# On Coupled Fixed Point Theorems for Nonlinear Contractions in Partially OrderedG-Metric Spaces

Authors: S. A. Mohiuddine; Abdullah Alotaibi
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

Two concepts—one of the coupled fixed point and the other of the generalized metric space—play a very active role in recent research on the fixed point theory. The definition of coupled fixed point was introduced by Bhaskar and Lakshmikantham (2006) while the generalized metric space was introduced by Mustafa and Sims (2006). In this work, we determine some coupled fixed point theorems for mixed monotone mapping satisfying nonlinear contraction in the framework of generalized metric space endowed with partial order. We also prove the uniqueness of the coupled fixed point for such mappings in this setup.


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## Body

## 1. Introduction

Fixed point theory is a very useful tool in solving variety of problems in the control theory, economic theory, nonlinear analysis and, global analysis. The Banach contraction principle [1] is the most famous, most simplest, and one of the most versatile elementary results in the fixed point theory. A huge amount of literature is witnessed on applications, generalizations, and extensions of this principle carried out by several authors in different directions, for example, by weakening the hypothesis, using different setups, and considering different mappings.Recently, the idea of generalized metric spaces was introduced and studied by Mustafa and Sims [2] originated from the concept of metric spaces. Some fixed point theorem in this setup was first determined by Mustafa et al. [3]; particularly, the Banach contraction principle was established in this work. Since then several fixed point, coupled fixed point, and triple fixed point theorems in the framework of generalized metric spaces have been investigated in [4–11].In the recent past, many authors obtained important fixed point theorems in partially ordered metric spaces, that is, metric spaces endowed with a partial ordering (see [12–24]).The aim of this paper is to determine some coupled fixed point theorems for nonlinear contractions in the framework of partially ordered generalized metric spaces.


## 2. Definitions and Preliminary Results

We will assume throughout this paper that the symbolℝ and ℕ will denote the set of real and natural numbers, respectively. In this section, we recall some definitions and preliminary results which we will use throughout the paper. Mustafa and Sims [2] have recently introduced the concept of generalized metric space as follows.LetX be a nonempty set and a mapping G:X×X×X→ℝ. Then G is called a generalized metric (for short, G-metric) on X and (X,G) a generalized metric space or simply G-metric space if the following conditions are satisfied:  (i)
G(x,y,z)=0 if x=y=z,(ii)
G(x,x,y)>0, for all x,y∈X and x≠y,(iii)
G(x,x,y)≤G(x,y,z), for all x,y,z∈X and y≠z,(iv)
G(x,y,z)=G(x,z,y)=G(y,z,x)=⋯ (symmetry in all three variables),(v)
G(x,y,z)≤G(x,a,a)+G(a,y,z), for all x,y,z,a∈X (rectangle inequality).We remark that everyG-metric on X defines a metric dG on X by dG(x,y)=G(x,y,y)+G(y,x,x), for all x,y∈X.Example 2.1 (see [2]).
Let(X,d) be a metric space. The function G:X×X×X→[0,∞) defined by
(2.1)G(x,y,z)=max{d(x,y),d(y,z),d(z,x)},
or
(2.2)G(x,y,z)=d(x,y)+d(y,z)+d(z,x),
for all x,y,z∈X, is a G-metric on X.
The concepts of convergence and Cauchy sequences and continuous functions inG-metric space are studied in [2].
Let(X,G) be a G-metric space. Then, a sequence (xn) is said to be convergent in (X,G) or simply G-convergent to x∈X if for every ϵ>0 there exists N∈ℕ such that G(xn,xm,x)<ϵ, for all n,m≥N.
Let(X,G) be a G-metric space. Then, (xn) is said to be Cauchy in (X,G) or simply G-Cauchy if for every ϵ>0 there exists N∈ℕ such that G(xn,xm,xk)<ϵ, for all n,m,k≥N. A  G-metric space (X,G) is said to be complete if every G-Cauchy sequence is G-convergent.
Let(X,G) be a G-metric space and f:X→X a mapping. Then, f is said to G-continuous at a point x∈X if and only if it is G-sequentially continuous at x; that is, whenever (xn) is G-convergent to x, we have (f(xn))G-convergent to f(x).Proposition 2.2 (see [2]).
Let(X,G) be a G-metric space and (xn) a sequence in X. Then, for all x∈X, the following statements are equivalent:  (i)
(xn) is G-convergent to x,(ii)
G(xn,xn,x)→0 as n→∞,(iii)
G(xn,x,x)→0 as n→∞,(iv)
G(xn,xm,x)→0 as n,m→∞.Proposition 2.3 (see [2]).
Let(X,G) be a G-metric space and (xn) a sequence in X. Then, the following statements are equivalent:  (i)
(xn) is G-Cauchy,(ii)
For everyϵ>0 there exists N∈ℕ such that G(xn,xm,xm)<ϵ, for all n,m≥N.Lemma 2.4 (see [2]).
If(X,G) is a G-metric space then G(x,y,y)≤2G(y,x,x) for all x,y∈X.Let(X,G) be a G-metric space and F:X×X→X a mapping. Then, a map F is said to be continuous [10] in (X,G) if for every G-convergent sequences xn→x and yn→y, (F(xn,yn)) is G-convergent to F(x,y).Quite recently, Bhaskar and Lakshmikantham [14] defined and studied the concepts of mixed monotone property and coupled fixed point in partially ordered metric space.Let(X,≤) be a partially ordered set and F:X×X→X a mapping. Then, a map F is said to have 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,
(2.3)x1,x2∈X,x1≤x2impliesF(x1,y)≤F(x2,y),y1,y2∈X,y1≤y2impliesF(x,y1)≥F(x,y2).An element(x,y)∈X×X is said to be a coupled fixed point of the mapping F:X×X→X if
(2.4)F(x,y)=x,F(y,x)=y.The following class of functions are considered in [25]. Denote with Φ the set of all functions φ:[0,∞)→[0,∞) which satisfy that(i)
φ is continuous and nondecreasing,(ii)
φ(t)=0 if and only if t=0,(iii)
φ(t+s)≤φ(t)+φ(s), for all t,s∈[0,∞).ByΨ we denote the set of all functions ψ:[0,∞)→(0,∞) which satisfy limt→rψ(t)>0, for all r>0 and limt→0+ψ(t)=0.For example, functionsφ1,φ2,φ3,φ3∈Φ, where  φ1(t)=kt(k>0), φ2(t)=t/t+1, φ3(t)=ln(t+1), and φ4(t)=min{t,1}, and the functions ψ1,ψ2,ψ3∈Ψ, where ψ1(t)=kt, ψ2(t)=ln(2t+1)/2, and
(2.5)ψ3(t)={1,ift=0,t=1,tt+1,if0<t<1,t2,ift>1.


## 3.    Main Results

In this section, we establish some coupled fixed point results by considering a map on generalized metric spaces endowed with a partial order.Theorem 3.1.
Let(X,⪯) be a partially ordered set, and let G be a G-metric on X such that (X,G) is a complete G-metric space. Suppose that there exist φ∈Φ, ψ∈Ψ, and a mapping F:X×X→X such that
(3.1)φ(G(F(x,y),F(u,v),F(s,t)))≤12φ(G(x,u,s)+G(y,v,t))-ψ(G(x,u,s)+G(y,v,t)2),
for all x,y,u,v,s,t∈X with x≽u≽s and y⪯v⪯t where either u≠s or v≠t. Suppose F has a mixed monotone property and also suppose that either  (a)
F is continuous or(b)
X has the following property:
(i)
if a nondecreasing sequence(xn) is G-convergent to x, then xn⪯x, for all n,(ii)
if a nonincreasing sequence(yn) is G-convergent to y, then yn≽y, for all n.If there exist x0,y0∈X such that x0⪯F(x0,y0) and y0≽F(y0,x0), then F has a coupled point; that is, there exist x,y∈X such that F(x,y)=x and F(y,x)=y.Proof.
Letx0,y0∈X be such that x0⪯F(x0,y0) and y0≽F(y0,x0). We can choose x1,y1∈X such that x1=F(x0,y0) and y1=F(y0,x0). Write
(3.2)xn+1=F(xn,yn),yn+1=F(yn,xn),
for all n≥1. Due to the mixed monotone property of F, we can find x2≽x1≽x0 and y2⪯y1⪯y0. By straightforward calculation, we obtain
(3.3)x0⪯x1⪯x2⪯⋯⪯xn+1⪯⋯,y0≽y1≽y2≽⋯≽yn+1≽⋯.
Using (3.1) and (3.2), we obtain
(3.4)φ(G(xn+1,xn+1,xn))=φ(G(F(xn,yn),F(xn,yn),F(xn-1,yn-1)))≤12φ(G(xn,xn,xn-1)+G(yn,yn,yn-1))-ψ(G(xn,xn,xn-1)+G(yn,yn,yn-1)2),
and similarly
(3.5)φ(G(yn+1,yn+1,yn))=φ(G(F(yn,xn),F(yn,xn),F(yn-1,xn-1)))≤12φ(G(yn,yn,yn-1)+G(xn,xn,xn-1))-ψ(G(yn,yn,yn-1)+G(xn,xn,xn-1)2).
Adding (3.4) and (3.5), we get
(3.6)φ(G(xn+1,xn+1,xn))+φ(G(yn+1,yn+1,yn))≤φ(G(xn,xn,xn-1)+G(yn,yn,yn-1))-2ψ(G(xn,xn,xn-1)+G(yn,yn,yn-1)2).
Using the property φ(t+s)≤φ(t)+φ(s) for all t,s∈[0,∞), we get
(3.7)φ(G(xn+1,xn+1,xn)+G(yn+1,yn+1,yn))≤φ(G(xn,xn,xn-1)+G(yn,yn,yn-1))-2ψ(G(xn,xn,xn-1)+G(yn,yn,yn-1)2),
which implies that
(3.8)φ(G(xn+1,xn+1,xn)+G(yn+1,yn+1,yn))≤φ(G(xn,xn,xn-1)+G(yn,yn,yn-1)).
Since φ is nondecreasing, we have
(3.9)G(xn+1,xn+1,xn)+G(yn+1,yn+1,yn)≤G(xn,xn,xn-1)+G(yn,yn,yn-1).
For all n∈ℕ, set
(3.10)sn=G(xn+1,xn+1,xn)+G(yn+1,yn+1,yn),
then a sequence (sn) is decreasing. Therefore, there exists some s≥0 such that
(3.11)limn→∞sn=limn→∞[G(xn+1,xn+1,xn)+G(yn+1,yn+1,yn)]=s.
Now we have to show that s=0. On the contrary, suppose that s>0. Letting n→∞ in (3.7) (equivalently, sn is G-convergent to s) and using the property of φ and ψ, we get
(3.12)φ(s)=limn→∞φ(sn)≤limn→∞{φ(sn-1)-2ψ(sn-12)}=φ(s)-2limsn-1→sψ(sn-12)<φ(s),
which is a contradiction. Thus s=0; from (3.11), we have
(3.13)limn→∞sn=limn→∞[G(xn+1,xn+1,xn)+G(yn+1,yn+1,yn)]=0.
Again, we have to show that (xn) and (yn) are Cauchy sequences in the G-metric space (X,G). On the contrary, suppose that at least one of (xn) or (yn) is not a Cauchy sequence in (X,G). Then there exists ϵ>0, for which we can find subsequences (xk(j)), (xl(j)) and (yk(j)), (yl(j)) of the sequences (xn) and (yn), respectively, with k(j)>l(j)≥j, for all j∈ℕ such that
(3.14)αj=G(xk(j),xk(j),xl(j))+G(yk(j),yk(j),yl(j))≥ϵ.
We may also assume that
(3.15)G(xk(j)-1,xk(j)-1,xl(j))+G(yk(j)-1,yk(j)-1,yl(j))<ϵ,
by choosing k(j) to be the smallest number exceeding l(j), for which (3.14) holds. From (3.14) and (3.15) and using the rectangle inequality, we obtain
(3.16)ϵ≤αj=G(xk(j),xk(j),xl(j))+G(yk(j),yk(j),yl(j))≤G(xk(j),xk(j),xk(j)-1)+G(xk(j)-1,xk(j)-1,xl(j))+G(yk(j),yk(j),yk(j)-1)+G(yk(j)-1,yk(j)-1,yl(j))<G(xk(j),xk(j),xk(j)-1)+G(yk(j),yk(j),yk(j)-1)+ϵ.
Letting j→∞ in the above inequality and using (3.13), we get
(3.17)limj→∞αj=limj→∞[G(xk(j),xk(j),xl(j))+G(yk(j),yk(j),yl(j))]=ϵ.
Again, by using rectangle inequality, we obtain
(3.18)αj=G(xk(j),xk(j),xl(j))+G(yk(j),yk(j),yl(j))≤G(xk(j),xk(j),xk(j)+1)+G(xk(j)+1,xk(j)+1,xl(j)+1)+G(xl(j)+1,xl(j)+1,xl(j))+G(yk(j),yk(j),yk(j)+1)+G(yk(j)+1,yk(j)+1,yl(j)+1)+G(yl(j)+1,yl(j)+1,yl(j))=sl(j)+G(xk(j),xk(j),xk(j)+1)+G(xk(j)+1,xk(j)+1,xl(j)+1)+G(yk(j),yk(j),yk(j)+1)+G(yk(j)+1,yk(j)+1,yl(j)+1).
By using Lemma 2.4, the above inequality becomes
(3.19)αj≤sl(j)+2G(xk(j)+1,xk(j)+1,xk(j))+2G(yk(j)+1,yk(j)+1,yk(j))+G(xk(j)+1,xk(j)+1,xl(j)+1)+G(yk(j)+1,yk(j)+1,yl(j)+1);
this implies that
(3.20)αj≤sl(j)+2sk(j)+G(xk(j)+1,xk(j)+1,xl(j)+1)+G(yk(j)+1,yk(j)+1,yl(j)+1).
Operating φ on both sides of the above inequality,
(3.21)φ(αj)≤φ(sl(j)+2sk(j)+G(xk(j)+1,xk(j)+1,xl(j)+1)+G(yk(j)+1,yk(j)+1,yl(j)+1))=φ(sl(j)+2sk(j))+φ(G(xk(j)+1,xk(j)+1,xl(j)+1))+φ(G(yk(j)+1,yk(j)+1,yl(j)+1)).
Now we find the expressions φ(G(xk(j)+1,xk(j)+1,xl(j)+1)) and φ(G(yk(j)+1,yk(j)+1,yl(j)+1)) in terms of φ and ψ by using (3.1) and (3.2); that is,
(3.22)φ(G(xk(j)+1,xk(j)+1,xl(j)+1))=φ(G(F(xk(j),yk(j)),F(xk(j),yk(j)),F(xl(j),yl(j))))≤12φ(G(xk(j),xk(j),xl(j))+G(yk(j),yk(j),yl(j)))-ψ(G(xk(j),xk(j),xl(j))+G(yk(j),yk(j),yl(j))2),(3.23)φ(G(yk(j)+1,yk(j)+1,yl(j)+1))=φ(G(F(yk(j),xk(j)),F(yk(j),xk(j)),F(yl(j),xl(j))))≤12φ(G(yk(j),yk(j),yl(j))+G(xk(j),xk(j),xl(j)))-ψ(G(yk(j),yk(j),yl(j))+G(xk(j),xk(j),xl(j))2).
Adding (3.22) and (3.23), we get
(3.24)φ(G(xk(j)+1,xk(j)+1,xl(j)+1))+φ(G(yk(j)+1,yk(j)+1,yl(j)+1))≤φ(αj)-2ψ(αj2).
From (3.21) and (3.24), we obtain
(3.25)φ(αj)≤φ(sl(j)+2sk(j))+φ(αj)-2ψ(αj2).
Taking limit as j→∞ on both sides of the above inequality, we get
(3.26)φ(ϵ)≤φ(0)+φ(ϵ)-2limj→∞ψ(αj2)=φ(ϵ)-2limαj→ϵψ(αj2)<φ(ϵ),
which is a contradiction, and hence (xn) and (yn) are Cauchy sequences in the G-metric space (X,G). Since (X,G) is complete G-metric space, hence (xn) and (yn) are G-convergent. Then, there exist x,y∈X such that (xn) and (yn) are G-convergent to x and y, respectively. Suppose that condition (a) holds. Letting n→∞ in (3.2), we get x=F(x,y) and y=F(y,x). Lastly, suppose that assumption (b) holds. Since a sequence (xn) is nondecreasing and G-convergent to x and also (yn) is nonincreasing sequence and G-convergent to y, by assumption (b), we have xn⪯x and yn≽y for all n. Using the rectangle inequality, write
(3.27)G(x,x,F(x,y))≤G(x,x,xn+1)+G(xn+1,xn+1,F(x,y))=G(x,x,xn+1)+G(F(xn,yn),F(xn,yn),F(x,y)).
Applying the function φ on both sides of the above equation and using (3.1), we have
(3.28)φ(G(x,x,F(x,y)))≤φ(G(x,x,xn+1))+φ(G(F(xn,yn),F(xn,yn),F(x,y)))≤φ(G(x,x,xn+1))+12φ(G(xn,xn,x)+G(yn,yn,y))-ψ(G(xn,xn,x)+G(yn,yn,y)2).
Letting n→∞, we get G(x,x,F(x,y))=0. Hence x=F(x,y). Similarly we obtain y=F(y,x). Thus, we conclude that F has a coupled fixed point.Corollary 3.2.
Let(X,⪯) be a partially ordered set, and let G be a G-metric on X such that (X,G) is a complete G-metric space. Suppose that F:X×X→X is a mapping having mixed monotone property. Assume that there exists ψ∈Ψ such that for all x,y,u,v,s,t∈X,
(3.29)G(F(x,y),F(u,v),F(s,t))≤G(x,u,s)+G(y,v,t)2-ψ(G(x,u,s)+G(y,v,t)2),
with x≽u≽s and y⪯v⪯t where either u≠s or v≠t. Suppose that either  (a)
F is continuous or(b)
X has the following property:
(i)
if a nondecreasing sequence(xn) is G-convergent to x, then xn⪯x, for all n,(ii)
if a nonincreasing sequence(yn) is G-convergent to y, then yn≽y, for all n.If there exist x0,y0∈X such that x0⪯F(x0,y0) and y0≽F(y0,x0), then there exist x,y∈X such that F(x,y)=x and F(y,x)=y; that is, F has a coupled point in X.Proof.
Takingφ(t)=t in Theorem 3.1 and proceeding the same lines as in this theorem, we get the desired result.Corollary 3.3.
Let(X,⪯) be a partially ordered set, and let G be a G-metric on X such that (X,G) is a complete G-metric space. Suppose F:X×X→X is a mapping having mixed monotone property and assume that there exists k∈[0,1) such that
(3.30)G(F(x,y),F(u,v),F(s,t))≤k2[G(x,u,s)+G(y,v,t)],
for all x,y,u,v,s,t∈X with x≽u≽s and y⪯v⪯t where either u≠s or v≠t. Suppose that either  (a)
F is continuous or(b)
X has the following property:
(i)
if a nondecreasing sequence(xn) is G-convergent to x, then xn⪯x, for all n,(ii)
if a nonincreasing sequence(yn) is G-convergent to y, then yn≽y, for all n.If there exist x0,y0∈X such that x0⪯F(x0,y0) and y0≽F(y0,x0), then F has a coupled point in X.Proof.
Takingφ(t)=t and ψ(t)=((1-k)/2)t in Theorem 3.1 and proceeding the same lines as in this theorem, we get the desired result.Remark 3.4.
To assure the uniqueness of a coupled fixed point, we shall consider the following condition. If(Y,⪯) is a partially ordered set, we endowed the product Y×Y with
(3.31)(x,y)⪯(u,v)iffx⪯u,y≽v,
for all (x,y),(u,v)∈Y×Y.Theorem 3.5.
In addition to the hypothesis of Theorem3.1, suppose that for all (x,y),(s,t)∈X×X, there exists (u,v)∈X×X such that (F(u,v),F(v,u)) is comparable with (F(x,y),F(y,x)) and (F(s,t),F(t,s)). Then, F has a unique coupled fixed point.Proof.
It follows from Theorem3.1 that the set of coupled fixed points is nonempty. Suppose (x,y) and (s,t) are coupled fixed points of the mappings F:X×X→X; that is, x=F(x,y), y=F(y,x), and s=F(s,t), t=F(t,s). By assumption there exists (u,v) in X×X such that (F(u,v),F(v,u)) is comparable to (F(x,y),F(y,x)) and (F(s,t),F(t,s)). Put u=u0 and v=v0 and choose u1,v1∈X such that u1=F(u1,v1) and v1=F(v1,u1). Thus, we can define two sequences (un) and (vn) as
(3.32)un+1=F(un,vn),vn+1=F(vn,un).
Since (u,v) is comparable to (x,y), we can assume that (x,y)≽(u,v)=(u0,v0). Then it is easy to show that (un,vn) and (x,y) are comparable; that is, (x,y)≽(un,vn), for all n. Thus, from (3.1), we have
(3.33)φ(G(uu+1,x,x))=φ(G(F(uu,vn),F(x,y),F(x,y)))≤12φ(G(un,x,x)+G(vn,y,y))-ψ(G(un,x,x)+G(vn,y,y)2),(3.34)φ(G(y,y,vu+1))=φ(G(F(y,x),F(y,x),F(vu,un)))≤12φ(G(y,y,vn)+G(x,x,un))-ψ(G(y,y,vn)+G(x,x,un)2).
Using the property of φ and adding (3.33) and (3.34), we get
(3.35)φ(G(uu+1,x,x)+G(vu+1,y,y))≤φ(G(un,x,x)+G(vn,y,y))-2ψ(G(un,x,x)+G(vn,y,y)2),
which implies that
(3.36)φ(G(uu+1,x,x)+G(vu+1,y,y))≤φ(G(un,x,x)+G(vn,y,y)).
Therefore,
(3.37)G(uu+1,x,x)+G(vu+1,y,y)≤G(un,x,x)+G(vn,y,y).
We see that the sequence (G(un,x,x)+G(vn,y,y)) is decreasing; there exists some ξ≥0 such that
(3.38)limn→∞[G(un,x,x)+G(vn,y,y)]=ξ.
Now we have to show that ξ=0. On the contrary, suppose that ξ>0. Letting n→∞ in (3.35), we get
(3.39)φ(ξ)≤φ(ξ)-2ψlimn→∞(G(un,x,x)+G(vn,y,y)2)<φ(ξ),
which is not possible. Hence ξ=0. Therefore, (3.38) becomes
(3.40)limn→∞[G(un,x,x)+G(vn,y,y)]=0,
which implies
(3.41)limn→∞G(un,x,x)=0=limn→∞G(vn,y,y).
Similarly, we can show that limn→∞G(un,s,s)=0=limn→∞G(vn,t,t). We conclude that x=s and y=t. Thus, F has a unique coupled fixed point.Theorem 3.6.
In addition to the hypothesis of Theorem3.1, suppose that x0 and y0 are comparable. Then, F has a unique fixed point.Proof.
Proceeding the same lines as in the proof of Theorem3.1, we know that a mapping F:X×X→X has a coupled fixed point (x,y). Now we need to show that x=y. Since x0 and y0 are comparable, we can assume that x0≽y0. It is easy to show that xn≽yn for all n≥0, where xn+1=F(xn,yn) and yn+1=F(yn,xn). Suppose that G(x,x,y)>0, for all x,y∈X with x≠y. Using the rectangle inequality, write
(3.42)G(x,x,y)≤G(x,x,xn+1)+G(xn+1,xn+1,yn+1)+G(yn+1,yn+1,y)=G(x,x,xn+1)+G(F(xn,yn),F(xn,yn),F(yn,xn))+G(yn+1,yn+1,y).
Operating φ on both sides of the above inequality, we get
(3.43)φ(G(x,x,y))≤φ(G(x,x,xn+1)+G(yn+1,yn+1,y))+φ(G(F(xn,yn),F(xn,yn),F(yn,xn))).
From (3.1), we have
(3.44)φ(G(x,x,y))≤φ(G(x,x,xn+1)+G(yn+1,yn+1,y))+12φ(G(xn,xn,yn)+G(yn,yn,xn))-ψ(G(xn,xn,yn)+G(yn,yn,xn)2).
Similarly, we obtain
(3.45)φ(G(y,y,x))≤φ(G(y,y,yn+1)+G(xn+1,xn+1,x))+12φ(G(yn,yn,xn)+G(xn,xn,yn))-ψ(G(yn,yn,xn)+G(xn,xn,yn)2).
Adding (3.44) and (3.45) and using the property of φ, we get
(3.46)φ(G(x,x,y)+G(y,y,x))≤φ(G(x,x,xn+1)+G(yn+1,yn+1,y))+φ(G(y,y,yn+1)+G(xn+1,xn+1,x))+φ(G(xn,xn,yn)+G(yn,yn,xn))-2ψ(G(xn,xn,yn)+G(yn,yn,xn)2).
Letting n→∞ in the above inequality, we get
(3.47)φ(G(x,x,y)+G(y,y,x))≤φ(0)+φ(0)+φ(G(x,x,y)+G(y,y,x))-2limn→∞ψ(G(xn,xn,yn)+G(yn,yn,xn)2),
which implies that
(3.48)φ(G(x,x,y)+G(y,y,x))<φ(G(x,x,y)+G(y,y,x)),
which is not possible, and hence G(x,x,y)=0. Thus x=y, whence the result.Theorem 3.7.
Let(X,⪯) be a partially ordered set, and let G be a G-metric on X such that (X,G) is a complete G-metric space. Let F:X×X→X be a mapping such that F has a mixed monotone property and F(x,y)⪯F(y,x) whenever x⪯y. Suppose that there exist φ∈Φ and ψ∈Ψ such that for all x,y,u,v,s,t∈X, (3.1) holds with x≽u≽s, y⪯v⪯t and x≺y where either u≠s or v≠t. Assume that either  (a)
F is continuous or(b)
X has the following property:
(i)
if a nondecreasing sequence(xn) is G-convergent to x, then xn⪯x, for all n,(ii)
if a nonincreasing sequence(yn) is G-convergent to y, then yn≽y, for all n.If there exist x0,y0∈X such that x0⪯y0,  x0⪯F(x0,y0) and y0≽F(y0,x0), then F has a coupled point; that is, there exist x,y∈X such that F(x,y)=x and F(y,x)=y.Proof.
Letx0,y0∈X be such that x0⪯F(x0,y0) and y0≽F(y0,x0). We can choose x1,y1∈X such that x1=F(x0,y0)≽x0 and y1=F(y0,x0)⪯y0. Since x0⪯y0, we have F(x0,y0)⪯F(y0,x0). Accordingly,
(3.49)x0⪯x1=F(x0,y0)⪯F(y0,x0)=y1⪯y0.
Continuing this process, we can construct two sequences (xn) and (yn) in X such that
(3.50)xn⪯F(xn,yn)=xn+1⪯yn+1=F(yn,xn)⪯yn,
for all n≥0. Therefore,
(3.51)x0⪯x1⪯x2⪯⋯⪯xn⪯xn+1⋯⪯yn+1⪯yn⪯⋯⪯y2⪯y1⪯y0.
The rest of the proof can be done on the same lines as in Theorem 3.1.


## 4.  Example and the Concluding Remark

In the following, we construct an example of aG-metric space involving the idea of coupled fixed point to see the applicability of our results.LetX=ℕ∪{0}. Define a mapping G from X3 to ℝ by
(4.1)G(x,y,z)={x+y+z,ifx,y,zarealldistinctanddifferentfromzero,x+z,ifx=y≠zandallaredifferentfromzero,y+z+1,ifx=0,y≠zandy,zaredifferentfromzero,y+2,ifx=0,y=z≠0,z+1,ifx=0=y,z≠0,0,ifx=y=z.
Then (X,G) is a complete G-metric space [10]. Let us consider a partial order ⪯ on X be such that x⪯y holds if x>y, 3 divides (x-y), and 3⪯1 and 0⪯1 hold, for all x,y∈X. Consider a mapping F:X×X→X defined by
(4.2)F(x,y)={1,ifx≺y,0,otherwise.
Suppose that s⪯u⪯x≺y⪯v⪯t holds. Therefore, we have F(x,y)=1=F(u,v)=F(s,t). It follows that G(F(x,y),F(u,v),F(s,t))=0.Now, applying the functionφ to this equality and then using the hypothesis of this function, we see that (3.1) is satisfied since the left-hand side of (3.1) becomes 0. For x0=81 and y0=0, Theorem 3.7 is applicable. In this case, the coupled fixed point is not unique. Hence (0,0) and (1,0) are two coupled fixed points of F.We remark that inequality (3.1) is not satisfied when s=u=x=y=3, v=0, and t=1, and hence Theorem 3.1 does not work for this example. We know that a G-metric naturally induces a metric dG given by dG(x,y)=G(x,y,y)+G(x,x,y) [2], but inequality (3.1) does not reduce to any metric inequality with the metric dG due to the condition that either u≠s or v≠t. Hence our theorems are more general, different from the classical results, and do not reduce to fixed point problems in the corresponding metric space (X,dG).



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*Source: 897198-2012-12-02.xml*