Abstract
In this article, we have designed two existence of fixed point theorems which are regarding to set-valued SU-type -contraction and -contraction via gauge function in the setting of metric spaces. An extensive set of nontrivial example will be given to justify our claim. At the end, we will give an application to prove the existence behavior for the system of functional equation in dynamical system and integral inclusion.
1. Introduction and Preliminaries
The most publicized famous result in nonlinear analysis is Banach contraction principle, which made clear a systematic rule to find the fixed point of a given mapping on a metric space. So far, numerous authors have studied this classical result to examine the existence and uniqueness of a solution for different forms of contractive structure.
In 2014, Jleli and Samet [1] introduced the concept of a new contraction known as the -contraction, which generalizes the Banach contraction principle in a beautiful way.
In 2015, Khojasteh et al. [2] introduced simulation function. Recently, many researchers have proved fixed point theorems for Suzuki-type (SU) mappings in metric space (see [3, 4]).
Let be a metric space. For and , let and denote the family of all nonempty closed subsets and the family of all nonempty closed bounded subsets of . Design the Pompeiu–Hausdorff metric induced by on asfor all and . A point is said to be a fixed point of , if . If, for , there exists a sequence in such that , then is said to be an orbit of . Mapping is said to be -orbitally lower semicontinuous (o.l.s.c), if a sequence in and .
A multivalued mapping is called a Nadler-contraction, if there exists such that
Nadler [5] obtained the multivalued version of the Banach contraction principle. Let be a complete metric space and be a Nadler-contraction. Then, has a fixed point. Recently, Vetro [6] proved the following result.to .
Theorem 1. Let be a complete metric space and be a multivalued mapping. Suppose that there exist and such thatwhere is the set of mapping satisfying :(i) is nondecreasing and right-continuous.(ii) For each in , .(iii) There exist and such that . Then, has at least one fixed point.
Remark 1. Let be a metric space. If is a multivalued mapping satisfying the above theorem, thenSince is nondecreasing, we obtain
Example 1. The functions , defined by and , are in .
Lemma 1. (see [6]). Let be a metric space and with . Then, for all and , there exists such that
Lemma 2. (see [6]). Let be a metric space, , and . Then, for each , there exists such that
Definition 1. (see [2]). Mapping is called a simulation function such that . for all . If such that , thenDue to , we have for all . Here, the set satisfies the conditions ()–().
Example 2. (see [2]). For , let be continuous functions such that . Functions () are in :
(i) for all , where .(ii) for all, where is a function such thatDefinition 2. (see [1]). Let be a metric space and be a nonempty subset of , and is known as -admissible, if there exists a mapping such thatfor all and .
Lemma 3. (see [7]). Suppose there is a point ( is a closed subset of ) that satisfiesand for some . Then, where denotes an interval on containing 0.
Definition 3. (see [7]). (inclusion ball) Suppose and . Then, for all iterates which belongs to , we define the closed-ball with center and radius , where is defined by (13).
Lemma 4. (see [7]). Suppose there is a point that satisfies and for some ; then, and .
Definition 4. (see [7]). Let , and is known as a gauge function of order on , if it satisfies the following conditions:(i) for all and .(ii) for all .Note that the first condition of Definition 4 is equivalent to and is nondecreasing on .
Definition 5. (see [7]). A gauge function is said to be a Bianchini–Grandolfi gauge function on ifNote that a Bianchini–Grandolfi gauge function also satisfies the following functional equation:
2. Set-Valued SU -Contraction
The first main definition of this exposition is as follows.
Definition 6. Let be a metric space, be a closed subset of , and be a Bianchini–Grandolfi gauge function on . A mapping is known as set-valued SU -contraction, if there exists such that for ,which implies thatwherefor all , with , where .
Theorem 2. Let be a complete metric space and be a multivalued SU -contraction. Suppose such that for some . Then, there exists an orbit of in and such that . In addition, is a fixed point of if and only if the function is -o.l.s.c at .
Proof. Choose . In the case that , is a fixed point of . Thus, we assume that . On the other hand, we haveDefine . From (13), we have . Hence, , and so . Since , from (15) and (17), it follows thatBy right continuity of , there exists a real number such thatFromby Lemma 1, there exists such that . Since is nondecreasing, by (19), this inequality gives thatwhereWe claim thatLet . If , we have , so we obtainwhich is a contradiction. Thus, we conclude that . We assume that ; otherwise, is a fixed point of . From Remark 1, we have , and so . Next, becauseAlso, sincefrom (15), we getSince is right-continuous, there exists a real number such thatNext, fromby Lemma 1, there exists such that . By (28), this inequality gives thatwhereWe claim thatLet . If , we have , so we obtainwhich is a contradiction. Thus, we conclude that . We assume that ; otherwise, is a fixed point of . From Remark 1, we have , and so . Also, we have , sinceContinuing this setup, we have two sequences and such that , with andfor all . Then,which gives thatand by , we haveNext, we prove that is a Cauchy sequence in . Setting , from (), there exist and such thatTake . From the definition of limit, there exists such thatUsing (36) and the above inequality,This implies thatHence, there exists such thatLet . Then, using the triangular inequality and (43), we getOwing to the convergence of the series , is a Cauchy sequence in . Since is closed in , there exists such that . Note that because . Now, we claim thatorfor all . Assume, on the contrary, that there exists such thatandBy (47), we havewhich implies thatwhich together with (38) givesSincefrom contractive condition (15), we havewhereWe claim thatLet . If . Since , we havewhich is a contradiction. Thus, we conclude that . From Remark 1, we haveFrom (38), (43), and (47), we obtainwhich is a contradiction. Hence, (45) holds true:Also, we know that for all n. Thus, from (15), we havewhereWe claim thatLet . If , we have , so we obtainwhich is a contradiction. Thus, we obtain . From Remark 1, we deduceTaking limit in (64),Since is -o.l.s.c at , thenSince is closed, we have . Conversely, if is a fixed point of , then , since .
Corollary 1. Let be a complete metric space, be a Bianchini–Grandolfi gauge function on an interval , and be a given set-valued mapping. If and exist,which implies thatfor all , with . Suppose such that for some . Then, there exists an orbit of in and such that . In addition, is a fixed point of if and only if the function is -o.l.s.c at point .
Corollary 2. Let be a complete metric space, be a Bianchini–Grandolfi gauge function on an interval , and be a given set-valued mapping. If and for exist,implies thatfor all , , and . In addition, suppose such that for some . Then, there exists an orbit of in , such that and is a fixed point of if and only if the function is -o.l.s.c at .
Corollary 3. Let be a complete metric space, be a Bianchini–Grandolfi gauge function on , and be a given set-valued mapping. If and exist, thenfor all , , and . Suppose that such that for some . Then, there exists an orbit of in that converges to the fixed point of .
Example 3. Let be an usual metric and let . Mapping is defined asClearly, if and only if . Let ; then, we have such that . Firstly, we claim that satisfies inequality (68) with setting , , and . For and , we obtainConsequently, the requirements of Corollary 1 are fulfilled and 0 is a fixed point of . For and ,for all and . Then, Corollary 1 cannot be satisfied.
3. Set-Valued SU-Type -Contraction
In this section, we prove the existence of fixed point in the class of metric space with respect to a simulation function.
Definition 7. Let be a metric space, be a closed subset of , and be a Bianchini–Grandolfi gauge function on . Mapping is known as set-valued SU-type -contraction, if there exists such that for ,which implies thatwherefor all , with .
Theorem 3. Let be a complete metric space and be a set-valued SU-type -contraction such that the following conditions are satisfied:(a) is -admissible.(b)There exists with for some such that . Then, there exists an orbit of in and such that . In addition, is a fixed point of if and only if the function is -o.l.s.c at .
Proof. By the hypothesis, there exists with for some such that . On the other hand, we haveIn the case that , is a fixed point of . Thus, we assume that . Define . From (13), we have . Hence, , and so . Since and , from (76) and (78), it follows thatwhich implies thatWe can choose such thatThus,It follows from Lemma 2 that there exists such thatFrom fd82(82) and (83), we infer thatwhereWe claim thatLet . If , we have , so we obtainwhich is a contradiction. Thus, we obtain . We assume that ; otherwise, is a fixed point of . Since , we deduce that . Next, becauseBecause is -admissible, . Also, sincefrom (76), we getThis implies thatNow choose such thatThus,It follows from Lemma 2 that there exists , such thatFrom (93) and (94),whereWe claim thatLet . If . Since , we havewhich is a contradiction. Thus, we have . We assume that ; otherwise, is a fixed point of . From (95), we have , and so . Also, we have , sinceContinuing this setup, we obtain a sequence such that , with and andFor with , by using the triangular inequality and (100), we getTo shows that is a Cauchy sequence in . Since is closed in , there exists an such that . Note that because . By same argument of Theorem 2, we haveAlso, we know that and for all n. Thus, from (76), we havewhich gives thatSince , from (100), we getwhereWe claim thatLet . If . Since , we havewhich is a contradiction. Thus, we have . Taking limit in (105), we obtainSince is -o.l.s.c at , thenSince is closed, . Conversely, if is a fixed point of , then , since .
Taking for all , in Theorem 3, we obtain the following theorem.
Corollary 4. Let be a complete metric space, be a Bianchini–Grandolfi gauge function on an interval , and be a given set-valued mapping. If , thenwhich implies thatfor all , , and , where is a function such that exists and for all such that the following holds:(a) is -admissible.(b)There exists with for some such that . Then, there exists an orbit of in and such that . In addition, is a fixed point of if and only if the function is -o.l.s.c at .
Corollary 5. Let be a complete metric space, be a Bianchini–Grandolfi gauge function on an interval , and be a given set-valued mapping. If exists, thenfor all , , and such that the following holds:(a) is -admissible.(b)There exists with for some such that . Then, there exists an orbit of in that converges to the fixed point of .
4. An Application to Dynamical System
Dynamical system is connected to a multistage operation reduced for solving the following functional equation:where
Assume that and are Banach spaces, is a state space, and is a decision space. For more details, see [3]. Let signify the set of all bounded real-valued functions on . Choose an arbitrary point defined as . endowed with the metric given byfor all , are BS. Define byfor all and . Also,
Consider such thatand we havefor all , where , and .
Theorem 4. Let be a l.s.c mapping as defined in (117) such that the following conditions are satisfied: and are continuous and bounded. There exists an orbit of and such that . is a fixed point iff is -o.l.s.c at .Then, functional (114) possesses a bounded solution.
Proof. Note that is a complete metric, where is the metric, as defined by (82). There exist , and such thatand we haveIt implies thatOwing to (123),By and with (124), we obtainFurthermore, are equivalent to of Theorem 3. So, there exists a fixed point in , which is a bounded solution of functional (117).
4.1. An Application to Integral Inclusion
In this section, we consider the following set-valued integral inclusion:where , is a bounded compact subset of , and is l.s.c. Let be the space of all continuous real-valued function and is complete w.r.t the metric , which defined by
Assume that there exists and is continuous onsuch that forwe havewhere , , and .
Moreover, let be a closed subspace of and the operator be defined by
Set . Note that
Consider
This implies that
Theorem 5. Let and be a l.s.c mapping. Suppose that the following assumptions hold:(i) is defined for all .(ii) is a CS of for all .Then, owing to (127)–(135), integral (126) has a solution on .
Proof. Let . Then, . Hence, we have . If , the integral equation in (132) exists. Since is continuous, is defined for all . Next, let . Then, for :Thus, for each . So, is a subset of . Now, let ; then, for . Since is compact, there exists subsequence such that is convergent to . Let ; then,Hence, is a CS of for all . Next,It yields thatBy , and with (139), we getFurthermore, - are equivalent to of Theorem 3. So, there exists a fixed point in , which is a bounded solution of (126).
5. Conclusion
The paper deals with the set-valued fixed point theorems satisfying SU-type contraction via Bianchini–Grandolfi gauge function in the context of metric spaces. Within this framework, we introduced two related fixed point results in metric space. An extensive set of nontrivial example is given to justify our claim. Also, we have proven the existence theorem for the system of functional equation and integral inclusion.
Data Availability
The data used to support the findings of this study are included within the article.
Conflicts of Interest
The authors declare that there are no conflicts of interest.
Authors’ Contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Acknowledgments
The authors are thankful to the Deanship, King Abdulaziz University, Jeddah, KSA, for funding this research.