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Some new common fixed point results for three pairs of mappings in generalized metric spaces

Abstract

In this paper, we use weakly commuting and weakly compatible conditions of self-mapping pairs, prove some new common fixed point theorems for three pairs of self-mappings in G-metric spaces. An example is provided to support our result. The results presented in this paper extend and improve several well-known comparable results.

MSC:47H10, 54H25, 54E50.

1 Introduction and preliminaries

The metric fixed point theory is very important and useful in mathematics. It can be applied in various areas, for instance, approximation theory, optimization and variational inequalities. Many authors have introduced the generalizations of metric spaces, for example, Gähler [1, 2] (called 2-metric spaces) and Dhage [3, 4] (called D-metric spaces). In 2003, Mustafa and Sims [5] found that most of the claims concerning the fundamental topological properties of D-metric spaces are incorrect. Therefore, they [6] introduced a new structure of generalized metric spaces, which are called G-metric spaces, as a generalization of metric spaces, to develop and introduce a new fixed point theory for various mappings in this new structure. Later, several fixed point and common fixed point theorems in G-metric spaces were obtained by [651].

The purpose of this paper is to use the concept of weakly commuting mappings and weakly compatible mappings to discuss some new common fixed point problem for six self-mappings in G-metric spaces. The results presented in this paper extend and improve the corresponding results of Abbas et al. [7], Mustafa and Sims [8], Abbas and Rhoades [9], Mustafa et al. [10], Mustafa et al. [11], Abbas et al. [12], Chugh and Kadian [13], Manro et al. [14], Vats et al. [15].

We now recall some definitions and properties in G-metric spaces.

Definition 1.1 [6]

Let X be a nonempty set and let G:X×X×X R + be a function satisfying the following properties:

  • ( G 1 ) G(x,y,z)=0 if x=y=z;

  • ( G 2 ) 0<G(x,x,y) for all x,yXwith xy;

  • ( G 3 ) G(x,x,y)G(x,y,z) for all x,y,zX with zy;

  • ( G 4 ) G(x,y,z)=G(x,z,y)=G(y,z,x)= , symmetry in all three variables;

  • ( G 5 ) G(x,y,z)G(x,a,a)+G(a,y,z) for all x,y,z,aX.

Then the function G is called a generalized metric, or, more specifically, a G-metric on X, and the pair (X,G) is called a G-metric space.

Definition 1.2 [6]

Let (X,G) be a G-metric space and let ( x n ) be a sequence of points of X. A point xX is said to be the limit of the sequence ( x n ) if lim n , m + G(x, x n , x m )=0, and we say that the sequence ( x n ) is G-convergent to x or ( x n ) G-convergent to x.

Thus, x n x in a G-metric space (X,G) if, for any ϵ>0, there exists kN such that G(x, x n , x m )<ϵ for all m,nk.

Proposition 1.1 [6]

Let (X,G) be a G-metric space, then the following are equivalent:

  1. 1.

    ( x n ) is G-convergent to x.

  2. 2.

    G( x n , x n ,x)0 as n+.

  3. 3.

    G( x n ,x,x)0 as n+.

  4. 4.

    G( x n , x m ,x)0 as n,m+.

Definition 1.3 [6]

Let (X,G) be a G-metric space. A sequence ( x n ) is called G-Cauchy if, for every ϵ>0, there is kN such that G( x n , x m , x l )<ϵ for all m,n,lk; that is, G( x n , x m , x l )0 as n,m,l+.

Proposition 1.2 [6]

Let (X,G) be a G-metric space. Then the following are equivalent:

  1. 1.

    The sequence ( x n ) is G-Cauchy.

  2. 2.

    For every ϵ>0, there is kN such that G( x n , x m , x m )<ϵ for all m,nk.

Definition 1.4 [6]

Let (X,G) and ( X , G ) be G-metric spaces and let f:(X,G)( X , G ) be a function. Then f is said to be G-continuous at a point aX if and only if, for every ϵ>0, there is δ>0 such that x,yX and G(a,x,y)<δ imply G (f(a),f(x),f(y))<ϵ. A function f is G-continuous at X if only if it is G-continuous at aX.

Proposition 1.3 [6]

Let (X,G) be a G-metric space. Then the function G(x,y,z) is jointly continuous in all three of its variables.

Definition 1.5 [6]

A G-metric space (X,G) is G-complete if every G-Cauchy sequence in (X,G) is G-convergent in X.

Definition 1.6 [16]

Two self-mappings f and g of a G-metric space (X,G) are said to be weakly commuting if G(fgx,gfx,gfx)G(fx,gx,gx) for all x in X.

Definition 1.7 [16]

Let f and g be two self-mappings from a G-metric space (X,G) into itself. Then the mappings f and g are said to be weakly compatible if G(fgx,gfx,gfx)=0 whenever G(fx,gx,gx)=0.

Proposition 1.4 [6]

Let (X,G) be a G-metric space. Then, for all x, y, z, a in X, it follows that:

  1. (i)

    If G(x,x,y)=0, then x=y=z;

  2. (ii)

    G(x,y,z)G(x,x,y)+G(x,x,z);

  3. (iii)

    G(x,y,y)2G(y,x,x);

  4. (iv)

    G(x,y,z)G(x,a,z)+G(a,y,z);

  5. (v)

    G(x,y,z) 2 3 (G(x,y,a)+G(x,a,z)+G(a,y,z));

  6. (vi)

    G(x,y,z)(G(x,a,a)+G(y,a,a)+G(z,a,a)).

2 Common fixed point theorems

Theorem 2.1 Let (X,G) be a complete G-metric space, and let f, g, h, A, B and C be six mappings of X into itself satisfying the following conditions:

  1. (i)

    f(X)B(X), g(X)C(X), h(X)A(X);

  2. (ii)

    x,y,zX,

    G(fx,gy,hz)kmax { G ( A x , B y , C z ) , G ( A x , f x , f x ) , G ( B y , g y , g y ) , G ( C z , h z , h z ) , G ( A x , g y , g y ) , G ( A x , h z , h z ) , G ( B y , f x , f x ) , G ( B y , h z , h z ) , G ( C z , f x , f x ) , G ( C z , g y , g y ) }
    (2.1)

    or

    G(fx,gy,hz)kmax { G ( A x , B y , C z ) , G ( A x , A x , f x ) , G ( B y , B y , g y ) , G ( C z , C z , h z ) , G ( A x , A x , g y ) , G ( A x , A x , h z ) , G ( B y , B y , f x ) , G ( B y , B y , h z ) , G ( C z , C z , f x ) , G ( C z , C z , g y ) } ,
    (2.2)

    where k[0, 1 2 ). If one of the following conditions is satisfied:

    1. (a)

      Either f or A is G-continuous, the pair (f,A) is weakly commuting, the pairs (g,B) and (h,C) are weakly compatible;

    2. (b)

      Either g or B is G-continuous, the pair (g,B) is weakly commuting, the pairs (f,A) and (h,C) are weakly compatible;

    3. (c)

      Either h or C is G-continuous, the pair (h,C) is weakly commuting, the pairs (f,A) and (g,B) are weakly compatible.

Then

  1. (I)

    one of the pairs (f,A), (g,B) and (h,C) has a coincidence point in X;

  2. (II)

    the mappings f, g, h, A, B and C have a unique common fixed point in X.

Proof Suppose that mappings f, g, h, A, B and C satisfy condition (2.1).

Let x 0 in X be an arbitrary point since f(X)B(X), g(X)C(X), h(X)A(X). There exist the sequences { x n } and { y n } in X such that

y 3 n =f x 3 n =B x 3 n + 1 , y 3 n + 1 =g x 3 n + 1 =C x 3 n + 2 , y 3 n + 2 =h x 3 n + 2 =A x 3 n + 3

for all n=0,1,2, .

If there exists n 0 N such that y n 0 = y n 0 + 1 , then the conclusion (I) of Theorem 2.1 holds. In fact, if there exists pN such that y 3 p + 2 = y 3 p + 3 , then fu=Au, where u= x 3 p + 3 . Hence the pair (f,A) has a coincidence point uX. If y 3 p = y 3 p + 1 , then gu=Bu, where u= x 3 p + 1 . Therefore, the pair (g,B) has a coincidence point uX. If y 3 p + 1 = y 3 p + 2 , then hu=Cu, where u= x 3 p + 2 . And so the pair (h,C) has a coincidence point uX.

On the other hand, if there exists n 0 N such that y n 0 = y n 0 + 1 = y n 0 + 2 , then y n = y n 0 for any n n 0 . This implies that { y n } is a G-Cauchy sequence.

Actually, if there exists pN such that y 3 p = y 3 p + 1 = y 3 p + 2 , then applying the contractive condition (2.1) with x= y 3 p + 3 , y= y 3 p + 1 and z= y 3 p + 2 , we get

G ( y 3 p + 1 , y 3 p + 2 , y 3 p + 3 ) = G ( f x 3 p + 3 , g x 3 p + 1 , h x 3 p + 2 ) k max { G ( A x 3 p + 3 , B x 3 p + 1 , C x 3 p + 2 ) , G ( A x 3 p + 3 , f x 3 p + 3 , f x 3 p + 3 ) , G ( B x 3 p + 1 , g x 3 p + 1 , g x 3 p + 1 ) , G ( C x 3 p + 2 , h x 3 p + 2 , h x 3 p + 2 ) , G ( A x 3 p + 3 , g x 3 p + 1 , g x 3 p + 1 ) , G ( A x 3 p + 3 , h x 3 p + 2 , h x 3 p + 2 ) , G ( B x 3 p + 1 , f x 3 p + 3 , f x 3 p + 3 ) , G ( B x 3 p + 1 , h x 3 p + 2 , h x 3 p + 2 ) , G ( C x 3 p + 2 , f x 3 p + 3 , f x 3 p + 3 ) , G ( C x 3 p + 2 , g x 3 p + 1 , g x 3 p + 1 ) } = k max { G ( y 3 p + 2 , y 3 p , y 3 p + 1 ) , G ( y 3 p + 2 , y 3 p + 3 , y 3 p + 3 ) , G ( y 3 p , y 3 p + 1 , y 3 p + 1 ) , G ( y 3 p + 1 , y 3 p + 2 , y 3 p + 2 ) , G ( y 3 p + 2 , y 3 p + 1 , y 3 p + 1 ) , G ( y 3 p + 2 , y 3 p + 2 , y 3 p + 2 ) , G ( y 3 p , y 3 p + 3 , y 3 p + 3 ) , G ( y 3 p , y 3 p + 2 , y 3 p + 2 ) , G ( y 3 p + 1 , y 3 p + 3 , y 3 p + 3 ) , G ( y 3 p + 1 , y 3 p + 1 , y 3 p + 1 ) } k max { G ( y 3 p + 2 , y 3 p + 3 , y 3 p + 3 ) , G ( y 3 p , y 3 p + 3 , y 3 p + 3 ) , G ( y 3 p + 1 , y 3 p + 3 , y 3 p + 3 ) } = k G ( y 3 p + 2 , y 3 p + 3 , y 3 p + 3 ) .

If y 3 p + 3 y 3 p + 1 , then from condition ( G 3 ) and Proposition 1.4(iii), we get

0<G( y 3 n + 1 , y 3 n + 2 , y 3 n + 3 )2kG( y 3 n + 2 , y 3 n + 2 , y 3 n + 3 )2kG( y 3 n + 1 , y 3 n + 2 , y 3 n + 3 ),

which implies that k 1 2 , that is a contradiction, since 0k< 1 2 . So, we find y n = y 3 p for any n3p. This implies that { y n } is a G-Cauchy sequence. The same conclusion holds if y 3 p + 1 = y 3 p + 2 = y 3 p + 3 , or y 3 p + 2 = y 3 p + 3 = y 3 p + 4 for some pN.

Assume for the rest of the paper that y n y m for any nm. Applying again (2.1) with x= y 3 n , y= y 3 n + 1 and z= y 3 n + 2 and using conditions ( G 3 ) and ( G 5 ), we get that

G ( y 3 n , y 3 n + 1 , y 3 n + 2 ) = G ( f x 3 n , g x 3 n + 1 , h x 3 n + 2 ) k max { G ( A x 3 n , B x 3 n + 1 , C x 3 n + 2 ) , G ( A x 3 n , f x 3 n , f x 3 n ) , G ( B x 3 n + 1 , g x 3 n + 1 , g x 3 n + 1 ) , G ( C x 3 n + 2 , h x 3 n + 2 , h x 3 n + 2 ) , G ( A x 3 n , g x 3 n + 1 , g x 3 n + 1 ) , G ( A x 3 n , h x 3 n + 2 , h x 3 n + 2 ) , G ( B x 3 n + 1 , f x 3 n , f x 3 n ) , G ( B x 3 n + 1 , h x 3 n + 2 , h x 3 n + 2 ) , G ( C x 3 n + 2 , f x 3 n , f x 3 n ) , G ( C x 3 n + 2 , g x 3 n + 1 , g x 3 n + 1 ) } = k max { G ( y 3 n 1 , y 3 n , y 3 n + 1 ) , G ( y 3 n 1 , y 3 n , y 3 n ) , G ( y 3 n , y 3 n + 1 , y 3 n + 1 ) , G ( y 3 n + 1 , y 3 n + 2 , y 3 n + 2 ) , G ( y 3 n 1 , y 3 n + 1 , y 3 n + 1 ) , G ( y 3 n 1 , y 3 n + 2 , y 3 n + 2 ) , G ( y 3 n , y 3 n , y 3 n ) , G ( y 3 n , y 3 n + 2 , y 3 n + 2 ) , G ( y 3 n + 1 , y 3 n , y 3 n ) , G ( y 3 n + 1 , y 3 n + 1 , y 3 n + 1 ) } k max { G ( y 3 n 1 , y 3 n , y 3 n + 1 ) , G ( y 3 n 1 , y 3 n , y 3 n + 1 ) , G ( y 3 n 1 , y 3 n , y 3 n + 1 ) , G ( y 3 n , y 3 n + 1 , y 3 n + 2 ) , G ( y 3 n 1 , y 3 n , y 3 n + 1 ) , G ( y 3 n 1 , y 3 n + 1 , y 3 n + 1 ) + G ( y 3 n + 1 , y 3 n + 2 , y 3 n + 2 ) , 0 , G ( y 3 n , y 3 n + 1 , y 3 n + 2 ) , G ( y 3 n 1 , y 3 n , y 3 n + 1 ) , 0 } k max { G ( y 3 n 1 , y 3 n , y 3 n + 1 ) , G ( y 3 n 1 , y 3 n , y 3 n + 1 ) , G ( y 3 n 1 , y 3 n , y 3 n + 1 ) , G ( y 3 n , y 3 n + 1 , y 3 n + 2 ) , G ( y 3 n 1 , y 3 n , y 3 n + 1 ) , G ( y 3 n 1 , y 3 n , y 3 n + 1 ) + G ( y 3 n , y 3 n + 1 , y 3 n + 2 ) , 0 , G ( y 3 n , y 3 n + 1 , y 3 n + 2 ) , G ( y 3 n 1 , y 3 n , y 3 n + 1 ) , 0 } = k max { G ( y 3 n 1 , y 3 n , y 3 n + 1 ) , G ( y 3 n , y 3 n + 1 , y 3 n + 2 ) , G ( y 3 n 1 , y 3 n , y 3 n + 1 ) + G ( y 3 n , y 3 n + 1 , y 3 n + 2 ) } = k [ G ( y 3 n 1 , y 3 n , y 3 n + 1 ) + G ( y 3 n , y 3 n + 1 , y 3 n + 2 ) ] .

From k[0, 1 2 ) we obtain

G( y 3 n , y 3 n + 1 , y 3 n + 2 )λG( y 3 n 1 , y 3 n , y 3 n + 1 ),
(2.3)

where λ= k 1 k [0,1). Similarly it can be shown that

G( y 3 n + 1 , y 3 n + 2 , y 3 n + 3 )λG( y 3 n , y 3 n + 1 , y 3 n + 2 )
(2.4)

and

G( y 3 n + 1 , y 3 n + 2 , y 3 n + 3 )λG( y 3 n , y 3 n + 1 , y 3 n + 2 ).
(2.5)

It follows from (2.3), (2.4) and (2.5) that, for all nN,

G( y n , y n + 1 , y n + 2 )λG( y n 1 , y n , y n + 1 ) λ 2 G( y n 2 , y n 1 , y n ) λ n G( y 0 , y 1 , y 2 ).

Therefore, for all n,mN, n<m, by ( G 3 ) and ( G 5 ), we have

G ( y n , y m , y m ) G ( y n , y n + 1 , y n + 1 ) + G ( y n + 1 , y n + 2 , y n + 2 ) + G ( y n + 2 , y n + 3 , y n + 3 ) + + G ( y m 1 , y m , y m ) G ( y n , y n + 1 , y n + 2 ) + G ( y n + 1 , y n + 2 , y n + 3 ) + + G ( y m 1 , y m , y m + 1 ) ( λ n + λ n + 1 + λ n + 2 + + λ m 1 ) G ( y 0 , y 1 , y 2 ) λ n 1 λ G ( y 0 , y 1 , y 2 ) 0 , as  n .

Hence { y n } is a G-Cauchy sequence in X. Since X is a complete G-metric space, there exists a point uX such that y n u (n).

Since the sequences {f x 3 n }={B x 3 n + 1 }, {g x 3 n + 1 }={C x 3 n + 2 } and {h x 3 n 1 }={A x 3 n } are all subsequences of { y n }, then they all converge to u

y 3 n = f x 3 n = B x 3 n + 1 u , y 3 n + 1 = g x 3 n + 1 = C x 3 n + 2 u , y 3 n 1 = h x 3 n 1 = A x 3 n u ( n ) .
(2.6)

Now we prove that u is a common fixed point of f, g, h, A, B and C under condition (a).

First, we suppose that A is continuous, the pair (f,A) is weakly commuting, the pairs (g,B) and (h,C) are weakly compatible.

Step 1. We prove that u=fu=Au.

By (2.6) and a weakly commuting of mapping pair (f,A), we have

G(fA x 3 n ,Af x 3 n ,Af x 3 n )G(f x 3 n ,A x 3 n ,A x 3 n )0(n).
(2.7)

Since A is continuous, then A 2 x 3 n Au (n), Af x 3 n Au (n). By (2.7) we know that fA x 3 n Au (n).

From condition (2.1) we know

G ( f A x 3 n , g x 3 n + 1 , h x 3 n + 2 ) k max { G ( A 2 x 3 n , B x 3 n + 1 , C x 3 n + 2 ) , G ( A 2 x 3 n , f A x 3 n , f A x 3 n ) , G ( B x 3 n + 1 , g x 3 n + 1 , g x 3 n + 1 ) , G ( C x 3 n + 2 , h x 3 n + 2 , h x 3 n + 2 ) , G ( A 2 x 3 n , g x 3 n + 1 , g x 3 n + 1 ) , G ( A 2 x 3 n , h x 3 n + 2 , h x 3 n + 2 ) , G ( B x 3 n + 1 , f A x 3 n , f A x 3 n ) , G ( B x 3 n + 1 , h x 3 n + 2 , h x 3 n + 2 ) , G ( C x 3 n + 2 , f x 3 n , f x 3 n ) , G ( C x 3 n + 2 , g x 3 n + 1 , g x 3 n + 1 ) } .

Letting n and using Proposition 1.4(iii), we have

G ( A u , u , u ) k max { G ( A u , u , u ) , G ( A u , A u , A u ) , G ( u , u , u ) , G ( u , u , u ) , G ( A u , u , u ) , G ( A u , u , u ) , G ( u , A u , A u ) , G ( u , u , u ) , G ( u , A u , A u ) , G ( u , u , u ) } = k max { G ( A u , u , u ) , G ( u , A u , A u ) } 2 k G ( A u , u , u ) ,

which implies that G(Au,u,u)=0, and so Au=u since 0k< 1 2 .

Again, by use of condition (2.1), we have

G(fu,g x 3 n + 1 ,h x 3 n + 2 )kmax { G ( A u , B x 3 n + 1 , C x 3 n + 2 ) , G ( A u , f u , f u ) , G ( B x 3 n + 1 , g x 3 n + 1 , g x 3 n + 1 ) , G ( C x 3 n + 2 , h x 3 n + 2 , h x 3 n + 2 ) , G ( A u , g x 3 n + 1 , g x 3 n + 1 ) , G ( A u , h x 3 n + 2 , h x 3 n + 2 ) , G ( B x 3 n + 1 , f u , f u ) , G ( B x 3 n + 1 , h x 3 n + 2 , h x 3 n + 2 ) , G ( C x 3 n + 2 , f u , f u ) , G ( C x 3 n + 2 , g x 3 n + 1 , g x 3 n + 1 ) } .

Letting n, using (2.6), u=Au and Proposition 1.4(iii), we obtain

G ( f u , u , u ) k max { G ( u , u , u ) , G ( u , f u , f u ) , G ( u , u , u ) , G ( u , u , u ) , G ( u , u , u ) , G ( u , u , u ) , G ( u , f u , f u ) , G ( u , u , u ) , G ( u , f u , f u ) , G ( u , u , u ) } = k G ( u , f u , f u ) 2 k G ( f u , u , u ) .

This implies that G(fu,u,u)=0 and so fu=u. Thus we have u=Au=fu.

Step 2. We prove that u=gu=Bu.

Since f(X)B(X) and u=fuf(X), there is a point vX such that u=fu=Bv. Again, by use of condition (2.1), we have

G(fu,gv,h x 3 n + 2 )kmax { G ( A u , B v , C x 3 n + 2 ) , G ( A u , f u , f u ) , G ( B v , g v , g v ) , G ( C x 3 n + 2 , h x 3 n + 2 , h x 3 n + 2 ) , G ( A u , g v , g v ) , G ( A u , h x 3 n + 2 , h x 3 n + 2 ) , G ( B v , f u , f u ) , G ( B v , h x 3 n + 2 , h x 3 n + 2 ) , G ( C x 3 n + 2 , f u , f u ) , G ( C x 3 n + 2 , g v , g v ) } .

Letting n, using u=Au=fu=Bv and Proposition 1.4(iii), we have

G ( u , g v , u ) k max { G ( u , u , u ) , G ( u , u , u ) , G ( u , g v , g v ) , G ( u , u , u ) , G ( u , g v , g v ) , G ( u , u , u ) , G ( u , u , u ) , G ( u , u , u ) , G ( u , u , u ) , G ( u , g v , g v ) } = k G ( u , g v , g v ) 2 k G ( u , g v , u ) ,

which implies that G(u,gv,u)=0, and so gv=u=Bv.

Since the pair (g,B) is weakly compatible, we have

gu=gBv=Bgv=Bu.

Again, by use of condition (2.1), we have

G(fu,gu,h x 3 n + 2 )kmax { G ( A u , B u , C x 3 n + 2 ) , G ( A u , f u , f u ) , G ( B u , g u , g u ) , G ( C x 3 n + 2 , h x 3 n + 2 , h x 3 n + 2 ) , G ( A u , g u , g u ) , G ( A u , h x 3 n + 2 , h x 3 n + 2 ) , G ( B u , f u , f u ) , G ( B u , h x 3 n + 2 , h x 3 n + 2 ) , G ( C x 3 n + 2 , f u , f u ) , G ( C x 3 n + 2 , g u , g u ) } .

Letting n, using u=Au=fu and gu=Bu and Proposition 1.4(iii), we have

G ( u , g u , u ) k max { G ( u , g u , u ) , G ( u , u , u ) , G ( g u , g u , g u ) , G ( u , u , u ) , G ( u , g u , g u ) , G ( u , u , u ) , G ( g u , u , u ) , G ( g u , u , u ) , G ( u , u , u ) , G ( u , g u , g u ) } = k max { G ( u , g u , u ) , G ( u , g u , g u ) } 2 k G ( u , g u , u ) .

This implies that G(u,gu,u)=0, and so u=gu=Bu.

Step 3. We prove that u=hu=Cu.

Since g(X)C(X) and u=gug(X), there is a point wX such that u=gu=Cw. Again, by use of condition (2.1), we have

G(fu,gu,hw)kmax { G ( A u , B u , C w ) , G ( A u , f u , f u ) , G ( B u , g u , g u ) , G ( C w , h w , h w ) , G ( A u , g u , g u ) , G ( A u , h w , h w ) , G ( B u , f u , f u ) , G ( B u , h w , h w ) , G ( C w , f u , f u ) , G ( C w , g u , g u ) } .

Using u=Au=fu, u=gu=Bu=Cw and Proposition 1.4(iii), we obtain

G ( u , u , h w ) k max { G ( u , u , u ) , G ( u , u , u ) , G ( u , u , u ) , G ( u , h w , h w ) , G ( u , u , u ) , G ( u , h w , h w ) , G ( u , u , u ) , G ( u , h w , h w ) , G ( u , u , u ) , G ( u , u , u ) } = k G ( u , h w , h w ) 2 k G ( u , u , h w ) .

Hence G(u,u,hw)=0, and so hw=u=Cw.

Since the pair (h,C) is weakly compatible, we have

hu=hCw=Chw=Cu.

Again, by use of condition (2.1), we have

G(fu,gu,hu)kmax { G ( A u , B u , C u ) , G ( A u , f u , f u ) , G ( B u , g u , g u ) , G ( C u , h u , h u ) , G ( A u , g u , g u ) , G ( A u , h u , h u ) , G ( B u , f u , f u ) , G ( B u , h u , h u ) , G ( C u , f u , f u ) , G ( C u , g u , g u ) } .

Using u=Au=fu, u=gu=Bu, Cu=hu and Proposition 1.4(iii), we have

G(u,u,hu)kmax { G ( u , u , h u ) , G ( u , h u , h u ) } 2kG(u,u,hu).

Thus G(u,u,hu)=0, and so u=hu=Cu.

Therefore u is the common fixed point of f, g, h, A, B and C when A is continuous and the pair (f,A) is weakly commuting, the pairs (g,B) and (h,C) are weakly compatible.

Next, we suppose that f is continuous, the pair (f,A) is weakly commuting, the pairs (g,B) and (h,C) are weakly compatible.

Step 1. We prove that u=fu.

By (2.6) and a weakly commuting mapping pair (f,A), we have

G(fA x 3 n ,Af x 3 n ,Af x 3 n )G(f x 3 n ,A x 3 n ,A x 3 n )0(n).
(2.8)

Since f is continuous, then f 2 x 3 n fu (n), fA x 3 n fu (n). By (2.6) we know Af x 3 n fu (n).

From condition (2.1) we know

G ( f 2 x 3 n , g x 3 n + 1 , h x 3 n + 2 ) k max { G ( A f x 3 n , B x 3 n + 1 , C x 3 n + 2 ) , G ( A f x 3 n , f 2 x 3 n , f 2 x 3 n ) , G ( B x 3 n + 1 , g x 3 n + 1 , g x 3 n + 1 ) , G ( C x 3 n + 2 , h x 3 n + 2 , h x 3 n + 2 ) , G ( A f x 3 n , g x 3 n + 1 , g x 3 n + 1 ) , G ( A f x 3 n , h x 3 n + 2 , h x 3 n + 2 ) , G ( B x 3 n + 1 , f 2 x 3 n , f 2 x 3 n ) , G ( B x 3 n + 1 , h x 3 n + 2 , h x 3 n + 2 ) , G ( C x 3 n + 2 , f 2 x 3 n , f 2 x 3 n ) , G ( C x 3 n + 2 , g x 3 n + 1 , g x 3 n + 1 ) } .

Letting n and noting Proposition 1.4(iii), we have

G ( f u , u , u ) k max { G ( f u , u , u ) , G ( f u , f u , f u ) , G ( u , u , u ) , G ( u , u , u ) , G ( f u , u , u ) , G ( f u , u , u ) , G ( u , f u , f u ) , G ( u , u , u ) , G ( u , f u , f u ) , G ( u , u , u ) } = k max { G ( f u , u , u ) , G ( u , f u , f u ) } 2 k G ( f u , u , u ) ,

which implies that G(fu,u,u)=0, and so fu=u.

Step 2. We prove that u=gu=Bu.

Since f(X)B(X) and u=fuf(X), there is a point zX such that u=fu=Bz. Again, by use of condition (2.1), we have

G ( f 2 x 3 n , g z , h x 3 n + 2 ) kmax { G ( A f x 3 n , B z , C x 3 n + 2 ) , G ( A f x 3 n , f 2 x 3 n , f 2 x 3 n ) , G ( B z , g z , g z ) , G ( C x 3 n + 2 , h x 3 n + 2 , h x 3 n + 2 ) , G ( A f x 3 n , g z , g z ) , G ( A f x 3 n , h x 3 n + 2 , h x 3 n + 2 ) , G ( B z , f 2 x 3 n , f 2 x 3 n ) , G ( B z , h x 3 n + 2 , h x 3 n + 2 ) , G ( C x 3 n + 2 , f 2 x 3 n , f 2 x 3 n ) , G ( C x 3 n + 2 , g z , g z ) } .

Letting n, using u=fu=Bz and Proposition 1.4(iii), we have

G ( u , g z , u ) k max { G ( u , u , u ) , G ( u , u , u ) , G ( u , g z , g z ) , G ( u , u , u ) , G ( u , g z , g z ) , G ( u , u , u ) , G ( u , u , u ) , G ( u , u , u ) , G ( u , u , u ) , G ( u , g z , g z ) } = k G ( u , g z , g z ) 2 k G ( u , g u , u ) .

This implies that G(u,gz,u)=0, and so gz=u=Bz.

Since the pair (g,B) is weakly compatible, we have

gu=gBz=Bgz=Bu.

Again, by use of condition (2.1), we have

G(f x 3 n ,gu,h x 3 n + 2 )kmax { G ( A x 3 n , B u , C x 3 n + 2 ) , G ( A x 3 n , f x 3 n , f x 3 n ) , G ( B u , g u , g u ) , G ( C x 3 n + 2 , h x 3 n + 2 , h x 3 n + 2 ) , G ( A x 3 n , g u , g u ) , G ( A x 3 n , h x 3 n + 2 , h x 3 n + 2 ) , G ( B u , f x 3 n , f x 3 n ) , G ( B u , h x 3 n + 2 , h x 3 n + 2 ) , G ( C x 3 n + 2 , f x 3 n , f x 3 n ) , G ( C x 3 n + 2 , g u , g u ) } .

Letting n, using u=fu, gu=Bu and Proposition 1.4(iii), we have

G ( u , g u , u ) k max { G ( u , g u , u ) , G ( u , u , u ) , G ( g u , g u , g u ) , G ( u , u , u ) , G ( u , g u , g u ) , G ( u , u , u ) , G ( g u , u , u ) , G ( g u , u , u ) , G ( u , u , u ) , G ( u , g u , g u ) } = k max { G ( u , g u , u ) , G ( u , g u , g u ) } 2 k G ( u , g u , u ) .

Therefore, G(u,gu,u)=0, and so gu=u=Bu.

Step 3. We prove that u=hu=Cu.

Since g(X)C(X) and u=gug(X), there is a point tX such that u=gu=Ct. Again, by use of condition (2.1), we have

G(f x 3 n ,gu,ht)kmax { G ( A x 3 n , B u , C t ) , G ( A x 3 n , f x 3 n , f x 3 n ) , G ( B u , g u , g u ) , G ( C t , h t , h t ) , G ( A x 3 n , g u , g u ) , G ( A x 3 n , h t , h t ) , G ( B u , f x 3 n , f x 3 n ) , G ( B u , h t , h t ) , G ( C t , f x 3 n , f x 3 n ) , G ( C t , g u , g u ) } .

Letting n, using u=gu=Bu=Ct and Proposition 1.4(iii), we obtain

G ( u , u , h t ) k max { G ( u , u , u ) , G ( u , u , u ) , G ( u , u , u ) , G ( u , h t , h t ) , G ( u , u , u ) , G ( u , h t , h t ) , G ( u , u , u ) , G ( u , h t , h t ) , G ( u , u , u ) , G ( u , u , u ) } = k G ( u , h t , h t ) 2 k G ( u , u , h t ) .

Thus G(u,u,ht)=0, and so ht=u=Ct.

Since the pair (h,C) is weakly compatible, we have

hu=hCt=Cht=Cu.

Again, by use of condition (2.1), we have

G(f x 3 n ,gu,hu)kmax { G ( A x 3 n , B u , C u ) , G ( A x 3 n , f x 3 n , f x 3 n ) , G ( B u , g u , g u ) , G ( C u , h u , h u ) , G ( A x 3 n , g u , g u ) , G ( A x 3 n , h u , h u ) , G ( B u , f x 3 n , f x 3 n ) , G ( B u , h u , h u ) , G ( C u , f x 3 n , f x 3 n ) , G ( C u , g u , g u ) } .

Letting n, using u=fu=gu=Bu, Cu=hu and Proposition 1.4(iii), we have

G ( u , u , h u ) k max { G ( u , u , h u ) , G ( u , u , u ) , G ( u , u , u ) , G ( h u , h u , h u ) , G ( u , u , u ) , G ( u , h u , h u ) , G ( u , u , u ) , G ( u , h u , h u ) , G ( h u , u , u ) , G ( h u , u , u ) } = k max { G ( u , u , h u ) , G ( u , h u , h u ) } 2 k G ( u , u , h u ) ,

which implies that G(u,u,hu)=0, and so hu=u=Cu.

Step 4. We prove that u=Au.

Since h(X)A(X) and u=huh(X), there is a point pX such that u=hu=Ap. Again, by use of condition (2.1), we have

G(fp,gu,hu)kmax { G ( A p , B u , C u ) , G ( A p , f p , f p ) , G ( B u , g u , g u ) , G ( C u , h u , h u ) , G ( A p , g u , g u ) , G ( A p , h u , h u ) , G ( B u , f p , f p ) , G ( B u , h u , h u ) , G ( C u , f p , f p ) , G ( C u , g u , g u ) } .

Using u=gu=Bu, u=hu=Cu and Proposition 1.4(iii), we obtain

G(fp,u,u)kG(u,fp,fp)2αG(fp,u,u).

Hence G(fp,u,u)=0, and so fp=u=Ap.

Since the pair (f,A) is weakly compatible, we have

fu=fAp=Afp=Au=u.

Therefore u is the common fixed point of f, g, h, A, B and C when S is continuous and the pair (f,A) is weakly commuting, the pairs (g,B) and (h,C) are weakly compatible.

Similarly we can prove the result that u is a common fixed point of f, g, h, A, B and C under the condition of (b) or (c).

Finally, we prove the uniqueness of a common fixed point u.

Let u and q be two common fixed points of f, g, h, A, B and C. By use of condition (2.1), we have

G ( q , u , u ) = G ( f q , g u , h u ) k max { G ( A q , B u , C u ) , G ( A q , f q , f q ) , G ( B u , g u , g u ) , G ( C u , h u , h u ) , G ( A q , g u , g u ) , G ( A q , h u , h u ) , G ( B u , f q , f q ) , G ( B u , h u , h u ) , G ( C u , f q , f q ) , G ( C u , g u , g u ) } = k max { G ( q , u , u ) , G ( u , q , q ) } 2 k G ( q , u , u ) .

This implies that G(q,u,u)=0, and so q=u. Thus the common fixed point is unique.

The proof using (2.2) is similar. This completes the proof. □

Now we introduce an example to support Theorem 2.1.

Example 2.2 Let X=[0,1] and let (X,G) be a G-metric space defined by G(x,y,z)=|xy|+|yz|+|zx| for all x,y,z in X. Let f, g, h, A, B and C be self-mappings defined by

f x = { 1 , x [ 0 , 1 2 ] , 5 6 , x ( 1 2 , 1 ] , g x = { 7 8 , x [ 0 , 1 2 ] , 5 6 , x ( 1 2 , 1 ] , h x = { 6 7 , x [ 0 , 1 2 ] , 5 6 , x ( 1 2 , 1 ] , A x = x , B x = { 1 , x [ 0 , 1 2 ] , 5 6 , x ( 1 2 , 1 ) , 0 , x = 1 , C x = { 1 , x [ 0 , 1 2 ] , 5 6 , x ( 1 2 , 1 ) , 7 8 , x = 1 .

Note that A is G-continuous in X, and f, g, h, B and C are not G-continuous in X.

  1. (i)

    Clearly we can get f(X)B(X), g(X)C(X), h(X)A(X).

Actually, because fX={ 5 6 ,1}, BX={0, 5 6 ,1}, gX={ 5 6 , 7 8 }, CX={0, 5 6 , 7 8 }, hX={ 5 6 , 6 7 }, AX=X=[0,1], so we know f(X)B(X), g(X)C(X) and h(X)A(X).

  1. (ii)

    By the definition of the mappings of f and A, for all x[0,1], G(fAx,Afx,Afx)=G(fx,fx,fx)=0G(fx,Ax,Ax), so we can get the pair (f,A) is weakly commuting.

By the definition of the mappings of g and B, only for x( 1 2 ,1), gx=Bx= 5 6 , at this time gBx=T( 5 6 )= 5 6 =B( 5 6 )=Bgx, so gBx=Bgx, so we can obtain the pair (g,B) is weakly compatible. Similarly we can prove the pair (h,C) is also weakly compatible.

  1. (iii)

    Now we prove the mappings f, g, h, A, B and C satisfy condition (2.1) of Theorem 2.1 with k= 2 5

    M(x,y,z)=max { G ( A x , B y , C z ) , G ( A x , f x , f x ) , G ( B y , g y , g y ) , G ( C z , h z , h z ) , G ( A x , g y , g y ) , G ( A x , h z , h z ) , G ( B y , f x , f x ) , G ( B y , h z , h z ) , G ( C z , f x , f x ) , G ( C z , g y , g y ) } .

Case 1. If x,y,z[0, 1 2 ], then

G ( f x , g y , h z ) = G ( 1 , 7 8 , 6 7 ) = 2 7 , G ( A x , f x , f x ) = G ( x , 1 , 1 ) = 2 | x 1 | 1 .

Thus we have

G(fx,gy,hz)= 2 7 < 2 5 ×1 2 5 G(Ax,fx,fx) 2 5 M(x,y,z).

Case 2. If x,y[0, 1 2 ], z( 1 2 ,1], then

G ( f x , g y , h z ) = G ( 1 , 7 8 , 5 6 ) = 1 3 , G ( A x , f x , f x ) = G ( x , 1 , 1 ) = 2 | x 1 | 1 .

Hence we get

G(fx,gy,hz)= 1 3 < 2 5 ×1 2 5 G(Ax,fx,fx) 2 5 M(x,y,z).

Case 3. If x,z[0, 1 2 ], y( 1 2 ,1], then

G ( f x , g y , h z ) = G ( 1 , 5 6 , 6 7 ) = 1 3 , G ( A x , f x , f x ) = G ( x , 1 , 1 ) = 2 | x 1 | 1 .

Therefore we obtain

G(fx,gy,hz)= 1 3 < 2 5 ×1 2 5 G(Ax,fx,fx) 2 5 M(x,y,z).

Case 4. If y,z[0, 1 2 ], x( 1 2 ,1], then

G ( f x , g y , h z ) = G ( 5 6 , 7 8 , 6 7 ) = 1 12 , G ( B y , g y , g y ) = G ( 1 , 7 8 , 7 8 ) = 1 4 .

Thus we have

G(fx,gy,hz)= 1 12 < 2 5 × 1 4 = 2 5 G(By,gy,gy) 2 5 M(x,y,z).

Case 5. If x[0, 1 2 ], y,z( 1 2 ,1], then

G ( f x , g y , h z ) = G ( 1 , 5 6 , 5 6 ) = 1 3 , G ( A x , f x , f x ) = G ( x , 1 , 1 ) = 2 | x 1 | 1 .

Hence we obtain

G(fx,gy,hz)= 1 3 < 2 5 ×1 2 5 G(Ax,fx,fx) 2 5 M(x,y,z).

Case 6. If y[0, 1 2 ], x,z( 1 2 ,1], then

G ( f x , g y , h z ) = G ( 5 6 , 7 8 , 5 6 ) = 1 12 , G ( B y , g y , g y ) = G ( 1 , 7 8 , 7 8 ) = 1 4 .

So we have

G(fx,gy,hz)= 1 12 < 2 5 × 1 4 = 2 5 G(By,gy,gy) 2 5 M(x,y,z).

Case 7. If z[0, 1 2 ], x,y( 1 2 ,1], then

G ( f x , g y , h z ) = G ( 5 6 , 5 6 , 6 7 ) = 1 21 , G ( C z , h z , h z ) = G ( 1 , 6 7 , 6 7 ) = 2 7 .

Thus we get

G(fx,gy,hz)= 1 21 < 2 5 × 2 7 = 2 5 G(Cz,hz,hz) 2 5 M(x,y,z).

Case 8. If x,y,z( 1 2 ,1], then

G(fx,gy,hz)=G ( 5 6 , 5 6 , 5 6 ) =0 2 5 M(x,y,z).

Then in all the above cases, the mappings f, g, h, A, B and C satisfy condition (2.1) of Theorem 2.1 with k= 2 5 . So that all the conditions of Theorem 2.1 are satisfied. Moreover, 5 6 is the unique common fixed point for all of the mappings f, g, h, A, B and C.

In Theorem 2.1, if we take A=B=C=I (I is identity mapping, the same below), then we have the following corollary.

Corollary 2.3 Let (X,G) be a complete G-metric space and let f, g and h be three mappings of X into itself satisfying the following conditions:

G ( f x , g y , h z ) k max { G ( x , y , z ) , G ( x , f x , f x ) , G ( y , g y , g y ) , G ( z , h z , h z ) , G ( x , g y , g y ) , G ( x , h z , h z ) , G ( y , f x , f x ) , G ( y , h z , h z ) , G ( z , f x , f x ) , G ( z , g y , g y ) }
(2.9)

or

G(fx,gy,hz)kmax { G ( x , y , z ) , G ( x , x , f x ) , G ( y , y , g y ) , G ( z , z , h z ) , G ( x , x , g y ) , G ( x , x , h z ) , G ( y , y , f x ) , G ( y , y , h z ) , G ( z , z , f x ) , G ( z , z , g y ) }
(2.10)

x,y,zX, where k[0, 1 2 ). Then f, g and h have a unique common fixed point in X.

Remark 2.4 Corollary 2.3 generalizes and extends the corresponding results in Abbas et al. [[7], Theorem 2.1].

Also, if we take f=g=h and A=B=C=I in Theorem 2.1, then we get the following.

Corollary 2.5 Let (X,G) be a complete G-metric space and let f be a mapping of X into itself satisfying the following conditions:

G(fx,fy,fz)kmax { G ( x , y , z ) , G ( x , f x , f x ) , G ( y , f y , f y ) , G ( z , f z , f z ) , G ( x , f y , f y ) , G ( x , f z , f z ) , G ( y , f x , f x ) , G ( y , f z , f z ) , G ( z , f x , f x ) , G ( z , f y , f y ) }
(2.11)

or

G(fx,fy,fz)kmax { G ( x , y , z ) , G ( x , x , f x ) , G ( y , y , f y ) , G ( z , z , f z ) , G ( x , x , f y ) , G ( x , x , f z ) , G ( y , y , f x ) , G ( y , y , f z ) , G ( z , z , f x ) , G ( z , z , f y ) }
(2.12)

x,y,zX, where k[0, 1 2 ). Then f has a unique fixed point in X.

Remark 2.6 Corollary 2.5 generalizes and extends the corresponding results in Mustafa and Sims [[8], Theorem 2.1].

Remark 2.7 Theorem 2.1, Corollaries 2.3 and 2.5 in this paper also improve and generalize the corresponding results of Abbas and Rhoades [[9], Theorems 2.4 and 2.5], Mustafa et al. [[10], Theorems 2.3, 2.5, 2.8 and Corollary 2.6], Mustafa et al. [[11], Theorem 2.5], Abbas et al. [[12], Theorem 2.1, Corollaries 2.3-2.6] and Chugh and Kadian [[13], Theorem 2.2].

Remark 2.8 In Theorem 2.1, we have taken: (1) f=g=h; (2) A=B=C; (3) g=h and B=C; (4) g=h, B=C=I, several new results can be obtained.

Theorem 2.9 Let (X,G) be a complete G-metric space and let f, g, h, A, B and C be six mappings of X into itself satisfying the following conditions:

  1. (i)

    f(X)B(X), g(X)C(X), h(X)A(X);

  2. (ii)

    The pairs (f,A), (g,B) and (h,C) are commuting mappings;

  3. (iii)

    x,y,zX,

    G ( f m x , g m y , h m z ) kmax { G ( A x , B y , C z ) , G ( A x , f m x , f m x ) , G ( B y , g m y , g m y ) , G ( C z , h m z , h m z ) , G ( A x , g m y , g m y ) , G ( A x , h m z , h m z ) , G ( B y , f m x , f m x ) , G ( B y , h m z , h m z ) , G ( C z , f m x , f m x ) , G ( C z , g m y , g m y ) }
    (2.13)

or

G ( f m x , g m y , h m z ) kmax { G ( A x , B y , C z ) , G ( A x , A x , f m x ) , G ( B y , B y , g m y ) , G ( C z , C z , h m z ) , G ( A x , A x , g m y ) , G ( A x , A x , h m z ) , G ( B y , B y , f m x ) , G ( B y , B y , h m z ) , G ( C z , C z , f m x ) , G ( C z , C z , g m y ) } ,
(2.14)

where α[0, 1 2 ), mN, then f, g, h, A, B and C have a unique common fixed point in X.

Proof Suppose that mappings f, g, h, A, B and C satisfy condition (2.13). Since f m X f m 1 XfX, fXBX, so that f m XBX. Similar, we can show that g m XCX and h m XAX. From Theorem 2.1, we see that f m , g m , h m , A, B and C have a unique common fixed point u.

It follows from (2.13) that

G ( f m f u , g m u , h m u ) αmax { G ( A f u , B u , C u ) , G ( A f u , f m f u , f m f u ) , G ( B u , g m u , g m u ) , G ( C u , h m u , h m u ) , G ( A f u , g m u , g m u ) , G ( A f u , h m u , h m u ) , G ( B u , f m f u , f m f u ) , G ( B u , h m u , h m u ) , G ( C u , f m f u , f m f u ) , G ( C u , g m u , g m u ) } .

By condition (ii) we have Afu=fAu=fu, note that fu=f( f m u)= f m + 1 u= f m (fu), and Proposition 1.4(iii), we obtain

G ( f u , u , u ) k max { G ( f u , u , u ) , G ( f u , f u , f u ) , G ( u , u , u ) , G ( u , u , u ) , G ( f u , u , u ) , G ( f u , u , u ) , G ( u , f u , f u ) , G ( u , u , u ) , G ( u , f u , f u ) , G ( u , u , u ) } = k max { G ( f u , u , u ) , G ( u , f u , f u ) } 2 k G ( f u , u , u ) ,

which implies that G(fu,u,u)=0, and so fu=u.

By the same argument, we can prove gu=u and hu=u. Thus we have u=fu=gu=hu=Au=Bu=Cu, so that f, g, h, A, B and C have a common fixed point u in X. Let v be any other common fixed point of f, g, h, A, B and C, then by use of condition (2.13) and Proposition 1.4(iii), we have

G ( u , u , v ) = G ( f m u , g m u , h m v ) k max { G ( A u , B u , C v ) , G ( A u , f m u , f m u ) , G ( B u , g m u , g m u ) , G ( C v , h m v , h m v ) , G ( A u , g m u , g m u ) , G ( A u , h m v , h m v ) , G ( B u , f m u , f m u ) , G ( B u , h m v , h m v ) , G ( C v , f m u , f m u ) , G ( C v , g m u , g m u ) } = k max { G ( u , u , v ) , G ( u , u , u ) , G ( u , u , u ) , G ( v , v , v ) , G ( u , u , u ) , G ( u , v , v ) , G ( u , u , u ) , G ( u , v , v ) , G ( v , u , u ) , G ( v , u , u ) } = k max { G ( u , u , v ) , G ( u , v , v ) } 2 k G ( u , u , v ) .

This implies that G(u,u,v)=0, and so u=v. Thus common fixed point is unique.

The proof using (2.14) is similar. This completes the proof. □

In Theorem 2.9, if we take A=B=C=I, then we have the following corollary.

Corollary 2.10 Let (X,G) be a complete G-metric space and let f, g and h be three mappings of X into itself satisfying the following conditions:

G ( f m x , g m y , h m z ) kmax { G ( x , y , z ) , G ( x , f m x , f m x ) , G ( y , g m y , g m y ) , G ( z , h m z , h m z ) , G ( x , g m y , g m y ) , G ( x , h m z , h m z ) , G ( y , f m x , f m x ) , G ( y , h m z , h m z ) , G ( z , f m x , f m x ) , G ( z , g m y , g m y ) }
(2.15)

or

G ( f m x , g m y , h m z ) kmax { G ( x , y , z ) , G ( x , x , f m x ) , G ( y , y , g m y ) , G ( z , z , h m z ) , G ( x , x , g m y ) , G ( x , x , h m z ) , G ( y , y , f m x ) , G ( y , y , h m z ) , G ( z , z , f m x ) , G ( z , z , g m y ) }
(2.16)

x,y,zX, where k[0, 1 2 ), mN, then f, g and h have a unique common fixed point in X.

Remark 2.11 Corollary 2.10 generalizes and extends the corresponding results in Abbas et al. [[7], Corollary 2.3].

Also, if we take f=g=h and A=B=C=I in Theorem 2.9, then we get the following.

Corollary 2.12 Let (X,G) be a complete G-metric space and let f be a mapping of X into itself satisfying the following conditions:

G ( f m x , f m y , f m z ) kmax { G ( x , y , z ) , G ( x , f m x , f m x ) , G ( y , f m y , f m y ) , G ( z , f m z , f m z ) , G ( x , f m y , f m y ) , G ( x , f m z , f m z ) , G ( y , f m x , f m x ) , G ( y , f m z , f m z ) , G ( z , f m x , f m x ) , G ( z , f m y , f m y ) }
(2.17)

or

G ( f m x , f m y , f m z ) kmax { G ( x , y , z ) , G ( x , x , f m x ) , G ( y , y , f m y ) , G ( z , z , f m z ) , G ( x , x , f m y ) , G ( x , x , f m z ) , G ( y , y , f m x ) , G ( y , y , f m z ) , G ( z , z , f m x ) , G ( z , z , f m y ) }
(2.18)

x,y,zX, where k[0, 1 2 ), mN, then f has a unique fixed point in X.

Remark 2.13 Corollary 2.12 generalizes and extends the corresponding results in Mustafa and Sims [[8], Corollary 2.3].

Remark 2.14 Theorem 2.9, Corollaries 2.10 and 2.12 generalize and extend the corresponding results in Mustafa et al. [[10], Corollaries 2.4 and 2.7].

Remark 2.15 In Theorem 2.9, we have taken: (1) f=g=h; (2) A=B=C; (3) g=h and B=C; (4) g=h, B=C=I, several new results can be obtained.

Theorem 2.16 Let (X,G) be a complete G-metric space and let f, g, h, A, B and C be six mappings of X into itself satisfying the following conditions:

  1. (i)

    f(X)B(X), g(X)C(X), h(X)A(X);

  2. (ii)

    x,y,zX,

    G ( f x , g y , h z ) a 1 G ( A x , B y , C z ) + a 2 G ( A x , f x , f x ) + a 3 G ( B y , g y , g y ) + a 4 G ( C z , h z , h z ) + a 5 G ( A x , g y , g y ) + a 6 G ( A x , h z , h z ) + a 7 G ( B y , f x , f x ) + a 8 G ( B y , h z , h z ) + a 9 G ( C z , f x , f x ) + a 10 G ( C z , g y , g y )
    (2.19)

    or

    G ( f x , g y , h z ) a 1 G ( A x , B y , C z ) + a 2 G ( A x , A x , f x ) + a 3 G ( B y , B y , g y ) + a 4 G ( C z , C z , h z ) + a 5 G ( A x , A x , g y ) + a 6 G ( A x , A x , h z ) + a 7 G ( B y , B y , f x ) + a 8 G ( B y , B y , h z ) + a 9 G ( C z , C z , f x ) + a 10 G ( C z , C z , g y ) ,
    (2.20)

    where a i 0 (i=1,2,3,,10) and 0 i = 1 10 a i < 1 2 . If one of the following conditions are satisfied:

    1. (a)

      Either f or A is G-continuous, the pair (f,A) is weakly commuting, the pairs (g,B) and (h,C) are weakly compatible;

    2. (b)

      Either g or B is G-continuous, the pair (g,B) is weakly commuting, the pairs (f,A) and (h,C) are weakly compatible;

    3. (c)

      Either h or C is G-continuous, the pair (h,C) is weakly commuting, the pairs (f,A) and (g,B) are weakly compatible.

Then

  1. (I)

    one of the pairs (f,A), (g,B) and (h,C) has a coincidence point in X;

  2. (II)

    the mappings f, g, h, A, B and C have a unique common fixed point in X.

Proof Suppose that mappings f, g, h, A, B and C satisfy condition (2.19). For x,y,zX, let

M(x,y,z)=max { G ( A x , B y , C z ) , G ( A x , f x , f x ) , G ( B y , g y , g y ) , G ( C z , h z , h z ) , G ( A x , g y , g y ) , G ( A x , h z , h z ) , G ( B y , f x , f x ) , G ( B y , h z , h z ) , G ( C z , f x , f x ) , G ( C z , g y , g y ) } .

Then

a 1 G ( A x , B y , C z ) + a 2 G ( A x , f x , f x ) + a 3 G ( B y , g y , g y ) + a 4 G ( C z , h z , h z ) + a 5 G ( A x , g y , g y ) + a 6 G ( A x , h z , h z ) + a 7 G ( B y , f x , f x ) + a 8 G ( B y , h z , h z ) + a 9 G ( C z , f x , f x ) + a 10 G ( C z , g y , g y ) ( i = 1 10 a i ) M ( x , y , z ) .

Therefore, it follows from (2.19) that

G(fx,gy,hz) ( i = 1 10 a i ) M(x,y,z).

Taking k= i = 1 10 a i in Theorem 2.1, the conclusion of Theorem 2.16 can be obtained from Theorem 2.1 immediately.

The proof using (2.20) is similar. This completes the proof. □

Remark 2.17 Theorem 2.16 generalizes and extends the corresponding results in Mustafa et al. [[10], Theorem 2.1], Mustafa et al. [[12], Theorem 2.5].

Remark 2.18 In Theorem 2.16, we have taken: (1) A=B=C=I; (2) f=g=h; (3) A=B=C; (4) g=h and B=C; (5) g=h, B=C=I, several new results can be obtained.

Corollary 2.19 Let (X,G) be a complete G-metric space and let f, g, h, A, B and C be six mappings of X into itself satisfying the following conditions:

  1. (i)

    f(X)B(X), g(X)C(X), h(X)A(X);

  2. (ii)

    The pairs (f,A), (g,B) and (h,C) are commuting mappings;

  3. (iii)

    x,y,zX,

    G ( f m x , g m y , h m z ) a 1 G ( A x , B y , C z ) + a 2 G ( A x , f m x , f m x ) + a 3 G ( B y , g m y , g m y ) + a 4 G ( C z , h m z , h m z ) + a 5 G ( A x , g m y , g m y ) + a 6 G ( A x , h m z , h m z ) + a 7 G ( B y , f m x , f m x ) + a 8 G ( B y , h m z , h m z ) + a 9 G ( C z , f m x , f m x ) + a 10 G ( C z , g m y , g m y )
    (2.21)

    or

    G ( f m x , g m y , h m z ) a 1 G ( A x , B y , C z ) + a 2 G ( A x , A x , f m x ) + a 3 G ( B y , B y , g m y ) + a 4 d G ( C z , B z , h m z ) + a 5 G (