# Common Fixed Points in a Partially Ordered Partial Metric Space

Authors: Daniela Paesano; Pasquale Vetro
Journal: International Journal of Analysis
 (2013)

---

## Abstract

In the first part of this paper, we prove some generalized versions of the result of Matthews in (Matthews, 1994) using different types of conditions in partially ordered partial metric spaces for dominated self-mappings or in partial metric spaces for self-mappings. In the second part, using our results, we deduce a characterization of partial metric 0-completeness in terms of fixed point theory. This result extends the Subrahmanyam characterization of metric completeness.


---

## Body

## 1. Introduction

In the mathematical field of domain theory, attempts were made in order to equip semantics domain with a notion of distance. In particular, Matthews [1] introduced the notion of a partial metric space as a part of the study of denotational semantics of data for networks, showing that the contraction mapping principle can be generalized to the partial metric context for applications in program verification. Moreover, the existence of several connections between partial metrics and topological aspects of domain theory has been lately pointed by other authors as O'Neill [2], Bukatin and Scott [3], Bukatin and Shorina [4], Romaguera and Schellekens [5], and others (see also [6–14] and the references therein).After the result of Matthews [1], the interest for fixed point theory developments in partial metric spaces has been constantly growing, and many authors presented significant contributions in the directions of establishing partial metric versions of well-known fixed point theorems for the existence of fixed points, common fixed points, and coupled fixed points in classical metric spaces (see e.g., [15, 16]). Obviously, we cannot cite all these papers but we give only a partial list [17–49].Recently, Romaguera [50] proved that a partial metric space (X,d) is 0-complete if and only if every ps-Caristi mapping on X has a fixed point. In particular, the result of Romaguera extended Kirk’s [51] characterization of metric completeness to a kind of complete partial metric spaces. Successively, Karapinar in [36] extended the result of Caristi and Kirk [52] to partial metric spaces.In the first part of this paper, following this research direction, we prove some generalized versions of the result of Matthews by using different types of conditions in ordered partial metric spaces for dominated self-mappings or in partial metric spaces for self-mappings. The notion of dominated mapping of economics, finance, trade, and industry is also applied to approximate the unique solution of nonlinear functional equations. In the second part, using the results obtained in the first part, we deduce a characterization of partial metric 0-completeness in terms of fixed point theory. This result extends the Subrahmanyam [53] characterization of metric completeness. For other characterizations of metric completeness in terms of fixed point theory, the reader can see, for example, [54, 55] and for partial metric completeness, [41].


## 2. Preliminaries

First, we recall some definitions and some properties of partial metric spaces that can be found in [1, 2, 40, 48, 50]. A partial metric on a nonempty set X is a function p:X×X→[0,+∞) such that for all x,y,z∈X(p1)x=y⇔p(x,x)=p(x,y)=p(y,y),(p2)p(x,x)≤p(x,y),(p3)p(x,y)=p(y,x),(p4)p(x,y)≤p(x,z)+p(z,y)-p(z,z).A partial metric space is a pair(X,p) such that X is a nonempty set and p is a partial metric on X. It is clear that if p(x,y)=0, then from (p1) and (p2), it follows that x=y. But if x=y, p(x,y) may not be 0. A basic example of a partial metric space is the pair ([0,+∞),p), where p(x,y)=max{x,y} for all x,y∈[0,+∞). Other examples of partial metric spaces which are interesting from a computational point of view can be found in [1].Each partial metricp on X generates a T0 topology τp on X which has as a base the family of open p-balls {Bp(x,ɛ):x∈X,ɛ>0}, where
(1)Bp(x,ɛ)={y∈X:p(x,y)<p(x,x)+ɛ}
for all x∈X and ɛ>0.Ifp is a partial metric on X, then the function ps:X×X→[0,+∞) given by
(2)ps(x,y)=2p(x,y)-p(x,x)-p(y,y)
is a metric on X.Let(X,p) be a partial metric space. A sequence {xn} in (X,p) converges to a point x∈X if and only if p(x,x)=limn→+∞p(x,xn).A sequence{xn} in (X,p) is called a Cauchy sequence if there exists (and is finite) limn,m→+∞p(xn,xm).A partial metric space(X,p) is said to be complete if every Cauchy sequence {xn} in X converges, with respect to τp, to a point x∈X such that p(x,x)=limn,m→+∞p(xn,xm).A sequence{xn} in (X,p) is called 0-Cauchy if limn,m→+∞p(xn,xm)=0. We say that (X,p) is 0-complete if every 0-Cauchy sequence in X converges, with respect to τp, to a point x∈X such that p(x,x)=0.On the other hand, the partial metric space(ℚ∩[0,+∞),p), where ℚ denotes the set of rational numbers and the partial metric p is given by p(x,y)=max{x,y}, provides an example of a 0-complete partial metric space which is not complete.It is easy to see that every closed subset of a complete partial metric space is complete.Lemma 1 (see [1, 40]).
Let(X,p) be a partial metric space. Then(a)
{xn} is a Cauchy sequence in (X,p) if and only if it is a Cauchy sequence in the metric space (X,ps).(b)
A partial metric space(X,p) is complete if and only if the metric space (X,ps) is complete. Furthermore, limn→+∞ps(xn,x)=0 if and only if
(3)p(x,x)=limn→+∞p(xn,x)=limn,m→+∞p(xn,xm).The following lemma is obvious.Lemma 2.
Let(X,p) be a partial metric space and {xn}⊂X. If xn→x∈X and p(x,x)=0, then limn→+∞p(xn,z)=p(x,z) for all z∈X.Definep(x,A)=inf{p(x,a):a∈A}. Then a∈A¯⇔p(a,A)=p(a,a), where A¯ denotes the closure of A (for details see [22, Lemma 1]). From
(4)ps(x,a)=2p(x,a)-p(x,x)-p(a,a)≤2p(x,a)
for every a∈A, we deduce that ps(x,A)≤2p(x,A).LetX be a nonempty set and T,f:X→X. The mappings T,f are said to be weakly compatible if they commute at their coincidence points (i.e., Tfx=fTx whenever Tx=fx). A point y∈X is called a point of coincidence of T and f if there exists a point x∈Xsuch that y=Tx=fx.Lemma 3 (see [56]).
LetX be a nonempty set and the mappings T,f:X→X have a unique pointof coincidence vinX.If  T and f are weakly compatible, then T and f have a unique common fixed point.LetX be a nonempty set. If (X,p) is a partial metric space and (X,⪯) is a partially ordered set, then (X,p,⪯) is called a partially ordered partial metric space. x,y∈X are called comparable if x⪯y or y⪯x holds. Let (X,⪯) be a partially ordered set and T,f:X→X two mappings. T is called an f-dominated mapping if Tx⪯fx for every x∈X.


## 3. Main Results

Let(X,p) be a partial metric space and T,f:X→X be such that TX⊂fX. For every x0∈X we consider the sequence {xn}⊂X defined by fxn=Txn-1 for all n∈ℕ and we say that {Txn} is a T-f-sequence of the initial point x0 (see [57]).Denote withΨ the family of non-decreasing functions ψ:[0,+∞)→[0,+∞) such that ψ(t)>0 and limn→+∞ψn(t)=0 for each t>0, where ψn is the nth iterate of ψ.Lemma 4.
For every functionψ∈Ψ, the following holds, if for each t>0, limn→+∞ψn(t)=0 then ψ(t)<t.The following theorem is one of our main results, and it ensures the existence of a common fixed point for two self-mappings in the setting of partially ordered partial metric spaces.Theorem 5.
Let(X,p,⪯) be a partially ordered partial metric space and T,f:X→X two mappings such that TX⊂fX. Assume that there exists ψ∈Ψ such that
(5)p(Tx,Ty)≤max{ψ(p(fx,fy)),ψ(p(fx,Tx)),ψ(p(fy,Ty))}
for all x,y∈X with fx and fy comparable. If the following conditions hold:  (i)
T is a f-dominated mapping,(ii)
eitherTX or fX is a 0-complete subspace of X,(iii)
for a non-increasing sequence{fxn}⊂X converging to fu∈X, we have fu⪯fxn for all n∈ℕ and ffu⪯fu,then T and f have a point of coincidence. Moreover, if   T and f are weakly compatible, then T and f have a common fixed point.Proof.
Letx0∈X be fixed and {Txn} be a T-f-sequence of the initial point x0. As fxn+1=Txn≤fxn for all n∈ℕ, then the sequence {Txn} is non-increasing.
IfTxn=Txn-1=fxn for some n∈ℕ, then y=Txn=fxn is a point of coincidence of T and f. Suppose that Txn≠Txn-1 for all n∈ℕ. Since fxn and fxn+1 are comparable for all n∈ℕ, we have
(6)p(Txn+1,Txn)≤max{ψ(p(fxn+1,fxn)),ψ(p(fxn+1,Txn+1)),ψ(p(fxn,Txn))}=max{ψ(p(Txn,Txn-1)),ψ(p(Txn+1,Txn))}.
Ifmax{ψ(p(Txn,Txn-1)),ψ(p(Txn+1,Txn))}=ψ(p(Txn+1,Txn)) from
(7)p(Txn+1,Txn)≤ψ(p(Txn+1,Txn))<p(Txn+1,Txn),
we obtain a contradiction and so
(8)max{ψ(p(Txn,Txn-1)),ψ(p(Txn+1,Txn))}=ψ(p(Txn,Txn-1)).
Then, we have
(9)p(Txn+1,Txn)≤ψn(p(Tx1,Tx0)),∀n∈ℕ
and hence
(10)limn→+∞p(Txn+1,Txn)=0.
Fix ɛ>0 and we choose n(ɛ)∈ℕ such that
(11)p(Txm,Txm+1)<ɛ-ψ(ɛ)
for all m≥n(ɛ). Let m≥n(ɛ) and we show that
(12)p(Txm,Txn+1)<ɛ,∀n≥m.
Clearly, (12) is true for n=m. Suppose that (12) holds for some n≥m, as fxi and fxj are comparable for all i,j∈ℕ, then
(13)p(Txm,Txn+2)≤p(Txm,Txm+1)+p(Txm+1,Txn+2)≤p(Txm,Txm+1)+max{ψ(p(Txm,Txn+1)),ψ(p(Txm,Txm+1)),ψ(p(Txn+1,Txn+2))}<ɛ-ψ(ɛ)+max{ψ(ɛ),ψ(ɛ-ψ(ɛ))}=ɛ.
This implies that (12) holds for n+1 and by induction, it holds for all n≥m. From (12), we deduce that there exists
(14)limn,m→+∞p(Txn,Txm)=0
and hence {Txn} is a 0-Cauchy sequence.
Suppose thatTX is a 0-complete subspace of (X,p), then there exists y∈TX⊂fX such that
(15)p(y,y)=limn→+∞p(Txn,y)=limn,m→+∞p(Txn,Txm)=0.
This holds also if fX is 0-complete with y∈fX.
Letu∈X be such that y=fu. We show that y is a point of coincidence of T and f. If not, we have p(fu,Tu)>0. This implies that there exists n0∈ℕ such that
(16)max{ψ(p(Txn-1,fu)),ψ(p(Txn-1,Txn)),ψ(p(fu,Tu))}=ψ(p(fu,Tu)),
for every n≥n0. By condition (iii), fxn and fu are comparable for every n∈ℕ and hence, by condition (5) with x=xn and y=u, we deduce that
(17)p(Txn,Tu)≤max{ψ(p(fxn,fu)),ψ(p(fxn,Txn)),ψ(p(fu,Tu))}=max{ψ(p(Txn-1,fu)),ψ(p(Txn-1,Txn)),ψ(p(fu,Tu))}=ψ(p(fu,Tu)),
for every n≥n0. Letting n→+∞ in the previous inequality and using Lemma 2, we obtain
(18)p(fu,Tu)≤ψ(p(fu,Tu))<p(fu,Tu),
which implies p(fu,Tu)=0, that is, Tu=fu. Thus, we have shown that y=fu=Tu is a point of coincidence of T and f. If T and f are weakly compatible, then Ty=Tfu=fTu=fy. By condition (iii), fy=ffu⪯fu, that is, fy and fu are comparable. Using the contractive condition (5), we get
(19)p(Tu,Ty)≤max{ψ(p(fu,fy)),ψ(p(fu,Tu)),ψ(p(fy,Ty))}=ψ(p(fu,fy))<p(fu,fy),
which implies Ty=Tu=y and hence y is a common fixed point of T and f.We shall give a sufficient condition for the uniqueness of the common fixed point in Theorem5.Theorem 6.
Let all the conditions of Theorem5 be satisfied. If the following condition holds:(iv)
for allx,y∈fX there exists v0∈X such that fv0⪯x,fv0⪯y and limn→+∞p(Tvn-1,Tvn)=0, where {Tvn} is the T-f-sequence of the initial point v0, then T and f have a unique common fixed point.Proof.
Letz,w be two common fixed points of T and f with z≠w. If z and w are comparable, then using the contractive condition (5), we deduce that z=w. If z and w are not comparable, then there exists v0∈X such that fv0⪯z=fz, fv0⪯w=fw. As T is a f-dominated mapping, we get that
(20)fv1=Tv0⪯fv0⪯z=fz.
To continue, we obtain
(21)fvn+1=Tvn⪯fvn⪯z=fz∀n∈ℕ
and hence fvn and fz are comparable.
Using the contractive condition (5) with x=vn and y=z, we get
(22)p(Tvn,Tz)≤max{ψ(p(fvn,fz)),ψ(p(fvn,Tvn)),ψ(p(fz,Tz))}=max{ψ(p(Tvn-1,Tz)),ψ(p(Tvn-1,Tvn)),ψ(p(Tz,Tz))}
for all n∈ℕ. Since, the contractive condition (5) ensures that p(Tz,Tz)=0, we have
(23)p(Tvn,Tz)≤max{ψ(p(Tvn-1,Tz)),ψ(p(Tvn-1,Tvn))}.
Now, by condition (iv), limn→+∞p(Tvn-1,Tvn)=0 and hence for n sufficiently large, we have
(24)p(Tvn,Tz)≤ψ(p(Tvn-1,Tz)).
Without loss of generality, assuming that (24) holds for all ∈ℕ, it follows that
(25)p(Tvn,Tz)≤ψn(p(Tv0,Tz)).
Now, letting n→+∞ in (25), we obtain
(26)limn→+∞p(Tvn,Tz)=0.
With similar arguments, we deduce thatlimn→+∞p(Tw,Tvn)=0. Hence
(27)0<p(w,z)=p(Tw,Tz)≤p(Tw,Tvn)+p(Tvn,Tz)→0
as n→+∞, which is a contradiction. Thus T and f have a unique common fixed point.As a consequence of Theorem5, we state the following result.Theorem 7.
Let(X,p) be a partial metric space and T,f:X→X two mappings such that TX⊂fX. Assume that there exists ψ∈Ψ such that
(28)p(Tx,Ty)≤max{ψ(p(fx,fy)),ψ(p(fx,Tx)),ψ(p(fy,Ty))}
for all x,y∈X. If TX or fX is a 0-complete subspace of X, then T and f have a unique point of coincidence. Moreover, if T and f are weakly compatible, then T and f have a unique common fixed point.Proof.
Proceeding as in the proof of Theorem5, we get that T and f have a unique point of coincidence and, by Lemma 3, T and f have a unique common fixed point.From Theorem7, we can deduce the following corollaries.Corollary 8 (Banach type).
Let(X,p) be a partial metric space and T,f:X→X two mappings such that TX⊂fX. Assume that
(29)p(Tx,Ty)≤kp(fx,fy)
for all x,y∈X, where 0≤k<1. If TX or fX is a 0-complete subspace of X, then T and f have a unique point of coincidence. Moreover, if T and f are weakly compatible, then T and f have a unique common fixed point.Corollary 9 (Bianchini type).
Let(X,p) be a partial metric space and let T,f:X→X be two mappings such that TX⊂fX. Assume that
(30)p(Tx,Ty)≤kmax{p(fx,Tx),p(fy,Ty)}
for all x,y∈X, where 0≤k<1. If TX or fX is a 0-complete subspace of X, then T and f have a unique point of coincidence. Moreover, if T and f are weakly compatible, then T and f have a unique common fixed point.Corollary 10 (Reich type [58]).
Let(X,p) be a partial metric space and let T,f:X→X be two mappings such that TX⊂fX. Assume that
(31)p(Tx,Ty)≤ap(fx,fy)+bp(fx,Tx)+cp(fy,Ty)
for all x,y∈X, where a,b,c≥0 and a+b+c<1. If   TX or fX is a 0-complete subspace of X, then T and f have a unique point of coincidence. Moreover, if T and f are weakly compatible, then T and f have a unique common fixed point.The following example shows that there exist mappings that satisfy the contractive condition (28), but are not quasi-contractions [59].Example 11.
Consider the setX={1,2,3} and the function p:X×X→[0,+∞) given by p(1,2)=p(2,3)=1,p(1,3)=3/2,p(1,1)=p(3,3)=1/2,p(2,2)=0, and p(x,y)=p(y,x). Obviously, p is a partial metric on X, but it is not a metric (since p(x,x)≠0 for x=1 and x=3). Clearly, (X,p) is a 0-complete partial metric space. Let T,f:X→X be defined by T1=2,T2=2,T3=1 and fx=x for every x∈X. Take ψ(t)=(2/3)t for every t≥0.
First, we will check thatT and f satisfy the contractive condition (28). If x,y∈{1,2}, then p(Tx,Ty)=p(2,2)=0 and (28) trivially holds. Let, for example, y=3, then we have the following three cases:
(32)p(T1,T3)=p(2,1)=1≤23·32=max{ψ(p(1,3)),ψ(p(1,2)),ψ(p(3,1))};p(T2,T3)=p(2,1)=1≤23·32=max{ψ(p(2,3)),ψ(p(2,2)),ψ(p(3,1))};p(T3,T3)=p(1,1)=12<23·32=max{ψ(p(3,3)),ψ(p(3,1))}.
Thus, all the conditions of Theorem7 are satisfied and the existence of a common fixed point of T and f (which is 2) follows. The same conclusion cannot be obtained by the main results from [59]. Indeed, using ps(a,b)=2p(a,b)-p(a,a)-p(b,b), and then taking ps instead of p, x=1,y=3 in (5), we obtain
(33)L=ps(T1,T3)=ps(2,1)=32,R=max{ψ(ps(1,3)),ψ(ps(1,T1)),ψ(ps(3,T3)),ψ(ps(1,T3)),ψ(ps(3,T1))}=23max{ps(1,3),ps(1,2),ps(3,1),ps(1,1),ps(3,2)}=23max{2,32,2,0,32}=43.
Since L>R, the conclusion follows.The following example shows that there exist mappings that satisfy the contractive condition (5), but do not satisfy the contractive condition (28).Example 12.
LetX=[0,2] be endowed with the partial metric
(34)p(x,y)={|x-y|ifx,y∈[0,1],max{x,y}if{x,y}∩(1,2]≠∅.
Clearly, (X,p) is a 0-complete partial metric space. Let T,f:X→X be defined by
(35)Tx={xifx∈[0,1],1+x2ifx∈(1,2]
and fx=x for each x∈X. As T and f have many common fixed points (each x∈[0,1] is a common fixed point), then it is immediate to show that T and f do not satisfy the contractive condition (28).
If(X,p) is ordered by
(36)x⪯y⟺(x=y)or(x,y∈(1,2],x≤y),
then T and f satisfy the contractive condition (5) where ψ:[0,+∞)→[0,+∞) is defined by
(37)ψ(t)={t2ift∈[0,1],1+t2ift∈(1,2].
Using Theorem 5, we deduce that T and f have a common fixed point.


## 4. Completeness in Partial Metric Spaces and Fixed Points

In this section, we characterize those partial metric spaces for which every Bianchini mapping has a fixed point in the style of Subrahmanyam characterization of metric completeness. This will be done by means of the notion of 0-completeness which was introduced by Romaguera in [50].Let(X,p) be a partial metric space and T:X→X be a mapping. We recall that T is a Bianchini [60] mapping if
(38)p(Tx,Ty)≤kmax{p(x,Tx),p(y,Ty)}
for all x,y∈X, where 0≤k<1.Theorem 13.
Let(X,p) be a partial metric space. If every mapping T:X→X satisfying the following conditions: (i)
p(Tx,Ty)≤λmax{p(x,Tx),p(y,Ty)} for all x,y∈X, for a fixed λ>0,(ii)
TX is countable has a fixed point, then (X,p) is 0-complete.Proof.
Suppose that there is a 0-Cauchy sequence{xn} of distinct points in (X,p) which is not convergent in (X,ps). We put A={xn} and we note that for every x∉A, we have ps(x,A)>0.
Now,0<ps(x,A)≤2p(x,A) implies that p(x,A)>0. Since {xn} is a 0-Cauchy sequence in (X,p), there exists a least positive integer N(x) such that
(39)p(xm,xn)<λp(x,A)≤λp(x,xl),l∈ℕ,∀m,n≥N(x).
In particular(40)p(xm,xN(x))<λp(x,A∖{x})≤λp(x,xN(x)),∀m≥N(x).
For fixedn∈ℕ, since ps(xn,A∖{xn})>0 and so p(xn,A∖{xn})>0, there is n′(n)>n such that
(41)p(xm,xn′)<λp(xn,A∖{xn})≤λp(xn,xn′),∀m≥n′.
Now, letT:X→X be defined by
(42)Tx={xN(x)ifx∉A,xn′ifx(=xn)∈A.
From the definition of T, we deduce that T satisfies the condition (ii). On the other hand, T satisfies also the condition (i). In fact, (i) is verified by assuming Tx=xn and Ty=xm, and noting that
(43)p(xm,xn)<{λp(y,A∖{y})≤λp(y,Ty)ifn≥m,λp(x,A∖{x})≤λp(x,Tx)ifn<m.
It is clear that T has not fixed points since xn′≠xn, n=1,2,…. Thus, the assumption that there is a 0-Cauchy sequence {xn} which is not convergent in (X,ps) leads to a contradiction to Theorem 13 and thereby establishes the same.If in Theorem13 we choose λ∈[0,1), by Corollary 9, we obtain the following characterization of 0-completeness for partial metric spaces.Theorem 14.
A partial metric space(X,p) is 0-complete if and only if every mapping T:X→X satisfying the following conditions: (i)
p(Tx,Ty)≤λmax{p(x,Tx),p(y,Ty)} for all x,y∈X, for a fixed λ∈[0,1),(ii)
TX is countable has a fixed point.In Theorem14, the class of mappings satisfying (i) and (ii) can be replaced by the class of mappings satisfying (ii) and the following condition:(i)
p(Tx,Ty)≤λ[p(x,Tx)+p(y,Ty)]forallx,y∈X,forafixedλ∈[0,1/2).



---

*Source: 428561-2013-01-13.xml*