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.
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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 such that for all :
(σ 1) if then ,
(σ 2) ,
(σ 3) ,
and the pair 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., [4–12]). Another interesting generalization of the Banach contraction mapping principle was given by Kirk et al. [13] via a cyclic mapping (see, e.g., [14–16]). 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 is called cyclic if and .
Theorem 1.2 (See [13])
Let A and B be two nonempty closed subsets of a complete metric space . Suppose that is cyclic and satisfies the following:
-
(C)
There exists a constant such that
Then T has a unique fixed point that belongs to .
Cyclic mappings and related fixed point theorems have been considered by many authors (see, e.g., [13–28]). 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 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 there exists such that . Also, is a σ-closed subset of X if is a σ-open subset of X.
Lemma 1.4 Let be a metric-like space and V be a σ-closed subset of X. Let be a sequence in V. If as , then .
Proof Let . By Definition 1.3, is a σ-open set. Then there exists such that . On the other hand, we have since as . Hence, there exists such that
for all . So, we conclude that for all . This is a contradiction since for all . □
Lemma 1.5 Let be a metric-like space and be a sequence in X such that as and . Then for all .
Proof From (σ 3) we have
Letting in the above inequalities, we get . □
Lemma 1.6 Let be a metric-like space. Then
-
(A)
if , then ;
-
(B)
if is a sequence such that , then we have
-
(C)
if , then ;
-
(D)
holds for all , where .
Proof We skip the proof (A) since it is evident.
(B) Due to the triangle inequality, we have . So, we find
Analogously, we derive
-
(C)
If and , then by (σ 1) we have , which is a contradiction.
-
(D)
Again from (σ 3) we get
where . Then we observe that
Hence, we derive that
□
At first, we define the class of Φ and Ψ by the following ways:
and
Definition 1.7 Let be a metric-like space, , let be σ-closed nonempty subsets of X and . We say that T is called a cyclic generalized ϕ-ψ-contractive mapping if
-
(1)
is a cyclic representation of Y with respect to T;
-
(2)
and
(1)
for any , , , where , , and
Let X be a nonempty set and be a given map. The set of all fixed points of T will be denoted by , that is, .
Theorem 1.8 Let be a complete metric-like space, , let be nonempty σ-closed subsets of X and . Suppose that is a cyclic generalized ϕ-ψ-contractive mapping. Then T has a fixed point in . Moreover, if for all , then T has a unique fixed point in .
Proof Let be an arbitrary point of Y. So, there exists some such that . Since , we conclude that . Thus, there exists in such that . Recursively, , where . Hence, for , there exists such that . In case for some , then it is clear that is a fixed point of T. Now assume that for all n. Hence, by Lemma 1.6(C) we have for all n. We shall show that the sequence is non-increasing where . Assume that there exists some such that
Hence
By taking and in condition (1) together with (2), we get
On the other hand, from Lemma 1.6(D) we have
and by (σ 3) we have
That is,
Then
Therefore from (3) we get
Now, if , then
a contradiction. Hence, we have
for all . By taking and in (4) and keeping (2) in mind, we deduce that
a contradiction. Hence, we conclude that holds for all . Thus, there exists such that . We shall show that by the method of reductio ad absurdum. For this purpose, we assume that . By (4), together with the properties of ϕ, ψ, we have
which yields that . This is a contradiction. Hence, we obtain that
We shall show that 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:
-
(K)
For every , there exists such that if with , then .
Suppose, on the contrary, that there exists such that for any , we can find with satisfying
Now, we take . Then, corresponding to , we can choose in such a way that it is the smallest integer with satisfying and . Therefore, . By using the triangular inequality, we obtain
Passing to the limit as in the last inequality and taking (5) into account, we obtain that
Again, by (σ 3), we derive that
Taking (5) and (7) into account, we get
By (σ 3), we have the following inequalities:
and
Letting in (9) and (10), we derive that
Again by (σ 3) we have
and
Letting in (12) and (13), we derive that
Since and lie in different adjacently labeled sets and for certain , by using (5), (7), (8), (11), (14) together with the fact that T is a generalized cyclic ϕ-ψ-contractive mapping, we find that
Regarding the properties of ϕ, ψ in the last inequality, we obtain that
a contradiction. Hence, the condition (K) is satisfied. Fix . By the claim, we find such that if with ,
Since , we also find such that
for any . Suppose that and . Then there exists such that . Therefore, for . So, we have for ,
By (15) and (16) and from the last inequality, we get
This proves that is a σ-Cauchy sequence. Since ε is arbitrary, is a Cauchy sequence. Since Y is σ-closed in , then is also complete, there exists such that in ; equivalently
In what follows, we prove that x is a fixed point of T. In fact, since and, as is a cyclic representation of Y with respect to T, the sequence has infinite terms in each for . Suppose that , , and we take a subsequence of with (the existence of this subsequence is guaranteed by the above-mentioned comment). By using the contractive condition, we can obtain
Passing to the limit as and using , lower semi-continuity of φ, we have
So, and, therefore, x is a fixed point of T. Finally, to prove the uniqueness of the fixed point, suppose that are two distinct fixed points of T. The cyclic character of T and the fact that are fixed points of T imply that . Suppose that and for all , . Using the contractive condition, we obtain
Then we have
which is a contradiction. Thus, we derive that , which finishes the proof. □
If in Theorem 1.8 we take for all , then we deduce the following theorem.
Theorem 1.9 Let be a complete metric-like space and T be a self-map on X. Assume that there exist and such that
for all , where
Then T has a fixed point. Moreover, if for all , then T has a unique fixed point.
If in Theorem 1.8 we take and , where , then we deduce the following corollary.
Corollary 1.10 Let be a complete metric-like space, , let be nonempty σ-closed subsets of X and . Suppose that is an operator such that
-
(i)
is a cyclic representation of X with respect to T;
-
(ii)
there exists such that
for any , , , where . Then T has a fixed point . Moreover, if for all , then T has a unique fixed point.
Example 1.11 Let with the metric-like for all . Suppose and and . Define by
It is clear that is a cyclic representation of Y with respect to T. Let and . Then
and so
Hence, the conditions of Corollary 1.10 (Theorem 1.8) hold and T has a fixed point in . Here, is a fixed point of T.
If in the above corollary we take for all , then we deduce the following corollary.
Corollary 1.12 Let be a complete metric-like space and T be a self-map on X. Assume that there exists such that
holds for all . Then T has a fixed point. Moreover, if for all , then T has a unique fixed point.
Example 1.13 Let with the metric-like for all . Let be defined by
Proof To show the existence and uniqueness point of T, we need to consider the following cases.
-
Let . Then
-
Let . Then
-
Let . Then
-
Let and . Then
-
Let and . Then
-
Let and . Then
and so
Hence, we conclude that all the conditions of Corollary 1.12 (Theorem 1.9) hold and hence T has a fixed point 0 in . □
By Corollary 1.10 we deduce the following result.
Corollary 1.14 Let be a complete metric-like space, , let be nonempty σ-closed subsets of X and . Suppose that is an operator such that
-
(i)
is a cyclic representation of X with respect to T;
-
(ii)
there exists such that
for any , , , where , and is a Lebesgue-integrable mapping satisfying for . Then T has a fixed point . Moreover, if for all , then T has a unique fixed point.
If in the above corollary we take for all , then we deduce the following corollary.
Corollary 1.15 Let be a complete metric-like space and let be a mapping such that for any ,
where is a Lebesgue-integrable mapping satisfying for and the constant . Then T has a unique fixed point.
Definition 1.16 Let and and . A mapping T is said to be a γ-ψ-subadmissible mapping if
Example 1.17 Let and be defined by and . Then T is a γ-ψ-subadmissible mapping where . Indeed, if , then , and hence . That is, .
Example 1.18 Let and be defined by and . Then T is a γ-ψ-subadmissible mapping where . Indeed, if , then , and hence . That is, .
Let Λ be the class of all the functions that are a continuous with the following property:
Definition 1.19 Let be a metric-like space, , let be σ-closed nonempty subsets of and . Assume that is a γ-ψ-subadmissible mapping where . Then T is called a ψ-cyclic generalized weakly C-contraction if
-
(1)
is a cyclic representation of Y with respect to T;
-
(2)
(18)
for any , , , where and .
Theorem 1.20 Let be a complete metric-like space, , let be nonempty σ-closed subsets of and . Suppose that is a ψ-cyclic generalized weakly C-contraction. If there exists such that , then T has a fixed point . Moreover, if , then z is unique.
Proof Let be such that . Since T is a sub ψ-admissible mapping with respect to , then . for all . Also, there exists some such that . Now implies that . Thus there exists in such that . Similarly, , where . Hence, for , there exists such that and . In case for some , then it is clear that is a fixed point of T. Now assume that for all n. Since is a cyclic generalized weak C-contraction, we have that for all ,
and so
On the other hand, from (σ 3) we have
and by Lemma 1.6(D) we have
Then by (19) we get
Therefore,
for any . Set . On the occasion of the facts above, is a non-increasing sequence of nonnegative real numbers. Consequently, there exists such that
We shall prove that . Since , then . Similarly, . Then
On the other hand, by taking limit as in (19), we have
which implies
Hence,
Now, from (18) we have
By taking limit as in the above inequality, we deduce
and so . Since , we get . Due to and , we have
We shall show that is a σ-Cauchy sequence. At first, we prove the following fact:
-
(K)
For every , there exists such that if with , then .
Suppose to the contrary that there exists such that for any , we can find with satisfying
Following the related lines of the proof of Theorem 1.8, we have
and
Since and lie in different adjacently labeled sets and for certain , using the fact that T is a ψ-cyclic generalized weakly C-contraction, we have
Now, by taking limit as in the above inequality, we derive that
which is a contradiction. Hence, condition (K) holds. We are ready to show that the sequence is Cauchy. Fix . By the claim, we find such that if with ,
Since , we also find such that
for any . Suppose that and . Then there exists such that . Therefore, for . So, we have, for , ,
By (25) and (26) and from the last inequality, we get
This proves that is a σ-Cauchy sequence.
Since Y is σ-closed in , then is also complete, there exists such that in ; equivalently
In what follows, we prove that x is a fixed point of T. In fact, since and, as is a cyclic representation of Y with respect to T, the sequence has infinite terms in each for . Suppose that , , and we take a subsequence of with (the existence of this subsequence is guaranteed by the above-mentioned comment). By using the contractive condition, we can obtain
Passing to the limit as and using , lower semi-continuity of φ, we have
So, and, therefore, x is a fixed point of T. Finally, to prove the uniqueness of the fixed point, suppose that are fixed points of T. The cyclic character of T and the fact that are fixed points of T imply that . Also, suppose that . By using the contractive condition, we derive that
Then
This gives us , that is, . This finishes the proof. □
Corollary 1.21 Let be a complete metric-like space, , let be nonempty σ-closed subsets of X and . Suppose that is an operator such that
-
(i)
is a cyclic representation of X with respect to T;
-
(ii)
there exists such that
(28)
for any , , , where . Then T has a fixed point .
Proof Let and . Here, it suffices to take the function defined by . Obviously, φ satisfies that if and only if , and . Then we apply Theorem 1.20 to finish the proof. □
Example 1.22 Let with the metric-like for all . Suppose and and . Define by
It is clear that is a cyclic representation of Y with respect to T.
Let and . Then
and so
Hence, the conditions of Corollary 1.21 (Theorem 1.20) hold and T has a fixed point in . Here, is a fixed point of T.
If in Theorem 1.20 we take for all , then we deduce the following theorem.
Theorem 1.23 Let be a complete metric-like space and let be a sub-ψ-admissible mapping such that
for any , where and . Then T has a unique fixed point in X.
Corollary 1.24 Let be a complete metric-like space and let be a sub-ψ-admissible mapping such that
for any , where . Then T has a unique fixed point in X.
Example 1.25 Let with the metric-like for all . Let be defined by
Proof To show the existence and uniqueness point of T, we investigate the following cases:
-
Let . Then we get
-
Let . So we have
-
Let and . Then we obtain
and hence
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 be a complete metric-like space, , let be nonempty σ-closed subsets of X and . Suppose that is an operator such that
-
(i)
is a cyclic representation of X with respect to T;
-
(ii)
there exists such that
for any , , , where , and is a Lebesgue-integrable mapping satisfying for . Then T has a unique fixed point .
If in Corollary 1.26, we take for , we obtain the following result.
Corollary 1.27 Let be a complete metric-like space and let be a mapping such that for any ,
where is a Lebesgue-integrable mapping satisfying for and the constant . Then T has a unique fixed point.
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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). https://doi.org/10.1186/1687-1812-2013-222
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DOI: https://doi.org/10.1186/1687-1812-2013-222