R E S E A R C H
Open Access
Common fixed point results for weak
contractive mappings in ordered
b-dislocated
metric spaces with applications
Nawab Hussain
1, Jamal Rezaei Roshan
2*, Vahid Parvaneh
3and Mujahid Abbas
4*Correspondence:
2Department of Mathematics,
Qaemshahr Branch, Islamic Azad University, Qaemshahr, Iran Full list of author information is available at the end of the article
Abstract
We first introduce a new concept ofb-dislocated metric space as a generalization of dislocated metric space and analyze different properties of such spaces.
A fundamental result for the convergence of sequences inb-dislocated metric spaces is established and is employed to prove some common fixed point results for four mappings satisfying the generalized weak contractive condition in partially ordered
b-dislocated metric spaces. Moreover, some examples and applications to integral equations are given here to illustrate the usability of the obtained results.
MSC: Primary 47H10; secondary 54H25
Keywords: coincidence point; common fixed point; dislocated metric space;
b-dislocated metric space; dominating and dominated maps; altering distance function
1 Introduction and preliminaries
The Banach contraction principle is one of the simplest and most applicable results of metric fixed point theory. It is a popular tool for proving the existence of solution of prob-lems in different fields of mathematics. There are several generalizations of the Banach contraction principle in literature on metric fixed point theory [–]. Hitzler and Seda [] introduced the concept of dislocated topologies and named their corresponding gen-eralized metric a dislocated metric. They have also established a fixed point theorem in complete dislocated metric spaces to generalize the celebrated Banach contraction prin-ciple. The notion of dislocated topologies has useful applications in the context of logic programming semantics (see []). Further useful results can be seen in [–].
Definition .[] LetXbe a nonempty set. A mappingdl:X×X→[,∞) is called a dislocated metric (or simplydl-metric) if the following conditions hold for anyx,y,z∈X:
(i) Ifdl(x,y) = , thenx=y;
(ii) dl(x,y) =dl(y,x);
(iii) dl(x,y)≤dl(x,z) +dl(z,y).
The pair (X,dl) is called a dislocated metric space or adl-metric space. Note that when
x=y,dl(x,y) may not be .
Example . IfX=R+∪ {}, thend
l(x,y) =x+ydefines a dislocated metric onX.
Definition .[] A sequence{xn}in adl-metric space is called: () a Cauchy sequence if, givenε> , there existsn∈Nsuch that for alln,m≥n, we havedl(xm,xn) <εor
limn,m→∞dl(xn,xm) = , () convergent with respect todl if there existsx∈Xsuch that
dl(xn,x)→ asn→ ∞. In this case,xis called the limit of{xn}and we writexn→x.
Adl-metric spaceX is called complete if every Cauchy sequence inXconverges to a point inX.
Definition . A nonempty setX is called an ordered dislocated metric space if it is
equipped with a partial orderingand there exists a dislocated metricdlonX.
Definition . Let (X,) be a partially ordered set. Thenx,y∈Xare called comparable ifxyoryxholds.
Definition .[] Let (X,) be a partially ordered set. A self-mappingf onXis called dominating ifxfxfor eachxinX.
Example .[] LetX= [, ] be endowed with the usual ordering, and letf :X→Xbe defined byfx=√nx. Sincex≤xn=fxfor allx∈X, thereforef is a dominating map.
Definition .[] Let (X,) be a partially ordered set. A self-mappingf onXis called dominated iffxxfor eachxinX.
Example .[] LetX= [, ] be endowed with the usual ordering, and letf:X→Xbe defined byfx=xnfor somen∈N. Sincefx=xn≤xfor allx∈X, thereforef is a dominated map.
In the following, we give the definition of ab-dislocated metric space.
Definition . LetX be a nonempty set. A mappingbd:X×X→[,∞) is called a
b-dislocated metric (or simplybd-metric) if the following conditions hold for anyx,y,z∈X ands≥:
(bd) Ifbd(x,y) = , thenx=y;
(bd) bd(x,y) =bd(y,x);
(bd) bd(x,y)≤s(bd(x,z) +bd(z,y)).
The pair (X,bd) is called ab-dislocated metric space or abd-metric space. It should be noted that the class ofbd-metric spaces is effectively larger than that ofdl-metric spaces, since abd-metric is abl-metric whens= .
Here, we present an example to show that in general ab-dislocated metric need not be abl-metric.
Example . Let (X,dl) be a dislocated metric space, andbd(x,y) = (bl(x,y))p, wherep>
If <p<∞, then the convexity of the functionf(x) =xp (x> ) implies that (a+b
)p≤
(ap+bp). Hence, (a+b)p≤p–(ap+bp) holds. Thus, for eachx,y,z∈X, we obtain that
bd(x,y) =
dl(x,y) p
≤dl(x,z) +dl(z,y) p
≤p–dl(x,z) p
+dl(z,y) p
= p–b
d(x,z) +bd(z,y)
.
So, condition (bd) of Definition . is also satisfied andbdis abd-metric.
However, if (X,dl) is a dislocated metric space, then (X,bd) is not necessarily a dislocated metric space. For example, ifX=Ris the set of real numbers, thendl(x,y) =|x|+|y|is a dislocated metric, andbd(x,y) = (|x|+|y|) is ab-dislocated metric onRwiths= , but not a dislocated metric onR.
Recently, Sarma and Kumari [] established the existence of a topology induced by a dislocated metric which is metrizable with a family of sets{B(x,ε)∪ {x}:x∈X,ε> }as a base, whereB(x,ε) ={y∈X:dl(x,y) <ε}for allx∈Xandε> . Also,B(x,ε) ={y∈X:
dl(x,y)≤ε}is a closed ball.
On the similar lines, we show that eachb-dislocated metric space onXgenerates a topol-ogyτbdwhose base is the family of openbd-balls
Bbd(x,ε) =
y∈X:bd(x,y) <ε
.
Definition . We say that a net (xα:α∈) inXconverges toxin (X,bd) and write
limα∈xα=xiflimα∈bd(xα,x) = .
Note that the limit of a net in (X,bd) is unique. ForA⊆X, we writeD(A) ={x∈X:
xis a limit of a net in (A,bd)}.
Proposition . If A,B⊆X,then (i) D(A) =∅ifA=∅,
(ii) D(A)⊆D(B)ifA⊆B, (iii) D(A∪B) =D(A)∪D(B), (iv) D(D(A))⊆D(A).
Proof To prove (i), (ii) and (iii), we refer to []. To prove (iv), letx∈D(D(A)). Suppose that for eachα in, (xαβ:β∈(α)) is a net inAsuch thatxα=limβ∈(α)xαβ. Thus, for
each positive integeri, there isαi∈such thatbd(xαi,x) <
is, andβi∈(αi) such that
bd(xαiβi,xαi) <
is. Takeαiβi=γifor eachi, then{γ,γ,γ, . . .}is a directed setγi<γjifi<j,
andbd(xγi,x)≤s(bd(xγi,xαi) +bd(xαi,x)) <
i. This implies thatx∈D(A).
As a corollary, we have the following.
Corollary . Let,for all A⊂X,A=A∪D(A).Then the operation A→A on P(X) sat-isfies Kuratowski’s closure axioms[]:
(i) ∅=∅,
Consequently, we have the following.
Theorem . Letϒ be the family of all subsets A of X for which A=A andτbdare the complements of members ofϒ.Then theτbd is a topology for X and theτbd-closure of a subset A of X is A.
Definition . The topologyτbdobtained in Theorem . is called the topology induced
bybdand simply referred to as thebd-topology ofX; and it is denoted by (X,bd,τbd).
Now we state some propositions and corollaries in (X,bd,τbd) which can be proved
fol-lowing similar arguments to those given in [].
Proposition . Let A⊆X.Then x∈D(A)iff for everyδ> ,Bδ(x)∩A=∅.
Corollary . x∈A⇐⇒x∈A or Bδ(x)∩A=∅,∀δ> .
Corollary . A set A⊆X is open in(X,bd,τbd)if and only if for every x∈A,there isδ> such that{x} ∪Bδ(x)⊆A.
Proposition . If x∈X andδ> ,then{x} ∪Bδ(x)is an open set in(X,bd,τbd).
Corollary . If x∈X and Vr(x) =Br(x)∪ {x}for r> ,then the collection{Vr(x)|x∈X}
is an open base at x in(X,bd,τbd).If bdis a b-metric and V=B(x),thenτbdcoincides with the metric topology.
Proposition . (X,bd,τbd)is a Hausdorff space.
Proof Ifx,y∈Xandbd(x,y)
s =r> , thenVr(x)∩Vr(y) =∅.
Corollary . If x∈X,then the collection{V(x)|x∈X}is an open base at x for(X,bd,τbd). Hence, (X,bd,τbd)is first countable.
Remark . The above corollary enables us to deal with sequences instead of nets.
Motivated by Proposition . in [], we have the following proposition for theb -dislo-cated metric space.
Proposition . Let(X,bd)be a b-dislocated metric space.The following three conditions
are equivalent:
(i) For allx∈X,we havebd(x,x) = .
(ii) bdis ab-metric.
(iii) For allx∈Xand allr> ,we haveBr(x)=∅.
Proof We show that (iii) implies (i). SinceBrs(x)=∅for allr> , there exists somey∈X
withbd(x,y) <rs. But for ally∈X, we havebd(x,x)≤sbd(x,y). Therefore,bd(x,x) <rfor allr> . Hence,bd(x,x) = .
Definition . A sequence{xn}in ab-dislocated metric space (X,bd) converges with respect tobd (bd-convergent) if there existsx∈Xsuch that bd(xn,x) converges to as
n→ ∞. In this case,xis called the limit of{xn}, and we writexn→x.
Proposition . Limit of a convergent sequence in a b-dislocated metric space is unique.
Proof Letxandybe limits of the sequence{xn}. By properties (bd) and (bd) of
Defini-tion ., it follows thatbd(x,y)≤s(bd(xn,x) +bd(xn,y))→. Hence,bd(x,y) = , and by property (bd) of Definition . it follows thatx=y.
Definition . A sequence{xn} in ab-dislocated metric space (X,bd) is called abd -Cauchy sequence if, givenε> , there exitsn∈N such that for alln,m≥n, we have
bd(xm,xn) <εorlimn,m→∞bd(xn,xm) = .
Proposition . Every convergent sequence in a b-dislocated space is bd-Cauchy.
Proof Let{xn}be a sequence which converges to some x, andε> . Then there exists
n∈Nwithbd(xn,x) <εsfor alln≥n. Form,n≥n, we obtainbd(xn,xm)≤s(bd(xn,x) +
bd(xm,x)) < sεs=ε. Hence,{xn}isbd-Cauchy.
Definition . Ab-dislocated metric space (X,bd) is called complete if everybd-Cauchy sequence inXisbd-convergent.
The following example shows that in general ab-dislocated metric is not continuous.
Example . LetX=N∪ {∞}andbd:X×X→Rbe defined by
bd(m,n) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
m+
n ifm,nare even ormn=∞, ifmandnare odd andm=n,
otherwise.
Then it is easy to see that for allm,n,p∈X, we have
bd(m,p)≤
bd(m,n) +bd(n,p)
.
Thus, (X,bd) is ab-dislocated metric space. Letxn= nfor eachn∈N. Then
bd(n,∞) =
n→ asn→ ∞,
that is,xn→ ∞, butbd(xn, ) = bd(∞, ) asn→ ∞.
We need the following simple lemma about thebd-convergent sequences in the proof of our main results.
Lemma . Let(X,bd)be a b-dislocated metric with parameter s≥.Suppose that{xn}
and{yn}are bd-convergent to x,y,respectively.Then we have
sbd(x,y)≤lim infn→∞ bd(xn,yn)≤lim supn→∞ bd(xn,yn)≤s
b
In particular,if bd(x,y) = , then we havelimn→∞bd(xn,yn) = =bd(x,y). Moreover,for
each z∈X,we have
sbd(x,z)≤lim infn→∞ bd(xn,z)≤lim supn→∞ bd(xn,z)≤sbd(x,z).
In particular,if bd(x,z) = ,then we havelimn→∞bd(xn,z) = =bd(x,z).
Proof Using the triangle inequality in ab-dislocated metric space, it is easy to see that
bd(x,y)≤sbd(x,xn) +sbd(xn,yn) +sbd(yn,y) and
bd(xn,yn)≤sbd(xn,x) +sbd(x,y) +sbd(y,yn).
Taking the lower limit asn→ ∞in the first inequality and the upper limit asn→ ∞in the second inequality, the result follows. Similarly, using again the triangle inequality, the
last assertion follows.
Definition .[] Letf andgbe two self-maps on a nonempty setX. Ifw=fx=gx, for somexinX, thenxis called a coincidence point off andg, wherewis called a point of coincidence off andg.
Definition .[] Letf andgbe two self-maps defined on a setX. Thenf andgare said to be weakly compatible if they commute at every coincidence point.
Definition . Let (X,bd) be ab-dislocated metric space. Then the pair (f,g) is said to be compatible if and only iflimn→∞bd(fgxn,gfxn) = , whenever{xn}is a sequence inXso thatlimn→∞fxn=limn→∞gxn=tfor somet∈X.
2 Common fixed point results Suppose that
=ψ: [,∞)→[,∞)|ψis a continuous non-decreasing function with
ψ(t) = ⇔t=
and
=ϕ: [,∞)→[,∞)|ϕis a lower semi-continuous function with
ϕ(t) = ⇔t= .
Theorem . Let(X,bd,)be an ordered complete b-dislocated metric space,and let f,
g,S and T be four self-maps on X such that(f,g)and(S,T)are dominated and dominat-ing maps,respectively,with fX⊆TX and gX⊆SX.Suppose that for all two comparable elements x,y∈X,
ψsbd(fx,gy)
≤ψMs(x,y)
–ϕMs(x,y)
is satisfied,where
Ms(x,y) =max
bd(Sx,Ty),bd(fx,Sx),bd(gy,Ty),
bd(Sx,gy) +bd(fx,Ty) s
, (.)
ψ∈andϕ∈.If for every non-increasing sequence{xn}and a sequence{yn}with yn
xn,for all n such that yn→u,we have uxnand either
(a) (f,S)are compatible,f orSis continuous and(g,T)is weakly compatible,or
(a) (g,T)are compatible,gorTis continuous and(f,S)is weakly compatible,
then f,g,S and T have a common fixed point.Moreover,the set of common fixed points of f,g,S and T is well ordered if and only if f,g,S and T have one and only one common fixed point.
Proof Letxbe an arbitrary point inX. We define inductively the sequences{xn}and{yn} inXby
yn+=fxn=Txn+, yn+=gxn+=Sxn+, n= , , , . . . .
This can be done asfX⊆TXandgX⊆SX. By given assumptions,xn+Txn+=fxn
xnandxnSxn=gxn–xn–. Thus, we havexn+xnfor alln≥. We will show that{yn}isbd-Cauchy. Suppose thatbd(yn,yn+) > for everyn. If not, then for somek,
bd(yk,yk+) = , and from (.), we obtain
ψbd(yk+,yk+)
≤ψsbd(yk+,yk+)
=ψsbd(fxk,gxk+)
≤ψMs(xk,xk+)
–ϕMs(xk,xk+)
, (.)
where
Ms(xk,xk+) = max
bd(Sxk,Txk+),bd(fxk,Sxk),bd(gxk+,Txk+),
bd(Sxk,gxk+) +bd(fxk,Txk+)
s
=max
bd(yk,yk+),bd(yk+,yk),bd(yk+,yk+),
bd(yk,yk+) +bd(yk+,yk+)
s
=max
, ,bd(yk+,yk+),
bd(yk,yk+) +bd(yk+,yk+)
s
=bd(yk+,yk+), (.)
since
bd(yk,yk+) +bd(yk+,yk+)
s ≤
sbd(yk,yk+) +sbd(yk+,yk+) + sbd(yk,yk+)
s
=sbd(yk+,yk+)
So, from (.) and (.), we obtain that
ψbd(yk+,yk+)
≤ψbd(yk+,yk+)
–ϕbd(yk+,yk+)
,
which givesϕ(bd(yk+,yk+))≤ and soyk+=yk+, which further implies thatyk+=
yk+. Thus,{yn}becomes a constant sequence, hence,ynis a Cauchy sequence.
Now, takebd(yn,yn+) > for eachn. Asxnandxn+are comparable, so from (.) we
have
ψbd(yn+,yn+)
≤ψsbd(yn+,yn+)
=ψsbd(fxn,gxn+)
≤ψMs(xn,xn+)
–ϕMs(xn,xn+)
≤ψMs(xn,xn+)
. (.)
Hence
bd(yn+,yn+)≤Ms(xn,xn+), (.)
where
Ms(xn,xn+) = max
bd(Sxn,Txn+),bd(fxn,Sxn),bd(gxn+,Txn+),
bd(Sxn,gxn+) +bd(fxn,Txn+)
s
=max
bd(yn,yn+),bd(yn+,yn),bd(yn+,yn+),
bd(yn,yn+) +bd(yn+,yn+)
s
≤max
bd(yn,yn+),bd(yn+,yn+),
sbd(yn,yn+) +sbd(yn+,yn+) + sbd(yn,yn+)
s
=max
bd(yn,yn+),bd(yn+,yn+),
sbd(yn,yn+) +sbd(yn+,yn+)
s
=maxbd(yn,yn+),bd(yn+,yn+)
.
If for some n, bd(yn+,yn+)≥bd(yn,yn+) > , then (.) gives that Ms(xn,xn+) =
bd(yn+,yn+) and from (.) we have
ψbd(yn+,yn+)
≤ψsbd(yn+,yn+)
≤ψMs(xn,xn+)
–ϕMs(xn,xn+)
=ψbd(yn+,yn+)
–ϕbd(yn+,yn+)
which yields thatϕ(bd(yn+,yn+))≤, or, equivalently,bd(yn+,yn+) = , a
contradic-tion.
Hence, Ms(xn,xn+)≤ bd(yn,yn+). Since Ms(xn,xn+) ≥bd(yn,yn+), therefore,
bd(yn+,yn+)≤Ms(xn,xn+) =bd(yn,yn+). Following similar arguments to those given
above, we have
bd(yn+,yn+)≤Ms(xn+,xn+) =bd(yn+,yn+). (.)
Therefore,{bd(yn,yn+)}is a non-increasing sequence and so there existsr≥ such that
lim
n→∞bd(yn–,yn) =nlim→∞Ms(xn,xn+) =r. Suppose thatr> . As
ψbd(yn+,yn+)
≤ψsbd(yn+,yn+)
≤ψMs(xn,xn+)
–ϕMs(xn,xn+)
,
by taking the upper limit asn→ ∞, we obtain
ψ(r)≤ψ(r) –lim inf
n→∞ ϕ
Ms(xn,xn+)
=ψ(r) –ϕ
lim inf
n→∞ Ms(xn,xn+)
=ψ(r) –ϕ(r),
a contradiction. Hence
lim
n→∞bd(yn–,yn) = . (.)
Now, we prove that{yn}is abd-Cauchy sequence. To do this, it is sufficient to show that the subsequence{yn}isbd-Cauchy inX. Assume on the contrary that{yn}is not abd-Cauchy sequence. Then there existsε> for which we can find subsequences{ymk}and{ynk}of
{yn}so thatnkis the smallest index for which nk> mk>k,
bd(ymk,ynk)≥ε (.)
and
bd(ymk,ynk–) <ε. (.)
Using the triangle inequality and (.), we obtain that
ε≤bd(ymk,ynk)≤sbd(ymk,ymk+) +sbd(ymk+,ynk).
Taking the upper limit ask→ ∞and using (.), we obtain
ε
Using the triangle inequality and (.), we have
ε≤bd(ymk,ynk)
≤sbd(ymk,ynk–) +s
b
d(ynk–,ynk–) +s
b
d(ynk–,ynk)
<εs+sbd(ynk–,ynk–) +s
b
d(ynk–,ynk).
Taking the upper limit ask→ ∞and using (.), we obtain
ε≤lim sup
k→∞ bd(ymk,ynk)≤εs. (.)
Also,
ε≤bd(ymk,ynk)≤sbd(ymk,ynk–) +sbd(ynk–,ynk).
Hence
ε
s ≤lim supk→∞ bd(ymk,ynk–).
On the other hand, we have
bd(ymk,ynk–)≤sbd(ymk,ynk) +sbd(ynk,ynk–).
So, from (.) and (.), we have
lim sup
k→∞ bd(ymk,ynk–)≤s
lim sup
k→∞ bd(ymk,ynk)≤εs
.
Consequently,
ε
s ≤lim supk→∞ bd(ymk,ynk–)≤εs
. (.)
Similarly,
ε
s ≤lim sup
k→∞
bd(ymk+,ynk–)≤εs
. (.)
Asxmk andxnk–are comparable, from (.) we have ψsbd(ymk+,ynk)
=ψsbd(fxmk,gxnk–)
≤ψMs(xmk,xnk–)
–ϕMs(xmk,xnk–)
,
where
Ms(xmk,xnk–) =max
bd(Sxmk,Txnk–),bd(fxmk,Sxmk),bd(gxnk–,Txnk–), bd(Sxmk,gxnk–) +bd(fxmk,Txnk–)
s
=max
bd(ymk,ynk–),bd(ymk+,ymk),bd(ynk,ynk–), bd(ymk,ynk) +bd(ymk+,ynk–)
s
.
Taking the upper limit and using (.) and (.)-(.), we get
ε+ ε
s s =min
ε s,
ε+ ε
s s
≤lim sup
k→∞ Ms(xmk,xnk–)
=max
lim sup
k→∞
bd(ymk,ynk–), , ,
lim supk→∞bd(ymk,ynk) +lim supk→∞bd(ymk+,ynk–)
s
≤max
εs,εs+εs
s
=εs.
Hence, we have
ε+ ε
s
s ≤lim supk→∞ Ms(xmk,xnk–)≤εs
. (.)
Similarly, we can obtain
ε+sε
s ≤lim infk→∞ Ms(xmk,xnk–)≤εs
. (.)
As
ψsbd(ymk+,ynk)
=ψsbd(fxmk,gxnk–)
≤ψMs(xmk,xnk–)
–ϕMs(xmk,xnk–)
,
so, by taking the upper limit ask→ ∞, and from (.) and (.), we obtain
ψεs=ψ
sε
s
≤ψ
slim sup
k→∞ bd(ymk+,ynk)
≤ψ
lim sup
k→∞ Ms(xmk,xnk–)
–lim inf
k→∞ ϕ
Ms(xmk,xnk–)
≤ψεs–ϕ
lim inf
k→∞ Ms(xmk,xnk–)
≤ψεs–ϕ
lim inf
k→∞ Ms(xmk,xnk–)
,
which implies that
ϕ
lim inf
k→∞ Ms(xmk,xnk–)
so lim infMs(xmk,xnk–) = , a contradiction to (.). Hence {yn} is a bd-Cauchy
se-quence inX. SinceXis complete, there existsy∈Xsuch that
lim
n→∞fxn=nlim→∞Txn+=nlim→∞gxn+=nlim→∞Sxn=y. Now, we show thatyis a common fixed point off,g,SandT.
Assume that (a) holds andSis continuous. Then
lim
n→∞S
x
n+=Sy and lim
n→∞Sfxn=Sy. Using the triangle inequality, we have
bd(fSxn,Sy)≤s
bd(fSxn,Sfxn) +bd(Sfxn,Sy)
.
Since the pair (f,S) is compatible,limn→∞bd(fSxn,Sfxn) = . So, by taking the limit when
n→ ∞in the above inequality, we have
lim
n→∞bd(fSxn,Sy)≤s
lim
n→∞bd(fSxn,Sfxn) +nlim→∞bd(Sfxn,Sy)
= .
Hence,limn→∞fSxn=Sy. AsSxn+=gxn+xn+, from (.) we obtain
ψsbd(fSxn+,gxn+)
≤ψMs(Sxn+,xn+)
–ϕMs(Sxn+,xn+)
, (.)
where
Ms(Sxn+,xn+)
=max
bd
Sxn+,Txn+
,bd
fSxn+,Sxn+
,bd(gxn+,Txn+),
bd(Sxn+,gxn+) +bd(fSxn+,Txn+)
s
.
Now, by using Lemma ., we get
lim sup
n→∞
Ms(Sxn+,xn+)≤max
sbd(Sy,y), , ,
sbd(Sy,y) +sbd(Sy,y) s
=sbd(Sy,y).
Hence, by taking the upper limit in (.) and using Lemma ., we obtain
ψsbd(Sy,y)
=ψ
s
sbd(Sy,y)
≤ψsbd(Sy,y)
–ϕsbd(Sy,y)
≤ψsbd(Sy,y)
–ϕsbd(Sy,y)
which givesϕ(sb
Now, sincegxn+xn+andgxn+→yasn→ ∞, thenyxn+and from (.) we
have
ψsbd(fy,gxn+)
≤ψMs(y,xn+)
–ϕMs(y,xn+)
, (.)
where
Ms(y,xn+) =max
bd(Sy,Txn+),bd(fy,Sy),bd(gxn+,Txn+),
bd(Sy,gxn+) +bd(fy,Txn+)
s
.
Taking the upper limit asn→ ∞in (.) and using Lemma ., we have
ψsbd(fy,y)
=ψ
s
sbd(fy,y)
≤ψbd(fy,y)
–ϕbd(fy,y)
≤ψsbd(fy,y)
–ϕbd(fy,y)
,
which implies thatϕ(bd(fy,y))≤, sofy=y.
Sincef(X)⊆T(X), there exists a pointv∈Xsuch thatfy=Tv. Suppose thatgv=Tv. SincevTv=fyy, from (.) we have
ψbd(Tv,gv)
=ψbd(fy,gv)
≤ψMs(y,v)
–ϕMs(y,v)
, (.)
where
Ms(y,v) =max
bd(Sy,Tv),bd(fy,Sy),bd(gv,Tv),
bd(Sy,gv) +bd(fy,Tv) s
=bd(gv,Tv).
So, from (.) we have
ψbd(Tv,gv)
≤ψbd(gv,Tv)
–ϕbd(gv,Tv)
,
a contradiction. Thereforegv=Tv. Since the pair (g,T) is weakly compatible,gy=gfy=
gTv=Tgv=Tfy=Tyandyis the coincidence point ofgandT. SinceSxnxnandSxn→
yasn→ ∞, it implies thatyxnand from (.) we obtain
ψsbd(fxn,gy)
≤ψMs(xn,y)
–ϕMs(xn,y)
, (.)
where
Ms(xn,y) =max
bd(Sxn,Ty),bd(fxn,Sxn),bd(gy,Ty),
bd(Sxn,gy) +bd(fxn,Ty) s
Taking the upper limit asn→ ∞in (.) and using Lemma ., we have
max
sbd(y,gy),bd(gy,Ty),
s
sbd(y,gy)
≤lim inf
n→∞ Ms(xn,y)
≤lim sup
n→∞ Ms(xn,y)
≤max
sbd(y,gy),bd(gy,Ty), s
sbd(y,gy)
=maxsbd(y,gy),bd(gy,gy)
≤maxsbd(y,gy), sbd(y,gy)
= sbd(y,gy). (.)
Taking the upper limit asn→ ∞in (.) and using Lemma . and (.), we have
ψsbd(y,gy)
=ψ
s
sbd(y,gy)
≤ψ
lim sup
n→∞
Ms(xn,y)
–lim inf
n→∞ ϕ
Ms(xn,y)
≤ψsbd(y,gy)
–ϕ
lim inf
n→∞ Ms(xn,y)
≤ψsbd(y,gy)
–ϕ
lim inf
n→∞ Ms(xn,y)
,
which implies thatlim infn→∞Ms(xn,y) = , so we havey=gy. Therefore,fy=gy=Sy=
Ty=y.
The proof is similar whenf is continuous. Similarly, if (a) holds, then the result follows.
Now, suppose that the set of common fixed points off,g,SandTis well ordered. We show that they have a unique common fixed point. Assume on the contrary thatfu=gu=
Su=Tu=uandfv=gv=Sv=Tv=v, butu=v. By assumption, we can apply (.) to obtain
ψsbd(u,v)
=ψsbd(fu,gv)
≤ψsbd(fu,gv)
≤ψMs(u,v)
–ϕMs(u,v)
,
where
Ms(u,v) =max
bd(Su,Tv),bd(fu,Su),bd(gv,Tv),
bd(Su,gv) +bd(fu,Tv) s
=max
bd(u,v),bd(u,u),bd(v,v),
bd(u,v) +bd(u,v) s
=maxbd(u,v),bd(u,u),bd(v,v)
≤maxbd(u,v), sbd(u,v), sbd(u,v)
Hence
ψsbd(u,v)
≤ψsbd(u,v)
–ϕMs(u,v)
.
So, we haveMs(u,v) = , a contradiction. Thereforeu=v. The converse is obvious.
In the following theorem, we omit the continuity assumption off,g,TandSand replace the compatibility of the pairs (f,S) and (g,T) by weak compatibility of the pairs, and we show thatf,g,S and T have a common fixed point on X.
Theorem . Let(X,bd,)be an ordered complete b-dislocated metric space,and f,g,S
and T be four self-maps on X such that(f,g)and(S,T)are dominated and dominating maps,respectively,with fX⊆TX and gX⊆SX,and TX and SX are bd-closed subsets of X.
Suppose that for all two comparable elements x,y∈X,
ψsd(fx,gy)≤ψMs(x,y)
–ϕMs(x,y)
(.)
is satisfied,where
Ms(x,y) =max
bd(Sx,Ty),bd(fx,Sx),bd(gy,Ty),
bd(Sx,gy) +bd(fx,Ty) s
,
ψ∈andϕ∈.If for every non-increasing sequence{xn}and a sequence{yn}with yn
xn,for all n such that yn→u,we have uxn,and the pairs(f,S)and(g,T)are weakly
compatible,then f,g,S and T have a common fixed point.Moreover,the set of common fixed points of f,g,S and T is well ordered if and only if f,g,S and T have one and only one common fixed point.
Proof Following the proof of Theorem ., there existsy∈Xsuch that
lim
k→∞bd(yk,y) = . (.)
SinceT(X) isbd-closed and{yn+} ⊆T(X), thereforey∈T(X). Hence, there existsu∈X
such thaty=Tuand
lim
n→∞bd(yn+,Tu) =nlim→∞bd(Txn+,Tu) = . (.) Similarly, there existsv∈Xsuch thaty=Tu=Svand
lim
n→∞bd(yn,Sv) =nlim→∞bd(Sxn,Sv) = . (.) Now we prove thatvis a coincidence point off andS.
SinceTxn+→y=Svasn→ ∞, so, by assumption,Txn+Sv. Therefore, from (.)
we have
ψsbd(fv,gxn+)
≤ψMs(v,xn+)
–ϕMs(v,xn+)
where
Ms(v,xn+) = max
bd(Sv,Txn+),bd(fv,Sv),bd(gxn+,Txn+),
bd(Sv,gxn+) +bd(fv,Txn+)
s
=max
bd(Tu,Txn+),bd(fv,y),bd(gxn+,Txn+),
bd(Sv,yn+) +bd(fv,Txn+)
s
.
Taking the upper limit asn→ ∞and using (.)-(.) and Lemma ., we obtain that
max
bd(fv,y),
sbd(y,y),
sbd(y,y)
≤lim inf
n→∞ Ms(v,xn+)
≤lim sup
n→∞ Ms(v,xn+)
≤max
,bd(fv,y),sbd(y,y), +sb
d(fv,y) s
≤maxbd(fv,y), sbd(fv,y)
= sbd(fv,y). (.)
Taking the upper limit asn→ ∞in (.) and using (.) and Lemma ., we obtain that
ψsd(fv,y)=ψ
s
sd(fv,y)
≤ψsbd(fv,y)
–ϕ
lim inf
n→∞ Ms(v,xn+)
≤ψsbd(fv,y)
–ϕ
lim inf
n→∞ Ms(v,xn+)
,
which implies thatlim infn→∞Ms(v,xn+) = , so from (.) we obtainfv=y=Sv.
Asf andSare weakly compatible, we havefy=fSv=Sfv=Sy. Thus,yis a coincidence point off andS.
Similarly, it can be shown thatyis a coincidence point of the pair (g,T). Now, we show thatfy=gy. From (.) we have
ψsd(fy,gy)≤ψMs(y,y)
–ϕMs(y,y)
,
where
Ms(y,y) =max
bd(Sy,Ty),bd(fy,Sy),bd(gy,Ty),
bd(Sy,gy) +bd(fy,Ty) s
=max
bd(fy,gy),bd(fy,fy),bd(gy,gy),
bd(fy,gy) +bd(fy,gy) s
=maxbd(fy,gy),bd(fy,fy),bd(gy,gy)
≤maxbd(fy,gy), sbd(fy,gy), sbd(fy,gy)
= sbd(fy,gy).
So, we have
ψsd(fy,gy)≤ψMs(y,y)
–ϕMs(y,y)
≤ψsbd(fy,gy)
–ϕMs(y,y)
≤ψsbd(fy,gy)
–ϕMs(y,y)
,
which implies thatMs(y,y) = , so we havefy=gy. Therefore,fy=gy=Sy=Ty.
Now, similar to the proof of Theorem ., indeed from (.)-(.), we havegy=y. Therefore,fy=gy=Sy=Ty=y, as required. The last conclusion follows similarly as in the
proof of Theorem ..
Now, we give an example to support our result.
Example . LetX= [,∞) be equipped with theb-dislocated metricbd(x,y) = (x+y) wheres= and suppose that ‘’ is the usual ordering≤onX. Obviously, (X,bd,≤) is an ordered completeb-dislocated metric space. Letf,g,S,T:X→Xbe defined as
f(x) =ln
+x
, g(x) =ln
+x
,
S(x) =ex– , T(x) =ex– . For eachx∈X, we have +x≤exand +x
≤ex, sof(x) =ln( +
x
)≤x,g(x) =ln( +
x
)≤x,
x≤ex– =S(x) andx≤ex– =T(x). Thus,f andg are dominated andT andSare dominating withf(X) =g(X) =S(X) =T(X) = [,∞). Also, the pair (g,T) is compatible,
gis continuous and (f,S) is weakly compatible. Let the control functionsψ,ϕ: [,∞)→ [,∞) be defined asψ(t) =btandϕ(t) = (b– )t, for allt∈[,∞), where <b≤ . Note that
ψsbd
f(x),g(y)= bf(x) +g(y) = b
ln
+x
+ln
+y
≤b
x
+
y
=
b(x+ y)
≤ex– +ey– =bd
S(x),T(y)
≤M(x,y) =ψ
M(x,y)
–ϕM(x,y)
, x,y∈X.
Corollary . Let(X,bd,)be an ordered complete b-dislocated metric space,and let f
and g be two dominated self-maps on X.Suppose that for every two comparable elements x,y∈X,
ψsbd(fx,gy)
≤ψMs(x,y)
–ϕMs(x,y)
is satisfied,where
Ms(x,y) =max
bd(x,y),bd(fx,x),bd(gy,y),
bd(x,gy) +bd(fx,y) s
,
ψ∈andϕ∈.If for every non-increasing sequence{xn}and a sequence{yn}with yn
xn,for all n such that yn→u,we have uxn,then f and g have a common fixed point.
Moreover,the set of common fixed points of f and g is well ordered if and only if f and g have one and only one common fixed point.
Proof TakingSandTas identity maps onX, the result follows from Theorem ..
Corollary . Let(X,bd,)be an ordered complete b-dislocated metric space.Let f and
g be dominated self-maps on X.Suppose that for every two comparable elements x,y∈X,
sbd(fx,gy)≤Ms(x,y) –ϕ
Ms(x,y)
is satisfied,where
Ms(x,y) =max
bd(x,y),bd(fx,x),bd(gy,y),
bd(x,gy) +bd(fx,y) s
,
andϕ∈.If for every non-increasing sequence{xn}and a sequence{yn}with ynxn,for
all n such that yn→u,it implies that uxn,then f and g have a common fixed point.
Moreover,the set of common fixed points of f and g is well ordered if and only if f and g have one and only one common fixed point.
Proof If we takeSandT as the identity maps onXandψ(t) =tfor allt∈[,∞), then from Theorem . it follows thatf andghave a common fixed point.
Remark . As corollaries we can state partial metric space as well asb-metric space ver-sions of our proved results in a similar way, which extends recent results in these settings.
3 Existence of a common solution for a system of integral equations Consider the following system of integral equations:
x(t) = b
a
K
t,r,x(r)dr,
x(t) = b
a
K
t,r,x(r)dr,
(.)
Here,K,K: [a,b]×[a,b]×R→R. The considered problem can be reformulated in
the following manner.
Letf,g:X→Xbe the mappings defined by
fx(t) = b
a
K
t,r,x(r)dr,
gx(t) = b
a
K
t,r,x(r)dr
for allx∈Xand for allt∈[a,b].
Then the existence of a solution to (.) is equivalent to the existence of a common fixed point off andg. According to Example .,Xequipped with
bd(u,v) = max t∈[a,b]
u(t)+v(t)p
for allu,v∈X, is a completeb-dislocated metric space withs= p–.
We endowXwith the partial orderinggiven by
xy ⇐⇒ x(t)≤y(t)
for allt∈[a,b]. Moreover, in [], it is proved that (X,) is regular. Now, we will prove the following result.
Theorem . Suppose that the following hypotheses hold:
(i) K,K: [a,b]×[a,b]×R→Rare continuous;
(ii) for allt,r∈[a,b]andx∈X,we have
x(t)≤min
b
a
K
t,r,x(r)dr, b
a
K
t,r,x(r)dr
;
(iii) for allr,t∈[a,b]andx,y∈Xwithxy,we have
K
t,r,x(r)+K
t,r,y(r)≤ξ(t,r)ln +x(r)+y(r)p,
whereξis a continuous function satisfying
sup
t∈[a,b]
b
a
ξ(t,r)pdr
< p–p
(b–a)p–.
Then the integral equations(.)have a common solution x∈X.
Now, letx,y∈Xbe such thatxy. From condition (iii), for allt∈[a,b], we have
p–fx(t)+gy(t)p ≤p–p
b
a K
t,r,x(r)+K
t,r,x(r)dr
p
≤p–p b
a qdr
q b
a K
t,r,x(r)+K
t,r,x(r)pdr
p
p
≤p–p(b–a)
p q
b
a
ξ(t,r)pln +x(r)+y(r)ppdr
≤p–p(b–a)pq
b
a
ξ(t,r)pln +bd(x,y) p
dr
≤p–p(b–a)
p q
b
a
ξ(t,r)pln +Ms(x,y) p
dr
= p–p(b–a)p– b
a
ξ(t,r)pdrln +Ms(x,y) p
<ln +Ms(x,y) p
=Ms(x,y)p–
Ms(x,y)p–
ln +Ms(x,y) p
.
Hence,
sbd(fx,gy) p
= s sup
t∈[a,b]
fx(t)+gy(t)p
≤Ms(x,y)p–
Ms(x,y)p–
ln +Ms(x,y) p
.
Takingψ(t) =tpandϕ(t) =tp– (ln( +t))pin Corollary ., there existsx∈X, a common fixed point off andg, that is,xis a solution for (.).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Author details
1Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia.2Department of
Mathematics, Qaemshahr Branch, Islamic Azad University, Qaemshahr, Iran.3Young Researchers and Elite Club,
Kermanshah Branch, Islamic Azad University, Kermanshah, Iran. 4Department of Mathematics and Applied Mathematics,
University of Pretoria, Lynwood road, Pretoria 0002, South Africa.
Acknowledgements
This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. Therefore, the first author acknowledges with thanks DSR, KAU for financial support.
Received: 6 July 2013 Accepted: 13 September 2013 Published:07 Nov 2013
References
1. Aghajani, A, Abbas, M, Roshan, JR: Common fixed point of generalized weak contractive mappings in partially orderedb-metric spaces. Math. Slovaca (2012, in press)
2. Kelle, JL: General Topology. Van Nostrand, New York (1960)
3. Jungck, G: Compatible mappings and common fixed points. Int. J. Math. Math. Sci.9(4), 771-779 (1986)
5. Abbas, M, Nemeth, SZ: Finding solutions of implicit complementarity problems by isotonicity of metric projection. Nonlinear Anal.75, 2349-2361 (2012)
6. Hussain, N, Kadelburg, Z, Radenovi´c, S, Al-Solamy, FR: Comparison functions and fixed point results in partial metric spaces. Abstr. Appl. Anal.2012, Article ID 605781 (2012)
7. Hussain, N, Ðori´c, D, Kadelburg, Z, Radenovi´c, S: Suzuki-type fixed point results in metric type spaces. Fixed Point Theory Appl.2012, 126 (2012)
8. Shah, MH, Simic, S, Hussain, N, Sretenovic, A, Radenovic, S: Common fixed points theorems for occasionally weakly compatible pairs on cone metric type spaces. J. Comput. Anal. Appl.14, 290-297 (2012)
9. Hussain, N, Shah, MH: KKM mappings in coneb-metric spaces. Comput. Math. Appl.62, 1677-1684 (2011) 10. Khamsi, MA, Hussain, N: KKM mappings in metric type spaces. Nonlinear Anal.73, 3123-3129 (2010) 11. Hitzler, P, Seda, AK: Dislocated topologies. J. Electr. Eng.51(12), 3-7 (2000)
12. Hitzler, P: Generalized metrics and topology in logic programming semantics. PhD thesis, National University of Ireland (University College, Cork) (2001)
13. Aage, CT, Salunke, JN: The results on fixed points in dislocated and dislocated quasi-metric space. Appl. Math. Sci. 2(59), 2941-2948 (2008)
14. Isufati, A: Fixed point theorems in dislocated quasi-metric space. Appl. Math. Sci.4(5), 217-233 (2010) 15. Sarma, IR, Kumari, PS: On dislocated metric spaces. Int. J. Math. Arch.3(1), 72-77 (2012)
16. Shrivastava, R, Ansari, ZK, Sharma, M: Some results on fixed points in dislocated and dislocated quasi-metric spaces. J. Adv. Stud. Topol.3(1), 25-31 (2012)
17. Shrivastava, M, Qureshi, K, Singh, AD: A fixed point theorem for continuous mapping in dislocated quasi-metric spaces. Int. J. Theor. Appl. Sci.4(1), 39-40 (2012)
18. Kumari, PS, Kumar, VV, Sarma, IR: Common fixed point theorems on weakly compatible maps on dislocated metric spaces. Math. Sci.6, 71 (2012)
19. Kumari, PS: Some fixed point theorems in generalized dislocated metric spaces. Math. Theory Model.1(4), 16-22 (2011)
20. Daheriya, RD, Jain, R, Ughade, M: Some fixed point theorem for expansive type mapping in dislocated metric space. ISRN Math. Anal.2012, Article ID 376832 (2012)
21. George, R: Cyclic contractions and fixed points in dislocated metric spaces. Int. J. Math. Anal.7(9), 403-411 (2013) 22. Jha, K: A common fixed point theorem in dislocated metric space. Appl. Math. Sci.6(91), 4497-4503 (2012) 23. Zoto, K: Some new results in dislocated and dislocated quasi-metric spaces. Appl. Math. Sci.6(71), 3519-3526 (2012)
10.1186/1029-242X-2013-486
