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

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.

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:DX be a map. We say f is contractive if there exists α[0,1) such that for all x,yD,

d(fx,fy)αd(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:XX on a metric space is called Φ-contraction if there exists an upper semi-continuous function Φ:[0,)[0,) such that

d(fx,fy)Φ ( d ( x , y ) ) for all x,yX.

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,zX,

(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):xX,γ>0}, where B p (x,γ)={yX:p(x,y)<p(x,x)+γ} for all xX 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)=2p(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 xX 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 xX 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 xX, then xA.

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, mN and f:XX 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, mN, 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(fx,fy)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 0ψ(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:XX 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, mN, A 1 , A 2 ,, A m be closed nonempty subsets of X and X= i = 1 m A i . An operator f:XX 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(fx,fy)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, mN, 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, mN, A 1 , A 2 ,, A m be nonempty subsets of X and Y= i = 1 m A i . An operator f:YY 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 ψΨ, φΘ, ϕΦ, and M(x,y)=max{p(x,y),p(x,fx),p(y,fy)}.

Theorem 3 Let (X,p) be a complete partial metric space, mN, A 1 , A 2 ,, A m be nonempty closed subsets of X and Y= i = 1 m A i . Let f:YY 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 n0, 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 nN,

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 η0. 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)2p(x,y)p(x,x)p(y,y) for all x,yX, 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 nN such that if r,qn with rq=1modm, then d p ( x r , x q )<ε.

Suppose the above statement is false. Then there exists ϵ>0 such that for any nN, there are r n , q n N with r n > q n n with r n q n =1modm satisfying

d p ( x q n , x r n )ϵ.

Now, we let n>2m. 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 =1modm 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)=2p(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 1im, 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 rq=1modm, 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,qmax{ n 1 , n 2 } and r>q. Then there exists k{1,2,,m} such that rq=kmodm. Therefore, rq+j=1modm for j=mk+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 νX 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)=2p(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,fν))=0, that is, p(ν,fν)=0. So, ν=fν.

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,yX,

and define the function f:XX by

f(x)= x 2 1 + x for all xX.

Now, we let φ,ϕ: R + R + and ψ: R + 3 R + be

φ(t)=2t,ϕ(t)= 2 t 5 ( 1 + t ) andψ(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, ABC is a cyclic representation of X with respect to f;

  2. (2)

    For xA and yB (or, xB and yC), without loss of generality, we may assume that xy, 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 xC and yA, without loss of generality, we may assume that xy, 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 ηt<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,yX with ηp(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, mN, A 1 , A 2 ,, A m be nonempty subsets of X and Y= i = 1 m A i . An operator f:YY 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(fx,fy)ϕ ( p ( x , y ) ) ψ ( p ( x , y ) ) ,
    (3.1)

where A m + 1 = A 1 , ϕΦ and ψΨ.

Theorem 4 Let (X,p) be a complete partial metric space, mN, A 1 , A 2 ,, A m be nonempty closed subsets of X and Y= i = 1 m A i . Let f:YY 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 n0, 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 η0. 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 ηp( 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)2p(x,y)p(x,x)p(y,y) for all x,yX, 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 nN such that if r,qn with rq=1modm, then d p ( x r , x q )<ε.

Suppose the above statement is false. Then there exists ϵ>0 such that for any nN, there are r n , q n N with r n > q n n with r n q n =1modm satisfying

d p ( x q n , x r n )ϵ.

Now, we let n>2m. 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 =1modm 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)=2p(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 1im, 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 rq=1modm, 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,qmax{ n 1 , n 2 } and r>q. Then there exists k{1,2,,m} such that rq=kmodm. Therefore, rq+j=1modm for j=mk+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 νX 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)=2p(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,fν)0,

and so ν=fν.

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,yX,

and define the function f:XX by

f(x)= x 2 1 + x for all xX.

Now, we let ψ,ϕ: R + R + be

ϕ(t)= 4 t 5 andψ(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, mN, A 1 , A 2 ,, A m be nonempty closed subsets of X, Y= i = 1 m A i and let f:YY. 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(fx,fy)ϕ ( p ( x , y ) ) ,

where A m + 1 = A 1 and ϕΦ.

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

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

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Chen, CM. Fixed point theory of cyclical generalized contractive conditions in partial metric spaces. Fixed Point Theory Appl 2013, 17 (2013). https://doi.org/10.1186/1687-1812-2013-17

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Keywords

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