On the Blended Solutions of Polynomial-Like Iterative Equation with Multivalued Functions

Abstract

The solutions are more general than the solutions for polynomial-like iterative equation with multivalued functions. In this paper we study the solutions of polynomial-like iterative equation with multivalued functions.

1. Introduction

The polynomial-like iterative equationis an interesting form of functional equations, where is a topological vector space, is a given function, and , stands for the th iteration of the unknown function , i.e., and . Since iteration is an important problem in many mathematical subjects, such as dynamical systems and numerical computation, and many fields of natural science, in recent years some attentions have been paid to (1) and its generalizations [1–7]. On the other hand, multivalued function is an important class of mappings, which has been extensively employed in control theory [8], stochastics [9], artificial intelligence [10], and economics [11]. Many nice results [12–16] were obtained on functional equations with multivalued functions. It is important to investigate (1) with multivalued functions, i.e., the equationwhere is an integer, are real constants, is a given multivalued function, denotes the family of all nonempty convex compact subsets of , and is an unknown multivalued function; the th iteration of the multivalued function is defined inductively as and for all . In 2004, Nikodem and Zhang [2] first studied (2) for with an increasing upper semicontinuous ( for short) multivalued function on . The result on the existence and uniqueness of solutions is given. As indicated in [2], the upper semicontinuity for multivalued functions is much weaker than the continuity for functions; the method used for continuous solutions [6, 7] is improved substantially in order to obtain solutions. In 2011, Xu, Nikodem, and Zhang [4] considered the general case of the equationwhich is a modified version of (2), for the multivalued functions defined in [4]. multivalued solutions of this equation were given in the inclusion sense. As defined in [4], a multivalued function is said to be if it satisfies on a sequence in ; that is, for each ; otherwise, is said to be blended on . Clearly, the blended requirement for multivalued functions is much weaker than the unblended requirement for multivalued functions. It is an interesting object to study the blended multivalued solutions of (2) in the inclusion sense. So far, we find no results on the multivalued solutions of (2) in the inclusion sense. In this paper, we investigate the existence of the multivalued solutions of (2) in the inclusion sense.

2. Preliminaries

The family endowed with the Hausdorff distance is defined bywhere is a complete metric space (cf., e.g., [17], Cor. 4.3.12).

A multivalued function is increasing (resp., strictly increasing) if for every with , we have (resp., ) (cf. [18], Def. 3.5.1). A multivalued function is at a point if for every open set with there exists a neighborhood of such that for every . is on if it is at every point in . For convenience, let where is the set of all multivalued functions . Some useful properties are summarized in the following Lemma (cf. [4, 15, 19]).

Lemma 1. For and for an arbitrary real , the following properties hold:(i),(ii),(iii)

Lemma 2. If , then .

As indicated in [4], if a function is not single-valued, there exists at least a point such that the cardinal of the set is more than 1; i.e., card . Actually, is a nontrivial interval because . Since is strictly increasing, there exist two small open intervals and such that is single-valued in both of them and satisfies , and . We call a jump-point of or a jump simply. For every , let denote the set of all jumps of . We easily see that each has at most countably infinite many jumps, i.e., the cardinal card . In fact, for each , the set is a nontrivial compact subinterval of . By the strict monotonicity, is a set of disjoint nonempty compact subintervals of . Choose a rational number for each . Then card . According to above the argument, was divided into two cases in [4]: one is unblended multivalued functions; the other is blended ones.

Since functions in are strictly increasing, it suffices to discuss multivalued functions in which satisfy either for all int or for all int . Let and denote the two classes of multivalued functions, respectively.

Lemma 3. Suppose that (resp., ) is on the strictly increasing (resp., decreasing) sequence . If and satisfies that and (resp., and ), then for each integer , (i) , and (ii) (resp., .

A function is said to be on with the sequence and constants if for each ,(C1),(C2),(C3),

where is a strictly monotonic sequence in such that int is a union of disjoint open intervals; i.e.,where each is either an element of or an endpoint of . For a strictly increasing sequence in such that and , define Similarly, for a strictly decreasing sequence in such that and define

Lemma 4. is a complete metric space equipped with the distance , defined by .

DefineClearly, , , , and are all closed subsets of .

Lemma 5. (resp., ) if (resp., ).

Lemma 6. If either or , then

3. Main Results

Theorem 7. Suppose that , and and that sequences and are strictly increasing sequences in such that , , and card of and that , where , such that ; then, for arbitrary constants such that (4) has a unique solution , which is a solution of (2) in the inclusion sense and is in .

Proof. Sets and are well defined since the sequences and are strictly increasing sequences in such that , , and card of . Define operator bywhere . We can deduce from hypotheses , , , and . Since and , we have which impliesBy in and card of , we can obtain that . In fact, Since , , and card of , then , with , and ; i.e., . From in , Lemma 5, and (13), we can infer thatwhich together with (15) yields Furthermore, for , , by (13) and Lemma 6 we haveIn view of Lemma 6,We know that operator is contract since . Making use of fixed point Theorem, has a unique fixed point in such thatand since , , and are three sets, according to sets operation relations we can get which implies is a unique solution of (2) in the inclusion sense. In the following, we shall prove that is in . By reduction to absurd, we suppose is in . Since is in , for , there exists some interval such that , which implies .    because is . Therefore, , which contradicts to the condition that is strictly increasing on ; thus is in .

Theorem 8. Suppose that , and and that sequences and are strictly decreasing sequences in such that , , and card of and that , where , such that ; then, for arbitrary constants satisfying (12), (4) has a unique solution , which is a solution of (2) in the inclusion sense and is in .

The proof is similar to Theorem 7, so we omit it.

Remark 9. Although we strengthen the condition of Theorem 4.1 in [4] in Theorem 7, more general solutions of (2) are given.

4. Example

In order to show the rationality conditions in Theorems, we consider the equationwhere , , and, obviously, , where with , for with ; one can check Taking , , we can see that conditions (12) hold. Thus, (2) has a solution by Theorem (12).

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This work is supported by the general item of Lingnan Normal University [grant ZL1505], KSP of Lingnan Normal University [number 1171518004], NSFC [grant number 11701476], the Fundamental Research Funds for the Central Universities [grant number 2682018CX63], and the young project of Zhanjiang Preschool Education College Scientific Research [grant number ZJYZQN201717].