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Fixed point theorems for weak C-contractions in partially ordered 2-metric spaces

Fixed Point Theory and Applications20132013:161

DOI: 10.1186/1687-1812-2013-161

Received: 10 February 2013

Accepted: 30 May 2013

Published: 21 June 2013

Abstract

The aim of this paper is to state some fixed point results for weak C-contractions in a partially ordered 2-metric space. Examples are given to illustrate the results.

1 Introduction and preliminaries

Chatterjea in [1] introduced the notion of a C-contraction.

Definition 1.1 [1]

Let ( X , d ) be a metric space and T : X X be a map. Then T is called a C-contraction if there exists α ( 0 , 1 2 ) such that for all x , y X , d ( T x , T y ) α [ d ( x , T y ) + d ( y , T x ) ] .

This notion was generalized to a weak C-contraction by Choudhury in [2].

Definition 1.2 ([2], Definition 1.3)

Let ( X , d ) be a metric space and T : X X be a map. Then T is called a weak C-contraction if there exists ψ : [ 0 , ) 2 [ 0 , ) which is continuous, and ψ ( s , t ) = 0 if and only if s = t = 0 such that
d ( T x , T y ) 1 2 [ d ( x , T y ) + d ( y , T x ) ] ψ ( d ( x , T y ) , d ( y , T x ) )
(1.1)

for all x , y X .

In [2], Choudhury proved that if X is a complete metric space, then every weak C-contraction has a unique fixed point; see [[2], Theorem 2.1]. This result was generalized to a complete, partially ordered metric space in [3]; see [[3], Theorems 2.1, 2.3 and 3.1].

There were some generalizations of a metric such as a 2-metric, a D-metric, a G-metric, a cone metric, and a complex-valued metric. The notion of a 2-metric has been introduced by Gähler in [4]. Note that a 2-metric is not a continuous function of its variables, whereas an ordinary metric is. This led Dhage to introduce the notion of a D-metric in [5]. But in [6] Mustafa and Sims showed that most of topological properties of D-metric were not correct. In [7] Mustafa and Sims introduced the notion of a G-metric to overcome flaws of a D-metric. After that, many fixed point theorems on G-metric spaces have been stated. However, it was shown in [8] and [9] that in several situations fixed point results in G-metric spaces can be in fact deduced from fixed point theorems in metric or quasi-metric spaces.

In [10] Huang and Zhang defined the notion of a cone metric. After that, many authors extended some fixed point theorems on metric spaces to cone metric spaces. However, it was shown later by various authors that in several cases the fixed point results in cone metric spaces can be obtained by reducing them to their standard metric counterparts; for example, see [1114]. In [15] Azam, Fisher and Khan have introduced the notion of a complex-valued metric and some fixed point theorems have been stated. But in [16] Sastry, Naidu and Bekeshie showed that some fixed point theorems recently generalized to complex-valued metric spaces are consequences of their counterparts in the setting of metric spaces and hence are redundant.

Note that in the above generalizations, only a 2-metric space has not been known to be topologically equivalent to an ordinary metric. Then there was no easy relationship between results obtained in 2-metric spaces and metric spaces. In particular, the fixed point theorems on 2-metric spaces and metric spaces may be unrelated easily. For the fixed point theorems on 2-metric spaces, the readers may refer to [1726].

The aim of this paper is to state some fixed point results for weak C-contractions in a partially ordered 2-metric space. Examples are given to illustrate the results.

First we recall some notions and lemmas which will be useful in what follows.

Definition 1.3 [4]

Let X be a non-empty set and let d : X × X × X R be a map satisfying the following conditions:
  1. 1.

    For every pair of distinct points x , y X , there exists a point z X such that d ( x , y , z ) 0 .

     
  2. 2.

    If at least two of three points x , y , z are the same, then d ( x , y , z ) = 0 .

     
  3. 3.

    The symmetry: d ( x , y , z ) = d ( x , z , y ) = d ( y , x , z ) = d ( y , z , x ) = d ( z , x , y ) = d ( z , y , x ) for all x , y , z X .

     
  4. 4.

    The rectangle inequality: d ( x , y , z ) d ( x , y , t ) + d ( y , z , t ) + d ( z , x , t ) for all x , y , z , t X .

     

Then d is called a 2-metric on X and ( X , d ) is called a 2-metric space which will be sometimes denoted by X if there is no confusion. Every member x X is called a point in X.

Definition 1.4 [4]

Let ( X , d ) be a 2-metric space and a , b X , r 0 . The set
B ( a , b , r ) = { x X : d ( a , b , x ) < r }

is called a 2-ball centered at a and b with radius r. The topology generated by the collection of all 2-balls as a subbasis is called a 2-metric topology on X.

Definition 1.5 [22]

Let { x n } be a sequence in a 2-metric space ( X , d ) .
  1. 1.

    { x n } is said to be convergent to x in ( X , d ) , written lim n x n = x , if for all a X , lim n d ( x n , x , a ) = 0 .

     
  2. 2.

    { x n } is said to be Cauchy in X if for all a X , lim n , m d ( x n , x m , a ) = 0 , that is, for each ε > 0 , there exists n 0 such that d ( x n , x m , a ) < ε for all n , m n 0 .

     
  3. 3.

    ( X , d ) is said to be complete if every Cauchy sequence is a convergent sequence.

     

Definition 1.6 ([24], Definition 8)

A 2-metric space ( X , d ) is said to be compact if every sequence in X has a convergent subsequence.

Lemma 1.7 ([24], Lemma 3)

Every 2-metric space is a T 1 -space.

Lemma 1.8 ([24], Lemma 4)

lim n x n = x in a 2-metric space ( X , d ) if and only if lim n x n = x in the 2-metric topological space X.

Lemma 1.9 ([24], Lemma 5)

If T : X Y is a continuous map from a 2-metric space X to a 2-metric space Y, then lim n x n = x in X implies lim n T x n = T x in Y.

Remark 1.10
  1. 1.

    It is straightforward from Definition 1.3 that every 2-metric is non-negative and every 2-metric space contains at least three distinct points.

     
  2. 2.

    A 2-metric d ( x , y , z ) is sequentially continuous in one argument. Moreover, if a 2-metric d ( x , y , z ) is sequentially continuous in two arguments, then it is sequentially continuous in all three arguments, see [[27], p.975].

     
  3. 3.

    A convergent sequence in a 2-metric space need not be a Cauchy sequence, see [[27], Remark 01 and Example 01].

     
  4. 4.

    In a 2-metric space ( X , d ) , every convergent sequence is a Cauchy sequence if d is continuous, see [[27], Remark 02].

     
  5. 5.

    There exists a 2-metric space ( X , d ) such that every convergent sequence is a Cauchy sequence but d is not continuous, see [[27], Remark 02 and Example 02].

     

2 Main results

First, we introduce the notion of a weak C-contraction on a partially ordered 2-metric space.

Definition 2.1 Let ( X , , d ) be a partially ordered 2-metric space and T : X X be a map. Then T is called a weak C-contraction if there exists ψ : [ 0 , ) 2 [ 0 , ) which is continuous, and ψ ( s , t ) = 0 if and only if s = t = 0 such that
d ( T x , T y , a ) 1 2 [ d ( x , T y , a ) + d ( y , T x , a ) ] ψ ( d ( x , T y , a ) , d ( y , T x , a ) )
(2.1)

for all x , y , a X and x y or y x .

The following example gives some examples of ψ in Definition 2.1. Note that in [[2], Example 2.1], Choudhury considered the function ψ ( a , b ) = 1 2 min { a , b } for all a , b [ 0 , ) . Unfortunately, for this function, we have ψ ( 0 , 1 ) = 0 , which is a contradiction to the condition that ψ ( s , t ) = 0 if and only if s = t = 0 in [[2], Definition 1.3], also in Definition 2.1.

Example 2.2
  1. 1.

    ψ ( a , b ) = a + b 2 for all a , b [ 0 , ) .

     
  2. 2.

    ψ ( a , b ) = max { a , b } 4 for all a , b [ 0 , ) .

     

The first result is a sufficient condition for the existence of a fixed point of a weak C-contraction on a 2-metric space. For the preceding one in metric spaces, see [[3], Theorem 2.1].

Theorem 2.3 Let ( X , , d ) be a complete, partially ordered 2-metric space and T : X X be a weak C-contraction such that:
  1. 1.

    T is continuous and non-decreasing.

     
  2. 2.

    There exists x 0 X with x 0 T x 0 .

     

Then T has a fixed point.

Proof If x 0 = T x 0 , then the proof is finished. Suppose now that x 0 T x 0 . Since T is a non-decreasing map, we have x 0 T x 0 T 2 x 0 T n x 0  . Put x n + 1 = T x n . Then, for all n 1 , from (2.1) and noting that x n 1 and x n are comparable, we get
d ( x n + 1 , x n , a ) = d ( T x n , T x n 1 , a ) 1 2 [ d ( x n , T x n 1 , a ) + d ( x n 1 , T x n , a ) ] ψ ( d ( x n , T x n 1 , a ) , d ( x n 1 , T x n , a ) ) = 1 2 [ d ( x n , x n , a ) + d ( x n 1 , x n + 1 , a ) ] ψ ( d ( x n , x n , a ) , d ( x n 1 , x n + 1 , a ) ) = 1 2 d ( x n 1 , x n + 1 , a ) ψ ( 0 , d ( x n 1 , x n + 1 , a ) ) 1 2 d ( x n 1 , x n + 1 , a )
(2.2)
for all a X . By choosing a = x n 1 in (2.2), we obtain d ( x n + 1 , x n , x n 1 ) 0 , that is,
d ( x n + 1 , x n , x n 1 ) = 0 .
(2.3)
It follows from (2.2) and (2.3) that
d ( x n + 1 , x n , a ) 1 2 d ( x n 1 , x n + 1 , a ) 1 2 [ d ( x n 1 , x n , a ) + d ( x n , x n + 1 , a ) + d ( x n 1 , x n , x n + 1 ) ] = 1 2 [ d ( x n 1 , x n , a ) + d ( x n , x n + 1 , a ) ] .
(2.4)
It implies that
d ( x n , x n + 1 , a ) d ( x n 1 , x n , a ) .
(2.5)
Thus { d ( x n , x n + 1 , a ) } is a decreasing sequence of non-negative real numbers and hence it is convergent. Let
lim n d ( x n , x n + 1 , a ) = r .
(2.6)
Taking the limit as n in (2.4) and using (2.6), we get
r lim n 1 2 d ( x n 1 , x n + 1 , a ) 1 2 ( r + r ) = r .
That is,
lim n d ( x n 1 , x n + 1 , a ) = 2 r .
(2.7)
Taking the limit as n in (2.2) and using (2.6), (2.7), we get r 1 2 2 r ψ ( 0 , 2 r ) 1 2 2 r = r . It implies that ψ ( 0 , 2 r ) = 0 , that is, r = 0 . Then (2.6) becomes
lim n d ( x n + 1 , x n , a ) = 0 .
(2.8)
From (2.5), we have if d ( x n 1 , x n , a ) = 0 , then d ( x n , x n + 1 , a ) = 0 . Since d ( x 0 , x 1 , x 0 ) = 0 , we have d ( x n , x n + 1 , x 0 ) = 0 for all n N . Since d ( x m 1 , x m , x m ) = 0 , we have
d ( x n , x n + 1 , x m ) = 0
(2.9)
for all n m 1 . For 0 n < m 1 , noting that m 1 n + 1 , from (2.9) we have
d ( x m 1 , x m , x n + 1 ) = d ( x m 1 , x m , x n ) = 0 .
It implies that
d ( x n , x n + 1 , x m ) d ( x n , x n + 1 , x m 1 ) + d ( x n + 1 , x m , x m 1 ) + d ( x m , x n , x m 1 ) = d ( x n , x n + 1 , x m 1 ) .
(2.10)
Since d ( x n , x n + 1 , x n + 1 ) = 0 , from (2.10) we have
d ( x n , x n + 1 , x m ) = 0
(2.11)

for all 0 n < m 1 . From (2.9) and (2.11), we have d ( x n , x n + 1 , x m ) = 0 for all n , m N .

Now, for all i , j , k N with i < j , we have d ( x j 1 , x j , x i ) = d ( x j 1 , x j , x k ) = 0 . Therefore,
d ( x i , x j , x k ) d ( x i , x j , x j 1 ) + d ( x j , x k , x j 1 ) + d ( x k , x i , x j 1 ) d ( x i , x j 1 , x k ) d ( x i , x i , x k ) = 0 .
This proves that for all i , j , k N
d ( x i , x j , x k ) = 0 .
(2.12)
In what follows, we will prove that { x n } is a Cauchy sequence. Suppose to the contrary that { x n } is not a Cauchy sequence. Then there exists ε > 0 for which we can find subsequences { x m ( k ) } and { x n ( k ) } where n ( k ) is the smallest integer such that n ( k ) > m ( k ) > k and
d ( x n ( k ) , x m ( k ) , a ) ε
(2.13)
for all k N . Therefore,
d ( x n ( k ) 1 , x m ( k ) , a ) < ε .
(2.14)
By using (2.12), (2.13) and (2.14), we have
ε d ( x n ( k ) , x m ( k ) , a ) d ( x n ( k ) , x n ( k ) 1 , a ) + d ( x n ( k ) 1 , x m ( k ) , a ) + d ( x n ( k ) , x m ( k ) , x n ( k ) 1 ) = d ( x n ( k ) , x n ( k ) 1 , a ) + d ( x n ( k ) 1 , x m ( k ) , a ) < d ( x n ( k ) , x n ( k ) 1 , a ) + ε .
(2.15)
Taking the limit as k in (2.15) and using (2.8), we have
lim k d ( x n ( k ) , x m ( k ) , a ) = lim k d ( x n ( k ) 1 , x m ( k ) , a ) = ε .
(2.16)
Also, from (2.12), we have
d ( x m ( k ) , x n ( k ) 1 , a ) d ( x m ( k ) , x m ( k ) 1 , a ) + d ( x m ( k ) 1 , x n ( k ) 1 , a ) + d ( x m ( k ) , x n ( k ) 1 , x m ( k ) 1 ) = d ( x m ( k ) , x m ( k ) 1 , a ) + d ( x m ( k ) 1 , x n ( k ) 1 , a ) d ( x m ( k ) , x m ( k ) 1 , a ) + d ( x m ( k ) 1 , x n ( k ) , a ) + d ( x n ( k ) 1 , x n ( k ) , a ) + d ( x m ( k ) 1 , x n ( k ) 1 , x n ( k ) ) = d ( x m ( k ) , x m ( k ) 1 , a ) + d ( x m ( k ) 1 , x n ( k ) , a ) + d ( x n ( k ) 1 , x n ( k ) , a )
(2.17)
and
d ( x m ( k ) 1 , x n ( k ) , a ) d ( x m ( k ) 1 , x m ( k ) , a ) + d ( x n ( k ) , x m ( k ) , a ) + d ( x m ( k ) 1 , x n ( k ) , x m ( k ) ) = d ( x m ( k ) 1 , x m ( k ) , a ) + d ( x n ( k ) , x m ( k ) , a ) .
(2.18)
Taking the limit as k in (2.17), (2.18) and using (2.8), (2.16), we obtain
lim k d ( x m ( k ) 1 , x n ( k ) , a ) = ε .
(2.19)
Since n ( k ) > m ( k ) and x n ( k ) 1 , x m ( k ) 1 are comparable, by using (2.1), we have
ε d ( x n ( k ) , x m ( k ) , a ) = d ( T x n ( k ) 1 , T x m ( k ) 1 , a ) 1 2 [ d ( x n ( k ) 1 , T x m ( k ) 1 , a ) + d ( x m ( k ) 1 , T x n ( k ) 1 , a ) ] ψ ( d ( x n ( k ) 1 , T x m ( k ) 1 , a ) , d ( x m ( k ) 1 , T x n ( k ) 1 , a ) ) = 1 2 [ d ( x n ( k ) 1 , x m ( k ) , a ) + d ( x m ( k ) 1 , x n ( k ) , a ) ] ψ ( d ( x n ( k ) 1 , x m ( k ) , a ) , d ( x m ( k ) 1 , x n ( k ) , a ) ) .
(2.20)
Taking the limit as k in (2.20) and using (2.16), (2.19) and the continuity of ψ, we have
ε 1 2 ( ε + ε ) ψ ( ε , ε ) = ε ψ ( ε , ε ) ε .

This proves that ψ ( ε , ε ) = 0 , that is, ε = 0 . It is a contradiction. This proves that { x n } is a Cauchy sequence. Since X is complete, there exists z X such that lim n x n = z . It follows from the continuity of T that z = lim n x n + 1 = lim n T x n = T z . Then z is a fixed point of T. □

The next result is another one for the existence of the fixed point of a weak C-contraction on a 2-metric space. For the preceding one in metric spaces, see [[3], Theorem 2.2].

Theorem 2.4 Let ( X , , d ) be a complete, partially ordered 2-metric space and T : X X be a weak C-contraction such that:
  1. 1.

    T is non-decreasing.

     
  2. 2.

    If { x n } is non-decreasing such that lim n x n = x , then x n x for all n N .

     
  3. 3.

    There exists x 0 X with x 0 T x 0 .

     

Then T has a fixed point.

Proof As in the proof of Theorem 2.3, we have a Cauchy sequence { x n } with lim n x n = z in X. We only have to prove that T z = z . Since { x n } is non-decreasing and lim n x n = z , we have x n z for all n N . It follows from (2.1) that
d ( x n + 1 , T z , a ) = d ( T x n , T z , a ) 1 2 [ d ( x n , T z , a ) + d ( z , T x n , a ) ] ψ ( d ( x n , T z , a ) , d ( z , T x n , a ) ) = 1 2 [ d ( x n , T z , a ) + d ( z , x n + 1 , a ) ] ψ ( d ( x n , T z , a ) , d ( z , x n + 1 , a ) ) .
(2.21)
Taking the limit as n in (2.21), we have
d ( z , T z , a ) 1 2 [ d ( z , T z , a ) + d ( z , z , a ) ] ψ ( d ( z , T z , a ) , d ( z , z , a ) ) 1 2 d ( z , T z , a ) ψ ( d ( z , T z , a ) , 0 ) 1 2 d ( z , T z , a ) .

It implies that d ( z , T z , a ) = 0 for all a X , that is, T z = z . □

In what follows, we prove a sufficient condition for the uniqueness of the fixed point in Theorem 2.3 and Theorem 2.4.

Theorem 2.5 Suppose that:
  1. 1.

    Either hypotheses of Theorem  2.3 or hypotheses of Theorem  2.4 hold.

     
  2. 2.

    For each x , y X , there exists z X that is comparable to x and y.

     

Then T has a unique fixed point.

Proof As in the proofs of Theorem 2.3 and Theorem 2.4, we see that T has a fixed point.

Now we prove the uniqueness of the fixed point of T. Let x, y be two fixed points of T. We consider the following two cases.

Case 1. x is comparable to y. Then T n x is comparable to T n y for all n N . For all a X , we have
d ( x , y , a ) = d ( T n x , T n y , a ) 1 2 [ d ( T n 1 x , T n y , a ) + d ( T n 1 y , T n x , a ) ] ψ ( d ( T n 1 x , T n y , a ) , d ( T n 1 y , T n x , a ) ) = 1 2 [ d ( x , y , a ) + d ( y , x , a ) ] ψ ( d ( x , y , a ) , d ( y , x , a ) ) d ( x , y , a ) .

This proves that ψ ( d ( x , y , a ) , d ( y , x , a ) ) = 0 , that is, d ( x , y , a ) = 0 for all a X . Then x = y .

Case 2. x is not comparable to y. Then there exists z X that is comparable to x and y. It implies that T n z is comparable to T n x = x and T n y = y . For all n N and a X , we have
d ( x , T n z , a ) = d ( T n x , T n z , a ) 1 2 [ d ( T n 1 x , T n z , a ) + d ( T n 1 z , T n x , a ) ] ψ ( d ( T n 1 x , T n z , a ) , d ( T n 1 z , T n x , a ) ) = 1 2 [ d ( x , T n z , a ) + d ( T n 1 z , x , a ) ] ψ ( d ( x , T n z , a ) , d ( T n 1 z , x , a ) ) 1 2 [ d ( x , T n z , a ) + d ( T n 1 z , x , a ) ] .
(2.22)

It implies that d ( x , T n z , a ) d ( x , T n 1 z , a ) . Then there exists lim n d ( x , T n z , a ) = l . Taking the limit as n in (2.22) and noting that ψ is continuous, we have l 1 2 ( l + l ) ψ ( l , l ) l . This proves that ψ ( l , l ) = 0 . Then l = 0 , that is, lim n T n z = x . Similarly, lim n T n z = y . By Lemma 1.7, we get x = y . □

Remark 2.6 Note that if ( X , ) is totally ordered, then the condition (2) in Theorem 2.5 is always satisfied.

The following result is an analogue of [[3], Theorem 3.1].

Theorem 2.7 Let ( X , , d ) be a complete, partially ordered 2-metric space and T : X X be a weak C-contraction such that:
  1. 1.

    For all x , y X , if x y then T x T y .

     
  2. 2.

    For each x , y X , there exists z X that is comparable to x and y.

     
  3. 3.

    There exists x 0 X with x 0 T x 0 or x 0 T x 0 .

     

Then, for all a X , inf { d ( x , T x , a ) : x X { a } } = 0 . In particular, inf { d ( x , T x , a ) : x X } = 0 .

Proof We consider the following two cases.

Case 1. x 0 T x 0 . By the hypothesis (1), consecutive terms of the sequence { T n x 0 } are comparable. It follows from (2.1) that for all a X ,
d ( T n + 1 x 0 , T n x 0 , a ) 1 2 [ d ( T n x 0 , T n x 0 , a ) + d ( T n 1 x 0 , T n + 1 x 0 , a ) ] ψ ( d ( T n x 0 , T n x 0 , a ) , d ( T n 1 x 0 , T n + 1 x 0 , a ) ) = 1 2 d ( T n 1 x 0 , T n + 1 x 0 , a ) ψ ( 0 , d ( T n 1 x 0 , T n + 1 x 0 , a ) ) 1 2 d ( T n 1 x 0 , T n + 1 x 0 , a ) 1 2 [ d ( T n 1 x 0 , T n x 0 , a ) + d ( T n x 0 , T n + 1 x 0 , a ) + d ( T n 1 x 0 , T n x 0 , T n + 1 x 0 ) ] .
(2.23)
As in the proof of (2.12) of Theorem 2.3, we have d ( x i , x j , x k ) = 0 for all i , j , k N . Then (2.23) implies
d ( T n + 1 x 0 , T n x 0 , a ) 1 2 [ d ( T n 1 x 0 , T n x 0 , a ) + d ( T n x 0 , T n + 1 x 0 , a ) ] .

That is, d ( T n + 1 x 0 , T n x 0 , a ) d ( T n x 0 , T n 1 x 0 , a ) . Then there exists lim n d ( T n + 1 x 0 , T n x 0 , a ) = r . As in the proof of Theorem 2.3, we get r = 0 . Then lim n d ( T n + 1 x 0 , T n x 0 , a ) = 0 . That is, inf { d ( x , T x , a ) : x X } = 0 .

Case 2. x 0 T x 0 . The same as in Case 1. □

For each a X , if d a ( x , y ) = d ( x , y , a ) for all x , y X is a metric on X, then the formula (2.1) becomes (1.1). Also, the above proofs may be similar to the method used in [2] and [3]. The following example guarantees that this fact is not true in general.

Example 2.8 There exists a 2-metric space ( X , d ) such that for each a X , the formula d a ( x , y ) = d ( x , y , a ) for all x , y X is not a metric on X.

Proof Let X = [ 0 , + ) and
d ( x , y , z ) = { x y + y z + z x if  x y z , 0 otherwise.
Then ( X , d ) is a 2-metric space. For each a X , we have
d a ( x , y ) = d ( x , y , a ) = { x y + y a + a x if  x y a , 0 otherwise.
If a = 0 , then we have
d 0 ( 1 , 6 ) = 6 > d 0 ( 1 , 1 6 ) + d 0 ( 1 6 , 6 ) = 1 6 + 1 = 7 6 .
If a 0 , then we have
d a ( 2 a , 3 a ) = 11 a 2 > d a ( 2 a , 0 ) + d a ( 0 , 3 a ) = 2 a 2 + 3 a 2 = 5 a 2 .

This proves that d a is not a metric on X for all a X . □

The following example shows that hypotheses in Theorem 2.3 and Theorem 2.4 do not guarantee the uniqueness of the fixed point.

Example 2.9 Let X = { ( 1 , 0 ) , ( 0 , 1 ) , ( 2 , 2 ) } R 2 with the order
( x , y ) ( z , t ) if and only if x z  and  y t .
Define a 2-metric d on X as follows:
d ( x , y , z ) = { 2 if  x y z , 0 otherwise.

Then ( X , , d ) is a partially ordered, complete 2-metric space whose different elements are not comparable. The identity map T ( x , y ) = ( x , y ) for all ( x , y ) X is continuous, non-decreasing, and contraction conditions in Theorem 2.3 and Theorem 2.4 are satisfied. Moreover, ( 1 , 0 ) T ( 1 , 0 ) and T has more than one fixed point.

The following example is an illustration of Theorem 2.4 and Theorem 2.7.

Example 2.10 Let X = { a , b , c } with the order x y if and only if x = y for all x , y X . Let d be a 2-metric on X defined by the symmetry of all three variables and
d ( x , y , z ) = { 1 if  x y z , 0 otherwise.

Let T : X X be defined by T a = b , T b = a , T c = c . It is easy to see that Theorem 2.4 and Theorem 2.7 are applicable to T and c is a unique fixed point of T. Moreover, the condition (2) in Theorem 2.5 does not hold, then it is not a necessary condition of the uniqueness of the fixed point.

Note that, in [[3], Theorem 3.1], if X is a compact metric space and T is continuous, then T has a unique fixed point. The following example shows that, in Theorem 2.7, if X is a compact 2-metric space and T is continuous, then T may not have a unique fixed point.

Example 2.11 Let ( X , d ) be a 2-metric space as in Example 2.10. Let T : X X be defined by
T a = a , T b = b , T c = c .

We see that all assumptions in Theorem 2.7 are satisfied but T has more than one fixed point.

Finally, Example 2.12 and Example 2.13 show that the above results can be used to prove the existence of a fixed point when standard arguments in metric spaces in [2] and [3] fail, even for trivial maps.

Example 2.12 Let X be the 2-metric space ( R , σ ) on [[4], p.145] with the usual order and T x = 0 for all x X . Then we have:
  1. 1.

    X is a complete, totally ordered 2-metric space.

     
  2. 2.

    X is not metrizable.

     
  3. 3.

    T is a C-weak contraction on the 2-metric space X.

     

Proof (1) and (2) See [[4], p.145].

(3) By choosing ψ ( a , b ) = a + b 2 for all a , b [ 0 , + ) , then the condition (2.1) holds. This proves that T is a C-weak contraction on the 2-metric space X. □

Example 2.13 Let X = { 0 , 1 , 2 , , n , } with the usual order,
d ( x , y , z ) = { 1 if  x y z  and there exists  n 1  with  { n , n + 1 } { x , y , z } , 0 otherwise
and T x = 0 for all x X . Then we have:
  1. 1.

    ( X , d ) is a complete, totally ordered 2-metric space.

     
  2. 2.

    ( X , d ) is not completely metrizable, that is, there does not exist any metric ρ on X such that the metric topology and the completeness on ( X , ρ ) are coincident with the 2-metric topology and the completeness on ( X , d ) , respectively.

     
  3. 3.

    T is a C-weak contraction on the 2-metric space ( X , d ) .

     

Proof (1) It is easy to see that ( X , d ) is a partially ordered 2-metric space.

Let { x n } be a Cauchy sequence in ( X , d ) . We have lim n , m d ( x n , x m , a ) = 0 for all a X . Then, for each a X , there exists n 0 ( a ) such that d ( x n , x m , a ) = 0 for all n , m n 0 ( a ) . We consider the following three cases.

Case 1. For all n , m n 0 ( a ) , x n = x m . This proves that { x n } is convergent.

Case 2. For all n n 0 ( a ) , x n = a . It is a contradiction because a is an arbitrary point of X.

Case 3. For all n , m n 0 ( a ) and all k 1 , x n x m a and { k , k + 1 } { x n , x m , a } . Then d ( x n , 0 , a ) = 0 for all n n 0 ( a ) . This proves that { x n } is convergent.

By the above three cases, the Cauchy sequence { x n } is convergent in ( X , d ) . This proves that ( X , d ) is complete.

(2) Since lim n d ( n , 0 , a ) = 0 , we have { n } is a convergent sequence in ( X , d ) . On the other hand, lim n d ( n , n + 1 , a ) = 1 , then { n } is not a Cauchy sequence in ( X , d ) . This proves that ( X , d ) is not completely metrizable.

(3) By choosing ψ ( a , b ) = a + b 2 , the condition (2.1) holds. This proves that T is a C-weak contraction on the 2-metric space ( X , d ) . □

Declarations

Acknowledgements

The authors would like to thank the referees for their valuable comments.

Authors’ Affiliations

(1)
Department of Mathematics, Dong Thap University
(2)
Journal of Science, Dong Thap University

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© Dung and Le Hang; licensee Springer. 2013

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