1 Introduction and preliminaries

Recently, Huang and Zhang [1] generalized the concept of a metric space, replacing the set of real numbers by ordered Banach space and obtained some fixed point theorems for mappings satisfying different contractive conditions. Subsequently, the study of fixed point theorems in such spaces is followed by some other mathematicians; see [224]. The study of fixed point theorems for non-self-mappings in metrically convex metric spaces was initiated by Assad and Kirk [25]. Utilizing the induction method of Assad and Kirk [25], many authors like Assad [26], Ćirić [2736], Ćirić et al. [3742], Kumam et al. [43], Hadžić [44], Hadžić and Gajić [45], Imdad and Kumar [46], Rhoades [47, 48] have obtained common fixed point in metrically convex spaces. Recently, Ćirić and Ume [49] defined a wide class of multi-valued non-self-mappings which satisfy a generalized contraction condition and proved a fixed point theorem which generalizes the results of Itoh [50] and Khan [51].

Very recently, Radenović and Rhoades [15] extended the fixed point theorem of Imdad and Kumar [46] for a pair of non-self-mappings to non-normal cone metric spaces. Huang et al. [6] proved a fixed point theorem for four non-self-mappings in cone metric spaces which generalizes the result of Radenović and Rhoades [15]. Janković et al. [9] proved new common fixed point results for a pair of non-self-mappings defined on a closed subset of metrically convex cone metric space which is not necessarily normal by adapting Assad-Kirk’s method. Sumitra et al. [20] generalized the fixed point theorems of Ćirić and Ume [49] for a pair of non-self-mappings to non-normal cone metric spaces. In the same time, Sumitra et al.’s [20] results extended the results of Radenović and Rhoades [15] and Janković et al. [9]. The aim of this paper is to prove a common fixed point theorem for two pairs of non-self-mappings on cone metric spaces in which the cone need not be normal and the condition is weaker. This result generalizes the result of Sumitra et al. [20] and Huang et al. [6].

Consistent with Huang and Zhang [1], the following definitions and results will be needed in the sequel.

Let E be a real Banach space. A subset P of E is called a cone if and only if:

  1. (a)

    P is closed, nonempty and P{θ};

  2. (b)

    a,bR, a,b0, x,yP implies ax+byP;

  3. (c)

    P(P)={θ}.

Given a cone PE, we define a partial ordering ⪯ with respect to P by xy if and only if yxP. A cone P is called normal if there is a number K>0 such that for all x,yE,

θxyimpliesxKy.

The least positive number satisfying the above inequality is called the normal constant of P, while xy stands for yxintP (interior of P).

Definition 1.1 [1]

Let X be a nonempty set. Suppose that the mapping d:X×XE satisfies:

(d1) θd(x,y) for all x,yX and d(x,y)=θ if and only if x=y;

(d2) d(x,y)=d(y,x) for all x,yX;

(d3) d(x,y)d(x,z)+d(z,y) for all x,y,zX.

Then d is called a cone metric on X and (X,d) is called a cone metric space.

The concept of a cone metric space is more general than that of a metric space.

Example 1.1 [1]

Let E= R 2 , P={(x,y)Ex,y0}, X=R, and d:X×XE be such that d(x,y)=(|xy|,α|xy|), where α0 is a constant. Then (X,d) is a cone metric space.

Definition 1.2 [1]

Let (X,d) be a cone metric space. We say that { x n } is:

  1. (e)

    a Cauchy sequence if for every cE with θc, there is an N such that for all n,m>N, d( x n , x m )c;

  2. (f)

    a convergent sequence if for every cE with θc, there is an N such that for all n>N, d( x n ,x)c for some fixed xX.

A cone metric space X is said to be complete if for every Cauchy sequence in X is convergent in X. It is well known that { x n } converges to xX if and only if d( x n ,x)θ as n. It is a Cauchy sequence if and only if d( x n , x m )θ (n,m).

Remark 1.1 [52]

Let E be an ordered Banach (normed) space. Then c is an interior point of P, if and only if [c,c] is a neighborhood of θ.

Corollary 1.1 [16]

  1. (1)

    If ab and bc, then ac.

    Indeed, ca=(cb)+(ba)cb implies [(ca),ca][(cb),cb].

  2. (2)

    If ab and bc, then ac.

Indeed, ca=(cb)+(ba)cb implies [(ca),ca][(cb),cb].

  1. (3)

    If θuc for each cintP then u=θ.

Remark 1.2 [10, 15]

If cintP, θ a n , and a n θ, then there exists an n 0 such that for all n> n 0 we have a n c.

Remark 1.3 [16, 17]

If E is a real Banach space with cone P and if aka where aP and 0<k<1, then a=θ.

Definition 1.3 [2]

Let f and g be self-maps on a set X (i.e., f,g:XX). If w=fx=gx for some x in X, then x is called a coincidence point of f and g, and w is called a point of coincidence of f and g. Self-maps f and g are said to be weakly compatible if they commute at their coincidence point; i.e., if fx=gx for some xX, then fgx=gfx.

2 Main results

The following theorem is Sumitra et al.’s [20] generalization of Ćirić and Ume’s [49] result in cone metric spaces.

Theorem 2.1 Let (X,d) be a complete cone metric space, C a nonempty closed subset of X such that for each xC and yC there exists a point zC (the boundary of C) such that

d(x,z)+d(z,y)=d(x,y).

Suppose that f,g:CX are two non-self-mappings satisfying, for all x,yC with xy,

d(gx,gy)αd(fx,fy)+βu+γv,
(2.1)

where u{d(fx,gx),d(fy,gy)}, v{d(fx,gx)+d(fy,gy),d(fx,gy)+d(fy,gx)}, and α, β, γ are nonnegative real numbers such that

α+2β+3γ+αγ<1.
(2.2)

Also assume that

  1. (i)

    CfC, gCCfC,

  2. (ii)

    fxC implies that gxC,

  3. (iii)

    fC is closed in X.

Then the pair (f,g) has a coincidence point in C. Moreover, if the pair (f,g) is weakly compatible, then f and g have a unique common fixed point in C.

Remark 2.1 From the proof of this theorem, it is easy to see that condition (2.2) can be weakened to α+β+2γ<1.

The purpose of this paper is to extend the above theorem for two pairs of non-self-mappings in cone metric spaces with a weaker condition.

We state and prove our main result as follows.

Theorem 2.2 Let (X,d) be a complete cone metric space, C a nonempty closed subset of X such that for each xC and yC there exists a point zC such that

d(x,z)+d(z,y)=d(x,y).

Suppose that F,G,S,T:CX are two pairs of non-self-mappings satisfying, for all x,yC with xy,

d(Fx,Gy)αd(Tx,Sy)+βu+γv,
(2.3)

where u{d(Tx,Fx),d(Sy,Gy)}, v{d(Tx,Fx)+d(Sy,Gy),d(Tx,Gy)+d(Sy,Fx)}, and α, β, γ are nonnegative real numbers such that

α+β+2γ<1.
(2.4)

Also assume that

  1. (I)

    CSCTC, FCCSC, GCCTC,

  2. (II)

    TxC implies that FxC, SxC implies that GxC,

  3. (III)

    SC and TC (or FC and GC) are closed in X.

Then

  1. (IV)

    (F,T) has a point of coincidence,

  2. (V)

    (G,S) has a point of coincidence.

Moreover, if (F,T) and (G,S) are weakly compatible pairs, then F, G, S, and T have a unique common fixed point.

Proof Firstly, we proceed to construct two sequences { x n } and { y n } in the following way.

Let xC be arbitrary. Then (due to CTC) there exists a point x 0 C such that x=T x 0 . Since TxC implies that FxC, one concludes that F x 0 FCCSC. Thus, there exists x 1 C such that y 1 =S x 1 =F x 0 C. Since y 1 =F x 0 there exists a point y 2 =G x 1 such that

d( y 1 , y 2 )=d(F x 0 ,G x 1 ).

Suppose y 2 C. Then y 2 GCCTC, which implies that there exists a point x 2 C such that y 2 =T x 2 . otherwise, if y 2 C, then there exists a point pC such that

d(S x 1 ,p)+d(p, y 2 )=d(S x 1 , y 2 ).

Since pCTC there exists a point x 2 C with p=T x 2 so that

d(S x 1 ,T x 2 )+d(T x 2 , y 2 )=d(S x 1 , y 2 ).

Let y 3 =F x 2 be such that d( y 2 , y 3 )=d(G x 1 ,F x 2 ). Thus, repeating the foregoing arguments, one obtains two sequences { x n } and { y n } such that

  1. (a)

    y 2 n =G x 2 n 1 , y 2 n + 1 =F x 2 n ,

  2. (b)

    y 2 n C implies that y 2 n =T x 2 n or y 2 n C implies that T x 2 n C and

    d(S x 2 n 1 ,T x 2 n )+d(T x 2 n , y 2 n )=d(S x 2 n 1 , y 2 n ),
  3. (c)

    y 2 n + 1 C implies that y 2 n + 1 =S x 2 n + 1 or y 2 n + 1 C implies that S x 2 n + 1 C and

    d(T x 2 n ,S x 2 n + 1 )+d(S x 2 n + 1 , y 2 n + 1 )=d(T x 2 n , y 2 n + 1 ).

We denote

P 0 = { T x 2 i { T x 2 n } : T x 2 i = y 2 i } , P 1 = { T x 2 i { T x 2 n } : T x 2 i y 2 i } , Q 0 = { S x 2 i + 1 { S x 2 n + 1 } : S x 2 i + 1 = y 2 i + 1 } , Q 1 = { S x 2 i + 1 { S x 2 n + 1 } : S x 2 i + 1 y 2 i + 1 } .

Note that (T x 2 n ,S x 2 n + 1 ) P 1 × Q 1 , as if T x 2 n P 1 , then y 2 n T x 2 n and one infers that T x 2 n C, which implies that y 2 n + 1 =F x 2 n C. Hence y 2 n + 1 =S x 2 n + 1 Q 0 . Similarly, one can argue that (S x 2 n 1 ,T x 2 n ) Q 1 × P 1 .

Now, we distinguish the following three cases.

Case 1. If (T x 2 n ,S x 2 n + 1 ) P 0 × Q 0 , then from (2.3)

d(T x 2 n ,S x 2 n + 1 )=d(F x 2 n ,G x 2 n 1 )αd(T x 2 n ,S x 2 n 1 )+β u 2 n +γ v 2 n ,

where

u 2 n { d ( T x 2 n , F x 2 n ) , d ( S x 2 n 1 , G x 2 n 1 ) } = { d ( T x 2 n , y 2 n + 1 ) , d ( S x 2 n 1 , y 2 n ) } , v 2 n { d ( T x 2 n , F x 2 n ) + d ( S x 2 n 1 , G x 2 n 1 ) , d ( T x 2 n , G x 2 n 1 ) + d ( S x 2 n 1 , F x 2 n ) } v 2 n = { d ( T x 2 n , y 2 n + 1 ) + d ( S x 2 n 1 , y 2 n ) , d ( S x 2 n 1 , y 2 n + 1 ) } .

Clearly, there are infinitely many n such that at least one of the following four cases holds:

  1. (1)

    If u 2 n =d(T x 2 n , y 2 n + 1 ) and v 2 n =d(T x 2 n , y 2 n + 1 )+d(S x 2 n 1 , y 2 n ), then

    d ( T x 2 n , S x 2 n + 1 ) α d ( T x 2 n , S x 2 n 1 ) + β d ( T x 2 n , y 2 n + 1 ) + γ ( d ( T x 2 n , y 2 n + 1 ) + d ( S x 2 n 1 , y 2 n ) ) = α d ( T x 2 n , S x 2 n 1 ) + β d ( T x 2 n , S x 2 n + 1 ) + γ d ( T x 2 n , S x 2 n + 1 ) + γ d ( S x 2 n 1 , T x 2 n ) .

    It implies that d(T x 2 n ,S x 2 n + 1 ) α + γ 1 β γ d(S x 2 n 1 ,T x 2 n ).

  2. (2)

    If u 2 n =d(T x 2 n , y 2 n + 1 ) and v 2 n =d(S x 2 n 1 , y 2 n + 1 ), then

    d ( T x 2 n , S x 2 n + 1 ) α d ( T x 2 n , S x 2 n 1 ) + β d ( T x 2 n , y 2 n + 1 ) + γ d ( S x 2 n 1 , y 2 n + 1 ) α d ( T x 2 n , S x 2 n 1 ) + β d ( T x 2 n , y 2 n + 1 ) + γ ( d ( S x 2 n 1 , y 2 n ) + d ( y 2 n , y 2 n + 1 ) ) = α d ( T x 2 n , S x 2 n 1 ) + β d ( T x 2 n , S x 2 n + 1 ) + γ d ( S x 2 n 1 , T x 2 n ) + γ d ( T x 2 n , S x 2 n + 1 ) .

It implies that d(T x 2 n ,S x 2 n + 1 ) α + γ 1 β γ d(S x 2 n 1 ,T x 2 n ).

  1. (3)

    If u 2 n =d(S x 2 n 1 , y 2 n ) and v 2 n =d(T x 2 n , y 2 n + 1 )+d(S x 2 n 1 , y 2 n ), then

    d ( T x 2 n , S x 2 n + 1 ) α d ( T x 2 n , S x 2 n 1 ) + β d ( S x 2 n 1 , y 2 n ) + γ ( d ( T x 2 n , y 2 n + 1 ) + d ( S x 2 n 1 , y 2 n ) ) = α d ( T x 2 n , S x 2 n 1 ) + β d ( S x 2 n 1 , T x 2 n ) + γ d ( T x 2 n , S x 2 n + 1 ) + γ d ( S x 2 n 1 , T x 2 n ) .

It implies that d(T x 2 n ,S x 2 n + 1 ) α + β + γ 1 γ d(S x 2 n 1 ,T x 2 n ).

  1. (4)

    If u 2 n =d(S x 2 n 1 , y 2 n ) and v 2 n =d(S x 2 n 1 , y 2 n + 1 ), then

    d ( T x 2 n , S x 2 n + 1 ) α d ( T x 2 n , S x 2 n 1 ) + β d ( S x 2 n 1 , y 2 n ) + γ d ( S x 2 n 1 , y 2 n + 1 ) α d ( T x 2 n , S x 2 n 1 ) + β d ( S x 2 n 1 , y 2 n ) + γ ( d ( S x 2 n 1 , y 2 n ) + d ( y 2 n , y 2 n + 1 ) ) = α d ( T x 2 n , S x 2 n 1 ) + β d ( S x 2 n 1 , T x 2 n ) + γ d ( S x 2 n 1 , T x 2 n ) + γ d ( T x 2 n , S x 2 n + 1 ) .

It implies that d(T x 2 n ,S x 2 n + 1 ) α + β + γ 1 γ d(S x 2 n 1 ,T x 2 n ).

From (1), (2), (3), (4) it follows that

d(T x 2 n ,S x 2 n + 1 )λd(S x 2 n 1 ,T x 2 n ),
(2.5)

where λ=max{ α + γ 1 β γ , α + β + γ 1 γ }<1 by (2.4).

Similarly, if (S x 2 n + 1 ,T x 2 n + 2 ) Q 0 × P 0 , we have

d(S x 2 n + 1 ,T x 2 n + 2 )=d(F x 2 n ,G x 2 n + 1 )λd(T x 2 n ,S x 2 n + 1 ).
(2.6)

If (S x 2 n 1 ,T x 2 n ) Q 0 × P 0 , we have

d(S x 2 n 1 ,T x 2 n )=d(F x 2 n 2 ,G x 2 n 1 )λd(T x 2 n 2 ,S x 2 n 1 ).
(2.7)

Case 2. If (T x 2 n ,S x 2 n + 1 ) P 0 × Q 1 , then S x 2 n + 1 Q 1 and

d(T x 2 n ,S x 2 n + 1 )+d(S x 2 n + 1 , y 2 n + 1 )=d(T x 2 n , y 2 n + 1 ),
(2.8)

which in turns yields

d(T x 2 n ,S x 2 n + 1 )d(T x 2 n , y 2 n + 1 )=d( y 2 n , y 2 n + 1 )
(2.9)

and hence

d(T x 2 n ,S x 2 n + 1 )d( y 2 n , y 2 n + 1 )=d(F x 2 n ,G x 2 n 1 ).
(2.10)

Now, proceeding as in Case 1, we see that (2.5) holds.

If (S x 2 n + 1 ,T x 2 n + 2 ) Q 1 × P 0 , then T x 2 n P 0 . We show that

d(S x 2 n + 1 ,T x 2 n + 2 )λd(T x 2 n ,S x 2 n 1 ).
(2.11)

Using (2.8), we get

d ( S x 2 n + 1 , T x 2 n + 2 ) d ( S x 2 n + 1 , y 2 n + 1 ) + d ( y 2 n + 1 , T x 2 n + 2 ) = d ( T x 2 n , y 2 n + 1 ) d ( T x 2 n , S x 2 n + 1 ) + d ( y 2 n + 1 , T x 2 n + 2 ) .
(2.12)

By noting that T x 2 n + 2 ,T x 2 n P 0 , one can conclude that

d( y 2 n + 1 ,T x 2 n + 2 )=d( y 2 n + 1 , y 2 n + 2 )=d(F x 2 n ,G x 2 n + 1 )λd(T x 2 n ,S x 2 n + 1 )
(2.13)

and

d(T x 2 n , y 2 n + 1 )=d( y 2 n , y 2 n + 1 )=d(F x 2 n ,G x 2 n 1 )λd(S x 2 n 1 ,T x 2 n ),
(2.14)

in view of Case 1.

Thus,

d(S x 2 n + 1 ,T x 2 n + 2 )λd(S x 2 n 1 ,T x 2 n )(1λ)d(T x 2 n ,S x 2 n + 1 )λd(S x 2 n 1 ,T x 2 n ),

and we proved (2.11).

Case 3. If (T x 2 n ,S x 2 n + 1 ) P 1 × Q 0 , then S x 2 n 1 Q 0 . We show that

d(T x 2 n ,S x 2 n + 1 )λd(S x 2 n 1 ,T x 2 n 2 ).
(2.15)

Since T x 2 n P 1 , then

d(S x 2 n 1 ,T x 2 n )+d(T x 2 n , y 2 n )=d(S x 2 n 1 , y 2 n ).
(2.16)

From this, we get

d ( T x 2 n , S x 2 n + 1 ) d ( T x 2 n , y 2 n ) + d ( y 2 n , S x 2 n + 1 ) = d ( S x 2 n 1 , y 2 n ) d ( S x 2 n 1 , T x 2 n ) + d ( y 2 n , S x 2 n + 1 ) .
(2.17)

By noting that S x 2 n + 1 ,S x 2 n 1 Q 0 , one can conclude that

d( y 2 n ,S x 2 n + 1 )=d( y 2 n , y 2 n + 1 )=d(F x 2 n ,G x 2 n 1 )λd(S x 2 n 1 ,T x 2 n )
(2.18)

and

d(S x 2 n 1 , y 2 n )=d( y 2 n 1 , y 2 n )=d(F x 2 n 2 ,G x 2 n 1 )λd(S x 2 n 1 ,T x 2 n 2 ),
(2.19)

in view of Case 1.

Thus,

d(T x 2 n ,S x 2 n + 1 )λd(S x 2 n 1 ,T x 2 n 2 )(1λ)d(S x 2 n 1 ,T x 2 n )λd(S x 2 n 1 ,T x 2 n 2 ),

and we proved (2.15).

Similarly, if (S x 2 n + 1 ,T x 2 n + 2 ) Q 0 × P 1 , then T x 2 n + 2 P 1 , and

d(S x 2 n + 1 ,T x 2 n + 2 )+d(T x 2 n + 2 , y 2 n + 2 )=d(S x 2 n + 1 , y 2 n + 2 ).

From this, we have

d ( S x 2 n + 1 , T x 2 n + 2 ) d ( S x 2 n + 1 , y 2 n + 2 ) + d ( y 2 n + 2 , T x 2 n + 2 ) d ( S x 2 n + 1 , T x 2 n + 2 ) d ( S x 2 n + 1 , y 2 n + 2 ) + d ( S x 2 n + 1 , y 2 n + 2 ) d ( S x 2 n + 1 , T x 2 n + 2 ) d ( S x 2 n + 1 , T x 2 n + 2 ) = 2 d ( S x 2 n + 1 , y 2 n + 2 ) d ( S x 2 n + 1 , T x 2 n + 2 ) d ( S x 2 n + 1 , T x 2 n + 2 ) d ( S x 2 n + 1 , y 2 n + 2 ) .

By noting that S x 2 n + 1 Q 0 , one can conclude that

d(S x 2 n + 1 ,T x 2 n + 2 )d(S x 2 n + 1 , y 2 n + 2 )=d(F x 2 n ,G x 2 n + 1 )λd(T x 2 n ,S x 2 n + 1 ),
(2.20)

in view of Case 1.

Thus, in all cases (1)-(3), there exists w 2 n {d(S x 2 n 1 ,T x 2 n ),d(T x 2 n 2 ,S x 2 n 1 )} such that

d(T x 2 n ,S x 2 n + 1 )λ w 2 n

and there exists w 2 n + 1 {d(S x 2 n 1 ,T x 2 n ),d(T x 2 n ,S x 2 n + 1 )} such that

d(S x 2 n + 1 ,T x 2 n + 2 )λ w 2 n + 1 .

Following the procedure of Assad and Kirk [25], it can easily be shown by induction that, for n1, there exists w 2 {d(T x 0 ,S x 1 ),d(S x 1 ,T x 2 )} such that

d(T x 2 n ,S x 2 n + 1 ) λ n 1 2 w 2 andd(S x 2 n + 1 ,T x 2 n + 2 ) λ n w 2 .
(2.21)

From (2.21) and by the triangle inequality, for n>m we have

d ( T x 2 n , S x 2 m + 1 ) d ( T x 2 n , S x 2 n 1 ) + d ( S x 2 n 1 , T x 2 n 2 ) + + d ( T x 2 m + 2 , S x 2 m + 1 ) ( λ m + λ m + 1 2 + + λ n 1 ) w 2 λ m 1 λ w 2 θ , as  m .

From Remark 1.2 and Corollary 1.1(1), it follows that d(T x 2 n ,S x 2 m + 1 )c.

Thus, the sequence {T x 0 ,S x 1 ,T x 2 ,S x 3 ,,S x 2 n 1 ,T x 2 n ,S x 2 n 1 ,} is a Cauchy sequence. Then, as noted in [45], there exists at least one subsequence, {T x 2 n k } or {S x 2 n k + 1 }, which is contained in P 0 or Q 0 , respectively, and one finds its limit zC. Furthermore, subsequences {T x 2 n k } and {S x 2 n k + 1 } both converge to zC as C is a closed subset of complete cone metric space (X,d). We assume that there exists a subsequence {T x 2 n k } P 0 for each kN, then T x 2 n k = y 2 n k =G x 2 n k 1 CGCTC Since TC as well as SC are closed in X and {T x 2 n k } is Cauchy in TC, it converges to a point zTC. Let w T 1 z, then Tw=z. Similarly, {T x 2 n k } and {S x 2 n k + 1 } being a subsequence of a Cauchy sequence, {T x 0 ,S x 1 ,T x 2 ,S x 3 ,,S x 2 n 1 ,T x 2 n ,S x 2 n 1 ,} also converges to z as SC is closed. Let θc, then d(z,S x 2 n k 1 ) c 2 α + γ 1 β γ , where α, β, γ are nonnegative real numbers with α+β+2γ<1.

Using (2.3), one can write

d ( F w , z ) d ( F w , G x 2 n k 1 ) + d ( G x 2 n k 1 , z ) α d ( T w , S x 2 n k 1 ) + β u w + γ v w + d ( G x 2 n k 1 , z ) = α d ( z , S x 2 n k 1 ) + β u w + γ v w + d ( G x 2 n k 1 , z ) ,

where

u w { d ( T w , F w ) , d ( S x 2 n k 1 , G x 2 n k 1 ) } = { d ( z , F w ) , d ( S x 2 n k 1 , G x 2 n k 1 ) } , v w { d ( T w , F w ) + d ( S x 2 n k 1 , G x 2 n k 1 ) , d ( T w , G x 2 n k 1 ) + d ( F w , S x 2 n k 1 ) } v w = { d ( z , F w ) + d ( S x 2 n k 1 , G x 2 n k 1 ) , d ( z , G x 2 n k 1 ) + d ( F w , S x 2 n k 1 ) } .

Let θc. Clearly at least one of the following four cases holds for infinitely many n:

  1. (1)

    If u w =d(z,Fw) and v w =d(z,Fw)+d(S x 2 n k 1 ,G x 2 n k 1 ), then

    d ( F w , z ) α d ( z , S x 2 n k 1 ) + β d ( z , F w ) d ( F w , z ) + γ ( d ( z , F w ) + d ( S x 2 n k 1 , G x 2 n k 1 ) ) + d ( G x 2 n k 1 , z ) d ( F w , z ) α d ( z , S x 2 n k 1 ) + β d ( z , F w ) + γ d ( z , F w ) d ( F w , z ) + γ ( d ( S x 2 n k 1 , z ) + d ( z , G x 2 n k 1 ) ) + d ( G x 2 n k 1 , z ) d ( F w , z ) = ( α + γ ) d ( z , S x 2 n k 1 ) + ( β + γ ) d ( z , F w ) + ( γ + 1 ) d ( G x 2 n k 1 , z ) d ( F w , z ) α + γ 1 β γ d ( z , S x 2 n k 1 ) + γ + 1 1 β γ d ( G x 2 n k 1 , z ) d ( F w , z ) α + γ 1 β γ c 2 α + γ 1 β γ + γ + 1 1 β γ c 2 γ + 1 1 β γ = c .
  2. (2)

    If u w =d(z,Fw) and v w =d(z,G x 2 n k 1 )+d(Fw,S x 2 n k 1 ), then

    d ( F w , z ) α d ( z , S x 2 n k 1 ) + β d ( z , F w ) d ( F w , z ) + γ ( d ( z , G x 2 n k 1 ) + d ( F w , S x 2 n k 1 ) ) + d ( G x 2 n k 1 , z ) d ( F w , z ) α d ( z , S x 2 n k 1 ) + β d ( z , F w ) + γ d ( z , G x 2 n k 1 ) d ( F w , z ) + γ ( d ( F w , z ) + d ( z , S x 2 n k 1 ) ) + d ( G x 2 n k 1 , z ) d ( F w , z ) = ( α + γ ) d ( z , S x 2 n k 1 ) + ( β + γ ) d ( z , F w ) + ( γ + 1 ) d ( G x 2 n k 1 , z ) d ( F w , z ) α + γ 1 β γ d ( z , S x 2 n k 1 ) + γ + 1 1 β γ d ( G x 2 n k 1 , z ) d ( F w , z ) α + γ 1 β γ c 2 α + γ 1 β γ + γ + 1 1 β γ c 2 γ + 1 1 β γ = c .
  3. (3)

    If u w =d(S x 2 n k 1 ,G x 2 n k 1 ) and v w =d(z,Fw)+d(S x 2 n k 1 ,G x 2 n k 1 ), then

    d ( F w , z ) α d ( z , S x 2 n k 1 ) + β d ( S x 2 n k 1 , G x 2 n k 1 ) d ( F w , z ) + γ ( d ( z , F w ) + d ( S x 2 n k 1 , G x 2 n k 1 ) ) + d ( G x 2 n k 1 , z ) d ( F w , z ) α d ( z , S x 2 n k 1 ) + β ( d ( S x 2 n k 1 , z ) + d ( z , G x 2 n k 1 ) ) + γ d ( z , F w ) d ( F w , z ) + γ ( d ( S x 2 n k 1 , z ) + d ( z , G x 2 n k 1 ) ) + d ( G x 2 n k 1 , z ) d ( F w , z ) = ( α + β + γ ) d ( z , S x 2 n k 1 ) + γ d ( z , F w ) + ( β + γ + 1 ) d ( G x 2 n k 1 , z ) d ( F w , z ) α + β + γ 1 γ d ( z , S x 2 n k 1 ) + β + γ + 1 1 γ d ( G x 2 n k 1 , z ) d ( F w , z ) α + β + γ 1 γ c 2 α + β + γ 1 γ + β + γ + 1 1 γ c 2 β + γ + 1 1 γ = c .
  4. (4)

    If u w =d(S x 2 n k 1 ,G x 2 n k 1 ) and v w =d(z,G x 2 n k 1 )+d(Fw,S x 2 n k 1 ), then

    d ( F w , z ) α d ( z , S x 2 n k 1 ) + β d ( S x 2 n k 1 , G x 2 n k 1 ) d ( F w , z ) + γ ( d ( z , G x 2 n k 1 ) + d ( F w , S x 2 n k 1 ) ) + d ( G x 2 n k 1 , z ) d ( F w , z ) α d ( z , S x 2 n k 1 ) + β ( d ( S x 2 n k 1 , z ) + d ( z , G x 2 n k 1 ) ) + γ d ( z , G x 2 n k 1 ) d ( F w , z ) + γ ( d ( F w , z ) + d ( z , S x 2 n k 1 ) ) + d ( G x 2 n k 1 , z ) d ( F w , z ) = ( α + β + γ ) d ( z , S x 2 n k 1 ) + γ d ( z , F w ) + ( β + γ + 1 ) d ( G x 2 n k 1 , z ) d ( F w , z ) α + β + γ 1 γ d ( z , S x 2 n k 1 ) + β + γ + 1 1 γ d ( G x 2 n k 1 , z ) d ( F w , z ) α + β + γ 1 γ c 2 α + β + γ 1 γ + β + γ + 1 1 γ c 2 β + γ + 1 1 γ = c .

In all cases we obtain d(Fw,z)c for each cintP. Using Corollary 1.1(3) it follows that d(Fw,z)=θ or Fw=z. Thus, Fw=z=Tw, that is, z is a coincidence point of F, T.

Further, since Cauchy sequence {T x 0 ,S x 1 ,T x 2 ,S x 3 ,,S x 2 n 1 ,T x 2 n ,S x 2 n 1 ,} converges to zC and z=Fw, zFCCSC, there exists bC such that Sb=z. Again using (2.3), we get

d(Sb,Gb)=d(z,Gb)=d(Fw,Gb)αd(Tw,Sb)+β u w +γ v w =β u w +γ v w ,

where

u w { d ( T w , F w ) , d ( S b , G b ) } = { θ , d ( S b , G b ) } , v w { d ( T w , F w ) + d ( S b , G b ) , d ( T w , G b ) + d ( S b , F w ) } v w = { d ( S b , G b ) , d ( z , G b ) } = { d ( S b , G b ) } .

Hence, we get the following cases:

d(Sb,Gb)βθ+γd(Sb,Gb)=γd(Sb,Gb)andd(Sb,Gb)(β+γ)d(Sb,Gb).

Since 0γβ+γ<1αγ1, using Remark 1.3 and Corollary 1.1(3), it follows that Sb=Gb, therefore, Sb=z=Gb, that is, z is a coincidence point of (G,S).

In case FC and GC are closed in X, then zFCCSC or zGCCTC. The analogous arguments establish (IV) and (V). If we assume that there exists a subsequence {S x 2 n k + 1 } Q 0 with TC as well SC are closed in X, then noting that {S x 2 n k + 1 } is a Cauchy sequence in SC, the foregoing arguments establish (IV) and (V).

Suppose now that (F,T) and (G,S) are weakly compatible pairs, then

z = F w = T w F z = F T w = T F w = T z and z = G b = S b G z = G S b = S G b = S z .

Then, from (2.3),

d(Fz,z)=d(Fz,Gb)αd(Tz,Sb)+βu+γv=αd(Fz,z)+βu+γv,

where

u { d ( T z , F z ) , d ( S b , G b ) } = { d ( F z , F z ) , d ( z , z ) } = { θ } , v { d ( T z , F z ) + d ( S b , G b ) , d ( T z , G b ) + d ( S b , F z ) } v = { θ , d ( F z , z ) + d ( z , F z ) } = { θ , 2 d ( F z , z ) } .

Hence, we get the following cases:

d(Fz,z)αd(Fz,z)andd(Fz,z)αd(Fz,z)+2γd(Fz,z)=(α+2γ)d(z,Fz).

Since 0αα+2γ<1β1, using Remark 1.3 and Corollary 1.1(3), it follows that Fz=z. Thus, Fz=z=Tz.

Similarly, we can prove Gz=z=Sz. Therefore z=Fz=Gz=Sz=Tz, that is, z is a common fixed point of F, G, S, and T.

Uniqueness of the common fixed point follows easily from (2.3). □

The following example shows that in general F, G, S, and T satisfying the hypotheses of Theorem 2.2 need not have a common coincidence justifying the two separate conclusions (IV) and (V).

Example 2.1 Let E= C 1 ([0,1],R), P={φE:φ(t)0,t[0,1]}, X=[0,+), C=[0,2], and d:X×XE defined by d(x,y)=|xy|φ, where φP is a fixed function, e.g., φ(t)= e t . Then (X,d) is a complete cone metric space with a non-normal cone having a nonempty interior. Define F, G, S, and T:CX as

Fx=x+ 4 5 ,Gx= x 2 + 4 5 ,Tx=5xandSx=5 x 2 ,xC.

Since C={0,2}. Clearly, for each xC and yC there exists a point z=2C such that d(x,z)+d(z,y)=d(x,y). Further, SCTC=[0,20][0,10]=[0,10]{0,2}=C, FCC=[ 4 5 , 14 5 ][0,2]=[ 4 5 ,2]SC, GCC=[ 4 5 , 24 5 ][0,2]=[ 4 5 ,2]TC, and, SC, TC, FC, and GC are closed in X.

Also,

T 0 = 0 C F 0 = 4 5 C , S 0 = 0 C G 0 = 4 5 C , T ( 2 5 ) = 2 C F ( 2 5 ) = 6 5 C , S ( 2 5 ) = 2 C G ( 2 5 ) = 6 5 C .

Moreover, for each x,yC,

d(Fx,Gy)=|x y 2 |φ= 1 5 d(Tx,Sy),

that is, (2.3) is satisfied with α= 1 5 , β=γ=0.

Obviously, 1=T( 1 5 )=F( 1 5 ) 1 5 and 1=S( 1 5 )=G( 1 5 ) 1 5 . Notice that two separate coincidence points are not common fixed points as FT( 1 5 )TF( 1 5 ) and SG( 1 5 )GS( 1 5 ), which shows the necessity of the weakly compatibility property in Theorem 2.2.

Next, we furnish an illustrate example in support of our result. In doing so, we are essentially inspired by Imdad and Kumar [46].

Example 2.2 Let E= C 1 ([0,1],R), P={φE:φ(t)0,t[0,1]}, X=[1,+), C=[1,3], and d:X×XE defined by d(x,y)=|xy|φ, where φP is a fixed function, e.g., φ(t)= e t . Then (X,d) is a complete cone metric space with a non-normal cone having the nonempty interior. Define F, G, S, and T:CX as

F x = { x 2 if  1 x 2 , 2 if  2 < x 3 , T x = { 4 x 4 3 if  1 x 2 , 13 if  2 < x 3 , G x = { x 3 if  1 x 2 , 2 if  2 < x 3 and S x = { 4 x 6 3 if  1 x 2 , 13 if  2 < x 3 .

Since C={1,3}. Clearly, for each xC and yC there exists a point z=3C such that d(x,z)+d(z,y)=d(x,y). Further, SCTC=[1,253][1,61]=[1,61]{1,3}=C, FCC=[1,4][1,3]=[1,3]SC, and GCC=[1,8][1,3]=[1,3]TC.

Also,

T 1 = 1 C F 1 = 1 C , S 1 = 1 C G 1 = 1 C , T ( 3 2 4 ) = 3 C F ( 3 2 4 ) = 3 2 C , S ( 3 2 6 ) = 3 C G ( 3 2 6 ) = 3 2 C .

Moreover, if x[1,2] and y[2,3], then

d(Fx,Gy)=| x 2 2|φ= | x 4 4 | | x 2 + 2 | φ= 4 | x 4 4 | 4 | x 2 + 2 | φ= 1 4 ( x 2 + 2 ) d(Tx,Sy).

Next, if x,y(2,3], then

d(Fx,Gy)=0=αd(Tx,Sy).

Finally, if x,y[1,2], then

d(Fx,Gy)=| x 2 y 3 |φ= | x 4 y 6 | | x 2 + y 3 | φ= 4 | x 4 y 6 | 4 | x 2 + y 3 | φ= 1 4 ( x 2 + y 3 ) d(Tx,Sy).

Therefore, condition (2.3) is satisfied if we choose α=max{ 1 4 ( x 2 + 2 ) , 1 4 ( x 2 + y 3 ) }(0, 1 4 ), β=γ=0. Moreover, 1 is a point of coincidence as T1=F1 as well as S1=G1, whereas both pairs (F,T) and (G,S) are weakly compatible as TF1=1=FT1 and SG1=1=GS1. Also, SC, TC, FC, and GC are closed in X. Thus, all the conditions of Theorem 2.2 are satisfied and 1 is the unique common fixed point of F, G, S, and T. One may note that 1 is also a point of coincidence for both pairs (F,T) and (G,S).

Remark 2.2 1. Setting G=F=g and T=S=f in Theorem 2.2, one deduces Theorem 2.1 due to Sumitra et al. [20] with weaker condition.

  1. 2.

    Setting G=F=g and T=S= I X in Theorem 2.2, we obtain the following result.

Corollary 2.1 Let (X,d) be a complete cone metric space, and C a nonempty closed subset of X such that for each xC and yC there exists a point zC such that

d(x,z)+d(z,y)=d(x,y).

Suppose that g:CX satisfies, for all x,yC with xy,

d(gx,gy)αd(x,y)+βu+γ,

where

u { d ( x , g x ) , d ( y , g y ) } ,v { d ( x , g x ) + d ( y , g y ) , d ( x , g y ) + d ( y , g x ) } ,

and α, β, γ are nonnegative real numbers such that α+β+2γ<1 and g has the additional property that if, for each xC, gxC, then g has a unique fixed point in C.

Corollary 2.2 Let (X,d) be a complete cone metric space, C a nonempty closed subset of X such that for each xC and yC there exists a point zC such that

d(x,z)+d(z,y)=d(x,y).

Let F,G,S,T:CX be such that

d(Fx,Gy)αd(Tx,Sy)
(2.22)

for some α(0,1) and for all x,yC with xy.

Suppose, further, that F, G, S, T, and C satisfy the following conditions:

  1. (I)

    CSCTC, FCCSC, GCCTC,

  2. (II)

    TxCFxC, SxCGxC,

  3. (III)

    SC and TC (or FC and GC) are closed in X.

Then

  1. (IV)

    (F,T) has a point of coincidence,

  2. (V)

    (G,S) has a point of coincidence.

Moreover, if (F,T) and (G,S) are weakly compatible pairs, then F, G, S, and T have a unique common fixed point.

Corollary 2.3 Let (X,d) be a complete cone metric space, C a nonempty closed subset of X such that for each xC and yC there exists a point zC such that

d(x,z)+d(z,y)=d(x,y).

Let F,G,S,T:CX be such that

d(Fx,Gy)γ ( d ( T x , F x ) + d ( S y , G y ) )
(2.23)

for some γ(0,1/2) and for all x,yC with xy.

Suppose, further, that F, G, S, T, and C satisfy the following conditions:

  1. (I)

    CSCTC, FCCSC, GCCTC,

  2. (II)

    TxCFxC, SxCGxC,

  3. (III)

    SC and TC (or FC and GC) are closed in X.

Then

  1. (IV)

    (F,T) has a point of coincidence,

  2. (V)

    (G,S) has a point of coincidence.

Moreover, if (F,T) and (G,S) are weakly compatible pairs, then F, G, S, and T have a unique common fixed point.

Corollary 2.4 Let (X,d) be a complete cone metric space, C a nonempty closed subset of X such that for each xC and yC there exists a point zC such that

d(x,z)+d(z,y)=d(x,y).

Let F,G,S,T:CX be such that

d(Fx,Gy)γ ( d ( T x , F y ) + d ( F x , S y ) )
(2.24)

for some γ(0,1/2) and for all x,yC with xy.

Suppose, further, that F, G, S, T, and C satisfy the following conditions:

  1. (I)

    CSCTC, FCCSC, GCCTC,

  2. (II)

    TxCFxC, SxCGxC,

  3. (III)

    SC and TC (or FC and GC) are closed in X.

Then

  1. (IV)

    (F,T) has a point of coincidence,

  2. (V)

    (G,S) has a point of coincidence.

Moreover, if (F,T) and (G,S) are weakly compatible pairs, then F, G, S, and T have a unique common fixed point.

Remark 2.3 Setting G=F=f and T=S=g in Corollaries 2.2-2.4, we obtain the following results.

Corollary 2.5 Let (X,d) be a complete cone metric space, C a nonempty closed subset of X such that for each xC and yC there exists a point zC such that

d(x,z)+d(z,y)=d(x,y).

Let f,g:CX be such that

d(fx,fy)αd(gx,gy)
(2.25)

for some α(0,1) and for all x,yC. Suppose, further, that f, g, and C satisfy the following conditions:

  1. (I)

    CgC, fCCgC,

  2. (II)

    gxCfxC,

  3. (III)

    gC is closed in X.

Then the pair (f,g) has a coincidence point in C. Moreover, if the pair (f,g) is weakly compatible, then f and g have a unique common fixed point in C.

Corollary 2.6 Let (X,d) be a complete cone metric space, C a nonempty closed subset of X such that for each xC and yC there exists a point zC (the boundary of C) such that

d(x,z)+d(z,y)=d(x,y).

Let f,g:CX be such that

d(fx,fy)γ ( d ( f x , g x ) + d ( f y , g y ) )
(2.26)

for some γ(0,1/2) and for all x,yC. Suppose, further, that f, g, and C satisfy the following conditions:

  1. (I)

    CgC, fCCgC,

  2. (II)

    gxCfxC,

  3. (III)

    gC is closed in X.

Then the pair (f,g) has a coincidence point in C. Moreover, if the pair (f,g) is weakly compatible, then f and g have a unique common fixed point in C.

Corollary 2.7 Let (X,d) be a complete cone metric space, C a nonempty closed subset of X such that for each xC and yC there exists a point zC (the boundary of C) such that

d(x,z)+d(z,y)=d(x,y).

Let f,g:CX be such that

d(fx,fy)γ ( d ( f x , g y ) + d ( f y , g x ) )
(2.27)

for some γ(0,1/2) and for all x,yC. Suppose, further, that f, g, and C satisfy the following conditions:

  1. (I)

    CgC, fCCgC,

  2. (II)

    gxCfxC,

  3. (III)

    gC is closed in X.

Then the pair (f,g) has a coincidence point in C. Moreover, if the pair (f,g) is weakly compatible, then f and g have a unique common fixed point in C.

Remark 2.4 Corollaries 2.5-2.7 are the corresponding theorems of Abbas and Jungck from [2] in the case that f, g are non-self-mappings.

There have been a number of papers written about fixed points for non-self-maps. One of the most general papers involving two maps is that in [30]. For cone metric spaces, the four-function analog of this result would have the contractive condition.

Suppose that F,G,S,T:CX are two pairs of non-self-mappings satisfying, for all x,yC with xy,

d(Fx,Gy)hw,
(2.28)

where w{ d ( T x , S y ) a ,d(Tx,Fx),d(Sy,Gy), d ( T x , G y ) + d ( S y , F x ) a + h }, a is a positive real number satisfying a1+ 2 h 2 1 + h and 0<h<1.

Note that if F, G, S, and T satisfy condition (2.28), then F, G, S, and T satisfy condition (2.3), but the implication is not reversible.

Indeed, there are four cases to consider:

  1. (1)

    If w= d ( T x , S y ) a in (2.28), then d(Fx,Gy) h a d(Tx,Sy). So setting α= h a h 1 + 2 h 2 1 + h = h 2 + h 2 h 2 + h + 1 = 1 2 1 h 2 ( 2 h 2 + h + 1 ) < 1 2 , β=γ=0 in (2.3), it follows that d(Fx,Gy)αd(Tx,Sy).

  2. (2)

    If w=d(Tx,Fx) in (2.28), then d(Fx,Gy)hd(Tx,Fx). So setting β=h<1, α=γ=0, u=d(Tx,Fx) in (2.3), it follows that d(Fx,Gy)βu, where u=d(Tx,Fx).

  3. (3)

    If w=d(Sy,Gy) in (2.28), then d(Fx,Gy)hd(Sy,Gy). So setting β=h<1, α=γ=0, u=d(Sy,Gy) in (2.3), it follows that d(Fx,Gy)βu, where u=d(Sy,Gy).

  4. (4)

    If w= d ( T x , G y ) + d ( S y , F x ) a + h in (2.28), then d(Fx,Gy) h a + h [d(Tx,Gy)+d(Sy,Fx)]. So setting γ= h ( a + h ) h 1 + 2 h 2 1 + h + h = h 2 + h 3 h 2 + 2 h + 1 = 1 3 1 h 3 ( 3 h 2 + 2 h + 1 ) < 1 3 , α=β=0, v=d(Tx,Gy)+d(Sy,Fx) in (2.3), it follows that d(Fx,Gy)γv, where v=d(Tx,Gy)+d(Sy,Fx).

Therefore, in all cases we find that F, G, S, and T satisfy condition (2.28), then F, G, S, and T satisfy condition (2.3).

Now, we give an example to show that condition (2.3) is more general than condition (2.28) above.

Example 2.3 Let E= C 1 ([0,1],R), P={φE:φ(t)0,t[0,1]}, X=[0,+), C=[ 3 5 ,1], and let d:X×XE be defined by d(x,y)=|xy|φ, where φP is a fixed function, e.g., φ(t)= e t . Then (X,d) is a complete cone metric space with a non-normal cone having a nonempty interior. Define F, G, S, and T:CX as

F(x)= 2 x 1 + x ,G(x)= 2 x 2 1 + x 2 ,T(x)=x,S(x)= x 2 ,xC.
(2.29)

Note that for all x,yC with xy,

d(Fx,Gy)=| 2 x 1 + x 2 y 2 1 + y 2 |φ= | 2 x 2 y 2 | φ ( 1 + x ) ( 1 + y 2 ) = d ( T x , S y ) 1 2 ( 1 + x ) ( 1 + y 2 ) .
(2.30)

Therefore, condition (2.3) is satisfied if we choose α= 1 1 2 ( 1 + x ) ( 1 + y 2 ) [ 1 2 , 125 144 ], β=γ=0.

Next, we shall see that the inequality (2.28) is not satisfied for all 0<h<1 and a1+ 2 h 2 1 + h by taking x=1 and y= 3 5 .

Indeed, d(F(1),G( 3 5 ))=|1 1 2 |φ= 1 2 φ and w{ d ( T ( 1 ) , S ( 3 5 ) ) a ,d(T(1),F(1)),d(S( 3 5 ),G( 3 5 )), d ( T ( 1 ) , G ( 3 5 ) ) + d ( S ( 3 5 ) , F ( 1 ) ) a + h }={ 1 a 16 25 φ,0, 7 50 φ, 1 a + h 57 50 φ}.

Since h a < 1 2 and h a + h < 1 3 , d(F(1),G( 3 5 ))hw for all possible cases of w, 0<h<1, and a1+ 2 h 2 1 + h , that is, (2.3) is more general than (2.28).

So, we can obtain the following corollary.

Corollary 2.8 Let (X,d) be a complete cone metric space, C a nonempty closed subset of X such that for each xC and yC there exists a point zC such that

d(x,z)+d(z,y)=d(x,y).

Suppose that F,G,S,T:CX are two pairs of non-self-mappings satisfying, for all x,yC with xy,

d(Fx,Gy)hw,
(2.31)

where w{ d ( T x , S y ) a ,d(Tx,Fx),d(Sy,Gy), d ( T x , G y ) + d ( S y , F x ) a + h }, a is a positive real number satisfying a1+ 2 h 2 1 + h and 0<h<1. Also assume that

  1. (I)

    CSCTC, FCCSC, GCCTC,

  2. (II)

    TxCFxC, SxCGxC,

  3. (III)

    SC and TC (or FC and GC) are closed in X.

Then

  1. (IV)

    (F,T) has a point of coincidence,

  2. (V)

    (G,S) has a point of coincidence.

Moreover, if (F,T) and (G,S) are weakly compatible pairs, then F, G, S, and T have a unique common fixed point.