Open Access

Common fixed points of Ćirić-type contractive mappings in two ordered generalized metric spaces

Fixed Point Theory and Applications20122012:139

https://doi.org/10.1186/1687-1812-2012-139

Received: 16 March 2012

Accepted: 22 August 2012

Published: 4 September 2012

Abstract

In this paper, using the setting of two ordered generalized metric spaces, a unique common fixed point of four mappings satisfying a generalized contractive condition is obtained. We also present an example to demonstrate the results presented herein.

MSC:54H25, 47H10, 54E50.

Keywords

weakly compatible mappings compatible mappings dominated mappings common fixed point partially ordered set generalized metric space

1 Introduction and preliminaries

The study of a unique common fixed point of given mappings satisfying certain contractive conditions has been at the center of rigorous research activity. Mustafa and Sims [1] generalized the concept of a metric in which a real number is assigned to every triplet of an arbitrary set. Based on the notion of generalized metric spaces, Mustafa et al. [25] obtained some fixed point theorems for some mappings satisfying different contractive conditions. The existence of common fixed points in generalized metric spaces was initiated by Abbas and Rhoades [6] (see also [7] and [8]). For further study of common fixed points in generalized metric spaces, we refer to [912] and references mentioned therein. Abbas et al. [13] showed the existence of coupled common fixed points in two generalized metric spaces (for more results on couple fixed points, see also [1421]).

The existence of fixed points in ordered metric spaces has been initiated in 2004 by Ran and Reurings [22] and further studied by Nieto and Lopez [23]. Subsequently, several interesting and valuable results have appeared in this direction [2430].

The aim of this paper is to study common fixed point of four mappings that satisfy the generalized contractive condition in two ordered generalized metric spaces.

In the sequel, R , R + and N denote the set of real numbers, the set of nonnegative integers and the set of positive integers respectively. The usual order on R (respectively, on R + ) will be indistinctly denoted by ≤ or by ≥.

In [1], Mustafa and Sims introduced the following definitions and results:

Definition 1.1 Let X be a nonempty set. Suppose that a mapping G : X × X × X R + satisfies the following conditions:
  1. (a)

    G ( x , y , z ) = 0 if x = y = z for all x , y , z X ;

     
  2. (b)

    0 < G ( x , y , z ) for all x , y , z X with x y ;

     
  3. (c)

    G ( x , x , y ) G ( x , y , z ) for all x , y , z X with y z ;

     
  4. (d)

    G ( x , y , z ) = G ( p { x , y , z } ) , where p is a permutation of x , y , z X (symmetry);

     
  5. (e)

    G ( x , y , z ) G ( x , a , a ) + G ( a , y , z ) for all x , y , z , a X .

     

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

Definition 1.2 A sequence { x n } in a G-metric space X is called:
  1. (1)

    a G-Cauchy sequence if, for any ε > 0 , there exists n 0 N (the set of natural numbers) such that, for all n , m , l n 0 , G ( x n , x m , x l ) < ε ;

     
  2. (2)

    G-convergent if, for any ε > 0 , there exist x X and n 0 N such that, for all n , m n 0 , G ( x , x n , x m ) < ε ;

     
  3. (3)

    A G-metric space X is said to be G-complete if every G-Cauchy sequence in X is G-convergent in X.

     

It is known that { x n } is G-convergent to a point x X if and only if G ( x m , x n , x ) 0 as n , m .

Proposition 1.3 [1]

Let X be a G-metric space. Then the following items are equivalent:
  1. (1)

    A sequence { x n } in X is G-convergent to a point x X ;

     
  2. (2)

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

     
  3. (3)

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

     
  4. (4)

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

     

Definition 1.4 A G-metric on X is said to be symmetric if G ( x , y , y ) = G ( y , x , x ) for all x , y X .

Proposition 1.5 Every G-metric on X defines a metric d G on X by
d G ( x , y ) = G ( x , y , y ) + G ( y , x , x )
(1.1)

for all x , y X .

For a symmetric G-metric, we have
d G ( x , y ) = 2 G ( x , y , y )
(1.2)
for all x , y X . However, if G is non-symmetric, then the following inequality holds:
3 2 G ( x , y , y ) d G ( x , y ) 3 G ( x , y , y )
(1.3)
for all x , y X . It is obvious that
G ( x , x , y ) 2 G ( x , y , y )

for all x , y X .

Now, we give an example of a non-symmetric G-metric.

Example 1.6 Let X = { 1 , 2 } and G : X × X × X R + be a mapping defined by Table 1.
Table 1

G -metric

(x,y,z)

G(x,y,z)

(1,1,1), (2,2,2)

0

(1,1,2), (1,2,1), (2,1,1)

0.5

(1,2,2), (2,1,2), (2,2,1)

1.0

Note that G satisfies all the axioms of a generalized metric, but G ( x , x , y ) G ( x , y , y ) for two distinct points x , y X .

Definition 1.7 Let f and g be self-mappings on a set X. If w = f x = g x for some x X , then the point x is called a coincidence point of f and g and w is called a point of coincidence of f and g.

Definition 1.8 [31]

Let f and g be self-mappings on a set X. Then f and g are said to be weakly compatible if they commute at every coincidence point.

Definition 1.9 [8]

Let X be a G-metric space and f, g be self-mappings on X. Then f and g are said to be R-weakly commuting if there exists a positive real number R such that G ( f g x , f g x , g f x ) R G ( f x , f x , g x ) for all x X .

The maps f and g are R-weakly commuting on X if and only if they commute at their coincidence points.

Recall that two mappings f and g on a G-metric space X are said to be compatible if, for a sequence { x n } in X such that { f x n } and { g x n } are G-convergent to some t X ,
lim n G ( f g x n , f g x n , g f x n ) = 0 .
Definition 1.10 Let X be a nonempty set. Then ( X , , G ) is called an ordered generalized metric space if the following conditions hold:
  1. (a)

    G is a generalized metric on X;

     
  2. (b)

    is a partial order on X.

     

Definition 1.11 Let ( X , ) be a partial ordered set. Then two points x , y X are said to be comparable if x y or y x .

Definition 1.12 [24]

Let ( X , ) be a partially ordered set. A self-mapping f on X is said to be dominating if x f x for all x X .

Example 1.13 [24]

Let X = [ 0 , 1 ] be endowed with usual ordering and f : X X be a mapping defined by f x = x n for some n N . Since x x 1 n = f x for all x X , f is a dominating mapping.

Definition 1.14 Let ( X , ) be a partially ordered set. A self-mapping f on X is said to be dominated if f x x for all x X .

Example 1.15 Let X = [ 0 , 1 ] be endowed with usual ordering and f : X X be a mapping defined by f x = x n for some n N . Since f x = x n x for all x X , f is a dominated mapping.

Definition 1.16 A subset K of a partially ordered set X is said to be well-ordered if every two elements of K are comparable.

2 Common fixed point theorems

In [32], Kannan proved a fixed point theorem for a single valued self-mapping T on a metric space X satisfying the following property:
d ( T x , T y ) h { d ( x , T x ) + d ( y , T y ) }
for all x , y X , where h [ 0 , 1 2 ) . If a self-mapping T on a metric space X satisfies the following property:
d ( T x , T y ) a d ( x , y ) + b d ( x , T x ) + c d ( y , T y ) + e [ d ( x , T y ) + d ( y , T x ) ]

for all x , y X , where a , b , c , e 0 with a + b + c + 2 e < 1 , then T has a unique fixed point provided that X is T-orbitally complete (for related definitions and results, we refer to [33]).

Afterwards, Ćirić [34] obtained a fixed point result for a mapping satisfying the following property:
d ( T x , T y ) q max { d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) + d ( y , T x ) 2 }

for all x , y X , where 0 q < 1 .

In this section, we show the existence of a unique common fixed point of four mappings satisfying Ćirić-type contractive condition in the framework of two ordered generalized metric spaces.

Now, we start with the following result:

Theorem 2.1 Let ( X , ) be a partially ordered set and G 1 , G 2 be two G-metrics on X such that G 2 ( x , y , z ) G 1 ( x , y , z ) for all x , y , z X with a complete metric G 1 on X. Suppose that f, g, S and T are self-mappings on X satisfying the following properties:
G 1 ( f x , f x , g y ) k max { G 2 ( S x , S x , T y ) , G 2 ( f x , f x , S x ) , G 2 ( g y , g y , T y ) , [ G 2 ( f x , f x , T y ) + G 2 ( g y , g y , S x ) ] / 2 }
(2.1)
and
G 1 ( f x , g y , g y ) k max { G 2 ( S x , T y , T y ) , G 2 ( f x , S x , S x ) , G 2 ( g y , T y , T y ) , [ G 2 ( f x , T y , T y ) + G 2 ( g y , S x , S x ) ] / 2 }
(2.2)
for all comparable x , y X , where k [ 0 , 1 ) . Suppose that f ( X ) T ( X ) and g ( X ) S ( X ) , f, g are dominated mappings and S, T are dominating mappings. If, for any nonincreasing sequence { x n } in X with y n x n for all n N , y n u implies that u x n and either
  1. (a)

    f, S are compatible, f or S is continuous and g, T are weakly compatible

     
or
  1. (b)

    g, T are compatible, g or T is continuous and f, S are weakly compatible,

     

then f, g, S and T have a common fixed point. Moreover, the set of common fixed points of f, g, S and T is well-ordered if and only if f, g, S and T have one and only one common fixed point.

Proof Let x 0 be an arbitrary point in X. Since f ( X ) T ( X ) and g ( X ) S ( X ) , we can define the sequences { x n } and { y n } in X by
y 2 n = g x 2 n = S x 2 n + 1 , y 2 n + 1 = f x 2 n + 1 = T x 2 n + 2
for all n 0 . By the given assumptions, we have
x 2 n + 2 T x 2 n + 2 = f x 2 n + 1 x 2 n + 1 , x 2 n + 1 S x 2 n + 1 = g x 2 n x 2 n .
Thus, for all n 0 , we have x n + 1 x n . Suppose that G 1 ( y 2 n , y 2 n + 1 , y 2 n + 1 ) > 0 for all n 0 . If not, then, for some m 0 , y m = y m + 1 . Indeed, if m = 2 k , then y 2 k = y 2 k + 1 and from (2.1), it follows that
(2.3)
Again, from (2.2), it follows that
(2.4)
Thus (2.3) and (2.4) imply that
G 1 ( y 2 k + 2 , y 2 k + 1 , y 2 k + 1 ) k 2 G 1 ( y 2 k + 2 , y 2 k + 1 , y 2 k + 1 )

and so y 2 k + 1 = y 2 k + 2 since k 2 < 1 .

Similarly, if m = 2 k + 1 , then one can easily obtain y 2 k + 2 = y 2 k + 3 . Thus { y n } becomes a constant sequence and y 2 n serves as the common fixed point of f, g, S and T.

Suppose that G 1 ( y 2 n , y 2 n + 1 , y 2 n + 1 ) > 0 for all n 0 .

If n N is even, then n = 2 k for some k N ; then it follows from (2.1) that
which implies that
G 1 ( y n + 1 , y n + 1 , y n ) k G 1 ( y n , y n , y n 1 ) .
If n N is odd, then n = 2 k + 1 for some k N . Again, it follows from (2.1) that
that is,
G 1 ( y n + 1 , y n + 1 , y n ) k G 1 ( y n , y n , y n 1 )
for all n N . Continuing the above process, we have
G 1 ( y n + 1 , y n + 1 , y n ) k n G 1 ( y 1 , y 1 , y 0 )
for all n N . Thus, for all n , m N with m > n , we have
G 1 ( y m , y m , y n ) G 1 ( y n , y n + 1 , y n + 1 ) + G 1 ( y n + 1 , y n + 2 , y n + 2 ) + + G 1 ( y m 1 , y m , y m ) k n G 1 ( y 0 , y 1 , y 1 ) + k n + 1 G 1 ( y 0 , y 1 , y 1 ) + + k m 1 G 1 ( y 0 , y 1 , y 1 ) = k n G 1 ( y 0 , y 1 , y 1 ) i = 0 m n 1 k i k n 1 k G 1 ( y 0 , y 1 , y 1 )
and so G 1 ( y n , y m , y m ) 0 as m , n . Hence { y n } is a G-Cauchy sequence in X. Since X is G 1 -complete, there exists a point z X such that lim n y n = z . Consequently, we have
lim n y 2 n + 1 = lim n f x 2 n + 1 = lim n T x 2 n + 2 = z
and
lim n y 2 n = lim n g x 2 n = lim n S x 2 n + 1 = z .
If S is continuous and { f , S } is compatible, then
lim n S 2 x 2 n + 1 = S z , lim n f S x 2 n + 1 = lim n S f x 2 n + 1 = S z .
Since S x 2 n + 1 = g x 2 n x 2 n , (2.1) gives
Taking the limit as n , we obtain
G 1 ( S z , S z , z ) k max { G 2 ( S z , S z , z ) , G 2 ( S z , S z , S z ) , G 2 ( z , z , z ) , [ G 2 ( S z , S z , z ) + G 2 ( z , z , S z ) ] / 2 } k max { G 1 ( S z , S z , z ) , [ G 1 ( S z , S z , z ) + G 1 ( z , z , S z ) ] / 2 } = k 2 [ G 1 ( S z , S z , z ) + G 1 ( z , z , S z ) ] ,
which further implies that
G 1 ( S z , S z , z ) h G 1 ( z , z , S z ) ,
(2.5)

where h = k 2 k . Obviously, 0 h < 1 .

Similarly, we obtain
G 1 ( S z , z , z ) h G 1 ( z , S z , S z ) .
(2.6)
From (2.5) and (2.6), we have
G 1 ( S z , S z , z ) h 2 G 1 ( z , S z , S z )
and so S z = z since 0 h 2 < 1 . Since g x 2 n x 2 n and g x 2 n z as n implies z x 2 n , it follows from (2.1) that
G 1 ( f z , f z , g x 2 n ) k max { G 2 ( S z , S z , T x 2 n ) , G 2 ( f z , f z , S z ) , G 2 ( g x 2 n , g x 2 n , T x 2 n ) , [ G 2 ( f z , f z , T x 2 n ) + G 2 ( g x 2 n , g x 2 n , S z ) ] / 2 } = k max { G 2 ( z , z , T x 2 n ) , G 2 ( f z , f z , z ) , G 2 ( g x 2 n , g x 2 n , T x 2 n ) , [ G 2 ( f z , f z , T x 2 n ) + G 2 ( g x 2 n , g x 2 n , z ) ] / 2 } ,
which, taking the limit as n , gives
G 1 ( f z , f z , z ) k max { G 2 ( z , z , z ) , G 2 ( f z , f z , z ) , G 2 ( z , z , z ) , [ G 2 ( f z , f z , z ) + G 2 ( z , z , z ) ] / 2 } k G 1 ( f z , f z , z ) .
(2.7)
Similarly, we obtain
G 1 ( f z , z , z ) k G 1 ( z , f z , f z ) .
(2.8)

Therefore, by using the above two inequalities, we have f z = z .

Since f ( X ) T ( X ) , there exists a point v X such that f z = T v . Since v T v = f z z , it follows from (2.1) that
G 1 ( f z , f z , g v ) k max { G 2 ( S z , S z , T v ) , G 2 ( f z , f z , S z ) , G 2 ( g v , g v , T v ) , [ G 2 ( f z , f z , T v ) + G 2 ( g v , g v , S z ) ] / 2 } = k max { G 2 ( f z , f z , f z ) , G 2 ( f z , f z , f z ) , G 2 ( g v , g v , f z ) , [ G 2 ( f z , f z , f z ) + G 2 ( g v , g v , f z ) ] / 2 } k G 1 ( f z , g v , g v ) .
(2.9)
Similarly, we get
G 1 ( f z , g v , g v ) k G 1 ( f z , f z , g v ) .
(2.10)

Thus (2.9) and (2.10) imply f z = g v . Since g and T are weakly compatible, we have g z = g f z = g T v = T g v = T f z = T z , and so z is the coincidence point of g and T.

Now, from (2.1), we have
G 1 ( z , z , g z ) = G 1 ( f z , f z , g z ) k max { G 2 ( S z , S z , T z ) , G 2 ( f z , f z , S z ) , G 2 ( g z , g z , T z ) , [ G 2 ( f z , f z , T z ) + G 2 ( g z , g z , S z ) ] / 2 } = k max { G 2 ( z , z , g z ) , G 2 ( z , z , z ) , G 2 ( g z , g z , g z ) , [ G 2 ( z , z , g z ) + G 2 ( g z , g z , z ) ] / 2 } = k max { G 2 ( z , z , g z ) , [ G 2 ( z , z , g z ) + G 2 ( g z , g z , z ) ] / 2 } k 2 [ G 1 ( z , z , g z ) + G 1 ( g z , g z , z ) ] ,
that is,
G 1 ( z , z , g z ) h G 1 ( g z , g z , z ) ,
(2.11)
where h = k 2 k . Obviously, 0 h < 1 . Using (2.2), we have
G 1 ( z , g z , g z ) h G 1 ( z , z , g z ) .
(2.12)
Combining the above two inequalities, we get
G 1 ( z , z , g z ) h 2 G 1 ( z , z , g z )

and so z = g z . Therefore, f z = g z = S z = T z = z . The proof is similar when f is continuous. Similarly, if (b) holds, then the result follows.

Now, suppose that the set of common fixed points of f, g, S and T is well ordered. We show that a common fixed point of f, g, S and T is unique. Let u be another common fixed point of f, g, S and T. Then, from (2.1), we have
G 1 ( z , z , u ) = G 1 ( f z , f z , g u ) k max { G 2 ( S z , S z , T u ) , G 2 ( f z , f z , S z ) , G 2 ( g u , g u , T u ) , [ G 2 ( f z , f z , T u ) + G 2 ( g u , g u , S z ) ] / 2 } = k max { G 2 ( z , z , u ) , G 2 ( z , z , z ) , G 2 ( u , u , u ) , [ G 2 ( z , z , u ) + G 2 ( u , u , z ) ] / 2 } = k 2 [ G 2 ( z , z , u ) + G 2 ( u , u , z ) ] 1 2 G 1 ( z , z , u ) + k 2 G 1 ( u , u , z ) ,
that is,
G 1 ( z , z , u ) k G 1 ( z , u , u ) .
Similarly, using (2.2), we obtain
G 1 ( z , u , u ) k G 1 ( z , z , u ) .
Combining the above two inequalities, we get
G 1 ( z , z , u ) k 2 G 1 ( z , z , u )

and hence z = u .

The converse follows immediately. This completes the proof. □

Example 2.2 Let X = { 0 , 1 , 2 , 3 } be endowed with the usual ordering and G 1 , G 2 be two G-metrics on X defined by Table 2. Then G 1 and G 2 are non-symmetric since G 1 ( 1 , 1 , 0 ) G 1 ( 1 , 0 , 0 ) and G 2 ( 1 , 1 , 0 ) G 2 ( 1 , 0 , 0 ) with G 2 ( x , y , z ) G 1 ( x , y , z ) for all x , y , z X . Let f , g , S , T : X X be the mappings defined by Table 3. Clearly, f ( X ) T ( X ) , g ( X ) S ( X ) , f, g are dominated mappings and S, T are dominating mappings, see Table 4.
Table 2

Two G -metrices

(x,y,z)

G 1 ( x , y , z )

G 2 ( x , y , z )

(0,0,0), (1,1,1), (2,2,2), (3,3,3),

0

0

(0,0,2), (0,2,0), (2,0,0), (0,2,2), (2,0,2), (2,2,0),

(0,0,1), (0,1,0), (1,0,0), (0,0,3), (0,3,0), (3,0,0),

4

3

(0,1,1), (1,0,1), (1,1,0), (0,3,3), (3,0,3), (3,3,0),

(1,1,2), (1,2,1), (2,1,1), (1,2,2), (2,1,2), (2,2,1),

(1,1,3), (1,3,1), (3,1,1), (1,3,3), (3,1,3), (3,3,1),

(2,2,3), (2,3,2), (3,2,2), (2,3,3), (3,2,3), (3,3,2),

8

6

(0,1,2), (0,1,3), (0,2,1), (0,2,3), (0,3,1), (0,3,2),

(1,0,2), (1,0,3), (1,2,0), (1,2,3), (1,3,0), (1,3,2),

(2,0,1), (2,0,3), (2,1,0), (2,1,3), (2,3,0), (2,3,1),

(3,0,1), (3,0,2), (3,1,0), (3,1,2), (3,2,0), (3,2,1),

8

6

Table 3

Self maps

x

f(x)

g(x)

S(x)

T(x)

0

0

0

0

0

1

0

0

2

2

2

0

2

2

3

3

0

0

3

3

Table 4

Dominated and dominating maps

xX

f is dominated

g is dominated

S is dominating

T is dominating

x = 0

f(0)=0

g(0)=0

0 = S(0)

0 = T(0)

x = 1

f(1)=0<1

g(1)=0<1

1<2 = S(1)

1<2 = T(1)

x = 2

f(2)=0<2

g(2)=2

2 = S(2)

2<3 = T(2)

x = 3

f(3)=0<3

g(3)=0<3

3 = S(3)

3 = T(3)

Now, we shall show that for all comparable x , y X , (2.1) and (2.2) are satisfied with k = 3 4 [ 0 , 1 ) . Note that for all x , y { 0 , 1 , 3 } , G ( f x , f x , g y ) = G ( f x , g y , g y ) = 0 and (2.1), (2.2) are satisfied obviously.
  1. (1)
    When x = 0 and y = 2 , then f x = 0 , g y = 2 , S x = 0 , T y = 3 and so
    G 1 ( f x , f x , g y ) = G 1 ( 0 , 0 , 2 ) = 4 < 3 4 ( 6 ) = 3 4 G 2 ( 2 , 2 , 3 ) = 3 4 G 2 ( g y , g y , T y ) k max { G 2 ( S x , S x , T y ) , G 2 ( f x , f x , S x ) , G 2 ( g y , g y , T y ) , [ G 2 ( f x , f x , T y ) + G 2 ( g y , g y , S x ) ] / 2 }
     
and
G 1 ( f x , g y , g y ) = G 1 ( 0 , 2 , 2 ) = 4 < 3 4 ( 6 ) = 3 4 G 2 ( 2 , 3 , 3 ) = 3 4 G 2 ( g y , T y , T y ) k max { G 2 ( S x , T y , T y ) , G 2 ( f x , S x , S x ) , G 2 ( g y , T y , T y ) , [ G 2 ( f x , T y , T y ) + G 2 ( g y , S x , S x ) ] / 2 } .
  1. (2)
    When x = 1 and y = 2 , then f x = 0 , g y = 2 , S x = 2 , T y = 3 and so
    G 1 ( f x , f x , g y ) = G 1 ( 0 , 0 , 2 ) = 4 < 3 4 ( 6 ) = 3 4 G 2 ( 2 , 2 , 3 ) = 3 4 G 2 ( g y , g y , T y ) k max { G 2 ( S x , S x , T y ) , G 2 ( f x , f x , S x ) , G 2 ( g y , g y , T y ) , [ G 2 ( f x , f x , T y ) + G 2 ( g y , g y , S x ) ] / 2 }
     
and
G 1 ( f x , g y , g y ) = G 1 ( 0 , 2 , 2 ) = 4 < 3 4 ( 6 ) = 3 4 G 2 ( 2 , 3 , 3 ) = 3 4 G 2 ( g y , T y , T y ) k max { G 2 ( S x , T y , T y ) , G 2 ( f x , S x , S x ) , G 2 ( g y , T y , T y ) , [ G 2 ( f x , T y , T y ) + G 2 ( g y , S x , S x ) ] / 2 } .
  1. (3)
    When x = 2 and y = 2 , then f x = 0 , g y = 2 , S x = 2 , T y = 3 and so
    G 1 ( f x , f x , g y ) = G 1 ( 0 , 0 , 2 ) = 4 < 3 4 ( 6 ) = 3 4 G 2 ( 2 , 2 , 3 ) = 3 4 G 2 ( S x , S x , T y ) k max { G 2 ( S x , S x , T y ) , G 2 ( f x , f x , S x ) , G 2 ( g y , g y , T y ) , [ G 2 ( f x , f x , T y ) + G 2 ( g y , g y , S x ) ] / 2 }
     
and
G 1 ( f x , g y , g y ) = G 1 ( 0 , 2 , 2 ) = 4 < 3 4 ( 6 ) = 3 4 G 2 ( 2 , 3 , 3 ) = 3 4 G 2 ( S x , T y , T y ) k max { G 2 ( S x , T y , T y ) , G 2 ( f x , S x , S x ) , G 2 ( g y , T y , T y ) , [ G 2 ( f x , T y , T y ) + G 2 ( g y , S x , S x ) ] / 2 } .
  1. (4)
    Finally, when x = 3 and y = 2 , then f x = 0 , g y = 2 , S x = 3 , T y = 3 and so
    G 1 ( f x , f x , g y ) = G 1 ( 0 , 0 , 2 ) = 4 < 3 4 ( 6 ) = 3 4 G 2 ( 2 , 2 , 3 ) = 3 4 G 2 ( g y , g y , T y ) k max { G 2 ( S x , S x , T y ) , G 2 ( f x , f x , S x ) , G 2 ( g y , g y , T y ) , [ G 2 ( f x , f x , T y ) + G 2 ( g y , g y , S x ) ] / 2 }
     
and
G 1 ( f x , g y , g y ) = G 1 ( 0 , 2 , 2 ) = 4 < 3 4 ( 6 ) = 3 4 G 2 ( 2 , 3 , 3 ) = 3 4 G 2 ( g y , T y , T y ) k max { G 2 ( S x , T y , T y ) , G 2 ( f x , S x , S x ) , G 2 ( g y , T y , T y ) , [ G 2 ( f x , T y , T y ) + G 2 ( g y , S x , S x ) ] / 2 } .

Thus, for all cases, the contractions (2.1) and (2.2) are satisfied. Hence all of the conditions of Theorem 2.1 are satisfied. Moreover, 0 is the unique common fixed point of f, g, S and g.

If we consider the same set equipped with two metrics given by d 1 ( x , y ) = | x y | and d 2 ( x , y ) = 1 2 | x y | for all x , y X , then for x = 1 and y = 2 , we have
d 1 ( f x , g y ) = d 1 ( 0 , 2 ) = 2 2 k k max { d 2 ( 2 , 3 ) , d 2 ( 0 , 2 ) , d 2 ( 2 , 3 ) , [ d 2 ( 0 , 3 ) + d 2 ( 2 , 2 ) ] / 2 } = k max { d 2 ( S x , T y ) , d 2 ( f x , S x ) , d 2 ( g y , T y ) , [ d 2 ( f x , T y ) + d 2 ( g y , S x ) ] / 2 }

for any k [ 0 , 1 ) . So corresponding results in ordinary metric spaces cannot be applied in this case.

Theorem 2.1 can be viewed as an extension of Theorem 2.1 of [8] to the case of two ordered G-metric spaces.

Since the class of weakly compatible mappings includes R-weakly commuting mappings, Theorem 2.1 generalizes the comparable results in [8].

Corollary 2.3 Let ( X , ) be a partially ordered set and G 1 , G 2 be two G-metrics on X such that G 2 ( x , y , z ) G 1 ( x , y , z ) for all x , y , z X with a complete metric G 1 on X. Suppose that f, g, S and T are self-mappings on X satisfying the following properties:
G 1 ( f x , f x , g y ) a 1 G 2 ( S x , S x , T y ) + a 2 G 2 ( S x , S x , f x ) + a 3 G 2 ( T y , T y , g y ) + a 4 [ G 2 ( S x , S x , g y ) + G 2 ( T y , T y , f x ) ]
(2.13)
and
G 1 ( f x , g y , g y ) a 1 G 2 ( S x , T y , T y ) + a 2 G 2 ( S x , f x , f x ) + a 3 G 2 ( T y , g y , g y ) + a 4 [ G 2 ( S x , g y , g y ) + G 2 ( T y , f x , f x ) ]
(2.14)
for all comparable x , y X , where a 1 + a 2 + a 3 + 2 a 4 < 1 . Suppose that f ( X ) T ( X ) , g ( X ) S ( X ) and f, g are dominated mappings and S, T are dominating mappings. If, for any nonincreasing sequence { x n } with y n x n for all n N , y n u implies that u x n and either
  1. (a)

    f, S are compatible, f or S is continuous and g, T are weakly compatible

     
or
  1. (b)

    g, T are compatible, g or T is continuous and f, S are weakly compatible,

     

then f, g, S and T have a common fixed point in X. Moreover, the set of common fixed points of f, g, S and T is well-ordered if and only if f, g, S and T have one and only one common fixed point in X.

Example 2.4 Let X = [ 0 , 1 ] be endowed with the usual ordering and G 1 , G 2 be two G-metrics on X given in [13]:
G 1 ( a , b , c ) = | a b | + | b c | + | c a | , G 2 ( a , b , c ) = 1 2 [ | a b | + | b c | + | c a | ] .
Define the mappings f , g , S , T : X X as
f x = x 12 , g x = { x 4 if  x [ 0 , 1 2 ) , x 6 if  x [ 1 2 , 1 ) , S ( x ) = 3 x 2 , T ( x ) = 5 x 2
for all x X . Clearly, f, g are dominated mappings and S, T are dominating mappings with f ( X ) T ( X ) and g ( X ) S ( X ) . Also, f, S are compatible, f is continuous and g, T are weakly compatible. Now, for all comparable x , y X , we check the following cases:
  1. (1)
    If x , y [ 0 , 1 2 ) , then we have
    G 1 ( f x , f x , g y ) = 1 12 | x 3 y | 1 12 ( x + 3 y ) 3 10 ( 17 12 x ) + 3 10 ( 9 4 y ) = a 2 G 2 ( f x , f x , S x ) + a 3 G 2 ( g y , g y , T y ) a 1 G 2 ( S x , S x , T y ) + a 2 G 2 ( f x , f x , S x ) + a 3 G 2 ( g y , g y , T y ) + a 4 [ G 2 ( f x , f x , T y ) + G 2 ( g y , g y , S x ) ] .
     
  2. (2)
    If x [ 0 , 1 2 ) and y [ 1 2 , 1 ] , then we have
    G 1 ( f x , f x , g y ) = 1 12 | x 2 y | 1 12 ( x + 2 y ) 3 10 ( 17 12 x ) + 3 10 ( 14 6 y ) = a 2 G 2 ( f x , f x , S x ) + a 3 G 2 ( g y , g y , T y ) a 1 G 2 ( S x , S x , T y ) + a 2 G 2 ( f x , f x , S x ) + a 3 G 2 ( g y , g y , T y ) + a 4 [ G 2 ( f x , f x , T y ) + G 2 ( g y , g y , S x ) ] .
     
  3. (3)
    If y [ 0 , 1 2 ) and x [ 1 2 , 1 ] , then we have
    G 1 ( f x , f x , g y ) = 1 12 | x 3 y | 1 12 ( x + 3 y ) 3 10 ( 17 12 x ) + 3 10 ( 9 4 y ) = a 2 G 2 ( f x , f x , S x ) + a 3 G 2 ( g y , g y , T y ) a 1 G 2 ( S x , S x , T y ) + a 2 G 2 ( f x , f x , S x ) + a 3 G 2 ( g y , g y , T y ) + a 4 [ G 2 ( f x , f x , T y ) + G 2 ( g y , g y , S x ) ] .
     
  4. (4)
    If x , y [ 1 2 , 1 ] , then we obtain
    G 1 ( f x , f x , g y ) = 1 12 | x 2 y | 1 12 ( x + 2 y ) 3 10 ( 17 12 x ) + 3 10 ( 14 6 y ) = a 2 G 2 ( f x , f x , S x ) + a 3 G 2 ( g y , g y , T y ) a 1 G 2 ( S x , S x , T y ) + a 2 G 2 ( f x , f x , S x ) + a 3 G 2 ( g y , g y , T y ) + a 4 [ G 2 ( f x , f x , T y ) + G 2 ( g y , g y , S x ) ] .
     

Thus (2.13) is satisfied with a 1 = a 4 = 1 10 and a 2 = a 3 = 3 10 , where a 1 + a 2 + a 3 + 2 a 4 < 1 . Similarly, (2.14) is satisfied. Thus all the conditions of Corollary 2.3 are satisfied. Moreover, 0 is the unique common fixed point of f and g.

3 Application

Let X = L 2 ( Ω ) , the set of comparable functions on Ω whose square is integrable on Ω where Ω = [ 0 , 1 ] , be a bounded set in R . We endow X with the partial ordered given by: x , y X , x y x ( t ) y ( t ) , for all t Ω . We consider the integral equations
x ( t ) = Ω q 1 ( t , s , x ( s ) ) d s c ( t ) , y ( t ) = Ω q 2 ( t , s , y ( s ) ) d s c ( t ) ,
(3.1)
where q 1 , q 2 : Ω × Ω × R R and c : Ω R + , to be given continuous mappings. Recently, Abbas et al. [35] obtained a common solution of integral equations (3.1) as an application of their results in the setup of ordered generalized metric spaces. Here we study a sufficient condition for the existence of a common solution of integral equations in the framework of two generalized metric spaces. Define G 1 , G 2 : X × X × X R + by
G 1 ( x , y , z ) = sup t Ω | x ( t ) y ( t ) | + sup t Ω | y ( t ) z ( t ) | + sup t Ω | z ( t ) x ( t ) | , G 2 ( x , y , z ) = 1 2 [ sup t Ω | x ( t ) y ( t ) | + sup t Ω | y ( t ) z ( t ) | + sup t Ω | z ( t ) x ( t ) | ] .
Obviously, G 2 ( x , y , z ) G 1 ( x , y , z ) for all x , y , z X . Suppose that the following hypotheses hold:
  1. (i)
    For each s , t Ω ,
    Ω q 1 ( t , s , u ( s ) ) d s u ( s )
     
and
Ω q 2 ( t , s , u ( s ) ) d s u ( s )
hold.
  1. (ii)
    There exists r : Ω Ω such that
    Ω | q 1 ( t , s , u ( t ) ) q 2 ( t , s , v ( t ) ) | d t r ( t ) | u ( t ) v ( t ) |
     

for each s , t Ω with sup t Ω r ( t ) k where k [ 0 , 1 ) .

Then the integral equations (3.1) have a common solution in L 2 ( Ω ) .

Proof Define f x ( t ) = Ω q 1 ( t , s , x ( t ) ) d t c ( t ) and g x ( t ) = Ω q 2 ( t , s , x ( t ) ) d t c ( t ) . As f x ( t ) x ( t ) and g x ( t ) x ( t ) , so f and g are dominated maps. Now, for all comparable x , y X ,
G 1 ( f x , f x , g y ) = 2 sup t Ω | f x ( t ) g y ( t ) | = 2 sup t Ω | Ω q 1 ( t , s , x ( t ) ) d t Ω q 2 ( t , s , y ( t ) ) d t | 2 sup t Ω Ω | q 1 ( t , s , x ( t ) ) q 2 ( t , s , y ( t ) ) | d t 2 sup t Ω r ( t ) | x ( t ) y ( t ) | 2 k sup t Ω | x ( t ) y ( t ) | = k G 2 ( x , y , y ) k max { G 2 ( x , x , y ) , G 2 ( f x , f x , x ) , G 2 ( g y , g y , y ) , [ G 2 ( f x , f x , y ) + G 2 ( g y , g y , x ) ] / 2 } .
Similarly,
G 1 ( f x , g y , g y ) k max { G 2 ( x , y , y ) , G 2 ( f x , x , x ) , G 2 ( g y , y , y ) , [ G 2 ( f x , y , y ) + G 2 ( g y , x , x ) ] / 2 }

is satisfied. Now we can apply Theorem 2.1 by taking S and T as identity maps to obtain the common solutions of integral equations (3.1) in L 2 ( Ω ) . □

Remarks (1) If we take f = g in Theorem 2.1, then it generalizes Corollary 2.3 in [8] to a more general class of commuting mappings in the setup of two ordered G-metric spaces.
  1. (2)

    If we take S = T in Theorem 2.1, then Corollary 2.4 in [8] is a special case of Theorem 2.1.

     
  2. (3)

    If S = T = I X (: the identity mapping on X) in Theorem 2.1, then we obtain Corollary 2.5 in [8] in a more general setup.

     
(4) Corollary 2.6 of [8] becomes a special case of Theorem 2.1 if we take f = g and S = T = I X .
  1. (5)

    A G-metric naturally induces a metric d G given by d G ( x , y ) = G ( x , y , y ) + G ( x , x , y ) . If the G-metric is not symmetric, then the inequalities (2.1), (2.2), (2.13) and (2.14) do not reduce to any metric inequality with the metric d G . Hence our results do not reduce to fixed point problems in the corresponding metric space ( X , , d G ) .

     

Declarations

Acknowledgements

The authors thank the referees for their appreciation and suggestions regarding this work.

Authors’ Affiliations

(1)
Department of Mathematics, Lahore University of Management Sciences
(2)
Department of Mathematics Education and the RINS, Gyeongsang National University

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