Open Access

Some new fixed point theorems for the Meir-Keeler contractions on partial Hausdorff metric spaces

Journal of Inequalities and Applications20152015:279

https://doi.org/10.1186/s13660-015-0803-4

Received: 14 July 2015

Accepted: 28 August 2015

Published: 17 September 2015

Abstract

The purpose of this paper is to study fixed point theorems for a multi-valued mapping concerning with three classes of Meir-Keeler contractions with respect to the partial Hausdorff metric \(\mathcal{H}\) in complete partial metric spaces. Our results generalize and improve many recent fixed point theorems for the partial Hausdorff metric in the literature.

Keywords

fixed pointMeir-Keeler contractionpartial Hausdorff metric space

MSC

47H1054C6054H2555M20

1 Introduction and preliminaries

Throughout this paper, by \(\mathbb{R}^{+}\), we denote the set of all nonnegative real numbers, while \(\mathbb{N}\) is the set of all natural numbers. Fixed point theory is one of the most crucial tools in nonlinear functional analysis and has application in distinct branches of mathematic. In 1922, Banach [1] introduced the most impressed fixed point result, and he concluded that each contraction has a unique fixed point in the complete metric space. Since then, this pioneer work has been generalized and extended in different abstract spaces. One of the interesting generalization of Banach fixed point theorem was given by Matthews [2] in 1994. In this paper, the author introduced the following notion of partial metric spaces and proved the Banach fixed point theorem in the context of complete partial metric space. We recall some basic definitions and fundamental results of partial metric spaces from the literature.

Definition 1

[2]

A partial metric on a nonempty set X is a function \(p:X\times X\rightarrow\mathbb{R^{+}}\) such that for all \(x,y,z\in X\)
(p1): 

\(x=y\) if and only if \(p(x,x)=p(x,y)=p(y,y)\);

(p2): 

\(p(x,x)\leq p(x,y)\);

(p3): 

\(p(x,y)=p(y,x)\);

(p4): 

\(p(x,y)\leq p(x,z)+p(z,y)-p(z,z)\).

A partial metric space is a pair \((X,p)\) such that X is a nonempty set and p is a partial metric on X.

Remark 1

[2]

It is clear that, if \(p(x,y)=0\), then from (p1) and (p2), we have \(x=y\). But, if \(x=y\), then the expression \(p(x,y)\) 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 \(\{ B_{p}(x,\gamma):x\in X, \gamma> 0\}\), where \(B_{p}(x,\gamma)=\{y\in X: p(x,y)< p(x,x)+\gamma\}\) for all \(x\in X\) and \(\gamma> 0\). If p is a partial metric on X, then the function \(d_{p}:X\times X\rightarrow \mathbb{R^{+}}\) given by
$$d_{p}(x,y)=2p(x,y)-p(x,x)-p(y,y) $$
is a metric on X.

Definition 2

[2]

Let \((X,p)\) be a partial metric space. Then
  1. (1)

    a sequence \(\{x_{n}\}\) in a partial metric space \((X,p)\) converges to \(x\in X\) if and only if \(p(x,x)=\lim_{n\rightarrow\infty }p(x,x_{n})\);

     
  2. (2)

    a sequence \(\{x_{n}\}\) in a partial metric space \((X,p)\) is called a Cauchy sequence if and only if \(\lim_{m,n\rightarrow\infty }p(x_{m},x_{n})\) exists (and is finite);

     
  3. (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 \(p(x,x)=\lim_{m,n\rightarrow\infty }p(x_{m},x_{n})\);

     
  4. (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 2

The limit in a partial metric space is not unique.

Lemma 1

[2, 3]

  1. (1)

    \(\{x_{n}\}\) is a Cauchy sequence in a partial metric space \((X,p)\) if and only if it is a Cauchy sequence in the metric space \((X,d_{p})\);

     
  2. (2)

    a partial metric space \((X,p)\) is complete if and only if the metric space \((X,d_{p})\) is complete. Furthermore, \(\lim_{n\rightarrow \infty}d_{p}(x_{n},x)=0\) if and only if \(p(x,x)=\lim_{n\rightarrow\infty }p(x_{n},x)=\lim_{n\rightarrow\infty}p(x_{n},x_{m})\).

     

Very recently Haghi et al. [4] proved that some fixed point results in partial metric space results are equivalent to results in the context of usual metric space. Recently, fixed point theory has developed rapidly on partial metric spaces; see e.g. [3, 512] and the references therein.

Let \((X,d)\) be a metric space and \(CB(X)\) denote the collection of all nonempty, closed and bounded subsets of X. For \(A,B\in CB(X)\), we define
$$\mathcal{H}(A,B):=\max\Bigl\{ \sup_{a\in A}d(a,B), \sup _{b\in B}d(b,A)\Bigr\} , $$
where \(d(x,B):=\inf\{d(x,b): b\in B\}\), and it is well known that \(\mathcal{H}\) is called the Hausdorff metric induced the metric d. A multi-valued mapping \(T:X\rightarrow CB(X)\) is called a contraction if
$$\mathcal{H}(Tx,Ty)\leq kd(x,y), $$
for all \(x,y\in X\) and \(k\in[0,1)\). The study of fixed points for multi-valued contractions using the Hausdorff metric was introduced in Nadler [13].

Theorem 1

[13]

Let \((X,d)\) be a complete metric space and \(T:X\rightarrow CB(X)\) be a multi-valued contraction. Then there exists \(x\in X\) such that \(x\in Tx\).

Very recently, Aydi et al. [14] established the notion of partial Hausdorff metric \(\mathcal{H}_{p}\) induced by the partial metric p. Let \((X,p)\) be a partial metric space and \(CB^{p}(X)\) be the collection of all nonempty, closed and bounded subset of the partial metric space \((X,p)\). Note that closedness is taken from \((X,\tau_{p})\) and boundedness is given as follows: A is a bounded subset in \((X,p)\) if there exist \(x_{0}\in X\) and \(M\in\mathbb{R}\) such that for all \(a\in A\), we have \(a\in B_{p}(x_{0},M)\), that is, \(p(x_{0},a)< p(a,a)+M\). For \(A,B\in CB^{p}(X)\) and \(x\in X\), they define
$$\begin{aligned}& p(x,A):=\inf\bigl\{ p(x,a):a\in A\bigr\} , \\& \delta_{p}(A,B):=\sup\bigl\{ p(a,B):a\in A\bigr\} , \\& \delta_{p}(B,A):=\sup\bigl\{ p(b,A):b\in B\bigr\} , \\& \mathcal{H}_{p}(A,B)=\max\bigl\{ \delta_{p}(A,B), \delta_{p}(B,A)\bigr\} . \end{aligned}$$
It is immediate to see that if \(p(x,A)=0\), then \(d_{p}(x,A)=0\) where \(d_{p}(x,A)=\inf\{d_{p}(x,a):a\in A\}\).

Remark 3

[14]

Let \((X,p)\) be a partial metric space and A a nonempty subset of X. Then
$$a\in\overline{A} \quad\mbox{if and only if} \quad p(a,A)= p(a,a). $$

Aydi et al. [14] also introduced the following properties of mappings \(\delta_{p}:CB^{p}(X)\times CB^{p}(X)\rightarrow \mathbb{R}\) and \(\mathcal{H}_{p}:CB^{p}(X)\times CB^{p}(X)\rightarrow\mathbb{R}\).

Proposition 1

[14]

Let \((X,p)\) be a partial metric space. For \(A,B\in CB^{p}(X)\), the following properties hold:
  1. (1)

    \(\delta_{p}(A,A)=\sup\{p(a,a):a \in A\}\);

     
  2. (2)

    \(\delta_{p}(A,A)\leq\delta_{p}(A,B)\);

     
  3. (3)

    \(\delta_{p}(A,B)=0\) implies that \(A\subset B\);

     
  4. (4)

    \(\delta_{p}(A,B)\leq\delta_{p}(A,C)+\delta_{p}(C,B)-\inf_{c\in C}p(c,c)\).

     

Proposition 2

[14]

Let \((X,p)\) be a partial metric space. For \(A,B\in CB^{p}(X)\), the following properties hold:
  1. (1)

    \(\mathcal{H}_{p}(A,A)\leq\mathcal{H}_{p}(A,B)\);

     
  2. (2)

    \(\mathcal{H}_{p}(A,B)=\mathcal{H}_{p}(B,A)\);

     
  3. (3)

    \(\mathcal{H}_{p}(A,B)\leq\mathcal{H}_{p}(A,C)+\mathcal {H}_{p}(C,B)-\inf_{c\in C}p(c,c)\);

     
  4. (4)

    \(\mathcal{H}_{p}(A,B)=0\) implies that \(A= B\).

     

And, Aydi et al. [14] proved the following important result.

Lemma 2

Let \((X,p)\) be a partial metric space, \(A,B\in CB^{p}(X)\) and \(h>1\). For any \(a\in A\), there exists \(b=b(a)\in B\) such that
$$p(a,b)\leq h\mathcal{H}_{p}(A,B). $$
In this study, we also recall the notion of Meir-Keeler-type function (see [15]). A function \(\xi:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}\) is said to be a Meir-Keeler-type function, if ξ satisfies the following condition:
$$\forall\eta>0\ \exists\delta>0\ \forall t\in\mathbb{R}^{+}\quad \bigl(\eta \leq t< \eta+ \delta\Rightarrow\xi(t)< \eta\bigr). $$

Remark 4

It is clear that if ξ is a Meir-Keeler-type function, then we have
$$\xi(t)< t, \quad\mbox{for all } t\in\mathbb{R}^{+}. $$

We first introduce the notion of stronger Meir-Keeler-type function, as follows:

Definition 3

A function \(\psi:\mathbb{R}^{+}\rightarrow[0,1)\) is said to be a stronger Meir-Keeler-type function, if ψ satisfies the following condition:
$$\forall\eta>0\ \exists\delta>0\ \exists\gamma_{\eta}\in(0,1)\ \forall t\in \mathbb{R}^{+}\quad \bigl(\eta\leq t< \eta+\delta\Rightarrow\psi (t)< \gamma_{\eta}\bigr). $$

In 1972, Reich introduced the following important function.

Definition 4

[16]

A function \(\psi:\mathbb{R}^{+}\rightarrow[0,1)\) is said to be a Reich function (\(\mathcal{R}\)-function, for short) if
$$\limsup_{s\rightarrow t^{+}} \psi(s)< 1, \quad\mbox{for all } t\in\mathbb{R}^{+}. $$

Remark 5

It is clear that, if the function \(\psi:\mathbb{R}^{+}\rightarrow[0,1)\) is a Reich function (\(\mathcal{R}\)-function), then ψ is also a stronger Meir-Keeler-type function.

We next introduce the notion of weaker Meir-Keeler function, as follows:

Definition 5

A function \(\varphi:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}\) is said to be a weaker Meir-Keeler-type function, if φ satisfies the following condition:
$$\forall\eta>0\ \exists\delta>0\ \forall t\in\mathbb{R}^{+} \quad\bigl(\eta \leq t< \eta+ \delta\Rightarrow\exists n_{0}\in\mathbb{N}, \varphi ^{n_{0}}(t)< \eta\bigr). $$

The purpose of this paper is to study fixed point theorems for a multi-valued mapping concerning three classes of Meir-Keeler contractions with respect to the partial Hausdorff metric \(\mathcal{H}\) in complete partial metric spaces. Our results generalize and improve many recent fixed point theorems for the partial Hausdorff metric in the literature.

2 Fixed point theorem (I)

In the sequel, we denote by Φ the class of functions \(\phi:\mathbb {R^{+}}^{4}\rightarrow\mathbb{R^{+}}\) satisfying the following conditions:
  1. (1)

    ϕ is an increasing and continuous function in each coordinate;

     
  2. (2)

    for \(t\in\mathbb{R^{+}}\backslash\{0\}\), \(\phi(t,t,t,t)\leq t\) and \(\phi(t_{1},t_{2},t_{3},t_{4})=0\) iff \(t_{1}=t_{2}=t_{3}=t_{4}=0\).

     

We now introduce the notion of \((\psi,\phi)\)-Meir-Keeler contraction on partial Hausdorff metric spaces.

Definition 6

Let \((X,p)\) be a partial metric space, \(\psi:\mathbb{R}^{+}\rightarrow [0,1)\) and \(\phi\in\Phi\). We call \(T:X\rightarrow CB^{p}(X)\) a \((\psi ,\phi)\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\), if the following conditions hold:
(c1): 

ψ is a stronger Meir-Keeler-type function;

(c2): 
for all \(x,y\in X\), we have
$$\mathcal{H}_{p}(Tx,Ty)\leq\psi\bigl(p(x,y)\bigr) \phi \biggl(p(x,y),p(x,Tx),p(y,Ty),\frac{p(x,Ty)+p(y,Tx)}{2} \biggr). $$

We state and prove the main fixed point result for the \((\psi,\phi )\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\).

Theorem 2

Let \((X,p)\) be a complete partial metric space. Suppose \(T:X\rightarrow CB^{p}(X)\) is a \((\psi,\phi)\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\). Then T has a fixed point in X, that is, there exists \(x^{*}\in X\) such that \(x^{*}\in Tx^{*}\).

Proof

Let \(x_{0}\in X\) be given and let \(x_{1}\in Tx_{0}\). Since \(T:X\rightarrow CB^{p}(X)\) is a \((\psi,\phi)\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\), we have
$$\begin{aligned} & \mathcal{H}_{p}(Tx_{0},Tx_{1}) \\ &\quad\leq\psi\bigl(p(x_{0},x_{1})\bigr)\phi \biggl(p(x_{0},x_{1}),p(x_{0},Tx_{0}),p(x_{1},Tx_{1}), \frac {p(x_{0},Tx_{1})+p(x_{1},Tx_{0})}{2} \biggr). \end{aligned}$$
(2.1)
Put \(p(x_{0},x_{1})=\eta_{0}>0\). Since ψ is a stronger Meir-Keeler-type function, there exists \(\gamma_{\eta_{0}}\in(0,1)\) such that
$$ \psi\bigl(p(x_{0},x_{1})\bigr)< \gamma_{\eta_{0}}. $$
(2.2)
From (2.1) and (2.2), we have
$$ \mathcal{H}_{p}(Tx_{0},Tx_{1})< \gamma_{\eta_{0}}\cdot\phi \biggl(p(x_{0},x_{1}),p(x_{0},Tx_{0}),p(x_{1},Tx_{1}), \frac {p(x_{0},Tx_{1})+p(x_{1},Tx_{0})}{2} \biggr). $$
(2.3)
From Lemma 2 with \(h=\frac{1}{\sqrt{\gamma_{\eta_{0}}}}>1\), there exists \(x_{2}\in Tx_{1}\) such that
$$ p(x_{1},x_{2})\leq\frac{1}{\sqrt{\gamma_{\eta_{0}}}}\mathcal {H}_{p}(Tx_{0},Tx_{1}). $$
(2.4)
Using (2.3) and (2.4), we obtain
$$\begin{aligned} p(x_{1},x_{2})&< \sqrt{\gamma_{\eta_{0}}} \cdot\phi \biggl(p(x_{0},x_{1}),p(x_{0},Tx_{0}),p(x_{1},Tx_{1}), \frac {p(x_{0},Tx_{1})+p(x_{1},Tx_{0})}{2} \biggr) \\ &\leq\sqrt{\gamma_{\eta_{0}}}\cdot\phi \biggl(p(x_{0},x_{1}),p(x_{0},x_{1}),p(x_{1},x_{2}), \frac{p(x_{0},x_{2})+p(x_{1},x_{1})}{2} \biggr) \\ &\leq\sqrt{\gamma_{\eta_{0}}}\cdot\phi \biggl(p(x_{0},x_{1}),p(x_{0},x_{1}),p(x_{1},x_{2}), \frac{p(x_{0},x_{1})+p(x_{1},x_{2})}{2} \biggr). \end{aligned}$$
(2.5)
If \(p(x_{0},x_{1})\leq p(x_{1},x_{2})\), then by the definition of the function ϕ, we have
$$\begin{aligned} p(x_{1},x_{2})&< \sqrt{\gamma_{\eta_{0}}}\cdot\phi \biggl(p(x_{0},x_{1}),p(x_{0},x_{1}),p(x_{1},x_{2}), \frac{p(x_{0},x_{1})+p(x_{1},x_{2})}{2} \biggr) \\ &\leq\sqrt{\gamma_{\eta_{0}}}\cdot\phi \bigl(p(x_{1},x_{2}),p(x_{1},x_{2}),p(x_{1},x_{2}),p(x_{1},x_{2}) \bigr) \\ &\leq\sqrt{\gamma_{\eta _{0}}}\cdotp(x_{1},x_{2}), \end{aligned}$$
which implies a contradiction, and hence \(p(x_{0},x_{1})> p(x_{1},x_{2})\). Therefore, we have
$$\begin{aligned} p(x_{1},x_{2})&< \sqrt{\gamma_{\eta_{0}}} \cdot\phi \biggl(p(x_{0},x_{1}),p(x_{0},x_{1}),p(x_{1},x_{2}), \frac{p(x_{0},x_{1})+p(x_{1},x_{2})}{2} \biggr) \\ &\leq\sqrt{\gamma_{\eta_{0}}}\cdot\phi \bigl(p(x_{0},x_{1}),p(x_{0},x_{1}),p(x_{0},x_{1}),p(x_{0},x_{1}) \bigr) \\ &\leq\sqrt{\gamma_{\eta _{0}}}\cdot p(x_{0},x_{1}). \end{aligned}$$
(2.6)
Put \(p(x_{1},x_{2})=\eta_{1}>0\). Since ψ is a stronger Meir-Keeler-type function, there exists \(\gamma_{\eta_{1}}\in(0,1)\) such that
$$ \psi\bigl(p(x_{1},x_{2})\bigr)< \gamma_{\eta_{1}}. $$
(2.7)
From Lemma 2 with \(h=\frac{1}{\sqrt{\gamma_{\eta_{1}}}}>1\), we have
$$ p(x_{2},x_{3})\leq\frac{1}{\sqrt{\gamma_{\eta_{1}}}}\mathcal {H}_{p}(Tx_{1},Tx_{2}). $$
(2.8)
Since \(T:X\rightarrow CB^{p}(X)\) is a \((\psi,\phi)\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal {H}_{p}\), we have
$$\begin{aligned} &\mathcal{H}_{p}(Tx_{1},Tx_{2}) \\ &\quad\leq\psi\bigl(p(x_{1},x_{2})\bigr)\phi \biggl(p(x_{1},x_{2}),p(x_{1},Tx_{1}),p(x_{2},Tx_{2}), \frac {p(x_{1},Tx_{2})+p(x_{2},Tx_{1})}{2} \biggr) \\ &\quad< \gamma_{\eta_{1}}\cdot\phi \biggl(p(x_{1},x_{2}),p(x_{1},x_{2}),p(x_{2},x_{3}), \frac{p(x_{1},x_{3})+p(x_{2},x_{2})}{2} \biggr) \\ &\quad\leq\gamma_{\eta_{1}}\cdot\phi \biggl(p(x_{1},x_{2}),p(x_{1},x_{2}),p(x_{2},x_{3}), \frac{p(x_{1},x_{2})+p(x_{2},x_{3})}{2} \biggr). \end{aligned}$$
(2.9)
Using (2.8) and (2.9), we obtain
$$\begin{aligned} p(x_{2},x_{3})< \sqrt{\gamma_{\eta_{1}}}\cdot\phi \biggl(p(x_{1},x_{2}),p(x_{1},x_{2}),p(x_{2},x_{3}), \frac{p(x_{1},x_{2})+p(x_{2},x_{3})}{2} \biggr). \end{aligned}$$
If \(p(x_{1},x_{2})\leq p(x_{2},x_{3})\), then
$$\begin{aligned}[b] p(x_{2},x_{3})&< \sqrt{\gamma_{\eta_{1}}}\cdot\phi \biggl(p(x_{1},x_{2}),p(x_{1},x_{2}),p(x_{2},x_{3}), \frac{p(x_{1},x_{2})+p(x_{2},x_{3})}{2} \biggr) \\ &\leq\sqrt{\gamma_{\eta_{1}}}\cdot\phi \bigl(p(x_{2},x_{3}),p(x_{2},3),p(x_{2},x_{3}),p(x_{2},x_{3}) \bigr) \\ &\leq\sqrt{\gamma_{\eta _{1}}}\cdot p(x_{2},x_{3}), \end{aligned} $$
which implies a contradiction, and hence \(p(x_{1},x_{2})> p(x_{2},x_{3})\). Therefore
$$ p(x_{2},x_{3})\leq\sqrt{\gamma_{\eta_{0}}}\sqrt{ \gamma_{\eta_{1}}}\cdot p(x_{0},x_{1}). $$
(2.10)
Continuing this process, we can obtain a sequence \(\{x_{n}\}\) of X recursively as follows:
$$x_{n}\in Tx_{n-1}, \quad \mbox{for all } n\in\mathbb{N}. $$
Put \(p(x_{n},x_{n+1})=\eta_{n}>0\). Since ψ is a stronger Meir-Keeler-type function, there exists \(\gamma_{\eta_{n}}\in(0,1)\) such that
$$ (x_{n},x_{n+1})< \gamma_{\eta_{n}}. $$
(2.11)
Since \(T:X\rightarrow CB^{p}(X)\) is a \((\psi,\phi)\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal {H}_{p}\), we have for all \(n\in\mathbb{N}\cup\{0\}\)
$$\begin{aligned} &\mathcal{H}_{p}(Tx_{n},Tx_{n+1}) \\ &\quad\leq \psi\bigl(p(x_{n},x_{n+1})\bigr)\phi \biggl(p(x_{n},x_{n+1}),p(x_{n},Tx_{n}),p(x_{n+1},Tx_{n+1}), \\ &\qquad{}\frac {1}{2}\bigl[p(x_{n},Tx_{n+1})+p(x_{n+1},Tx_{n}) \bigr]\biggr) \\ &\quad< \gamma_{\eta_{n}}\phi \biggl(p(x_{n},x_{n+1}),p(x_{n},Tx_{n}),p(x_{n+1},Tx_{n+1}), \frac {1}{2}\bigl[p(x_{n},Tx_{n+1})+p(x_{n+1},Tx_{n}) \bigr]\biggr) \\ &\quad< \gamma_{\eta_{n}}\phi \biggl(p(x_{n},x_{n+1}),p(x_{n},x_{n+1}),p(x_{n+1},x_{n+2}), \frac {1}{2}\bigl[p(x_{n},x_{n+1})+p(x_{n+1},x_{n+2}) \bigr]\biggr). \end{aligned}$$
(2.12)
From Lemma 2 with \(h=\frac{1}{\sqrt{\gamma_{\eta_{n}}}}>1\), we have
$$ p(x_{n+1},x_{n+2})\leq\frac{1}{\sqrt{\gamma_{\eta_{n}}}}\mathcal {H}_{p}(Tx_{n},Tx_{n+1}), \quad n\in\mathbb{N}\cup\{0 \}. $$
(2.13)
Using (2.12) and (2.13), we obtain
$$\begin{aligned} &p(x_{n+1},x_{n+2})\\ &\quad< \sqrt{\gamma_{\eta_{n}}}\phi \biggl(p(x_{n},x_{n+1}),p(x_{n},x_{n+1}),p(x_{n+1},x_{n+2}), \frac {1}{2}\bigl[p(x_{n},x_{n+1})+p(x_{n+1},x_{n+2}) \bigr]\biggr). \end{aligned}$$
If \(p(x_{n},x_{n+1})\leq p(x_{n+1},x_{n+2})\), then
$$\begin{aligned} p(x_{n+1},x_{n+2}) &< \sqrt{\gamma_{\eta_{n}}}\phi \bigl(p(x_{n+1},x_{n+2}),p(x_{n+1},x_{n+2}),p(x_{n+1},x_{n+2}),p(x_{n+1},x_{n+2}) \bigr) \\ &\leq\sqrt{\gamma_{\eta_{n}}} p(x_{n+1},x_{n+2}), \end{aligned}$$
which implies a contradiction, and hence \(p(x_{0},x_{1})> p(x_{1},x_{2})\). Therefore, we have
$$ p(x_{n+1},x_{n+2})\leq\sqrt{\gamma_{\eta_{n}}}p(x_{n},x_{n+1}). $$
(2.14)
By the mathematical induction, we obtain
$$\begin{aligned} p(x_{n+1},x_{n+2})&\leq\sqrt{ \gamma_{\eta_{n}}}p(x_{n},x_{n+1}) \\ &\leq\sqrt {\gamma_{\eta_{n}}}\sqrt{\gamma_{\eta_{n-1}}}p(x_{n-1},x_{n}) \\ &\leq\cdots \\ &\leq\sqrt{\gamma_{\eta_{n}}}\sqrt{\gamma_{\eta_{n-1}}}\cdots\sqrt { \gamma_{\eta_{0}}} p(x_{0},x_{1}). \end{aligned}$$
(2.15)
Put
$$ \overline{k}=\max\bigl\{ \sqrt{\gamma_{\eta_{n}}}: n\in\mathbb{N}\cup\{0\} \bigr\} . $$
(2.16)
Using (2.15) and (2.16), we obtain
$$ p(x_{n+1},x_{n+2})\leq(\overline{k})^{n+1} p(x_{0},x_{1}), \quad\mbox{for all } n\in\mathbb{N}\cup\{0 \}. $$
(2.17)
Let \(n\rightarrow\infty\) in (2.17). Then
$$ \lim_{n\rightarrow\infty}p(x_{n},x_{n+1})=0. $$
(2.18)
By the property (p2) of a partial metric and using (2.18), we have
$$ \lim_{n\rightarrow\infty}p(x_{n},x_{n})=0. $$
(2.19)
Using (2.17) and the property (p4) of a partial metric, for any \(m\in\mathbb{N}\), we have
$$\begin{aligned} p(x_{n},x_{n+m})&\leq\sum_{i=1}^{m} p(x_{n+i-1},x_{n+i})-\sum_{i=1}^{m-1} p(x_{n+i},x_{n+i}) \\ &\leq\sum_{i=1}^{m} (\overline {k})^{n+i-1}p(x_{0},x_{1})-\sum _{i=1}^{m-1} p(x_{n+i},x_{n+i}) \\ &\leq \frac{(\overline{k})^{n}}{(1-\overline{k})} p(x_{0},x_{1})-\sum _{i=1}^{m-1} p(x_{n+i},x_{n+i}). \end{aligned}$$
(2.20)
Using (2.19) and (2.20), we get
$$\lim_{n\rightarrow\infty}p(x_{n},x_{n+m})=0. $$
By the definition of \(d_{p}\), we see that, for any \(m\in\mathbb{N}\),
$$ \lim_{n\rightarrow\infty}d_{p}(x_{n},x_{n+m}) \leq\lim_{n\rightarrow \infty}2p(x_{n},x_{n+m})= 0. $$
(2.21)
This shows that \(\{x_{n}\}\) is a Cauchy sequence in \((X,d_{p})\). Since \((X,p)\) is complete, from Lemma 1, \((X,d_{p})\) is a complete metric space. Therefore, \(\{x_{n}\}\) converges to some \(x^{*}\in X\) with respect to the metric \(d_{p}\), and we also have
$$ p\bigl(x^{*},x^{*}\bigr)=\lim_{n\rightarrow\infty}p\bigl(x_{n},x^{*} \bigr)=\lim_{n\rightarrow \infty}p(x_{n},x_{m})=0. $$
(2.22)
Since \(T:X\rightarrow CB^{p}(X)\) is a \((\psi,\phi)\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal {H}_{p}\), we have
$$\begin{aligned} &\mathcal{H}_{p}\bigl(Tx_{n},Tx^{*}\bigr) \\ &\quad\leq\psi\bigl(p\bigl(x_{n},x^{*}\bigr)\bigr)\phi \biggl(p \bigl(x_{n},x^{*}\bigr),p(x_{n},Tx_{n}),p \bigl(x^{*},Tx^{*}\bigr),\frac{p(x_{n},Tx^{*})+p(x^{*},Tx_{n})}{2}\biggr) \\ &\quad\leq\psi\bigl(p\bigl(x_{n},x^{*}\bigr)\bigr)\phi\biggl(p \bigl(x_{n},x^{*}\bigr),p(x_{n},x_{n+1}),p\bigl(x^{*},Tx^{*} \bigr),\frac {p(x_{n},Tx^{*})+p(x^{*},x_{n+1})}{2}\biggr). \end{aligned}$$
By the definition of the mapping ψ and using (2.22), we get
$$\begin{aligned} \lim_{n\rightarrow\infty}\mathcal{H}_{p}\bigl(Tx_{n},Tx^{*} \bigr)&< \phi \biggl(0,0,p\bigl(x^{*},Tx^{*}\bigr),\frac{1}{2}p\bigl(x^{*},Tx^{*} \bigr)\biggr) \\ &\leq p\bigl(x^{*},Tx^{*}\bigr). \end{aligned}$$
(2.23)
Now \(x_{n+1}\in Tx_{n}\) shows
$$p\bigl(x_{n+1},Tx^{*}\bigr)\leq\delta_{p}\bigl(Tx_{n},Tx^{*} \bigr)\leq\mathcal{H}_{p}\bigl(Tx_{n},Tx^{*}\bigr). $$
Using (2.23), we get
$$p\bigl(x^{*},Tx^{*}\bigr)=\lim_{n\rightarrow\infty}p\bigl(x_{n+1},Tx^{*} \bigr)< p\bigl(x^{*},Tx^{*}\bigr), $$
a contradiction. So, we have
$$p\bigl(x^{*},Tx^{*}\bigr)=0. $$
Therefore, from (2.22), \(p(x^{*},x^{*})=0\), we obtain
$$p\bigl(x^{*},x^{*}\bigr)=p\bigl(x^{*},Tx^{*}\bigr), $$
which implies \(x^{*}\in Tx^{*}\) by Remark 3. □

Using the Reich function and stronger Meir-Keeler function, we establish the following notion of \((\psi,\phi)\)-Reich’s contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\).

Definition 7

Let \((X,p)\) be a partial metric space, \(\psi:\mathbb{R}^{+}\rightarrow [0,1)\), and \(\phi\in\Phi\). We call \(T:X\rightarrow CB^{p}(X)\) a \((\psi ,\phi)\)-Reich’s contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\) if the following conditions hold:
  1. (1)

    ψ is a Reich function (\(\mathcal{R}\)-function);

     
  2. (2)
    for all \(x,y\in X\), we have
    $$\mathcal{H}_{p}(Tx,Ty)\leq\psi\bigl(p(x,y)\bigr) \phi \biggl(p(x,y),p(x,Tx),p(y,Ty),\frac{p(x,Ty)+p(y,Tx)}{2} \biggr). $$
     

Apply above Remark 5, Definition 7, and Theorem 2, we are easy to get the following theorem.

Theorem 3

Let \((X,p)\) be a complete partial metric space. Suppose \(T:X\rightarrow CB^{p}(X)\) is a \((\psi,\phi)\)-Reich’s contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\). Then T has a fixed point in X, that is, there exists \(x^{*}\in X\) such that \(x^{*}\in Tx^{*}\).

3 Fixed point theorem (II)

In this section, we let Ξ be the class of all non-decreasing function \(\varphi:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}\) satisfying the following conditions:
  1. (1)

    φ is a weaker Meir-Keeler-type function;

     
  2. (2)

    for all \(t\in(0,\infty)\), \(\{\varphi^{n}(t)\}_{n\in\mathbb {N}}\) is decreasing;

     
  3. (3)

    \(\varphi(t)>0\) for \(t>0\) and \(\varphi(0)=0\),

     
  4. (4)

    for \(t>0\), if \(\lim_{n\rightarrow\infty}\varphi^{n}(t)=0\), then \(\lim_{n\rightarrow\infty}\sum_{i=n}^{m}\varphi^{i}(t)=0\), where \(m>n\);

     
  5. (5)

    for \(t_{n}\in\mathbb{R}^{+}\), if \(\lim_{n\rightarrow\infty }t_{n}=0\), then \(\lim_{n\rightarrow\infty}\varphi(t_{n})=0\).

     

We recall the notion of α-admissible function that was introduced in [17].

Definition 8

[17]

Let \(f:X\rightarrow X\) be a self-mapping of a set X and \(\alpha :X\times X\rightarrow\mathbb{R}^{+}\). Then f is called a α-admissible if
$$x,y\in X, \quad\alpha(x,y)\geq1\quad\Longrightarrow\quad\alpha(fx,fy)\geq1. $$

In [18], the authors introduced the following notion of strictly α-admissible.

Definition 9

Let \((X,p)\) be a partial metric space, \(T:X\rightarrow CB^{p}(X)\) and \(\alpha:X\times X\rightarrow\mathbb{R}^{+}\diagdown\{0\}\). We say that T is strictly α-admissible if
$$\alpha(x,y)>1 \quad\mbox{implies} \quad \alpha(y,z)>1,\quad x\in X, y\in Tx, z\in Ty. $$

We now introduce the notion of \((\alpha,\varphi)\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal {H}_{p}\), as follows:

Definition 10

Let \((X,p)\) be a partial metric space, \(\varphi\in\Xi\), and \(\alpha :X\times X\rightarrow\mathbb{R}^{+}\diagdown\{0\}\). We call \(T:X\rightarrow CB^{p}(X)\) a \((\alpha,\varphi)\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\) if the following conditions hold:
(c1): 

T is strictly α-admissible;

(c2): 
for each \(x,y\in X\),
$$\alpha(x,y)\mathcal{H}_{p}(Tx,Ty)\leq\varphi\bigl(p(x,y)\bigr). $$

We now state and prove our main result for the \((\alpha,\varphi )\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\).

Theorem 4

Let \((X,p)\) be a complete partial metric space. Suppose \(T:X\rightarrow CB^{p}(X)\) is a \((\alpha,\varphi)\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\). Suppose also that
  1. (i)

    there exists \(x_{0}\in X\) such that \(\alpha(x_{0},y)>1\) for all \(y\in Tx_{0}\);

     
  2. (ii)

    if \(\{x_{n}\}\) is a sequence in X such that \(\alpha (x_{n},x_{n+1})\geq1\) for all n and \(x_{n} \rightarrow x\in X\) as \(n\rightarrow\infty\), then \(\alpha(x_{n},x)\geq1\) for all n.

     
Then T has a fixed point in X (that is, there exists \(x^{*}\in X\) such that \(x^{*}\in Tx^{*}\)).

Proof

Let \(x_{1}\in Tx_{0}\). Since \(T:X\rightarrow CB^{p}(X)\) is a \((\alpha,\varphi )\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\), we have
$$ \alpha(x_{0},x_{1})\mathcal{H}_{p}(Tx_{0},Tx_{1}) \leq\varphi\bigl(p(x_{0},x_{1})\bigr). $$
(3.1)
Put \(\alpha(x_{0},x_{1})=h_{0}>1\). From Lemma 2 with \(h=h_{0}\), there exists \(x_{2}\in Tx_{1}\) such that
$$ p(x_{1},x_{2})\leq h_{0}\mathcal{H}_{p}(Tx_{0},Tx_{1}). $$
(3.2)
Using (3.1) and (3.2), we obtain
$$ p(x_{1},x_{2})\leq\varphi\bigl(p(x_{0},x_{1}) \bigr). $$
(3.3)
Continuing this process, we can obtain a sequence \(\{x_{n}\}\) of X recursively as follows:
$$x_{n}\in Tx_{n-1}, \quad \mbox{for all } n\in\mathbb{N}. $$
Since T is strictly α-admissible, we deduce that \(\alpha (x_{1},x_{2})=h_{1}>1\). Continuing this process, we have
$$ \alpha(x_{n},x_{n+1})=h_{n}>1, \quad\mbox{for all } n\in\mathbb{N}\cup\{0\}. $$
(3.4)
And, since \(T:X\rightarrow CB^{p}(X)\) is a \((\alpha,\varphi)\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal {H}_{p}\), we have
$$ \alpha(x_{n},x_{n+1})\mathcal{H}_{p}(Tx_{n},Tx_{n+1}) \leq\varphi \bigl(p(x_{n},x_{n+1})\bigr), \quad \mbox{for all } n \in\mathbb{N}\cup\{0\}. $$
(3.5)
From Lemma 2 with \(h_{n}\), we have
$$ p(x_{n+1},x_{n+2})\leq h_{n}\mathcal{H}_{p}(Tx_{n},Tx_{n+1}), \quad\mbox{for all } n\in\mathbb{N}\cup\{0\}. $$
(3.6)
Using (3.5) and (3.6), we obtain
$$ p(x_{n+1},x_{n+2})\leq\varphi\bigl(p(x_{n},x_{n+1}) \bigr), \quad\mbox{for all } n\in\mathbb{N}\cup\{0\}. $$
(3.7)
Therefore, we conclude that
$$\begin{aligned} p(x_{n+1},x_{n+2})&\leq\varphi \bigl(p(x_{n},x_{n+1})\bigr) \\ &\leq\varphi ^{2}\bigl(p(x_{n-1},x_{n})\bigr) \\ &\leq\cdots \\ &\leq\varphi^{n+1}\bigl(p(x_{0},x_{1})\bigr). \end{aligned}$$
(3.8)
By the condition (\(\varphi_{2}\)), \(\{\varphi^{n}(p(x_{0},x_{1}))\}_{n\in\mathbb {N}}\) is decreasing, it must converges to some \(\eta\geq0\). We claim that \(\eta= 0\). On the contrary, assume that \(\eta> 0\). Then by the definition of the weaker Meir-Keeler-type function, there exists \(\delta> 0\) such that for \(x_{0},x_{1}\in X\) with \(\eta\leq p(x_{0},x_{1})< \delta+\eta\), there exists \(n_{0}\in\mathbb{N}\) such that \(\varphi ^{n_{0}}(p(x_{0},x_{1}))<\eta\). Since \(\lim_{n\rightarrow\infty}\varphi ^{n}(p(x_{0},x_{1}))=\eta\), there exists \(m_{0}\in\mathbb{N}\) such that \(\eta \leq\varphi^{m}(p(x_{0},x_{1}))< \delta+\eta\), for all \(m\geq m_{0}\). Thus, we conclude that \(\varphi^{m_{0}+n_{0}}(p(x_{0},x_{1}))< \eta\). So we get a contradiction. So \(\lim_{n\rightarrow\infty}\varphi^{n}(p(x_{0},x_{1}))=0\), and so
$$ \lim_{n\rightarrow\infty}p(x_{n},x_{n+1})=0. $$
(3.9)
By the property (p2) of a partial metric and using (3.9), we have
$$ \lim_{n\rightarrow\infty}p(x_{n},x_{n})=0. $$
(3.10)
We will prove that the sequence \(\{x_{n}\}\) is a Cauchy sequence. Using (3.8), we have
$$\begin{aligned} p(x_{n},x_{n+m})={}& p(x_{n},x_{n+1})+p(x_{n+1},x_{n+m})-p(x_{n+1},x_{n+1}) \\ \leq{}&p(x_{n},x_{n+1})+p(x_{n+1},x_{n+2})+p(x_{n+2},x_{n+m}) \\ &{}-p(x_{n+1},x_{n+1})-p(x_{n+2},x_{n+2}) \\ \leq{}&\sum_{i=1}^{k}p(x_{n+i-1},x_{n+i})- \sum_{i=1}^{k-1}p(x_{n+i},x_{n+i}) \\ \leq{}&\sum_{i=1}^{m}\varphi^{n+i-1} p(x_{0},x_{1})-\sum_{i=1}^{m-1}p(x_{n+i},x_{n+i}) \\ \leq{}&\sum_{i=1}^{m}\varphi^{n+i-1} p(x_{0},x_{1}). \end{aligned}$$
Letting \(n\rightarrow\infty\), then, by using the condition (\(\varphi _{4}\)), we have
$$ \lim_{n\rightarrow\infty}p(x_{n},x_{n+m})=0. $$
(3.11)
By the definition of \(d_{p}\), we see that, for any \(m\in\mathbb{N}\),
$$ \lim_{n\rightarrow\infty}d_{p}(x_{n},x_{n+m}) \leq\lim_{n\rightarrow \infty}2p(x_{n},x_{n+m})= 0. $$
(3.12)
This shows that \(\{x_{n}\}\) is a Cauchy sequence in \((X,d_{p})\). Since \((X,p)\) is complete, from Lemma 1, \((X,d_{p})\) is a complete metric space. Therefore, \(\{x_{n}\}\) converges to some \(x^{*}\in X\) with respect to the metric \(d_{p}\), and we also have
$$ p\bigl(x^{*},x^{*}\bigr)=\lim_{n\rightarrow\infty}p\bigl(x_{n},x^{*} \bigr)=\lim_{n\rightarrow \infty}p(x_{n},x_{m})=0. $$
(3.13)
Since \(T:X\rightarrow CB^{p}(X)\) is a \((\alpha,\varphi)\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal {H}_{p}\), we have
$$\alpha\bigl(x_{n},x^{*}\bigr)\mathcal{H}_{p} \bigl(Tx_{n},Tx^{*}\bigr)\leq\varphi\bigl(p\bigl(x_{n},x^{*}\bigr) \bigr). $$
By the definition of the mapping α, we have \(\alpha(x_{n},x^{*})>0\). By the condition (\(\varphi_{5}\)) and using (3.13), we get
$$ \lim_{n\rightarrow\infty}\mathcal{H}_{p}\bigl(Tx_{n},Tx^{*} \bigr)=0. $$
(3.14)
Now \(x_{n+1}\in Tx_{n}\) shows
$$p\bigl(x_{n+1},Tx^{*}\bigr)\leq\delta_{p}\bigl(Tx_{n},Tx^{*} \bigr)\leq\mathcal{H}_{p}\bigl(Tx_{n},Tx^{*}\bigr). $$
Using (3.14), we get
$$p\bigl(x^{*},Tx^{*}\bigr)=\lim_{n\rightarrow\infty}p\bigl(x_{n+1},Tx^{*} \bigr)=0. $$
Therefore, from (3.13), \(p(x^{*},x^{*})=0\), we obtain
$$p\bigl(x^{*},x^{*}\bigr)=p\bigl(x^{*},Tx^{*}\bigr), $$
which implies \(x^{*}\in Tx^{*}\) by Remark 3. □

4 Fixed point theorem (III)

In this section, we consider the family
$$\Omega=\bigl\{ (\xi_{1},\xi_{2},\xi_{3}, \xi_{4})|\xi_{i}:\mathbb{R}^{+}\rightarrow\mathbb {R}^{+},i=1,2,3,4\bigr\} $$
such that:
  1. (1)

    \(\xi_{1}(t),\xi_{2}(t),\xi_{3}(t)\leq\xi_{4}(t)\) for all \(t>0\);

     
  2. (2)

    \(\xi_{1}(t)\), \(\xi_{2}(t)\), \(\xi_{3}(t)\), \(\xi_{4}(t)\) are continuous;

     
  3. (3)

    \(\xi_{1}(t_{1})=\xi_{2}(t_{2})=\xi_{3}(t_{3})=\xi_{4}(t_{4})=0\) if and only if \(t_{1}=t_{2}=t_{3}=t_{4}=0\);

     
  4. (4)

    \(\xi_{4}\) is a Meir-Keeler-type function;

     
  5. (5)

    \(\xi_{4}(t_{1}+t_{2})\leq\xi_{4}(t_{1})+\xi_{4}(t_{2})\) for all \(t_{1},t_{2}>0\).

     

We now introduce the notion of \((\alpha,\phi,\xi_{1},\xi_{2},\xi_{3},\xi _{4})\)-Meir-Keeler contraction on partial Hausdorff metric spaces.

Definition 11

Let \((X,p)\) be a partial metric space, \(\phi\in\Phi\), \((\xi_{1},\xi_{2},\xi _{3},\xi_{4})\in\Omega\), and \(\alpha:X\times X\rightarrow\mathbb {R}^{+}\diagdown\{0\}\). We call \(T:X\rightarrow CB^{p}(X)\) a \((\alpha,\phi ,\xi_{1},\xi_{2},\xi_{3},\xi_{4})\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\) if the following conditions hold:
  1. (1)

    T is strictly α-admissible;

     
  2. (2)
    for all \(x,y\in X\), we have
    $$\alpha(x,y)\mathcal{H}_{p}(Tx,Ty)\leq\phi \biggl(\xi_{1} \bigl(p(x,y)\bigr),\xi _{2}\bigl(p(x,Tx)\bigr),\xi_{3} \bigl(p(y,Ty)\bigr),\frac{\xi_{4}(p(x,Ty)+p(y,Tx))}{2} \biggr). $$
     

We now state and prove our main result for the \((\alpha,\phi,\xi_{1},\xi _{2},\xi_{3},\xi_{4})\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\).

Theorem 5

Let \((X,p)\) be a complete partial metric space. Suppose \(T:X\rightarrow CB^{p}(X)\) is a \((\alpha,\phi,\xi_{1},\xi_{2},\xi_{3},\xi_{4})\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal {H}_{p}\). Suppose also that
  1. (i)

    there exists \(x_{0}\in X\) such that \(\alpha(x_{0},y)>1\) for all \(y\in Tx_{0}\);

     
  2. (ii)

    if \(\{x_{n}\}\) is a sequence in X such that \(\alpha (x_{n},x_{n+1})\geq1\) for all n and \(x_{n} \rightarrow x\in X\) as \(n\rightarrow\infty\), then \(\alpha(x_{n},x)\geq1\) for all n.

     
Then T has a fixed point in X (that is, there exists \(x^{*}\in X\) such that \(x^{*}\in Tx^{*}\)).

Proof

Let \(x_{1}\in Tx_{0}\). Since \(T:X\rightarrow CB^{p}(X)\) is a \((\alpha,\phi,\xi _{1},\xi_{2},\xi_{3},\xi_{4})\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\), we have
$$\begin{aligned} &\alpha(x_{0},x_{1})\mathcal{H}_{p}(Tx_{0},Tx_{1}) \\ &\quad\leq\phi \biggl(\xi _{1}\bigl(p(x_{0},x_{1}) \bigr),\xi_{2}\bigl(p(x_{0},Tx_{0})\bigr), \xi_{3}\bigl(p(x_{1},Tx_{1})\bigr), \frac{\xi _{4}(p(x_{0},Tx_{1})+p(x_{1},Tx_{0}))}{2} \biggr). \end{aligned}$$
(4.1)
By the condition (i), we have \(\alpha(x_{0},x_{1})>1\). Put \(\alpha(x_{0},x_{1})=h_{0}\). Note \(h_{0}>1\). From Lemma 2 with \(h=\sqrt{h_{0}}\), there exists \(x_{2}\in Tx_{1}\) such that
$$ p(x_{1},x_{2})\leq\sqrt{h_{0}} \mathcal{H}_{p}(Tx_{0},Tx_{1}). $$
(4.2)
Using (4.1) and (4.2), we have
$$\begin{aligned} 0&\leq p(x_{1},x_{2}) \\ &\leq\frac{1}{\sqrt{h_{0}}}\phi \biggl(\xi _{1}\bigl(p(x_{0},x_{1}) \bigr),\xi_{2}\bigl(p(x_{0},Tx_{0})\bigr), \xi_{3}\bigl(p(x_{1},Tx_{1})\bigr), \frac{\xi _{4}(p(x_{0},Tx_{1})+p(x_{1},Tx_{0}))}{2} \biggr) \\ &\leq\frac{1}{\sqrt{h_{0}}}\phi \biggl(\xi_{1}\bigl(p(x_{0},x_{1}) \bigr),\xi_{2}\bigl(p(x_{0},x_{1})\bigr), \xi_{3}\bigl(p(x_{1},x_{2})\bigr), \frac{\xi _{4}(p(x_{0},x_{1})+p(x_{1},x_{2}))}{2} \biggr) \\ &\leq\frac{1}{\sqrt{h_{0}}}\phi \biggl(\xi_{1}\bigl(p(x_{0},x_{1}) \bigr),\xi_{2}\bigl(p(x_{0},x_{1})\bigr), \xi_{3}\bigl(p(x_{1},x_{2})\bigr), \frac{\xi _{4}(p(x_{0},x_{1}))+\xi_{4}(p(x_{1},x_{2}))}{2} \biggr). \end{aligned}$$
(4.3)
Now, if \(\xi_{4}(p(x_{0},x_{1}))\leq\xi_{4}(p(x_{1},x_{2}))\), then by using (4.3) and since \(\phi\in\Phi\), \((\xi_{1},\xi_{2},\xi_{3},\xi_{4})\in\Omega \), we have
$$\begin{aligned} 0&\leq p(x_{1},x_{2}) \\ &\leq\frac{1}{\sqrt{h_{0}}}\phi \biggl(\xi _{4}\bigl(p(x_{0},x_{1}) \bigr),\xi_{4}\bigl(p(x_{0},x_{1})\bigr), \xi_{4}\bigl(p(x_{1},x_{2})\bigr), \frac{\xi _{4}(p(x_{0},x_{1})+p(x_{1},x_{2}))}{2} \biggr) \\ &\leq\frac{1}{\sqrt{h_{0}}}\phi \bigl(\xi_{4}\bigl(p(x_{1},x_{2}) \bigr),\xi_{4}\bigl(p(x_{1},x_{2})\bigr), \xi_{4}\bigl(p(x_{1},x_{2})\bigr),\xi _{4}\bigl(p(x_{1},x_{2})\bigr) \bigr) \\ &\leq\frac{1}{\sqrt{h_{0}}}\xi_{4}\bigl(p(x_{1},x_{2}) \bigr). \end{aligned}$$
By Remark 4, we also have
$$p(x_{1},x_{2})\leq\frac{1}{\sqrt{h_{0}}}\xi_{4} \bigl(p(x_{1},x_{2})\bigr)< p(x_{1},x_{2}), $$
which implies a contradiction. Therefore, we have
$$p(x_{1},x_{2})\leq\frac{1}{\sqrt{h_{0}}}\xi_{4} \bigl(p(x_{0},x_{1})\bigr)\leq\frac{1}{\sqrt {h_{0}}}p(x_{0},x_{1}). $$
Since \(T:X\rightarrow CB^{p}(X)\) is a \((\alpha,\phi,\xi_{1},\xi_{2},\xi_{3},\xi _{4})\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\), we have
$$\begin{aligned} &\alpha(x_{1},x_{2})\mathcal{H}_{p}(Tx_{1},Tx_{2}) \\ &\quad\leq\phi \biggl(\xi _{1}\bigl(p(x_{1},x_{2}) \bigr),\xi_{2}\bigl(p(x_{1},Tx_{1})\bigr), \xi_{3}\bigl(p(x_{2},Tx_{2})\bigr), \frac{\xi _{4}(p(x_{1},Tx_{2})+p(x_{2},Tx_{1}))}{2} \biggr). \end{aligned}$$
(4.4)
Since T is strictly α-admissible, we have \(\alpha(x_{1},x_{2})>1\). Put \(\alpha(x_{1},x_{2})=h_{1}\). Note \(h_{1}>1\). From Lemma 2 with \(h=\sqrt{h_{1}}\), there exists \(x_{3}\in Tx_{2}\) such that
$$ p(x_{2},x_{3})\leq\sqrt{h_{1}} \mathcal{H}_{p}(Tx_{1},Tx_{2}). $$
(4.5)
Using (4.4) and (4.5), we have
$$\begin{aligned} 0&\leq p(x_{2},x_{3}) \\ &\leq\frac{1}{\sqrt{h_{1}}}\phi \biggl(\xi _{1}\bigl(p(x_{1},x_{2}) \bigr),\xi_{2}\bigl(p(x_{1},Tx_{1})\bigr), \xi_{3}\bigl(p(x_{2},Tx_{2})\bigr), \frac{\xi _{4}(p(x_{1},Tx_{2})+p(x_{2},Tx_{1}))}{2} \biggr) \\ &\leq\frac{1}{\sqrt{h_{1}}}\phi \biggl(\xi_{1}\bigl(p(x_{1},x_{2}) \bigr),\xi_{2}\bigl(p(x_{1},x_{2})\bigr), \xi_{3}\bigl(p(x_{2},x_{3})\bigr), \frac{\xi _{4}(p(x_{1},x_{2})+p(x_{2},x_{3}))}{2} \biggr) \\ &\leq\frac{1}{\sqrt{h_{1}}}\phi \biggl(\xi_{1}\bigl(p(x_{1},x_{2}) \bigr),\xi_{2}\bigl(p(x_{1},x_{2})\bigr), \xi_{3}\bigl(p(x_{2},x_{3})\bigr), \frac{\xi _{4}(p(x_{1},x_{2}))+\xi_{4}(p(x_{2},x_{3}))}{2} \biggr). \end{aligned}$$
(4.6)
Now, if \(\xi_{4}(p(x_{1},x_{2}))\leq\xi_{4}(p(x_{2},x_{3}))\), then by using (4.6) and since \(\phi\in\Phi\), \((\xi_{1},\xi_{2},\xi_{3},\xi_{4})\in\Omega \), we have
$$\begin{aligned} 0&\leq p(x_{2},x_{3}) \\ &\leq\frac{1}{\sqrt{h_{1}}}\phi \biggl(\xi _{4}\bigl(p(x_{1},x_{2}) \bigr),\xi_{4}\bigl(p(x_{1},x_{2})\bigr), \xi_{4}\bigl(p(x_{2},x_{3})\bigr), \frac{\xi _{4}(p(x_{1},x_{2})+p(x_{2},x_{3}))}{2} \biggr) \\ &\leq\frac{1}{\sqrt{h_{1}}}\phi \bigl(\xi_{4}\bigl(p(x_{2},x_{3}) \bigr),\xi_{4}\bigl(p(x_{2},x_{3})\bigr), \xi_{4}\bigl(p(x_{2},x_{3})\bigr),\xi _{4}\bigl(p(x_{2},x_{3})\bigr) \bigr) \\ &\leq\frac{1}{\sqrt{h_{1}}}\xi_{4}\bigl(p(x_{2},x_{3}) \bigr). \end{aligned}$$
By Remark 4,we also have
$$p(x_{2},x_{3})\leq\frac{1}{\sqrt{h_{1}}}\xi_{4} \bigl(p(x_{2},x_{3})\bigr)< p(x_{2},x_{3}), $$
which implies a contradiction. Therefore, we have
$$p(x_{2},x_{3})\leq\frac{1}{\sqrt{h_{1}}}\xi_{4} \bigl(p(x_{1},x_{2})\bigr)\leq\frac{1}{\sqrt {h_{1}}}p(x_{1},x_{2}) \leq\frac{1}{\sqrt{h_{0}}}\frac{1}{\sqrt{h_{1}}}p(x_{0},x_{1}). $$
Continuing this process, we can obtain a sequence \(\{x_{n}\}\) of X recursively as follows:
$$x_{n}\in Tx_{n-1}, \quad\mbox{for all } n\in\mathbb{N}. $$
Since \(T:X\rightarrow CB^{p}(X)\) is a \((\alpha,\phi,\xi_{1},\xi_{2},\xi_{3},\xi _{4})\)-Meir-Keeler contraction with respect to the partial Hausdorff metric \(\mathcal{H}_{p}\), we have
$$\begin{aligned} &\alpha(x_{n},x_{n+1})\mathcal{H}_{p}(Tx_{n},Tx_{n+1}) \\ &\quad\leq\phi \biggl(\xi_{1}\bigl(p(x_{n},x_{n+1}) \bigr),\xi_{2}\bigl(p(x_{n},Tx_{n})\bigr),\xi _{3}\bigl(p(x_{n+1},Tx_{n+1})\bigr), \\ &\qquad{}\frac{\xi _{4}(p(x_{n},Tx_{n+1})+p(x_{n+1},Tx_{n}))}{2} \biggr). \end{aligned}$$
(4.7)
Since T is strictly α-admissible, we have \(\alpha (x_{n},x_{n+1})>1\) for all \(n\in\mathbb{N}\). Put \(\alpha(x_{n},x_{n+1})=h_{n}\). From Lemma 2 with \(h=\sqrt{h_{n}}\), there exists \(x_{n+2}\in Tx_{n+1}\) such that
$$ p(x_{n+1},x_{n+2})\leq\sqrt{h_{n}} \mathcal{H}_{p}(Tx_{n},Tx_{n+1}). $$
(4.8)
Using (4.7) and (4.8), we have
$$\begin{aligned} 0\leq{}& p(x_{n+1},x_{n+2}) \\ \leq{}&\frac{1}{\sqrt{h_{n}}}\phi \biggl(\xi _{1}\bigl(p(x_{n},x_{n+1}) \bigr),\xi_{2}\bigl(p(x_{n},Tx_{n})\bigr),\xi _{3}\bigl(p(x_{n+1},Tx_{n+1})\bigr), \\ &{}\frac{\xi _{4}(p(x_{n},Tx_{n+1})+p(x_{n+1},Tx_{n}))}{2} \biggr) \\ \leq{}&\frac{1}{\sqrt {h_{n}}}\phi \biggl(\xi_{1}\bigl(p(x_{n},x_{n+1}) \bigr),\xi_{2}\bigl(p(x_{n},x_{n+1})\bigr),\xi _{3}\bigl(p(x_{n+1},x_{n+2})\bigr), \\ &{}\frac{\xi _{4}(p(x_{n},x_{n+1})+p(x_{n+1},x_{n+2}))}{2} \biggr) \\ \leq{}&\frac{1}{\sqrt {h_{n}}}\phi \biggl(\xi_{1}\bigl(p(x_{n},x_{n+1}) \bigr),\xi_{2}\bigl(p(x_{n},x_{n+1})\bigr),\xi _{3}\bigl(p(x_{n+1},x_{n+2})\bigr), \\ &{}\frac{\xi_{4}(p(x_{n},x_{n+1}))+\xi _{4}(p(x_{n+1},x_{n+2}))}{2} \biggr). \end{aligned}$$
(4.9)
Now, if \(\xi_{4}(p(x_{n},x_{n+1}))\leq\xi_{4}(p(x_{n+1},x_{n+2}))\), then by using (4.9) and since \(\phi\in\Phi\), \((\xi_{1},\xi_{2}, \xi_{3},\xi _{4})\in\Omega\), we have
$$\begin{aligned} 0\leq{}& p(x_{n+1},x_{n+2}) \\ \leq{}&\frac{1}{\sqrt{h_{n}}}\phi \biggl(\xi _{4}\bigl(p(x_{n},x_{n+1}) \bigr),\xi_{4}\bigl(p(x_{n},x_{n+1})\bigr),\xi _{4}\bigl(p(x_{n+1},x_{n+2})\bigr),\\ &{}\frac{\xi _{4}(p(x_{n},x_{n+1})+p(x_{n+1},x_{n+2}))}{2} \biggr) \\ \leq{}&\frac{1}{\sqrt {h_{n}}}\phi \bigl(\xi_{4}\bigl(p(x_{n+1},x_{n+2}) \bigr),\xi_{4}\bigl(p(x_{n+1},x_{n+2})\bigr),\xi _{4}\bigl(p(x_{n+1},x_{n+2})\bigr),\xi_{4} \bigl(p(x_{n+1},x_{n+2})\bigr) \bigr) \\ \leq{}&\frac {1}{\sqrt{h_{n}}}\xi_{4}\bigl(p(x_{n+1},x_{n+2}) \bigr). \end{aligned}$$
By Remark 4,we also have
$$p(x_{n+1},x_{n+2})\leq\frac{1}{\sqrt{h_{n}}}\xi_{4} \bigl(p(x_{n+1},x_{n+2})\bigr)< p(x_{n+1},x_{n+2}), $$
which implies a contradiction. Therefore, we have
$$\begin{aligned} p(x_{n+1},x_{n+2})&\leq\frac{1}{\sqrt{h_{n}}} \xi_{4}\bigl(p(x_{n},x_{n+1})\bigr) \\ &\leq\frac{1}{\sqrt{h_{n}}}p(x_{n},x_{n+1}) \\ &\leq\cdots \\ &\leq\frac {1}{\sqrt{h_{n}}}\frac{1}{\sqrt{h_{n-1}}}\cdots\frac{1}{\sqrt {h_{0}}}p(x_{0},x_{1}). \end{aligned}$$
(4.10)
Since \(h_{n}>1\) for all \(n\in\mathbb{N}\cup\{0\}\), we get
$$\frac{1}{\sqrt{h_{n}}}< 1, \quad\mbox{for all } n\in\mathbb{N}\cup\{0\}. $$
Put
$$ \overline{k}=\max\biggl\{ \frac{1}{\sqrt{h_{n}}}: n\in\mathbb{N}\cup\{0\}\biggr\} . $$
(4.11)
Using (4.10) and (4.11), we obtain
$$ p(x_{n+1},x_{n+2})\leq(\overline{k})^{n+1} p(x_{0},x_{1}),\quad \mbox{for all } n\in\mathbb{N}\cup\{0 \}. $$
(4.12)
Let \(n\rightarrow\infty\) in (4.12). Then
$$ \lim_{n\rightarrow\infty}p(x_{n},x_{n+1})=0. $$
(4.13)
By the property (p2) of a partial metric and using (4.13), we have
$$ \lim_{n\rightarrow\infty}p(x_{n},x_{n})=0. $$
(4.14)
Using (4.12) and the property (p4) of a partial metric, for any \(m\in\mathbb{N}\), we have
$$\begin{aligned} p(x_{n},x_{n+m})&\leq\sum _{i=1}^{m} p(x_{n+i-1},x_{n+i})-\sum _{i=1}^{m-1} p(x_{n+i},x_{n+i}) \\ &\leq\sum_{i=1}^{m} (\overline {k})^{n+i-1}p(x_{0},x_{1})-\sum _{i=1}^{m-1} p(x_{n+i},x_{n+i}) \\ &\leq \frac{(\overline{k})^{n}}{(1-\overline{k})} p(x_{0},x_{1})-\sum _{i=1}^{m-1} p(x_{n+i},x_{n+i}). \end{aligned}$$
(4.15)
Using (4.14) and (4.15), we get
$$\lim_{n\rightarrow\infty}p(x_{n},x_{n+m})=0. $$
By the definition of \(d_{p}\), we see that, for any \(m\in\mathbb{N}\),
$$ \lim_{n\rightarrow\infty}d_{p}(x_{n},x_{n+m}) \leq\lim_{n\rightarrow \infty}2p(x_{n},x_{n+m})= 0. $$
(4.16)
This shows that \(\{x_{n}\}\) is a Cauchy sequence in \((X,d_{p})\). Since \((X,p)\) is complete, from Lemma 1, \((X,d_{p})\) is a complete metric space. Therefore, \(\{x_{n}\}\) converges to some \(x^{*}\in X\) with respect to the metric \(d_{p}\), and we also have
$$ p\bigl(x^{*},x^{*}\bigr)=\lim_{n\rightarrow\infty}p\bigl(x_{n},x^{*} \bigr)=\lim_{n\rightarrow \infty}p(x_{n},x_{m})=0. $$
(4.17)
By the definition of the mapping α, we have \(\alpha(x_{n},x^{*})>0\). By using (4.17) together with the properties of the auxiliary functions ϕ, \(\xi_{1}\), \(\xi_{2}\), \(\xi_{3}\), \(\xi_{4}\), and the condition (ii), we get
$$\begin{aligned} & \mathcal{H}_{p}\bigl(Tx_{n},Tx^{*}\bigr) \\ &\quad\leq\alpha\bigl(x_{n},x^{*}\bigr)\mathcal {H}_{p} \bigl(Tx_{n},Tx^{*}\bigr) \\ &\quad\leq\phi\biggl(\xi_{1}\bigl(p\bigl(x_{n},x^{*}\bigr) \bigr),\xi_{2}\bigl(p(x_{n},Tx_{n})\bigr),\xi _{3}\bigl(p\bigl(x^{*},Tx^{*}\bigr)\bigr),\frac{\xi_{4}(p(x_{n},Tx^{*})+p(x^{*},Tx_{n}))}{2}\biggr) \\ &\quad\leq\phi \biggl(\xi_{1}\bigl(p\bigl(x_{n},x^{*}\bigr) \bigr),\xi_{2}\bigl(p(x_{n},x_{n+1})\bigr), \xi_{3}\bigl(p\bigl(x^{*},Tx^{*}\bigr)\bigr), \\ &\qquad{}\frac{\xi _{4}(p(x_{n},x^{*})+p(x^{*},Tx^{*})-p(x^{*},x^{*})+p(x^{*},x_{n+1}))}{2}\biggr) \\ &\quad\leq\phi \biggl(\xi_{4}\bigl(p\bigl(x_{n},x^{*}\bigr) \bigr),\xi_{4}\bigl(p(x_{n},x_{n+1})\bigr), \xi_{4}\bigl(p\bigl(x^{*},Tx^{*}\bigr)\bigr), \\ &\qquad{}\frac{\xi _{4}(p(x_{n},x^{*})+p(x^{*},Tx^{*})-p(x^{*},x^{*})+p(x^{*},x_{n+1}))}{2}\biggr). \end{aligned}$$
(4.18)
Let \(n\rightarrow\infty\) in (4.18). By Remark 4,
$$\begin{aligned} &\lim_{n\rightarrow\infty}\mathcal{H}_{p} \bigl(Tx_{n},Tx^{*}\bigr) \\ &\quad\leq \phi(0,0,\xi_{4}\bigl(p\bigl(x^{*},Tx^{*}\bigr)\bigr), \frac{1}{2}\xi_{4}\bigl(p\bigl(x^{*},Tx^{*}\bigr)\bigr) \\ &\quad\leq\xi _{4}\bigl(p\bigl(x^{*},Tx^{*}\bigr)\bigr) \\ &\quad< p\bigl(x^{*},Tx^{*}\bigr). \end{aligned}$$
(4.19)
Now \(x_{n+1}\in Tx_{n}\) shows
$$ p\bigl(x_{n+1},Tx^{*}\bigr)\leq\delta_{p}\bigl(Tx_{n},Tx^{*} \bigr)\leq\mathcal {H}_{p}\bigl(Tx_{n},Tx^{*} \bigr). $$
(4.20)
Using (4.18), (4.19), and (4.20), we get
$$p\bigl(x^{*},Tx^{*}\bigr)=\lim_{n\rightarrow\infty}p\bigl(x_{n+1},Tx^{*} \bigr)< p\bigl(x^{*},Tx^{*}\bigr), $$
a contradiction. So, we have
$$p\bigl(x^{*},Tx^{*}\bigr)=0. $$
Therefore, from (4.17), \(p(x^{*},x^{*})=0\), we obtain
$$p\bigl(x^{*},x^{*}\bigr)=p\bigl(x^{*},Tx^{*}\bigr), $$
which implies \(x^{*}\in Tx^{*}\) by Remark 3. □

Declarations

Acknowledgements

This research was supported by the Ministry of Science and Technology of the Republic of China.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
General Education Center, St. John’s University
(2)
Department of Applied Mathematics, National Hsinchu University of Education

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© Jen et al. 2015