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Existence of a tripled coincidence point in ordered G b -metric spaces and applications to a system of integral equations

Abstract

In this paper, tripled coincidence points of mappings satisfying some nonlinear contractive conditions in the framework of partially ordered G b -metric spaces are obtained. Our results extend the results of Aydi et al. (Fixed Point Theory Appl., 2012:101, 2012, doi:10.1186/1687-1812-2012-101). Moreover, some examples of the main result are given. Finally, some tripled coincidence point results for mappings satisfying some contractive conditions of integral type in complete partially ordered G b -metric spaces are deduced.

MSC: 47H10, 54H25.

1 Introduction and preliminaries

The concepts of mixed monotone mapping and coupled fixed point were introduced in [1] by Bhaskar and Lakshmikantham. Also, they established some coupled fixed point theorems for a mixed monotone mapping in partially ordered metric spaces. For more details on coupled fixed point theorems and related topics in different metric spaces, we refer the reader to [213] and [1425].

Also, Berinde and Borcut [26] introduced a new concept of tripled fixed point and obtained some tripled fixed point theorems for contractive-type mappings in partially ordered metric spaces. For a survey of tripled fixed point theorems and related topics, we refer the reader to [2632].

Definition 1.1 [26]

An element (x,y,z) X 3 is called a tripled fixed point of F: X 3 X if F(x,y,z)=x, F(y,x,y)=y and F(z,y,x)=z.

Definition 1.2 [27]

An element (x,y,z) X 3 is called a tripled coincidence point of the mappings F: X 3 X and g:XX if F(x,y,z)=g(x), F(y,x,y)=gy and F(z,y,x)=gz.

Definition 1.3 [27]

An element (x,y,z) X 3 is called a tripled common fixed point of F: X 3 X and g:XX if x=g(x)=F(x,y,z), y=g(y)=F(y,x,y) and z=g(z)=F(z,y,x).

Definition 1.4 [29]

Let X be a nonempty set. We say that the mappings F: X 3 X and g:XX are commutative if g(F(x,y,z))=F(gx,gy,gz) for all x,y,zX.

The notion of altering distance function was introduced by Khan et al. [10] as follows.

Definition 1.5 The function ψ:[0,)[0,) is called an altering distance function if

  1. 1.

    ψ is continuous and nondecreasing.

  2. 2.

    ψ(t)=0 if and only if t=0.

The concept of generalized metric space, or G-metric space, was introduced by Mustafa and Sims [33]. Mustafa and others studied several fixed point theorems for mappings satisfying different contractive conditions (see [3345]).

Definition 1.6 (G-metric space, [33])

Let X be a nonempty set and G: X 3 R + be a function satisfying the following properties:

  • (G1) G(x,y,z)=0 iff x=y=z;

  • (G2) 0<G(x,x,y) for all x,y?X with x?y;

  • (G3) G(x,x,y)=G(x,y,z) for all x,y,z?X with z?y;

  • (G4) G(x,y,z)=G(x,z,y)=G(y,z,x)=? (symmetry in all three variables);

  • (G5) G(x,y,z)=G(x,a,a)+G(a,y,z) for all x,y,z,a?X (rectangle inequality).

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

Example 1.7 If we think that G(x,y,z) is measuring the perimeter of the triangle with vertices at x, y and z, then (G5) can be interpreted as

[x,y]+[x,z]+[y,z]2[x,a]+[a,y]+[a,z]+[y,z],

where [x,y] is the ‘length’ of the side x, y. If we take y=z, we have

2[x,y]2[x,a]+2[a,y].

Thus, (G5) embodies the triangle inequality. And so (G5) can be sharp.

In [46], Aydi et al. established some tripled coincidence point results for mappings F: X 3 X and g:XX involving nonlinear contractions in the setting of ordered G-metric spaces.

Theorem 1.8 [46]

Let (X,) be a partially ordered set and (X,G) be a G-metric space such that (X,G) is G-complete. Let F: X 3 X and g:XX. Assume that there exist ψ,ϕ:[0,)[0,) such that ψ is an altering distance function and ϕ is a lower-semicontinuous and nondecreasing function with ϕ(t)=0 if and only if t=0 and for all x,y,z,u,v,w,r,s,tX, with gxgugr, gygvgs and gzgwgt, we have

ψ ( G ( F ( x , y , z ) , F ( u , v , w ) , F ( r , s , t ) ) ) ψ ( max { G ( g x , g u , g r ) , G ( g y , g v , g s ) , G ( g z , g w , g t ) } ) ϕ ( max { G ( g x , g u , g r ) , G ( g y , g v , g s ) , G ( g z , g w , g t ) } ) .

Assume that F and g satisfy the following conditions:

  1. (1)

    F( X 3 )g(X),

  2. (2)

    F has the mixed g-monotone property,

  3. (3)

    F is continuous,

  4. (4)

    g is continuous and commutes with F.

Let there exist x 0 , y 0 , z 0 X such that g x 0 F( x 0 , y 0 , z 0 ), g y 0 F( y 0 , x 0 , y 0 ) and g z 0 F( z 0 , y 0 , x 0 ). Then F and g have a tripled coincidence point in X, i.e., there exist x,y,zX such that F(x,y,z)=gx, F(y,x,y)=gy and F(z,y,x)=gz.

Also, they proved that the above theorem is still valid for F not necessarily continuous assuming the following hypothesis (see Theorem 19 of [46]).

  1. (I)

    If { x n } is a nondecreasing sequence with x n x, then x n x for all nN.

  2. (II)

    If { y n } is a nonincreasing sequence with y n y, then y n y for all nN.

A partially ordered G-metric space (X,G) with the above properties is called regular.

In this paper, we obtain some tripled coincidence point theorems for nonlinear (ψ,φ)-weakly contractive mappings in partially ordered G b -metric spaces. This results generalize and modify several comparable results in the literature. First, we recall the concept of generalized b-metric spaces, or G b -metric spaces.

Definition 1.9 [47]

Let X be a nonempty set and s1 be a given real number. Suppose that a mapping G: X 3 R + satisfies:

  • (G b 1) G(x,y,z)=0 if x=y=z,

  • (G b 2) 0<G(x,x,y) for all x,y?X with x?y,

  • (G b 3) G(x,x,y)=G(x,y,z) for all x,y,z?X with y?z,

  • (G b 4) G(x,y,z)=G(p{x,y,z}), where p is a permutation of x, y, z (symmetry),

  • (G b 5) G(x,y,z)=s[G(x,a,a)+G(a,y,z)] for all x,y,z,a?X (rectangle inequality).

Then G is called a generalized b-metric and the pair (X,G) is called a generalized b-metric space or a G b -metric space.

Obviously, each G-metric space is a G b -metric space with s=1. But the following example shows that a G b -metric on X need not be a G-metric on X (see also [48]).

Example 1.10 If we think that G b (x,y,z) is the maximum of the squares of length sides of a triangle with vertices at x, y and z such that:

If xyz, then G b (x,y,z)=max{ ( [ x , y ] ) 2 , ( [ y , z ] ) 2 , ( [ z , x ] ) 2 }.

If xy=z, then G b (x,y,y)= ( [ x , y ] ) 2 ,

where [x,y] is the ‘length’ of the side x, y. Then it is easy to see that G b (x,y,z) is a G b function with s=2.

Since by the triangle inequality we have

[x,y][x,a]+[a,y],[z,x][z,a]+[a,x],

hence

G b ( x , y , z ) = max { ( [ x , y ] ) 2 , ( [ y , z ] ) 2 , ( [ z , x ] ) 2 } max { ( [ x , a ] + [ a , y ] ) 2 , ( [ y , z ] ) 2 , ( [ z , a ] + [ a , x ] ) 2 } max { 2 ( ( [ x , a ] ) 2 + ( [ a , y ] ) 2 ) , ( [ y , z ] ) 2 , 2 ( ( [ z , a ] ) 2 + ( [ a , x ] ) 2 ) } 2 ( [ x , a ] ) 2 + max { 2 ( [ a , y ] ) 2 , ( [ y , z ] ) 2 , 2 ( [ z , a ] ) 2 } 2 ( [ x , a ] ) 2 + max { 2 ( [ a , y ] ) 2 , 2 ( [ y , z ] ) 2 , 2 ( [ z , a ] ) 2 } = 2 ( G b ( x , a , a ) + G b ( a , y , z ) ) .

Example 1.11 [47]

Let (X,G) be a G-metric space and G (x,y,z)=G ( x , y , z ) p , where p>1 is a real number. Then G is a G b -metric with s= 2 p 1 .

Also, in the above example, (X, G ) is not necessarily a G-metric space. For example, let X=R and G-metric G be defined by

G(x,y,z)= 1 3 ( | x y | + | y z | + | x z | )

for all x,y,zR (see [33]). Then G (x,y,z)=G ( x , y , z ) 2 = 1 9 ( | x y | + | y z | + | x z | ) 2 is a G b -metric on with s= 2 2 1 =2, but it is not a G-metric on .

Example 1.12 [47]

Let X=R and d(x,y)=|xy | 2 . We know that (X,d) is a b-metric space with s=2. Let G(x,y,z)=d(x,y)+d(y,z)+d(z,x), then (X,G) is not a G b -metric space. Indeed, (G b 3) is not true for x=0, y=2 and z=1. To see this, we have

G(0,0,2)=d(0,0)+d(0,2)+d(2,0)=2d(0,2)=8

and

G(0,2,1)=d(0,2)+d(2,1)+d(1,0)=4+1+1=6.

So, G(0,0,2)>G(0,2,1).

However, G(x,y,z)=max{d(x,y),d(y,z),d(z,x)} is a G b -metric on with s=2. Similarly, if d(x,y)=|xy | p is selected with p1, then G(x,y,z)=max{d(x,y),d(y,z),d(z,x)} is a G b -metric on with s= 2 p 1 .

Now we present some definitions and propositions in a G b -metric space.

Definition 1.13 [47]

A G b -metric G is said to be symmetric if G(x,y,y)=G(y,x,x) for all x,yX.

Definition 1.14 Let (X,G) be a G b -metric space. Then, for x 0 X and r>0, the G b -ball with center x 0 and radius r is

B G ( x 0 ,r)= { y X G ( x 0 , y , y ) < r } .

By some straight forward calculations, we can establish the following.

Proposition 1.15 [47]

Let X be a G b -metric space. Then, for each x,y,z,aX, it follows that:

  1. (1)

    if G(x,y,z)=0, then x=y=z,

  2. (2)

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

  3. (3)

    G(x,y,y)2sG(y,x,x),

  4. (4)

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

Definition 1.16 [47]

Let X be a G b -metric space. We define d G (x,y)=G(x,y,y)+G(x,x,y) for all x,yX. It is easy to see that d G defines a b-metric d on X, which we call the b-metric associated with G.

Proposition 1.17 [47]

Let X be a G b -metric space. Then, for any x 0 X and r>0, if y B G ( x 0 ,r), then there exists δ>0 such that B G (y,δ) B G ( x 0 ,r).

From the above proposition, the family of all G b -balls

Ϝ= { B G ( x , r ) x X , r > 0 }

is a base of a topology τ(G) on X, which we call the G b -metric topology.

Now, we generalize Proposition 5 in [34] for a G b -metric space as follows.

Proposition 1.18 [47]

Let X be a G b -metric space. Then, for any x 0 X and r>0, we have

B G ( x 0 , r 2 s + 1 ) B d G ( x 0 ,r) B G ( x 0 ,r).

Thus every G b -metric space is topologically equivalent to a b-metric space. This allows us to readily transport many concepts and results from b-metric spaces into G b -metric space setting.

Definition 1.19 [47]

Let X be a G b -metric space. A sequence { x n } in X is said to be:

  1. (1)

    G b -Cauchy if for each ε>0, there exists a positive integer n 0 such that, for all m,n,l n 0 , G( x n , x m , x l )<ε;

  2. (2)

    G b -convergent to a point xX if for each ε>0, there exists a positive integer n 0 such that, for all m,n n 0 , G( x n , x m ,x)<ε.

Proposition 1.20 [47]

Let X be a G b -metric space. Then the following are equivalent:

  1. (1)

    the sequence { x n } is G b -Cauchy;

  2. (2)

    for any ε>0, there exists n 0 N such that G( x n , x m , x m )<ε for all m,n n 0 .

Proposition 1.21 [47]

Let X be a G b -metric space. The following are equivalent:

  1. (1)

    { x n } is G b -convergent to x;

  2. (2)

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

  3. (3)

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

Definition 1.22 [47]

A G b -metric space X is called complete if every G b -Cauchy sequence is G b -convergent in X.

Definition 1.23 [47]

Let (X,G) and ( X , G ) be two G b -metric spaces. Then a function f:X X is G b -continuous at a point xX if and only if it is G b -sequentially continuous at x, that is, whenever { x n } is G b -convergent to x, {f( x n )} is G b -convergent to f(x).

Mustafa and Sims proved that each G-metric function G(x,y,z) is jointly continuous in all three of its variables (see Proposition 8 in [33]). But, in general, a G b -metric function G(x,y,z) for s>1 is not jointly continuous in all its variables. Now, we recall an example of a discontinuous G b -metric.

Example 1.24 [49]

Let X=N{} and let D:X×XR be defined by

D(m,n)={ 0 if  m = n , | 1 m 1 n | if one of  m , n  is even and the other is even or  , 5 if one of  m , n  is odd and the other is odd  ( and  m n )  or  , 2 otherwise .

Then it is easy to see that for all m,n,pX, we have

D(m,p) 5 2 ( D ( m , n ) + D ( n , p ) ) .

Thus, (X,D) is a b-metric space with s= 5 2 (see corrected Example 3 in [9]).

Let G(x,y,z)=max{D(x,y),D(y,z),D(z,x)}. It is easy to see that G is a G b -metric with s=5/2. In [49], it is proved that G(x,y,z) is not a continuous function.

So, from the above discussion, we need the following simple lemma about the G b -convergent sequences in the proof of our main result.

Lemma 1.25 [49]

Let (X,G) be a G b -metric space with s>1 and suppose that { x n }, { y n } and { z n } are G b -convergent to x, y and z, respectively. Then we have

1 s 3 G ( x , y , z ) lim inf n G ( x n , y n , z n ) lim sup n G ( x n , y n , z n ) s 3 G ( x , y , z ) .

In particular, if x=y=z, then we have lim n G( x n , y n , z n )=0.

In this paper, we present some tripled coincidence point results in ordered G b -metric spaces. Our results extend and generalize the results in [46].

2 Main results

Let (X,,G) be an ordered G b -metric space and F: X 3 X and g:XX. In the rest of this paper, unless otherwise stated, for all x,y,z,u,v,w,r,s,tX, let

M F ( x , y , z , u , v , w , r , s , t ) = max { G ( F ( x , y , z ) , F ( u , v , w ) , F ( r , s , t ) ) , G ( F ( y , x , y ) , F ( v , u , v ) , F ( s , r , s ) ) , G ( F ( z , y , x ) , F ( w , v , u ) , F ( t , s , r ) ) }

and

M g (x,y,z,u,v,w,r,s,t)=max { G ( g x , g u , g r ) , G ( g y , g v , g s ) , G ( g z , g w , g t ) } .

Now, the main result is presented as follows.

Theorem 2.1 Let (X,,G) be a partially ordered G b -metric space and F: X 3 X and g:XX be such that F( X 3 )g(X). Assume that

ψ ( s M F ( x , y , z , u , v , w , r , s , t ) ) ψ ( M g ( x , y , z , u , v , w , r , s , t ) ) φ ( M g ( x , y , z , u , v , w , r , s , t ) )
(2.1)

for every x,y,z,u,v,w,r,s,tX with gxgugr, gygvgs and gzgwgt, or grgugx, gsgvgy and gtgwgz, where ψ,φ:[0,)[0,) are altering distance functions.

Assume that

  1. (1)

    F has the mixed g-monotone property.

  2. (2)

    g is G b -continuous and commutes with F.

Also suppose that

  1. (a)

    either F is G b -continuous and (X,G) is G b -complete, or

  2. (b)

    (X,G) is regular and (g(X),G) is G b -complete.

If there exist x 0 , y 0 , z 0 X such that g x 0 F( x 0 , y 0 , z 0 ), g y 0 F( y 0 , x 0 , y 0 ) and g z 0 F( z 0 , y 0 , x 0 ), then F and g have a tripled coincidence point in X.

Proof Let x 0 , y 0 , z 0 X be such that g x 0 F( x 0 , y 0 , z 0 ), g y 0 F( y 0 , x 0 , y 0 ) and g z 0 F( z 0 , y 0 , x 0 ). Define x 1 , y 1 , z 1 X such that g x 1 =F( x 0 , y 0 , z 0 ), g y 1 =F( y 0 , x 0 , y 0 ) and g z 1 =F( z 0 , y 0 , x 0 ). Then g x 0 g x 1 , g y 0 g y 1 and g z 0 g z 1 . Similarly, define g x 2 =F( x 1 , y 1 , z 1 ), g y 2 =F( y 1 , x 1 , y 1 ) and g z 2 =F( z 1 , y 1 , x 1 ). Since F has the mixed g-monotone property, we have g x 0 g x 1 g x 2 , g y 0 g y 1 g y 2 and g z 0 g z 1 g z 2 .

In this way, we construct the sequences { a n }, { b n } and { c n } as

a n = g x n = F ( x n 1 , y n 1 , z n 1 ) , b n = g y n = F ( y n 1 , x n 1 , y n 1 )

and

c n =g z n =F( z n 1 , y n 1 , x n 1 )

for all n1.

We will finish the proof in two steps.

Step I. We shall show that { a n }, { b n } and { c n } are G b -Cauchy.

Let

δ n =max { G ( a n 1 , a n , a n ) , G ( b n 1 , b n , b n ) , G ( c n 1 , c n , c n ) } .

So, we have

δ n = M F ( x n 2 , y n 2 , z n 2 , x n 1 , y n 1 , z n 1 , x n 1 , y n 1 , z n 1 )

and

δ n = M g ( x n 1 , y n 1 , z n 1 , x n , y n , z n , x n , y n , z n ).

As g x n 1 g x n , g y n 1 g y n and g z n 1 g z n , using (2.1) we obtain that

ψ ( s δ n + 1 ) = ψ ( s M F ( x n 1 , y n 1 , z n 1 , x n , y n , z n , x n , y n , z n ) ) ψ ( M g ( x n 1 , y n 1 , z n 1 , x n , y n , z n , x n , y n , z n ) ) φ ( M g ( x n 1 , y n 1 , z n 1 , x n , y n , z n , x n , y n , z n ) ) = ψ ( δ n ) φ ( δ n ) ψ ( s δ n ) φ ( δ n ) .
(2.2)

Since ψ is an altering distance function, by (2.2) we deduce that

δ n + 1 δ n ,

that is, { δ n } is a nonincreasing sequence of nonnegative real numbers. Thus, there is r0 such that

lim n δ n =r.

Letting n in (2.2), from the continuity of ψ and φ, we obtain that

ψ(sr)ψ(sr)φ(r),

which implies that φ(r)=0 and hence r=0.

Next, we claim that { a n }, { b n } and { c n } are G b -Cauchy.

We shall show that for every ε>0, there exists kN such that if m,nk,

max { G ( a m , a n , a n ) , G ( b m , b n , b n ) , G ( c m , c n , c n ) } <ε.

Suppose that the above statement is false. Then there exists ε>0 for which we can find subsequences { a m ( k ) } and { a n ( k ) } of { a n }, { b m ( k ) } and { b n ( k ) } of { b n } and { c m ( k ) } and { c n ( k ) } of { c n } such that n(k)>m(k)>k and

max { G ( a m ( k ) , a n ( k ) , a n ( k ) ) , G ( b m ( k ) , b n ( k ) , b n ( k ) ) , G ( c m ( k ) , c n ( k ) , c n ( k ) ) } ε,
(2.3)

where n(k) is the smallest index with this property, i.e.,

max { G ( a m ( k ) , a n ( k ) 1 , a n ( k ) 1 ) , G ( b m ( k ) , b n ( k ) 1 , b n ( k ) 1 ) , G ( c m ( k ) , c n ( k ) 1 , c n ( k ) 1 ) } <ε.
(2.4)

From (2.4), we have

lim sup k max { G ( a m ( k ) , a n ( k ) 1 , a n ( k ) 1 ) , G ( b m ( k ) , b n ( k ) 1 , b n ( k ) 1 ) , G ( c m ( k ) , c n ( k ) 1 , c n ( k ) 1 ) } ε .
(2.5)

From the rectangle inequality,

G( a m ( k ) , a n ( k ) , a n ( k ) )s [ G ( a m ( k ) , a n ( k ) 1 , a n ( k ) 1 ) + G ( a n ( k ) 1 , a n ( k ) , a n ( k ) ) ] .
(2.6)

Similarly,

G( b m ( k ) , b n ( k ) , b n ( k ) )s [ G ( b m ( k ) , b n ( k ) 1 , b n ( k ) 1 ) + G ( b n ( k ) 1 , b n ( k ) , b n ( k ) ) ]
(2.7)

and

G( c m ( k ) , c n ( k ) , c n ( k ) )s [ G ( c m ( k ) , c n ( k ) 1 , c n ( k ) 1 ) + G ( c n ( k ) 1 , c n ( k ) , c n ( k ) ) ] .
(2.8)

So,

max { G ( a m ( k ) , a n ( k ) , a n ( k ) ) , G ( b m ( k ) , b n ( k ) , b n ( k ) ) , G ( c m ( k ) , c n ( k ) , c n ( k ) ) } s max { G ( a m ( k ) , a n ( k ) 1 , a n ( k ) 1 ) , G ( b m ( k ) , b n ( k ) 1 , b n ( k ) 1 ) , G ( c m ( k ) , c n ( k ) 1 , c n ( k ) 1 ) } + s max { G ( a n ( k ) 1 , a n ( k ) , a n ( k ) ) , G ( b n ( k ) 1 , b n ( k ) , b n ( k ) ) , G ( c n ( k ) 1 , c n ( k ) , c n ( k ) ) } .
(2.9)

Letting k as lim n δ n =0, by (2.3) and (2.4), we can conclude that

ε s lim inf k max { G ( a m ( k ) , a n ( k ) 1 , a n ( k ) 1 ) , G ( b m ( k ) , b n ( k ) 1 , b n ( k ) 1 ) , G ( c m ( k ) , c n ( k ) 1 , c n ( k ) 1 ) } .
(2.10)

Since

G( a m ( k ) , a n ( k ) , a n ( k ) )sG( a m ( k ) , a m ( k ) + 1 , a m ( k ) + 1 )+sG( a m ( k ) + 1 , a n ( k ) , a n ( k ) )
(2.11)

and

G( b m ( k ) , b n ( k ) , b n ( k ) )sG( b m ( k ) , b m ( k ) + 1 , b m ( k ) + 1 )+sG( b m ( k ) + 1 , b n ( k ) , b n ( k ) ),
(2.12)

and

G( c m ( k ) , c n ( k ) , c n ( k ) )sG( c m ( k ) , c m ( k ) + 1 , c m ( k ) + 1 )+sG( c m ( k ) + 1 , c n ( k ) , c n ( k ) ),
(2.13)

we obtain that

max { G ( a m ( k ) , a n ( k ) , a n ( k ) ) , G ( b m ( k ) , b n ( k ) , b n ( k ) ) , G ( c m ( k ) , c n ( k ) , c n ( k ) ) } s max { G ( a m ( k ) , a m ( k ) + 1 , a m ( k ) + 1 ) , G ( b m ( k ) , b m ( k ) + 1 , b m ( k ) + 1 ) , G ( c m ( k ) , c m ( k ) + 1 , c m ( k ) + 1 ) } + s max { G ( a m ( k ) + 1 , a n ( k ) , a n ( k ) ) , G ( b m ( k ) + 1 , b n ( k ) , b n ( k ) ) , G ( c m ( k ) + 1 , c n ( k ) , c n ( k ) ) } .
(2.14)

If in the above inequality k as lim n δ n =0, from (2.3) we have

ε s lim sup k max { G ( a m ( k ) + 1 , a n ( k ) , a n ( k ) ) , G ( b m ( k ) + 1 , b n ( k ) , b n ( k ) ) , G ( c m ( k ) + 1 , c n ( k ) , c n ( k ) ) } .
(2.15)

As n(k)>m(k), we have g x m ( k ) g x n ( k ) 1 , g y m ( k ) g y n ( k ) 1 and g z m ( k ) g z n ( k ) 1 . Putting x= x m ( k ) , y= y m ( k ) , z= z m ( k ) , u= x n ( k ) 1 , v= y n ( k ) 1 , w= z n ( k ) 1 , r= x n ( k ) 1 , s= y n ( k ) 1 and t= z n ( k ) 1 in (2.1), we have

ψ ( s max { G ( a m ( k ) + 1 , a n ( k ) , a n ( k ) ) , G ( b m ( k ) + 1 , b n ( k ) , b n ( k ) ) , G ( c m ( k ) + 1 , c n ( k ) , c n ( k ) ) } ) = ψ ( s M F ( x m ( k ) , y m ( k ) , z m ( k ) , x n ( k ) 1 , y n ( k ) 1 , z n ( k ) 1 , x n ( k ) 1 , y n ( k ) 1 , z n ( k ) 1 ) ) ψ ( M g ( x m ( k ) , y m ( k ) , z m ( k ) , x n ( k ) 1 , y n ( k ) 1 , z n ( k ) 1 , x n ( k ) 1 , y n ( k ) 1 , z n ( k ) 1 ) ) φ ( M g ( x m ( k ) , y m ( k ) , z m ( k ) , x n ( k ) 1 , y n ( k ) 1 , z n ( k ) 1 , x n ( k ) 1 , y n ( k ) 1 , z n ( k ) 1 ) ) = ψ ( max { G ( g x m ( k ) , g x n ( k ) 1 , g x n ( k ) 1 ) , G ( g y m ( k ) , g y n ( k ) 1 , g y n ( k ) 1 ) , G ( g z m ( k ) , g z n ( k ) 1 , g z n ( k ) 1 ) } ) φ ( max { G ( g x m ( k ) , g x n ( k ) 1 , g x n ( k ) 1 ) , G ( g y m ( k ) , g y n ( k ) 1 , g y n ( k ) 1 ) , G ( g z m ( k ) , g z n ( k ) 1 , g z n ( k ) 1 ) } ) = ψ ( max { G ( a m ( k ) , a n ( k ) 1 , a n ( k ) 1 ) , G ( b m ( k ) , b n ( k ) 1 , b n ( k ) 1 ) , G ( c m ( k ) , c n ( k ) 1 , c n ( k ) 1 ) } ) φ ( max { G ( a m ( k ) , a n ( k ) 1 , a n ( k ) 1 ) , G ( b m ( k ) , b n ( k ) 1 , b n ( k ) 1 ) , G ( c m ( k ) , c n ( k ) 1 , c n ( k ) 1 ) } ) .
(2.16)

Letting k in (2.16),

ψ ( s ε s ) ψ ( s lim sup k max { G ( a m ( k ) + 1 , a n ( k ) , a n ( k ) ) , G ( b m ( k ) + 1 , b n ( k ) , b n ( k ) ) , G ( c m ( k ) + 1 , c n ( k ) , c n ( k ) ) } ) ψ ( lim sup k max { G ( a m ( k ) , a n ( k ) 1 , a n ( k ) 1 ) , G ( b m ( k ) , b n ( k ) 1 , b n ( k ) 1 ) , G ( c m ( k ) , c n ( k ) 1 , c n ( k ) 1 ) } ) φ ( lim inf k max { G ( a m ( k ) , a n ( k ) 1 , a n ( k ) 1 ) , G ( b m ( k ) , b n ( k ) 1 , b n ( k ) 1 ) , G ( c m ( k ) , c n ( k ) 1 , c n ( k ) 1 ) } ) ψ ( ε ) φ ( lim inf k max { G ( a m ( k ) , a n ( k ) 1 , a n ( k ) 1 ) , G ( b m ( k ) , b n ( k ) 1 , b n ( k ) 1 ) , G ( c m ( k ) , c n ( k ) 1 , c n ( k ) 1 ) } ) .
(2.17)

From (2.17), we have

φ ( lim inf k max { G ( a m ( k ) , a n ( k ) 1 , a n ( k ) 1 ) , G ( b m ( k ) , b n ( k ) 1 , b n ( k ) 1 ) , G ( c m ( k ) , c n ( k ) 1 , c n ( k ) 1 ) } ) 0 .

Therefore,

lim inf k max { G ( a m ( k ) , a n ( k ) 1 , a n ( k ) 1 ) , G ( b m ( k ) , b n ( k ) 1 , b n ( k ) 1 ) , G ( c m ( k ) , c n ( k ) 1 , c n ( k ) 1 ) } =0,

which is a contradiction to (2.10). Consequently, { a n }, { b n } and { c n } are G b -Cauchy.

Step II. We shall show that F and g have a tripled coincidence point.

First, let (a) hold, that is, F is G b -continuous and (X,G) is G b -complete.

Since X is G b -complete and { a n } is G b -Cauchy, there exists aX such that

lim n G( a n , a n ,a)= lim n G(g x n ,g x n ,a)=0.
(2.18)

Similarly, there exist b,cX such that

lim n G( b n , b n ,b)= lim n G(g y n ,g y n ,b)=0
(2.19)

and

lim n G( c n , c n ,c)= lim n G(g z n ,g z n ,c)=0.
(2.20)

Now, we prove that (a,b,c) is a tripled coincidence point of F and g.

Continuity of g and Lemma 1.25 yields that

0 = 1 s 3 G ( g a , g a , g a ) lim inf n G ( g ( g x n ) , g ( g x n ) , g a ) lim sup n G ( g ( g x n ) , g ( g x n ) , g a ) s 3 G ( g a , g a , g a ) = 0 .

Hence,

lim n G ( g ( g x n ) , g ( g x n ) , g a ) =0
(2.21)

and similarly,

lim n G ( g ( g y n ) , g ( g y n ) , g b ) =0
(2.22)

and

lim n G ( g ( g z n ) , g ( g z n ) , g c ) =0.
(2.23)

Since g x n + 1 =F( x n , y n , z n ), g y n + 1 =F( y n , x n , y n ) and g z n + 1 =F( z n , y n , x n ), the commutativity of F and g yields that

g(g x n + 1 )=g ( F ( x n , y n , z n ) ) =F(g x n ,g y n ,g z n ),
(2.24)
g(g y n + 1 )=g ( F ( y n , x n , y n ) ) =F(g y n ,g x n ,g y n )
(2.25)

and

g(g z n + 1 )=g ( F ( z n , y n , x n ) ) =F(g z n ,g y n ,g x n ).
(2.26)

From the continuity of F and (2.24), (2.25) and (2.26) and Lemma 1.25, {g(g x n + 1 )} is G b -convergent to F(a,b,c), {g(g y n + 1 )} is G b -convergent to F(b,a,b) and {g(g z n + 1 )} is G b -convergent to F(c,b,a). From (2.21), (2.22) and (2.23) and uniqueness of the limit, we have F(a,b,c)=ga, F(b,a,b)=gb and F(c,b,a)=gc, that is, g and F have a tripled coincidence point.

In what follows, suppose that assumption (b) holds.

Following the proof of the previous step, there exist u,v,wX such that

lim n G(g x n ,g x n ,gu)=0,
(2.27)
lim n G(g y n ,g y n ,gv)=0
(2.28)

and

lim n G(g z n ,g z n ,gw)=0,
(2.29)

as (g(X),G) is G b -complete.

Now, we prove that F(u,v,w)=gu, F(v,u,v)=gv and F(w,v,u)=gw. From regularity of X and using (2.1), we have

ψ ( s M F ( x n , y n , z n , u , v , w , u , v , w ) ) ψ ( max { G ( g x n , g u , g u ) , G ( g y n , g v , g v ) , G ( g z n , g w , g w ) } ) φ ( max { G ( g x n , g u , g u ) , G ( g y n , g v , g v ) , G ( g z n , g w , g w ) } ) .
(2.30)

As {g x n } is G b -convergent to gu, from Lemma 1.25, we have lim n G(g x n ,gu,gu)=0. Analogously, lim n G(g y n ,gv,gv)= lim n G(g z n ,gw,gw)=0.

As ψ and φ are continuous, from (2.30) we have

lim n M F ( x n , y n , z n ,u,v,w,u,v,w)=0,

or, equivalently,

lim n G ( g x n + 1 , F ( u , v , w ) , F ( u , v , w ) ) =0.
(2.31)

Similarly,

lim n G ( g y n + 1 , F ( v , u , v ) , F ( v , u , v ) ) = lim n G ( g z n + 1 , F ( w , v , u ) , F ( w , v , u ) ) =0.
(2.32)

On the other hand,

G ( g u , F ( u , v , w ) , F ( u , v , w ) s G ( g u , g x n + 1 , g x n + 1 ) + s G ( g x n + 1 , F ( u , v , w ) , F ( u , v , w ) .
(2.33)

Taking the limit when n and using (2.27) and (2.31), we get

G ( g u , F ( u , v , w ) , F ( u , v , w ) ) s lim n G ( g u , g x n + 1 , g x n + 1 ) + s lim n G ( g x n + 1 , F ( u , v , w ) , F ( u , v , w ) = 0 ,
(2.34)

that is, gu=F(u,v,w).

Analogously, we can show that gv=F(v,u,v) and gw=F(w,v,u).

Thus, we have proved that g and F have a tripled coincidence point. This completes the proof of the theorem. □

Let

M(x,y,z,u,v,w,r,s,t)=max { G ( x , u , r ) , G ( y , v , s ) , G ( z , w , t ) } .

Taking g= I X (the identity mapping on X) in Theorem 2.1, we obtain the following tripled fixed point result.

Corollary 2.2 Let (X,,G) be a G b -complete partially ordered G b -metric space, and let F: X 3 X be a mapping with the mixed monotone property. Assume that

ψ ( s M F ( x , y , z , u , v , w , r , s , t ) ) ψ ( M ( x , y , z , u , v , w , r , s , t ) ) φ ( M ( x , y , z , u , v , w , r , s , t ) )
(2.35)

for every x,y,z,u,v,w,r,s,tX with xur, yvs and zwt, or rux, svy and twz, where ψ,φ:[0,)[0,) are altering distance functions.

Also suppose that

  1. (a)

    either F is G b -continuous, or

  2. (b)

    (X,G) is regular.

If there exist x 0 , y 0 , z 0 X such that x 0 F( x 0 , y 0 , z 0 ), y 0 F( y 0 , x 0 , y 0 ) and z 0 F( z 0 , y 0 , x 0 ), then F has a tripled fixed point in X.

Taking ψ(t)=t and φ(t)= t 2 1 + t for all t[0,) in Corollary 2.2, we obtain the following tripled fixed point result.

Corollary 2.3 Let (X,,G) be a G b -complete partially ordered G b -metric space and F: X 3 X with the mixed monotone property. Assume that

s M F (x,y,z,u,v,w,r,s,t) M ( x , y , z , u , v , w , r , s , t ) 1 + M ( x , y , z , u , v , w , r , s , t )
(2.36)

for every x,y,z,u,v,w,r,s,tX with xur, yvs and zwt, or rux, svy and twz.

Also suppose that

  1. (a)

    either F is G b -continuous, or

  2. (b)

    (X,G) is regular.

If there exist x 0 , y 0 , z 0 X such that x 0 F( x 0 , y 0 , z 0 ), y 0 F( y 0 , x 0 , y 0 ) and z 0 F( z 0 , y 0 , x 0 ), then F has a tripled fixed point in X.

Remark 2.4 Theorem 1.8 is a special case of Theorem 2.1.

Remark 2.5 Theorem 2.1 part (a) holds if we replace the commutativity assumption of F and g by compatibility assumption (also see Remark 2.2 of [30]).

The following corollary can be deduced from our previously obtained results.

Corollary 2.6 Let (X,) be a partially ordered set and (X,G) be a G b -complete G b -metric space. Let F: X 3 X be a mapping with the mixed monotone property such that

ψ ( s M F ( x , y , z , u , v , w , r , s , t ) ) ψ ( G ( x , u , r ) + G ( y , v , s ) + G ( z , w , t ) 3 ) φ ( max { G ( x , u , r ) , G ( y , v , s ) , G ( z , w , t ) } )
(2.37)

for every x,y,z,u,v,w,r,s,tX with xur, yvs and zwt, or rux, svy and twz.

Also suppose that

  1. (a)

    either F is G b -continuous, or

  2. (b)

    (X,G) is regular.

If there exist x 0 , y 0 , z 0 X such that x 0 F( x 0 , y 0 , z 0 ), y 0 F( y 0 , x 0 , y 0 ) and z 0 F( z 0 , y 0 , x 0 ), then F has a tripled fixed point in X.

Proof If F satisfies (2.37), then F satisfies (2.35). So, the result follows from Theorem 2.1. □

In Theorem 2.1, if we take ψ(t)=t and φ(t)=(1k)t for all t[0,), where k[0,1), we obtain the following result.

Corollary 2.7 Let (X,) be a partially ordered set and (X,G) be a G b -complete G b -metric space. Let F: X 3 X be a mapping having the mixed monotone property and

M F (x,y,z,u,v,w,r,s,t) k s M(x,y,z,u,v,w,r,s,t)
(2.38)

for every x,y,z,u,v,w,r,s,tX with xur, yvs and zwt, or rux, svy and twz.

Also suppose that

  1. (a)

    either F is G b -continuous, or

  2. (b)

    (X,G) is regular.

If there exist x 0 , y 0 , z 0 X such that x 0 F( x 0 , y 0 , z 0 ), y 0 F( y 0 , x 0 , y 0 ) and z 0 F( z 0 , y 0 , x 0 ), then F has a tripled fixed point in X.

Corollary 2.8 Let (X,) be a partially ordered set and (X,G) be a G b -complete G b -metric space. Let F: X 3 X be a mapping with the mixed monotone property such that

M F (x,y,z,u,v,w,r,s,t) k 3 s [ G ( x , u , r ) + G ( y , v , s ) + G ( z , w , t ) ]
(2.39)

for every x,y,z,u,v,w,r,s,tX with xur, yvs and zwt, or rux, svy and twz.

Also suppose that

  1. (a)

    either F is G b -continuous, or

  2. (b)

    (X,G) is regular.

If there exist x 0 , y 0 , z 0 X such that x 0 F( x 0 , y 0 , z 0 ), y 0 F( y 0 , x 0 , y 0 ) and z 0 F( z 0 , y 0 , x 0 ), then F has a tripled fixed point in X.

Proof If F satisfies (2.39), then F satisfies (2.38). □

Note that if (X,) is a partially ordered set, then we can endow X 3 with the following partial order relation:

(x,y,z)(u,v,w)xu,yv,zw

for all (x,y,z),(u,v,w) X 3 (see [26]).

In the following theorem, we give a sufficient condition for the uniqueness of the common tripled fixed point (also see, e.g., [4, 46, 50] and [51]).

Theorem 2.9 In addition to the hypotheses of Theorem 2.1, suppose that for every (x,y,z) and ( x , y , z )X×X×X, there exists (u,v,w) X 3 such that (F(u,v,w),F(v,u,v),F(w,v,u)) is comparable with (F(x,y,z),F(y,x,y),F(z,y,x)) and (F( x , y , z ),F( y , x , y ),F( z , y , x )). Then F and g have a unique common tripled fixed point.

Proof From Theorem 2.1 the set of tripled coincidence points of F and g is nonempty. We shall show that if (x,y,z) and ( x , y , z ) are tripled coincidence points, that is,

g(x)=F(x,y,z),g(y)=F(y,x,y),g(z)=F(z,y,x)

and

g ( x ) =F ( x , y , z ) ,g ( y ) =F ( y , x , y ) ,g ( z ) =F ( z , y , x ) ,

then gx=g x and gy=g y and gz=g z .

Choose an element (u,v,w) X 3 such that (F(u,v,w),F(v,u,v),F(w,v,u)) is comparable with

( F ( x , y , z ) , F ( y , x , y ) , F ( z , y , x ) )

and

( F ( x , y , z ) , F ( y , x , y ) , F ( z , y , x ) ) .

Let u 0 =u, v 0 =v and w 0 =w and choose u 1 , v 1 and w 1 X so that g u 1 =F( u 0 , v 0 , w 0 ), g v 1 =F( v 0 , u 0 , v 0 ) and g w 1 =F( w 0 , v 0 , u 0 ). Then, similarly as in the proof of Theorem 2.1, we can inductively define sequences {g u n }, {g v n } and {g w n } such that g u n + 1 =F( u n , v n , w n ), g v n + 1 =F( v n , u n , v n ) and g w n + 1 =F( w n , v n , u n ). Since (gx,gy,gz)=(F(x,y,z),F(y,x,y),F(w,y,x)) and (F(u,v,w),F(v,u,v),F(w,v,u))=(g u 1 ,g v 1 ,g w 1 ) are comparable, we may assume that (gx,gy,gz)(g u 1 ,g v 1 ,g w 1 ). Then gxg u 1 , gyg v 1 and gzg w 1 . Using the mathematical induction, it is easy to prove that gxg u n , gyg v n and gzg w n for all n0.

Applying (2.1), as gxg u n , gyg v n and gzg w n , one obtains that

ψ ( s max { G ( g x , g u n + 1 , g u n + 1 ) , G ( g y , g v n + 1 , g v n + 1 ) , G ( g z , g w n + 1 , g w n + 1 ) } ) = ψ ( s M F ( x , y , z , u n , v n , w n , u n , v n , w n ) ) ψ ( M g ( x , y , z , u n , v n , w n , u n , v n , w n ) ) φ ( M g ( x , y , z , u n , v n , w n , u n , v n , w n ) ) = ψ ( max { G ( g x , g u n , g u n ) , G ( g y , g v n , g v n ) , G ( g z , g w n , g w n ) } ) φ ( max { G ( g x , g u n , g u n ) , G ( g y , g v n , g v n ) , G ( g z , g w n , g w n ) } ) .
(2.40)

From the properties of ψ, we deduce that

{ max { G ( g x , g u n , g u n ) , G ( g y , g v n , g v n ) , G ( g z , g w n , g w n ) } }

is nonincreasing.

Hence, if we proceed as in Theorem 2.1, we can show that

lim n max { G ( g x , g u n , g u n ) , G ( g y , g v n , g v n ) , G ( g z , g w n , g w n ) } =0,

that is, {g u n }, {g v n } and {g w n } are G b -convergent to gx, gy and gz, respectively.

Similarly, we can show that

lim n max { G ( g x , g u n , g u n ) , G ( g y , g v n , g v n ) , G ( g z , g w n , g w n ) } =0,

that is, {g u n }, {g v n } and {g w n } are G b -convergent to g x , g y and g z , respectively. Finally, since the limit is unique, gx=g x , gy=g y and gz=g z .

Since gx=F(x,y,z), gy=F(y,x,y) and gz=F(z,y,x), by commutativity of F and g, we have g(gx)=g(F(x,y,z))=F(gx,gy,gz), g(gy)=g(F(y,x,y))=F(gy,gx,gy) and g(gz)=g(F(z,y,x))=F(gz,gy,gx). Let gx=a, gy=b and g(z)=c. Then ga=F(a,b,c), gb=F(b,a,b) and gc=F(c,b,a). Thus, (a,b,c) is another tripled coincidence point of F and g. Then a=gx=ga, b=gy=gb and c=gz=gc. Therefore, (a,b,c) is a tripled common fixed point of F and g.

To prove the uniqueness, assume that (p,q,r) is another tripled common fixed point of F and g. Then p=gp=F(p,q,r), q=gq=F(q,p,q) and r=gr=F(r,p,q). Since (p,q,r) is a tripled coincidence point of F and g, we have gp=gx, gq=gy and gr=gz. Thus, p=gp=ga=a, q=gq=gb=b and r=gr=gc=c. Hence, the tripled common fixed point is unique. □

3 Examples

The following examples support our results.

Example 3.1 Let X=(,) be endowed with the usual ordering and the G b -complete G b -metric

G(x,y,z)= ( | x y | + | y z | + | z x | ) 2 ,

where s=2.

Define F: X 3 X as

F(x,y,z)= x 2 y + 4 z 96

for all x,y,zX and g:XX with g(x)= x 2 for all xX.

Let φ:[0,)[0,) be defined by φ(t)=ln(t+1), and let ψ:[0,)[0,) be defined by ψ(t)=ln( 4 t + 4 t + 4 ).

Now, from the fact that for α,β,γ0, ( α + β + γ ) p 2 2 p 2 α p + 2 2 p 2 β p + 2 p 1 γ p , we have

ψ ( s G ( F ( x , y , z ) , F ( u , v , w ) , F ( r , s , t ) ) ) = ln ( 2 ( 1 96 [ | ( x 2 y + 4 z ) ( u 2 v + 4 w ) | ] + 1 96 [ | ( u 2 v + 4 w ) ( r 2 s + 4 t ) | ] + 1 96 [ | ( r 2 s + 4 t ) ( x 2 y + 4 z ) | ] ) 2 + 1 ) ln ( 2 ( 1 48 | x 2 u 2 | + 1 24 | y 2 v 2 | + 1 12 | z 2 w 2 | + 1 48 | u 2 r 2 | + 1 24 | v 2 s 2 | + 1 12 | w 2 t 2 | + 1 48 | r 2 x 2 | + 1 24 | s 2 y 2 | + 1 12 | t 2 z 2 | ) 2 + 1 ) = ln ( 2 ( 1 48 [ | x 2 u 2 | + | u 2 r 2 | + | r 2 x 2 | ] + 1 24 [ | y 2 v 2 | + | v 2 s 2 | + | s 2 y 2 | ] + 1 12 [ | z 2 w 2 | + | w 2 t 2 | + | t 2 z 2 | ] ) 2 + 1 ) ln ( 8 48 2 ( [ | x 2 u 2 | + | u 2 r 2 | + | r 2 x 2 | ] 2 + 8 24 2 [ | y 2 v 2 | + | v 2 s 2 | + | s 2 y 2 | ] 2 + 4 12 2 [ | z 2 w 2 | + | w 2 t 2 | + | t 2 z 2 | ] 2 ) + 1 ) ln ( 1 12 ( [ | x 2 u 2 | + | u 2 r 2 | + | r 2 x 2 | ] 2 + 1 12 [ | y 2 v 2 | + | v 2 s 2 | + | s 2 y 2 | ] 2 + 1 12 [ | z 2 w 2 | + | w 2 t 2 | + | t 2 z 2 | ] 2 ) + 1 ) ln ( 1 4 max { [ | x 2 u 2 | + | u 2 r 2 | + | r 2 x 2 | ] 2 , [ | y 2 v 2 | + | v 2 s 2 | + | s 2 y 2 | ] 2 , [ | z 2 w 2 | + | w 2 t 2 | + | t 2 z 2 | ] 2 } + 1 ) ln ( 1 4 max { G ( g x , g u , g r ) , G ( g y , g v , g s ) , G ( g z , g w , g t ) } + 1 ) = ln ( max { G ( g x , g u , g r ) , G ( g y , g v , g s ) , G ( g z , g w , g t ) } + 1 ) ln ( 4 max { G ( g x , g u , g r ) , G ( g y , g v , g s ) , G ( g z , g w , g t ) } + 4 max { G ( g x , g u , g r ) , G ( g y , g v , g s ) , G ( g z , g w , g t ) } + 4 ) = ψ ( max { G ( g x , g u , g r ) , G ( g y , g v , g s ) , G ( g z , g w , g t ) } ) φ ( max { G ( g x , g u , g r ) , G ( g y , g v , g s ) , G ( g z , g w , g t ) } ) .

Analogously, we can show that

ψ ( G ( F ( y , x , y ) , F ( v , u , v ) , F ( s , r , s ) ) ) ψ ( max { G ( g x , g u , g r ) , G ( g y , g v , g s ) , G ( g z , g w , g t ) } ) φ ( max { G ( g x , g u , g r ) , G ( g y , g v , g s ) , G ( g z , g w , g t ) } )

and

ψ ( G ( F ( z , y , x ) , F ( w , v , u ) , F ( t , s , r ) ) ) ψ ( max { G ( g x , g u , g r ) , G ( g y , g v , g s ) , G ( g z , g w , g t ) } ) φ ( max { G ( g x , g u , g r ) , G ( g y , g v , g s ) , G ( g z , g w , g t ) } ) .

Thus,

ψ ( s M F ( x , y , z , u , v , w , r , s , t ) ) ψ ( M g ( x , y , z , u , v , w , r , s , t ) ) φ ( M g ( x , y , z , u , v , w , r , s , t ) ) .

Hence, all of the conditions of Theorem 2.1 are satisfied. Moreover, (0,0,0) is the unique common tripled fixed point of F and g.

The following example has been constructed according to Example 2.12 of [2].

Example 3.2 Let X={(x,0,x)}{(0,x,0)} R 3 , where x[0,] with the order defined as

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

Let d be given as

d(x,y)=max { | x 1 x 2 | 2 , | y 1 y 2 | 2 , | z 1 z 2 | 2 }

and

G(x,y,z)=max { d ( x , y ) , d ( y , z ) , d ( z , x ) } ,

where x=( x 1 , y 1 , z 1 ) and y=( x 2 , y 2 , z 2 ). (X,G) is, clearly, a G b -complete G b -metric space.

Let g:XX and F: X 3 X be defined as follows:

F(x,y,z)=x

and

g ( ( x , 0 , x ) ) =(0,x,0)andg ( ( 0 , x , 0 ) ) =(x,0,x).

Let ψ,φ:[0,)[0,) be as in the above example.

According to the order on X and the definition of g, we see that for any element xX, g(x) is comparable only with itself.

By a careful computation, it is easy to see that all of the conditions of Theorem 2.1 are satisfied. Finally, Theorem 2.1 guarantees the existence of a unique common tripled fixed point for F and g, i.e., the point ((0,0,0),(0,0,0),(0,0,0)).

4 Applications

In this section, we obtain some tripled coincidence point theorems for a mapping satisfying a contractive condition of integral type in a complete ordered G b -metric space.

We denote by Λ the set of all functions μ:[0,+)[0,+) verifying the following conditions:

  1. (I)

    μ is a positive Lebesgue integrable mapping on each compact subset of [0,+).

  2. (II)

    For all ε>0, 0 ε μ(t)dt>0.

Corollary 4.1 Replace the contractive condition (2.1) of Theorem 2.1 by the following condition:

There exists μΛ such that

0 ψ ( s M F ( x , y , z , u , v , w , r , s , t ) ) μ ( t ) d t 0 ψ ( M g ( x , y , z , u , v , w , r , s , t ) ) μ ( t ) d t 0 φ ( M g ( x , y , z , u , v , w , r , s , t ) ) μ ( t ) d t .
(4.1)

If the other conditions of Theorem 2.1 are satisfied, then F and g have a tripled coincidence point.

Proof Consider the function Γ(x)= 0 x μ(t)dt. Then (4.1) becomes

Γ ( ψ ( s M F ( x , y , z , u , v , w , r , s , t ) ) ) Γ ( ψ ( M g ( x , y , z , u , v , w , r , s , t ) ) ) Γ ( φ ( M g ( x , y , z , u , v , w , r , s , t ) ) ) .

Taking ψ 1 =Γoψ and φ 1 =Γoφ and applying Theorem 2.1, we obtain the proof (it is easy to verify that ψ 1 and φ 1 are altering distance functions). □

Corollary 4.2 Substitute the contractive condition (2.1) of Theorem 2.1 by the following condition:

There exists μΛ such that

ψ ( 0 s M F ( x , y , z , u , v , w , r , s , t ) μ ( t ) d t ) ψ ( 0 M g ( x , y , z , u , v , w , r , s , t ) μ ( t ) d t ) φ ( 0 M g ( x , y , z , u , v , w , r , s , t ) μ ( t ) d t ) .
(4.2)

If the other conditions of Theorem 2.1 are satisfied, then F and g have a tripled coincidence point.

Proof Again, as in Corollary 4.1, define the function Γ(x)= 0 x μ(t)dt. Then (4.2) changes to

ψ ( Γ ( s M F ( x , y , z , u , v , w , r , s , t ) ) ) ψ ( Γ ( M g ( x , y , z , u , v , w , r , s , t ) ) ) φ ( Γ ( M g ( x , y , z , u , v , w , r , s , t ) ) ) .

Now, if we define ψ 1 =ψoΓ and φ 1 =φoΓ and apply Theorem 2.1, then the proof is completed. □

Corollary 4.3 Replace the contractive condition (2.1) of Theorem 2.1 by the following condition:

There exists μΛ such that

ψ 1 ( 0 ψ 2 ( s M F ( x , y , z , u , v , w , r , s , t ) ) μ ( t ) d t ) ψ 1 ( 0 ψ 2 ( M g ( x , y , z , u , v , w , r , s , t ) ) μ ( t ) d t ) φ 1 ( 0 φ 2 ( M g ( x , y , z , u , v , w , r , s , t ) ) μ ( t ) d t )
(4.3)

for altering distance functions ψ 1 , ψ 2 , φ 1 and φ 2 . If the other conditions of Theorem 2.1 are satisfied, then F and g have a tripled coincidence point.

Similar to [52], let NN be fixed. Let { μ i } 1 i N be a family of N functions which belong to Λ. For all t0, we define

I 1 ( t ) = 0 t μ 1 ( s ) d s , I 2 ( t ) = 0 I 1 t μ 2 ( s ) d s = 0 0 t μ 1 ( s ) d s μ 2 ( s ) d s , I 3 ( t ) = 0 I 2 t μ 3 ( s ) d s = 0 0 0 t μ 1 ( s ) d s μ 2 ( s ) d s μ 3 ( s ) d s , , I N ( t ) = 0 I ( N 1 ) t μ N ( s ) d s .

We have the following result.

Corollary 4.4 Replace inequality (2.1) of Theorem 2.1 by the following condition:

ψ ( I N ( s M F ( x , y , z , u , v , w , r , s , t ) ) ) ψ ( I N ( M g ( x , y , z , u , v , w , r , s , t ) ) ) φ ( I N ( M g ( x , y , z , u , v , w , r , s , t ) ) ) .
(4.4)

If the other conditions of Theorem 2.1 are satisfied, then F and g have a tripled coincidence point.

Proof Consider Ψ ˆ =ψo I N and Φ ˆ =φo I N . Then the above inequality becomes

Ψ ˆ ( s M F ( x , y , z , u , v , w , r , s , t ) ) Ψ ˆ ( M g ( x , y , z , u , v , w , r , s , t ) ) Φ ˆ ( M g ( x , y , z , u , v ,