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Coupled common fixed point theorems for generalized nonlinear contraction mappings with the mixed monotone property in partially ordered metric spaces

Abstract

The purpose of this paper is to establish some coupled coincidence point theorems for generalized nonlinear contraction mappings with the mixed g-monotone property in the framework of metric spaces endowed with partial order. The theorems presented in this paper are generalizations and improvements of the several well-known results in the literature.

MSC:46N40, 47H10, 54H25, 46T99.

1 Introduction and preliminaries

The Banach contraction principle is one of very popular tools in solving the existence in many problems of mathematical analysis. Due to its simplicity and usefulness, there are a lot of generalizations of this principle in the literature. Ran and Reurings [1] extended the Banach contraction principle in partially ordered sets with some applications to linear and nonlinear matrix equations. While Nieto and López [2] extended the result of Ran and Reurings and applied their main theorems to obtain a unique solution for a first-order ordinary differential equation with periodic boundary conditions. Bhaskar and Lakshmikantham [3] introduced the concept of mixed monotone mappings and obtained some coupled fixed point results. Also, they applied their results to a first-order differential equation with periodic boundary conditions. Recently, many researchers have obtained fixed point, common fixed point, coupled fixed point and coupled common fixed point results in cone metric spaces, partially ordered metric spaces and others (see [1–27]).

Definition 1 Let (X,d) be a metric space and F:X×X→X and g:X→X, F and g are said to commute if F(gx,gy)=g(F(x,y)) for all x,y∈X.

Definition 2 Let (X,⪯) be a partially ordered set and F:X×X→X. The mapping F is said to be non-decreasing if for x,y∈X, x⪯y implies F(x)⪯F(y) and non-increasing if for x,y∈X, x⪯y implies F(x)⪰F(y).

Definition 3 Let (X,⪯) be a partially ordered set and F:X×X→X and g:X→X. The mapping F is said to have the mixed g-monotone property if F(x,y) is monotone g-non-decreasing in x and monotone g-non-increasing in y, 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),

and

y 1 , y 2 ∈X,g y 1 ⪯g y 2 ⇒F(x, y 1 )⪰F(x, y 2 ).

If g=identity mapping in Definition 3, then the mapping F is said to have the mixed monotone property.

Definition 4 An element (x,y)∈X×X is called a coupled coincidence point of the mappings F:X×X→X and g:X→X if F(x,y)=gx and F(y,x)=gy.

If g is the identity mapping in Definition 4, then (x,y)∈X×X is called a coupled fixed point.

Geraghty [16] introduced an extension of the Banach contraction principle in which the contraction constant was replaced by a function having some specified properties.

Definition 5 Let Θ be the class of functions β: R + →[0,1) with

  1. (i)

    R + ={R/t>0};

  2. (ii)

    β( t n )→1 implies t n →0.

The method applied by Geraghty [16] was utilized to obtain further new fixed point result works like [6, 7, 15].

The purpose of this paper is to establish some coupled coincidence point results for a pair of mappings with the mixed g-monotone property satisfying a generalized contractive condition by using the ideas of Geraghty [16] in partially ordered metric spaces. Also we give some examples to illustrate the main results.

2 Main results

Theorem 6 Let (X,⪯) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Suppose that F:X×X→X and g:X→X are self-mappings on X such that F has the mixed g-monotone property on X such that there exist two elements x 0 , y 0 ∈X with g( x 0 )⪯F( x 0 , y 0 ) and g( y 0 )⪰F( y 0 , x 0 ). Suppose that there exists θ∈Θ such that

d ( F ( x , y ) , F ( u , v ) ) ≤θ ( d ( g x , g u ) + d ( g y , g v ) 2 ) ( d ( g x , g u ) + d ( g y , g v ) 2 )
(2.1)

for all x,y,u,v∈X with gx⪰gu and gy⪯gv. Further suppose that F(X×X)⊆g(X), g is continuous and commutes with F, and also suppose that either

  1. (a)

    F is continuous, or

  2. (b)

    X has the following properties:

  3. (i)

    if {g( x n )}⊂X is a non-decreasing sequence with g x n →gx in g(X), then g x n ⪯gx for every n;

  4. (ii)

    if {g( y n )}⊂X is a non-increasing sequence with g y n →gy in g(X), then g y n ⪰gy for every n.

Then there exist two elements x,y∈X such that F(x,y)=g(x) and gy=F(y,x), that is, F and g have a coupled coincidence point (x,y)∈X×X.

Proof Let x 0 , y 0 ∈X be such that g x 0 ⪯F( x 0 , y 0 ) and g y 0 ⪰F( y 0 , x 0 ). Since F(X×X)⊆g(X), we can construct sequences { x n } and { y n } in X such that

g x n + 1 =F( x n , y n )andg y n + 1 =F( y n , x n ),∀n≥0.
(2.2)

We claim that for all n≥0,

g x n ⪯g x n + 1 ,
(2.3)

and

g y n ⪰g y n + 1 .
(2.4)

We use the mathematical induction. Let n=0. Since g x 0 ⪯F( x 0 , y 0 ) and g y 0 ⪰F( y 0 , x 0 ), in view of g x 1 =F( x 0 , y 0 ) and g y 1 =F( y 0 , x 0 ), we have g x 0 ⪯g x 1 and g y 0 ⪰g y 1 , that is, (2.3) and (2.4) hold for n=0. Suppose that (2.3) and (2.4) hold for some n>0. As F has the mixed g-monotone property and g x n ⪯g x n + 1 and g y n ⪰g y n + 1 , from (2.2), we get

g x n + 1 =F( x n , y n )⪯F( x n + 1 , y n )⪯F( x n + 1 , y n + 1 )=g x n + 2
(2.5)

and

g y n + 1 =F( y n , x n )⪰F( y n + 1 , x n )⪰F( y n + 1 , x n + 1 )=g y n + 2 .
(2.6)

Now from (2.5) and (2.6), we obtain that g x n + 1 ⪯g x n + 2 and g y n + 1 ⪰g y n + 2 . Thus, by the mathematical induction, we conclude that (2.3) and (2.4) hold for all n≥0. Therefore

g x 0 ⪯g x 1 ⪯g x 2 ⪯⋯⪯g x n ⪯g x n + 1 ⪯⋯
(2.7)

and

g y 0 ⪰g y 1 ⪰g y 2 ⪰⋯⪰g y n ⪰g y n + 1 ⪰⋯.
(2.8)

Assume that there is some r∈N such that d(g x r ,g x r − 1 )+d(g y r ,g y r − 1 )=0, that is, g x r =g x r − 1 and g y r =g y r − 1 . Then g x r − 1 =F( x r − 1 , y r − 1 ) and g y r − 1 =F( y r − 1 , x r − 1 ), and hence we get the result.

For simplicity, let t n + 1 :=d(g x n + 1 ,g x n )+d(g y n + 1 ,g y n ). Now, we assume that

t n =d(g x n ,g x n − 1 )+d(g y n ,g y n − 1 )≠0

for all n. Since g x n ⪰g x n − 1 and g y n ⪯g y n − 1 , from (2.1) and (2.2), we have

t n + 1 = d ( g x n + 1 , g x n ) + d ( g y n + 1 , g y n ) = d ( F ( x n , y n ) , F ( x n − 1 , y n − 1 ) ) + d ( F ( y n , x n ) , F ( y n − 1 , x n − 1 ) ) ≤ θ ( d ( g x n − 1 , g x n ) + d ( g y n − 1 , g y n ) 2 ) ( d ( g x n − 1 , g x n ) + d ( g y n − 1 , g y n ) 2 ) + θ ( d ( g x n − 1 , g x n ) + d ( g y n − 1 , g y n ) 2 ) ( d ( g x n − 1 , g x n ) + d ( g y n − 1 , g y n ) 2 ) = θ ( d ( g x n − 1 , g x n ) + d ( g y n − 1 , g y n ) 2 ) ( d ( g x n − 1 , g x n ) + d ( g y n − 1 , g y n ) ) = θ ( d ( g x n − 1 , g x n ) + d ( g y n − 1 , g y n ) 2 ) ( t n ) ≤ t n ,
(2.9)

which implies that t n + 1 ≤ t n . It follows that { t n } is a monotone decreasing sequence of non-negative real numbers. Therefore, there is some t≥0 such that lim n → ∞ t n =t.

Now, we show that t=0. Assume to the contrary that t>0, then from (2.9) we have

t n + 1 t n ≤θ ( d ( g x n − 1 , g x n ) + d ( g y n − 1 , g y n ) 2 ) <1,

which yields that lim n → ∞ θ( t n 2 )=1. This implies that d(g x n − 1 ,g x n )→0 and d(g y n − 1 ,g y n )→0. Therefore t=0, that is,

lim n → ∞ t n = lim n → ∞ [ d ( g x n + 1 , g x n ) + d ( g y n + 1 , g y n ) ] =0.
(2.10)

Next, we prove that {g x n } and {g y n } are Cauchy sequences. On the contrary, assume that at least one of {g x n } or {g y n } is not a Cauchy sequence. Then there exists an ϵ>0 for which we can find subsequences {g x m ( k ) } and {g x n ( k ) } of {g x n }, and {g y m ( k ) } and {g y n ( k ) } of {g y n } with n(k)>m(k)>k such that for every k,

d(g x m ( k ) ,g x n ( k ) )+d(g y m ( k ) ,g y n ( k ) )≥ϵ.
(2.11)

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 (2.11). Then

d(g x n ( k ) − 1 ,g x m ( k ) )+d(g y n ( k ) − 1 ,g y m ( k ) )<ϵ.
(2.12)

Using (2.11) and (2.12), we have

ϵ ≤ r k : = d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) ≤ 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 ) ) < ϵ + t n ( k ) .

Letting k→∞ and using (2.10), we have

lim r k =lim [ d ( g x m ( k ) , g x n ( k ) ) + d ( g y m ( k ) , g y n ( k ) ) ] =ϵ.
(2.13)

Also, by the triangle inequality, we have

r k = d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) ≤ 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 ) ) + 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 ) ) = t n ( k ) + t m ( k ) + d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + d ( g y n ( k ) + 1 , g x m ( k ) + 1 ) .

Since n(k)>m(k), g x n ( k ) ⪰g x m ( k ) and g y n ( k ) ⪯g y m ( k ) , from (2.1) and (2.2), we have

d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) = d ( F ( x n ( k ) , y n ( k ) ) , F ( x m ( k ) , y m ( k ) ) ) + d ( F ( y n ( k ) , x n ( k ) ) , F ( y m ( k ) , x m ( k ) ) ) ≤ θ ( d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) 2 ) × ( d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) ) = θ ( d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) 2 ) r k .

Therefore, we have

r k ≤ t n ( k ) + t m ( k ) +θ ( d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) 2 ) r k .

This implies that

r k − t n ( k ) − t m ( k ) r k ≤θ ( d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) 2 ) <1.

Taking k→∞, we get

θ ( d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) 2 ) =1.

Since θ∈Θ, we get

lim n → ∞ d(g x n ( k ) ,g x m ( k ) )= lim n → ∞ d(g y n ( k ) ,g y m ( k ) )=0,

which is a contradiction. This implies that {g x n } and {g y n } are Cauchy sequences in X. Since X is a complete metric space, there is (x,y)∈X×X such that g x n →x and g y n →y. Since g is continuous, g(g x n )→gx and g(g y n )→gy.

First, suppose that F is continuous. Then F(g x n ,g y n )→F(x,y) and F(g y n ,g x n )→F(y,x). As F commutes with g, we have F(g x n ,g y n )=gF( x n , y n )=g(g x n + 1 )→gx and F(g y n ,g x n )=gF( y n , x n )=g(g y n + 1 )→gy. By the uniqueness of the limit, we get gx=F(x,y) and gy=F(y,x).

Second, suppose that (b) holds. Since {g x n } is a non-decreasing sequence such that g x n →x and {g y n } is a non-increasing sequence such that g y n →y, and g is a non-increasing function, we get g(g x n )⪯gx and g(g y n )⪰gy hold for all n∈N. Hence, by (2.1), we have

d ( g ( g x n + 1 ) , F ( x , y ) ) + d ( g ( g y n + 1 ) , F ( y , x ) ) = d ( F ( g x n , g y n ) , F ( x , y ) ) + d ( F ( g y n , g x n ) F ( y , x ) ) ≤ θ ( d ( g ( g x n ) , g x ) + d ( g ( g y n ) , g y ) 2 ) ( d ( g ( g x n ) , g x ) + d ( g ( g y n ) , g y ) ) .

Taking n→∞, we get d(gx+F(x,y))+d(gy,F(y,x))=0, and hence gx=F(x,y) and gy=F(y,x). Thus F and g have a coupled coincidence point. □

If θ(t)=k, where k∈[0,1), then we have the following result.

Corollary 7 Let (X,⪯) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Suppose that F:X×X→X and g:X→X are self-mappings on X such that F has the mixed g-monotone property on X such that there exist two elements x 0 , y 0 ∈X with g( x 0 )⪯F( x 0 , y 0 ) and g( y 0 )⪰F( y 0 , x 0 ). Suppose that there exists k∈[0,1) such that

d ( F ( x , y ) , F ( u , v ) ) ≤k ( d ( g x , g u ) + d ( g y , g v ) 2 )
(2.14)

satisfies for all x,y,u,v∈X with gx⪰gu and gy⪯gv. Further suppose that F(X×X)⊆g(X), g is continuous non-decreasing and commutes with F and also suppose that either

  1. (a)

    F is continuous, or

  2. (b)

    X has the following properties:

  3. (i)

    if {g( x n )}⊂X is a non-decreasing sequence with g x n →gx in g(X), then g x n ⪯gx for every n;

  4. (ii)

    if {g( y n )}⊂X is a non-increasing sequence with g y n →gy in g(X), then g y n ⪰gy for every n.

Then there exist two elements x,y∈X such that F(x,y)=g(x) and gy=F(y,x), that is, F and g have a coupled coincidence point (x,y)∈X×X.

If g is an identity mapping, we have the following result of Bhaskar and Lakshmikantham [3].

Corollary 8 Let (X,⪯) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a metric space. Suppose that F:X×X→X is a mapping on X and has the mixed monotone property on X such that there exist two elements x 0 , y 0 ∈X with x 0 ⪯F( x 0 , y 0 ) and y 0 ⪰F( y 0 , x 0 ). Suppose that there exists k∈[0,1) such that

d ( F ( x , y ) , F ( u , v ) ) ≤ k 2 ( d ( x , u ) + d ( y , v ) )
(2.15)

for all x,y,u,v∈X with x⪰u and y⪯v. Further suppose that either

  1. (a)

    F is continuous, or

  2. (b)

    X has the following properties:

  3. (i)

    if { x n }⊂X is a non-decreasing sequence with x n →x in X, then x n ⪯x for each n≥1;

  4. (ii)

    if { y n }⊂X is a non-increasing sequence with y n →y in X, then y n ⪰y for each n≥1.

Then there exist two elements x,y∈X such that F(x,y)=x and y=F(y,x), that is, F has a coupled fixed point (x,y)∈X×X.

Example 9 Let X=[0,1]. Then (X,≤) is a partially ordered set with the natural ordering of real numbers. Let d(x,y)=|x−y| for all x,y∈X. Define a mapping g:X→X by g(x)=x and a mapping F:X×X→X by

F(x,y)={ x − y 4 , x ≥ y ; 0 , x < y .

Then it is easy to prove that (X,d) is a complete metric space, g(X) is complete, F:X×X→X⊆g(X)=X, X satisfies conditions (1) and (2) of Theorem 6 and F has the g-monotone property. Let θ:(0,∞)→[0,1) be defined by

θ(t)={ 1 − t 2 , t ≤ 1 ; α < 1 , t > 1 .

Now, we verify the inequality (2.1) of Theorem 6 for all x,y,u,v∈X with gx⪰gu and gy⪯gv.

Now, we consider the following cases.

Case 1. (x,y)=(0,0), (u,v)=(0,1) or (x,y)=(1,1), (u,v)=(0,1), we have

d ( F ( x , y ) , F ( u , v ) ) =0.

Hence inequality (2.1) holds.

Case 2. (x,y)=(1,0), (u,v)=(0,0), we have

d ( F ( x , y ) , F ( u , v ) ) =d ( F ( 1 , 0 ) , F ( 0 , 0 ) ) = 1 4

and

θ ( d ( g x , g u ) + d ( g y , g v ) 2 ) ( d ( g x , g u ) + d ( g y , g v ) 2 ) = θ ( d ( 1 , 0 ) + d ( 0 , 0 ) 2 ) ( d ( 1 , 0 ) + d ( 0 , 0 ) 2 ) = θ ( 1 2 ) ( 1 2 ) = ( 1 − 1 4 ) 1 2 .

Hence inequality (2.1) holds.

Case 3. (x,y)=(1,0), (u,v)=(0,1), we have

d ( F ( x , y ) , F ( u , v ) ) =d ( F ( 1 , 0 ) , F ( 0 , 1 ) ) = 1 4

and

θ ( d ( g x , g u ) + d ( g y , g v ) 2 ) ( d ( g x , g u ) + d ( g y , g v ) 2 ) = θ ( d ( 1 , 0 ) + d ( 0 , 1 ) 2 ) ( d ( 1 , 0 ) + d ( 0 , 1 ) 2 ) = θ ( 1 ) ( 1 ) = ( 1 − 1 2 ) .

Hence inequality (2.1) holds.

Case 4. (x,y)=(1,0), (u,v)=(1,1), we have

d ( F ( x , y ) , F ( u , v ) ) =d ( F ( 1 , 0 ) , F ( 1 , 1 ) ) = 1 4

and

θ ( d ( g x , g u ) + d ( g y , g v ) 2 ) ( d ( g x , g u ) + d ( g y , g v ) 2 ) = θ ( d ( 1 , 1 ) + d ( 0 , 1 ) 2 ) ( d ( 1 , 1 ) + d ( 0 , 1 ) 2 ) = θ ( 1 2 ) ( 1 2 ) = ( 1 − 1 4 ) 1 2 .

Hence inequality (2.1) holds.

Thus, in all the cases, inequality (2.1) of Theorem 6 is satisfied. Hence, by Theorem 6, (0,0) is a coupled coincidence point of F and g.

Theorem 10 Let (X,⪯) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Suppose that F:X×X→X and g:X→X are self-mappings on X such that F has the mixed g-monotone property on X such that there exist two elements x 0 , y 0 ∈X with g( x 0 )⪯F( x 0 , y 0 ) and g( y 0 )⪰F( y 0 , x 0 ). Suppose that there exists θ∈Θ such that

d ( F ( x , y ) , F ( u , v ) ) ≤θ ( M ( x , y , u , v ) ) ( M ( x , y , u , v ) ) ,
(2.16)

where

M(x,y,u,v)= d ( g x , F ( x , y ) ) + d ( g y , F ( y , x ) ) + d ( g u , F ( u , v ) ) + d ( g v , F ( v , u ) ) 4

for all x,y,u,v∈X with gx⪰gu and gy⪯gv. Further suppose that F(X×X)⊆g(X), g is continuous non-decreasing and commutes with F and also suppose that either

  1. (a)

    F is continuous, or

  2. (b)

    X has the following properties:

  3. (i)

    if {g( x n )}⊂X is a non-decreasing sequence with g x n →gx in g(X), then g x n ⪯gx for every n;

  4. (ii)

    if {g( y n )}⊂X is a non-increasing sequence with g y n →gy in g(X), then g y n ⪰gy for every n.

Then there exist two elements x,y∈X such that F(x,y)=g(x) and gy=F(y,x), that is, F and g have a coupled coincidence point (x,y)∈X×X.

Proof Following the proof of Theorem 6, we have an increasing sequence { x n } and a decreasing sequence { y n } in X. Now, we assume that

t n =d(g x n ,g x n − 1 )+d(g y n ,g y n − 1 )≠0

for all n.

Since g x n ⪰g x n − 1 and g y n ⪯g y n − 1 , from (2.16) and (2.2), we have

t n + 1 = d ( g x n + 1 , g x n ) + d ( g y n + 1 , g y n ) t n + 1 = d ( F ( x n , y n ) , F ( x n − 1 , y n − 1 ) ) + d ( F ( y n , x n ) , F ( y n − 1 , x n − 1 ) ) t n + 1 ≤ θ ( d ( g x n , F ( x n , y n ) ) + d ( g y n , F ( y n , x n ) ) + d ( g x n − 1 , F ( x n − 1 , y n − 1 ) ) + d ( g y n − 1 , F ( y n − 1 , x n − 1 ) ) 4 ) t n + 1 = × ( d ( g x n , F ( x n , y n ) ) + d ( g y n , F ( y n , x n ) ) + d ( g x n − 1 , F ( x n − 1 , y n − 1 ) ) + d ( g y n − 1 , F ( y n − 1 , x n − 1 ) ) 4 ) t n + 1 = + θ ( d ( g y n , F ( y n , x n ) ) + d ( g x n , F ( x n , y n ) ) + d ( g y n − 1 , F ( y n − 1 , x n − 1 ) ) + d ( g x n − 1 , F ( x n − 1 , y n − 1 ) ) 4 ) t n + 1 = × ( d ( g y n , F ( y n , x n ) ) + d ( g x n , F ( x n , y n ) ) + d ( g y n − 1 , F ( y n − 1 , x n − 1 ) ) + d ( g x n − 1 , F ( x n − 1 , y n − 1 ) ) 4 ) t n + 1 = θ ( d ( g x n , F ( x n , y n ) ) + d ( g y n , F ( y n , x n ) ) + d ( g x n − 1 , F ( x n − 1 , y n − 1 ) ) + d ( g y n − 1 , F ( y n − 1 , x n − 1 ) ) 4 ) t n + 1 = × ( d ( g x n , F ( x n , y n ) ) + d ( g y n , F ( y n , x n ) ) + d ( g x n − 1 , F ( x n − 1 , y n − 1 ) ) + d ( g y n − 1 , F ( y n − 1 , x n − 1 ) ) 2 ) t n + 1 = θ ( t n + t n + 1 4 ) ( t n + t n + 1 2 ) t n + 1 ≤ t n .
(2.17)

It follows that { t n } is a monotone decreasing sequence of non-negative real numbers. Therefore, there is some t≥0 such that lim n → ∞ t n =t.

Next, we show that t=0. Assume to the contrary that t>0, then from (2.17) we have

t n + 1 t n + t n + 1 2 ≤θ ( t n + t n + 1 4 ) <1,

which yields that lim n → ∞ θ( t n + t n + 1 4 )=1. This implies that d(g x n − 1 ,g x n )→0 and d(g y n − 1 ,g y n )→0. Therefore t=0, that is,

lim n → ∞ t n = lim n → ∞ [ d ( g x n + 1 , g x n ) + d ( g y n + 1 , g y n ) ] =0.
(2.18)

Now, we prove that {g x n } and {g y n } are Cauchy sequences. On the contrary, assume that at least one of {g x n } or {g y n } is not a Cauchy sequence. Then there exists an ϵ>0 for which we can find subsequences {g x m ( k ) } and {g x n ( k ) } of {g x n } and {g y m ( k ) } and {g y n ( k ) } of {g y n } with n(k)>m(k)>k such that for every k,

r k =d(g x m ( k ) ,g x n ( k ) )+d(g y m ( k ) ,g y n ( k ) )≥ϵ.
(2.19)

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 (2.19). Then

d(g x n ( k ) − 1 ,g x m ( k ) )+d(g y n ( k ) − 1 ,g y m ( k ) )<ϵ.
(2.20)

Using (2.19) and (2.20), we have

ϵ ≤ r k : = d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) ≤ 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 ) ) < ϵ + t n ( k ) .

Letting k→∞ and using (2.18), we have

lim r k =lim [ d ( g x m ( k ) , g x n ( k ) ) + d ( g y m ( k ) , g y n ( k ) ) ] =ϵ.
(2.21)

Also, by the triangle inequality, we have

r k = d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) ≤ 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 ) ) + 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 ) ) = t n ( k ) + t m ( k ) + d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + d ( g y n ( k ) + 1 , g x m ( k ) + 1 ) .

Since n(k)>m(k), g x n ( k ) ⪰g x m ( k ) and g y n ( k ) ⪯g y m ( k ) , from (2.16) and (2.2), we have

d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) = d ( F ( x n ( k ) , y n ( k ) ) , F ( x m ( k ) , y m ( k ) ) ) ≤ θ ( d ( g x n ( k ) , g x n ( k ) + 1 ) + d ( g y n ( k ) , g y n ( k ) + 1 ) + d ( g x m ( k ) , g x m ( k ) + 1 ) + d ( g y m ( k ) , g y m ( k ) + 1 ) 4 ) × ( d ( g x n ( k ) , g x n ( k ) + 1 ) + d ( g y n ( k ) , g y n ( k ) + 1 ) + d ( g x m ( k ) , g x m ( k ) + 1 ) + d ( g y m ( k ) , g y m ( k ) + 1 ) 4 ) ,

and similarly,

d ( g y n ( k ) + 1 , g x m ( k ) + 1 ) = d ( F ( y n ( k ) , x n ( k ) ) , F ( y m ( k ) , x m ( k ) ) ) ≤ θ ( d ( g x n ( k ) , g x n ( k ) + 1 ) + d ( g y n ( k ) , g y n ( k ) + 1 ) + d ( g x m ( k ) , g x m ( k ) + 1 ) + d ( g y m ( k ) , g y m ( k ) + 1 ) 4 ) × ( d ( g x n ( k ) , g x n ( k ) + 1 ) + d ( g y n ( k ) , g y n ( k ) + 1 ) + d ( g x m ( k ) , g x m ( k ) + 1 ) + d ( g y m ( k ) , g y m ( k ) + 1 ) 4 ) .

Therefore, we have

r k ≤ t n ( k ) + t m ( k ) r k ≤ + 2 θ ( d ( g x n ( k ) , g x n ( k ) + 1 ) + d ( g y n ( k ) , g y n ( k ) + 1 ) + d ( g x m ( k ) , g x m ( k ) + 1 ) + d ( g y m ( k ) , g y m ( k ) + 1 ) 4 ) r k ≤ × ( d ( g x n ( k ) , g x n ( k ) + 1 ) + d ( g y n ( k ) , g y n ( k ) + 1 ) + d ( g x m ( k ) , g x m ( k ) + 1 ) + d ( g y m ( k ) , g y m ( k ) + 1 ) 4 ) .

Taking k→∞, we have t n ( k ) , t m ( k ) →0, and

d ( g x n ( k ) , g x n ( k ) + 1 ) + d ( g y n ( k ) , g y n ( k ) + 1 ) + d ( g x m ( k ) , g x m ( k ) + 1 ) + d ( g y m ( k ) , g y m ( k ) + 1 ) 4 →0.

Hence, we get r k =ϵ=0, which is a contradiction. This implies that {g x n } and {g y n } are Cauchy sequences in g(X).

Since X is a complete metric space, there is (x,y)∈X×X such that g x n →x and g y n →y. Since g is continuous, g(g x n )→gx and g(g y n )→gy.

First, suppose that F is continuous. Then F(g x n ,g y n )→F(x,y) and F(g y n ,g x n )→F(y,x). As F commutes with g, we have

F(g x n ,g y n )=gF( x n , y n )=g(g x n + 1 )→gx

and

F(g y n ,g x n )=gF( y n , x n )=g(g y n + 1 )→gy.

By the uniqueness of the limit, we get gx=F(x,y) and gy=F(y,x).

Second, suppose that (b) holds. Since {g x n } is a non-decreasing sequence such that g x n →x and {g y n } is a non-increasing sequence such that g y n →y, and g is a non-decreasing function, we get that g(g x n )⪯gx and g(g y n )⪰gy hold for all n∈N. Hence, by (2.16), we have

d ( g ( g x n + 1 ) , F ( x , y ) ) = d ( F ( g x n , g y n ) , F ( x , y ) ) ≤ θ ( d ( g ( g x n ) , F ( g x n , g y n ) ) + d ( g ( g y n ) , F ( g y n , g x n ) ) + d ( g x , F ( x , y ) ) + d ( g y , F ( y , x ) ) 4 ) × ( d ( g ( g x n ) , F ( g x n , g y n ) ) + d ( g ( g y n ) , F ( g y n , g x n ) ) + d ( g x , F ( x , y ) ) + d ( g y , F ( y , x ) ) 4 ) = θ ( d ( g ( g x n ) , g ( g x n + 1 ) ) + d ( g ( g y n ) , g ( g y n + 1 ) ) + d ( g x , F ( x , y ) ) + d ( g y , F ( y , x ) ) 4 ) × ( d ( g ( g x n ) , g ( g x n + 1 ) ) + d ( g ( g y n ) , g ( g y n + 1 ) ) + d ( g x , F ( x , y ) ) + d ( g y , F ( y , x ) ) 4 )

and

d ( g ( g y n + 1 ) , F ( y , x ) ) = d ( F ( g y n , g x n ) , F ( y , x ) ) ≤ θ ( d ( g ( g y n ) , F ( g y n , g x n ) ) + d ( g ( g x n ) , F ( g x n , g y n ) ) + d ( g y , F ( y , x ) ) + d ( g x , F ( x , y ) ) 4 ) × ( d ( g ( g y n ) , F ( g y n , g x n ) ) + d ( g ( g x n ) , F ( g x n , g y n ) ) + d ( g y , F ( y , x ) ) + d ( g x , F ( x , y ) ) 4 ) = θ ( d ( g ( g y n ) , g ( g y n + 1 ) ) + d ( g ( g x n ) , g ( g x n + 1 ) ) + d ( g y , F ( y , x ) ) + d ( g x , F ( x , y ) ) 4 ) × ( d ( g ( g y n ) , g ( g y n + 1 ) ) + d ( g ( g x n ) , g ( g x n + 1 ) ) + d ( g y , F ( y , x ) ) + d ( g x , F ( x , y ) ) 4 ) .

Taking n→∞, we get d(gx+F(x,y))+d(gy,F(y,x))=0, and hence gx=F(x,y) and gy=F(y,x). Thus F and g have a coupled coincidence point. □

Example 11 Let X=[0,1]. Then (X,≤) is a partially ordered set with the natural ordering of real numbers. Let d(x,y)=|x−y| for all x,y∈X. Define a mapping g:X→X by g(x)=x and a mapping F:X×X→X by

F(x,y)={ x − y 16 , x ≥ y ; 0 , x < y .

Then it is easy to prove that (X,d) is a complete metric space, g(X) is complete, F:X×X→X⊆g(X)=X, X satisfies conditions (1) and (2) of Theorem 10 and F has the g-monotone property. Let θ:(0,∞)→[0,1) be defined as

θ(t)={ 1 − t , t ≤ 1 ; α < 1 , t > 1 .

Now, we verify inequality (2.16) of Theorem 10 for all x,y,u,v∈X with gx⪰gu and gy⪯gv.

Now, we consider the following cases.

Case 1. (x,y)=(0,0), (u,v)=(0,1) or (x,y)=(1,1), (u,v)=(0,1), we have

d ( F ( x , y ) , F ( u , v ) ) =0.

Hence, inequality (2.16) holds.

Case 2. (x,y)=(1,0), (u,v)=(0,0), we have

d ( F ( x , y ) , F ( u , v ) ) =d ( F ( 1 , 0 ) , F ( 0 , 0 ) ) = 1 16

and

θ ( d ( g x , F ( x , y ) ) + d ( g y , F ( y , x ) ) + d ( g u , F ( u , v ) ) + d ( g v , F ( v , u ) ) 4 ) × ( d ( g x , F ( x , y ) ) + d ( g y , F ( y , x ) ) + d ( g u , F ( u , v ) ) + d ( g v , F ( v , u ) ) 4 ) = θ ( d ( 1 , F ( 1 , 0 ) ) + d ( 0 , F ( 0 , 1 ) ) + 0 + 0 4 ) ( d ( 1 , F ( 1 , 0 ) ) + d ( 0 , F ( 0 , 1 ) ) + 0 + 0 4 ) = θ ( 15 64 ) ( 15 64 ) = ( 1 − 15 64 ) 15 64 .

Hence, inequality (2.16) holds.

Case 3. (x,y)=(1,0), (u,v)=(0,1), we have

d ( F ( x , y ) , F ( u , v ) ) =d ( F ( 1 , 0 ) , F ( 0 , 1 ) ) = 1 16

and

θ ( d ( g x , F ( x , y ) ) + d ( g y , F ( y , x ) ) + d ( g u , F ( u , v ) ) + d ( g v , F ( v , u ) ) 4 ) × ( d ( g x , F ( x , y ) ) + d ( g y , F ( y , x ) ) + d ( g u , F ( u , v ) ) + d ( g v , F ( v , u ) ) 4 ) = θ ( d ( 1 , F ( 1 , 0 ) ) + d ( 0 , F ( 0 , 1 ) ) + d ( 0 , F ( 0 , 1 ) ) + d ( 1 , F ( 1 , 0 ) ) 4 ) × ( d ( 1 , F ( 1 , 0 ) ) + d ( 0 , F ( 0 , 1 ) ) + d ( 0 , F ( 0 , 1 ) ) + d ( 1 , F ( 1 , 0 ) ) 4 ) = θ ( 30 64 ) ( 30 64 ) = ( 1 − 30 64 ) 30 64 .

Hence, inequality (2.16) holds.

Case 4. (x,y)=(1,0), (u,v)=(1,1), we have

d ( F ( x , y ) , F ( u , v ) ) =d ( F ( 1 , 0 ) , F ( 1 , 1 ) ) = 1 16 ;

and

θ ( d ( g x , F ( x , y ) ) + d ( g y , F ( y , x ) ) + d ( g u , F ( u , v ) ) + d ( g v , F ( v , u ) ) 4 ) × ( d ( g x , F ( x , y ) ) + d ( g y , F ( y , x ) ) + d ( g u , F ( u , v ) ) + d ( g v , F ( v , u ) ) 4 ) = θ ( d ( 1 , F ( 1 , 0 ) ) + d ( 0 , F ( 0 , 1 ) ) + d ( 1 , F ( 1 , 1 ) ) + d ( 1 , F ( 1 , 1 ) ) 4 ) × ( d ( 1 , F ( 1 , 0 ) ) + d ( 0 , F ( 0 , 1 ) ) + d ( 1 , F ( 1 , 1 ) ) + d ( 1 , F ( 1 , 1 ) ) 4 ) = θ ( 47 64 ) ( 47 64 ) = ( 1 − 47 64 ) 47 64 .

Hence, inequality (2.16) holds.

Thus, in all the cases, inequality (2.16) of Theorem 10 is satisfied. Hence, by Theorem 10, (0,0) is a coupled coincidence point of F and g.

Next, we prove the existence of a coupled coincidence point theorem, where we do not require that F and g are commuting.

The following lemma proved by Haghi et al. [17] is useful for our results.

Lemma 12 [17]

Let X be a nonempty set, and let g:X→X be a mapping. Then there exists a subset E⊆X such that g(E)=g(X) and g:E→X is one-to-one.

Theorem 13 Let (X,⪯) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a metric space. Suppose that F:X×X→X and g:X→X are self-mappings on X such that F has the mixed g-monotone property on X such that there exist two elements x 0 , y 0 ∈X with g( x 0 )⪯F( x 0 , y 0 ) and g( y 0 )⪰F( y 0 , x 0 ). Suppose that there exists θ∈Θ such that

d ( F ( x , y ) , F ( u , v ) ) ≤θ ( d ( g x , g u ) + d ( g y , g v ) 2 ) ( d ( g x , g u ) + d ( g y , g v ) 2 )
(2.22)

for all x,y,u,v∈X with gx⪰gu and gy⪯gv. Further suppose that F(X×X)⊆g(X) and g(X) is a complete subspace of X. Also assume that either

  1. (a)

    F is continuous, or

  2. (b)

    X has the following properties:

  3. (i)

    if {g( x n )}⊂X is a non-decreasing sequence with g x n →gx in g(X), then g x n ⪯gx for every n;

  4. (ii)

    if {g( y n )}⊂X is a non-increasing sequence with g y n →gy in g(X), then g y n ⪰gy for every n.

Then there exist two elements x,y∈X such that F(x,y)=g(x) and gy=F(y,x), that is, F and g have a coupled coincidence point (x,y)∈X×X.

Proof Using Lemma 12, there exists E⊆X such that g(E)=g(X) and g:E→X is one-to-one. We define a mapping A:g(E)×g(E)→X by

A(gx,gy)=F(x,y)for every gx,gy∈g(E).
(2.23)

As g is one-to-one on g(E), so A is well defined.

Since F has the mixed g-monotone property, for all x,y∈X, we have

x 1 , x 2 ∈X,g x 1 ⪯g x 2 ⇔F( x 1 ,y)⪯F( x 2 ,y)
(2.24)

and

y 1 , y 2 ∈X,g y 1 ⪰g y 2 ⇔F(x, y 1 )⪰F(x, y 2 ).
(2.25)

Thus, it follows from (2.23), (2.24) and (2.25) that for all gx,gy∈g(E),

g x 1 ,g x 2 ∈g(E),g x 1 ⪯g x 2 ⇔A(g x 1 ,gy)⪯A(g x 2 ,gy)
(2.26)

and

g y 1 ,g y 2 ∈g(E),g y 1 ⪰g y 2 ⇔A(gx,g y 1 )⪰A(gx,g y 2 ),
(2.27)

which implies that A has the mixed monotone property.

Suppose that the assumption (a) holds. Since F is continuous, A is also continuous. Using Theorem 2.1 of [15] with the mapping A, it follows that A has a coupled fixed point (u,v)∈g(X)×g(X).

Suppose that the assumption (b) holds. We can conclude similarly to the proof of Theorem 2.1 of [15] that the mapping A has a coupled fixed point (u,v)∈g(X)×g(X).

Finally, we prove that F and g have a coupled coincidence point in X. Since (u,v) is a coupled fixed point of A, we get

u=A(u,v),v=A(v,u).
(2.28)

Since (u,v)∈g(X)×g(X), there exists a point ( u ˆ , v ˆ )∈X×X such that

u=g u ˆ ,v=g v ˆ .
(2.29)

Thus, it follows from (2.28) and (2.29) that

g u ˆ =A(g u ˆ ,g v ˆ ),g v ˆ =A(g v ˆ ,g u ˆ ).
(2.30)

Also, from (2.23) and (2.30), we get

g u ˆ =F( u ˆ , v ˆ ),g v ˆ =F( v ˆ , u ˆ ).
(2.31)

Therefore, ( u ˆ , v ˆ ) is a coupled coincidence point of F and g. This completes the proof. □

Theorem 14 Let (X,⪯) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a metric space. Suppose that F:X×X→X and g:X→X are self-mappings on X such that F has the mixed g-monotone property on X such that there exist two elements x 0 , y 0 ∈X with g( x 0 )⪯F( x 0 , y 0 ) and g( y 0 )⪰F( y 0 , x 0 ). Suppose that there exists θ∈Θ such that

d ( F ( x , y ) , F ( u , v ) ) ≤θ ( M ( x , y , u , v ) ) ( M ( x , y , u , v ) ) ,
(2.32)

where

M(x,y,u,v)= d ( g x , F ( x , y ) ) + d ( g y , F ( y , x ) ) + d ( g u , F ( u , v ) ) + d ( g v , F ( v , u ) ) 4

for all x,y,u,v∈X with gx⪰gu and gy⪯gv. Further suppose that F(X×X)⊆g(X) and g(X) is a complete subspace of X. Also assume that either

  1. (a)

    F is continuous, or

  2. (b)

    X has the following properties:

  3. (i)

    if {g( x n )}⊂X is a non-decreasing sequence with g x n →gx in g(X), then g x n ⪯gx for every n;

  4. (ii)

    if {g( y n )}⊂X is a non-increasing sequence with g y n →gy in g(X), then g y n ⪰gy for every n.

Then there exist two elements x,y∈X such that F(x,y)=g(x) and gy=F(y,x), that is, F and g have a coupled coincidence point (x,y)∈X×X.

Proof Following similar arguments to those in Theorem 13 and using Theorem 2.2 of [15], we get the result. □

Remark 15 Although Theorem 6 and Theorem 10 are an essential tool in the partially ordered metric spaces to claim the existence of coupled coincidence points of two mappings, some mappings do not have the commutative property. For example, see the following.

Example 16 Let X=[0,1]. Then (X,⪯) is a partially ordered set with the natural ordering of real numbers. Let d(x,y)=x−y for all x,y∈X. Define mappings F:X×X→X and g:X→X by F(x,y)=1 for all (x,y)∈X×X and g(x)=x−1 for each x∈X. Since g(F(x,y))=g(1)=0≠1=F(gx,gy) for all x,y∈X, the mappings F and g do not satisfy the commutative condition. Hence, the above two theorems cannot be applied to this example. But, by a simple calculation, we see that F(X×X)⊆g(X), g and F are continuous and F has the mixed g-monotone property. Moreover, there exist x 0 =1 and y 0 =3 with g(1)=0⪯1=F(1,3) and g(3)=2⪰1=F(3,1).

Therefore, it is very interesting to use Theorems 13 and 14 as another auxiliary tool to claim the existence of a coupled coincidence point.

References

  1. Ran ACM, Reurings MCB: A fixed point theorem in partially ordered sets and some applications to matrix equations. Proc. Am. Math. Soc. 2004, 132(5):1435–1443. 10.1090/S0002-9939-03-07220-4

    Article  MathSciNet  Google Scholar 

  2. Nieto JJ, López RR: Contractive mapping theorems in partially ordered sets and applications to ordinary differential equations. Order 2005, 22: 223–239. 10.1007/s11083-005-9018-5

    Article  MathSciNet  Google Scholar 

  3. Bhaskar TG, Lakshmikantham V: Fixed point theorems in partially ordered metric spaces and applications. Nonlinear Anal. 2006, 65: 1379–1393. 10.1016/j.na.2005.10.017

    Article  MathSciNet  Google Scholar 

  4. Abbas M, Sintunavarat W, Kumam P: Coupled fixed point of generalized contractive mappings on partially ordered G -metric spaces. Fixed Point Theory Appl. 2012., 2012: Article ID 31

    Google Scholar 

  5. Agarwal RP, El-Gebeily MA, O’Regan D: Generalized contractions in partially ordered metric spaces. Appl. Anal. 2008, 87: 1–8. 10.1080/00036810701714164

    Article  MathSciNet  Google Scholar 

  6. Amini-Harandi A, Emami H: A fixed point theorem for contraction type maps in partially ordered metric spaces and application to ordinary differential equations. Nonlinear Anal. TMA 2010, 72(5):2238–2242. 10.1016/j.na.2009.10.023

    Article  MathSciNet  Google Scholar 

  7. Caballero J, Harjani J, Sadarangani K: Contractive-like mapping principles in ordered metric spaces and application to ordinary differential equations. Fixed Point Theory Appl. 2010., 2010: Article ID 916064 10.1155/2010/916064

    Google Scholar 

  8. Chandok S: Some common fixed point theorems for generalized f -weakly contractive mappings. J. Appl. Math. Inform. 2011, 29: 257–265.

    MathSciNet  Google Scholar 

  9. Chandok S: Some common fixed point theorems for generalized nonlinear contractive mappings. Comput. Math. Appl. 2011, 62: 3692–3699. 10.1016/j.camwa.2011.09.009

    Article  MathSciNet  Google Scholar 

  10. Chandok S: Common fixed points, invariant approximation and generalized weak contractions. Int. J. Math. Math. Sci. 2012., 2012: Article ID 102980

    Google Scholar 

  11. Chandok S, Khan MS, Rao KPR: Some coupled common fixed point theorems for a pair of mappings satisfying a contractive condition of rational type without monotonicity. Int. J. Math. Anal. 2013, 7(9):433–440.

    MathSciNet  Google Scholar 

  12. Chandok S, Kim JK: Fixed point theorem in ordered metric spaces for generalized contractions mappings satisfying rational type expressions. Nonlinear Funct. Anal. Appl. 2012, 17: 301–306.

    Google Scholar 

  13. Chandok S, Narang TD, Taoudi MA: Some common fixed point results in partially ordered metric spaces for generalized rational type contraction mappings. Vietnam J. Math. 2013. 10.1007/s10013-013-0024-4

    Google Scholar 

  14. Chandok S, Sintunavarat W, Kumam P: Some coupled common fixed points for a pair of mappings in partially ordered G -metric spaces. Math. Sci. 2013., 7: Article ID 24 10.1186/2251-7456-7-24

    Google Scholar 

  15. Choudhury BS, Kundu A: On coupled generalised Banach and Kannan type contractions. J. Nonlinear Sci. Appl. 2012, 5: 259–270.

    MathSciNet  Google Scholar 

  16. Geraghty MA: On contractive mappings. Proc. Am. Math. Soc. 1973, 40: 604–608. 10.1090/S0002-9939-1973-0334176-5

    Article  MathSciNet  Google Scholar 

  17. Haghi RH, Rezapour S, Shahzad N: Some fixed point generalizations are not real generalizations. Nonlinear Anal. 2011, 74: 1799–1803. 10.1016/j.na.2010.10.052

    Article  MathSciNet  Google Scholar 

  18. Janos L: On mappings contractive in the sense of Kannan. Proc. Am. Math. Soc. 1976, 61(1):171–175. 10.1090/S0002-9939-1976-0425936-3

    Article  MathSciNet  Google Scholar 

  19. Karapinar E, Kumam P, Sintunavarat W: Coupled fixed point theorems in cone metric spaces with a c -distance and applications. Fixed Point Theory Appl. 2012., 2012: Article ID 194

    Google Scholar 

  20. Lakshmikantham V, Ciric L: Coupled fixed point theorems for nonlinear contractions in partially ordered metric spaces. Nonlinear Anal. 2009, 70: 4341–4349. 10.1016/j.na.2008.09.020

    Article  MathSciNet  Google Scholar 

  21. Sintunavarat W, Cho YJ, Kumam P: Coupled fixed point theorems for weak contraction mapping under F -invariant set. Abstr. Appl. Anal. 2012., 2012: Article ID 324874

    Google Scholar 

  22. Sintunavarat W, Kumam P: Coupled best proximity point theorem in metric spaces. Fixed Point Theory Appl. 2012., 2012: Article ID 93

    Google Scholar 

  23. Sintunavarat W, Kumam P: Coupled coincidence and coupled common fixed point theorems in partially ordered metric spaces. Thai J. Math. 2012, 10(3):551–563.

    MathSciNet  Google Scholar 

  24. Sintunavarat W, Cho YJ, Kumam P: Coupled fixed-point theorems for contraction mapping induced by cone ball-metric in partially ordered spaces. Fixed Point Theory Appl. 2012., 2012: Article ID 128

    Google Scholar 

  25. Sintunavarat W, Petrusel A, Kumam P:Common coupled fixed point theorems for w ∗ -compatible mappings without mixed monotone property. Rend. Circ. Mat. Palermo 2012, 61: 361–383. 10.1007/s12215-012-0096-0

    Article  MathSciNet  Google Scholar 

  26. Sintunavarat W, Kumam P, Cho YJ: Coupled fixed point theorems for nonlinear contractions without mixed monotone property. Fixed Point Theory Appl. 2012., 2012: Article ID 170

    Google Scholar 

  27. Sintunavarat W, Radenovic S, Golubovic Z, Kumam P: Coupled fixed point theorems for F -invariant set. Appl. Math. Inform. Sci. 2013, 7(1):247–255. 10.12785/amis/070131

    Article  MathSciNet  Google Scholar 

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Acknowledgements

This work was supported by the Kyungnam University Research Fund 2013. The authors are thankful to the referees for valuable suggestions.

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The main idea of this paper was proposed by JKK. JKK and SC prepared the manuscript initially and performed all the steps of the proof in this research. All authors read and approved the final manuscript.

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Kim, J.K., Chandok, S. Coupled common fixed point theorems for generalized nonlinear contraction mappings with the mixed monotone property in partially ordered metric spaces. Fixed Point Theory Appl 2013, 307 (2013). https://doi.org/10.1186/1687-1812-2013-307

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