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# Fixed point theory of cyclical generalized contractive conditions in partial metric spaces

Fixed Point Theory and Applications20132013:17

https://doi.org/10.1186/1687-1812-2013-17

• Accepted: 13 January 2013
• Published:

## Abstract

The purpose of this paper is to study fixed point theorems for a mapping satisfying the cyclical generalized contractive conditions in complete partial metric spaces. Our results generalize or improve many recent fixed point theorems in the literature.

MSC:47H10, 54C60, 54H25, 55M20.

## Keywords

• fixed point
• cyclic $\mathcal{CW}$-contraction
• cyclic $\mathcal{MK}$-contraction
• partial metric space

## 1 Introduction and preliminaries

Throughout this paper, by ${\mathbb{R}}^{+}$, we denote the set of all nonnegative real numbers, while is the set of all natural numbers. Let $\left(X,d\right)$ be a metric space, D be a subset of X and $f:D\to X$ be a map. We say f is contractive if there exists $\alpha \in \left[0,1\right)$ such that for all $x,y\in D$,
$d\left(fx,fy\right)\le \alpha \cdot d\left(x,y\right).$
The well-known Banach fixed point theorem asserts that if $D=X$, f is contractive and $\left(X,d\right)$ is complete, then f has a unique fixed point in X. It is well known that the Banach contraction principle  is a very useful and classical tool in nonlinear analysis. Also, this principle has many generalizations. For instance, in 1969, Boyd and Wong  introduced the notion of Φ-contraction. A mapping $f:X\to X$ on a metric space is called Φ-contraction if there exists an upper semi-continuous function $\mathrm{\Phi }:\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ such that

In 1994, Mattews  introduced the following notion of partial metric spaces.

Definition 1  A partial metric on a nonempty set X is a function $p:X×X\to {\mathbb{R}}^{+}$ such that for all $x,y,z\in X$,

(p1) $x=y$ if and only if $p\left(x,x\right)=p\left(x,y\right)=p\left(y,y\right)$;

(p2) $p\left(x,x\right)\le p\left(x,y\right)$;

(p3) $p\left(x,y\right)=p\left(y,x\right)$;

(p4) $p\left(x,y\right)\le p\left(x,z\right)+p\left(z,y\right)-p\left(z,z\right)$.

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.

Remark 1 It is clear that if $p\left(x,y\right)=0$, then from (p1) and (p2), $x=y$. But if $x=y$, $p\left(x,y\right)$ may not be 0.

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

We recall some definitions of a partial metric space as follows.

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 $x\in X$ 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 if and only if ${lim}_{m,n\to \mathrm{\infty }}p\left({x}_{m},{x}_{n}\right)$ 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 $p\left(x,x\right)={lim}_{m,n\to \mathrm{\infty }}p\left({x}_{m},{x}_{n}\right)$;

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 2 The limit in a partial metric space is not unique.

Lemma 1 [3, 4]

1. (a)

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

In 2003, Kirk, Srinivasan and Veeramani  introduced the following notion of the cyclic representation.

Definition 3 

Let X be a nonempty set, $m\in \mathbb{N}$ and $f:X\to X$ be an operator. Then $X={\bigcup }_{i=1}^{m}{A}_{i}$ is called a cyclic representation of X with respect to f if
1. (1)

${A}_{i}$, $i=1,2,\dots ,m$ are nonempty subsets of X;

2. (2)

$f\left({A}_{1}\right)\subset {A}_{2},f\left({A}_{2}\right)\subset {A}_{3},\dots ,f\left({A}_{m-1}\right)\subset {A}_{m},f\left({A}_{m}\right)\subset {A}_{1}$.

Kirk, Srinivasan and Veeramani  also proved the following theorem.

Theorem 1 

Let $\left(X,d\right)$ be a complete metric space, $m\in \mathbb{N}$, ${A}_{1},{A}_{2},\dots ,{A}_{m}$, be closed nonempty subsets of X and $X={\bigcup }_{i=1}^{m}{A}_{i}$. Suppose that f satisfies the following condition:
$d\left(fx,fy\right)\le \psi \left(d\left(x,y\right)\right),\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}x\in {A}_{i},y\in {A}_{i+1},i\in \left\{1,2,\dots ,m\right\},$

where $\psi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ is upper semi-continuous from the right and $0\le \psi \left(t\right) for $t>0$. Then f has a fixed point $z\in {\bigcap }_{i=1}^{n}{A}_{i}$.

Recently, the fixed theorems for an operator $f:X\to X$ defined on a metric space X with a cyclic representation of X with respect to f have appeared in the literature (see, e.g., ). In 2010, Pǎcurar and Rus  introduced the following notion of a cyclic weaker φ-contraction.

Definition 4 

Let $\left(X,d\right)$ be a metric space, $m\in \mathbb{N}$, ${A}_{1},{A}_{2},\dots ,{A}_{m}$ be closed nonempty subsets of X and $X={\bigcup }_{i=1}^{m}{A}_{i}$. An operator $f:X\to X$ is called a cyclic weaker φ-contraction if
1. (1)

$X={\bigcup }_{i=1}^{m}{A}_{i}$ is a cyclic representation of X with respect to f;

2. (2)
there exists a continuous, non-decreasing function $\phi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ with $\phi \left(t\right)>0$ for $t\in \left(0,\mathrm{\infty }\right)$ and $\phi \left(0\right)=0$ such that
$d\left(fx,fy\right)\le d\left(x,y\right)-\phi \left(d\left(x,y\right)\right)$

for any $x\in {A}_{i}$, $y\in {A}_{i+1}$, $i=1,2,\dots ,m$, where ${A}_{m+1}={A}_{1}$.

And Pǎcurar and Rus  proved the following main theorem.

Theorem 2 

Let $\left(X,d\right)$ be a complete metric space, $m\in \mathbb{N}$, ${A}_{1},{A}_{2},\dots ,{A}_{m}$ be closed nonempty subsets of X and $X={\bigcup }_{i=1}^{m}{A}_{i}$. Suppose that f is a cyclic weaker φ-contraction. Then f has a fixed point $z\in {\bigcap }_{i=1}^{n}{A}_{i}$.

In the recent years, fixed point theory has developed rapidly on cyclic contraction mappings, see .

The purpose of this paper is to study fixed point theorems for a mapping satisfying the cyclical generalized contractive conditions in complete partial metric spaces. Our results generalize or improve many recent fixed point theorems in the literature.

## 2 Fixed point theorems (I)

In the section, we denote by Ψ the class of functions $\psi :{{\mathbb{R}}^{+}}^{3}\to {\mathbb{R}}^{+}$ satisfying the following conditions:

(${\psi }_{1}$) ψ is an increasing and continuous function in each coordinate;

(${\psi }_{2}$) for $t\in {\mathbb{R}}^{+}$, $\psi \left(t,t,t\right)\le t$, $\psi \left(t,0,0\right)\le t$ and $\psi \left(0,0,t\right)\le t$.

Next, we denote by Θ the class of functions $\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ satisfying the following conditions:

(${\phi }_{1}$) φ is continuous and non-decreasing;

(${\phi }_{2}$) for $t>0$, $\phi \left(t\right)>0$ and $\phi \left(0\right)=0$.

And we denote by Φ the class of functions $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ satisfying the following conditions:

(${\varphi }_{1}$) ϕ is continuous;

(${\varphi }_{2}$) for $t>0$, $\varphi \left(t\right)>0$ and $\varphi \left(0\right)=0$.

We now state a new notion of cyclic $\mathcal{CW}$-contractions in partial metric spaces as follows.

Definition 5 Let $\left(X,p\right)$ be a partial metric space, $m\in \mathbb{N}$, ${A}_{1},{A}_{2},\dots ,{A}_{m}$ be nonempty subsets of X and $Y={\bigcup }_{i=1}^{m}{A}_{i}$. An operator $f:Y\to Y$ is called a cyclic $\mathcal{CW}$-contraction if
1. (1)

${\bigcup }_{i=1}^{m}{A}_{i}$ is a cyclic representation of Y with respect to f;

2. (2)
for any $x\in {A}_{i}$, $y\in {A}_{i+1}$, $i=1,2,\dots ,m$,
$\phi \left(p\left(fx,fy\right)\right)\le \psi \left(\phi \left(p\left(x,y\right)\right),\phi \left(p\left(x,fx\right)\right),\phi \left(p\left(y,fy\right)\right)\right)-\varphi \left(M\left(x,y\right)\right),$
(2.1)

where $\psi \in \mathrm{\Psi }$, $\phi \in \mathrm{\Theta }$, $\varphi \in \mathrm{\Phi }$, and $M\left(x,y\right)=max\left\{p\left(x,y\right),p\left(x,fx\right),p\left(y,fy\right)\right\}$.

Theorem 3 Let $\left(X,p\right)$ be a complete partial metric space, $m\in \mathbb{N}$, ${A}_{1},{A}_{2},\dots ,{A}_{m}$ be nonempty closed subsets of X and $Y={\bigcup }_{i=1}^{m}{A}_{i}$. Let $f:Y\to Y$ be a cyclic $\mathcal{CW}$-contraction. Then f has a unique fixed point $z\in {\bigcap }_{i=1}^{m}{A}_{i}$.

Proof Given ${x}_{0}$ and let ${x}_{n+1}=f{x}_{n}={f}^{n}{x}_{0}$ for $n=0,1,2,\dots$ . If there exists ${n}_{0}\in \mathbb{N}$ such that ${x}_{{n}_{0}+1}={x}_{{n}_{0}}$, then we finished the proof. Suppose that ${x}_{n+1}\ne {x}_{n}$ for any $n=0,1,2,\dots$ . Notice that for any $n\ge 0$, there exists ${i}_{n}\in \left\{1,2,\dots ,m\right\}$ such that ${x}_{n}\in {A}_{{i}_{n}}$ and ${x}_{n+1}\in {A}_{{i}_{n}+1}$.

Step 1. We will prove that
$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{n+1}\right)=0,\phantom{\rule{1em}{0ex}}\text{that is},\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{n},{x}_{n+1}\right)=0.$
Using (2.1), we have
$\begin{array}{rcl}\phi \left(p\left({x}_{n},{x}_{n+1}\right)\right)& =& \phi \left(p\left(f{x}_{n-1},f{x}_{n}\right)\right)\\ \le & \psi \left(\phi \left(p\left({x}_{n-1},{x}_{n}\right)\right),\phi \left(p\left({x}_{n-1},f{x}_{n-1}\right)\right),\phi \left(p\left({x}_{n},f{x}_{n}\right)\right)\right)-\varphi \left(M\left({x}_{n-1},{x}_{n}\right)\right)\\ =& \psi \left(\phi \left(p\left({x}_{n-1},{x}_{n}\right)\right),\phi \left(p\left({x}_{n-1},{x}_{n}\right)\right),\phi \left(p\left({x}_{n},{x}_{n+1}\right)\right)\right)-\varphi \left(M\left({x}_{n-1},{x}_{n}\right)\right),\end{array}$
where
$\begin{array}{rl}M\left({x}_{n-1},{x}_{n}\right)& =max\left\{p\left({x}_{n-1},{x}_{n}\right),p\left({x}_{n-1},f{x}_{n-1}\right),p\left({x}_{n},f{x}_{n}\right)\right\}\\ =max\left\{p\left({x}_{n-1},{x}_{n}\right),p\left({x}_{n-1},{x}_{n}\right),p\left({x}_{n},{x}_{n+1}\right)\right\}.\end{array}$
If $M\left({x}_{n-1},{x}_{n}\right)=p\left({x}_{n},{x}_{n+1}\right)$, then
$\begin{array}{rcl}\phi \left(p\left({x}_{n},{x}_{n+1}\right)\right)& \le & \psi \left(\phi \left(p\left({x}_{n},{x}_{n+1}\right)\right),\phi \left(p\left({x}_{n},{x}_{n+1}\right)\right),\phi \left(p\left({x}_{n},{x}_{n+1}\right)\right)\right)-\varphi \left(p\left({x}_{n},{x}_{n+1}\right)\right)\\ \le & \phi \left(p\left({x}_{n},{x}_{n+1}\right)\right)-\varphi \left(p\left({x}_{n},{x}_{n+1}\right)\right),\end{array}$

which implies that $\varphi \left(p\left({x}_{n},{x}_{n+1}\right)\right)=0$, and hence $p\left({x}_{n},{x}_{n+1}\right)=0$. This contradicts our initial assumption.

From the above argument, we have that for each $n\in \mathbb{N}$,
$\phi \left(p\left({x}_{n},{x}_{n+1}\right)\right)\le \phi \left(p\left({x}_{n-1},{x}_{n}\right)\right)-\varphi \left(p\left({x}_{n-1},{x}_{n}\right)\right),$
(2.2)
and
$p\left({x}_{n},{x}_{n+1}\right)
And since the sequence $\left\{p\left({x}_{n},{x}_{n+1}\right)\right\}$ is decreasing, it must converge to some $\eta \ge 0$. Taking limit as $n\to \mathrm{\infty }$ in (2.2) and by the continuity of φ and ϕ, we get
$\phi \left(\eta \right)\le \phi \left(\eta \right)-\varphi \left(\eta \right),$
and so we conclude that $\varphi \left(\eta \right)=0$ and $\eta =0$. Thus, we have
$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{n+1}\right)=0.$
(2.3)
By (p2), we also have
$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{n}\right)=0.$
(2.4)
Since ${d}_{p}\left(x,y\right)\le 2p\left(x,y\right)-p\left(x,x\right)-p\left(y,y\right)$ for all $x,y\in X$, using (2.3) and (2.4), we obtain that
$\underset{n\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{n},{x}_{n+1}\right)=0.$
(2.5)

Step 2. We show that $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the metric space $\left(Y,{d}_{p}\right)$. We claim that the following result holds.

Claim For every $\epsilon >0$, there exists $n\in \mathbb{N}$ such that if $r,q\ge n$ with $r-q=1modm$, then ${d}_{p}\left({x}_{r},{x}_{q}\right)<\epsilon$.

Suppose the above statement is false. Then there exists $ϵ>0$ such that for any $n\in \mathbb{N}$, there are ${r}_{n},{q}_{n}\in \mathbb{N}$ with ${r}_{n}>{q}_{n}\ge n$ with ${r}_{n}-{q}_{n}=1modm$ satisfying
${d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)\ge ϵ.$
Now, we let $n>2m$. Then corresponding to ${q}_{n}\ge n$ use, we can choose ${r}_{n}$ in such a way it is the smallest integer with ${r}_{n}>{q}_{n}\ge n$ satisfying ${r}_{n}-{q}_{n}=1modm$ and ${d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)\ge ϵ$. Therefore, ${d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}-m}\right)\le ϵ$ and
$\begin{array}{rl}ϵ& \le {d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)\\ \le {d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}-m}\right)+\sum _{i=1}^{m}{d}_{p}\left({x}_{{r}_{n-i}},{x}_{{r}_{n-i+1}}\right)\\ <ϵ+\sum _{i=1}^{m}{d}_{p}\left({x}_{{r}_{n-i}},{x}_{{r}_{n-i+1}}\right).\end{array}$
Letting $n\to \mathrm{\infty }$, we obtain that
$\underset{n\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)=ϵ.$
(2.6)
On the other hand, we can conclude that
$\begin{array}{rl}ϵ& \le {d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)\\ \le {d}_{p}\left({x}_{{q}_{n}},{x}_{{q}_{n+1}}\right)+{d}_{p}\left({x}_{{q}_{n+1}},{x}_{{r}_{n+1}}\right)+{d}_{p}\left({x}_{{r}_{n+1}},{x}_{{r}_{n}}\right)\\ \le {d}_{p}\left({x}_{{q}_{n}},{x}_{{q}_{n+1}}\right)+{d}_{p}\left({x}_{{q}_{n+1}},{x}_{{q}_{n}}\right)+{d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)+{d}_{p}\left({x}_{{r}_{n}},{x}_{{r}_{n+1}}\right)+{d}_{p}\left({x}_{{r}_{n+1}},{x}_{{r}_{n}}\right).\end{array}$
Letting $n\to \mathrm{\infty }$, we obtain that
$\underset{n\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{{q}_{n+1}},{x}_{{r}_{n+1}}\right)=ϵ.$
(2.7)
Since ${d}_{p}\left(x,y\right)=2p\left(x,y\right)-p\left(x,x\right)-p\left(y,y\right)$ and using (2.4), (2.6) and (2.7), we have that
$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)=\frac{ϵ}{2},$
(2.8)
and
$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{{q}_{n+1}},{x}_{{r}_{n+1}}\right)=\frac{ϵ}{2}.$
(2.9)
Since ${x}_{{q}_{n}}$ and ${x}_{{r}_{n}}$ lie in different adjacently labeled sets ${A}_{i}$ and ${A}_{i+1}$ for certain $1\le i\le m$, by using the fact that f is a cyclic $\mathcal{CW}$-contraction, we have
$\begin{array}{rcl}\phi \left(p\left(f{x}_{{q}_{n}+1},f{x}_{{r}_{n}+1}\right)\right)& =& \phi \left(p\left(f{x}_{{q}_{n}},f{x}_{{r}_{n}}\right)\right)\\ \le & \psi \left(\phi \left(p\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)\right),\phi \left(p\left({x}_{{q}_{n}},f{x}_{{q}_{n}}\right)\right),\phi \left(p\left({x}_{{r}_{n}},f{x}_{{r}_{n}}\right)\right)\right)\\ -\varphi \left(M\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)\right)\\ =& \psi \left(\phi \left(p\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)\right),\phi \left(p\left({x}_{{q}_{n}},{x}_{{q}_{n}+1}\right)\right),\phi \left(p\left({x}_{{r}_{n}},{x}_{{r}_{n}+1}\right)\right)\right)\\ -\varphi \left(M\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)\right),\end{array}$
where
$M\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)=max\left\{p\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right),p\left({x}_{{q}_{n}},{x}_{{q}_{n}+1}\right),p\left({x}_{{r}_{n}},{x}_{{r}_{n}+1}\right)\right\}.$
Thus, letting $n\to \mathrm{\infty }$, we can conclude that
$\phi \left(\frac{ϵ}{2}\right)\le \psi \left(\phi \left(\frac{ϵ}{2}\right),\phi \left(0\right),\phi \left(0\right)\right)-\varphi \left(\frac{ϵ}{2}\right)\le \phi \left(\frac{ϵ}{2}\right)-\varphi \left(\frac{ϵ}{2}\right),$

which implies $\varphi \left(\frac{ϵ}{2}\right)=0$, that is, $ϵ=0$. So, we get a contradiction. Therefore, our claim is proved.

In the sequel, we will show that $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the metric space $\left(Y,{d}_{p}\right)$. Let $\epsilon >0$ be given. By our claim, there exists ${n}_{1}\in \mathbb{N}$ such that if $r,q\ge {n}_{1}$ with $r-q=1modm$, then
${d}_{p}\left({x}_{r},{x}_{q}\right)\le \frac{\epsilon }{2}.$
Since ${lim}_{n\to \mathrm{\infty }}{d}_{p}\left({x}_{n},{x}_{n+1}\right)=0$, there exists ${n}_{2}\in \mathbb{N}$ such that
${d}_{p}\left({x}_{n},{x}_{n+1}\right)\le \frac{\epsilon }{2m}$

for any $n\ge {n}_{2}$.

Let $r,q\ge max\left\{{n}_{1},{n}_{2}\right\}$ and $r>q$. Then there exists $k\in \left\{1,2,\dots ,m\right\}$ such that $r-q=kmodm$. Therefore, $r-q+j=1modm$ for $j=m-k+1$, and so we have
$\begin{array}{rl}{d}_{p}\left({x}_{q},{x}_{r}\right)& \le {d}_{p}\left({x}_{q},{x}_{r+j}\right)+{d}_{p}\left({x}_{r+j},{x}_{r+j-1}\right)+\cdots +{d}_{p}\left({x}_{r-1},{x}_{r}\right)\\ \le \frac{\epsilon }{2}+j×\frac{\epsilon }{2m}\\ \le \frac{\epsilon }{2}+m×\frac{\epsilon }{2m}=\epsilon .\end{array}$

Thus, $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the metric space $\left(Y,{d}_{p}\right)$.

Step 3. We show that f has a fixed point ν in ${\bigcap }_{i=1}^{m}{A}_{i}$.

Since Y is closed, the subspace $\left(Y,p\right)$ is complete. Then from Lemma 1, we have that $\left(Y,{d}_{p}\right)$ is complete. Thus, there exists $\nu \in X$ such that
$\underset{n\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{n},\nu \right)=0.$
And it follows from Lemma 1 that we have
$p\left(\nu ,\nu \right)=\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},\nu \right)=\underset{n,m\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{m}\right).$
(2.10)
On the other hand, since the sequence $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the metric space $\left(Y,{d}_{p}\right)$, we also have
$\underset{n\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{n},{x}_{m}\right)=0.$
Since ${d}_{p}\left(x,y\right)=2p\left(x,y\right)-p\left(x,x\right)-p\left(y,y\right)$, we can deduce that
$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{m}\right)=0.$
(2.11)
Since $Y={\bigcup }_{i=1}^{m}{A}_{i}$ is a cyclic representation of X with respect to f, the sequence $\left\{{x}_{n}\right\}$ has infinite terms in each ${A}_{i}$ for $i\in \left\{1,2,\dots ,m\right\}$. Now, for all $i=1,2,\dots ,m$, we may take a subsequence $\left\{{x}_{{n}_{k}}\right\}$ of $\left\{{x}_{n}\right\}$ with ${x}_{{n}_{k}}\in {A}_{i-1}$ and also all converge to ν. Using (2.10) and (2.11), we have
$p\left(\nu ,\nu \right)=\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},\nu \right)=\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{{n}_{k}},\nu \right)=0.$
By (2.1),
$\begin{array}{rl}\phi \left(p\left({x}_{{n}_{k+1}},f\nu \right)\right)& =\phi \left(p\left(f{x}_{{n}_{k}},f\nu \right)\right)\\ \le \psi \left(\phi \left(p\left({x}_{{n}_{k}},\nu \right)\right),\phi \left(p\left({x}_{{n}_{k}},f{x}_{{n}_{k}}\right)\right),\phi \left(p\left(\nu ,f\nu \right)\right)\right)-\varphi \left(M\left({x}_{{n}_{k}},\nu \right)\right)\\ =\psi \left(\phi \left(p\left({x}_{{n}_{k}},\nu \right)\right),\phi \left(p\left({x}_{{n}_{k}},{x}_{{n}_{k}+1}\right)\right),\phi \left(p\left(\nu ,f\nu \right)\right)\right)-\varphi \left(M\left({x}_{{n}_{k}},\nu \right)\right),\end{array}$
where
$M\left({x}_{{n}_{k}},\nu \right)=max\left\{p\left({x}_{{n}_{k}},\nu \right),p\left({x}_{{n}_{k}},{x}_{{n}_{k}+1}\right),p\left(\nu ,f\nu \right)\right\}.$
Letting $k\to \mathrm{\infty }$, we have
$\begin{array}{rl}\phi \left(p\left(\nu ,f\nu \right)\right)& \le \psi \left(\phi \left(0\right),\phi \left(0\right),\phi \left(p\left(\nu ,f\nu \right)\right)\right)-\varphi \left(p\left(\nu ,f\nu \right)\right)\\ \le \phi \left(p\left(\nu ,f\nu \right)\right)-\varphi \left(p\left(\nu ,f\nu \right)\right),\end{array}$

which implies $\varphi \left(p\left(\nu ,f\nu \right)\right)=0$, that is, $p\left(\nu ,f\nu \right)=0$. So, $\nu =f\nu$.

Step 4. Finally, to prove the uniqueness of the fixed point, suppose that μ, ν are fixed points of f. Then using the inequality (2.1), we obtain that
$\begin{array}{rcl}\phi \left(p\left(\mu ,\nu \right)\right)& =& \phi \left(p\left(f\mu ,f\nu \right)\right)\le \psi \left(\phi \left(p\left(\mu ,\nu \right)\right),\phi \left(p\left(\mu ,f\mu \right)\right),\phi \left(p\left(\nu ,f\nu \right)\right)\right)\\ -\varphi \left(M\left(\mu ,\nu \right)\right),\end{array}$
where
$M\left(\mu ,\nu \right)=max\left\{p\left(\mu ,\nu \right),p\left(\mu ,f\mu \right),p\left(\nu ,f\nu \right)\right\}=p\left(\mu ,\nu \right).$
So, we also deduce that
$\begin{array}{rl}\phi \left(p\left(\mu ,\nu \right)\right)& \le \psi \left(\phi \left(p\left(\mu ,\nu \right),0,0\right)\right)\\ \le \phi \left(p\left(\mu ,\nu \right)\right)-\varphi \left(p\left(\mu ,\nu \right)\right),\end{array}$

which implies that $\varphi \left(p\left(\mu ,\nu \right)\right)=0$, and hence $p\left(\mu ,\nu \right)=0$, that is, $\mu =\nu$. So, we complete the proof.  □

The following provides an example for Theorem 3.

Example 1 Let $X=\left[0,1\right]$ and $A=\left[0,1\right]$, $B=\left[0,\frac{1}{2}\right]$, $C=\left[0,\frac{1}{4}\right]$. We define the partial metric p on X by
and define the function $f:X\to X$ by
Now, we let $\phi ,\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ and $\psi :{{\mathbb{R}}^{+}}^{3}\to {\mathbb{R}}^{+}$ be
$\phi \left(t\right)=2t,\phantom{\rule{2em}{0ex}}\varphi \left(t\right)=\frac{2t}{5\left(1+t\right)}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\psi \left(t\right)=\frac{4}{5}\cdot max\left\{{t}_{1},{t}_{2},{t}_{3}\right\}.$

Then f is a cyclic $\mathcal{CW}$-contraction and 0 is the unique fixed point.

Proof We claim that f is a cyclic $\mathcal{CW}$-contraction.
1. (1)

Note that $f\left(A\right)=\left[0,\frac{1}{2}\right]\subset B$, $f\left(B\right)=\left[0,\frac{1}{6}\right]\subset C$ and $f\left(C\right)=\left[0,\frac{1}{20}\right]\subset A$. Thus, $A\cup B\cup C$ is a cyclic representation of X with respect to f;

2. (2)
For $x\in A$ and $y\in B$ (or, $x\in B$ and $y\in C$), without loss of generality, we may assume that $x\ge y$, then we have

Since
$\frac{2{x}^{2}}{1+x}\le \frac{8x}{5}-\frac{2x}{5\left(1+x\right)},$
we have
$\begin{array}{rcl}\phi \left(p\left(fx,fy\right)\right)& \le & \psi \left(\phi \left(p\left(x,y\right)\right),\phi \left(p\left(x,fx\right)\right),\phi \left(p\left(y,fy\right)\right)\right)\\ -\varphi \left(max\left\{p\left(x,y\right),p\left(x,fx\right),p\left(y,fy\right)\right\}\right).\end{array}$

On the other hand, for $x\in C$ and $y\in A$, without loss of generality, we may assume that $x\le y$, then it is easy to get the above inequality.

Note that Example 1 satisfies all of the hypotheses of Theorem 3, and we get that 0 is the unique fixed point. □

## 3 Fixed point theorems (II)

In this article, we also recall the notion of a Meir-Keeler function (see ). A function $\varphi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ is said to be a Meir-Keeler function if for each $\eta >0$, there exists $\delta >0$ such that for $t\in \left[0,\mathrm{\infty }\right)$ with $\eta \le t<\eta +\delta$, we have $\varphi \left(t\right)<\eta$. We now introduce a new notion of a weaker Meir-Keeler function $\varphi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ in a partial metric space $\left(X,p\right)$ as follows.

Definition 6 Let $\left(X,p\right)$ be a partial metric space. We call $\varphi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ a weaker Meir-Keeler function in X if for each $\eta >0$, there exists $\delta >0$ such that for $x,y\in X$ with $\eta \le p\left(x,y\right)<\eta +\delta$, there exists ${n}_{0}\in \mathbb{N}$ such that ${\varphi }^{{n}_{0}}\left(p\left(x,y\right)\right)<\eta$.

In the section, we denote by Φ the class of weaker Meir-Keeler functions $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ in a partial metric space in $\left(X,p\right)$ satisfying the following conditions:

(${\varphi }_{1}$) $\varphi \left(t\right)>0$ for $t>0$, $\varphi \left(0\right)=0$;

(${\varphi }_{2}$) ${\left\{{\varphi }^{n}\left(t\right)\right\}}_{n\in \mathbb{N}}$ is decreasing;

(${\varphi }_{3}$) for ${t}_{n}\in \left[0,\mathrm{\infty }\right)$,
1. (a)

if ${lim}_{n\to \mathrm{\infty }}{t}_{n}=\gamma >0$, then ${lim}_{n\to \mathrm{\infty }}\varphi \left({t}_{n}\right)<\gamma$ and

2. (b)

if ${lim}_{n\to \mathrm{\infty }}{t}_{n}=0$, then ${lim}_{n\to \mathrm{\infty }}\varphi \left({t}_{n}\right)=0$.

And we denote by the class Ψ of functions $\psi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ a continuous function satisfying $\psi \left(t\right)>0$ for $t>0$, $\psi \left(0\right)=0$.

First, we state a new notion of cyclic $\mathcal{MK}$-contractions in partial metric spaces as follows.

Definition 7 Let $\left(X,p\right)$ be a partial metric space, $m\in \mathbb{N}$, ${A}_{1},{A}_{2},\dots ,{A}_{m}$ be nonempty subsets of X and $Y={\bigcup }_{i=1}^{m}{A}_{i}$. An operator $f:Y\to Y$ is called a cyclic $\mathcal{MK}$-contraction if
1. (1)

${\bigcup }_{i=1}^{m}{A}_{i}$ is a cyclic representation of Y with respect to f;

2. (2)
for any $x\in {A}_{i}$, $y\in {A}_{i+1}$, $i=1,2,\dots ,m$,
$p\left(fx,fy\right)\le \varphi \left(p\left(x,y\right)\right)-\psi \left(p\left(x,y\right)\right),$
(3.1)

where ${A}_{m+1}={A}_{1}$, $\varphi \in \mathrm{\Phi }$ and $\psi \in \mathrm{\Psi }$.

Theorem 4 Let $\left(X,p\right)$ be a complete partial metric space, $m\in \mathbb{N}$, ${A}_{1},{A}_{2},\dots ,{A}_{m}$ be nonempty closed subsets of X and $Y={\bigcup }_{i=1}^{m}{A}_{i}$. Let $f:Y\to Y$ be a cyclic $\mathcal{MK}$-contraction. Then f has a unique fixed point $z\in {\bigcap }_{i=1}^{m}{A}_{i}$.

Proof Given ${x}_{0}$ and let ${x}_{n+1}=f{x}_{n}={f}^{n}{x}_{0}$, for $n=0,1,2,\dots$ . If there exists ${n}_{0}\in \mathbb{N}$ such that ${x}_{{n}_{0}+1}={x}_{{n}_{0}}$, then we finished the proof. Suppose that ${x}_{n+1}\ne {x}_{n}$ for any $n=0,1,2,\dots$ . Notice that for any $n\ge 0$, there exists ${i}_{n}\in \left\{1,2,\dots ,m\right\}$ such that ${x}_{n}\in {A}_{{i}_{n}}$ and ${x}_{n+1}\in {A}_{{i}_{n}+1}$. Then by (3.1), we have
$p\left({x}_{n},{x}_{n+1}\right)=p\left(f{x}_{n-1},f{x}_{n}\right)\le \varphi \left(p\left({x}_{n-1},{x}_{n}\right)\right)-\psi \left(p\left({x}_{n-1},{x}_{n}\right)\right).$
Step 1. We will prove that
$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{n+1}\right)=0,\phantom{\rule{1em}{0ex}}\text{that is},\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{n},{x}_{n+1}\right)=0.$
Since f is a cyclic $\mathcal{MK}$-contraction, we can conclude that
$\begin{array}{rl}p\left({x}_{n},{x}_{n+1}\right)& \le \varphi \left(p\left({x}_{n-1},{x}_{n}\right)\right)\\ \le \varphi \left(\varphi \left(p\left({x}_{n-2},{x}_{n-1}\right)\right)={\varphi }^{2}\left(p\left({x}_{n-2},{x}_{n-1}\right)\right)\\ \le \cdots \\ \le {\varphi }^{n}\left(p\left({x}_{0},{x}_{1}\right)\right).\end{array}$
Since ${\left\{{\varphi }^{n}\left(p\left({x}_{0},{x}_{1}\right)\right)\right\}}_{n\in \mathbb{N}}$ is decreasing, it must converge to some $\eta \ge 0$. We claim that $\eta =0$. On the contrary, assume that $0<\eta$. Then by the definition of a weaker Meir-Keeler function ϕ, there exists $\delta >0$ such that for ${x}_{0},{x}_{1}\in X$ with $\eta \le p\left({x}_{0},{x}_{1}\right)<\delta +\eta$, there exists ${n}_{0}\in \mathbb{N}$ such that ${\varphi }^{{n}_{0}}\left(p\left({x}_{0},{x}_{1}\right)\right)<\eta$. Since ${lim}_{n\to \mathrm{\infty }}{\varphi }^{n}\left(p\left({x}_{0},{x}_{1}\right)\right)=\eta$, there exists ${k}_{0}\in \mathbb{N}$ such that $\eta \le {\varphi }^{k}\left(p\left({x}_{0},{x}_{1}\right)\right)<\delta +\eta$, for all $k\ge {k}_{0}$. Thus, we conclude that ${\varphi }^{{k}_{0}+{n}_{0}}\left(p\left({x}_{0},{x}_{1}\right)\right)<\eta$. So, we get a contradiction. Therefore, ${lim}_{n\to \mathrm{\infty }}{\varphi }^{n}\left(p\left({x}_{0},{x}_{1}\right)\right)=0$, and so we have
$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{n+1}\right)=0.$
(3.2)
By (p2), we also have
$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{n}\right)=0.$
(3.3)
Since ${d}_{p}\left(x,y\right)\le 2p\left(x,y\right)-p\left(x,x\right)-p\left(y,y\right)$ for all $x,y\in X$, using (3.2) and (3.3), we obtain that
$\underset{n\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{n},{x}_{n+1}\right)=0.$
(3.4)

Step 2. We show that $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the metric space $\left(Y,{d}_{p}\right)$. We claim that the following result holds.

Claim For every $\epsilon >0$, there exists $n\in \mathbb{N}$ such that if $r,q\ge n$ with $r-q=1modm$, then ${d}_{p}\left({x}_{r},{x}_{q}\right)<\epsilon$.

Suppose the above statement is false. Then there exists $ϵ>0$ such that for any $n\in \mathbb{N}$, there are ${r}_{n},{q}_{n}\in \mathbb{N}$ with ${r}_{n}>{q}_{n}\ge n$ with ${r}_{n}-{q}_{n}=1modm$ satisfying
${d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)\ge ϵ.$
Now, we let $n>2m$. Then corresponding to ${q}_{n}\ge n$ use, we can choose ${r}_{n}$ in such a way it is the smallest integer with ${r}_{n}>{q}_{n}\ge n$ satisfying ${r}_{n}-{q}_{n}=1modm$ and ${d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)\ge ϵ$. Therefore, ${d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}-m}\right)\le ϵ$ and
$\begin{array}{rl}ϵ& \le {d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)\\ \le {d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}-m}\right)+\sum _{i=1}^{m}{d}_{p}\left({x}_{{r}_{n-i}},{x}_{{r}_{n-i+1}}\right)\\ <ϵ+\sum _{i=1}^{m}{d}_{p}\left({x}_{{r}_{n-i}},{x}_{{r}_{n-i+1}}\right).\end{array}$
Letting $n\to \mathrm{\infty }$, we obtain that
$\underset{n\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)=ϵ.$
(3.5)
On the other hand, we can conclude that
$\begin{array}{rl}ϵ& \le {d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)\\ \le {d}_{p}\left({x}_{{q}_{n}},{x}_{{q}_{n+1}}\right)+{d}_{p}\left({x}_{{q}_{n+1}},{x}_{{r}_{n+1}}\right)+{d}_{p}\left({x}_{{r}_{n+1}},{x}_{{r}_{n}}\right)\\ \le {d}_{p}\left({x}_{{q}_{n}},{x}_{{q}_{n+1}}\right)+{d}_{p}\left({x}_{{q}_{n+1}},{x}_{{q}_{n}}\right)+{d}_{p}\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)+{d}_{p}\left({x}_{{r}_{n}},{x}_{{r}_{n+1}}\right)+{d}_{p}\left({x}_{{r}_{n+1}},{x}_{{r}_{n}}\right).\end{array}$
Letting $n\to \mathrm{\infty }$, we obtain that
$\underset{n\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{{q}_{n+1}},{x}_{{r}_{n+1}}\right)=ϵ.$
(3.6)
Since ${d}_{p}\left(x,y\right)=2p\left(x,y\right)-p\left(x,x\right)-p\left(y,y\right)$ and using (3.5) and (3.6), we have that
$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)=\frac{ϵ}{2},$
(3.7)
and
$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{{q}_{n+1}},{x}_{{r}_{n+1}}\right)=\frac{ϵ}{2}.$
(3.8)
Since ${x}_{{q}_{n}}$ and ${x}_{{r}_{n}}$ lie in different adjacently labeled sets ${A}_{i}$ and ${A}_{i+1}$ for certain $1\le i\le m$, by using the fact that f is a cyclic $\mathcal{MK}$-contraction, we have
$p\left({x}_{{q}_{n+1}},{x}_{{r}_{n+1}}\right)=p\left(f{x}_{{q}_{n}},f{x}_{{r}_{n}}\right)\le \varphi \left(p\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)\right)-\psi \left(p\left({x}_{{q}_{n}},{x}_{{r}_{n}}\right)\right).$
Letting $n\to \mathrm{\infty }$, by using the condition ${\varphi }_{3}$ of the function ϕ, we obtain that
$\frac{ϵ}{2}\le \frac{ϵ}{2}-\psi \left(\frac{ϵ}{2}\right),$

and consequently, $\psi \left(\frac{ϵ}{2}\right)=0$. By the definition of a function ψ, we get $ϵ=0$ which is a contraction. Therefore, our claim is proved.

In the sequel, we will show that $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the metric space $\left(Y,{d}_{p}\right)$. Let $\epsilon >0$ be given. By our claim, there exists ${n}_{1}\in \mathbb{N}$ such that if $r,q\ge {n}_{1}$ with $r-q=1modm$, then
${d}_{p}\left({x}_{r},{x}_{q}\right)\le \frac{\epsilon }{2}.$
Since ${lim}_{n\to \mathrm{\infty }}{d}_{p}\left({x}_{n},{x}_{n+1}\right)=0$, there exists ${n}_{2}\in \mathbb{N}$ such that
${d}_{p}\left({x}_{n},{x}_{n+1}\right)\le \frac{\epsilon }{2m}$

for any $n\ge {n}_{2}$.

Let $r,q\ge max\left\{{n}_{1},{n}_{2}\right\}$ and $r>q$. Then there exists $k\in \left\{1,2,\dots ,m\right\}$ such that $r-q=kmodm$. Therefore, $r-q+j=1modm$ for $j=m-k+1$, and so we have
$\begin{array}{rl}{d}_{p}\left({x}_{q},{x}_{r}\right)& \le {d}_{p}\left({x}_{q},{x}_{r+j}\right)+{d}_{p}\left({x}_{r+j},{x}_{r+j-1}\right)+\cdots +{d}_{p}\left({x}_{r-1},{x}_{r}\right)\\ \le \frac{\epsilon }{2}+j×\frac{\epsilon }{2m}\\ \le \frac{\epsilon }{2}+m×\frac{\epsilon }{2m}=\epsilon .\end{array}$

Thus, $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the metric space $\left(Y,{d}_{p}\right)$.

Step 3. We show that f has a fixed point ν in ${\bigcap }_{i=1}^{m}{A}_{i}$.

Since Y is closed, the subspace $\left(Y,p\right)$ is complete. Then from Lemma 1, we have that $\left(Y,{d}_{p}\right)$ is complete. Thus, there exists $\nu \in X$ such that
$\underset{n\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{n},\nu \right)=0.$
And it follows from Lemma 1 that we have
$p\left(\nu ,\nu \right)=\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},\nu \right)=\underset{n,m\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{m}\right).$
(3.9)
On the other hand, since the sequence $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the metric space $\left(Y,{d}_{p}\right)$, we also have
$\underset{n\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{n},{x}_{m}\right)=0.$
Since ${d}_{p}\left(x,y\right)=2p\left(x,y\right)-p\left(x,x\right)-p\left(y,y\right)$, we can deduce that
$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{m}\right)=0.$
(3.10)
Since $Y={\bigcup }_{i=1}^{m}{A}_{i}$ is a cyclic representation of X with respect to f, the sequence $\left\{{x}_{n}\right\}$ has infinite terms in each ${A}_{i}$ for $i\in \left\{1,2,\dots ,m\right\}$. Now, for all $i=1,2,\dots ,m$, we may take a subsequence $\left\{{x}_{{n}_{k}}\right\}$ of $\left\{{x}_{n}\right\}$ with ${x}_{{n}_{k}}\in {A}_{i-1}$ and also all converge to ν. Using (3.9) and (3.10), we have
$p\left(\nu ,\nu \right)=\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},\nu \right)=\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{{n}_{k}},\nu \right)=0.$
By (3.1),
$\begin{array}{rl}p\left({x}_{{n}_{k+1}},f\nu \right)& =p\left(f{x}_{{n}_{k}},f\nu \right)\\ \le \varphi \left(p\left({x}_{{n}_{k}},\nu \right)\right)-\psi \left(p\left({x}_{{n}_{k}},\nu \right)\right)\\ \le \varphi \left(p\left({x}_{{n}_{k}},\nu \right)\right).\end{array}$
Letting $k\to \mathrm{\infty }$, we have
$p\left(\nu ,f\nu \right)\le 0,$

and so $\nu =f\nu$.

Step 4. Finally, to prove the uniqueness of the fixed point, let μ be another fixed point of f in ${\bigcap }_{i=1}^{m}{A}_{i}$. By the cyclic character of f, we have $\mu ,\nu \in {\bigcap }_{i=1}^{n}{A}_{i}$. Since f is a cyclic weaker $\mathcal{MK}$-contraction, we have
$\begin{array}{rl}p\left(\nu ,\mu \right)& =p\left(\nu ,f\mu \right)\\ =\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{{n}_{k+1}},f\mu \right)\\ =\underset{n\to \mathrm{\infty }}{lim}p\left(f{x}_{{n}_{k}},f\mu \right)\\ \le \underset{n\to \mathrm{\infty }}{lim}\left[\varphi \left(p\left({x}_{{n}_{k}},\mu \right)\right)-\psi \left(p\left({x}_{{n}_{k}},\mu \right)\right)\right]\\ \le p\left(\nu ,\mu \right)-\psi \left(p\left(\nu ,\mu \right)\right),\end{array}$
and we can conclude that
$\psi \left(p\left(\nu ,\mu \right)\right)=0,$

which implies $p\left(\nu ,\mu \right)=0$. So, we have $\mu =\nu$. We complete the proof.  □

The following provides an example for Theorem 4.

Example 2 Let $X=\left[0,1\right]$ and $A=\left[0,1\right]$, $B=\left[0,\frac{1}{2}\right]$, $C=\left[0,\frac{1}{4}\right]$. We define the partial metric p on X by
and define the function $f:X\to X$ by
Now, we let $\psi ,\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ be
$\varphi \left(t\right)=\frac{4t}{5}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\psi \left(t\right)=\frac{t}{5\left(1+t\right)}.$

Then f is a cyclic $\mathcal{MK}$-contraction and 0 is the unique fixed point.

By Theorem 4, it is easy to get the following corollary.

Corollary 1 Let $\left(X,p\right)$ be a complete partial metric space, $m\in \mathbb{N}$, ${A}_{1},{A}_{2},\dots ,{A}_{m}$ be nonempty closed subsets of X, $Y={\bigcup }_{i=1}^{m}{A}_{i}$ and let $f:Y\to Y$. Assume that
1. (1)

${\bigcup }_{i=1}^{m}{A}_{i}$ is a cyclic representation of Y with respect to f;

2. (2)
for any $x\in {A}_{i}$, $y\in {A}_{i+1}$, $i=1,2,\dots ,m$,
$p\left(fx,fy\right)\le \varphi \left(p\left(x,y\right)\right),$

where ${A}_{m+1}={A}_{1}$ and $\varphi \in \mathrm{\Phi }$.

Then f has a unique fixed point $z\in {\bigcap }_{i=1}^{m}{A}_{i}$.

## Declarations

### Acknowledgements

The authors would like to thank referee(s) for many useful comments and suggestions for the improvement of the paper.

## Authors’ Affiliations

(1)
Department of Applied Mathematics, National Hsinchu University of Education, Hsinchu, Taiwan

## References 