{1}{2^k}}x\right)-{\displaystyle \frac{1}{2^{k+1}}}f\left({\displaystyle \frac{1}{2^{k+1}}}x\right),2{\gamma }^{k+1}t\right)\right\}\hfill \\ \hfill & \ge N\left(\varphi (x,x),t\right),\hfill \end{array}$$

which yields

$$N\left(f\left(x\right)-{\displaystyle \frac{1}{2^n}}f\left({\displaystyle \frac{1}{2^n}}x\right),2\sum _{k=0}^{n-1}{\gamma }^{k+1}t\right)\ge N\left(\varphi (x,x),t\right)$$

(9)

for all $x\in {\mathbb{R}}^{*}$, $n\ge 1$, and $t\in {\mathbb{R}}_{+}$.

Replacing x by ${\displaystyle \frac{1}{2^m}}x$ and dividing the left-hand side by ${\displaystyle \frac{1}{2^m}}$ in inequality (9), we obtain

$$N\left({\displaystyle \frac{1}{2^m}}f\left({\displaystyle \frac{1}{2^m}}x\right)-{\displaystyle \frac{1}{2^{n+m}}}f\left({\displaystyle \frac{1}{2^{n+m}}}x\right),2\sum _{k=0}^{n-1}{\gamma }^{k+1}{\displaystyle \frac{1}{2^m}}t\right)\ge N\left(\varphi (x,x),{\displaystyle \frac{1}{{\left(2\gamma \right)}^m}}t\right)$$

for all $x\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$.

Finally, replacing t by ${\left(2\gamma \right)}^{m}t$ in the last inequality gives

$$N\left({\displaystyle \frac{1}{2^m}}f\left({\displaystyle \frac{1}{2^m}}x\right)-{\displaystyle \frac{1}{2^{n+m}}}f\left({\displaystyle \frac{1}{2^{n+m}}}x\right),2\sum _{k=m}^{n+m-1}{\gamma }^{k+1}t\right)\ge N\left(\varphi (x,x),t\right)$$

for all $n,m\ge 0$, $x\in {\mathbb{R}}^{*}$, and $t\in {\mathbb{R}}_{+}$. The previous inequality yields

$$N\left({\displaystyle \frac{1}{2^m}}f\left({\displaystyle \frac{1}{2^m}}x\right)-{\displaystyle \frac{1}{2^{n+m}}}f\left({\displaystyle \frac{x}{2^{n+m}}}\right),t\right)\ge N\left(\varphi (x,x),{\displaystyle \frac{1}{2{\sum }_{k=m}^{n+m-1}{\gamma }^{k+1}}}t\right)$$

(10)

for all $n,m\ge 0$, $x\in {\mathbb{R}}^{*}$, and $t\in {\mathbb{R}}_{+}$.

Since ${\sum }_{k=0}^{\infty }{\gamma }^k<\infty$, it follows from inequality (10) and property (4) of Definition 3 that the sequence

$$\left\{{\displaystyle \frac{1}{2^n}}f\left({\displaystyle \frac{x}{2^n}}\right)\right\}$$

is Cauchy in the fuzzy Banach space $(\mathbb{R},N)$ for each $x\in {\mathbb{R}}^{*}$. Consequently, this sequence converges to a point $R\left(x\right)\in \mathbb{R}$ for each $x\in {\mathbb{R}}^{*}$. Thus, we define a mapping $R:{\mathbb{R}}^{*}\to \mathbb{R}$ by

$$R\left(x\right)=\underset{n\to \infty }{lim}{\displaystyle \frac{1}{2^n}}f\left({\displaystyle \frac{x}{2^n}}\right),x\in {\mathbb{R}}^{*}.$$

Moreover, taking $m=0$ in inequality (10) gives

$$\begin{array}{cc}\hfill N\left(f\right(x)-R(x),max\{t,s\left\}\right)& \ge min\left\{N\left(f\left(x\right)-{\displaystyle \frac{1}{2^n}}f\left({\displaystyle \frac{x}{2^n}}\right),t\right),N\left(R\left(x\right)-{\displaystyle \frac{1}{2^n}}f\left({\displaystyle \frac{x}{2^n}}\right),s\right)\right\}\hfill \\ \hfill & \ge min\left\{N\left(\varphi (x,x),{\displaystyle \frac{1}{2{\sum }_{k=0}^{n-1}{\gamma }^{k+1}}}t\right),N\left(R\left(x\right)-{\displaystyle \frac{1}{2^n}}f\left({\displaystyle \frac{x}{2^n}}\right),s\right)\right\}\hfill \end{array}$$

for all $x\in {\mathbb{R}}^{*}$, $s,t\in {\mathbb{R}}_{+}$, and integers $n\ge 1$. By taking $n\to \infty$ in the previous inequality, we obtain

$$N(f\left(x\right)-R\left(x\right),max\{t,s\})\ge N\left(\varphi (x,x),{\displaystyle \frac{1-\gamma }{2\gamma }}t\right)$$

for all $x\in {\mathbb{R}}^{*}$ and $s,t\in {\mathbb{R}}_{+}$. Now, taking $s\to 0$, we recover inequality (5). Next, we show that the mapping R is the unique reciprocal mapping on ${\mathbb{R}}^{*}$. For all $x\in {\mathbb{R}}^{*}$, $t\in {\mathbb{R}}_{+}$, and integers $n\ge 1$, we have

$$\begin{array}{ccc}\hfill N\left(DR\right(x,y),t)& \ge & min\left\{N\left(DR(x,y)-{\displaystyle \frac{1}{2^n}}Df\left({\displaystyle \frac{x}{2^n}},{\displaystyle \frac{y}{2^n}}\right),t\right),N\left({\displaystyle \frac{1}{2^n}}Df\left({\displaystyle \frac{x}{2^n}},{\displaystyle \frac{y}{2^n}}\right),t\right)\right\}\hfill \\ & \ge & min\left\{N\left(DR(x,y)-{\displaystyle \frac{1}{2^n}}Df\left({\displaystyle \frac{x}{2^n}},{\displaystyle \frac{y}{2^n}}\right),t\right),N\left(\varphi \left({\displaystyle \frac{x}{2^n}},{\displaystyle \frac{x}{2^n}}\right),2^{n}t\right)\right\}.\hfill \end{array}$$

Since inequality (3) holds and ${lim}_{n\to \infty }{\displaystyle \frac{1}{{\gamma }^n}}=\infty$, it follows that

$$N\left(\varphi \left({\displaystyle \frac{x}{2^n}},{\displaystyle \frac{x}{2^n}}\right),2^{n}t\right)\ge N\left(\varphi (x,x),{\displaystyle \frac{1}{{\gamma }^n}}t\right)=1$$

(11)

for all $x,y\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$. Also, since

$$\underset{n\to \infty }{lim}N\left(DR(x,y)-{\displaystyle \frac{1}{2^n}}Df\left({\displaystyle \frac{x}{2^n}},{\displaystyle \frac{y}{2^n}}\right),t\right)=1,$$

it follows that

$$N\left(DR\right(x,y),t)=1$$

for all $x,y\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$. Hence, $DR(x,y)=0$, that is, the mapping R is a reciprocal mapping on ${\mathbb{R}}^{*}$. We now prove the uniqueness of R. Assume that $S:{\mathbb{R}}^{*}\to \mathbb{R}$ is another reciprocal mapping. It is easy to check that

$$R\left(x\right)={\displaystyle \frac{1}{2^n}}R\left({\displaystyle \frac{x}{2^n}}\right)\mathrm{and}S\left(x\right)={\displaystyle \frac{1}{2^n}}S\left({\displaystyle \frac{x}{2^n}}\right)$$

for all $x\in {\mathbb{R}}^{*}$ and integers $n\ge 0$. Hence,

$$\begin{array}{ccc}\hfill N\left(R\right(x)-S(x),t)& =& N\left({\displaystyle \frac{1}{2^n}}R\left({\displaystyle \frac{x}{2^n}}\right)-{\displaystyle \frac{1}{2^n}}S\left({\displaystyle \frac{x}{2^n}}\right),t\right)\hfill \\ & \ge & min\left\{N\left({\displaystyle \frac{1}{2^n}}R\left({\displaystyle \frac{x}{2^n}}\right)-{\displaystyle \frac{1}{2^n}}f\left({\displaystyle \frac{x}{2^n}}\right),t\right),N\left({\displaystyle \frac{1}{2^n}}S\left({\displaystyle \frac{x}{2^n}}\right)-{\displaystyle \frac{1}{2^n}}f\left({\displaystyle \frac{x}{2^n}}\right),t\right)\right\}\hfill \\ & \ge & N\left(\varphi \left({\displaystyle \frac{x}{2^n}},{\displaystyle \frac{x}{2^n}}\right),2^{n}t\right)\hfill \\ & \ge & N\left(\varphi (x,x),{\displaystyle \frac{1}{{\gamma }^n}}t\right)\hfill \end{array}$$

for all $x\in {\mathbb{R}}^{*}$, $t\in {\mathbb{R}}_{+}$, and integers $n\ge 0$. From Equation (11), it follows that

$$N\left(R\right(x)-S(x),t)=1$$

for all $x\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$. Thus, $R\left(x\right)=S\left(x\right)$ for all $x\in {\mathbb{R}}^{*}$. Therefore, the reciprocal mapping R is unique, as desired. □

3. Fixed Point Alternative Method

We will first present a generalized metric and the theorems of the fixed point alternative in a generalized metric space [17].

Definition 5.

Let X be a set. A function $d:X\times X\to [0,\infty ]$ is called a generalized metric on X if d satisfies

 (1) 

$d(x,y)=0$ if and only if $x=y;$

 (2) 

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

 (3) 

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

Theorem 2

(The alternative of fixed point [17,18]). Suppose that we are given a complete generalized metric space $(X,d)$ and a strictly contractive mapping $T:X\to X$ with Lipschitz constant $L<1,$ that is,

$$d(Tx,Ty)\le Ld(x,y)forallx,y\in X.$$

Then for each given $x\in X,$ either

$$d(T^{n}x,T^{n+1}x)=\infty foralln\ge 0$$

or

$$d(T^{n}x,T^{n+1}x)<\infty foralln\ge n_0$$

for some natural number $n_0.$ Moreover, if the second alternative holds then

 1. 

The sequence $\left\{T^{n}x\right\}$ is convergent to a fixed point $y^{*}$ of $T;$

 2. 

$y^{*}$ is the unique fixed point of T in the set

$$Y=\{y\in X|d(T^{n_0}x,y)<\infty \};$$

 3. 

$d(y,y^{*})\le {\displaystyle \frac{1}{1-L}}d(y,Ty)$ for all $y\in Y.$

Now, we will analyze stability using the fixed point alternative method on a non-Archimedean fuzzy norm space for the mapping (2).

Theorem 3.

Let $f:{\mathbb{R}}^{*}\to \mathbb{R}$ be a function for which there exists a function $\varphi :{\mathbb{R}}^{*}\times {\mathbb{R}}^{*}\to {\mathbb{R}}_{+}$ and a constant L with $0<L<1$, such that the following inequalities hold for all $x,y\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$:

$$N\left(Df\right(x,y),t)\ge N\left(\varphi \right(x,y),t),$$

(12)

$$N\left({\displaystyle \frac{1}{2}}\varphi \left({\displaystyle \frac{x}{2}},{\displaystyle \frac{y}{2}}\right),t\right)\ge N(L\varphi (x,y),t).$$

Then there exists a unique reciprocal function $R:{\mathbb{R}}^{*}\to \mathbb{R}$, defined by

$$R\left(x\right)=\underset{n\to \infty }{lim}{\displaystyle \frac{1}{2^n}}f\left({\displaystyle \frac{x}{2^n}}\right),$$

such that

$$N(f\left(x\right)-R\left(x\right),t)\ge N\left({\displaystyle \frac{2L}{1-L}}\varphi (x,x),t\right)$$

(13)

for all $x\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$.

Proof. 

Consider the set

$$\mathrm{\Omega }=\{g:{\mathbb{R}}^{*}\to \mathbb{R}\}$$

and introduce the generalized metric on $\mathrm{\Omega }$ by

$$d(g,h)=inf\left\{c\in (0,\infty )|N(g\left(x\right)-h\left(x\right),t)\ge N(c\varphi (x,x),t),x\in {\mathbb{R}}^{*},t\in {\mathbb{R}}_{+}\right\}.$$

It is easy to see that $(\mathrm{\Omega },d)$ is a complete generalized metric space (see [19]). Now, we define a mapping $T:\mathrm{\Omega }\to \mathrm{\Omega }$ by

$$\left(Tg\right)\left(x\right)={\displaystyle \frac{1}{2}}g\left({\displaystyle \frac{x}{2}}\right),g\in \mathrm{\Omega },x\in {\mathbb{R}}^{*}.$$

We next check that T is a strictly contractive mapping on $\mathrm{\Omega }$. Let $g,h\in \mathrm{\Omega }$. Then, by the definition of $d(g,h)$, there exists a constant $c\in (0,\infty )$ with $d(g,h)<c$. Thus, for all $x\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$, we have

$$N\left(g\right(x)-h(x),2t)\ge N\left(c\varphi \right(x,x),2t).$$

Replacing x with $\frac{x}{2}$ in the above inequality yields

$$N\left({\displaystyle \frac{1}{2}}g\left({\displaystyle \frac{x}{2}}\right)-{\displaystyle \frac{1}{2}}h\left({\displaystyle \frac{x}{2}}\right),t\right)\ge N\left({\displaystyle \frac{c}{2}}\varphi \left(\frac{x}{2},\frac{x}{2}\right),t\right).$$

Since $\varphi$ satisfies the structural inequality

$$\varphi \left(\frac{x}{2},\frac{x}{2}\right)\le L\varphi (x,x),\mathrm{for}\mathrm{some}L\in (0,1),$$

it follows that

$$N\left({\displaystyle \frac{1}{2}}g\left({\displaystyle \frac{x}{2}}\right)-{\displaystyle \frac{1}{2}}h\left({\displaystyle \frac{x}{2}}\right),t\right)\ge N\left(Lc\varphi (x,x),t\right),$$

for all $x\in {\mathbb{R}}^{*}$ and $t>0$. Hence, we have

$$d(Tg,Th)\le Ld(g,h)$$

for all $g,h\in \mathrm{\Omega }$, that is, T is a strictly contractive mapping on $\mathrm{\Omega }$ with Lipschitz constant L. By setting $x=\frac{x}{2}$ and $y=\frac{x}{2}$ in the inequality (12), we obtain

$$N\left(f\left(x\right)-\frac{1}{2}f\left(\frac{x}{2}\right),t\right)\ge N\left(L\varphi (x,x),\frac{t}{2}\right)=N\left(2L\varphi (x,x),t\right)$$

(14)

for all $x\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$. That is, $d(Tf,f)\le 2L<\infty$. Therefore, by the fixed point alternative Theorem 2, there exists a fixed point R of T in $\mathrm{\Omega }$ such that

• ${lim}_{n\to \infty }d(T^{n}f,R)=0$;
• R is the unique fixed point of T in the set $Y=\{g\in \mathrm{\Omega }\mid d(T^{n_0}f,g)<\infty \}$;
• $d(f,R)\le {\displaystyle \frac{1}{1-L}}d(Tf,f)$.

Since ${lim}_{n\to \infty }d(T^{n}f,R)=0$, we may write $d(T^{n}f,R)={ϵ}_n$ with ${lim}_{n\to \infty }{ϵ}_n=0$. Then, for all $x\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$,

$$N(T^{n}f\left(x\right)-f\left(x\right),t)\ge N({ϵ}_n\varphi (x,x),t).$$

Since ${lim}_{n\to \infty }{ϵ}_n=0$, we infer that

$$R\left(x\right)=\underset{n\to \infty }{lim}{\displaystyle \frac{1}{2^n}}f\left({\displaystyle \frac{x}{2^n}}\right)$$

(15)

for all $x\in {\mathbb{R}}^{*}$. Letting $x=\frac{1}{2^n}x$, $y=\frac{1}{2^n}y$, and $t=2^{n}t$ in the inequality (12), we obtain

$$N\left(Df\left(\frac{x}{2^n},\frac{y}{2^n}\right),2^{n}t\right)\ge N\left(\varphi \left(\frac{x}{2^n},\frac{y}{2^n}\right),2^{n}t\right).$$

That is,

$$N\left(\frac{1}{2^n}Df\left(\frac{x}{2^n},\frac{y}{2^n}\right),t\right)\ge N\left(\frac{1}{2^n}\varphi \left(\frac{x}{2^n},\frac{y}{2^n}\right),t\right)\ge N\left(\varphi (x,y),\frac{1}{L^n}t\right)$$

for all $x,y\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$.

As $n\to \infty$, we have $N(\varphi (x,y),\frac{1}{L^n}t)\to 1$. Hence, the limiting function R satisfies the reciprocal functional Equation (1). Moreover, by the fixed point alternative method, such a function R is the unique solution of (1).

Again, using part (4) of the fixed point alternative theorem, we obtain

$$d(f,R)\le {\displaystyle \frac{1}{1-L}}d(Tf,f).$$

Therefore, we may conclude that

$$N(f\left(x\right)-R\left(x\right),t)\ge N\left({\displaystyle \frac{2L}{1-L}}\varphi (x,x),t\right)$$

for all $x\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$. This inequality coincides with (13), as desired. □

Corollary 1.

Let θ and L be positive real numbers with $0<L<1$, and let $f:{\mathbb{R}}^{*}\to \mathbb{R}$ be a mapping such that

$$N(Df(x,y),t)\ge N\left(\theta (\parallel x\parallel +\parallel y\parallel ),t\right)$$

(16)

for all $x,y\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$. Then there exists a unique solution $R:{\mathbb{R}}^{*}\to \mathbb{R}$ satisfying

$$N(f\left(x\right)-R\left(x\right),t)\ge N\left({\displaystyle \frac{4L\theta }{1-L}}\parallel x\parallel ,t\right)$$

for all $x\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$.

Proof. 

On taking $\varphi (x,y)=\theta (\parallel x\parallel +\parallel y\parallel )$ for all $x,y\in {\mathbb{R}}^{*}$, it is straightforward to verify that inequality (16) holds. In particular, by an argument similar to the proof of Theorem 3, we obtain

$$\begin{array}{ccc}\hfill N\left(f\right(x)-R(x),t)& \ge & N\left({\displaystyle \frac{2L}{1-L}}\varphi (x,x),t\right)\hfill \\ & =& N\left({\displaystyle \frac{4L\theta }{1-L}}\parallel x\parallel ,t\right)\hfill \end{array}$$

for all $x\in {\mathbb{R}}^{*}$ and $t\in {\mathbb{R}}_{+}$. □

4. Brzdȩk’s Fixed Point Method

In this section, we investigate the reciprocal functional Equation (1) using Brzdȩk’s fixed point method. To apply this method, we first reformulate the equation in the following form:

$${\displaystyle \frac{1}{f(x+y)}}={\displaystyle \frac{1}{f\left(x\right)}}+{\displaystyle \frac{1}{f\left(y\right)}}.$$

(17)

We begin by recalling Brzdȩk’s fixed point method. In their work, Brzdȩk and Ciepliński [8] established an existence theorem for fixed points of nonlinear operators in metric spaces and subsequently applied this result to prove stability results for functional equations in non-Archimedean metric spaces. Their framework also yields fixed point results in arbitrary metric spaces. Specifically, Brzdȩk’s fixed point method can be viewed as a direct consequence of the theorems presented in [8,20].

Throughout this section, we denote by $Y^X$ the set of all functions from a set X to a set Y.

Theorem 4

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

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

(18)

and

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

(19)

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

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

(20)

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 $x\in X.$

We now investigate the stability of the reciprocal functional Equation (17) using Brzdȩk’s fixed point method in the setting of non-Archimedean fuzzy metric spaces and non-Archimedean fuzzy normed spaces.

Theorem 5. 

Let $(\mathbb{R},M)$ be a non-Archimedean fuzzy metric space. Suppose that the metric M is invariant, i.e., for all $x,y,z\in \mathbb{R}$ and $t\in \mathbb{R}$,

$$M(x+z,y+z,t)=M(x,y,t),$$

and suppose there exists a function $h:\mathbb{R}\to {\mathbb{R}}_{+}$ such that the set

$$L_0:=\{m\in \mathbb{N}\mid s\left(m\right)+s(1+m)<1\}$$

(21)

is non-empty, where

$$s\left(m\right):=inf\left\{c\in \mathbb{R}\mid {\displaystyle \frac{1}{h\left(mx\right)}}\le c{\displaystyle \frac{1}{h\left(x\right)}}forallx\in \mathbb{R}\right\}.$$

Assume that for all $x,y\in \mathbb{R}$, $n,m\in \mathbb{N}$, and $t\in {\mathbb{R}}_{+}$, the following inequality holds:

$$N\left({\displaystyle \frac{1}{h\left(nx\right)}}+{\displaystyle \frac{1}{h\left(my\right)}},t\right)\ge N\left({\displaystyle \frac{s\left(n\right)}{h\left(x\right)}}+{\displaystyle \frac{s\left(m\right)}{h\left(y\right)}},t\right).$$

(22)

Additionally, suppose there exists a function $f:\mathbb{R}\to {\mathbb{R}}^{*}$ such that

$$M\left({\displaystyle \frac{1}{f(x+y)}},{\displaystyle \frac{1}{f\left(x\right)}}+{\displaystyle \frac{1}{f\left(y\right)}},t\right)\ge N\left({\displaystyle \frac{1}{h\left(x\right)}}+{\displaystyle \frac{1}{h\left(y\right)}},t\right)$$

(23)

for all $x,y\in \mathbb{R}$ and $t\in {\mathbb{R}}_{+}$.

Then there exists a unique solution $R:\mathbb{R}\to {\mathbb{R}}^{*}$ for the Equation (17) such that, for all $x\in \mathbb{R}$ and $t\in {\mathbb{R}}_{+}$,

$$M\left({\displaystyle \frac{1}{f\left(x\right)}},{\displaystyle \frac{1}{R\left(x\right)}},t\right)\ge N\left(S_0{\displaystyle \frac{1}{h\left(x\right)}},t\right),$$

(24)

where

$$S_0:=inf\left\{{\displaystyle \frac{1+s\left(m\right)}{1-s\left(m\right)-s(1+m)}}\mid m\in L_0\right\}.$$

(25)

Proof. 

Replacing y with $mx,$ where $x\in \mathbb{R}$ and $m\in \mathbb{N},$ in the inequality (23), we obtain

$$M\left({\displaystyle \frac{1}{f\left(\right(1+m\left)x\right)}},{\displaystyle \frac{1}{f\left(x\right)}}+{\displaystyle \frac{1}{f\left(mx\right)}},t\right)\ge N\left((1+s\left(m\right)){\displaystyle \frac{1}{h\left(x\right)}},t\right)$$

(26)

for all $t\in {\mathbb{R}}_{+}.$ For each $m\in \mathbb{N},$ we define the operators ${\mathcal{T}}_m:{\left({\mathbb{R}}^{*}\right)}^{\mathbb{R}}\to {\left({\mathbb{R}}^{*}\right)}^{\mathbb{R}}$ and ${\mathrm{\Lambda }}_m:{{\mathbb{R}}_{+}}^{\mathbb{R}}\to {{\mathbb{R}}_{+}}^{\mathbb{R}}$ by

$${\mathcal{T}}_m{\displaystyle \frac{1}{\xi \left(x\right)}}:={\displaystyle \frac{1}{\xi \left(\right(1+m\left)x\right)}}-{\displaystyle \frac{1}{\xi \left(mx\right)}},$$

(27)

$${\mathrm{\Lambda }}_m{\displaystyle \frac{1}{\mu \left(x\right)}}:={\displaystyle \frac{1}{\mu \left(\right(1+m\left)x\right)}}+{\displaystyle \frac{1}{\mu \left(mx\right)}},$$

(28)

for all $x\in \mathbb{R},\xi \in {\left({\mathbb{R}}^{*}\right)}^{\mathbb{R}},\mu \in {{\mathbb{R}}_{+}}^{\mathbb{R}}.$ For each $m\in \mathbb{N}$, $\mathrm{\Lambda }:={\mathrm{\Lambda }}_m$ has the form described in (19) in Theorem 4 with $f_1\left(x\right)=(1+m)x$ and $f_2\left(x\right)=mx$. Since M is invariant, we have

$$M\left({\mathcal{T}}_m{\displaystyle \frac{1}{f\left(x\right)}},{\displaystyle \frac{1}{f\left(x\right)}},t\right)\ge N\left({\epsilon }_m\left(x\right),t\right),$$

and

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

where

$${\displaystyle \frac{1}{{\epsilon }_m\left(x\right)}}:=(1+s\left(m\right)){\displaystyle \frac{1}{h\left(x\right)}},x\in \mathbb{R},t\in {\mathbb{R}}_{+}.$$

Hence, for each $m\in \mathbb{N}$, $\mathcal{T}:={\mathcal{T}}_m$ satisfies the inequalities (18) in Theorem 4. It can be shown by mathematical induction on k that

$${\mathrm{\Lambda }}_m^k{\displaystyle \frac{1}{{\epsilon }_m\left(x\right)}}={[s(1+m)+s\left(m\right)]}^k(1+s\left(m\right)){\displaystyle \frac{1}{h\left(x\right)}}$$

for all $x\in \mathbb{R}$ and $m\in L_0$.

For each $m\in L_0$ and $x\in \mathbb{R}$, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{{\epsilon }_m^{*}\left(x\right)}}:=\sum _{j=0}^{\infty }{\mathrm{\Lambda }}_m^j{\displaystyle \frac{1}{{\epsilon }_m\left(x\right)}}& \le (1+s\left(m\right)){\displaystyle \frac{1}{h\left(x\right)}}\sum _{j=0}^{\infty }{(s\left(m\right)+s(1+m))}^j\hfill \\ \hfill & ={\displaystyle \frac{1+s\left(m\right)}{1-s(1+m)-s\left(m\right)}}{\displaystyle \frac{1}{h\left(x\right)}},\hfill \end{array}$$

where ${\mathrm{\Lambda }}_m^0{\displaystyle \frac{1}{{\epsilon }_m\left(x\right)}}={\displaystyle \frac{1}{{\epsilon }_m\left(x\right)}}$. Now, we can use Brzdȩk’s fixed point method as in Theorem 4. Hence, we obtain the limit

$${\displaystyle \frac{1}{R_m\left(x\right)}}:=\underset{k\to \infty }{lim}{\mathcal{T}}_m^k{\displaystyle \frac{1}{f\left(x\right)}},$$

which exists for each $m\in L_0$ and $x\in \mathbb{R}$, and satisfies

$$M\left({\displaystyle \frac{1}{f\left(x\right)}},{\displaystyle \frac{1}{R_m\left(x\right)}},t\right)\ge N\left({\displaystyle \frac{1}{{\epsilon }_m^{*}\left(x\right)}},t\right),t\in {\mathbb{R}}_{+}.$$

Next, we show that

$$M\left({\mathcal{T}}_m^n{\displaystyle \frac{1}{f(x+y)}},{\mathcal{T}}_m^n{\displaystyle \frac{1}{f\left(x\right)}}+{\mathcal{T}}_m^n{\displaystyle \frac{1}{f\left(y\right)}},t\right)\ge N\left({(s(1+m)+s\left(m\right))}^n\left({\displaystyle \frac{1}{h\left(x\right)}}+{\displaystyle \frac{1}{h\left(y\right)}}\right),t\right)$$

for all non-negative integers n, $x\in \mathbb{R}$, $t\in {\mathbb{R}}_{+}$, and $m\in L_0$. We note that ${\mathcal{T}}_m^0{\displaystyle \frac{1}{f\left(x\right)}}={\displaystyle \frac{1}{f\left(x\right)}}$. The case $n=0$ is covered by inequality (23). Assume that the inequality holds for $n=k$. Then, for $n=k+1$, we have

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

for all $x\in \mathbb{R}$, $t\in {\mathbb{R}}_{+}$, and $m\in L_0$.

As $n\to \infty$, we have

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

for $m\in L_0$. Hence, we obtain the following equality

$${\displaystyle \frac{1}{R_m(x+y)}}={\displaystyle \frac{1}{R_m\left(x\right)}}+{\displaystyle \frac{1}{R_m\left(y\right)}},$$

for all $x,y\in \mathbb{R}$ and $m\in L_0$. That is, for each $m\in L_0$, the mapping $R_m$ is defined in this way and is a solution of the Equation (17).

Let $L>0$ be a constant. We will prove that each mapping $R:\mathbb{R}\to {\mathbb{R}}^{*}$ satisfying the Equation (17) and

$$M\left({\displaystyle \frac{1}{f\left(x\right)}},{\displaystyle \frac{1}{R\left(x\right)}},t\right)\ge N\left(L{\displaystyle \frac{1}{h\left(x\right)}},t\right)$$

coincides with $R_m$ for each $m\in L_0$. Fix $m_0\in L_0$. Since $N(x,·)$ is non-decreasing, we have

$$\begin{array}{cc}\hfill & M\left({\displaystyle \frac{1}{R\left(x\right)}},{\displaystyle \frac{1}{R_{m_0}\left(x\right)}},t\right)\hfill \\ \hfill & \ge min\left\{M\left({\displaystyle \frac{1}{R\left(x\right)}},{\displaystyle \frac{1}{f\left(x\right)}},t\right),M\left({\displaystyle \frac{1}{f\left(x\right)}},{\displaystyle \frac{1}{R_{m_0}\left(x\right)}},t\right)\right\}\hfill \\ \hfill & \ge min\left\{N\left(L{\displaystyle \frac{1}{h\left(x\right)}},t\right),N\left({\displaystyle \frac{1+s\left(m_0\right)}{1-s(1+m_0)-s\left(m_0\right)}}{\displaystyle \frac{1}{h\left(x\right)}},t\right)\right\}\hfill \\ \hfill & \ge N\left(S_0\sum _{k=0}^{\infty }{(s(1+m_0)+s\left(m_0\right))}^k{\displaystyle \frac{1}{h\left(x\right)}},t\right),\hfill \end{array}$$

where

$$S_0:=\left((1+s\left(m_0\right))+(1-s(1+m_0)-s\left(m_0\right))L\right).$$

Hence, we obtain

$$M\left({\displaystyle \frac{1}{R\left(x\right)}},{\displaystyle \frac{1}{R_{m_0}\left(x\right)}},t\right)\ge N\left(S_0\sum _{k=0}^{\infty }{(s(1+m_0)+s\left(m_0\right))}^k{\displaystyle \frac{1}{h\left(x\right)}},t\right)$$

(29)

for all $x\in \mathbb{R}$ and $t\in {\mathbb{R}}_{+}$. For $l\in {\mathbb{N}}_0$, we will show that

$$M\left({\displaystyle \frac{1}{R\left(x\right)}},{\displaystyle \frac{1}{R_{m_0}\left(x\right)}},t\right)\ge N\left(S_0\sum _{k=l}^{\infty }{(s(1+m_0)+s\left(m_0\right))}^k{\displaystyle \frac{1}{h\left(x\right)}},t\right)$$

(30)

for all $x\in \mathbb{R}$ and $t\in {\mathbb{R}}_{+}$. The case $l=0$ corresponds to inequality (29). Assume that inequality (30) holds for $l=n$. Then, we have

$$\begin{array}{cccc}\hfill & M\left({\displaystyle \frac{1}{R\left(x\right)}},{\displaystyle \frac{1}{R_{m_0}\left(x\right)}},t\right)\hfill & \hfill & \\ \hfill & =M\left({\displaystyle \frac{1}{R\left((1+m_0)x\right)}}-{\displaystyle \frac{1}{R\left(m_{0}x\right)}},{\displaystyle \frac{1}{R_{m_0}\left((1+m_0)x\right)}}-{\displaystyle \frac{1}{R_{m_0}\left(m_{0}x\right)}},t\right)\hfill & \hfill & \\ \hfill & \ge min\{N\left(s(1+m_0)S_0{\displaystyle \frac{1}{h\left(x\right)}}·\sum _{k=n}^{\infty }{(s(1+m_0)+s\left(m_0\right))}^k,t\right),\hfill & \hfill & \\ \hfill & N\left(s\left(m_0\right)S_0{\displaystyle \frac{1}{h\left(x\right)}}·\sum _{k=n}^{\infty }{(s(1+m_0)+s\left(m_0\right))}^k,t\right)\}\hfill & \hfill & \\ \hfill & \ge N\left(S_0{\displaystyle \frac{1}{h\left(x\right)}}·\sum _{k=n+1}^{\infty }{(s(1+m_0)+s\left(m_0\right))}^k,t\right),\hfill & \hfill \end{array}$$

for all $x\in \mathbb{R}$ and $t\in {\mathbb{R}}_{+}$. As $l\to \infty$, we have

$$N\left(S_0{\displaystyle \frac{1}{h\left(x\right)}}·\sum _{k=l+1}^{\infty }{(s(1+m_0)+s\left(m_0\right))}^k,t\right)\to 1.$$

Hence, we conclude that

$${\displaystyle \frac{1}{R}}={\displaystyle \frac{1}{R_{m_0}}}$$

for all $m_0\in L_0$. Thus, for each $m_0\in L_0$, it follows that

$${\displaystyle \frac{1}{R_m}}={\displaystyle \frac{1}{R_{m_0}}},$$

as desired. □

The following corollary is a direct consequence of Theorem 5.

Corollary 2.

Let M be as in Theorem 5, and let $h:\mathbb{R}\to (0,\infty )$ be a mapping satisfying

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

(31)

Suppose that $f:\mathbb{R}\to {\mathbb{R}}^{*}$ satisfies the inequality (23). Then there exists a unique reciprocal mapping $R:\mathbb{R}\to {\mathbb{R}}^{*}$ such that

$$M\left({\displaystyle \frac{1}{f\left(x\right)}},{\displaystyle \frac{1}{R\left(x\right)}},t\right)\ge N\left({\displaystyle \frac{1}{h\left(x\right)}},t\right)$$

(32)

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

Proof. 

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

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

The assumption of Theorem 5 implies

$$s(1+n)\le \underset{x\in \mathbb{R}}{sup}{\displaystyle \frac{h\left(x\right)}{h\left(\right(1+n\left)x\right)}}\le a_n,s\left(n\right)\le \underset{x\in \mathbb{R}}{sup}{\displaystyle \frac{h\left(x\right)}{h\left(nx\right)}}\le a_n.$$

Hence, we obtain

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

(33)

By the assumption that the sequence $\left\{a_n\right\}$ has a subsequence $\left\{a_{n_k}\right\}$ such that ${lim}_{k\to \infty }{a}_{n_k}=0$, we get

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

(34)

From inequalities (33) and (34), we deduce

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

which implies

$$\underset{k\to \infty }{lim}s(1+n_k)=0\mathrm{and}\underset{k\to \infty }{lim}s\left(n_k\right)=0.$$

Thus, we conclude

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

Finally, by letting $S_0=1$ as in Theorem 5, the inequality (32) follows from inequality (24). □

Remark 1.

It should be emphasized that reciprocal-type functional equations inherently possess a symmetric property. This symmetry can be illustrated, for instance, by the substitutions

Case (1) $x={\displaystyle \frac{1}{2}}x,y={\displaystyle \frac{1}{2}}x$ or case (2) $y={\displaystyle \frac{1}{2}}y,x={\displaystyle \frac{1}{2}}y$ in the direct method and the fixed point alternative method, or

Case (1) $x=x,y=mx$ or case (2) $y=y,x=m$ in the Brzdȩk’s fixed point method.

These substitutions highlight the underlying symmetric structure of the original functional equation. However, each case considered separately does not, in fact, preserve symmetry on its own. As a result, when stability methods such as the direct method, the fixed point alternative, or Brzdȩk’s fixed point approach are applied, the symmetric property intrinsic to the original equation is no longer maintained within the stability framework.

Remark 2.

In this work, we have investigated the direct method alongside two fixed point frameworks—the classical fixed point alternative and Brzdȩk’s fixed point method—in studying stability problems. These approaches were selected to provide a more comprehensive analysis by leveraging their complementary strengths.

The direct method constructs explicit approximating functions for a given functional equation and estimates the deviation from exact solutions without relying on abstract fixed point arguments. While it offers an intuitive and constructive framework, Brzdȩk’s fixed point method guarantees existence and uniqueness from a more general theoretical framework.

The classical fixed point alternative typically requires a complete generalized metric space and restrictive Lipschitz-type conditions. In contrast, Brzdȩk’s theorem introduces a general control function, allowing generalized contractive conditions that drive the operator toward a fixed point even when conventional Lipschitz continuity fails. This generalization significantly extends the applicability of fixed point techniques to nonlinear functional equations.

For the reciprocal functional equation, the relation $y=x$ plays a central role in both the direct and classical fixed point approaches. In Brzdȩk’s method, the linear relation $y=mx$ not only aids in establishing stability results but also enhances the effectiveness of analyzing a slightly modified reciprocal equation. Stability estimates further indicate that the underlying non-Archimedean fuzzy metric must satisfy an invariance property.

By integrating these methods, we obtain specific stability results under restrictive assumptions and more general results under relaxed conditions, thereby strengthening the completeness of the analysis.

An open question remains for future research: Can Brzdȩk’s fixed point method be systematically extended to more general classes of functional equations and more general normed or metric spaces?