# Fixed Points of Discontinuous Multivalued Operators in Ordered Spaces with Applications

- ShihuangH Hong
^{1}Email author and - Zheyong Qiu
^{1}

**2010**:745769

**DOI: **10.1155/2010/745769

© S. H. Hong and Z. Qiu. 2010

**Received: **24 September 2009

**Accepted: **3 March 2010

**Published: **15 March 2010

## Abstract

Existence theorems of fixed points for multivalued increasing operators in partially ordered spaces are presented. Here neither the continuity nor compactness is assumed for multivalued operators. As an application, we lead to the existence principles for integral inclusions of Hammerstein type multivalued maps.

## 1. Introduction

The influence of fixed point theorems for contractive and nonexpansive mappings (see [1, 2]) on fixed point theory is so huge that there are many results dealing with fixed points of mappings satisfying various types of contractive and nonexpansive conditions. On the other hand, it is also huge that well-known Brouwer's and Schauder's fixed point theorems for set-contractive mappings exert an influence on this theory. However, if a mapping is not completely continuous, in general, it is difficult to verify that the mapping satisfies the set-contractive condition. In 1980, Mönch [3] has obtained the following important fixed point theorem which avoids the above mentioned difficulty.

Theorem 1.1.

Let be a Banach space, a closed convex subset. Suppose that (single) operator is continuous and satisfies that

(i)there exists such that if is countable, then is relatively compact,

It has been observed that continuity is an ideal and important property in the above cited works, while in some applications the mapping under consideration may not be continuous, yet at the same time it may be "not very discontinuous". this idea has motivated many authors to study corresponding problems, for instance, the stability of Brouwer's fixed point theorem [4], similar result for nonexpansive mappings [5], and existence and approximation of the synthetic approaches to fixed point theorems [6]. Recently, fixed point theory for discontinuous multivalued mappings has attracted much attention and many authors studied the existence of fixed points for such mappings. We refer to [7–11]. For example, Hong [8] has extended Mönch [3] to discontinuous multivalued operators in ordered Banach spaces by using a quite weak compactness condition; that is, assuming the following condition is satisfied.

(H)If is a countable totally ordered set and , then is weakly relatively compact. Here is a multivalued operator and denotes the weak closure of the set .

The purpose of this paper is to present some results on fixed point theorems of Mönch type of multivalued increasing operators for which neither the continuity nor the compactness is assumed in ordered topological spaces. However, we will use the following hypothesis.

(H1)If is a countable totally ordered set and , then has a supremum.

is a topological vector space endowed with partial ordering " ", stands for the closure of the set , and with is a given ordered set of .

This paper is organized as follows. In Section 2, we introduce some definitions and preliminary facts from partially ordered theory and multivalued analysis which are used later. In especial, we introduce a new partial ordering of sets which forms a basis to our main results. In Section 3, we state and prove existence of fixed points, also, maximal and minimal fixed point theorem is presented for discontinuous multivalued increasing operators which are our main results. To illustrate the applicability of our theory, in Section 4, we discuss the existence of solutions to the Hammerstein integral inclusions of the form

## 2. Preliminaries

Let be a partially ordered topological vector space. By the notation " " we always mean that and . Let stand for the collection of all nonempty subsets of . Take and let be a given ordered set of . The ordered interval of is written as .

For two subsets of , we write (or ) if

Given a nonempty subsets of we say that is increasing upwards if , , and imply that there exists such that . is increasing downwards if , , and imply an existence of such that . If is increasing upwards and downwards we say that is increasing.

Let be nonempty. The element is called an upper (lower) bound of if ( ) whenever . is called upper (lower) bounded with respect to the ordering if its upper (lower) bounds exist. The element is called a supremum of , written as , if is an upper bound and as long as is another upper bound of . Similarly, we can define the infimum of .

Throughout this paper, unless otherwise mentioned, the partial ordering of always introduced by a closed cone if is a Banach space. The following lemmas will be used in after sections.

Lemma 2.1 (see [12]).

Let be an ordered Banach space and a totally ordered and weakly relatively compact subset of , then there exists such that foe all .

An ordered topological vector space is said to have the limit ordinal property if with for , and for imply . By an analogy of the proof of Lemma 1.1.2 in [12], we have the following.

Lemma 2.2.

If has the limit ordinal property and is a relatively compact monotone sequence of , then is convergent. Moreover, if is increasing and if is decreasing for Here .

Remark 2.3.

Under the assumptions of Lemma 2.2, it is evident that is the supremum (infimum) of increasing (decreasing) sequence .

Lemma 2.4.

Let the increasing sequence have the supremum . If is a infinity subsequence of , then has the supremum , too.

Proof.

Evidently, is an upper bound of . Let be the other one, then for For any given , since is infinity, there exists such that , which implies that for all . From the definition of supremums it follows that , that is, is the supremum of .

Lemma 2.5.

Suppose that every countable totally ordered subset of the partially ordered set has a supremum in . Let the operator satisfy for all , then there exists such that .

Proof.

If , then the conclusion of the lemma is proved. Otherwise, take for Again, the set has the supremum . Denote . If , then the conclusion of the lemma is proved. Otherwise, take for , and let with . In general, having defined with and , where and , if , which completes the proof. Otherwise, repeating this process, either the conclusion of the lemma is proved, or we can obtain a set sequence satisfying

In fact, for any , there exists such that . There exists such that . If for some nature number , then which yields . Otherwise, we have . This implies that . Consequently, . From the arbitrariness of it follows that (2.4) is satisfied.

Finally, combining and we see easily that is totally ordered. Our hypothesis guarantees that has a supremum, written as . Note that (2.4) guarantees , we have . On the other hand, the definition of ensures that . Hence . This proof is completed.

Let be a nonempty subset of . In this section we impose the following hypotheses on the increasing upwards multivalued operator . Set

where, and is given as follows: since , there exists such that . In virtue of the fact that is increasing upwards, there exists such that . On the analogy of this process, there exists such that for Obviously, , thus, the condition (H1) guarantees that the supremum of exists.

Remark 2.6.

In general, the sequences of these kinds, , may not be unique, that is, every corresponds to , moreover, corresponds to . For given , we denote with and the families of and as above, respectively.

In addition, if has the limit ordinal property, is a closed set for any . In fact, let be any infinity subsequence of for which

observing that is increasing, by Lemma 2.2 we get that is a supremum of and by Lemma 2.4 we get .

Definition 2.7.

A set is said to be sup-closed if the supremum of each countable subset of (provided that it exists) belongs to . A multivalued operator is said to have sup-closed values if is sup-closed for each .

Defining

Lemma 2.8.

Let be an ordered topological space, a nonempty subset of with ; let have sup-closed values and satisfy hypothesis (H1). Moreover, assume that

(H2) is increasing upwards and satisfies ,

Proof.

It is clear that has the supremum . For any , from and there exists such that . We can assume that the sequence is increasing. Indeed, if for , our purpose is reached. Otherwise, there exists such that , then we take instead of . Let , then . Condition (H1) guarantees that has a supremum . Clearly, . By virtue of the fact that has sup-closed values, we have . This proof is complete.

For the sake of convenience, in this paper, by we always stand for the supremum of . For given , let be a set consisting of all given as in Lemma 2.8, then is an increasing map. Now for any Lemma 2.4 shows , thus, . Define

It is obvious that . Hence, is nonempty. A relation on Z is defined as follows (it is easy to see that is a partially ordered set):

Remark 2.9.

It is clear we may assume that, for any , there exists such that if .

Let us assume that there exists some such that

Define

Obviously, , that is, is nonempty if is increasing upwards. Now we denote as a relation on defined by, for any ,

(II) , there exists such that and

(b)there exists a countable at most and totally ordered subset such that

() is a totally ordered set and satisfies . is called a link of linking with .

Remark 2.10.

( ) may be satisfied. In fact, we can take empty set as a link of linking and . Thus, implies that for any we can find such that . In this case, we take . Besides, can be a finite set, for example, with then . ( ) and the condition (H1) ensure to exist the supremum, so, from Lemma 2.2 the element satisfying ( ) exists.

Lemma 2.11.

The relation " " satisfies that

Therefore, is a partially ordered set.

- (i)

Hence, is a countable totally ordered subset and satisfies (b ).

Next we will prove that satisfies (b ). It is clear that the following consequences are true, that is, and . It is easy to see that for all by and for all . Also, for all .

Finally, we prove that satisfies (b ). , there exist with (because is totally ordered) and such that . If (or ), then and are ordered by (b ). If , from (b ) and (b ) it follows that , which shows that . To conclude, is totally ordered. Noting that both and have supremums, by the definition of , we have

This shows that satisfies (b ). Consequently, which completes this proof.

## 3. Main Results

Now we can state and prove our main results.

Definition 3.1.

is said to be a fixed point of the multivalued operator if . The fixed point of is said to be a maximal fixed point of if whenever and . If is a fixed point and if whenever and , we say that is a minimal fixed point of .

Theorem 3.2.

Assume that is an ordered topological space. Let , be nonempty and the multivalued operator have sup-closed values such that hypotheses (H1)–(H3) hold. Then admits at least one fixed point in .

Proof.

If has a maximal element , then is a fixed point of . In fact, since , we can find such that . From the definition of we can let . This implies . We claim that . Suppose that , then and . Take empty set as a link of linking with , we have , which contradicts the definition of maximal element.

To prove the existence of maximal element of , by Zorn's lemma, is thus sufficient to show that every totally ordered subset of has an upper bound. Let be any such a subset of . To this purpose, we consider the set . Obviously, . We claim that is totally ordered. Indeed, for any there exist and such that . If , then is ordered. Otherwise, we can assume that , thus, from Lemma 2.8 and (b ) it follows that . Conclusively, is a totally ordered subset.

We will prove that any countable totally ordered subset of has a supremum. It is enough to prove that any given strictly monotone sequence of there is a supremum. From the definition of , there exist and such that for For any , from the definition of , it follows has a supremum. Moreover, Lemma 2.4 guarantees that has a supremum if with for some given . It is suffices to consider the fact that there exists a subsequence of (without loss of generality, we may assume that it is itself) such that .

Case 1.

If it is contrary, there exists some such that . It is easy to know that . (b ) implies that , therefore, . This contradicts increasing. The claim follows.

Taking as the link of linking with for Let

If with , then, by (3.3), . If with , then from condition (b ) it follows that . To sum up, , which, combining the condition (H1), yields that has a supremum. Hence, by Lemma 2.4, has a supremum.

Case 2.

It is clear that has a supremum when is decreasing.

Now, we prove that has a maximal element. Suppose, on the contrary, for any , that there exists such that and . Let , then is an operator mapping into and satisfies and for every . In virtue of Lemma 2.5, there exists such that . On the other hand, by the definition of , we have and , a contradiction. Therefore, has a maximal element, that is, there exists such that for all .

Finally, we shall prove that is an upper bound of . Since , there exists and such that , which implies that On the other hand, since , we have . This compels . Taking empty set as a link of linking with , we have that . Given , in virtue of being totally ordered, or which implies ; or , which, applying (b ), yields . Therefore, . Noting that , we have that . Conclusively, , so, by (a) we have . This shows that is an upper bound of . This proof is completed.

Remark 3.3.

We observe that the result of Theorem 3.2 is true under assumptions of Theorem 3.2 if all are written as The following corollary shows that Theorem 3.2 extends and improves the results of [8].

Corollary 3.4.

Let be an ordered Banach space, be a multivalued operator having nonempty and weakly closed values. Assume that there exists such that conditions (H2), (H3) and (H) hold, then has at least a fixed point.

Proof.

Lemma 2.1 shows that there exists such that for all . By means of Eberlein's theorem and Lemma 2.2 we have that is the supremum of , that is, (H1) is satisfied. Moreover, this implies that , that is, is upper sequentially order closed in the sense of "weak." Since has weakly closed values, has sup-closed values. From Remark 3.3 has a fixed point.

Corollary 3.5.

Let be a weakly sequently completed ordered Banach space, a normal cone. If the operator is bounded and satisfies conditions (H2) and (H3), then has at least one fixed point in .

Proof.

It is suffice to prove that condition (H1) holds. Under these hypotheses, every bounded subset is weakly relatively compact (see [4]), which implies that (H1) is true.

In what follows, we shall consider the existence of maximal and minimal fixed points.

Theorem 3.6.

Under assumptions of Theorem 3.2, has a minimal fixed point in .

Proof.

Let denote the set consisting of fixed points of . From Theorem 3.2 it follows that is nonempty. Set for . Clearly, and is a partially ordered set. By the same methods as to prove Theorem 3.2, we can prove that has a maximal element and is a fixed point of in . It is easy to see that is minimal fixed point of . This completes the proof of Theorem 3.6.

The next result is dual to that of Theorem 3.6.

Theorem 3.7.

Assume that is an ordered topological space. Let , be nonempty and the multivalued operator have inf-closed values such that the following hypotheses are satisfied.

(h1)If is a countable totally ordered set and , then has a infimum.

(h2) is increasing downwards and .

(h3) , where stands for a set which consists of all infimums of (its definition is similar to ).

Then admits at least one fixed point in .

Theorem 3.8.

Assume that the operator is increasing and satisfies conditions (H1)–(H3) and (h1)–(h3), then has maximal and minimal fixed points on .

Remark 3.9.

If has the limit ordinal property, is increasing and has nonempty closed values. Assume that is relatively sequentially compact and conditions (H3) and (h3) hold, then has sup-closed and inf-closed values and satisfies conditions (H1) and (h1) on . Thereby, has maximal and minimal fixed points on . In this sense, we extend and improve the corresponding results of Theorem 2.1 in [10].

Corollary 3.10.

Let be a partially ordered Banach space. If there exist with such that . Assume that is increasing, has nonempty closed values, and satisfies one of the following hypotheses, then has maximal and minimal fixed points on .

(s2) If is countable totally ordered subset and then is relatively compact subset.

(s3) is a weakly relatively compact set.

(s4) is a bounded ordered interval, and for any countable noncompact subset with , one has , where denotes Kuratowskii's noncompactness measure.

Proof.

(s2) implies (H1) and (h1) holds. The rest is clear.

Remark 3.11.

(s2) is main condition of [12] for single-valued operators, (s4) is main condition of [13]. Hence the results presented here extend and improve the corresponding results of the above mentioned papers.

## 4. Application

In this section we assume that is a Banach space with partial ordering derived by the continuous bounded function as follows (see [14]):

To illustrate the ideas involved in Theorem 3.8 we discuss the Hammerstein integral inclusions of the form

Here is a real single-valued function, while is a multivalued map with nonempty closed values.

Let and be the conjugate exponent of , that is, . Let denote the norm of the space . For stipulate that if and only if with all .

In order to prove the existence of solutions to (4.2) in we assume the following.

(S1)The function satisfies that and belongs to .

(S2) is increasing with regard to for fixed .

(S3)There exist with such that and for every .

(S4) has a strongly measurable selection on for each .

Theorem 4.1.

Assume that conditions (S1)–(S5) hold, then (4.2) has maximal and minimal solutions in .

Proof.

whenever . This shows that is a Cauchy sequence, thereby is convergent. Lemma 2.2 guarantees , which yields . From the arbitrariness of it follows that is supremum of . From (4.4) and the dominated convergence theorem, it follows that . Moreover, for . Consequently, . Similarly, we have . This shows that (H3) and (h3) are satisfied for . (S2) guarantees that is increasing. It is easy to see that has closed values. This yields that has sup-closed and inf-closed values.

Finally, we check conditions (H1) and (h1). Suppose that the set is countable, totally ordered, and satisfies for all . We have to prove that the set has a supremum. Since is countable totally ordered, we can assume with for This implies that the sequence is decreasing. In the same way, we can prove that the sequence is convergent. Again, Lemma 2.2 guarantees that has a supremum, which implies that condition (H1) is satisfied. Similarly, we can prove that condition (h1) holds. All conditions of Theorem 3.8 are satisfied, consequently, the operator has minimum and maximum fixed points in and this proof is completed.

Remark 4.2.

By comparing the results of Theorem 4.2 in [15] in which Couchouron and Precup have proved that (4.2) has at least one solution, we omit the conditions that is continuous and has compact values in Theorem 4.1.

## Declarations

### Acknowledgments

This work was supported by the Natural Science Foundation of Zhejiang Province (Y607178) and the Natural Science Foundation of China (10771048).

## Authors’ Affiliations

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