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An original coupled coincidence point result for a pair of mappings without MMP

Abstract

The purpose of this paper is to establish a coupled coincidence point theorem for a pair of mappings without MMP (mixed monotone property) in metric spaces endowed with partial order, which is not an immediate consequence of a well-known theorem in the literature. Also, we present a result on the existence and uniqueness of coupled common fixed points. The results presented in the paper generalize and extend some of the results of Bhaskar and Lakshmikantham (Nonlinear Anal. 65:1379-1393, 2006), Choudhury, Metiya and Kundu (Ann. Univ. Ferrara 57:1-16, 2011), Harjani, Lopez and Sadarangani (Nonlinear Anal. 74:1749-1760, 2011) and of Luong and Thuan (Bull. Math. Anal. Appl. 2:16-24, 2010) for the mappings having no MMP. We introduce an example that there exists a common coupled fixed point of the mappings g and F such that F does not satisfy the g-mixed monotone property, and also g and F do not commute.

MSC:41A50, 47H10, 54H25.

1 Introduction and preliminaries

Fixed point theory is one of the famous and traditional theories in mathematics and has a large number of applications. The Banach contraction mapping is one of the pivotal results of analysis. It is a very popular tool for solving existence problems in many different fields of mathematics. There are a lot of generalizations of the Banach contraction 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 Rodŕiguez-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 on 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, fuzzy metric spaces, intuitionistic fuzzy normed spaces, partially ordered metric spaces and others (see [125]).

Definition 1.1 Let (X,d) be a metric space and F:X×XX and g:XX, F and g are said to commute if F(gx,gy)=g(F(x,y)) for all x,yX.

Definition 1.2 Let (X,d) be a metric space and let g:XX, F:X×XX. The mappings g and F are said to be compatible if lim n d(gF( x n , y n ),F(g x n ,g y n ))=0 and lim n d(gF( y n , x n ),F(g y n ,g x n ))=0 hold whenever { x n } and { y n } are sequences in X such that lim n F( x n , y n )= lim n g x n and lim n F( y n , x n )= lim n g y n .

Definition 1.3 Let (X,) be a partially ordered set and F:X×XX. The mapping F is said to be non-decreasing if for x,yX, xy implies F(x)F(y) and non-increasing if for x,yX, xy implies F(x)F(y).

Definition 1.4 Let (X,) be a partially ordered set and F:X×XX and g:XX. 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,yX,

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 1.4, then the mapping F is said to have the mixed monotone property.

Recently, Ðoric et al. [12] showed that the mixed monotone property in coupled fixed point results for mappings in ordered metric spaces can be replaced by another property which is often easy to check. In particular, it is automatically satisfied in the case of a totally ordered space, the case which is important in applications. Hence, these results can be applied in a much wider class of problems.

If elements x, y of a partially ordered set (X,) are comparable (i.e., xy or yx holds) we will write xy. Let g:XX and F:X×XX. We will consider the following condition:

if x,y,u,vX are such that gxF(x,y)=gu, then F(x,y)F(u,v).

If g is an identity mapping, for all x, y, v, if xF(x,y) then F(x,y)F(F(x,y),v).

Ðoric et al. [12] gave some examples that these conditions may be satisfied when F does not have the g-mixed monotone property.

Definition 1.5 An element (x,y)X×X is called a coupled coincidence point of the mappings F:X×XX and g:XX if F(x,y)=gx and F(y,x)=gy.

If g= identity mapping in Definition 1.5, then (x,y)X×X is called a coupled fixed point.

The purpose of this paper is to establish some coupled coincidence point results in partially ordered metric spaces for a pair of mappings without mixed monotone property satisfying a contractive condition. Also, we present a result on the existence and uniqueness of coupled common fixed points. Also, we give an example to illustrate the main result in this paper. The results proved generalize some of the results of Bhaskar and Lakshmikantham [3], Choudhury et al. [10], Luong and Thuan [17] and Harjani et al. [13] for the mappings having no mixed monotone property.

2 Main results

2.1 Coupled common fixed point theorems

In this section, we prove some coupled common fixed point theorems in the context of ordered metric spaces.

We denote by Φ the set of functions ϕ:[0,)[0,) satisfying:

  1. (i)

    ϕ is continuous;

  2. (ii)

    ϕ(t)<t for all t>0 and ϕ(t)=0 if and only if t=0.

Theorem 2.1 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×XX and g:XX are self-mappings on X such that the following conditions hold:

  1. (i)

    g is continuous and g(X) is closed;

  2. (ii)

    F(X×X)g(X) and g and F are compatible;

  3. (iii)

    for all x,y,u,vX, if g(x)F(x,y)=gu, then F(x,y)F(u,v);

  4. (iv)

    there exist x 0 , y 0 X such that g x 0 F( x 0 , y 0 ) and g y 0 F( y 0 , x 0 );

  5. (v)

    there exists a non-negative real number L such that

    d ( F ( x , y ) , F ( u , v ) ) ϕ ( max { d ( g x , g u ) , d ( g y , g v ) } ) + L min { d ( F ( x , y ) , g u ) , d ( F ( u , v ) , g x ) , d ( F ( x , y ) , g x ) , d ( F ( u , v ) , g u ) }
    (2.1)

for all x,y,u,vX with gxgu and gygv, where ϕΦ;

  1. (vi)

    (a) F is continuous or (b) x n x, when n in X, then x n x for sufficiently large n.

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

Proof Using conditions (ii) and (iv), construct sequences { x n } and { y n } in X satisfying g x n =F( x n 1 , y n 1 ) and g y n =F( y n 1 , x n 1 ) for n=1,2, .

By (iv), g x 0 F( x 0 , y 0 )=g x 1 and condition (iii) implies that g x 1 =F( x 0 , y 0 )F( x 1 , y 1 )=g x 2 . Proceeding by induction, we get that g x n 1 g x n , and similarly, g y n 1 g y n for each nN.

Now from the contractive condition (2.1), we have

d ( g x n + 1 , g x n ) = d ( F ( x n , y n ) , F ( x n 1 , y n 1 ) ) ϕ ( max { d ( g x n , g x n 1 ) , d ( g y n , g y n 1 ) } ) + L min { d ( F ( x n , y n ) , g x n 1 ) , d ( F ( x n 1 , y n 1 ) , g x n ) , d ( F ( x n , y n ) , g x n ) , d ( F ( x n 1 , y n 1 ) , g x n 1 ) } ,
(2.2)

which implies that d(g x n + 1 ,g x n )ϕ(max{d(g x n ,g x n 1 ),d(g y n ,g y n 1 )}).

Similarly, we have d(g y n + 1 ,g y n )ϕ(max{d(g y n ,g y n 1 ),d(g x n ,g x n 1 )}).

Therefore, from the above two inequalities we have

max { d ( g x n + 1 , g x n ) , d ( g y n + 1 , g y n ) } ϕ ( max { d ( g x n , g x n 1 ) , d ( g y n , g y n 1 ) } ) .
(2.3)

Since ϕ(t)<t for all t>0 and ϕ(t)=0 if and only if t=0, from (2.3) we have

max { d ( g x n + 1 , g x n ) , d ( g y n + 1 , g y n ) } max { d ( g x n , g x n 1 ) , d ( g y n , g y n 1 ) } .

Set ϱ n :=max{d(g x n + 1 ,g x n ),d(g y n + 1 ,g y n )}, then { ϱ n } is a non-increasing sequence of positive real numbers. Thus, there is d0 such lim n ϱ n =d.

Suppose that d>0, letting n two sides of (2.3) and using the properties of ϕ, we have

d= lim n d(g y n ,g y n + 1 ) lim n ϕ ( max { d ( g y n + 1 , g y n ) , d ( g x n + 1 , g x n ) } ) =ϕ(d)<d,
(2.4)

which is a contradiction. Hence d=0, i.e.,

lim n ϱ n = lim n max { d ( g x n + 1 , g x n ) , d ( g y n + 1 , g y n ) } =0.
(2.5)

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

max { d ( g x n ( k ) , g x m ( k ) ) , d ( g y n ( k ) , g y m ( k ) ) } ϵ.
(2.6)

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 (2.6). Then

max { d ( g x n ( k ) 1 , g x m ( k ) ) , d ( g y n ( k ) 1 , g y m ( k ) ) } <ϵ.
(2.7)

Using the triangle inequality and (2.7), we have

d(g x n ( k ) ,g x m ( k ) )d(g x n ( k ) ,g x n ( k ) 1 )+d(g x n ( k ) 1 ,g x m ( k ) )<d(g x n ( k ) ,g x n ( k ) 1 )+ϵ
(2.8)

and

d(g y n ( k ) ,g y m ( k ) )d(g y n ( k ) ,g y n ( k ) 1 )+d(g y n ( k ) 1 ,g y m ( k ) )<d(g y n ( k ) ,g y n ( k ) 1 )+ϵ.
(2.9)

From (2.6), (2.8) and (2.9), we have

ϵ max { d ( g x n ( k ) , g x m ( k ) ) , d ( g y n ( k ) , g y m ( k ) ) } < max { d ( g x n ( k ) , g x n ( k ) 1 ) , d ( g y n ( k ) , g y n ( k ) 1 ) } + ϵ .

Letting k in the inequalities above and using (2.5), we get

lim k max { d ( g x n ( k ) , g x m ( k ) ) , d ( g y n ( k ) , g y m ( k ) ) } =ϵ.
(2.10)

By the triangle inequality,

d(g x n ( k ) ,g x m ( k ) )d(g x n ( k ) ,g x n ( k ) 1 )+d(g x n ( k ) 1 ,g x n ( k ) 1 )+d(g x m ( k ) 1 ,g x m ( k ) )

and

d(g y n ( k ) ,g y 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 ) ).

From the last two inequalities and (2.6), we have

ϵ max { d ( g x n ( k ) , g x m ( k ) ) , d ( g y n ( k ) , g y m ( k ) ) } max { d ( g x n ( k ) , g x n ( k ) 1 ) , d ( g y n ( k ) , g y n ( k ) 1 ) } + max { d ( g x m ( k ) 1 , g x m ( k ) ) , d ( g y m ( k ) 1 , g y m ( k ) ) } + max { d ( g x n ( k ) 1 , g x m ( k ) 1 ) , d ( g y n ( k ) 1 , g y m ( k ) 1 ) } .
(2.11)

Again, by the triangle inequality,

d ( g x n ( k ) 1 , g x m ( k ) 1 ) d ( g x n ( k ) 1 , g x m ( k ) ) + d ( g x m ( k ) , g x m ( k ) 1 ) < d ( g x m ( k ) , g x m ( k ) 1 ) + ϵ ,

and

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

Therefore,

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

Taking k in (2.11) and (2.12) and using (2.5), (2.10), we have

lim k max { d ( g x n ( k ) 1 , g x m ( k ) 1 ) , d ( g y n ( k ) 1 , g y m ( k ) 1 ) } =ϵ.
(2.13)

Since n(k)>m(k), g x n ( k ) 1 g x m ( k ) 1 and g y n ( k ) 1 g y m ( k ) 1 . Then from (2.1) we have

d ( g x n ( k ) , g x m ( k ) ) = d ( F ( x n ( k ) 1 , y n ( k ) 1 ) , F ( x m ( k ) 1 , y m ( k ) 1 ) ) ϕ ( max { d ( g x n ( k ) 1 , g x m ( k ) 1 ) , d ( g y n ( k ) 1 , g y m ( k ) 1 ) } ) + L min { d ( F ( x n ( k ) 1 , y n ( k ) 1 ) , g x m ( k ) 1 ) , d ( F ( x m ( k ) 1 , y m ( k ) 1 ) , g x n ( k ) 1 ) , d ( F ( x n ( k ) 1 , y n ( k ) 1 ) , g x n ( k ) 1 ) , d ( F ( x m ( k ) 1 , y m ( k ) 1 ) , g x m ( k ) 1 ) } ϕ ( max { d ( g x n ( k ) 1 , g x m ( k ) 1 ) , d ( g y n ( k ) 1 , g y m ( k ) 1 ) } ) + L min { d ( g x n ( k ) , g x n ( k ) 1 ) , d ( g x m ( k ) , g x m ( k ) 1 ) } .
(2.14)

Similarly,

d ( g y m ( k ) , g y n ( k ) ) ϕ ( max { d ( g x n ( k ) 1 , g x m ( k ) 1 ) , d ( g y n ( k ) 1 , g y m ( k ) 1 ) } ) + L min { d ( g y m ( k ) , g y m ( k ) 1 ) , d ( g y n ( k ) , g y n ( k ) 1 ) } .
(2.15)

From (2.14) and (2.15), we have

max { d ( g x n ( k ) , g x m ( k ) ) , d ( g y n ( k ) , g y m ( k ) ) } ϕ ( max { d ( g x n ( k ) 1 , g x m ( k ) 1 ) , d ( g y n ( k ) 1 , g y m ( k ) 1 ) } ) + L min { d ( g x n ( k ) , g x n ( k ) 1 ) , d ( g x m ( k ) , g x m ( k ) 1 ) } + L min { d ( g y m ( k ) , g y m ( k ) 1 ) , d ( g y n ( k ) , g y n ( k ) 1 ) } .

Letting n in the above inequality and using (2.5), (2.10), (2.13) and the properties of ϕ, we have

ϵϕ(ϵ)+2Lmin{0,0}<ϵ,

which is a contradiction. Therefore, {g x n } and {g y n } are Cauchy sequences and since g(X) is closed in a complete metric space (condition (i)), there exist x,yg(X) such that lim n g x n = lim n F( x n 1 , y n 1 )=x and lim n g y n = lim n F( y n 1 , x n 1 )=y.

Compatibility of F and g (condition (ii)) implies 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.

Consider the two possibilities given in condition (vi).

  1. (a)

    Suppose that F is continuous. Using the triangle inequality, we get that

    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 ) ) .

By taking limit n and using the continuity of F and g, we have d(gx,F(x,y))=0, i.e., gx=F(x,y) and, in a similar way, we have gy=F(y,x). Thus F and g have a coupled coincidence point.

  1. (b)

    In this case g x n u=gx and g y n v=gy for some x,yX and n sufficiently large. For such n, using (2.1) we get

    d ( F ( x , y ) , g x ) d ( F ( x , y ) , g x n + 1 ) + d ( g x n + 1 , g x ) = d ( F ( x , y ) , F ( x n , y n ) ) + d ( g x n + 1 , g x ) ϕ ( max { d ( g x , g x n ) , d ( g y , g y n ) } ) + L min { d ( F ( x , y ) , g x n ) , d ( F ( x n , y n ) , g x n ) , d ( F ( x , y ) , g x ) , d ( F ( x n , y n ) , g x n ) } + d ( g x n + 1 , g x ) .

Taking n in the above inequality and using the compatibility of F and g and the properties of ϕ, we have d(F(x,y),gx)ϕ(max{0,0})+0+0=0. Hence F(x,y)=gx. Similarly, one can show that F(y,x)=gy. Hence the result. □

Remark 2.1 Very recently, using the equivalence of the three basic metrics, Samet et al. [19] show that many of the coupled fixed point theorems are immediate consequences of well-known fixed point theorems in the literature.

In our Theorem 2.1, it is easy to see that if L0 there is no equivalence and this theorem is not a consequence of a known fixed point theorem.

Remark 2.2 In the above theorem, condition (iii) is a substitution for the mixed monotone property that has been used in most of the coupled fixed point results so far. Note that this condition is trivially satisfied if the order on X is total, which is the case in most of the examples in articles mentioned in the references.

If g is an identity mapping in the above theorem, we have the following result.

Corollary 2.2 Let (X,d,) be a complete partially ordered metric space and let F:X×XX. Suppose that the following hold:

  1. (i)

    for all x,y,vX, if xF(x,y), then F(x,y)F(F(x,y),v);

  2. (ii)

    there exist x 0 , y 0 X such that x 0 F( x 0 , y 0 ) and y 0 F( y 0 , x 0 );

  3. (iii)

    there exists a non-negative real number L such that

    d ( F ( x , y ) , F ( u , v ) ) ϕ ( max { d ( x , u ) , d ( y , v ) } ) + L min { d ( F ( x , y ) , u ) , d ( F ( u , v ) , x ) , d ( F ( x , y ) , x ) , d ( F ( u , v ) , u ) }

for all x,y,u,vX with xu and yv, where ϕΦ;

  1. (iv)

    (a) F is continuous or (b) x n x, when n in X, then x n x for sufficiently large n.

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

Remark 2.3 Letting L=0, in inequality (2.1), for all x,y,u,vX, α,β0, α+β<1, we have

αd(x,u)+βd(y,v)(α+β)max { d ( x , u ) , d ( y , v ) } =ϕ ( max { d ( x , u ) , d ( y , v ) } ) ,

where ϕ(t)=(α+β)(t) for all t0 is in Φ. Hence Theorem 2.1 generalizes the corresponding coupled fixed point results of Bhaskar and Lakshmikantham [3], Choudhury et al. [10], Luong and Thuan [17] and Harjani et al. [13] for the mappings having no mixed monotone property.

Taking L=0, we have the following result.

Corollary 2.3 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×XX and g:XX are self-mappings on X such that the following conditions hold:

  1. (i)

    g is continuous and g(X) is closed;

  2. (ii)

    F(X×X)g(X) and g and F are compatible;

  3. (iii)

    for all x,y,u,vX, if g(x)F(x,y)=gu, then F(x,y)F(u,v);

  4. (iv)

    there exist x 0 , y 0 X such that g x 0 F( x 0 , y 0 ) and g y 0 F( y 0 , x 0 );

  5. (v)

    F and g satisfy

    d ( F ( x , y ) , F ( u , v ) ) ϕ ( max { d ( g x , g u ) , d ( g y , g v ) } )
    (2.16)

for all x,y,u,vX with gxgu and gygv, where ϕΦ;

  1. (vi)

    (a) F is continuous or (b) x n x, when n in X, then x n x for sufficiently large n.

Then there exist x,yX 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.

Corollary 2.4 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×XX and g:XX are self-mappings on X such that following conditions hold:

  1. (i)

    g is continuous and g(X) is closed;

  2. (ii)

    F(X×X)g(X) and g and F are compatible;

  3. (iii)

    for all x,y,u,vX, if g(x)F(x,y)=gu, then F(x,y)F(u,v);

  4. (iv)

    there exist x 0 , y 0 X such that g x 0 F( x 0 , y 0 ) and g y 0 F( y 0 , x 0 );

  5. (v)

    there exist non-negative real numbers α, β with α+β<1 such that

    d ( F ( x , y ) , F ( u , v ) ) αd(gx,gu)+βd(gy,gv)
    (2.17)

for all x,y,u,vX with gxgu and gygv;

  1. (vi)

    (a) F is continuous or (b) x n x, when n in X, then x n x for sufficiently large n.

Then there exist x,yX 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.

Taking α=β=k[0,1), we have the following result.

Corollary 2.5 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×XX and g:XX are self-mappings on X such that following conditions hold:

  1. (i)

    g is continuous and g(X) is closed;

  2. (ii)

    F(X×X)g(X) and g and F are compatible;

  3. (iii)

    for all x,y,u,vX, if g(x)F(x,y)=gu, then F(x,y)F(u,v);

  4. (iv)

    there exist x 0 , y 0 X such that g x 0 F( x 0 , y 0 ) and g y 0 F( y 0 , x 0 );

  5. (v)

    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 ) ]
    (2.18)

for all x,y,u,vX with gxgu and gygv;

  1. (vi)

    (a) F is continuous or (b) x n x, when n in X, then x n x for sufficiently large n.

Then there exist x,yX 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.

Now, we shall prove the existence and uniqueness of a coupled common fixed point. Note that if (X,) is a partially ordered set, then we endow the product space X×X with the following partial order relation:

for (x,y),(u,v)X×X,(u,v)(x,y)xu,yv.

Theorem 2.6 In addition to hypotheses of Theorem 2.1, suppose that

  1. (vii)

    for every (x,y),(u,v)X×X, there exists (w,z)X×X such that (F(w,z),F(z,w)) is comparable to (F(x,y),F(y,x)) and (F(u,v),F(v,u)).

Then F and g have a unique coupled common fixed point, that is, there exists a unique (p,q)X×X such that p=gp=F(p,q) and q=gq=F(q,p).

Proof From Theorem 2.1, there exists (x,y)X×X such that gx=F(x,y) and gy=F(y,x). Suppose that there is also (u,v)X×X such that gu=F(u,v) and gv=F(v,u). We will prove that gx=gu and gy=gv. Condition (vii) implies that there exists (w,z)X×X such that (F(w,z),F(z,w)) is comparable to both (F(x,y),F(y,x)) and (F(u,v),F(v,u)). Put w 0 =w, z 0 =z and, analogously to the proof of Theorem 2.1, choose sequences { w n }, { z n } satisfying

g w n =F( w n 1 , z n 1 )andg z n =F( z n 1 , w n 1 )

for nN. Starting from x 0 =x, y 0 =y and u 0 =u, v 0 =v, choose sequences { x n }, { y n } and { u n }, { v n }, satisfying g x n =F( x n 1 , y n 1 ), g y n =F( y n 1 , x n 1 ) and g u n =F( u n 1 , v n 1 ), g v n =F( v n 1 , u n 1 ) for nN, taking into account properties of coincidence points, it is easy to see that this can be done so that x n =x, y n =y and u n =u, v n =v, i.e.,

g x n =F(x,y),g y n =F(y,x)andg u n =F(u,v),g v n =F(v,u)for nN.

Since (F(x,y),F(y,x))=(gx,gy) and (F(w,z),F(z,w))=(g w 1 ,g z 1 ) are comparable, then gxg w 1 and gyg z 1 , and, in a similar way, we have gxg w n and gyg z n . Thus from (2.1) we have

d ( g x , g w n + 1 ) = d ( F ( x , y ) , F ( w n , z n ) ) ϕ ( max { d ( g x , g w n ) , d ( g y , g z n ) } ) + L min { d ( F ( x , y ) , g w n ) , d ( F ( w n , z n ) , g x ) , d ( F ( x , y ) , g x ) , d ( F ( w n , z n ) , g w n ) } ,
(2.19)

which implies that d(gx,g w n + 1 )ϕ(max{d(gx,g w n ),d(gy,g z n )}).

Similarly, we can prove that d(gy,g z n + 1 )ϕ(max{d(gx,g w n ),d(gy,g z n )}).

Therefore, from the above two inequalities we have

max { d ( g x , g w n + 1 ) , d ( g y , g z n + 1 ) } ϕ ( max { d ( g x , g w n ) , d ( g y , g z n ) } ) .
(2.20)

Since ϕ(t)t for all t0, from (2.20) we have

max { d ( g x , g w n + 1 ) , d ( g y , g z n + 1 ) } max { d ( g x , g w n ) , d ( g y , g z n ) } .

Hence the sequence { δ n } defined by δ n :=max{d(gx,g w n + 1 ),d(gy,g z n + 1 )} is non-negative and decreasing and so lim n δ n =δ for some δ0.

Now, we show that δ=0. Assume that δ>0, letting n two sides of (2.20) and using the properties of ϕ, we have

δ= lim n d(gy,g z n + 1 ) lim n ϕ ( max { d ( g x , g w n ) , d ( g y , g z n ) } ) =ϕ(δ)<δ,
(2.21)

which is a contradiction. Hence d=0, i.e.,

lim n d(gx,g w n + 1 )=0= lim n d(gy,g z n + 1 )=0.
(2.22)

Similarly, we can prove that

lim n d(gu,g w n + 1 )=0= lim n d(gv,g z n + 1 ).
(2.23)

Using relations (2.22) and (2.23), together with the triangle inequality, we have d(gx,gu)=0 and d(gy,gv)=0 and so gx=gu and gy=gv.

Denote gx=p and gy=q. So, we have that

gp=g(gx)=gF(x,y)andgq=g(gy)=gF(y,x).
(2.24)

By definition of the sequences { x n } and { y n } we have

g x n =F(x,y)=F( x n 1 , y n 1 )andg y n =F(y,x)=F( y n 1 , x n 1 ),

and so

F( x n 1 , y n 1 )F(x,y)andg x n F(x,y),

as well as

F( y n 1 , x n 1 )F(y,x)andg y n F(y,x).

Compatibility of g and F implies that

d ( g F ( x n , y n ) , F ( g x n , g y n ) ) 0,n,

i.e., gF(x,y)=F(gx,gy). This together with (2.24) implies that gp=F(p,q) and, in a similar way, gq=F(q,p). Thus, we have another coincidence, and by the property we have just proved, it follows that gp=gx=p and gq=gy=q. In other words, p=gp=F(p,q) and q=gq=F(q,p), and (p,q) is a common coupled fixed point of g and F.

To prove the uniqueness, assume that (r,s) is another coupled common fixed point. Then by (2.24) we have r=gr=gp=p and s=gs=gq=q. Hence we get the result. □

Example 2.7 Let X=[0,1]. Then (X,) is a partially ordered set with the natural ordering of real numbers. Let d(x,y)=|xy| for all x,yX. Define a mapping g:XX by g(x)= x 2 and a mapping F:X×XX by

F(x,y)={ x 2 2 y 2 4 x y , 0 x < y .

Then it is easy to check all the conditions of Theorems 2.1 and 2.6. In particular, we will check that g and F are compatible.

Let { x n } and { y n } be two sequences in X such that

lim n g x n = lim n F( x n , y n )=aand lim n g y n = lim n F( y n , x n )=b.

Then a 2 b 4 =a and b 2 a 4 =b, where from it follows that a=b=0. Then

d ( g F ( x n , y n ) , F ( g x n , g y n ) ) = | ( x n 2 2 y n 2 4 ) 2 x n 4 2 y n 4 4 | 0(n),

and similarly d(gF( y n , x n ),F(g y n ,g x n ))0.

Now, we verify inequality (2.1) of Theorem 2.1 for ϕ(t)= 3 4 t, t>0 and L=[0,) for all x,y,u,vX with gxgu and gygv.

d ( F ( x , y ) , F ( u , v ) ) = | x 2 2 y 2 4 u 2 2 v 2 4 | 1 4 | x 2 u 2 | + 2 4 | y 2 v 2 | 3 4 max { | x 2 u 2 | , | y 2 v 2 | } ϕ ( max { d ( g x , g u ) , d ( g y , g v ) } ) + L min { d ( F ( x , y ) , g u ) , d ( F ( u , v ) , g x ) , d ( F ( x , y ) , g x ) , d ( F ( u , v ) , g u ) } .

Thus there exists a common coupled fixed point (0,0) of the mappings g and F. Note that F does not satisfy the g-mixed monotone property. Also, g and F do not commute.

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Correspondence to Sumit Chandok.

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Chandok, S., Tas, K. An original coupled coincidence point result for a pair of mappings without MMP. J Inequal Appl 2014, 61 (2014). https://doi.org/10.1186/1029-242X-2014-61

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Keywords

  • coupled fixed point
  • mixed monotone property
  • ordered metric spaces