Fixed point results for cyclic $\left(\psi ,\varphi ,A,B\right)$-contraction in partial metric spaces

Abstract

Very recently, Agarwal et al. (Fixed Point Theory Appl. 2012:40, 2012) initiated the study of fixed point theorems for mappings satisfying cyclical generalized contractive conditions in complete partial metric spaces. In the present paper, we study some fixed point theorems for a mapping satisfying a cyclical generalized contractive condition based on a pair of altering distance functions in complete partial metric spaces. Also, we introduce an example and an application to support the usability of our paper.

MSC:54H25, 47H10.

1 Introduction

The existence and uniqueness of fixed and common fixed point theorems of operators has been a subject of great interest since Banach [1] proved the Banach contraction principle in 1922. Many authors generalized the Banach contraction principle in various spaces such as quasi-metric spaces, generalized metric spaces, cone metric spaces and fuzzy metric spaces. Matthews [2] introduced the notion of partial metric spaces in such a way that each object does not necessarily have to have a zero distance from itself and proved a modified version of the Banach contraction principle. Afterwards, many authors proved many existing fixed point theorems in partial metric spaces (see [321] for examples).

We recall below the definition of partial metric space and some of its properties.

Definition 1 [2]

A partial metric on a nonempty set X is a function $p:X×X\to {\mathbb{R}}^{+}$ such that for all $x,y,z\in X$:

(${\mathrm{p}}_{1}$) $x=y⟺p\left(x,x\right)=p\left(x,y\right)=p\left(y,y\right)$,

(${\mathrm{p}}_{2}$) $p\left(x,x\right)\le p\left(x,y\right)$,

(${\mathrm{p}}_{3}$) $p\left(x,y\right)=p\left(y,x\right)$,

(${\mathrm{p}}_{4}$) $p\left(x,y\right)\le p\left(x,z\right)+p\left(z,y\right)-p\left(z,z\right)$.

A partial metric space is a pair $\left(X,p\right)$ such that X is a nonempty set and p is a partial metric on X. It is clear that, if $p\left(x,y\right)=0$, then from (${\mathrm{p}}_{1}$) and (${\mathrm{p}}_{2}$), $x=y$. But if $x=y$, $p\left(x,y\right)$ may not be 0. The function $p\left(x,y\right)=max\left\{x,y\right\}$ for all $x,y\in {\mathbb{R}}^{+}$ defines a partial metric on ${\mathbb{R}}^{+}$.

Each partial metric p on X generates a ${T}_{0}$ topology ${\tau }_{p}$ on X which has as a base the family of open p-balls $\left\{{B}_{p}\left(x,\epsilon \right):x\in X,\epsilon >0\right\}$, where ${B}_{p}\left(x,\epsilon \right)=\left\{y\in X:p\left(x,y\right) for all $x\in X$ and $\epsilon >0$.

If p is a partial metric on X, then the function ${d}_{p}:X×X\to {\mathbb{R}}^{+}$ given by

${d}_{p}\left(x,y\right)=2p\left(x,y\right)-p\left(x,x\right)-p\left(y,y\right)$

is a metric on X.

Definition 2 Let $\left(X,p\right)$ be a partial metric space. Then

1. (1)

A sequence $\left\{{x}_{n}\right\}$ in a partial metric space $\left(X,p\right)$ converges to a point $x\in X$ if and only if $p\left(x,x\right)={lim}_{n\to \mathrm{\infty }}p\left(x,{x}_{n}\right)$.

2. (2)

A sequence $\left\{{x}_{n}\right\}$ in a partial metric space $\left(X,p\right)$ is called a Cauchy sequence iff ${lim}_{n,m\to \mathrm{\infty }}p\left({x}_{n},{x}_{m}\right)$ exists (and is finite).

3. (3)

A partial metric space $\left(X,p\right)$ is said to be complete if every Cauchy sequence $\left\{{x}_{n}\right\}$ in X converges, with respect to ${\tau }_{p}$, to a point $x\in X$ such that $p\left(x,x\right)={lim}_{n,m\to \mathrm{\infty }}p\left({x}_{n},{x}_{m}\right)$.

4. (4)

A subset A of a partial metric space $\left(X,p\right)$ is closed if whenever $\left\{{x}_{n}\right\}$ is a sequence in A such that $\left\{{x}_{n}\right\}$ converges to some $x\in X$, then $x\in A$.

Remark 1 The limit in a partial metric space is not unique.

Lemma 1 ([2, 17])

Let $\left(X,p\right)$ be a partial metric space.

1. (a)

$\left\{{x}_{n}\right\}$ is a Cauchy sequence in $\left(X,p\right)$ if and only if it is a Cauchy sequence in the metric space $\left(X,{d}_{p}\right)$.

2. (b)

A partial metric space $\left(X,p\right)$ is complete if and only if the metric space $\left(X,{d}_{p}\right)$ is complete. Furthermore, ${lim}_{n\to \mathrm{\infty }}{d}_{p}\left({x}_{n},x\right)=0$ if and only if

$p\left(x,x\right)=\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},x\right)=\underset{n,m\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{m}\right).$

Now, we define the cyclic map.

Definition 3 Let A and B be nonempty subsets of a metric space $\left(X,d\right)$ and $T:A\cup B\to A\cup B$. Then T is called a cyclic map if $T\left(A\right)\subseteq B$ and $T\left(B\right)\subseteq A$.

In 2003, Kirk et al. [22] gave the following fixed point theorem for a cyclic map.

Theorem 1 [22]

Let A and B be nonempty closed subsets of a complete metric space $\left(X,d\right)$. Suppose that $T:A\cup B\to A\cup B$ is a cyclic map such that

$d\left(Tx,Ty\right)\le kd\left(x,y\right)\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in A,\mathrm{\forall }y\in B.$

If $k\in \left[0,1\right)$, then T has a unique fixed point in $A\cap B$.

Karapınar and Erhan [23] introduced the following types of cyclic contractions:

Definition 4 [23]

Let A and B be nonempty closed subsets of a metric space $\left(X,d\right)$. A cyclic map $T:A\cup B\to A\cup B$ is said to be a Kannan type cyclic contraction if there exists $k\in \left(0,\frac{1}{2}\right)$ such that

$d\left(Tx,Ty\right)\le k\left(d\left(Tx,x\right)+d\left(Ty,y\right)\right)\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in A,\mathrm{\forall }y\in B.$

Definition 5 [23]

Let A and B be nonempty closed subsets of a metric space $\left(X,d\right)$. A cyclic map $T:A\cup B\to A\cup B$ is said to be a Reich type cyclic contraction if there exists $k\in \left(0,\frac{1}{3}\right)$ such that

$d\left(Tx,Ty\right)\le k\left(d\left(x,y\right)+d\left(Tx,x\right)+d\left(Ty,y\right)\right)\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in A,\mathrm{\forall }y\in B.$

Definition 6 [23]

Let A and B be nonempty closed subsets of a metric space $\left(X,d\right)$. A cyclic map $T:A\cup B\to A\cup B$ is said to be a Ćirić type cyclic contraction if there exists $k\in \left(0,\frac{1}{3}\right)$ such that

$d\left(Tx,Ty\right)\le kmax\left\{d\left(x,y\right),d\left(Tx,x\right),d\left(Ty,y\right)\right\}\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in A,\mathrm{\forall }y\in B.$

Moreover, Karapınar and Erhan [23] obtained the following results:

Theorem 2 [23]

Let A and B be nonempty closed subsets of a complete metric space $\left(X,d\right)$, and let $T:A\cup B\to A\cup B$ be a Kannan type cyclic contraction. Then T has a unique fixed point in $A\cap B$.

Theorem 3 [23]

Let A and B be nonempty closed subsets of a complete metric space $\left(X,d\right)$, and let $T:A\cup B\to A\cup B$ be a Reich type cyclic contraction. Then T has a unique fixed point in $A\cap B$.

Theorem 4 [23]

Let A and B be nonempty closed subsets of a complete metric space $\left(X,d\right)$, and let $T:A\cup B\to A\cup B$ be a Ćirić type cyclic contraction. Then T has a unique fixed point in $A\cap B$.

For more results on cyclic contraction mappings, see [24, 25].

Very recently, Agarwal et al. [26] initiated the study of fixed point theorems for mappings satisfying cyclical generalized contractive conditions in complete partial metric spaces.

Khan et al. [27] introduced the notion of altering distance function as follows.

Definition 7 (Altering distance function [27])

The function $\varphi :\left[0,+\mathrm{\infty }\right)\to \left[0,+\mathrm{\infty }\right)$ is called an altering distance function if the following properties are satisfied:

1. (1)

ϕ is continuous and nondecreasing.

2. (2)

$\varphi \left(t\right)=0$ if and only if $t=0$.

For some work on altering distance function, we refer the reader to [2833].

The purpose of this paper is to study some fixed point theorems for a mapping satisfying a cyclical generalized contractive condition based on a pair of altering distance functions in partial metric spaces.

2 Main result

Definition 8 Let $\left(X,p\right)$ be a partial metric space and A, B be nonempty closed subsets of X. A mapping $T:X\to X$ is called a cyclic $\left(\psi ,\varphi ,A,B\right)$-contraction if

1. (1)

ψ and ϕ are altering distance functions;

2. (2)

$A\cup B$ has a cyclic representation w.r.t. T; that is, $T\left(A\right)\subseteq B$ and $T\left(B\right)\subseteq A$; and(3)

$\begin{array}{rcl}\psi \left(p\left(Tx,Ty\right)\right)& \le & \psi \left(max\left\{p\left(x,y\right),p\left(x,Tx\right),p\left(y,Ty\right),\frac{1}{2}\left(p\left(x,Ty\right)+p\left(Tx,y\right)\right)\right\}\right)\\ -\varphi \left(max\left\{p\left(x,y\right),p\left(y,Ty\right)\right\}\right)\end{array}$
(2.1)

for all $x\in A$ and $y\in B$.

From now on, by ψ and ϕ we mean altering distance functions unless otherwise stated.

In the rest of this paper, N stands for the set of nonnegative integer numbers.

Theorem 5 Let A and B be nonempty closed subsets of a complete partial metric space $\left(X,p\right)$. If $T:X\to X$ is a cyclic $\left(\psi ,\varphi ,A,B\right)$-contraction, then T has a unique fixed point $u\in A\cap B$.

Proof Let ${x}_{0}\in A$. Since $TA\subseteq B$, we choose ${x}_{1}\in B$ such that $T{x}_{0}={x}_{1}$. Also, since $TB\subseteq A$, we choose ${x}_{2}\in A$ such that $T{x}_{1}={x}_{2}$. Continuing this process, we can construct sequences $\left\{{x}_{n}\right\}$ in X such that ${x}_{2n}\in A$, ${x}_{2n+1}\in B$, ${x}_{2n+1}=T{x}_{2n}$ and ${x}_{2n+2}=T{x}_{2n+1}$. If ${x}_{2{n}_{0}+1}={x}_{2{n}_{0}+2}$ for some $n\in \mathbf{N}$, then ${x}_{2{n}_{0}+1}=T{x}_{2{n}_{0}+1}$. Thus, ${x}_{2{n}_{0}+1}$ is a fixed point of T in $A\cap B$. Thus, we may assume that ${x}_{2n+1}\ne {x}_{2n+2}$ for all $n\in \mathbf{N}$.

Given $n\in \mathbf{N}$. If n is even, then $n=2t$ for some $t\in \mathbf{N}$. By (2.1), we have

$\begin{array}{c}\psi \left(p\left({x}_{n+1},{x}_{n+2}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\psi \left(p\left({x}_{2t+1},{x}_{2t+2}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\psi \left(p\left(T{x}_{2t},T{x}_{2t+1}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \psi \left(max\left\{p\left({x}_{2t},{x}_{2t+1}\right),p\left(T{x}_{2t},{x}_{2t}\right),p\left(T{x}_{2t+1},{x}_{2t+1}\right),\hfill \\ \phantom{\rule{2em}{0ex}}\frac{1}{2}\left(p\left({x}_{2t},T{x}_{2t+1}\right)+p\left(T{x}_{2t},{x}_{2t+1}\right)\right)\right\}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\varphi \left(max\left\{p\left({x}_{2t},{x}_{2t+1}\right),p\left(T{x}_{2t+1},{x}_{2t+1}\right)\right\}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\psi \left(max\left\{p\left({x}_{2t},{x}_{2t+1}\right),p\left({x}_{2t+2},{x}_{2t+1}\right),\hfill \\ \phantom{\rule{2em}{0ex}}\frac{1}{2}\left(p\left({x}_{2t},{x}_{2t+2}\right)+p\left({x}_{2t+1},{x}_{2t+1}\right)\right)\right\}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\varphi \left(max\left\{p\left({x}_{2t},{x}_{2t+1}\right),p\left({x}_{2t+2},{x}_{2t+1}\right)\right\}\right).\hfill \end{array}$

By (${\mathrm{p}}_{4}$), we have

$\begin{array}{c}\psi \left(p\left({x}_{n+1},{x}_{n+2}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\psi \left(p\left({x}_{2t+1},{x}_{2t+2}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \psi \left(max\left\{p\left({x}_{2t},{x}_{2t+1}\right),p\left({x}_{2t+2},{x}_{2t+1}\right),\frac{1}{2}\left(p\left({x}_{2t},{x}_{2t+1}\right)+p\left({x}_{2t+1},{x}_{2t+2}\right)\right)\right\}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\varphi \left(max\left\{p\left({x}_{2t},{x}_{2t+1}\right),p\left({x}_{2t+2},{x}_{2t+1}\right)\right\}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \psi \left(max\left\{p\left({x}_{2t},{x}_{2t+1}\right),p\left({x}_{2t+2},{x}_{2t+1}\right)\right\}\right)-\varphi \left(max\left\{p\left({x}_{2t},{x}_{2t+1}\right),p\left({x}_{2t+2},{x}_{2t+1}\right)\right\}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \psi \left(max\left\{p\left({x}_{2t},{x}_{2t+1}\right),p\left({x}_{2t+2},{x}_{2t+1}\right)\right\}\right).\hfill \end{array}$

If

$max\left\{p\left({x}_{2t},{x}_{2t+1}\right),p\left({x}_{2t+2},{x}_{2t+1}\right)\right\}=p\left({x}_{2t+2},{x}_{2t+1}\right),$

then

$\psi \left(p\left({x}_{2t+1},{x}_{2t+2}\right)\right)\le \psi \left(p\left({x}_{2t+2},{x}_{2t+1}\right)\right)-\varphi \left(p\left({x}_{2t+2},{x}_{2t+1}\right)\right).$

Therefore, $\varphi \left(p\left({x}_{2t+1},{x}_{2t+2}\right)\right)=0$, and hence $p\left({x}_{2t+1},{x}_{2t+2}\right)=0$. By (${\mathrm{p}}_{1}$) and (${\mathrm{p}}_{2}$), we have ${x}_{2t+1}={x}_{2t+2}$, which is a contradiction. Therefore,

$max\left\{p\left({x}_{2t},{x}_{2t+1}\right),p\left({x}_{2t+2},{x}_{2t+1}\right)\right\}=p\left({x}_{2t},{x}_{2t+1}\right).$

Hence,

$p\left({x}_{n+1},{x}_{n+2}\right)=p\left({x}_{2t+2},{x}_{2t+1}\right)\le p\left({x}_{2t},{x}_{2t+1}\right)=p\left({x}_{n},{x}_{n+1}\right)$
(2.2)

and

$\psi \left(p\left({x}_{n+1},{x}_{n+2}\right)\right)\le \psi \left(p\left({x}_{n},{x}_{n+1}\right)\right)-\varphi \left(p\left({x}_{n},{x}_{n+1}\right)\right).$
(2.3)

If n is odd, then $n=2t+1$ for some $t\in \mathbf{N}$. By (2.1), we have

$\begin{array}{c}\psi \left(p\left({x}_{n+1},{x}_{n+2}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\psi \left(p\left({x}_{2t+2},{x}_{2t+3}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\psi \left(p\left({x}_{2t+3},{x}_{2t+2}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\psi \left(p\left(T{x}_{2t+2},T{x}_{2t+1}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \psi \left(max\left\{p\left({x}_{2t+2},{x}_{2t+1}\right),p\left(T{x}_{2t+2},{x}_{2t+2}\right),p\left(T{x}_{2t+1},{x}_{2t+1}\right),\hfill \\ \phantom{\rule{2em}{0ex}}\frac{1}{2}\left(p\left({x}_{2t+2},T{x}_{2t+1}\right)+p\left(T{x}_{2t+2},{x}_{2t+1}\right)\right)\right\}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\varphi \left(max\left\{p\left({x}_{2t+2},{x}_{2t+1}\right),p\left(T{x}_{2t+1},{x}_{2t+1}\right)\right\}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\psi \left(max\left\{p\left({x}_{2t+3},{x}_{2t+2}\right),p\left({x}_{2t+2},{x}_{2t+1}\right),\frac{1}{2}\left(p\left({x}_{2t+2},{x}_{2t+2}\right)+p\left({x}_{2t+3},{x}_{2t+1}\right)\right)\right\}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\varphi \left(max\left\{p\left({x}_{2t+2},{x}_{2t+1}\right),p\left({x}_{2t+2},{x}_{2t+1}\right)\right\}\right).\hfill \end{array}$

By (${\mathrm{p}}_{4}$), we have

$\begin{array}{c}\psi \left(p\left({x}_{n+1},{x}_{n+2}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\psi \left(p\left({x}_{2t+3},{x}_{2t+2}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \psi \left(max\left\{p\left({x}_{2t+3},{x}_{2t+2}\right),p\left({x}_{2t+2},{x}_{2t+1}\right),\frac{1}{2}\left(p\left({x}_{2t+3},{x}_{2t+2}\right)+p\left({x}_{2t+2},{x}_{2t+1}\right)\right)\right\}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\varphi \left(max\left\{p\left({x}_{2t+2},{x}_{2t+1}\right),p\left(T{x}_{2t+1},{x}_{2t+1}\right)\right\}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \psi \left(max\left\{p\left({x}_{2t+3},{x}_{2t+2}\right),p\left({x}_{2t+2},{x}_{2t+1}\right)\right\}\right)-\varphi \left(max\left\{p\left({x}_{2t+2},{x}_{2t+1}\right),p\left(T{x}_{2t+1},{x}_{2t+1}\right)\right\}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \psi \left(max\left\{p\left({x}_{2t+3},{x}_{2t+2}\right),p\left({x}_{2t+2},{x}_{2t+1}\right)\right\}\right).\hfill \end{array}$

If

$max\left\{p\left({x}_{2t+3},{x}_{2t+2}\right),p\left({x}_{2t+2},{x}_{2t+1}\right)\right\}=p\left({x}_{2t+3},{x}_{2t+2}\right),$

then

$\varphi \left(p\left({x}_{2t+3},{x}_{2t+2}\right)\right)\le \psi \left(p\left({x}_{2t+3},{x}_{2t+2}\right)\right)-\varphi \left(p\left({x}_{2t+2},{x}_{2t+1}\right)\right).$

Therefore, $\varphi \left(p\left({x}_{2t+2},{x}_{2t+1}\right)\right)=0$, and hence $p\left({x}_{2t+3},{x}_{2t+2}\right)=0$. By (${p}_{1}$) and (${p}_{2}$), we have ${x}_{2t+2}={x}_{2t+1}$, which is a contradiction. Therefore,

$max\left\{p\left({x}_{2t+3},{x}_{2t+2}\right),p\left({x}_{2t+2},{x}_{2t+1}\right)\right\}=p\left({x}_{2t+2},{x}_{2t+1}\right).$

Hence,

(2.4)
(2.5)

From (2.2) and (2.4), we have $\left\{p\left({x}_{n+1},{x}_{n}\right):n\in \mathbf{N}\right\}$ is a nonincreasing sequence and hence there exists $r\ge 0$ such that

$\underset{n\to +\mathrm{\infty }}{lim}p\left({x}_{n},{x}_{n+1}\right)=r.$

Also, from (2.3) and (2.5), we have

$\psi \left(p\left({x}_{n+2},{x}_{n+1}\right)\right)\le \psi \left(p\left({x}_{n},{x}_{n+1}\right)\right)-\varphi \left(p\left({x}_{n},{x}_{n+1}\right)\right)\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbf{N}.$
(2.6)

Letting $n\to +\mathrm{\infty }$ in (2.6) and using the fact that ψ and ϕ are continuous, we get that

$\psi \left(r\right)\le \psi \left(r\right)-\varphi \left(r\right).$

Therefore, $\varphi \left(r\right)=0$ and hence $r=0$. Thus

$\underset{n\to +\mathrm{\infty }}{lim}p\left({x}_{n},{x}_{n+1}\right)=0.$
(2.7)

By (${\mathrm{p}}_{2}$), we get that

$\underset{n\to +\mathrm{\infty }}{lim}p\left({x}_{n},{x}_{n}\right)=0.$
(2.8)

Since ${d}_{p}\left(x,y\right)\le 2p\left(x,y\right)$ for all $x,y\in X$, we get that

$\underset{n\to +\mathrm{\infty }}{lim}{d}_{p}\left({x}_{n},{x}_{n+1}\right)=0.$
(2.9)

Next, we show that $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the metric space $\left(A\cup B,{d}_{p}\right)$. It is sufficient to show that $\left\{{x}_{2n}\right\}$ is a Cauchy sequence in $\left(A\cup B,{d}_{p}\right)$. Suppose the contrary; that is, $\left\{{x}_{2n}\right\}$ is not a Cauchy sequence in $\left(A\cup B,{d}_{p}\right)$. Then there exists $ϵ>0$ for which we can find two subsequences $\left\{{x}_{2m\left(i\right)}\right\}$ and $\left\{{x}_{2n\left(i\right)}\right\}$ of $\left\{{x}_{2n}\right\}$ such that $n\left(i\right)$ is the smallest index for which

$n\left(i\right)>m\left(i\right)>i,\phantom{\rule{1em}{0ex}}{d}_{p}\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)}\right)\ge ϵ.$
(2.10)

This means that

${d}_{p}\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)-2}\right)<ϵ.$
(2.11)

From (2.10), (2.11) and the triangular inequality, we get that

$\begin{array}{rcl}ϵ& \le & {d}_{p}\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)}\right)\\ \le & {d}_{p}\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)-2}\right)+{d}_{p}\left({x}_{2n\left(i\right)-2},{x}_{2n\left(i\right)-1}\right)\\ +{d}_{p}\left({x}_{2n\left(i\right)-1},{x}_{2n\left(i\right)}\right)\\ <& ϵ+{d}_{p}\left({x}_{2n\left(i\right)-2},{x}_{2n\left(i\right)-1}\right)+{d}_{p}\left({x}_{2n\left(i\right)-1},{x}_{2n\left(i\right)}\right).\end{array}$

On letting $i\to +\mathrm{\infty }$ in the above inequalities and using (2.9), we have

$\underset{i\to +\mathrm{\infty }}{lim}{d}_{p}\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)}\right)=ϵ.$
(2.12)

Again, from (2.10) and the triangular inequality, we get that

$\begin{array}{rcl}ϵ& \le & {d}_{p}\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)}\right)\\ \le & {d}_{p}\left({x}_{2n\left(i\right)},{x}_{2n\left(i\right)-1}\right)+{d}_{p}\left({x}_{2n\left(i\right)-1},{x}_{2m\left(i\right)}\right)\\ \le & {d}_{p}\left({x}_{2n\left(i\right)},{x}_{2n\left(i\right)-1}\right)+{d}_{p}\left({x}_{2n\left(i\right)-1},{x}_{2m\left(i\right)+1}\right)+{d}_{p}\left({x}_{2m\left(i\right)+1},{x}_{2m\left(i\right)}\right)\\ \le & {d}_{p}\left({x}_{2n\left(i\right)},{x}_{2n\left(i\right)-1}\right)+{d}_{p}\left({x}_{2n\left(i\right)-1},{x}_{2m\left(i\right)}\right)+2{d}_{p}\left({x}_{2m\left(i\right)+1},{x}_{2m\left(i\right)}\right)\\ \le & 2{d}_{p}\left({x}_{2n\left(i\right)},{x}_{2n\left(i\right)-1}\right)+{d}_{p}\left({x}_{2n\left(i\right)},{x}_{2m\left(i\right)}\right)+2{d}_{p}\left({x}_{2m\left(i\right)+1},{x}_{2m\left(i\right)}\right).\end{array}$

Letting $i\to +\mathrm{\infty }$ in the above inequalities and using (2.9) and (2.12), we get that

$\begin{array}{rcl}\underset{i\to +\mathrm{\infty }}{lim}{d}_{p}\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)}\right)& =& \underset{i\to +\mathrm{\infty }}{lim}{d}_{p}\left({x}_{2m\left(i\right)+1},{x}_{2n\left(i\right)-1}\right)\\ =& \underset{i\to +\mathrm{\infty }}{lim}{d}_{p}\left({x}_{2m\left(i\right)+1},{x}_{2n\left(i\right)}\right)\\ =& \underset{i\to +\mathrm{\infty }}{lim}{d}_{p}\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)-1}\right)\\ =& ϵ.\end{array}$

Since

${d}_{p}\left(x,y\right)=2p\left(x,y\right)-p\left(x,x\right)-p\left(y,y\right)$

for all $x,y\in X$, then

$\begin{array}{rcl}\underset{i\to +\mathrm{\infty }}{lim}p\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)}\right)& =& \underset{i\to +\mathrm{\infty }}{lim}p\left({x}_{2m\left(i\right)+1},{x}_{2n\left(i\right)-1}\right)\\ =& \underset{i\to +\mathrm{\infty }}{lim}p\left({x}_{2m\left(i\right)+1},{x}_{2n\left(i\right)}\right)\\ =& \underset{i\to +\mathrm{\infty }}{lim}p\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)-1}\right)\\ =& \frac{ϵ}{2}.\end{array}$

By (2.1), we have

$\begin{array}{rcl}\psi \left(p\left({x}_{2m\left(i\right)+1},{x}_{2n\left(i\right)}\right)\right)& =& \psi \left(p\left(T{x}_{2m\left(i\right)},T{x}_{2n\left(i\right)-1}\right)\right)\\ \le & \psi \left(max\left\{p\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)-1}\right),p\left({x}_{2m\left(i\right)},T{x}_{2m\left(i\right)}\right),p\left({x}_{2n\left(i\right)-1},T{x}_{2n\left(i\right)-1}\right),\\ \frac{1}{2}\left(p\left({x}_{2m\left(i\right)},T{x}_{2n\left(i\right)-1}\right)+p\left({x}_{2n\left(i\right)-1},T{x}_{2m\left(i\right)}\right)\right)\right\}\right)\\ -\varphi \left(max\left\{p\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)-1}\right),p\left({x}_{2n\left(i\right)-1},T{x}_{2n\left(i\right)-1}\right)\right\}\right)\\ =& \psi \left(max\left\{p\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)-1}\right),p\left({x}_{2m\left(i\right)},{x}_{2m\left(i\right)+1}\right),p\left({x}_{2n\left(i\right)-1},{x}_{2n\left(i\right)}\right),\\ \frac{1}{2}\left(p\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)}\right)+p\left({x}_{2n\left(i\right)-1},{x}_{2m\left(i\right)+1}\right)\right)\right\}\right)\\ -\varphi \left(max\left\{p\left({x}_{2m\left(i\right)},{x}_{2n\left(i\right)-1}\right),p\left({x}_{2n\left(i\right)-1},{x}_{2n\left(i\right)}\right)\right\}\right).\end{array}$

Letting $i\to +\mathrm{\infty }$ and using the continuity of ϕ and ψ, we get that

$\psi \left(\frac{ϵ}{2}\right)\le \psi \left(\frac{ϵ}{2}\right)-\varphi \left(\frac{ϵ}{2}\right).$

Therefore, we get that $\varphi \left(\frac{ϵ}{2}\right)=0$. Hence, $ϵ=0$ is a contradiction. Thus $\left\{{x}_{n}\right\}$ is a Cauchy sequence in $\left(A\cup B,{d}_{p}\right)$. Since $\left(X,p\right)$ is complete and $A\cup B$ is a closed subspace of $\left(X,p\right)$, then we have $\left(A\cup B,p\right)$ is complete. From Lemma 1, the sequence $\left\{{x}_{n}\right\}$ converges in the metric space $\left(A\cup B,{d}_{p}\right)$, say ${lim}_{n\to \mathrm{\infty }}{d}_{p}\left({x}_{n},u\right)=0$. Again from Lemma 1, we have

$p\left(u,u\right)=\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},u\right)=\underset{n,m\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{m}\right).$
(2.13)

Moreover, since $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the metric space $\left(A\cup B,{d}_{p}\right)$, we have

$\underset{n,m\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{n},{x}_{m}\right)=0.$
(2.14)

From the definition of ${d}_{p}$ we have

${d}_{p}\left({x}_{n},{x}_{m}\right)=2p\left({x}_{n},{x}_{m}\right)-p\left({x}_{n},{x}_{n}\right)-p\left({x}_{m},{x}_{m}\right).$

Letting $n,m\to +\mathrm{\infty }$ in the above equality and using (2.8) and (2.14), we get

$\underset{n,m\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{m}\right)=0.$

Thus by (2.13), we have

$\underset{n\to +\mathrm{\infty }}{lim}p\left({x}_{n},u\right)=p\left(u,u\right)=0.$
(2.15)

Since $p\left({x}_{2n},u\right)\to 0=p\left(u,u\right)$, $\left\{{x}_{2n}\right\}$ is a sequence in A, and A is closed in $\left(X,p\right)$, we have $u\in A$. Similarly, we have $u\in B$, that is $u\in A\cap B$. Again, from the definition of p, we have

$\begin{array}{rcl}p\left({x}_{n},Tu\right)& \le & p\left({x}_{n},u\right)+p\left(u,Tu\right)-p\left(u,u\right)\\ \le & p\left({x}_{n},u\right)+p\left(u,{x}_{n}\right)+p\left({x}_{n},Tu\right)-p\left({x}_{n},{x}_{n}\right)-p\left(u,u\right).\end{array}$

Letting $n\to +\mathrm{\infty }$ in the above inequalities and using (2.9) and (2.15), we get that

$\underset{n\to +\mathrm{\infty }}{lim}p\left({x}_{n},Tu\right)=p\left(u,Tu\right).$

Now, we claim that $Tu=u$.

Since ${x}_{2n}\in A$ and $u\in B$, by (2.1) we have

$\begin{array}{rcl}\psi \left(p\left({x}_{2n+1},Tu\right)\right)& =& \psi \left(p\left(T{x}_{2n},Tu\right)\right)\\ \le & \psi \left(max\left\{p\left({x}_{2n},u\right),p\left(T{x}_{2n},{x}_{2n}\right),p\left(Tu,u\right),\\ \frac{1}{2}\left(p\left({x}_{2n},Tu\right)+p\left(u,T{x}_{2n}\right)\right)\right\}\right)\\ -\varphi \left(max\left\{p\left({x}_{2n},u\right),p\left(Tu,u\right)\right\}\right)\\ =& \psi \left(max\left\{p\left({x}_{2n},u\right),p\left({x}_{2n},{x}_{2n+1}\right),p\left(Tu,u\right),\\ \frac{1}{2}\left(p\left({x}_{2n},Tu\right)+p\left(u,{x}_{2n+1}\right)\right)\right\}\right)\\ -\varphi \left(max\left\{p\left({x}_{2n},u\right),p\left(u,Tu\right)\right\}\right).\end{array}$

Letting $n\to +\mathrm{\infty }$, we get that

$\psi \left(p\left(u,Tu\right)\right)\le \psi \left(p\left(u,Tu\right)\right)-\varphi \left(p\left(u,Tu\right)\right).$

Therefore, $\varphi \left(p\left(u,Tu\right)\right)=0$. Since ϕ is an altering distance function, $p\left(u,Tu\right)=0$, that is, $u=Tu$.

Therefore, u is a fixed point of T. To prove the uniqueness of the fixed point, we let v be any other fixed point of T in $A\cap B$. It is an easy matter to prove that $p\left(v,v\right)=0$. Now, we prove that $u=v$. Since $u\in A\cap B\subseteq A$ and $v\in A\cap B\subseteq B$, we have

$\begin{array}{rcl}\psi \left(p\left(u,v\right)\right)& =& \psi \left(p\left(Tu,Tv\right)\right)\\ \le & \psi \left(max\left\{p\left(u,v\right),p\left(u,u\right),p\left(v,v\right)\right\}\right)-\varphi \left(max\left\{p\left(u,v\right),p\left(v,v\right)\right\}\right)\\ =& \psi \left(p\left(u,v\right)\right)-\varphi \left(p\left(u,v\right)\right).\end{array}$

Thus $\varphi \left(p\left(u,v\right)\right)=0$ and hence $p\left(u,v\right)=0$. Therefore, $u=v$. □

Taking $\psi ={I}_{\left[0,+\mathrm{\infty }\right)}$ (the identity function) in Theorem 5, we have the following result.

Corollary 1 Let A and B be nonempty closed subsets of a complete partial metric space $\left(X,p\right)$. Let $T:X\to X$ be a mapping such that $A\cup B$ has a cyclic representation w.r.t. T. Suppose there exists an altering distance function ϕ such that

$\begin{array}{rcl}p\left(Tx,Ty\right)& \le & max\left\{p\left(x,y\right),p\left(x,Tx\right),p\left(y,Ty\right),\frac{1}{2}\left(p\left(x,Ty\right)+p\left(Tx,y\right)\right)\right\}\\ -\varphi \left(max\left\{p\left(x,y\right),p\left(y,Ty\right)\right\}\right)\end{array}$

for all $x\in A$ and $y\in B$. Then T has a unique fixed point $u\in A\cap B$.

Corollary 2 Let A and B be nonempty closed subsets of a complete partial metric space $\left(X,p\right)$. Let $T:X\to X$ be a mapping such that $A\cup B$ has a cyclic representation w.r.t. T. Suppose there exists an altering distance function ϕ such that

$p\left(Tx,Ty\right)\le max\left\{p\left(x,y\right),p\left(x,Tx\right),p\left(y,Ty\right)\right\}-\varphi \left(max\left\{p\left(x,y\right),p\left(x,Tx\right),p\left(y,Ty\right)\right\}\right)$

for all $x\in A$ and $y\in B$. Then T has a unique fixed point $u\in A\cap B$.

Now, we introduce an example to support the usability of our results.

Example 1 Let $X=\left[0,1\right]$. Define the partial metric p on X by

Also, define the mapping $T:X\to X$ by $T\left(x\right)=\frac{{x}^{2}}{1+x}$ and the functions $\psi ,\varphi :\left[0,+\mathrm{\infty }\right)\to \left[0,+\mathrm{\infty }\right)$ by $\psi \left(t\right)=2t$ and $\varphi \left(t\right)=\frac{t}{1+2t}$. Take $A=\left[0,\frac{1}{2}\right]$ and $B=\left[0,1\right]$. Then

1. (1)

$\left(X,p\right)$ is a complete partial metric space.

2. (2)

$A\cup B$ has a cyclic representation w.r.t. T.

3. (3)

For all $x\in A$ and $y\in B$, we have

$\begin{array}{rcl}\psi \left(p\left(Tx,Ty\right)\right)& \le & \psi \left(max\left\{p\left(x,y\right),p\left(x,Tx\right),p\left(y,Ty\right),\frac{1}{2}\left(p\left(x,Ty\right)+p\left(Tx,y\right)\right)\right\}\right)\\ -\varphi \left(max\left\{p\left(x,y\right),p\left(y,Ty\right)\right\}\right).\end{array}$

Proof Note that $TA=\left[0,\frac{1}{6}\right]\subseteq B$ and $TB=\left[0,\frac{1}{2}\right]\subseteq A$. Thus $A\cup B$ has a cyclic representation of T. To prove (3), given $x\in A$ and $y\in B$, without loss of generality, we may assume that $x\le y$. So,

$\begin{array}{c}\psi \left(p\left(Tx,Ty\right)\right)=\psi \left(p\left(\frac{{x}^{2}}{1+x},\frac{{y}^{2}}{1+y}\right)\right)=\psi \left(\frac{{y}^{2}}{1+y}\right)=\frac{2{y}^{2}}{1+y},\hfill \\ \psi \left(max\left\{p\left(x,y\right),p\left(x,Tx\right),p\left(y,Ty\right),\frac{1}{2}\left(P\left(x,Ty\right)+p\left(Tx,y\right)\right)\right\}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\psi \left(max\left\{y,p\left(x,\frac{{x}^{2}}{1+x}\right),p\left(y,\frac{{y}^{2}}{1+y}\right),\frac{1}{2}\left(p\left(x,\frac{{y}^{2}}{1+y}\right)+p\left(\frac{{x}^{2}}{1+x},y\right)\right)\right\}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \psi \left(y\right)=2y,\hfill \end{array}$

and

$\varphi \left(max\left\{p\left(x,y\right),p\left(y,Ty\right)\right\}\right)=\varphi \left(max\left\{y,p\left(y,\frac{{y}^{2}}{1+y}\right)\right\}\right)=\varphi \left(y\right)=\frac{y}{1+2y}.$

Since

$\frac{2{y}^{2}}{1+y}\le 2y-\frac{y}{1+2y},$

we have

$\begin{array}{rcl}\psi \left(p\left(Tx,Ty\right)\right)& \le & \psi \left(max\left\{p\left(x,y\right),p\left(x,Tx\right),p\left(y,Ty\right),\frac{1}{2}\left(p\left(x,Ty\right)+p\left(Tx,y\right)\right)\right\}\right)\\ -\varphi \left(max\left\{p\left(x,y\right),p\left(y,Ty\right)\right\}\right).\end{array}$

□

Note that Example 1 satisfies all the hypotheses of Theorem 5.

3 Application

Denote by Λ the set of functions $\mu :\left[0,+\mathrm{\infty }\right)\to \left[0,+\mathrm{\infty }\right)$ satisfying the following hypotheses:

(h1) μ is a Lebesgue-integrable mapping on each compact of $\left[0,+\mathrm{\infty }\right)$.

(h2) For every $ϵ>0$, we have

${\int }_{0}^{ϵ}\mu \left(t\right)\phantom{\rule{0.2em}{0ex}}dt>0.$

Theorem 6 Let A and B be nonempty closed subsets of a complete partial metric space $\left(X,p\right)$. Let $T:X\to X$ be a mapping such that $A\cup B$ has a cyclic representation w.r.t. T. Suppose that for $x\in A$ and $y\in B$, we have

$\begin{array}{rcl}{\int }_{0}^{p\left(Tx,Ty\right)}{\mu }_{1}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt& \le & {\int }_{0}^{max\left\{p\left(x,y\right),p\left(x,Tx\right),p\left(y,Ty\right),\frac{1}{2}\left(p\left(x,Ty\right)+p\left(Tx,y\right)\right)\right\}}{\mu }_{1}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\\ -{\int }_{0}^{max\left\{p\left(x,y\right),p\left(y,Ty\right)\right\}}{\mu }_{2}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt,\end{array}$

where ${\mu }_{1},{\mu }_{2}\in \mathrm{\Lambda }$. Then T has a unique fixed point $u\in A\cap B$.

Proof Follows from Theorem 5 by defining $\psi ,\varphi :\left[0,+\mathrm{\infty }\right)\to \left[0,+\mathrm{\infty }\right)$ via $\psi \left(t\right)={\int }_{0}^{t}{\mu }_{1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds$ and $\varphi \left(t\right)={\int }_{0}^{t}{\mu }_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds$ and noting that ψ, ϕ are altering distance functions. □

Remark 2 Theorem 2.1 of [23] is a special case of Corollary 2.

Remark 3 Theorem 2.3 of [23] is a special case of Corollary 2.

Remark 4 Theorem 2.4 of [23] is a special case of Corollary 2.

Remark 5 Theorem 1.1 of [22] is a special case of Corollary 2.

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Shatanawi, W., Manro, S. Fixed point results for cyclic $\left(\psi ,\varphi ,A,B\right)$-contraction in partial metric spaces. Fixed Point Theory Appl 2012, 165 (2012). https://doi.org/10.1186/1687-1812-2012-165