Contractive multivalued maps in terms of Qfunctions on complete quasimetric spaces
 Erdal Karapınar^{1, 2},
 Salvador Romaguera^{3} and
 Pedro Tirado^{3}Email author
https://doi.org/10.1186/16871812201453
© Karapinar et al.; licensee Springer. 2014
Received: 17 December 2013
Accepted: 18 February 2014
Published: 27 February 2014
Abstract
In this paper we prove the existence of a fixed point for multivalued maps satisfying a contraction condition in terms of Qfunctions, and via BianchiniGrandolfi gauge functions, for complete ${T}_{0}$quasipseudometric spaces. Our results extend, improve, and generalize some recent results in the literature. We present some examples to validate and illustrate our results.
MSC:54H25, 47H10, 54E50.
Keywords
1 Introduction and preliminaries
The notion of metric space, introduced by Fréchet [1], is one of the cornerstones of both applied and pure mathematics. The metric space is indispensable in many branches of mathematics. For example, in these days, one of the core topics in group theory is to construct a metric on a given group under the certain conditions. Due to its wide application areas in all quantitative sciences, this notion has been generalized and extended in various way, such as quasimetrics, symmetrics, bmetrics, Gmetrics, fuzzy metrics, etc. Among all, we attract attention to the notion of Qfunction, introduced by AlHomidan et al. [2] in the framework of quasimetric space as an extension of the concept of wdistance defined by Kada et al. [3]. In fact, the authors of [2] proved, among other results, a quasimetric version of the celebrated Nadler fixed point theorem [4]. Recently, Marín et al. [5] generalized some results of [2] by using BianchiniGrandolfi gauge functions. Almost simultaneously, Latif and AlMezel [6] obtained a quasimetric generalization of a wellknown fixed point theorem of Mizoguchi and Takahashi [[7], Theorem 5] (see also [8, 9]) for multivalued maps on complete metric spaces.
In this paper we prove the existence of fixed point for a lower semicontinuous multivalued map satisfying certain contraction condition in terms of Qfunctions via BianchiniGrandolfi gauge functions on a complete ${T}_{0}$quasipseudometric space. We also prove a weaker version of that theorem by removing the lower semicontinuity assumption. We state some examples to show the validity of the conditions and to indicate our generalizations have worth, and finally give applications to the case of contractive multivalued maps on complete partial metric spaces. Our results improve, generalize, and extend several known results in this direction.
Let ℕ denote the set of positive integer numbers, while ω denotes the set of nonnegative integer numbers.
For the sake of completeness of the paper, we recall several pertinent notions and fundamental results.
Let X be nonempty set and $d:X\times X\to [0,\mathrm{\infty})$ be a function such that
(qpm_{1}) $d(x,y)=d(y,x)=0\iff x=y$, and
(qpm_{2}) $d(x,z)\le d(x,y)+d(y,z)$,
for all $x,y,z\in X$. Then d is called a ${T}_{0}$quasipseudometric on a set X. The pair $(X,d)$ is said to be a ${T}_{0}$quasipseudometric space.
If one replaces the condition (qpm_{1}) with the stronger condition
${({\mathrm{qpm}}_{1})}^{\ast}$ $d(x,y)=0\iff x=y$,
then d is called a quasimetric on X. In this case, the pair $(X,d)$ is said to be a quasimetric space.
In the sequel we will use the abbreviation ${T}_{0}$qpm (respectively, ${T}_{0}$qpm space) instead of ${T}_{0}$quasipseudometric (respectively, ${T}_{0}$quasipseudometric space).
Given a ${T}_{0}$qpm d on a set X, the function ${d}^{1}$ defined by ${d}^{1}(x,y)=d(y,x)$ is also a ${T}_{0}$qpm, called the conjugate of d. It is clear that the function ${d}^{s}$ defined by ${d}^{s}(x,y)=max\{{d}^{1}(x,y),d(x,y)\}$ is a metric on X. (Note that if d is a metric on X, then $d={d}^{s}$.)
Consequently, every ${T}_{0}$qpm d on X induces three topologies defined as follows.
(${\tau}_{1}$) The first topology, ${\tau}_{d}$ which has as a base the family of open balls $\{{B}_{d}(x,\epsilon ):x\in X\text{and}\epsilon 0\}$, where ${B}_{d}(x,\epsilon )=\{y\in X:d(x,y)<\epsilon \}$ for all $x\in X$ and $\epsilon >0$.
(${\tau}_{2}$) The second topology, ${\tau}_{{d}^{1}}$ which has as a base the family of open balls $\{{B}_{{d}^{1}}(x,\epsilon ):x\in X\text{and}\epsilon 0\}$, where ${B}_{{d}^{1}}(x,\epsilon )=\{y\in X:{d}^{1}(x,y)<\epsilon \}$ for all $x\in X$ and $\epsilon >0$.
(${\tau}_{3}$) The last topology induced by the metric ${d}^{s}$ and denoted by ${\tau}_{{d}^{s}}$.
Notice that both ${\tau}_{d}$ and ${\tau}_{{d}^{1}}$ are ${T}_{0}$ topologies on X. Furthermore, if d is a quasimetric on X, then ${d}^{1}$ is also a quasimetric on X and hence, both ${\tau}_{d}$ and ${\tau}_{{d}^{1}}$ are ${T}_{1}$ topologies on X.
It immediately follows that a sequence ${({x}_{n})}_{n\in \mathbb{N}}$ in a ${T}_{0}$qpm space $(X,d)$ is ${\tau}_{d}$convergent to $x\in X$ if and only if ${lim}_{n\to \mathrm{\infty}}d(x,{x}_{n})=0$. Analogously, a sequence ${({x}_{n})}_{n\in \mathbb{N}}$ in a ${T}_{0}$qpm space $(X,d)$ is ${\tau}_{{d}^{1}}$convergent to $x\in X$ if and only if ${lim}_{n\to \mathrm{\infty}}d({x}_{n},{x}_{n})=0$.
In the literature, the notion of completeness for quasimetric spaces can be varied; see e.g. [5, 10, 11]. In the context of our paper we shall use the following very general notion: A ${T}_{0}$qpm space $(X,d)$ is said to be complete if every Cauchy sequence in the metric space $(X,{d}^{s})$ is ${\tau}_{{d}^{1}}$convergent.
Now, we recall the definition of Qfunction, as introduced by AlHomidanAnsariYao [2].
Definition 1 Let $(X,d)$ be a ${T}_{0}$qpm space and $q:X\times X\to [0,\mathrm{\infty})$ be a function which satisfies
(Q_{1}) $q(x,z)\le q(x,y)+q(y,z)$, for all $x,y,z\in X$,
(Q_{2}) if $x\in X$, $M>0$ and ${({y}_{n})}_{n\in \mathbb{N}}$ is a sequence in X that ${\tau}_{{d}^{1}}$converges to a point $y\in X$, and satisfies $q(x,{y}_{n})\le M$, for all $n\in \mathbb{N}$, then $q(x,y)\le M$,
(Q_{3}) for each $\epsilon >0$ there exists $\delta >0$ such that $q(x,y)\le \delta $ and $q(x,z)\le \delta $ imply $d(y,z)\le \epsilon $.
Then q is called a Qfunction on $(X,d)$.
If q satisfies conditions (Q_{1}) and (Q_{3}), and
(${\mathrm{Q}}_{2}^{\prime}$) for each $x,y\in X$ the function $q(x,\cdot ):X\to [0,\mathrm{\infty})$ is ${\tau}_{{d}^{1}}$lower semicontinuous on $(X,d)$,
then q is called a wdistance on $(X,d)$. Note that every wdistance is a Qfunction.
Remark 1 It is evident that d is a wdistance on $(X,d)$ if d is a metric on X. Note also that if $(X,d)$ is a ${T}_{0}$qpm space then d is not necessarily a Qfunction on $(X,d)$ [[2], Example 2.3] (see also [[5], Proposition 2.3]).
We conclude this section with the following simple fact which will be useful in the rest of the paper.
Lemma 1 [5]
Let q be a Qfunction on a ${T}_{0}$qpm space $(X,d)$, let $\epsilon >0$ and let $\delta =\delta (\epsilon )>0$ for which condition (Q_{3}) holds. If $q(x,y)\le \delta $ and $q(x,z)\le \delta $ then ${d}^{s}(y,z)\le \epsilon $.
2 Main results
Let $(X,d)$ be a ${T}_{0}$qpm space. The collection of all nonempty subsets (respectively, ${\tau}_{{d}^{s}}$closed subsets) of X will be denoted by ${2}^{X}$ (respectively, ${Cl}_{{d}^{s}}(X)$).
Let Ψ be the family of functions $\phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty})$ satisfying the following conditions:
(${\phi}_{1}$) φ is nondecreasing;
(${\phi}_{2}$) ${\sum}_{n=1}^{+\mathrm{\infty}}{\phi}^{n}(t)<\mathrm{\infty}$ for all $t>0$, where ${\phi}^{n}$ is the n th iterate of φ.
These functions are known in the literature as BianchiniGrandolfi gauge functions in some sources (see e.g. [12–14]) and as $(c)$comparison functions in some other sources (see e.g. [15]). It is easily proved that if $\phi \in \mathrm{\Psi}$, then $\phi (t)<t$ for any $t>0$ (see e.g. [15]).
The following lemma will be crucial to prove our first theorem.
 (a)
${x}_{n+1}\in T{x}_{n}$ for all $n\in \omega $.
 (b)
For each $\delta >0$ there exists ${n}_{\delta}\in \mathbb{N}$ such that $q({x}_{n},{x}_{m})<\delta $ whenever $m>n\ge {n}_{\delta}$.
 (c)
${({x}_{n})}_{n\in \omega}$ is a Cauchy sequence in the metric space $(X,{d}^{s})$.
for all $n\in \omega $.
Now we distinguish two cases.
Case 1. There exists $k\in \omega $ such that $q({x}_{k},{x}_{k+1})=0$. Then, by condition (2) and the fact that $\phi (t)<t$ for all $t>0$, we deduce that $q({x}_{k+1},{x}_{k+2})=0$. Repeating this argument, we obtain $q({x}_{k+j},{x}_{k+j+1})=0$ for all $j\in \omega $, so, by condition (Q_{1}), $q({x}_{n},{x}_{m})=0$ whenever $m>n\ge k$. It follows from Lemma 1 that for each $\epsilon >0$, ${d}^{s}({x}_{n},{x}_{m})\le \epsilon $ whenever $n,m>k$, and thus ${({x}_{n})}_{n\in \omega}$ is a Cauchy sequence in $(X,{d}^{s})$. Thus we have shown that conditions (a), (b), and (c) are satisfied.
In particular, $q({x}_{{n}_{\delta}},{x}_{n})\le \delta $ and $q({x}_{{n}_{\delta}},{x}_{m})\le \delta $ whenever $n,m>{n}_{\delta}$. Thus, by Lemma 1, ${d}^{s}({x}_{n},{x}_{m})\le \epsilon $ whenever $n,m>{n}_{\delta}$. Hence ${({x}_{n})}_{n\in \omega}$ is a Cauchy sequence in $(X,{d}^{s})$. This concludes the proof. □
We also need the following notion.
Definition 2 Let q be a Qfunction on a ${T}_{0}$qpm space $(X,d)$. We say that a multivalued map $T:X\to {2}^{X}$ is qlower semicontinuous (ql.s.c. in short) if the function $x\mapsto q(x,Tx)$ is lower semicontinuous on the metric space $(X,{d}^{s})$, where $q(x,Tx)=inf\{q(x,y):y\in Tx\}$.
Remark 2 An antecedent of the above concept can be found in Theorem 3.3 of the paper by Daffer and Kaneko [8], where they proved a generalization of Nadler’s fixed point theorem for a multivalued map T on a complete metric space $(X,d)$ by assuming that the function $x\mapsto d(x,Tx)$ is lower semicontinuous on $(X,d)$.
Before establishing our first fixed point result we recall that a point $z\in X$ is said to be a fixed point of a multivalued map $T:X\to {2}^{X}$ if $z\in Tz$.
Then T has a fixed point $z\in X$ such that $q(z,z)=0$.
Proof Fix ${x}_{0}\in X$. Then there is a sequence ${({x}_{n})}_{n\in \omega}$ satisfying the three conditions (a), (b) and (c) of Lemma 2. Since $(X,d)$ is complete, there exists $z\in X$ such that ${lim}_{n\to \mathrm{\infty}}d({x}_{n},z)=0$.
We shall prove that $z\in Tz$. To this end, first we prove that ${lim}_{n\to \mathrm{\infty}}q({x}_{n},z)=0$. Indeed, given $\epsilon >0$ take $\delta =\delta (\epsilon )<\epsilon $ for which condition (Q_{3}) holds. Fix $n\ge {n}_{\delta}$. By condition (b), we have $q({x}_{n},{x}_{m})\le \delta $ whenever $m>n$, so from condition (Q_{2}) we deduce that $q({x}_{n},z)\le \delta <\epsilon $ whenever $n\ge {n}_{\delta}$.
Next we show that ${lim}_{n\to \mathrm{\infty}}{d}^{s}({x}_{n},z)=0$. Indeed, given $\epsilon >0$ take $\delta =\delta (\epsilon )<\epsilon $ for which condition (Q_{3}) holds. Since $q({x}_{{n}_{\delta}},z)\le \delta $ and $q({x}_{{n}_{\delta}},{x}_{n})\le \delta $ whenever $n>{n}_{\delta}$, it follows from Lemma 1 that ${d}^{s}(z,{x}_{n})\le \epsilon $ whenever $n>{n}_{\delta}$.
Consequently $z\in {Cl}_{{d}^{s}}Tz=Tz$. Finally, $q(z,z)=0$ by (7), (8), and condition (Q_{2}). □
Next we give an example which shows that qlower semicontinuity of T cannot be omitted in Theorem 1 not even when $(X,d)$ is a complete metric space.
Example 1 Let $X=\{0\}\cup \mathbb{N}\cup A$, where $A=\{1/n:n\in \mathbb{N}\mathrm{\setminus}\{1\}\}$, and let d be the restriction to X of the usual metric on the set of real numbers. It is clear that $(X,d)$ is a complete metric space.
Now let $q:X\times X\to [0,\mathrm{\infty})$ be given by
$q(x,x)=0$ for all $x\in X$,
$q(0,x)=2$ for all $x\in \mathbb{N}\cup A$,
$q(x,y)=q(y,x)=1$ for all $x\in \mathbb{N}$, $y\in A$,
$q(x,y)=1$ for all $x,y\in \mathbb{N}$,
$q(x,0)=1$ for all $x\in \mathbb{N}$,
$q(x,y)=xy$ for all $x,y\in A$, and
$q(x,0)=x$ for all $x\in A$.
It is easy to check that q is a Qfunction (actually it is a wdistance) on $(X,d)$.
Define $T:X\to {Cl}_{d}(X)$ as
$T0=\mathbb{N}$,
$Tx=1/2x$ for all $x\in \mathbb{N}$, and
$Tx=x/2$ for all $x\in A$.
Since $q(0,T0)=2$ and $q(x,Tx)=x/2$ for all $x\in A$, we deduce that T is not ql.s.c. Moreover, it is obvious that T has no fixed point. However, we shall show that the contraction condition (6) is satisfied for the BianchiniGrandolfi gauge function φ defined as $\phi (t)=t/2$ for all $t\ge 0$.
The case that $x\in A$ and $y\in \mathbb{N}$ is similar, and hence it is omitted.
Our next fixed point result shows that qlower semicontinuity of T can be removed if the contraction condition (6) is replaced with $q(u,v)\le \phi (max\{q(x,y),q(x,u)\})$.
Then T has a fixed point.
Proof Fix ${x}_{0}\in X$. Then there is a sequence ${({x}_{n})}_{n\in \omega}$ satisfying the three conditions (a), (b), and (c) of Lemma 2. Since $(X,d)$ is complete, there exists $z\in X$ such that ${lim}_{n\to \mathrm{\infty}}d({x}_{n},z)=0$.
Now, as in the proof of Theorem 1, we obtain ${lim}_{n\to \mathrm{\infty}}q({x}_{n},z)=0$.
so, by (10) and the fact that $\phi (t)<t$ for all $t>0$, we deduce that $q({x}_{n},{z}_{n})<\epsilon $.
by Lemma 1. We conclude that $z\in Tz$. □
The following consequences of Theorem 2, which are also illustrated by Example 4 below, improve and generalize in several directions the Banach contraction principle.
Then T has a fixed point.
If we take $\phi (t)=rt$ where $r\in [0,1)$ we get one of the main results in [2].
Then T has a fixed point.
Corollary 1 was proved in [[5], Theorem 3.3]. In fact, it was showed that there is a fixed point $z\in X$ of T such that $q(z,z)=0$. This suggests the following question that remains open: Under the conditions of Theorem 2, is there is a fixed point $z\in X$ of T such that $q(z,z)=0$?
Remark 3 Example 1 shows that Theorem 2 is not true when the contraction condition (9) is replaced with $q(u,v)\le \phi (max\{q(x,y),q(y,v)\})$. Indeed, take in that example, $x\in A$, $y=0$ and $u\in Tx$. Then we have $u=x/2$, and hence $q(u,v)=1>x=max\{q(x,y),q(x,u)\}$.
Theorems 1 and 2 are independent from each other. The following two examples show this fact.
Example 2 Let $X=\omega $, i.e., $X=\{0\}\cup \mathbb{N}$, and let d be the quasimetric on X defined as
$d(x,x)=0$ for all $x\in X$,
$d(x,y)=x$ if $x>y$, and
$d(x,y)=x+y$ if $x<y$.
Clearly $(X,d)$ is a complete quasimetric space and ${\tau}_{d}$ is the discrete topology on X, so ${\tau}_{d}={\tau}_{{d}^{s}}$. Furthermore, it is almost obvious that d is a wdistance on $(X,d)$.
Now let $T:X\to {Cl}_{{d}^{s}}(X)$ given as
$T0=0$,
$T1=\{x\in \mathbb{N}:x>1\}$, and
$Tx=\{0\}\cup \{y\in \mathbb{N}:y>x\}$ for all $x\in \mathbb{N}\mathrm{\setminus}\{1\}$.
Since ${\tau}_{{d}^{s}}$ is the discrete topology on X it immediately follows that T is dl.s.c.
Consider the BianchiniGrandolfi gauge function φ given by
$\phi (t)=t/2$ if $0\le t<2$, and
$\phi (t)=n$ if $t\in [n+1,n+2)$, $n\in \mathbb{N}$.
An easy computation of the different cases shows that the contraction condition (6) is satisfied. Indeed, let $x,y\in X$ and $u\in Tx$. In the cases where for $u=0$ we can choose $v=0\in Ty$, the conclusion is obvious. Therefore, we briefly discuss the rest of the cases.
Example 3 Let $X=\{0,1\}\cup A$, where $A=\{11/n:n\in \mathbb{N}\mathrm{\setminus}\{1\}\}$, and let d be the restriction to X of the usual metric on the set of real numbers. It is clear that $(X,d)$ is a complete metric space.
Now let $q:X\times X\to [0,\mathrm{\infty})$ be given by
$q(x,x)=0$ for all $x\in X\mathrm{\setminus}\{1\}$,
$q(1,1)=1$,
$q(0,x)=q(x,0)=1/2$ for all $x\in X\mathrm{\setminus}\{0\}$,
$q(1,x)=x$ for all $x\in A$,
$q(x,1)=1x$ for all $x\in A$, and
$q(x,y)=xy$ for all $x,y\in A$.
It is not difficult to check that q is a wdistance on $(X,d)$.
Define $T:X\to {Cl}_{d}(X)$ as
$T0=0$,
$T1=\{0,1\}$, and
$Tx=\{0,(1+x)/2\}$ for all $x\in A$,
and let φ be the BianchiniGrandolfi gauge function given by $\phi (t)=t/2$ for all $t\ge 0$.
Hence, we cannot apply Theorem 1 to this example. We show that, nevertheless, the contraction condition (9) is satisfied and consequently the conditions of Theorem 2 hold.
Indeed, let $x,y\in X$ and $u\in Tx$. In the cases where for $u=0$ we can choose $v=0\in Ty$, the conclusion is obvious. Therefore we discuss the rest of the cases.
If $x=1$, $y=0$ and $u=1$, we have $v=0$, and, as in the preceding case, $q(u,v)=\phi (q(x,u))$.
If $x=1$, $y\in A$ and $u=1$, take $v=0$, and, as in the preceding case, $q(u,v)=\phi (q(x,u))$.
We conclude this section with an example where the conditions of Theorems 1 and 2 are satisfied, but for which we cannot apply Corollary 1.
Example 4 Let $X=\{0,1/4,2\}\cup [5/12,3/4]$ and let d be the ${T}_{0}$qpm on X defined as
$d(x,x)=0$ for all $x\in X$,
$d(x,0)=0$ for all $x\in X$, and
$d(x,y)=1$ otherwise.
It is clear that d is complete. In fact, ${d}^{s}$ is the discrete metric on X.
for all $x,y\in X$, is a wdistance on $(X,d)$.
Now let $T:X\to {Cl}_{{d}^{s}}(X)$ given as
$T0=T\frac{1}{4}=\{0\}$,
$Tx=\{0,1/4\}$ if $x\in [5/12,3/4]$, and
$T2=[5/12,3/4]$,
and let $\phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty})$ given as $\phi (t)=t/(2+t)$ if $0\le t<1$, and $\phi (t)=t/3$ if $t\ge 1$.
Clearly φ is a BianchiniGrandolfi gauge function (note that for $0<t<1$ and $n\in \mathbb{N}$ we have ${\phi}^{n}(t)<t/{2}^{n}$).
We shall show that the conditions of Theorem 2 are satisfied. Note that it suffices to check (9). To this end, and due to the facts that $T0=T1/4=\{0\}$, $q(0,0)=0$, and that q is symmetric we only consider the following cases.

Case 1. $5/12\le x\le 3/4$, $y\in \{0,1/4\}$ and $u\in Tx$. Take $v=0\in Ty$. If $u=0$, $q(u,v)=0$, and the inequality (1) is trivially satisfied. If $u=1/4$ we deduce that$q(u,v)=\frac{1}{4}\le \frac{x+u}{2+x+u}=\phi (x+u)=\phi (q(x,u)).$

Case 2. $5/12\le x,y\le 1$ and $u\in Tx$. Take $v=0\in Ty$. Then the conclusion follows exactly as in Case 1.

Case 3. $x\in \{0,1/4\}$, $y=2$ and $u\in Tx$. Then $u=0$. Taking $v=5/12\in Ty$, we deduce that$q(u,v)=\frac{5}{12}<\frac{2}{3}=\phi (2)\le \phi (q(x,y)).$

Case 4. $5/12\le x\le 3/4$, $y=2$ and $u\in Tx$. Taking $v=5/12\in Ty$, we deduce that$q(u,v)\le \frac{1}{4}+\frac{5}{12}=\frac{2}{3}=\phi (2)\le \phi (q(x,y)).$

Case 5. $x=2$, $y\in \{0,1/4\}$ and $u\in Tx$. Take $v=0\in Ty$. Then$q(u,v)=u<\frac{x+u}{3}=\phi (q(x,u)).$

Case 6. $x=2$, $5/12\le y\le 3/4$ and $u\in Tx$. Taking $v=0\in Ty$, we deduce that$q(u,v)=u\le \frac{3}{4}\le \frac{2+5/12}{3}\le \frac{x+u}{3}=\phi (q(x,u)).$

Case 7. $x=y=2$ and $u\in Tx$. Taking $v=5/12\in Ty$, we deduce that$q(u,v)=u+v\le \frac{3}{4}+\frac{5}{12}<\frac{4}{3}=\phi (q(x,y)).$
Moreover, the conditions of Theorem 1 are also satisfied because T is trivially ql.s.c.
Finally, note that the preceding relations also show that condition (9) does not follow for the ${T}_{0}$qpm d.
3 Application to partial metric spaces
Matthews introduced in [16] (see also [17]) the ‘equivalent’ notions of a weightable ${T}_{0}$qpm space and of a partial metric space in order to construct a consistent topological model for certain programming languages.
for all $x,y\in X$. In this case, we say that the pair $(X,d)$ is a weightable ${T}_{0}$qpm space. The function w is called a weighting function for $(X,d)$.
Note that Matthews used the term ‘weightable quasimetric spaces’ for such spaces.
Now, we state the definition of partial metric space as given by Matthews [16, 17].
Definition 3 A partial metric on a set X is a function $p:X\times X\to [0,\mathrm{\infty})$ satisfying the following conditions for all $x,y,z\in X$:
(P1) $x=y\iff p(x,x)=p(y,y)=p(x,y)$,
(P2) $p(x,x)\le p(x,y)$,
(P3) $p(x,y)=p(y,x)$,
(P4) $p(x,y)\le p(x,z)+p(z,y)p(z,z)$.
Then the pair $(X,p)$ is called a partial metric space.
Observe that from (P1) and (P2) it follows that if $p(x,y)=0$ then $x=y$.
Each partial metric p on a set X induces a ${T}_{0}$ topology ${\tau}_{p}$ on X, which has as a base the family of open pballs $\{{B}_{p}(x,\epsilon ):x\in X,\epsilon >0\}$ where ${B}_{p}(x,\epsilon )=\{y\in X:p(x,y)<p(x,x)+\epsilon \}$ for all $x\in X$ and $\epsilon >0$.
A typical example of a partial metric space is the pair $(X,p)$ where $X=[0,\mathrm{\infty})$ and p is the partial metric on X given by $p(x,y)=max\{x,y\}$ for all $x,y\in X$.
A partial metric p on a set X induces, in a natural way, three metrics on X, denoted by ${p}^{S}$, ${p}_{M}$ and ${p}_{0}$, respectively, that are defined, for each $x,y\in X$, as follows:
${p}^{S}(x,y)=2p(x,y)p(x,x)p(y,y)$,
${p}_{M}(x,y)=p(x,y)min\{p(x,x),p(y,y)\}$, and
${p}_{0}(x,x)=0$, and ${p}_{0}(x,y)=p(x,y)$ if $x\ne y$.
It is easy to show (see e.g. [18]) that ${\tau}_{p}\subseteq {\tau}_{{p}^{S}}={\tau}_{{p}_{M}}\subseteq {\tau}_{{p}_{0}}$.
Matthews proved [[17], Theorems 4.1 and 4.2] that the concepts of weightable ${T}_{0}$qpm space and partial metric space are equivalent in the following sense.
Proposition 1 (i) Let $(X,d)$ be a weightable ${T}_{0}$qpm space with weighting function w. Then the function ${p}_{d}:X\times X\to [0,\mathrm{\infty})$ defined by ${p}_{d}(x,y)=d(x,y)+w(x)$ for all $x,y\in X$ is a partial metric on X. Furthermore, ${\tau}_{d}={\tau}_{{p}_{d}}$.
(ii) Conversely, let $(X,p)$ be a partial metric space. Then the function ${d}_{p}:X\times X\to [0,\mathrm{\infty})$ defined by ${d}_{p}(x,y)=p(x,y)p(x,x)$ for all $x,y\in X$ is a weightable ${T}_{0}$qpm space on X. Furthermore, ${\tau}_{d}={\tau}_{{d}_{p}}$.
It is clear from the above proposition that for each partial metric p on X one has $p={p}_{{d}_{p}}$, and that for each weightable ${T}_{0}$qpm on X one has $d={d}_{{p}_{d}}$.
Since Matthews proved in [[17], Theorem 5.3] his wellknown partial metric generalization of the Banach contraction principle several authors have investigated the problem of obtaining fixed points for a variety of contractive conditions for selfmaps and multivalued maps on partial metric spaces. This research has been specially intensive in the last five years (see e.g. [19, 20] and the bibliographic references contained in them). In connection with our approach it is interesting to note that the partial metric ${p}_{d}$ induced by a weightable ${T}_{0}$qpm space $(X,d)$ allows us to construct some nice Qfunctions on $(X,d)$. This is stated in the next two lemmas. The first one was proved in [[5], Proposition 2.10].
Lemma 3 [5]
Let $(X,d)$ be a weightable ${T}_{0}$qpm space with weighting function w. Then the induced partial metric ${p}_{d}$ is a Qfunction on $(X,d)$.
A slight modification of the proof of the above lemma allows us to state the following.
Lemma 4 Let $(X,d)$ be a weightable ${T}_{0}$qpm space with weighting function w. Then the function ${q}_{d}:X\times X\to [0,\mathrm{\infty})$ defined by ${q}_{d}(x,y)={p}_{d}(x,y)+{p}_{d}(y,y)$ for all $x,y\in X$, is a Qfunction on $(X,d)$.
Then, and as a natural consequence of Theorems 1 and 2, we obtain the following results that generalize and improve, among other results, [[5], Theorem 3.3] and [[17], Theorem 3.5].
 (A)T is pl.s.c. and for each $x,y\in X$ and $u\in Tx$, there is $v\in T(y)$ satisfying$p(u,v)\le \phi (max\{p(x,y),p(x,u),p(y,v)\}).$
 (B)T is ${q}_{{d}_{p}}$l.s.c. and for each $x,y\in X$ and $u\in Tx$, there is $v\in T(y)$ satisfying$p(u,v)+p(v,v)\le \phi (max\{p(x,y)+p(y,y),p(x,u)+p(u,u),p(y,v)+p(v,v)\}).$
 (A)For each $x,y\in X$ and $u\in Tx$, there is $v\in T(y)$ satisfying$p(u,v)\le \phi (max\{p(x,y),p(x,u)\}).$
 (B)For each $x,y\in X$ and $u\in Tx$, there is $v\in T(y)$ satisfying$p(u,v)+p(v,v)\le \phi (max\{p(x,y)+p(y,y),p(x,u)+p(u,u)\}).$
 (A)For each $x,y\in X$ and $u\in Tx$, there is $v\in T(y)$ satisfying$p(u,v)\le \phi (p(x,y)).$
 (B)For each $x,y\in X$ and $u\in Tx$, there is $v\in T(y)$ satisfying$p(u,v)+p(v,v)\le \phi (p(x,y)+p(y,y)).$
 (A)For each $x,y\in X$ and $u\in Tx$, there is $v\in T(y)$ satisfying$p(u,v)\le rp(x,y).$
 (B)For each $x,y\in X$ and $u\in Tx$, there is $v\in T(y)$ satisfying$p(u,v)+p(v,v)\le r(p(x,y)+p(y,y)).$
We conclude the paper with a simple example where we can apply the part (B) of the above corollaries but not the part (A) of them.
Example 5 Let $X=\{0,1\}$ and let p be the partial metric on X given by $p(x,y)=max\{x,y\}$ for all $x,y\in X$. Obviously ${d}_{p}$ is a complete ${T}_{0}$qpm on X. Let $T:X\to {Cl}_{{({d}_{p})}^{s}}(X)$ such that $T0=X$ and $T1=0$. Since ${({d}_{p})}^{s}$ is the discrete metric on X, it follows that T is pl.s.c. on $(X,{d}_{p})$.
Now we show that we can apply Corollary 6(B), and hence Corollaries 5(B), 4(B), and 3(B).
Let $r=1/2$, $x,y\in X$ and $u\in Tx$. In the cases where for $u=0$ we can choose $v=0\in Ty$, the conclusion is obvious. Therefore we only consider the following two cases.
Case 2. $x=0$, $y=1$ and $u=1$. Then $v=0$, and as in Case 1, $p(u,v)+p(v,v)=r(p(x,u)+p(u,u))$.
Declarations
Acknowledgements
The second and third named authors thank the supports of the Universitat Politècnica de València, grant PAID0612SP20120471, and the Ministry of Economy and Competitiveness of Spain, grant MTM201237894C0201.
Authors’ Affiliations
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