Open Access

Coupled coincidence points for monotone operators in partially ordered metric spaces

Fixed Point Theory and Applications20112011:44

https://doi.org/10.1186/1687-1812-2011-44

Received: 18 March 2011

Accepted: 30 August 2011

Published: 30 August 2011

Abstract

Using the notion of compatible mappings in the setting of a partially ordered metric space, we prove the existence and uniqueness of coupled coincidence points involving a (ϕ, ψ)-contractive condition for a mappings having the mixed g-monotone property. We illustrate our results with the help of an example.

Keywords

coupled coincidence pointpartially ordered metric spacemixed g-monotone property

1 Introduction

The Banach contraction principle is the most celebrated fixed point theorem. Afterward many authors obtained many important extensions of this principle (cf. [116]). Recently Bhaskar and Lakshmikantham [5], Nieto and Lopez [12, 13], Ran and Reurings [14] and Agarwal et al. [3] presented some new results for contractions in partially ordered metric spaces. Bhaskar and Lakshmikantham [5] noted that their theorem can be used to investigate a large class of problems and have discussed the existence and uniqueness of solution for a periodic boundary value problem.

Recently, Luong and Thuan [11] presented some coupled fixed point theorems for a mixed monotone mapping in a partially ordered metric space which are generalizations of the results of Bhaskar and Lakshmikantham [5]. In this paper, we establish the existence and uniqueness of coupled coincidence point involving a (ϕ,ψ)-contractive condition for mappings having the mixed g-monotone property. We also illustrate our results with the help of an example.

2 Preliminaries

A partial order is a binary relation over a set X which is reflexive, antisymmetric, and transitive. Now, let us recall the definition of the monotonic function f : XX in the partially order set (X, ). We say that f is non-decreasing if for x, y X, x y, we have fx fy. Similarly, we say that f is non-increasing if for x, y X, x y, we have fx fy. Any one could read on [9] for more details on fixed point theory.

Definition 2.1 [10] (Mixed g-Monotone Property)

Let (X, ) be a partially ordered set and F : X × XX. We say that the mapping F has the mixed g-monotone property if F is monotone g-non-decreasing in its first argument and is monotone g-non-increasing in its second argument. That is, for any x, y X,
x 1 , x 2 X , g x 1 g x 2 F ( x 1 , y ) F ( x 2 , y )
(1)
and
y 1 , y 2 X , g y 1 g y 2 F ( x , y 1 ) F ( x , y 2 ) .
(2)

Definition 2.2 [10] (Coupled Coincidence Point)

Let (x, y) X × X, F : X × XX and g : XX. We say that (x, y) is a coupled coincidence point of F and g if F(x, y) = gx and F(y, x) = gy for x, y X.

Definition 2.3 [10] Let X be a non-empty set and let F : X × XX and g : XX. We say F and g are commutative if, for all x, y X,
g ( F ( x , y ) ) = F ( g ( x ) , g ( y ) ) .
Definition 2.4 [6] The mapping F and g where F : X × XX and g : XX, are said to be compatible if
lim n d ( g ( F ( x n , y n ) ) , F ( g x n , g y n ) ) = 0
and
lim n d ( g ( F ( y n , x n ) ) , F ( g y n , g x n ) ) = 0 ,

whenever {x n } and {y n } are sequences in X, such that lim n →∞ F (x n , y n ) = lim n →∞ gx n = x and lim n →∞ F (y n , x n ) = lim n →∞ gy n = y, for all x, y X are satisfied.

3 Existence of coupled coincidence points

As in [11], let ϕ denote all functions ϕ : [0, ∞) → [0, ∞) which satisfy
  1. 1.

    ϕ is continuous and non-decreasing,

     
  2. 2.

    ϕ (t) = 0 if and only if t = 0,

     
  3. 3.

    ϕ (t + s) ≤ ϕ (t) + ϕ (s), t, s [0, ∞)

     

and let ψ denote all the functions ψ : [0, ∞) → (0, ∞) which satisfy lim t r ψ (t) > 0 for all r > 0 and lim t 0 + ψ ( t ) = 0 .

For example [11], functions ϕ1(t) = kt where k > 0, ϕ 2 t = t t + 1 , ϕ3(t) = ln(t + 1), and ϕ4(t) = min{t, 1} are in Φ; ψ1(t) = kt where k > 0, ψ 2 t = ln 2 t + 1 2 , and
ψ 3 ( t ) = 1 , t = 0 t t + 1 , 0 < t < 1 1 , t = 1 1 2 t , t > 1

are in Ψ,

Now, let us start proving our main results.

Theorem 3.1 Let (X, ) be a partially ordered set and suppose there is a metric d on X such that (X, d) is a complete metric space. Let F : X × XX be a mapping having the mixed g-monotone property on X such that there exist two elements x0, y0 X with
g x 0 F ( x 0 , y 0 ) a n d g y 0 F ( y 0 , x 0 ) .
Suppose there exist ϕ Φ and ψ Ψ such that
ϕ ( d ( F ( x , y ) , F ( u , v ) ) ) 1 2 ϕ ( d ( g x , g u ) + d ( g y , g v ) ) ψ ( d ( g x , g u ) + d ( g y , g v ) ) 2 )
(3)

for all x, y, u, v X with gx gu and gy gv. Suppose F(X × X) g(X), g is continuous and compatible with F and also suppose either

(a) F is continuous or

(b) X has the following property:

(i) if a non-decreasing sequence {x n } → x, then x n x, for all n,

(ii) if a non-increasing sequence {y n } → y, then y y n , for all n.

Then there exists x, y X such that
g x = F ( x , y ) a n d g y = F ( y , x ) ,

i.e., F and g have a coupled coincidence point in X.

Proof. Let x0, y0 X be such that gx0 F (x0, y0) and gy0 F (y0, x0).

Using F(X × X) g(X), we construct sequences {x n } and {y n } in X as
g x n + 1 = F ( x n , y n ) and g y n + 1 = F ( y n , x n ) for all  n 0 .
(4)
We are going to prove that
g x n g x n + 1 for all  n 0
(5)
and
g y n g y n + 1 for all  n 0 .
(6)

To prove these, we are going to use the mathematical induction.

Let n = 0. Since gx0 F(x0, y0) and gy0 F(y0, x0) and as gx1 = F(x0, y0) and gy1 = F (y0, x0), we have gx0 gx1 and gy0 gy1. Thus (5) and (6) hold for n = 0.

Suppose now that (5) and (6) hold for some fixed n ≥ 0, Then, since gx n gx n +1 and gy n gy n +1 , and by mixed g-monotone property of F, we have
g x n + 2 = F ( x n + 1 , y n + 1 ) F ( x n , y n + 1 ) F ( x n , y n ) = g x n + 1
(7)
and
g y n + 2 = F ( y n + 1 , x n + 1 ) F ( y n , x n + 1 ) F ( y n , x n ) = g y n + 1 .
(8)
Using (7) and (8), we get
g x n + 1 g x n + 2 and g y n + 1 g y n + 2 .
Hence by the mathematical induction we conclude that (5) and (6) hold for all n ≥ 0. Therefore,
g x 0 g x 1 g x 2 g x n g x n + 1
(9)
and
g y 0 g y 1 g y 2 g y n g y n + 1 .
(10)
Since gx n gx n - 1 and gy n gy n - 1 , using (3) and (4), we have
ϕ ( d ( g x n + 1 , g x n ) ) = ϕ ( d ( F ( x n , y n ) , F ( x n - 1 , y n - 1 ) ) ) (1) 1 2 ϕ ( d ( g x n , g x n - 1 ) + d ( g y n , g y n - 1 ) ) (2) - ψ d ( g x n , g x n - 1 ) + d ( g y n , g y n - 1 ) 2 . (3) (4)
(11)
Similarly, since gy n - 1 gy n and gx n - 1 gx n , using (3) and (4), we also have
ϕ ( d ( g y n , g y n + 1 ) ) = ϕ ( d ( F ( y n 1 , x n 1 ) , F ( y n , x n ) ) ) 1 2 ϕ ( d ( g y n 1 , g y n ) + d ( g x n 1 , g x n ) ) ψ ( d ( ( g y n 1 , g y n ) + d ( g x n 1 , g x n ) 2 ) .
(12)
Using (11) and (12), we have
ϕ ( d ( g x n + 1 , g x n ) ) + ϕ ( d ( g y n + 1 , g y n ) ) ϕ ( d ( g x n , g x n 1 ) + d ( g y n , g y n 1 ) ) 2 ψ ( d ( g x n , g x n 1 ) + d ( g y n , g y n 1 ) 2 ) .
(13)
By property (iii) of ϕ, we have
ϕ ( d ( g x n + 1 , g x n ) + d ( g y n + 1 , g y n ) ) ϕ ( d ( g x n + 1 , g x n ) ) + ϕ ( d ( g y n + 1 , g y n ) ) .
(14)
Using (13) and (14), we have
ϕ ( d ( g x n + 1 , g x n ) + d ( g y n + 1 , g y n ) ) ϕ ( d ( g x n , g x n - 1 ) + d ( g y n , g y n - 1 ) ) (1) - 2 ψ d ( g x n , g x n - 1 ) + d ( g y n , g y n - 1 ) 2 (2) (3)
(15)
which implies, since ψ is a non-negative function,
ϕ ( d ( g x n + 1 , g x n ) + d ( g y n + 1 , g y n ) ) ϕ ( d ( g x n , g x n - 1 ) + d ( g y n , g y n - 1 ) ) .
Using the fact that ϕ is non-decreasing, we get
d ( g x n + 1 , g x n ) + d ( g y n + 1 , g y n ) d ( g x n , g x n - 1 ) + d ( g y n , g y n - 1 ) .
Set
δ n = d ( g x n + 1 , g x n ) + d ( g y n + 1 , g y n ) .
Now we would like to show that δ n → 0 as n → ∞. It is clear that the sequence {δ n } is decreasing. Therefore, there is some δ ≥ 0 such that
lim n δ n = lim n [ d ( g x n + 1 , g x n ) + d ( g y n + 1 , g y n ) ] = δ .
(16)
We shall show that δ = 0. Suppose, to the contrary, that δ > 0. Then taking the limit as n → ∞ (equivalently, δ n δ) of both sides of (15) and remembering lim t r ψ(t) > 0 for all r > 0 and ϕ is continuous, we have
ϕ ( δ ) = lim n ϕ ( δ n ) lim n ϕ ( δ n - 1 ) - 2 ψ δ n - 1 2 (1) = ϕ ( δ ) - 2 lim δ n - 1 δ ψ δ n - 1 2 < ϕ ( δ ) (2) (3)
a contradiction. Thus δ = 0, that is
lim n δ n = lim n [ d ( g x n + 1 , g x n ) + d ( g y n + 1 , g y n ) ] = 0 .
(17)
Now, we will prove that {gx n } and {gy n } are Cauchy sequences. Suppose, to the contrary, that at least one of {gx n } or {gy n } is not Cauchy sequence. Then there exists an ε > 0 for which we can find subsequences {gx n ( k )}, {gx m ( k )} of {gx n } and {gy n ( k )}, {gy m ( k )} of {gy n } with n(k) > m(k) ≥ k such that
d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) ε .
(18)
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) and satisfying (18). Then
d ( g x n ( k ) - 1 , g x m ( k ) ) + d ( g y n ( k ) - 1 , g y m ( k ) ) < ε .
(19)
Using (18), (19) and the triangle inequality, we have
ε r k : = d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) (1) d ( g x n ( k ) , g x n ( k ) - 1 ) + d ( g x n ( k ) - 1 , g x m ( k ) ) + d ( g y n ( k ) , g y n ( k ) - 1 ) + d ( g y n ( k ) - 1 , g y m ( k ) ) (2) d ( g x n ( k ) , g x n ( k ) - 1 ) + d ( g y n ( k ) , g y n ( k ) - 1 ) + ε . (3) (4) 
Letting k → ∞ and using (17), we get
lim k r k = lim k [ d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ] = ε .
(20)
By the triangle inequality
r k = d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) (1) d ( g x n ( k ) , g x n ( k ) + 1 ) + d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + d ( g x m ( k ) + 1 , g x m ( k ) ) (2) + d ( g y n ( k ) , g y n ( k ) + 1 ) + d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) + d ( g y m ( k ) + 1 , g y m ( k ) ) (3) = δ n ( k ) + δ m ( k ) + d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) . (4) (5)
Using the property of ϕ, we have
ϕ ( r k ) = ϕ ( δ n ( k ) + δ m ( k ) + d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) ) (1) ϕ ( δ n ( k ) + δ m ( k ) ) + ϕ ( d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) ) (2) + ϕ ( d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) ) . (3) (4)
(21)

Since n(k) > m(k), hence gx n ( k ) gx m ( k ) and gy n ( k ) gy m ( k ). Using (3) and

(4), we get
ϕ ( d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) ) = ϕ ( d ( F ( x n ( k ) , y n ( k ) ) , F ( x m ( k ) , y m ( k ) ) ) ) (1) 1 2 ϕ ( d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) ) (2) - ψ d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) 2 (3) = 1 2 ϕ ( r k ) - ψ r k 2 . (4) (5)
(22)
By the same way, we also have
ϕ ( d ( g y m ( k ) + 1 , g y n ( k ) + 1 ) ) = ϕ ( d ( F ( y m ( k ) , x m ( k ) ) , F ( y n ( k ) , x n ( k ) ) ) ) \ (1) 1 2 ϕ ( d ( g y m ( k ) , g y n ( k ) ) + d ( g x m ( k ) , g x n ( k ) ) ) (2) - ψ d ( g y m ( k ) , g y n ( k ) ) + d ( g x m ( k ) , g x n ( k ) ) 2 (3) = 1 2 ϕ ( r k ) - ψ r k 2 . (4) (5)
(23)
Inserting (22) and (23) in (21), we have
ϕ ( r k ) ϕ ( δ n ( k ) + δ m ( k ) ) + ϕ ( r k ) - 2 ψ r k 2 .
Letting k → ∞ and using (17) and (20), we get
ϕ ( ε ) ϕ ( 0 ) + ϕ ( ε ) - 2 lim k ψ r k 2 = ϕ ( ε ) - 2 lim r k ε ψ r k 2 < ϕ ( ε )

a contradiction. This shows that {gx n } and {gy n } are Cauchy sequences.

Since X is a complete metric space, there exist x, y X such that
lim n F ( x n , y n ) = lim n g x n = x a n d lim n F ( y n , x n ) = lim n g y n = y .
(24)
Since F and g are compatible mappings, we have
lim n d ( g ( F ( x n , y n ) ) , F ( g x n , g y n ) ) = 0
(25)
and
lim n d ( g ( F ( y n , x n ) ) , F ( g y n , g x n ) ) = 0
(26)
We now show that gx = F(x, y) and gy = F(y, x). Suppose that the assumption (a) holds. For all n ≥ 0, we have,
d ( g x , F ( g x n , g y n ) ) d ( g x , g ( F ( x n , y n ) ) ) + d ( g ( F ( x n , y n ) ) , F ( g x n , g y n ) .

Taking the limit as n → ∞, using (4), (24), (25) and the fact that F and g

are continuous, we have d(gx, F(x, y)) = 0.

Similarly, using (4), (24), (26) and the fact that F and g are continuous, we have d(gy, F(y, x)) = 0.

Combining the above two results we get
g x = F ( x , y ) a n d g y = F ( y , x ) .
Finally, suppose that (b) holds. By (5), (6) and (24), we have {gx n } is a non-decreasing sequence, gx n x and {gy n } is a non-increasing sequence, gy n y as n → ∞. Hence, by assumption (b), we have for all n ≥ 0,
g x n x a n d g y n y .
(27)

Since F and g are compatible mappings and g is continuous, by (25) and (26)

we have
lim n g ( g x n ) = g x = lim n g ( F ( x n , y n ) ) = lim n F ( g x n , g y n )
(28)
and,
lim n g ( g y n ) = g y = lim n g ( F ( y n , x n ) ) = lim n F ( g y n , g x n ) .
(29)
Now we have
d ( g x , F ( x , y ) ) d ( g x , g ( g x n + 1 ) ) + d ( g ( g x n + 1 ) , F ( x , y ) ) .
Taking n → ∞ in the above inequality, using (4) and (21) we have,
d ( g x , F ( x , y ) ) lim n d ( g x , g ( g x n + 1 ) ) + lim n d ( g ( F ( x n , y n ) ) , F ( x , y ) ) lim n d ( F ( g x n , g y n ) ) , F ( x , y ) )
(30)
Using the property of ϕ, we get
ϕ ( d ( g x , F ( x , y ) ) ) lim n ϕ ( d ( F ( g x n , g y n ) ) , F ( x , y ) ) )
Since the mapping g is monotone increasing, using (3), (27) and (30), we have for all n ≥ 0,
ϕ ( d ( g x , F ( x , y ) ) ) lim n 1 2 ϕ ( d ( g g x n , g x ) + d ( g y n , g g y ) ) (1) - lim n ψ d ( g g x n , g x ) + d ( g g y n , g y ) 2 . (2) (3)

Using the above inequality, using (24) and the property of ψ, we get ϕ(d(gx, F(x, y))) = 0, thus d(gx, F(x, y)) = 0. Hence gx = F(x, y).

Similarly, we can show that gy = F(y, x). Thus we proved that F and g have a coupled coincidence point.

Corollary 3.1 [11] Let (X, ) be a partially ordered set and suppose there is a metric d on X such that (X, d) is a complete metric space. Let F : X × XX be a mapping having the mixed monotone property on X such that there exist two elements x0, y0 X with
x 0 F ( x 0 , y 0 ) a n d y 0 F ( y 0 , x 0 ) .
Suppose there exist ϕ Φ and ψ Ψ such that
ϕ ( d ( F ( x , y ) , F ( u , v ) ) ) 1 2 ϕ ( d ( x , u ) + d ( y , v ) ) ψ ( d ( x , u ) + d ( y , v ) ) 2 )

for all x, y, u, v X with xu and yv. Suppose either

(a) F is continuous or

(b) X has the following property.

(i) if a non-decreasing sequence {x n } → x, then x n x, for all n,

(ii) if a non-increasing sequence {y n } → y, then y y n , for all n,

then there exist x, y X such that
x = F ( x , y ) a n d y = F ( y , x )

that is, F has a coupled fixed point in X.

Corollary 3.2 [11] Let (X, ) be a partially ordered set and suppose there is a metric d on X such that (X, d) is a complete metric space. Let F : X × XX be a mapping having the mixed monotone property on X such that there exist two elements x0, y0 X with
x 0 F ( x 0 , y 0 ) a n d y 0 F ( y 0 , x 0 ) .
Suppose there exists ψ Ψ such that
d ( F ( x , y ) , F ( u , v ) ) d ( x , u ) + d ( y , v ) 2 - ψ d ( x , u ) + d ( y , v ) 2

for all x, y, u, v X with xu and yv. Suppose either

(a) F is continuous or

(b) X has the following property:

(i) if a non-decreasing sequence {x n } → x, then x n x, for all n,

(ii) if a non-increasing sequence {y n } → y, then y y n , for all n,

then there exist x, y X such that
x = F ( x , y ) a n d y = F ( y , x )

that is, F has a coupled fixed point in X.

Proof. Take ϕ(t) = t in Corollary 3.1, we get Corollary 3.2.

Corollary 3.3 [5] eses of Corollary 3.1, suppose that for Let (X, ) be a partially ordered set and suppose there is a metric d on X such that (X, d) is a complete metric space. Let F : X × XX be a mapping having the mixed monotone property on X such that there exist two elements x0, y0 X with
x 0 F ( x 0 , y 0 ) a n d y 0 F ( y 0 , x 0 ) .
Suppose there exists a real number k [0, 1) such that
d ( F ( x , y ) , F ( u , v ) ) k 2 [ d ( x , u ) + d ( y , v ) ]

for all x, y, u, v X with xu and yv. Suppose either

(a) F is continuous or

(b) X has the following property.

(i) if a non-decreasing sequence {x n } → x, then x n x, for all n,

(ii) if a non-increasing sequence {y n } → y, then y y n , for all n,

then there exist x, y X such that
x = F ( x , y ) a n d y = F ( y , x )

that is, F has a coupled fixed point in X.

Proof. Taking ψ t = 1 - k 2 t in Corollary 3.2.

4 Uniqueness of coupled coincidence point

In this section, we will prove the uniqueness of the coupled coincidence point. Note that if (X, ) is a partially ordered set, then we endow the product X × X with the following partial order relation, for all (x, y), (u, v) X × X,
( x , y ) ( u , v ) x u , y v .

Theorem 4.1 In addition to hypotheses of Theorem 3.1, suppose that for every (x, y), (z, t) in X × X, if there exists a (u, v) in X ×X that is comparable to (x, y) and (z, t), then F has a unique coupled coincidence point.

Proof. From Theorem 3.1, the set of coupled coincidence points of F and g is non-empty. Suppose (x, y) and (z, t) are coupled coincidence points of F and g, that is gx = F(x, y), gy = F(y, x), gz = F(z, t) and gt = F(t, z). We are going to show that gx = gz and gy = gt. By assumption, there exists (u, v) X × X that is comparable to (x, y) and (z, t). We define sequences {gu n }, {gv n } as follows
u 0 = u v 0 = v . g u u + 1 = F u n , v n and g v n + 1 = F ( v n , u n ) for all  n .
Since (u, v) is comparable with (x, y), we may assume that (x, y) (u, v) = (u0, v0). Using the mathematical induction, it is easy to prove that
( x , y ) ( u n , v n ) for all  n .
(31)
Using (3) and (31), we have
φ ( d ( g x , g u n + 1 ) ) = φ ( d ( F ( x , y ) , F ( u n , v n ) ) ) (1) < 1 2 φ ( d ( x , u n ) + d ( y , v n ) ) - ψ d ( x , u n ) + d ( y , v n ) 2 (2) (3)
(32)
Similarly
φ ( d ( g v n + 1 , g y ) ) = φ ( d ( F ( v n , u n ) , F ( y , x ) ) ) (1) < 1 2 φ ( d ( v n , y ) + d ( u n , x ) ) - ψ d ( v n , y ) + d ( u n , x ) 2 (2) (3)
(33)
Using (32), (33) and the property of φ, we have
φ ( d ( g x , g u n + 1 ) + d ( g y , g v n + 1 ) ) φ ( d ( g x , g u n + 1 ) ) + φ ( d ( g y , g v n + 1 ) ) (1) φ ( d ( g x , g u n ) + d ( g y , g v n ) ) (2) - 2 ψ d ( g x , g u n ) + d ( g y , g v n ) 2 . (3) (4)
(34)
which implies, using the property of ψ,
φ ( d ( g x , g u n + 1 ) + d ( g y , g v n + 1 ) ) φ ( d ( g x , g u n ) + d ( g y , g v n ) ) .
Thus, using the property of ϕ,
d ( g x , g u n + 1 ) + d ( g y , g v n + 1 ) d ( g x , g u n ) + d ( g y , g v n ) .
That is the sequence {d(gx, gu n )+ d(gy, gv n )} is decreasing. Therefore, there exists α ≥ 0 such that
lim n [ d ( g x , g u n ) + d ( g y , g v n ) ] = α .
(35)
We will show that α = 0. Suppose, to the contrary, that α > 0. Taking the limit as n → ∞ in (34), we have, using the property of ψ,
φ ( α ) φ ( α ) - 2 lim n ψ d ( g x , g u n ) + d ( g y , g v n ) 2 < φ ( α )
a contradiction. Thus. α = 0, that is,
lim n [ d ( g x , g u n ) + d ( g y , g v n ) ] = 0 .
It implies
lim n d ( g x , g u n ) = lim n d ( g y , g v n ) = 0 .
(36)
Similarly, we show that
lim n d ( g z , g u n ) = lim n d ( g t , g v n ) = 0 .
(37)

Using (36) and (37) we have gx = gz and gy = gt.

Corollary 4.1 [11] In addition to hypotheses of Corollary 3.1, suppose that for every (x, y), (z, t) in X × X, if there exists a (u, v) in X × X that is comparable to (x, y) and (z, t), then F has a unique coupled fixed point.

5 Example

Example 5.1 Let X = [0, 1]. Then (X, ≤) is a partially ordered set with the natural ordering of real numbers. Let
d ( x , y ) = x - y f o r x , y [ 0 , 1 ] .

Then (X, d) is a complete metric space.

Let g : XX be defined as
g x = x 2 , f o r a l l x X ,
and let F : X × XX be defined as
F ( x , y ) = x 2 - y 2 3 , i f x y , 0 , i f x < y .

F obeys the mixed g-monotone property.

Let ϕ : [0, ∞) → [0, ∞) be defined as
ϕ ( t ) = 3 4 t , f o r t [ 0 , ) .
and let ψ : [0, ∞) → [0, ∞) be defined as
ψ ( t ) = 1 4 t , f o r t [ 0 , ) .
Let {x n } and {y n } be two sequences in X such that lim n →∞ F (x n , y n ) = a, lim n →∞ gx n = a, lim n →∞ F (y n , x n ) = b and lim n →∞ gy n = b Then obviously, a = 0 and b = 0. Now, for all n ≥ 0,
g x n = x n 2 , g y n = y n 2 , F ( x n , y n ) = x n 2 - y n 2 3 , i f x n y n , 0 , i f x n < y n .
and
F ( y n , x n ) = y n 2 - x n 2 3 , i f y n x n , 0 , i f y n < x n .
Then it follows that,
lim n d ( g ( F ( x n , y n ) ) , F ( g x n , g y n ) ) = 0
and
lim n d ( g ( F ( y n , x n ) ) , F ( g y n , g x n ) ) = 0 ,
Hence, the mappings F and g are compatible in X. Also, x0 = 0 and y0 = c(> 0) are two points in X such that
g x 0 = g ( 0 ) = 0 = F ( 0 , c ) = F ( x 0 , y 0 )
and
g y 0 = g ( c ) = c 2 c 2 3 = F ( c , 0 ) = F ( y 0 , x 0 ) .

We next verify the contraction (3). We take x, y, u, v, X, such that gxgu and gygv, that is, x2u2 and y2v2.

We consider the following cases:

Case 1. xy, uv. Then,
ϕ ( d ( F ( x , y ) , F ( u , v ) ) ) = 3 4 [ d ( F ( x , y ) , F ( u , v ) ] = 3 4 [ d ( x 2 y 2 3 , u 2 v 2 3 ) ] = 3 4 | ( x 2 y 2 ) ( u 2 v 2 ) 3 | = 3 4 | x 2 u 2 | + | y 2 v 2 | 3 = 1 2 ( d ( g x , g u ) + d ( g y , g v ) 2 ) = 3 4 ( d ( g x , g u ) + d ( g y , g v ) 2 ) 1 4 ( d ( g x , g u ) + d ( g y , g v ) 2 ) = 3 8 ( d ( g x , g u ) + d ( g y , g v ) ) 1 4 ( d ( g x , g u ) + d ( g y , g v ) 2 ) = 1 2 ϕ ( d ( g x , g u ) + d ( g y , g v ) ) ψ ( d ( g x , g u ) + d ( g y , g v ) 2 )
Case 2. xy, u < v.Then
ϕ ( d ( F ( x , y ) , F ( u , v ) ) = 3 4 [ d ( F ( x , y ) , F ( u , v ) ] = 3 4 [ d ( x 2 y 2 3 , 0 ) ] = 3 4 | x 2 y 2 | 3 = 3 4 | v 2 + x 2 y 2 u 2 | 3 = 3 4 | ( v 2 y 2 ) ( u 2 x 2 ) | 3 3 4 | v 2 y 2 | + | u 2 x 2 | 3 = 3 4 ( | u 2 x 2 | + | y 2 v 2 | 3 ) = 1 2 ( | u 2 x 2 | + | y 2 v 2 | 2 ) = 1 2 ( d ( g x , g u ) + d ( g y , g v ) 2 ) = 3 4 ( d ( g x , g u ) + d ( g y , g v ) 2 ) 1 4 ( d ( g x , g u ) + d ( g y , g v ) 2 ) = 3 8 ( d ( g x , g u ) + d ( g y , g v ) ) 1 4 ( d ( g x , g u ) + d ( g y , g v ) 2 = 1 2 ϕ ( d ( g x , g u ) + d ( g y , g v ) ) ψ ( d ( g x , g u ) + d ( g y , g v ) 2 )
Case 3. x < y and uv. Then
ϕ ( d ( F ( x , y ) , F ( u , v ) ) ) = 3 4 d ( 0 , u 2 - v 2 3 ) (1) = 3 4 | u 2 - v 2 | 3 (2) = 3 4 | u 2 + x 2 - v 2 - x 2 | 3 (3) = 3 4 | ( x 2 - v 2 ) + ( u 2 - x 2 ) | 3 ( s i n c e y > x ) (4) 3 4 | y 2 - v 2 | + | u 2 - x 2 | 3 (5) = 1 2 | u 2 - x 2 | + | y 2 - v 2 | 2 (6) = 1 2 d ( g x , g u ) + d ( g y , g v ) 2 (7) = 3 4 d ( g x , g u ) + d ( g y , g v ) 2 (8) - 1 4 d ( g x , g u ) + d ( g y , g v ) 2 (9) = 3 8 ( d ( g x , g u ) + d ( g y , g v ) ) (10) - 1 4 d ( g x , g u ) + d ( g y , g v ) 2 (11) = 1 2 ϕ ( d ( g x , g u ) + d ( g y , g v ) ) (12) - ψ d ( g x , g u ) + d ( g y , g v ) 2 (13) (14)
Case 4. x < y and u < v with x2u2 and y2v2. Then, F(x, y) = 0 and F(u, v) = 0, that is,
ϕ ( d ( F ( x , y ) , F ( u , v ) ) ) = ϕ ( d ( 0 , 0 ) ) = ϕ ( 0 ) = 0 .

Therefore all conditions of Theorem 3.1 are satisfied. Thus the conclusion follows.

Declarations

Acknowledgements

The authors would like to thank the referees for the invaluable comments that improved this paper.

Authors’ Affiliations

(1)
Department of Mathematics, King Abdulaziz University

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© Alotaibi and Alsulami; licensee Springer. 2011

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