R E S E A R C H
Open Access
Coincidence and fixed points for contractions
and cyclical contractions in partial metric
spaces
Hassen Aydi
1, Calogero Vetro
2, Wutiphol Sintunavarat
3and Poom Kumam
3**Correspondence: poom.kum@kmutt.ac.th 3Department of Mathematics, Faculty of Science, King Mongkut’s University of Technology Thonburi (KMUTT), Bangkok, 10140, Thailand Full list of author information is available at the end of the article
Abstract
We prove some coincidence and common fixed point results for three mappings satisfying a generalized weak contractive condition in ordered partial metric spaces. As application of the presented results, we give a unique fixed point result for a mapping satisfying a weak cyclical contractive condition. We also provide some illustrative examples.
MSC: 47H10; 54H25
Keywords: coincidence point; common fixed point; compatible mappings; cyclic weak (
ψ
,ϕ
)-contraction; partial metric space; weakly increasing mappings1 Introduction and preliminaries
In the last decades, several authors have worked on domain theory in order to equip se-mantics domain with a notion of distance. In , Matthews [] introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks and showed that the Banach contraction principle [] can be generalized to the partial metric context for applications in program verification. Later on, many researchers stud-ied fixed point theorems in partial metric spaces as well as ordered partial metric spaces. For more details, see [, , –, , , , , ].
Recently, there have been so many exciting developments in the field of existence of fixed points in partially ordered sets. For instance, Ran and Reurings [] extended the Banach contraction principle in partially ordered sets with some applications to matrix equations. For more details on fixed point theory in partially ordered sets, we refer the reader to [–, , , , , , , –, , ] and the references cited therein.
In this paper, we establish some coincidence and common fixed point results for three self-mappings on an ordered partial metric space satisfying a generalized weak contractive condition. The presented theorems extend some recent results in the literature. Moreover, as application, we give a unique fixed point theorem for a mapping satisfying a weak cycli-cal contractive condition.
Throughout this paper,R+will denote the set of all non-negative real numbers. First, we
start by recalling some known definitions and properties of partial metric spaces.
Definition .([]) A partial metric on a nonempty setXis a functionp:X×X→R+
such that for allx,y,z∈X:
(p) x=y⇐⇒p(x,x) =p(x,y) =p(y,y), (p) p(x,x)≤p(x,y),
(p) p(x,y) =p(y,x),
(p) p(x,y)≤p(x,z) +p(z,y) –p(z,z).
A partial metric space is a pair (X,p) such thatXis a nonempty set andpis a partial metric onX.
It is clear that, ifp(x,y) = , then from (p) and (p),x=y; but ifx=y,p(x,y) may not be . A basic example of a partial metric space is the pair (R+,p), wherep(x,y) =max{x,y}
for allx,y∈R+.
Other examples of partial metric spaces which are interesting from a computational point of view may be found in [, ].
Each partial metric ponXgenerates a Ttopologyτp onX which has as a base the
family of openp-balls{Bp(x,ε),x∈X,ε> }, whereBp(x,ε) ={y∈X:p(x,y) <p(x,x) +ε}
for allx∈Xandε> .
Ifpis a partial metric onX, then the functionps:X×X→R+given by
ps(x,y) = p(x,y) –p(x,x) –p(y,y) (.)
is a metric onX.
Definition .([]) Let{xn}be a sequence inX. Then
(i) {xn}converges to a pointx∈Xif and only ifp(x,x) =limn→+∞p(x,xn). We may write this asxn→x.
(ii) {xn}is called a Cauchy sequence iflimn,m→+∞p(xn,xm)exists and is finite. (iii) (X,p)is said to be complete if every Cauchy sequence{xn}inXconverges, with
respect toτp, to a pointx∈X, such thatp(x,x) =limn,m→+∞p(xn,xm).
Lemma .([]) 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) = if and only if
p(x,x) = lim
n→+∞p(xn,x) =n,mlim→+∞p(xn,xm).
Definition .([]) Let (X,p) be a partial metric space andT:X→Xbe a given map-ping. We say thatT is continuous atx∈X, if for everyε> , there existsη> such that
T(Bp(x,η))⊆Bp(Tx,ε).
Lemma .(Sequential characterization of continuity) Let(X,p)be a partial metric space
and T:X→X be a given mapping. T:X→X is continuous at x∈X if it is sequentially
continuous at x, that is, if and only if
∀{xn} ⊂X: lim
LetXbe a nonempty set andR:X→Xbe a given mapping. For everyx∈X, we denote byR–(x) the subset ofXdefined by
R–(x) :={u∈X|Ru=x}.
Definition . LetXbe a nonempty set. Then (X,,p) is called an ordered partial metric space if and only if
(i) (X,p)is a partial metric space, (ii) (X,)is a partially ordered set.
Definition . Let (X,) be a partially ordered set. Thenx,y∈Xare called comparable ifxyoryxholds.
Definition .([]) Let (X,) be a partially ordered set andT,S,R:X→Xbe given mappings such thatTX⊆RXandSX⊆RX. We say thatSandT are weakly increasing with respect toRif and only if, for allx∈X, we have
TxSy, ∀y∈R–(Tx)
and
SxTy, ∀y∈R–(Sx).
Remark . IfR:X→Xis the identity mapping (Rx=xfor allx∈X, shortly R=IX),
then the fact thatSandT are weakly increasing with respect toRimplies thatSandT
are weakly increasing mappings, that is,SxTSxandTxSTxfor allx∈X. Finally, a mappingT:X→Xis weakly increasing if and only ifTxTTxfor allx∈X.
Example . ConsiderX=R+ endowed with the usual ordering of real numbers and
defineT,S,R:X→Xby
Tx= for allx∈X, Sx= ⎧ ⎨ ⎩
x ifx∈[, ],
ifx> and Rx=x for allx∈X.
Now,R–(Tx) ={}andR–(Sx) =Sx, thenSandTare weakly increasing with respect toR.
Definition . Let (X,,p) be an ordered partial metric space. We say thatXis regular
if and only if the following hypothesis holds:{zn}is a non-decreasing sequence inXwith
respect tosuch thatzn→zasn→+∞, thenznzfor alln∈N.
Finally, we recall the following definition of partial-compatibility introduced by Samet
et al.[].
Definition . Let (X,p) be a partial metric space andT,R:X→Xbe given mappings. We say that the pair{T,R}is partial-compatible if the following conditions hold:
(b) limn→+∞p(TRxn,RTxn) = , whenever{xn}is a sequence inXsuch thatTxn→t andRxn→tfor somet∈X.
Note that Definition . extends and generalizes the notion of compatibility introduced by Jungck [].
2 Main results
We start this section with some auxiliary results (see also []).
Lemma . Let (X,d) be a metric space and let {xn} be a sequence in X such that
{d(xn+,xn)}is non-increasing and
lim
n→+∞d(xn+,xn) = .
If{xn}is not a Cauchy sequence, then there existε> and two sequences{mk}and{nk} of positive integers such that mk>nk>k and the following four sequences tend toεwhen k→+∞:
d(xmk,xnk)
, d(xmk,xnk+)
, d(xmk–,xnk)
, d(xmk–,xnk+)
.
As a corollary, applying Lemma . to the associated metricpsof a partial metricp, and
using Lemma ., we obtain the following lemma (see also []).
Lemma . Let(X,p)be a partial metric space and let{xn}be a sequence in X such that
{p(xn+,xn)}is non-increasing and
lim
n→+∞p(xn+,xn) = .
If{xn}is not a Cauchy sequence, then there existε> and two sequences{mk}and{nk} of positive integers such that mk>nk>k and the following four sequences tend toεwhen k→+∞:
p(xmk,xnk)
, p(xmk,xnk+)
, p(xmk–,xnk)
, p(xmk–,xnk+)
.
In the sequel, letbe the set of functionsψ:R+→R+such thatψis continuous, strictly
increasing andψ(t) = if and only ift= . Also, letbe the set of functionsϕ:R+→R+
such thatϕis lower semi-continuous andϕ(t) = if and only ift= . Suchψ andϕare called control functions.
Our first main result is the following.
Theorem . Let(X,)be a partially ordered set. Suppose that there exists a partial met-ric p on X such that the partial metmet-ric space(X,p)is complete. Let T,S,R:X→X be given mappings satisfying
(a) T,SandRare continuous,
Suppose that for every(x,y)∈X×X such that Rx and Ry are comparable, we have
ψp(Tx,Sy)≤ψ p(Tx,Rx) +p(Sy,Ry)
–ϕp(Rx,Ry), (.)
whereψ ∈andϕ∈. Then T , S and R have a coincidence point u∈X, that is, Tu=
Su=Ru.
Proof By Definition ., it follows thatTX∪SX⊆RX. Letxbe an arbitrary point inX.
SinceTX⊆RX, there existsx∈Xsuch thatRx=Tx. SinceSX⊆RX, there existsx∈X
such thatRx=Sx. Continuing this process, we can construct a sequence{xn}inXdefined
by
Rxn+=Txn, Rxn+=Sxn+, ∀n∈N. (.)
By construction, we havex∈R–(Tx) andx∈R–(Sx). Then using the fact thatSand
T are weakly increasing with respect toR, we obtain
Rx=TxSx=RxTx=Rx.
We continue this process to get
RxRx · · · Rxn+Rxn+ · · ·. (.)
We claim that{Rxn}is a Cauchy sequence in the partial metric space (X,p). To this aim,
we distinguish the following two cases.
Case . We suppose that there existsk∈Nsuch thatp(Rxk,Rxk+) = , so thatRxk= Rxk+. By (.), applying (.) withx=xkandy=xk+, we get
ψp(Rxk+,Rxk+)
=ψp(Txk,Sxk+)
≤ψ p(Txk,Rxk) +p(Sxk+,Rxk+)
–ϕp(Rxk,Rxk+)
=ψ p(Rxk+,Rxk) +p(Rxk+,Rxk+)
–ϕp(Rxk,Rxk+)
=ψ p(Rxk+,Rxk+)
.
Sinceψis strictly increasing, we have
p(Rxk+,Rxk+)≤
p(Rxk+,Rxk+).
This implies thatp(Rxk+,Rxk+) = . Continuing this process, we obtainp(Rxn,Rxk) =
for alln≥k. This implies thatRxn=Rxk, therefore{Rxn}is Cauchy in (X,p). The same
conclusion holds ifRxk+=Rxk+for somek∈N.
Case . Now, we suppose that
Here, we havep(Rxn,Rxn+)= for alln≥. Thanks to (.),RxnandRxn+are
compa-rable, then using (.) and takingx=xn+andy=xn+in (.), we get
ψp(Rxn+,Rxn+)
=ψp(Txn+,Sxn+)
≤ψ p(Txn+,Rxn+) +p(Sxn+,Rxn+)
–ϕp(Rxn+,Rxn+)
=ψ p(Rxn+,Rxn+) +p(Rxn+,Rxn+)
–ϕp(Rxn+,Rxn+)
≤ψ p(Rxn+,Rxn+) +p(Rxn+,Rxn+)
.
Sinceψis strictly increasing, the above inequality implies that
p(Rxn+,Rxn+)≤p(Rxn+,Rxn+). (.)
Now, takingx=xnandy=xn+in (.), we have
ψp(Rxn+,Rxn+)
=ψp(Txn,Sxn+)
≤ψ p(Txn,Rxn) +p(Sxn+,Rxn+)
–ϕp(Rxn,Rxn+)
=ψ p(Rxn+,Rxn) +p(Rxn+,Rxn+)
–ϕp(Rxn,Rxn+)
≤ψ p(Rxn+,Rxn) +p(Rxn+,Rxn+)
, (.)
which implies that
p(Rxn+,Rxn+)≤p(Rxn+,Rxn). (.)
Combining (.) and (.), we get
p(Rxn+,Rxn+)≤p(Rxn,Rxn+) for alln≥. (.)
It follows that the sequence{p(Rxn,Rxn+)} is non-increasing and bounded below by .
Hence, there existsr≥ such that
p(Rxn,Rxn+)→r asn→+∞.
We claim thatr= . Suppose thatr> . Taking thelim supasn→+∞in (.) and using the properties of the functionsψ andϕ, we have
This implies thatϕ(r) = , and by a property of the functionϕ, we haver= , that is a contradiction. We deduce thatr= ,i.e.,
p(Rxn,Rxn+)→ asn→+∞. (.)
We shall show that{Rxn}is a Cauchy sequence in the partial metric space (X,p). For this,
it is sufficient to prove that{Rxn}is Cauchy in (X,p). Suppose to the contrary that{Rxn}
is not a Cauchy sequence. Then, having in mind that{p(Rxn,Rxn+)}is non-increasing and
(.), it follows by Lemma . that there existε> and two sequences{mk}and{nk}of
positive integers such thatmk>nk>kand the following four sequences tend toεwhen k→+∞:
p(Rxmk,Rxnk)
, p(Rxmk,Rxnk+)
,
p(Rxmk–,Rxnk)
, p(Rxmk–,Rxnk+)
.
Applying (.) withx=xnk andy=xmk–, we get
ψp(Rxnk+,Rxmk)
=ψp(Txnk,Sxmk–)
≤ψ p(Txnk,Rxnk) +p(Sxmk–,Rxmk–)
–ϕp(Rxnk,Rxmk–)
=ψ p(Rxnk+,Rxn(k)) +p(Rxm(k),Rxmk–)
–ϕp(Rxnk,Rxmk–)
.
Takinglim supk→+∞in the above inequality and using the continuity ofψand the lower semi-continuity ofϕ, we obtain
ψ(ε)≤ψ +
–ϕ(ε) = –ϕ(ε), (.)
from which a contradiction follows sinceε> . Then, we deduce that{Rxn}is a Cauchy
sequence in the partial metric space (X,p), which is complete, so{Rxn}converges to some u∈X, that is, from (p) and Definition .,
p(u,u) = lim
n→+∞p(Rxn,u) =n→lim+∞p(Rxn,Rxn).
But from (.) and condition (p), we have
lim
n→+∞p(Rxn,Rxn) = ,
therefore, it follows that
p(u,u) = lim
n→+∞p(Rxn,u) =n→lim+∞p(Rxn,Rxn) = . (.)
From (.) and the continuity ofR, we get
lim n→+∞p
R(Rxn),Ru
The triangular inequality yields
p(Ru,Tu)≤pRu,R(Rxn+)
+pR(Txn),T(Rxn)
+pT(Rxn),Tu
. (.)
By (.) and (.), we have
Rxn→u, Txn=Rxn+→u asn→+∞. (.)
Having in mind that the pair{R,T}is partial-compatible, then
pR(Txn),T(Rxn)
→ asn→+∞. (.)
Also, sincep(u,u) = , then we havep(Tu,Tu) = . The continuity ofTtogether with (.) give us
pT(Rxn),Tu
→p(Tu,Tu) = . (.)
Combining (.) and (.) together with (.) and lettingn→+∞in (.), we obtain
p(Ru,Tu)≤p(Ru,Ru). (.)
By condition (p) and (.), one can write
p(Ru,Ru) =p(Ru,Tu). (.)
Similarly, by triangular inequality, we get
p(Ru,Su)≤pRu,R(Rxn+)
+pR(Sxn+),S(Rxn+)
+pS(Rxn+),Su
. (.)
By (.) and (.), we have
Rxn+→u, Sxn+→u asn→+∞. (.)
Since the pair{S,R}is partial-compatible, then
pR(Sxn+),S(Rxn+)
→ asn→+∞. (.)
Also, sincep(u,u) = , it followsp(Ru,Ru) = . Thus, from (.),p(Ru,Tu) =p(Ru,Ru) = and soRu=Tu.
The continuity ofSand (.) give us
pS(Rxn+),Su
→p(Su,Su) asn→+∞. (.)
Combining (.) and (.) together with (.) and lettingn→+∞in (.), we obtain
By condition (p) and (.), we get
p(Ru,Su) =p(Su,Su). (.)
Applying (.) withx=y=u, we get
ψp(Tu,Su)≤ψ p(Tu,Ru) +p(Su,Ru)
–ϕp(Ru,Ru)
=ψ p(Su,Tu)
–ϕ()
=ψ p(Tu,Su)
.
This implies that
p(Tu,Su)≤
p(Tu,Su),
and so it followsp(Tu,Su) = , that isTu=Su. Thus, we have obtained
Ru=Tu=Su,
that is,uis a coincidence point ofT,SandR.
Remark . We point out that the order in which the mappings in condition (b) of The-orem . are considered is crucial. Trivially, TheThe-orem . remains true if we assume that the partial-compatible pairs are{T,R}and{R,S}.
Example . LetX= [,] be endowed with the partial metricp(x,y) =max{x,y}and the order given as follows:
xy ⇐⇒ x≥y.
Consider the mappingsT,S,R:X→Xdefined byTx=Sx=xandRx=xfor allx∈X. Also, define the functionsψ,ϕ:R+→R+byψ(t) =tandϕ(t) = t, for allt≥. Clearly,
condition (.) is satisfied. In fact, for every (x,y)∈X×Xwithx≥y, we get
ψp(Tx,Sy)=x≤x
≤
x+y
–
x
=ψ p(Tx,Rx) +p(Sy,Ry)
–ϕp(Rx,Ry).
All the other hypotheses of Theorem . are satisfied andT,SandRhave a coincidence pointu= . (Moreover,u= is the unique common fixed point ofT,SandR.)
Note that Theorem . is not applicable in respect of the usual order of real numbers becauseT is not weakly increasing. It follows that the partial order may be fundamental.
Theorem . Let(X,)be a partially ordered set. Suppose that there exists a partial met-ric p on X such that(X,p)is complete. Let T,S,R:X→X be given mappings satisfying
(a) RXis a closed subspace of(X,p),
(b) T andSare weakly increasing with respect toR, (c) Xis regular.
Suppose that for every(x,y)∈X×X such that Rx and Ry are comparable, we have
ψp(Tx,Sy)≤ψ p(Tx,Rx) +p(Sy,Ry)
–ϕp(Rx,Ry), (.)
whereψ∈ andϕ∈. Then, T , S and R have a coincidence point u∈X, that is, Tu=
Su=Ru.
Proof Following the proof of Theorem ., we have that{Rxn}is a Cauchy sequence in the
closed subspaceRX, then there existsv=Ru, withu∈X, such that
Rxn→v=Ru asn→+∞. (.)
Thanks to (.),{Rxn}is a non-decreasing sequence, and so, since it converges tov=Ru,
from the regularity ofX, we get
RxnRu, ∀n∈N.
Therefore,RxnandRuare comparable. Puttingx=xnandy=uin (.) and using (.),
we get
ψp(Rxn+,Su)
=ψp(Txn,Su)
≤ψ p(Txn,Rxn) +p(Su,Ru)
–ϕp(Rxn,Ru)
=ψ p(Rxn+,Rxn) +p(Su,Ru)
–ϕp(Rxn,Ru)
.
Takinglim supn→+∞ in the above inequality, using (.) and the properties ofϕ andψ, we obtain
ψp(Ru,Su)≤ψ p(Su,Ru)
–ϕp(Ru,Ru)
≤ψ p(Su,Ru)
.
This implies that
p(Ru,Su)≤
p(Su,Ru),
Analogously, puttingx=uandy=xn+in (.), we have
ψ(p(Tu,Rxn+) =ψ
p(Tu,Sxn+)
≤ψ p(Tu,Ru) +p(Sxn+,Rxn+)
–ϕp(Ru,Rxn+)
=ψ p(Rxn+,Rxn+) +p(Tu,Ru)
–ϕp(Ru,Rxn+)
.
Takinglim supn→+∞ in the above inequality, using (.) and the properties ofϕ andψ, we obtain
p(Tu,Ru)≤
p(Tu,Ru),
which yields that
Tu=Ru.
We conclude thatuis a coincidence point ofT,SandR.
IfR:X→Xis the identity mappingIX, by Theorem ., we obtain the following
com-mon fixed point result involving two mappings.
Corollary . Let(X,)be a partially ordered set. Suppose that there exists a partial metric p on X such that the partial metric space(X,p)is complete. Let X be regular and T,S:X→X be given mappings such that T and S are weakly increasing. Suppose that for every(x,y)∈X×X such that x and y are comparable, we have
ψp(Tx,Sy)≤ψ p(Tx,x) +p(Sy,y)
–ϕp(x,y), (.)
whereψ∈ andϕ∈. Then, T and S have a common fixed point u∈X, that is, Tu=
Su=u.
The following example shows that the hypothesis ‘TandSare weakly increasing (with respect toR)’ has a key role for the validity of our results.
Example . LetX= [, ] be endowed with the partial metricp(x,y) =max{x,y}and the ordergiven as follows:
xy ⇐⇒ x≥y.
Consider the mappingsT,S:X→X defined byTx= x andSx=x, for allx∈X. Also, define the functionsψ,ϕ:R+→R+byψ(t) =tandϕ(t) = t, for allt≥. It is easy to show
thatSxTSxandTxSTx, for allx∈X, that is,TandSare weakly increasing. Now, take
xandycomparable and, without loss of generality, assumeyx, so thatx≤y. It is easy to show that (.) holds and all the other hypotheses of Corollary . are satisfied. Then,
Note that Corollary . is not applicable in respect of the usual order of real numbers becauseT andSare not weakly increasing.
Now, we shall prove the existence and uniqueness of a common fixed point for three mappings.
Theorem . In addition to the hypotheses of Theorem ., suppose that for any(x,y)∈
X×X, there exists z∈X such that TxTz and TyTz. Then, T , S and R have a unique common fixed point, that is, there exists a unique u∈X such that u=Ru=Tu=Su.
Proof Referring to Theorem ., the set of coincidence points ofT,SandRis nonempty. Now, we shall show that ifx*andy*are coincidence points ofT,SandR, that is,Rx*=
Tx*=Sx*andRy*=Ty*=Sy*, then
pRx*,Ry*= . (.)
For the coincidence pointsx*andy*, Theorem . gives us that
pRx*,Tx*=pTx*,Sx*= =pRy*,Ty*=pTy*,Sy*.
By assumption, there existsz∈Xsuch that
Tx*Tz, Ty*Tz. (.)
Now, proceeding similarly to the proof of Theorem ., we can immediately define a se-quence{Rzn}as follows:
Rzn+=Tzn, Rzn+=Szn+, ∀n∈N. (.)
SinceTandSare weakly increasing with respect toR, we have
Tx*=Rx*Rzn, Ty*=Ry*Rzn, ∀n∈N. (.)
Puttingx=znandy=x*in (.) and using (.), we get
ψpRzn+,Rx*
=ψpTzn,Sx*
≤ψ p(Tzn,Rzn) +p(Sx
*,Rx*)
–ϕpRzn,Rx*
=ψ p(Rzn+,Rzn)
–ϕpRzn,Rx*
≤ψ p(Rzn+,Rzn)
.
Sinceψis strictly increasing, we have
pRzn+,Rx*
≤
p(Rzn+,Rzn)≤ p
Rzn+,Rx*
+
p
Rx*,Rzn
This gives us
pRzn+,Rx*
≤pRzn,Rx*
. (.)
Puttingx=x*andy=z
nin (.), then similarly to the above, one can find pRzn+,Rx*
≤pRzn+,Rx*
. (.)
We combine (.) and (.) to remark that
pRzn+,Rx*
≤pRzn,Rx*
, ∀n∈N. (.)
Then, the sequence{p(Rzn,Rx*)} is non-increasing and bounded below, so there exists r≥ such that
pRzn,Rx*
→r asn→+∞.
Adopting the strategy used in the proof of Theorem ., one can show thatr= ,i.e.,
pRzn,Rx*
→ asn→+∞. (.)
The same idea yields
pRzn,Ry*
→ asn→+∞. (.)
Now,p(Rx*,Ry*)≤p(Rx*,Rz
n) +p(Rzn,Ry*) and from (.), (.), we obtainp(Rx*,Ry*) =
, and so (.) holds.
Thanks to (.) and (.), one can write
Tzn→Rx*=Ry*, Szn+→Rx*=Ry* asn→+∞. (.)
From partial-compatibility of the pairs{R,T}and{S,R}, using (.) and (.), we obtain
pR(Tzn),T(Rzn)
→, pR(Szn+),S(Rzn+)
→ asn→+∞. (.)
Denote
u=Rx*.
Sincep(u,u) =p(Rx*,Ry*) = , so again by partial-compatibility of the pairs{R,T} and
{S,R}, we get
p(Tu,Tu) =p(Ru,Ru) = . (.)
By triangular inequality, we have
p(Ru,Tu)≤pRu,R(Tzn)
+pR(Tzn),T(Rzn)
+pT(Rzn),Tu
Using (.), (.), (.), the continuity ofTand lettingn→+∞in the above inequality, we get
p(Ru,Tu)≤p(Ru,Ru) +p(Tu,Tu) = ,
that is,Ru=Tuanduis a coincidence point ofT andR. Analogously, the triangular inequality gives us
p(Ru,Su)≤pRu,R(Szn+)
+pR(Szn+),S(Rzn+)
+pS(Rzn+),Su
.
Using (.), (.), (.), the continuity ofSand lettingn→+∞in the above inequality, we get
p(Ru,Su)≤p(Ru,Ru) +p(Su,Su) =p(Su,Su).
By condition (p), it follows immediately
p(Ru,Su) =p(Su,Su).
Now, applying (.) withx=y=u, we have
ψp(Tu,Su)≤ψ p(Tu,Ru) +p(Su,Ru)
–ϕp(Ru,Ru)
=ψ p(Su,Tu)
–ϕ()
=ψ p(Tu,Su)
.
This implies that
p(Tu,Su)≤
p(Tu,Su),
then we deduce thatp(Tu,Su) = , and soTu=Su. Until now, we have obtained
Ru=Tu=Su.
Withy*=uand from (.), we have
u=Rx*=Ru=Tu=Su.
This proves thatuis a common fixed point of the mappingsT,SandR.
Now our purpose is to check that such a point is unique. Suppose to the contrary that there is another common fixed point ofT,SandR, sayq. Then, applying (.) withx=
y=q, we obtain easily that p(q,Tq) =p(q,Sq) =p(q,Rq) = . It is immediate thatq is a coincidence point ofT,SandR. From (.), this implies that
Hence, we get
q=Rq=Ru=u,
which yields the uniqueness of the common fixed point ofT,SandR. This completes the
proof.
Remark . We leave, as exercise for the reader, to verify that our results hold even if we replace condition (.) by the following
ψp(Tx,Sy)≤ψ p(Tx,Ry) +p(Sy,Rx)
–ϕp(Rx,Ry)
for allx,y∈Xsuch thatRxandRyare comparable.
3 Application to cyclical contractions
In this section we use the previous results to prove a fixed point theorem for a mapping satisfying a weak cyclical contractive condition. In , Kirket al.[] studied existence and uniqueness of a fixed point for mappings satisfying cyclical contractive conditions in complete metric spaces.
Definition . Let (X,d) be a metric space,ma positive integer andY, . . . ,Ymnonempty
subsets ofX. A mappingT onmi=Yi is called am-cyclic mapping ifT(Yi)⊂Yi+,i=
, . . . ,m, whereYm+=Y.
Later on, Pacurar and Rus [] introduced the following notion, suggested by the con-siderations in [].
Definition . LetYbe a nonempty set,ma positive integer andT:Y→Yan operator. By definition,Y=mi=Yiis a cyclic representation ofYwith respect toTifTis am-cyclic
mapping andYiare nonempty sets.
Example . Let X=R. Assume Y =Y= [–, ] and Y =Y= [, ], so that Y =
i=Yi = [–, ]. Define T :Y →Y such that Tx= –x, for all x∈Y. It is clear that Y=i=Yiis a cyclic representation ofY.
Inspired by Karapinar [] and Gopalet al.[], we present the notion of a cyclic weak (ψ,ϕ)-contraction in partial metric spaces.
Definition . Let (X,,p) be an ordered partial metric space,Y,Y, . . . ,Ym be closed
subsets of X and Y =mi=Yi. An operatorT :Y →Y is called a cyclic weak (ψ,
ϕ)-contraction if the following conditions hold:
(i) Y=mi=Yiis a cyclic representation ofYwith respect toT,
(ii) there existψ∈andϕ∈such that
ψp(Tx,Ty)≤ψ p(Tx,x) +p(Ty,y)
–ϕp(x,y), (.)
Now, we state and prove the following result.
Theorem . Let(X,)be a partially ordered set. Suppose that there exists a partial
met-ric p on X such that the partial metmet-ric space(X,p)is complete. Let T:mi=Yi→
m i=Yi be a given mapping satisfying
(a) T is a cyclic weak(ψ,ϕ)-contraction, (b) T is weakly increasing and continuous, (c) the pair{Ix,T}is partial-compatible,
(d) for any(x,y)∈X×X, there existsz∈Xsuch thatTxTzandTyTz.
Then, T has a unique fixed point u∈mi=Yi, that is, Tu=u.
Proof Letx∈Y=
m
i=Yiand set
xn+=Txn, ∀n∈N. (.)
For anyn∈N, there isin∈ {, . . . ,m}such thatxn∈Yin andxn+∈Yin+. Then, following
the lines of the proof of Theorem ., it is easy to show that{xn}is a Cauchy sequence in
the partial metric space (Y,p), which is complete, so{xn}converges to somey∈Y. On the
other hand, by condition (i) of Definition ., it follows that the iterative sequence{xn}has
an infinite number of terms inYifor eachi= , , . . . ,m. Since (Y,p) is complete, from each Yi,i= , , . . . ,m, one can extract a subsequence of{xn}that converges toy. In virtue of the
fact that eachYi,i= , , . . . ,m, is closed, we conclude thaty∈
m
i=Yiand thus
m i=Yi=∅.
Obviously,mi=Yiis closed and complete. Now, consider the restriction ofTon
m i=Yi,
that isT|mi=Yi:
m i=Yi→
m
i=Yiwhich satisfies the assumptions of Theorem . and
thus,T|mi=Yihas a unique fixed point in
m
i=Yi, sayu, which is obtained by iteration
from the starting pointx∈Y. To conclude, we have to show that, for any initial value
x∈Y, we get the same limit pointu∈mi=Yi. Due to condition (c) and using the analogous
ideas of the proof of Theorem ., it can be obtained that, for any initial valuex∈Y,xn→u
asn→+∞. This completes the proof.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly to writing this paper. All authors read and approved the final manuscript.
Author details
1Institut Supérieur d’Informatique et des Technologies de Communication de Hammam Sousse, Université de Sousse, Route GP1-4011, H. Sousse, Tunisia.2Dipartimento di Matematica e Informatica, Università degli Studi di Palermo, Via Archirafi 34, Palermo, 90123, Italy. 3Department of Mathematics, Faculty of Science, King Mongkut’s University of Technology Thonburi (KMUTT), Bangkok, 10140, Thailand.
Acknowledgements
The authors are really thankful to the anonymous referee for his/her precious suggestions useful to improve the quality of the paper. The third author would like to thank the Research Professional Development Project under the Science Achievement Scholarship of Thailand (SAST) and the forth author would like to thank the Commission on Higher Education and the Thailand Research Fund under Grant MRG no. 5380044 for financial support during the preparation of this manuscript.
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doi:10.1186/1687-1812-2012-124