Some Noncontractive Maps on Incomplete Metric Spaces Have Also Fixed Points

Abstract

In this article, we discuss a new version of metric fixed point theory. The application of this newly introduced concept is to find some fixed point results where many well-known results in literature cannot be applied. We give some examples to illustrate the given concepts and obtained results.

1. Introduction

The fixed point theory has a long history. After the Banach contraction principle [1], there has been a huge development in metric fixed point theory. This principle has been generalized and extended by many researchers, either by changing the contraction condition or the underlying space. For more details, see [2–11]. The theory of fixed points in ordered sets was started first by Turinici [12]. In 2004, Ran and Reurings [13] have generalized the Banach contraction principle in the setting of ordered sets. The key feature in Ran-Reurings theorem is that the contractive condition on the nonlinear map is only assumed to hold on the comparable elements instead of the whole space as in Banach contraction principle. In 2005, Nieto, Rodríguez-López [14] proved a fixed point theorem by relaxing some conditions in Ran-Reurings [13]. In 2008, Suzuki [15] proved a fixed point theorem by assuming contraction condition on those elements which satisfy the given condition. Besides all these results, there exist various maps on metric spaces which possess a fixed point. This is because either the underlying metric space is not complete or the contraction condition is not satisfied. In this paper, we have tackled both the problems in setting of ordered metric spaces.

First, we recall some well-known results.

Theorem 1 ((Banach contraction principle) [1]). Let be a complete metric space and be a self-mapping such that for all ,where . Then has a unique fixed point in .

Theorem 2 ((Ran-Reurings theorem) [13]). Let be a partially ordered complete metric space such that every pair has a lower bound and an upper bound. Suppose is a continuous and monotone map (, either order-preserving or order-reversing ) such that(1) for all , where ;(2)there exists such that or . Then has a unique fixed point, say . Moreover, for every ,

Theorem 3 ((Suzuki) [15]). Let be a complete metric space and be a self-map. Define by Assume there exists such thatfor all . Then has a unique fixed point in .

Definition 4 (see [12]). A sequence in an ordered set is said to be increasing or ascending (resp. strictly increasing) if for , (resp. and . We denote it by ).

Definition 5 (see [16]). An ordered metric space is said to be -complete, if every increasing Cauchy sequence in converges in . In an ordered metric space, completeness implies -completeness.

In this paper, we introduce a new contraction condition which is assumed to hold for comparable elements of a subset of whole space. Our result guarantees the existence of a fixed point in such cases where neither Banach contraction principle nor Ran-Reurings and other theorems can be applied. We prove that, under certain conditions, noncontractive maps on incomplete metric spaces have also fixed points. We give examples to illustrate our concepts and obtained results. We also discuss some classes of contraction maps.

2. Main Results

First, we present the following definitions along with some examples.

Definition 6. Let be an ordered set and be a self-map. A subset is said to be a -subset of with respect to if

Example 7. Let be equipped with the natural ordering . Define by . The following are -subsets of with respect to :
;
.

Example 8. Let . Equip with the ordering defined as iff for each and define by . Then , is a -subset of with respect to .

Definition 9. Let be a partially ordered metric space and be a self-map. For any subset of , is said to be a -contraction with respect to if for all with , we havewhere .

The following examples illustrate Definition 9.

Example 10. Let us take . Endow with the usual metric of and the natural ordering . Let us consider the subset defined by . Then . Define by For any with , we have , and . Take . For such , we get Hence is a -contraction with respect to .

Example 11. Let be endowed with the usual metric of and the natural ordering . Define by and take . Clearly, is a -contraction with respect to .

Our first main result is as follows.

Theorem 12. Let be an -complete ordered metric space and be a self-map. Let be any nonempty -subset with respect to . Suppose that (a) is a -contraction with respect to ;(b) is continuous;(c). Then has at least one fixed point in .

Proof. The subset is nonempty. Let , so . If , the proof is completed. Otherwise, choose . By assumption (c), we have . By definition of a -set, . If , the proof is completed. Otherwise, choose . Therefore, . Continuing in this process, we get a strictly increasing sequence such thatAs with , then by (6) we haveAgain, as with , we haveUsing (10) in (11), we haveContinuing in this way, we getNow, we show that is a Cauchy sequence. For , by using triangular inequality, (9) and (13), we get This shows that is an increasing Cauchy sequence in and hence in . But is -complete; therefore, there exists such thatSince is continuous, we haveTaking in (13), we haveSo . By using (15) and (16), Thus, is a fixed point of in .

Now, we present an example illustrating Theorem 12, where Banach contraction principle, Ran-Reurings theorem, Suzuki theorem, and other results cannot be applied.

Example 13. Let be endowed with the usual metric and the natural ordering . Obviously, is an -complete ordered metric space. Define by Thus is a continuous mapping on . Considering the subset as , then . Clearly, is a -subset of with respect to and . Now, we show that is a -contraction with respect to . For any with , we have , , and . Set . For such , we get Hence is a -contraction with respect to . Thus all the conditions of Theorem 12 are satisfied, and has a fixed point.

Our second main result is as follows.

Theorem 14. Let be a -complete ordered metric space and be a self-map. Let . Suppose that(i);(ii) is nondecreasing on ;(iii) is continuous on ;(iv)there exists such that ;(v) is a -contraction with respect to . Then has at least one fixed point in .

Proof. As with . If , the proof is completed. Otherwise, choose such that . Continuing this process and using monotonicity of in , we get a strictly increasing Cauchy sequence in such that . As , using (6), we haveAgain as with , we haveUsing (20) in (21), we get Continuing in this way, we getAs Theorem 12, is an increasing Cauchy sequence in and hence in . But is -complete, so there exists such thatSince is continuous on ,By taking in (23) and using continuity of metric , we getThenBy using (24) and (25), we get , and so is a fixed point of in .

The following examples show that Theorem 12 cannot be applied, while the existence of a fixed point can be obtained using Theorem 14.

Example 15. Let be endowed with the usual metric and the natural ordering . Then is an -complete metric space. Let us define by Clearly, is a continuous function. Note that neither Banach contraction principle, nor Ran-Reurings theorem, nor Suzuki result can be applied. Now, we show that Theorem 14 can work in this case. Choose the subset such thatThen . Clearly, is not a -subset with respect to . Thus, Theorem 12 cannot be applied. Now, and is nondecreasing in . So, it remains to prove that (6) is satisfied. For any with , we have and . Choose . We haveThus, all the conditions of Theorem 14 are satisfied, and so has a fixed point in .

Example 16. Let be endowed with the usual metric and the natural ordering . Then is an -complete metric space. Let us define byMention that is a continuous function on . Choose the subset such thatThen . Clearly, and is nondecreasing in . Now, it remains to prove that (6) is satisfied. For any with , we have and . Choose . We have Thus, all the conditions of Theorem 14 are satisfied, so has a fixed point in .

Corollary 17. If all the conditions of Theorem 12 are satisfied, then the fixed point of exists and lies in (closure of ).

Proof. Theorem 12 guarantees the existence of a fixed point in . From (15), is a sequence in converging to ; therefore, .

Theorem 18. Let be an ordered metric space not necessarily complete and be a continuous self-map such that is -complete and for all . Suppose thatwhere . Then has at least one fixed point in . Also, every strict upper bound of fixed points of in is also a fixed point of .

Proof. Let . By assumption, . If , the proof is completed. Otherwise, let . Then and we have . If , again the proof is completed. Otherwise, continuing in the process, we get an increasing sequence in such that Now, with ; then, by (33), we haveAgain, as with , we have Using (34), we getContinuing in this way, we getAs Theorem 12, is an increasing Cauchy sequence in , which is -complete, so there exists such thatThe continuity of implies thatLetting in (37) and using (38) and (39), we getHence . We deduce that . Thus is a fixed point of in .
Now, let be any strict upper bound of in , , such that . By (33), we have This implies that , so is also a fixed point of in .

Example 19. Let be equipped with the natural ordering and the usual metric . Let be a self-map defined byClearly, is continuous and is an -complete metric space. It can be seen that satisfies all the assumptions of Theorem 18, and so there exists a fixed point of in . It can be seen that the existing known classical fixed point results in literature cannot be applied.

Corollary 20. If, in Theorem 18, we replace the -completeness of by the fact that (closure of ) is complete, then has a fixed point.

Proof. From (15), we have proved that is a Cauchy sequence in and hence in . Since is complete, converges to . We have proved in Theorem 12 that is a fixed point.

Remarks. Let be a partially ordered metric space. ConsiderandWe have . Further, let be a -subset of such that . Define We shall show that provided that exists.

For this, let . Suppose there exists such that and is a -subset with respect to . Then for all with . In particular,As , and . Also, implies that . Since (46) holds for all such that , we can replace by . From (46), we get This shows that . Hence . It should be noted that the converse is not true in general.

3. Conclusion

In this article, we have introduced new contraction type mappings assumed to hold only on comparable elements of a subset of whole space. By using this concept, we have guaranteed the existence of a fixed point in such cases where Banach contraction principle, Ran-Reurings theorem, Suzuki theorem, and others remain silent. The field for applying this result is not restricted to only contractive maps. This result can be applied to expansive and nonexpansive maps too. Further, it is clear that the fixed point is unique if the -subset of with respect to is connected and either bounded above or bounded below, but not both. However, proof for uniqueness needs further explorations.

Inclosing, we want to bring to the reader attention the following open questions.

Question 1. Is it possible to replace the continuity hypothesis in Theorems 12 and 18, by a weaker condition?

Question 2. In Theorems 12 and 18, under what condition we will have uniqueness of the fixed point? If such condition exists, does it give us uniqueness in an orbit or in the whole space?

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Authors’ Contributions

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

Acknowledgments

The third author would like to thank Prince Sultan University for funding this work through research group Nonlinear Analysis Methods in Applied Mathematics (NAMAM) group number RG-DES-2017-01-17.