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Dislocated metric space to metric spaces with some fixed point theorems

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Abstract

In this paper, we notice the notions metric-like space and dislocated metric space are exactly the same. After this historical remark, we discuss the existence and uniqueness of a fixed point of a cyclic mapping in the context of metric-like spaces. We consider some examples to illustrate the validity of the derived results of this paper.

MSC:47H10, 54H25.

1 Introduction and preliminaries

Fixed point theory is one of the most dynamic research subjects in nonlinear sciences. Regarding the feasibility of application of it to the various disciplines, a number of authors have contributed to this theory with a number of publications. The most impressing result in this direction was given by Banach, called the Banach contraction mapping principle: Every contraction in a complete metric space has a unique fixed point. In fact, Banach demonstrated how to find the desired fixed point by offering a smart and plain technique. This elementary technique leads to increasing of the possibility of solving various problems in different research fields. This celebrated result has been generalized in many abstract spaces for distinct operators. In particular, Hitzler [1] obtained one of interesting characterizations of the Banach contraction mapping principle by introducing dislocated metric spaces, which is rediscovered by Amini-Harandi [2].

Definition 1.1 A dislocated (metric-like) on a nonempty set X is a function σ:X×X[0,+) such that for all x,y,zX:

(σ 1) if σ(x,y)=0 then x=y,

(σ 2) σ(x,y)=σ(y,x),

(σ 3) σ(x,y)σ(x,z)+σ(z,y),

and the pair (X,σ) is called a dislocated (metric-like) space.

The motivation of defining this new notion is to get better results in logic programming semantics (see, e.g., [1, 3]). Following these initial reports, many authors paid attention to the subject and have published several papers (see, e.g., [412]). Another interesting generalization of the Banach contraction mapping principle was given by Kirk et al. [13] via a cyclic mapping (see, e.g., [1416]). In this remarkable paper, the mappings, for which the existence and uniqueness of a fixed point were discussed, do not need to be continuous.

A mapping T:ABAB is called cyclic if T(A)B and T(B)A.

Theorem 1.2 (See [13])

Let A and B be two nonempty closed subsets of a complete metric space (X,d). Suppose that T:ABAB is cyclic and satisfies the following:

  1. (C)

    There exists a constant k(0,1) such that

    d(Tx,Ty)kd(x,y)for allxA,yB.

Then T has a unique fixed point that belongs to AB.

Cyclic mappings and related fixed point theorems have been considered by many authors (see, e.g., [1328]). In this paper, we discuss the existence and uniqueness of fixed point theory of a cyclic mapping with certain properties in the context of metric-like spaces.

We recall some basic definitions and crucial results on the topic. In this paper, we follow the notations of Amini-Harandi [2].

Definition 1.3 (See [2])

Let (X,σ) be a metric-like space and U be a subset of X. We say U is a σ-open subset of X if for all xX there exists r>0 such that B σ (x,r)U. Also, VX is a σ-closed subset of X if (XV) is a σ-open subset of X.

Lemma 1.4 Let (X,σ) be a metric-like space and V be a σ-closed subset of X. Let { x n } be a sequence in V. If x n x as n, then xV.

Proof Let xV. By Definition 1.3, (XV) is a σ-open set. Then there exists r>0 such that B σ (x,r)XV. On the other hand, we have lim n |σ( x n ,x)σ(x,x)|=0 since x n x as n. Hence, there exists n 0 N such that

| σ ( x n , x ) σ ( x , x ) | <r

for all n n 0 . So, we conclude that { x n } B σ (x,r)XV for all n n 0 . This is a contradiction since { x n }V for all nN. □

Lemma 1.5 Let (X,σ) be a metric-like space and { x n } be a sequence in X such that x n x as n and σ(x,x)=0. Then lim n σ( x n ,y)=σ(x,y) for all yX.

Proof From (σ 3) we have

σ(x,y)σ( x n ,x)σ( x n ,y)σ( x n ,x)+σ(x,y).

Letting n in the above inequalities, we get lim n σ( x n ,y)=σ(x,y). □

Lemma 1.6 Let (X,σ) be a metric-like space. Then

  1. (A)

    if σ(x,y)=0, then σ(x,x)=σ(y,y)=0;

  2. (B)

    if { x n } is a sequence such that lim n σ( x n , x n + 1 )=0, then we have

    lim n σ( x n , x n )= lim n σ( x n + 1 , x n + 1 )=0;
  3. (C)

    if xy, then σ(x,y)>0;

  4. (D)

    σ(x,x) 2 n i = 1 i = n σ(x, x i ) holds for all x i ,xX, where 1in.

Proof We skip the proof (A) since it is evident.

(B) Due to the triangle inequality, we have σ( x n , x n )σ( x n , x n + 1 )+σ( x n + 1 , x n )=2σ( x n + 1 , x n ). So, we find

0 lim n σ( x n , x n )2 lim n σ( x n , x n + 1 )=0.

Analogously, we derive

0 lim n σ( x n + 1 , x n + 1 )2 lim n σ( x n , x n + 1 )=0.
  1. (C)

    If xy and σ(x,y)=0, then by (σ 1) we have x=y, which is a contradiction.

  2. (D)

    Again from (σ 3) we get

    σ(x,x)2σ(x, x i ),

where 1in. Then we observe that

i = 1 i = n σ(x,x)2 i = 1 i = n σ(x, x i ).

Hence, we derive that

σ(x,x) 2 n i = 1 i = n σ(x, x i ).

 □

At first, we define the class of Φ and Ψ by the following ways:

Ψ= { ψ : [ 0 , ) [ 0 , )  such that  ψ  is non-decreasing and continuous }

and

Φ= { ϕ : [ 0 , ) [ 0 , )  such that  ϕ  is lower semicontinuous } .

Definition 1.7 Let (X,σ) be a metric-like space, mN, let A 1 , A 2 ,, A m be σ-closed nonempty subsets of X and Y= i = 1 m A i . We say that T is called a cyclic generalized ϕ-ψ-contractive mapping if

  1. (1)

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

  2. (2)
    ψ(t)ψ(s)+ϕ(s)>0for all t>0 and s=t or s=0

    and

    ψ ( σ ( T x , T y ) ) ψ ( M σ ( x , y ) ) ϕ ( M σ ( x , y ) )
    (1)

for any x A i , y A i + 1 , i=1,2,,m, where A m + 1 = A 1 , ϕΦ, ψΨ and

M σ (x,y)=max { σ ( x , y ) , σ ( x , T x ) , σ ( y , T y ) , σ ( x , T y ) + σ ( y , T x ) 4 } .

Let X be a nonempty set and T:XX be a given map. The set of all fixed points of T will be denoted by Fix(T), that is, Fix(T)={xX;x=Tx}.

Theorem 1.8 Let (X,σ) be a complete metric-like space, mN, let A 1 , A 2 ,, A m be nonempty σ-closed subsets of X and Y= i = 1 m A i . Suppose that T:YY is a cyclic generalized ϕ-ψ-contractive mapping. Then T has a fixed point in i = 1 n A i . Moreover, if σ(x,y)σ(x,x) for all x,yFix(T), then T has a unique fixed point in i = 1 n A i .

Proof Let x 0 be an arbitrary point of Y. So, there exists some i 0 such that x 0 A i 0 . Since T( A i 0 ) A i 0 + 1 , we conclude that T x 0 A i 0 + 1 . Thus, there exists x 1 in A i 0 + 1 such that T x 0 = x 1 . Recursively, T x n = x n + 1 , where x n A i n . Hence, for n0, there exists i n {1,2,,m} such that x n A i n . In case x n 0 = x n 0 + 1 for some n 0 =0,1,2, , then it is clear that x n 0 is a fixed point of T. Now assume that x n x n + 1 for all n. Hence, by Lemma 1.6(C) we have σ( x n 1 , x n )>0 for all n. We shall show that the sequence { σ n } is non-increasing where σ n =σ( x n , x n + 1 ). Assume that there exists some n 0 N such that

σ( x n 0 1 , x n 0 )σ( x n 0 , x n 0 + 1 ).

Hence

ψ ( σ ( x n 0 1 , x n 0 ) ) ψ ( σ ( x n 0 , x n 0 + 1 ) ) .
(2)

By taking x= x n 1 and y= x n in condition (1) together with (2), we get

ψ ( σ ( x n , x n + 1 ) ) = ψ ( σ ( T x n 1 , T x n ) ) ψ ( max { σ ( x n 1 , x n ) , σ ( x n 1 , T x n 1 ) , σ ( x n , T x n ) , σ ( x n 1 , T x n ) + σ ( x n , T x n 1 ) 4 } ) ϕ ( max { σ ( x n 1 , x n ) , σ ( x n 1 , T x n 1 ) , σ ( x n , T x n ) , σ ( x n 1 , T x n ) + σ ( x n , T x n 1 ) 4 } ) ψ ( max { σ ( x n 1 , x n ) , σ ( x n , x n + 1 ) , σ ( x n 1 , x n + 1 ) + σ ( x n , x n ) 4 } ) ϕ ( max { σ ( x n 1 , x n ) , σ ( x n , x n + 1 ) , σ ( x n 1 , x n + 1 ) + σ ( x n , x n ) 4 } ) .
(3)

On the other hand, from Lemma 1.6(D) we have

σ( x n , x n )σ( x n 1 , x n )+σ( x n , x n + 1 ),

and by (σ 3) we have

σ( x n 1 , x n + 1 )σ( x n 1 , x n )+σ( x n , x n + 1 ).

That is,

max { σ ( x n 1 , x n ) , σ ( x n , x n + 1 ) } max { σ ( x n 1 , x n ) , σ ( x n , x n + 1 ) , σ ( x n 1 , x n + 1 ) + σ ( x n , x n ) 4 } max { σ ( x n 1 , x n ) , σ ( x n , x n + 1 ) , σ ( x n 1 , x n ) + σ ( x n , x n + 1 ) 2 } = max { σ ( x n 1 , x n ) , σ ( x n , x n + 1 ) } .

Then

max { σ ( x n 1 , x n ) , σ ( x n , x n + 1 ) , σ ( x n 1 , x n + 1 ) + σ ( x n , x n ) 4 } =max { σ ( x n 1 , x n ) , σ ( x n , x n + 1 ) } .

Therefore from (3) we get

ψ ( σ ( x n , x n + 1 ) ) ψ ( max { σ ( x n 1 , x n ) , σ ( x n , x n + 1 ) } ) ϕ ( max { σ ( x n 1 , x n ) , σ ( x n , x n + 1 ) } ) .

Now, if max{σ( x n 1 , x n ),σ( x n , x n + 1 )}=σ( x n , x n + 1 ), then

ψ ( σ ( x n , x n + 1 ) ) α ( σ ( x n , x n + 1 ) ) β ( σ ( x n , x n + 1 ) ) ,

a contradiction. Hence, we have

ψ ( σ ( x n , x n + 1 ) ) ψ ( σ ( x n 1 , x n ) ) ϕ ( σ ( x n 1 , x n ) )
(4)

for all nN. By taking x= x n 0 1 and y= x n 0 in (4) and keeping (2) in mind, we deduce that

ψ ( σ ( x n 0 1 , x n 0 ) ) ψ ( σ ( x n 0 1 , x n 0 ) ) ϕ ( σ ( x n 0 1 , x n 0 ) ) ,

a contradiction. Hence, we conclude that σ n < σ n 1 holds for all nN. Thus, there exists r0 such that lim n σ n =r. We shall show that r=0 by the method of reductio ad absurdum. For this purpose, we assume that r>0. By (4), together with the properties of ϕ, ψ, we have

ψ(r)= lim sup n ψ( σ n ) lim sup n [ ψ ( σ n 1 ) ϕ ( σ n 1 ) ] ψ(r)ϕ(r),

which yields that ϕ(r)0. This is a contradiction. Hence, we obtain that

lim n σ n = lim n σ( x n , x n + 1 )=0.
(5)

We shall show that { x n } is a σ-Cauchy sequence. To reach this goal, we shall follow the standard techniques that can be found in, e.g., [22]. For the sake of completeness, we shall adopt the techniques used in [22]. First, we prove the following claim:

  1. (K)

    For every ε>0, there exists nN such that if r,qn with rq1(m), then σ( x r , x q )<ε.

Suppose, on the contrary, that there exists ε>0 such that for any nN, we can find r n > q n n with r n q n 1(m) satisfying

σ( x q n , x r n )ε.
(6)

Now, we take n>2m. Then, corresponding to q n n, we can choose r n in such a way that it is the smallest integer with r n > q n satisfying r n q n 1(m) and σ( x q n , x r n )ε. Therefore, σ( x q n , x r n m )ε. By using the triangular inequality, we obtain

εσ( x q n , x r n )σ( x q n , x r n m )+ i = 1 m σ( x r n i , x r n i + 1 )<ε+ i = 1 m p( x r n i , x r n i + 1 ).

Passing to the limit as n in the last inequality and taking (5) into account, we obtain that

lim n σ( x q n , x r n )=ε.
(7)

Again, by (σ 3), we derive that

ε σ ( x q n , x r n ) σ ( x q n , x q n + 1 ) + σ ( x q n + 1 , x r n + 1 ) + σ ( x r n + 1 , x r n ) σ ( x q n , x q n + 1 ) + σ ( x q n + 1 , x q n ) + σ ( x q n , x r n ) + σ ( x r n , x r n + 1 ) + σ ( x r n + 1 , x r n ) = 2 σ ( x q n , x q n + 1 ) + σ ( x q n , x r n ) + 2 σ ( x r n , x r n + 1 ) .

Taking (5) and (7) into account, we get

lim n σ( x q n + 1 , x r n + 1 )=ε.
(8)

By (σ 3), we have the following inequalities:

σ( x q n , x r n + 1 )σ( x q n , x r n )+σ( x r n , x r n + 1 )
(9)

and

σ( x q n , x r n )σ( x q n , x r n + 1 )+σ( x r n , x r n + 1 ).
(10)

Letting n in (9) and (10), we derive that

lim n σ( x q n , x r n + 1 )=ε.
(11)

Again by (σ 3) we have

σ( x r n , x q n + 1 )σ( x r n , x r n + 1 )+σ( x r n + 1 , x q n + 1 )
(12)

and

σ( x r n + 1 , x q n + 1 )σ( x r n + 1 , x r n )+σ( x r n , x q n + 1 ).
(13)

Letting n in (12) and (13), we derive that

lim n σ( x r n , x q n + 1 )=ε.
(14)

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 (5), (7), (8), (11), (14) together with the fact that T is a generalized cyclic ϕ-ψ-contractive mapping, we find that

ψ ( σ ( x q n + 1 , x r n + 1 ) ) = ψ ( σ ( T x q n , T x r n ) ) ψ ( max { σ ( x q n , x r n ) , σ ( x q n , T x q n ) , σ ( x r n , T x r n ) , σ ( x q n , T x r n ) + σ ( x r n , T x q n ) 4 } ) ϕ ( max { σ ( x q n , x r n ) , σ ( x q n , T x q n ) , σ ( x r n , T x r n ) , σ ( x q n , T x r n ) + σ ( x r n , T x q n ) 4 } ) = ψ ( max { σ ( x q n , x r n ) , σ ( x q n , x q n + 1 ) , σ ( x r n , x r n + 1 ) , σ ( x q n , x r n + 1 ) + σ ( x r n , x q n + 1 ) 4 } ) ϕ ( max { σ ( x q n , x r n ) , σ ( x q n , x q n + 1 ) , σ ( x r n , x r n + 1 ) , σ ( x q n , x r n + 1 ) + σ ( x r n , x q n + 1 ) 4 } ) .

Regarding the properties of ϕ, ψ in the last inequality, we obtain that

ψ(ε)ψ(ε)ϕ(ε),

a contradiction. Hence, the condition (K) is satisfied. Fix ε>0. By the claim, we find n 0 N such that if r,q n 0 with rq1(m),

σ( x r , x q ) ε 2 .
(15)

Since lim n σ( x n , x n + 1 )=0, we also find n 1 N such that

σ( x n , x n + 1 ) ε 2 m
(16)

for any n n 1 . Suppose that r,smax{ n 0 , n 1 } and s>r. Then there exists k{1,2,,m} such that srk(m). Therefore, sr+φ1(m) for φ=mk+1. So, we have for j{1,,m}, s+jr1(m)

σ( x r , x s )σ( x r , x s + j )+σ( x s + j , x s + j 1 )++σ( x s + 1 , x s ).

By (15) and (16) and from the last inequality, we get

σ( x r , x s ) ε 2 +j× ε 2 m ε 2 +m× ε 2 m =ε.

This proves that { x n } is a σ-Cauchy sequence. Since ε is arbitrary, { x n } is a Cauchy sequence. Since Y is σ-closed in (X,σ), then (Y,σ) is also complete, there exists xY= i = 1 m A i such that lim n x n =x in (Y,σ); equivalently

σ(x,x)= lim n σ(x, x n )= lim n , m σ( x n , x m )=0.
(17)

In what follows, we prove that x is a fixed point of T. In fact, since lim n x n =x and, as Y= i = 1 m A i is a cyclic representation of Y with respect to T, the sequence { x n } has infinite terms in each A i for i{1,2,,m}. Suppose that x A i , Tx A i + 1 , and we take a subsequence x n k of { x n } with x n k A i 1 (the existence of this subsequence is guaranteed by the above-mentioned comment). By using the contractive condition, we can obtain

ψ ( σ ( T x , T x n k ) ) ψ ( max { σ ( x , x n k ) , σ ( x , T x ) , σ ( x n k , T x n k ) , σ ( x , T x n k ) + σ ( x n k , T x ) 4 } ) ϕ ( max { σ ( x , x n k ) , σ ( x , T x ) , σ ( x n k , T x n k ) , σ ( x , T x n k ) + σ ( x n k , T x ) 4 } ) .

Passing to the limit as n and using x n k x, lower semi-continuity of φ, we have

ψ ( σ ( x , T x ) ) ψ ( σ ( x , T x ) ) ϕ ( σ ( x , T x ) ) .

So, σ(x,Tx)=0 and, therefore, x is a fixed point of T. Finally, to prove the uniqueness of the fixed point, suppose that y,zX are two distinct fixed points of T. The cyclic character of T and the fact that y,zX are fixed points of T imply that x,y i = 1 m A i . Suppose that xy and for all u,wFix(T), σ(u,w)σ(u,u). Using the contractive condition, we obtain

ψ ( σ ( T x , T y ) ) ψ ( max { σ ( x , y ) , σ ( x , T x ) , σ ( y , T y ) , σ ( x , T y ) + σ ( y , T x ) 4 } ) ϕ ( max { σ ( x , y ) , σ ( x , T x ) , σ ( y , T y ) , σ ( x , T y ) + σ ( y , T x ) 4 } ) .

Then we have

ψ ( σ ( x , y ) ) ψ ( σ ( x , y ) ) ϕ ( σ ( x , y ) ) ,

which is a contradiction. Thus, we derive that σ(y,z)=0y=z, which finishes the proof. □

If in Theorem 1.8 we take A i =X for all 0im, then we deduce the following theorem.

Theorem 1.9 Let (X,σ) be a complete metric-like space and T be a self-map on X. Assume that there exist ϕΦ and ψΨ such that

ψ ( σ ( T x , T y ) ) ψ ( M σ ( x , y ) ) ϕ ( M σ ( x , y ) )

for all x,yX, where

M σ (x,y)=max { σ ( x , y ) , σ ( x , T x ) , σ ( y , T y ) , σ ( x , T y ) + σ ( y , T x ) 4 } .

Then T has a fixed point. Moreover, if σ(x,y)σ(x,x) for all x,yFix(T), then T has a unique fixed point.

If in Theorem 1.8 we take ψ(t)=t and ϕ(t)=(1r)t, where r[0,1), then we deduce the following corollary.

Corollary 1.10 Let (X,σ) be a complete metric-like space, mN, let A 1 , A 2 ,, A m be nonempty σ-closed subsets of X and Y= i = 1 m A i . Suppose that T:YY is an operator such that

  1. (i)

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

  2. (ii)

    there exists r[0,1) such that

    σ(Tx,Ty)rmax { σ ( x , y ) , σ ( x , T x ) , σ ( y , T y ) , σ ( x , T y ) + σ ( y , T x ) 4 }

for any x A i , y A i + 1 , i=1,2,,m, where A m + 1 = A 1 . Then T has a fixed point z i = 1 m A i . Moreover, if σ(x,y)σ(x,x) for all x,yFix(T), then T has a unique fixed point.

Example 1.11 Let X=R with the metric-like σ(x,y)=max{|x|,|y|} for all x,yX. Suppose A 1 =[1,0] and A 2 =[0,1] and Y= i = 1 2 A i . Define T:YY by

Tx={ x 3 if  x [ 1 , 0 ] , x 3 2 if  x [ 0 , 1 ] .

It is clear that i = 1 2 A i is a cyclic representation of Y with respect to T. Let x A 1 =[1,0] and y A 2 =[0,1]. Then

σ ( T x , T y ) = max { | x 3 | , | y 3 2 | } = max { x 3 , y 3 2 } max { x 2 , y 2 } = 1 2 max { x , y } = 1 2 σ ( x , y ) ,

and so

σ(Tx,Ty) 1 2 max { σ ( x , y ) , σ ( x , T x ) , σ ( y , T y ) , σ ( x , T y ) + σ ( y , T x ) 4 } .

Hence, the conditions of Corollary 1.10 (Theorem 1.8) hold and T has a fixed point in A 1 A 2 . Here, x=0 is a fixed point of T.

If in the above corollary we take A i =X for all 0im, then we deduce the following corollary.

Corollary 1.12 Let (X,σ) be a complete metric-like space and T be a self-map on X. Assume that there exists r[0,1) such that

σ(Tx,Ty)rmax { σ ( x , y ) , σ ( x , T x ) , σ ( y , T y ) , σ ( x , T y ) + σ ( y , T x ) 4 }

holds for all x,yX. Then T has a fixed point. Moreover, if σ(x,y)σ(x,x) for all x,yFix(T), then T has a unique fixed point.

Example 1.13 Let X=R with the metric-like σ(x,y)=max{x,y} for all x,yX. Let T:XX be defined by

Tx={ 1 5 x 2 if  0 x < 1 / 3 , ( 1 x ) / 2 if  1 / 3 x 1 , 1 6 x if  x > 1 .

Proof To show the existence and uniqueness point of T, we need to consider the following cases.

  • Let 0x,y<1/3. Then

    σ(Tx,Ty)= 1 5 max { x 2 , y 2 } 1 2 max{x,y}= 1 2 σ(x,y).
  • Let 1/3x,y1. Then

    σ(Tx,Ty)= 1 2 max{1x,1y} 1 2 max{x,y}= 1 2 σ(x,y).
  • Let x,y>1. Then

    σ(Tx,Ty)= 1 6 max{x,y} 1 2 max{x,y}= 1 2 σ(x,y).
  • Let 0x<1/3 and 1/3y1. Then

    σ(Tx,Ty)=max { 1 5 x 2 , ( 1 y ) / 2 } 1 2 max{x,y}= 1 2 σ(x,y).
  • Let 0x<1/3 and y>1. Then

    σ(Tx,Ty)=max { 1 5 x 2 , 1 6 y } 1 2 max{x,y}= 1 2 σ(x,y).
  • Let 1/3x1 and y>1. Then

    σ(Tx,Ty)=max { ( 1 x ) / 2 , 1 6 y } 1 2 max{x,y}= 1 2 σ(x,y),

and so

σ(Tx,Ty) 1 2 max { σ ( x , y ) , σ ( x , T x ) , σ ( y , T y ) , σ ( x , T y ) + σ ( y , T x ) 4 } .

Hence, we conclude that all the conditions of Corollary 1.12 (Theorem 1.9) hold and hence T has a fixed point 0 in [0,). □

By Corollary 1.10 we deduce the following result.

Corollary 1.14 Let (X,σ) be a complete metric-like space, mN, let A 1 , A 2 ,, A m be nonempty σ-closed subsets of X and Y= i = 1 m A i . Suppose that T:YY is an operator such that

  1. (i)

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

  2. (ii)

    there exists r[0,1) such that

    0 σ ( T x , T y ) ρ(t)dtr 0 max { σ ( x , y ) , σ ( x , T x ) , σ ( y , T y ) , σ ( x , T y ) + σ ( y , T x ) 4 } ρ(t)dt

for any x A i , y A i + 1 , i=1,2,,m, where A m + 1 = A 1 , and ρ:[0,)[0,) is a Lebesgue-integrable mapping satisfying 0 ε ρ(t)dt>0 for ε>0. Then T has a fixed point z i = 1 m A i . Moreover, if σ(x,y)σ(x,x) for all x,yFix(T), then T has a unique fixed point.

If in the above corollary we take A i =X for all 0im, then we deduce the following corollary.

Corollary 1.15 Let (X,σ) be a complete metric-like space and let T:XX be a mapping such that for any x,yX,

0 σ ( T x , T y ) ρ(t)dtr 0 max { σ ( x , y ) , σ ( x , T x ) , σ ( y , T y ) , σ ( x , T y ) + σ ( y , T x ) 4 } ρ(t)dt,

where ρ:[0,)[0,) is a Lebesgue-integrable mapping satisfying 0 ε ρ(t)dt for ε>0 and the constant β[0, 1 4 ). Then T has a unique fixed point.

Definition 1.16 Let T:XX and ψ:X[0,) and γ[0,1]. A mapping T is said to be a γ-ψ-subadmissible mapping if

ψ(x)γimpliesψ(Tx)γ,xX.

Example 1.17 Let T:RR and ψ:R R + be defined by Tx= x 3 and ψ(x)= 1 2 e x . Then T is a γ-ψ-subadmissible mapping where γ= 1 6 . Indeed, if ψ(x)= 1 6 e x 1 6 , then x0, and hence Tx0. That is, ψ(Tx)= 1 6 e T x 1 6 .

Example 1.18 Let T:[π,π][π,π] and ψ:[π,π] R + be defined by Tx= π 2 sin(x) and ψ(x)=|x 1 2 π|+ 1 2 . Then T is a γ-ψ-subadmissible mapping where γ= 1 2 . Indeed, if ψ(x)=|x 1 2 π|+ 1 2 1 2 , then x= 1 2 π, and hence Tx= 1 2 π. That is, ψ(Tx)= 1 2 .

Let Λ be the class of all the functions φ: [ 0 , + ) 3 [0,+) that are a continuous with the following property:

φ(x,y,z)=0if and only ifx=y=z=0.

Definition 1.19 Let (X,σ) be a metric-like space, mN, let A 1 , A 2 ,, A m be σ-closed nonempty subsets of (X, d p ) and Y= i = 1 m A i . Assume that T:YY is a γ-ψ-subadmissible mapping where γ= 1 6 . Then T is called a ψ-cyclic generalized weakly C-contraction if

  1. (1)

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

  2. (2)
    σ ( T x , T y ) ψ ( x ) σ ( x , T x ) + ψ ( T x ) σ ( y , T y ) + ψ ( T 2 x ) σ ( x , T y ) + ψ ( T 3 x ) σ ( y , T x ) φ ( σ ( x , T x ) , σ ( x , T y ) , 1 2 [ σ ( x , T y ) + σ ( y , T x ) ] )
    (18)

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

Theorem 1.20 Let (X,σ) be a complete metric-like space, mN, let A 1 , A 2 ,, A m be nonempty σ-closed subsets of (X,p) and Y= i = 1 m A i . Suppose that T:YY is a ψ-cyclic generalized weakly C-contraction. If there exists x 0 Y such that ψ( x 0 ) 1 6 , then T has a fixed point z i = 1 n A i . Moreover, if ψ(z) 1 6 , then z is unique.

Proof Let x 0 Y be such that ψ( x 0 ) 1 6 . Since T is a sub ψ-admissible mapping with respect to 1 6 , then ψ(T x 0 ) 1 6 . ψ( T n x 0 ) 1 6 for all nN0. Also, there exists some i 0 such that x 0 A i 0 . Now T( A i 0 ) A i 0 + 1 implies that T x 0 A i 0 + 1 . Thus there exists x 1 in A i 0 + 1 such that T x 0 = x 1 . Similarly, T x n = x n + 1 , where x n A i n . Hence, for n0, there exists i n {1,2,,m} such that x n A i n and x n + 1 A i n + 1 . In case x n 0 = x n 0 + 1 for some n 0 =0,1,2, , then it is clear that x n 0 is a fixed point of T. Now assume that x n x n + 1 for all n. Since T:YY is a cyclic generalized weak C-contraction, we have that for all n N ,

σ ( x n , x n + 1 ) = σ ( T x n 1 , T x n ) ψ ( x n 1 ) σ ( x n 1 , T x n 1 ) + ψ ( T x n 1 ) σ ( x n , T x n ) + ψ ( T 2 x n 1 ) σ ( x n 1 , T x n ) + ψ ( T 3 x n 1 ) σ ( x n , T x n 1 ) φ ( σ ( x n 1 , T x n 1 ) , σ ( x n , T x n ) , 1 2 [ σ ( x n 1 , T x n ) + σ ( x n , T x n 1 ) ] ) = ψ ( x n 1 ) σ ( x n 1 , x n ) + ψ ( x n ) σ ( x n , x n + 1 ) + ψ ( x n + 1 ) σ ( x n 1 , x n + 1 ) + ψ ( x n + 2 ) σ ( x n , x n ) φ ( σ ( x n 1 , x n ) , σ ( x n , x n + 1 ) , 1 2 [ σ ( x n 1 , x n + 1 ) + σ ( x n , x n ) ] ) 1 6 [ σ ( x n 1 , x n ) + σ ( x n , x n + 1 ) + σ ( x n 1 , x n + 1 ) + σ ( x n , x n ) ] φ ( σ ( x n 1 , x n ) , σ ( x n , x n + 1 ) , 1 2 [ σ ( x n 1 , x n + 1 ) + σ ( x n , x n ) ] ) ,

and so

σ( x n , x n + 1 ) 1 6 [ σ ( x n 1 , x n ) + σ ( x n , x n + 1 ) + σ ( x n 1 , x n + 1 ) + σ ( x n , x n ) ] .
(19)

On the other hand, from (σ 3) we have

σ( x n 1 , x n + 1 )σ( x n 1 , x n )+σ( x n , x n + 1 ),

and by Lemma 1.6(D) we have

σ( x n , x n )σ( x n 1 , x n )+σ( x n , x n + 1 ).

Then by (19) we get

σ( x n , x n + 1 ) 1 2 [ σ ( x n 1 , x n ) + σ ( x n , x n + 1 ) ] .

Therefore,

σ( x n , x n + 1 )σ( x n 1 , x n )
(20)

for any n1. Set t n =φ( x n , x n 1 ). On the occasion of the facts above, { t n } is a non-increasing sequence of nonnegative real numbers. Consequently, there exists L0 such that

lim n σ( x n , x n + 1 )=L.
(21)

We shall prove that L=0. Since σ( x n , x n )2φ( x n , x n + 1 ), then lim n σ( x n , x n )2L. Similarly, lim n σ( x n 1 , x n + 1 )2L. Then

lim n [ σ ( x n , x n ) + σ ( x n 1 , x n + 1 ) ] 4L.

On the other hand, by taking limit as n in (19), we have

L 1 6 [ 2 L + lim n [ σ ( x n , x n ) + σ ( x n 1 , x n + 1 ) ] ] ,

which implies

4L lim n [ σ ( x n , x n ) + σ ( x n 1 , x n + 1 ) ] .

Hence,

lim n [ σ ( x n , x n ) + σ ( x n 1 , x n + 1 ) ] =4L.

Now, from (18) we have

t n + 1 ψ ( x n 1 ) t n + ψ ( x n ) t n + 1 + ψ ( x n + 1 ) σ ( x n 1 , x n + 1 ) + ψ ( x n + 2 ) σ ( x n , x n ) φ ( t n , t n + 1 , 1 2 [ σ ( x n 1 , x n + 1 ) + σ ( x n , x n ) ] ) 1 6 [ t n + t n + 1 + σ ( x n 1 , x n + 1 ) + σ ( x n , x n ) ] φ ( t n , t n + 1 , 1 2 [ σ ( x n 1 , x n + 1 ) + σ ( x n , x n ) ] ) .

By taking limit as n in the above inequality, we deduce

LLφ(L,L,2L),

and so φ(L,L,2L)=0. Since φ(x,y,z)=0x=y=z=0, we get L=0. Due to lim n σ( x n , x n )2L and lim n σ( x n 1 , x n + 1 )2L, we have

lim n σ( x n , x n )= lim n σ( x n 1 , x n + 1 )= lim n σ( x n , x n + 1 )=0.
(22)

We shall show that { x n } is a σ-Cauchy sequence. At first, we prove the following fact:

  1. (K)

    For every ε>0, there exists nN such that if r,qn with rq1(m), then σ( x r , x q )<ε.

Suppose to the contrary that there exists ε>0 such that for any nN, we can find r n > q n n with r n q n 1(m) satisfying

σ( x q n , x r n )ε.
(23)

Following the related lines of the proof of Theorem 1.8, we have

lim n σ ( x q n , x r n ) = ε ; lim n σ ( x q n + 1 , x r n + 1 ) = ε ; lim n σ ( x q n , x r n + 1 ) = ε

and

lim n σ( x r n , x q n + 1 )=ε.
(24)

Since x q n and x r n lie in different adjacently labeled sets A i and A i + 1 for certain 1im, using the fact that T is a ψ-cyclic generalized weakly C-contraction, we have

σ ( x q n + 1 , x r n + 1 ) = σ ( T x q n , T x r n ) ψ ( x q n ) σ ( x q n , T x q n ) + ψ ( T x q n ) σ ( x r n , T x r n ) + ψ ( T 2 x q n ) σ ( x q n , T x r n ) + ψ ( T 3 x q n ) σ ( x r n , T x q n ) φ ( σ ( x q n , T x q n ) , σ ( x r n , T x r n ) , 1 2 [ σ ( x q n , T x r n ) + σ ( x r n , T x q n ) ] ) 1 6 [ σ ( x q n , x q n + 1 ) + σ ( x r n , x r n + 1 ) + σ ( x q n , x r n + 1 ) + σ ( x r n , x q n + 1 ) ] φ ( σ ( x q n , x q n + 1 ) , σ ( x r n , x r n + 1 ) , 1 2 [ σ ( x q n , x r n + 1 ) + σ ( x r n , x q n + 1 ) ] ) .

Now, by taking limit as n in the above inequality, we derive that

ε 1 6 [0+0+ε+ε]φ(0,0,ε) 1 3 ε,

which is a contradiction. Hence, condition (K) holds. We are ready to show that the sequence { x n } is Cauchy. Fix ε>0. By the claim, we find n 0 N such that if r,q n 0 with rq1(m),

σ( x r , x q ) ε 2 .
(25)

Since lim n σ( x n , x n + 1 )=0, we also find n 1 N such that

σ( x n , x n + 1 ) ε 2 m
(26)

for any n n 1 . Suppose that r,smax{ n 0 , n 1 } and s>r. Then there exists k{1,2,,m} such that srk(m). Therefore, sr+φ1(m) for φ=mk+1. So, we have, for j{1,,m}, s+jr1(m),

σ( x r , x s )σ( x r , x s + j )+σ( x s + j , x s + j 1 )++σ( x s + 1 , x s ).

By (25) and (26) and from the last inequality, we get

σ( x r , x s ) ε 2 +j× ε 2 m ε 2 +m× ε 2 m =ε.

This proves that { x n } is a σ-Cauchy sequence.

Since Y is σ-closed in (X,σ), then (Y,σ) is also complete, there exists zY= i = 1 m A i such that lim n x n =z in (Y,p); equivalently

σ(z,z)= lim n σ(z, x n )= lim n , m σ( x n , x m )=0.
(27)

In what follows, we prove that x is a fixed point of T. In fact, since lim n x n =z and, as Y= i = 1 m A i is a cyclic representation of Y with respect to T, the sequence ( x n ) has infinite terms in each A i for i{1,2,,m}. Suppose that x A i , Tx A i + 1 , and we take a subsequence x n k of ( x n ) with x n k A i 1 (the existence of this subsequence is guaranteed by the above-mentioned comment). By using the contractive condition, we can obtain

σ ( x n k + 1 , T x ) = σ ( T x n k , T x ) ψ ( x n k ) σ ( x n k , T x n k ) + ψ ( T x n k ) σ ( x , T x ) + ψ ( T 2 x n k ) σ ( x n k , T x ) + ψ ( T 3 x n k ) σ ( x , T x n k ) φ ( σ ( x n k , T x n k ) , σ ( x , T x ) , 1 2 [ σ ( x n k , T x ) + σ ( x , T x n k ) ] ) 1 6 [ σ ( x n k , x n k + 1 ) + σ ( x , T x ) + σ ( x n k , T x ) + σ ( x , x n k + 1 ) ] φ ( σ ( x n k , x n k + 1 ) , σ ( x , T x ) , 1 2 [ σ ( x n k , T x ) + σ ( x , x n k + 1 ) ] ) .

Passing to the limit as n and using x n k x, lower semi-continuity of φ, we have

σ(x,Tx) 1 3 σ(x,Tx)φ ( 0 , σ ( x , T x ) , 1 2 σ ( x , T x ) ) 1 3 σ(x,Tx).

So, σ(x,Tx)=0 and, therefore, x is a fixed point of T. Finally, to prove the uniqueness of the fixed point, suppose that y,zX are fixed points of T. The cyclic character of T and the fact that y,zX are fixed points of T imply that y,z i = 1 m A i . Also, suppose that ψ(y) 1 6 . By using the contractive condition, we derive that

σ ( y , z ) = σ ( T y , T x ) ψ ( y ) σ ( y , T y ) + ψ ( T y ) σ ( z , T z ) + ψ ( T 2 y ) σ ( y , T z ) + ψ ( T 2 y ) σ ( z , T y ) φ ( σ ( y , T y ) , σ ( z , T z ) , 1 2 [ σ ( y , T z ) + σ ( z , T y ) ] ) .

Then

σ ( y , z ) 1 6 [ 2 σ ( y , z ) + σ ( y , y ) ] φ ( 0 , 0 , 1 2 [ σ ( y , z ) + σ ( z , y ) ] ) 1 6 [ 2 σ ( y , z ) + 2 σ ( y , z ) ] φ ( 0 , 0 , 1 2 [ σ ( y , z ) + σ ( z , y ) ] ) = 2 3 σ ( y , z ) φ ( 0 , 0 , 1 2 [ σ ( y , z ) + σ ( z , y ) ] ) 2 3 σ ( y , z ) .

This gives us σ(y,z)=0, that is, y=z. This finishes the proof. □

Corollary 1.21 Let (X,σ) be a complete metric-like space, mN, let A 1 , A 2 ,, A m be nonempty σ-closed subsets of X and Y= i = 1 m A i . Suppose that T:YY is an operator such that

  1. (i)

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

  2. (ii)

    there exists β[0, 1 6 ) such that

    σ(Tx,Ty)β [ σ ( x , T x ) + σ ( y , T y ) + σ ( x , T y ) + σ ( y , T x ) ]
    (28)

for any x A i , y A i + 1 , i=1,2,,m, where A m + 1 = A 1 . Then T has a fixed point z i = 1 n A i .

Proof Let ψ(t)= 1 6 and β[0, 1 6 ). Here, it suffices to take the function φ: [ 0 , + ) 4 [0,+) defined by φ(a,b,c,e)=( 1 6 β)(a+b+c+e). Obviously, φ satisfies that φ(a,b,c,e)=0 if and only if a=b=c=e=0, and φ(x,y,z,t)=( 1 6 β)(x+y+z+t)=φ(x+y+z+t,0). Then we apply Theorem 1.20 to finish the proof. □

Example 1.22 Let X=R with the metric-like σ(x,y)=max{|x|,|y|} for all x,yX. Suppose A 1 =[1,0] and A 2 =[0,1] and Y= i = 1 2 A i . Define T:YY by

Tx={ 1 24 x if  x [ 1 , 0 ] , 1 12 x if  x [ 0 , 1 ] .

It is clear that i = 1 2 A i is a cyclic representation of Y with respect to T.

Let x A 1 =[1,0] and y A 2 =[0,1]. Then

σ ( T x , T y ) = max { | 1 24 x | , | 1 12 y | } = max { 1 24 x , 1 12 y } max { x 12 , y 12 } = 1 12 max { x , y } = 1 12 σ ( x , y ) ,

and so

σ(Tx,Ty) 1 12 [ σ ( x , T x ) + σ ( y , T y ) + σ ( x , T y ) + σ ( y , T x ) ] .

Hence, the conditions of Corollary 1.21 (Theorem 1.20) hold and T has a fixed point in A 1 A 2 . Here, x=0 is a fixed point of T.

If in Theorem 1.20 we take A i =X for all 0im, then we deduce the following theorem.

Theorem 1.23 Let (X,σ) be a complete metric-like space and let T:XX be a sub-ψ-admissible mapping such that

σ ( T x , T y ) ψ ( x ) σ ( x , T x ) + ψ ( T x ) σ ( y , T y ) + ψ ( T 2 x ) σ ( x , T y ) + ψ ( T 3 x ) σ ( y , T x ) φ ( σ ( x , T x ) , σ ( x , T y ) , 1 2 [ σ ( x , T y ) + σ ( y , T x ) ] )

for any x,yX, where ψΨ and φΛ. Then T has a unique fixed point in X.

Corollary 1.24 Let (X,σ) be a complete metric-like space and let T:XX be a sub-ψ-admissible mapping such that

σ(Tx,Ty)β [ σ ( x , T x ) + σ ( y , T y ) + σ ( x , T y ) + σ ( y , T x ) ]

for any x,yX, where β[0, 1 6 ). Then T has a unique fixed point in X.

Example 1.25 Let X= R + with the metric-like σ(x,y)=max{x,y} for all x,yX. Let T:XX be defined by

Tx={ 1 14 ( x 3 + x ) if  0 x < 1 , 1 10 x 2 if  x 1 .

Proof To show the existence and uniqueness point of T, we investigate the following cases:

  • Let 0x,y<1. Then we get

    σ(Tx,Ty)=max { 1 14 ( x 3 + x ) , 1 14 ( y 3 + y ) } 1 7 max{x,y}= 1 7 σ(x,y).
  • Let x,y1. So we have

    σ(Tx,Ty)= 1 10 max { x 2 , y 2 } 1 10 max{x,y} 1 7 max{x,y}= 1 7 σ(x,y).
  • Let 0x<1 and y1. Then we obtain

    σ(Tx,Ty)=max { 1 14 ( x 2 + x ) , 1 10 y 2 } max { 1 7 x , 1 10 y } 1 7 max{x,y}= 1 7 σ(x,y),

and hence

σ(Tx,Ty) 1 7 [ σ ( x , T x ) + σ ( y , T y ) + σ ( x , T y ) + σ ( y , T x ) ] .

Then all the conditions of Corollary 1.24 (Theorem 1.23) are satisfied. Thus, T has a unique fixed point X. Indeed, 0 is the unique fixed point of T. □

Corollary 1.26 Let (X,σ) be a complete metric-like space, mN, let A 1 , A 2 ,, A m be nonempty σ-closed subsets of X and Y= i = 1 m A i . Suppose that T:YY is an operator such that

  1. (i)

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

  2. (ii)

    there exists β[0, 1 6 ) such that

    0 σ ( T x , T y ) ρ(t)dtβ 0 σ ( x , T x ) + σ ( y , T y ) + σ ( x , T y ) + σ ( y , T x ) ρ(t)dt

for any x A i , y A i + 1 , i=1,2,,m, where A m + 1 = A 1 , and ρ:[0,)[0,) is a Lebesgue-integrable mapping satisfying 0 ε ρ(t)dt for ε>0. Then T has a unique fixed point z i = 1 m A i .

If in Corollary 1.26, we take A i =X for i=1,2,,m, we obtain the following result.

Corollary 1.27 Let (X,σ) be a complete metric-like space and let T:XX be a mapping such that for any x,yX,

0 σ ( T x , T y ) ρ(t)dtβ 0 σ ( x , T x ) + σ ( y , T y ) + σ ( x , T y ) + σ ( y , T x ) ρ(t)dt,

where ρ:[0,)[0,) is a Lebesgue-integrable mapping satisfying 0 ε ρ(t)dt for ε>0 and the constant β[0, 1 6 ). Then T has a unique fixed point.

References

  1. 1.

    Hitzler, P: Generalized metrics and topology in logic programming semantics. PhD thesis, School of Mathematics, Applied Mathematics and Statistics, National University Ireland, University College Cork (2001)

  2. 2.

    Amini-Harandi A: Metric-like spaces, partial metric spaces and fixed points. Fixed Point Theory Appl. 2012., 2012: Article ID 204

  3. 3.

    Hitzler P, Seda AK: Dislocated topologies. J. Electr. Eng. 2000, 51(12):3–7.

  4. 4.

    Aage CT, Salunke JN: The results on fixed points in dislocated and dislocated quasi-metric space. Appl. Math. Sci. 2008, 2(59):2941–2948.

  5. 5.

    Aage CT, Salunke JN: Some results of fixed point theorem in dislocated quasi-metric spaces. Bull. Marathwada Math. Soc. 2008, 9: 1–5.

  6. 6.

    Daheriya RD, Jain R, Ughade M: Some fixed point theorem for expansive type mapping in dislocated metric space. ISRN Math. Anal. 2012., 2012: Article ID 376832

  7. 7.

    Sarma IR, Kumari PS: On dislocated metric spaces. Int. J. Math. Arch. 2012, 3(1):72–77.

  8. 8.

    Shrivastava R, Ansari ZK, Sharma M: Some results on fixed points in dislocated and dislocated quasi-metric spaces. J. Adv. Stud. Topol. 2012, 3(1):25–31.

  9. 9.

    Zeyada FM, Hassan GH, Ahmed MA: A generalization of a fixed point theorem due to Hitzler and Seda in dislocated quasi-metric spaces. Arab. J. Sci. Eng., Sect. A 2005, 31(1):111–114.

  10. 10.

    Zoto K, Hoxha E, Isufati A: Some new results in dislocated and dislocated quasi-metric spaces. Appl. Math. Sci. 2012, 6(71):3519–3526.

  11. 11.

    Zoto K, Hoxha E: Fixed point theorems for ψ -contractive type mappings in dislocated quasi-metric spaces. Int. Math. Forum 2012, 7(51):2503–2508.

  12. 12.

    Zoto K, Hoxha E: Fixed point theorems in dislocated and dislocated quasi-metric spaces. J. Adv. Stud. Topol. 2012, 3(4):119–124.

  13. 13.

    Kirk WA, Srinavasan PS, Veeramani P: Fixed points for mapping satisfying cyclical contractive conditions. Fixed Point Theory 2003, 4: 79–89.

  14. 14.

    Sintunavarat W, Kumam P: Common fixed point theorem for cyclic generalized multi-valued contraction mappings. Appl. Math. Lett. 2012, 25(11):1849–1855. 10.1016/j.aml.2012.02.045

  15. 15.

    Nashine HK, Sintunavarat W, Kumam P: Cyclic generalized contractions and fixed point results with applications to integral equation. Fixed Point Theory Appl. 2012., 2012: Article ID 217

  16. 16.

    Mongkolkeha C, Kumam P: Best proximity point theorems for generalized cyclic contractions in ordered metric spaces. J. Optim. Theory Appl. 2012. 10.1007/s10957-012-9991-y

  17. 17.

    Eldered AA, Veeramani P: Convergence and existence for best proximity points. J. Math. Anal. Appl. 2006, 323: 1001–1006. 10.1016/j.jmaa.2005.10.081

  18. 18.

    Eldered AA, Veeramani P: Proximal pointwise contraction. Topol. Appl. 2009, 156: 2942–2948. 10.1016/j.topol.2009.01.017

  19. 19.

    Chen CM: Fixed point theorems for cyclic Meir-Keeler type mappings in complete metric spaces. Fixed Point Theory Appl. 2012., 2012: Article ID 41

  20. 20.

    Karapınar E: Fixed point theory for cyclic weak ϕ -contraction. Appl. Math. Lett. 2011, 24: 822–825. 10.1016/j.aml.2010.12.016

  21. 21.

    Karapınar E, Erhan IM, Ulus AY: Fixed point theorem for cyclic maps on partial metric spaces. Appl. Math. Inf. Sci. 2012, 6(1):239–244.

  22. 22.

    Karapınar E, Sadarangani K:Fixed point theory for cyclic (ψϕ) contractions. Fixed Point Theory Appl. 2011., 2011: Article ID 69

  23. 23.

    Karpagam S, Agrawal S: Best proximity points theorems for cyclic Meir-Keeler contraction maps. Nonlinear Anal. 2011, 74: 1040–1046. 10.1016/j.na.2010.07.026

  24. 24.

    Karpagam S, Agrawal S: Existence of best proximity points of p -cyclic contractions. Fixed Point Theory 2012, 13(1):99–105.

  25. 25.

    Păcurar M, Rus IA: Fixed point theory for cyclic φ -contractions. Nonlinear Anal. 2010, 72(3–4):1181–1187. 10.1016/j.na.2009.08.002

  26. 26.

    Petruşel G: Cyclic representations and periodic points. Stud. Univ. Babeş-Bolyai, Math. 2005, 50: 107–112.

  27. 27.

    Rezapour S, Derafshpour M, Shahzad N: Best proximity point of cyclic φ -contractions in ordered metric spaces. Topol. Methods Nonlinear Anal. 2011, 37: 193–202.

  28. 28.

    Rus IA: Cyclic representations and fixed points. Ann. Tiberiu Popoviciu Semin. Funct. Equ. Approx. Convexity 2005, 3: 171–178.

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The authors thanks to anonymous referee for his remarkable comments, suggestion and ideas that helps to improve this paper.

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Karapınar, E., Salimi, P. Dislocated metric space to metric spaces with some fixed point theorems. Fixed Point Theory Appl 2013, 222 (2013) doi:10.1186/1687-1812-2013-222

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Keywords

  • dislocated metric spaces
  • metric-like spaces
  • fixed point