e-Distance in Menger PGM Spaces with an Application

Abstract

The main purpose of the present paper is to define the concept of an e-distance (as a generalization of r-distance) on a Menger PGM space and to introduce some of its properties. Moreover, some coupled fixed point results, in terms of this distance on a complete PGM space, are proved. To support our definitions and main results, several examples and an application are considered.

1. Introduction and Preliminaries

In 1942, Menger [1] introduced Menger probabilistic metric spaces as an extension of metric spaces. After that, Sehgal and Bharucha-Reid [2,3] studied some fixed point results for different classes of probabilistic contractions (also, see and references in the citation). Moreover, in 2009, Saadati et al. [4] introduced the concept of r-distance on this space.

Throughout this paper, the set of all Menger distance distribution functions are denoted by $D^{+}$.

Definition 1

([5], page 1). A binary mapping $\mathcal{T}:[0,1]\times [0,1]\to [0,1]$ is called t-norm if the following propertied are held:

(a) 

$\mathcal{T}$ is commutative and associative;

(b) 

$\mathcal{T}$ is continuous;

(c) 

$\mathcal{T}(a,1)=a$ if $a\in [0,1]$;

(d) 

$\mathcal{T}(a,b)\le \mathcal{T}(c,d)$ if $a\le c$ and $b\le d$ for every $a,b,c,d\in [0,1]$.

Definition 2

([4]). A t-norm $\mathcal{T}$ is called an H-type I if for $ϵ\in (0,1)$, there exist $\delta \in (0,1)$ so that ${\mathcal{T}}^m(1-\delta ,\dots ,1-\delta )>1-ϵ$ for each $m\in \mathbb{N}$, where ${\mathcal{T}}^m$ recursively defined by ${\mathcal{T}}^1=\mathcal{T}$ and ${\mathcal{T}}^m(t_1,t_2,\dots ,t_{m+1})=\mathcal{T}({\mathcal{T}}^{m-1}(t_1,t_2,\dots ,t_m),t_{m+1})$ for $m=2,3,\cdots$ and $t_i\in [0,1]$.

All t-norms in the sequel are from class of H-type I.

From another point of view, Mustafa and Sims [6] defined G-metric spaces as another extension of metric spaces, analyzed the structure of this space, and continued the theory of fixed point in such spaces. In 2014, Zhou et al. [7], by combining Menger $PM$-spaces and G-metric spaces, defined Menger probabilistic generalized metric space (shortly, Menger PGM space). Other researchers extended several fixed point theorems in [8,9,10] and references contained therein.

Definition 3

([7]). Assume that $\mathcal{X}$ is a nonempty set, $\mathcal{T}$ is a continuous t-norm and $G:{\mathcal{X}}^3\to D^{+}$ is a mapping satisfying the following properties for all $x,y,z,a\in \mathcal{X}$ and $s,t>0$:

• (PG1) $G_{x,y,z}\left(t\right)=1$ if and only if $x=y=z$;
• (PG2) $G_{x,x,y}\left(t\right)\ge G_{x,y,z}\left(t\right)$, where $z\ne y$;
• (PG3) $G_{x,y,z}\left(t\right)=G_{x,z,y}\left(t\right)=G_{y,x,z}\left(t\right)=\cdots$;
• (PG4) $G_{x,y,z}(t+s)\ge \mathcal{T}(G_{x,a,a}\left(s\right),G_{a,y,z}\left(t\right))$.
• Then $(\mathcal{X},G,\mathcal{T})$ is named a Menger PGM space.

For the definitions of convergent, completeness, closedness and some theorems by regarding these concepts in such spaces, one can see [7]. In 2004, Ran and Reurings [11] discussed on fixed point results for comparable elements of a metric space $(\mathcal{X},d)$ provided with a partial order. Then, Bhaskar and Lakshmikantham [12] presented several fixed point results for a mapping having mixed monotone property in such spaces (see [13,14]).

Definition 4

([12]). Consider a ordered set $(\mathcal{X},⪯)$ and a mapping $F:{\mathcal{X}}^2\to \mathcal{X}$. The mapping F is told to be have mixed monotone property if

$$\begin{array}{cc}\hfill & x_1⪯x_{2}implies thatF(x_1,y)⪯F(x_2,y)\forall x_1,x_2\in \mathcal{X},\hfill \\ \hfill & y_1⪯y_{2}implies thatF(x,y_1)⪰F(x,y_2)\forall y_1,y_2\in \mathcal{X}.\hfill \end{array}$$

for every $x,y\in \mathcal{X}$.

Here we introduce an e-distance on Menger PGM spaces and some of its properties. Then we obtain some coupled fixed point results in the quasi-ordered version of such spaces. The subject of the paper offers novelties compared to the related background literature since a new distance in Menger spaces is defined while some of its properties are revisited and extended.

2. Main Results

Here, we consider an e-distance on a Menger PGM space, which is an extension of r-distance introduced by Saadati et al. [4].

Definition 5.

Consider a Menger PGM space $(\mathcal{X},G,\mathcal{T})$. Then the function $g:{\mathcal{X}}^3\times [0,\infty ]\to [0,1]$ is called an e-distance, if for all $x,y,z,a\in \mathcal{X}$ and $s,t\ge 0$ the following are held:

(r1) 

$g_{x,y,z}(t+s)\ge \mathcal{T}(g_{x,a,a}\left(s\right),g_{a,y,z}\left(t\right))$;

(r2) 

$g_{x,y,.}\left(t\right)$ and $g_{x,.,y}\left(t\right)$ are continuous;

(r3)  

for each $ϵ>0$, there exists $\delta >0$ provided that $g_{a,y,z}\left(t\right)\ge 1-\delta$ and $g_{x,a,a}\left(s\right)\ge 1-\delta$ conclude that $G_{x,y,z}(t+s)\ge 1-ϵ$.

Lemma 1.

Each Menger PGM is an e-distance on $\mathcal{X}$.

Proof. 

Clearly, (r1) and (r2) are true. Only, we prove that (r3) is true. Assume $ϵ>0$ and select $\delta >0$ so that $\mathcal{T}(1-\delta ,1-\delta )\ge 1-ϵ.$ Then, for $G_{a,y,z}\left(t\right)\ge 1-\delta$ and $G_{x,a,a}\left(s\right)\ge 1-\delta$, we get

$G_{x,y,z}(t+s)\ge \mathcal{T}(G_{a,y,z}\left(t\right),G_{x,a,a}\left(s\right))\ge \mathcal{T}(1-\delta ,1-\delta )\ge 1-ϵ.$

□

Example 1.

Assume $(\mathcal{X},G,\mathcal{T})$ is a Menger PGM space. Define a function $g:{\mathcal{X}}^3\times [0,\infty ]\to [0,1]$ by $g_{x,y,z}\left(t\right)=1-c$ for each $x,y,z\in \mathcal{X}$ and $t>0$ with $c\in (0,1)$. Then g is an e-distance.

Lemma 2.

Consider a Menger PGM space with a continuous mapping A on $\mathcal{X}$ and a function $g:{\mathcal{X}}^3\times [0,\infty ]\to [0,1]$ by $g_{x,y,z}\left(t\right)=min\{G_{x,y,z}\left(t\right),G_{Ax,Ay,Az}\left(t\right)\}$ for each $x,y,z\in \mathcal{X}$ and $t>0$. Then g is an e-distance on $\mathcal{X}$.

Proof. 

The condition (r2) is clearly established. To prove (r1), consider $x,y,z,a\in \mathcal{X}$ and $t,s>0$. Then, we have two following cases:

  Case 1: if $G_{x,y,z}\left(t\right)=min\{G_{x,y,z}\left(t\right),G_{Ax,Ay,Az}\left(t\right)\}$, then

$$\begin{array}{cc}\hfill g_{x,y,z}(t+s)& =G_{x,y,z}(t+s)\hfill \\ \hfill & \ge \mathcal{T}(G_{x,a,a}\left(t\right),G_{a,y,z}\left(s\right))\hfill \\ \hfill & \ge \mathcal{T}(min\{G_{x,a,a}\left(t\right),G_{Ax,Aa,Aa}\left(t\right)\},min\{G_{a,y,z}\left(s\right),G_{Aa,Ay,Az}\left(s\right)\})\hfill \\ \hfill & \ge \mathcal{T}(g_{x,a,a}\left(t\right),g_{a,y,z}\left(s\right)).\hfill \end{array}$$

  Case 2: if $G_{Ax,Ay,Az}\left(t\right)=min\{G_{x,y,z}\left(t\right),G_{Ax,Ay,Az}\left(t\right)\}$, then

$$\begin{array}{cc}\hfill g_{x,y,z}(t+s)& =G_{Ax,Ay,Az}(t+s)\hfill \\ \hfill & \ge \mathcal{T}(G_{Ax,Aa,Aa}\left(t\right),G_{Aa,Ay,Az}\left(s\right))\hfill \\ \hfill & \ge \mathcal{T}(min\{G_{x,a,a}\left(t\right),G_{Ax,Aa,Aa}\left(t\right)\},min\{G_{a,y,z}\left(s\right),G_{Aa,Ay,Az}\left(s\right)\})\hfill \\ \hfill & \ge \mathcal{T}(g_{x,a,a}\left(t\right),g_{a,y,z}\left(s\right)).\hfill \end{array}$$

Therefore, (r1) is established. Now, assume $ϵ>0$ and select $\delta >0$ so that $\mathcal{T}(1-\delta ,1-\delta )\ge 1-ϵ$. Using $g_{x,a,a}\left(t\right)\ge 1-\delta$ and $g_{a,y,z}\left(s\right)\ge 1-\delta$, we get

$$\begin{array}{cc}\hfill & min\{G_{x,a,a}\left(t\right),G_{Ax,Aa,Aa}\left(t\right)\}=g_{x,a,a}\left(t\right)\ge 1-\delta ,\hfill \\ \hfill & min\{G_{a,y,z}\left(s\right),G_{Aa,Ay,Az}\left(s\right)\}=g_{a,y,z}\left(s\right)\ge 1-\delta ,\hfill \end{array}$$

which induces that

$$\begin{array}{cc}\hfill G_{x,y,z}(t+s)& \ge \mathcal{T}(G_{x,a,a}\left(t\right),G_{a,y,z}\left(s\right))\hfill \\ \hfill & \ge \mathcal{T}(min\{G_{x,a,a}\left(t\right),G_{Ax,Aa,Aa}\left(t\right)\},min\{G_{a,y,z}\left(s\right),G_{Aa,Ay,Az}\left(s\right)\})\hfill \\ \hfill & =\mathcal{T}(g_{x,a,a}\left(t\right),g_{a,y,z}\left(s\right))\ge \mathcal{T}(1-\delta ,1-\delta )\ge 1-ϵ.\hfill \end{array}$$

Thus, (r3) is established. This completes the proof. □

Lemma 3.

Consider an e-distance g on $(\mathcal{X},G,\mathcal{T})$ with two sequences $\left\{x_n\right\}$ and $\left\{y_n\right\}$ in $\mathcal{X}$. Suppose that $\left\{{\alpha }_n\right\}$ and $\left\{{\beta }_n\right\}$ are two non-negative sequences converging to 0. Then for $x,y,z\in \mathcal{X}$ and $t,s>0$ the following assertions are established:

(i) 

$g_{z,y,x_n}\left(t\right)\ge 1-{\alpha }_n$ and $g_{x,x_n,x_n}\left(t\right)\ge 1-{\beta }_n$ for any $n\in \mathbb{N}$ imply $x=y=z$. Specially, $g_{x,a,a}\left(t\right)=1$ and $g_{a,y,z}\left(s\right)=1$ imply $x=y=z$;

(ii) 

$g_{y_n,x_n,x_n}\left(t\right)\ge 1-{\alpha }_n$ and $g_{x_n,y_m,z}\left(t\right)\ge 1-{\beta }_n$ for all $m>n$ with $m,n\in \mathbb{N}$ imply $G_{y_n,y_m,z}(t+s)\to 1$ as $n\to \infty$;

(iii) 

let $g_{x_n,x_m,x_l}\left(t\right)\ge 1-{\alpha }_n$ for all $n,m,l\in \mathbb{N}$, where $l>m>n$. Then $\left\{x_n\right\}$ is a Cauchy sequence;

(iv) 

let $g_{y,y,x_l}\left(t\right)\ge 1-{\alpha }_n$ for all $n\in \mathbb{N}$. Then $\left\{x_n\right\}$ is a Cauchy sequence.

Proof. 

To prove (ii), assume $ϵ>0$. By applying the definition of e-distance, there exists $\delta >0$ so that $g_{a,y,z}\left(t\right)\ge 1-\delta$ and $g_{x,a,a}\left(s\right)\ge 1-\delta$ induce $G_{x,y,z}(t+s)\ge 1-ϵ$. Select $n_0\in \mathbb{N}$ provided that ${\alpha }_n\le \delta$ and ${\beta }_n\le \delta$ for each $n\ge n_0$. Then $g_{y_n,x_n,x_n}\left(t\right)\ge 1-{\alpha }_n\ge 1-\delta$ and $g_{x_n,y_m,z}\left(t\right)\ge 1-{\beta }_n\ge 1-\delta$ for any $n\ge n_0$ and hence $G_{y_n,y_m,z}(t+s)\ge 1-ϵ$. Therefore, $\left\{y_n\right\}$ converges to z. Now, using (ii), (i) is established. To prove (iii), assume $ϵ>0$. Similar to the proof of (ii), select $\delta >0$ and $n_0\in \mathbb{N}$. Then, for all $n,m,l\ge n_0+1$, we get $g_{x_n,x_{n_0},x_{n_0}}\left(t\right)\ge 1-{\alpha }_{n_0}\ge 1-\delta$ and $g_{x_{n_0},x_l,x_m}\left(t\right)\ge 1-{\alpha }_{n_0}\ge 1-\delta$. Therefore, $G_{x_n,x_m,x_l}\left(t\right)\ge 1-ϵ$. Hence, $\left\{x_n\right\}$ is a Cauchy sequence. Now, it follows from (iii) that (iv) is true. □

Lemma 4.

Consider an e-distance g on $(\mathcal{X},G,\mathcal{T})$. Suppose that $E_{\lambda ,g}:{\mathcal{X}}^3\to {\mathbb{R}}^{+}\cup \left\{0\right\}$ is introduced by $E_{\lambda ,g}(x,y,z)=inf\{t>0:g_{x,y,z}\left(t\right)>1-\lambda \}$ for any $x,y,z\in \mathcal{X}$ and $\lambda \in (0,1)$. Then

(1) 

for all $\mu \in (0,1)$, there exists $\lambda \in (0,1)$ so that

$$E_{\mu ,g}(x_1,x_1,x_n)\le E_{\lambda ,g}(x_1,x_1,x_2)+E_{\lambda ,g}(x_2,x_2,x_3)+\cdots +E_{\lambda ,g}(x_{n-1},x_{n-1},x_n)$$

for each $x_1,\cdots ,x_n\in \mathcal{X}$;

(2) 

for every sequence $\left\{x_n\right\}$ in $\mathcal{X}$, $g_{x_n,x,x}\left(t\right)\to 1$ iff $E_{\lambda ,g}(x_n,x,x)\to 0$. Further, the sequence $\left\{x_n\right\}$ is Cauchy w.r.t. g iff it is Cauchy with $E_{\lambda ,g}$.

Proof. 

(1)

For every $\mu \in (0,1)$, we can gain $\lambda \in (0,1)$ provided that ${\mathcal{T}}^{n-1}(1-\lambda ,\dots ,1-\lambda )\ge 1-\mu$. Now, for every $\delta >0$, we have

$$\begin{array}{cc}\hfill & g_{x_1,x_1,x_n}(E_{\lambda ,g}(x_1,x_1,x_2)+E_{\lambda ,g}(x_2,x_2,x_3)+\cdots +E_{\lambda ,g}(x_{n-1},x_{n-1},x_n)+n\delta )\hfill \\ \hfill & \ge {\mathcal{T}}^{n-1}(g_{x_1,x_1,x_2}(E_{\lambda ,g}(x_1,x_1,x_2)+\delta ),g_{x_2,x_2,x_3}(E_{\lambda ,g}(x_2,x_2,x_3)+\delta )\hfill \\ \hfill & ,\cdots ,g_{x_{n-1},x_{n-1},x_n}(E_{\lambda ,g}(x_{n-1},x_{n-1},x_n)+\delta ))\hfill \\ \hfill & \ge {\mathcal{T}}^{n-1}(1-\lambda ,\dots ,1-\lambda )\ge 1-\mu \hfill \end{array}$$

which induces that

$$E_{\mu ,g}(x_1,x_1,x_n)\le E_{\lambda ,g}(x_1,x_1,x_2)+E_{\lambda ,g}(x_2,x_2,x_3)+\cdots +E_{\lambda ,g}(x_{n-1},x_{n-1},x_n)+n\delta .$$

Since $\delta >0$ is optional, we obtain

$$E_{\mu ,g}(x_1,x_1,x_n)\le E_{\lambda ,g}(x_1,x_1,x_2)+E_{\lambda ,g}(x_2,x_2,x_3)+\cdots +E_{\lambda ,g}(x_{n-1},x_{n-1},x_n).$$

(2)

Note that $g_{x_n,x,x}\left(\eta \right)\to 1-\lambda$ as $n\to \infty$ iff $E_{\lambda ,g}(x_n,x,x)<\eta$ for each $n\in \mathbb{N}$ and $\eta >0$.

□

In the sequel, we establish some coupled fixed point theorems by regarding an e-distance on a quasi-ordered complete PGM space.

Theorem 1.

Let $(\mathcal{X},G,\mathcal{T},⪯)$ be a quasi-ordered complete Menger PGM space with $\mathcal{T}$ of Hadzić-type I, g be an e-distance and $f:{\mathcal{X}}^2\to \mathcal{X}$ be a mapping having the mixed monotone property on $\mathcal{X}$. Assume that there exists a $k\in [0,1)$ such that

$$g_{f(x,y),f(u,v),f(w,z)}\left(t\right)\ge \frac{1}{2}(g_{x,u,w}\left(\frac{t}{k}\right)+g_{y,v,z}\left(\frac{t}{k}\right))$$

(1)

for all $x,y,z,u,v,w\in \mathcal{X}$ with $x⪰u⪰w$ and $y⪯v⪯z$, where either $u\ne w$ or $v\ne z$ and

$$sup\{\mathcal{T}(g_{x,y,z}\left(t\right),g_{x,y,f(x,y)}\left(t\right)):x,y\in \mathcal{X}\}<1.$$

(2)

for all $z\in \mathcal{X}$, where $z\ne f(z,q)$ for all $q\in \mathcal{X}$. If there exist $x_0,y_0\in \mathcal{X}$ so that $x_0⪯f(x_0,y_0)$ and $y_0⪰f(y_0,x_0)$, then f have a coupled fixed point in ${\mathcal{X}}^2$.

Proof. 

Since there exist $x_0,y_0\in \mathcal{X}$ with $x_0⪯f(x_0,y_0)$ and $y_0⪰f(y_0,x_0)$, and f has the mixed monotone property, we can construct Bhaskar-Lakshmikantham type iterative as follow:

$$\begin{array}{c}\hfill x_0⪯x_1⪯x_2⪯\cdots ⪯x_{n+1}⪯\cdots ,y_0⪰y_1⪰y_2⪰\cdots ⪰y_{n+1}⪰\cdots \end{array}$$

for all $n\ge 0$, where

$$\begin{array}{cc}\hfill & x_{n+1}=f^{n+1}(x_0,y_0)=f(f^n(x_0,y_0),f^n(y_0,x_0)),\hfill \\ \hfill & y_{n+1}=f^{n+1}(y_0,x_0)=f(f^n(y_0,x_0),f^n(x_0,y_0)).\hfill \end{array}$$

If $(x_{n+1},y_{n+1})=(x_n,y_n)$, then f has a coupled fixed point. Otherwise, assume $(x_{n+1},y_{n+1})\ne (x_n,y_n)$ for each $n\ge 0$; that is, either $x_{n+1}=f(x_n,y_n)\ne x_n$ or $y_{n+1}=f(y_n,x_n)\ne y_n$. Now, by induction and (1), we obtain

$$\begin{array}{cc}\hfill g_{x_n,x_n,x_{n+1}}\left(t\right)& \ge \frac{1}{2}(g_{x_0,x_0,x_1}(\frac{t}{k^n})+g_{y_0,y_0,y_1}(\frac{t}{k^n})),\hfill \\ \hfill g_{y_n,y_n,y_{n+1}}\left(t\right)& \ge \frac{1}{2}(g_{y_0,y_0,y_1}(\frac{t}{k^n})+g_{x_0,x_0,x_1}(\frac{t}{k^n})),\hfill \end{array}$$

for each $n\ge 0$ which induces that $g_{x_n,x_n,x_{n+1}}\left(t\right)\ge \frac{1}{2}{g}_{x_0,x_0,x_1}(\frac{t}{k^n})$ and $g_{y_n,y_n,y_{n+1}}\left(t\right)\ge \frac{1}{2}{g}_{y_0,y_0,y_1}(\frac{t}{k^n})$. Therefore,

$$\begin{array}{cc}\hfill E_{\lambda ,g}(x_n,x_n,x_{n+1})& =inf\{t>0:g_{x_n,x_n,x_{n+1}}\left(t\right)>1-\lambda \}\hfill \\ \hfill & \le inf\{t>0:\frac{1}{2}{g}_{x_0,x_0,x_1}(\frac{t}{k^n})>1-\lambda \}\hfill \\ \hfill & =2k^n{E}_{\lambda ,g}(x_0,x_0,x_1).\hfill \end{array}$$

Thus, for $m>n$ and $\lambda \in (0,1)$, there exists $\gamma \in (0,1)$ so that

$$\begin{array}{c}\hfill E_{\lambda ,g}(x_n,x_n,x_m)\le E_{\gamma ,g}(x_n,x_n,x_{n+1})+\cdots +E_{\gamma ,g}(x_{m-1},x_{m-1},x_m)\le \frac{2k^n}{1-k}{E}_{\gamma ,g}(x_0,x_0,x_1).\end{array}$$

Now, there exists $n_0\in \mathbb{N}$ so that for each $n>n_0$, $E_{\lambda ,g}(x_n,x_n,x_m)\to 0$. By Lemmas 3 and 4, $\left\{x_n\right\}$ is a Cauchy sequence. Thus, using Lemma 4 (ii), there exit $n_1\in \mathbb{N}$ and a sequence ${\delta }_n\to 0$ so that $g_{x_n,x_n,x_m}\left(t\right)\ge 1-{\delta }_n$ for $n\ge max\{n_0,n_1\}$. Since $\mathcal{X}$ is complete, $\left\{x_n\right\}$ converges to a point $p\in \mathcal{X}$. Similarly, $\left\{y_n\right\}$ is convergent to a point $q\in \mathcal{X}$. By (r2), we obtain $g_{x_n,x_n,p}\left(t\right)={lim}_{m\to \infty }{g}_{x_n,x_n,x_m}\left(t\right)\ge 1-{\delta }_n$ for $n\ge max\{n_0,n_1\}$. Moreover, we get $g_{x_n,x_{n+1},x_{n+1}}\left(t\right)\ge 1-{\delta }_n$. Now, we show that f has a coupled fixed point. Let $p\ne f(p,q)$. Then, by (2), we obtain

$$\begin{array}{cc}\hfill 1& >sup\{\mathcal{T}(g_{x,y,p}\left(t\right),g_{x,y,f(x,y)}\left(t\right)):x,y\in \mathcal{X}\}\hfill \\ \hfill & \ge sup\{\mathcal{T}(g_{x_n,x_n,p}\left(t\right),g_{x_n,x_{n+1},x_{n+1}}\left(t\right)):n\in \mathbb{N}\}\hfill \\ \hfill & \ge sup\{\mathcal{T}(1-{\delta }_n,1-{\delta }_n):n\in \mathbb{N}\}=1,\hfill \end{array}$$

which is a contradiction. Consequently, we get $p=f(p,q)$. Similarly, we obtain $f(q,p)=q$. Here, the proof ends. □

Theorem 2.

Assume the assumptions of Theorem 1 are held and consider the continuity of f instead of relation (2) . Then f has a coupled fixed point.

Proof. 

As in the proof of Theorem 1, construct $\left\{x_n\right\}$ and $\left\{y_n\right\}$, where $x_n\to p$, $y_n\to q$, $x_{n+1}=f(x_n,y_n)$. Now, by the continuity of f and by taking the limit as $n\to \infty$, we get $f(p,q)=p$. Analogously, we can obtain $f(q,p)=q$. Therefore, $(p,q)$ is a coupled fixed point of f. □

Example 2.

Assume that $\mathcal{X}=[0,\infty )$, $‘‘⪯"$ is a quasi-ordered on $\mathcal{X}$ and $\mathcal{T}(a,b)=min\{a,b\}$. Define a constant function $f:{\mathcal{X}}^2\to \mathcal{X}$ by $f(a,b)=p$ and $G:{\mathcal{X}}^3\to D^{+}$ by $G_{x,y,z}\left(t\right)=\frac{t}{t+G^{*}(x,y,z)}$ with $G^{*}(x,y,z)=|x-y|+|x-z|+|y-z|$ for each $x,y,z\in \mathcal{X}$. Clearly, G satisfies (PG1)-(PG4). Consider $g_{x,y,z}\left(t\right)=1-c$, where $c\in (0,1)$. Then g is an e-distance on $\mathcal{X}$. Clearly, for all $x,y,z,u,v,w\in \mathcal{X}$ and for any $t>0$, we have $g_{f(x,y),f(u,v),f(w,z)}\left(t\right)\ge \frac{1}{2}(g_{x,u,w}\left(\frac{t}{k}\right)+g_{y,v,z}\left(\frac{t}{k}\right))$. Moreover, there exist $x_0=0$ and $y_0=1$ so that $0=x_0⪯f(x_0,y_0)$ and $1=y_0⪰f(y_0,x_0)=1$. Therefore, all of the hypothesis of Theorem 2 are held. Clearly, $(p,p)$ is a coupled fixed point the function f.

3. Application

Consider the following system of integral equations:

$$\left\{\begin{array}{c}x\left(t\right)={\int }_a^{b}M(t,s)K(s,x\left(s\right),y\left(s\right))ds,\hfill \\ y\left(t\right)={\int }_a^{b}M(t,s)K(s,y\left(s\right),x\left(s\right))ds,\hfill \end{array}\right.$$

(3)

for all $t\in I=[a,b]$, where $b>a$, $M\in C(I\times I,[0,\infty \left)\right)$ and $K\in C(I\times \mathbb{R}\times \mathbb{R},\mathbb{R})$.

Let $C(I,\mathbb{R})$ be the Banach space of every real continuous functions on I with ${\left|\right|x\left|\right|}_{\infty }={max}_{t\in I}\left|x\left(t\right)\right|$ for all $x\in C(I,\mathbb{R})$ and $C(I\times I\times C(I,\mathbb{R}),\mathbb{R})$ be the space of every continuous functions on $I\times I\times C(I,\mathbb{R})$. Define a mapping $G:C(I,\mathbb{R})\times C(I,\mathbb{R})\to D^{+}$ by $G_{x,y,z}\left(t\right)=\chi (\frac{t}{2}{-(\parallel x-y\parallel }_{\infty }{+\parallel x-z\parallel }_{\infty }{+\parallel y-z\parallel }_{\infty }\left)\right)$ for all $x,y,z\in C(I,\mathbb{R})$ and $t>0$, where

$$\chi \left(t\right)=\left\{\begin{array}{c}0ift\le 0\hfill \\ 1ift>0\hfill \end{array}\right.$$

Then, $\left(C\right(I,\mathbb{R}),G,\mathcal{T})$ with $\mathcal{T}(a,b)=min\{a,b\}$ is a complete Menger PGM space ([7]). Consider an e-distance on $\mathcal{X}$ by $g_{x,y,z}\left(t\right)=min\{G_{x,y,z}\left(t\right),G_{Ax,Ay,Az}\left(t\right)\},$ where $A:C(I,\mathbb{R})\to C(I,\mathbb{R})$ and $Ax=\frac{x}{2}$. Moreover, we define the relation $‘‘⪯"$ on $C(I,\mathbb{R})$ by $x⪯y\iff {\left|\right|x\left|\right|}_{\infty }\le {\left|\right|y\left|\right|}_{\infty }$ for all $x,y\in C(I,\mathbb{R})$. Clearly the relation $‘‘⪯"$ is a quasi-order relation on $C(I,\mathbb{R})$ and $\left(C\right(I,\mathbb{R}),G,\mathcal{T},⪯)$ is a quasi-ordered complete PGM space.

Theorem 3.

Let $\left(C\right(I,\mathbb{R}),G,\mathcal{T},⪯)$ be a quasi-ordered complete Menger PGM space and $f:C(I,\mathbb{R})\times C(I,\mathbb{R})\to C(I,\mathbb{R})$ be a operator defined by $f(x,y)\left(t\right)={\int }_a^{b}M(t,s)K(s,x\left(s\right),y\left(s\right))ds$, where $M\in C(I\times I,[0,\infty \left)\right)$ and $K\in C(I\times \mathbb{R}\times \mathbb{R},\mathbb{R})$ are two operators. Assume the following properties are held:

(i) 

${\left|\right|K\left|\right|}_{\infty }={sup}_{s\in I,x,y\in C(I,\mathbb{R})}\left|K(s,x\left(s\right),y\left(s\right))\right|<\infty$;

(ii) 

for every $x,y\in C(I,\mathbb{R})$ and every $t,s\in I$, we have

$$\left|\right|K(s,x\left(s\right),y\left(s\right))-{K(s,u\left(s\right),v\left(s\right))\left|\right|}_{\infty }\le \frac{1}{4}(max|x\left(s\right)-u\left(s\right)|+max|y\left(s\right)-v\left(s\right)\left|\right);$$

(iii) 

${max}_{t\in I}{\int }_a^{b}M(t,s)ds<1$.

Then, the system (3) have a solution in $C(I,\mathbb{R})\times C(I,\mathbb{R})$.

Proof. 

For all $x,y\in C(I,\mathbb{R})$, let ${\parallel x-y\parallel }_{\infty }={max}_{t\in I}\left(\right|x\left(t\right)-y\left(t\right)\left|\right)$. Then, for all $x,y,z,u,v,w\in C(I,\mathbb{R})$, we have

$$\begin{array}{cc}\hfill {\parallel f(x,y)-f(u,v)\parallel }_{\infty }& \le \underset{t\in I}{max}{\int }_a^{b}M(t,s)|K(s,x\left(s\right)y\left(s\right))-K(s,u\left(s\right),v\left(s\right))|ds\hfill \\ \hfill & \le max(\frac{1}{4}\left(\right|x\left(s\right)-u\left(s\right)|+|y\left(s\right)-v\left(s\right)\left|\right))\underset{t\in I}{max}{\int }_a^{b}M(t,s)ds\hfill \\ \hfill & \le max(\frac{1}{4}\left(\right|x\left(s\right)-u\left(s\right)|+|y\left(s\right)-v\left(s\right)\left|\right)).\hfill \end{array}$$

We consider two following cases:

  Case 1. Let

$$\begin{array}{cc}\hfill g_{f(x,y),f(u,v),f(w,z)}\left(t\right)& =min\{G_{f(x,y),f(u,v),f(w,z)}\left(t\right),G_{Af(x,y),Af(u,v),Af(w,z)}\left(t\right)\}\hfill \\ \hfill & =G_{f(x,y),f(u,v),f(w,z)}\left(t\right).\hfill \end{array}$$

Then, we obtain

$$\begin{array}{cc}\hfill g_{f(x,y),f(u,v),f(w,z)}\left(t\right)& =G_{f(x,y),f(u,v),f(w,z)}\left(t\right)\hfill \\ \hfill & =\chi (\frac{t}{2}{-(\parallel f(x,y)-f(u,v)\parallel }_{\infty }{+\parallel f(x,y)-f(w,z)\parallel }_{\infty }{+\parallel f(u,v)-f(w,z)\parallel }_{\infty }\left)\right)\hfill \\ \hfill & \ge \chi (\frac{t}{2}-(max(\frac{1}{4}\left(\right|x\left(s\right)-u\left(s\right)|+|y\left(s\right)-v\left(s\right)\left|\right))\hfill \\ \hfill & +max(\frac{1}{4}\left(\right|x\left(s\right)-w\left(s\right)|+|y\left(s\right)-z\left(s\right)\left|\right))\hfill \\ \hfill & +max(\frac{1}{4}\left(\right|u\left(s\right)-w\left(s\right)|+|v\left(s\right)-z\left(s\right)\left|\right)\left)\right))\hfill \\ \hfill & =\chi (t-\frac{1}{2}(max(\left(\right|x\left(s\right)-u\left(s\right)|+|y\left(s\right)-v\left(s\right)\left|\right))\hfill \\ \hfill & +max\left(\right(\left|x\right(s)-w(s\left)\right|+\left|y\right(s)-z(s\left)\right|\left)\right)+max\left(\right(\left|u\right(s)-w(s\left)\right|+\left|v\right(s)-z(s\left)\right|\left)\right)\left)\right)\hfill \\ \hfill & \ge \frac{1}{2}\left(\chi \right(t-(max(|x\left(s\right)-u\left(s\right)|+|x\left(s\right)-w\left(s\right)|+|u\left(s\right)-w\left(s\right)|\left)\right))\hfill \\ \hfill & +\chi (t-(max\left(\right|y\left(s\right)-v\left(s\right)|+|y\left(s\right)-z\left(s\right)|+|v\left(s\right)-z\left(s\right)\left|\right)\left)\right))\hfill \\ \hfill & =\frac{1}{2}(G_{x,u,w}\left(2t\right)+G_{y,v,z}\left(2t\right))\ge \frac{1}{2}(g_{x,u,w}\left(2t\right)+g_{y,v,z}\left(2t\right)).\hfill \end{array}$$

  Case 2. Let

$$\begin{array}{cc}\hfill g_{f(x,y),f(u,v),f(w,z)}\left(t\right)& =min\{G_{f(x,y),f(u,v),f(w,z)}\left(t\right),G_{Af(x,y),Af(u,v),Af(w,z)}\left(t\right)\}\hfill \\ \hfill & =G_{Af(x,y),Af(u,v),Af(w,z)}\left(t\right).\hfill \end{array}$$

By $Ax=\frac{x}{2}$, we have

$$\begin{array}{cc}\hfill g_{f(x,y),f(u,v),f(w,z)}\left(t\right)& =G_{Af(x,y),Af(u,v),Af(w,z)}\left(t\right)\hfill \\ \hfill & =\chi (\frac{t}{2}-\frac{1}{2}{(\parallel f(x,y)-f(u,v)\parallel }_{\infty }{+\parallel f(x,y)-f(w,z)\parallel }_{\infty }{+\parallel f(u,v)-f(w,z)\parallel }_{\infty }\left)\right)\hfill \\ \hfill & \ge \chi (\frac{t}{2}{-(\parallel f(x,y)-f(u,v)\parallel }_{\infty }{+\parallel f(x,y)-f(w,z)\parallel }_{\infty }{+\parallel f(u,v)-f(w,z)\parallel }_{\infty }\left)\right)\hfill \\ \hfill & \ge \chi (\frac{t}{2}-(max(\frac{1}{4}\left(\right|x\left(s\right)-u\left(s\right)|+|y\left(s\right)-v\left(s\right)\left|\right))\hfill \\ \hfill & +max(\frac{1}{4}\left(\right|x\left(s\right)-w\left(s\right)|+|y\left(s\right)-z\left(s\right)\left|\right))\hfill \\ \hfill & +max(\frac{1}{4}\left(\right|u\left(s\right)-w\left(s\right)|+|v\left(s\right)-z\left(s\right)\left|\right)\left)\right))\hfill \\ \hfill & =\chi (t-\frac{1}{2}(max(\left(\right|x\left(s\right)-u\left(s\right)|+|y\left(s\right)-v\left(s\right)\left|\right))\hfill \\ \hfill & +max\left(\right(\left|x\right(s)-w(s\left)\right|+\left|y\right(s)-z(s\left)\right|\left)\right)+max\left(\right(\left|u\right(s)-w(s\left)\right|+\left|v\right(s)-z(s\left)\right|\left)\right)\left)\right)\hfill \\ \hfill & \ge \frac{1}{2}\left(\chi \right(t-(max(|x\left(s\right)-u\left(s\right)|+|x\left(s\right)-w\left(s\right)|+|u\left(s\right)-w\left(s\right)|\left)\right))\hfill \\ \hfill & +\chi (t-(max\left(\right|y\left(s\right)-v\left(s\right)|+|y\left(s\right)-z\left(s\right)|+|v\left(s\right)-z\left(s\right)\left|\right)\left)\right))\hfill \\ \hfill & =\frac{1}{2}(G_{x,u,w}\left(2t\right)+G_{y,v,z}\left(2t\right))\ge \frac{1}{2}(g_{x,u,w}\left(2t\right)+g_{y,v,z}\left(2t\right))\hfill \end{array}$$

for all $x,y,z,u,v,w\in C(I,\mathbb{R})$. Therefore, by Theorem 2 with $k=\frac{1}{2}$ for all $x,y,z,u,v,w\in C(I,\mathbb{R})$ and $t>0$, we deduce that the operator f has a coupled fixed point which is the solution of the system of the integral equations. □

4. Conclusions

The new concept of e-distance, which is a generalization of r-distance in PGM space has been introduced. Moreover, some of properties of e-distance have been discussed. In addition, we obtained several new coupled fixed point results. Ultimately, to illustrate the usability of the main theorem, the existence of a solution for a system of integral equations is proved.