Fixed Points for Discontinuous Monotone Operators

Abstract

We obtain some new existence theorems of the maximal and minimal fixed points for discontinuous monotone operator on an order interval in an ordered normed space. Moreover, the maximal and minimal fixed points can be achieved by monotone iterative method under some conditions. As an example of the application of our results, we show the existence of extremal solutions to a class of discontinuous initial value problems.

1. Introduction

Let be a Banach space. A nonempty convex closed set is said to be a cone if it satisfies the following two conditions: (i) , implies ; (ii) , implies , where denotes the zero element. The cone defines an ordering in given by if and only if . Let be an ordering interval in , and an increasing operator such that , . It is a common knowledge that fixed point theorems on increasing operators are used widely in nonlinear differential equations and other fields in mathematics ([1–7]).

But in most well-known documents, it is assumed generally that increasing operators possess stronger continuity and compactness. Recently, there have been some papers that considered the existence of fixed points of discontinuous operators. For example, Krasnosel'skii and Lusnikov [8] and Chen [9] discussed the fixed point problems for discontinuous monotonically compact operator. They called an operator A to be a monotonically compact operator if () implies that converges to some in norm and that (). A monotonically compact operator is referred to as an MMC-operator. A is said to be -monotone if implies , where , , and . They proved the following theorem.

Theorem 1.1 (see [8]).

Let be an -monotone MMC-operator with . Then has at least one fixed point possessing the property of -continuity.

Motivated by the results of [3, 8, 9], in this paper we study the existence of the minimal and maximal fixed points of a discontinuous operator , which is expressed as the form . We do not assume any continuity on . It is only required that (or ) is an MMC-operator and (or ) possesses the quasiseparability, which are satisfied naturally in some spaces. As an example for application, we applied our theorem to study first order discontinuous nonlinear differential equation to conclude our paper.

We give the following definitions.

Definition 1.2 (see [3]).

Let be an Hausdorff topological space with an ordering structure. is called an ordered topological space if for any two sequences and in , and , imply .

Definition 1.3 (see [3]).

Let be an ordered topological space, is said to be a quasi-separable set in if for any totally ordered set in , there exists a countable set such that is dense in (i.e., for any , there exists such that ).

Obviously, the separability implies the quasi-separability.

Definition 1.4 (see [3]).

Let be two ordered topological spaces. An operator is said to be a monotonically compact operator if () implies that converges to some in norm and that .

Remark 1.5.

The definition of the MMC-operator is slightly different from that of [8, 9].

2. Main Results

Theorem 2.1.

Let be an ordered topological space, and an order interval in . Let be an operator. Assume that

(i)there exist ordered topological space , increasing operator , and increasing operator such that ;

(ii) is quasiseparable and is an MMC-operator;

(iii), .

Then has at least one fixed point in .

Proof.

It follows from the monotonicity of and condition (iii) that . Set . Since , is nonempty. Suppose that is a totally ordered set in . We now show that has an upper bound in .

Since , by condition (ii) there exists a countable subset of such that is dense in . Consider the sequence

(2.1)

Since is a totally ordered set, makes sense and

(2.2)

By condition (ii), and Definition 1.4, there exists such that

(2.3)

(2.4)

and hence make sense.

Set

(2.5)

Using (2.1) and (2.2), we have

(2.6)

Since is dense in , for any there exists a subsequence of such that (). By (2.6) and Definition 1.2, we get

(2.7)

Hence , therefore is an upper bound of .

Now we show . By virtue of (2.4) and condition (iii)

(2.8)

Thus and hence . By (2.7) and condition (ii), we get and hence . By (2.3) and Definition 1.2, we get and

(2.9)

Hence , and therefore .

This shows that is an upper bound of in . It follows from Zorn's lemma that has maximal element . Thus . And so , which implies that and . As is a maximal element of , ; that is, is a fixed point of .

Theorem 2.2.

Let be an ordered topological space, and an order interval in . Let be an operator. Assume that

(i)there exist ordered topological space , increasing operator , and increasing operator such that ;

(ii) is quasiseparable and is an MMC-operator;

(iii), .

Then has at least one fixed point in .

Proof.

Let , . By the conditions (i) and (iii), we have

(2.10)

Since is increasing, for any , we get

(2.11)

that is, ; therefore the quasiseparability of implies that is quasiseparable. Applying Theorem 2.1, the operator has at least one fixed point in , that is,

(2.12)

Set . Since is increasing, by (2.12), we have

(2.13)

that is, is a fixed point of the operator in .

Theorem 2.3.

If the conditions in Theorem 2.1 are satisfied, then has the minimal fixed point and the maximal fixed point in ; that is, and are fixed points of , and for any fixed point of in , one has .

Proof.

Set

(2.14)

By Theorem 2.1, . Set

(2.15)

Since is increasing, for any , we have

(2.16)

and hence

(2.17)

therefore , and thus . An order of is defined by the inclusion relation, that is, for any , , and if , then we define . We show that has a minimal element. Let be a totally subset of and . Obviously, is a totally ordered set in . Since is quasiseparable, it follows from that there exists a countable subset of such that is dense in . Let

(2.18)

Since is a totally ordered set, makes sense and

(2.19)

Then there exists such that

(2.20)

Using the same method as in Theorem 2.1, we can prove that makes sense, (where ) is an upper bound of , and

(2.21)

Since (for all ), for any , we have , for all . Since , . By (2.20), , and hence , for all , and therefore

(2.22)

Consider . Similarly, we can prove that there exists such that is a lower bound of and

(2.23)

By (2.22) and (2.23), . Set . By virtue of (2.21), (2.22), and (2.23), . It is easy to see that is a lower bound of in . It follows from Zorn's lemma that has a minimal element.

Let be a minimal element of . Therefore, , , and . Obviously, is a fixed point of . In fact, on the contrary, and . Hence

(2.24)

Since is an increasing operator, this implies that and includes properly . This contradicts that is the minimal element of . Similarly, is a fixed point of . Since , is the minimal fixed point of and is the maximal fixed point of .

Theorem 2.4.

If the conditions in Theorem 2.2 are satisfied, then has the minimal fixed point and the maximal fixed point in ; that is, and are fixed points of , and for any fixed point of in , one has .

Proof.

It is similar to the proof of Theorem 2.4; so we omit it.

Theorem 2.5.

Let be an ordered topological space, and an order interval in . Let be an operator. Assume that

(i)there exist ordered topological space , increasing operator , and increasing operator such that ;

(ii) is an continuous operator;

(iii) is a demicontinuous MMC-operator;

(iv), .

Then has both the minimal fixed point and the maximal fixed point in , and and can be obtained via monotone iterates:

(2.25)

with and .

Proof.

We define the sequences

(2.26)

and conclude from the monotonicity of operator and the condition (iv) that

(2.27)

Let

(2.28)

Since is increasing, by (2.27). By the condition (iii), we get

(2.29)

By (2.29) and Definition 1.2, we have

(2.30)

and hence makes sense. Set , then . Since is continuous,

(2.31)

By the condition (iii), , that is, . Note that ; we have ; hence ; that is, is a fixed point of . Similarly, there exists such that and is a fixed point of . By the routine standard proof, it is easy to prove that is the minimal fixed point of and is the maximal fixed point of in .

3. Applications

As some simple applications of Theorem 2.5, we consider the existence of extremal solutions for a class of discontinuous scalar differential equations.

In the following, stands for the set of real numbers and a compact real interval. Let be the class of continuous functions on . is a normed linear space with the maximum norm and partially ordered by the cone . is a normal cone in .

For any , set

(3.1)

Then is a Banach space by the norm .

A function is said to be a Carathéodory function if is measurable as a function of for each fixed and continuous as a function of for a.a. (almost all) .

We list for convenience the following assumptions.

(H1), ,

(3.2)

(H2) is a Carathéodory function.

(H3)There exists such that

(3.3)

(H4)There exists such that is nondecreasing for a.a. .

Consider the differential equation

(3.4)

where . It is a common knowledge that the initial value problem (3.4) is equivalent to the equation

(3.5)

if is continuous. Therefore, when is not continuous, we define the solution of the integral equation (3.5) as the solution of the equation (3.4).

Theorem 3.1.

Under the hypotheses (H1)–(H4), the IVP (3.4) has the minimal solution and maximal solution in . Moreover, there exist monotone iteration sequences such that

(3.6)

where and satisfy

(3.7)

Proof.

For any , we consider the linear integral equation:

(3.8)

where . Obviously, is a linear completely continuous operator. By direct computation, the operator equation has only zero solution; then by Fredholm theorem, for any , the operator equation (3.8) has a unique solution in . We definition the mapping by

(3.9)

where is the unique solution of (3.8) corresponding to . Obviously is a linear continuous operator; now we show that is increasing. Suppose that , . Set . By the definition of the operator we get

(3.10)

This integral inequality implies (for all ); that is, is an increasing operator. Set

(3.11)

Obviously, is an increasing continuous operator. Set

(3.12)

By (H2), maps element of into measurable functions. For any , by (H3) and (H4) we get

(3.13)

This implies . Hence maps into and is an increasing operator. Set

(3.14)

By above discussions we know that and are all increasing. Thus conditions (i) and (ii) in Theorem 2.5 are satisfied.

Let such that in ; by (H2) we have

(3.15)

For any (), by (2.29), we have

(3.16)

and hence

(3.17)

where . By (H3), ; thus

(3.18)

where . Applying the Lebesgue dominated convergence theorem, we have

(3.19)

This implies that in ; that is, is a demicontinuous operator. Since the cone in is regular, it is easy to see that is an MMC-operator. Thus condition (iii) in Theorem 2.5 is satisfied.

We now show that condition (iv) in Theorem 2.5 is fulfilled. By (H1) and (3.5), and noting the definition of operator , we get

(3.20)

This implies that for all , that is, . Similarly we can show that .

Since all conditions in Theorem 2.5 are satisfied, by Theorem 2.5, has the maximal fixed point and the minimal fixed point in . Observing that fixed point of is equivalent to solutions of (3.5), and (3.5) is equivalent to (3.4), the conclusions of Theorem 3.1 hold.

Remark 3.2.

In the proof of Theorem 3.1, we obtain the uniformly convergence of the monotone sequences without the compactness condition.