# Fixed point results for Meir-Keeler-type ϕ-α-contractions on partial metric spaces

## Abstract

The purpose of this paper is to study fixed point theorems for a mapping satisfying the generalized Meir-Keeler-type ϕ-α-contractions in complete partial metric spaces. Our results generalize or improve many recent fixed point theorems in the literature.

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

## 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. In 1994, Mattews [1] introduced the following notion of partial metric spaces.

Definition 1 [1]

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

(${p}_{1}$) $x=y$ if and only if $p\left(x,x\right)=p\left(x,y\right)=p\left(y,y\right)$;

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

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

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

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.

Remark 1 It is clear that if $p\left(x,y\right)=0$, then from (${p}_{1}$) and (${p}_{2}$), $x=y$. 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 [1]

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 [1, 2]

1. (1)

$\left\{{x}_{n}\right\}$ 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. (2)

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 recent years, fixed point theory has developed rapidly on partial metric spaces, see [210].

In this study, we also recall the Meir-Keeler-type contraction [11] and α-admissible one [12]. In 1969, Meir and Keeler [11] introduced the following notion of Meir-Keeler-type contraction in a metric space $\left(X,d\right)$.

Definition 3 Let $\left(X,d\right)$ be a metric space, $f:X\to X$. Then f is called a Meir-Keeler-type contraction whenever, for each $\eta >0$, there exists $\gamma >0$ such that

$\eta \le d\left(x,y\right)<\eta +\gamma \phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}d\left(fx,fy\right)<\eta .$

The following definition was introduced in [12].

Definition 4 Let $f:X\to X$ be a self-mapping of a set X and $\alpha :X×X\to {\mathbb{R}}^{+}$. Then f is called α-admissible if

$x,y\in X,\phantom{\rule{1em}{0ex}}\alpha \left(x,y\right)\ge 1\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}\alpha \left(fx,fy\right)\ge 1.$

The purpose of this paper is to study fixed point theorems for a mapping satisfying the generalized Meir-Keeler-type ϕ-α-contractions in complete partial metric spaces. Our results generalize or improve many recent fixed point theorems in the literature.

## 2 Main results

In the article, we denote by Φ the class of functions $\varphi :{{\mathbb{R}}^{+}}^{4}\to {\mathbb{R}}^{+}$ satisfying the following conditions:

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

(${\varphi }_{2}$) for $t\in {\mathbb{R}}^{+}\mathrm{\setminus }\left\{0\right\}$, $\varphi \left(t,t,t,t\right)\le t$, $\varphi \left(t,0,0,t\right)\le t$, $\varphi \left(0,0,t,\frac{t}{2}\right)\le t$; and $\varphi \left({t}_{1},{t}_{2},{t}_{3},{t}_{4}\right)=0$ iff ${t}_{1}={t}_{2}={t}_{3}={t}_{4}=0$.

We now state the new notions of generalized Meir-Keeler-type ϕ-contractions and generalized Meir-Keeler-type ϕ-α-contractions in partial metric spaces as follows.

Definition 5 Let $\left(X,p\right)$ be a partial metric space, $f:X\to X$ and $\varphi \in \mathrm{\Phi }$. Then f is called a generalized Meir-Keeler-type ϕ-contraction whenever, for each $\eta >0$, there exists $\delta >0$ such that

$\begin{array}{r}\eta \le \varphi \left(p\left(x,y\right),p\left(x,fx\right),p\left(y,fy\right),\frac{1}{2}\left[p\left(x,fy\right)+p\left(y,fx\right)\right]\right)<\eta +\delta \\ \phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}p\left(fx,fy\right)<\eta .\end{array}$

Definition 6 Let $\left(X,p\right)$ be a partial metric space, $f:X\to X$, $\varphi \in \mathrm{\Phi }$ and $\alpha :X×X\to {\mathbb{R}}^{+}$. Then f is called a generalized Meir-Keeler-type ϕ-α-contraction if the following conditions hold:

1. (1)

2. (2)

for each $\eta >0$, there exists $\delta >0$ such that

$\begin{array}{r}\eta \le \varphi \left(p\left(x,y\right),p\left(x,fx\right),p\left(y,fy\right),\frac{1}{2}\left[p\left(x,fy\right)+p\left(y,fx\right)\right]\right)<\eta +\delta \\ \phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}\alpha \left(x,x\right)\alpha \left(y,y\right)p\left(fx,fy\right)<\eta .\end{array}$
(2.1)

Remark 3 Note that if f is a generalized Meir-Keeler-type ϕ-α-contraction, then we have that for all $x,y\in X$,

$\begin{array}{r}\alpha \left(x,x\right)\alpha \left(y,y\right)p\left(fx,fy\right)\\ \phantom{\rule{1em}{0ex}}\le \varphi \left(p\left(x,y\right),p\left(x,fx\right),p\left(y,fy\right),\frac{1}{2}\left[p\left(x,fy\right)+p\left(y,fx\right)\right]\right).\end{array}$

Further, if $\varphi \left(p\left(x,y\right),p\left(x,fx\right),p\left(y,fy\right),\frac{1}{2}\left[p\left(x,fy\right)+p\left(y,fx\right)\right]\right)=0$, then $p\left(fx,fy\right)=0$. On the other hand, if $\varphi \left(p\left(x,y\right),p\left(x,fx\right),p\left(y,fy\right),\frac{1}{2}\left[p\left(x,fy\right)+p\left(y,fx\right)\right]\right)>0$, then $\alpha \left(x,x\right)\alpha \left(y,y\right)p\left(fx,fy\right)<\varphi \left(p\left(x,y\right),p\left(x,fx\right),p\left(y,fy\right),\frac{1}{2}\left[p\left(x,fy\right)+p\left(y,fx\right)\right]\right)$.

We now state our main result for the generalized Meir-Keeler-type ϕ-α-contraction as follows.

Theorem 1 Let $\left(X,p\right)$ be a complete partial metric space, and $\varphi \in \mathrm{\Phi }$. If $\alpha :X×X\to {\mathbb{R}}^{+}$ satisfies the following conditions:

(${\alpha }_{1}$) there exists ${x}_{0}\in X$ such that $\alpha \left({x}_{0},{x}_{0}\right)\ge 1$;

(${\alpha }_{2}$) if $\alpha \left({x}_{n},{x}_{n}\right)\ge 1$ for all $n\in \mathbb{N}$, then ${lim}_{n\to \mathrm{\infty }}\alpha \left({x}_{n},{x}_{n}\right)\ge 1$;

(${\alpha }_{3}$) $\alpha :X×X\to {\mathbb{R}}^{+}$ is a continuous function in each coordinate.

Suppose that $f:X\to X$ is a generalized Meir-Keeler-type ϕ-α-contraction. Then f has a fixed point in X.

Proof Let ${x}_{0}$ and let ${x}_{n+1}=f{x}_{n}={f}^{n}{x}_{0}$ for $n=0,1,2,\dots$ . Since f is α-admissible and $\alpha \left({x}_{0},{x}_{0}\right)\ge 1$, we have

$\alpha \left(f{x}_{0},f{x}_{0}\right)=\alpha \left({x}_{1},{x}_{1}\right)\ge 1.$

By continuing this process, we get

(2.2)

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 $n=0,1,2,\dots$ . By the definition of the function ϕ, we have $\varphi \left(p\left({x}_{n},{x}_{n+1}\right),p\left({x}_{n},f{x}_{n}\right),p\left({x}_{n+1},f{x}_{n+1}\right),\frac{1}{2}\left[p\left({x}_{n},f{x}_{n+1}\right)+p\left({x}_{n+1},f{x}_{n}\right)\right]\right)>0$ for all $n\in \mathbb{N}\cup \left\{0\right\}$.

Step 1. We shall prove that

$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{n+1}\right)=0,\phantom{\rule{1em}{0ex}}\text{that is}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{n},{x}_{n+1}\right)=0.$

By Remark 3 and (${p}_{4}$), using (2.2), we have

$\begin{array}{r}p\left({x}_{n+1},{x}_{n+2}\right)\\ \phantom{\rule{1em}{0ex}}=p\left(f{x}_{n},f{x}_{n+1}\right)\\ \phantom{\rule{1em}{0ex}}\le \alpha \left({x}_{n},{x}_{n}\right)\alpha \left({x}_{n+1},{x}_{n+1}\right)p\left(f{x}_{n},f{x}_{n+1}\right)\\ \phantom{\rule{1em}{0ex}}<\varphi \left(p\left({x}_{n},{x}_{n+1}\right),p\left({x}_{n},f{x}_{n}\right),p\left({x}_{n+1},f{x}_{n+1}\right),\frac{1}{2}\left[p\left({x}_{n},f{x}_{n+1}\right)+p\left({x}_{n+1},f{x}_{n}\right)\right]\right)\\ \phantom{\rule{1em}{0ex}}=\varphi \left(p\left({x}_{n},{x}_{n+1}\right),p\left({x}_{n},{x}_{n+1}\right),p\left({x}_{n+1},{x}_{n+2}\right),\frac{1}{2}\left[p\left({x}_{n},{x}_{n+2}\right)+p\left({x}_{n+1},{x}_{n+1}\right)\right]\right)\\ \phantom{\rule{1em}{0ex}}\le \varphi \left(p\left({x}_{n},{x}_{n+1}\right),p\left({x}_{n},{x}_{n+1}\right),p\left({x}_{n+1},{x}_{n+2}\right),\frac{1}{2}\left[p\left({x}_{n},{x}_{n+1}\right)+p\left({x}_{n+1},{x}_{n+2}\right)\right]\right).\end{array}$
(2.3)

If $p\left({x}_{n},{x}_{n+1}\right)\le p\left({x}_{n+1},{x}_{n+2}\right)$, then

$\begin{array}{rl}p\left({x}_{n+1},{x}_{n+2}\right)& =p\left(f{x}_{n},f{x}_{n+1}\right)\\ <\varphi \left(p\left({x}_{n+1},{x}_{n+2}\right),p\left({x}_{n+1},{x}_{n+2}\right),p\left({x}_{n+1},{x}_{n+2}\right),p\left({x}_{n+1},{x}_{n+2}\right)\right)\\ \le p\left({x}_{n+1},{x}_{n+2}\right),\end{array}$

which implies a contradiction, and hence $p\left({x}_{n},{x}_{n+1}\right). From the argument above, we also have that for each $n\in \mathbb{N}$,

$\begin{array}{rl}p\left({x}_{n+1},{x}_{n+2}\right)& =p\left(f{x}_{n},f{x}_{n+1}\right)\\ <\varphi \left(p\left({x}_{n},{x}_{n+1}\right),p\left({x}_{n},{x}_{n+1}\right),p\left({x}_{n},{x}_{n+1}\right),p\left({x}_{n},{x}_{n+1}\right)\right)\\ \le p\left({x}_{n},{x}_{n+1}\right).\end{array}$
(2.4)

Since the sequence $\left\{p\left({x}_{n},{x}_{n+1}\right)\right\}$ is decreasing, it must converge to some $\eta \ge 0$, that is,

$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{n+1}\right)=\eta .$
(2.5)

It follows from (2.4) and (2.5) that

$\underset{n\to \mathrm{\infty }}{lim}\varphi \left(p\left({x}_{n},{x}_{n+1}\right),p\left({x}_{n},{x}_{n+1}\right),p\left({x}_{n},{x}_{n+1}\right),p\left({x}_{n},{x}_{n+1}\right)\right)=\eta .$
(2.6)

Notice that $\eta =inf\left\{p\left({x}_{n},{x}_{n+1}\right):n\in \mathbb{N}\right\}$. We claim that $\eta =0$. Suppose, to the contrary, that $\eta >0$. Since f is a generalized Meir-Keeler-type ϕ-contraction, corresponding to η use, and taking into account the above inequality (2.6), there exist $\delta >0$ and a natural number k such that

$\begin{array}{r}\eta \le \varphi \left(p\left({x}_{k},{x}_{k+1}\right),p\left({x}_{k},{x}_{k+1}\right),p\left({x}_{k},{x}_{k+1}\right),p\left({x}_{k},{x}_{k+1}\right)\right)<\eta +\delta \\ \phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}\alpha \left({x}_{k},{x}_{k}\right)\alpha \left({x}_{k+1},{x}_{k+1}\right)p\left(f{x}_{k},f{x}_{k+1}\right)<\eta ,\end{array}$

which implies

$p\left({x}_{k+1},{x}_{k+2}\right)=p\left(f{x}_{k},f{x}_{k+1}\right)\le \alpha \left({x}_{k},{x}_{k}\right)\alpha \left({x}_{k+1},{x}_{k+1}\right)p\left(f{x}_{k},f{x}_{k+1}\right)<\eta .$

So, we get a contradiction since $\eta =inf\left\{p\left({x}_{n},{x}_{n+1}\right):n\in \mathbb{N}\right\}$. Thus we have that

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

By (${p}_{2}$), we also have

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

Since ${d}_{p}\left(x,y\right)=2p\left(x,y\right)-p\left(x,x\right)-p\left(y,y\right)$ for all $x,y\in X$, using (2.7) and (2.8), we obtain that

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

Step 2. We show that $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the partial metric space $\left(X,p\right)$, that is, it is sufficient to show that $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the metric space $\left(X,{d}_{p}\right)$.

Suppose that the above statement is false. Then there exists $ϵ>0$ such that for any $k\in \mathbb{N}$, there are ${n}_{k},{m}_{k}\in \mathbb{N}$ with ${n}_{k}>{m}_{k}\ge k$ satisfying

${d}_{p}\left({x}_{{m}_{k}},{x}_{{n}_{k}}\right)\ge ϵ.$
(2.10)

Further, corresponding to ${m}_{k}\ge k$, we can choose ${n}_{k}$ in such a way that it is the smallest integer with ${n}_{k}>{m}_{k}\ge k$ and $d\left({x}_{2{m}_{k}},{x}_{2{n}_{k}}\right)\ge ϵ$. Therefore

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

Now we have that for all $k\in \mathbb{N}$,

$\begin{array}{rl}ϵ& \le {d}_{p}\left({x}_{{m}_{k}},{x}_{{n}_{k}}\right)\\ \le {d}_{p}\left({x}_{{m}_{k}},{x}_{{n}_{k}-2}\right)+{d}_{p}\left({x}_{{n}_{k}-2},{x}_{{n}_{k}-1}\right)+{d}_{p}\left({x}_{{n}_{k}-1},{x}_{{n}_{k}}\right)\\ <ϵ+{d}_{p}\left({x}_{{n}_{k}-2},{x}_{{n}_{k}-1}\right)+{d}_{p}\left({x}_{{n}_{k}-1},{x}_{{n}_{k}}\right).\end{array}$
(2.12)

Letting $k\to \mathrm{\infty }$ in the above inequality and using (2.12), we get

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

On the other hand, we have

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

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

$\underset{n\to \mathrm{\infty }}{lim}{d}_{p}\left({x}_{{m}_{k+1}},{x}_{{n}_{k+1}}\right)=ϵ.$
(2.14)

Since ${d}_{p}\left(x,y\right)=2p\left(x,y\right)-p\left(x,x\right)-p\left(y,y\right)$ and using (2.13) and (2.14), we have that

$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{{m}_{k}},{x}_{{n}_{k}}\right)=\frac{ϵ}{2}$
(2.15)

and

$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{{m}_{k+1}},{x}_{{n}_{k+1}}\right)=\frac{ϵ}{2}$
(2.16)

By Remark 3 and (${p}_{4}$), we have

$\begin{array}{r}p\left({x}_{{m}_{k+1}},{x}_{{n}_{k+1}}\right)\\ \phantom{\rule{1em}{0ex}}=p\left(f{x}_{{m}_{k}},f{x}_{{n}_{k}}\right)\\ \phantom{\rule{1em}{0ex}}\le \alpha \left({x}_{{m}_{k}},{x}_{{m}_{k}}\right)\alpha \left({x}_{{n}_{k}},{x}_{{n}_{k}}\right)p\left(f{x}_{{m}_{k}},f{x}_{{n}_{k}}\right)\\ \phantom{\rule{1em}{0ex}}<\varphi \left(p\left({x}_{{m}_{k}},{x}_{{n}_{k}}\right),p\left({x}_{{m}_{k}},f{x}_{{m}_{k}}\right),p\left({x}_{{n}_{k}},f{x}_{{n}_{k}}\right),\frac{1}{2}\left[p\left({x}_{{m}_{k}},f{x}_{{n}_{k}}\right)+p\left({x}_{{n}_{k}},f{x}_{{m}_{k}}\right)\right]\right)\\ \phantom{\rule{1em}{0ex}}=\varphi \left(p\left({x}_{{m}_{k}},{x}_{{n}_{k}}\right),p\left({x}_{{m}_{k}},{x}_{{m}_{k}+1}\right),p\left({x}_{{n}_{k}},{x}_{{n}_{k+1}}\right),\\ \phantom{\rule{2em}{0ex}}\frac{1}{2}\left[p\left({x}_{{m}_{k}},{x}_{{n}_{k}+1}\right)+p\left({x}_{{n}_{k}},{x}_{{m}_{k}+1}\right)\right]\right).\end{array}$
(2.17)

Since

$p\left({x}_{{m}_{k}},{x}_{{n}_{k}+1}\right)\le p\left({x}_{{m}_{k}},{x}_{{m}_{k}+1}\right)+p\left({x}_{{m}_{k}+1},{x}_{{n}_{k}+1}\right)-p\left({x}_{{m}_{k}+1},{x}_{{m}_{k}+1}\right)$
(2.18)

and

$p\left({x}_{{n}_{k}},{x}_{{m}_{k}+1}\right)\le p\left({x}_{{n}_{k}},{x}_{{n}_{k}+1}\right)+p\left({x}_{{n}_{k}+1},{x}_{{m}_{k}+1}\right)-p\left({x}_{{n}_{k}+1},{x}_{{n}_{k}+1}\right).$
(2.19)

Taking into account the above inequalities (2.8), (2.17), (2.18) and (2.19), letting $k\to \mathrm{\infty }$, we have

$\frac{ϵ}{2}<\varphi \left(\frac{ϵ}{2},0,0,\frac{ϵ}{2}\right)\le \frac{ϵ}{2},$

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

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

Since $\left(X,p\right)$ is complete, then from Lemma 1, we have that $\left(X,{d}_{p}\right)$ is complete. Thus, there exists $\nu \in X$ such that

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

Moreover, it follows from Lemma 1 that

$p\left(\nu ,\nu \right)=\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},\nu \right)=\underset{n,m\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{m}\right).$
(2.20)

On the other hand, since the sequence $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the metric space $\left(X,{d}_{p}\right)$, 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

$\underset{n\to \mathrm{\infty }}{lim}p\left({x}_{n},{x}_{m}\right)=0.$
(2.21)

Using (2.20) and (2.21), we have

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

Again, by Remark 3, (${p}_{4}$), and the conditions of the mapping α, we have

$\begin{array}{rl}p\left({x}_{n+1},f\nu \right)& =p\left(f{x}_{n},f\nu \right)\\ \le \alpha \left({x}_{n},{x}_{n}\right)\alpha \left(\nu ,\nu \right)p\left(f{x}_{n},f\nu \right)\\ <\varphi \left(p\left({x}_{n},\nu \right),p\left({x}_{n},f{x}_{n}\right),p\left(\nu ,f\nu \right),\frac{1}{2}\left[p\left({x}_{n},f\nu \right)+p\left(\nu ,f{x}_{n}\right)\right]\right)\\ =\varphi \left(p\left({x}_{n},\nu \right),p\left({x}_{n},{x}_{n+1}\right),p\left(\nu ,f\nu \right),\frac{1}{2}\left[p\left({x}_{n},f\nu \right)+p\left(\nu ,{x}_{n+1}\right)\right]\right).\end{array}$
(2.22)

Letting $n\to \mathrm{\infty }$ in (2.22), we get

$p\left(\nu ,f\nu \right)<\varphi \left(0,0,p\left(\nu ,f\nu \right),\frac{1}{2}p\left(\nu ,f\nu \right)\right)\le p\left(\nu ,f\nu \right),$

a contradiction. So, we have $p\left(\nu ,f\nu \right)=0$, that is, $f\nu =\nu$. □

We give the following example to illustrate Theorem 2.

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

$p\left(x,y\right)=max\left\{x,y\right\}.$

Let $\alpha :\left[0,1\right]×\left[0,1\right]\to {\mathbb{R}}^{+}$ be defined as

$\alpha \left(x,y\right)=1+x+y,$

let $f:X\to X$ be defined as

$f\left(x\right)=\frac{1}{16}{x}^{2},$

and, let $\varphi :{{\mathbb{R}}^{+}}^{4}\to {\mathbb{R}}^{+}$ denote

$\psi \left({t}_{1},{t}_{2},{t}_{3},{t}_{4}\right)=\frac{1}{2}\cdot max\left\{{t}_{1},{t}_{2},{t}_{3},\frac{1}{2}{t}_{4}\right\}.$

Without loss of generality, we assume that $x>y$ and verify the inequality (2.1). For all $x,y\in \left[0,1\right]$ with $x>y$, we have

$\begin{array}{c}\alpha \left(x,x\right)\alpha \left(y,y\right)p\left(fx,fy\right)\ge \frac{1}{16}{x}^{2},\hfill \\ p\left(x,y\right)=x,\phantom{\rule{2em}{0ex}}p\left(x,fx\right)=x,\phantom{\rule{2em}{0ex}}p\left(y,fy\right)=y\phantom{\rule{1em}{0ex}}\text{and}\hfill \\ \begin{array}{rl}\frac{1}{2}\left[p\left(x,fy\right)+p\left(y,fx\right)\right]& =\frac{1}{2}\left[max\left\{x,{y}^{2}\right\}+max\left\{y,{x}^{2}\right\}\right]\\ \le \frac{1}{2}\left[max\left\{x,y\right\}+max\left\{y,x\right\}\right]\\

and hence $\varphi \left(p\left(x,y\right),p\left(x,fx\right),p\left(y,fy\right),\frac{1}{2}\left[p\left(x,fy\right)+p\left(y,fx\right)\right]\right)=\frac{1}{2}x$. Therefore, all the conditions of Theorem 1 are satisfied, and we obtained that 0 is a fixed point of f.

If we let

then it is easy to get the following theorem.

Theorem 2 Let $\left(X,p\right)$ be a complete partial metric space and $\varphi \in \mathrm{\Phi }$. Suppose that $f:X\to X$ is a generalized Meir-Keeler-type ϕ-contraction. Then f has a fixed point in X.

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## Acknowledgements

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

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Correspondence to Chi-Ming Chen.

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All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

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Chen, CH., Chen, CM. Fixed point results for Meir-Keeler-type ϕ-α-contractions on partial metric spaces. J Inequal Appl 2013, 341 (2013). https://doi.org/10.1186/1029-242X-2013-341