Open Access

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

Journal of Inequalities and Applications20132013:453

https://doi.org/10.1186/1029-242X-2013-453

Received: 14 January 2013

Accepted: 2 October 2013

Published: 7 November 2013

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.

Keywords

tripled fixed pointgeneralized weakly contractiongeneralized metric spacespartially ordered set

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 : X X if F ( x , y , z ) = g ( x ) , F ( y , x , y ) = g y and F ( z , y , x ) = g z .

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 : X X 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 : X X are commutative if g ( F ( x , y , z ) ) = F ( g x , g y , g z ) for all x , y , z X .

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 : X X 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 : X X . 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 , t X , with g x g u g r , g y g v g s and g z g w g t , 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 , z X such that F ( x , y , z ) = g x , F ( y , x , y ) = g y and F ( z , y , x ) = g z .

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 n N .

     
  2. (II)

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

     

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 s 1 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 x y z , then G b ( x , y , z ) = max { ( [ x , y ] ) 2 , ( [ y , z ] ) 2 , ( [ z , x ] ) 2 } .

If x y = 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 , z R (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 ) = | x y | 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 ) = 2 d ( 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 ) = | x y | p is selected with p 1 , 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 , y X .

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 , a X , 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 ) 2 s G ( 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 , y X . 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 x X 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 x X 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 × X R 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 , p X , 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 : X X . In the rest of this paper, unless otherwise stated, for all x , y , z , u , v , w , r , s , t X , 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 : X X 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 , t X with g x g u g r , g y g v g s and g z g w g t , or g r g u g x , g s g v g y and g t g w g z , 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 n 1 .

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 r 0 such that
lim n δ n = r .
Letting n in (2.2), from the continuity of ψ and φ, we obtain that
ψ ( s r ) ψ ( s r ) 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 k N such that if m , n k ,
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 ) ) s G ( a m ( k ) , a m ( k ) + 1 , a m ( k ) + 1 ) + s G ( a m ( k ) + 1 , a n ( k ) , a n ( k ) )
(2.11)
and
G ( b m ( k ) , b n ( k ) , b n ( k ) ) s G ( b m ( k ) , b m ( k ) + 1 , b m ( k ) + 1 ) + s G ( b m ( k ) + 1 , b n ( k ) , b n ( k ) ) ,
(2.12)
and
G ( c m ( k ) , c n ( k ) , c n ( k ) ) s G ( c m ( k ) , c m ( k ) + 1 , c m ( k ) + 1 ) + s G ( 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 a X 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 , c X 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 ) = g a , F ( b , a , b ) = g b and F ( c , b , a ) = g c , 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 , w X such that
lim n G ( g x n , g x n , g u ) = 0 ,
(2.27)
lim n G ( g y n , g y n , g v ) = 0
(2.28)
and
lim n G ( g z n , g z n , g w ) = 0 ,
(2.29)

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

Now, we prove that F ( u , v , w ) = g u , F ( v , u , v ) = g v and F ( w , v , u ) = g w . 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 , g u , g u ) = 0 . Analogously, lim n G ( g y n , g v , g v ) = lim n G ( g z n , g w , g w ) = 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, g u = F ( u , v , w ) .

Analogously, we can show that g v = F ( v , u , v ) and g w = 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 , t X with x u r , y v s and z w t , or r u x , s v y and t w z , 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 , t X with x u r , y v s and z w t , or r u x , s v y and t w z .

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 , t X with x u r , y v s and z w t , or r u x , s v y and t w z .

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 ) = ( 1 k ) 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 , t X with x u r , y v s and z w t , or r u x , s v y and t w z .

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 , t X with x u r , y v s and z w t , or r u x , s v y and t w z .

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 ) x u , y v , z w

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 g x = g x and g y = g y and g z = 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 ( g x , g y , g z ) = ( 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 ( g x , g y , g z ) ( g u 1 , g v 1 , g w 1 ) . Then g x g u 1 , g y g v 1 and g z g w 1 . Using the mathematical induction, it is easy to prove that g x g u n , g y g v n and g z g w n for all n 0 .

Applying (2.1), as g x g u n , g y g v n and g z g 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, g x = g x , g y = g y and g z = g z .

Since g x = F ( x , y , z ) , g y = F ( y , x , y ) and g z = F ( z , y , x ) , by commutativity of F and g, we have g ( g x ) = g ( F ( x , y , z ) ) = F ( g x , g y , g z ) , g ( g y ) = g ( F ( y , x , y ) ) = F ( g y , g x , g y ) and g ( g z ) = g ( F ( z , y , x ) ) = F ( g z , g y , g x ) . Let g x = a , g y = b and g ( z ) = c . Then g a = F ( a , b , c ) , g b = F ( b , a , b ) and g c = F ( c , b , a ) . Thus, ( a , b , c ) is another tripled coincidence point of F and g. Then a = g x = g a , b = g y = g b and c = g z = g c . 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 = g p = F ( p , q , r ) , q = g q = F ( q , p , q ) and r = g r = F ( r , p , q ) . Since ( p , q , r ) is a tripled coincidence point of F and g, we have g p = g x , g q = g y and g r = g z . Thus, p = g p = g a = a , q = g q = g b = b and r = g r = g c = 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 , z X and g : X X with g ( x ) = x 2 for all x X .

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 , γ 0 , ( 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 : X X and F : X 3 X be defined as follows:
F ( x , y , z ) = x
and
g ( ( x , 0 , x ) ) = ( 0 , x , 0 ) and g ( ( 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 x X , 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 ) d t > 0 .

     

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

There exists μ Λ 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 ) d t . 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 = Γ o ψ and φ 1 = Γ o φ 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 μ Λ 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 ) d t . 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 = ψ o Γ and φ 1 = φ o Γ 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 μ Λ 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 N N be fixed. Let { μ i } 1 i N be a family of N functions which belong to Λ. For all t 0 , 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 = ψ o I N and Φ ˆ = φ o 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 , w , r , s , t ) ) .

Applying Theorem 2.1, we obtain the desired result (it is easy to verify that Ψ ˆ and Φ ˆ are altering distance functions). □

Another consequence of the main theorem is the following result.

Corollary 4.5 Substitute contractive condition (2.1) of Theorem 2.1 by the following condition:

There exist μ 1 , μ 2 Λ such that
0 s M F ( x , y , z , u , v , w , r , s , t ) μ 1 ( t ) d t 0 M g ( x , y , z , u , v , w , r , s , t ) μ 1 ( t ) d t 0 M g ( x , y , z , u , v , w , r , s , t ) μ 2 ( t ) d t .

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

Proof It is clear that the function s 0 s μ i ( t ) d t for i = 1 , 2 is an altering distance function. □

Motivated by [46], we study the existence of solutions for nonlinear integral equations using the results proved in the previous section.

Consider the integral equations in the following system.
x ( t ) = ω ( t ) + 0 T S ( t , r ) [ f ( r , x ( r ) ) + k ( r , y ( r ) ) + h ( r , z ( r ) ) ] d r , y ( t ) = ω ( t ) + 0 T S ( t , r ) [ f ( r , y ( r ) ) + k ( r , x ( r ) ) + h ( r , y ( r ) ) ] d r , z ( t ) = ω ( t ) + 0 T S ( t , r ) [ f ( λ , z ( r ) ) + k ( r , y ( r ) ) + h ( r , x ( r ) ) ] d r .
(4.5)
We will consider system (4.5) under the following assumptions:
  1. (i)

    f , k , h : [ 0 , T ] × R R are continuous,

     
  2. (ii)

    ω : [ 0 , T ] R is continuous,

     
  3. (iii)

    S : [ 0 , T ] × R [ 0 , ) is continuous,

     
  4. (iv)
    there exists q > 0 such that for all x , y R ,
    0 f ( r , y ) f ( r , x ) q ( y x ) , 0 k ( r , x ) k ( r , y ) q ( y x )
    and
    0 h ( r , y ) h ( r , x ) q ( y x ) .
     
  5. (v)
    We suppose that
    2 3 p 3 3 q p sup t [ 0 , T ] ( 0 T | S ( t , r ) | d r ) p < 1 .
     
  6. (vi)
    There exist continuous functions α , β , γ : [ 0 , T ] R such that
    α ( t ) ω ( t ) + 0 T S ( t , r ) [ f ( r , α ( r ) ) + k ( r , β ( r ) ) + h ( r , γ ( r ) ) ] d r , β ( t ) ω ( t ) + 0 T S ( t , r ) [ f ( r , β ( r ) ) + k ( r , α ( r ) ) + h ( r , β ( r ) ) ] d r
     
and
γ ( t ) ω ( t ) + 0 T S ( t , r ) [ f ( r , γ ( r ) ) + k ( r , β ( r ) ) + h ( r , α ( r ) ) ] d r .
We consider the space X = C ( [ 0 , T ] , R ) of continuous functions defined on [ 0 , T ] endowed with the G b -metric given by
G ( θ , φ , ψ ) = ( max t [ 0 , T ] | θ ( t ) φ ( t ) | p , max t [ 0 , T ] | φ ( t ) ψ ( t ) | p , max t [ 0 , T ] | ψ ( t ) θ ( t ) | p )

for all θ , φ , ψ X , where s = 2 p 1 and p 1 (see Example 1.12).

We endow X with the partial ordered given by
x y x ( t ) y ( t ) , for all  t [ 0 , T ] .

On the other hand, ( X , d ) is regular [53].

Our result is the following.

Theorem 4.6 Under assumptions (i)-(vi), system (4.5) has a solution in X 3 where X = ( C [ 0 , T ] , R ) .

Proof As in [46], we consider the operators F : X 3 X and g : X X defined by
F ( x 1 , x 2 , x 3 ) ( t ) = ω ( t ) + 0 T S ( t , r ) [ f ( r , x 1 ( r ) ) + k ( r , x 2 ( r ) ) + h ( r , x 3 ( r ) ) ] d r
and
g ( x ) = x

for all t [ 0 , T ] and x 1 , x 2 , x 3 , x X .

F has the mixed monotone property (see Theorem 25 of [46]).

Let x , y , z , u , v , w X be such that x u , y v and z w . Since F has the mixed monotone property, we have
F ( u , v , w ) F ( x , y , z ) .
On the other hand,
G ( F ( x , y , z ) , F ( u , v , w ) , F ( a , b , c ) ) = max { max t [ 0 , T ] | F ( x , y , z ) ( t ) F ( u , v , w ) ( t ) | p , max t [ 0 , T ] | F ( u , v , w ) ( t ) F ( a , b , c ) ( t ) | p , max t [ 0 , T ] | F ( a , b , c ) ( t ) F ( x , y , z ) ( t ) |