Abstract
In this paper, using the concept of -admissibility, we prove some fixed point results for interpolate Ćirić-Reich-Rus-type contraction mappings. We also present some consequences and a useful example.
MSC:
46T99; 47H10; 54H25
1. Introduction and Preliminaries
In [1], the notion of an interpolative Kannan-type contraction was introduced and the following fixed point theorem was stated:
A self-mapping T on a complete metric space such that:
where and , and with , has a unique fixed point in X.
Very recently, the authors in [2] (see also [3]) pointed out a gap in [1], that is the guaranteed fixed point in the theorem above need not be unique.
The next theorem and its invariants were considered and proven independently by L.B. Ćirić (Serbia), S. Reich (Israel), and I.A. Rus (Romania); see, e.g., [4,5,6,7,8,9,10,11]. Regarding the contributions of these authors, we shall call the following result the Ćirić-Reich-Rus theorem, by which our main result is inspired.
Ćirić-Reich-Rus theorem: A self-mapping T on a complete metric space such that:
for all , where , possesses a unique fixed point.
Denote by the set of all nondecreasing self-mappings on such that:
Note that for , we have and for each ; see, e.g., [10,12].
The notion of -orbital admissible maps was introduced by Popescu as a refinement of the concept of -admissible maps of Samet et al. [13].
Definition 1
([14]). Let be a mapping and . A self-mapping is said to be an ω-orbital admissible if for all , we have:
Many papers used and generalized this above concept in order to prove variant (common) fixed point results (see, for instance, [15,16,17,18,19,20,21,22,23,24]). In this setting, the following condition has often been considered in order to avoid the continuity of the involved contractive mappings.
- If is a sequence in X such that for each n and as , then there exists from such that for each k.
In this paper, using the notion of -admissibility, we initiate the idea of -interpolative Ćirić-Reich-Rus-type contraction. We also present some consequences and an example in support of our obtained result.
2. Main Results
First, we initiate the concept of -interpolative Ćirić-Reich-Rus-type contractions.
Definition 2.
Let be a metric space. The map is said to be an ω-interpolative Ćirić-Reich-Rus-type contraction if there exist , and positive reals , verifying , such that:
for all , where denotes the set of all fixed points of T (that is, points such that ).
The essential main result is given as follows.
Theorem 1.
Suppose a continuous self-mapping is ω-orbital admissible and forms an ω-interpolative Ćirić-Reich-Rus-type contraction on a complete metric space . If there exists such that , then T possesses a fixed point in X.
Proof.
Let be a point such that . Let be the sequence defined by , . If for some , we have , then is a fixed point of T, which ends the proof. Otherwise, for each . We have . Since T is -orbital admissible,
Continuing as above, we obtain that:
Taking and in (4), we find that:
In particular, as for each ,
We derive:
Therefore:
Hence, the positive sequence is decreasing. Eventually, there is a real in order that . Taking into account (9),
so (6) together with the nondecreasing character of lead to:
By repeating this argument, we get:
Taking in (10) and using the fact for each , we deduce that , that is,
We assert that is a Cauchy sequence, that is for all . On account of the triangle inequality together with (10), we find:
Letting in the inequality above, we conclude that the right-hand side tends to zero. Thus, is a Cauchy sequence. Regarding the completeness of the metric space , we deduce that there is some so that:
Since T is continuous, we have . ☐
In what follows, we replace the continuity criteria by a weakened condition .
Theorem 2.
Suppose a self-mapping is ω-orbital admissible and forms an ω-interpolative Ćirić-Reich-Rus-type contraction on a complete metric space . Suppose also that the condition is fulfilled. If there exists such that , then T possesses a fixed point in X.
Proof.
By the proof of Theorem 1 verbatim, we conclude that the constructed sequence is Cauchy and (11) holds. Suppose the condition holds. We argue by contradiction by assuming that . Recall that for each . Due to , there is a partial subsequence of such that for all k. Since , and , there is such that, for each ,
Taking and in (4), we get that:
As is nondecreasing, it follows from (12) that:
Letting , we find that:
which is a contradiction. Thus, . ☐
In what follows, we introduce the notion of -interpolative Kannan-type contractions.
Definition 3.
The self-mapping T on the metric space is called an ω-interpolative Kannan-type contraction if there exist , and such that:
for all .
The following one is our second main result.
Theorem 3.
Let be an ω-orbital admissible and ω-interpolative Kannan-type contraction mapping on a complete metric space . Assume also that either T is continuous on or holds. If there exists so that , then there exists a fixed point of T in X.
We skipped the proof due to the verbatim proof of Theorem 1.
By considering in Theorem 1, we state the following.
Corollary 1.
Let T be a self-mapping on a complete metric space such that:
for all , where are positive reals satisfying . Then, T admits a fixed point.
Corollary 2.
Let T be a self-mapping on a complete metric space such that:
for all , where . Then, T admits a fixed point in X.
Taking (where ) in Corollary 1, we state:
Corollary 3.
Let T be a self-mapping on a complete metric space such that:
for all , where are positive reals verifying and . Then, T has a fixed point in X.
Taking (where ) in Corollary 2, we state:
Corollary 4.
Let T be a self-mapping on a complete metric space such that:
for all , where and . Then, there exists a fixed point of T.
Remark 1.
Corollary 3 corresponds to Corollary 2.1 in [2].
Let be a partially-ordered metric space. Let us consider the following condition.
- If is a sequence in X such that for each n and as , then there exists a subsequence of such that for each k.
Following [23], we may state the following consequences of Theorem 1.
Corollary 5.
Let be a complete partially-ordered metric space. Let be the mapping such that:
for all with , where and are positive reals such that . Assume that:
- (i)
- T is nondecreasing with respect to ⪯;
- (ii)
- there exists such that ;
- (iii)
- either T is continuous on or holds.
Then, T has a fixed point in X.
Proof.
It suffices to take, in Theorem 1,
☐
Corollary 6.
Let be a complete partially-ordered metric space and be a given mapping satisfying:
for all with , where and . Assume that:
- (i)
- T is nondecreasing with respect to ⪯;
- (ii)
- there exists such that ;
- (iii)
- either T is continuous on or holds.
Then, T has a fixed point in X.
Proof.
We take in Theorem 3,
☐
Corollary 7.
Suppose that the subsets and of a complete metric space are closed. Suppose also that satisfies:
for all and , such that , where and are positive reals such that . If and , then there exists a fixed point of T in .
Proof.
It suffices to take, in Theorem 1,
☐
Corollary 8.
Let and be two nonempty closed subsets of a complete metric space . Suppose that satisfies:
for all and such that , where and . If and , then there exists a fixed point of T in .
Proof.
It suffices to take, in Theorem 3,
☐
Theorem 1 is supported by the following.
Example 1.
Let us consider the set endowed with . Let T be a self-mapping on X defined by:
Take:
Let be such that , and . Then, and . We have . Hence, (4) holds. For , we have:
Now, let be such that . It yields that , so . Hence, , that is T is ω-orbital admissible. Notice that T is not continuous. We shall show that holds. Let be a sequence in X such that for each . Then, . If as , we have as . Hence, , and so, . All conditions of Theorem 1 hold. Note that and are two fixed points of T.
Author Contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Funding
This research received no external funding.
Conflicts of Interest
The authors declare no conflict of interest.
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