Fixed point results for cyclic $(\psi ,\varphi ,A,B)$-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 [3–21] 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 $p:X\times X\to {\mathbb{R}}^{+}$ such that for all $x,y,z\in X$:

(${\mathrm{p}}_1$) $x=y⟺p(x,x)=p(x,y)=p(y,y)$,

(${\mathrm{p}}_2$) $p(x,x)\le p(x,y)$,

(${\mathrm{p}}_3$) $p(x,y)=p(y,x)$,

(${\mathrm{p}}_4$) $p(x,y)\le p(x,z)+p(z,y)-p(z,z)$.

A partial metric space is a pair $(X,p)$ such that X is a nonempty set and p is a partial metric on X. It is clear that, if $p(x,y)=0$, then from (${\mathrm{p}}_1$) and (${\mathrm{p}}_2$), $x=y$. But if $x=y$, $p(x,y)$ may not be 0. The function $p(x,y)=max\{x,y\}$ 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 $\{B_p(x,\epsilon ):x\in X,\epsilon >0\}$, where $B_p(x,\epsilon )=\{y\in X:p(x,y)<p(x,x)+\epsilon \}$ for all $x\in X$ and $\epsilon >0$.

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

$$d_p(x,y)=2p(x,y)-p(x,x)-p(y,y)$$

is a metric on X.

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

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

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

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

• (4) A subset A of a partial metric space $(X,p)$ is closed if whenever $\{x_n\}$ is a sequence in A such that $\{x_n\}$ 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 $(X,p)$ be a partial metric space.

• (a) $\{x_n\}$ is a Cauchy sequence in $(X,p)$ if and only if it is a Cauchy sequence in the metric space $(X,d_p)$.

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

  $$p(x,x)=\underset{n\to \mathrm{\infty }}{lim}p(x_n,x)=\underset{n,m\to \mathrm{\infty }}{lim}p(x_n,x_m).$$

Now, we define the cyclic map.

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

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 $(X,d)$. Suppose that $T:A\cup B\to A\cup B$ is a cyclic map such that

$$d(Tx,Ty)\le kd(x,y)\mathrm{\forall }x\in A,\mathrm{\forall }y\in B.$$

If $k\in [0,1)$, 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 $(X,d)$. A cyclic map $T:A\cup B\to A\cup B$ is said to be a Kannan type cyclic contraction if there exists $k\in (0,\frac{1}{2})$ such that

$$d(Tx,Ty)\le k(d(Tx,x)+d(Ty,y))\mathrm{\forall }x\in A,\mathrm{\forall }y\in B.$$

Definition 5 [23]

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

$$d(Tx,Ty)\le k(d(x,y)+d(Tx,x)+d(Ty,y))\mathrm{\forall }x\in A,\mathrm{\forall }y\in B.$$

Definition 6 [23]

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

$$d(Tx,Ty)\le kmax\{d(x,y),d(Tx,x),d(Ty,y)\}\mathrm{\forall }x\in A,\mathrm{\forall }y\in B.$$

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

• (1) ϕ is continuous and nondecreasing.

• (2) $\varphi (t)=0$ if and only if $t=0$.

For some work on altering distance function, we refer the reader to [28–33].

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

We start with the following definition.

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

• (1) ψ and ϕ are altering distance functions;

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

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

  (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 $(X,p)$. If $T:X\to X$ is a cyclic $(\psi ,\varphi ,A,B)$-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 $\{x_n\}$ 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 $x_{2n_0+1}=x_{2n_0+2}$ for some $n\in \mathbf{N}$, then $x_{2n_0+1}=T{x}_{2n_0+1}$. Thus, $x_{2n_0+1}$ 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

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

By (${\mathrm{p}}_4$), we have

$$\begin{array}{c}\psi (p(x_{n+1},x_{n+2}))\hfill \\ =\psi (p(x_{2t+1},x_{2t+2}))\hfill \\ \le \psi (max\{p(x_{2t},x_{2t+1}),p(x_{2t+2},x_{2t+1}),\frac{1}{2}(p(x_{2t},x_{2t+1})+p(x_{2t+1},x_{2t+2}))\})\hfill \\ -\varphi (max\{p(x_{2t},x_{2t+1}),p(x_{2t+2},x_{2t+1})\})\hfill \\ \le \psi (max\{p(x_{2t},x_{2t+1}),p(x_{2t+2},x_{2t+1})\})-\varphi (max\{p(x_{2t},x_{2t+1}),p(x_{2t+2},x_{2t+1})\})\hfill \\ \le \psi (max\{p(x_{2t},x_{2t+1}),p(x_{2t+2},x_{2t+1})\}).\hfill \end{array}$$

If

$$max\{p(x_{2t},x_{2t+1}),p(x_{2t+2},x_{2t+1})\}=p(x_{2t+2},x_{2t+1}),$$

then

$$\psi (p(x_{2t+1},x_{2t+2}))\le \psi (p(x_{2t+2},x_{2t+1}))-\varphi (p(x_{2t+2},x_{2t+1})).$$

Therefore, $\varphi (p(x_{2t+1},x_{2t+2}))=0$, and hence $p(x_{2t+1},x_{2t+2})=0$. By (${\mathrm{p}}_1$) and (${\mathrm{p}}_2$), we have $x_{2t+1}=x_{2t+2}$, which is a contradiction. Therefore,

$$max\{p(x_{2t},x_{2t+1}),p(x_{2t+2},x_{2t+1})\}=p(x_{2t},x_{2t+1}).$$

Hence,

$$p(x_{n+1},x_{n+2})=p(x_{2t+2},x_{2t+1})\le p(x_{2t},x_{2t+1})=p(x_n,x_{n+1})$$

(2.2)

and

$$\psi (p(x_{n+1},x_{n+2}))\le \psi (p(x_n,x_{n+1}))-\varphi (p(x_n,x_{n+1})).$$

(2.3)

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

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

By (${\mathrm{p}}_4$), we have

$$\begin{array}{c}\psi (p(x_{n+1},x_{n+2}))\hfill \\ =\psi (p(x_{2t+3},x_{2t+2}))\hfill \\ \le \psi (max\{p(x_{2t+3},x_{2t+2}),p(x_{2t+2},x_{2t+1}),\frac{1}{2}(p(x_{2t+3},x_{2t+2})+p(x_{2t+2},x_{2t+1}))\})\hfill \\ -\varphi (max\{p(x_{2t+2},x_{2t+1}),p(T{x}_{2t+1},x_{2t+1})\})\hfill \\ \le \psi (max\{p(x_{2t+3},x_{2t+2}),p(x_{2t+2},x_{2t+1})\})-\varphi (max\{p(x_{2t+2},x_{2t+1}),p(T{x}_{2t+1},x_{2t+1})\})\hfill \\ \le \psi (max\{p(x_{2t+3},x_{2t+2}),p(x_{2t+2},x_{2t+1})\}).\hfill \end{array}$$

If

$$max\{p(x_{2t+3},x_{2t+2}),p(x_{2t+2},x_{2t+1})\}=p(x_{2t+3},x_{2t+2}),$$

then

$$\varphi (p(x_{2t+3},x_{2t+2}))\le \psi (p(x_{2t+3},x_{2t+2}))-\varphi (p(x_{2t+2},x_{2t+1})).$$

Therefore, $\varphi (p(x_{2t+2},x_{2t+1}))=0$, and hence $p(x_{2t+3},x_{2t+2})=0$. By ($p_1$) and ($p_2$), we have $x_{2t+2}=x_{2t+1}$, which is a contradiction. Therefore,

$$max\{p(x_{2t+3},x_{2t+2}),p(x_{2t+2},x_{2t+1})\}=p(x_{2t+2},x_{2t+1}).$$

Hence,

(2.4)

(2.5)

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

$$\underset{n\to +\mathrm{\infty }}{lim}p(x_n,x_{n+1})=r.$$

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

$$\psi (p(x_{n+2},x_{n+1}))\le \psi (p(x_n,x_{n+1}))-\varphi (p(x_n,x_{n+1}))\mathrm{\forall }n\in \mathbf{N}.$$

(2.6)

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

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

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

$$\underset{n\to +\mathrm{\infty }}{lim}p(x_n,x_{n+1})=0.$$

(2.7)

By (${\mathrm{p}}_2$), we get that

$$\underset{n\to +\mathrm{\infty }}{lim}p(x_n,x_n)=0.$$

(2.8)

Since $d_p(x,y)\le 2p(x,y)$ for all $x,y\in X$, we get that

$$\underset{n\to +\mathrm{\infty }}{lim}{d}_p(x_n,x_{n+1})=0.$$

(2.9)

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

$$n(i)>m(i)>i,d_p(x_{2m(i)},x_{2n(i)})\ge ϵ.$$

(2.10)

This means that

$$d_p(x_{2m(i)},x_{2n(i)-2})<ϵ.$$

(2.11)

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

$$\begin{array}{rcl}ϵ& \le & d_p(x_{2m(i)},x_{2n(i)})\\ \le & d_p(x_{2m(i)},x_{2n(i)-2})+d_p(x_{2n(i)-2},x_{2n(i)-1})\\ +d_p(x_{2n(i)-1},x_{2n(i)})\\ <& ϵ+d_p(x_{2n(i)-2},x_{2n(i)-1})+d_p(x_{2n(i)-1},x_{2n(i)}).\end{array}$$

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

$$\underset{i\to +\mathrm{\infty }}{lim}{d}_p(x_{2m(i)},x_{2n(i)})=ϵ.$$

(2.12)

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

$$\begin{array}{rcl}ϵ& \le & d_p(x_{2m(i)},x_{2n(i)})\\ \le & d_p(x_{2n(i)},x_{2n(i)-1})+d_p(x_{2n(i)-1},x_{2m(i)})\\ \le & d_p(x_{2n(i)},x_{2n(i)-1})+d_p(x_{2n(i)-1},x_{2m(i)+1})+d_p(x_{2m(i)+1},x_{2m(i)})\\ \le & d_p(x_{2n(i)},x_{2n(i)-1})+d_p(x_{2n(i)-1},x_{2m(i)})+2d_p(x_{2m(i)+1},x_{2m(i)})\\ \le & 2d_p(x_{2n(i)},x_{2n(i)-1})+d_p(x_{2n(i)},x_{2m(i)})+2d_p(x_{2m(i)+1},x_{2m(i)}).\end{array}$$

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

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

Since

$$d_p(x,y)=2p(x,y)-p(x,x)-p(y,y)$$

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

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

By (2.1), we have

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

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

$$p(u,u)=\underset{n\to \mathrm{\infty }}{lim}p(x_n,u)=\underset{n,m\to \mathrm{\infty }}{lim}p(x_n,x_m).$$

(2.13)

Moreover, since $\{x_n\}$ is a Cauchy sequence in the metric space $(A\cup B,d_p)$, we have

$$\underset{n,m\to \mathrm{\infty }}{lim}{d}_p(x_n,x_m)=0.$$

(2.14)

From the definition of $d_p$ we have

$$d_p(x_n,x_m)=2p(x_n,x_m)-p(x_n,x_n)-p(x_m,x_m).$$

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

$$\underset{n,m\to \mathrm{\infty }}{lim}p(x_n,x_m)=0.$$

Thus by (2.13), we have

$$\underset{n\to +\mathrm{\infty }}{lim}p(x_n,u)=p(u,u)=0.$$

(2.15)

Since $p(x_{2n},u)\to 0=p(u,u)$, $\{x_{2n}\}$ is a sequence in A, and A is closed in $(X,p)$, 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

$$\begin{array}{rcl}p(x_n,Tu)& \le & p(x_n,u)+p(u,Tu)-p(u,u)\\ \le & p(x_n,u)+p(u,x_n)+p(x_n,Tu)-p(x_n,x_n)-p(u,u).\end{array}$$

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

$$\underset{n\to +\mathrm{\infty }}{lim}p(x_n,Tu)=p(u,Tu).$$

Now, we claim that $Tu=u$.

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

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

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

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

Therefore, $\varphi (p(u,Tu))=0$. Since ϕ is an altering distance function, $p(u,Tu)=0$, 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 $p(v,v)=0$. Now, we prove that $u=v$. Since $u\in A\cap B\subseteq A$ and $v\in A\cap B\subseteq B$, we have

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

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

Taking $\psi =I_{[0,+\mathrm{\infty })}$ (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 $(X,p)$. 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

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

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 $(X,p)$. 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

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

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 $X=[0,1]$. Define the partial metric p on X by

$$p(x,y)=\{\begin{array}{cc}0,\hfill & \text{if }x=y;\hfill \\ max\{x,y\},\hfill & \text{if }x\ne y.\hfill \end{array}$$

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

• (1) $(X,p)$ is a complete partial metric space.

• (2) $A\cup B$ has a cyclic representation w.r.t. T.

• (3) For all $x\in A$ and $y\in B$, we have

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

Proof Note that $TA=[0,\frac{1}{6}]\subseteq B$ and $TB=[0,\frac{1}{2}]\subseteq A$. 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,

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

and

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

Since

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

we have

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

 □

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

3 Application

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

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

(h2) For every $ϵ>0$, we have

$${\int }_0^{ϵ}\mu (t)dt>0.$$

Theorem 6 Let A and B be nonempty closed subsets of a complete partial metric space $(X,p)$. 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

$$\begin{array}{rcl}{\int }_0^{p(Tx,Ty)}{\mu }_1(t)dt& \le & {\int }_0^{max\{p(x,y),p(x,Tx),p(y,Ty),\frac{1}{2}(p(x,Ty)+p(Tx,y))\}}{\mu }_1(t)dt\\ -{\int }_0^{max\{p(x,y),p(y,Ty)\}}{\mu }_2(t)dt,\end{array}$$

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 $\psi ,\varphi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ via $\psi (t)={\int }_0^t{\mu }_1(s)ds$ and $\varphi (t)={\int }_0^t{\mu }_2(s)ds$ 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.