A Solution to the Non-Cooperative Equilibrium Problem for Two and Three Players Using the Fixed-Point Technique

Abstract

The aims of this paper are (a) to introduce the concept of the 0-complete m-metric spaces, (b) to obtain the results for $m^w$-Caristi mapping using Kirk’s approach, (c) to investigate the problem of non-cooperative equilibrium (abbreviated as NCE) in two- and three-person games in the structure of game theory and find the solution by employing coupled and tripled fixed-point results within the framework of 0-complete m-metric spaces (m-metric spaces, respectively), and (d) to establish some coupled fixed-point results which extend the scope of metric fixed point theory. We provide some examples to support the concepts and results presented in this paper. As an application of our results in this paper, we obtain the existence of a solution for a nonlinear integral equation.

1. Introduction and Preliminaries

It is widely acknowledged that real-world problems can be represented using mathematical equations. Metrical fixed-point theory, contraction mapping theorem, and theoretical monotone iterative method are very convenient instruments for solving a variety of problems in nonlinear analysis, control theory, game theory, and economic theory. Banach’s contraction principle [1] is one of the basic results of a metric fixed-point theory which states that if f is a self mapping on a complete metric space $(X,d)$ and there exists a constant $k\in [0,1)$ such that $d(fx,fy)\le kd(x,y)$ holds for all $x,y\in X$, then f has a unique fixed-point in $X.$ Furthermore, for any initial guess $x_0\in X$, the sequence of successive approximations $\{x_0,f{x}_0,f^2{x}_0,\dots \}$ converges to a fixed-point of f. Due to its significance and simplicity, several authors have extended Banach’s contraction principle in various directions; see [2,3,4,5,6,7]. However, the domain of metric fixed-point theory extends beyond pure mathematics and finds application in various quantitative sciences, including engineering, economics, operations research, network theory, game theory, and many more. Within the realm of economics, game theory has utilized techniques and approaches from fixed-point theory to address its complexities.

Game theory is a formal mathematical framework used to analyze games systematically. In fact, games can be viewed as conflicts in which a set of individuals (known as players) participate, and each player aims to maximize their utility within the scope of this conflict. The games can be classified in different ways. However, in this paper, we will focus on cooperative games, where players have the option to collaborate, and non-cooperative games, where players are not permitted to cooperate.

In the subsequent sections, we illustrate the relationship between the existence of equilibria and the presence of a fixed point. In this work, we adopt the concepts and notations introduced in [8]. Let us recall some fundamental notions. In the context of normal form, a two-person game G is defined by the following data:

(i)

The topological spaces $Z_1$ and $Z_2$ are regarded as the strategies for the first player and second player, respectively;

(ii)

A topological subspace $\mathrm{\Xi }$ of the product space $Z_1$$\times Z_2$ represents valid strategy pairs;

(iii)

We define a bi-loss operator as follows: $\pounds :$$\mathrm{\Xi }\to {\mathbb{R}}^2:\left(z_1\times z_2\right)\mapsto \left({\pounds }_1\left(z_1,z_2\right);{\pounds }_2\left(z_1,z_2\right)\right)$, where ${\pounds }_j\left(z_1,z_2\right)$ represents the loss acquired by player j when strategies $z_1$ and $z_2$ are employed. A pair $\left(z_1^{\P },z_2^{\P }\right)\in \mathrm{\Xi }$ is stated as a NCE if

$$\begin{array}{ccc}\hfill {\pounds }_1\left(z_1^{\P },z_2^{\P }\right)& \le & {\pounds }_1\left(z_1,z_2^{\P }\right)\mathrm{for}\mathrm{all}{z}_1\mathrm{in}{Z}_1,\hfill \\ \hfill {\pounds }_1\left(z_1^{\P },z_2^{\P }\right)& \le & {\pounds }_1\left(z_1^{\P },z_2\right)\mathrm{for}\mathrm{all}{z}_2\mathrm{in}{Z}_1.\hfill \end{array}$$

Hence,

$$\begin{array}{ccc}\hfill {\pounds }_1\left(z_1^{\P },z_2^{\P }\right)& \le & \underset{z_1\in Z_1}{min}{\pounds }_1\left(z_1,z_2^{\P }\right),\hfill \\ \hfill {\pounds }_2\left(z_1^{\P },z_2^{\P }\right)& \le & \underset{z_2\in Z_2}{min}{\pounds }_2\left(z_1^{\P },z_2\right).\hfill \end{array}$$

Assume that the following maps exist:

$$\begin{array}{ccc}\hfill {\mathrm{\yen }}_1& :& Z_2\to Z_1\hfill \\ \hfill {\mathrm{\yen }}_2& :& Z_1\to Z_2\hfill \end{array}$$

such that the following equations hold:

$$\begin{array}{ccc}\hfill {\pounds }_1\left({\mathrm{\yen }}_1\left(z_2\right),z_2\right)& =& \underset{z_1\in Z_1}{min}{\pounds }_1\left(z_1,z_2\right),\hfill \\ \hfill {\pounds }_2\left(z_1,{\mathrm{\yen }}_2\left(z_1\right)\right)& =& \underset{z_2\in Z_2.}{min}{\pounds }_2\left(z_1,z_2\right).\hfill \end{array}$$

The mappings ${\mathrm{\yen }}_1$ and ${\mathrm{\yen }}_2$ satisfying the above properties are called optimal decision rules. Any solution $\left(z_1^{\P },z_2^{\P }\right)$ to the system

$$\begin{array}{ccc}\hfill {\mathrm{\yen }}_1\left(z_2^{\P }\right)& =& z_1^{\P },\hfill \\ \hfill {\mathrm{\yen }}_2\left(z_1^{\P }\right)& =& z_2^{\P },\hfill \end{array}$$

is a NCE. Let $\mathrm{{\rm Y}}$ represent the function

$$\begin{array}{c}\mathrm{{\rm Y}}:Z_1\times Z_2\to Z_1\times Z_2\hfill \\ \left(z_1^{\P },z_2^{\P }\right)\mapsto \left({\mathrm{\yen }}_1\left(z_2^{\P }\right);{\mathrm{\yen }}_2\left(z_1^{\P }\right)\right).\hfill \end{array}$$

A fixed point $\left(z_1^{\P },z_2^{\P }\right)$ of $\mathrm{{\rm Y}}$ is indeed a NCE. Therefore, the existence of a solution for a NCE is identified as a pair of fixed-points. Further insights into game theory can be found in [8,9].

However, if we take $Z_1=Z_2=Z$ and ${\mathrm{\yen }}_1={\mathrm{\yen }}_2=\mathrm{\yen }$ and define $\mathrm{{\rm Y}}:Z\times Z\to Z$ by $\left(z_1^{\P },z_2^{\P }\right)\mapsto \mathrm{\yen }\left(z_2^{\P }\right)$, a coupled fixed-point of $\mathrm{{\rm Y}}$ then becomes a non-cooperative equilibrium point.

In 1995, Matthews [10] introduced the concept of a partial metric space and gave an interesting generalization of the Banach contraction principal by replacing an ordinary metric space with a partial metric space. Asadi et al. [11] proposed the notion of an m-metric space and studied its topological properties. They also obtained some fixed-point results, which extended the scope of the classic Banach and Kannan fixed-point theorems. Many authors then extended these results by introducing more general contractive conditions (see [12,13,14,15,16]).

Throughout this paper, we use the symbol $\mathbb{N}$ to indicate the set of positive integers, and ${\mathbb{N}}_0$ to indicate the set of nonnegative integers. Similarly, we denote the set of real numbers and the set of positive real numbers as $\mathbb{R}$ and ${\mathbb{R}}_0^{+}=$${\mathbb{R}}^{+}\cup \left\{0\right\}$, respectively.

Definition 1

([11]). Consider a non-empty set ξ. Then, a mapping $m:\xi \times \xi \to {\mathbb{R}}^{+}\cup \left\{0\right\}$ is called the m-metric (or $M.M$) on ξ if for all $k,s,l\in \xi$, the following conditions hold:

(i) 

$m\left(k,k\right)=m\left(s,s\right)=m\left(k,s\right)⟺k=s;$

(ii) 

$m_{k,s}\le m(k,s);$

(iii) 

$m\left(k,s\right)=m\left(s,k\right);$

(iv) 

$m\left(k,s\right)-m_{k,s}\le (m\left(k,l\right)-m)+(m\left(l,s\right)-m_{l,s});$ where $m_{k,s}=min\left\{m\left(k,k\right),m\left(s,s\right)\right\}$, and $M_{k,s}=max\left\{m\left(k,s\right),m\left(s,s\right)\right\}.$

Then, the pair $\left(\xi ,m\right)$ is known as the m-metric space (or $M.M$-space).

Every ($M.M)$-metric m on $\xi$ generates a $T_0$ topology ${\tau }_m\left(\mathrm{say}\right)$ on $\xi$ which has a base of collection of m-open balls

$$\left\{B_m\left(k,ϵ\right):k\in \xi ,ϵ>0\right\},$$

where

$$B_m\left(k,ϵ\right)=\{s\in \xi :m\left(k,s\right)<m_{k,s}+ϵ\}\mathrm{for}\mathrm{all}k\in \xi ,\epsilon >0.$$

If m is an $M.M$ on $\xi$, then the functions $m^w$, $m^s:\xi \times \xi \to {\mathbb{R}}^{+}\cup \left\{0\right\}$ given by:

$$m^w\left(k,s\right)=m\left(k,s\right)-2m_{k,s}+M_{k,s},$$

and

$$m^s\left(k,s\right)=\left\{\begin{array}{c}m\left(k,s\right)-m_{k,s},\mathrm{if}k\ne s,\hfill \\ 0,\mathrm{if}k=s.\hfill \end{array}\right.$$

are ordinary metrics on $\xi$. It is easy to see that $m^w$ and $m^s$ are equivalent metrics on $\xi .$

In the following example, we set the following notations:

(i)

Let $M_1$ and $M_2$ be two players, each choosing strategies from their respective strategy set $Z_1$ and $Z_2.$

(ii)

The strategy space $\zeta =Z_1\times Z_2$ is equipped with $M.M$-space by defining the distance function m as follows:

$$m\left(\left(z_1,z_2\right),\left(z_1^{\P },z_2^{\P }\right)\right)=\left\{\begin{array}{cc}0\hfill & \mathrm{if}\left(z_1,z_2\right)=\left(z_1^{\P },z_2^{\P }\right),\hfill \\ 1\hfill & \mathrm{if}{z}_1\ne z_1^{\P }\mathrm{or}{z}_2\ne z_2^{\P }.\hfill \end{array}\right.$$

This space models the closedness of the strategies.

(iii)

The payoff functions are ${\pounds }_1:Z_1\times Z_2\to \mathbb{R}$ for player $M_1$ and ${\pounds }_2:Z_1\times Z_1\to \mathbb{R}$ for player $M_2$

(iv)

A Nash equilibrium occurs at the strategy pair $\left(z_1^{\P },z_2^{\P }\right)\in Z_1\times Z_2$ such that no player can improve their payoff by unilaterally changing their strategy:

$$\begin{array}{ccc}\hfill {\pounds }_1\left(z_1^{\P },z_2^{\P }\right)& \le & {\pounds }_1\left(z_1,z_2^{\P }\right)\mathrm{for}\mathrm{all}{z}_1\mathrm{in}{Z}_1\hfill \\ \hfill {\pounds }_2\left(z_1^{\P },z_2^{\P }\right)& \le & {\pounds }_2\left(z_1^{\P },z_2\right)\mathrm{for}\mathrm{all}{z}_2\mathrm{in}{Z}_2.\hfill \end{array}$$

The challenge of finding the Nash equilibrium (or NCE) can be reformulated as a fixed-point problem in $MM$-space, where the concept of distance is generalized, allowing for non-zero self-distance and modified triangle inequality ([11]).

Example 1.

Consider a simple coordination game between two players where they have the following strategy set:

1 

$Z_1=\left\{\xi ,\zeta \right\}$ for player $M_1$

2 

$Z_2=\left\{A,B\right\}$ for player $M_{2.}$

Their payoffs are described in the following Table 1, where each cell contains the payoff pair $\left({\pounds }_1,{\pounds }_2\right)$ corresponding to the strategies chosen by the two players.

Table 1. In this table shows the payoffs in a coordination game where Player $M_1$ and Player $M_2$ each choose from two strategies, and the goal is to coordinate on the same strategy for the best outcome.

	$Z_2=A$	$Z_2=B$
$Z_1=\xi$	$\left(3,3\right)$	$\left(0,2\right)$
$Z_1=\zeta$	$\left(0,2\right)$	$\left(1,1\right)$

Payoff Analysis

・

If both players choose $\left(\xi ,A\right)$, they both obtain a payoff of $\left(3,3\right)$ which is a mutually beneficial outcome.

・

If $M_1$ chooses $\xi$ and $M_2$ chooses B, $M_1$, gets 0 and $M_2$ gets $2.$

・

The other combinations yield lower payoffs for at least one player.

Best Response Dynamics

・

Best Response of Player $M_1$:

・

Given $Z_2=A$, the best response of $M_1$ is $Z_1=\xi$ (since $3>2$).

・

Given $Z_2=B$, the best response of $M_1$ is $Z_1=\xi$ (since $0>1$).

・

Best Response of Player $M_2$

・

Given $Z_1=\xi$, the best response of player $M_2$ is $Z_2=A$ (since $3>2$).

・

Given $Z_1=\zeta$, the best response of player $M_2$ is $Z_2=B$ (since $1>0$).

Thus, the strategy pair $\left(\xi ,A\right)$ is a Nash equilibrium (or NCE), as neither player can improve their payoff by unilaterally deviating.

The Nash equilibrium (or non-cooperative equilibrium) corresponds to the fixed-point of the best response function $\mathrm{{\rm Y}}$, where $\mathrm{{\rm Y}}\left(\left(\zeta ,A\right)\right)=\left(\zeta ,A\right)$, which implies that the strategy pair $\left(\zeta ,A\right)$ is a self-consistent solution (i.e., the Nash equilibrium or NCE).

Definition 2 

([11]). Let $\left(\xi ,m\right)$ be a $M.M$-space. Then:

(i) 

A sequence $\left\{k_{\mu }\right\}$ in $M.M$-space $\left(\xi ,m\right)$ converges with respect to ${\tau }_m$ to k if and only if

$$\underset{\mu \to \infty }{lim}\left(m\left(k_{\mu },k\right)-m_{k_{\mu },k}\right)=0,forall\mu \in \mathbb{N}.$$

(ii) 

A sequence $\left\{k_{\mu }\right\}$ in $M.M$-space $\left(\xi ,m\right)$ is called m-Cauchy if

$$\underset{\mu ,\nu \to \infty }{lim}\left(m\left(k_{\mu },k_{\nu }\right)-m_{k_{\mu },k_{\nu }}\right)and\underset{\mu ,\nu \to \infty }{lim}\left(M_{k_{\mu },k_{\nu }}-m_{k_{\mu ,}{k}_{\nu }}\right),forall\mu ,\nu \in \mathbb{N}$$

exist (and are finite).

(iii) 

A space $\left(\xi ,m\right)$ is said to be complete if every m-Cauchy $\left\{k_{\mu }\right\}$ in ξ is m-convergent with respect to ${\tau }_m$ to k in ξ such that

$$\underset{\mu \to \infty }{lim}m\left(k_{\mu },k\right)-m_{k_{\mu },k}=0,and\underset{\mu \to \infty }{lim}\left(M_{k_{\mu },k}-m_{k_{\mu },k}\right)=0.$$

(iv) 

A sequence $\left\{k_{\mu }\right\}$ is a Cauchy sequence in $\left(\xi ,m\right)$ if and only if it is a Cauchy sequence in the metric space $\left(\xi ,m^w\right).$

(v) 

A space $\left(\xi ,m\right)$ is complete if and only if $\left(\xi ,m^w\right)$ is complete.

Very recently, Mohanta et al. [17] introduced the notion of 0-complete $M.M$-spaces as follows:

Consistent with the concepts in [17], let us recall the following:

Definition 3

([17]). Let $\left(\xi ,m\right)$ be a $M.M$-space, then

(ii) 

A sequence $\left\{k_{\mu }\right\}$ in $\left(\xi ,m\right)$ is called 0-Cauchy sequence if

$$\underset{\mu ,\nu \to \infty }{lim}\left(m\left(k_{\mu },k_{\nu }\right)-m_{k_{\mu },k_{\nu }}\right)=0and\underset{\mu ,\nu \to \infty }{lim}\left(M_{k_{\mu },k_{\nu }}-m_{k_{\mu ,}{k}_{\nu }}\right)=0.$$

(iii) 

A space $\left(\xi ,m\right)$ is said to be 0-complete if every 0-Cauchy sequence $\left\{k_{\mu }\right\}$ in ξ is convergent with respect to ${\tau }_m$ to k in ξ such that

$$\underset{\mu \to \infty }{lim}m\left(k_{\mu },k\right)-m_{k_{\mu },k}=0,and\underset{\mu \to \infty }{lim}\left(M_{k_{\mu },k}-m_{k_{\mu },k}\right)=0,$$

We now present the following two alternative ways to define the Caristi mapping within the context of $M.M$-spaces:

Definition 4.

(i) 

A self-mapping $\mathrm{{\rm Y}}$on a $M.M$-space is called a m-Caristi mapping if there is a function $\mathrm{\Psi }:$ ξ$\to [0,\infty )$ with lower semicontinuity in the setup of $\left(\xi ,m\right)$, and it satisfies the inequality

$$m(k,\mathrm{{\rm Y}}(k\left)\right)\le \mathrm{\Psi }\left(k\right)-\mathrm{\Psi }\left(\mathrm{{\rm Y}}\right(k\left)\right)forallk\in \xi .$$

(ii) 

Self-mapping $\mathrm{{\rm Y}}$ on an $M.M$-space is called a m-Caristi mapping if there is a function $\mathrm{\Psi }:$ ξ$\to [0,\infty )$ with lower semicontinuity in $\left(\xi ,m^w\right)$, and it satisfies the inequality

$$m(k,\mathrm{{\rm Y}}(k\left)\right)\le \mathrm{\Psi }\left(k\right)-\mathrm{\Psi }\left(\mathrm{{\rm Y}}\right(k\left)\right)forallk\in \xi .$$

In an initial endeavor to extend Kirk’s characterization of metric completeness and partial metric completeness to the $M.M$ setup, one might speculate that an $M.M$-space $\left(\xi ,m\right)$ is complete if and only if every m-Caristi mapping on $\xi$ possesses a fixed-point. However, the subsequent straightforward example demonstrates the falsity of this speculation.

Example 2.

Define the m-metric m the set $\mathbb{N}$ of natural numbers as follows $m\left(\mu ,\nu \right)={\displaystyle \frac{\frac{1}{\mu }+\frac{1}{\nu }}{2}}.$

Observing that $\left(\mathbb{N},m\right)$ is not complete, as the metric $m^w$ induces the discrete topology on $\mathbb{N}$ and the sequence ${\left\{k_{\mu }\right\}}_{\mu \in \mathbb{N}}$ is a Cauchy sequence in $\left(\mathbb{N},m^w\right)$. However, we establish that there are no m-Caristi mappings on $\mathbb{N}$, as demonstrated in the following analysis.

Certainly, consider $\mathrm{{\rm Y}}:\mathbb{N}\to \mathbb{N}$ and assume the existence of a lower semicontinuous function $\mathrm{\Psi }:\left(\mathbb{N},t_m\right)\to {\mathbb{R}}^{+}\cup \left\{0\right\}$ such that

$$m(\mu ,\mathrm{{\rm Y}}(\mu \left)\right)\le \mathrm{\Psi }\left(\mu \right)-\mathrm{\Psi }\left(\mathrm{{\rm Y}}\right(\mu \left)\right)\mathrm{for}\mathrm{all}\mu \in \mathbb{N}.$$

Let $1=\mathrm{{\rm Y}}\left(1\right)$, we obtain

$$\begin{array}{ccc}\hfill m(1,\mathrm{{\rm Y}}(1\left)\right)& =& 1,\mathrm{and}m\left(1,1\right)={\displaystyle \frac{\frac{1}{1}+\frac{1}{1}}{2}}={\displaystyle \frac{2}{2}}=1\hfill \\ & \Rightarrow & m(1,\mathrm{{\rm Y}}\left(1\right))=1=m\left(1,1\right).\hfill \end{array}$$

which means that $\mathrm{{\rm Y}}\left(1\right)\in B_m\left(1,ϵ\right)$ for any $ϵ>0.$ Therefore, $1=\mathrm{{\rm Y}}\left(1\right)$, which contradicts the condition

$$m(\mu ,\mathrm{{\rm Y}}(\mu \left)\right)\le \mathrm{\Psi }\left(\mu \right)-\mathrm{\Psi }\left(f\right(\mu \left)\right)\mathrm{for}\mathrm{all}\mu \in \mathbb{N}.$$

Hence, $\mathrm{{\rm Y}}$ is not m-Caristi mappings on $\mathbb{N}.$

Following the ideas of Ran and Reurings [18] and Nieto and Rodriguez-Lopez [19,20], many authors have explored the existence and uniqueness of fixed-points in partially ordered metric spaces for contractive-type mappings. Moreover, the idea of a coupled fixed-point was proposed by Guo and Lakshmikantham in [21].

Let us recall that a pair $(k,s)$ belonging to $\xi \times \xi$ is denoted as a coupled fixed-point for the mapping $\mathrm{{\rm Y}}:\xi \times \xi \to \xi$ when it fulfills the conditions $\mathrm{{\rm Y}}(k,s)$$=k$ and $\mathrm{{\rm Y}}(s,k)=s$ (see, e.g., [21]). Bhaskar et.al [22] introduced the concept of mixed monotone properties and obtained coupled fixed-point theorems in the context of partially ordered metric spaces.

Definition 5

([22]). Consider a partially ordered set $(\xi ,\le )$ and a mapping $\mathrm{{\rm Y}}:\xi \times \xi \to \xi .$ The mapping $\mathrm{{\rm Y}}$ is said to have the mixed monotone property if $\mathrm{{\rm Y}}$ is monotone non-decreasing in first argument and monotone non-increasing in second argument; that is, for any $k,s\in \xi$, we have the following:

$$\begin{array}{ccc}\hfill k_1,k_2& \in & \xi ,k_1\le k_{2}implies\mathrm{{\rm Y}}(k_1,s)\le \mathrm{{\rm Y}}(k_2,s),\hfill \\ \hfill s_1,s_2& \in & \xi ,s_1\le s_{2}implies\mathrm{{\rm Y}}(k,s_1)\ge \mathrm{{\rm Y}}(k,s_2).\hfill \end{array}$$

For more results in this direction, we refer to [23], and the relevant findings and coupled fixed-point theorems can be deduced from [24,25].

Wardowski [26] presented the novel concept of an F-contraction and provided proof for fixed-point theorems within the standard framework of metric spaces. After that, many authors extended and generalized the concept of F-contraction and proved some fixed-point results (see [27,28,29,30] for more details).

Definition 6

([31]). Consider a mapping $F:{\mathbb{R}}_{+}\to \mathbb{R}$ that fulfills the following:

• ($F_1$) F is strictly increasing and continuous.
• ($F_2$) For any sequence $\left\{k_{\mu }\right\}\subset {\mathbb{R}}_{+}$,

$$\underset{\mu \to \infty }{lim}{k}_{\mu }=0\iff \underset{\mu \to \infty }{lim}F\left(k_{\mu }\right)=-\infty .$$

Here, $\mathrm{\Gamma }\left(F\right)$ denotes the set of all functions F that meet the criteria defined in ($F_1$)-($F_2$). Note that $F\left(k\right)=ln\left(k\right)$ and $F^{*}\left(k\right)=lnk+k$ for all k∈${\mathbb{R}}_{+}$ belong to $\mathrm{\Gamma }\left(F\right).$

Note that the manifestation of a fixed point for every $m^w$-Caristi mapping on $M.M$-space $\left(\xi ,m\right)$ does not serve as a characterization for the completeness of $\left(\xi ,m\right)$, as given in the following section of this paper.

2. Main Results

In this section, drawing inspiration from Kirk’s approach to metric completeness and using the idea in [32], we aim to characterize $M.M$-space where each $m^w$-Caristi mapping possesses a fixed-point. To achieve this, we introduce the concept of a 0-complete $M.M$-space, defined as follows.

Definition 7.

Consider a sequence ${\left\{k_{\mu }\right\}}_{\mu \in \mathbb{N}}$ in $M.M$-space $\left(\xi ,m\right)$ is called 0-Cauchy if

$$\underset{\mu ,\nu }{lim}m\left(k_{\mu },s_{\nu }\right)=0and\underset{\mu ,\nu }{lim}{M}_{k_{\mu },s_{\nu }}=0.$$

We say that $\left(\xi ,m\right)$ is 0-complete if every 0-Cauchy sequence in ξ converges to a point $k\in \xi$ such that $m_{k,s}=0.$

Remark 1.

Our assumption is strong from Remark 2.14 [17]. Note that each 0-Cauchy sequence in $\left(\xi ,m\right)$ is a Cauchy sequence in $\left(\xi ,m^w\right)$ and each complete $M.M$-space is 0-complete.

Next, we provide the non-trivial example of 0-complete $M.M$-space, which is not complete.

Example 3.

Consider the $M.M$-space $\left(Q\cap {\mathbb{R}}_0^{+},m\right)$, where Q represents the set of rational numbers and the m-metric m is defined as $m\left(k,s\right)={\displaystyle \frac{k+s}{2}}$. This serves as an illustrative example of a 0-complete $M.M$-space that is not complete.

Proposition 1.

Let $\left(\xi ,m\right)$ be a $M.M$-space. Then, the function $d:\xi \times \xi \to {\mathbb{R}}^{+}\cup \left\{0\right\}$ given by

$$d\left(k,s\right)=\left\{\begin{array}{cc}0\hfill & \mathrm{if}k=s,\hfill \\ m\left(k,s\right)\hfill & \mathrm{if}k\ne s.\hfill \end{array}\right.$$

is a metric on ξ such that ${\tau }_m\subseteq {\tau }_d.$ Moreover, $\left(\xi ,m\right)$ is complete if and only if $\left(\xi ,m\right)$ is 0-complete.

Proof. 

It is clear that

$$d\left(k,s\right)=0\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}k=s,$$

(1)

and $d\left(k,s\right)=d\left(k,s\right)$ for all $k,s\in \xi .$ Let $k,s,l\in \xi .$ Note that $d\left(k,s\right)\le m\left(k,s\right)-m_{k,s}\le m\left(k,l\right)-m_{k,l}+m\left(l,s\right)-m_{l,s}$, from the Equation (1), $m_{k,s}=0.$ If $k\ne s$ and $k=l$, then

$$\begin{array}{ccc}\hfill d\left(k,s\right)& \le & m\left(k,k\right)-min\left(\left(k,k\right),m\left(l,l\right)\right)+m\left(l,s\right)-min\left(\left(l,l\right),m\left(s,s\right)\right)-min\left(\left(k,k\right),m\left(s,s\right)\right)\hfill \\ & =& m\left(k,k\right)-\left(k,k\right)+m\left(l,s\right)\hfill \\ & =& m\left(l,s\right)\hfill \\ & =& d\left(k,s\right)\hfill \end{array}$$

If $k\ne s$ and $s=l$, then we can easily verify that $d\left(k,s\right)=d\left(k,s\right).$ If $k=s$, then

$$d\left(k,s\right)=0\le d\left(k,l\right)+d\left(l,s\right)$$

Thus, $\left(d,\xi \right)$ is a metric space. According to Definition 7, clearly, $\left(\xi ,m\right)$ is complete if and only if $\left(\xi ,m\right)$ is 0-complete. □

Lemma 1.

Let $\left(\xi ,m\right)$ be a $M.M$-space. Then, for each $k\in \xi$, the function ${\mathrm{{\rm Y}}}_k:$ $\xi \to \left[0,\infty \right)$ given by ${\mathrm{{\rm Y}}}_k\left(s\right)=m\left(k,s\right)-m_{k,s}$ is lower semicontinuous for $\left(\xi ,m^w\right)$.

Proof. 

Assume that

$$\underset{n}{lim}{m}^w\left(s,s_n\right)=0,$$

then,

$${\mathrm{{\rm Y}}}_k\left(s\right)\le {\mathrm{{\rm Y}}}_k\left(s\right)+m\left(s,s_n\right)-m_{s,s_n}={\mathrm{{\rm Y}}}_k\left(s\right)+m^w\left(s,s_n\right)-m\left(s,s_n\right)+M_{s,s}.$$

This yield $lim{inf}_n{\mathrm{{\rm Y}}}_k\left(s_n\right)\ge {\mathrm{{\rm Y}}}_k\left(s\right)$ because $m_{s,s_n}\le m\left(s,s_n\right).$ □

Theorem 1.

A $M.M$-space $\left(\xi ,m\right)$ is 0-complete if and only if every $m^w$-Caristi mapping $\mathrm{{\rm Y}}$ on ξ has a fixed-point.

Proof. 

Assume that $\mathrm{{\rm Y}}$ is a $m^w$-Caristi mapping on ξ, then there is a function $\mathrm{\Psi }:\xi \to \left[0,\infty \right)$ which is lower semicontinuous in the setup of $\left(\xi ,m^w\right)$ and

$$m\left(k,\mathrm{{\rm Y}}\left(k\right)\right)\le \mathrm{\Psi }\left(k\right)-\mathrm{\Psi }\left(\mathrm{{\rm Y}}\left(k\right)\right),\mathrm{holds}\mathrm{for}\mathrm{all}k\in \xi .$$

Now, for every $k\in \xi$, consider a set given by

$${\mathrm{\Lambda }}_k=\left\{k\in \xi :m\left(k,s\right)\le \mathrm{\Psi }\left(k\right)-\mathrm{\Psi }\left(s\right)\right\}.$$

Observe that ${\mathrm{\Lambda }}_k$ is nonempty because $\mathrm{{\rm Y}}\left(k\right)$ in ${\mathrm{\Lambda }}_k.$ Moreover, ${\mathrm{\Lambda }}_k$ is closed in $\left(\xi ,m^w\right)$ since $s\mapsto m\left(k,s\right)+\mathrm{\Psi }\left(k\right)$ is lower semicontinuous for $\left(\xi ,m^w\right).$ Fix $k_0$ in ${\mathrm{\Lambda }}_{k_0}$, take $k_1$ in ${\mathrm{\Lambda }}_k$ such that

$$\mathrm{\Psi }\left(k_1\right)<\underset{s\in {\mathrm{\Lambda }}_{k_0}}{inf}\mathrm{\Psi }\left(s\right)+2^{-1}.$$

Clearly, ${\mathrm{\Lambda }}_{k_1}\subseteq {\mathrm{\Lambda }}_{k_0.}$ Hence, for every $k\in {\mathrm{\Lambda }}_{k_1}$, we have

$$\begin{array}{ccc}\hfill m\left(k_1,k\right)& \le & \mathrm{\Psi }\left(k\right)-\mathrm{\Psi }\left(k\right)\hfill \\ & <& \underset{s\in {\mathrm{\Lambda }}_{k_0}}{inf}\mathrm{\Psi }\left(s\right)+2^{-1}-\mathrm{\Psi }\left(k\right)\le \mathrm{\Psi }\left(k\right)+2^{-1}-\mathrm{\Psi }\left(k\right).\hfill \end{array}$$

By following the arguments similar to those given above, we generate a sequence ${\left\{k_{\mu }\right\}}_{\mu \in \mathbb{N}}$ in set ξ, ensuring its corresponding sequence is constructed ${\left\{{\mathrm{\Lambda }}_{k_{\mu }}\right\}}_{\mu \in \mathbb{N}}$ of closed subsets in $\left(\xi ,m^w\right)$ and satisfies

(i)

${\mathrm{\Lambda }}_{k_{\mu +1}}\subseteq {\mathrm{\Lambda }}_{k_{\mu }}$, $k_{\mu +1}\in {\mathrm{\Lambda }}_{k_{\mu }}$ for all $\mu \in \mathbb{N}$

(ii)

$m\left(k_{\mu },k\right)<2^{-\mu }$∀$k\in {\mathrm{\Lambda }}_{k_{\mu }}$, for all $\mu \in \mathbb{N}.$

As

$$m_{k_{\mu },k_{n+1}}\le m\left(k_{\mu },k_{\mu +1}\right),$$

based on (i) and (ii), we have

$$m\left(k_{\mu },k_{\nu }\right)<2^{-\mu }\forall ,\mathrm{for}\mathrm{all}\mu >\nu .$$

If follows that

$$\underset{\mu ,\nu }{lim}m\left(k_{\mu },k\right)=0.$$

There exist $q\in \xi$ such that

$$\underset{\mu }{lim}m\left(k_{\mu },q\right)=m_{k_{\mu },q}=0,$$

and thus

$$\underset{\mu }{lim}m\left(k_{\mu },q\right)=0,$$

and

$$\underset{\mu ,\nu }{lim}{M}_{k_{\mu },k}=0.$$

Therefore, ${\left\{k_{\mu }\right\}}_{\mu \in \mathbb{N}}$ is a 0-Cauchy sequence in $\left(\xi ,m\right)$. Therefore,

$$q\in {\cap }_{\mu \in \mathbb{N}}{\mathrm{\Lambda }}_{k_{\mu }}.$$

In conclusion, we demonstrate that $q=\mathrm{{\rm Y}}\left(q\right)$. To achieve this, observe that

$$\begin{array}{ccc}\hfill m\left(k_{\mu },\mathrm{{\rm Y}}\left(q\right)\right)-m_{k_{\mu },\mathrm{{\rm Y}}\left(q\right)}& \le & m\left(k_{\mu },q\right)-m_{k_{\mu },q}+m\left(q,\mathrm{{\rm Y}}\left(q\right)\right)-m_{q,\mathrm{{\rm Y}}\left(q\right)}\hfill \\ & \le & m\left(k_{\mu },q\right)+m\left(q,\mathrm{{\rm Y}}\left(q\right)\right).\hfill \end{array}$$

Since $m_{k_{\mu },\mathrm{{\rm Y}}\left(q\right)}=0$ and for all $\mu \in \mathbb{N}.$ Consequently, $\mathrm{{\rm Y}}\left(q\right)\in {\cap }_{\mu \in \mathbb{N}}{\mathrm{\Lambda }}_{k_{\mu }}$, so based on $\left(ii\right)$

$$m\left(k_{\mu },\mathrm{{\rm Y}}\left(q\right)\right)<2^{-\mu }\mathrm{for}\mathrm{all}\mu \in \mathbb{N}.$$

Since

$$\begin{array}{ccc}\hfill m\left(q,\mathrm{{\rm Y}}\left(q\right)\right)-m_{q,\mathrm{{\rm Y}}\left(q\right)}& \le & m\left(q,k_n\right)-m_{q,k_n}+m\left(k_n,\mathrm{{\rm Y}}\left(q\right)\right)-m_{k_n,\mathrm{{\rm Y}}\left(q\right)}\hfill \\ & \le & m\left(q,k_n\right)+m\left(k_n,\mathrm{{\rm Y}}\left(q\right)\right).\hfill \end{array}$$

and ${lim}_{\mu }m\left(q,k_n\right)=0$. Also, $m_{q,\mathrm{{\rm Y}}\left(q\right)}=0$ implies that $m\left(q,\mathrm{{\rm Y}}\left(q\right)\right)=0.$ Hence, $m^w\left(q,\mathrm{{\rm Y}}\left(q\right)\right)=0.$ Now, $m^w\left(q,\mathrm{{\rm Y}}\left(q\right)\right)\le m\left(q,\mathrm{{\rm Y}}\left(q\right)\right)-2m_{q,\mathrm{{\rm Y}}\left(q\right)}+M_{q,\mathrm{{\rm Y}}\left(q\right)}$ implies that $m^w\left(q,\mathrm{{\rm Y}}\left(q\right)\right)\le m\left(q,\mathrm{{\rm Y}}\left(q\right)\right).$ Thus, $q=\mathrm{{\rm Y}}\left(q\right).$

Conversely, assume that there is a 0-Cauchy sequence ${\left\{k_{\mu }\right\}}_{\mu \in \mathbb{N}}$ of distinct points in $\left(\xi ,m\right)$, which is not convergent in $\left(\xi ,m^w\right).$ Generate a subsequence ${\left\{s_{\mu }\right\}}_{\mu \in \mathbb{N}}$ from the sequence ${\left\{s_{\mu }\right\}}_{\mu \in \mathbb{N}}$ such that

$$m\left(s_{\mu },s_{\mu +1}\right)<2^{-\left(\mu +1\right)}\mathrm{for}\mathrm{all}\mu \in \mathbb{N}.$$

Put

$$\mathrm{\Lambda }=\left\{s_{\mu }:\mu \in \mathbb{N}\right\},$$

and define a map $\mathrm{{\rm Y}}:\xi \to \xi$ by $\mathrm{{\rm Y}}\left(k\right)=s_0$ if k in $\xi \setminus \mathrm{\Lambda }$, and $\mathrm{{\rm Y}}\left(s_{\mu }\right)=$$s_{\mu +1}$ for all $\mu \in \mathbb{N}.$ Observe that Λ is closed in $m^w\left(\xi ,m\right).$ Now, define $\mathrm{\Psi }:\xi \to \left[0,\infty \right)$ as

$$\mathrm{\Psi }\left(k\right)=m\left(k,s_0\right)+1\mathrm{if}k\in \xi \setminus \mathrm{\Lambda },$$

and $\mathrm{\Psi }\left(s_n\right)=2^{-\mu }$ for all $\mu \in \mathbb{N}.$ Note that $\mathrm{\Psi }\left(s_{\mu +1}\right)<\mathrm{\Psi }\left(s_{\mu +1}\right)$ for all $\mu \in \mathbb{N}$ and so $\mathrm{\Psi }\left(s_0\right)<\mathrm{\Psi }\left(k\right)$ for all k in $\xi \setminus \mathrm{\Lambda }.$ We now conclude that Ψ is a lower semicontinuity in $\left(\xi ,m^w\right).$ Moreover, for every k in $\xi \setminus \mathrm{\Lambda }$, we obtain

$$m\left(k,\mathrm{{\rm Y}}k\right)=m\left(k,s_0\right)=\mathrm{\Psi }\left(k\right)-\mathrm{\Psi }\left(s_0\right)=\mathrm{\Psi }\left(k\right)-\mathrm{\Psi }\left(\mathrm{{\rm Y}}\left(k\right)\right),$$

and for every $s_{\mu }$ in Λ,

$$\begin{array}{ccc}\hfill m\left(s_{\mu },\mathrm{{\rm Y}}\left(s_{\mu }\right)\right)& =& m\left(s_{\mu },s_{\mu +1}\right)<2^{-\left(\mu +1\right)}=\mathrm{\Psi }\left(s_{\mu }\right)-\mathrm{\Psi }\left(s_{\mu +1}\right)\hfill \\ & =& \mathrm{\Psi }\left(s_{\mu }\right)-\mathrm{\Psi }\left(\mathrm{{\rm Y}}\left(s_{\mu }\right)\right).\hfill \end{array}$$

Thus, $\mathrm{{\rm Y}}$ is a Caristi $m^w$-mapping on ξ lacking a fixed point, which leads to a contradiction. □

3. Coupled Fixed-Point Results in M-Metric Spaces

In this section, we prove some new coupled fixed-point results in 0-complete $M.M$-spaces, which are then employed to demonstrate the existence of a NCE in a two-person game.

Throughout this section, let $\left(\xi ,\le \right)$ be a partially ordered set and m be a $M.M$-metric on $\xi$ such that $\left(\xi ,m\right)$ is complete $M.M$-spaces and 0-complete $M.M$-spaces. Further, the product spaces $\xi \times \xi$ satisfy the following:

$$(a,b)\le (k,s)\iff a\le k,s\le b;\mathrm{for}\mathrm{all}(k,s),(a,b)\in \xi \times \xi .$$

Theorem 2.

Let $\left(\xi ,\le \right)$ be a partially ordered set, $\left(\xi ,m\right)$ a 0-complete $M.M$-space, and $\mathrm{{\rm Y}}:\xi \times \xi \to \xi$ a continuous mapping having a mixed monotone property on $\xi .$ Assume that there exist some $F\in \mathrm{\Gamma }\left(F\right)$ and $\chi >0$ with the following:

(i) 

$$\chi +F\left(m\left(\mathrm{{\rm Y}}\left(k,s\right),\mathrm{{\rm Y}}\left(a,b\right)\right)\right)\le F\left(max\left\{m\left(k,a\right),m\left(s,b\right)\right\}\right)forallk\le a,s\le b,$$

(ii) There are $k_0,s_0$ in ξ such that $k_0\le \mathrm{{\rm Y}}\left(k_0,s_0\right)$, $k_0\ge \mathrm{{\rm Y}}\left(k_0,s_0\right).$

Then, there exists $k,s$ in ξ, such that $k=\mathrm{{\rm Y}}\left(k,s\right)$ and $k=\mathrm{{\rm Y}}\left(k,s\right).$

Proof. 

Assume $k_0,s_0$ in $\xi$ such that $k_0\le \mathrm{{\rm Y}}\left(k_0,s_0\right)$, $s_0\ge \mathrm{{\rm Y}}\left(s_0,k_0\right).$ Let

$$k_1=\mathrm{{\rm Y}}\left(k_0,s_0\right)\mathrm{and}{s}_1=\mathrm{{\rm Y}}\left(s_0,k_0\right).$$

Then, $s_0\le s_1$ and $r_0\ge r_1.$ Again, let

$$k_2=\mathrm{{\rm Y}}\left(k_1,s_1\right)\mathrm{and}{s}_2=\mathrm{{\rm Y}}\left(s_1,k_1\right).$$

Using the mixed monotone property of $\mathrm{{\rm Y}}$, we have $k_1\le k_1$ and $s_1\ge s_1.$ Continuing in a similar manner, we have two sequences $\left\{k_n\right\}$ and $\left\{s_n\right\}$ such that

$$k_{n+1}=\mathrm{{\rm Y}}\left(k_n,s_n\right)\mathrm{and}{s}_{n+1}=\mathrm{{\rm Y}}\left(s_n,k_n\right),$$

and

$$\begin{array}{ccc}\hfill k_0& \le & k_1\le k_2\le \dots k_n\le k_{n+1}\dots ,\hfill \\ \hfill s_0& \ge & s_1\ge s_2\ge \dots s_n\ge s_{n+1}\dots \hfill \end{array}$$

For, each $\mu =0,1,2\dots$, from (i), we have the following:

$$\chi +F\left(m\left(k_{\mu },k_{\mu +1}\right)\right)=\chi +F\left(m\left(\mathrm{{\rm Y}}\left(k_{\mu -1},s_{\mu -1}\right),\mathrm{{\rm Y}}\left(k_{\mu },s_{\mu }\right)\right)\right)\le F\left(max\left\{m\left(k_{\mu -1},k_{\mu }\right),m\left(s_{\mu -1},s_{\mu }\right)\right\}\right),$$

(2)

and

$$\chi +F\left(m\left(s_{\mu },s_{\mu +1}\right)\right)=\chi +F\left(m\left(\mathrm{{\rm Y}}\left(s_{\mu -1},k_{\mu -1}\right),\mathrm{{\rm Y}}\left(s_{\mu },k_{\mu }\right)\right)\right)\le F\left(max\left\{m\left(s_{\mu -1},s_{\mu }\right),m\left(k_{\mu -1},k_{\mu }\right)\right\}\right).$$

(3)

Since (2) and (3) hold, we obtain that

$$\chi +F\left(max\left\{m\left(k_{\mu },k_{\mu +1}\right),m\left(s_{\mu },s_{\mu +1}\right)\right\}\right)\le F\left(max\left\{m\left(k_{\mu -1},k_{\mu }\right),\left(s_{\mu -1},s_{\mu }\right)\right\}\right),$$

(4)

and by $\left(F_1\right)$, we have

$$max\left\{m\left(k_{\mu },k_{\mu +1}\right),m\left(s_{\mu },s_{\mu +1}\right)\right\}\le max\left\{m\left(k_{\mu -1},k_{\mu }\right),\left(s_{\mu -1},s_{\mu }\right)\right\}\mathrm{for}\mathrm{all}\mu =1,2,\dots$$

It follows that the sequence $\left\{Z_{\mu }=max\left\{m\left(k_{\mu },k_{\mu +1}\right),m\left(s_{\mu },s_{\mu +1}\right)\right\}\right\}$ is monotone decreasing. Therefore, there is some $Z\ge 0$ such that

$$\underset{\mu \to \infty }{lim}{Z}_{\mu }=Z.$$

Since, F is continuous, on taking the limit on both sides of (4), we determine that

$$\chi +F\left(Z\right)\le F\left(Z\right).$$

As $\chi >0$, using $\left(F_2\right)$, we have

$$\underset{\mu \to \infty }{lim}F\left(Z\right)=-\infty \Rightarrow Z=0.$$

Therefore,

$$\underset{\mu \to \infty }{lim}max\left\{m\left(k_{\mu },k_{\mu +1}\right),m\left(s_{\mu },s_{\mu +1}\right)\right\}=0.$$

(5)

Then,

$$m\left(k_{\mu },k_{\mu +1}\right)\to 0\mathrm{and}m\left(s_{\mu },s_{\mu +1}\right)\to 0.$$

(6)

Also, we have

$$0\le m_{k_{\mu },k_{\mu +1}}\le m\left(k_{\mu },k_{\mu +1}\right)\Rightarrow \underset{\mu \to \infty }{lim}{m}_{k_{\mu },k_{\mu +1}}=0,$$

and

$$m_{k_{\mu },k_{\mu +1}}=min\left\{m\left(k_{\mu },k_{\mu }\right),m\left(k_{\mu +1},k_{\mu +1}\right)\right\}\Rightarrow \underset{\mu \to \infty }{lim}m\left(k_{\mu },k_{\mu }\right)=0.$$

It follows that

$$m_{k_{\mu },k_m}=min\left\{m\left(k_{\mu },k_{\mu }\right),m\left(k_m,k_m\right)\right\}\Rightarrow \underset{\mu \to \infty }{lim}{m}_{k_{\mu },k_m}=0,$$

therefore,

$$\underset{m,\mu \to \infty }{lim}{M}_{k_{\mu },k_m}-m_{k_{\mu },k_m}=0.$$

We now show that

$$\underset{m,\mu \to \infty }{lim}\left(m\left(k_{\mu },k_m\right)-m_{k_{\mu },k_m}\right)=0.$$

Define

$$M\left(k,s\right)=m\left(k,s\right)-m_{k,s}.$$

Based on $\left(m_2\right)$, we have

$$m_{k_{\mu },k_{\mu }}\to 0,m_{s_{\mu },s_{\mu }}\to 0.$$

(7)

Now, we have to prove that ${lim}_{m,\mu \to \infty }m\left(k_m,k_{\mu }\right)={lim}_{m,\mu \to \infty }m\left(s_{\mu },s_m\right)=0.$ Suppose on the contrary that it is not true; there exist $\epsilon >0$ and two sequences $\left\{k_{m_l}\right\}$ and $\left\{k_{{\mu }_l}\right\}$ of positive integers such that for $\left\{{\mu }_l\right\}>\left\{m_l\right\}>l$, we get

$$H_l=max\left\{M\left(k_{m_l},k_{{\mu }_l}\right),M\left(s_{m_l},s_{{\mu }_l}\right)\right\}\ge \epsilon .$$

(8)

Without any loss of generality, let ${\mu }_l$ be the smallest such integer. Then,

$$max\left\{M\left(k_{m_l},k_{{\mu }_{l-1}}\right),M\left(s_{m_l},s_{{\mu }_{l-1}}\right)\right\}\le \epsilon .$$

(9)

Using (6) and (7), and based on the last property of $M.M$-space, for $H_l=M\left(k_{m_l},k_{{\mu }_l}\right)$, we obtain the following:

$$\begin{array}{ccc}\hfill \epsilon & \le & H_l\le M\left(k_{m_l},k_{{\mu }_{l-1}}\right)+M\left(k_{{\mu }_{l-1}},k_{m_l}\right)+M\left(k_{{\mu }_{l-1}},k_{{\mu }_{l-1}}\right)\hfill \\ & \le & \epsilon +M\left(k_{m_l},k_{{\mu }_{l-1}}\right)+M\left(k_{{\mu }_{l-1}},k_{m_l}\right)\hfill \\ \hfill \epsilon & \le & H_l\le M\left(k_{m_l},k_{{\mu }_{l-1}}\right)+M\left(k_{{\mu }_{l-1}},k_{m_l}\right)+M\left(s_{m_l},s_{{\mu }_{l-1}}\right)+M\left(s_{{\mu }_{l-1}},s_{m_l}\right)\hfill \\ & & +M\left(k_{{\mu }_{l-1}},k_{{\mu }_{l-1}}\right)+M\left(s_{{\mu }_{l-1}},s_{{\mu }_{l-1}}\right)\hfill \\ & <& \epsilon +M\left(k_{{\mu }_{l-1}},k_{m_l}\right)+M\left(s_{{\mu }_{l-1}},s_{m_l}\right)+M\left(k_{{\mu }_{l-1}},k_{{\mu }_{l-1}}\right)+M\left(s_{{\mu }_{l-1}},s_{{\mu }_{l-1}}\right)\hfill \end{array}$$

Letting $l\to \infty$ in the above inequality and using (6) and (7), and $\left(m_2\right)$, we have ${lim}_{l\to \infty }{H}_l=\epsilon .$ If for any ${\mu }_0\in \mathbb{N}$ and for all $l\ge$${\mu }_0$, $H_l=0$, we obtain $\epsilon =0$. Assume $H_l>0$ for infinitely many $l.$ Now, we obtain

$$H_l=M\left(k_{m_l},k_{{\mu }_l}\right)\le M\left(k_{m_l},k_{m_{l+1}}\right)+M\left(k_{m_{l+1}},k_{{\mu }_{l+1}}\right)+M\left(k_{{\mu }_{l+1}},k_{{\mu }_l}\right).$$

Put

$$U_l=M\left(k_{m_l},k_{m_{l+1}}\right)+M\left(k_{{\mu }_{l+1}},k_{{\mu }_l}\right).$$

Obviously,

$$\underset{l\to \infty }{lim}{U}_l=0.$$

Therefore,

$$H_l\le U_l+M\left(k_{m_{l+1}},k_{{\mu }_{l+1}}\right).$$

(10)

Similarly, $H_l=M\left(s_{m_l},s_{{\mu }_l}\right)$, we have

$$H_l\le U_l+M\left(s_{m_{l+1}},s_{{\mu }_{l+1}}\right),$$

(11)

using (10) and (11); hence, we have

$$\underset{l\to \infty }{lim}max\left\{U_l+M\left(s_{m_{l+1}},s_{m_{\mu l+1}}\right),U_l+M\left(k_{m_{l+1}},k_{{\mu }_{l+1}}\right)\right\}=H_l=l.$$

(12)

Next, since we have $k_{m_l}\le k_{{\mu }_l}$ and $s_{m_l}\ge s_{m_l}$, using the condition $\left(i\right)$, we obtain that

$$\begin{array}{ccc}\hfill \chi +F\left(m\left(k_{m_{l+1}},k_{{\mu }_{l+1}}\right)\right)& =& \chi +F\left(m\left(\mathrm{{\rm Y}}\left(k_{m_l},s_{m_l}\right),\mathrm{{\rm Y}}\left(k_{{\mu }_l},s_{{\mu }_l}\right)\right)\right)\hfill \\ & \le & F\left(max\left\{m\left(k_{m_l},k_{{\mu }_l}\right),m\left(s_{m_l},s_{{\mu }_l}\right)\right\}\right)\hfill \\ & \le & F\left(max\left\{U_l+m_{k_{m_l},k_{{\mu }_l}},U_l+m_{s_{m_l},s_{{\mu }_l}}\right\}\right)\hfill \end{array}$$

Similarly,

$$\begin{array}{ccc}\hfill \chi +F\left(m\left(s_{m_{l+1}},s_{{\mu }_{l+1}}\right)\right)& =& \chi +F\left(m\left(\mathrm{{\rm Y}}\left(s_{m_l},k_{m_l}\right),\mathrm{{\rm Y}}\left(s_{{\mu }_l},k_{{\mu }_l}\right)\right)\right)\hfill \\ & \le & F\left(max\left\{m\left(k_{m_l},k_{{\mu }_l}\right),m\left(s_{m_l},s_{{\mu }_l}\right)\right\}\right)\hfill \\ & \le & F\left(max\left\{U_l+m_{k_{m_l},k_{{\mu }_l}},U_l+m_{s_{m_l},s_{{\mu }_l}}\right\}\right).\hfill \end{array}$$

Therefore,

$$\chi +max\left\{F\left(m\left(k_{m_{l+1}},k_{{\mu }_{l+1}}\right)\right),F\left(m\left(s_{m_{l+1}},k_{{\mu }_{l+1}}\right)\right)\right\}\le F\left(max\left\{U_l+m_{k_{m_l},k_{{\mu }_l}},U_l+m_{s_{m_l},s_{{\mu }_l}}\right\}\right).$$

From (12), letting $l\to \infty$, we have

$$\chi +F\left(ϵ\right)\le F\left(ϵ\right).$$

This leads to $ϵ=0$, a contradiction. Therefore, $\left\{k_{\mu }\right\}$ and $\left\{s_{\mu }\right\}$ are M-Cauchy sequences. Since $\left(\xi ,m\right)$ is 0-complete $M.M$-space, $m\left(k_{\mu },a\right)\to 0$ and $m\left(s_{\mu },b\right)\to 0$ as $\mu \to \infty$ for some $a,b\in \xi$. Therefore,

$$\begin{array}{ccc}\hfill \underset{\mu \to \infty }{lim}\left(m\left(k_{\mu },a\right)-m_{k_{\mu },a}\right)& =& 0,\hfill \\ \hfill \underset{\mu \to \infty }{lim}\left(m\left(s_{\mu },b\right)-m_{s_{\mu },b}\right)& =& 0.\hfill \end{array}$$

Since ${lim}_{\mu \to \infty }{m}_{k_{\mu },a}=0$ implies that $m\left(a,a\right)=0.$ Similarly, ${lim}_{\mu \to \infty }{m}_{s_{\mu },b}=0$ implies that $m\left(b,b\right)=0$. Now, we prove that $a=\mathrm{{\rm Y}}\left(a,b\right)$ and $b=\mathrm{{\rm Y}}\left(b,a\right).$ Indeed, since $a\le a$ and $b\ge b$, based on the definition of F, we obtain

$$\begin{array}{ccc}\hfill \chi +F\left(m\left(\mathrm{{\rm Y}}\left(a,b\right),\mathrm{{\rm Y}}\left(a,b\right)\right)\right)& \le & F\left(max\left\{m\left(a,b\right),m\left(a,b\right)\right\}\right)\hfill \\ & =& F\left(0\right)=-\infty .\hfill \end{array}$$

This implies that

$$m\left(\mathrm{{\rm Y}}\left(a,b\right),\mathrm{{\rm Y}}\left(a,b\right)\right)=0.$$

Since $k_{\mu }\to a$, $s_{\mu }\to b$ as $\mu \to \infty$ in $\left(\xi ,m\right)$ and $\mathrm{{\rm Y}}$ is m-continuous, we have $\mathrm{{\rm Y}}\left(k_{\mu },s_{\mu }\right)\to \mathrm{{\rm Y}}\left(a,b\right).$ Therefore,

$$\begin{array}{ccc}& & \underset{\mu \to \infty }{lim}m\left(k_{\mu +1},\mathrm{{\rm Y}}\left(a,b\right)\right)\hfill \\ & =& \underset{\mu \to \infty }{lim}m\left(\mathrm{{\rm Y}}\left(k_{\mu +1},s_{\mu +1}\right),\mathrm{{\rm Y}}\left(a,b\right)\right)\hfill \\ & =& 0.\hfill \end{array}$$

Now, we obtain

$$\begin{array}{ccc}\hfill m\left(a,\mathrm{{\rm Y}}\left(a,b\right)\right)-m_{a,\mathrm{{\rm Y}}\left(a,b\right)}& \le & m\left(a,k_{\mu +1}\right)-m_{a,k_{\mu +1}}\hfill \\ & & +m\left(k_{\mu +1},\mathrm{{\rm Y}}\left(a,b\right)\right)-m_{m\left(k_{\mu +1},\mathrm{{\rm Y}}\left(a,b\right)\right)}\hfill \\ & \le & m\left(a,k_{\mu +1}\right)+m\left(k_{\mu +1},\mathrm{{\rm Y}}\left(a,b\right)\right).\hfill \end{array}$$

Letting $\mu \to \infty$, since $m_{a,\mathrm{{\rm Y}}\left(a,b\right)}=0$, we determine that $m\left(a,\mathrm{{\rm Y}}\left(a,b\right)\right)=0$, and therefore $a=\mathrm{{\rm Y}}\left(a,b\right).$ Similarly, $m\left(b,\mathrm{{\rm Y}}\left(b,a\right)\right)=0$ and hence $b=\mathrm{{\rm Y}}\left(b,a\right).$ □

Our next results are obtained by dropping the continuity condition of $\mathrm{{\rm Y}}.$

Theorem 3.

Let $\left(\xi ,\le \right)$be a partially ordered set and $\left(\xi ,m\right)$ a 0-complete $M.M$-space. Suppose that ξ has the following properties:

(a) 

If a $\left\{k_{\mu }\right\}$ is a nondecreasing sequence in ξ such that $k_{\mu }\le k_{\mu +1}$ for all $\mu =1,2,3\dots$, and $k_{\mu }\to k$, then $k_{\mu }\le k.$

(b) 

If a $\left\{k_{\mu }\right\}$ is a nonincreasing sequence in ξ such that $k_{\mu }\ge k_{\mu +1}$ for all $\mu =1,2,3\dots$ and $k_{\mu }\to k$, then $k_{\mu }\ge k.$

Also, assume $\mathrm{{\rm Y}}:\xi \times \xi \to \xi$ is a mapping with a mixed monotone property on $\xi .$ Assume there exists some F $\in \mathrm{\Gamma }\left(F\right)$ and $\chi >0$ such that:

(i) 

$$\chi +F\left(m\left(\mathrm{{\rm Y}}\left(k,s\right),\mathrm{{\rm Y}}\left(a,b\right)\right)\right)\le F\left(max\left\{m\left(k,a\right),m\left(s,b\right)\right\}\right)\mathrm{for}\mathrm{all}k\le a,s\le b.$$

(ii) There are $k_0,s_0$ in ξ such that $k_0\le \mathrm{{\rm Y}}\left(k_0,s_0\right)$, $s_0\ge \mathrm{{\rm Y}}\left(s_0,k_0\right).$

Then, there exists $k,s$ in ξ such that $k=\mathrm{{\rm Y}}\left(k,s\right)$ and $s=\mathrm{{\rm Y}}\left(s,k\right).$

Proof. 

Following arguments similar to those given in the proof of Theorem 2, we now prove that

$$\underset{\mu \to \infty }{lim}m\left(k_{\mu +1},\mathrm{{\rm Y}}\left(a,b\right)\right)=\underset{\mu \to \infty }{lim}m\left(\mathrm{{\rm Y}}\left(k_{\mu },s_{\mu }\right),\mathrm{{\rm Y}}\left(a,b\right)\right)=0.$$

From assumptions (i) and (ii), we get $k_{\mu }\le a$ and $s_{\mu }\le b$ for all $\mu$. Using assumption (i), we have

$$\begin{array}{ccc}\hfill \chi +F\left(m\left(k_{\mu +1},\mathrm{{\rm Y}}\left(a,b\right)\right)\right)& =& F\left(m\left(\mathrm{{\rm Y}}\left(k_{\mu },s_{\mu }\right),\mathrm{{\rm Y}}\left(a,b\right)\right)\right)\hfill \\ & \le & F\left(max\left\{m\left(s_{\mu },a\right),m\left(r_{\mu },b\right)\right\}\right).\hfill \end{array}$$

Letting $\mu \to \infty$, we deduce that ${lim}_{\mu \to \infty }F\left(m\left(k_{\mu +1},\mathrm{{\rm Y}}\left(a,b\right)\right)\right)=-\infty .$ Hence,

$$\underset{\mu \to \infty }{lim}m\left(k_{\mu +1},\mathrm{{\rm Y}}\left(a,b\right)\right)=\underset{\mu \to \infty }{lim}m\left(\mathrm{{\rm Y}}\left(k_{\mu },s_{\mu }\right),\mathrm{{\rm Y}}\left(a,b\right)\right)=0.$$

The proof is over. □

We straightforwardly develop the following corollaries:

Corollary 1.

Consider the partially ordered set $\left(\xi ,\le \right)$, and $\left(\xi ,m\right)$ is a 0-complete $M.M$-space. Let $\mathrm{{\rm Y}}:\xi \times \xi \to \xi$ be a mapping having a mixed monotone property on $\xi .$ Assume that there exist some $F\in \mathrm{\Gamma }\left(F\right)$ and $\chi >0$ such that:

(i) 

$$\chi +F\left(m\left(\mathrm{{\rm Y}}\left(k,s\right),\mathrm{{\rm Y}}\left(a,b\right)\right)\right)\le F\left({\displaystyle \frac{m\left(k,a\right)+m\left(s,b\right)}{2}}\right)forallk\le a,s\le b,$$

(13)

(ii) There are $k_0,s_0$ in ξ such that $k_0\le \mathrm{{\rm Y}}\left(k_0,s_0\right)$, $s_0\ge \mathrm{{\rm Y}}\left(s_0,k_0\right),$

Also, assume that either:

(1) $\mathrm{{\rm Y}}$ is continuous, or (2) ξ possesses the following conditions:

(a) If $\left\{k_{\mu }\right\}$ is a non-decreasing sequence in ξ such that $k_{\mu }\le k_{\mu +1}$ for all $\mu =1,2,3\dots$ and $k_{\mu }\to k$ then $k_{\mu }\le k$,

(b) If $\left\{k_{\mu }\right\}$ is a non-increasing sequence in ξ such that $k_{\mu }\ge k_{\mu +1}$ for all $\mu =1,2,3\dots$ and $k_{\mu }\to k$, then $k_{\mu }\ge k.$

Then, there exists $k,s$ in ξ such that $k=\mathrm{{\rm Y}}\left(k,s\right)$ and $s=\mathrm{{\rm Y}}\left(s,k\right).$

Proof. 

Since

$${\displaystyle \frac{m\left(k,a\right)+m\left(s,b\right)}{2}}\le max\left\{m\left(k,a\right),m\left(s,b\right)\right\}\mathrm{for}\mathrm{all}k,s,a,b\in \xi ,$$

the condition (13) implies the first property of Theorem 2. Hence, the desired result can be deduced from Theorems 2 and 3. □

Corollary 2.

In addition to the assumptions of Corollary 1, if $k_0$ and $s_0$ are comparable, then there exists k in ξ such that $k=\mathrm{{\rm Y}}\left(k,k\right)$ is a unique fixed-point of $\mathrm{{\rm Y}}.$

Proof. 

Consider the assumption (ii) of Corollary 1

$$k_0\le \mathrm{{\rm Y}}\left(k_0,s_0\right)\mathrm{and}{s}_0\ge \mathrm{{\rm Y}}\left(s_0,k_0\right).$$

Since $k_0$ and $s_0$ are comparable, we conclude that:

$$k_0\le s_0\mathrm{or}{s}_0\ge k_0.$$

Consider the first case. Then, the mixed monotone property of $\mathrm{{\rm Y}}$ implies that

$$k_1=\mathrm{{\rm Y}}\left(k_0,s_0\right)\ge \mathrm{{\rm Y}}\left(s_0,s_0\right)\ge \mathrm{{\rm Y}}\left(k_0,s_0\right)=s_1,$$

hence, through induction, one gets $k_{\mu }\ge s_{\mu }$, $\mu \ge 0.$ Now, based on the continuity of distance function m, we determine that

$$k=\underset{\mu \to \infty }{lim}{k}_{\mu +1}=s=\underset{\mu \to \infty }{lim}{s}_{\mu +1},$$

we have

$$m\left(k,s\right)=\underset{\mu \to \infty }{lim}m\left(k_{\mu +1},s_{\mu +1}\right).$$

Alternatively, we obtain

$$\begin{array}{ccc}\hfill \chi +F\left(m\left(k_{\mu +1},s_{\mu +1}\right)\right)& \le & F\left(m\left(\mathrm{{\rm Y}}\left(k_{\mu },s_{\mu }\right),\mathrm{{\rm Y}}\left(k_{\mu },s_{\mu }\right)\right)\right)\hfill \\ & \le & F\left(max\left\{m\left(k_{\mu },k_{\mu }\right),m\left(s_{\mu },s_{\mu }\right)\right\}\right),\hfill \end{array}$$

(14)

Therefore, based on $\left(F_2\right)$, we deduce that

$$\underset{\mu ,m\to \infty }{lim}max\left\{m\left(k_{\mu },k_{\mu }\right),m\left(s_{\mu },s_{\mu }\right)\right\}=0.$$

This means

$$\underset{\mu ,m\to \infty }{lim}m\left(k_{\mu },k_{\mu }\right)=\underset{\mu ,m\to \infty }{lim}m\left(s_{\mu },s_{\mu }\right)=0.$$

Letting limit in (14), we reach at

$$\underset{\mu \to \infty }{lim}m\left(k_{\mu +1},s_{\mu +1}\right)=0.$$

Therefore, $m\left(k,s\right)=0$, or $k=s.$ Hence, $k=\mathrm{{\rm Y}}\left(k,k\right).$ □

4. Solution of Some Non-Cooperative Equilibrium Problems of Two Persons

In this section, we utilize the findings from Section 2 to demonstrate the existence of a NCE in a two-person game. For more detailed investigation of the concepts of two-person games, we refer to [8]. Consider $\left(\xi ,\le \right)$ a partial set and assume that there exists m-metric such that $\left(Z,m\right)$ be a 0-complete $M.M$-space. Consistent with [9], let G be a normal-form game that admits the following data:

(i)

$Z=Z_1$ and $Z_2=Z$ represent strategies for the first and second players, respectively;

(ii)

The $\mathrm{\Xi }=$$Z_1\times Z_2$ denotes the set of allowed strategy pairs;

(iii)

The biloss operator is as follows:

$\pounds :$ $\mathrm{\Xi }\to {\mathbb{R}}^2$

$\left(z_1\times z_2\right)\mapsto \left({\pounds }_1\left(z_1,z_2\right);{\pounds }_2\left(z_1,z_2\right)\right)$,

where ${\pounds }_j\left(z_1,z_2\right)$ represents the loss acquired by player j when strategies $z_1$ and $z_2$ are employed, assuming that there exist maps ${\mathrm{\yen }}_1$ and ${\mathrm{\yen }}_2$, which are optimal decision rules.

As mentioned before, any solution $\left(z_1^{\P },z_2^{\P }\right)$ to the system

$$\begin{array}{ccc}\hfill {\mathrm{\yen }}_1\left(z_2^{\P }\right)& =& z_1^{\P },\hfill \\ \hfill {\mathrm{\yen }}_2\left(z_1^{\P }\right)& =& z_2^{\P },\hfill \end{array}$$

is a NCE. Let $\mathrm{{\rm Y}}$ represent the following function:

$$\begin{array}{c}\mathrm{{\rm Y}}:Z_1\times Z_2\to Z_1\times Z_2\hfill \\ \left(z_1^{\P },z_2^{\P }\right)\mapsto \left({\mathrm{\yen }}_1\left(z_2^{\P }\right);{\mathrm{\yen }}_2\left(z_1^{\P }\right)\right).\hfill \end{array}$$

A fixed-point $\left(z_1^{\P },z_2^{\P }\right)$ of $\mathrm{{\rm Y}}$ is indeed a NCE. Therefore, exploring the existence of a solution for a NCE is identified as a pair of fixed points. However, if we take $Z_1=Z_2=Z$ and ${\mathrm{\yen }}_1={\mathrm{\yen }}_2=\mathrm{\yen }$ and define $\mathrm{{\rm Y}}:Z\times Z\to Z$ by $\left(z_1^{\P },z_2^{\P }\right)\mapsto \mathrm{\yen }\left(z_2^{\P }\right)$, a coupled fixed point of $\mathrm{{\rm Y}}$ then becomes a non-cooperative equilibrium point.

Theorem 4.

Let G be a normal-form game and ξ a strategy for players. Assume that $\left(\xi ,m\right)$ is an 0-complete $M.M$-space and the optimal decision rule $\mathrm{\yen }:\xi \to \xi$ is a monotone continuous operator that meets the following conditions:

(i) 

$$\chi +F\left(m\left(\mathrm{\yen }\left(k\right),\mathrm{\yen }\left(s\right)\right)\right)\le F\left(m\left(k,s\right)\right),$$

for all $k,s\in \xi$ and $s\ge k$ for any $F\in \mathrm{\Gamma }\left(F\right)$ and $\chi >0$,

(ii) There are $k_0,s_0\in {\mathbb{R}}_{+}\cup \left\{0\right\}$ so that $k_0\le \mathrm{\yen }\left(s_0\right)$, $s_0\ge$ $\mathrm{\yen }\left(k_0\right).$

Then, the two-person game G possesses a NCE.

Proof. 

Suppose a mapping $\mathrm{{\rm Y}}:\xi \times \xi \to$$\mathbb{R}$ is given by

$$\mathrm{{\rm Y}}\left(k,s\right)=\mathrm{\yen }\left(s\right)\mathrm{for}\mathrm{all}k,s\in \xi .$$

Based on the continuity and monotonicity of $\mathrm{\yen }$, $\mathrm{{\rm Y}}$ is also continuous and has a mixed monotone property. For all $k,s,a,b\in {\mathbb{R}}_{+}\cup \left\{0\right\}$, with $k\le a$, $s\le b$, we obtain

$$m\left(\mathrm{{\rm Y}}\left(k,s\right),\mathrm{{\rm Y}}\left(a,b\right)\right)=m\left(\mathrm{\yen }\left(r\right),\mathrm{\yen }\left(b\right)\right).$$

Therefore, the assumption (i) reduces to the following:

$$F\left(m\left(\mathrm{\yen }\left(s\right),\mathrm{\yen }\left(b\right)\right)\right)\le F\left(max\left\{m\left(k,a\right),m\left(s,b\right)\right\}\right)$$

(15)

for each $k\le a$, $s\le b.$ Since

$$max\left\{m\left(k,a\right),\left(s,b\right)\right\}\ge m\left(s,b\right),$$

and due to $\left(F_1\right)$, we see that the assumption (i) implies (15), and following the argument in Theorem 2, we determine that $\mathrm{{\rm Y}}$ has a couple fixed-point. Hence, the two-person game G has a NCE. □

Our next results also hold for complete $M.M$-space because every complete $M.M$-space is considered a 0-complete $M.M$-space.

Corollary 3.

Let G be a normal-form game and ξ be a strategy for players. Assume that $\left(\xi ,m\right)$ is a complete $M.M$-space and the optimal decision rule $\mathrm{\yen }:\xi \to \xi$ is a monotone continuous operator that meets the following conditions:

(i) 

$$\chi +F\left(m\left(\mathrm{\yen }\left(k\right),\mathrm{\yen }\left(s\right)\right)\right)\le F\left(m\left(k,s\right)\right)$$

for all $k,s\in \xi$ and $k<s$ for any $F\in \mathrm{\Gamma }\left(F\right)$ and $\chi >0$;

(ii) There are $k_0,s_0\in {\mathbb{R}}_{+}\cup \left\{0\right\}$ so that $k_0\le H\left(s_0\right)$, $s_0\ge$ $H\left(k_0\right).$

Then, the two-person game G possesses a NCE.

Example 4.

Let $\xi ={\mathbb{R}}_{+}\cup \left\{0\right\}$ and $m\left(k,s\right)=\left|{\displaystyle \frac{k+s}{2}}\right|$ for all $k,s\in \xi .$ Suppose that G is a non-normal two-person game with the following biloss operator:

$$\begin{array}{ccc}\hfill {\pounds }_1\left(z_1,z_2\right)& =& z_1^2\left(1+z_2\right)e^{-\chi }-z_1,\hfill \\ \hfill {\pounds }_2\left(z_1,z_2\right)& =& z_2^2\left(1+z_1\right)e^{-\chi }-z_2,\hfill \end{array}$$

for every $z_1,z_2\in {\mathbb{R}}_{+}\cup \left\{0\right\}$, and $\chi >0.$ Suppose ${\mathrm{\yen }}_1$ and ${\mathrm{\yen }}_2$ are the optimal decision rules. We can easily compute for G, and $z_1,z_2$ are the strategies of ${\mathrm{\yen }}_1$ and ${\mathrm{\yen }}_2.$ We have

$${\mathrm{\yen }}_1\left(z_2\right)={\displaystyle \frac{e^{-\chi }}{2\left(1+z_2\right)}}$$

and

$${\mathrm{\yen }}_2\left(z_1\right)={\displaystyle \frac{e^{-\chi }}{2\left(1+z_1\right)}},$$

where $z_1,z_2\in {\mathbb{R}}_{+}\cup \left\{0\right\}.$ As we know, ${\mathrm{\yen }}_1\left(z\right)={\mathrm{\yen }}_2\left(z\right)$ for all $z\in {\mathbb{R}}_{+}\cup \left\{0\right\}.$ In Table 2, we find the solution for the NCE. Table 2 shows the iterative process of calculating the NCE for the two-person non-normal game using optimal strategies for each player, where $\chi =0.5.$

Table 2. This table illustrates an exponential decay process for both players’ strategies over 20 iterations.

Iteration Player ${\mathit{z}}_1$	Iteration Player ${\mathit{z}}_2$	Player 1 Strategy $\left({\mathit{z}}_1\right)$	Player 2 Strategy $\left({\mathit{z}}_2\right)$
0	1	$0.367879$	$0.367879$
1	2	$0.275335$	$0.275335$
2	3	$0.213061$	$0.213061$
3	4	$0.169013$	$0.169013$
4	5	$0.136863$	$0.136863$
5	6	$0.112832$	$0.112832$
6	7	$0.094517$	$0.094517$
…,	…,	…,	…,
17	18	$0.031034$	$0.031034$
18	19	$0.029394$	$0.029394$
19	20	$0.027943$	$0.027943={\displaystyle \frac{e^{-\chi }}{2\left(1+z_2\right)}}$

In Figure 1, we use the iteration for the player $z_1$ and find the strategy value of the first player.

Figure 1. Strategies of player 1.

In Figure 2, we use the iteration for the player $z_2$ and find the strategy value of the second player.

Figure 2. Strategies of player 2.

In Figure 3, the values for both players’ strategies converge to approximately 0.027943 by iteration 19, which represents the NCE of the game. Therefore, $\mathrm{\yen }$ is continuous. To prove that $\mathrm{\yen }$ fulfills all assumptions of Corollary 3, we have

$$\begin{array}{ccc}\hfill m\left(\mathrm{\yen }\left(k\right),\mathrm{\yen }\left(s\right)\right)& =& e^{-\chi }\left|{\displaystyle \frac{\frac{1}{2\left(1+k\right)}+\frac{1}{2\left(1+s\right)}}{2}}\right|\hfill \\ & \le & e^{-\chi }\left|{\displaystyle \frac{k+s}{2}}\right|\hfill \\ & =& e^{-\chi }m\left(k,s\right),\hfill \end{array}$$

for all $k,s\in {\mathbb{R}}_{+}\cup \left\{0\right\}$. Taking the logarithm of the above inequality, we have

$$\chi +ln\left(m\left(\left(\mathrm{\yen }\left(k\right),\mathrm{\yen }\left(s\right)\right)\right)\right)\le ln\left(m\left(k,s\right)\right)\mathrm{for}\mathrm{all}k\ne s.$$

As we know,

$$F\left(s\right)=ln\left(s\right)\in \mathrm{\Gamma }\left(F\right).$$

Therefore, the first assumption of Corollary 3 holds. Take $k_0=0$, we deduce $\mathrm{\yen }\left(k_0\right)=e^{-\xi }$ and chose $s_0=1.$ We obtain $s_0\ge \mathrm{\yen }\left(k_0\right)$ and

$$k_0=0\le \mathrm{\yen }\left(k_0\right)={\displaystyle \frac{e^{-\xi }}{2}}.$$

Therefore, operator $\mathrm{\yen }$ satisfies all conditions of Corollary 3. Consequently, the two-person game G possesses a NCE.

Figure 3. Strategies of both players.

5. Tripled Fixed-Points in M-Metric Spaces

Definition 8.

Let $\mathrm{{\rm Y}}:$$\xi \times \xi \times \xi \to \xi .$ An element $\left(k,s,l\right)\in \xi \times \xi \times \xi$ is called a tripled fixed-point of $\mathrm{{\rm Y}}$ if $\mathrm{{\rm Y}}\left(k,s,l\right)=k$, $\mathrm{{\rm Y}}\left(s,k,s\right)=s$, $\mathrm{{\rm Y}}\left(l,s,k\right)=l.$

Definition 9.

Let $(\xi ,\le )$ be a partially ordered set and $\mathrm{{\rm Y}}:\xi \times \xi \to \xi$ be a mapping. Then the map $\mathrm{{\rm Y}}$ is said to have the mixed monotone property if $\mathrm{{\rm Y}}(k,s,l)$ is monotone non-decreasing in k and l, and monotone non-increasing in s; that is, for any $k,s,l\in \xi$

$$\begin{array}{ccc}\hfill k_1,k_2& \in & \xi ,k_1\le k_{2}implies\mathrm{{\rm Y}}(k_1,s,l)\le \mathrm{{\rm Y}}(k_2,s,l),\hfill \\ \hfill s_1,s_2& \in & \xi ,s_1\le s_{2}implies\mathrm{{\rm Y}}(k,s_1,l)\ge \mathrm{{\rm Y}}(k,s_2,l),\hfill \\ \hfill l_1,l_2& \in & \xi ,l_1\le l_{2}implies\mathrm{{\rm Y}}(k,s,l_1)\le \mathrm{{\rm Y}}(k,s,l_2).\hfill \end{array}$$

Theorem 5.

Consider the partially ordered set $(\xi ,\le )$. $\left(\xi ,m\right)$ is a 0-complete $M.M$-space, and $\mathrm{{\rm Y}}:\xi \times \xi \to \xi$ is a continuous mapping with the mixed monotone property on ξ. Assume that there exist some $F\in \mathrm{\Gamma }\left(F\right)$ and $\chi >0$ with the following:

(i) 

$$\begin{array}{ccc}& & F+Ϝ\left(m\left(\mathrm{{\rm Y}}\left(k,s,l\right),\mathrm{{\rm Y}}\left(a,b,l\right)\right)\right)\hfill \\ & \le & F\left(max\left\{m\left(k,a\right),m\left(s,b\right),m\left(l,c\right)\right\}\right)\hfill \end{array}$$

(16)

for all $k\le a,s\ge b,l\le c.$

(ii) There exists $k_0$, $s_0$, $l_0\in \xi$ such that $k_0\le$ $\mathrm{{\rm Y}}(k_0,s_0,l_0)$, $s_0\ge \mathrm{{\rm Y}}(s_0,k_0,s_0),l_0\le$ $\mathrm{{\rm Y}}(l_0,s_0,k_0).$

Then, $\mathrm{{\rm Y}}$ possesses a tripled fixed point; that is, there exists $k_0$, $s_0$, $l_0\in \xi$ such that

$$k_0=\mathrm{{\rm Y}}(k_0,s_0,l_0),s_0=\mathrm{{\rm Y}}(s_0,k_0,s_0),l_0=\mathrm{{\rm Y}}(l_0,s_0,k_0).$$

Proof. 

Assume $k_0$, $s_0$ in ξ such that $k_0\le \mathrm{{\rm Y}}(k_0,s_0,l_0)$, $s_0\ge \mathrm{{\rm Y}}(s_0,k_0,s_0)$, $l_0\le \mathrm{{\rm Y}}(l_0,s_0,k_0).$ Let

$$k_1=\mathrm{{\rm Y}}\left(k_0,s_0,l_0\right)\mathrm{and}{s}_1=\mathrm{{\rm Y}}\left(s_0,k_0,s_0\right),\mathrm{and}{l}_0=\mathrm{{\rm Y}}\left(l_0,s_0,k_0\right).$$

Then, $k_0\le$ $k_1$, $s_0\ge s_1$ and $l_0\le l_1.$ Again, let

$$k_2=\mathrm{{\rm Y}}\left(k_1,s_1,l_1\right)\mathrm{and}{s}_1=\mathrm{{\rm Y}}\left(s_1,k_1,s_1\right),\mathrm{and}{l}_2=\mathrm{{\rm Y}}\left(l_1,s_1,k_1\right).$$

Owing to the mixed monotone property of $\mathrm{{\rm Y}}$, we deduce that $k_0\le$ $k_1$, $s_0\ge s_1$ and $l_0\le l_1.$ Continuing in a similar fashion, we obtain three sequences $\left\{k_{\mu }\right\}$, $\left\{s_{\mu }\right\}$ and $\left\{l_{\mu }\right\}$ within ξ where each term is defined as $k_{\mu +1}=\mathrm{{\rm Y}}\left(k_{\mu },s_{\mu },l_{\mu }\right)$, $s_{\mu +1}=\mathrm{{\rm Y}}\left(s_{\mu },k_{\mu },s_{\mu }\right)$, and $l_{\mu +1}=\mathrm{{\rm Y}}\left(l_{\mu },s_{\mu },k_{\mu },\right)$ and

$$\begin{array}{ccc}\hfill k_0& \le & k_1\le k_2\le k_3\le \dots k_{\mu }\le k_{\mu +1}\dots \hfill \\ \hfill s_0& \ge & s_1\ge s_2\ge s_3\ge \dots s_{\mu }\ge s_{\mu +1}\dots \hfill \\ \hfill l_0& \le & l_1\le l_2\le l_3\le \dots l_{\mu }\le l_{\mu +1}\dots \hfill \end{array}$$

Now, for every $\mu =0,1,2$, and so on, we find that

$$\begin{array}{ccc}\hfill \chi +F\left(m\left(k_{\mu },k_{\mu +1}\right)\right)& =& \chi +F\left(m\left(\mathrm{{\rm Y}}\left(k_{\mu -1},s_{\mu -1},l_{\mu -1}\right),\mathrm{{\rm Y}}\left(k_{\mu },s_{\mu },l_{\mu }\right)\right)\right)\hfill \\ & \le & F\left(max\left\{m\left(k_{\mu -1},k_{\mu }\right),m\left(s_{\mu -1},s_{\mu }\right),m\left(l_{\mu -1},l_{\mu }\right)\right\}\right)\hfill \end{array}$$

(17)

$$\begin{array}{ccc}& & \chi +F\left(m\left(s_{\mu },s_{\mu +1}\right)\right)\hfill \\ & =& \chi +F\left(m\left(\mathrm{{\rm Y}}\left(s_{\mu -1},k_{\mu -1},s_{\mu }\right),\mathrm{{\rm Y}}\left(s_{\mu },k_{\mu },s_{\mu }\right)\right)\right)\hfill \\ & \le & F\left(max\left\{m\left(s_{\mu -1},s_{\mu }\right),m\left(k_{\mu -1},k_{\mu }\right),m\left(s_{\mu -1},s_{\mu -1}\right)\right\}\right)\hfill \end{array}$$

(18)

$$\begin{array}{ccc}& & F+Ϝ\left(m\left(l_{\mu },l_{\mu +1}\right)\right)\hfill \\ & =& \chi +F\left(m\left(\mathrm{{\rm Y}}\left(l_{\mu -1},s_{\mu -1},k_{\mu -1}\right),\mathrm{{\rm Y}}\left(l_{\mu },s_{\mu },k_{\mu }\right)\right)\right)\hfill \\ & \le & F\left(max\left\{m\left(l_{\mu -1},l_{\mu }\right),m\left(s_{\mu -1},s_{\mu }\right),m\left(k_{\mu -1},k_{\mu -1}\right)\right\}\right)\hfill \end{array}$$

(19)

Considering Equations (17)–(19), and given that F is increasing, we conclude that

$$\begin{array}{ccc}& & \chi +F\left(max\left\{m\left(k_{\mu },k_{\mu +1}\right),m\left(s_{\mu },s_{\mu +1}\right),m\left(l_{\mu },l_{\mu +1}\right)\right\}\right)\hfill \\ & \le & F\left(max\left\{m\left(k_{\mu -1},k_{\mu }\right),m\left(s_{\mu -1},s_{\mu }\right),m\left(l_{\mu -1},l_{\mu }\right)\right\}\right).\hfill \end{array}$$

(20)

Consequently, we can deduce that

$$\begin{array}{ccc}& & max\left\{m\left(k_{\mu },k_{\mu +1}\right),m\left(s_{\mu },s_{\mu +1}\right),m\left(l_{\mu },l_{\mu +1}\right)\right\}\hfill \\ & \le & max\left\{m\left(k_{\mu -1},k_{\mu }\right),m\left(s_{\mu -1},s_{\mu }\right),m\left(l_{\mu -1},l_{\mu }\right)\right\},\hfill \end{array}$$

for all $\mu =1,2$, and so on, it follows that the sequence $\left\{B_{\mu }=max\left\{m\left(k_{\mu },k_{\mu +1}\right),m\left(s_{\mu },s_{\mu +1}\right),m\left(l_{\mu },l_{\mu +1}\right)\right\}\right\}$ is non-increasing. Hence, there exists $B_{\mu }\ge 0$ such that ${lim}_{\mu \to \infty }{B}_{\mu }=B$. As F is continuous, by allowing μ to approach infinity in Equation (20), we reach the following:

$$\chi +F\left(B\right)\le F\left(B\right).$$

Because $\chi >0$, and considering the definition of F, we can conclude that $F\left(B\right)=-\infty .$ This implies that B $= 0$. Consequently,

$$\underset{\mu \to \infty }{lim}max\left\{m\left(k_{\mu },k_{\mu +1}\right),m\left(s_{\mu },s_{\mu +1}\right),m\left(l_{\mu },l_{\mu +1}\right)\right\}=0.$$

(21)

By following the same approach as demonstrated in the proof of Theorem 2, we complete the proof. Hence, $\mathrm{{\rm Y}}$ has tripled fixed-point; that is,

$$a=\mathrm{{\rm Y}}(a,b,c),b=\mathrm{{\rm Y}}(b,a,b),c=\mathrm{{\rm Y}}(c,b,c).$$

□

Remark 2.

It is important to highlight that the coupled and tripled fixed-point results presented in this article can be derived from the fixed-point results of a single mapping using the arguments in [25,33].

6. Certain Non-Cooperative Equilibrium Problems Involving Three Players

In this section, we will utilize tripled fixed-point theorems to establish the existence of a in NCE in a three-person game. For a more comprehensive understanding of the general concepts related to three-person games, interested readers are referred to [8]. Consider $(Z,m)$ as a 0-complete $M.M$-space, with Z having a partially ordered relation ≤. We now examine a three-person game, denoted as G in normal form, which is defined by the following data:

(i)

$Z_1=Z$, $Z_2=Z$, $Z_3=Z$ represent strategies for the first, second, and third players, respectively;

(ii)

The $\mathrm{\Xi }=$$Z_1\times Z_2\times Z_3$ denotes the set of allowed strategy pairs;

(iii)

We define a triloss operator as follows:

$$\pounds :\mathrm{\Xi }\to {\mathbb{R}}^3$$

$$\left(z_1\times z_2\times z_3\right)\mapsto \left({\pounds }_1\left(z_1,z_2,z_3\right);{\pounds }_2\left(z_1,z_2,z_3\right);{\pounds }_3\left(z_1,z_2,z_3\right)\right),$$

where ${\pounds }_j\left(z_1,z_2,z_3\right)$ represents the loss acquired by player j when strategies $z_1$, $z_2$ and $z_3$ are employed. A pair $\left(z_1^{\P },z_2^{\P },z_3^{\P }\right)\in \mathrm{\Xi }$ is stated as a NCE if

$$\begin{array}{ccc}\hfill {\pounds }_1\left(z_1^{\P },z_2^{\P },z_3^{\P }\right)& \le & {\pounds }_1\left(z_1^{\P },z_2^{\P },z_3^{\P }\right),\mathrm{for}\mathrm{all}{z}_1\in Z_1\hfill \\ \hfill {\pounds }_1\left(z_1^{\P },z_2^{\P },z_3^{\P }\right)& \le & {\pounds }_1\left(z_1^{\P },z_2^{\P },z_3^{\P }\right),\mathrm{for}\mathrm{all}{z}_2\in Z_2\hfill \\ \hfill {\pounds }_1\left(z_1^{\P },z_2^{\P },z_3^{\P }\right)& \le & {\pounds }_1\left(z_1^{\P },z_2^{\P },z_3^{\P }\right),\mathrm{for}\mathrm{all}{z}_3\in Z_3.\hfill \end{array}$$

This implies that

$$\begin{array}{ccc}\hfill {\pounds }_1\left(z_1^{\P },z_2^{\P },z_3^{\P }\right)& =& \underset{z_1\in Z_1}{min}{\pounds }_1\left(z_1^{\P },z_2^{\P },z_3^{\P }\right)\hfill \\ \hfill {\pounds }_1\left(z_1^{\P },z_2^{\P },z_3^{\P }\right)& =& \underset{z_2\in Z_2}{min}{\pounds }_1\left(z_1^{\P },z_2^{\P },z_3^{\P }\right)\hfill \\ \hfill {\pounds }_1\left(z_1^{\P },z_2^{\P },z_3^{\P }\right)& =& \underset{z_3\in Z_3}{min}{\pounds }_1\left(z_1^{\P },z_2^{\P },z_3^{\P }\right).\hfill \end{array}$$

To determine the strategy pairs that succeed as non-cooperative equilibria, we examine the optimal decision rules ${\mathrm{\yen }}_1,{\mathrm{\yen }}_2$ and ${\mathrm{\yen }}_3$, defined as follows:

$$\begin{array}{ccc}\hfill {\pounds }_1\left({\mathrm{\yen }}_1\left(z_1\right),z_2,z_3\right)& =& \underset{z_1\in Z_1}{min}{\pounds }_1\left(z_1,z_2,z_3\right)\hfill \\ \hfill {\pounds }_1\left(z_1,{\yen }_2\left(z_2\right),z_3\right)& =& \underset{z_2\in Z_2}{min}{\pounds }_1\left(z_1,z_2,z_3\right)\hfill \\ \hfill {\pounds }_1\left(z_1,z_2,{\yen }_3\left(z_3\right)\right)& =& \underset{z_3\in Z_3}{min}{\pounds }_1\left(z_1,z_2,z_3\right).\hfill \end{array}$$

Consider the any fixed-point mapping

$$\left(z_1,z_2,z_3\right)⟼\left({\mathrm{\yen }}_1\left(z_1\right),{\mathrm{\yen }}_2\left(z_2\right),{\mathrm{\yen }}_3\left(z_3\right)\right)$$

is a NCE.

In this section, we will assume that ${\mathrm{\yen }}_1\left(z_1\right)={\mathrm{\yen }}_2\left(z_2\right)={\mathrm{\yen }}_3\left(z_3\right)$ for all $z\in Z$. It is straightforward to observe that if ${\mathrm{\yen }}_1\left(z_1,z_2,z_3\right)={\mathrm{\yen }}_2\left(z_1,z_2,z_3\right)={\mathrm{\yen }}_3\left(z_1,z_2,z_3\right)$ for all $\left(z_1,z_2,z_3\right)\in {\mathrm{\yen }}_1\times {\mathrm{\yen }}_2\times {\mathrm{\yen }}_3$, then ${\mathrm{\yen }}_1\left(z\right)={\mathrm{\yen }}_2\left(z\right)={\mathrm{\yen }}_3\left(z\right)$. Moreover, it is straightforward to provide an example where ${\mathrm{\yen }}_1\left(z\right)={\mathrm{\yen }}_2\left(z\right)={\mathrm{\yen }}_3\left(z\right)$ when the condition ${\mathrm{\yen }}_1\left(z_1,z_2,z_3\right)\ne {\mathrm{\yen }}_2\left(z_1,z_2,z_3\right)\ne {\mathrm{\yen }}_3\left(z_1,z_2,z_3\right)$ holds. Let $\mathrm{{\rm Y}}:Z_1\times Z_2\times Z_3$ be defined by

$$\mathrm{{\rm Y}}\left(k,s,l\right)=\mathrm{\yen }\left(l\right),$$

for all $k,s,l\in Z.$ Assume that $\mathrm{{\rm Y}}$ possesses a tripled fixed-point $(a,b,c)\in \mathbb{R}$. Consequently, we can deduce that:

$$\begin{array}{ccc}\hfill a& =& \mathrm{{\rm Y}}\left(a,b,c\right)=\mathrm{\yen }\left(a\right)\hfill \\ \hfill b& =& \mathrm{{\rm Y}}\left(b,a,c\right)=\mathrm{\yen }\left(b\right)\hfill \\ \hfill c& =& \mathrm{{\rm Y}}\left(c,a,b\right)=\mathrm{\yen }\left(c\right)\hfill \end{array}$$

and $\left(c,b,a\right)$ is a fixed-point of the map $\left(z_1,z_2,z_3\right)⟼\left(\mathrm{\yen }\left(z_1\right),\mathrm{\yen }\left(z_2\right),\mathrm{\yen }\left(z_3\right)\right).$ Thus, the occurrence of a triple fixed-point for $\mathrm{{\rm Y}}$ implies the existence of a NCE.

Corollary 4.

Consider Z and G as given above. Consider $(Z,m)$ as a 0-complete $M.M$-space, and suppose that the optimal decision rule is represented by monotone continuous functions $\mathrm{\yen }$ that meet the following criteria:

(i) 

$$\chi +F\left(m\left(\mathrm{\yen }\left(k\right),\mathrm{\yen }\left(s\right)\right)\right)\le F\left(m\left(k,s\right)\right)$$

(22)

for all $k,s\in Z$ and $s\ge k$, for some $F\in \mathrm{\Gamma }\left(F\right)$ and $\chi >0.$

(ii) There are $k_0$, $s_0$, $l_0\in {\mathbb{R}}_{+}\cup \left\{0\right\}$ such that $k_0\le \mathrm{\yen }\left(s_0\right),$ $s_0\ge \mathrm{\yen }\left(k_0\right).$

Then, the three-person game G possesses a NCE.

Proof. 

Let $\mathrm{{\rm Y}}:Z\times Z\times Z\to \mathbb{R}$ be defined by

$$\mathrm{{\rm Y}}\left(k,s,l\right)=\mathrm{\yen }\left(l\right)\mathrm{for}\mathrm{all}k,s,l\in Z.$$

As $\mathrm{\yen }$ is a continuous function, $\mathrm{{\rm Y}}$ is also continuous. Furthermore, given that $\mathrm{\yen }$ is monotone. Now, to prove $\mathrm{{\rm Y}}$ satisfies the mixed monotone property on Z. For all $k,s,l,a,b,c\in {\mathbb{R}}_{+}\cup \left\{0\right\}$, with $k\le a,s\ge b$, $l\le c$, we have

$$m\left(\mathrm{{\rm Y}}\left(k,s,l\right),\mathrm{{\rm Y}}\left(a,b,c\right)\right)=m\left(\mathrm{\yen }\left(l\right),\mathrm{\yen }\left(c\right)\right).$$

Thus, condition(22) reduces to

$$\chi +F\left(m\left(\mathrm{\yen }\left(k\right),\mathrm{\yen }\left(s\right)\right)\right)\le F\left(max\left\{m\left(k,a\right),m\left(s,b\right),m\left(l,c\right)\right\}\right),$$

(23)

for all $k\le a,s\ge b,l\le c.$ Since F is increasing, we determine that the condition $\left(i\right)$⇒(23). Applying Theorem 5 leads us to conclude that a tripled fixed-point exists. Consequently, this implies the existence of a NCE in the three-person game $G.$ □

Note As every complete $M.M$-space can be considered a 0-complete $M.M$-space, we can readily derive the following corollary.

Corollary 5.

Let G be a normal-form game and ξ be a strategy for players. Suppose $\left(Z,m\right)$ is a complete $M.M$-space, and the optimal decision rule is represented by monotonously continuous functions $\mathrm{\yen }$, which obeys to the following criteria for some $F\in \mathrm{\Gamma }\left(F\right)$ and $\chi >0$:

(i) 

$$\chi +F\left(m\left(\mathrm{\yen }\left(k\right),\mathrm{\yen }\left(s\right)\right)\right)\le F\left(m\left(k,s\right)\right),forallk,s\in Z.$$

(ii) There exists $k_0,s_0,l_0$$\in {\mathbb{R}}_{+}\cup \left\{0\right\}$ such that $k_0\le \mathrm{\yen }\left(s_0\right)$, $s_0\ge \mathrm{\yen }\left(k_0\right).$

Then, the three-person game G possesses a NCE.

Example 5.

Let Z $= {\mathbb{R}}_{+}\cup \left\{0\right\}$ equipped with the m-metric $m\left(k,s\right)={\displaystyle \frac{1}{2}}\left|k+s\right|$ for all $k,s\in Z$. Now, consider G to be a three-person game with a triloss operator.

$$\begin{array}{ccc}\hfill {\mathrm{\yen }}_1\left(z_1,z_2,z_3\right)& =& z_1^2\left(1+z_2+z_3\right)e^{\chi }-z_1\hfill \\ \hfill {\mathrm{\yen }}_1\left(z_1,z_2,z_3\right)& =& z_2^2\left(1+z_2+z_3\right)e^{\chi }-z_2\hfill \\ \hfill {\mathrm{\yen }}_1\left(z_1,z_2,z_3\right)& =& z_3^2\left(1+z_2+z_3\right)e^{\chi }-z_3,\hfill \end{array}$$

where $\left(z_1,z_2,z_3\right)\in {\mathbb{R}}_{+}^3\cup \left\{0\right\}$ and, since $\chi >0$, it is straightforward to calculate the optimal decision rules ${\mathrm{\yen }}_1,{\mathrm{\yen }}_2$ and ${\mathrm{\yen }}_3$ for the game G.

$${\mathrm{\yen }}_1\left(z_1\right)={\displaystyle \frac{e^{\chi }}{3\left(1+z_1\right)}},{\mathrm{\yen }}_1\left(z_2\right)={\displaystyle \frac{e^{\chi }}{3\left(1+z_2\right)}},{\mathrm{\yen }}_1\left(z_2\right)={\displaystyle \frac{e^{\chi }}{3\left(1+z_3\right)}}.$$

Given $z_1,z_2,z_3\in {\mathbb{R}}_{+}\cup \left\{0\right\}$, it holds that

$${\mathrm{\yen }}_1\left(z\right)={\mathrm{\yen }}_2\left(z\right)={\mathrm{\yen }}_2\left(c\right)forallz\in {\mathbb{R}}_{+}\cup \left\{0\right\},$$

and $\mathrm{\yen }$ is a continuous map. To establish that $\mathrm{\yen }$ fulfills all the conditions of Corollary 5, we observe that

$$\begin{array}{ccc}\hfill m\left(\mathrm{\yen }\left(k\right),\mathrm{\yen }\left(s\right)\right)& \le & {\displaystyle \frac{1}{2}}{e}^{-\chi }\left|{\displaystyle \frac{1}{3\left(1+k\right)}}+{\displaystyle \frac{1}{3\left(1+s\right)}}\right|\hfill \\ & \le & e^{-\chi }{\displaystyle \frac{1}{2}}\left|k+s\right|\hfill \\ & =& m\left(k,s\right)\hfill \end{array}$$

for each $k,s$ in ${\mathbb{R}}_{+}\cup \left\{0\right\}.$ By taking the logarithm of both sides, we reach the following expression:

$$\chi +lnm\left(\mathrm{\yen }\left(k\right),\mathrm{\yen }\left(s\right)\right)\le lnm\left(k,s\right),\forall k\ne s.$$

Since,

$$F\left(k\right)=lnk\in \mathrm{\Gamma }\left(F\right).$$

Hence, we can deduce that $\mathrm{\yen }$ satisfies the initial condition of Corollary 5. When we choose $k_0=0$, we find that

$$\mathrm{\yen }\left(k_0\right)=e^{-\chi }.$$

Now, if we set $s_0=1$, we find that

$$s_0\ge \mathrm{\yen }\left(k_0\right).$$

On the other hand,

$$k_0=0\le \mathrm{\yen }\left(s_0\right)={\displaystyle \frac{e^{-\chi }}{2}}.$$

Therefore, $\mathrm{\yen }$ fulfills the all assumptions of the Corollary 5. Utilizing this corollary leads to the inference that the three-person game G possesses a NCE.

7. Solution of an Integral Equation

In this section, we apply Corollary 2 to study the existence and uniqueness o f solution of the following integral equation ([34]).

Let us examine the following integral equation:

$$k\left(\alpha \right)=\varpi \left(\alpha \right)+{\int }_0^{\mathrm{\Lambda }}\left[{\mathrm{\Gamma }}_1\left(\alpha ,\sigma \right)+{\mathrm{\Gamma }}_1\left(\alpha ,\sigma \right)\right]\left(\lambda \left(\sigma ,k\left(\alpha \right)\right)+\beta \left(\sigma ,k\left(\alpha \right)\right)\right)d\sigma$$

(24)

for all $\alpha \in \left[0,\mathrm{\Lambda }\right].$ We will analyze Equation (24) under the following assumptions as in reference [34].

(i)

The unknown function k is real-valued,

(ii)

$\lambda ,\beta :\left[0,\mathrm{\Lambda }\right]\times \mathbb{R}\to \mathbb{R}$ are increasing and decreasing functions, respectively, where:

(iii)

$\varpi :$$\left[0,\mathrm{\Lambda }\right]\to \mathbb{R}$ is a continuous function,

(iv)

${\mathrm{\Gamma }}_j:\left[0,\mathrm{\Lambda }\right]\times \left[0,\mathrm{\Lambda }\right]\to \mathbb{R}$, where $j=1,2$, ${\mathrm{\Gamma }}_{j=1}\left(\alpha ,q\right)\ge 0$ and ${\mathrm{\Gamma }}_{j=1}\left(\alpha ,\sigma \right)\le 0$ and for all $\sigma ,\alpha \in \left[0,\mathrm{\Lambda }\right]$ such that

$${\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_j\left(\alpha ,\sigma \right)d\sigma \le \mathrm{\Lambda },j=1,2,\mathrm{\Lambda }>0,$$

(v)

there exists $\chi \in \left[1,\infty \right)$ such that

$$0\le {\displaystyle \frac{1}{2}}\lambda \left(\alpha ,\sigma \right)-\lambda \left(\alpha ,\sigma \right)\le e^{-\chi }{\displaystyle \frac{k-s}{2}}\mathrm{for}\mathrm{all}k\ge s,$$

and

$$-\xi e^{-\chi }{\displaystyle \frac{k-s}{2}}\le \beta \left(\alpha ,\sigma \right)+\beta \left(\alpha ,\sigma \right)\le 0\mathrm{for\; all}k\le s.$$

Definition 10.

A pair $\left(\rho ,\varrho \right)\in \xi \times \xi$ is called a couple lower–upper solution to Equation (24) if $\rho \le \varrho$ then

$$\rho \le \mathrm{{\rm Y}}\left(\rho ,\varrho \right)\mathrm{and}\varrho \ge \mathrm{{\rm Y}}\left(\rho ,\varrho \right).$$

Now, we formulate the major result of this part:

Theorem 6.

Considering the presence of coupled lower–upper solutions under assumptions (i)–(v), the Equation (24) admits a solution within the space $\xi =C\left(\left[0,1\right],\mathbb{R}\right).$

Proof. 

Consider the mapping $\mathrm{{\rm Y}}:\xi \times \xi \to \xi$ defined by $\mathrm{{\rm Y}}\left(k,s\right)\left(\alpha \right)=\varpi \left(\alpha \right)+$${\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_1\left(\alpha ,\sigma \right)\lambda \left(\sigma ,k\left(\sigma \right)\right)+\beta \left(\alpha ,s\left(\sigma \right)\right)d\sigma +{\mathrm{\Gamma }}_2\left(\alpha ,\sigma \right)\left(\lambda \left(\sigma ,s\left(\sigma \right)\right)+\beta \left(\sigma ,k\left(\sigma \right)\right)\right)d\sigma .$ Assume that for a monotone non-decreasing sequence $\left\{k_{\mu }\right\}\in \xi$, and $k_{\mu }\to k\in \xi .$ Then, for any t in $\left[0,\mathrm{\Lambda }\right]$,

$$k_1\left(\alpha \right)\le k_2\left(\alpha \right)\dots k_{\mu }\left(\alpha \right)\dots ,$$

the sequence of real numbers converges to $k\left(\alpha \right).$ Therefore, for all $\alpha$ in $\left[0,\mathrm{\Lambda }\right]$, $\mu \ge 0$, $k_{\mu }\left(\alpha \right)\le k\left(\alpha \right)$, i.e., $k_{\mu }\le k.$ In the same way, we can easily see that the limit $s\left(\alpha \right)$ of an abstract monotone non-increasing sequence $s_{\mu }\left(\alpha \right)\in \xi$ is a lower-bound for all members in the sequence, i.e., $s\le s_{\mu }\left(\alpha \right)$ for all $\mu$. Therefore, assumption (ii) holds for Corollary 2.

For arbitrary k in $\xi$, define ${∥k∥}_{\chi }={max}_{t\in \left[0,\mathrm{\Lambda }\right]}\left\{\left|k\right|e^{-\chi }\right\}$, where $\chi \ge 1.$ Note that ${∥.∥}_{\chi }$ is a norm equivalent to the maximum norm and $\left(\xi ,{∥.∥}_{\chi }\right)$ equipped with metric $d_{\xi }$, as stated by

$$d_{\xi }\left(k,s\right)={∥k-s∥}_{\chi }=\underset{\alpha \in \left[0,\mathrm{\Lambda }\right]}{max}\left\{\left|k\left(\alpha \right)-s\left(\alpha \right)\right|e^{-\chi \alpha }\right\}\mathrm{for}\mathrm{all}k,s\in \xi ,$$

is a Banach space ( [31]).

Now, we define a $M.M$-space on $\xi$ as given by

$$m_{\xi }\left(s,r\right)=\left\{\begin{array}{cc}{∥s-r∥}_{\chi }\hfill & \mathrm{if}{∥s∥}_{\chi },{∥s∥}_{\chi }\le 1,\hfill \\ \chi +{∥{\displaystyle \frac{s+r}{2}}∥}_{\chi }\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Clearly, $\left(\xi ,m\right)$ is 0-complete but not m-complete. To prove $\mathrm{{\rm Y}}$ is a mixed monotone mapping. Actually, for $k_1,k_2\in \xi$ and $k_1\le k_2$, i.e., $k_1\left(\alpha \right)\le k_2\left(\alpha \right)$, we have

$$\begin{array}{ccc}& & \mathrm{{\rm Y}}\left(k_1,s\right)\left(\alpha \right)-\mathrm{{\rm Y}}\left(k_2,s\right)\left(\alpha \right)\hfill \\ & =& \varpi \left(\alpha \right)+{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_1\left(\alpha ,\sigma \right)\left[\lambda \left(\sigma ,k_1\left(\sigma \right)\right)+\beta \left(\sigma ,s\left(\sigma \right)\right)\right]d\sigma \hfill \\ & & +{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_2\left(\alpha ,\sigma \right)\left[\lambda \left(\sigma ,s\left(\sigma \right)\right)+\beta \left(\sigma ,k_1\left(\sigma \right)\right)\right]d\sigma \hfill \\ & & -{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_1\left(\alpha ,\sigma \right)\left[\lambda \left(\sigma ,k_2\left(\sigma \right)\right)+\beta \left(\sigma ,s\left(\sigma \right)\right)\right]d\sigma \hfill \\ & & -{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_2\left(\alpha ,\sigma \right)\left[\lambda \left(\sigma ,s\left(\sigma \right)\right)+\beta \left(\sigma ,k_2\left(\sigma \right)\right)\right]d\sigma \hfill \\ & & -\varpi \left(\alpha \right)\hfill \\ & =& {\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_1\left(\alpha ,\sigma \right)\left[\lambda \left(\sigma ,k_1\left(\sigma \right)\right)+\lambda \left(\sigma ,k_2\left(\sigma \right)\right)\right]d\sigma \hfill \\ & & -{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_2\left(\alpha ,\sigma \right)\left[\beta \left(\sigma ,k_1\left(\sigma \right)\right)+\beta \left(\sigma ,k_2\left(\sigma \right)\right)\right]d\sigma \hfill \\ & \le & 0,\hfill \end{array}$$

where $\alpha$ in $\left[0,1\right]$ form assumptions (iv). This produces

$$\mathrm{{\rm Y}}\left(k_1,s\right)\left(\alpha \right)\le \mathrm{{\rm Y}}\left(k_2,s\right)\left(\alpha \right)$$

for every $\sigma$ in $\left[0,1\right]$, that is

$$\mathrm{{\rm Y}}\left(k_1,s\right)\le \mathrm{{\rm Y}}\left(k_2,s\right).$$

Similarly, we can reach at

$$\mathrm{{\rm Y}}\left(k,s_1\right)\le \mathrm{{\rm Y}}\left(k,s_2\right)\mathrm{if}{s}_1\le s_2.$$

Therefore, using the mixed monotone property of $.$ Now, for $k\ge a$ and $s\le b$, we deduce that

$$\left|\left(k,s\right)\left(\alpha \right)-\left(a,b\right)\right|=\left|\right.\left[\right.\varpi \left(\alpha \right)+{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_1\left(\alpha ,\sigma \right)\left(\lambda \left(\sigma ,k\left(\sigma \right)\right)+\beta \left(\sigma ,s\left(\sigma \right)\right)\right)d\sigma$$

$$\begin{array}{ccc}& & +{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_2\left(\alpha ,\sigma \right)\left(\lambda \left(\sigma ,s\left(\sigma \right)\right)+\beta \left(\sigma ,k\left(\sigma \right)\right)\right)d\sigma \left.\right]\hfill \\ & & -\left[\right.\varpi \left(\alpha \right)+{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_1\left(\alpha ,\sigma \right)\left(\lambda \left(\sigma ,a\left(\alpha \right)\right)+\beta \left(\sigma ,b\left(\alpha \right)\right)\right)d\sigma \hfill \\ & & +{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_2\left(\alpha ,q\right)\left(\lambda \left(\sigma ,b\left(\alpha \right)\right)+\beta \left(\sigma ,a\left(\alpha \right)\right)\right)d\sigma \left.+\varpi \left(\alpha \right)\right]d\sigma \left.\right|\hfill \\ & =& \left|\right.{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_1\left(\alpha ,q\right)\left[\left(\lambda \left(\sigma ,k\left(\sigma \right)\right)-\lambda \left(\sigma ,a\left(\sigma \right)\right)\right)+\left(\beta \left(\sigma ,s\left(\sigma \right)\right)-\beta \left(\sigma ,b\left(\sigma \right)\right)\right)\right]d\sigma \hfill \\ & & +{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_2\left(\alpha ,\sigma \right)\left[\left(\lambda \left(\sigma ,s\left(\sigma \right)\right)-\lambda \left(\sigma ,b\left(\sigma \right)\right)\right)+\left(\beta \left(\sigma ,k\left(\sigma \right)\right)-\beta \left(\sigma ,a\left(\sigma \right)\right)\right)\right]d\sigma \left.\right|\hfill \\ & =& \left|\right.{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_1\left(\alpha ,\sigma \right)\left[\left(\lambda \left(\sigma ,k\left(\sigma \right)\right)-\lambda \left(\sigma ,a\left(\sigma \right)\right)\right)-\left(\beta \left(\sigma ,b\left(\sigma \right)\right)-\beta \left(\sigma ,s\left(\sigma \right)\right)\right)\right]d\sigma \hfill \\ & & -{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_2\left(\alpha ,\sigma \right)\left[\left(\lambda \left(\sigma ,b\left(\sigma \right)\right)-\lambda \left(\sigma ,s\left(\sigma \right)\right)\right)-\left(\beta \left(\sigma ,k\left(\sigma \right)\right)-\beta \left(\sigma ,a\left(\sigma \right)\right)\right)\right]d\sigma \left.\right|\hfill \end{array}$$

$$\begin{array}{ccc}& \le & \left|\right.{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_1\left(\alpha ,\sigma \right)\xi e^{-\xi }\left[{\displaystyle \frac{k\left(\sigma \right)-a\left(\sigma \right)}{2}}+{\displaystyle \frac{b\left(\sigma \right)-s\left(\sigma \right)}{2}}\right]d\sigma \hfill \\ & & -{\int }_0^{\mathrm{\Lambda }}{\mathrm{\Gamma }}_2\left(\alpha ,q\right)\xi e^{-\xi }\left[{\displaystyle \frac{b\left(\sigma \right)-s\left(\sigma \right)}{2}}+{\displaystyle \frac{k\left(\sigma \right)-a\left(\sigma \right)}{2}}\right]d\sigma \left.\right|\hfill \\ & \le & \xi e^{-\chi }{\int }_0^{\mathrm{\Lambda }}\left|\right.\left[{\mathrm{\Gamma }}_1\left(\alpha ,\sigma \right)-{\mathrm{\Gamma }}_2\left(\alpha ,\sigma \right)\right]\left[{\displaystyle \frac{b\left(\sigma \right)-r\left(\sigma \right)}{2}}+{\displaystyle \frac{s\left(\sigma \right)-a\left(\sigma \right)}{2}}\right]d\sigma \left.\right|\hfill \\ & \le & \xi e^{-\chi }{\int }_0^{\mathrm{\Lambda }}\left|{\mathrm{\Gamma }}_1\left(\alpha ,\sigma \right)-{\mathrm{\Gamma }}_2\left(\alpha ,\sigma \right)\right|e^{\xi q}\left[{\displaystyle \frac{\left|b\left(\sigma \right)-r\left(\sigma \right)\right|e^{-\xi \sigma }}{2}}+{\displaystyle \frac{\left|s\left(\sigma \right)-a\left(\sigma \right)\right|e^{-\xi \sigma }}{2}}\right]d\sigma \hfill \\ & \le & \xi e^{-\chi }{\int }_0^{\mathrm{\Lambda }}\underset{\alpha ,\sigma \in \left[0,1\right]}{max}\left|{\mathrm{\Gamma }}_1\left(\alpha ,\sigma \right)-{\mathrm{\Gamma }}_2\left(\alpha ,\sigma \right)\right|e^{\xi \sigma }\left[{\displaystyle \frac{{∥s-a∥}_{\xi }}{2}}+{\displaystyle \frac{{∥r-b∥}_{\xi }}{2}}\right]d\sigma \hfill \\ & \le & \xi e^{-\chi }{\displaystyle \frac{e^{\alpha \chi }}{\xi }}\left[{\displaystyle \frac{{∥k-a∥}_{\chi }}{2}}+{\displaystyle \frac{{∥s-b∥}_{\chi }}{2}}\right].\hfill \end{array}$$

It implies that

$$\begin{array}{ccc}& & \left|\mathrm{{\rm Y}}\left(k,s\right)\left(t\right)-\mathrm{{\rm Y}}\left(a,b\right)\right|e^{-\alpha \xi }\hfill \\ & \le & e^{-\xi }\left[{\displaystyle \frac{{∥k-a∥}_{\xi }}{2}}+{\displaystyle \frac{{∥s-b∥}_{\xi }}{2}}\right].\hfill \end{array}$$

Therefore, for each $k,s,a,b\in \xi$ such that $k\ge a$ and $s\le b$, since

$${∥k∥}_{\chi },{∥s∥}_{\chi }{∥a∥}_{\chi },{∥b∥}_{\chi }\le 1,$$

we have

$$\begin{array}{ccc}& & m_{\chi }\left(\mathrm{{\rm Y}}\left(k,s\right),\mathrm{{\rm Y}}\left(a,b\right)\right)\hfill \\ & \le & e^{-\chi }{\displaystyle \frac{1}{2}}\left[m_{\chi }\left(k,a\right)+m_{\chi }\left(s,b\right)\right].\hfill \end{array}$$

Taking the logarithms, we reach the following:

$$\begin{array}{ccc}& & \chi +ln{m}_{\chi }\left(\mathrm{{\rm Y}}\left(k,s\right),\mathrm{{\rm Y}}\left(a,b\right)\right)\hfill \\ & \le & ln\left({\displaystyle \frac{m_{\chi }\left(k,a\right)+m_{\chi }\left(s,b\right)}{2}}\right).\hfill \end{array}$$

This shows that $\mathrm{{\rm Y}}$ satisfies the condition. Now, let the lower–upper solution of the integral Equation (24) be a pair $\left(\rho ,\varrho \right)\in \xi \times \xi$, and we obtain $\rho \le \varrho$

$$\rho \le \mathrm{{\rm Y}}\left(\rho ,\varrho \right)\mathrm{and}\varrho \ge \mathrm{{\rm Y}}\left(\rho ,\varrho \right).$$

Therefore, all assumptions of Corollary 2 hold. Hence, $\mathrm{{\rm Y}}$ has a couple fixed-point say $s.$ Thus, $\mathrm{{\rm Y}}\left(s,s\right)=s$ is the unique solution of Equation (24). □

8. Conclusions

In this paper, we established some coupled and tripled fixed points in the context of 0-complete $M.M$-spaces and also discussed a solution to the NCE problem of two and three-person games within the framework of game theory. Additionally, as an application, we use these results to prove the existence of a solution for the class of nonlinear integral equations. Our proposed work can be extended in many directions and solve the NCE problem of two, three, and N-person games within the framework of game theory under generalized distance spaces. In essence, this research contributes valuable insights into the field of non-linear contraction principles and opens up new avenues for future exploration in various related domains.