Open Access Research

Coupled coincidence points for monotone operators in partially ordered metric spaces

Abdullah Alotaibi* and Saud M Alsulami

Author Affiliations

Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia

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Fixed Point Theory and Applications 2011, 2011:44  doi:10.1186/1687-1812-2011-44


The electronic version of this article is the complete one and can be found online at: http://www.fixedpointtheoryandapplications.com/content/2011/1/44


Received: 18 March 2011
Accepted: 30 August 2011
Published: 30 August 2011

© 2011 Alotaibi and Alsulami; licensee Springer.

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 point; partially ordered metric space; mixed 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. [1-16]). 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 : X X 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 × X X. 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 × X X and g : X X. 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 × X X and g : X X. 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 × X X and g : X X, 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 {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.

3 Existence of coupled coincidence points

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

1. ϕ is continuous and non-decreasing,

2. ϕ (t) = 0 if and only if t = 0,

3. ϕ (t + s) ≤ ϕ (t) + ϕ (s), ∀t, s ∈ [0, ∞)

and let ψ denote all the functions ψ : [0, ∞) → (0, ∞) which satisfy limtr 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 × X X 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 {xn} → x, then xn x, for all n,

(ii) if a non-increasing sequence {yn} → y, then y yn, 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 {xn} and {yn} 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 gxn gxn+1 and gyn gyn+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 gxn gxn - 1 and gyn gyn - 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 gyn - 1 gyn and gxn - 1 gxn, 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 δn → 0 as n → ∞. It is clear that the sequence {δn} 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, δn δ) of both sides of (15) and remembering limtr 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 {gxn} and {gyn} are Cauchy sequences. Suppose, to the contrary, that at least one of {gxn} or {gyn} is not Cauchy sequence. Then there exists an ε > 0 for which 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

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 gxn(k) ≽ gxm(k) and gyn(k) ≽ gym(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 {gxn} and {gyn} 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 {gxn} is a non-decreasing sequence, gxn x and {gyn} is a non-increasing sequence, gyn 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 × X X 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 x u and y v. Suppose either

(a) F is continuous or

(b) X has the following property.

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

(ii) if a non-increasing sequence {yn} → y, then y yn, 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 × X X 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 x u and y v. Suppose either

(a) F is continuous or

(b) X has the following property:

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

(ii) if a non-increasing sequence {yn} → y, then y yn, 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 × X X 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 x u and y v. Suppose either

(a) F is continuous or

(b) X has the following property.

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

(ii) if a non-increasing sequence {yn} → y, then y yn, 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 {gun}, {gvn} 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, gun)+ d(gy, gvn)} 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 : X X be defined as

g x = x 2 , f o r a l l x X ,

and let F : X × X X 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 {xn} and {yn} be two sequences in X such that limn→∞ F (xn, yn) = a, limn→∞ gxn = a, limn→∞ F (yn, xn) = b and limn→∞ gyn = 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 gx gu and gy gv, that is, x2 u2 and y2 v2.

We consider the following cases:

Case 1. x y, 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 , 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. x y, 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 u v. 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 x2 u2 and y2 v2. 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.

Competing interests

The authors declare that they have no competing interests.

Authors' contributions

The authors have been working together on each step of the paper such as the literature review, results and examples.

Acknowledgements

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

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