Open Access

Fixed point theory of cyclical generalized contractive conditions in partial metric spaces

Fixed Point Theory and Applications20132013:17

https://doi.org/10.1186/1687-1812-2013-17

Received: 6 November 2012

Accepted: 13 January 2013

Published: 28 January 2013

Abstract

The purpose of this paper is to study fixed point theorems for a mapping satisfying the cyclical generalized contractive conditions in complete partial metric spaces. Our results generalize or improve many recent fixed point theorems in the literature.

MSC:47H10, 54C60, 54H25, 55M20.

Keywords

fixed point cyclic CW -contraction cyclic MK -contraction partial metric space

1 Introduction and preliminaries

Throughout this paper, by R + , we denote the set of all nonnegative real numbers, while is the set of all natural numbers. Let ( X , d ) be a metric space, D be a subset of X and f : D X be a map. We say f is contractive if there exists α [ 0 , 1 ) such that for all x , y D ,
d ( f x , f y ) α d ( x , y ) .
The well-known Banach fixed point theorem asserts that if D = X , f is contractive and ( X , d ) is complete, then f has a unique fixed point in X. It is well known that the Banach contraction principle [1] is a very useful and classical tool in nonlinear analysis. Also, this principle has many generalizations. For instance, in 1969, Boyd and Wong [2] introduced the notion of Φ-contraction. A mapping f : X X on a metric space is called Φ-contraction if there exists an upper semi-continuous function Φ : [ 0 , ) [ 0 , ) such that
d ( f x , f y ) Φ ( d ( x , y ) ) for all  x , y X .

In 1994, Mattews [3] introduced the following notion of partial metric spaces.

Definition 1 [3] A partial metric on a nonempty set X is a function p : X × X R + such that for all x , y , z X ,

(p1) x = y if and only if p ( x , x ) = p ( x , y ) = p ( y , y ) ;

(p2) p ( x , x ) p ( x , y ) ;

(p3) p ( x , y ) = p ( y , x ) ;

(p4) p ( x , y ) p ( x , z ) + p ( z , y ) p ( z , z ) .

A partial metric space is a pair ( X , p ) such that X is a nonempty set and p is a partial metric on X.

Remark 1 It is clear that if p ( x , y ) = 0 , then from (p1) and (p2), x = y . But if x = y , p ( x , y ) may not be 0.

Each partial metric p on X generates a T 0 topology τ p on X which has as a base the family of open p-balls { B p ( x , γ ) : x X , γ > 0 } , where B p ( x , γ ) = { y X : p ( x , y ) < p ( x , x ) + γ } for all x X and γ > 0 . If p is a partial metric on X, then the function d p : X × X R + given by
d p ( x , y ) = 2 p ( x , y ) p ( x , x ) p ( y , y )

is a metric on X.

We recall some definitions of a partial metric space as follows.

Definition 2 [3]

Let ( X , p ) be a partial metric space. Then
  1. (1)

    a sequence { x n } in a partial metric space ( X , p ) converges to x X if and only if p ( x , x ) = lim n p ( x , x n ) ;

     
  2. (2)

    a sequence { x n } in a partial metric space ( X , p ) is called a Cauchy sequence if and only if lim m , n p ( x m , x n ) exists (and is finite);

     
  3. (3)

    a partial metric space ( X , p ) is said to be complete if every Cauchy sequence { x n } in X converges, with respect to τ p , to a point x X such that p ( x , x ) = lim m , n p ( x m , x n ) ;

     
  4. (4)

    a subset A of a partial metric space ( X , p ) is closed if whenever { x n } is a sequence in A such that { x n } converges to some x X , then x A .

     

Remark 2 The limit in a partial metric space is not unique.

Lemma 1 [3, 4]

  1. (a)

    { x n } is a Cauchy sequence in a partial metric space ( X , p ) if and only if it is a Cauchy sequence in the metric space ( x , d p ) ;

     
  2. (b)

    a partial metric space ( X , p ) is complete if and only if the metric space ( X , d p ) is complete. Furthermore, lim n d p ( x n , x ) = 0 if and only if p ( x , x ) = lim n p ( x n , x ) = lim n p ( x n , x m ) .

     

In 2003, Kirk, Srinivasan and Veeramani [5] introduced the following notion of the cyclic representation.

Definition 3 [5]

Let X be a nonempty set, m N and f : X X be an operator. Then X = i = 1 m A i is called a cyclic representation of X with respect to f if
  1. (1)

    A i , i = 1 , 2 , , m are nonempty subsets of X;

     
  2. (2)

    f ( A 1 ) A 2 , f ( A 2 ) A 3 , , f ( A m 1 ) A m , f ( A m ) A 1 .

     

Kirk, Srinivasan and Veeramani [5] also proved the following theorem.

Theorem 1 [5]

Let ( X , d ) be a complete metric space, m N , A 1 , A 2 , , A m , be closed nonempty subsets of X and X = i = 1 m A i . Suppose that f satisfies the following condition:
d ( f x , f y ) ψ ( d ( x , y ) ) , for all x A i , y A i + 1 , i { 1 , 2 , , m } ,

where ψ : [ 0 , ) [ 0 , ) is upper semi-continuous from the right and 0 ψ ( t ) < t for t > 0 . Then f has a fixed point z i = 1 n A i .

Recently, the fixed theorems for an operator f : X X defined on a metric space X with a cyclic representation of X with respect to f have appeared in the literature (see, e.g., [68]). In 2010, Pǎcurar and Rus [7] introduced the following notion of a cyclic weaker φ-contraction.

Definition 4 [7]

Let ( X , d ) be a metric space, m N , A 1 , A 2 , , A m be closed nonempty subsets of X and X = i = 1 m A i . An operator f : X X is called a cyclic weaker φ-contraction if
  1. (1)

    X = i = 1 m A i is a cyclic representation of X with respect to f;

     
  2. (2)
    there exists a continuous, non-decreasing function φ : [ 0 , ) [ 0 , ) with φ ( t ) > 0 for t ( 0 , ) and φ ( 0 ) = 0 such that
    d ( f x , f y ) d ( x , y ) φ ( d ( x , y ) )
     

for any x A i , y A i + 1 , i = 1 , 2 , , m , where A m + 1 = A 1 .

And Pǎcurar and Rus [7] proved the following main theorem.

Theorem 2 [7]

Let ( X , d ) be a complete metric space, m N , A 1 , A 2 , , A m be closed nonempty subsets of X and X = i = 1 m A i . Suppose that f is a cyclic weaker φ-contraction. Then f has a fixed point z i = 1 n A i .

In the recent years, fixed point theory has developed rapidly on cyclic contraction mappings, see [915].

The purpose of this paper is to study fixed point theorems for a mapping satisfying the cyclical generalized contractive conditions in complete partial metric spaces. Our results generalize or improve many recent fixed point theorems in the literature.

2 Fixed point theorems (I)

In the section, we denote by Ψ the class of functions ψ : R + 3 R + satisfying the following conditions:

( ψ 1 ) ψ is an increasing and continuous function in each coordinate;

( ψ 2 ) for t R + , ψ ( t , t , t ) t , ψ ( t , 0 , 0 ) t and ψ ( 0 , 0 , t ) t .

Next, we denote by Θ the class of functions φ : R + R + satisfying the following conditions:

( φ 1 ) φ is continuous and non-decreasing;

( φ 2 ) for t > 0 , φ ( t ) > 0 and φ ( 0 ) = 0 .

And we denote by Φ the class of functions ϕ : R + R + satisfying the following conditions:

( ϕ 1 ) ϕ is continuous;

( ϕ 2 ) for t > 0 , ϕ ( t ) > 0 and ϕ ( 0 ) = 0 .

We now state a new notion of cyclic CW -contractions in partial metric spaces as follows.

Definition 5 Let ( X , p ) be a partial metric space, m N , A 1 , A 2 , , A m be nonempty subsets of X and Y = i = 1 m A i . An operator f : Y Y is called a cyclic CW -contraction if
  1. (1)

    i = 1 m A i is a cyclic representation of Y with respect to f;

     
  2. (2)
    for any x A i , y A i + 1 , i = 1 , 2 , , m ,
    φ ( p ( f x , f y ) ) ψ ( φ ( p ( x , y ) ) , φ ( p ( x , f x ) ) , φ ( p ( y , f y ) ) ) ϕ ( M ( x , y ) ) ,
    (2.1)
     

where ψ Ψ , φ Θ , ϕ Φ , and M ( x , y ) = max { p ( x , y ) , p ( x , f x ) , p ( y , f y ) } .

Theorem 3 Let ( X , p ) be a complete partial metric space, m N , A 1 , A 2 , , A m be nonempty closed subsets of X and Y = i = 1 m A i . Let f : Y Y be a cyclic CW -contraction. Then f has a unique fixed point z i = 1 m A i .

Proof Given x 0 and let x n + 1 = f x n = f n x 0 for n = 0 , 1 , 2 ,  . If there exists n 0 N such that x n 0 + 1 = x n 0 , then we finished the proof. Suppose that x n + 1 x n for any n = 0 , 1 , 2 ,  . Notice that for any n 0 , there exists i n { 1 , 2 , , m } such that x n A i n and x n + 1 A i n + 1 .

Step 1. We will prove that
lim n p ( x n , x n + 1 ) = 0 , that is , lim n d p ( x n , x n + 1 ) = 0 .
Using (2.1), we have
φ ( p ( x n , x n + 1 ) ) = φ ( p ( f x n 1 , f x n ) ) ψ ( φ ( p ( x n 1 , x n ) ) , φ ( p ( x n 1 , f x n 1 ) ) , φ ( p ( x n , f x n ) ) ) ϕ ( M ( x n 1 , x n ) ) = ψ ( φ ( p ( x n 1 , x n ) ) , φ ( p ( x n 1 , x n ) ) , φ ( p ( x n , x n + 1 ) ) ) ϕ ( M ( x n 1 , x n ) ) ,
where
M ( x n 1 , x n ) = max { p ( x n 1 , x n ) , p ( x n 1 , f x n 1 ) , p ( x n , f x n ) } = max { p ( x n 1 , x n ) , p ( x n 1 , x n ) , p ( x n , x n + 1 ) } .
If M ( x n 1 , x n ) = p ( x n , x n + 1 ) , then
φ ( p ( x n , x n + 1 ) ) ψ ( φ ( p ( x n , x n + 1 ) ) , φ ( p ( x n , x n + 1 ) ) , φ ( p ( x n , x n + 1 ) ) ) ϕ ( p ( x n , x n + 1 ) ) φ ( p ( x n , x n + 1 ) ) ϕ ( p ( x n , x n + 1 ) ) ,

which implies that ϕ ( p ( x n , x n + 1 ) ) = 0 , and hence p ( x n , x n + 1 ) = 0 . This contradicts our initial assumption.

From the above argument, we have that for each n N ,
φ ( p ( x n , x n + 1 ) ) φ ( p ( x n 1 , x n ) ) ϕ ( p ( x n 1 , x n ) ) ,
(2.2)
and
p ( x n , x n + 1 ) < p ( x n 1 , x n ) .
And since the sequence { p ( x n , x n + 1 ) } is decreasing, it must converge to some η 0 . Taking limit as n in (2.2) and by the continuity of φ and ϕ, we get
φ ( η ) φ ( η ) ϕ ( η ) ,
and so we conclude that ϕ ( η ) = 0 and η = 0 . Thus, we have
lim n p ( x n , x n + 1 ) = 0 .
(2.3)
By (p2), we also have
lim n p ( x n , x n ) = 0 .
(2.4)
Since d p ( x , y ) 2 p ( x , y ) p ( x , x ) p ( y , y ) for all x , y X , using (2.3) and (2.4), we obtain that
lim n d p ( x n , x n + 1 ) = 0 .
(2.5)

Step 2. We show that { x n } is a Cauchy sequence in the metric space ( Y , d p ) . We claim that the following result holds.

Claim For every ε > 0 , there exists n N such that if r , q n with r q = 1 mod m , then d p ( x r , x q ) < ε .

Suppose the above statement is false. Then there exists ϵ > 0 such that for any n N , there are r n , q n N with r n > q n n with r n q n = 1 mod m satisfying
d p ( x q n , x r n ) ϵ .
Now, we let n > 2 m . Then corresponding to q n n use, we can choose r n in such a way it is the smallest integer with r n > q n n satisfying r n q n = 1 mod m and d p ( x q n , x r n ) ϵ . Therefore, d p ( x q n , x r n m ) ϵ and
ϵ d p ( x q n , x r n ) d p ( x q n , x r n m ) + i = 1 m d p ( x r n i , x r n i + 1 ) < ϵ + i = 1 m d p ( x r n i , x r n i + 1 ) .
Letting n , we obtain that
lim n d p ( x q n , x r n ) = ϵ .
(2.6)
On the other hand, we can conclude that
ϵ d p ( x q n , x r n ) d p ( x q n , x q n + 1 ) + d p ( x q n + 1 , x r n + 1 ) + d p ( x r n + 1 , x r n ) d p ( x q n , x q n + 1 ) + d p ( x q n + 1 , x q n ) + d p ( x q n , x r n ) + d p ( x r n , x r n + 1 ) + d p ( x r n + 1 , x r n ) .
Letting n , we obtain that
lim n d p ( x q n + 1 , x r n + 1 ) = ϵ .
(2.7)
Since d p ( x , y ) = 2 p ( x , y ) p ( x , x ) p ( y , y ) and using (2.4), (2.6) and (2.7), we have that
lim n p ( x q n , x r n ) = ϵ 2 ,
(2.8)
and
lim n p ( x q n + 1 , x r n + 1 ) = ϵ 2 .
(2.9)
Since x q n and x r n lie in different adjacently labeled sets A i and A i + 1 for certain 1 i m , by using the fact that f is a cyclic CW -contraction, we have
φ ( p ( f x q n + 1 , f x r n + 1 ) ) = φ ( p ( f x q n , f x r n ) ) ψ ( φ ( p ( x q n , x r n ) ) , φ ( p ( x q n , f x q n ) ) , φ ( p ( x r n , f x r n ) ) ) ϕ ( M ( x q n , x r n ) ) = ψ ( φ ( p ( x q n , x r n ) ) , φ ( p ( x q n , x q n + 1 ) ) , φ ( p ( x r n , x r n + 1 ) ) ) ϕ ( M ( x q n , x r n ) ) ,
where
M ( x q n , x r n ) = max { p ( x q n , x r n ) , p ( x q n , x q n + 1 ) , p ( x r n , x r n + 1 ) } .
Thus, letting n , we can conclude that
φ ( ϵ 2 ) ψ ( φ ( ϵ 2 ) , φ ( 0 ) , φ ( 0 ) ) ϕ ( ϵ 2 ) φ ( ϵ 2 ) ϕ ( ϵ 2 ) ,

which implies ϕ ( ϵ 2 ) = 0 , that is, ϵ = 0 . So, we get a contradiction. Therefore, our claim is proved.

In the sequel, we will show that { x n } is a Cauchy sequence in the metric space ( Y , d p ) . Let ε > 0 be given. By our claim, there exists n 1 N such that if r , q n 1 with r q = 1 mod m , then
d p ( x r , x q ) ε 2 .
Since lim n d p ( x n , x n + 1 ) = 0 , there exists n 2 N such that
d p ( x n , x n + 1 ) ε 2 m

for any n n 2 .

Let r , q max { n 1 , n 2 } and r > q . Then there exists k { 1 , 2 , , m } such that r q = k mod m . Therefore, r q + j = 1 mod m for j = m k + 1 , and so we have
d p ( x q , x r ) d p ( x q , x r + j ) + d p ( x r + j , x r + j 1 ) + + d p ( x r 1 , x r ) ε 2 + j × ε 2 m ε 2 + m × ε 2 m = ε .

Thus, { x n } is a Cauchy sequence in the metric space ( Y , d p ) .

Step 3. We show that f has a fixed point ν in i = 1 m A i .

Since Y is closed, the subspace ( Y , p ) is complete. Then from Lemma 1, we have that ( Y , d p ) is complete. Thus, there exists ν X such that
lim n d p ( x n , ν ) = 0 .
And it follows from Lemma 1 that we have
p ( ν , ν ) = lim n p ( x n , ν ) = lim n , m p ( x n , x m ) .
(2.10)
On the other hand, since the sequence { x n } is a Cauchy sequence in the metric space ( Y , d p ) , we also have
lim n d p ( x n , x m ) = 0 .
Since d p ( x , y ) = 2 p ( x , y ) p ( x , x ) p ( y , y ) , we can deduce that
lim n p ( x n , x m ) = 0 .
(2.11)
Since Y = i = 1 m A i is a cyclic representation of X with respect to f, the sequence { x n } has infinite terms in each A i for i { 1 , 2 , , m } . Now, for all i = 1 , 2 , , m , we may take a subsequence { x n k } of { x n } with x n k A i 1 and also all converge to ν. Using (2.10) and (2.11), we have
p ( ν , ν ) = lim n p ( x n , ν ) = lim n p ( x n k , ν ) = 0 .
By (2.1),
φ ( p ( x n k + 1 , f ν ) ) = φ ( p ( f x n k , f ν ) ) ψ ( φ ( p ( x n k , ν ) ) , φ ( p ( x n k , f x n k ) ) , φ ( p ( ν , f ν ) ) ) ϕ ( M ( x n k , ν ) ) = ψ ( φ ( p ( x n k , ν ) ) , φ ( p ( x n k , x n k + 1 ) ) , φ ( p ( ν , f ν ) ) ) ϕ ( M ( x n k , ν ) ) ,
where
M ( x n k , ν ) = max { p ( x n k , ν ) , p ( x n k , x n k + 1 ) , p ( ν , f ν ) } .
Letting k , we have
φ ( p ( ν , f ν ) ) ψ ( φ ( 0 ) , φ ( 0 ) , φ ( p ( ν , f ν ) ) ) ϕ ( p ( ν , f ν ) ) φ ( p ( ν , f ν ) ) ϕ ( p ( ν , f ν ) ) ,

which implies ϕ ( p ( ν , f ν ) ) = 0 , that is, p ( ν , f ν ) = 0 . So, ν = f ν .

Step 4. Finally, to prove the uniqueness of the fixed point, suppose that μ, ν are fixed points of f. Then using the inequality (2.1), we obtain that
φ ( p ( μ , ν ) ) = φ ( p ( f μ , f ν ) ) ψ ( φ ( p ( μ , ν ) ) , φ ( p ( μ , f μ ) ) , φ ( p ( ν , f ν ) ) ) ϕ ( M ( μ , ν ) ) ,
where
M ( μ , ν ) = max { p ( μ , ν ) , p ( μ , f μ ) , p ( ν , f ν ) } = p ( μ , ν ) .
So, we also deduce that
φ ( p ( μ , ν ) ) ψ ( φ ( p ( μ , ν ) , 0 , 0 ) ) φ ( p ( μ , ν ) ) ϕ ( p ( μ , ν ) ) ,

which implies that ϕ ( p ( μ , ν ) ) = 0 , and hence p ( μ , ν ) = 0 , that is, μ = ν . So, we complete the proof.  □

The following provides an example for Theorem 3.

Example 1 Let X = [ 0 , 1 ] and A = [ 0 , 1 ] , B = [ 0 , 1 2 ] , C = [ 0 , 1 4 ] . We define the partial metric p on X by
p ( x , y ) = max { x , y } for all  x , y X ,
and define the function f : X X by
f ( x ) = x 2 1 + x for all  x X .
Now, we let φ , ϕ : R + R + and ψ : R + 3 R + be
φ ( t ) = 2 t , ϕ ( t ) = 2 t 5 ( 1 + t ) and ψ ( t ) = 4 5 max { t 1 , t 2 , t 3 } .

Then f is a cyclic CW -contraction and 0 is the unique fixed point.

Proof We claim that f is a cyclic CW -contraction.
  1. (1)

    Note that f ( A ) = [ 0 , 1 2 ] B , f ( B ) = [ 0 , 1 6 ] C and f ( C ) = [ 0 , 1 20 ] A . Thus, A B C is a cyclic representation of X with respect to f;

     
  2. (2)
    For x A and y B (or, x B and y C ), without loss of generality, we may assume that x y , then we have
     
and
Since
2 x 2 1 + x 8 x 5 2 x 5 ( 1 + x ) ,
we have
φ ( p ( f x , f y ) ) ψ ( φ ( p ( x , y ) ) , φ ( p ( x , f x ) ) , φ ( p ( y , f y ) ) ) ϕ ( max { p ( x , y ) , p ( x , f x ) , p ( y , f y ) } ) .

On the other hand, for x C and y A , without loss of generality, we may assume that x y , then it is easy to get the above inequality.

Note that Example 1 satisfies all of the hypotheses of Theorem 3, and we get that 0 is the unique fixed point. □

3 Fixed point theorems (II)

In this article, we also recall the notion of a Meir-Keeler function (see [16]). A function ϕ : [ 0 , ) [ 0 , ) is said to be a Meir-Keeler function if for each η > 0 , there exists δ > 0 such that for t [ 0 , ) with η t < η + δ , we have ϕ ( t ) < η . We now introduce a new notion of a weaker Meir-Keeler function ϕ : [ 0 , ) [ 0 , ) in a partial metric space ( X , p ) as follows.

Definition 6 Let ( X , p ) be a partial metric space. We call ϕ : [ 0 , ) [ 0 , ) a weaker Meir-Keeler function in X if for each η > 0 , there exists δ > 0 such that for x , y X with η p ( x , y ) < η + δ , there exists n 0 N such that ϕ n 0 ( p ( x , y ) ) < η .

In the section, we denote by Φ the class of weaker Meir-Keeler functions ϕ : R + R + in a partial metric space in ( X , p ) satisfying the following conditions:

( ϕ 1 ) ϕ ( t ) > 0 for t > 0 , ϕ ( 0 ) = 0 ;

( ϕ 2 ) { ϕ n ( t ) } n N is decreasing;

( ϕ 3 ) for t n [ 0 , ) ,
  1. (a)

    if lim n t n = γ > 0 , then lim n ϕ ( t n ) < γ and

     
  2. (b)

    if lim n t n = 0 , then lim n ϕ ( t n ) = 0 .

     

And we denote by the class Ψ of functions ψ : R + R + a continuous function satisfying ψ ( t ) > 0 for t > 0 , ψ ( 0 ) = 0 .

First, we state a new notion of cyclic MK -contractions in partial metric spaces as follows.

Definition 7 Let ( X , p ) be a partial metric space, m N , A 1 , A 2 , , A m be nonempty subsets of X and Y = i = 1 m A i . An operator f : Y Y is called a cyclic MK -contraction if
  1. (1)

    i = 1 m A i is a cyclic representation of Y with respect to f;

     
  2. (2)
    for any x A i , y A i + 1 , i = 1 , 2 , , m ,
    p ( f x , f y ) ϕ ( p ( x , y ) ) ψ ( p ( x , y ) ) ,
    (3.1)
     

where A m + 1 = A 1 , ϕ Φ and ψ Ψ .

Theorem 4 Let ( X , p ) be a complete partial metric space, m N , A 1 , A 2 , , A m be nonempty closed subsets of X and Y = i = 1 m A i . Let f : Y Y be a cyclic MK -contraction. Then f has a unique fixed point z i = 1 m A i .

Proof Given x 0 and let x n + 1 = f x n = f n x 0 , for n = 0 , 1 , 2 ,  . If there exists n 0 N such that x n 0 + 1 = x n 0 , then we finished the proof. Suppose that x n + 1 x n for any n = 0 , 1 , 2 ,  . Notice that for any n 0 , there exists i n { 1 , 2 , , m } such that x n A i n and x n + 1 A i n + 1 . Then by (3.1), we have
p ( x n , x n + 1 ) = p ( f x n 1 , f x n ) ϕ ( p ( x n 1 , x n ) ) ψ ( p ( x n 1 , x n ) ) .
Step 1. We will prove that
lim n p ( x n , x n + 1 ) = 0 , that is , lim n d p ( x n , x n + 1 ) = 0 .
Since f is a cyclic MK -contraction, we can conclude that
p ( x n , x n + 1 ) ϕ ( p ( x n 1 , x n ) ) ϕ ( ϕ ( p ( x n 2 , x n 1 ) ) = ϕ 2 ( p ( x n 2 , x n 1 ) ) ϕ n ( p ( x 0 , x 1 ) ) .
Since { ϕ n ( p ( x 0 , x 1 ) ) } n N is decreasing, it must converge to some η 0 . We claim that η = 0 . On the contrary, assume that 0 < η . Then by the definition of a weaker Meir-Keeler function ϕ, there exists δ > 0 such that for x 0 , x 1 X with η p ( x 0 , x 1 ) < δ + η , there exists n 0 N such that ϕ n 0 ( p ( x 0 , x 1 ) ) < η . Since lim n ϕ n ( p ( x 0 , x 1 ) ) = η , there exists k 0 N such that η ϕ k ( p ( x 0 , x 1 ) ) < δ + η , for all k k 0 . Thus, we conclude that ϕ k 0 + n 0 ( p ( x 0 , x 1 ) ) < η . So, we get a contradiction. Therefore, lim n ϕ n ( p ( x 0 , x 1 ) ) = 0 , and so we have
lim n p ( x n , x n + 1 ) = 0 .
(3.2)
By (p2), we also have
lim n p ( x n , x n ) = 0 .
(3.3)
Since d p ( x , y ) 2 p ( x , y ) p ( x , x ) p ( y , y ) for all x , y X , using (3.2) and (3.3), we obtain that
lim n d p ( x n , x n + 1 ) = 0 .
(3.4)

Step 2. We show that { x n } is a Cauchy sequence in the metric space ( Y , d p ) . We claim that the following result holds.

Claim For every ε > 0 , there exists n N such that if r , q n with r q = 1 mod m , then d p ( x r , x q ) < ε .

Suppose the above statement is false. Then there exists ϵ > 0 such that for any n N , there are r n , q n N with r n > q n n with r n q n = 1 mod m satisfying
d p ( x q n , x r n ) ϵ .
Now, we let n > 2 m . Then corresponding to q n n use, we can choose r n in such a way it is the smallest integer with r n > q n n satisfying r n q n = 1 mod m and d p ( x q n , x r n ) ϵ . Therefore, d p ( x q n , x r n m ) ϵ and
ϵ d p ( x q n , x r n ) d p ( x q n , x r n m ) + i = 1 m d p ( x r n i , x r n i + 1 ) < ϵ + i = 1 m d p ( x r n i , x r n i + 1 ) .
Letting n , we obtain that
lim n d p ( x q n , x r n ) = ϵ .
(3.5)
On the other hand, we can conclude that
ϵ d p ( x q n , x r n ) d p ( x q n , x q n + 1 ) + d p ( x q n + 1 , x r n + 1 ) + d p ( x r n + 1 , x r n ) d p ( x q n , x q n + 1 ) + d p ( x q n + 1 , x q n ) + d p ( x q n , x r n ) + d p ( x r n , x r n + 1 ) + d p ( x r n + 1 , x r n ) .
Letting n , we obtain that
lim n d p ( x q n + 1 , x r n + 1 ) = ϵ .
(3.6)
Since d p ( x , y ) = 2 p ( x , y ) p ( x , x ) p ( y , y ) and using (3.5) and (3.6), we have that
lim n p ( x q n , x r n ) = ϵ 2 ,
(3.7)
and
lim n p ( x q n + 1 , x r n + 1 ) = ϵ 2 .
(3.8)
Since x q n and x r n lie in different adjacently labeled sets A i and A i + 1 for certain 1 i m , by using the fact that f is a cyclic MK -contraction, we have
p ( x q n + 1 , x r n + 1 ) = p ( f x q n , f x r n ) ϕ ( p ( x q n , x r n ) ) ψ ( p ( x q n , x r n ) ) .
Letting n , by using the condition ϕ 3 of the function ϕ, we obtain that
ϵ 2 ϵ 2 ψ ( ϵ 2 ) ,

and consequently, ψ ( ϵ 2 ) = 0 . By the definition of a function ψ, we get ϵ = 0 which is a contraction. Therefore, our claim is proved.

In the sequel, we will show that { x n } is a Cauchy sequence in the metric space ( Y , d p ) . Let ε > 0 be given. By our claim, there exists n 1 N such that if r , q n 1 with r q = 1 mod m , then
d p ( x r , x q ) ε 2 .
Since lim n d p ( x n , x n + 1 ) = 0 , there exists n 2 N such that
d p ( x n , x n + 1 ) ε 2 m

for any n n 2 .

Let r , q max { n 1 , n 2 } and r > q . Then there exists k { 1 , 2 , , m } such that r q = k mod m . Therefore, r q + j = 1 mod m for j = m k + 1 , and so we have
d p ( x q , x r ) d p ( x q , x r + j ) + d p ( x r + j , x r + j 1 ) + + d p ( x r 1 , x r ) ε 2 + j × ε 2 m ε 2 + m × ε 2 m = ε .

Thus, { x n } is a Cauchy sequence in the metric space ( Y , d p ) .

Step 3. We show that f has a fixed point ν in i = 1 m A i .

Since Y is closed, the subspace ( Y , p ) is complete. Then from Lemma 1, we have that ( Y , d p ) is complete. Thus, there exists ν X such that
lim n d p ( x n , ν ) = 0 .
And it follows from Lemma 1 that we have
p ( ν , ν ) = lim n p ( x n , ν ) = lim n , m p ( x n , x m ) .
(3.9)
On the other hand, since the sequence { x n } is a Cauchy sequence in the metric space ( Y , d p ) , we also have
lim n d p ( x n , x m ) = 0 .
Since d p ( x , y ) = 2 p ( x , y ) p ( x , x ) p ( y , y ) , we can deduce that
lim n p ( x n , x m ) = 0 .
(3.10)
Since Y = i = 1 m A i is a cyclic representation of X with respect to f, the sequence { x n } has infinite terms in each A i for i { 1 , 2 , , m } . Now, for all i = 1 , 2 , , m , we may take a subsequence { x n k } of { x n } with x n k A i 1 and also all converge to ν. Using (3.9) and (3.10), we have
p ( ν , ν ) = lim n p ( x n , ν ) = lim n p ( x n k , ν ) = 0 .
By (3.1),
p ( x n k + 1 , f ν ) = p ( f x n k , f ν ) ϕ ( p ( x n k , ν ) ) ψ ( p ( x n k , ν ) ) ϕ ( p ( x n k , ν ) ) .
Letting k , we have
p ( ν , f ν ) 0 ,

and so ν = f ν .

Step 4. Finally, to prove the uniqueness of the fixed point, let μ be another fixed point of f in i = 1 m A i . By the cyclic character of f, we have μ , ν i = 1 n A i . Since f is a cyclic weaker MK -contraction, we have
p ( ν , μ ) = p ( ν , f μ ) = lim n p ( x n k + 1 , f μ ) = lim n p ( f x n k , f μ ) lim n [ ϕ ( p ( x n k , μ ) ) ψ ( p ( x n k , μ ) ) ] p ( ν , μ ) ψ ( p ( ν , μ ) ) ,
and we can conclude that
ψ ( p ( ν , μ ) ) = 0 ,

which implies p ( ν , μ ) = 0 . So, we have μ = ν . We complete the proof.  □

The following provides an example for Theorem 4.

Example 2 Let X = [ 0 , 1 ] and A = [ 0 , 1 ] , B = [ 0 , 1 2 ] , C = [ 0 , 1 4 ] . We define the partial metric p on X by
p ( x , y ) = max { x , y } for all  x , y X ,
and define the function f : X X by
f ( x ) = x 2 1 + x for all  x X .
Now, we let ψ , ϕ : R + R + be
ϕ ( t ) = 4 t 5 and ψ ( t ) = t 5 ( 1 + t ) .

Then f is a cyclic MK -contraction and 0 is the unique fixed point.

By Theorem 4, it is easy to get the following corollary.

Corollary 1 Let ( X , p ) be a complete partial metric space, m N , A 1 , A 2 , , A m be nonempty closed subsets of X, Y = i = 1 m A i and let f : Y Y . Assume that
  1. (1)

    i = 1 m A i is a cyclic representation of Y with respect to f;

     
  2. (2)
    for any x A i , y A i + 1 , i = 1 , 2 , , m ,
    p ( f x , f y ) ϕ ( p ( x , y ) ) ,
     

where A m + 1 = A 1 and ϕ Φ .

Then f has a unique fixed point z i = 1 m A i .

Declarations

Acknowledgements

The authors would like to thank referee(s) for many useful comments and suggestions for the improvement of the paper.

Authors’ Affiliations

(1)
Department of Applied Mathematics, National Hsinchu University of Education

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© Chen; licensee Springer. 2013

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