1 Introduction

In this paper we prove certain coincidence point results in partially ordered metric spaces for functions which satisfy a certain inequality involving three control functions. Two of the control functions are discontinuous. Fixed point theory in partially ordered metric spaces is of relatively recent origin. An early result in this direction is due to Turinici [1], in which fixed point problems were studied in partially ordered uniform spaces. Later, this branch of fixed point theory has developed through a number of works, some of which are in [26].

Weak contraction was studied in partially ordered metric spaces by Harjani et al. [3]. In a recent result by Choudhury et al. [2], a generalization of the above result to a coincidence point theorem has been done using three control functions. Here we prove coincidence point results by assuming a weak contraction inequality with three control functions, two of which are not continuous. The results are obtained under two sets of additional conditions. A fixed point theorem is also established. There are several corollaries and two examples. One of the examples shows that the corollaries are properly contained in their respective theorem. The corollaries are generalizations of several existing works.

We apply our result to obtain some coupled coincidence point results. Coupled fixed theorems and coupled coincidence point theorems have appeared prominently in recent literature. Although the concept of coupled fixed points was introduced by Guo et al. [7], starting with the work of Gnana Bhaskar and Lakshmikantham [8], where they established a coupled contraction principle, this line of research has developed rapidly in partially ordered metric spaces. References [920] are some examples of these works. There is a viewpoint from which coupled fixed and coincidence point theorems can be considered as problems in product spaces [21]. We adopt this approach here. Specifically, we apply our theorem to a product of two metric spaces on which a metric is defined from the metric of the original spaces. We establish a generalization of several results. We also discuss an example which shows that our result is an actual improvement over the results it generalizes.

2 Mathematical preliminaries

Let (X,) be a partially ordered set and T:XX. The mapping T is said to be nondecreasing if for all x 1 , x 2 X, x 1 x 2 implies T x 1 T x 2 and nonincreasing if for all x 1 , x 2 X, x 1 x 2 implies T x 1 T x 2 .

Definition 2.1 ([22])

Let (X,) be a partially ordered set and T:XX and G:XX. The mapping T is said to be G-nondecreasing if for all x,yX, GxGy implies TxTy and G-nonincreasing if for all x,yX, GxGy implies TxTy.

Definition 2.2 Two self-mappings G and T of a nonempty set X are said to be commutative if GTx=TGx for all xX.

Definition 2.3 ([23])

Two self-mappings G and T of a metric space (X,d) are said to be compatible if the following relation holds:

lim n d(GT x n ,TG x n )=0,

whenever { x n } is a sequence in X such that lim n G x n = lim n T x n =x for some xX is satisfied.

Definition 2.4 ([24])

Two self-mappings G and T of a nonempty set X are said to be weakly compatible if they commute at their coincidence points; that is, if Gx=Tx for some xX, then GTx=TGx.

Definition 2.5 ([8])

Let (X,) be a partially ordered set and F:X×XX. The mapping F is said to have the mixed monotone property if F is monotone nondecreasing in its first argument and is monotone nonincreasing in its second argument; that is, if

x 1 , x 2 X, x 1 x 2 F( x 1 ,y)F( x 2 ,y)for all yX;

and

y 1 , y 2 X, y 1 y 2 F(x, y 1 )F(x, y 2 )for all xX.

Definition 2.6 ([17])

Let (X,) be a partially ordered set, F:X×XX and g:XX. We say that F has the mixed g-monotone property if

x 1 , x 2 X,g x 1 g x 2 F( x 1 ,y)F( x 2 ,y)for all yX;

and

y 1 , y 2 X,g y 1 g y 2 F(x, y 1 )F(x, y 2 )for all xX.

Definition 2.7 ([8])

An element (x,y)X×X is called a coupled fixed point of the mapping F:X×XX if F(x,y)=x and F(y,x)=y.

Definition 2.8 ([17])

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.

Definition 2.9 ([17])

Let X be a nonempty set. The mappings g and F, where g:XX and F:X×XX, are said to be commutative if gF(x,y)=F(gx,gy) for all x,yX.

Definition 2.10 ([12])

Let (X,d) be a metric space. The mappings g and F, where g:XX and F:X×XX, are said to be compatible if the following relations hold:

lim n d ( g F ( x n , y n ) , F ( g x n , g y n ) ) =0, 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 g x n = lim n F( x n , y n )=x and lim n g y n = lim n F( y n , x n )=y for some x,yX are satisfied.

Definition 2.11 Let X be a nonempty set. The mappings g and F, where g:XX and F:X×XX, are said to be weakly compatible if they commute at their coupled coincidence points, that is, if F(x,y)=gx and F(y,x)=gy for some (x,y)X×X, then gF(x,y)=F(gx,gy) and gF(y,x)=F(gy,gx).

Definition 2.12 ([25])

A function ψ:[0,)[0,) is called an altering distance function if the following properties are satisfied:

  1. (i)

    ψ is monotone increasing and continuous;

  2. (ii)

    ψ(t)=0 if and only if t=0.

In our results in the following sections, we use the following classes of functions.

We denote by Ψ the set of all functions ψ:[0,)[0,) satisfying

(i ψ ) ψ is continuous and monotone non-decreasing,

(ii ψ ) ψ(t)=0 if and only if t=0;

and by Θ we denote the set of all functions α:[0,)[0,) such that

(i α ) α is bounded on any bounded interval in [0,),

(ii α ) α is continuous at 0 and α(0)=0.

3 Main results

Let (X,,d) be an ordered metric space. X is called regular if it has the following properties:

  1. (i)

    if a nondecreasing sequence { x n }x in (X,d), then x n x for all n0;

  2. (ii)

    if a nonincreasing sequence { y n }y in (X,d), then y y n for all n0.

Theorem 3.1 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T,G:XX be two mappings such that G is continuous and nondecreasing, T(X)G(X), T is G-nondecreasing with respect toand the pair (G,T) is compatible. Suppose that there exist ψΨ and φ,θΘ such that

ψ(x)φ(y)xy,
(3.1)

for any sequence { x n } in [0,) with x n t>0,

(3.2)

and for all x,yX with GxGy

ψ ( d ( T x , T y ) ) φ ( d ( G x , G y ) ) θ ( d ( G x , G y ) ) .
(3.3)

Also, suppose that

  1. (a)

    T is continuous, or

  2. (b)

    X is regular.

If there exists x 0 X such that G x 0 T x 0 , then G and T have a coincidence point in X.

Proof Let x 0 X be such that G x 0 T x 0 . Since T(X)G(X), we can choose x 1 X such that G x 1 =T x 0 . Again, we can choose x 2 X such that G x 2 =T x 1 . Continuing this process, we construct a sequence { x n } in X such that

G x n + 1 =T x n for all n0.
(3.4)

Since G x 0 T x 0 and G x 1 =T x 0 , we have G x 0 G x 1 , which implies that T x 0 T x 1 . Now, T x 0 T x 1 , that is, G x 1 G x 2 implies that T x 1 T x 2 . Again, T x 1 T x 2 , that is, G x 2 G x 3 implies that T x 2 T x 3 . Continuing this process, we have

G x 0 G x 1 G x 2 G x 3 G x n G x n + 1 ,
(3.5)

and

T x 0 T x 1 T x 2 T x 3 T x n T x n + 1 .
(3.6)

Let R n =d(G x n + 1 ,G x n ) for all n0.

Since G x n + 1 G x n , from (3.3) and (3.4), we have

ψ ( d ( G x n + 2 , G x n + 1 ) ) =ψ ( d ( T x n + 1 , T x n ) ) φ ( d ( G x n + 1 , G x n ) ) θ ( d ( G x n + 1 , G x n ) ) ,

that is,

ψ( R n + 1 )φ( R n )θ( R n ),
(3.7)

which, in view of the fact that θ0, yields ψ( R n + 1 )φ( R n ), which by (3.1) implies that R n + 1 R n for all positive integer n, that is, { R n } is a monotone decreasing sequence. Hence there exists an r0 such that

R n =d(G x n + 1 ,G x n )ras n.
(3.8)

Taking limit supremum on both sides of (3.7), using (3.8), the property (i α ) of φ and θ, and the continuity of ψ, we obtain

Since , it follows that

that is,

which by (3.2) is a contradiction unless r=0. Therefore,

R n =d(G x n + 1 ,G x n )0as n.
(3.9)

Next we show that {G x n } is a Cauchy sequence.

Suppose that {G x n } is not a Cauchy sequence. Then there exists an ϵ>0 for which we can find two sequences of positive integers {m(k)} and {n(k)} such that for all positive integers k, n(k)>m(k)>k and d(G x m ( k ) ,G x n ( k ) )ϵ. Assuming that n(k) is the smallest such positive integer, we get

n ( k ) > m ( k ) > k , d ( G x m ( k ) , G x n ( k ) ) ϵ and d ( G x m ( k ) , G x n ( k ) 1 ) < ϵ .

Now,

ϵd(G x m ( k ) ,G x n ( k ) )d(G x m ( k ) ,G x n ( k ) 1 )+d(G x n ( k ) 1 ,G x n ( k ) ),

that is,

ϵd(G x m ( k ) ,G x n ( k ) )ϵ+d(G x n ( k ) 1 ,G x n ( k ) ).

Letting k in the above inequality and using (3.9), we have

lim k d(G x m ( k ) ,G x n ( k ) )=ϵ.
(3.10)

Again,

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

and

d(G x m ( k ) ,G x n ( k ) )d(G x m ( k ) + 1 ,G x m ( k ) )+d(G x m ( k ) + 1 ,G x n ( k ) + 1 )+d(G x n ( k ) + 1 ,G x n ( k ) ).

Letting k in the above inequalities, using (3.9) and (3.10), we have

lim k d(G x m ( k ) + 1 ,G x n ( k ) + 1 )=ϵ.
(3.11)

As n(k)>m(k), G x n ( k ) G x m ( k ) , from (3.3) and (3.4), we have

ψ ( d ( G x n ( k ) + 1 , G x m ( k ) + 1 ) ) = ψ ( d ( T x n ( k ) , T x m ( k ) ) ) φ ( d ( G x n ( k ) , G x m ( k ) ) ) θ ( d ( G x n ( k ) , G x m ( k ) ) ) .

Taking limit supremum on both sides of the above inequality, using (3.10), (3.11), the property (i α ) of φ and θ, and the continuity of ψ, we obtain

Since , it follows that

that is,

which is a contradiction by (3.2). Therefore, {G x n } is a Cauchy sequence in X. From the completeness of X, there exists xX such that

lim n T x n = lim n G x n =x.
(3.12)

Since the pair (G,T) is compatible, from (3.12), we have

lim n d(GT x n ,TG x n )=0.
(3.13)

Let the condition (a) hold.

By the triangular inequality, we have

d(Gx,TG x n )d(Gx,GT x n )+d(GT x n ,TG x n ).

Taking n in the above inequality, using (3.12), (3.13) and the continuities of T and G, we have d(Gx,Tx)=0, that is, Gx=Tx, that is, x is a coincidence point of the mappings G and T.

Next we suppose that the condition (b) holds.

By (3.5) and (3.12), we have G x n x for all n0. Using the monotone property of G, we obtain

GG x n Gx.
(3.14)

As G is continuous and the pair (G,T) is compatible, by (3.12) and (3.13), we have

lim n GG x n =Gx= lim n GT x n = lim n TG x n .
(3.15)

Then

d(Tx,Gx)= lim n d(Tx,TG x n ).

Since ψ is continuous, from the above inequality, we obtain

ψ ( d ( T x , G x ) ) =ψ ( lim n d ( T x , T G x n ) ) = lim n ψ ( d ( T x , T G x n ) ) ,

which, by (3.3) and (3.14), implies that

ψ ( d ( T x , G x ) ) lim n [ φ ( d ( G x , G G x n ) ) θ ( d ( G x , G G x n ) ) ] .

Using (3.15) and the property (ii α ) of φ and θ, we have

ψ ( d ( T x , G x ) ) =0,

which, by the property of ψ, implies that d(Tx,Gx)=0, that is, Gx=Tx, that is, x is a coincidence point of the mappings G and T. □

Next we discuss some corollaries of Theorem 3.1. By an example, we show that Theorem 3.1 properly contains all its corollaries.

Every commuting pair (G,T) is also a compatible pair. Then considering (G,T) to be the commuting pair in Theorem 3.1, we have the following corollary.

Corollary 3.1 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T,G:XX be two mappings such that G is continuous and nondecreasing, T(X)G(X), T is G-nondecreasing with respect toand the pair (G,T) is commutative. Suppose that there exist ψΨ and φ,θΘ such that (3.1), (3.2) and (3.3) are satisfied. Also, suppose that

  1. (a)

    T is continuous, or

  2. (b)

    X is regular.

If there exists x 0 X such that G x 0 T x 0 , then G and T have a coincidence point in X.

Considering ψ to be the identity mapping and θ(t)=0 for all t[0,) in Theorem 3.1, we have the following corollary.

Corollary 3.2 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T,G:XX be two mappings such that G is continuous and nondecreasing, T(X)G(X), T is G-nondecreasing with respect toand the pair (G,T) is compatible. Suppose that there exists φΘ such that for any sequence { x n } in [0,) with x n t>0,

(3.16)

and for all x,yX with GxGy,

d(Tx,Ty)φ ( d ( G x , G y ) ) .
(3.17)

Also, suppose that

  1. (a)

    T is continuous, or

  2. (b)

    X is regular.

If there exists x 0 X such that G x 0 T x 0 , then G and T have a coincidence point in X.

Considering θ(t)=0 for all t[0,) and φ(t)=kψ(t) with k[0,1) in Theorem 3.1, we have the following corollary.

Corollary 3.3 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T,G:XX be two mappings such that G is continuous and nondecreasing, T(X)G(X), T is G-nondecreasing with respect toand the pair (G,T) is compatible. Suppose that there exists ψΨ and k[0,1) such that for all x,yX with GxGy,

ψ ( d ( T x , T y ) ) kψ ( d ( G x , G y ) ) .
(3.18)

Also, suppose that

  1. (a)

    T is continuous, or

  2. (b)

    X is regular.

If there exists x 0 X such that G x 0 T x 0 , then G and T have a coincidence point in X.

Considering φ to be the function ψ in Theorem 3.1, we have the following corollary.

Corollary 3.4 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T,G:XX be two mappings such that G is continuous and nondecreasing, T(X)G(X), T is G-nondecreasing with respect toand the pair (G,T) is compatible. Suppose that there exist ψΨ and θΘ such that for any sequence { x n } in [0,) with x n t>0,

(3.19)

and for all x,yX with GxGy,

ψ ( d ( T x , T y ) ) ψ ( d ( G x , G y ) ) θ ( d ( G x , G y ) ) .
(3.20)

Also, suppose that

  1. (a)

    T is continuous or

  2. (b)

    X is regular.

If there exists x 0 X such that G x 0 T x 0 , then G and T have a coincidence point in X.

If ψ and φ are the identity mappings in Theorem 3.1, we have the following corollary.

Corollary 3.5 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T,G:XX be two mappings such that G is continuous and nondecreasing, T(X)G(X), T is G-nondecreasing with respect toand the pair (G,T) is compatible. Suppose that there exists θΘ such that for any sequence { x n } in [0,) with x n t>0, and for all x,yX with GxGy,

d(Tx,Ty)d(Gx,Gy)θ ( d ( G x , G y ) ) .
(3.21)

Also, suppose that

  1. (a)

    T is continuous, or

  2. (b)

    X is regular.

If there exists x 0 X such that G x 0 T x 0 , then G and T have a coincidence point in X.

Considering ψ and φ to be the identity mappings and θ(t)=(1k)t, where 0k<1 in Theorem 3.1, we have the following corollary.

Corollary 3.6 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T,G:XX be two mappings such that G is continuous and nondecreasing, T(X)G(X), T is G-nondecreasing with respect toand the pair (G,T) is compatible. Assume that there exists k[0,1) such that for all x,yX with GxGy,

d(Tx,Ty)kd(Gx,Gy).
(3.22)

Also, suppose that

  1. (a)

    T is continuous, or

  2. (b)

    X is regular.

If there exists x 0 X such that G x 0 T x 0 , then G and T have a coincidence point in X.

The condition (i), the continuity and the monotone property of the function G, and (ii), the compatibility condition of the pairs (G,T), which were considered in Theorem 3.1, are relaxed in our next theorem by taking G(X) to be closed in (X,d).

Theorem 3.2 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T,G:XX be two mappings such that T(X)G(X) and T is G-nondecreasing with respect toand G(X) is closed in X. Suppose that there exist ψΨ and φ,θΘ such that (3.1), (3.2) and (3.3) are satisfied. Also, suppose that X is regular.

If there exists x 0 X such that G x 0 T x 0 , then G and T have a coincidence point in X.

Proof We take the same sequence { x n } as in the proof of Theorem 3.1. Then we have (3.12), that is,

lim n T x n = lim n G x n =x.

Since {G x n } is a sequence in G(X) and G(X) is closed in X, xG(X). As xG(X), there exists zX such that x=Gz. Then

lim n T x n = lim n G x n =Gz.
(3.23)

Now, {G x n } is nondecreasing and converges to Gz. So, by the order condition of the metric space X, we have

G x n Gz.
(3.24)

Putting x=z and y= x n in (3.3), by the virtue of (3.24), we get

ψ ( d ( T z , T x n ) ) φ ( d ( G z , G x n ) ) θ ( d ( G z , G x n ) ) .

Taking n in the above inequality, using (3.23), the property (ii α ) of φ and θ and the continuity of ψ, we have

ψ ( d ( T z , G z ) ) =0,

which, by the property of ψ, implies that d(Tz,Gz)=0, that is, Tz=Gz, that is, z is a coincidence point of the mappings G and T. □

In the following, our aim is to prove the existence and uniqueness of the common fixed point in Theorems 3.1 and 3.2.

Theorem 3.3 In addition to the hypotheses of Theorems  3.1 and 3.2, in both of the theorems, suppose that for every x,yX there exists uX such that Tu is comparable to Tx and Ty, and also the pair (G,T) is weakly compatible. Then G and T have a unique common fixed point.

Proof From Theorem 3.1 or Theorem 3.2, the set of coincidence points of G and T is non-empty. Suppose x and y are coincidence points of G and T, that is, Gx=Tx and Gy=Ty. Now, we show

Gx=Gy.
(3.25)

By the assumption, there exists uX such that Tu is comparable with Tx and Ty.

Put u 0 =u and choose u 1 X so that G u 1 =T u 0 . Then, similarly to the proof of Theorem 3.1, we can inductively define sequences {G u n } where G u n + 1 =T u n for all n0. Hence Tx=Gx and Tu=T u 0 =G u 1 are comparable.

Suppose that G u 1 Gx (the proof is similar to that in the other case).

We claim that G u n Gx for each nN.

In fact, we will use mathematical induction. Since G u 1 Gx, our claim is true for n=1. We presume that G u n Gx holds for some n>1. Since T is G-nondecreasing with respect to ⪯, we get

G u n + 1 =T u n Tx=Gx;

and this proves our claim.

Let R n =d(Gx,G u n ). Since G u n Gx, using the contractive condition (3.3), for all n1, we have

ψ ( d ( G x , G u n + 1 ) ) =ψ ( d ( T x , T u n ) ) φ ( d ( G x , G u n ) ) θ ( d ( G x , G u n ) ) ,

that is, ψ( R n + 1 )φ( R n )θ( R n ), which, in view of the fact that θ0, yields ψ( R n + 1 )φ( R n ), which by (3.1) implies that R n + 1 R n for all positive integer n, that is, { R n } is a monotone decreasing sequence.

Then, as in the proof of Theorem 3.1, we have

lim n R n = lim n d(Gx,G u n )=0.
(3.26)

Similarly, we show that

lim n d(Gy,G u n )=0.
(3.27)

By the triangle inequality, using (3.26) and (3.27), we have

d(Gx,Gy) [ d ( G x , G u n ) + d ( G u n , G y ) ] 0as n.

Hence Gx=Gy. Thus (3.25) is proved.

Since Gx=Tx, by weak compatibility of G and T, we have

GGx=GTx=TGx.
(3.28)

Denote

Gx=z.
(3.29)

Then from (3.28) we have

Gz=Tz.

Thus z is a coincidence point of G and T. Then from (3.25) with y=z it follows that

Gx=Gz.

By (3.29) it follows that

z=Gz.
(3.30)

From (3.29) and (3.30), we get z=Gz=Tz.

Therefore, z is a common fixed point of G and T.

To prove the uniqueness, assume that r is another common fixed point of G and T. Then by (3.25) we have r=Gr=Gz=z. Hence the common fixed point of G and T is unique. □

Example 3.1 Let X=[0,). Then (X,) is a partially ordered set with the natural ordering of real numbers. Let d(x,y)=|xy| for x,yX. Then (X,d) is a complete metric space.

Let T,G:XX be given respectively by the formulas Tx= 1 3 x 2 and Gx= x 2 for all xX. Then T and G satisfy all the properties mentioned in Theorem 3.1.

Let ψ,φ,θ:[0,)[0,) be given respectively by the formulas

ψ(t)= t 2 ,φ(t)= { 1 2 [ t ] 2 if  3 < t < 4 , 4 9 t 2 otherwise , θ(t)= { 1 9 [ t ] 2 if  3 < t < 4 , 0 otherwise .

Then ψ, φ and θ have the properties mentioned in Theorem 3.1.

It can be verified that (3.3) is satisfied for all x,yX with GxGy. Hence the required conditions of Theorem 3.1 are satisfied and it is seen that 0 is a coincidence point of G and T. Also, the conditions of Theorem 3.3 are satisfied and it is seen that 0 is the unique common fixed point of G and T.

Remark 3.1 In the above example, the pair (G,T) is compatible but not commuting so that Corollary 3.1 is not applicable to this example and hence Theorem 3.1 properly contains its Corollary 3.1.

Remark 3.2 In the above example, ψ is not the identity mapping and θ(t)0 for all t in [0,). Let us consider the sequence { x n } in [0,), where x n =3 for all n. Then φ( x n )=4 for all n. Now x n t=3>0, but . Therefore, Corollary 3.2 is not applicable to this example, and hence Theorem 3.1 properly contains its Corollary 3.2.

Remark 3.3 The above example θ(t)0 for all t[0,), and hence Corollary 3.3 is not applicable to the example, and so Theorem 3.1 properly contains its Corollary 3.3.

Remark 3.4 In the above example, φ is not identical to the function ψ, and also for any sequence { x n } in [0,) with x n t>0. Therefore, Corollaries 3.4 and 3.5 are not applicable to this example, and hence Theorem 3.1 properly contains its Corollaries 3.4 and 3.5.

Remark 3.5 In the above example, ψ and φ are not the identity functions and θ(t)(1k)t with 0k<1. Therefore, Corollary 3.6 is not applicable to the above example. Hence Theorem 3.1 properly contains its Corollary 3.6.

Remark 3.6 Theorem 3.1 generalizes the results in [24, 6, 2529].

Example 3.2 Let X=[0,1]. Then (X,) is a partially ordered set with the natural ordering of real numbers. Let d(x,y)=|xy| for x,yX. Then (X,d) is a metric space with the required properties of Theorem 3.2.

Let T,G:XX be given respectively by the formulas

Tx=1for all xX,Gx= { 1 if  x  is rational , 1 2 if  x  is irrational .

Then T and G have the properties mentioned in Theorem 3.2.

Let ψ,φ,θ:[0,)[0,) be given respectively by the formulas

ψ(t)= t 2 ,φ(t)= { 1 2 [ t ] 2 if  3 < t < 4 , 4 9 t 2 otherwise , θ(t)= { 1 9 [ t ] 2 if  3 < t < 4 , 0 otherwise .

Then ψ, φ and θ have the properties mentioned in Theorem 3.2.

All the required conditions of Theorem 3.2 are satisfied. It is seen that every rational number xX is a coincidence point of G and T. Also, the conditions of Theorem 3.3 are satisfied and it is seen that 1 is the unique common fixed point of G and T.

Remark 3.7 In the above example, the function g is not continuous. Therefore, Theorem 3.1 is not applicable to the above example.

4 Applications to coupled coincidence point results

In this section, we use the results of the previous section to establish new coupled coincidence point results in partially ordered metric spaces. Our results are extensions of some existing results.

Let (X,) be a partially ordered set. Now, we endow the product space X×X with the following partial order:

for (x,y),(u,v)X×X,(u,v)(x,y)xu,yv.

Let (X,d) be a metric space. Then d 1 given by the law

d 1 ( ( x , y ) , ( u , v ) ) =d(x,u)+d(y,v)for (x,y),(u,v)X×X

is a metric on X×X.

Let g:XX and F:X×XX be two mappings. Then we define two functions G:X×XX×X and T:X×XX×X respectively as follows:

G(x,y)=(gx,gy)for x,yX,T(x,y)= ( F ( x , y ) , F ( y , x ) ) for x,yX.

Lemma 4.1 Let (X,) be a partially ordered set, F:X×XX and g:XX. If F has the mixed g-monotone property, then T is G-nondecreasing.

Proof Let ( x 1 , y 1 ),( x 2 , y 2 )X×X such that G( x 1 , y 1 )G( x 2 , y 2 ). Then, by the definition of G, it follows that (g x 1 ,g y 1 )(g x 2 ,g y 2 ), that is, g x 1 g x 2 and g y 1 g y 2 . Since F has the mixed g-monotone property, we have

F( x 1 , y 1 )F( x 2 , y 1 )F( x 2 , y 2 ),

and

F( y 1 , x 1 )F( y 2 , x 1 )F( y 2 , x 2 ).

It follows that (F( x 1 , y 1 ),F( y 1 , x 1 ))(F( x 2 , y 2 ),F( y 2 , x 2 )), that is, T( x 1 , y 1 )T( x 2 , y 2 ). Therefore, T is G-nondecreasing. □

Lemma 4.2 Let X be a nonempty set, g:XX and F:X×XX. If g and F are commutative, then the mappings G and T are also commutative.

Proof Let (x,y)X×X. Since g and F are commutative, by the definition of G and T, we have

G T ( x , y ) = G ( F ( x , y ) , F ( y , x ) ) = ( g F ( x , y ) , g F ( y , x ) ) = ( F ( g x , g y ) , F ( g y , g x ) ) = T ( g x , g y ) = T G ( x , y ) ,

which shows that G and T are commutative. □

Lemma 4.3 Let (X,d) be metric space and g:XX, F:X×XX. If g and F are compatible, then the mappings G and T are also compatible.

Proof Let {( x n , y n )} be a sequence in X×X such that lim n G( x n , y n )= lim n T( x n , y n )=(x,y) for some (x,y)X×X. By the definition of G and T, we have lim n (g x n ,g y n )= lim n (F( x n , y n ),F( y n , y n ))=(x,y), which implies that

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

Now

d 1 ( G T ( x n , y n ) , T G ( x n , y n ) ) = d 1 ( G ( F ( x n , y n ) , F ( y n , x n ) ) , T ( g x n , g y n ) ) = d 1 ( ( g F ( x n , y n ) , g F ( y n , x n ) ) , ( F ( g x n , g y n ) , F ( g y n , g x n ) ) ) = d ( g F ( x n , y n ) , F ( g x n , g y n ) ) + d ( g F ( y n , x n ) , F ( g y n , g x n ) ) .

Since g and F are compatible, we have

lim n d 1 ( G T ( x n , y n ) , T G ( x n , y n ) ) =0.

It follows that G and T are compatible. □

Lemma 4.4 Let X be a nonempty set, g:XX and F:X×XX. If g and F are weak compatible, then the mappings G and T are also weak compatible.

Proof Let (x,y)X×X be a coincidence point G and T. Then G(x,y)=T(x,y), that is, (gx,gy)=(F(x,y),F(y,x)), that is, gx=F(x,y) and gy=F(y,x). Since g and F are weak compatible, by the definition of G and T, we have

G T ( x , y ) = G ( F ( x , y ) , F ( y , x ) ) = ( g F ( x , y ) , g F ( y , x ) ) = ( F ( g x , g y ) , F ( g y , g x ) ) = T ( g x , g y ) = T G ( x , y ) ,

which shows that G and T commute at their coincidence point, that is, G and T are weak compatible. □

Theorem 4.1 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let F:X×XX and g:XX be two mappings such that g is continuous and nondecreasing, F(X×X)g(X), F has the mixed g-monotone property on X and the pair (g,F) is compatible. Suppose that there exist ψΨ and φ,θΘ such that (3.1) and (3.2) are satisfied and for all x,y,u,vX with gxgu and gygv,

ψ ( d ( F ( x , y ) , F ( u , v ) ) + d ( F ( y , x ) , F ( v , u ) ) ) φ ( d ( g x , g u ) + d ( g y , g v ) ) θ ( d ( g x , g u ) + d ( g y , g v ) ) .
(4.1)

Also, suppose that

  1. (a)

    F is continuous, or

  2. (b)

    X is regular.

If 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 ), then there exist x,yX such that gx=F(x,y) and gy=F(y,x); that is, g and F have a coupled coincidence point in X.

Proof We consider the product space (X×X,), the metric d 1 on X×X and the functions G:X×XX×X and T:X×XX×X as mentioned above. Denote S=X×X. Then (S, d 1 ) is a complete metric space. By the definition of G and T, we have that

  1. (i)

    G is continuous and nondecreasing; and T is continuous,

  2. (ii)

    T(S)G(S),

  3. (iii)

    T is G-nondecreasing with respect to ⪯,

  4. (iv)

    the pair (G,T) is compatible.

Let p=(x,y),q=(u,v)S=X×X such that gxgu and gygv, that is, (gx,gy)(gu,gv), that is, GpGq. Then (4.1) reduces to

ψ ( d 1 ( T p , T q ) ) φ ( d 1 ( G p , G q ) ) θ ( d 1 ( G p , G q ) ) .

Now, the existence of x 0 , y 0 X such that g x 0 F( x 0 , y 0 ) and g y 0 F( y 0 , x 0 ) implies the existence of a point p 0 =( x 0 , y 0 )S such that (g x 0 ,g y 0 )(F( x 0 , y 0 ),F( y 0 , x 0 )), that is, G p 0 T p 0 . Therefore, the theorem reduces to Theorem 3.1, and hence there exists w=(x,y)S=X×X such that Gw=Tw, that is, G(x,y)=T(x,y), that is, (gx,gy)=(F(x,y),F(y,x)), that is, gx=F(x,y) and gy=F(y,x), that is, (x,y) is a coupled coincidence point of g and F. □

The following corollary is a consequence of Corollary 3.1.

Corollary 4.1 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let F:X×XX and g:XX be two mappings such that g is continuous and nondecreasing, F(X×X)g(X), F has the mixed g-monotone property on X and the pair (g,F) is commutative. Suppose that there exist ψΨ, φ,θΘ such that (3.1), (3.2) and (4.1) are satisfied. Also, suppose that

  1. (a)

    F is continuous, or

  2. (b)

    X is regular.

If 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 ), then there exist x,yX such that gx=F(x,y) and gy=F(y,x); that is, g and F have a coupled coincidence point in X.

The following corollary is a consequence of Corollary 3.2.

Corollary 4.2 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let F:X×XX and g:XX be two mappings such that g is continuous and nondecreasing, F(X×X)g(X), F has the mixed g-monotone property on X and the pair (g,F) is compatible. Suppose that there exists φΘ such that for any sequence { x n } in [0,) with x n t>0,

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

d ( F ( x , y ) , F ( u , v ) ) +d ( F ( y , x ) , F ( v , u ) ) φ ( d ( g x , g u ) + d ( g y , g v ) ) .

Also, suppose that

  1. (a)

    F is continuous, or

  2. (b)

    X is regular.

If 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 ), then there exist x,yX such that gx=F(x,y) and gy=F(y,x); that is, g and F have a coupled coincidence point in X.

The following corollary is a consequence of Corollary 3.3.

Corollary 4.3 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let F:X×XX and g:XX be two mappings such that g is continuous and nondecreasing, F(X×X)g(X), F has the mixed g-monotone property on X and the pair (g,F) is compatible. Suppose that there exist ψΨ and k[0,1) such that for all x,y,u,vX with gxgu and gygv,

ψ ( d ( F ( x , y ) , F ( u , v ) ) + d ( F ( y , x ) , F ( v , u ) ) ) kψ ( d ( g x , g u ) + d ( g y , g v ) ) .

Also, suppose that

  1. (a)

    F is continuous, or

  2. (b)

    X is regular.

If 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 ), then there exist x,yX such that gx=F(x,y) and gy=F(y,x); that is, g and F have a coupled coincidence point in X.

The following corollary is a consequence of Corollary 3.4.

Corollary 4.4 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let F:X×XX and g:XX be two mappings such that g is continuous and nondecreasing, F(X×X)g(X), F has the mixed g-monotone property on X and the pair (g,F) is compatible. Suppose that there exist ψΨ and θΘ such that for any sequence { x n } in [0,) with x n t>0, and for all x,y,u,vX with gxgu and gygv,

ψ ( d ( F ( x , y ) , F ( u , v ) ) + d ( F ( y , x ) , F ( v , u ) ) ) ψ ( d ( g x , g u ) + d ( g y , g v ) ) θ ( d ( g x , g u ) + d ( g y , g v ) ) .
(4.2)

Also, suppose that

  1. (a)

    F is continuous, or

  2. (b)

    X is regular.

If 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 ), then there exist x,yX such that gx=F(x,y) and gy=F(y,x); that is, g and F have a coupled coincidence point in X.

Remark 4.1 The above result is also true if the arguments of ψ and θ in (4.2) are replaced by their half values, that is, when (4.2) is replaced by

ψ ( d ( F ( x , y ) , F ( u , v ) ) + d ( F ( y , x ) , F ( v , u ) ) 2 ) ψ ( d ( g x , g u ) + d ( g y , g v ) 2 ) θ ( d ( g x , g u ) + d ( g y , g v ) 2 ) .

In this case, we can write ψ 1 (t)=ψ( t 2 ) and θ 1 (t)=θ( t 2 ) and proceed with the same proof by replacing ψ, θ by ψ 1 , θ 1 respectively. Then we obtain a generalization of Theorem 2 of Berinde in [10].

The following corollary is a consequence of Corollary 3.5.

Corollary 4.5 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let F:X×XX and g:XX be two mappings such that g is continuous and nondecreasing, F(X×X)g(X), F has the mixed g-monotone property on X and the pair (g,F) is compatible. Suppose that there exists θΘ such that for any sequence { x n } in [0,) with x n t>0, and for all x,y,u,vX with gxgu and gygv,

d ( F ( x , y ) , F ( u , v ) ) +d ( F ( y , x ) , F ( v , u ) ) d(gx,gu)+d(gy,gv)θ ( d ( g x , g u ) + d ( g y , g v ) ) .

Also, suppose that

  1. (a)

    F is continuous, or

  2. (b)

    X is regular.

If 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 ), then there exist x,yX such that gx=F(x,y) and gy=F(y,x); that is, g and F have a coupled coincidence point in X.

The following corollary is a consequence of Corollary 3.6.

Corollary 4.6 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let F:X×XX and g:XX be two mappings such that g is continuous and nondecreasing, F(X×X)g(X), F has the mixed g-monotone property on X and the pair (g,F) is compatible. Assume that there exists k[0,1) such that for all x,y,u,vX with gxgu,gygv,

d ( F ( x , y ) , F ( u , v ) ) +d ( F ( y , x ) , F ( v , u ) ) k [ d ( g x , g u ) + d ( g y , g v ) ] .

Also, suppose that

  1. (a)

    F is continuous, or

  2. (b)

    X is regular.

If 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 ), then there exist x,yX such that gx=F(x,y) and gy=F(y,x); that is, g and F have a coupled coincidence point in X.

The following theorem is a consequence of Theorem 3.2.

Theorem 4.2 Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Consider the mappings F:X×XX and g:XX such that F(X×X)g(X), F has the mixed g-monotone property on X and G(X) is closed in X. Suppose that there exist ψΨ and φ,θΘ such that (3.1), (3.2) and (4.1) are satisfied. Also, suppose that X is regular.

If 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 ), then there exist x,yX such that gx=F(x,y) and gy=F(y,x); that is, g and F have a coupled coincidence point in X.

The following theorem is a consequence of Theorem 3.3.

Theorem 4.3 In addition to the hypotheses of Theorems  4.1 and 4.2, in both of the theorems, suppose that for every (x,y),( x , y )X×X, there exists a (u,v)X×X such that (F(u,v),F(v,u)) is comparable to (F(x,y),F(y,x)) and (F( x , y ),F( y , x )), and also the pair (g,F) is weakly compatible. Then g and F have a unique coupled common fixed point; that is, there exists a unique (x,y)X×X such that x=gx=F(x,y) and y=gy=F(y,x).

Example 4.1 Let X=[0,). Then (X,) is a partially ordered set with the natural ordering of real numbers. Let d(x,y)=|xy| for x,yX. Then (X,d) is a complete metric space.

Let g:XX be given by gx= x 2 for all xX. Also, consider

F:X×XX,F(x,y)= { 1 3 ( x 2 y 2 ) if  x , y X , x y , 0 if  x y ,

which obeys the mixed g-monotone property.

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

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

Then, obviously, a=0 and b=0.

Now, for all n0, g x n = x n 2 , g y n = y n 2 , while

F( x n , y n )= { 1 3 ( x n 2 y n 2 ) if  x n y n , 0 if  x n y n , F( y n , x n )= { 1 3 ( y n 2 x n 2 ) if  y n x n , 0 if  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, lim n d ( g F ( y n , x n ) , F ( g y n , g x n ) ) =0.

Hence, the pair (g,F) is compatible in X.

Let x 0 =0 and y 0 =c(>0) be two points in X. Then

g( x 0 )=g(0)=0=F(0,c)=F( x 0 , y 0 ),g( y 0 )=g(c)= c 2 c 2 3 =F(c,0)=F( y 0 , x 0 ).

Let ψ,φ,θ:[0,)[0,) be given respectively by the formulas

ψ(t)= t 2 ,φ(t)= { 1 2 [ t ] 2 if  3 < t < 4 , 4 9 t 2 otherwise , θ(t)= { 1 9 [ t ] 2 if  3 < t < 4 , 0 otherwise .

Then ψ, φ and θ have the properties mentioned in Theorem 4.1.

We now verify inequality (4.1) of Theorem 4.1.

We take x,y,u,vX such that gxgu and gygv, that is, x 2 u 2 and y 2 v 2 .

Let M=d(gx,gu)+d(gy,gv)=| x 2 u 2 |+| y 2 v 2 |=( x 2 u 2 )+( v 2 y 2 ).

The following are the four possible cases.

Case-1: xy and uv. Then

d ( F ( x , y ) , F ( u , v ) ) + d ( F ( y , x ) , F ( v , u ) ) = d ( x 2 y 2 3 , u 2 v 2 3 ) + d ( 0 , 0 ) = | ( x 2 y 2 ) ( u 2 v 2 ) 3 | = | ( x 2 u 2 ) + ( v 2 y 2 ) 3 | = ( x 2 u 2 ) + ( v 2 y 2 ) 3 = 1 3 M .

Case-2: x<y and u<v. Then

d ( F ( x , y ) , F ( u , v ) ) + d ( F ( y , x ) , F ( v , u ) ) = d ( 0 , 0 ) + d ( y 2 x 2 3 , v 2 u 2 3 ) = | ( y 2 x 2 ) ( v 2 u 2 ) 3 | = | ( x 2 u 2 ) + ( v 2 y 2 ) 3 | = ( x 2 u 2 ) + ( v 2 y 2 ) 3 = 1 3 M .

Case-3: xy and uv. Then

d ( F ( x , y ) , F ( u , v ) ) + d ( F ( y , x ) , F ( v , u ) ) = d ( x 2 y 2 3 , 0 ) + d ( 0 , v 2 u 2 3 ) = x 2 y 2 3 + v 2 u 2 3 = ( x 2 u 2 ) + ( v 2 y 2 ) 3 = 1 3 M .

Case-4: The case ‘x<y and u>v’ is not possible. Under this condition, x 2 < y 2 and u 2 > v 2 . Then by the condition y 2 v 2 , we have x 2 < y 2 v 2 < u 2 , which contradicts that x 2 u 2 .

In all the above cases, it can be verified that (4.1) is satisfied. Hence the required conditions of Theorem 4.1 are satisfied, and it is seen that (0,0) is a coupled coincidence point of g and F in X. Also, the conditions of Theorem 4.3 are satisfied, and it is seen that (0,0) is the unique coupled common fixed point of g and F in X.

Remark 4.2 In the above example, the pair (g,F) is compatible but not commuting so that Corollary 4.1 is not applicable to this example, and hence Theorem 4.1 properly contains its Corollary 4.1.

Remark 4.3 As discussed in Remarks 3.2-3.5, Theorem 4.1 properly contains its Corollaries 4.2-4.6.

Remark 4.4 Theorem 4.1 properly contains its Corollary 4.6, which is an extension of Theorem 3 of Berinde [9], and Theorems 2.1 and 2.2 of Bhaskar and Lakshmikantham [8]. Therefore, Theorem 4.1 is an actual extension over Theorem 3 of Berinde [9] and Theorems 2.1 and 2.2 of Bhaskar and Lakshmikantham [8].

Example 4.2 Let X=[0,1]. Then (X,) is a partially ordered set with the natural ordering of real numbers. Let d(x,y)=|xy| for x,yX. Then (X,d) is a metric space with the required properties of Theorem 4.2.

Let F:X×XX, F(x,y)=1 for (x,y)X×X. Let g:XX be given by the formula

gx= { 1 if  x  is rational , 1 2 if  x  is irrational .

Then F and g have the properties mentioned in Theorems 4.2.

Let ψ,φ,θ:[0,)[0,) be given respectively by the formulas

ψ(t)= t 2 ,φ(t)= { 1 2 [ t ] 2 if  3 < t < 4 , 4 9 t 2 otherwise , θ(t)= { 1 9 [ t ] 2 if  3 < t < 4 , 0 otherwise .

Then ψ, φ and θ have the properties mentioned in Theorem 4.2.

All the required conditions of Theorem 4.2 are satisfied. It is seen that every (x,y)X×X, where both x and y are rational, is a coupled coincidence point of g and F in X. Also, the conditions of Theorem 4.3 are satisfied and it is seen that (1,1) is the unique coupled common fixed point of g and F in X.

Remark 4.5 In the above example, the function g is not continuous. Therefore, Theorem 4.1 is not applicable to the above example.

Remark 4.6 In some recent papers [30, 31] it has been proved that some of the contractive conditions involving continuous control functions are equivalent. Here two of our control functions are discontinuous. Therefore, the contraction we use here is not included in the class of contractions addressed by Aydi et al. [30] and Jachymski [31].