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Fixed Points of Discontinuous Multivalued Operators in Ordered Spaces with Applications
Fixed Point Theory and Applications volume 2010, Article number: 745769 (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,
then has a fixed point in
.
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ1_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ2_HTML.gif)
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.
Take any fixed and let
for
, then
that is,
is increasing. From our assumption it follows that
has a supremum denoted by
. Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ3_HTML.gif)
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
(i) with
and
,
;
(ii) for
;
(iii), and
.
Let , then
is a countable subset and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ4_HTML.gif)
We claim that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ5_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ6_HTML.gif)
and for any define that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ7_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ8_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ9_HTML.gif)
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
,
then for any ,
has the supremum
which belongs to
that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ10_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ11_HTML.gif)
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):
;
and
.
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
(H3).
Define
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ12_HTML.gif)
Obviously, , that is,
is nonempty if
is increasing upwards. Now we denote
as a relation on
defined by, for any
,
(I)
(II), there exists
such that
and
(b)there exists a countable at most and totally ordered subset such that
() for any
.
() There exists such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ13_HTML.gif)
() 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
(i);
(ii) and
implies
;
(iii) implies
.
Therefore, is a partially ordered set.
Proof.
-
(i)
and (ii) are satisfied. Trivial by (I) and (II)(a). To prove (iii), for any given
we take
such that
and we can find
such that
. It is sufficient to assume that at least one of the above equalities does not hold. The definition of
guarantees that
(2.13)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ15_HTML.gif)
and at least one strictly inequality in (2.13) holds. The links linking with
and linking
with
are written, respectively, as
and
. Let
, for any
, if none of equalities in (2.13) holds, then by (b
) we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ16_HTML.gif)
If at least one equality in (2.13) holds, for instance, , then (b
) and (b
) show that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ17_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ18_HTML.gif)
Therefore,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ19_HTML.gif)
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 is strictly increasing, then
. We claim that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ20_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ21_HTML.gif)
(b) shows that
is countable totally ordered. For any
, there exists
such that
. If
, by means of (b
) and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ22_HTML.gif)
we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ23_HTML.gif)
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
.
(s1) is a regular cone.
(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]):
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ24_HTML.gif)
To illustrate the ideas involved in Theorem 3.8 we discuss the Hammerstein integral inclusions of the form
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ25_HTML.gif)
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
.
(S5) a.e. on
for all
. Here
.
Theorem 4.1.
Assume that conditions (S1)–(S5) hold, then (4.2) has maximal and minimal solutions in .
Proof.
Define a multivalued operator as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ26_HTML.gif)
(S4) guarantees that makes sense. For any
with
, there exists
such that
. From (
) and Hölder inequality, it follows that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ27_HTML.gif)
This implies that , that is,
maps
into itself. We seek to apply Theorem 3.8. Note that (S1) and (S3) guarantee that
. For every given
and any
, set
with
for
Thus,
is a decreasing sequence. Note that
is a bounded function, we obtain that the sequence
is convergent. Hence, for any
, there exists a natural number
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F745769/MediaObjects/13663_2009_Article_1336_Equ28_HTML.gif)
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.
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This work was supported by the Natural Science Foundation of Zhejiang Province (Y607178) and the Natural Science Foundation of China (10771048).
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Hong, S., Qiu, Z. Fixed Points of Discontinuous Multivalued Operators in Ordered Spaces with Applications. Fixed Point Theory Appl 2010, 745769 (2010). https://doi.org/10.1155/2010/745769
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DOI: https://doi.org/10.1155/2010/745769