# Coupled Coincidence Points of Mappings in Ordered Partial Metric Spaces

Authors: Zorana Golubović; Zoran Kadelburg; Stojan Radenović
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

New coupled coincidence point and coupled fixed point results in
ordered partial metric spaces under the contractive conditions of Geraghty,
Rakotch, and Branciari types are obtained. Examples show that these results
are distinct from the known ones.


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## Body

## 1. Introduction

In recent years many authors have worked on domain theory 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 dataflow networks. He showed that the Banach contraction mapping theorem can be generalized to the partial metric context for applications in program verification. Subsequently, several authors (see, e.g., [2–9]) have proved fixed point theorems in partial metric spaces.Many generalizations of Banach's contractive condition have been introduced in order to obtain more general fixed point results in metric spaces and their generalizations. We mention here conditions introduced by Geraghty [10], Rakotch [11], and Branciari [12].The notion of a coupled fixed point was introduced and studied by Bhaskar and Lakshmikantham in [13]. In subsequent papers several authors proved various coupled and common coupled fixed point theorems in partially ordered metric spaces (e.g., [14–17]). These results were applied for investigation of solutions of differential and integral equations. In a recent paper [18], Berinde presented a method of reducing coupled fixed point results in ordered metric spaces to the respective results for mappings with one variable.In this paper, we further develop the method of Berinde and obtain new coupled coincidence and coupled fixed point results in ordered partial metric spaces, under the contractive conditions of Geraghty, Rakotch, and Branciari types. Examples show that these results are distinct from the known ones. In particular, they show that using the order and/or the partial metric enables conclusions which cannot be obtained in the classical case.


## 2. Notation and Preliminary Results

### 2.1. Partial Metric Spaces

The following definitions and details can be seen in [1–9].Definition 2.1.
A partial metric on a nonempty setX is a function p:X×X→ℝ+ 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). The pair (X,p) is called a partial metric space.It is clear that ifp(x,y)=0, then from (P1) and (P2) x=y. But if x=y, p(x,y) may not be 0.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 Bp(x,ɛ)={y∈X:p(x,y)<p(x,x)+ɛ} for all x∈X and ɛ>0. A sequence {xn} in (X,p) converges to a point x∈X, with respect to τp, if limn→∞p(x,xn)=p(x,x). This will be denoted as xn→x, n→∞, or limn→∞xn=x.Ifp is a partial metric on X, then the function ps:X×X→ℝ+ given by(2.1)ps(x,y)=2p(x,y)-p(x,x)-p(y,y)
is metric on X. Furthermore, limn→∞ps(xn,x)=0 if and only if(2.2)p(x,x)=lim⁡n→∞p(xn,x)=lim⁡n,m→∞p(xn,xm).A basic example of a partial metric space is the pair(ℝ+,p), where p(x,y)=max{x,y} for all x,y∈ℝ+. The corresponding metric is(2.3)ps(x,y)=2max⁡{x,y}-x-y=|x-y|.
Other examples of partial metric spaces which are interesting from a computational point of view may be found in [1, 19].Remark 2.2.
Clearly, a limit of a sequence in a partial metric space need not be unique. Moreover, the functionp(·,·) need not be continuous in the sense that xn→x and yn→y implies p(xn,yn)→p(x,y).Definition 2.3.
Let(X,p) be a partial metric space. Then,(1)
a sequence{xn} in (X,p) is called a Cauchy sequence if limn,m→∞p(xn,xm) exists (and is finite);(2)
the 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).Lemma 2.4.
Let(X,p) be a partial metric space.(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)
The space(X,p) is complete if and only if the metric space (X,ps) is complete.Definition 2.5.
LetX be a nonempty set. Then (X,⪯,p) is called an ordered partial metric space if (i) (X,⪯) is a partially ordered set, and (ii) (X,p) is a partial metric space.
We will say that the space(X,⪯,p) satisfies the ordered-regular condition (abr. (ORC)) if the following holds: if {xn} is a nondecreasing sequence in X with respect to ⪯ such that xn→x∈X as n→∞, then xn⪯x for all n∈ℕ.Definition 2.6 (see [13, 14]).
Let(X,⪯) be a partially ordered set, F:X×X→X, and g:X→X.(1)
F is said to have g-mixed monotone property if the following two conditions are satisfied:
(2.4)(∀x1,x2,y∈X)gx1⪯gx2⟹F(x1,y)⪯F(x2,y),(∀x,y1,y2∈X)gy1⪯gy2⟹F(x,y1)⪰F(x,y2).
If g=iX (the identity map), we say that F has the mixed monotone property.(2)
A point(x,y)∈X×X is said to be a coupled coincidence point of F and g if F(x,y)=gx and F(y,x)=gy and their common coupled fixed point if F(x,y)=gx=x and F(y,x)=gy=y.Definition 2.7 (see [20]).
Let(X,d) be a metric space, and let F:X×X→X and g:X→X. The pair (F,g) is said to be compatible if
(2.5)lim⁡n→∞d(gF(xn,yn),F(gxn,gyn))=0,lim⁡n→∞d(gF(yn,xn),F(gyn,gxn))=0
whenever {xn} and {yn} are sequences in X such that limn→∞F(xn,yn)=limn→∞gxn=x and limn→∞F(yn,xn)=limn→∞gyn=y for some x,y∈X.


### 2.2. Some Auxiliary Results

Lemma 2.8.
(i) Let(X,⪯,p) be an ordered partial metric space. If relation ⊑ is defined on X2 by
(2.6)Y⊑V⟺x⪯u∧y⪰v,Y=(x,y),V=(u,v)∈X2,
and P:X2×X2→ℝ+ is given by
(2.7)P(Y,V)=p(x,u)+p(y,v),Y=(x,y),V=(u,v)∈X2,
then (X2,⊑,P) is an ordered partial metric space. The space (X2,⊑,P) is complete iff (X,⪯,p) is complete.
(ii) IfF:X×X→X and g:X→X, and F has the g-mixed monotone property, then the mapping TF:X2→X2 given by
(2.8)TFY=(F(x,y),F(y,x)),Y=(x,y)∈X2
is Tg-nondecreasing with respect to ⊑, that is,
(2.9)TgY⊑TgV⟹TFY⊑TFV,
where TgY=Tg(x,y)=(gx,gy).
(iii) Ifg is continuous in (X,p) (i.e., with respect to τp), then Tg is continuous in (X2,P) (i.e., with respect to τP). If F is continuous from (X2,P) to (X,p) (i.e., xn→x and yn→y imply F(xn,yn)→F(x,y)), then TF is continuous in (X2,P).Proof.
(i) Relation⊑ is obviously a partial order on X2. To prove that P is a partial metric on X2, only conditions (P1) and (P4) are nontrivial.
(P1) If Y=V∈X2, then obviously p(Y,Y)=p(Y,V)=p(V,V) holds. Conversely, let p(Y,Y)=p(Y,V)=p(V,V), that is,
(2.10)p(x,x)+p(y,y)=p(x,u)+p(y,v)=p(u,u)+p(v,v).
We know by (P2) that p(x,x)≤p(x,u) and p(y,y)≤p(y,v). Adding up, we obtain that p(x,x)+p(y,y)≤p(x,u)+p(y,v), and since in fact equality holds, we conclude that p(x,x)=p(x,u) and p(y,y)=p(y,v). Similarly, we get that p(u,u)=p(x,u) and p(v,v)=p(y,v). Hence,
(2.11)p(x,x)=p(x,u)=p(u,u),p(y,y)=p(y,v)=p(v,v),
and applying property (P1) of partial metric p, we get that x=u and y=v, that is, Y=V.
(P4) Let Y=(x,y),V=(u,v),Z=(w,z)∈X2. Then
(2.12)P(Y,V)=p(x,u)+p(y,v)≤(p(x,w)+p(w,u)-p(w,w))+(p(y,z)+p(z,v)-p(z,z))=p(x,w)+p(y,z)+p(w,u)+p(z,v)-[p(w,w)+p(z,z)]=P(Y,Z)+P(Z,V)-P(Z,Z).
(ii, iii) The proofs of these assertions are straightforward.Remark 2.9.
Letps be the metric associated with the partial metric p as in (2.1). It is easy to see that, with notation as in the previous lemma,
(2.13)Ps(Y,V)=ps(x,u)+ps(y,v),Y=(x,y),V=(u,v)∈X2
is the associated metric to the partial metric P on X2. We note, however, that when we speak about continuity of mappings, we always assume continuity in the sense of the partial metric p, that is, in the sense of the respective topology τp. This should not be confused with the approach given by O'Neill in [3] where both p- and ps-continuity were assumed.It is easy to see that (using notation as in the previous lemma), the mappingsF and g are ps-compatible (in the sense of Definition 2.7) if and only if the mappings TF and Tg are Ps-compatible in the usual sense (i.e., limn→∞d(TFTgYn,TgTFYn)=0, whenever {Yn} is a sequence in X2 such that limn→∞TFYn=limn→∞TgYn).Assertions similar to the following lemma (see, e.g., [21]) were used (and proved) in the course of proofs of several fixed point results in various papers.Lemma 2.10.
Let(X,d) be a metric space, and let {xn} be a sequence in X such that {d(xn+1,xn)} is decreasing and
(2.14)lim⁡n→∞d(xn+1,xn)=0.
If {x2n} is not a Cauchy sequence, then there exist ɛ>0 and two sequences {mk} and {nk} of positive integers such that the following four sequences tend to ɛ when k→∞:
(2.15)d(x2mk,x2nk),d(x2mk,x2nk+1),d(x2mk-1,x2nk),d(x2mk-1,x2nk+1).As a corollary (applying Lemma2.10 to the associated metric ps of a partial metric p, and using Lemma 2.4) we obtain the following.Lemma 2.11.
Let(X,p) be a partial metric space, and let {xn} be a sequence in X such that {p(xn+1,xn)} is decreasing and
(2.16)lim⁡n→∞p(xn+1,xn)=0.
If {x2n} is not a Cauchy sequence in (X,p), then there exist ɛ>0 and two sequences {mk} and {nk} of positive integers such that the following four sequences tend to ɛ when k→∞:
(2.17)p(x2mk,x2nk),p(x2mk,x2nk+1),p(x2mk-1,x2nk),p(x2mk-1,x2nk+1).


## 2.1. Partial Metric Spaces

The following definitions and details can be seen in [1–9].Definition 2.1.
A partial metric on a nonempty setX is a function p:X×X→ℝ+ 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). The pair (X,p) is called a partial metric space.It is clear that ifp(x,y)=0, then from (P1) and (P2) x=y. But if x=y, p(x,y) may not be 0.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 Bp(x,ɛ)={y∈X:p(x,y)<p(x,x)+ɛ} for all x∈X and ɛ>0. A sequence {xn} in (X,p) converges to a point x∈X, with respect to τp, if limn→∞p(x,xn)=p(x,x). This will be denoted as xn→x, n→∞, or limn→∞xn=x.Ifp is a partial metric on X, then the function ps:X×X→ℝ+ given by(2.1)ps(x,y)=2p(x,y)-p(x,x)-p(y,y)
is metric on X. Furthermore, limn→∞ps(xn,x)=0 if and only if(2.2)p(x,x)=lim⁡n→∞p(xn,x)=lim⁡n,m→∞p(xn,xm).A basic example of a partial metric space is the pair(ℝ+,p), where p(x,y)=max{x,y} for all x,y∈ℝ+. The corresponding metric is(2.3)ps(x,y)=2max⁡{x,y}-x-y=|x-y|.
Other examples of partial metric spaces which are interesting from a computational point of view may be found in [1, 19].Remark 2.2.
Clearly, a limit of a sequence in a partial metric space need not be unique. Moreover, the functionp(·,·) need not be continuous in the sense that xn→x and yn→y implies p(xn,yn)→p(x,y).Definition 2.3.
Let(X,p) be a partial metric space. Then,(1)
a sequence{xn} in (X,p) is called a Cauchy sequence if limn,m→∞p(xn,xm) exists (and is finite);(2)
the 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).Lemma 2.4.
Let(X,p) be a partial metric space.(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)
The space(X,p) is complete if and only if the metric space (X,ps) is complete.Definition 2.5.
LetX be a nonempty set. Then (X,⪯,p) is called an ordered partial metric space if (i) (X,⪯) is a partially ordered set, and (ii) (X,p) is a partial metric space.
We will say that the space(X,⪯,p) satisfies the ordered-regular condition (abr. (ORC)) if the following holds: if {xn} is a nondecreasing sequence in X with respect to ⪯ such that xn→x∈X as n→∞, then xn⪯x for all n∈ℕ.Definition 2.6 (see [13, 14]).
Let(X,⪯) be a partially ordered set, F:X×X→X, and g:X→X.(1)
F is said to have g-mixed monotone property if the following two conditions are satisfied:
(2.4)(∀x1,x2,y∈X)gx1⪯gx2⟹F(x1,y)⪯F(x2,y),(∀x,y1,y2∈X)gy1⪯gy2⟹F(x,y1)⪰F(x,y2).
If g=iX (the identity map), we say that F has the mixed monotone property.(2)
A point(x,y)∈X×X is said to be a coupled coincidence point of F and g if F(x,y)=gx and F(y,x)=gy and their common coupled fixed point if F(x,y)=gx=x and F(y,x)=gy=y.Definition 2.7 (see [20]).
Let(X,d) be a metric space, and let F:X×X→X and g:X→X. The pair (F,g) is said to be compatible if
(2.5)lim⁡n→∞d(gF(xn,yn),F(gxn,gyn))=0,lim⁡n→∞d(gF(yn,xn),F(gyn,gxn))=0
whenever {xn} and {yn} are sequences in X such that limn→∞F(xn,yn)=limn→∞gxn=x and limn→∞F(yn,xn)=limn→∞gyn=y for some x,y∈X.


## 2.2. Some Auxiliary Results

Lemma 2.8.
(i) Let(X,⪯,p) be an ordered partial metric space. If relation ⊑ is defined on X2 by
(2.6)Y⊑V⟺x⪯u∧y⪰v,Y=(x,y),V=(u,v)∈X2,
and P:X2×X2→ℝ+ is given by
(2.7)P(Y,V)=p(x,u)+p(y,v),Y=(x,y),V=(u,v)∈X2,
then (X2,⊑,P) is an ordered partial metric space. The space (X2,⊑,P) is complete iff (X,⪯,p) is complete.
(ii) IfF:X×X→X and g:X→X, and F has the g-mixed monotone property, then the mapping TF:X2→X2 given by
(2.8)TFY=(F(x,y),F(y,x)),Y=(x,y)∈X2
is Tg-nondecreasing with respect to ⊑, that is,
(2.9)TgY⊑TgV⟹TFY⊑TFV,
where TgY=Tg(x,y)=(gx,gy).
(iii) Ifg is continuous in (X,p) (i.e., with respect to τp), then Tg is continuous in (X2,P) (i.e., with respect to τP). If F is continuous from (X2,P) to (X,p) (i.e., xn→x and yn→y imply F(xn,yn)→F(x,y)), then TF is continuous in (X2,P).Proof.
(i) Relation⊑ is obviously a partial order on X2. To prove that P is a partial metric on X2, only conditions (P1) and (P4) are nontrivial.
(P1) If Y=V∈X2, then obviously p(Y,Y)=p(Y,V)=p(V,V) holds. Conversely, let p(Y,Y)=p(Y,V)=p(V,V), that is,
(2.10)p(x,x)+p(y,y)=p(x,u)+p(y,v)=p(u,u)+p(v,v).
We know by (P2) that p(x,x)≤p(x,u) and p(y,y)≤p(y,v). Adding up, we obtain that p(x,x)+p(y,y)≤p(x,u)+p(y,v), and since in fact equality holds, we conclude that p(x,x)=p(x,u) and p(y,y)=p(y,v). Similarly, we get that p(u,u)=p(x,u) and p(v,v)=p(y,v). Hence,
(2.11)p(x,x)=p(x,u)=p(u,u),p(y,y)=p(y,v)=p(v,v),
and applying property (P1) of partial metric p, we get that x=u and y=v, that is, Y=V.
(P4) Let Y=(x,y),V=(u,v),Z=(w,z)∈X2. Then
(2.12)P(Y,V)=p(x,u)+p(y,v)≤(p(x,w)+p(w,u)-p(w,w))+(p(y,z)+p(z,v)-p(z,z))=p(x,w)+p(y,z)+p(w,u)+p(z,v)-[p(w,w)+p(z,z)]=P(Y,Z)+P(Z,V)-P(Z,Z).
(ii, iii) The proofs of these assertions are straightforward.Remark 2.9.
Letps be the metric associated with the partial metric p as in (2.1). It is easy to see that, with notation as in the previous lemma,
(2.13)Ps(Y,V)=ps(x,u)+ps(y,v),Y=(x,y),V=(u,v)∈X2
is the associated metric to the partial metric P on X2. We note, however, that when we speak about continuity of mappings, we always assume continuity in the sense of the partial metric p, that is, in the sense of the respective topology τp. This should not be confused with the approach given by O'Neill in [3] where both p- and ps-continuity were assumed.It is easy to see that (using notation as in the previous lemma), the mappingsF and g are ps-compatible (in the sense of Definition 2.7) if and only if the mappings TF and Tg are Ps-compatible in the usual sense (i.e., limn→∞d(TFTgYn,TgTFYn)=0, whenever {Yn} is a sequence in X2 such that limn→∞TFYn=limn→∞TgYn).Assertions similar to the following lemma (see, e.g., [21]) were used (and proved) in the course of proofs of several fixed point results in various papers.Lemma 2.10.
Let(X,d) be a metric space, and let {xn} be a sequence in X such that {d(xn+1,xn)} is decreasing and
(2.14)lim⁡n→∞d(xn+1,xn)=0.
If {x2n} is not a Cauchy sequence, then there exist ɛ>0 and two sequences {mk} and {nk} of positive integers such that the following four sequences tend to ɛ when k→∞:
(2.15)d(x2mk,x2nk),d(x2mk,x2nk+1),d(x2mk-1,x2nk),d(x2mk-1,x2nk+1).As a corollary (applying Lemma2.10 to the associated metric ps of a partial metric p, and using Lemma 2.4) we obtain the following.Lemma 2.11.
Let(X,p) be a partial metric space, and let {xn} be a sequence in X such that {p(xn+1,xn)} is decreasing and
(2.16)lim⁡n→∞p(xn+1,xn)=0.
If {x2n} is not a Cauchy sequence in (X,p), then there exist ɛ>0 and two sequences {mk} and {nk} of positive integers such that the following four sequences tend to ɛ when k→∞:
(2.17)p(x2mk,x2nk),p(x2mk,x2nk+1),p(x2mk-1,x2nk),p(x2mk-1,x2nk+1).


## 3. Coupled Coincidence and Fixed Points under Geraghty-Type  Conditions

Let𝒢 denote the class of real functions γ:[0,+∞)→[0,1) satisfying the condition (3.1)γ(tn)⟶1impliestn⟶0.
An example of a function in 𝒢 may be given by γ(t)=e-2t for t>0 and γ(0)∈[0,1). In an attempt to generalize the Banach contraction principle, Geraghty proved in 1973 the following.Theorem 3.1 (see [10]).
Let(X,d) be a complete metric space, and let T:X→X be a self-map. Suppose that there exists γ∈𝒢 such that
(3.2)d(Tx,Ty)≤γ(d(x,y))d(x,y)
holds for all x,y∈X. Then T has a unique fixed point z∈X and for each x∈X the Picard sequence {Tnx} converges to z when n→∞.Subsequently, several authors proved such results, including the very recent paper of Ðukić et al. [22].We begin with the following auxiliary result.Lemma 3.2.
Let(X,⪯,p) be an ordered partial metric space which is complete. Let T,S:X→X be self-maps such that S is continuous, TX⊂SX, and one of these two subsets of X is closed. Suppose that T is S-nondecreasing (with respect to ⪯) and there exists x0∈X with Sx0⪯Tx0 or Tx0⪯Sx0. Assume also that there exists γ∈𝒢 such that
(3.3)p(Tx,Ty)≤γ(p(Sx,Sy))p(Sx,Sy)
holds for all x,y∈X such that Sx and Sy are comparable. Assume that either 1°: T is continuous and the pair (T,S) is ps- compatible or 2°: X satisfies (ORC). Then, T and S have a coincidence point in X.Proof.
The proof follows the lines of proof of [22, Theorems 3.1 and 3.5].
Takex0∈X with, say, Sx0⪯Tx0, and using that T is S-nondecreasing and that TX⊂SX form the sequence {xn} satisfying Txn=Sxn+1, n=0,1,2,…, and
(3.4)Sx0⪯Tx0=Sx1⪯Tx1=Sx2⪯⋯⪯Txn⪯Sxn+1⪯⋯.
Since Sxn-1 and Sxn are comparable, we can apply the contractive condition to obtain
(3.5)p(Sxn+1,Sxn)=p(Txn,Txn-1)≤γ(p(Sxn-1,Sxn))p(Sxn-1,Sxn)≤p(Sxn-1,Sxn).
Consider the following two cases:(1)
p(Sxn0+1,Sxn0)=0 for some n0∈ℕ;(2)
p(Sxn+1,Sxn)>0 for each n∈ℕ.Case 1.
Under this assumption, we get that(3.6)p(Sxn0+2,Sxn0+1)=p(Txn0+1,Txn0)≤γ(p(Sxn0+1,Sxn0))p(Sxn0+1,Sxn0)=γ(0)⋅0=0,
and it follows that p(Sxn0+2,Sxn0+1)=0. By induction, we obtain that p(Sxn+1,Sxn)=0 for all n≥n0 and so Sxn=Sxn0 for all n≥n0. Hence, {Sxn} is a Cauchy sequence, converging to Sxn0, and xn0 is a coincidence point of S and T.Case 2.
We will prove first that in this case the sequence{p(Sxn+1,Sxn)} is strictly decreasing and tends to 0 as n→∞.
For eachn∈ℕ we have that
(3.7)0<p(Sxn+2,Sxn+1)=p(Txn+1,Txn)≤γ(p(Sxn+1,Sxn))p(Sxn+1,Sxn)<p(Sxn+1,Sxn).
Hence, p(Sxn+1,Sxn) is strictly decreasing and bounded from below, thus converging to some q≥0. Suppose that q>0. Then, it follows from (3.7) that
(3.8)p(Sxn+2,Sxn+1)p(Sxn+1,Sxn)≤γ(p(Sxn+1,Sxn))<1,
wherefrom, passing to the limit when n→∞, we get that limn→∞γ(p(Sxn+1,Sxn))=1. Using property (3.1) of the function γ, we conclude that limn→∞p(Sxn+1,Sxn)=0, that is, q=0, a contradiction. Hence, limn→∞p(Sxn+1,Sxn)=0 is proved.
In order to prove that{Sxn} is a Cauchy sequence in (X,p), suppose the contrary. As was already proved, p(Sxn+1,Sxn)→0 as n→∞, and so, using (P2), p(Sxn,Sxn)→0 as n→∞. Hence, using (2.1), we get that ps(Sxn+1,Sxn)→0 as n→∞. Using Lemma 2.11, we obtain that there exist ɛ>0 and two sequences {mk} and {nk} of positive integers such that the following four sequences tend to ɛ when k→∞:
(3.9)p(Sx2mk,Sx2nk),p(Sx2mk,Sx2nk+1),p(Sx2mk-1,Sx2nk),p(Sx2mk-1,Sx2nk+1).
Putting in the contractive condition x=x2mk-1 and y=x2nk, it follows that
(3.10)p(Sx2mk,Sx2nk+1)≤γ(p(Sx2mk-1,Sx2nk))p(Sx2mk-1,Sx2nk)<p(Sx2mk-1,Sx2nk).
Hence,
(3.11)p(Sx2mk,Sx2nk+1)p(Sx2mk-1,Sx2nk)≤γ(p(Sx2mk-1,Sx2nk))<1
and limk→∞γ(p(Sx2mk-1,Sx2nk))=1. Since γ∈𝒢, it follows that limk→∞p(Sx2mk-1,Sx2nk)=0, which is in contradiction with ɛ>0.
Thus,{Sxn} is a Cauchy sequence, both in (X,p) and in (X,ps). Hence, it converges (in p and in ps) to a point Sz∈SX (we suppose SX to be closed, that is, complete; the case when TX is closed is treated similarly) such that
(3.12)p(Sz,Sz)=lim⁡n→∞p(Sxn,Sz)=lim⁡n,m→∞p(Sxn,Sxm).
Also, it follows easily that
(3.13)lim⁡n→∞p(Sxn,Sz)=p(Sz,Sz)=0.
We will prove that S and T have a coincidence point.
(i) Suppose thatT:(X,p)→(X,p) is continuous and that (T,S) is a ps-compatible pair. We have that
(3.14)ps(TSz,SSz)≤ps(TSz,TSxn)+ps(TSxn,STxn)+ps(STxn,SSz)⟶ps(TSz,TSz)+0+ps(SSz,SSz)=0,asn⟶∞.
It follows that T(Sz)=S(Sz) and Sz is a coincidence point of T and S.
(ii) If(X,p) satisfies (ORC), since {Sxn} is an increasing sequence tending to Sz, we have that Sxn⪯Sz for each n∈ℕ. So we can apply (P4) and the contractive condition to obtain
(3.15)p(Sz,Tz)≤p(Sz,Sxn+1)+p(Txn,Tz)≤p(Sz,Sxn+1)+γ(p(Sxn,Sz))p(Sxn,Sz)≤p(Sz,Sxn+1)+p(Sxn,Sz).
Letting n→∞ we get p(Sz,Tz)=0. Hence, we obtain that Tz=Sz and z is a coincidence point.Now, we are in the position to prove the main result of this section.Theorem 3.3.
Let(X,⪯,p) be an ordered partial metric space which is complete, and let g:X→X and F:X×X→X be such that F has the g-mixed monotone property. Suppose that(i)
there existsγ∈𝒢 such that
(3.16)p(F(x,y),F(u,v))+p(F(y,x),F(v,u))≤γ(p(gx,gu)+p(gy,gv))(p(gx,gu)+p(gy,gv))
holds for all x,y,u,v∈X satisfying (gx⪯gu and gy≽gv) or (gx≽gu and gy⪯gv);(ii)
g is continuous, F(X×X)⊂g(X), and one of these two subsets of X is closed;(iii)
there existx0,y0∈X such that (gx0⪯F(x0,y0) and gy0≽F(y0,x0)) or (gx0≽F(x0,y0) and gy0⪯F(y0,x0));(iv)
F is continuous and (F,g) is compatible in the sense of Definition 2.7, or(iv′)
(X,⪯,p) satisfies (ORC). Then there exist x¯,y¯∈X such that
(3.17)gx¯=F(x¯,y¯),gy¯=F(y¯,x¯),
that is, g and F have a coupled coincidence point.Proof.
Let relation⊑, partial metric P, and mappings TF, Tg on X2 be defined as in Lemma 2.8. Then (X2,⊑,P) is an ordered partial metric space which is complete and TF is a Tg-nondecreasing self-map on X2. Moreover,(i)
there existsγ∈𝒢 such that
(3.18)P(TFY,TFV)≤γ(P(TgY,TgV))P(TgY,TgV)
holds for all ⊑-comparable Y,V∈X2;(ii)
Tg is continuous, TF(X2)⊂Tg(X2), and one of these two subsets of X2 is closed;(iii)
there existsY0∈X2 such that TgY0 and TFY0 are comparable;(iv)
TF is continuous and the pair (TF,Tg) is Ps-compatible, or(iv′)
(X2,⊑,P) satisfies (ORC).Thus, all the conditions of Lemma 3.2 are satisfied (with T=TF and S=Tg). Hence, there exists Y¯∈X2 such that TgY¯=TFY¯, that is, there exist x¯,y¯∈X such that gx¯=F(x¯,y¯) and gy¯=F(y¯,x¯). Therefore, g and F have a coupled coincidence point.Puttingg=iX (the identity map) in Theorem 3.3, we obtain the following.Corollary 3.4.
Let(X,⪯,p) be an ordered partial metric space which is complete, and let F:X×X→X have the mixed monotone property. Suppose that(i)
there existsγ∈𝒢 such that
(3.19)p(F(x,y),F(u,v))+p(F(y,x),F(v,u))≤γ(p(x,u)+p(y,v))(p(x,u)+p(y,v))
holds for all x,y,u,v∈X satisfying (x⪯u and y≽v) or (x≽u and y⪯v);(ii)
there existx0,y0∈X such that (x0⪯F(x0,y0) and y0≽F(y0,x0)) or (x0≽F(x0,y0) and y0⪯F(y0,x0));(iii)
F is continuous, or(iii′)
(X,⪯,p) satisfies (ORC).Then there exist x¯,y¯∈X such that
(3.20)x¯=F(x¯,y¯),y¯=F(y¯,x¯),
that is, F has a coupled fixed point.Ifp=d is a standard metric, this reduces to [15, Corollary 2.3]. The following example shows how Corollary 3.4 can be used.Example 3.5.
LetX=[0,1/2] be ordered by the standard relation ≤. Consider the partial metric p on X given by p(x,y)=max{x,y}. Let F:X×X→X be given as
(3.21)F(x,y)={16(x-y),x≥y,0,x<y.
Finally, take γ∈𝒢 given as γ(t)=e-t/(t+1) for t>0 and γ(0)∈[0,1). We will show that conditions of Corollary 3.4 hold true.Take arbitraryx,y,u,v∈X satisfying x≥u and y≤v (the other possible case is treated symmetrically), and denote L=p(F(x,y),F(u,v))+p(F(y,x),F(v,u)) and R=γ(p(x,u)+p(y,v))(p(x,u)+p(y,v)). Consider the six possible cases: 1°: 0≤y≤u≤v≤x≤1/2, 2°: 0≤y≤v≤u≤x≤1/2, 3°: 0≤u≤y≤x≤v≤1/2, 4°: 0≤u≤x≤y≤v≤1/2, 5°: 0≤y≤u≤x≤v≤1/2, and 6°: 0≤u≤y≤v≤x≤1/2. It is easy to check that in all these cases L≤(1/6)(x+v) and that R=(e-(x+v)/(x+v+1))(x+v). Since(3.22)16t≤e-tt+1⋅t
holds for each t∈[0,1] (as far as it holds for t=1), we obtain that L≤R. The conditions of Corollary 3.4 are satisfied and F has a coupled fixed point (which is (0,0)).


## 4. Coupled Coincidence and Fixed Points under Rakotch-Type Conditions

Letℛ denote the class of real functions ρ:[0,+∞)→[0,1) satisfying the condition(4.1)limsup⁡s→tρ(s)<1foreacht>0.
Rakotch proved in 1962 the following.Theorem 4.1 (see [11]).
Let(X,d) be a complete metric space, and let T:X→X be a self-map. Suppose that there exists ρ∈ℛ such that
(4.2)d(Tx,Ty)≤ρ(d(x,y))d(x,y)
holds for all x,y∈X. Then T has a unique fixed point z∈X.We will prove the respective result for the existence of a coupled coincidence point in the frame of partial metric spaces. We begin with the following auxiliary result, which may be of interest on its own.Lemma 4.2.
Let(X,⪯,p) be an ordered partial metric space which is complete. Let T,S:X→X be self-maps such that S is continuous, TX⊂SX, and one of these two subsets of X is closed. Suppose that T is S-nondecreasing (with respect to ⪯) and there exists x0∈X with Sx0⪯Tx0 or Tx0⪯Sx0. Assume also that there exists ρ∈ℛ such that
(4.3)p(Tx,Ty)≤ρ(p(Sx,Sy))p(Sx,Sy)
holds for all x,y∈X such that Sx and Sy are comparable. Assume that either 1°: T is continuous and the pair (T,S) is ps-compatible or 2°: X satisfies (ORC). Then, T and S have a coincidence point in X.Proof.
The proof is similar to the proof of Lemma3.2, so we note only the basic step.
Withγ replaced by ρ, (3.7) shows that p(Sxn+1,Sxn) is strictly decreasing, thus converging to some q≥0. Suppose that q>0. Then, it follows that
(4.4)q≤limsup⁡n→∞ρ(p(Sxn+1,Sxn))<q,
a contradiction. Hence, limn→∞p(Sxn+1,Sxn)=0 is proved. The rest of the proof is identical to that of Lemma 3.2.The following example shows that the existence of order may be crucial.Example 4.3.
LetX=[0,2] be endowed with the partial metric defined by
(4.5)p(x,y)={|x-y|,x,y∈[0,1],max⁡{x,y},x∈(1,2]∨y∈(1,2].
The order ⪯ is given by
(4.6)x⪯y⟺x=y∨(x,y∈(1,2]∧x≤y).
Then it is easy to check that (X,⪯,p) is an ordered partial metric space which is complete and satisfies (ORC). Consider the mappings S,T:X→X given as
(4.7)Sx=x,Tx={1-x,x∈[0,1],x2,x∈(1,2].
Take the function ρ∈ℛ given by ρ(t)=1/2. It is easy to check that the contractive condition (4.3) holds. Indeed, if Sx=x and Sy=y∈X are comparable, say x≽y, then either x=y∈[0,1] or (x,y∈(1,2] and x≥y). In the first case, p(Sx,Sy)=p(x,x)=0 and p(Tx,Ty)=|(1-x)-(1-x)|=0; hence, p(Tx,Ty)≤ρ(0)·0 trivially holds (whichever function ρ∈ℛ is chosen). In the second case, p(Sx,Sy)=p(x,y)=x and
(4.8)p(Tx,Ty)=p(x2,y2)=x-y2<x2
and p(Tx,Ty)≤(1/2)·x=ρ(p(Sx,Sy))p(Sx,Sy). All the conditions of Lemma 4.2 are fulfilled and mappings T and S have a coincidence point (x=1/2).On the other hand, consider the same problem, but without order. Then the contractive condition does not hold and the conclusion about the coincidence point cannot be obtained in this way. Indeed, take anyx,y∈[0,1] with x≠y. Then, p(Sx,Sy)=p(x,y)=|x-y| and p(Tx,Ty)=|(1-x)-(1-y)|=|x-y|. Hence, if p(Tx,Ty)≤ρ(p(Sx,Sy))p(Sx,Sy), then ρ(p(Sx,Sy))≥1 and ρ∉ℛ.Now, we are in the position to giveTheorem 4.4.
Let(X,⪯,p) be an ordered partial metric space which is complete, and let g:X→X and F:X×X→X be such that F has the g-mixed monotone property. Suppose that(i)
there existsρ∈ℛ such that
(4.9)p(F(x,y),F(u,v))+p(F(y,x),F(v,u))≤ρ(p(gx,gu)+p(gy,gv))(p(gx,gu)+p(gy,gv))
holds for all x,y,u,v∈X satisfying (gx⪯gu and gy≽gv) or (gx≽gu and gy⪯gv) and conditions (ii), (iii), (iv), respectively, (iv′) of Theorem 3.3 hold true. Then there exist x¯,y¯∈X such that
(4.10)gx¯=F(x¯,y¯),gy¯=F(y¯,x¯),
that is, g and F have a coupled coincidence point.Proof.
The proof is analogous to the proof of Theorem3.3, and so is omitted.Puttingg=iX (the identity map) in Theorem 4.4, we obtain the following.Corollary 4.5.
Let(X,⪯,p) be an ordered partial metric space which is complete, and let F:X×X→X have the mixed monotone property. Suppose that(i)
there existsρ∈ℛ such that
(4.11)p(F(x,y),F(u,v))+p(F(y,x),F(v,u))≤ρ(p(x,u)+p(y,v))(p(x,u)+p(y,v))
holds for all x,y,u,v∈X satisfying (x⪯u and y≽v) or (x≽u and y⪯v) and conditions (ii), (iii), respectively, (iii′) of Corollary 3.4 hold true. Then there exist x¯,y¯∈X such that
(4.12)x¯=F(x¯,y¯),y¯=F(y¯,x¯),
that is, F has a coupled fixed point.


## 5. Coupled Coincidence and Fixed Points under Integral Conditions

Denote byΦ the set of functions φ:ℝ+→ℝ+ that are Lebesgue integrable and summable (having finite integral) on each compact subset of ℝ+, and satisfying condition ∫0ɛφ(t)dt>0 for each ɛ>0. Branciari [12] was the first to use integral-type contractive conditions in order to obtain fixed point results. He proved the following.Theorem 5.1 (see [12]).
Let(X,d) be a complete metric space, and let T:X→X be a self-map satisfying
(5.1)∫0d(Tx,Ty)φ(t)dt≤c∫0d(x,y)φ(t)dt
for some c∈(0,1) and some φ∈Φ. Then T has a unique fixed point z∈X and for each x∈X the Picard sequence {Tnx} converges to z when n→∞.Subsequently, several authors proved such results, including the very recent paper of Liu et al. [23].We begin with the following.Lemma 5.2.
Let(X,⪯,p) be an ordered partial metric space which is complete. Let T,S:X→X be self-maps such that S is continuous, TX⊂SX, and one of these two subsets of X is closed. Suppose that T is S-nondecreasing (with respect to ⪯) and there exists x0∈X with Sx0⪯Tx0 or Tx0⪯Sx0. Assume also that there exists ρ∈ℛ and φ∈Φ such that
(5.2)∫0p(Tx,Ty)φ(t)dt≤ρ(p(Sx,Sy))∫0p(Sx,Sy)φ(t)dt
holds for all x,y∈X such that Sx and Sy are comparable. Assume that either 1°:T is continuous and the pair (T,S) is ps-compatible or 2°:X satisfies (ORC). Then, T and S have a coincidence point in X.Proof.
Takex0∈X with, say, Sx0⪯Tx0, and using that T is S-nondecreasing and that TX⊂SX form the sequence {xn} satisfying Txn=Sxn+1, n=0,1,2,…, and
(5.3)Sx0⪯Tx0=Sx1⪯Tx1=Sx2⪯⋯⪯Txn⪯Sxn+1⪯⋯.
Since Sxn-1 and Sxn are comparable, we can apply the contractive condition (5.2) to obtain
(5.4)∫0p(Sxn+1,Sxn)φ(t)dt=∫0p(Txn,Txn-1)φ(t)dt≤ρ(p(Sxn,Sxn-1))∫0p(Sxn,Sxn-1)φ(t)dt≤∫0p(Sxn,Sxn-1)φ(t)dt.
If p(Sxn0,Sxn0-1)=0 for some n0∈ℕ, then it easily follows that Sxn0-1=Sxn0=Sxn0+1=⋯ and xn0 is a point of coincidence of S and T. Suppose that p(Sxn,Sxn+1)>0 for each n∈ℕ (and, hence, the strict inequality holds in (5.4)). We will prove that {p(Sxn,Sxn-1)} is a nonincreasing sequence. Suppose, to the contrary, that p(Sxn0+1,Sxn0)>p(Sxn0,Sxn0-1) for some n0∈ℕ. Then, since φ∈Φ, we get that
(5.5)0<∫0p(Sxn0,Sxn0-1)φ(t)dt≤∫0p(Sxn0+1,Sxn0)φ(t)dt=∫0p(Txn0,Txn0-1)φ(t)dt≤ρ(p(Sxn0,Sxn0-1))∫0p(Sxn0,Sxn0-1)φ(t)dt<∫0p(Sxn0,Sxn0-1)φ(t)dt,
a contradiction.
Denote byq≥0 the limit of the nonincreasing sequence {p(Sxn,Sxn-1)} of positive numbers. Suppose that q>0. Then, using Lemmas 2.1 and 2.2 from [23], the contractive condition (5.2), and property (4.1) of function ρ, we get that
(5.6)0<∫0qφ(t)dt=limsup⁡n→∞∫0p(Txn,Txn-1)φ(t)dt≤limsup⁡n→∞(ρ(p(Sxn,Sxn-1))∫0p(Sxn,Sxn-1)φ(t)dt)≤limsup⁡n→∞ρ(p(Sxn,Sxn-1))limsup⁡n→∞∫0p(Sxn,Sxn-1)φ(t)dt≤(limsup⁡s→qρ(s))∫0qφ(t)dt<∫0qφ(t)dt,
a contradiction. We conclude that limn→∞p(Sxn+1,Sxn)=0.
Now we prove that{Sxn} is a Cauchy sequence in (X,p) (and in (X,ps)). If this were not the case, as in the proof of Lemma 3.2, using Lemma 2.11, we would get that there exist ɛ>0 and two sequences {mk} and {nk} of positive integers such that the following four sequences tend to ɛ when k→∞:
(5.7)p(Sx2mk,Sx2nk),p(Sx2mk,Sx2nk+1),p(Sx2mk-1,Sx2nk),p(Sx2mk-1,Sx2nk+1).
Putting in the contractive condition x=x2mk-1 and y=x2nk, it follows that
(5.8)∫0p(Tx2mk-1,Tx2nk)φ(t)dt≤ρ(p(Sx2mk-1,Sx2nk))∫0p(Sx2mk-1,Sx2nk).
Passing to the upper limit as k→∞ and using properties of functions φ∈Φ, ρ∈ℛ as well as [23, Lemma 2.1], we get a contradiction
(5.9)0<∫0ɛφ(t)dt≤limsup⁡n→∞ρ(p(Sx2mk-1,Sx2nk))∫0ɛφ(t)dt<∫0ɛφ(t)dt.
Thus, {Sxn} is a Cauchy sequence, converging to some Sz∈SX (which we suppose to be closed in X) such that p(Sz,Sz)=0. We will prove that T and S have a coincidence point.
(i) Suppose thatT:(X,p)→(X,p) is continuous and (T,S) is ps-compatible. As in the proof of Lemma 3.2, we get that T(Sz)=S(Sz) and Sz is a coincidence point of T and S.
(ii) If(X,p) satisfies (ORC), since {Sxn} is an increasing sequence tending to Sz, we have that Sxn⪯Sz for each n∈ℕ. So we can apply (P4) and the contractive condition to obtain
(5.10)p(Sz,Tz)≤p(Sz,Sxn+1)+p(Txn,Tz)≤p(Sz,Sxn+1)+ρ(p(Sxn,Sz))∫0p(Sxn,Sz)φ(t)dt.
Passing to the upper limit as n→∞, using properties of functions ρ∈ℛ, φ∈Φ, as well as [23, Lemma 2.1], we get that p(Sz,Tz)=0. Hence, we obtain that Tz=Sz.PuttingS=iX and ρ(t)=c∈(0,1) in Lemma 5.2, we obtain an ordered partial metric extension of Branciari's Theorem 5.1. The following example shows that this extension is proper.Example 5.3.
LetX=[0,1] be endowed with the standard metric and order, and consider T,S:X→X defined by Tx=x/(1+x), and Sx=x, x∈X. Take φ∈Φ given by φ(t)=1. Then condition (5.1) does not hold. Indeed,
(5.11)∫0d(Tx,Ty)1⋅dt≤c∫0d(x,y)1⋅dt
would imply that
(5.12)|x-y|(1+x)(1+y)≤c|x-y|
for all x,y∈X. But taking x≠y and x,y→0+ would give that 1≤c, a contradiction.On the other hand, consider the same problem in an (ordered) partial metric space, with the partial metric given byp(x,y)=max{x,y}. Then, condition (5.2) reduces to(5.13)∫0p(Tx,Ty)1⋅dt≤c∫0p(x,y)1⋅dt,
which is, for x≥y, equivalent to x/(1+x)≤cx and holds true (for each x∈X) if c∈[1/2,1). Obviously, T has a (unique) fixed point x=0.The following theorem is obtained from Lemma5.2 in a similar way as Theorems 3.3 and 4.4 from respective lemmas.Theorem 5.4.
Let(X,⪯,p) be an ordered partial metric space which is complete, and let g:X→X and F:X×X→X be such that F has the g-mixed monotone property. Suppose that(i)
there existsρ∈ℛ and φ∈Φ such that
(5.14)∫0p(F(x,y),F(u,v))+p(F(y,x),F(v,u))φ(t)dt≤ρ(p(gx,gu)+p(gy,gv))∫0p(gx,gu)+p(gy,gv)φ(t)dt
holds for all x,y,u,v∈X satisfying (gx⪯gu and gy≽gv) or (gx≽gu and gy⪯gv); and conditions (ii), (iii), (iv), respectively, (iv′) of Theorem 3.3 hold true. Then there exist x¯,y¯∈X such that
(5.15)gx¯=F(x¯,y¯),gy¯=F(y¯,x¯),
that is, g and F have a coupled coincidence point.Puttingg=iX in Theorem 5.4, we obtain the following.Corollary 5.5.
Let(X,⪯,p) be an ordered partial metric space which is complete, and let F:X×X→X have the mixed monotone property. Suppose that(i)
there existsρ∈ℛ and φ∈Φ such that
(5.16)∫0p(F(x,y),F(u,v))+p(F(y,x),F(v,u))φ(t)dt≤ρ(p(x,u)+p(y,v))∫0p(x,u)+p(y,v)φ(t)dt
holds for all x,y,u,v∈X satisfying (x⪯u and y≽v) or (x≽u and y⪯v) and conditions (ii), (iii), respectively, (iii′) of Corollary 3.4 hold true. Then there exist x¯,y¯∈X such that
(5.17)x¯=F(x¯,y¯),y¯=F(y¯,x¯),
that is, F has a coupled fixed point.



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*Source: 192581-2012-03-20.xml*