Common Fixed Point and Endpoint Theorems for a Countable Family of Multi-Valued Mappings

Abstract

We prove some common fixed point and endpoint theorems for a countable infinite family of multi-valued mappings, as well as Allahyari et al. (2015) did for self-mappings. An example and an application to a system of integral equations are given to show the usability of the results.

1. Introduction

The study of common fixed point for a family of contraction mappings was initiated by Ćirić in [1]. Recently, in 2015, Allahyari et al. [2] introduced some new type of contractions for a countable family of contraction self-mappings and studied common fixed point for them.

On the other hand, existence of a fixed point for multi-valued mappings has been important for many mathematicians. In 1969, Nadler [3] extended the Banach contraction principle to multi-valued mappings. After that, many authors generalized Nadler’s result in different ways (see, for instance [4,5,6,7,8]).

In 2012, Samet et al. [9] introduced the notion of $\alpha$-admisssible mappings and a new type of contraction to a mapping $\mathsf{T}:\mathsf{X}\to \mathsf{X}$ called $\alpha$-$\psi$-contractive mapping, that is, $\alpha (x,y)d(\mathsf{T}x,\mathsf{T}y)\le \psi (d(x,y))$ for all $x,y\in \mathsf{X}$. This result generalized and improved many existing fixed point results. In the last few years, some authors have extended the notion of $\alpha$-admisssibility and $\alpha$-$\psi$-contraction to multi-valued mappings (see, [10,11]). In addition, common fixed point for a finite family or countable family of multi-valued mappings has been studied by some researchers (see, for example [12,13,14,15,16]).

The aim of this paper is to extend the new type of common contractivity for a family of mappings, introduced by Allahyari et al. (2015), to $\alpha$-admisssible multi-valued mappings.

Let $(\mathsf{X},d)$ be a metric space, $2^{\mathsf{X}}$ the set of all nonempty subsets of $\mathsf{X}$, and $\mathcal{CL}(\mathsf{X})$ the set of all nonempty closed subsets of $\mathsf{X}$. Assume that $\mathcal{H}$ is the generalized Hausdorff metric on $\mathcal{CL}(\mathsf{X})$ defined by

$$\begin{array}{c}\hfill \mathcal{H}(A,B)=\left\{\begin{array}{cc}max\{\underset{x\in A}{sup}\mathsf{D}(x,B),\underset{y\in B}{sup}\mathsf{D}(y,A)\},\hfill & \mathrm{if} \mathrm{it} \mathrm{exists},\hfill \\ \infty ,\hfill & \mathrm{otherwise},\hfill \end{array}\right.\end{array}$$

(1)

for all $A,B\in \mathcal{CL}(\mathsf{X})$, where $\mathsf{D}(x,B)={inf}_{y\in B}d(x,y)$. Let $\mathsf{T}:\mathsf{X}\to 2^{\mathsf{X}}$ is a multi-valued mapping. An element $x\in \mathsf{X}$ is said to be a fixed point of $\mathsf{T}$ if $x\in \mathsf{T}x$, and x is called an endpoint of $\mathsf{T}$ whenever $\mathsf{T}x=\left\{x\right\}$.

2. Main Results

Now, we are ready to state and prove the main results of this study.

Definition 1.

Let $\mathsf{X}$ be an arbitrary space and $\alpha :\mathsf{X}\times \mathsf{X}\to [0,\infty )$ be a function. Assume that ${\mathsf{T}}_n:\mathsf{X}\to 2^{\mathsf{X}}$ (n = 1,2,...) is a family of multi-valued mappings. We say that $\left\{{\mathsf{T}}_n\right\}$ is α-admissible whenever for each $x\in \mathsf{X}$ and $y\in {\mathsf{T}}_{n}x$ with $\alpha (x,y)\ge 1$, we have $\alpha (y,\mathsf{z})\ge 1$ for all $\mathsf{z}\in {\mathsf{T}}_{n+1}y$.

Theorem 1.

Let $(\mathsf{X},d)$ be a complete metric space and $0<a_{i,j}(i,j=1,2,\dots )$ with $a_{i,i+1}\ne 1$ for all $i=1,2,\dots$ satisfy:

(i) 

for each j, ${\overline{lim}}_{i\to \infty }{a}_{i,j}<1;$

(ii) 

${\sum }_{n=1}^{\infty }{A}_n<\infty$, where $A_n={\prod }_{i=1}^n\frac{a_{i,i+1}}{1-a_{i,i+1}}$.

Let $\alpha :\mathsf{X}\times \mathsf{X}\to [0,\infty )$ be a given function and $\left\{{\mathsf{T}}_n\right\}$ be a sequence of multi-valued operators ${\mathsf{T}}_n:\mathsf{X}\to \mathcal{CL}(\mathsf{X})$ (n = 1,2,...) such that

$$\begin{array}{c}\hfill \alpha (x,y)\mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{j}y)\le a_{i,j}[\mathsf{D}(x,{\mathsf{T}}_{j}y)+\mathsf{D}(y,{\mathsf{T}}_{i}x)],\end{array}$$

(2)

for all $x,y\in \mathsf{X};$ $i,j=1,2,\dots$ with $x\ne y$ and $i\ne j$. Moreover, assume that the following assertions hold:

(iii) 

there exist $x_0\in \mathsf{X}$ and $x_1\in {\mathsf{T}}_1{x}_0$ with $x_0\ne x_1$ and $\alpha (x_0,x_1)\ge 1;$

(iv) 

$\left\{{\mathsf{T}}_n\right\}$ is α-admissible;

(v) 

for each sequence $\left\{x_n\right\}$ in $\mathsf{X}$ with $\alpha (x_n,x_{n+1})\ge 1$ for all n and $x_n\to x$, we have $\alpha (x_n,x)\ge 1$ for all n.

Then each ${\mathsf{T}}_n$ have a common fixed point in $\mathsf{X}$.

Proof. 

Using (iii) and (2), we have

$$\begin{array}{ccc}\mathsf{D}(x_1,{\mathsf{T}}_2{x}_1)\hfill & \le \hfill & \alpha (x_0,x_1)\mathcal{H}({\mathsf{T}}_1{x}_0,{\mathsf{T}}_2{x}_1)\hfill \\ & \le \hfill & a_{1,2}[\mathsf{D}(x_0,{\mathsf{T}}_2{x}_1)+\mathsf{D}(x_1,{\mathsf{T}}_1{x}_0)]\hfill \\ & =\hfill & a_{1,2}\mathsf{D}(x_0,{\mathsf{T}}_2{x}_1)\hfill \\ & \le \hfill & a_{1,2}[d(x_0,x_1)+\mathsf{D}(x_1,{\mathsf{T}}_2{x}_1)],\hfill \end{array}$$

which implies

$$\mathsf{D}(x_1,{\mathsf{T}}_2{x}_1)\le \frac{a_{1,2}}{1-a_{1,2}}d(x_0,x_1)<\frac{a_{1,2}}{1-a_{1,2}}pd(x_0,x_1),$$

where $p>1$ is a fixed number. From the above inequality, there exists $x_2\in {\mathsf{T}}_2{x}_1$ such that $d(x_1,x_2)<\frac{a_{1,2}}{1-a_{1,2}}pd(x_0,x_1)$. Since $\left\{{\mathsf{T}}_n\right\}$ is $\alpha$-admissible, we have $\alpha (x_1,x_2)\ge 1$. Similarly,

$$\mathsf{D}(x_2,{\mathsf{T}}_3{x}_2)\le \frac{a_{2,3}}{1-a_{2,3}}d(x_1,x_2)<\frac{a_{2,3}}{1-a_{2,3}}\frac{a_{1,2}}{1-a_{1,2}}pd(x_0,x_1),$$

and so there exists $x_3\in {\mathsf{T}}_3{x}_2$ such that $d(x_2,x_3)<\frac{a_{2,3}}{1-a_{2,3}}\frac{a_{1,2}}{1-a_{1,2}}pd(x_0,x_1)$. Continuing this process, we obtain a sequence $\left\{x_n\right\}$ in $\mathsf{X}$ such that $x_{n+1}\in {\mathsf{T}}_{n+1}{x}_n$, $\alpha (x_n,x_{n+1})\ge 1$, and

$$\begin{array}{c}\hfill d(x_n,x_{n+1})<A_{n}pd(x_0,x_1),\mathrm{for}\mathrm{all}n=1,2,\dots .\end{array}$$

(3)

For any $n,m\in \mathbb{N}$ with $n<m$, from triangle inequality, we get

$$d(x_n,x_m)\le \sum _{k=n}^{m-1}d(x_k,x_{k+1})\le \sum _{k=n}^{m-1}{A}_{k}pd(x_0,x_1)\to 0$$

as $n,m\to \infty$. Therefore, we have shown that $\left\{x_n\right\}$ is a Cauchy sequence. Since $(\mathsf{X},d)$ is complete, there exists $x\in \mathsf{X}$ such that $x_n\to x$. From (v), we get $\alpha (x_n,x)\ge 1$ for all n. Now, we shall show that x is a common fixed point of ${\mathsf{T}}_n$. Let m be an arbitrary positive integer. Then, for any $n\in \mathbb{N}$, we have

$$\begin{array}{ccc}\mathsf{D}(x,{\mathsf{T}}_{m}x)\hfill & \le \hfill & d(x,x_n)+\mathsf{D}(x_n,{\mathsf{T}}_{m}x)\hfill \\ & \le \hfill & d(x,x_n)+\alpha (x_{n-1},x)\mathcal{H}({\mathsf{T}}_n{x}_{n-1},{\mathsf{T}}_{m}x)\hfill \\ & \le \hfill & d(x,x_n)+a_{n,m}[\mathsf{D}(x_{n-1},{\mathsf{T}}_{m}x)+\mathsf{D}(x,{\mathsf{T}}_n{x}_{n-1})]\hfill \\ & \le \hfill & d(x,x_n)+a_{n,m}[\mathsf{D}(x_{n-1},{\mathsf{T}}_{m}x)+d(x,x_n)].\hfill \end{array}$$

Taking $\overline{lim}$ in both sides of the above inequality, as $n\to \infty$, we get

$$\mathsf{D}(x,{\mathsf{T}}_{m}x)\le ({\overline{lim}}_{n\to \infty }{a}_{n,m})\mathsf{D}(x,{\mathsf{T}}_{m}x),$$

which implies $\mathsf{D}(x,{\mathsf{T}}_{m}x)=0$ and so $x\in {\mathsf{T}}_{m}x$. □

Theorem 2.

Let $(\mathsf{X},d)$ be a complete metric space and $0<a_{i,j}(i,j=1,2,\dots )$ with $a_{i,i+1}\ne 1$ for all $i=1,2,\dots$ satisfy:

(i) 

for each (j), ${\overline{lim}}_{i\to \infty }{a}_{i,j}<1;$

(ii) 

${\sum }_{n=1}^{\infty }{A}_n<\infty$ where $A_n={\prod }_{i=1}^n\frac{a_{i,i+1}}{1-a_{i,i+1}}$.

Let $\alpha :\mathsf{X}\times \mathsf{X}\to [0,\infty )$ be a given function and $\left\{{\mathsf{T}}_n\right\}$ be a sequence of multi-valued operators ${\mathsf{T}}_n:\mathsf{X}\to \mathcal{CL}(\mathsf{X})$ (n = 1,2,...) such that

$$\begin{array}{c}\hfill \alpha (x,y)\mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{j}y)\le a_{i,j}max\{d(x,y),\mathsf{D}(x,{\mathsf{T}}_{i}x),\mathsf{D}(y,{\mathsf{T}}_{j}y),\mathsf{D}(x,{\mathsf{T}}_{j}y),\mathsf{D}(y,{\mathsf{T}}_{i}x)\},\end{array}$$

(4)

for all $x,y\in \mathsf{X};$ $i,j=1,2,\dots$ with $x\ne y$ and $i\ne j$. Moreover, assume that the following assertions hold:

(iii) 

there exist $x_0\in \mathsf{X}$ and $x_1\in {\mathsf{T}}_1{x}_0$ with $x_0\ne x_1$ and $\alpha (x_0,x_1)\ge 1;$

(iv) 

$\left\{{\mathsf{T}}_n\right\}$ is α-admissible;

(v) 

for each sequence $\left\{x_n\right\}$ in $\mathsf{X}$ with $\alpha (x_n,x_{n+1})\ge 1$ for all n and $x_n\to x$, we have $\alpha (x_n,x)\ge 1$ for all n.

Then each ${\mathsf{T}}_n$ have a common fixed point in $\mathsf{X}$.

Proof. 

By (iii) and (4), we have

$$\begin{array}{ccc}\mathsf{D}(x_1,{\mathsf{T}}_2{x}_1)\hfill & \le \hfill & \alpha (x_0,x_1)\mathcal{H}({\mathsf{T}}_1{x}_0,{\mathsf{T}}_2{x}_1)\hfill \\ & \le \hfill & a_{1,2}max\{d(x_0,x_1),\mathsf{D}(x_0,{\mathsf{T}}_1{x}_0),\mathsf{D}(x_1,{\mathsf{T}}_2{x}_1),\mathsf{D}(x_0,{\mathsf{T}}_2{x}_1),\mathsf{D}(x_1,{\mathsf{T}}_1{x}_0)\}\hfill \\ & \le \hfill & a_{1,2}[d(x_0,x_1)+\mathsf{D}(x_1,{\mathsf{T}}_2{x}_1)],\hfill \end{array}$$

which implies

$$\mathsf{D}(x_1,{\mathsf{T}}_2{x}_1)\le \frac{a_{1,2}}{1-a_{1,2}}d(x_0,x_1)<\frac{a_{1,2}}{1-a_{1,2}}pd(x_0,x_1),$$

which $p>1$ is a fixed number. From the above inequality, there exists $x_2\in {\mathsf{T}}_2{x}_1$ such that $d(x_1,x_2)<\frac{a_{1,2}}{1-a_{1,2}}pd(x_0,x_1)$. Continuing in this manner and as in proof of Theorem 1, we obtain a sequence $\left\{x_n\right\}$ with $\alpha (x_n,x_{n+1})\ge 1$ and $x\in \mathsf{X}$ such that $x_n\to x$. Using (v), we get $\alpha (x_n,x)\ge 1$ for all n. Next, we show that x is a common fixed point of ${\mathsf{T}}_n$. Let m be an arbitrary positive integer. Then, for any $n\in \mathbb{N}$, we have

$$\begin{array}{ccc}\mathsf{D}(x,{\mathsf{T}}_{m}x)\hfill & \le \hfill & d(x,x_n)+\mathsf{D}(x_n,{\mathsf{T}}_{m}x)\hfill \\ & \le \hfill & d(x,x_n)+\alpha (x_{n-1},x)\mathcal{H}({\mathsf{T}}_n{x}_{n-1},{\mathsf{T}}_{m}x)\hfill \\ & \le \hfill & d(x,x_n)+a_{n,m}max\{d(x_{n-1},x),\mathsf{D}(x_{n-1},{\mathsf{T}}_n{x}_{n-1}),\mathsf{D}(x,{\mathsf{T}}_{m}x),\hfill \\ & & \mathsf{D}(x_{n-1},{\mathsf{T}}_{m}x),\mathsf{D}(x,{\mathsf{T}}_n{x}_{n-1})\}\hfill \\ & \le \hfill & d(x,x_n)+a_{n,m}max\{d(x_{n-1},x),d(x_{n-1},x_n),\mathsf{D}(x,{\mathsf{T}}_{m}x),\mathsf{D}(x_{n-1},{\mathsf{T}}_{m}x),d(x,x_n)\}.\hfill \end{array}$$

Taking $\overline{lim}$ as $n\to \infty$, we obtain $\mathsf{D}(x,{\mathsf{T}}_{m}x)\le ({\overline{lim}}_{n\to \infty }{a}_{n,m})\mathsf{D}(x,{\mathsf{T}}_{m}x),$ which implies $\mathsf{D}(x,{\mathsf{T}}_{m}x)=0$. This means that $x\in {\mathsf{T}}_{m}x$ and the proof is complete. □

Theorem 3.

Let $(\mathsf{X},d)$ be a complete metric space and $0\le a_{i,j},0<b_{i,j}(i,j=1,2,\dots )$ with $a_{i,i+1}\ne 1$ for all $i=1,2,\dots$ satisfy:

(i) 

for each j, ${\overline{lim}}_{i\to \infty }{a}_{i,j}<1$ and ${\overline{lim}}_{i\to \infty }{b}_{i,j}<\infty ;$

(ii) 

${\sum }_{n=1}^{\infty }{A}_n<\infty$ where $A_n={\prod }_{i=1}^n\frac{b_{i,i+1}}{1-a_{i,i+1}}$.

Let $\alpha :\mathsf{X}\times \mathsf{X}\to [0,\infty )$ be a given function and $\left\{{\mathsf{T}}_n\right\}$ be a sequence of multi-valued operators ${\mathsf{T}}_n:\mathsf{X}\to \mathcal{CL}(\mathsf{X})$ (n = 1,2,...) such that

$$\begin{array}{c}\hfill \alpha (x,y)\mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{j}y)\le a_{i,j}\mathsf{D}(y,{\mathsf{T}}_{j}y)\phi (\mathsf{D}(x,{\mathsf{T}}_{i}x),d(x,y))+b_{i,j}d(x,y),\end{array}$$

(5)

for all $x,y\in \mathsf{X};$ $i,j=1,2,\dots$ with $x\ne y$ and $i\ne j$, where $\phi :[0,\infty )\times [0,\infty )\to [0,\infty )$ is a continuous function such that $\phi (t,t)=1$ for all $t\in [0,\infty )$ and for any $t_1,s_1,t_2,s_2\in [0,\infty ),$

$$t_1\le t_2,s_1=s_2⟹\phi (t_1,s_1)\le \phi (t_2,s_2).$$

Moreover, assume that the following assertions hold:

(iii) 

there exist $x_0\in \mathsf{X}$ and $x_1\in {\mathsf{T}}_1{x}_0$ with $x_0\ne x_1$ and $\alpha (x_0,x_1)\ge 1;$

(iv) 

$\left\{{\mathsf{T}}_n\right\}$ is α-admissible;

(v) 

for each sequence $\left\{x_n\right\}$ in $\mathsf{X}$ with $\alpha (x_n,x_{n+1})\ge 1$ for all n and $x_n\to x$, we have $\alpha (x_n,x)\ge 1$ for all n.

Then each ${\mathsf{T}}_n$ have a common fixed point in $\mathsf{X}$.

Proof. 

By (iii) and (5), we have

$$\begin{array}{ccc}\mathsf{D}(x_1,{\mathsf{T}}_2{x}_1)\hfill & \le \hfill & \alpha (x_0,x_1)\mathcal{H}({\mathsf{T}}_1{x}_0,{\mathsf{T}}_2{x}_1)\hfill \\ & \le \hfill & a_{1,2}\mathsf{D}(x_1,{\mathsf{T}}_2{x}_1)\phi (\mathsf{D}(x_0,{\mathsf{T}}_1{x}_0),d(x_0,x_1))+b_{1,2}d(x_0,x_1)\hfill \\ & \le \hfill & a_{1,2}\mathsf{D}(x_1,{\mathsf{T}}_2{x}_1)\phi (d(x_0,x_1),d(x_0,x_1))+b_{1,2}d(x_0,x_1)\hfill \\ & \le \hfill & a_{1,2}\mathsf{D}(x_1,{\mathsf{T}}_2{x}_1)+b_{1,2}d(x_0,x_1),\hfill \end{array}$$

which gives us

$$\mathsf{D}(x_1,{\mathsf{T}}_2{x}_1)\le \frac{b_{1,2}}{1-a_{1,2}}d(x_0,x_1)<\frac{b_{1,2}}{1-a_{1,2}}pd(x_0,x_1),$$

where $p>1$ is a fixed number. From the above inequality, there exists $x_2\in {\mathsf{T}}_2{x}_1$ such that $d(x_1,x_2)<\frac{b_{1,2}}{1-a_{1,2}}pd(x_0,x_1)$. Similarly,

$$\mathsf{D}(x_2,{\mathsf{T}}_3{x}_2)\le \frac{b_{2,3}}{1-a_{2,3}}d(x_1,x_2)<\frac{b_{2,3}}{1-a_{2,3}}\frac{b_{1,2}}{1-a_{1,2}}pd(x_0,x_1),$$

and so there exists $x_3\in {\mathsf{T}}_3{x}_2$ such that $d(x_2,x_3)<\frac{b_{2,3}}{1-a_{2,3}}\frac{b_{1,2}}{1-a_{1,2}}pd(x_0,x_1)$. Continuing this process, we obtain a sequence $\left\{x_n\right\}$ in $\mathsf{X}$ such that $x_{n+1}\in {\mathsf{T}}_{n+1}{x}_n$, $\alpha (x_n,x_{n+1})\ge 1$, and

$$\begin{array}{c}\hfill d(x_n,x_{n+1})<A_{n}pd(x_0,x_1),\mathrm{for}\mathrm{all}n=1,2,\dots .\end{array}$$

(6)

Again, as in the proof of Theorem 1, we conclude that $\left\{x_n\right\}$ is a Cauchy sequence, and so there exists $x\in \mathsf{X}$ such that $x_n\to x$. From the assumption (v), we get $\alpha (x_n,x)\ge 1$ for all n. To show that x is a common fixed point of ${\mathsf{T}}_n$, let m be an arbitrary positive integer. Then, for any $n\in \mathbb{N}$, we have

$$\begin{array}{ccc}\mathsf{D}(x,{\mathsf{T}}_{m}x)\hfill & \le \hfill & d(x,x_n)+\mathsf{D}(x_n,{\mathsf{T}}_{m}x)\le d(x,x_n)+\alpha (x_{n-1},x)\mathcal{H}({\mathsf{T}}_n{x}_{n-1},{\mathsf{T}}_{m}x)\hfill \\ & \le \hfill & d(x,x_n)+a_{n,m}\mathsf{D}(x,{\mathsf{T}}_{m}x)\phi (\mathsf{D}(x_{n-1},{\mathsf{T}}_n{x}_{n-1}),d(x_{n-1},x))+b_{n,m}d(x_{n-1},x)\hfill \\ & \le \hfill & d(x,x_n)+a_{n,m}\mathsf{D}(x,{\mathsf{T}}_{m}x)\phi (d(x_{n-1},x_n),d(x_{n-1},x))+b_{n,m}d(x_{n-1},x).\hfill \end{array}$$

Taking $\overline{lim}$ in both sides of the above inequality, as $n\to \infty$, we obtain

$$\mathsf{D}(x,{\mathsf{T}}_{m}x)\le ({\overline{lim}}_{n\to \infty }{a}_{n,m})\mathsf{D}(x,{\mathsf{T}}_{m}x).$$

We conclude $\mathsf{D}(x,{\mathsf{T}}_{m}x)=0$ and thus $x\in {\mathsf{T}}_{m}x$. □

3. Common Endpoint Theorems

The notion of endpoints of multi-valued mappings has been studied by some researchers in the last decade (see for instance, [17,18,19]). In current section, we state and prove some common endpoint theorems for a sequence of multi-valued mappings with the contractions mentioned in Section 2. We need the following definition.

Definition 2.

Let ${\mathsf{T}}_n:\mathsf{X}\to \mathcal{CL}(\mathsf{X})$ (n = 1,2,...) be a sequence of multi-valued mappings. We say that $\left\{{\mathsf{T}}_n\right\}$ has (HS) property whenever for each $x\in \mathsf{X}$ there exists $y\in {\mathsf{T}}_{n}x$ such that $\mathcal{H}({\mathsf{T}}_{n}x,{\mathsf{T}}_{n+1}y)\ge {sup}_{b\in {\mathsf{T}}_{n+1}y}d(y,b)$.

Theorem 4.

Let $(\mathsf{X},d)$ be a complete metric space and $0\le a_{i,j}(i,j=1,2,\dots )$ with $a_{i,i+1}\ne 1$ for all $i=1,2,\dots$ satisfy:

(i) 

for each (j), ${\overline{lim}}_{i\to \infty }{a}_{i,j}<1;$

(ii) 

${\sum }_{n=1}^{\infty }{A}_n<\infty$ where $A_n={\prod }_{i=1}^n\frac{a_{i,i+1}}{1-a_{i,i+1}}$.

Let $\alpha :\mathsf{X}\times \mathsf{X}\to [0,\infty )$ be a given function and $\left\{{\mathsf{T}}_n\right\}$ be a sequence of multi-valued operators ${\mathsf{T}}_n:\mathsf{X}\to \mathcal{CL}(\mathsf{X})$ (n = 1,2,...) satisfying (HS) property such that

$$\begin{array}{c}\hfill \alpha (x,y)\mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{j}y)\le a_{i,j}[\mathsf{D}(x,{\mathsf{T}}_{j}y)+\mathsf{D}(y,{\mathsf{T}}_{i}x)],\end{array}$$

(7)

for all $x,y\in \mathsf{X};$ $i,j=1,2,\dots$ with $x\ne y$ and $i\ne j$. Moreover, assume that the following assertions hold:

(iii) 

there exists $x_0\in \mathsf{X}$ such that for any $x\in {\mathsf{T}}_1{x}_0,$ we have $\alpha (x_0,x)\ge 1;$

(iv) 

$\left\{{\mathsf{T}}_n\right\}$ is α-admissible;

(v) 

for each sequence $\left\{x_n\right\}$ in $\mathsf{X}$ with $\alpha (x_n,x_{n+1})\ge 1$ for all n and $x_n\to x$, we have $\alpha (x_n,x)\ge 1$ for all n.

Then each ${\mathsf{T}}_n$ have a common endpoint in $\mathsf{X}$.

Proof. 

Since $\left\{{\mathsf{T}}_n\right\}$ has (HS) property, there exists $x_1\in {\mathsf{T}}_1{x}_0$ such that $\mathcal{H}({\mathsf{T}}_1{x}_0,{\mathsf{T}}_2{x}_1)\ge {sup}_{b\in {\mathsf{T}}_2{x}_1}d(x_1,b)$. From (iii), we have $\alpha (x_0,x_1)\ge 1$. Similarly, there exists $x_2\in {\mathsf{T}}_2{x}_1$ such that $\mathcal{H}({\mathsf{T}}_2{x}_1,{\mathsf{T}}_3{x}_2)\ge {sup}_{b\in {\mathsf{T}}_3{x}_2}d(x_2,b)$. Since $\left\{{\mathsf{T}}_n\right\}$ is $\alpha$-admissible, so $\alpha (x_1,x_2)\ge 1$. If we continue this process, we obtain a sequence $\left\{x_n\right\}$ in $\mathsf{X}$ such that $x_n\in {\mathsf{T}}_n{x}_{n-1}$, $\alpha (x_{n-1},x_n)\ge 1$, and

$$\begin{array}{c}\hfill \mathcal{H}({\mathsf{T}}_n{x}_{n-1},{\mathsf{T}}_{n+1}{x}_n)\ge \underset{b\in {\mathsf{T}}_{n+1}{x}_n}{sup}d(x_n,b),\end{array}$$

(8)

for all $n\ge 1$. Then we have

$$\begin{array}{ccc}d(x_n,x_{n+1})\hfill & \le \hfill & \underset{b\in {\mathsf{T}}_{n+1}{x}_n}{sup}d(x_n,b)\le \alpha (x_{n-1},x_n)\mathcal{H}({\mathsf{T}}_n{x}_{n-1},{\mathsf{T}}_{n+1}{x}_n)\hfill \\ & \le \hfill & a_{n,n+1}[\mathsf{D}(x_{n-1},{\mathsf{T}}_{n+1}{x}_n)+\mathsf{D}(x_n,{\mathsf{T}}_n{x}_{n-1})]\hfill \\ & \le \hfill & a_{n,n+1}\left[d(x_{n-1},x_{n+1})\right]\le a_{n,n+1}[d(x_{n-1},x_n)+d(x_n,x_{n+1})].\hfill \end{array}$$

From the above inequality, we get

$$d(x_n,x_{n+1})\le \frac{a_{n,n+1}}{1-a_{n,n+1}}d(x_{n-1},x_n)\le \dots \le A_{n}d(x_0,x_1).$$

Hence $\left\{x_n\right\}$ is a Cauchy sequence, and so there exists $x\in \mathsf{X}$ such that $x_n\to x$. From (v) we deduce $\alpha (x_n,x)\ge 1$ for all n. Now we show that x is a common endpoint of ${\mathsf{T}}_n$. Let $m\in \mathbb{N}$ be arbitrary. Then, for any $n\in \mathbb{N}$, we have

$$\begin{array}{ccc}\mathcal{H}(\left\{x\right\},{\mathsf{T}}_{m}x)\hfill & \le \hfill & d(x,x_n)+\mathcal{H}(\left\{x_n\right\},{\mathsf{T}}_{n+1}{x}_n)+\alpha (x_n,x)\mathcal{H}({\mathsf{T}}_{n+1}{x}_n,{\mathsf{T}}_{m}x)\hfill \\ & \le \hfill & d(x,x_n)+\alpha (x_{n-1},x_n)\mathcal{H}({\mathsf{T}}_n{x}_{n-1},{\mathsf{T}}_{n+1}{x}_n)+\alpha (x_n,x)\mathcal{H}({\mathsf{T}}_{n+1}{x}_n,{\mathsf{T}}_{m}x)\hfill \\ & \le \hfill & d(x,x_n)+a_{n,n+1}[\mathsf{D}(x_{n-1},{\mathsf{T}}_{n+1}{x}_n)+\mathsf{D}(x_n,{\mathsf{T}}_n{x}_{n-1})]\hfill \\ & & +a_{n+1,m}[\mathsf{D}(x_n,{\mathsf{T}}_{m}x)+\mathsf{D}(x,{\mathsf{T}}_{n+1}{x}_n)]\hfill \\ & \le \hfill & d(x,x_n)+a_{n,n+1}\left[d(x_{n-1},x_{n+1})\right]+a_{n+1,m}[\mathsf{D}(x_n,{\mathsf{T}}_{m}x)+d(x,x_{n+1})].\hfill \end{array}$$

Taking $\overline{lim}$ as $n\to \infty$, we obtain

$$\mathcal{H}(\left\{x\right\},{\mathsf{T}}_{m}x)\le ({\overline{lim}}_{n\to \infty }{a}_{n+1,m})\mathsf{D}(x,{\mathsf{T}}_{m}x)\le ({\overline{lim}}_{n\to \infty }{a}_{n+1,m})\mathcal{H}(\left\{x\right\},{\mathsf{T}}_{m}x),$$

which implies $\mathcal{H}(\left\{x\right\},{\mathsf{T}}_{m}x)=0$ and so ${\mathsf{T}}_{m}x=\left\{x\right\}$. Since m was arbitrary, the proof is complete. □

Theorem 5.

In the statement of Theorem 4, if we add the extra condition $\alpha (x,y)\ge 1$ for any common endpoints $x,y$ of ${\mathsf{T}}_n$, then the common endpoint of ${\mathsf{T}}_n$ is unique.

Proof. 

Let $x,y$ be two common endpoints of ${\mathsf{T}}_n$. Since ${\sum }_{n=1}^{\infty }{A}_n<\infty$, there exists $i_0\in \mathbb{N}$ such that $\frac{a_{i_0,i_0+1}}{1-a_{i_0,i_0+1}}<1$, which implies $a_{i_0,i_0+1}<{\displaystyle \frac{1}{2}}$. Then, using (7), we get

$$\begin{array}{ccc}d(x,y)\hfill & =\hfill & \mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{i_0+1}y)\hfill \\ & \le \hfill & \alpha (x,y)\mathcal{H}({\mathsf{T}}_{i_0}x,{\mathsf{T}}_{i_0+1}y)\hfill \\ & \le \hfill & a_{i_0,i_0+1}[\mathsf{D}(x,{\mathsf{T}}_{i_0+1}y)+\mathsf{D}(y,{\mathsf{T}}_{i_0}x)]\hfill \\ & =\hfill & 2a_{i_0,i_0+1}d(x,y),\hfill \end{array}$$

which implies $d(x,y)=0$ and so $x=y$. □

Example 1.

Consider the space $\mathsf{X}=[0,1]$ with the usual metric $d(x,y)=|x-y|$. Define a sequence of mappings ${\mathsf{T}}_n:\mathsf{X}\to \mathcal{CL}(\mathsf{X})$ by

$${\mathsf{T}}_n(x)=\left\{\begin{array}{cc}\left\{1\right\},\hfill & \frac{1}{2}\le x\le 1,\hfill \\ \left[\frac{2}{3}+\frac{1}{n+2},1\right],\hfill & x=0,\hfill \\ \left\{0\right\},\hfill & 0<x<\frac{1}{2}.\hfill \end{array}\right.$$

Also consider the constants $a_{i,j}=\frac{1}{3}+\frac{1}{|i-j|+6}$. Then ${\overline{lim}}_{i\to \infty }{a}_{i,j}=\frac{1}{3}<1$, for all $j\in \mathbb{N}$. $A_n={\prod }_{i=1}^n\frac{a_{i,i+1}}{1-a_{i,i+1}}={(\frac{10}{11})}^n$. Thus ${\sum }_{n=1}^{\infty }{A}_n={\sum }_{n=1}^{\infty }{(\frac{10}{11})}^n<\infty$. Also let

$$\alpha (x,y)=\left\{\begin{array}{cc}1,\hfill & x,y\in \left\{0\right\}\cup \left[\frac{1}{2},1\right],\hfill \\ 0,\hfill & otherwise.\hfill \end{array}\right.$$

Now we show that $\alpha (x,y)\mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{j}y)\le a_{i,j}[\mathsf{D}(x,{\mathsf{T}}_{j}y)+\mathsf{D}(y,{\mathsf{T}}_{i}x)]$, for all $x,y\in X$. If $0<x<\frac{1}{2}$ or $0<y<\frac{1}{2}$, then $\alpha (x,y)=0$ and we have nothing to prove. Therefore, we may assume $x,y\in \left\{0\right\}\cup \left[\frac{1}{2},1\right]$. We consider the following cases:

(1) 

$x,y\in \left[\frac{1}{2},1\right]$. In this case we have $\alpha (x,y)\mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{j}y)=\mathcal{H}(\left\{1\right\},\left\{1\right\})=0\le a_{i,j}[\mathsf{D}(x,{\mathsf{T}}_{j}y)+\mathsf{D}(y,{\mathsf{T}}_{i}x)]$, for all $x,y\in \mathsf{X}$.

(2) 

$x\in \left[\frac{1}{2},1\right]$ and $y=0$. In this case we have

$$\begin{array}{cc}\hfill \alpha (x,y)\mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{j}y)& =\mathcal{H}(\left\{1\right\},\left[\frac{2}{3}+\frac{1}{j+2},1\right])\hfill \\ \hfill & =|1-(\frac{2}{3}+\frac{1}{j+2})|=\frac{1}{3}-\frac{1}{j+2}\le \frac{1}{3}\hfill \\ \hfill & \le (\frac{1}{3}+\frac{1}{|i-j|+6})(|x-(\frac{2}{3}+\frac{1}{j+2})|+|0-1|)\hfill \\ \hfill & =a_{i,j}[\mathsf{D}(x,{\mathsf{T}}_{j}y)+\mathsf{D}(y,{\mathsf{T}}_{i}x).\hfill \end{array}$$

(3) 

$x=y=0,i<j$. Then

$$\begin{array}{cc}\hfill \alpha (x,y)\mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{j}y)& =|\frac{2}{3}+\frac{1}{j+2}-(\frac{2}{3}+\frac{1}{i+2})|=\frac{1}{i+2}-\frac{1}{j+2}\le \frac{1}{i+2}\hfill \\ \hfill & \le (\frac{1}{3}+\frac{1}{|i-j|+6})(\frac{2}{3}+\frac{1}{i+2}+(\frac{2}{3}+\frac{1}{j+2}))\hfill \\ \hfill & =a_{i,j}[\mathsf{D}(x,{\mathsf{T}}_{j}y)+\mathsf{D}(y,{\mathsf{T}}_{i}x)].\hfill \end{array}$$

Also for $x_0=0$ and $x_1=1$, we have $x_1\in \left\{1\right\}=\left[\frac{2}{3}+\frac{1}{1+2},1\right]={\mathsf{T}}_1{x}_0$ and $\alpha (x,y)=1\ge 1$. It is easy to check that $\left\{{\mathsf{T}}_n\right\}$ is α-admissible. Also, for any common endpoints $x,y$, we have $\alpha (x,y)\ge 1$. Thus, all of the conditions of Theorem 4 and Theorem 5 are satisfied. Therefore, the mappings ${\mathsf{T}}_n$ have a unique common endpoint. Here $x=1$ is the unique common endpoint of ${\mathsf{T}}_n$.

Theorem 6.

Let $(\mathsf{X},d)$ be a complete metric space and $0\le a_{i,j}(i,j=1,2,\dots )$ with $a_{i,i+1}\ne 1$ for all $i=1,2,\dots$ satisfy:

(i) 

for each (j), ${\overline{lim}}_{i\to \infty }{a}_{i,j}<1;$

(ii) 

${\sum }_{n=1}^{\infty }{A}_n<\infty$ where $A_n={\prod }_{i=1}^n\frac{a_{i,i+1}}{1-a_{i,i+1}}$.

Let $\alpha :\mathsf{X}\times \mathsf{X}\to [0,\infty )$ be a given function and $\left\{{\mathsf{T}}_n\right\}$ be a sequence of multi-valued operators ${\mathsf{T}}_n:\mathsf{X}\to \mathcal{CL}(\mathsf{X})$ (n = 1,2,...) satisfying (HS) property such that

$$\begin{array}{c}\hfill \alpha (x,y)\mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{j}y)\le a_{i,j}max\{d(x,y),\mathsf{D}(x,{\mathsf{T}}_{i}x),\mathsf{D}(y,{\mathsf{T}}_{j}y),\mathsf{D}(x,{\mathsf{T}}_{j}y),\mathsf{D}(y,{\mathsf{T}}_{i}x)\},\end{array}$$

(9)

for all $x,y\in \mathsf{X};$ $i,j=1,2,\dots$ with $x\ne y$ and $i\ne j$. Moreover, assume that the following assertions hold:

(iii) 

there exists $x_0\in \mathsf{X}$ such that for any $x\in {\mathsf{T}}_1{x}_0,$ we have $\alpha (x_0,x)\ge 1$;

(iv) 

$\left\{{\mathsf{T}}_n\right\}$ is α-admissible;

(v) 

for each sequence $\left\{x_n\right\}$ in $\mathsf{X}$ with $\alpha (x_n,x_{n+1})\ge 1$ for all n and $x_n\to x$, we have $\alpha (x_n,x)\ge 1$ for all n.

Then each ${\mathsf{T}}_n$ have a common endpoint in $\mathsf{X}$.

Proof. 

As in the proof of Theorem 4, there exists a sequence $\left\{x_n\right\}$ in $\mathsf{X}$ such that $x_n\in {\mathsf{T}}_n{x}_{n-1}$, $\alpha (x_{n-1},x_n)\ge 1$, and

$$\mathcal{H}({\mathsf{T}}_n{x}_{n-1},{\mathsf{T}}_{n+1}{x}_n)\ge \underset{b\in {\mathsf{T}}_{n+1}{x}_n}{sup}d(x_n,b),$$

for all $n\ge 1$. Then we have

$$\begin{array}{ccc}d(x_n,x_{n+1})\hfill & \le \hfill & \underset{b\in {\mathsf{T}}_{n+1}{x}_n}{sup}d(x_n,b)\le \alpha (x_{n-1},x_n)\mathcal{H}({\mathsf{T}}_n{x}_{n-1},{\mathsf{T}}_{n+1}{x}_n)\hfill \\ & \le \hfill & a_{n,n+1}max\{d(x_{n-1},x_n),\mathsf{D}(x_{n-1},{\mathsf{T}}_n{x}_{n-1}),\mathsf{D}(x_n,{\mathsf{T}}_{n+1}{x}_n),\hfill \\ & & \mathsf{D}(x_{n-1},{\mathsf{T}}_{n+1}{x}_n),\mathsf{D}(x_n,{\mathsf{T}}_n{x}_{n-1})\}\hfill \\ & \le \hfill & a_{n,n+1}[d(x_{n-1},x_n)+d(x_n,x_{n+1})].\hfill \end{array}$$

From the above inequality, we get

$$d(x_n,x_{n+1})\le \frac{a_{n,n+1}}{1-a_{n,n+1}}d(x_{n-1},x_n)\le \dots \le A_{n}d(x_0,x_1).$$

Thus, $\left\{x_n\right\}$ is a Cauchy sequence and so there exists $x\in \mathsf{X}$ such that $x_n\to x$ and $\alpha (x_n,x)\ge 1$ for all n. Now, we show that x is a common endpoint of ${\mathsf{T}}_n$. Let $m\in \mathbb{N}$ be arbitrary. Then, for any $n\in \mathbb{N}$, we have

$$\begin{array}{ccc}\mathcal{H}(\left\{x\right\},{\mathsf{T}}_{m}x)\hfill & \le \hfill & d(x,x_n)+\mathcal{H}(\left\{x_n\right\},{\mathsf{T}}_{n+1}{x}_n)+\alpha (x_n,x)\mathcal{H}({\mathsf{T}}_{n+1}{x}_n,{\mathsf{T}}_{m}x)\hfill \\ & \le \hfill & d(x,x_n)+\alpha (x_{n-1},x_n)\mathcal{H}({\mathsf{T}}_n{x}_{n-1},{\mathsf{T}}_{n+1}{x}_n)+\alpha (x_n,x)\mathcal{H}({\mathsf{T}}_{n+1}{x}_n,{\mathsf{T}}_{m}x)\hfill \\ & \le \hfill & d(x,x_n)+a_{n,n+1}[d(x_{n-1},x_n)+d(x_n,x_{n+1})]\hfill \\ & & +a_{n+1,m}max\{d(x_n,x),\mathsf{D}(x_n,{\mathsf{T}}_{n+1}{x}_n),\mathsf{D}(x,{\mathsf{T}}_{m}x),\mathsf{D}(x_n,{\mathsf{T}}_{m}x),\mathsf{D}(x,{\mathsf{T}}_{n+1}{x}_n)\}\hfill \\ & \le \hfill & d(x,x_n)+a_{n,n+1}[d(x_{n-1},x_n)+d(x_n,x_{n+1})]\hfill \\ & & +a_{n+1,m}max\{d(x_n,x),\mathsf{D}(x_n,x_{n+1}),\mathsf{D}(x,{\mathsf{T}}_{m}x),\mathsf{D}(x_n,{\mathsf{T}}_{m}x),\mathsf{D}(x,x_{n+1})\}.\hfill \end{array}$$

Taking $\overline{lim}$ in both sides of the above inequality, as $n\to \infty$, we obtain

$$\mathcal{H}(\left\{x\right\},{\mathsf{T}}_{m}x)\le ({\overline{lim}}_{n\to \infty }{a}_{n+1,m})\mathsf{D}(x,{\mathsf{T}}_{m}x)\le ({\overline{lim}}_{n\to \infty }{a}_{n+1,m})\mathcal{H}(\left\{x\right\},{\mathsf{T}}_{m}x),$$

which implies $\mathcal{H}(\left\{x\right\},{\mathsf{T}}_{m}x)=0$ and so ${\mathsf{T}}_{m}x=\left\{x\right\}$. □

Theorem 7.

With the conditions of Theorem 6, if we add the extra condition $\alpha (x,y)\ge 1$ for any common endpoints $x,y$ of ${\mathsf{T}}_n$, then the common endpoint of ${\mathsf{T}}_n$ is unique.

Proof. 

Let $x,y$ be two common endpoints of ${\mathsf{T}}_n$. Using (9), we get

$$\begin{array}{ccc}d(x,y)\hfill & =\hfill & \mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{j}y)\le \alpha (x,y)\mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{j}y)\hfill \\ & \le \hfill & a_{i,j}max\{d(x,y),\mathsf{D}(x,{\mathsf{T}}_{i}x),\mathsf{D}(y,{\mathsf{T}}_{j}y),\mathsf{D}(x,{\mathsf{T}}_{j}y),\mathsf{D}(y,{\mathsf{T}}_{i}x)\}\hfill \\ & =\hfill & a_{i,j}d(x,y).\hfill \end{array}$$

Thus, $d(x,y)\le {\overline{lim}}_{i\to \infty }{a}_{i,j}d(x,y)$, which means that $d(x,y)=0$ and hence $x=y$. □

Theorem 8.

Let $(\mathsf{X},d)$ be a complete metric space and $0\le a_{i,j},0\le b_{i,j}(i,j=1,2,\dots )$ with $a_{i,i+1}\ne 1$ for all $i=1,2,\dots$ satisfy:

(i) 

for each (j), ${\overline{lim}}_{i\to \infty }{a}_{i,j}<1$, ${\overline{lim}}_{i\to \infty }{b}_{i,j}<1;$

(ii) 

${\sum }_{n=1}^{\infty }{A}_n<\infty$ where $A_n={\prod }_{i=1}^n\frac{b_{i,i+1}}{1-a_{i,i+1}}$.

Let $\alpha :\mathsf{X}\times \mathsf{X}\to [0,\infty )$ be a given function and $\left\{{\mathsf{T}}_n\right\}$ be a sequence of multi-valued operators ${\mathsf{T}}_n:\mathsf{X}\to \mathcal{CL}(\mathsf{X})$ (n = 1,2,...) satisfying (HS) property such that

$$\begin{array}{c}\hfill \alpha (x,y)\mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{j}y)\le a_{i,j}\mathsf{D}(y,{\mathsf{T}}_{j}y)\phi (\mathsf{D}(x,{\mathsf{T}}_{i}x),d(x,y))+b_{i,j}d(x,y),\end{array}$$

(10)

for all $x,y\in \mathsf{X};$ $i,j=1,2,\dots$ with $x\ne y$ and $i\ne j$, where φ is as in Theorem 3. Moreover, assume that the following assertions hold:

(iii) 

there exists $x_0\in \mathsf{X}$ such that for any $x\in {\mathsf{T}}_1{x}_0,$ we have $\alpha (x_0,x)\ge 1;$

(iv) 

$\left\{{\mathsf{T}}_n\right\}$ is α-admissible;

(v) 

for each sequence $\left\{x_n\right\}$ in $\mathsf{X}$ with $\alpha (x_n,x_{n+1})\ge 1$ for all n and $x_n\to x$, we have $\alpha (x_n,x)\ge 1$ for all n.

Then each ${\mathsf{T}}_n$ have a common endpoint in $\mathsf{X}$.

Proof. 

As in the proof of Theorem 4, there exists a sequence $\left\{x_n\right\}$ in $\mathsf{X}$ such that $x_n\in {\mathsf{T}}_n{x}_{n-1}$, $\alpha (x_{n-1},x_n)\ge 1$, and

$$\mathcal{H}({\mathsf{T}}_n{x}_{n-1},{\mathsf{T}}_{n+1}{x}_n)\ge \underset{b\in {\mathsf{T}}_{n+1}{x}_n}{sup}d(x_n,b),$$

for all $n\ge 1$. Then we have

$$\begin{array}{ccc}d(x_n,x_{n+1})\hfill & \le \hfill & \underset{b\in {\mathsf{T}}_{n+1}{x}_n}{sup}d(x_n,b)\hfill \\ & \le \hfill & \alpha (x_{n-1},x_n)\mathcal{H}({\mathsf{T}}_n{x}_{n-1},{\mathsf{T}}_{n+1}{x}_n)\hfill \\ & \le \hfill & a_{n,n+1}\mathsf{D}(x_n,{\mathsf{T}}_{n+1}{x}_n)\phi (\mathsf{D}(x_{n-1},{\mathsf{T}}_n{x}_{n-1}),d(x_{n-1},x_n))+b_{n,n+1}d(x_{n-1},x_n)\hfill \\ & \le \hfill & a_{n,n+1}d(x_n,x_{n+1})+b_{n,n+1}d(x_{n-1},x_n).\hfill \end{array}$$

From the above inequality, we get

$$d(x_n,x_{n+1})\le \frac{b_{n,n+1}}{1-a_{n,n+1}}d(x_{n-1},x_n)\le \dots \le A_{n}d(x_0,x_1).$$

As in proof of Theorem 1, we conclude that $\left\{x_n\right\}$ is a Cauchy sequence, and so there exists $x\in \mathsf{X}$ such that $x_n\to x$ and $\alpha (x_n,x)\ge 1$ for all n. To show that x is a common endpoint of ${\mathsf{T}}_n$, consider an arbitrary natural number m. Then, for any $n\in \mathbb{N}$, we have

$$\begin{array}{ccc}\mathcal{H}(\left\{x\right\},{\mathsf{T}}_{m}x)\hfill & \le \hfill & d(x,x_n)+\mathcal{H}(\left\{x_n\right\},{\mathsf{T}}_{n+1}{x}_n)+\alpha (x_n,x)\mathcal{H}({\mathsf{T}}_{n+1}{x}_n,{\mathsf{T}}_{m}x)\hfill \\ & \le \hfill & d(x,x_n)+\alpha (x_{n-1},x_n)\mathcal{H}({\mathsf{T}}_n{x}_{n-1},{\mathsf{T}}_{n+1}{x}_n)+\alpha (x_n,x)\mathcal{H}({\mathsf{T}}_{n+1}{x}_n,{\mathsf{T}}_{m}x)\hfill \\ & \le \hfill & d(x,x_n)+a_{n,n+1}\mathsf{D}(x_n,{\mathsf{T}}_{n+1}{x}_n)\phi (\mathsf{D}(x_{n-1},{\mathsf{T}}_n{x}_{n-1}),d(x_{n-1},x_n))\hfill \\ & & +b_{n,n+1}d(x_{n-1},x_n)\hfill \\ & & +a_{n+1,m}\mathsf{D}(x,{\mathsf{T}}_{m}x)\phi (\mathsf{D}(x_n,{\mathsf{T}}_{n+1}{x}_n),d(x_n,x))+b_{n+1,m}d(x_n,x)\hfill \\ & \le \hfill & d(x,x_n)+a_{n,n+1}d(x_n,x_{n+1})+b_{n,n+1}d(x_{n-1},x_n)\hfill \\ & & +a_{n+1,m}\mathsf{D}(x,{\mathsf{T}}_{m}x)\phi (d(x_n,x_{n+1}),d(x_n,x))+b_{n+1,m}d(x_n,x).\hfill \end{array}$$

Taking $\overline{lim}$ as $n\to \infty$, we obtain

$$\begin{array}{ccc}\mathcal{H}(\left\{x\right\},{\mathsf{T}}_{m}x)\hfill & \le \hfill & ({\overline{lim}}_{n\to \infty }{a}_{n+1,m})\mathsf{D}(x,{\mathsf{T}}_{m}x)\hfill \\ & \le \hfill & ({\overline{lim}}_{n\to \infty }{a}_{n+1,m})\mathcal{H}(\left\{x\right\},{\mathsf{T}}_{m}x),\hfill \end{array}$$

which shows $\mathcal{H}(\left\{x\right\},{\mathsf{T}}_{m}x)=0$. Thus $T_{m}x=\left\{x\right\}$. □

Theorem 9.

In the statement of Theorem 8, if we add the extra condition $\alpha (x,y)\ge 1$ for any common endpoints $x,y$ of ${\mathsf{T}}_n$, then the common endpoint of ${\mathsf{T}}_n$ is unique.

Proof. 

Let $x,y$ be two common endpoints of ${\mathsf{T}}_n$. Using (10), we have

$$\begin{array}{ccc}d(x,y)\hfill & =\hfill & \mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{j}y)\hfill \\ & \le \hfill & \alpha (x,y)\mathcal{H}({\mathsf{T}}_{i}x,{\mathsf{T}}_{j}y)\hfill \\ & \le \hfill & a_{i,j}\mathsf{D}(y,{\mathsf{T}}_{j}y)\phi (\mathsf{D}(x,{\mathsf{T}}_{i}x),d(x,y))+b_{i,j}d(x,y)\hfill \\ & =\hfill & b_{i,j}d(x,y).\hfill \end{array}$$

Therefore, $d(x,y)\le {\overline{lim}}_{i\to \infty }{b}_{i,j}d(x,y)$. Hence $d(x,y)=0$, which means that $x=y$. □

4. Application to Integral Equations

Take $\mathsf{I}=[0,T]$. Let $\mathsf{X}:=C(\mathsf{I},\mathbb{R})$ be the set of all real valued continuous functions with domain $\mathsf{I}$. Define the meric d on $\mathsf{X}$ with

$$d(x,y)=\underset{t\in \mathsf{I}}{sup}(|x(t)-y(t)|)=\left|\right|x-y\left|\right|.$$

Consider the system of integral equation:

$$x(t)=p(t)+{\int }_0^{T}G(t,s)F_n(s,x(s))ds,t\in \mathsf{I},n=1,2,3,\dots .$$

(11)

Our hypotheses on the data are the following:

(A)

$p:\mathsf{I}\to \mathbb{R}$ and $F_n:\mathsf{I}\times \mathbb{R}\to \mathbb{R}$ are continuous, for all $n\in \mathbb{N}$;

(B)

$G:\mathsf{I}\times \mathsf{I}\to \mathbb{R}$ is continuous and measurable at $s\in \mathsf{I}$ for all $t\in \mathsf{I}$;

(C)

$G(t,s)\ge 0$ for all $t,s\in \mathsf{I}$ and ${\int }_0^{T}G(t,s)ds\le 1$ for all $t\in \mathsf{I}$;

(D)

there exists $x_0\in \mathsf{X}$ such that $x_0(t)\le {\int }_0^{T}G(t,s)F_1(s,x_0(s))ds$, for all $t\in \mathsf{I}$;

(E)

for any $x\in \mathsf{X}$ with $x(t)\le {\int }_0^{T}G(t,s)F_n(s,x(s))ds,$ for all $t\in \mathsf{I}$, then we have ${\int }_0^{T}G(t,s)F_n(s,x(s))ds\le {\int }_0^{T}G(t,s)F_{n+1}(s,{\int }_0^{T}G(s,\tau )F_n(\tau ,x(\tau ))d\tau )ds,$ for all $t\in \mathsf{I}$.

Let $0\le a_{i,j}(i,j=1,2,\dots )$ with $a_{i,i+1}\ne 1$ for all $i=1,2,\dots$ satisfy:

(F)

for each (j), ${\overline{lim}}_{i\to \infty }{a}_{i,j}<1;$

(G)

${\Sigma }_{n=1}^{\infty }{A}_n<\infty$ where $A_n={\prod }_{i=1}^n\frac{a_{i,i+1}}{1-a_{i,i+1}}$;

(H)

for each $t\in \mathsf{I}$, $x,y\in \mathsf{X}$ with $x\le y,$ and $i\ne j$, we have

$$\begin{array}{cc}|F_i(t,x(t))-F_j(t,y(t))|\hfill & \le a_{i,j}(|x(t)-{\int }_0^{T}G(t,s)F_j(s,y(s))ds|\hfill \\ \hfill & +|y(t)-{\int }_0^{T}G(t,s)F_i(s,x(s))ds|).\hfill \end{array}$$

(12)

Theorem 10.

Under the assumptions $(A)$–$(H),$ the system of integral Equation (11) has a solution in $\mathsf{X}$.

Proof. 

Define ${\mathrm{Y}}_n:\mathsf{X}\to \mathsf{X}$ as

$$({\mathrm{Y}}_{n}x)(t)=p(t)+{\int }_0^{T}G(t,s)F_n(s,x(s))ds,t\in \mathsf{I}$$

for all $n\in \mathbb{N}$. In addition, define $\alpha :\mathsf{X}\times \mathsf{X}\to [0,\infty )$ by

$$\alpha (x,y)=\left\{\begin{array}{cc}1,\hfill & x(t)\le y(t)forallt\in \mathsf{I},\hfill \\ 0,\hfill & otherwise.\hfill \end{array}\right.$$

Let $x,y$ be two arbitrary elements of $\mathsf{X}$. If $x\nleqq y$, then $\alpha (x,y)=0$ and so inequality (2) holds, obviously. Now, let $x\le y$. Then

$$\begin{array}{cc}\hfill |({\mathrm{Y}}_{i}x)(t)-({\mathrm{Y}}_{j}y)(t)|& =|{\int }_0^{T}G(t,s)(F_i(s,x(s))-F_j(s,y(s))ds|\hfill \\ \hfill & \le {\int }_0^{T}G(t,s)|F_i(s,x(s))-F_j(s,y(s))|ds\hfill \\ \hfill & \le {\int }_0^{T}G(t,s)a_{i,j}(|x(s)-{\int }_0^{T}G(s,\tau )F_j(\tau ,y(\tau ))d\tau |\hfill \\ \hfill & +|y(s)-{\int }_0^{T}G(s,\tau )F_i(\tau ,x(\tau ))d\tau |)ds\hfill \\ \hfill & \le {\int }_0^{T}G(t,s)a_{i,j}(|x(s)-({\mathrm{Y}}_{j}y)(s)|+|y(s)-{\mathrm{Y}}_{i}x)(s)|)ds\hfill \\ \hfill & \le {\int }_0^{T}G(t,s)a_{i,j}(||x-({\mathrm{Y}}_{j}y)\left|\right|+\left|\right|y-{\mathrm{Y}}_{i}x)(s)\left|\right|)ds\hfill \\ \hfill & \le a_{i,j}(||x-{\mathrm{Y}}_{j}y\left|\right|+\left|\right|y-{\mathrm{Y}}_{i}x\left|\right|)\hfill \end{array}$$

for every $t\in \mathsf{I}$. Take sup in the above inequality to find that

$$\begin{array}{cc}\hfill \alpha (x,y)d({\mathrm{Y}}_{i}x,{\mathrm{Y}}_{j}y)& =\left|\right|\mathrm{Y}x-\mathrm{Y}y\left|\right|\hfill \\ \hfill & \le a_{i,j}(||x-{\mathrm{Y}}_{j}y\left|\right|+\left|\right|y-{\mathrm{Y}}_{i}x\left|\right|)=a_{i,j}(d(x,{\mathrm{Y}}_{j}y)+d(y,{\mathrm{Y}}_{i}x)).\hfill \end{array}$$

The properties $(D)$ and $(E)$ yield that properties $(iii)$ and $(iv)$ of Theorem 1 are satisfied. Obviously, the property $(v)$ of Theorem 1 holds. Thus, by that theorem, $\left\{{\mathrm{Y}}_n\right\}$ have a common fixed point, that is, the system of integral Equation (11) having a solution. □