Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2012, Article ID 897198, 15 pages
doi:10.1155/2012/897198
Research Article
On Coupled Fixed Point Theorems for Nonlinear
Contractions in Partially Ordered G-Metric Spaces
S. A. Mohiuddine and Abdullah Alotaibi
Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203,
Jeddah 21589, Saudi Arabia
Correspondence should be addressed to S. A. Mohiuddine, mohiuddine@gmail.com
Received 31 July 2012; Revised 2 October 2012; Accepted 2 October 2012
Academic Editor: Simeon Reich
Copyright q 2012 S. A. Mohiuddine and A. Alotaibi. This is an open access article distributed
under the Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
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.
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.
2
Abstract and Applied Analysis
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 symbol R 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.
Let X 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 Gx, y, z 0 if x y z,
ii Gx, x, y > 0, for all x, y ∈ X and x / y,
iii Gx, x, y ≤ Gx, y, z, for all x, y, z ∈ X and y /
z,
iv Gx, y, z Gx, z, y Gy, z, x · · · symmetry in all three variables,
v Gx, y, z ≤ Gx, a, a Ga, y, z, for all x, y, z, a ∈ X rectangle inequality.
We remark that every G-metric on X defines a metric dG on X by dG x, y Gx, y, y Gy, 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
G x, y, z max d x, y , d y, z , dz, x ,
2.1
G x, y, z d x, y d y, z dz, x,
2.2
or
for all x, y, z ∈ X, is a G-metric on X.
The concepts of convergence and Cauchy sequences and continuous functions in Gmetric 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 > 0 there exists N ∈ N such that Gxn , xm , x < ,
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 GCauchy if for every > 0 there exists N ∈ N such that Gxn , xm , xk < , 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 fxn G-convergent to fx.
Abstract and Applied Analysis
3
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 Gxn , xn , x → 0 as n → ∞,
iii Gxn , x, x → 0 as n → ∞,
iv Gxn , 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 every > 0 there exists N ∈ N such that Gxn , xm , xm < , for all n, m ≥ N.
Lemma 2.4 see 2. If X, G is a G-metric space then Gx, y, y ≤ 2Gy, 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,
Fxn , yn is G-convergent to Fx, 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 Fx, y is monotone nondecreasing in x and is
monotone nonincreasing in y; that is, for any x, y ∈ X,
x1 , x2 ∈ X,
x1 ≤ x2
y1 , y2 ∈ X,
y1 ≤ y2
implies F x1 , y ≤ F x2 , y ,
implies F x, y1 ≥ F x, y2 .
2.3
An element x, y ∈ X × X is said to be a coupled fixed point of the mapping F : X × X →
X if
F x, y x,
F y, x y.
2.4
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 ϕt 0 if and only if t 0,
iii ϕt s ≤ ϕt ϕs, for all t, s ∈ 0, ∞.
By Ψ we denote the set of all functions ψ : 0, ∞ → 0, ∞ which satisfy limt → r ψt >
0, for all r > 0 and limt → 0 ψt 0.
4
Abstract and Applied Analysis
For example, functions ϕ1 , ϕ2 , ϕ3 , ϕ3 ∈ Φ, where ϕ1 t kt k > 0, ϕ2 t t/t 1,
ϕ3 t lnt 1, and ϕ4 t min{t, 1}, and the functions ψ1 , ψ2 , ψ3 ∈ Ψ, where ψ1 t kt,
ψ2 t ln2t 1/2, and
⎧
1,
⎪
⎪
⎪
⎪
⎪
⎨ t
ψ3 t t 1 ,
⎪
⎪
⎪
⎪
⎪
⎩t,
2
if t 0, t 1,
if 0 < t < 1,
2.5
if t > 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
1 ϕ G F x, y , Fu, v, Fs, t ≤ ϕ Gx, u, s G y, v, t − ψ
2
Gx, u, s G y, v, t
,
2
3.1
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 Fx0 , y0 and y0 Fy0 , x0 , then F has a coupled point; that
is, there exist x, y ∈ X such that Fx, y x and Fy, x y.
Proof. Let x0 , y0 ∈ X be such that x0 Fx0 , y0 and y0 Fy0 , x0 . We can choose x1 , y1 ∈ X
such that x1 Fx0 , y0 and y1 Fy0 , x0 . Write
xn1 F xn , yn ,
yn1 F yn , xn ,
3.2
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
x0 x1 x2 · · · xn1 · · · ,
y0 y1 y2 · · · yn1 · · · .
3.3
Abstract and Applied Analysis
5
Using 3.1 and 3.2, we obtain
ϕGxn1 , xn1 , xn ϕ G F xn , yn , F xn , yn , F xn−1 , yn−1
Gxn , xn , xn−1 G yn , yn , yn−1
1 ,
≤ ϕ Gxn , xn , xn−1 G yn , yn , yn−1 − ψ
2
2
3.4
and similarly
ϕ G yn1 , yn1 , yn
ϕ G F yn , xn , F yn , xn , F yn−1 , xn−1
G yn , yn , yn−1 Gxn , xn , xn−1 1 .
≤ ϕ G yn , yn , yn−1 Gxn , xn , xn−1 − ψ
2
2
3.5
Adding 3.4 and 3.5, we get
ϕGxn1 , xn1 , xn ϕ G yn1 , yn1 , yn
≤ ϕ Gxn , xn , xn−1 G yn , yn , yn−1
Gxn , xn , xn−1 G yn , yn , yn−1
.
− 2ψ
2
3.6
Using the property ϕt s ≤ ϕt ϕs for all t, s ∈ 0, ∞, we get
ϕ Gxn1 , xn1 , xn G yn1 , yn1 , yn
≤ ϕ Gxn , xn , xn−1 G yn , yn , yn−1 − 2ψ
Gxn , xn , xn−1 G yn , yn , yn−1
,
2
3.7
which implies that
ϕ Gxn1 , xn1 , xn G yn1 , yn1 , yn ≤ ϕ Gxn , xn , xn−1 G yn , yn , yn−1 .
3.8
Since ϕ is nondecreasing, we have
Gxn1 , xn1 , xn G yn1 , yn1 , yn ≤ Gxn , xn , xn−1 G yn , yn , yn−1 .
3.9
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Abstract and Applied Analysis
For all n ∈ N, set
sn Gxn1 , xn1 , xn G yn1 , yn1 , yn ,
3.10
then a sequence sn is decreasing. Therefore, there exists some s ≥ 0 such that
lim sn lim Gxn1 , xn1 , xn G yn1 , yn1 , yn s.
n→∞
n→∞
3.11
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
s s n−1
n−1
ϕs − 2 lim ψ
< ϕs,
ϕs lim ϕsn ≤ lim ϕsn−1 − 2ψ
n→∞
n→∞
sn−1 → s
2
2
3.12
which is a contradiction. Thus s 0; from 3.11, we have
lim sn lim Gxn1 , xn1 , xn G yn1 , yn1 , yn 0.
n→∞
n→∞
3.13
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 > 0, for which we can find subsequences xkj , xlj and ykj , ylj of the sequences xn and yn , respectively, with kj > lj ≥ j, for all j ∈ N such that
αj G xkj , xkj , xlj G ykj , ykj , ylj ≥ .
3.14
We may also assume that
G xkj−1 , xkj−1 , xlj G ykj−1 , ykj−1 , ylj < ,
3.15
by choosing kj to be the smallest number exceeding lj, for which 3.14 holds. From 3.14
and 3.15 and using the rectangle inequality, we obtain
≤ αj G xkj , xkj , xlj G ykj , ykj , ylj
≤ G xkj , xkj , xkj−1 G xkj−1 , xkj−1 , xlj
G ykj , ykj , ykj−1 G ykj−1 , ykj−1 , ylj
< G xkj , xkj , xkj−1 G ykj , ykj , ykj−1 .
3.16
Letting j → ∞ in the above inequality and using 3.13, we get
lim αj lim G xkj , xkj , xlj G ykj , ykj , ylj .
j →∞
j →∞
3.17
Abstract and Applied Analysis
7
Again, by using rectangle inequality, we obtain
αj G xkj , xkj , xlj G ykj , ykj , ylj
≤ G xkj , xkj , xkj1 G xkj1 , xkj1 , xlj1 G xlj1 , xlj1 , xlj
G ykj , ykj , ykj1 G ykj1 , ykj1 , ylj1 G ylj1 , ylj1 , ylj
3.18
slj G xkj , xkj , xkj1 G xkj1 , xkj1 , xlj1
G ykj , ykj , ykj1 G ykj1 , ykj1 , ylj1 .
By using Lemma 2.4, the above inequality becomes
αj ≤ slj 2G xkj1 , xkj1 , xkj 2G ykj1 , ykj1 , ykj
G xkj1 , xkj1 , xlj1 G ykj1 , ykj1 , ylj1 ;
3.19
this implies that
αj ≤ slj 2skj G xkj1 , xkj1 , xlj1 G ykj1 , ykj1 , ylj1 .
3.20
Operating ϕ on both sides of the above inequality,
ϕ αj ≤ ϕ slj 2skj G xkj1 , xkj1 , xlj1 G ykj1 , ykj1 , ylj1
ϕ slj 2skj ϕ G xkj1 , xkj1 , xlj1 ϕ G ykj1 , ykj1 , ylj1 .
3.21
Now we find the expressions ϕGxkj1 , xkj1 , xlj1 and ϕGykj1 , ykj1 , ylj1 in
terms of ϕ and ψ by using 3.1 and 3.2; that is,
ϕ G xkj1 , xkj1 , xlj1 ϕ G F xkj , ykj , F xkj , ykj , F xlj , ylj
1 ϕ G xkj , xkj , xlj G ykj , ykj , ylj
2
G xkj , xkj , xlj G ykj , ykj , ylj
,
−ψ
2
ϕ G ykj1 , ykj1 , ylj1 ϕ G F ykj , xkj , F ykj , xkj , F ylj , xlj
≤
≤
1 ϕ G ykj , ykj , ylj G xkj , xkj , xlj
2
G ykj , ykj , ylj G xkj , xkj , xlj
.
−ψ
2
3.22
3.23
8
Abstract and Applied Analysis
Adding 3.22 and 3.23, we get
αj
.
ϕ G xkj1 , xkj1 , xlj1 ϕ G ykj1 , ykj1 , ylj1 ≤ ϕ αj − 2ψ
2
3.24
From 3.21 and 3.24, we obtain
αj
ϕ αj ≤ ϕ slj 2skj ϕ αj − 2ψ
.
2
3.25
Taking limit as j → ∞ on both sides of the above inequality, we get
ϕ ≤ ϕ0 ϕ − 2 lim ψ
j →∞
αj
2
ϕ − 2 lim ψ
αj → αj
2
< ϕ,
3.26
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 Fx, y and y Fy, 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
G x, x, F x, y ≤ Gx, x, xn1 G xn1 , xn1 , F x, y
Gx, x, xn1 G F xn , yn , F xn , yn , F x, y .
3.27
Applying the function ϕ on both sides of the above equation and using 3.1, we have
ϕ G x, x, F x, y
≤ ϕGx, x, xn1 ϕ G F xn , yn , F xn , yn , F x, y
1 ≤ ϕGx, x, xn1 ϕ Gxn , xn , x G yn , yn , y
2
Gxn , xn , x G yn , yn , y
.
−ψ
2
3.28
Letting n → ∞, we get Gx, x, Fx, y 0. Hence x Fx, y. Similarly we obtain y Fy, 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,
Gx, u, s G y, v, t
Gx, u, s G y, v, t
−ψ
,
G F x, y , Fu, v, Fs, t ≤
2
2
3.29
Abstract and Applied Analysis
9
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 Fx0 , y0 and y0 Fy0 , x0 , then there exist x, y ∈ X such
that Fx, y x and Fy, x y; that is, F has a coupled point in X.
Proof. Taking ϕt 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
k
Gx, u, s G y, v, t ,
G F x, y , Fu, v, Fs, t ≤
2
3.30
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 Fx0 , y0 and y0 Fy0 , x0 , then F has a coupled point in
X.
Proof. Taking ϕt t and ψt 1 − k/2t 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
x, y u, v iff x u, y v,
3.31
for all x, y, u, v ∈ Y × Y .
Theorem 3.5. In addition to the hypothesis of Theorem 3.1, suppose that for all x, y, s, t ∈ X × X,
there exists u, v ∈ X × X such that Fu, v, Fv, u is comparable with Fx, y, Fy, x and
Fs, t, Ft, s. Then, F has a unique coupled fixed point.
Proof. It follows from Theorem 3.1 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 Fx, y,
y Fy, x, and s Fs, t, t Ft, s. By assumption there exists u, v in X × X such
10
Abstract and Applied Analysis
that Fu, v, Fv, u is comparable to Fx, y, Fy, x and Fs, t, Ft, s. Put u u0 and
v v0 and choose u1 , v1 ∈ X such that u1 Fu1 , v1 and v1 Fv1 , u1 . Thus, we can define
two sequences un and vn as
un1 Fun , vn ,
vn1 Fvn , un .
3.32
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
ϕGuu1 , x, x ϕ G Fuu , vn , F x, y , F x, y
Gun , x, x G vn , y, y
1 ,
≤ ϕ Gun , x, x G vn , y, y − ψ
2
2
3.33
ϕ G y, y, vu1 ϕ G F y, x , F y, x , Fvu , un G y, y, vn Gx, x, un 1 .
≤ ϕ G y, y, vn Gx, x, un − ψ
2
2
3.34
Using the property of ϕ and adding 3.33 and 3.34, we get
ϕ Guu1 , x, x G vu1 , y, y ≤ ϕ Gun , x, x G vn , y, y
Gun , x, x G vn , y, y
− 2ψ
,
2
3.35
which implies that
ϕ Guu1 , x, x G vu1 , y, y ≤ ϕ Gun , x, x G vn , y, y .
3.36
Guu1 , x, x G vu1 , y, y ≤ Gun , x, x G vn , y, y .
3.37
Therefore,
We see that the sequence Gun , x, x Gvn , y, y is decreasing; there exists some ξ ≥ 0 such
that
lim Gun , x, x G vn , y, y ξ.
n→∞
3.38
Abstract and Applied Analysis
11
Now we have to show that ξ 0. On the contrary, suppose that ξ > 0. Letting n → ∞ in
3.35, we get
ϕξ ≤ ϕξ − 2ψ lim
n→∞
Gun , x, x G vn , y, y
< ϕξ,
2
3.39
which is not possible. Hence ξ 0. Therefore, 3.38 becomes
lim Gun , x, x G vn , y, y 0,
3.40
lim Gun , x, x 0 lim G vn , y, y .
3.41
n→∞
which implies
n→∞
n→∞
Similarly, we can show that limn → ∞ Gun , s, s 0 limn → ∞ Gvn , 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 Theorem 3.1, suppose that x0 and y0 are comparable.
Then, F has a unique fixed point.
Proof. Proceeding the same lines as in the proof of Theorem 3.1, 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 xn1 Fxn , yn and yn1 Fyn , xn . Suppose that Gx, x, y > 0, for all
x, y ∈ X with x /
y. Using the rectangle inequality, write
G x, x, y ≤ Gx, x, xn1 G xn1 , xn1 , yn1 G yn1 , yn1 , y
Gx, x, xn1 G F xn , yn , F xn , yn , F yn , xn G yn1 , yn1 , y .
3.42
Operating ϕ on both sides of the above inequality, we get
.
ϕ G x, x, y ≤ ϕ Gx, x, xn1 G yn1 , yn1 , y ϕ G F xn , yn , F xn , yn , F yn , xn
3.43
From 3.1, we have
ϕ G x, x, y ≤ ϕ Gx, x, xn1 G yn1 , yn1 , y
1 ϕ G xn , xn , yn G yn , yn , xn − ψ
2
G xn , xn , yn G yn , yn , xn
.
2
3.44
12
Abstract and Applied Analysis
Similarly, we obtain
ϕ G y, y, x ≤ ϕ G y, y, yn1 Gxn1 , xn1 , x
1 ϕ G yn , yn , xn G xn , xn , yn − ψ
2
G yn , yn , xn G xn , xn , yn
.
2
3.45
Adding 3.44 and 3.45 and using the property of ϕ, we get
ϕ G x, x, y G y, y, x
≤ ϕ Gx, x, xn1 G yn1 , yn1 , y ϕ G y, y, yn1 Gxn1 , xn1 , x
G xn , xn , yn G yn , yn , xn
ϕ G xn , xn , yn G yn , yn , xn − 2ψ
.
2
3.46
Letting n → ∞ in the above inequality, we get
ϕ G x, x, y G y, y, x ≤ ϕ0 ϕ0 ϕ G x, x, y G y, y, x
G xn , xn , yn G yn , yn , xn
− 2 lim ψ
,
n→∞
2
3.47
which implies that
ϕ G x, x, y G y, y, x < ϕ G x, x, y G y, y, x ,
3.48
which is not possible, and hence Gx, 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 Fx, y Fy, 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 Fx0 , y0 and y0 Fy0 , x0 , then F has a coupled
point; that is, there exist x, y ∈ X such that Fx, y x and Fy, x y.
Abstract and Applied Analysis
13
Proof. Let x0 , y0 ∈ X be such that x0 Fx0 , y0 and y0 Fy0 , x0 . We can choose x1 , y1 ∈ X
such that x1 Fx0 , y0 x0 and y1 Fy0 , x0 y0 . Since x0 y0 , we have Fx0 , y0 Fy0 , x0 . Accordingly,
x0 x1 F x0 , y0 F y0 , x0 y1 y0 .
3.49
Continuing this process, we can construct two sequences xn and yn in X such that
xn F xn , yn xn1 yn1 F yn , xn yn ,
3.50
x0 x1 x2 · · · xn xn1 · · · yn1 yn · · · y2 y1 y0 .
3.51
for all n ≥ 0. Therefore,
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 a G-metric space involving the idea of coupled
fixed point to see the applicability of our results.
Let X N ∪ {0}. Define a mapping G from X 3 to R by
⎧
x y z,
⎪
⎪
⎪
⎪
⎪
⎪
x z,
⎪
⎪
⎪
⎨y z 1,
G x, y, z ⎪
y 2,
⎪
⎪
⎪
⎪
⎪
z 1,
⎪
⎪
⎪
⎩
0,
if x, y, z are all distinct and different from zero,
if x y /
z and all are different from zero,
if x 0, y /
z and y, z are different from zero,
if x 0, y z / 0,
if x 0 y, z /
0,
4.1
if x 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
F x, y 1,
0,
if x ≺ y,
otherwise.
4.2
Suppose that s u x ≺ y v t holds. Therefore, we have Fx, y 1 Fu, v Fs, t.
It follows that GFx, y, Fu, v, Fs, t 0.
Now, applying the function ϕ 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
14
Abstract and Applied Analysis
induces a metric dG given by dG x, y Gx, y, y Gx, x, y 2, but inequality 3.1 does
s
not reduce to any metric inequality with the metric dG due to the condition that either u /
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 .
Acknowledgment
The authors have benefited from the reports of the anonymous referees and they are thankful
for their valuable comments on the first draft of this paper which improved the presentation
and readability.
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