Existence of Fixed Points for a Class of Decreasing Operators with Parameter and Applications

Abstract

This paper studies the existence of fixed points for a class of decreasing operators with parameter in real Banach spaces. The existence theorems of fixed point are obtained as when the parameter is increasing, there will still be a large fixed point. These results have reduced the requirements of convexity, compactness, and lattice structure of spaces. By this new method, the existence of solutions for a class of second-order differential equations with parameter in infinite intervals is studied.

1. Introduction and Preliminaries

The solvability of nonlinear equations can be transformed into the existence of fixed points for operators. The existence of fixed points has been considered widely [1–11]. There are many conclusions about fixed points for increasing operators, but less for decreasing operators [6–11]. The partial-order method is subject to many constraints to verify the convergence of decreasing operators, and it is also difficult to construct iterative conditions.

The existence theory of fixed points for decreasing operators has been developed rapidly in the field of microeconomics [12–16] and has been found having various applications to economic models. In [15], the existence of game equilibrium for alternative strategy is proved. However, the above results require convexity, compactness, or lattice structure in player’s strategy space.

Inspired by [16], this paper studies a class of decreasing operators with parameter in real Banach spaces by cone theory, and the existence theorems of fixed points for decreasing operators with parameter are obtained. It is proved that when the parameter is increasing, there will still be a larger fixed point. This result reduces the requirements for convexity, compactness, and lattice structure of spaces. As an application of the results in this paper, we study the existence of solutions to the initial value problem of a class of second-order initial value differential equations in infinite intervals. It is proved that the solution of the problem is monotonically increasing with respect to the parameters.

The result of this study can also be used as a mathematical tool to solve the comparative static analysis of the existence of Nash equilibrium in noncooperative alternative strategy games. It provides a mathematical means for us to study the phenomenon of diminishing returns, mutual substitution, incomplete competition, and various modern economic phenomena characterized by “difference.”

Here some concepts and theorems are given, which are related to incomplete preference.

Let be a nonempty set. An ordering relation on may satisfy the following axioms:(i)Reflexive:, for any ;(ii)Symmetry: for any , and imply (iii)Transitive: for any , and imply (iv)Complete: if , for any , there is at least one inequality to be established

Definition 1 (see [17]). Let be a nonempty set. An order relation defined among certain elements of is said to be partial order if the order relation satisfies reflexive, transitivity, and antisymmetry axioms. Then, is called a poset.

Definition 2 (see [17]). Let E be the Banach space and be a zero element of E. Let P be a nonempty convex closed sets of E. If the following conditions are satisfied,(1)(2); then, P is called a cone

Definition 3 (see [17]). A cone is called normal if there is a constant , such that implies , for all . The least positive constant satisfying the above inequality is called the normal constant of . We call P a regular cone. If every increasing sequence in E has an upper bound in order, it must have a limit. If, for arbitrary , and exist, the cone is called minimal.

Definition 4 (see [18]). Let be a partially ordered set and . The operator is said to be a decreasing operator if, for all implies .

Definition 5 (see [18]). Let and the operator be continuous and bounded. The operator is said to be a condensing operator if, for any nonrelatively compact and bounded set, implies , where means noncompact measurement in .

Remark 1. If the operator satisfies continuity and compactness, then it is condensing.

Lemma 1 (see [11]). Let be a normal cone in a real Banach space and be decreasing. Suppose that there is which satisfiesIf the operator is condensing, then has a fixed point in .

Lemma 2 (see [18]). Let be a normal cone in a real Banach space and be an increasing and condensing operator such that . Then, has a maximal fixed point and a minimal fixed point in ; moreover,

Lemma 3 (Sadovskii [18]). Let be a real Banach space, be nonempty, closed bounded convex, and be condensing or completely continuous. Then, has a fixed point in .

2. Main Results and Proof

Let be a Banach space and be a cone of . We define a partial ordering with respect to by if .

Theorem 1. Let be a normal cone in a real Banach space , be a nonempty partially ordered set, and . Assume that, for any is a condensing operator which satisfies the following conditions  For any is increasing on , and for any is decreasing on   For any , there exists a , such that Then, the following is true:(1)For any has a fixed point in ; if has another fixed point in , then and are incomparable(2)For any , there exists a such that

Proof. (1)For any , Lemma 1 implies that the operator has a fixed point in  Next, it will be proved that any two unequal fixed points of operator in cannot be compared. If and are any two unequal fixed points of operator in , then . Suppose that and can be compared. Without loss of generality, let and , that is, . Since is decreasing on , we know that . And because , we have . Yet this is in contradiction to . So, any two unequal fixed points of operator in cannot be compared.(2)For any and , it is easy to know that by the condition that is increasing on . Noticing that is decreasing on , we know thatwhere . is increasing on . So, from condition , the following conclusion can be obtained:(3) and (4) show thatLetBy (3) and (5), we haveSince is an increasing and condensing operator on , which satisfies the conditions (7) and (8), Lemma 2 implies has a maximal fixed point and a minimal fixed point in . Therefore, we obtainLet be a fixed point of . Then,and thus, is also a fixed point of .
We distinguish two cases to discuss:(i)If , that is, has only one fixed point, then . This proves that must have a fixed point in the ordered interval , and .(ii)If , (10) implies that and are fixed points of . Since is decreasing on , we haveFor , since is decreasing on , we have . By (11), we have . Since is a normal cone of real , we have that closed bounded convex ordered interval. So, satisfies the conditions in Lemma 3. Thus, Lemma 3 implies that the operator has a fixed point in , and .
The proof of Theorem 1 is completed.

Remark 2. For operator , the requirements of lattice structure, continuity, and compactness are weaker.

Theorem 2. Let be a regular cone in a real Banach space , be a nonempty partially ordered set, and . Assume that, for any is a condensing operator which satisfies the following conditions:  For any is increasing in , and for any is decreasing in   For any , there exists a , such that Then, the following is true.(1)For any has a fixed point in ; if has another fixed point in , then and are incomparable.(2)For any , there exists such that .

Proof. Similar to the proof of Theorem 1, by Lemma 3, we can get the result.
In order to support the main results, let us consider the following a nontrivial example that shows that, for any is increasing in , and for any is decreasing in .

Example 1. For any , we define . With the standard product order, it is easy to check that, for any is increasing in , and for any is decreasing in ; for any , , such that . Then, the equation , has a nontrivial solution , and is increasing in .

3. Applications

In this section, we use Theorem 1 to show the existence of unique solution for the second-order initial value problem:where are nonnegative continuous functions in , and is nonnegative continuous function, for all , . Then, problem (12) has a positive solution , and is increasing on .

Proof. Supposing that is function, for all , ifis substituted intothen the following conclusion holds:So, the nonnegative solution of the initial value problem (12) for all is equivalent to the fixed point of the integral operator so that is the nonnegative solution of the initial value problem (12) which is equivalent to that which is the nonnegative continuous solution of the following integral equation:Let . It is easy to know that is a Banach space with maximum norm . Let , , and be a normal cone in .
Consider the operator:For any fixed , is a completely continuous operator from into , and thus, is condensing. For any fixed , is also a decreasing operator from into and for any , is increasing on .
Letting in (17), we haveFor ,So, the condition of Theorem 1 holds. By Theorem 1, we get that has a fixed point in . Then, problem (12) has a positive solution , and is increasing on .
The problem in this section is proved.

Data Availability

No data were used to support the results of the study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

The first and second authors were supported financially by the National Foundation of Social Science of China (13CJL006).