Open Access

Some fixed point results in ordered G p -metric spaces

Fixed Point Theory and Applications20132013:317

https://doi.org/10.1186/1687-1812-2013-317

Received: 9 August 2013

Accepted: 25 October 2013

Published: 25 November 2013

Abstract

In this paper, first we present some coincidence point results for six mappings satisfying the generalized ( ψ , φ ) -weakly contractive condition in the framework of partially ordered G p -metric spaces. Secondly, we consider α-admissible mappings in the setup of G p -metric spaces. An example is also provided to support our results.

MSC:47H10, 54H25.

Keywords

coincidence point common fixed point generalized weak contraction generalized metric space partially weakly increasing mapping altering distance function

1 Introduction and mathematical preliminaries

Recently, Zand and Nezhad [1] have introduced a new generalized metric space, a G p -metric space, as a generalization of both partial metric spaces [2] and G-metric spaces [3].

We will use the following definition of a G p -metric space.

Definition 1.1 [4]

Let X be a nonempty set. Suppose that a mapping G p : X × X × X R + satisfies:

( G p 1 ) x = y = z if G p ( x , y , z ) = G p ( z , z , z ) = G p ( y , y , y ) = G p ( x , x , x ) ;

( G p 2 ) G p ( x , x , x ) G p ( x , x , y ) G p ( x , y , z ) for all x , y , z X with z y ;

( G p 3 ) G p ( x , y , z ) = G p ( p { x , y , z } ) , where p is any permutation of x, y, z (symmetry in all three variables);

( G p 4 ) G p ( x , y , z ) G p ( x , a , a ) + G p ( a , y , z ) G p ( a , a , a ) for all x , y , z , a X (rectangle inequality).

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

The G p -metric G p is called symmetric if
G p ( x , x , y ) = G p ( x , y , y )
(1)

holds for all x , y X . Otherwise, G p is an asymmetric G p -metric.

Remark 1 In [1] (see also [5]), instead of ( G p 2 ), the following condition was used:

( G p 2 ) G p ( x , x , x ) G p ( x , x , y ) G p ( x , y , z ) for all x , y , z X .

However, with this assumption, it is very easy to obtain that (1) holds for all x , y X , i.e., the respective space is symmetric. On the other hand, there are a lot of examples of non-symmetric G-metric spaces. Hence, the conclusion stated in [1, 5] that each G-metric space is a G p -metric space (satisfying ( G p 2 )) does not hold. With our assumption ( G p 2 ), this conclusion holds true.

The following are some easy examples of G p -metric spaces.

Example 1.1 Let X = [ 0 , + ) , and let G p : X 3 R + be given by G p ( x , y , z ) = max { x , y , z } . Obviously, ( X , G p ) is a symmetric G p -metric space which is not a G-metric space.

Example 1.2 Let X = { 0 , 1 , 2 , 3 , } . Define G p : X 3 X by
G p ( x , y , z ) = { x + y + z + 1 , x y z , x + z + 1 , y = z x , y + z + 1 , x = z y , x + z + 1 , x = y z , 1 , x = y = z .

It is easy to see that ( X , G p ) is a symmetric G p -metric space.

Example 1.3 [4]

Let X = { 0 , 1 , 2 , 3 } . Let
A = { ( 1 , 0 , 0 ) , ( 0 , 1 , 0 ) , ( 0 , 0 , 1 ) , ( 2 , 0 , 0 ) , ( 0 , 2 , 0 ) , ( 0 , 0 , 2 ) , ( 3 , 0 , 0 ) , ( 0 , 3 , 0 ) , ( 0 , 0 , 3 ) , ( 1 , 2 , 2 ) , ( 2 , 1 , 2 ) , ( 2 , 2 , 1 ) , ( 2 , 3 , 3 ) , ( 3 , 2 , 3 ) , ( 3 , 3 , 2 ) } , B = { ( 0 , 1 , 1 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) , ( 0 , 2 , 2 ) , ( 2 , 0 , 2 ) , ( 2 , 2 , 0 ) , ( 0 , 3 , 3 ) , ( 3 , 0 , 3 ) , ( 3 , 3 , 0 ) , ( 2 , 1 , 1 ) , ( 1 , 2 , 1 ) , ( 1 , 1 , 2 ) , ( 3 , 2 , 2 ) , ( 2 , 3 , 2 ) , ( 2 , 2 , 3 ) } .
Define G p : X 3 R + by
G ( x , y , z ) = { 1 if  x = y = z 2 , 0 if  x = y = z = 2 , 2 if  ( x , y , z ) A , 5 2 if  ( x , y , z ) B , 3 if  x y z .

It is easy to see that ( X , G p ) is an asymmetric G p -metric space.

Proposition 1.1 [1]

Every G p -metric space ( X , G p ) defines a metric space ( X , d G p ) where
d G p ( x , y ) = G p ( x , y , y ) + G p ( y , x , x ) G p ( x , x , x ) G p ( y , y , y )

for all x , y X .

Proposition 1.2 [1]

Let X be a G p -metric space. Then, for each x , y , z , a X , it follows that:
  1. (1)

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

     
  2. (2)

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

     
  3. (3)

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

     
  4. (4)

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

     

Definition 1.2 [1]

Let ( X , G p ) be a G p -metric space. Let { x n } be a sequence of points of X.
  1. 1.

    A point x X is said to be a limit of the sequence { x n } , denoted by x n x , if lim n , m G p ( x , x n , x m ) = G p ( x , x , x ) .

     
  2. 2.

    { x n } is said to be a G p -Cauchy sequence if lim n , m G p ( x n , x m , x m ) exists (and is finite).

     
  3. 3.

    ( X , G p ) is said to be G p -complete if every G p -Cauchy sequence in X is G p -convergent to x X .

     

Using the above definitions, one can easily prove the following proposition.

Proposition 1.3 [1]

Let ( X , G p ) be a G p -metric space. Then, for any sequence { x n } in X and a point x X , the following are equivalent:
  1. (1)

    { x n } is G p -convergent to x.

     
  2. (2)

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

     
  3. (3)

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

     

Lemma 1.1 [4]

If G p is a G p -metric on X, then the mappings d G p , d G p : X × X R + , given by
d G p ( x , y ) = G p ( x , y , y ) + G p ( y , x , x ) G p ( x , x , x ) G p ( y , y , y )
and
d G p ( x , y ) = max { G p ( x , y , y ) G p ( x , x , x ) , G p ( y , x , x ) G p ( y , y , y ) } ,

define equivalent metrics on X.

Proof a + b 2 max { a , b } a + b for all nonnegative real numbers a, b. □

Based on Lemma 2.2 of [6], Parvaneh et al. have proved the following essential lemma.

Lemma 1.2 [4]
  1. (1)

    A sequence { x n } is a G p -Cauchy sequence in a G p -metric space ( X , G p ) if and only if it is a Cauchy sequence in the metric space ( X , d G p ) .

     
  2. (2)
    A G p -metric space ( X , G p ) is G p -complete if and only if the metric space ( X , d G p ) is complete. Moreover, lim n d G p ( x , x n ) = 0 if and only if
    lim n G p ( x , x n , x n ) = lim n G p ( x n , x , x ) = lim n , m G p ( x n , x n , x m ) = lim n , m G p ( x n , x m , x m ) = G p ( x , x , x ) .
     

Lemma 1.3 [4]

Assume that x n x as n in a G p -metric space ( X , G p ) such that G p ( x , x , x ) = 0 . Then, for every y X ,
  1. (i)

    lim n G p ( x n , y , y ) = G p ( x , y , y ) ,

     
  2. (ii)

    lim n G p ( x n , x n , y ) = G p ( x , x , y ) .

     

Lemma 1.4 [4]

Assume that { x n } , { y n } and { z n } are three sequences in a G p -metric space X such that
lim n G p ( x n , x , x ) = lim n G p ( x n , x n , x n ) = G p ( x , x , x ) , lim n G p ( y n , y , y ) = lim n G p ( y n , y n , y n ) = G p ( y , y , y )
and
lim n G p ( z n , z , z ) = lim n G p ( z n , z n , z n ) = G p ( z , z , z ) .
Then
  1. (i)

    lim n G p ( x n , y n , z n ) = G p ( x , y , z ) and

     
  2. (ii)

    lim n G p ( x n , x n , y ) = G p ( x , x , y )

     

for every y , z X .

Lemma 1.5 [5]

Let ( X , G p ) be a G p -metric space. Then
  1. (A)

    If G p ( x , y , z ) = 0 , then x = y = z .

     
  2. (B)

    If x y , then G p ( x , y , y ) > 0 .

     

Definition 1.3 [1]

Let ( X 1 , G 1 ) and ( X 2 , G 2 ) be two G p -metric spaces, and let f : ( X 1 , G 1 ) ( X 2 , G 2 ) be a mapping. Then f is said to be G p -continuous at a point a X 1 if for a given ε > 0 , there exists δ > 0 such that x , y X 1 and G 1 ( a , x , y ) < δ + G 1 ( a , a , a ) imply that G 2 ( f ( a ) , f ( x ) , f ( y ) ) < ε + G 2 ( f ( a ) , f ( a ) , f ( a ) ) . The mapping f is G p -continuous on X 1 if it is G p -continuous at all a X 1 .

Proposition 1.4 [1]

Let ( X 1 , G 1 ) and ( X 2 , G 2 ) be two G p -metric spaces. Then a mapping f : X 1 X 2 is G p -continuous at a point x X 1 if and only if it is G p -sequentially continuous at x; that is, whenever { x n } is G p -convergent to x, { f ( x n ) } is G p -convergent to f ( x ) .

The concept of an altering distance function was introduced by Khan et al. [7] as follows.

Definition 1.4 The function ψ : [ 0 , ) [ 0 , ) is called an altering distance function if the following properties are satisfied:
  1. 1.

    ψ is continuous and nondecreasing.

     
  2. 2.

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

     
A self-mapping f on X is called a weak contraction if the following contractive condition is satisfied:
d ( f x , f y ) d ( x , y ) φ ( d ( x , y ) ) ,

for all x , y X , where φ is an altering distance function.

The concept of a weakly contractive mapping was introduced by Alber and Guerre-Delabrere [8] in the setup of Hilbert spaces. Rhoades [9] considered this class of mappings in the setup of metric spaces and proved that a weakly contractive mapping defined on a complete metric space has a unique fixed point.

Zhang and Song [10] introduced the concept of a generalized φ-weakly contractive mapping as follows.

Definition 1.5 Self-mappings f and g on a metric space X are called generalized φ-weak contractions if there exists a lower semicontinuous function φ : [ 0 , ) [ 0 , ) with φ ( 0 ) = 0 and φ ( t ) > 0 for all t > 0 such that for all x , y X ,
d ( f x , g y ) N ( x , y ) φ ( N ( x , y ) ) ,
where
N ( x , y ) = max { d ( x , y ) , d ( x , f x ) , d ( y , g y ) , 1 2 [ d ( x , g y ) + d ( y , f x ) ] } .

Based on the above definition, they proved the following common fixed point result.

Theorem 1.1 [10]

Let ( X , d ) be a complete metric space. If f , g : X X are generalized φ-weakly contractive mappings, then there exists a unique point u X such that u = f u = g u .

So far, many authors extended Theorem 1.1 (see [1113] and [14]). Moreover, Ðorić [12] generalized it by the definition of generalized ( ψ , φ ) -weak contractions.

Definition 1.6 Two mappings f , g : X X are called generalized ( ψ , φ ) -weakly contractive if there exist two maps ψ , φ : [ 0 , ) [ 0 , ) such that
ψ ( d ( f x , g y ) ) ψ ( N ( x , y ) ) φ ( N ( x , y ) ) ,

for all x , y X , where N and φ are as in Definition 1.5 and ψ : [ 0 , ) [ 0 , ) is an altering distance function.

Theorem 1.2 [12]

Let ( X , d ) be a complete metric space, and let f , g : X X be generalized ( ψ , φ ) -weakly contractive self-mappings. Then there exists a unique point u X such that u = f u = g u .

Recently, many researchers have focused on different contractive conditions in various metric spaces endowed with a partial order and studied fixed point theory in the so-called bi-structured spaces. For more details on fixed point results, their applications, comparison of different contractive conditions and related results in ordered various metric spaces, we refer the reader to [1529] and the references mentioned therein.

Let X be a nonempty set and f : X X be a given mapping. For every x X , let f 1 ( x ) = { u X : f u = x } .

Definition 1.7 [24]

Let ( X , ) be a partially ordered set, and let f , g , h : X X be given mappings such that f X h X and g X h X . We say that f and g are weakly increasing with respect to h if for all x X , we have
f x g y for all  y h 1 ( f x )
and
g x f y for all  y h 1 ( g x ) .

If f = g , we say that f is weakly increasing with respect to h.

If h = I (the identity mapping on X), then the above definition reduces to that of a weakly increasing mapping [30] (see also [24, 31]).

Definition 1.8 A partially ordered G p -metric space ( X , , G p ) is said to have the sequential limit comparison property if for every nondecreasing sequence (nonincreasing sequence) { x n } in X, x n x implies that x n x ( x x n ).

The aim of this paper is to prove some coincidence and common fixed point theorems for weakly ( ψ , φ ) -contractive mappings in partially ordered G p -metric spaces.

2 Main results

Let ( X , , G p ) be an ordered G p -metric space and f , g , h , R , S , T : X X be six self-mappings. Throughout this paper, unless otherwise stated, for all x , y , z X , let
M ( x , y , z ) = max { G p ( T x , R y , S z ) , G p ( T x , f x , f x ) , G p ( R y , g y , g y ) , G p ( S z , h z , h z ) , G p ( T x , T x , f x ) + G p ( R y , R y , g y ) + G p ( S z , S z , h z ) 3 } .
Theorem 2.1 Let ( X , , G p ) be a partially ordered G p -metric space with the sequential limit comparison property. Let f , g , h , R , S , T : X X be six mappings such that f ( X ) R ( X ) , g ( X ) S ( X ) and h ( X ) T ( X ) , and RX, SX and TX are G p -complete subsets of X. Suppose that for comparable elements T x , R y , S z X , we have
ψ ( 2 G p ( f x , g y , h z ) ) ψ ( M ( x , y , z ) ) φ ( M ( x , y , z ) ) ,
(2)

where ψ , φ : [ 0 , ) [ 0 , ) are altering distance functions. Then the pairs ( f , T ) , ( g , R ) and ( h , S ) have a coincidence point z in X provided that the pairs ( f , T ) , ( g , R ) and ( h , S ) are weakly compatible and the pairs ( f , g ) , ( g , h ) and ( h , f ) are partially weakly increasing with respect to R, S and T, respectively. Moreover, if R z , S z and T z are comparable, then z X is a coincidence point of f, g, h, R, S and T.

Proof Let x 0 be an arbitrary point of X. Choose x 1 X such that f x 0 = R x 1 , x 2 X such that g x 1 = S x 2 and x 3 X such that h x 2 = T x 3 . This can be done as f ( X ) R ( X ) , g ( X ) S ( X ) and h ( X ) T ( X ) .

Continuing this way, construct a sequence { z n } defined by z 3 n + 1 = R x 3 n + 1 = f x 3 n , z 3 n + 2 = S x 3 n + 2 = g x 3 n + 1 and z 3 n + 3 = T x 3 n + 3 = h x 3 n + 2 for all n 0 . The sequence { z n } in X is said to be a Jungck-type iterative sequence with initial guess x 0 .

As x 1 R 1 ( f x 0 ) , x 2 S 1 ( g x 1 ) and x 3 T 1 ( h x 2 ) and the pairs ( f , g ) , ( g , h ) and ( h , f ) are partially weakly increasing with respect to R, S and T, respectively, we have
R x 1 = f x 0 g x 1 = S x 2 h x 2 = T x 3 f x 3 = R x 4 .

Continuing this process, we obtain R x 3 n + 1 S x 3 n + 2 T x 3 n + 3 for all n 0 .

We will complete the proof in three steps.

Step I. We will prove that { z n } is a G p -Cauchy sequence. First, we show that lim k G p ( z k , z k + 1 , z k + 2 ) = 0 .

Define G p k = G p ( z k , z k + 1 , z k + 2 ) . Suppose G p k 0 = 0 for some k 0 . Then z k 0 = z k 0 + 1 = z k 0 + 2 . In the case that k 0 = 3 n , then z 3 n = z 3 n + 1 = z 3 n + 2 gives z 3 n + 1 = z 3 n + 2 = z 3 n + 3 . Indeed,
ψ ( 2 G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) ) = ψ ( 2 G p ( f x 3 n , g x 3 n + 1 , h x 3 n + 2 ) ) ψ ( M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) ) φ ( M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) ) ,
where
M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) = max { G p ( T x 3 n , R x 3 n + 1 , S x 3 n + 2 ) , G p ( T x 3 n , f x 3 n , f x 3 n ) , G p ( R x 3 n + 1 , g x 3 n + 1 , g x 3 n + 1 ) , G p ( S x 3 n + 2 , h x 3 n + 2 , h x 3 n + 2 ) , G p ( T x 3 n , T x 3 n , f x 3 n ) + G p ( R x 3 n + 1 , R x 3 n + 1 , g x 3 n + 1 ) + G p ( S x 3 n + 2 , S x 3 n + 2 , h x 3 n + 2 ) 3 } = max { G p ( z 3 n , z 3 n + 1 , z 3 n + 2 ) , G p ( z 3 n , z 3 n + 1 , z 3 n + 1 ) , G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 2 ) , G p ( z 3 n + 2 , z 3 n + 3 , z 3 n + 3 ) , G p ( z 3 n , z 3 n , z 3 n + 1 ) + G p ( z 3 n + 1 , z 3 n + 1 , z 3 n + 2 ) + G p ( z 3 n + 2 , z 3 n + 2 , z 3 n + 3 ) 3 } = max { 0 , 0 , 0 , G p ( z 3 n + 2 , z 3 n + 3 , z 3 n + 3 ) , 0 + 0 + G p ( z 3 n + 2 , z 3 n + 2 , z 3 n + 3 ) 3 } = G p ( z 3 n + 2 , z 3 n + 3 , z 3 n + 3 ) 2 G p ( z 3 n + 2 , z 3 n + 2 , z 3 n + 3 ) = 2 G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) .
Thus
ψ ( 2 G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) ) ψ ( 2 G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) ) φ ( G p ( z 3 n + 2 , z 3 n + 3 , z 3 n + 3 ) )

implies that φ ( G p ( z 3 n + 2 , z 3 n + 3 , z 3 n + 3 ) ) = 0 , that is, z 3 n + 1 = z 3 n + 2 = z 3 n + 3 . Similarly, if k 0 = 3 n + 1 , then z 3 n + 1 = z 3 n + 2 = z 3 n + 3 gives z 3 n + 2 = z 3 n + 3 = z 3 n + 4 . Also, if k 0 = 3 n + 2 , then z 3 n + 2 = z 3 n + 3 = z 3 n + 4 implies that z 3 n + 3 = z 3 n + 4 = z 3 n + 5 . Consequently, the sequence { z k } becomes constant for k k 0 , hence { z k } is G p -Cauchy.

Suppose that
z k z k + 1 z k + 2
(3)
for each k. We now claim that the following inequality holds:
G p ( z k + 1 , z k + 2 , z k + 3 ) G p ( z k , z k + 1 , z k + 2 ) = M ( x k , x k + 1 , x k + 2 )
(4)

for each k = 1 , 2 , 3 ,  .

Let k = 3 n and for n 0 , G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) > G p ( z 3 n , z 3 n + 1 , z 3 n + 2 ) > 0 . Then, as T x 3 n R x 3 n + 1 S x 3 n + 2 , using (2) we obtain that
ψ ( G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) ) ψ ( 2 G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) ) = ψ ( 2 G p ( f x 3 n , g x 3 n + 1 , h x 3 n + 2 ) ) ψ ( M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) ) φ ( M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) ) ,
(5)
where
M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) = max { G p ( T x 3 n , R x 3 n + 1 , S x 3 n + 2 ) , G p ( T x 3 n , f x 3 n , f x 3 n ) , G p ( R x 3 n + 1 , g x 3 n + 1 , g x 3 n + 1 ) , G p ( S x 3 n + 2 , h x 3 n + 2 , h x 3 n + 2 ) , G p ( T x 3 n , T x 3 n , f x 3 n ) + G p ( R x 3 n + 1 , R x 3 n + 1 , g x 3 n + 1 ) + G p ( S x 3 n + 2 , S x 3 n + 2 , h x 3 n + 2 ) 3 } = max { G p ( z 3 n , z 3 n + 1 , z 3 n + 2 ) , G p ( z 3 n , z 3 n + 1 , z 3 n + 1 ) , G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 2 ) , G p ( z 3 n + 2 , z 3 n + 3 , z 3 n + 3 ) , G p ( z 3 n , z 3 n , z 3 n + 1 ) + G p ( z 3 n + 1 , z 3 n + 1 , z 3 n + 2 ) + G p ( z 3 n + 2 , z 3 n + 2 , z 3 n + 3 ) 3 } max { G p ( z 3 n , z 3 n + 1 , z 3 n + 2 ) , G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) , 2 G p ( z 3 n , z 3 n + 1 , z 3 n + 2 ) + G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) 3 } = G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) .
Hence (5) implies that
ψ ( G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) ) ψ ( G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) ) φ ( M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) ) ,
which is possible only if M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) = 0 , that is, G p ( z 3 n , z 3 n + 1 , z 3 n + 2 ) = 0 . A contradiction to (3). Hence, G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) G p ( z 3 n , z 3 n + 1 , z 3 n + 2 ) and
M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) = G p ( z 3 n , z 3 n + 1 , z 3 n + 2 ) .

Therefore, (4) is proved for k = 3 n .

Similarly, it can be shown that
G p ( z 3 n + 2 , z 3 n + 3 , z 3 n + 4 ) G p ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) = M ( x 3 n + 1 , x 3 n + 2 , x 3 n + 3 )
(6)
and
G p ( z 3 n + 3 , z 3 n + 4 , z 3 n + 5 ) G p ( z 3 n + 2 , z 3 n + 3 , z 3 n + 4 ) = M ( x 3 n + 2 , x 3 n + 3 , x 3 n + 4 ) .
(7)
Hence, { G p ( z k , z k + 1 , z k + 2 ) } is a nonincreasing sequence of nonnegative real numbers. Therefore, there is r 0 such that
lim k G p ( z k , z k + 1 , z k + 2 ) = r .
(8)
Since
G p ( z k + 1 , z k + 2 , z k + 3 ) M ( x k , x k + 1 , x k + 2 ) G p ( z k , z k + 1 , z k + 2 ) ,
(9)
taking the limit as k in (9), we obtain
lim k M ( x k , x k + 1 , x k + 2 ) = r .
(10)
Taking the limit as n in (5), using (8), (10) and the continuity of ψ and φ, we have ψ ( r ) ψ ( r ) φ ( r ) . Therefore, φ ( r ) = 0 . Hence
lim k G p ( z k , z k + 1 , z k + 2 ) = 0
(11)
from our assumptions about φ. Also, from Definition 1.1, part ( G p 2 ), we have
lim k G p ( z k , z k + 1 , z k + 1 ) = 0 ,
(12)
and since G p ( x , y , y ) 2 G p ( x , x , y ) for all x , y X , we have
lim k G p ( z k , z k , z k + 1 ) = 0 .
(13)
Step II. We now show that { z n } is a G p -Cauchy sequence in X. Therefore, we will show that
lim m , n G p ( z m , z n , z n ) = 0 .
Because of (11), (12) and (13), it is sufficient to show that
lim m , n G p ( z 3 m , z 3 n , z 3 n ) = 0 ,

i.e., we prove that { z 3 n } is G p -Cauchy.

Suppose the opposite. Then there exists ε > 0 for which we can find subsequences { z 3 m ( k ) } and { z 3 n ( k ) } of { z 3 n } such that n ( k ) > m ( k ) k and
G p ( z 3 m ( k ) , z 3 n ( k ) , z 3 n ( k ) ) ε ,
(14)
and n ( k ) is the smallest number such that the above statement holds; i.e.,
G p ( z 3 m ( k ) , z 3 n ( k ) 3 , z 3 n ( k ) 3 ) < ε .
(15)
From the rectangle inequality and (15), we have
G p ( z 3 m ( k ) , z 3 n ( k ) , z 3 n ( k ) ) G p ( z 3 m ( k ) , z 3 n ( k ) 3 , z 3 n ( k ) 3 ) + G p ( z 3 n ( k ) 3 , z 3 n ( k ) , z 3 n ( k ) ) < ε + G p ( z 3 n ( k ) 3 , z 3 n ( k ) , z 3 n ( k ) ) < ε + G p ( z 3 n ( k ) 3 , z 3 n ( k ) 2 , z 3 n ( k ) 2 ) + G p ( z 3 n ( k ) 2 , z 3 n ( k ) 1 , z 3 n ( k ) 1 ) + G p ( z 3 n ( k ) 1 , z 3 n ( k ) , z 3 n ( k ) ) .
(16)
Taking limit as k in (16), from (12) and (14) we obtain that
lim k G p ( z 3 m ( k ) , z 3 n ( k ) , z 3 n ( k ) ) = ε .
(17)
Using the rectangle inequality and ( G p 2 ), we have
G p ( z 3 m ( k ) , z 3 n ( k ) + 1 , z 3 n ( k ) + 2 ) G p ( z 3 m ( k ) , z 3 n ( k ) , z 3 n ( k ) ) + G p ( z 3 n ( k ) , z 3 n ( k ) + 1 , z 3 n ( k ) + 2 ) G p ( z 3 m ( k ) , z 3 n ( k ) + 1 , z 3 n ( k ) + 1 ) + G p ( z 3 n ( k ) + 1 , z 3 n ( k ) , z 3 n ( k ) ) + G p ( z 3 n ( k ) , z 3 n ( k ) + 1 , z 3 n ( k ) + 2 ) G p ( z 3 m ( k ) , z 3 n ( k ) + 1 , z 3 n ( k ) + 2 ) + G p ( z 3 n ( k ) + 1 , z 3 n ( k ) + 2 , z 3 n ( k ) + 2 ) + G p ( z 3 n ( k ) + 1 , z 3 n ( k ) , z 3 n ( k ) ) + G p ( z 3 n ( k ) , z 3 n ( k ) + 1 , z 3 n ( k ) + 2 ) .
(18)
Taking limit as k , we have
lim k G p ( z 3 m ( k ) , z 3 n ( k ) + 1 , z 3 n ( k ) + 2 ) ε lim k G p ( z 3 m ( k ) , z 3 n ( k ) + 1 , z 3 n ( k ) + 2 ) .
Finally,
G p ( z 3 m ( k ) + 1 , z 3 n ( k ) + 2 , z 3 n ( k ) + 3 ) G p ( z 3 m ( k ) + 1 , z 3 m ( k ) , z 3 m ( k ) ) + G p ( z 3 m ( k ) , z 3 n ( k ) + 2 , z 3 n ( k ) + 3 ) G p ( z 3 m ( k ) + 1 , z 3 m ( k ) , z 3 m ( k ) ) + G p ( z 3 m ( k ) , z 3 n ( k ) , z 3 n ( k ) ) + G p ( z 3 n ( k ) , z 3 n ( k ) + 2 , z 3 n ( k ) + 3 ) .
(19)
Taking limit as k and using (17), we have
lim k G p ( z 3 m ( k ) + 1 , z 3 n ( k ) + 2 , z 3 n ( k ) + 3 ) ε .
Consider,
G p ( z 3 m ( k ) , z 3 n ( k ) + 1 , z 3 n ( k ) + 2 ) G p ( z 3 m ( k ) , z 3 m ( k ) + 1 , z 3 m ( k ) + 1 ) + G p ( z 3 m ( k ) + 1 , z 3 n ( k ) + 1 , z 3 n ( k ) + 2 ) G p ( z 3 m ( k ) , z 3 m ( k ) + 1 , z 3 m ( k ) + 1 ) + G p ( z 3 m ( k ) + 1 , z 3 n ( k ) + 3 , z 3 n ( k ) + 3 ) + G p ( z 3 n ( k ) + 3 , z 3 n ( k ) + 1 , z 3 n ( k ) + 2 ) G p ( z 3 m ( k ) , z 3 m ( k ) + 1 , z 3 m ( k ) + 1 ) + G p ( z 3 m ( k ) + 1 , z 3 n ( k ) + 2 , z 3 n ( k ) + 3 ) + G p ( z 3 n ( k ) + 1 , z 3 n ( k ) + 2 , z 3 n ( k ) + 3 ) .
(20)
Taking limit as k and using (11), (12) and (13), we have
ε lim k G p ( z 3 m ( k ) + 1 , z 3 n ( k ) + 2 , z 3 n ( k ) + 3 ) .
Therefore,
lim k G p ( z 3 m ( k ) + 1 , z 3 n ( k ) + 2 , z 3 n ( k ) + 3 ) = ε .
(21)
As T x m ( k ) R x n ( k ) + 1 S x n ( k ) + 2 , so from (2) we have
ψ ( G p ( z 3 m ( k ) + 1 , z 3 n ( k ) + 2 , z 3 n ( k ) + 3 ) ) = ψ ( G p ( f x 3 m ( k ) , g x 3 n ( k ) + 1 , h x 3 n ( k ) + 2 ) ) ψ ( M ( x 3 m ( k ) , x 3 n ( k ) + 1 , x 3 n ( k ) + 2 ) ) φ ( M ( x 3 m ( k ) , x 3 n ( k ) + 1 , x 3 n ( k ) + 2 ) ) ,
(22)
where
M ( x 3 m ( k ) , x 3 n ( k ) + 1 , x 3 n ( k ) + 2 ) = max { G p ( T x 3 m ( k ) , R x 3 n ( k ) + 1 , S x 3 n ( k ) + 2 ) , G p ( T x 3 m ( k ) , f x 3 m ( k ) , f x 3 m ( k ) ) , G p ( R x 3 n ( k ) + 1 , g x 3 n ( k ) + 1 , g x 3 n ( k ) + 1 ) , G p ( S x 3 n ( k ) + 2 , h x 3 n ( k ) + 2 , h x 3 n ( k ) + 2 ) , G p ( T x 3 m ( k ) , T x 3 m ( k ) , f x 3 m ( k ) ) + G p ( R x 3 n ( k ) + 1 , R x 3 n ( k ) + 1 , g x 3 n ( k ) + 1 ) + G p ( S x 3 n ( k ) + 2 , S x 3 n ( k ) + 2 , h x 3 n ( k ) + 2 ) 3 } = max { G p ( z 3 m ( k ) , z 3 n ( k ) + 1 , z 3 n ( k ) + 2 ) , G p ( z 3 m ( k ) , z 3 m ( k ) + 1 , z 3 m ( k ) + 1 ) , G p ( z 3 n ( k ) + 1 , z 3 n ( k ) + 2 , z 3 n ( k ) + 2 ) , G p ( z 3 n ( k ) + 2 , z 3 n ( k ) + 3 , z 3 n ( k ) + 3 ) , G p ( z 3 m ( k ) , z 3 m ( k ) , z 3 m ( k ) + 1 ) + G p ( z 3 n ( k ) + 1 , z 3 n ( k ) + 1 , z 3 n ( k ) + 2 ) + G p ( z 3 n ( k ) + 2 , z 3 n ( k ) + 2 , z 3 n ( k ) + 3 ) 3 } .
Taking limit as k and using (12), (13), (17), (21) in (22), we have
ψ ( ε ) ψ ( ε ) φ ( ε ) < ψ ( ε ) ,

a contradiction. Hence, { z n } is a G p -Cauchy sequence.

Step III. We will show that f, g, h, R, S and T have a coincidence point.

Since { z n } is a G p -Cauchy sequence in the complete G p -metric space X, from Lemma 1.2, { z n } is a Cauchy sequence in the metric space ( X , d G p ) . Completeness of ( X , G p ) yields that ( X , d G p ) is also complete. Then there exists z X such that
lim n d G p ( z n , z ) = 0 .
(23)

Now, since lim m , n G p ( z m , z n , z n ) = 0 , (23) and part (2) of Lemma 1.2 yield that G p ( z , z , z ) = 0 .

Since R ( X ) is G p -complete and { z 3 n + 1 } R ( X ) , there exists u X such that z = R u and
lim n G p ( z 3 n + 1 , z 3 n + 1 , R u ) = lim n G p ( R x 3 n + 1 , R x 3 n + 1 , R u ) = lim n G p ( f x 3 n , f x 3 n , R u ) = G ( R u , R u , R u ) = 0 .
(24)
By similar arguments, there exist v , w X such that z = S v = T w and
lim n G p ( z 3 n + 2 , z 3 n + 2 , z ) = lim n G p ( S x 3 n + 2 , S x 3 n + 2 , z ) = lim n G p ( g x 3 n + 1 , g x 3 n + 1 , z ) = G ( z , z , z ) = 0
(25)
and
lim n G p ( z 3 n + 3 , z 3 n + 3 , z ) = lim n G p ( T x 3 n + 3 , T x 3 n + 3 , z ) = lim n G p ( h x 3 n + 2 , h x 3 n + 2 , z ) = G ( z , z , z ) = 0 .
(26)

Now, we prove that w is a coincidence point of f and T.

Since S x 3 n + 2 z = T w = R u as n , so S x 3 n + 2 T w = R u . Therefore, from (2), we have
ψ ( G p ( f w , g u , h x 3 n + 2 ) ) ψ ( M ( w , u , x 3 n + 2 ) ) φ ( M ( w , u , x 3 n + 2 ) ) ,
(27)
where
M ( w , u , x 3 n + 2 ) = max { G p ( T w , R u , S x 3 n + 2 ) , G ( T w , f w , f w ) , G p ( R u , g u , g u ) , G ( S x 3 n + 2 , h x 3 n + 2 , h x 3 n + 2 ) , G p ( T w , T w , f w ) + G ( R u , R u , g u ) + G p ( S x 3 n + 2 , S x 3 n + 2 , h x 3 n + 2 ) 3 } .
Taking limit as n in (27), as G ( z , z , z ) = 0 , from Lemma 1.3, we obtain that
ψ ( G p ( f w , g u , z ) ) ψ ( G p ( f w , g u , z ) ) φ ( max { G p ( z , f w , f w ) , G p ( z , g u , g u ) , G p ( z , z , f w ) + G p ( z , z , g u ) 3 } ) ,

which implies that g u = f w = z = T w = R u .

As f and T are weakly compatible, we have f z = f T w = T f w = T z . Thus z is a coincidence point of f and T.

Similarly it can be shown that z is a coincidence point of the pairs ( g , R ) and ( h , S ) .

Now, let R z , S z and T z be comparable. By (2) we have
ψ ( G p ( f z , g z , h z ) ) ψ ( M ( z , z , z ) ) φ ( M ( z , z , z ) ) ,
(28)
where
M ( z , z , z ) = max { G p ( T z , R z , S z ) , G p ( T z , f z , f z ) , G p ( R z , g z , g z ) , G p ( S z , h z , h z ) , G p ( T z , T z , f z ) + G p ( R z , R z , g z ) + G p ( S z , S z , h z ) 3 } = G p ( T z , R z , S z ) = G p ( f z , g z , h z ) .
Hence (28) gives
ψ ( G p ( f z , g z , h z ) ) ψ ( G p ( f z , g z , h z ) ) φ ( G p ( f z , g z , h z ) ) = 0 .

Therefore f z = g z = h z = T z = R z = S z . □

Theorem 2.2 Let ( X , , G p ) be a partially ordered complete G p -metric space. Let f , g , h : X X be three mappings. Suppose that for every three comparable elements x , y , z X , we have
ψ ( 2 G p ( f x , g y , h z ) ) ψ ( M ( x , y , z ) ) φ ( M ( x , y , z ) ) ,
(29)
where
M ( x , y , z ) = max { G p ( x , y , z ) , G p ( x , f x , f x ) , G p ( y , g y , g y ) , G p ( z , h z , h z ) , G p ( x , x , f x ) + G p ( y , y , g y ) + G p ( z , z , h z ) 3 }

and ψ , φ : [ 0 , ) [ 0 , ) are altering distance functions. Let f, g, h be continuous and the pairs ( f , g ) , ( g , h ) and ( h , f ) be partially weakly increasing. Then f, g and h have a common fixed point z in X.

Proof Let x 0 be an arbitrary point and x 3 n + 1 = f x 3 n , x 3 n + 2 = g x 3 n + 1 and x 3 n + 3 = h x 3 n + 2 for all n 0 .

Following the proof of the previous theorem, we can show that there exists z X such that
G p ( z , z , z ) = 0
(30)
and
lim n G p ( x 3 n , x 3 n , z ) = 0 .
(31)
Continuity of f yields that
lim n G p ( f x 3 n , f x 3 n , f z ) = G p ( f z , f z , f z ) .
(32)
By the rectangle inequality, we have
G p ( f z , z , z ) G p ( f z , f x 3 n , f x 3 n ) + G p ( x 3 n + 1 , z , z )
(33)
and
G p ( f z , f z , z ) G p ( z , f x 3 n , f x 3 n ) + G p ( f x 3 n , f z , f z ) .
(34)
Taking limit as n in (33) and (34), from (30) we obtain
G p ( f z , z , z ) G p ( f z , f z , f z )
and
G p ( f z , f z , z ) G p ( f z , f z , f z ) .

Similar inequalities are obtained for g and h.

On the other hand, as z z z , using (29) we obtain that
ψ ( G p ( f z , g z , h z ) ) ψ ( 2 G p ( f z , g z , h z ) ) ψ ( M ( z , z , z ) ) φ ( M ( z , z , z ) ) ,
(35)
where
M ( z , z , z ) = max { G p ( z , z , z ) , G p ( z , f z , f z ) , G p ( z , g z , g z ) , G p ( z , h z , h z ) , G p ( z , z , f z ) + G p ( z , z , g z ) + G p ( z , z , h z ) 3 } max { G p ( f z , f z , f z ) , G p ( g z , g z , g z ) , G p ( h z , h z , h z ) } .
(36)
We consider three cases as follows:
  1. 1.

    f z = g z = h z .

     
  2. 2.

    f z g z h z .

     
  3. 3.

    a. f z = g z h z , or b. f z g z = h z .

     

For case 1, by (36), M ( z , z , z ) G p ( f z , g z , h z ) .

For case 2, by ( G p 2 ), M ( z , z , z ) G p ( f z , g z , h z ) .

Now, from (35),
ψ ( G p ( f z , g z , h z ) ) ψ ( G p ( f z , g z , h z ) ) φ ( M ( z , z , z ) ) ,
(37)

hence M ( z , z , z ) = 0 . Therefore, z = f z = g z = h z .

On the other hand, for case 3, part a, by ( G p 2 ), M ( z , z , z ) 2 G p ( f z , g z , h z ) and hence from (35), we have
ψ ( 2 G p ( f z , g z , h z ) ) ψ ( 2 G p ( f z , g z , h z ) ) φ ( M ( z , z , z ) ) ,
(38)

hence M ( z , z , z ) = 0 . Therefore, z = f z = g z = h z .

Now, let x and y as two fixed points of f, g and h be comparable. So, from (29) we have
ψ ( 2 G p ( x , x , y ) ) = ψ ( 2 G p ( f x , g x , h y ) ) ψ ( M ( x , x , y ) ) φ ( M ( x , x , y ) ) ,
(39)
where
M ( x , x , y ) = max { G p ( x , x , y ) , G p ( x , f x , f x ) , G p ( x , g x , g x ) , G p ( y , h y , h y ) , G p ( x , x , f x ) + G p ( x , x , g x ) + G p ( y , y , h y ) 3 } 2 G p ( x , x , y ) .
Hence (39) gives
ψ ( 2 G p ( x , x , y ) ) ψ ( 2 G p ( x , x , y ) ) φ ( M ( x , x , y ) ) .

Therefore, φ ( M ( x , x , y ) ) = 0 and hence x = y . □

The following corollaries are special cases of the above results.

Corollary 2.1 Let ( X , , G p ) be a partially ordered complete G p -metric space. Let f : X X be a mapping such that for every three comparable elements x , y , z X , we have
ψ ( 2 G p ( f x , f y , f z ) ) ψ ( M ( x , y , z ) ) φ ( M ( x , y , z ) ) ,
(40)
where
M ( x , y , z ) = max { G p ( x , y , z ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) , G p ( z , f z , f z ) , G p ( x , x , f x ) + G p ( y , y , f y ) + G p ( z , z , f z ) 3 }
and ψ , φ : [ 0 , ) [ 0 , ) are altering distance functions. Then f has a fixed point in X provided that f x f ( f x ) for all x X and either
  1. a.

    f is continuous, or

     
  2. b.

    X has the sequential limit comparison property.

     

Moreover, f has a unique fixed point provided that the fixed points of f are comparable.

Taking y = z in Corollary 2.1, we obtain the following common fixed point result.

Corollary 2.2 Let ( X , , G p ) be a partially ordered complete G p -metric space, and let f be a self-mapping on X such that for every comparable elements x , y X ,
ψ ( 2 G p ( f x , f y , f y ) ) ψ ( M ( x , y , y ) ) φ ( M ( x , y , y ) ) ,
(41)
where
M ( x , y , y ) = max { G p ( x , y , y ) , G ( x , f x , f x ) , G p ( y , f y , f y ) , G p ( x , x , f x ) + 2 G p ( y , y , f y ) 3 } ,
and ψ , φ : [ 0 , ) [ 0 , ) are altering distance functions. Then f has a fixed point in X provided that f x f ( f x ) for all x X and either
  1. a.

    f is continuous, or

     
  2. b.

    X has the sequential limit comparison property.

     

3 Fixed point results via an α-admissible mapping with respect to η in G p -metric spaces

Samet et al. [32] defined the notion of α-admissible mappings and proved the following result.

Definition 3.1 Let T be a self-mapping on X and α : X × X [ 0 , + ) be a function. We say that T is an α-admissible mapping if
x , y X , α ( x , y ) 1 α ( T x , T y ) 1 .

Denote with Ψ the family of all nondecreasing functions ψ : [ 0 , + ) [ 0 , + ) such that n = 1 ψ n ( t ) < + for all t > 0 , where ψ n is the n th iterate of ψ.

Theorem 3.1 Let ( X , d ) be a complete metric space and T be an α-admissible mapping. Assume that
α ( x , y ) d ( T x , T y ) ψ ( d ( x , y ) ) ,
(42)
where ψ Ψ . Also suppose that the following assertions hold:
  1. (i)

    there exists x 0 X such that α ( x 0 , T x 0 ) 1 ;

     
  2. (ii)

    either T is continuous or for any sequence { x n } in X with α ( x n , x n + 1 ) 1 for all n N { 0 } and x n x as n + , we have α ( x n , x ) 1 for all n N { 0 } .

     

Then T has a fixed point.

For more details on α-admissible mappings, we refer the reader to [3337].

Very recently, Salimi et al. [38] modified and generalized the notions of α-ψ-contractive mappings and α-admissible mappings as follows.

Definition 3.2 [38]

Let T be a self-mapping on X and α , η : X × X [ 0 , + ) be two functions. We say that T is an α-admissible mapping with respect to η if
x , y X , α ( x , y ) η ( x , y ) α ( T x , T y ) η ( T x , T y ) .

Note that if we take η ( x , y ) = 1 , then this definition reduces to Definition 3.1. Also, if we take α ( x , y ) = 1 , then we say that T is an η-subadmissible mapping.

The following result properly contains Theorem 3.1 and Theorems 2.3 and 2.4 of [37].

Theorem 3.2 [38]

Let ( X , d ) be a complete metric space and T be an α-admissible mapping with respect to η. Assume that
x , y X , α ( x , y ) η ( x , y ) d ( T x , T y ) ψ ( M ( x , y ) ) ,
(43)
where ψ Ψ and
M ( x , y ) = max { d ( x , y ) , d ( x , T x ) + d ( y , T y ) 2 , d ( x , T y ) + d ( y , T x ) 2 } .
Also, suppose that the following assertions hold:
  1. (i)

    there exists x 0 X such that α ( x 0 , T x 0 ) η ( x 0 , T x 0 ) ;

     
  2. (ii)

    either T is continuous or for any sequence { x n } in X with α ( x n , x n + 1 ) η ( x n , x n + 1 ) for all n N { 0 } and x n x as n + , we have α ( x n , x ) η ( x n , x ) for all n N { 0 } .

     

Then T has a fixed point.

In fact, the Banach contraction principle and Theorem 3.2 hold for the following example, but Theorem 3.1 does not hold.

Example 3.1 [38]

Let X = [ 0 , ) be endowed with the usual metric d ( x , y ) = | x y | for all x , y X , and let T : X X be defined by T x = 1 4 x . Also, define α : X 2 [ 0 , ) by α ( x , y ) = 3 and ψ : [ 0 , ) [ 0 , ) by ψ ( t ) = 1 2 t .

Theorem 3.3 Let ( X , G p ) be a G p -complete G p -metric space, f be a continuous α-admissible mapping with respect to η on X, there exists x 0 X such that α ( x 0 , f x 0 ) η ( x 0 , f x 0 ) and if any sequence { x n } in X converges to a point x, then we have α ( x , x ) η ( x , x ) . Assume that
α ( x , y ) η ( x , y ) G p ( f x , f y , f y ) r max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) }
(44)

for all x , y X , where 0 r < 1 . Then f has a fixed point.

Proof Let x 0 X and define a sequence { x n } by x n = f n x 0 for all n N . Since f is an α-admissible mapping with respect to η and α ( x 0 , x 1 ) = α ( x 0 , f x 0 ) η ( x 0 , f x 0 ) = η ( x 0 , x 1 ) , we deduce that α ( x 1 , x 2 ) = α ( f x 0 , f x 1 ) η ( f x 0 , f x 1 ) = η ( x 1 , x 2 ) . Continuing this process, we get α ( x n , x n + 1 ) η ( x n , x n + 1 ) for all n N { 0 } . Now, from (44) we have
G p ( f f n x 0 , f 2 f n x 0 , f 2 f n x 0 ) r max { G p ( f n x 0 , f f n x 0 , f f n x 0 ) , G p ( f f n x 0 , f 2 f n x 0 , f 2 f n x 0 ) } ,
which implies
G p ( f n + 1 x 0 , f n + 2 x 0 , f n + 2 x 0 ) r G p ( f n x 0 , f n + 1 x 0 , f n + 1 x 0 ) .
(45)
Continuing the above process, we can obtain
G p ( f n x 0 , f n + 1 x 0 , f n + 1 x 0 ) r G p ( f n 1 x 0 , f n x 0 , f n x 0 ) r n G p ( x 0 , f x 0 , f x 0 ) .
(46)
Then, for any m > n , by (46) we get
G p ( f n x 0 , f m x 0 , f m x 0 ) G p ( f n x 0 , f n + 1 x 0 , f n + 1 x 0 ) + G p ( f n + 1 x 0 , f m x 0 , f m x 0 ) G p ( f n x 0 , f n + 1 x 0 , f n + 1 x 0 ) + G p ( f n + 1 x 0 , f n + 2 x 0 , f n + 2 x 0 ) + G p ( f n + 2 x 0 , f m x 0 , f m x 0 ) G ( f n x 0 , f n + 1 x 0 , f n + 1 x 0 ) + G p ( f n + 1 x 0 , f n + 2 x 0 , f n + 2 x 0 ) + G p ( f n + 2 x 0 , f n + 3 x 0 , f n + 3 x 0 ) + + G p ( f m 1 x 0 , f m x 0 , f m x 0 ) r n 1 r G p ( x 0 , f x 0 , f x 0 ) .

This implies that lim m , n + G p ( f n x 0 , f m x 0 , f m x 0 ) = 0 , that is, { x n } is a G p -Cauchy sequence.

Since { x n } is a G p -Cauchy sequence in the complete G p -metric space X, from Lemma 1.2, { x n } is a Cauchy sequence in the metric space ( X , d G p ) . Completeness of ( X , G p ) yields that ( X , d G p ) is also complete. Then there exists z X such that
lim n d G p ( x n , z ) = 0 .
(47)
Since lim m , n + G p ( x n , x m , x m ) = 0 , from Lemma 1.2 we get
lim n + G p ( x n , z , z ) = lim n + G p ( x n , x n , z ) = G p ( z , z , z ) = 0 .
(48)
From the continuity of f, we have
lim n + G p ( x n + 1 , f z , f z ) = G p ( f z , f z , f z ) ,
and hence we get
G p ( z , f z , f z ) lim n + G ( z , x n + 1 , x n + 1 ) + lim n + G ( x n + 1 , f z , f z ) = G p ( f z , f z , f z ) .
So, we get that G p ( z , f z , f z ) G p ( f z , f z , f z ) . Since the opposite inequality always holds, we get that
G p ( z , f z , f z ) = G p ( f z , f z , f z ) .
As α ( z , z ) η ( z , z ) we have
G p ( z , f z , f z ) = G p ( f z , f z , f z ) r max { G p ( z , z , z ) , G p ( z , f z , f z ) , G p ( z , f z , f z ) } ,
(49)

where 0 r < 1 . Hence, G p ( z , f z , f z ) r G p ( z , f z , f z ) . Thus, G p ( z , f z , f z ) = 0 , that is, z = f z . □

If in Theorem 3.3 we take η ( x , y ) = 1 , then we deduce the following corollary.

Corollary 3.1 Let ( X , G p ) be a G p -complete G p -metric space, f be a continuous α-admissible mapping on X, and there exists x 0 X such that α ( x 0 , f x 0 ) 1 . Assume that
α ( x , y ) 1 G p ( f x , f y , f y ) r max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) }

for all x , y X , where 0 r < 1 , and if any sequence { x n } in X converges to a point x, then we have α ( x , x ) 1 . Then f has a fixed point.

If in Theorem 3.3 we take α ( x , y ) = 1 , then we deduce the following corollary.

Corollary 3.2 Let ( X , G p ) be a G p -complete G p -metric space, f be a continuous η-subadmissible mapping on X, and there exists x 0 X such that η ( x 0 , f x 0 ) 1 . Assume that
η ( x , y ) 1 G p ( f x , f y , f y ) r max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) }
(50)

for all x , y X , where 0 r < 1 , and if any sequence { x n } in X converges to a point x, then we have 1 η ( x , x ) . Then f has a fixed point.

In the following theorem, we omit the continuity of the mapping f.

Theorem 3.4 Let ( X , G p ) be a G p -complete G p -metric space and f be an α-admissible mapping with respect to η on X such that
α ( x , y ) η ( x , y ) G p ( f x , f y , f y ) r max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) }
(51)
for all x , y X , where 0 r < 1 . Assume that the following conditions hold:
  1. (i)

    there exists x 0 X such that α ( x 0 , f x 0 ) η ( x 0 , f x 0 ) ;

     
  2. (ii)

    if { x n } is a sequence in X such that α ( x n , x n + 1 ) η ( x n , x n + 1 ) for all n and x n x as n + , then α ( x n , x ) η ( x n , x ) for all n N { 0 } .

     

Then f has a fixed point.

Proof Let x 0 X be such that α ( x 0 , f x 0 ) η ( x 0 , f x 0 ) and define a sequence { x n } in X by x n = f n x 0 = f x n 1 for all n N . Following the proof of Theorem 3.1, we have α ( x n , x n + 1 ) η ( x n , x n + 1 ) for all n N { 0 } and there exists x X such that x n x as n + . Hence, from (ii) we deduce that α ( x n , x ) η ( x n , x ) for all n N { 0 } .

Hence, by (51), it follows that for all n,
G p ( x n + 1 , f x , f x ) r max { G p ( x n , x , x ) , G p ( x n , x n + 1 , x n + 1 ) , G p ( x , f x , f x ) } .

Taking the limit as n + in the above inequality, from Lemma 1.3 we obtain ( 1 r ) G ( x , f x , f x ) 0 , which implies that x = f x . □

Corollary 3.3 Let ( X , G p ) be a G p -complete G p -metric space and f be an α-admissible mapping on X such that
α ( x , y ) 1 G p ( f x , f y , f y ) r max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) }
(52)
for all x , y X , where 0 r < 1 . Assume that the following conditions hold:
  1. (i)

    there exists x 0 X such that α ( x 0 , f x 0 ) 1 ;

     
  2. (ii)

    if { x n } is a sequence in X such that α ( x n , x n + 1 ) 1 for all n and x n x as n + , then α ( x n , x ) 1 for all n N { 0 } .

     

Then f has a fixed point.

Example 3.2 Let X = [ 0 , + ) and G p ( x , y , z ) = max { x , y , z } be a G p -metric on X. Define f : X X by
f x = { x 24 if  x [ 0 , 1 ] { 2 } = U , 37 / 12 if  x = 3 , ( 1 + x ) x if  x [ 0 , + ) ( [ 0 , 1 ] { 2 , 3 } ) = V ,
and α : X × X [ 0 , + ) by
α ( x , y ) = { 1 if  x , y [ 0 , 1 ] , 1 / 8 if  x = 2  and  y = 3 , 0 otherwise .

Now, we prove that all the hypotheses of Corollary 3.3 are satisfied and hence f has a fixed point.

Let x , y X , if α ( x , y ) 1 , then x , y [ 0 , 1 ] . On the other hand, for all x [ 0 , 1 ] , we have f x 1 and hence α ( f x , f y ) 1 . This implies that f is an α-admissible mapping on X. Obviously, α ( 0 , f 0 ) 1 .

Now, if { x n } is a sequence in X such that α ( x n , x n + 1 ) 1 for all n N { 0 } and x n x as n + , then { x n } [ 0 , 1 ] and hence x [ 0 , 1 ] . This implies that α ( x n , x ) 1 for all n N { 0 } .

If α ( x , y ) 1 , then x , y [ 0 , 1 ] . Hence,
G p ( f x , f y , f y ) = max { f x , f y } = max { x 24 , y 24 } 1 12 max { x , y } 1 12 max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) } .

Thus, all the conditions of Corollary 3.3 are satisfied and therefore f has a fixed point ( x = 0 ).

Corollary 3.4 Let ( X , G p ) be a G p -complete G p -metric space and f be an η-subadmissible mapping on X such that
η ( x , y ) 1 G p ( f x , f y , f y ) r max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) }
for all x , y X , where 0 r < 1 . Assume that the following conditions hold:
  1. (i)

    there exists x 0 X such that η ( x 0 , f x 0 ) 1 ;

     
  2. (ii)

    if { x n } is a sequence in X such that η ( x n , x n + 1 ) 1 for all n and x n x as n + , then η ( x n , x ) 1 for all n N { 0 } .

     

Then f has a fixed point.

4 Consequences

Theorem 4.1 Let ( X , G p ) be a G p -complete G p -metric space, f be a continuous α-admissible mapping on X, and there exists x 0 X such that α ( x 0 , f x 0 ) 1 . Assume that
α ( x , y ) G p ( f x , f y , f y ) r max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) }
(53)

for all x , y X , where 0 r < 1 and if any sequence { x n } in X converges to a point x, then we have α ( x , x ) η ( x , x ) . Then f has a fixed point.

Proof Assume that α ( x , y ) 1 , then from (53) we get
G p ( f x , f y , f y ) r max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) } .
That is,
α ( x , y ) 1 G p ( f x , f y , f y ) r max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) } .

Hence all the conditions of Corollary 3.1 hold and f has a fixed point. □

Similarly, we can deduce the following results.

Theorem 4.2 Let ( X , G p ) be a G p -complete G p -metric space, f be a continuous α-admissible mapping on X, and there exists x 0 X such that α ( x 0 , f x 0 ) 1 . Assume that
( G p ( f x , f y , f y ) + ) α ( x , y ) r max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) } +

for all x , y X , where 0 r < 1 and 1 , and if any sequence { x n } in X converges to a point x, then we have α ( x , x ) 1 . Then f has a fixed point.

Theorem 4.3 Let ( X , G p ) be a G p -complete G p -metric space, f be a continuous α-admissible mapping on X, and there exists x 0 X such that α ( x 0 , f x 0 ) 1 . Assume that
( α ( x , y ) + ) G p ( f x , f y , f y ) ( 1 + ) r max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) }
(54)

for all x , y X , where 0 r < 1 and > 0 , and if any sequence { x n } in X converges to a point x, then we have α ( x , x ) 1 . Then f has a fixed point.

Theorem 4.4 Let ( X , G p ) be a G p -complete G p -metric space, f be a continuous η-subadmissible mapping on X, and there exists x 0 X such that η ( x 0 , f x 0 ) 1 . Assume that
G p ( f x , f y , f y ) r η ( x , y ) max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) }
(55)

for all x , y X , where 0 r < 1 , and if any sequence { x n } in X converges to a point x, then we have 1 η ( x , x ) . Then f has a fixed point.

Theorem 4.5 Let ( X , G p ) be a G p -complete G p -metric space, f be a continuous η-subadmissible mapping on X, and there exists x 0 X such that η ( x 0 , f x 0 ) 1 . Assume that
G p ( f x , f y , f y ) + ( r max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) } + ) η ( x , y )

for all x , y X , where 0 r < 1 and 1 , and if any sequence { x n } in X converges to a point x, then we have 1 η ( x , x ) . Then f has a fixed point.

Theorem 4.6 Let ( X , G p ) be a G p -complete G p -metric space, f be a continuous η-subadmissible mapping on X, and there exists x 0 X such that η ( x 0 , f x 0 ) 1 . Assume that
( 1 + ) G p ( f x , f y , f y ) ( η ( x , y ) + ) r max { G p ( x , y , y ) , G p ( x , f x , f x ) , G p ( y , f y , f y ) }
(56)

for all x , y X , where 0 r < 1 and > 0 , and if any sequence { x n } in X converges to a point x, then we have 1 η ( x , x ) . Then f has a fixed point.

Theorem 4.7 Let ( X , G p ) be a G p -complete G p -metric space, f be an α-admissible mapping on X, and there exists x 0 X such that α <