R E S E A R C H
Open Access
Coupled common fixed point results
involving a
(
ϕ
,
ψ
)
-contractive condition for
mixed
g
-monotone operators in partially
ordered metric spaces
Manish Jain
1, Kenan Tas
2*, Sanjay Kumar
3and Neetu Gupta
4*Correspondence:
2Department of Mathematics and
Computer Sciences, Cankaya University, Ankara, Turkey Full list of author information is available at the end of the article
Abstract
In the setting of partially ordered metric spaces, using the notion of compatible mappings, we establish the existence and uniqueness of coupled common fixed points involving a (
ϕ
,ψ
)-contractive condition for mixedg-monotone operators. Our results extend and generalize the well-known results of Berinde (Nonlinear Anal. TMA 74:7347-7355, 2011; Nonlinear Anal. TMA 75:3218-3228, 2012) and weaken the contractive conditions involved in the results of Alotaibiet al.(Fixed Point Theory Appl. 2011:44, 2011), Bhaskaret al.(Nonlinear Anal. TMA 65:1379-1393, 2006), and Luonget al.(Nonlinear Anal. TMA 74:983-992, 2011). The effectiveness of the presented work is validated with the help of suitable examples.MSC: 54H10; 54H25
Keywords: partially ordered set; compatible mappings;g-mixed monotone mappings; coupled coincidence point; coupled common fixed point
1 Introduction and preliminaries
Bhaskar and Lakshmikantham [] introduced the notion of coupled fixed points and proved some coupled fixed point theorems for a mapping with the mixed monotone prop-erty in the setting of partially ordered metric spaces. These concepts are defined as follows.
Definition .[] Let (X,≤) be a partially ordered set andF:X×X→X. The mapping
Fis said to have the mixed monotone property ifF(x,y) is monotone non-decreasing inx
and monotone non-increasing iny; that is, for anyx,y∈X,
x,x∈X, x≤x implies F(x,y)≤F(x,y)
and
y,y∈X, y≤y implies F(x,y)≥F(x,y).
Definition .[] An element (x,y)∈X×Xis called a coupled fixed point of the mapping
F:X×X→XifF(x,y) =xandF(y,x) =y.
Bhaskar and Lakshmikantham [] proved the following results.
Theorem .[] Let(X,≤)be a partially ordered set and suppose there exists a metric d on X such that(X,d)is a complete metric space.Let F:X×X→X be a continuous mapping having the mixed monotone property on X.Assume that there exists a k∈[, )with
dF(x,y),F(u,v)≤k
d(x,u) +d(y,v) (.)
for all x≥u and y≤v.
If there exist two elements x,y∈X with x≤F(x,y)and y≥F(y,x),then there
exist x,y∈X such that x=F(x,y)and y=F(y,x).
Theorem .[] Let(X,≤)be a partially ordered set and suppose there exists a metric d
on X such that(X,d)is a complete metric space.Assume that X has the following property:
(i) if a non-decreasing sequence{xn} →x,thenxn≤xfor alln, (ii) if a non-increasing sequence{yn} →y,theny≤ynfor alln.
Let F :X×X→X be a mapping having the mixed monotone property on X.Assume that there exists a k∈[, )with the condition(.).If there exist two elements x,y∈X
with x≤F(x,y)and y≥F(y,x),then there exist x,y∈X such that x=F(x,y)and
y=F(y,x).
Lakshmikantham and Ćirić [] extended the notion of mixed monotone property to mixedg-monotone property and generalized the results of Bhaskar and Lakshmikantham [] by establishing the existence of coupled coincidence point results using a pair of com-mutative maps.
Definition .[] Let (X,≤) be a partially ordered set andF:X×X→Xandg:X→X.
We sayFhas the mixedg-monotone property ifF(x,y) is monotoneg-nondecreasing in its first argument and is monotoneg-nonincreasing in its second argument; that is, for anyx,y∈X,
x,x∈X, gx≤gx implies F(x,y)≤F(x,y)
and
y,y∈X, gy≤gy implies F(x,y)≥F(x,y).
Definition .[] An element (x,y)∈X×Xis called a coupled coincidence point of the mappingsF:X×X→Xandg:X→XifF(x,y) =gxandF(y,x) =gy.
Definition .[] An element (x,y)∈X×Xis called a coupled common fixed point of
the mappingsF:X×X→Xandg:X→Xifx=gx=F(x,y) andy=gy=F(y,x).
Definition .[] LetXbe a non-empty set andF:X×X→Xandg:X→X. We sayF
Later, Choudhury and Kundu [] introduced the notion of compatibility in the context of coupled coincidence point problems and used the notion to improve the results of Lak-shmikantham and Ćirić [].
Definition .[] The mappingsF:X×X→Xandg:X→Xare said to be compatible if
lim
n→∞d
gF(xn,yn),F(gxn,gyn)
= and lim
n→∞d
gF(yn,xn),F(gyn,gxn)
=
whenever{xn}and{yn}are sequences inXsuch thatlimn→∞F(xn,yn) =limn→∞gxn=x
andlimn→∞F(yn,xn) =limn→∞gyn=yfor somex,y∈X.
In recent years, following Bhaskar and Lakhsmikantham [], the existence and unique-ness of coupled fixed points under more general contractive conditions were established by various authors. One can refer to [, –].
In order to generalize the results of Bhaskar and Lakshmikantham [], Luong and Thuan [] considered the following class of control functions.
Definition .[] Letdenote the class of functionsϕ: [,∞)→[,∞) which satisfy
(ϕ) ϕis continuous and non-decreasing;
(ϕ) ϕ(t) = if and only ift= ;
(ϕ) ϕ(t+s)≤ϕ(t) +ϕ(s), for allt,s∈[,∞).
Definition .[] Letdenote the class of functionsψ: [,∞)→[,∞) which satisfy
(iψ) limt→rψ(t) > for allr> andlimt→+ψ(t) = .
The contractive condition considered by Luong and Thuan [] is given below:
ϕdF(x,y),F(u,v)≤ ϕ
d(x,u) +d(y,v)–ψ
d(x,u) +d(y,v)
, (.)
whereϕ∈,ψ∈andx≥u,y≤v.
On the other hand, Alotaibi and Alsulami [] extended the results of Luong and Thuan [] for a compatible pair (F,g), whereF:X×X→Xandg:X→Xare the maps satisfying the following contractive condition:
ϕdF(x,y),F(u,v)≤ ϕ
d(gx,gu) +d(gy,gv)–ψ
d(gx,gu) +d(gy,gv)
, (.)
withϕ∈,ψ∈andgx≥gu,gy≤gv.
We consider the classredefined by Berinde [] as follows.
Definition .[] Letdenote the class of functionsϕ: [,∞)→[,∞) which satisfy
(iϕ) ϕis continuous and (strictly) increasing;
(iiϕ) ϕ(t) <tfor allt> ;
(iiiϕ) ϕ(t+s)≤ϕ(t) +ϕ(s)for allt,s∈[,∞).
Berinde [] weakened the contractive conditions (.) and (.) by considering the more general one
ϕ
d(F(x,y),F(u,v)) +d(F(y,x),F(v,u))
≤ϕ
d(x,u) +d(y,v)
–ψ
d(x,u) +d(y,v)
(.)
for a mixed monotone mappingF:X×X→X,x≥u,y≤v, whereϕ∈andψ∈. The present work extends and generalizes several results presented in the literature of fixed point theory. Our theorems directly derive the main results of Berinde [, ]. We give suitable examples to show how our results extend the well-known results of Alotaibi
et al.[], Bhaskaret al.[] and Luonget al.[] by significantly weakening the involved contractive condition.
2 Main results
Theorem . Let(X,≤)be a partially ordered set and suppose there exists a metric d on X such that(X,d)is a complete metric space.Let F:X×X→X,g:X→X be two maps with F having the mixed g-monotone property on X such that there exist two elements x,y∈X
with gx≤F(x,y)and gy≥F(y,x).Suppose there existϕ∈andψ∈such that
ϕ
d(F(x,y),F(u,v)) +d(F(y,x),F(v,u))
≤ϕ
d(gx,gu) +d(gy,gv)
–ψ
d(gx,gu) +d(gy,gv)
(.)
for all x,y,u,v∈X with gx≥gu and gy≤gv.
Suppose that F(X×X)⊆g(X),g is continuous and the pair(F,g)is compatible.
Also suppose either (a) Fis continuous,or
(b) Xhas the following property:
(i) if a non-decreasing sequence{xn} →x,thengxn≤gxfor alln; (ii) if a non-increasing sequence{yn} →y,thengy≤gynfor alln.
Then there exist x,y∈X such that gx=F(x,y)and gy=F(y,x);that is,F and g have a coupled coincidence point in X.
Proof Letx,y∈Xsuch thatgx≤F(x,y) andgy≥F(y,x). SinceF(X×X)⊆g(X), we can choosex,y∈Xsuch thatgx=F(x,y),gy=F(y,x). Again, we can choose
x,y∈Xsuch thatgx=F(x,y),gy=F(y,x).
Continuing this process, we can construct sequences{gxn}and{gyn}inXsuch that
gxn+=F(xn,yn), gyn+=F(yn,xn) for alln≥. (.)
We shall prove, for alln≥, that
gxn≤gxn+, (.)
Sincegx≤F(x,y) andgy≥F(y,x),gx=F(x,y),gy=F(y,x), we havegx≤
gx,gy≥gy; that is, (.) and (.) hold forn= .
Suppose that (.) and (.) hold for somen> ,i.e.,gxn≤gxn+,gyn≥gyn+. AsFhas
the mixedg-monotone property, by (.), we have
gxn+=F(xn,yn)≤F(xn+,yn)≤F(xn+,yn+) =gxn+,
and
gyn+=F(yn,xn)≥F(yn+,xn)≥F(yn+,xn+) =gyn+;
that is,
gxn+≤gxn+ and gyn+≥gyn+.
Then, by mathematical induction, it follows that (.) and (.) hold for alln≥. If, for somen≥, we have (gxn+,gyn+) = (gxn,gyn), thenF(xn,yn) =gxnandF(yn,xn) = gyn; that is,Fandghave a coincidence point. So, now onwards, we suppose (gxn+,gyn+)= (gxn,gyn) for alln≥; that is, we suppose that eithergxn+=F(xn,yn)=gxn orgyn+=
F(yn,xn)=gyn.
Sincegxn≥gxn–andgyn≤gyn–, by (.) and (.), we have, for alln≥, that
ϕ
d(gxn+,gxn) +d(gyn+,gyn)
=ϕ
d(F(xn,yn),F(xn–,yn–)) +d(F(yn,xn),F(yn–,xn–))
≤ϕ
d(gxn,gxn–) +d(gyn,gyn–)
–ψ
d(gxn,gxn–) +d(gyn,gyn–)
. (.)
Sinceψis non-negative, we have
ϕ
d(gxn+,gxn) +d(gyn+,gyn)
≤ϕ
d(gxn,gxn–) +d(gyn,gyn–)
.
By the monotonicity ofϕ, we have
d(gxn+,gxn) +d(gyn+,gyn)
≤
d(gxn,gxn–) +d(gyn,gyn–)
.
Let Rn = d(gxn+,gxn)+d(gyn+,gyn), then {Rn} is a monotone decreasing sequence of
non-negative real numbers. Therefore, there exists someR≥ such that
lim
n→∞Rn=nlim→∞
d(gxn+,gxn) +d(gyn+,gyn)
=R. (.)
We claim thatR= .
Taking limit asn→ ∞on both sides of (.) and using the properties ofϕ andψ, we have
ϕ(R) = lim
n→∞ϕ(Rn)≤nlim→∞
ϕ(Rn–) –ψ(Rn–)
=ϕ(R) – lim
Rn–→R
ψ(Rn–) <ϕ(R),
a contradiction. Thus,R= ; that is,
lim
n→∞Rn=nlim→∞
d(gxn+,gxn) +d(gyn+,gyn)
= . (.)
Next, we shall show that{gxn}and{gyn}are Cauchy sequences.
If possible, suppose that at least one of{gxn}and{gyn}is not a Cauchy sequence. Then
there exists an ε> for which we can find subsequences{gxn(k)}, {gxm(k)}of{gxn}and {gyn(k)},{gym(k)}of{gyn}withn(k) >m(k)≥ksuch that
rk=
d(gxn(k),gxm(k)) +d(gyn(k),gym(k))
≥ε. (.)
Further, corresponding tom(k), we can choosen(k) in such a way that it is the smallest integer withn(k) >m(k)≥kand satisfies (.). Then
d(gxn(k)–,gxm(k)) +d(gyn(k)–,gym(k))
<ε. (.)
By (.), (.) and the triangle inequality, we have
ε≤rk=
d(gxn(k),gxm(k)) +d(gyn(k),gym(k))
≤{d(gxn(k),gxn(k)–) +d(gxn(k)–,gxm(k)) +d(gyn(k),gyn(k)–) +d(gyn(k)–,gym(k))}
< d(gxn(k),gxn(k)–) +d(gyn(k),gyn(k)–)
+ε.
Lettingk→ ∞and using (.) in the last inequality, we have
lim
k→∞rk=klim→∞
d(gxn(k),gxm(k)) +d(gyn(k),gym(k))
=ε. (.)
Again, by the triangle inequality
rk =
d(gxn(k),gxm(k)) +d(gyn(k),gym(k))
≤
d(gxn(k),gxn(k)+) +d(gxn(k)+,gxm(k)+) +d(gxm(k)+,gxm(k))
+d(gyn(k),gyn(k)+) +d(gyn(k)+,gym(k)+) +d(gym(k)+,gym(k))
=Rn(k)+Rm(k)+
d(gxn(k)+,gxm(k)+) +d(gyn(k)+,gym(k)+)
By the monotonicity ofϕand the property (iiiϕ), we have
ϕ(rk)≤ϕ(Rn(k)) +ϕ(Rm(k)) +ϕ
d(gxn(k)+,gxm(k)+) +d(gyn(k)+,gym(k)+)
. (.)
Sincen(k) >m(k),gxn(k)≥gxm(k)andgyn(k)≤gym(k). Then by (.) and (.), we have
ϕ
d(gxn(k)+,gxm(k)+) +d(gyn(k)+,gym(k)+)
=ϕ
d(F(xn(k),yn(k)),F(xm(k),ym(k))) +d(F(yn(k),xn(k)),F(ym(k),xm(k)))
≤ϕ
d(gxn(k),gxm(k)) +d(gyn(k),gym(k))
–ψ
d(gxn(k),gxm(k)) +d(gyn(k),gym(k))
=ϕ(rk) –ψ(rk). (.)
By (.) and (.), we have
ϕ(rk)≤ϕ(Rn(k)) +ϕ(Rm(k)) +ϕ(rk) –ψ(rk).
Lettingk→ ∞, using (.), (.) and the properties ofϕandψin the last inequality, we have
ϕ(ε)≤ϕ() +ϕ() +ϕ(ε) – lim
k→∞ψ(rk) =ϕ(ε) – lim
rk→ε
ψ(rk) <ϕ(ε),
a contradiction.
Therefore, both{gxn}and{gyn}are Cauchy sequences inX. By the completeness ofX,
there existx,y∈Xsuch that
lim
n→∞F(xn,yn) =nlim→∞gxn=x and nlim→∞F(yn,xn) =nlim→∞gyn=y. (.)
SinceFandgare compatible mappings, we have from (.)
lim
n→∞d
gF(xn,yn),F(gxn,gyn)
= , (.)
lim
n→∞d
gF(yn,xn),F(gyn,gxn)
= . (.)
Let the condition (a) hold. For alln≥, we have
dgx,F(gxn,gyn)
≤dgx,gF(xn,yn)
+dgF(xn,yn),F(gxn,gyn)
Takingn→ ∞in the last inequality, using the inequalities (.), (.) and the continu-ities ofFandg, we haved(gx,F(x,y)) = ; that is,gx=F(x,y). Again, for alln≥,
dgy,F(gyn,gxn)
≤dgy,gF(yn,xn)
+dgF(yn,xn),F(gyn,gxn)
.
Takingn→ ∞ in the last inequality, using the inequalities (.), (.) and the con-tinuities of F andg, we havedgy,F(y,x)= ; that is,gy=F(y,x). Hence, the element (x,y)∈X×Xis a coupled coincidence point of the mappingsF:X×X→Xandg:X→X.
Next, we suppose that the condition (b) holds.
By (.), (.) and (.), we have{gxn}is a non-decreasing sequence,gxn→xand{gyn}
is a non-increasing sequence,gyn→yasn→ ∞. Hence, by the assumption (b), we have
for alln≥,
ggxn≤gx and ggyn≥gy. (.)
SinceFandgare compatible mappings andgis continuous, by inequalities (.)-(.), we have
lim
n→∞ggxn=gx=nlim→∞g
F(xn,yn)
= lim
n→∞F(gxn,gyn), (.)
and
lim
n→∞ggyn=gy=nlim→∞g
F(yn,xn)
= lim
n→∞F(gyn,gxn). (.)
Now,
dF(x,y),gx≤dF(x,y),ggxn+
+d(ggxn+,gx);
that is,
dF(x,y),gx≤dF(x,y),gF(xn,yn)
+d(ggxn+,gx).
Takingn→ ∞in the last inequality and using (.), we have
dF(x,y),gx≤ lim
n→∞d
F(x,y),gF(xn,yn)
+ lim
n→∞d(ggxn+,gx) ≤ lim
n→∞d
F(x,y),F(gxn,gyn)
. (.)
Similarly,
dF(y,x),gy≤ lim
n→∞d
F(y,x),F(gyn,gxn)
. (.)
By (.), (.) and the property (iϕ), we have
ϕ
d(F(x,y),gx) +d(F(y,x),gy)
≤ lim
n→∞ϕ
d(F(x,y),F(gxn,gyn)) +d(F(y,x),F(gyn,gxn))
By (.) and (.), we have
ϕ
d(F(x,y),F(gxn,gyn)) +d(F(y,x),F(gyn,gxn))
≤ϕ
d(gx,ggxn) +d(gy,ggyn)
–ψ
d(gx,ggxn) +d(gy,ggyn)
. (.)
Inserting (.) in (.), we have
ϕ
d(F(x,y),gx) +d(F(y,x),gy)
≤ lim
n→∞
ϕ
d(gx,ggxn) +d(gy,ggyn)
–ψ
d(gx,ggxn) +d(gy,ggyn)
= lim
n→∞ϕ
d(gx,ggxn) +d(gy,ggyn)
– lim
n→∞ψ
d(gx,ggxn) +d(gy,ggyn)
.
By (.), (.), the continuity ofϕandlimt→+ψ(t) = , we get
ϕ
d(F(x,y),gx) +d(F(y,x),gy)
≤ lim
n→∞ϕ
d(gx,ggxn) +d(gy,ggyn)
=ϕ() = .
Sinceϕis non-negative andϕ() = , we have
dF(x,y),gx= and dF(y,x),gy= ;
that is,
F(x,y) =gx and F(y,x) =gy.
Hence, the element (x,y)∈X×Xis a coupled coincidence point of the mappingsF:X×
X→Xandg:X→X.
Now, we give an example in support of Theorem ..
Example . LetX= [, ]. Then (X,≤) is a partially ordered set with the natural ordering of real numbers.
Letd(x,y) =|x–y|forx,y∈X.
Then (X,d) is a complete metric space. Let :X→Xbe defined as
g(x) =x, for allx∈X.
LetF:X×X→Xbe defined as
F(x,y) = ⎧ ⎨ ⎩
x–y
Let{xn}and{yn}be two sequences inXsuch that
lim
n→∞F(xn,yn) =a, nlim→∞g(xn) =a,
lim
n→∞F(yn,xn) =b and nlim→∞g(yn) =b.
Now, for alln≥,
g(xn) =xn, g(yn) =yn,
F(xn,yn) =
⎧ ⎨ ⎩
xn–yn
, ifx,y∈[, ],xn≥yn, , ifxn<yn,
and
F(yn,xn) =
⎧ ⎨ ⎩
yn–xn
, ifx,y∈[, ],yn≥xn, , ifyn<xn.
Obviously,a= andb= . Then it follows that
dgF(xn,yn),F(gxn,gyn)
→ asn→ ∞,
and
dgF(yn,xn),F(gyn,gxn)
→ asn→ ∞.
Hence, the mappingsFandgare compatible inX. Clearly,Fobeys the mixedg-monotone property. Also,F(X×X)⊆g(X).
Letϕ,ψ: [,∞)→[,∞) be defined asϕ(t) = t
,ψ(t) =
t
, fort∈[,∞).
Also,x= andy=c(>) are two points inXsuch thatg(x) =g() = =F(,c) =
F(x,y) andg(y) =g(c) =c≥c =F(c, ) =F(y,x).
Next, we verify inequality (.) of Theorem .. We takex,y,u,v∈Xsuch thatgx≥gu
andgy≤gv; that is,x≥uandy≤ v. We discuss the following cases. Case :x≥y,u≥v.
Then
ϕ
d(F(x,y),F(u,v)) +d(F(y,x),F(v,u))
=
d(F(x,y),F(u,v)) +d(, )
=
d
x–y ,
u–v
=
x–y –u –v
=
(x–u) + ( v–y)=
(x–u) +
(v–y)
≤
(x–u) + (v–y)
=
d(gx,gu) +d(gv,gy)
=ϕ
d(gx,gu) +d(gv,gy)
–ψ
d(gx,gu) +d(gv,gy)
Case :x≥y,u<v. Then
ϕ
d(F(x,y),F(u,v)) +d(F(y,x),F(v,u))
=
d(F(x,y),F(u,v)) +d(F(y,x),F(v,u)) = d
x–y ,
+d
,v
–u
=
x–y
+
v–u
=
x–u
+
v–y
≤
x–u
+
v–y
=
(x–u) + (v–y)
=
d(gx,gu) +d(gv,gy)
=ϕ
d(gx,gu) +d(gv,gy)
–ψ
d(gx,gu) +d(gv,gy)
.
Case :x<y,u≥v. Then
ϕ
d(F(x,y),F(u,v)) +d(F(y,x),F(v,u))
=
d(F(x,y),F(u,v)) +d(F(y,x),F(v,u)) = d ,u
–v
+d
y–x ,
=
u–v
+
y–x
=
–(x–u) – (v–y)
≤
(x–u) + (v–y)
≤
(x–u) + (v–y)
=
d(gx,gu) +d(gv,gy)
=ϕ
d(gx,gu) +d(gv,gy)
–ψ
d(gx,gu) +d(gv,gy)
.
Case :x<y,u<v. Then
ϕ
d(F(x,y),F(u,v)) +d(F(y,x),F(v,u))
=
d(, ) +d(F(y,x),F(v,u))
=
d
y–x ,
v–u
=
y–x –v –u
=
–(x–u) – ( v–y)=
|
(x–u) + (v–y)|
=
(x–u) + (v–y)
≤
(x–u) + (v–y)
=
d(gx,gu) +d(gv,gy)
=ϕ
d(gx,gu) +d(gv,gy)
–ψ
d(gx,gu) +d(gv,gy)
.
Hence, the inequality (.) of Theorem . is satisfied.
Thus, all the conditions of Theorem . are satisfied, and it can be easily seen that (, ) is the required coupled coincidence point ofFandginX.
Remark . If we choose the functionsϕ(t) =t/ andψ(t) =t/, fort∈[,∞), then with
this choice of functions, we can obtain the already existing contractive condition. Sinceϕ andψare actually contractions, this will be cleared in Corollary .. But if we chooseϕ(t) =
t/(t+ ) andψ(t) =t/, fort∈[,∞), then with this choice ofϕandψ, the contractive condition (.) does not turn to the existing contractive condition.
The next example shows that Theorem . is more general than Theorem . in [] since the contractive condition (.) is more general than (.).
Example . LetX=R. Then (X,≤) is a partially ordered set with the natural ordering of real numbers. Letd:X×X→R+be defined by
d(x,y) =|x–y| forx,y∈X.
Then (X,d) is a complete metric space.
DefineF:X×X→XbyF(x,y) =x–y, (x,y)∈X×Xandg:X→Xbyg(x) = x,x∈X. Clearly,F(X×X)⊆g(X),Fis continuous and has the mixedg-monotone property, the pair (F,g) is compatible and satisfies the condition (.) but does not satisfy the condition (.). Assume, to the contrary, that there existϕ∈(in accordance with Definition .) andψ∈such that (.) holds. Then we must have
ϕx– y –
u– v
≤
ϕ
x–u +y
–
v
–ψ
|x
–
u
|+|
y
–
v
|
= ϕ
|
x–u|+|y–v|
–ψ
|
x–u|+|y–v|
for allx≥uandy≤v. Takex=u,y=vin the last inequality and letρ=|y–v|, we obtain
ϕ(ρ)≤
ϕ(ρ) –ψ(ρ), ρ> .
But by (ϕ) we have ϕ(ρ)≤ϕ(ρ) and hence we deduce that, for allρ> ,ψ(ρ)≤, that is,ψ(ρ) = , which contradicts (iψ). This shows thatFdoes not satisfy (.).
Now, we prove that (.) holds. Indeed, forx≥uandy≤v, we have
x– y–u– v
≤
and
y– x–v– u
≤
|y–v|+ |x–u|.
By summing up the last two inequalities, we get exactly (.) withϕ(t) = t,ψ(t) = t. Also,x= –,y= are the two points inXsuch thatgx≤F(x,y) andgy≥F(y,x).
F,g,ϕ,ψ satisfy all the conditions of Theorem .. So, by Theorem ., we obtain thatF
andghave a coupled coincidence point (, ), but Theorem . in [] cannot be applied toFandgin this example.
The following Corollary . is Theorem in [].
Corollary .[] Let(X,≤)be a partially ordered set and suppose there exists a metric d on X such that(X,d)is a complete metric space.Let F:X×X→X be a mapping having the mixed monotone property on X such that there exist two elements x,y∈X with x≤
F(x,y)and y≥F(y,x).Suppose there existϕ∈andψ∈such that
ϕ
d(F(x,y),F(u,v)) +d(F(y,x),F(v,u))
≤ϕ
d(x,u) +d(y,v)
–ψ
d(x,u) +d(y,v)
(.)
for all x,y,u,v∈X with x≥u and y≤v.Suppose either (a) Fis continuous,or
(b) Xhas the following property:
(i) if a non-decreasing sequence{xn} →x,thenxn≤xfor alln; (ii) if a non-increasing sequence{yn} →y,theny≤ynfor all n. Then there exist x,y∈X such that x=F(x,y)and y=F(y,x).
Proof Takinggto be an identity mapping in Theorem ., we obtain Corollary .. The following example shows that Corollary . is more general than Theorem . (i.e., Theorem . in []) and Theorem . in [], since the contractive condition (.) is more general than (.) and (.).
Example . LetX=R. Then (X,≤) is a partially ordered set with the natural ordering of real numbers. Letd:X×X→R+be defined by
d(x,y) =|x–y| forx,y∈X.
Then (X,d) is a complete metric space.
DefineF:X×X→XbyF(x,y) =x–y, (x,y)∈X×X.
Then F is continuous, has the mixed monotone property and satisfies the condition (.) but does not satisfy either the condition (.) or the condition (.). Indeed, assume there existsk∈[, ) such that (.) holds. Then we must have
x– y–u– v
≤k
by which, forx=u, we get
|y–v| ≤k|y–v|, y≤v,
which fory<vimplies ≤k, a contradiction, sincek∈[, ). Hence,F does not satisfy (.).
Further, (.) is also not satisfied. Assume, to the contrary, that there existϕ∈(in accordance with Definition .) andψ∈such that (.) holds. Then we must have
ϕx– y –
u– v
≤
ϕ
|x–u|+|y–v|–ψ |
x–u|+|y–v|
,
for allx≥uandy≤v. Takex=u,y=vin the last inequality and letα=|y–v|, we obtain
ϕ(α)≤
ϕ(α) –ψ(α), α> .
But by (ϕ), we haveϕ(α)≤ϕ(α) and hence we deduce that, for allα> ,ψ(α)≤, that is,ψ(α) = , which contradicts (iψ). This shows thatFdoes not satisfy (.).
Now, we prove that (.) holds. Indeed, forx≥uandy≤v, we have
x– y–u– v
≤
|x–u|+ |y–v|,
and y– x
–
v– u
≤
|y–v|+ |x–u|.
By summing up the last two inequalities, we get exactly (.) withϕ(t) =t,ψ(t) = t. Also,x= –,y= are the two points inXsuch thatx≤F(x,y) andy≥F(y,x).
So, by Corollary ., we obtain thatFhas a coupled fixed point (, ) but neither Theo-rem . in [] nor TheoTheo-rem . in [] can be applied toFin this example.
The following Corollary . is Corollary in [].
Corollary .[] Let(X,≤)be a partially ordered set and suppose there exists a metric d on X such that(X,d)is a complete metric space.Let F:X×X→X be a mapping having the mixed monotone property on X such that there exist two elements x,y∈X with x≤
F(x,y)and y≥F(y,x).Suppose there existsψ∈such that
dF(x,y),F(u,v)+dF(y,x),F(v,u)
≤d(x,u) +d(y,v) – ψ
d(x,u) +d(y,v)
(.)
for all x,y,u,v∈X with x≥u and y≤v.Suppose either (a) Fis continuous,or
(i) if a non-decreasing sequence{xn} →x,thenxn≤xfor alln; (ii) if a non-increasing sequence{yn} →y,theny≤ynfor alln. Then F has a coupled fixed point in X.
Proof Note that ifψ ∈, then for allr> ,rψ ∈. Now divide (.) by and take ϕ(t) =t,t∈[,∞), then the condition (.) reduces to (.) withψ=ψandg(x) =x; and hence by Theorem ., we obtain Corollary .. Corollary . Let(X,≤)be a partially ordered set and suppose there exists a metric d on X such that(X,d)is a complete metric space.Let F:X×X→X,g:X→X be two maps with F having the mixed g-monotone property on X such that there exist two elements x,y∈X
with gx≤F(x,y)and gy≥F(y,x).Suppose there exists a real number k∈[, )such
that
dF(x,y),F(u,v)+dF(y,x),F(v,u)≤kd(gx,gu) +d(gy,gv) (.)
for all x,y,u,v∈X with x≥u,y≤v.Suppose either (a) Fis continuous,or
(b) Xhas the following property:
(i) if a non-decreasing sequence{xn} →x,thengxn≤gxfor alln; (ii) if a non-increasing sequence{yn} →y,thengy≤gynfor alln.
Suppose that F(X×X)⊆g(X),g is continuous and the pair(F,g)is compatible,then there exist x,y∈X such that gx=F(x,y)and gy=F(y,x).
Proof Takingϕ(t) = t andψ(t) = ( –k)t, ≤k< , in Theorem ., we obtain
Corol-lary ..
Remark . (i) Corollary . is an extension of the recent coupled fixed point result of
Berinde (Theorem in []) to a coupled coincidence point theorem for a pair of compat-ible mappings having the mixedg-monotone property.
(ii) Again, the choice of functionsFandgin Example . shows that Corollary . is more general than Theorem . in [], since the contractive condition (.) is more general than (.). Indeed, the contractive condition (.) does not hold for the choice of functions
Fandg, but (.) holds exactly fork= withx= – andy= and yields (, ) as the coupled coincidence point ofFandg.
Corollary . Let(X,≤)be a partially ordered set and suppose there exists a metric d on X such that(X,d)is a complete metric space.Let F:X×X→X,be a mapping having the mixed monotone property on X such that there exist two elements x,y∈X with x≤
F(x,y)and y≥F(y,x).Suppose there exists a real number k∈[, )such that
dF(x,y),F(u,v)+dF(y,x),F(v,u)≤kd(x,u) +d(y,v) (.)
for all x,y,u,v∈X with x≥u,y≤v.Suppose either (a) Fis continuous,or
(i) if a non-decreasing sequence{xn} →x,thenxn≤xfor alln; (ii) If a non-increasing sequence{yn} →y,theny≤ynfor alln. Then F has a coupled fixed point in X.
Proof Takinggto be the identity mapping in Corollary ., we obtain Corollary .. Remark . (i) By considering the condition of continuity of F in Corollary ., we obtain Theorem in [].
(ii) Again, the choice of the functionFin Example . shows that Corollary . is more general than Theorem . (i.e., Theorem . in []) and Theorem . in [], since the tractive condition (.) is more general than (.) and (.). Indeed, the contractive con-ditions (.) and (.) do not hold for the choice of the functionF, but (.) holds exactly fork=withx= – andy= and yields (, ) as the coupled fixed point ofF.
Now, in order to prove the existence and uniqueness of the coupled common fixed point for our main results, we need the following lemma.
Lemma . Let F :X×X→X and g:X→X be compatible maps and let an element
(x,y)∈X×X such that gx=F(x,y)and gy=F(y,x)exist, then gF(x,y) =F(gx,gy)and gF(y,x) =F(gy,gx).
Proof Since the pair (F,g) is compatible, it follows that
lim
n→∞d
gF(xn,yn),F
g(xn),g(yn)
= ,
lim
n→∞d
gF(yn, xn),F
g(yn),g(xn)
= ,
whenever{xn}and{yn}are sequences inXsuch thatlimn→∞F(xn,yn) =limn→∞g(xn) =a,
limn→∞F(yn,xn) =limn→∞g(yn) =b for some a,b∈X. Takingxn=x,yn=yand using gx=F(x,y),gy=F(y,x), it follows that
dgF(x,y),F(gx,gy)= and dgF(y,x),F(gy,gx)= .
Hence,gF(x,y) =F(gx,gy) andgF(y,x) =F(gy,gx). Theorem . In addition to the hypothesis of Theorem ., suppose that for every
(x,y), (x*,y*)∈X×X,there exists a(u,v)∈X×X such that(F(u,v),F(v,u))is comparable
to(F(x,y),F(y,x))and(F(x*,y*),F(y*,x*)).Then F and g have a unique coupled common fixed point;that is, there exists a unique(x,y)∈X×X such that x=g(x) =F(x,y)and y=g(y) =F(y,x).
Proof By Theorem ., the set of coupled coincidences is non-empty. In order to prove the theorem, we shall first show that if (x,y) and (x*,y*) are coupled coincidence points, that is, ifgx=F(x,y),gy=F(y,x) andgx*=F(x*,y*),gy*=F(y*,x*), then
By assumption, there is (u,v)∈X× X such that (F(u,v),F(v,u)) is comparable with (F(x,y),F(y,x)) and (F(x*,y*),F(y*,x*)). Putu
=u,v=vand chooseu,v ∈X so that
gu=F(u,v),gv=F(v,u).
Then, similarly as in the proof of Theorem ., we can inductively define sequences{gun}
and{gvn}such thatgun+=F(un,vn) andgvn+=F(vn,un).
Further, set x=x,y=y,x*=x*,y*=y* and, in the same way, define the sequences
{gxn},{gyn}and{gx*n},{gy*n}. Then it is easy to show that
gxn+=F(xn,yn), gyn+=F(yn,xn)
and
gx*n+=Fx*n,y*n, gy*n+=Fy*n,x*n for alln≥.
Since (F(u,v),F(v,u)) = (gu,gv) and (F(x,y),F(y,x)) = (gx,gy) = (gx,gy) are comparable, thengu≥gxandgv≤gy. It is easy to show that (gun,gvn) and (gx,gy) are comparable,
that is,gun≥gxandgvn≤gyfor alln≥. Thus by (.),
ϕ
d(gun+,gx) +d(gvn+,gy)
=ϕ
d(F(un,vn),F(x, y)) +d(F(vn,un),F(y,x))
≤ϕ
d(gun,gx) +d(gvn,gy)
–ψ
d(gun,gx) +d(gvn,gy)
. (.)
Sinceψis non-negative, we have
ϕ
d(gun+,gx) +d(gvn+,gy)
≤ϕ
d(gun,gx) +d(gvn,gy)
.
By the monotonicity ofϕ, we have
d(gun+,gx) +d(gvn+,gy)
≤
d(gun,gx) +d(gvn,gy)
. (.)
Thus, the sequence{dn}defined bydn=d(gun,gx)+d(gvn,gy), is a monotonically decreasing
se-quence of non-negative real numbers, so there exists somed≥ such thatlimn→∞dn=d.
We shall show thatd= . Suppose, to the contrary, thatd> . Then taking limit as
n→ ∞, in (.) and using the continuity ofϕ, we have
ϕ(d)≤ϕ(d) – lim
dn→d
ψ(dn) <ϕ(d),
a contradiction. Thus,d= ; that is,limn→∞dn= .
Hence, it follows thatgun→gx,gvn→gy.
Similarly, one can show thatgun→gx*,gvn→gy*.
Sincegx=F(x,y), gy=F(y,x) and the pair (F,g) is compatible, then by Lemma ., it follows that
ggx=gF(x,y) =F(gx,gy) and ggy=gF(y,x) =F(gy,gx). (.)
Denotegx=z,gy=w. Then by (.),
gz=F(z,w) and gw=F(w,z). (.)
Thus, (z,w) is a coupled coincidence point.
Then by (.) withx*=zandy*=w, it follows thatgz=gxandgw=gy; that is,
gz=z, gw=w. (.)
By (.) and (.),
z=gz=F(z,w) and w=gw=F(w,z).
Therefore, (z,w) is the coupled common fixed point ofFandg.
To prove the uniqueness, assume that (p,q) is another coupled common fixed point of
Fandg. Then by (.), we havep=gp=gz=zandq=gq=gw=w. Corollary . In addition to the hypothesis of Corollary ., suppose that for every
(x,y), (x*,y*)∈X×X,there exists a(u,v)∈X×X such that(F(u,v),F(v,u))is comparable to(F(x,y),F(y,x))and(F(x*,y*),F(y*,x*)).Then F and g have a unique coupled common
fixed point;that is, there exists a unique(x,y)∈X×X such that x=g(x) =F(x,y)and y=g(y) =F(y,x).
Proof Takingϕ(t) = t andψ(t) = ( –k)t, ≤k< in Theorem ., we obtain
Corol-lary ..
Remark . Indeed, (, ) is the unique coupled common fixed point of the mapsFand
gin Example . in view of Theorem . and Corollary ..
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors read and approved the final manuscript.
Author details
1Department of Mathematics, Ahir College, Rewari, 123401, India.2Department of Mathematics and Computer Sciences,
Cankaya University, Ankara, Turkey.3Department of Mathematics, DCRUST, Murthal, Sonepat, India. 4HAS Department,
YMCAUST, Faridabad, India.
Acknowledgements
The authors are thankful to the referees for their appreciation and valuable suggestions regarding this work.
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Cite this article as:Jain et al.:Coupled common fixed point results involving a(ϕ,ψ)-contractive condition for mixed