Abstract

We admonish to be careful on studying coupled fixed point theorems since most of the reported fixed point results can be easily derived from the existing corresponding theorems in the literature. In particular, we notice that the recent paper [Semwal and Dimri (2014)] has gaps and the announced result is false. The authors claimed that their result generalized the main result in [Ðoric and Lazović (2011)] but, in fact, the contrary case is true. Finally, we present a fixed point theorem for Suzuki type (, r)-admissible contractions.

1. Introduction and Preliminaries

Throughout this note, we follow the notions and notations given in [1, 2]. Let be the mapping defined, for all , by Let be a metric space. We denote by (or by when it is convenient to clarify the involved metric) the class of all nonempty closed and bounded subsets of . For every , , let where for all and all . It is well known that is a metric on .

Very recently, Semwal and Dimri [1] announced the following result.

Theorem 1 (Semwal and Dimri [1], Theorem 2.1). Let be a complete metric space and let be mapping from into . Assume that there exists such that implies for all , , , . Then there exist , such that and .

Semwal and Dimri [1] claimed that it was a generalization of Đoric and Lazović’s recent result (see [2]), which is the following theorem.

Theorem 2 (Đoric and Lazović [2], Theorem 2.1). Let be a complete metric space and let be mapping from into . Assume that there exists such that implies for all . Then, there exists such that .

This note is devoted to the following three aims. We will show that the proof of the main result of Semwal and Dimri [1] is incorrect; in fact, it is possible to fix the glitch of the given proof in [1]. By modifying its contractivity condition, we obtain a correct version of Theorem 1 but, in such a case, we realize that the obtained result is a simple consequence of Theorem 2. Finally, we present a generalization of Theorem 2 for Suzuki type -admissible contractions.

2. Main Gaps

Let us review the lines of their proof. First at all, notice that, in general, if is a metric space, we know that, for all , , , all and all : The contrary inequalities can be false.

The authors took , arbitrarily and, later, they chose Taking into account that and using (7), the authors wrote the following (see [1], page 3, line 17): However, this inequality is not strong enough to apply the contractivity condition given in Theorem 1 because in the antecedent condition cannot appear the points and in the second member if they are not in the first member. Therefore, the contractivity condition that we found in Theorem 1 is not applicable.

Furthermore, assume that we would have been able to apply the mentioned contractivity condition. In this case, the authors wrote the following (see [1], page 3, lines 19–22): Immediately, they deduced that However, this inequality is based on which, in general, is false. Then, this argument is not correct.

The same mistake occurred when the authors tried to upper bound the terms and (see [1], page 4).

3. A Correct Version of Theorem 1

If we want to modify the contractivity condition given in Theorem 1 in order that (9) can be applied, then antecedent condition (3) must be replaced by the following one: for all , , , . In such a case, we obtain the following result.

Theorem 3. Let be a complete metric space and let be a mapping. Assume that there exists such that implies for all , , , . Then there exist , such that and .

However, we claim that this result is not a proper generalization of Theorem 2, but it is an immediate consequence of such theorem. To prove it, we need some preliminaries.

Lemma 4 (see, e.g., [3, 4]). Given a metric on , define , for all , by Then is a metric on . In addition to this, if is complete and then is also complete.

Remark 5. Notice that .

Given a mapping , denote by the mapping If there exists a point such that , then there exist two points such that and . This is precisely the thesis of Theorem 3. Therefore, we only have to prove that has a fixed point .

Notice that antecedent condition (14) can be written as Moreover, the second member in (15) is Notice that, associated to the metric , we can also consider given, for all , , by In such a case,

Theorem 6. Theorem 3 is a consequence of Theorem 2.

Proof. It is evident from (17) that (14) and (15) are equivalent to (5) and (6). Regarding Lemma 4, Remark 5, and the observations above, we conclude that, under the conditions of Theorem 3, all hypotheses of Theorem 2 are satisfied.

4. A Fixed Point Theorem for Suzuki Type -Admissible Contractions

In this section, we introduce a generalization of Theorem 2 using a slightly different kind of contractivity condition. We use the following preliminaries. Let be a metric space. Given a mapping , let be the mapping for all , . We say that the mapping is transitive if and implies that .

Definition 7 (see [5]). Let be a metric space and let be a mapping. We say that a mapping is -admissible if for all , such that .
We say that the metric space is -regular if for all provided that is a sequence such that and for all .

Remark 8. If for all , then is transitive, any metric space is -regular and any mapping is -admissible.

Definition 9. Let be a metric space, let be a multivalued mapping, let be a mapping, and let . One says that is a Suzuki type -admissible contraction if implies for all , .

In the following theorem, we will use the following condition, which can be verified for .

: if is a sequence such that verifying for all and , then there exist and such that

We must clarify that this condition is always satisfied when .

Lemma 10. If is a metric space and verifies for all , then condition holds for all and all .

Proof. Since and is closed, then . If , there is nothing to prove. Assume that and let be any positive real number in the interval As is an infimum, there exists such that . Therefore

Theorem 11. Let be a complete metric space and let be a Suzuki type -admissible multi-valued contraction from into . Suppose also that(i)is -admissible;(ii)there exist and such that ;(iii)at least, one of the following properties holds:(iii.1) is continuous, or(iii.2) is -regular and
Then has, at least, a fixed point; that is, there exists such that .

Taking into account Remark 8, this result admits Theorem 2 as a particularization to the case in which for all . Notice that the following proof is a slightly modified version of the proof of Theorem 2.1 in [2] using .

Proof. We follow the lines of the proof of Theorem 1.2 in [2], doing slight changes due to mapping . Let be arbitrary.
Step 1. There exists a sequence such that, for all , Starting from and such that , we notice that because is -admissible. If , then , so is a fixed point of and the proof is finished. On the contrary, assume that . As we can apply contractivity condition (24) and we deduce that As , and . Moreover, Hence, If , then the maximum is , and we have . As , we deduce . Therefore, , and as is closed, we conclude and the proof is finished. Suppose, on the contrary, that . In such a case, (33) means that Since is an infimum and , there exists such that Furthermore, Repeating this argument, either there exists such that (in this case, and the proof is finished) or there exists a sequence verifying (29).
Step 2. There exists such that . This fact is a consequence of being . Following a classical argument, it is easy to prove that is Cauchy in and, therefore, by the completeness, it is convergent.
Step 3. Assume that is continuous. In such a case, we have that ; that is, . By (7), it follows that for all (because ), and, taking limit as , we deduce that . Therefore, , and as is closed, we conclude and the proof is finished.
Step 4. Assume that is -regular and condition (28) holds.
In this case, using that is -regular, we have that and taking into account that is -admissible, we also have that If , the proof is also finished in this case. Therefore, we assume that and we will get a contradiction.
Next, we are going to show the following claim: Let be such that for all . As is -admissible, As and , there exists such that Therefore, for all , As we can apply contractivity condition (24), we obtain that, for all , Letting , we deduce that Assume that the maximum value is the last term. In that case, This proves that if the maximum in (46) is , then it is also true that . Therefore, in any case, we have that and it follows that (41) holds.
Next we distinguish between the cases and .
Case 4.1. Assume that and condition holds. Then, there exists and such that If , the proof is finished. On the contrary, assume that ; that is, . As and for all , property (41) guarantees that Since , we have that , and we can use contractivity condition (24). Notice that, as , , and therefore As and , If we suppose that , then , and the previous inequality leads to , so , which is false. Then and (50) implies that However, in this case, using the last two inequalities, (7) and (49), which is a contradiction. Then, we must admit that and the proof is finished in this case.
Case 4.2. Assume that and (or ) is transitive. We claim that, in this case, for all such that , we have that If , there is nothing to prove. Assume that . As is -admissible, Using (38) and the transitivity of , and, as is -admissible, then (The same conclusion is valid if is transitive.) By (41) and (58), we have that
Given , as and is an infimum, there exists a sequence such that Therefore, by (60) and (61), Letting , we deduce that
Two cases can be considered. If , then the previous inequality means that Therefore Hence, we can use contractivity condition (24), which means, using (57), that which guarantees that (56) holds.
On the contrary, assume that . Hence, inequality (63) yields that is, As we can also apply contractivity condition (24), reasoning as in (66), we deduce that, in this case, (56) holds.
In any case, using (56), we deduce that As , it follows that , so , which is the desired contradiction.

If we take for all , then we have the following result (using Remark 8).

Corollary 12. Theorem 2 follows from Theorem 11.

In the next example we show that Theorem 11 improves Theorem 2.

Example 13. Let be provided with the Euclidean metric for all . Define and by Clearly, is continuous and, if we identify , then Let and . Then Any number verifies . However, condition is false. Therefore, Theorem 2 cannot be applied to deduce that has a fixed point because is not contractive in the sense of Theorem 2. However, let . We will show that Theorem 11 is applicable.
Indeed, let , be such that . If , then there is nothing to prove. Assume that . In this case, , which means that , , and . Taking into account that then condition is obvious. Therefore, Theorem 11 guarantees that has a fixed point.

Conflict of Interests

The authors declare that they have no competing interests regarding the publication of this paper.

Authors’ Contribution

All authors contributed equally and significantly in writing this paper. All authors read and approved the final paper.

Acknowledgments

This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, under Grant no. 55-130-35-HiCi. The authors, therefore, acknowledge technical and financial support of KAU. The fourth author has been partially supported by Junta de Andaluca by Project FQM-268 of the Andalusian CICYE.