Open Access

A new approach to G-metric and related fixed point theorems

Journal of Inequalities and Applications20132013:454

https://doi.org/10.1186/1029-242X-2013-454

Received: 17 July 2013

Accepted: 16 September 2013

Published: 7 November 2013

Abstract

Very recently, Samet et al. and Jleli and Samet reported that most of fixed point results in the context of G-metric space, defined by Sims and Zead, can be derived from the usual fixed point theorems on the usual metric space. In this paper, we state and prove some fixed point theorems in the framework of G-metric space that cannot be obtained from the existence results in the context of associated metric space.

1 Introduction and preliminaries

In 2007, Mustafa and Sims introduced the notion of G-metric and investigated the topology of such spaces. The authors also characterized some celebrated fixed point results in the context of G-metric space. Following this initial paper, a number of authors have published many fixed point results on the setting of G-metric space (see, e.g., [133] and the references therein). Samet et al. [24] and Jleli and Samet [25] reported that some published results can be considered as a straight consequence of the existence theorem in the setting of the usual metric space. More precisely, the authors of these two papers noticed that p ( x , y ) = p G ( x , y ) = G ( x , y , y ) is a quasi-metric whenever G : X × X × X [ 0 , ) is a G-metric. It is evident that each quasi-metric induces a metric. In particular, if the pair ( X , p ) is a quasi-metric space, then the function defined by
d ( x , y ) = d G ( x , y ) = max { p ( x , y ) , p ( y , x ) } , for all  x , y X ,

forms a metric on X.

The object of this paper is to get some fixed point results in the context of G-metric space that cannot be concluded from the existence results. This paper can be considered as a continuation of [27], which was inspired by [26].

First, we recollect some necessary definitions and results in this direction. The notion of G-metric spaces is defined as follows.

Definition 1.1 (See [1])

Let X be a non-empty set, G : X × X × X R + be a function satisfying the following properties:
  1. (G1)

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

     
  2. (G2)

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

     
  3. (G3)

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

     
  4. (G4)

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

     
  5. (G5)

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

     

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

Note that every G-metric on X induces a metric d G on X defined by
d G ( x , y ) = G ( x , y , y ) + G ( y , x , x ) , for all  x , y X .
(1)

For a better understanding of the subject, we give the following examples of G-metrics.

Example 1.1 Let ( X , d ) be a metric space. Function G : X × X × X [ 0 , + ) , defined by
G ( x , y , z ) = max { d ( x , y ) , d ( y , z ) , d ( z , x ) } ,

for all x , y , z X , is a G-metric on X.

Example 1.2 (See, e.g., [1])

Let X = [ 0 , ) . Function G : X × X × X [ 0 , + ) , defined by
G ( x , y , z ) = | x y | + | y z | + | z x | ,

for all x , y , z X , is a G-metric on X.

In their initial paper, Mustafa and Sims [1] also defined the basic topological concepts in G-metric spaces as follows.

Definition 1.2 (See [1])

Let ( X , G ) be a G-metric space, and let { x n } be a sequence of points of X. We say that { x n } is G-convergent to x X if
lim n , m + G ( x , x n , x m ) = 0 ,

that is, for any ε > 0 , there exists N N such that G ( x , x n , x m ) < ε for all n , m N . We call x the limit of the sequence and write x n x or lim n + x n = x .

Proposition 1.1 (See [1])

Let ( X , G ) be a G-metric space. The following are equivalent:
  1. (1)

    { x n } is G-convergent to x,

     
  2. (2)

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

     
  3. (3)

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

     
  4. (4)

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

     

Definition 1.3 (See [1])

Let ( X , G ) be a G-metric space. Sequence { x n } is called a G-Cauchy sequence if, for any ε > 0 , there exists N N such that G ( x n , x m , x l ) < ε for all m , n , l N , that is, G ( x n , x m , x l ) 0 as n , m , l + .

Proposition 1.2 (See [1])

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

    sequence { x n } is G-Cauchy,

     
  2. (2)

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

     

Definition 1.4 (See [1])

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

Definition 1.5 Let ( X , G ) be a G-metric space. Mapping F : X × X × X X is said to be continuous if for any three G-convergent sequences { x n } , { y n } and { z n } converging to x, y and z, respectively, { F ( x n , y n , z n ) } is G-convergent to F ( x , y , z ) .

Mustafa [4] extended the well-known Banach [34] contraction principle mapping in the framework of G-metric spaces as follows.

Theorem 1.1 (See [4])

Let ( X , G ) be a complete G-metric space and T : X X be a mapping satisfying the following condition for all x , y , z X :
G ( T x , T y , T z ) k G ( x , y , z ) ,
(2)

where k [ 0 , 1 ) . Then T has a unique fixed point.

Theorem 1.2 (See [4])

Let ( X , G ) be a complete G-metric space and T : X X be a mapping satisfying the following condition for all x , y X :
G ( T x , T y , T y ) k G ( x , y , y ) ,
(3)

where k [ 0 , 1 ) . Then T has a unique fixed point.

Remark 1.1 We notice that condition (2) implies condition (3). The converse is true only if k [ 0 , 1 2 ) . For details see [4].

Lemma 1.1 [4]

By the rectangle inequality (G5) together with the symmetry (G4), we have
G ( x , y , y ) = G ( y , y , x ) G ( y , x , x ) + G ( x , y , x ) = 2 G ( y , x , x ) .
(4)

2 Main results

Theorem 2.1 Let ( X , G ) be a complete G-metric space and T : X X be a mapping satisfying the following condition for all x , y X :
G ( T x , T y , T y ) k G ( x , T x , y ) ,
(5)

where k [ 0 , 1 ) . Then T has a unique fixed point.

Proof Let x 0 X be an arbitrary point, and define the sequence x n by x n = T n ( x 0 ) . By (5), we have
G ( x n , x n + 1 , x n + 1 ) k G ( x n 1 , x n , x n ) .
(6)
Continuing in the same argument, we will get
G ( x n , x n + 1 , x n + 1 ) k n G ( x 0 , x 1 , x 1 ) .
(7)
Moreover, for all n , m N ; n < m , we have by rectangle inequality that
G ( x n , x m , x m ) G ( x n , x n + 1 , x n + 1 ) + G ( x n + 1 , x n + 2 , x n + 2 ) + G ( x n + 2 , x n + 3 , x n + 3 ) + + G ( x m 1 , x m , x m ) ( k n + k n + 1 + k n + 2 + + k m 1 ) G ( x 0 , x 1 , x 1 ) k n 1 k G ( x 0 , x 1 , x 1 ) ,
(8)

and so, lim G ( x n , x m , x m ) = 0 , as n , m . Thus, { x n } is G-Cauchy sequence. Due to the completeness of ( X , G ) , there exists u X such that { x n } is G-convergent to u.

Suppose that T u u , then
G ( x n , T u , T u ) k G ( x n 1 , x n , u ) ,
(9)
taking the limit as n , and using the fact that function G is continuous, then
G ( u , T u , T u ) k G ( u , u , u ) .
(10)

This contradiction implies that u = T u .

To prove uniqueness, suppose that u v such that T v = v , and use Lemma 1.1, then
G ( u , u , v ) = G ( T u , T u , T v ) k G ( u , T u , v ) = k G ( u , u , v ) ,
(11)

which implies that u = v . □

Example 2.1 Let X = [ 0 , ) and
G ( x , y , z ) = { 0 , if  x = y = z , max { x , y , z } , otherwise
be a G-metric on X. Define T : X X by T x = 1 5 x . Then the condition of Theorem 2.1 holds. In fact,
G ( T x , T y , T y ) = 1 5 max { x , y }
and
G ( x , T x , y ) = max { x , y } ,
and so,
G ( T x , T y , T y ) 1 4 G ( x , T x , y ) .

That is, conditions of Theorem 2.1 hold for this example.

Corollary 2.1 Let ( X , G ) be a complete G-metric space and T : X X be a mapping satisfying the following condition for all x , y , z X :
G ( T x , T y , T z ) a G ( x , T x , z ) + b G ( x , T x , y ) ,

where 0 a + b < 1 . Then T has a unique fixed point.

Theorem 2.2 Let ( X , G ) be a complete G-metric space and T : X X be a mapping satisfying the following condition for all x , y X , where a + b + c + d < 1
G ( T x , T y , T 2 y ) a G ( x , T x , T 2 x ) + b G ( y , T y , T 2 y ) + c G ( x , T x , T y ) + c G ( y , T y , T 3 x ) .
(12)

Then T has a unique fixed point.

Proof Take x 0 X . We construct sequence { x n } n = 0 of points in X in the following way:
x n + 1 = T x n for all  n = 0 , 1 , 2 , .
Notice that if x n = x n + 1 for some n N , then obviously T has a fixed point. Thus, we suppose that
x n x n + 1

for all n N .

That is, we have
G ( x n , x n + 1 , x n + 2 ) > 0 .
From (12), with x = x n 1 and y = x n , we have
G ( T x n 1 , T x n , T 2 x n ) a G ( x n 1 , T x n 1 , T 2 x n 1 ) + b G ( x n , T x n , T 2 x n ) + c G ( x n 1 , T x n 1 , T x n ) + d G ( x n , T x n , T 3 x n 1 ) ,
which implies that
G ( x n , x n + 1 , x n + 2 ) a G ( x n 1 , x n , x n + 1 ) + b G ( x n , x n + 1 , x n + 2 ) + c G ( x n 1 , x n , x n + 1 ) + d G ( x n , x n + 1 , x n + 2 ) ,
and so,
G ( x n , x n + 1 , x n + 2 ) k G ( x n 1 , x n , x n + 1 ) ,
where k = a + c 1 b d < 1 . Then
G ( x n , x n + 1 , x n + 2 ) k n G ( x 0 , x 1 , x 2 )
(13)
for all n N . Note that from (G3), we know that
G ( x n , x n , x n + 1 ) G ( x n , x n + 1 , x n + 2 )
with x n x n + 1 , and by Lemma 1.1, we know that
G ( x n + 1 , x n + 1 , x n ) 2 G ( x n , x n , x n + 1 ) .
Then by (13), we have
G ( x n + 1 , x n + 1 , x n ) 2 k n G ( x 0 , x 1 , x 2 ) .
Moreover, for all n , m N ; n < m , we have by rectangle inequality that
G ( x m , x m , x n ) G ( x n , x n + 1 , x n + 1 ) + G ( x n + 1 , x n + 2 , x n + 2 ) + G ( x n + 2 , x n + 3 , x n + 3 ) + + G ( x m 1 , x m , x m ) 2 ( k n + k n + 1 + k n + 2 + + k m 1 ) G ( x 0 , x 1 , x 2 ) 2 k n 1 k G ( x 0 , x 1 , x 2 ) ,
(14)
and so, lim G ( x n , x m , x m ) = 0 , as n , m . Thus, { x n } is G-Cauchy sequence. Due to the completeness of ( X , G ) , there exists u X such that { x n } is G-convergent to z. From (12), with x = x n and y = z , we have
G ( T x n , T z , T 2 z ) a G ( x n , T x n , T 2 x n ) + b G ( z , T z , T 2 z ) + c G ( x n , T x n , T z ) + d G ( z , T z , T 3 x n ) .
Then
G ( x n + 1 , T z , T 2 z ) a G ( x n , x n + 1 , x n + 2 ) + b G ( z , T z , T 2 z ) + c G ( x n , x n + 1 , T z ) + d G ( z , T z , x n + 3 ) .
Taking limit as n in the inequality above, we have
G ( z , T z , T 2 z ) ( c + d ) 1 b G ( z , z , T z ) .
Now, if T z = T 2 z , then T has a fixed point. Hence, we assume that T z T 2 z . Therefore, by (G3), we get
G ( z , T z , T 2 z ) ( c + d ) 1 b G ( z , z , T z ) ( c + d ) 1 b G ( z , T z , T 2 z ) ,

which implies that G ( z , T z , T 2 z ) = 0 , i.e., z = T z = T 2 z . □

At first, we assume that
Ψ 1 = { ψ : [ 0 , ) [ 0 , )  such that  ψ  is non-decreasing and continuous }
and
Φ = { φ : [ 0 , ) [ 0 , )  such that  φ  is lower semicontinuous } ,

where ψ ( t ) = ϕ ( t ) = 0 if and only if t = 0 .

Theorem 2.3 Let ( X , G ) be a complete G-metric space and T : X X be a mapping satisfying the following condition for all x , y X , where ψ Ψ and ϕ Φ holds
ψ ( G ( T x , T 2 x , T y ) ) ψ ( G ( x , T x , y ) ) ϕ ( G ( x , T x , y ) ) .
(15)

Then T has a unique fixed point.

Proof Take x 0 X . We construct sequence { x n } n = 0 of points in X in the following way:
x n + 1 = T x n for all  n = 0 , 1 , 2 , .
Notice that if x n = x n + 1 for some n N , then obviously T has a fixed point. Thus, we suppose that
x n x n + 1

for all n N .

By (G2), we have
G ( x n , x n + 1 , x n + 1 ) > 0 .
From (15), with x = x n 1 and y = x n , we have
ψ ( G ( T x n 1 , T 2 x n 1 , T x n ) ) ψ ( G ( x n 1 , T x n 1 , x n ) ) ϕ ( G ( x n 1 , T x n 1 , x n ) ) ,
which implies that
ψ ( G ( x n , x n + 1 , x n + 1 ) ) ψ ( G ( x n 1 , x n , x n ) ) ϕ ( G ( x n 1 , x n , x n ) )
(16)
ψ ( G ( x n 1 , x n , x n ) ) ,
(17)
then G ( x n , x n + 1 , x n + 1 ) G ( x n 1 , x n , x n ) . So sequence { G ( x n , x n + 1 , x n + 1 ) } is a decreasing sequence in R + , and thus, it is convergent, say t R + . We claim that t = 0 . Suppose, to the contrary, that t > 0 . Taking limit as n in (16), we get
ψ ( t ) ψ ( t ) ϕ ( t ) ,
which implies ϕ ( t ) = 0 . That is, t = 0 , which is a contrary. Hence, t = 0 , i.e.,
lim n G ( x n , x n + 1 , x n + 1 ) = 0 .
(18)
We shall show that { x n } n = 0 is a G-Cauchy sequence. Suppose, to the contrary, that there exists ε > 0 , and sequence x n ( k ) of x n such that
G ( x m ( k ) , T x m ( k ) , x n ( k ) ) ε
(19)
with n ( k ) m ( k ) > k . Further, corresponding to m ( k ) , we can choose n ( k ) in such a way that it is the smallest integer with n ( k ) > m ( k ) satisfying (19). Hence,
G ( x m ( k ) , T x m ( k ) , x n ( k ) 1 ) < ε .
(20)
By Lemma 1.1 and (G5), we have
ε G ( x m ( k ) , T x m ( k ) , x n ( k ) ) = G ( x n ( k ) , x m ( k ) , T x m ( k ) ) G ( x n ( k ) , x n ( k ) 1 , x n ( k ) 1 ) + G ( x n ( k ) 1 , T x m ( k ) , x m ( k ) ) G ( x m ( k ) , T x m ( k ) , x n ( k ) 1 ) + 2 s n ( k ) 1 ε + 2 s n ( k ) 1 ,
(21)
where s n ( k ) 1 = G ( x n ( k ) 1 , x n ( k ) , x n ( k ) ) . Letting k in (21), we derive that
lim k G ( x m ( k ) , T x m ( k ) , x n ( k ) ) = ε .
(22)
Also, by Lemma 1.1 and (G5), we obtain the following inequalities:
G ( x m ( k ) , T x m ( k ) , x n ( k ) ) G ( x m ( k ) , x m ( k ) 1 , x m ( k ) 1 ) + G ( x m ( k ) 1 , T x m ( k ) , x n ( k ) ) = G ( x m ( k ) , x m ( k ) 1 , x m ( k ) 1 ) + G ( x n ( k ) , x m ( k ) 1 , T x m ( k ) ) G ( x m ( k ) , x m ( k ) 1 , x m ( k ) 1 ) + G ( x n ( k ) , x n ( k ) 1 , x n ( k ) 1 ) + G ( x n ( k ) 1 , x m ( k ) 1 , T x m ( k ) ) 2 s m ( k ) 1 + 2 s n ( k ) 1 + G ( x n ( k ) 1 , x m ( k ) 1 , T x m ( k ) )
(23)
and
G ( x n ( k ) 1 , x m ( k ) 1 , T x m ( k ) ) G ( x n ( k ) 1 , x n ( k ) , x n ( k ) ) + G ( x n ( k ) , x m ( k ) 1 , T x m ( k ) ) = G ( x n ( k ) 1 , x n ( k ) , x n ( k ) ) + G ( x m ( k ) 1 , T x m ( k ) , x n ( k ) ) G ( x n ( k ) 1 , x n ( k ) , x n ( k ) ) + G ( x m ( k ) 1 , x m ( k ) , x m ( k ) ) + G ( x m ( k ) , T x m ( k ) , x n ( k ) ) = s n ( k ) 1 + s m ( k ) 1 + G ( x m ( k ) , T x m ( k ) , x n ( k ) ) .
(24)
Letting k in (23) and (24) and applying (22), we find that
lim k G ( x n ( k ) 1 , x m ( k ) 1 , T x m ( k ) ) = ε .
(25)
Again, by Lemma 1.1 and (G5), we have
G ( x n ( k ) 1 , x m ( k ) 1 , T x m ( k ) ) = G ( T x m ( k ) , x m ( k ) 1 , x n ( k ) 1 ) = G ( x m ( k ) + 1 , x m ( k ) 1 , x n ( k ) 1 ) G ( x m ( k ) + 1 , x m ( k ) , x m ( k ) ) + G ( x m ( k ) , x m ( k ) 1 , x n ( k ) 1 ) = G ( x m ( k ) + 1 , x m ( k ) , x m ( k ) ) + G ( x m ( k ) 1 , x m ( k ) , x n ( k ) 1 ) 2 s m ( k ) + G ( x m ( k ) 1 , T x m ( k ) 1 , x n ( k ) 1 ) ,
(26)
and
G ( x m ( k ) 1 , T x m ( k ) 1 , x n ( k ) 1 ) = G ( x m ( k ) 1 , x m ( k ) , x n ( k ) 1 ) G ( x m ( k ) 1 , x m ( k ) + 1 , x m ( k ) + 1 ) + G ( x m ( k ) + 1 , x m ( k ) , x n ( k ) 1 ) G ( x m ( k ) 1 , x m ( k ) , x m ( k ) ) + G ( x m ( k ) , x m ( k ) + 1 , x m ( k ) + 1 ) + G ( x m ( k ) + 1 , x m ( k ) , x n ( k ) 1 ) = s m ( k ) 1 + s m ( k ) + G ( x m ( k ) + 1 , x m ( k ) , x n ( k ) 1 ) = s m ( k ) 1 + s m ( k ) + G ( x m ( k ) , T x m ( k ) , x n ( k ) 1 ) < s m ( k ) 1 + s m ( k ) + ε .
(27)
Taking limit as n in (26) and (27) and applying (25), we have
lim k G ( x m ( k ) 1 , T x m ( k ) 1 , x n ( k ) 1 ) = ε .
(28)
By (15), with x = x m ( k ) 1 and y = x n ( k ) 1 , we have
ψ ( G ( x m ( k ) , T x m ( k ) , x n ( k ) ) ) = ψ ( G ( T x m ( k ) 1 , T 2 x m ( k ) 1 , T x n ( k ) 1 ) ) ψ ( G ( x m ( k ) 1 , T x m ( k ) 1 , x n ( k ) 1 ) ) ϕ ( G ( x m ( k ) 1 , T x m ( k ) 1 , x n ( k ) 1 ) ) .
Taking limit as k in the inequality above and applying, we have
ψ ( ε ) ψ ( ε ) ϕ ( ε ) ,
which implies ε = 0 , which is a contradiction. Then
lim m , n G ( x m , T x m , x n ) = lim m , n G ( x m , x m + 1 , x n ) = 0 .
That is, { x n } 0 is a Cauchy sequence. Since ( X , G ) is a G-complete, then there exist z X such that x n z as n . From (15), with x = x n and y = z , we have
ψ ( G ( x n + 1 , x n + 2 , T z ) ) = ψ ( G ( T x n , T 2 x n , T z ) ) ψ ( G ( x n , T x n , z ) ) ϕ ( G ( x n , T x n , z ) ) = ψ ( G ( x n , x n + 1 , z ) ) ϕ ( G ( x n , x n + 1 , z ) ) .
Taking limit as n , we get
ψ ( G ( z , z , T z ) ) ψ ( 0 ) ϕ ( 0 ) = 0 .
Then G ( z , z , T z ) = 0 , i.e., z = T z . To prove uniqueness, suppose that z u , such that T u = u . Now, by (15), we get
ψ ( G ( T z , T 2 z , T u ) ) ψ ( G ( z , T z , u ) ) ϕ ( G ( z , T z , u ) ) ,
(29)

which implies that ϕ ( G ( z , T z , u ) ) = 0 , i.e., z = u . □

If we take ψ ( t ) = t and ϕ ( t ) = ( 1 r ) t in Theorem 2.3, where 0 r < 1 , then we deduce the following corollary.

Corollary 2.2 Let ( X , G ) be a complete G-metric space and T : X X be a mapping satisfying the following condition for all x , y X , where 0 r < 1 holds
G ( T x , T 2 x , T y ) r G ( x , T x , y ) .

Then T has a unique fixed point.

Example 2.2 Let X = [ 0 , ) and
G ( x , y , z ) = { 0 , if  x = y = z , max { x , y } + max { y , z } + max { x , z } , otherwise
be a G-metric on X. Define T : X X by T x = 1 4 x . Then all the conditions of Corollary 2.2 (Theorem 2.3) hold. Indeed,
G ( T x , T 2 x , T y ) = 1 4 x + 1 4 max { 1 4 x , y } + 1 4 max { x , y }
and
G ( x , T x , y ) = x + max { 1 4 x , y } + max { x , y } ,
and so,
G ( T x , T 2 x , T y ) 1 2 G ( x , T x , y )

That is, the conditions of Corollary 2.2 (Theorem 2.3) hold for this example.

Corollary 2.3 Let ( X , G ) be a complete G-metric space and T : X X be a mapping satisfying the following condition for all x , y , z X , where 0 a + b < 2 holds
G ( T x , T 2 x , T y ) + G ( T x , T 2 x , T z ) a G ( x , T x , y ) + b G ( x , T x , z ) .

Then T has a unique fixed point.

Proof By taking y = z , we get
G ( T x , T 2 x , T y ) ( a + b ) 2 G ( x , T x , y ) ,

where 0 ( a + b ) 2 < 1 . That is, conditions of Theorem 2.3 hold, and T has a unique fixed point. □

3 Fixed point results for expansive mappings

In this section, we establish some fixed point results for expansive mappings.

Theorem 3.1 Let ( X , G ) be a complete G-metric space and T : X X be an onto mapping satisfying the following condition for all x , y X , where α > 1 holds
G ( T x , T 2 x , T y ) α G ( x , T x , y ) .
(30)

Then T has a unique fixed point.

Proof Let x 0 X , since T is onto, then there exists x 1 X such that x 0 = T x 1 . By continuing this process, we get x n = T x n + 1 for all n N 0 . In case x n 0 = x n 0 + 1 , for some n 0 N 0 , then it is clear that x n 0 is a fixed point of T. Now, assume that x n x n + 1 for all n. From (30), with x = x n + 1 and y = x n , we have
G ( x n , x n 1 , x n 1 ) = G ( T x n + 1 , T 2 x n + 1 , T x n ) α G ( x n + 1 , T x n + 1 , x n ) = α G ( x n + 1 , x n , x n ) ,
which implies that
G ( x n + 1 , x n , x n ) h G ( x n , x n 1 , x n 1 ) ,
(31)
where h = 1 α < 1 . Then we have
G ( x n + 1 , x n , x n ) h n G ( x 0 , x 1 , x 1 ) .
(32)
By Lemma 1.1, we get
G ( x n , x n + 1 , x n + 1 ) 2 G ( x n + 1 , x n , x n ) 2 h n G ( x 0 , x 1 , x 1 ) .
(33)
Following the lines of the proof of Theorem 2.1, we derive that { x n } is a Cauchy sequence. Since ( X , G ) is complete, then there exists z X such that x n z as n . Consequently, since T is onto, then there exists w X such that z = T w . From (30), with x = x n + 1 and y = w , we have
G ( x n , x n 1 , z ) = G ( T x n + 1 , T 2 x n + 1 , T w ) α G ( x n + 1 , T x n + 1 , w ) = α G ( x n + 1 , x n , w ) .
Taking limit as n in the inequality above, we get
G ( z , z , w ) = lim n G ( x n , x n 1 , z ) = 0 .
That is, z = w . Then z = T w = T z . To prove uniqueness, suppose that u v such that T v = v and T u = u . Now by (30), we get
G ( u , u , v ) = G ( T u , T 2 u , T v ) α G ( u , T u , v ) α G ( u , u , v ) > G ( u , u , v ) ,

which is a contradiction. Hence, u = v . □

Theorem 3.2 Let ( X , G ) be a complete G-metric space and T : X X be a mapping satisfying the following condition for all x , y X , where a > 1
G ( T x , T y , T 2 y ) α G ( x , T x , T 2 x ) .
(34)

Then T has a unique fixed point.

Proof Let x 0 X , since T is onto, then there exists x 1 X such that x 0 = T x 1 . By continuing this process, we get x n = T x n + 1 for all n N 0 . In case x n 0 = x n 0 + 1 , for some n 0 N 0 , then it is clear that x n 0 is a fixed point of T. Now, assume that x n x n + 1 for all n. From (34), with x = x n + 1 and y = x n , we have
G ( T x n + 1 , T x n , T 2 x n ) α G ( x n + 1 , T x n + 1 , T 2 x n + 1 ) ,
which implies that
G ( x n , x n 1 , x n 2 ) α G ( x n + 1 , x n , x n 1 ) ,
and so,
G ( x n + 1 , x n , x n 1 ) h G ( x n , x n 1 , x n 2 ) ,
where h = 1 α < 1 . By the mimic of the proof of Theorem 2.1, we can show that { x n } is a Cauchy sequence. Since ( X , G ) is a complete G-metric space, then there exists z X such that x n z as n . Consequently, since T is onto, then there exists w X such that z = T w . From (34), with x = w and y = x n + 1 , we have
G ( z , x n , x n 1 ) = G ( T w , T x n + 1 , T 2 x n + 1 ) α G ( w , T w , T 2 w ) .

Taking limit as n in the inequality above, we have G ( w , T w , T 2 w ) = 0 . That is, w = T w = T 2 w . To prove the uniqueness, suppose that u v such that T v = v and T u = u . □

Declarations

Acknowledgements

The authors thank to anonymous referees for their remarkable comments, suggestions and ideas that helped to improve this paper.

Authors’ Affiliations

(1)
Department of Mathematics, Zanjan Branch, Islamic Azad University
(2)
Atilim University, Incek
(3)
Young Researchers and Elite Club, Rasht Branch, Islamic Azad University

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