Abstract

In this paper we utilize the concept of manageable functions to define multivalued manageable contractions and prove fixed point theorems for such contractions. As applications we deduce certain fixed point theorems which generalize and improve Nadler’s fixed point theorem, Mizoguchi-Takahashi’s fixed point theorem, and some other well-known results in the literature. Also, we give an illustrating example showing that our results are a proper generalization of Nadler’s theorem and provide an application to integral equations.

1. Introduction and Preliminaries

The Banach contraction principle [1] is an elementary result in metric fixed point theory. This golden principle has been broadened in several directions by different authors (see [118]). An interesting generalization is the elongation of the Banach contraction principle to multivalued maps, known as Nadler’s fixed point theorem [19] and Mizoguchi-Takahashi’s fixed point theorem [20]. In 2012, Samet et al. [18] defined -contractive and -admissible mappings and then Salimi et al. [17] generalized this idea by introducing function and established fixed point theorems. Further Hasanzade Asl et al. [13] extended these notions to multivalued functions by introducing the concepts of -contractive and -admissible for multivalued mappings and proved some fixed point results.

Hussain et al. [14] modified the notions of -admissible as follows.

Definition 1 (see [14]). Let be a multifunction on a metric space and be two functions, where is bounded; then is an -admissible mapping with respect to if where

Further, Ali et al. [3] generalized the results of Hussain et al. and introduced the following definition.

Definition 2 (see [3]). Let be a closed valued mapping on a metric space and be two functions. We say that is generalized -admissible mapping with respect to if

Very recently, Ali et al. [2] modified Definition 2 for the sequence of multivalued functions as follows.

Definition 3 (see [2]). Let be a sequence of closed valued maps on a metric space and be two functions; then the sequence is -admissible mapping with respect to if for each . If, for all   , the sequence is called -subadmissible and, for , the sequence is called -admissible.

Recently, Du and Khojasteh [10] initiated the concept of manageable functions and proved some fixed point theorems. In this paper, we introduce multivalued manageable contraction and prove certain fixed point results. We also prove common fixed point theorem for multivalued contraction. The investigated results of this paper conclude several existing fixed point results including Nadler’s theorem.

Throughout this paper, denotes the family of all nonempty closed subsets of a metric space . The Hausdorff metric is defined on by where . In the sequel, denotes the set of all fixed points of mapping , denotes the set of all common fixed points of mappings , denotes the class of all functions fulfilling ,  for all , denotes the set of all functions such that exists and ,   for each , denotes the class of all nondecreasing functions such that   for all , and denotes the set of all -functions. Recall that a function is said to be an -function if and   for all and for every there exists such that   for [21].

2. Fixed Point and Common Fixed Point Results for Multivalued Contractions via Manageable Function

Consistent with Du and Khojasteh [10], we denote by the set of all manageable functions fulfilling the following conditions: for all ;for any bounded sequence and any nondecreasing sequence , it holds that

Example 4 (see [10]). Let . Then defined by is a manageable function.

Example 5. Let defined by where and is any function. Then . Indeed, by using Lemma  1 of [12], we have, for any ,so, holds. Let be a bounded sequence and let be a nonincreasing sequence. Then for some ; we getso, is also satisfied.

Definition 6. Let be a metric space and be a closed valued mapping. Let be two functions and . Then is called a multivalued -manageable contraction with respect to if for all

Now we state and prove the main result of this section.

Theorem 7. Let be a complete metric space and let be a closed valued map satisfying the following conditions:(1) is -admissible map with respect to ;(2) is manageable contraction with respect to ;(3)there exists and such that ;(4)for a sequence , and , for all , implies for all .Then .

Proof. Let be such that . Since is -admissible map with respect to , then . Therefore, from (11), we have If , then ; also if , then . So, we adopt that and . Thus . Define by By , we know that Also note that if , then So, from (12) and (14), we get LetSince . So, by using (16), we get and This implies that there exists such that Note that (since ). Now if , then is a fixed point of . Otherwise, . Also, since , , and , then . So, , so, from (11), we get By takingthere exists with such thatHence, by induction, we form a sequence in satisfying for each , , , , and by takingBy using (14), (15), (23), and (25), we get for each This implies that is a bounded sequence. By combining (25) and (27), for each , we get Which means that is a monotonically decreasing sequence of nonnegative reals and so it must be convergent. So, let From , we getNow, if, in (29), , then, by taking in (28) and using (30), we have This contradiction shows that . Hence, Next, we prove that is a Cauchy sequence in . Let, for each , Then, from (16), we have . By (28), we obtain Equation (30) implies that , so there exists and , such that For any , since for all and , (34) and (35) imply that Put , . For with , we have from (36) thatSince , . Hence This shows that is a Cauchy sequence in . Completeness of ensures the existence of such that as . Now, since for all , , and so, from (11), we have Then, from (14) and (15), we have Since , so, by using (40), we get Letting limit in the above inequality, we get Hence .

Example 8. Let with usual metric . Then is a complete metric space. Define , and by , for all and , where . Then is a manageable function. Indeed, for any , we have so, holds. Let be a bounded sequence and let be a nonincreasing sequence. Then which means that holds. Hence .
Since when , this implies that Hence is -admissible mapping with respect to .
Let ; then . This implies thatThus, all conditions of Theorem 7 are satisfied and is a fixed point of .
On the other hand, for and , we have This implies that is not a multivalued contraction, so we cannot apply Nadler’s theorem [19] with this example.

On bearing in Theorem 7, we get the following corollary.

Corollary 9. Let be a complete metric space and let be an -admissible and closed valued map enjoying the following:(1) implies ;(2)there exists and such that ;(3)for a sequence , , and , for all , one has for all ,for all and . Then .

By taking in Theorem 7, we get the following corollary.

Corollary 10. Let be a complete metric space and let be an -admissible and closed valued map enjoying the following conditions:(1) implies ;(2)there exists and such that ;(3)for a sequence , , and , for all , one has for all ,for all and . Then .

Theorem 11. Let be a complete metric space and let the sequence of closed valued mappings enjoy the following with condition (4) of Theorem 7: (1) is -admissible with respect to ;(2) implies  ;(3)there exists and for each such that ;for all , , and . Then for each .

Proof. Let be such that ; then from (11) we haveIf , for each , then . Adopt that . Thus From (50), we get where is defined in (13). Let Since ,  . So, by using (51), we get and This implies that there exists such that Note that (since ). Now if , for each , then . Let ; then Since the sequence is -admissible with respect to , so we have and, from condition (11), we get By taking there exists with such that Hence, by induction, we can establish a sequence in satisfying for each , , , for each , and by taking By using (14), (15), (59), and (61), we get for each Which means that is a bounded sequence. By combining (61) and (63), for each , we get Which means that is a monotonically decreasing sequence of nonnegative reals and so it must be convergent. So, let From , we getNow, if, in (65), , then, by taking in (64) and using (66), we have This contradiction shows that . Hence,Next, by following the same procedure as in proof of Theorem 7, we show that sequence is a Cauchy sequence in . Completeness of guarantees that there is such that as . Now, since for all , and from condition (11), we have Then, from (14) and (15), we have Since , by using (70), we get Letting limit in the above inequality, we get Hence .

Let us take for each ; then Theorem 11 reduces to the following.

Theorem 12. Let be a complete metric space and let the sequence be a closed valued mapping enjoying the following with conditions (3) and (4) of Theorem 7:(1) is generalized -admissible with respect to ;(2);for all and . Then .

As an application of Theorems 7 and 11, we can deduce the following result.

Theorem 13. Let be a complete metric space and let be a closed valued mapping satisfying for all where . Then .

Proof. Define by for all . So that . This implies that for all . That is, all the conditions of Theorem 7 hold true. Hence has a fixed point.

In the following corollaries, we obtain some known and some new results in literature via manageable functions.

Corollary 14 (see [19]). Let be a complete metric space and let be a closed valued mapping satisfying for all where . Then .

Proof. Define by Then , by Example 4. Therefore the result follows by taking in Theorem 13.

Corollary 15 (see [20]). Let be a complete metric space and let be a closed valued mapping satisfying for all where . Then .

Proof. Define by where and be any function. Then (see Example C in [10]). Therefore the result follows by taking in Theorem 13.

Corollary 16. Let be a complete metric space and let be a closed valued mapping satisfying for all where . Then .

Proof. Define by where and be any function. Then for we have Now let be a bounded sequence and let be a nonincreasing sequence. Then for some and Hence . Therefore the result follows by taking in Theorem 13.

Corollary 17. Let be a complete metric space and let be a closed valued mapping satisfying for all where . Then .

Proof. Define by where and be any function. Then . Indeed, by using Lemma  10 of [22], we have for so, holds. Let be a bounded sequence and let be a nonincreasing sequence. Then for some and So, is also satisfied. Therefore the result follows by taking in Theorem 13.

Remark 18. Since is a Meir-Keeler multivalued mapping of a metric space if and only if there exists (nondecreasing, right continuous) mapping such that (see [21], Theorem  2), therefore, from Corollary 17, we get the main result of [23].

By taking in Theorem 12, as defined in Corollary 15, we get the following.

Corollary 19 (see [3]). Let be a complete metric space and let be a closed valued mapping enjoying the following with conditions (3) and (4) of Theorem 7.(1) is generalized -admissible with respect to ;(2)  implies  ,for all and . Then .

On consideringwhere and is any function, in Theorems 7, 11, and 12, we get the following existing theorems, respectively.

Corollary 20 (see [14]). Let be a complete metric space and let be a closed valued mapping fulfilling conditions (1), (3), and (4) withfor all and . Then .

Corollary 21 (see [2]). Let be a complete metric space and let the sequence be -admissible with respect to fulfilling condition (4) of Theorem 7 and (3) of Theorem 11 with for all , and . Then .

Corollary 22 (see [2]). Let be a complete metric space and let be a closed valued mapping and generalized -admissible with respect to fulfilling conditions (3) and (4) of Theorem 7 with for all and . Then .

3. Fixed Point Results in Partially Ordered Metric Spaces and an Application to Integral Equations

Let be a partially ordered metric space. Recall that is monotone increasing if for all , for which (see [8]). There are many applications in differential and integral equations of monotone mappings in ordered metric spaces (see [15, 2426] and references therein). In this section, from Theorems 713, we derive the following new results in partially ordered metric spaces and give an example to integral equations.

Theorem 23. Let be a complete partially ordered metric space and let be a closed valued mapping satisfying the following assertions for all with : (1) is monotone increasing;(2);(3)there exists and such that ;(4)for a sequence , , and for all , one has for all .Then .

Proof. Define by Then for with , implies and otherwise. Thus, all the conditions of Theorem 7 are satisfied and hence has a fixed point.

Theorem 24. Let be a complete partially ordered metric space and let the sequence of closed valued mappings enjoy the following assertions, for all with and for each :(1);(2)if , then for each ;(3)there exists and for each such that ;(4)for a sequence , , and , for all , we have for all .Then for all .

Proof. By defining as in Theorem 23 and by using Theorem 11, we get the required result.

Similar to the arguments of Theorems 23 and 24, we conclude the following result and omit its proof.

Theorem 25. Let be a complete partial ordered metric space and let be a closed valued mapping satisfying (73) of Theorem 13 for all with . Then .

In case of single valued mapping, Theorems 2325 reduced to the following.

Theorem 26. Let be a complete partially ordered metric space and let be a self-map fulfilling the following assertions:(1) is monotone increasing;(2);(3)there exists and such that ;(4)for a sequence , , and , for all , we have for all ,for all with and . Then .

Theorem 27. Let be a complete partially ordered metric space and let the sequence of self-mappings fulfill the following assertions: (1);(2) is nondecreasing for each ;(3)there exists and for each such that ;(4)for a sequence , and for all , we have for all ,for all with , and . Then .

Theorem 28. Let be a complete partial ordered metric space and let be a self-map satisfying for all with where . Then .

Now we give an application of our results and establish the existence of solution of the integral equation.

Let be a partial order relation on . Define by

Theorem 29. Let with the usual supremum norm. Suppose that (1) and are continuous;(2)there exists a continuous function such that for each and with ;(3);(4)there exists and such that ;(5)for a sequence , , and , for all , one has for all .Then the integral equation (93) has a solution in .

Proof. Let and , for . Consider a partial order defined on by Then is a complete partial ordered metric space and for any increasing sequence in converging to , we have for any (see [27]). By using (94) and conditions (2) and (3) and taking for all with , we obtain This implies that So for all with . Hence all the conditions of Theorem 26 are satisfied. Therefore has a fixed point; consequently, integral equation (93) has a solution in .

Competing Interests

The authors declare that they have no competing interests.

Acknowledgments

This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. Therefore, the authors acknowledge with thanks DSR, KAU, for financial support.