# Coupled Coincidence Point Theorem in Partially Ordered Metric Spaces via Implicit Relation

Authors: Nguyen Manh Hung; Erdal Karapınar; Nguyen Van Luong
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We prove a coupled coincidence point theorem for mappingsF : X×X→X and g : X→X, where F has the mixed g-monotone property, in partially ordered metric spaces via implicit relations. Our result extends and improves several results in the literature. Examples are also given to illustrate our work.


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## Body

## 1. Introduction and Preliminaries

The notion of coupled fixed point was introduced by Guo and Lakshmikantham [1] in 1987. Later, Bhaskar and Lakshmikantham [2] defined the notions of mixed monotone mapping and proved some coupled fixed point theorems for the mixed monotone mappings. In this pioneer paper [2], they also discussed the existence and uniqueness of solution for a periodic boundary value problem. We start with recalling these basic concepts.Definition 1.1 (see [2]).
Let(X,⪯) be a partially ordered set and F:X×X→X. The mapping F is said to have the mixed monotone property if F(x,y) is monotone nondecreasing in x and is monotone nonincreasing in y, that is, for any x,y∈X, (1.1)x1,x2∈X,x1⪯x2⟹F(x1,y)⪯F(x2,y),y1,y2∈X,y1⪯y2⟹F(x,y1)⪰F(x,y2).Definition 1.2 (see [2]).
An element(x,y)∈X×X is called a coupled fixed point of the mapping F:X×X→X if
(1.2)x=F(x,y),y=F(y,x).The main results of Bhaskar and Lakshmikantham in [2] are the following theorems.Theorem 1.3 (see [2]).
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∈[0,1) with
(1.3)d(F(x,y),F(u,v))≤k2[d(x,u)+d(y,v)],
for all x≽u and y⪯v. If there exist two elements x0,y0∈X with
(1.4)x0⪯F(x0,y0),y0⪰F(y0,x0),
then there exist x,y∈X such that
(1.5)x=F(x,y),y=F(y,x).Theorem 1.4 (see [2]).
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 nondecreasing sequence{xn}→x, then xn⪯x for all n,(ii)
if a nonincreasing sequence{yn}→y, then y⪯yn for all n.
LetF:X×X→X be a mapping having the mixed monotone property on X. Assume that there exists a k∈[0,1) with
(1.6)d(F(x,y),F(u,v))≤k2[d(x,u)+d(y,v)],
for all x≽u and y⪯v. If there exist two elements x0,y0∈X with
(1.7)x0⪯F(x0,y0),y0⪰F(y0,x0),
then there exist x,y∈X such that
(1.8)x=F(x,y),y=F(y,x).Afterwards, a number of coupled coincidence/fixed point theorems and their application to integral equations, matrix equations, and periodic boundary value problem have been established (e.g., see [3–28] and references therein). In particular, Lakshmikantham and Ćirić [7] established coupled coincidence and coupled fixed point theorems for two mappings F:X×X→X and g:X→X, where F has the mixed g-monotone property and the functions F and g commute, as an extension of the fixed point results in [2]. Choudhury and Kundu in [15] introduced the concept of compatibility and proved the result established in [7] under a different set of conditions. Precisely, they established their result by assuming that F and g are compatible mappings. For the sake of completeness, we remind these characterizations.Definition 1.5 (see [7]).
Let(X,⪯) be a partially ordered set and let F:X×X→X and g:X→X are two mappings. We say F has the mixed g-monotone property if F(x,y) is g-nondecreasing in its first argument and is g-nonincreasing in its second argument, that is, for any x,y∈X,
(1.9)x1,x2∈X,gx1⪯gx2⟹F(x1,y)⪯F(x2,y),y1,y2∈X,gy1⪯gy2⟹F(x,y1)⪰F(x,y2).Definition 1.6 (see [7]).
An element(x,y)∈X×X is called a coupled coincident point of the mappings F:X×X→X and g:X→X if
(1.10)gx=F(x,y),gy=F(y,x).Definition 1.7 (see [15]).
The mappingsF and g where F:X×X→X,g:X→X are said to be compatible if
(1.11)lim⁡n→∞d(gF(xn,yn),F(gxn,gyn))=0,lim⁡n→∞d(gF(yn,xn),F(gyn,gxn))=0,
where {xn} and {yn} are sequences in X such that limn→∞F(xn,yn)=limn→∞gxn=x and limn→∞F(yn,xn)=limn→∞gyn=y for all x,y∈X are satisfied.Luong and Thuan [11] slightly extended the concept of compatible mappings into the context of partially ordered metric spaces, namely, O-compatible mappings and proved some coupled coincidence point theorems for such mappings in partially ordered generalized metric spaces.The concept ofO-compatible mappings is stated as follows.Definition 1.8 (cf. [11]).
Let(X,⪯,d) be a partially ordered metric space. The mappings F:X×X→X and g:X→X are said to be O-compatible if
(1.12)lim⁡n→∞d(gF(xn,yn),F(gxn,gyn))=0,lim⁡n→∞d(gF(yn,xn),F(gyn,gxn))=0,
where {xn} and {yn} are sequences in X such that {gxn}, {gyn} are monotone and
(1.13)lim⁡n→∞F(xn,yn)=lim⁡n→∞gxn=x,lim⁡n→∞F(yn,xn)=lim⁡n→∞gyn=y,
for all x,y∈X are satisfied.Let(X,⪯,d) be a partially metric space. If F:X×X→X and g:X→X are compatible then they are O-compatible. However, the converse is not true. The following example shows that there exist mappings that are O-compatible but not compatible.Example 1.9 (see [11]).
LetX={0}∪[1/2,2] with the usual metric d(x,y)=|x-y|, for all x,y∈X. We consider the following order relation on X:
(1.14)x,y∈Xx⪯y⟺x=yor(x,y)∈{(0,0),(0,1),(1,1)}.
Let F:X×X→X be given by
(1.15)F(x,y)={0,ifx,y∈{0}∪[12,1],1,otherwise,
and g:X→X be defined by
(1.16)gx={0,ifx=0,1,if12≤x≤1,2-x,if1<x≤32,12,if32<x≤2.
Then F and g are O-compatible but not compatible.Indeed, let{xn},{yn} in X such that {gxn},{gyn} are monotone and
(1.17)lim⁡n→∞F(xn,yn)=lim⁡n→∞gxn=x,lim⁡n→∞F(yn,xn)=lim⁡n→∞gyn=y,
for some x,y∈X. Since F(xn,yn)=F(yn,xn)∈{0,1} for all n, x=y∈{0,1}. The case x=y=1 is impossible. In fact, if x=y=1. Then since {gxn},{gyn} are monotone, gxn=gyn=1 for all n≥n1, for some n1. That is xn,yn∈[1/2,1] for all n≥n1. This implies F(xn,yn)=F(yn,xn)=0, for all n≥n1, which is a contradiction. Thus x=y=0. That implies gxn=gyn=0 for all n≥n2, for some n2. That is xn=yn=0 for all n≥n2. Thus, for all n≥n2,
(1.18)d(gF(xn,yn),F(gxn,gyn))=d(gF(yn,xn),F(gyn,gxn))=0.
Hence
(1.19)lim⁡n→∞d(gF(xn,yn),F(gxn,gyn))=0,lim⁡n→∞d(gF(yn,xn),F(gyn,gxn))=0
hold. Therefore F and g are O-compatible.Now let{xn},{yn} in X be defined by
(1.20)xn=yn=1+1n+1,n=1,2,3,….
We have
(1.21)F(xn,yn)=F(yn,xn)=F(1+1n+1,1+1n+1)=1,gxn=gyn=g(1+1n+1)=1-1n+1⟶1asn⟶∞,
but
(1.22)d(gF(xn,yn),F(gxn,gyn))=d(F(1-1n+1,1-1n+1),g1)=d(0,1)=1↛0asn⟶∞.
Thus, F and g are not compatible.Implicit relation on metric spaces has been used in many articles (see, e.g., [29–31] and references therein). In this paper, we use the following implicit relation to prove a coupled coincidence point theorem for mappings F:X×X→X and g:X→X, where F has the mixed g-monotone property and F,g are O-compatible.LetΦ denote all functions φ:ℝ+→ℝ+ which satisfy(i)
φ is continuous,(ii)
φ(t)<t for each t>0.Obviously, ifφ∈Φ then φ(0)=0.Letℍ denote all continuous functions H:(ℝ+)5→ℝ which satisfy(H1)
H(t1,t2,t3,t4,t5) is nonincreasing in t2 and t5,(H2)
there exists a functionφ∈Φ such that
(1.23)H(u,u+v,v,w,u+v)≤0impliesu≤φ(max⁡{v,w}).It is easy to check that the following functions are inℍ:(i)
H1(t1,t2,t3,t4,t5)=t1-αt2-βt3-γt4-θt5, where α,β,γ,θ are nonnegative real numbers satisfying 2α+β+γ+2θ<1;(ii)
H2(t1,t2,t3,t4,t5)=t1-αmax{t2/2,t3,t4,t5/2}, where α∈(0,1);(iii)
H3(t1,t2,t3,t4,t5)=t1-φ(max{t3,t4}), where φ∈Φ.In this paper, we prove a coupled coincidence point theorem for mappings satisfying such implicit relations.


## 2. Coupled Coincidence Point Theorem

Now we are going to prove our main result.Theorem 2.1.
Let(X,d,⪯) be a partially ordered complete metric space. Suppose  F:X×X→X and  g:X→X are mappings such that  F has the mixed  g-monotone property. Assume that there exists  H∈ℍ such that
(2.1)H(d(F(x,y),F(u,v)),d(F(x,y),gx)+d(F(u,v),gu),d(gx,gu),d(gy,gv),d(F(x,y),gu)+d(F(u,v),gx))≤0,
for all x,y,u,v∈X with gx≽gu and gy⪯gv. Suppose F(X×X)⊆g(X), g is continuous and g is O-compatible with F. Suppose either(a)
F is continuous or;(b)
X has the following property:
(i)
if a nondecreasing sequence{xn}→x, then gxn⪯gx for all n,(ii)
if a nonincreasing sequence{yn}→y, then gy⪯gyn for all n.
If there exist two elementsx0,y0∈X with
(2.2)gx0⪯F(x0,y0),gy0⪰F(y0,x0),
then F and g have a coupled coincidence point in X.Proof.
Letx0,y0∈X be such that gx0⪯F(x0,y0) and gy0≽F(y0,x0). Since F(X×X)⊆g(X), we construct the sequences {xn} and {yn} in X as follows:
(2.3)gxn+1=F(xn,yn),gyn+1=F(yn,xn)∀n≥0.
By using the mathematical induction and the mixed g-monotone property of F, we can show that
(2.4)gxn⪯gxn+1,gyn⪰gyn+1,∀n≥0.
If there is somen0∈ℕ* such that gxn0=gxn0+1 and gyn0=gyn0+1 then
(2.5)gxn0=gxn0+1=F(xn0,yn0),gyn0=gyn0+1=F(yn0,xn0),
that means (xn0,yn0) is a coupled coincidence point of F and g. Thus we may assume that max{d(gxn+1,gxn),d(gyn+1,gyn)}>0 for all n.
Sincegxn+1≽gxn and gyn+1⪯gyn, from (2.1), we have
(2.6)H(d(F(xn+1,yn+1),F(xn,yn)),d(F(xn+1,yn+1),gxn+1)+d(F(xn,yn),gxn),d(gxn+1,gxn),d(gyn+1,gyn),d(F(xn+1,yn+1),gxn)+d(F(xn,yn),gxn+1))≤0
or
(2.7)H(d(gxn+2,gxn+1),d(gxn+2,gxn+1)+d(gxn+1,gxn),d(gxn+1,gxn),d(gyn+1,gyn),d(gxn+2,gxn))≤0.
By the properties of H, we have
(2.8)H(d(gxn+2,gxn+1),d(gxn+2,gxn+1)+d(gxn+1,gxn),d(gxn+1,gxn),d(gyn+1,gyn),d(gxn+2,gxn+1)+d(gxn+1,gxn))≤0,
which implies that
(2.9)d(gxn+2,gxn+1)≤φ(max⁡{d(gxn+1,gxn),d(gyn+1,gyn)}).
Similarly, one can show that
(2.10)d(gyn+2,gyn+1)≤φ(max⁡{d(gxn+1,gxn),d(gyn+1,gyn)}).
From (2.9) and (2.10), we have
(2.11)max⁡{d(gxn+2,gxn+1),d(gyn+2,gyn+1)}≤φ(max⁡{d(gxn+1,gxn),d(gyn+1,gyn)}),
which implies
(2.12)max⁡{d(gxn+2,gxn+1),d(gyn+2,gyn+1)}<max⁡{d(gxn+1,gxn),d(gyn+1,gyn)}.
This means that {dn:=max{d(gxn+1,gxn),d(gyn+1,gyn)}} is a decreasing sequence of positive real numbers. So there is a d≥0 such that
(2.13)lim⁡n→∞dn=lim⁡n→∞max⁡{d(gxn+1,gxn),d(gyn+1,gyn)}=d.
We will show that d=0. Assume, to the contrary, that d>0. Taking n→∞ in (2.11), we have
(2.14)d≤lim⁡n→∞φ(dn)=φ(d)<d,
which is a contradiction. Thus d=0.
In what follows, we will show that{gxn} and {gyn} are Cauchy sequences. Suppose, to the contrary that at least one of {gxn} or {gyn} is not a Cauchy sequence. This means that there exists an ε>0 for wich we can find subsequences {gxn(k)},{gxm(k)} of {gxn} and {gyn(k)},{gym(k)} of {gyn} with n(k)>m(k)≥k such that
(2.15)max⁡{d(gxn(k),gxm(k)),d(gyn(k),gym(k))}≥ε.
Further, corresponding to m(k), we can choose n(k) in such a way that it is the smallest integer with n(k)>m(k)≥k and satisfies (2.15). Then
(2.16)max⁡{d(gxn(k)-1,gxm(k)),d(gyn(k)-1,gym(k))}<ε.
Using the triangle inequality and (2.16), we have
(2.17)d(gxn(k),gxm(k))≤d(gxn(k),gxn(k)-1)+d(gxn(k)-1,gxm(k))<d(gxn(k),gxn(k)-1)+ε,d(gyn(k),gym(k))≤d(gyn(k),gyn(k)-1)+d(gyn(k)-1,gym(k))<d(gyn(k),gyn(k)-1)+ε.
From (2.15) and (2.17), we have
(2.18)ε≤max⁡{d(gxn(k),gxm(k)),d(gyn(k),gym(k))}<max⁡{d(gxn(k),gxn(k)-1),d(gyn(k),gyn(k)-1)}+ε.
Letting k→∞ in the inequalities above and using (2.13) we get
(2.19)lim⁡k→∞max⁡{d(gxn(k),gxm(k)),d(gyn(k),gym(k))}=ε.
By the triangle inequality
(2.20)d(gxn(k),gxm(k))≤d(gxn(k),gxn(k)-1)+d(gxn(k)-1,gxm(k)-1)+d(gxm(k)-1,gxm(k)),d(gyn(k),gym(k))≤d(gyn(k),gyn(k)-1)+d(gyn(k)-1,gym(k)-1)+d(gym(k-1,gym(k)).
From the last two inequalities and (2.15), we have
(2.21)ε≤max⁡{d(gxn(k),gxm(k)),d(gyn(k),gym(k))}≤max⁡{d(gxn(k),gxn(k)-1),d(gyn(k),gyn(k)-1)}+max⁡{d(gxm(k)-1,gxm(k)),d(gym(k)-1,gym(k))}+max⁡{d(gxn(k)-1,gxm(k)-1),d(gyn(k)-1,gym(k)-1)}.
Again, by the triangle inequality,
(2.22)d(gxn(k)-1,gxm(k)-1)≤d(gxn(k)-1,gxm(k))+d(gxm(k),gxm(k)-1)<d(gxm(k),gxm(k)-1)+ε,d(gyn(k)-1,gym(k)-1)≤d(gyn(k)-1,gym(k))+d(gym(k),gym(k)-1)<d(gym(k),gym(k)-1)+ε.
Therefore,
(2.23)max⁡{d(gxn(k)-1,gxm(k)-1),d(gyn(k)-1,gym(k)-1)}<max⁡{d(gxm(k),gxm(k)-1),d(gym(k),gym(k)-1)}+ε.
From (2.21) and (2.23), we have
(2.24)ε-max⁡{d(gxn(k),gxn(k)-1),d(gyn(k),gyn(k)-1)}-max⁡{d(gxm(k)-1,gxm(k)),d(gym(k)-1,gym(k))}≤max⁡{d(gxn(k)-1,gxm(k)-1),d(gyn(k)-1,gym(k)-1)}<max⁡{d(gxm(k),gxm(k)-1),d(gym(k),gym(k)-1)}+ε.
Taking k→∞ in the inequalities above and using (2.13), we get
(2.25)lim⁡k→∞max⁡{d(gxn(k)-1,gxm(k)-1),d(gyn(k)-1,gym(k)-1)}=ε.
From (2.19) and (2.25), the sequences {d(gxn(k),gxm(k))}, {d(gyn(k),gym(k))}, {d(gxn(k)-1,gxm(k)-1)}, and {d(gyn(k)-1,gym(k)-1)} have subsequences converging to ε1, ε2, ε3 and ε4, respectively, and max{ε1,ε2}=max{ε3,ε4}=ε>0. We may assume that
(2.26)lim⁡k→∞d(gxn(k),gxm(k))=ε1,lim⁡k→∞d(gyn(k),gym(k))=ε2,lim⁡k→∞d(gxn(k)-1,gxm(k)-1)=ε3,lim⁡k→∞d(gyn(k)-1,gym(k)-1)=ε4.
We first assume thatε1=max{ε1,ε2}=ε. Since n(k)>m(k), gxn(k)-1≽gxm(k)-1 and gyn(k)-1⪯gym(k)-1. From (2.1), we have
(2.27)H(d(F(xn(k)-1,yn(k)-1),F(xm(k)-1,ym(k)-1)),d(F(xn(k)-1,yn(k)-1),gxn(k)-1)+d(F(xm(k)-1,ym(k)-1),gxm(k)-1),d(gxn(k)-1,gxm(k)-1),d(gyn(k)-1,gym(k)-1),d(F(xn(k)-1,yn(k)-1),gxm(k)-1)+d(F(xm(k)-1,ym(k)-1),gxn(k)-1))≤0
or
(2.28)H(d(gxn(k),gxm(k)),d(gxn(k),gxn(k)-1)+d(gxm(k),gxm(k)-1),d(gxn(k)-1,gxm(k)-1),d(gyn(k)-1,gym(k)-1),d(gxn(k),gxm(k)-1)+d(gxm(k),gxn(k)-1))≤0
or
(2.29)H(d(gxn(k),gxm(k)),d(gxn(k),gxn(k)-1)+d(gxm(k),gxm(k)-1),d(gxn(k)-1,gxm(k)-1),d(gyn(k)-1,gym(k)-1),d(gxn(k),gxm(k))+d(gxm(k),gxm(k)-1)+d(gxm(k),gxm(k)-1)+d(gxm(k)-1,gxn(k)-1))≤0.
Letting k→∞, we have
(2.30)H(ε1,0,ε3,ε4,ε1+ε3)≤0.
Thus,
(2.31)H(ε1,ε1+ε3,ε3,ε4,ε1+ε3)≤0,
which implies ε=ε1≤φ(max{ε3,ε4})=φ(ε)<ε. That is a contradiction.
Using the same argument as above for the caseε2=max{ε1,ε2}=ε, we also get a contradiction. Thus {gxn} and {gyn} are Cauchy sequences. Since X is complete, there exist x,y∈X such that
(2.32)lim⁡n→∞gxn=x,lim⁡n→∞gyn=y.
Thus
(2.33)lim⁡n→∞F(xn,yn)=lim⁡n→∞gxn=x,lim⁡n→∞F(yn,xn)=lim⁡n→∞gyn=y.
Since F and g are O-compatible, from (2.33), we have
(2.34)lim⁡n→∞d(gF(xn,yn),F(gxn,gyn))=0,(2.35)lim⁡n→∞d(gF(yn,xn),F(gyn,gxn))=0.
Now, suppose that assumption (a) holds. We have
(2.36)d(gx,F(g(xn),g(yn)))≤d(gx,gF(xn,yn))+d(gF(xn,yn),F(gxn,gyn)).
Taking the limit as n→∞ in (2.36) and by (2.32), (2.34) and the continuity of F and g we get d(gx,F(x,y))=0.
Similarly, we can show thatd(gy,F(y,x))=0. Therefore, gx=F(x,y) and gy=F(y,x).
Finally, suppose that assumption (b) holds. Since{gxn} is nondecreasing sequence and gxn→x and {gyn} is nonincreasing sequence and gyn→y, by the assumption, we have ggxn⪯gx and ggyn≽gy for all n.
Sinceg is continuous, from (2.32), (2.34), and (2.35) we have
(2.37)lim⁡n→∞ggxn=gx=lim⁡n→∞gF(xn,yn)=lim⁡n→∞F(gxn,gyn),lim⁡n→∞ggyn=gy=lim⁡n→∞gF(yn,xn)=lim⁡n→∞F(gyn,gxn).
We have
(2.38)H(d(F(gxn,gyn),F(x,y)),d(F(gxn,gyn),ggxn)+d(F(x,y),gx)d(ggxn,gx),d(ggyn,gy),d(F(gxn,gyn),gx)+d(F(x,y),ggxn))≤0.
Letting n→∞ and using (2.37), we have
(2.39)H(d(gx,F(x,y)),d(gx,F(x,y)),0,0,d(gx,F(x,y)))≤0,
which implies that d(gx,F(x,y))≤φ(max{0,0})=0. Hence gx=F(x,y). Similarly, one can show that gy=F(y,x).
Thus proved thatF and g have a coupled coincidence point in X.Example 2.2 (see, e.g., [11]).
Let(X,d,⪯),F and g be defined as in Example 1.9. Then(i)
X is complete and X has the property
(a)
if a nondecreasing sequence{xn}→x, then gxn⪯gx for all n,(b)
if a nonincreasing sequence{yn}→y, then gy⪯gyn for all n;(ii)
F(X×X)={0,1}⊂{0}∪[1/2,1]=g(X);(iii)
g is continuous and g and F are O-compatible;(iv)
there existx0=0,y0=1 such that gx0⪯F(x0,y0) and gy0≽F(y0,x0);(v)
F has the mixed g-monotone property. Indeed, for every y∈X, let x1,x2∈X such that gx1⪯gx2
(a)
ifgx1=gx2 then x1,x2=0 or x1,x2∈[1/2,1] or x1,x2∈(1,3/2] or x1,x2∈(3/2,2]. Thus,
(2.40)F(x1,y)=0=F(x2,y)ify∈{0}∪[12,1],x1,x2=0orx1,x2∈[12,1],
otherwise F(x1,y)=1=F(x2,y),(b)
ifgx1≺gx2, then gx1=0 and gx2=1, that is, x1=0 and x2∈[1/2,1]. Thus
(2.41)F(x1,y)=0=F(x2,y)ify∈{0}∪[12,1],F(x1,y)=1=F(x2,y)ify∈(1,2],
therefore, F is the g-nondecreasing in its first argument. Similarly, F is the g-nonincreasing in its second argument;(vi)
forx,y,u,v∈X, if gx≽gu and gy⪯gv then d(F(x,y),F(u,v))=0. Indeed,
(a)
ifgx≻gu and gy≺gv then y=u=0 and x,v∈[1/2,1]. Thus d(F(x,y),F(u,v))=d(0,0)=0,(b)
ifgx=gu and gy≺gv then y=0 and v∈[1/2,1]. Thus if x=u=0 or x,u∈[1/2,1] then d(F(x,y),F(u,v))=d(0,0)=0, otherwise d(F(x,y),F(u,v))=d(1,1)=0. Similarly, if gx≻gu and gy=gv then d(F(x,y),F(u,v))=0,(c)
ifgx=gu and gy=gv then both x,u are in one of the sets {0}, [1/2,1], (1,3/2] or (3/2,2] and both y,v are also in one of the sets {0}, [1/2,1], (1,3/2] or (3/2,2]. Thus d(F(x,y),F(u,v))=d(0,0)=0 if x=u=0 or x,u∈[1/2,1] and y=v=0 or y,v∈[1/2,1], otherwise, d(F(x,y),F(u,v))=d(1,1)=0.
Therefore, all the conditions of Theorem2.1 are satisfied with H(t1,t2,t3,t4,t5)=t1-max{t3,t4}/2. Applying Theorem 2.1, we conclude that F and g have a coupled coincidence point.
Note that, we cannot apply the result of Choudhury and Kundu [15], the result of Choudhury et al. [32] as well as the result of Lakshmikantham and Ćirić [7] to this example.



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*Source: 796964-2012-07-08.xml*