Journal of Function Spaces
Volume 2020 (2020), Article ID 5393521, 5 pages
https://doi.org/10.1155/2020/5393521

Research Article

New Results for Set-Valued Mappings on Ordered Metric Spaces

F. Sabetghadam,¹ H. P. Masiha,¹ and H. Aydi2,3,4

¹Faculty of Mathematics, K. N. Toosi University of Technology, P.O. Box 16315-1618, Tehran, Iran
²Nonlinear Analysis Research Group, Ton Duc Thang University, Ho Chi Minh City, Vietnam
³Faculty of Mathematics and Statistics, Ton Duc Thang University, Ho Chi Minh City, Vietnam
⁴China Medical University Hospital, China Medical University, Taichung 40402, Taiwan

Correspondence should be addressed to H. Aydi; hassen.aydi@tdtu.edu.vn

Received 6 February 2020; Accepted 22 August 2020; Published 24 September 2020

Academic Editor: Lars E. Persson

Copyright © 2020 F. Sabetghadam et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

We establish some new fixed point results for order-closed multivalued mappings in complete metric spaces endowed with a partial order.

1. Introduction

In 1976, an order relation was defined in metric spaces [1]. Later, many researchers proved various fixed point results in this setting (see [2–11]), while the authors in [4, 5, 9] considered coupled questions for so-called monotone conditions. Zhang [12] proved some (coupled) fixed point theorems for multivalued mappings with monotone conditions in metric spaces with a partial order. Since then, Agarwal and Khamsi [13] extended Caristi’s fixed point to vector-valued metric spaces. Also, Chung [14] considered nonlinear contraction mappings. Nadler [15] defined multivalued contractions, and Assad and Kirk [16] proved fixed point theorems for set-valued mappings (see also the related works [17–22]).

In this paper, we present some fixed point results in ordered metric spaces for order-closed multivalued operators. First, we need some facts.

Lemma 1. [23] Let be a metric space and be a functional. Consider a nondecreasing, continuous, and subadditive function so that Take the relation “” on given as

Then, “” is a partial order relation on . Apparently, then .

Here, we state some definitions. Let be a topological space. Denote by the family of nonempty subsets of . Let be a partial order on .

Definition 2. [24] Given two nonempty subsets of .

Note that and are different relations between and (see Remark 114 of [24]). Also, , , and are not partial orders on (see Remark 2 of [24]).

Definition 3. If satisfies or , then is called a monotone sequence.

Definition 4. A multivalued operator is said to be order-closed if, for monotone sequences and , we have , , and imply .

Definition 5. A function is said to be order upper (lower) semicontinuous, if, for a monotone sequence with , we have

Note that an upper (lower) semicontinuous function is an order upper (lower) semicontinuous. But the converse is not true (see Remark 3 of [24]).

2. Main Results

Let be a metric space. For , define the partial order “” on induced by and in Lemma 1.

Theorem 6. Let be a complete ordered metric space and be a bounded below function. Suppose that is order-closed with respect to “” so that

Then, there is a monotone sequence , so that for , and converges to , which is a fixed point of . If in addition, is order lower semicontinuous on , then for each .

Proof. By the condition , take . From , there is so that . Again, from , there is with . Continuing this procedure, we have an increasing sequence so that . Therefore,

That is, the real sequence is decreasing, so it is a convergent sequence because is bounded from below on .
Using Remark 3 of [25] and Remark 2 of [26], we find that

provided that is subadditive. Thus,

By (7), there are and so that

Since is nondecreasing, we have for each .
Let . Then

i.e.,

Therefore, we have

which together with (8) implies that

Note that is convergent; then, there is so that for all ,

Moreover, by (12), we get

The convergence of implies that is a Cauchy sequence. By the completeness of , there is so that as . Since is order-closed, is monotone and . Consequently,

i.e., is a fixed point of .
If is order lower semicontinuous on , by definition of “,” then we have for each

It yields that . The proof is completed.

The proof of the following theorem carries over in the same manner as for Theorem 6.

Theorem 7. Let be a complete ordered metric space and be a bounded below function. Suppose that is order-closed with respect to “.” Assume that

Then, there is a monotone sequence , , , such that converges to , which is a fixed point of . If, in addition, is order upper semicontinuous on , then for all .

Theorem 8. Let be a complete ordered metric space and be bounded below function. Suppose that is order-closed with respect to “.” Assume that

(i)for each with (ii)there exists such that, .

Then, has a fixed point , and there is a sequence with for , such that . Moreover, if is order lower semicontinuous, then for all .

Proof. Since , by , we can choose so that . This implies that , by definition of , there is so that . Continuing this procedure, we can find an increasing sequence such that . The rest of the proof is similar as in Theorem 6.

The following supports Theorem 8.

Example 9. Let and , for . The metric space is complete. Consider the multivalued mapping given as

where is the set of rational numbers. Let and . Note that is bounded from below. Take the order induced by , that is, given as follows:

Mention that verifies the following assertions:

(a)for each with , we have (b)(c) is order-closed on .

Hence, has a fixed point on , which is .

Theorem 10. Let be a complete ordered metric space and be a bounded below function. Suppose is order-closed with respect to “.” If the following conditions hold:

(i)for each with (ii)there exists such that,

then has a fixed point . Also, there is a sequence with for any such that . Moreover, if is order upper semicontinuous, then for all .
The following example illustrates Theorem 10.

Example 11. Let , , for and for each . Here, is a complete metric space, and is a bounded below function. Consider the order induced by :

Clearly, this partial order is the usual order on . Define by . It is obvious that satisfies the following:

(a)for each with (b)(c) is order-closed on .

Hence, has a fixed point on , which is .

Now, a multivalued version of Theorem 2 of [27] may be obtained. Here, the considered multivalued mapping is not necessarily continuous.

Theorem 12. Let be a complete ordered metric space and be a continuous function bounded below. Let be a multivalued mapping. Suppose that

Then, has a fixed point.

Proof. Set

We claim that has a maximum element. For a directed set , let be a totally ordered subset in . For with , the fact that yields . Due to the fact that is bounded below, is a convergent set in . Consider

As in proof of Theorem 6, is Cauchy in , which is complete, so converges to in . For ,

Hence, for each . By , for each , there is so that . By the compactness of , there is a convergence subset of . Assume that converges to . Take such that implies . One writes

So for all . Also,

So and . Thus, has an upper bound in .
By Zorn’s Lemma, there is a maximum element . also, there is so that . Using , there is so that . Hence, . The element is maximum in , so and . That is, is a fixed point of .

Theorem 13. Let be a complete ordered metric space and be a continuous function bounded below. Let be a multivalued mapping. Assume that

Then, has a fixed point.

Remark 14. If is a continuous single-valued mapping in Theorem 6 (resp., Theorem 7), we can replace the condition by

and we can obtain the same result.

Remark 15. If in Theorem 8 (resp., Theorem 10), is assumed to be a continuous single-valued mapping, then we get the same result when replacing the conditions and by

(i) is monotone increasing (resp., decreasing), that is, for , we have (ii)there exists with (resp., ).

If in addition is order upper (resp., lower) semicontinuous on , then is the smallest (resp., largest) fixed point of in (resp., ).

Proof. Let be a fixed point of in , i.e., . Since , we get . Hence, , i.e., . Suppose , then , i.e., . By a mathematical induction, we have for all , then

That is, .
Note that if we omit the continuity of and add above condition on , the results remain true.

Data Availability

The data used to support the findings of this study are available from the corresponding author upon request.

Conflicts of Interest

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

Authors’ Contributions

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

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