R E S E A R C H
Open Access
Coupled fixed point theorems for partially
contractive mappings
Thabet Abdeljawad
**Correspondence:
[email protected] Mathematics and Computer Science Department, Çankaya University, Eski¸sehir Yolu, Yenimahalle, Ankara, 06810, Turkey
Abstract
Recently, some authors have started to generalize fixed point theorems for contractive mappings in a class of generalized metric spaces in which the self-distance need not be zero. These spaces, partial metric spaces, were first introduced by Matthews in 1994. The proved fixed point theorems have been obtained for mappings satisfying contraction type conditions empty of the self-distance. In this article, we prove some coupled fixed point theorems for mappings satisfying contractive conditions allowing the appearance of self-distance terms. These partially contractive mappings do reflect the structure of the partial metric space, and hence their coupled fixed theorems generalize the previously obtained by (Aydi in Int. J. Math. Sci. 2011:Article ID 647091, 2011). Some examples are given to support our claims.
MSC: 47H10; 54H25
Keywords: partial metric space; coupled fixed point; partially contractive map; strong partially contractive; 0-complete
1 Introduction and preliminaries
The Banach contraction mapping principle is considered to be the soul of many extended fixed point theorems. It has widespread applications in many branches of mathematics, engineering and computer science. During the last decades many authors were able to gen-eralize this principle [–]. After the appearance of partial metric spaces as a place for dis-tinct research work into flow analysis, non-symmetric topology and domain theory [, ], many authors started to generalize this principle to these spaces (see [–]. However, the contraction type conditions used in those generalizations do not reflect the structure of a partial metric space apparently. Later, the authors in [] proved a more reasonable con-traction principle in the partial metric space in which they used self-distance terms. On the other hand, the theory of coupled fixed point theorems has recently attracted some authors (see [–]). Also, Meir-Keeler type common and tripled fixed point theorems have been recently considered over partial metric spaces [, ]. In this article, we prove a coupled partial contraction principle generalizing the recently published coupled fixed point theorems in []. An example is presented to show that our coupled partial contrac-tion principle is worthy of investigacontrac-tion.
A partial metric space (PMS) (see,e.g., [, ]) is a pair (X,p:X×X→R+) (whereR+
denotes the set of all nonnegative real numbers) such that (P) p(x,y) =p(y,x)(symmetry)
(P) If≤p(x,x) =p(x,y) =p(y,y)thenx=y(equality) (P) p(x,x)≤p(x,y)(small self-distances)
(P) p(x,z) +p(y,y)≤p(x,y) +p(y,z)(triangularity) for allx,y,z∈X.
For a partial metricponX, the functiondp:X×X→R+given by
dp(x,y) = p(x,y) –p(x,x) –p(y,y) ()
is a (usual) metric onX. Each partial metricponXgenerates aTtopologyτponXwith a
base of 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> .
Definition (see,e.g., [, , ])
(i) A sequence{xn}in a PMS(X,p)converges tox∈Xif and only if p(x,x) =limn→∞p(x,xn);
(ii) A sequence{xn}in a PMS(X,p)is called Cauchy if and only iflimn,m→∞p(xn,xm)
exists (and is finite);
(iii) A PMS(X,p)is said to be complete if every Cauchy sequence{xn}inXconverges, with respect toτp, to a pointx∈Xsuch thatp(x,x) =limn,m→∞p(xn,xm);
(iv) A mappingf :X→Xis said to be continuous atx∈X, if for everyε> , there
existsδ> such thatf(Bp(x,δ))⊂Bp(f(x),ε).
Lemma (see,e.g., [, , ])
(A) A sequence{xn}is Cauchy in a PMS(X,p)if and only if{xn}is Cauchy in a metric space(X,dp);
(B) A PMS(X,p)is complete if and only if the metric space(X,dp)is complete. Moreover,
lim
n→∞dp(x,xn) = ⇔ p(x,x) =n→∞lim p(x,xn) =n,mlim→∞p(xn,xm). ()
A sequence{xn}is called -Cauchy [] iflimm,np(xn,xm) = . The partial metric space
(X,p) is called -complete [, ] if every -Cauchy sequence inxconverges to a point x∈X with respect topandp(x,x) = . Clearly, every complete partial metric space is -complete. The converse need not be true.
Example (see []) LetX=Q∩[,∞) with the partial metricp(x,y) =max{x,y}. Then (X,p) is a -complete partial metric space which is not complete.
Letρp=inf{p(x,y) :x,y∈X}and defineXp={x∈X:p(x,x) =ρp}.
The following theorem was presented in [].
Theorem Let(X,p)be a complete metric space,α∈[, )and T:X→X a given map-ping. Suppose that for each x,y∈X the following condition holds:
p(x,y)≤maxαp(x,y),p(x,x),p(y,y).
Then
() there is a uniqueu∈Xpsuch thatTu=u;
() for eachx∈Xpthe sequence{Tnx}n≥converges touwith respect to the metricdp.
If (X,p) is a partial metric space, then clearly (p×p) :X×X→Rdefined by (p× p)((x,y), (u,v)) =p(x,u) +p(y,v) is a partial metric, and hence (X×X, (p×p)) is a par-tial metric space. Clearly, if (X,p) is complete (-complete) then so is (X×X, (p×p)). The minimum self partial distance in (X×X, (p×p)) is defined by
ρp×p:=inf(p×p)(x,y), (x,y)=p(x,x) +p(y,y) :x,y∈X.
The set of all points inX×Xwith self partial distanceρp×pis denoted by
(X×X)p×p=(x,y)∈X×X: (p×p)(x,y), (x,y)=ρp×p.
Example LetX= [, ] and provideXwith the partial metricp(x,y) =|x–y|if bothx,y∈ [, ) and p(x,y) =max{x,y} otherwise. Then clearly (X,p) is a complete partial metric space,ρp×p= and (X×X)p×p= [, )×[, ).
Definition Let (X,p) be a partial metric space andF:X×X→Xbe a mapping. Then Fis called
(a) partially contractive if there exist constants≤k,l< such that for allx,y,u,v∈X,
pF(x,y),F(u,v)
≤
max
kp(x,u) +lp(y,v), (p×p)(x,y), (x,y), (p×p)(u,v), (u,v), ()
(b) strong partially contractive if there exist constants≤k,l< such that for all x,y,u,v∈X,
pF(x,y),F(u,v)
≤
max
kp(x,u) +lp(y,v),(p×p)((x,y), (x,y)) + (p×p)((u,v), (u,v))
. ()
Aydi [] proved the following coupled fixed point theorems in partial metric spaces:
Theorem Let(X,p)be a complete partial metric space. Suppose that the mapping F: X×X→X satisfies the following contractive condition for all x,y,u,v∈X:
pF(x,y),F(u,v)≤kp(x,u) +lp(y,v),
where≤k,l< with k+l< . Then F has a unique coupled fixed point. That is, there exists unique(x,y)∈X×X such that F(x,y) =x and F(y,x) =y.
Theorem Let(X,p)be a complete partial metric space. Suppose that the mapping F: X×X→X satisfies the following contractive condition for all x,y,u,v∈X:
pF(x,y),F(u,v)≤kpF(x,y),x+lpF(u,v),u,
2 Main results
Theorem Let(X,p)be a complete partial metric space and F:X×X→X be a partially contractive mapping. Then
(a) (X×X)p×p=∅,
(b) there exists a unique(u,v)∈(X×X)p×psuch that
F(u,v) =u and F(v,u) =v; ()
(c) for each(x,y)∈(X×X)p×pthe sequence{(xn,yn)} ∈X×Xdefined by
x=F(x,y), y=F(y,x), xn+=F(xn,yn), yn+=F(yn,xn), n≥,
converges to(u,v)with respect todp×dp. That is
lim
n→∞(dp×dp)
(xn,yn), (u,v)
= lim
n→∞
dp(xn,u) +dp(yn,v)
= .
Proof If (x,y)∈X×X, define the sequence{(xn,yn)} ∈X×Xby
x=F(x,y), y=F(y,x), xn+=F(xn,yn), yn+=F(yn,xn), n≥.
We divide the proof of the claim (a) into the following steps:
Step I: The sequence{(p×p)((xn,yn)(xn,yn))}={p(xn,xn) +p(yn,yn)}is non-increasing.
From the condition () we have
Tn=p
F(xn,yn),F(xn,yn)
+pF(yn,xn),F(yn,xn)
≤
p(xn,xn) +p(yn,yn) +
p(xn,xn) +p(yn,yn)
=p(xn,xn) +p(yn,yn) = (p×p)
(xn,yn), (xn,yn)
,
whereTn= (p×p)((xn+,yn+), (xn+,yn+)). As a result of Step I, there existsr(x,y)≥
such that
lim
n→∞(p×p)
(xn,yn), (xn,yn)
=inf
n(p×p)
(xn,yn), (xn,yn)
=r(x,y).
Step II: For eachn≥, we show that
(p×p)(xn,yn), (x,y)
≤M(x,y) :=
–τ(p×p)
(x,y), (x,y)
+ (p×p)(x,y), (x,y)
, ()
whereτ=max{k,l}. We follow by induction. Forn= , , it is clear. Assume () is true for n≤r, and let us prove () forn=r+ ≥. Then by the help of (), the triangle inequality and Step I, we have
E = (p×p)(xr+,yr+), (x,y)
=p(x,x) +p(y,y) +p
F(xr,yr),F(x,y)
+pF(yr,xr),F(y,x)
≤(p×p)(x,y), (x,y)
+ max
kp(xr,x) +lp(yr,y), (p×p)
(x,y), (x,y)
+ max
lp(xr,x) +kp(yr,y), (p×p)
(x,y), (x,y)
≤(p×p)(x,y), (x,y)
+maxτ(p×p)(xr,yr), (x,y)
, (p×p)(x,y), (x,y)
≤(p×p)(x,y), (x,y)
+maxτM(x,y), (p×p)
(x,y), (x,y)
≤M(x,y),
whereE=p(xr+,x) +p(yr+,y).
Step III: We show that
lim
m,n→∞(p×p)
(xn,yn), (xm,ym)
= lim
m,n→∞
p(xn.xm) +p(yn,ym) =r(x,y). ()
Let> . Choosensuch that (p×p)((xn,yn), (xn,yn)) <r(x,y)+and M(x,y)τ
n<
r(x,y) +. Then form,n≥n, we have
S =pF(xn–,yn–),F(xm–,ym–)
+pF(yn–,xn–),F(ym–,xm–)
≤
max
kp(xn–,xm–) +lp(yn–,ym–), (p×p)
(xn–,yn–),
(xn–,yn–)
, (p×p)(xm–,ym–), (xm–,ym–)
+ max
lp(xn–,xm–) +kp(yn–,ym–), (p×p)
(xn–,yn–),
(xn–,yn–)
, (p×p)(xm–,ym–), (xm–,ym–)
≤maxτp(xn–,xm–) +p(yn–,ym–) , (p×p)
(xn–,yn–),
(xn–,yn–)
, (p×p)(xm–,ym–), (xm–,ym–)
≤ · · · ≤maxτnp(x
n–n,xm–n) +p(yn–n,ym–n), (p×p)
(xn–n,yn–n), (xn–n,yn–n)
, (p×p)(xm–n,ym–n), (xm–n,ym–n)
≤maxr(x,y) +,r(x,y) +, M(x,y)τn
≤r(x,y) +, ()
whereS=p(xn,xm) +p(yn,ym). On the other hand, (P) impliesr(x,y) –≤p(xn,xn) +
p(yn,yn)≤p(xn,xm) +p(yn,ym) and so () is obtained.
As a result of () and completeness of (X,p), there exists (x.,y.)∈X×Xsuch that r(x,y) = lim
m,n→∞(p×p)
(xn,yn), (xm,ym)
= lim
n→∞(p×p)
(xn,yn),
x.,y.= (p×p)x.,y.,x.,y.. () Step IV: We show that for eachn≥ the following holds:
First, note that by (P) applied toporp×p, we have
(p×p)x.,y.,x.,y.≤(p×p)x.,y.,Fx.,y.,Fy.,x.. () By the help of the triangle inequality applied toporp×p, for eachn≥, we have
T ≤(p×p)x.,y., (xn+,yn+)
+ (p×p)(xn+,yn+),
Fx.,y.,Fy.,x. – (p×p)(xn+,yn+), (xn+,yn+)
, ()
whereT= (p×p)((x.
,y.), (F(x.,y.),F(y.,x.))). By the condition () and definition ofp×p,
we have
L≤ max
kpxn,x.
+lpyn,y.
,
(p×p)(xn,yn), (xn,yn)
, (p×p)x.,y.,x.,y. +
max
kpyn,y.
+lpxn,x.
,
(p×p)(yn,xn), (yn,xn)
, (p×p)y.,x.,y.,x.
≤maxτpxn,x.
+pyn,y. ,
(p×p)(xn,yn), (xn,yn)
, (p×p)x.,y.,x.,y.,
whereL= (p×p)((xn+,yn+), (F(x.,y.),F(y.,x.))). Without loss of generality (by passing
to subsequence if necessary), we may assume that for eachn≥, we have one of the fol-lowing:
(p×p)(xn+,yn+),
Fx.,y.,Fy.,x.≤τpxn,x.
+pyn,y. ,
or
(p×p)(xn+,yn+),
Fx.,y.,Fy.,x.≤(p×p)(xn,yn), (xn,yn)
,
or
(p×p)(xn+,yn+),
Fx.,y.,Fy.,x.≤(p×p)x.,y.,x.,y..
Substituting each of these cases in (), lettingn→ ∞and using () together with (), we arrive at ().
Step V: We prove that (X×X)p×p=∅.
For eachk∈N, pick (xk,yk)∈X×Xwith
(p×p)(xk,yk), (xk,yk)
<ρp×p+
k. ()
We show that
lim
m,n→∞(p×p)
Given> putn= [(–τ)] + . Ifk≥nthen by () we have
ρp×p≤(p×p)Fx.k,y.k
,Fy.k,x.k,Fxk.,y.k,Fy.k,x.k =pFx.k,y.k,Fx.k,y.k+pFy.k,x.k,Fy.k,x.k
≤px.k,x.k+py.k,y.k
= (p×p)x.k,y.k,x.k,y.k=r(xk,yk) ≤(p×p)(xk,yk), (xk,yk)
<ρp×p+
k
≤ρp×p+ n
<ρp×p+( –τ)
. ()
Hence, we conclude that for allk≥n,
Uk:= (p×p)
x.k,y.k,x.k,yk.– (p×p)Fx.k,y.k,Fy.k,x.k,Fx.k,y.k,Fy.k,x.k
≤ ( –τ)
+
ρp×p– (p×p)Fx.k,y.k,Fy.k,xk.,Fx.k,yk.,Fy.k,x.k < ( –τ)
()
and
(p×p)x.k,y.k,x.k,y.k≤ρp×p+( –τ)
. ()
Now, ifm,n≥n, then by the triangle inequality applied to (p×p), Step IV and (), we
have
(p×p)x.n,y.n,x.m,y.m =Un+Um+p
Fx.n,y.n,Fx.m,y.m+pFy.n,x.n,Fy.m,x.m
≤Un+Um+max
τ(p×p)x.n,yn.,x.m,y.m ,
(p×p)x.n,y.n,x.n,y.n, (p×p)x.m,y.m,x.m,y.m. ()
Hence, using () and (), we obtain
ρp×p≤(p×p)x.n,y.n,x.m,y.m
≤max
,
( –τ)
+ (p×p)
x.n,y.n,x.n,y.n, ( –τ)
+ (p×p)
x.m,y.m,x.m,y.m
≤max
,( –τ) +ρp×p
This shows () and so{(x.
n,y.n)}is Cauchy in the complete partial metric space (X×X,p×
p). Therefore, there exists (x,y)∈X×Xsuch that
(p×p)(x,y), (x,y)
= lim
n→∞(p×p)
x.n,y.n,x.n,y.n= lim
n→∞(p×p)
x.n,y.n,x.m,y.m=ρp×p. ()
In particular (x,y)∈(X×X)p×pand so (X×X)p×p=∅.
Now, let (x,y)∈(X×X)p×pbe arbitrary. Then by (), (P) applied top×pand (),
we have
ρp×p≤(p×p)x.,y.,x.,y.
≤(p×p)x.,y.,Fx.,y.,Fy.,x.= (p×p)x.,y.,x.,y.=r(x,y)
≤(p×p)(x,y), (x,y)
=ρp×p. ()
Therefore, (P) applied top×pimplies that
(p×p)x.,y.,x.,y.= (p×p)x.,y.,Fx.,y.,Fy.,x.
and so (x.,y.) is a coupled fixed point ofF. Clearly, () implies that{(xn,yn)}converges
to the coupled fixed point (x.,y.) with respect todp×dpwhich proves (c). To complete
the proof of (b), assume (x,y)∈(X×X)p×pand (u,v)∈(X×X)p×pare both coupled fixed
points ofF. Then by () we have
(p×p)(x,y), (u,v)= (p×p)F(x,y),F(y,x),F(u,v),F(v,u)
≤maxτ(p×p)(x,y), (u,v), (p×p)(x,y), (x,y),
(p×p)(u,v), (u,v). ()
From which it follows either (p×p)((x,y), (u,v))≤τ(p× p)((x,y), (u,v)) and so (p× p)((x,y), (u,v)) = implies that (x,y) = (u,v), or ρp×p ≤ (p × p)((x,y), (u,v))≤ (p× p)((x,y), (x,y)) = (p×p)((u,v), (u,v)) = ρp×p, and hence (P) applied to p× p implies
(x,y) = (u,v).
Remark Although Theorem does not imply the uniqueness of the fixed point, it is easy to see that, under the assumptions made, if (x,y) and (u,v) are both coupled fixed points forFsatisfying (p×p)((x,y), (x,y)) = (p×p)((u,v), (u,v)), then (x,y) = (u,v). If the partially contractive condition () is replaced by the somewhat stronger condition below, the uniqueness of the coupled fixed point is guaranteed.
Theorem Let(X,p)be a complete partial metric space,≤k,l< and F:X×X→X be a strong partially contractive mapping. Then there exists a unique coupled fixed point (x,y). Furthermore, (x,y)∈(X×X)p×p and for each(x,y)∈(X×X)p×p, the sequence {(xn,yn)} ∈X×X defined by
converges to(x,y)with respect to dp×dp. That is
lim
n→∞(dp×dp)
(xn,yn), (x,y)
= lim
n→∞
dp(xn,x) +dp(yn,y)
= .
Proof By Theorem , we only need to prove the uniqueness of the coupled fixed point. If (x,y) and (u,v) are two coupled fixed points, then
G=p(x,u) +p(y,v) =pF(x,y),F(u,v)+pF(y,x),F(v,u)
≤
max
kp(x,u) +lp(y,v),(p×p)((x,y), (x,y)) + (p×p)((u,v), (u,v))
+ max
kp(x,u) +lp(y,v),(p×p)((x,y), (x,y)) + (p×p)((u,v), (u,v))
≤max
τp(x,u) +p(y,v),(p×p)((x,y), (x,y)) + (p×p)((u,v), (u,v))
, ()
whereG= (p×p)((x,y), (u,v)) andτ =max{k,l}. From () we distinguish the following cases:
Case : If (p×p)((x,y), (u,v))≤τ(p(x,u) +p(y,v)) =τ(p×p)((x,y), (u,v)), then (p× p)((x,y), (u,v)) = and so (x,y) = (u,v).
Case : If (p×p)((x,y), (u,v))≤(p×p)((x,y),(x,y))+(p×p)((u,v),(u,v)), then (dp×dp)((x,y), (u,v)) =
and so (x,y) = (u,v).
As a corollary, we obtain the already mentioned result [] stated in Theorem . As well, let us remark that the result of Aydi in Theorem is valid also for -complete partial metric spaces.
Corollary Let(X,p)be a-complete partial metric space. Suppose that the mapping F:X×X→X satisfies the following contractive condition for all x,y,u,v∈X:
pF(x,y),F(u,v)≤kp(x,u) +lp(y,v), ()
where≤k,l< with k+l< . Then F has a unique coupled fixed point(u,v)such that (p×p)((u,v), (u,v)) = . Also for each(x,y)∈X×X the sequence{(xn+,yn+)}converges
to(u,v)with respect to the metric dp×dp.
Proof The condition () implies that (p× p)((xn+,yn+), (xn+,yn+)) ≤ (k + l)n(p×
p)((x,y), (x,y)), which, in turn, by () implies (p×p)((x.,y.)) = . But then by ()
we have (p×p)((x.,y.), (F(x.,y.),F(y.,x.))) = so the dp ×dp-limit of the sequence {(xn+,yn+)}is actually the unique coupled fixed point (x.,y.).
Example Define p: [, ]×[, ]→[, ] byp(x,y) =max{x,y}. Then ([, ],p) is a complete partial metric space. LetF: [, ]×[, ]→[, ] be defined by
F(x,y) =|x–y| .
(a) There are no ≤k,l< withk+l< such thatp(F(x,y),F(u,v))≤kp(x,u) +lp(y,v), for allx,y,u,v∈[, ]. That is,Fdoes not verify the assumptions of Theorem . Indeed, if we assume that there exist ≤k,l< withk+l< such that for allx,y,u,v∈X= [, ], we have
pF(x,y),F(u,v)≤kp(x,u) +lp(y,v),
then
pF(, ),F(, )=
≤kp(, ) +lp(, ) =k, ()
and
pF(, ),F(, )=
≤kp(, ) +lp(, ) =l, ()
then we havek+l≥, which is a contradiction.
(b) There are no ≤k,l< withk+l< such thatp(F(x,y),F(u,v))≤kp(F(x,y),x) + lp(F(u,v),u). That is,Fdoes not verify the assumptions of Theorem . Indeed, if we assume that there exist ≤k,l< withk+l< such that for allx,y,u,v∈X= [, ], we have
pF(x,y),F(u,v)≤kpF(x,y),x+lpF(u,v),u. ()
Then we conclude that
pF(, ),F(, )= ≤kp
,
+p(, ) =k <
,
which is a contradiction.
(c)Fis partially contractive. That isFverifies the assumptions of Theorem . Hence, it has the coupled fixed point (, ). The condition () is clearly satisfied since|x–y| ≤x+y for allx,y∈X= [, ].
(d) Note that (X×X)p×p={(, )}and hence the coupled Picard sequence{(xn,yn)}
con-verges for (x,y) = (, ) and for (x,y)∈(X×X)\{(, )}may converge or not depending
on the particular choice of (x,y).
(e) IfFis replaced byH(x,y) =|x–y|, then it is strong partially contractive, and hence by Theorem , we guarantee the uniqueness of the coupled fixed point (, ).
Competing interests
The author declares that they have no competing interests.
Received: 14 May 2012 Accepted: 24 August 2012 Published: 12 September 2012 References
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