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Some Results for Integral Inclusions of Volterra Type in Banach Spaces
Advances in Difference Equations volume 2010, Article number: 798067 (2010)
Abstract
We first present several existence results and compactness of solutions set for the following Volterra type integral inclusions of the form: , where , is the infinitesimal generator of an integral resolvent family on a separable Banach space , and is a setvalued map. Then the Filippov's theorem and a FilippovWażewski result are proved.
1. Introduction
In the past few years, several papers have been devoted to the study of integral equations on real compact intervals under different conditions on the kernel (see, e.g., [1–4]) and references therein. However very few results are available for integral inclusions on compact intervals, see [5–7]. Topological structure of the solution set of integral inclusions of Volterra type is studied in [8].
In this paper we present some results on the existence of solutions, the compactness of set of solutions, Filippov's theorem, and relaxation for linear and semilinear integral inclusions of Volterra type of the form
where and is the generator of an integral resolvent family defined on a complex Banach space , and is a multivalued map.
In 1980, Da Prato and Iannelli introduced the concept of resolvent families, which can be regarded as an extension of semigroups in the study of a class of integrodifferential equations [9]. It is well known that the following abstract Volterra equation
where is a continuous function, is wellposed if and only if it admits a resolvent family, that is, there is a strongly continuous family , , of bounded linear operators defined in , which commutes with and satisfies the resolvent equation
The study of diverse properties of resolvent families such as the regularity, positivity, periodicity, approximation, uniform continuity, compactness, and others are studied by several authors under different conditions on the kernel and the operator (see [10–24]). An important kernel is given by
where
is the RiemannLiouville kernel. In this case (1.1) and (1.2) can be represented in the form of fractional differential equations and inclusions or abstract fractional differential equations and inclusions. Also in the case where , and is a RiemanLiouville kernel, (1.1) and (1.2) can be represented in the form of fractional differential equations and inclusions, see for instants [25–27].
Our goal in this paper is to complement and extend some recent results to the case of infinitedimensional spaces; moreover the righthand side nonlinearity may be either convex or nonconvex. Some auxiliary results from multivalued analysis, resolvent family theory, and so forth, are gathered together in Sections 2 and 3. In the first part of this work, we prove some existence results based on the nonlinear alternative of LeraySchauder type (in the convex case), on BressanColombo selection theorem and on the Covitz combined the nonlinear alternative of LeraySchauder type for singlevalued operators, and CovitzNadler fixed point theorem for contraction multivalued maps in a generalized metric space (in the nonconvex case). Some topological ingredients including some notions of measure of noncompactness are recalled and employed to prove the compactness of the solution set in Section 4.2. Section 5 is concerned with Filippov's theorem for the problem (1.1). In Section 6, we discuss the relaxed problem, namely, the density of the solution set of problem (1.1) in that of the convexified problem.
2. Preliminaries
In this section, we recall from the literature some notations, definitions, and auxiliary results which will be used throughout this paper. Let be a separable Banach space, an interval in and the Banach space of all continuous functions from into with the norm
refers to the Banach space of linear bounded operators from into with norm
A function is called measurable provided for every open subset , the set is Lebesgue measurable. A measurable function is Bochner integrable if is Lebesgue integrable. For properties of the Bochner integral, see, for example, Yosida [28]. In what follows, denotes the Banach space of functions , which are Bochner integrable with norm
Denote by , closed}, bounded}, convex}, compact.
2.1. Multivalued Analysis
Let and be two metric spaces and be a multivalued map. A singlevalued map is said to be a selection of and we write whenever for every .
is called upper semicontinuous (u.s.c. for short) on if for each the set is a nonempty, closed subset of , and if for each open set of containing , there exists an open neighborhood of such that . That is, if the set is closed for any closed set in . Equivalently, is u.s.c. if the set is open for any open set in .
The following two results are easily deduced from the limit properties.
Lemma 2.1 (see, e.g., [29, Theorem 1.4.13]).
If is u.s.c., then for any ,
Lemma 2.2 (see, e.g., [29, Lemma 1.1.9]).
Let be a sequence of subsets where is compact in the separable Banach space . Then
where refers to the closure of the convex hull of .
is said to be completely continuous if it is u.s.c. and, for every bounded subset , is relatively compact, that is, there exists a relatively compact set such that . is compact if is relatively compact. It is called locally compact if, for each , there exists such that is relatively compact. is quasicompact if, for each subset , is relatively compact.
Definition 2.3.
A multivalued map is said measurable provided for every open , the set is Lebesgue measurable.
We have
The mapping is measurable if and only if for each , the function defined by
is Lebesgue measurable.
The following two lemmas are needed in this paper. The first one is the celebrated KuratowskiRyllNardzewski selection theorem.
Lemma 2.5 (see [31, Theorem 19.7]).
Let be a separable metric space and a measurable multivalued map with nonempty closed values. Then has a measurable selection.
Lemma 2.6 (see [32, Lemma 3.2]).
Let be a measurable multivalued map and a measurable function. Then for any measurable , there exists a measurable selection of such that for a.e. ,
Corollary 2.7.
Let be a measurable multivalued map and a measurable function. Then there exists a measurable selection of such that for a.e. ,
2.1.1. Closed Graphs
We denote the graph of to be the set .
Definition 2.8.
is closed if is a closed subset of , that is, for every sequences and , if , as with, then .
We recall the following two results; the first one is classical.
Lemma 2.9 (see [33, Proposition 1.2]).
If is u.s.c., then is a closed subset of . Conversely, if is locally compact and has nonempty compact values and a closed graph, then it is u.s.c.
Lemma 2.10.
If is quasicompact and has a closed graph, then is u.s.c.
Given a separable Banach space , for a multivalued map , denote
Definition 2.11.
A multivalued map is called a Carathéodory function if
(a)the function is measurable for each ;
(b)for a.e. , the map is upper semicontinuous.
Furthermore, is Carathéodory if it is locally integrably bounded, that is, for each positive , there exists such that
For each , the set
is known as the set of selection functions.
Remark 2.12.

(a)
For each , the set is closed whenever has closed values. It is convex if and only if is convex for .

(b)
From [34] (see also [35] when is finitedimensional), we know that is nonempty if and only if the mapping belongs to . It is bounded if and only if the mapping belongs to ; this particularly holds true when is Carathéodory. For the sake of completeness, we refer also to Theorem 1.3.5 in [36] which states that contains a measurable selection whenever is measurable and is a Carathéodory function.
Lemma 2.13 (see [35]).
Given a Banach space , let be an Carathéodory multivalued map, and let be a linear continuous mapping from into . Then the operator
has a closed graph in .
For further readings and details on multivalued analysis, we refer to the books by Andres and Górniewicz [37], Aubin and Cellina [38], Aubin and Frankowska [29], Deimling [33], Górniewicz [31], Hu and Papageorgiou [34], Kamenskii et al. [36], and Tolstonogov [39].
2.2. Semicompactness in
Definition 2.14.
A sequence is said to be semicompact if

(a)
it is integrably bounded, that is, there exists such that
(2.13) 
(b)
the image sequence is relatively compact in for a.e. .
We recall two fundamental results. The first one follows from the DunfordPettis theorem (see [36, Proposition 4.2.1]). This result is of particular importance if is reflexive in which case (a) implies (b) in Definition 2.14.
Lemma 2.15.
Every semicompact sequence is weakly compact in .
The second one is due to Mazur, 1933.
Lemma 2.16 (Mazur's Lemma, [28]).
Let be a normed space and be a sequence weakly converging to a limit . Then there exists a sequence of convex combinations with for and , which converges strongly to .
3. Resolvent Family
The Laplace transformation of a function is defined by
if the integral is absolutely convergent for . In order to defined the mild solution of the problems (1.1) we recall the following definition.
Definition 3.1.
Let be a closed and linear operator with domain defined on a Banach space . We call the generator of an integral resolvent if there exists and a strongly continuous function such that
In this case, is called the integral resolvent family generated by .
The following result is a direct consequence of ([16, Proposition 3.1 and Lemma 2.2]).
Proposition 3.2.
Let be an integral resolvent family with generator . Then the following conditions are satisfied

(a)
is strongly continuous for and ;

(b)
and for all ;

(c)
for every and ,
(3.3) 
(d)
let . Then , and
(3.4)
In particular, .
Remark 3.3.
The uniqueness of resolvent is well known (see Prüss [24]).
If an operator with domain is the infinitesimal generator of an integral resolvent family and is a continuous, positive and nondecreasing function which satisfies , then for all we have
(see [22, Theorem 2.1]). For example, the case corresponds to the generator of a semigroup and actually corresponds to the generator of a sine family; see [40]. A characterization of generators of integral resolvent families, analogous to the HilleYosida Theorem for semigroups, can be directly deduced from [22, Theorem 3.4]. More information on the semigroups and sine families can be found in [41–43].
Definition 3.4.
A resolvent family of bounded linear operators, , is called uniformly continuous if
Definition 3.5.
The solution operator is called exponentially bounded if there are constants and such that
4. Existence Results
4.1. Mild Solutions
In order to define mild solutions for problem (1.1), we proof the following auxiliary lemma.
Lemma 4.1.
Let . Assume that generates an integral resolvent family on , which is in addition integrable and . Let be a continuous function (or ), then the unique bounded solution of the problem
is given by
Proof.
Let be a solution of the integral equation (4.2), then
Using the fact that is solution operator and Fubini's theorem we obtain
Thus
This lemma leads us to the definition of a mild solution of the problem (1.1).
Definition 4.2.
A function is said to be a mild solution of problem (1.1) if there exists such that a.e. on such that
Consider the following assumptions.
(B_{1}) The operator solution is compact for .
(B_{2})There exist a function and a continuous nondecreasing function such that
with
(B_{3}) For every , is uniformly continuous.
In all the sequel we assume that is exponentially bounded. Our first main existence result is the following.
Theorem 4.3.
Assume is a Carathéodory map satisfying or . Then problem (1.1) has at least one solution. If further is a reflexive space, then the solution set is compact in .
The following socalled nonlinear alternatives of LeraySchauder type will be needed in the proof (see [31, 44]).
Lemma 4.4.
Let be a normed space and a compact, u.s.c. multivalued map. Then either one of the following conditions holds.

(a)
has at least one fixed point,

(b)
the set is unbounded.
The singlevalued version may be stated as follows.
Lemma 4.5.
Let be a Banach space and a nonempty bounded, closed, convex subset. Assume is an open subset of with and let be a a continuous compact map. Then

(a)
either there is a point and with ,

(b)
or has a fixed point in .
Proof of Theorem 4.3.
We have the following parts.
Part 1: Existence of Solutions
It is clear that all solutions of problem (1.1) are fixed points of the multivalued operator defined by
where
Notice that the set is nonempty (see Remark 2.12,(b)). Since, for each , the nonlinearity takes convex values, the selection set is convex and therefore has convex values.
Step 1 ( is completely continuous). (a) sends bounded sets into bounded sets in. Let , be a bounded set in , and . Then for each , there exists such that
Thus for each ,
(b) maps bounded sets into equicontinuous sets of. Let , and be a bounded set of as in (a). Let ; then for each
The righthand side tends to zero as since is uniformly continuous.
(c) As a consequence of parts (a) and (b) together with the ArzélaAscoli theorem, it suffices to show that maps into a precompact set in . Let and let . For , define
Then
which tends to 0 as . Therefore, there are precompact sets arbitrarily close to the set . This set is then precompact in .
Step 2 ( has a closed graph).
Let and. We will prove that . means that there exists such that for each
First, we have
Now, consider the linear continuous operator defined by
From the definition of , we know that
Since and is a closed graph operator by Lemma 2.13, then there exists such that
Hence , proving our claim. Lemma 2.10 implies that is u.s.c.
Step 3 (a priori bounds on solutions).
Let be such that . Then there exists such that
Then
Set
then and for a.e. we have
Thus
Using a change of variable we get
From ) there exists such that
Let
and consider the operator . From the choice of , there is no such that for some . As a consequence of the LeraySchauder nonlinear alternative (Lemma 4.4), we deduce that has a fixed point in which is a mild solution of problem (1.1).
Part 2: Compactness of the Solution Set
Let
From Part 1, and there exists such that for every , . Since is completely continuous, then is relatively compact in . Let ; then and . It remains to prove that is closed set in . Let such that converge to . For every , there exists a.e. such that
implies that , hence is integrably bounded. Note this still remains true when holds for is a bounded set. Since is reflexive, is semicompact. By Lemma 2.15, there exists a subsequence, still denoted , which converges weakly to some limit . Moreover, the mapping defined by
is a continuous linear operator. Then it remains continuous if these spaces are endowed with their weak topologies. Therefore for a.e. , the sequence converges to , it follows that
It remains to prove that , for a.e.. Lemma 2.16 yields the existence of , such that and the sequence of convex combinaisons converges strongly to in . Since takes convex values, using Lemma 2.2, we obtain that
Since is u.s.c. with compact values, then by Lemma 2.1, we have
This with (4.33) imply that . Since has closed, convex values, we deduce that , for a.e., as claimed. Hence which yields that is closed, hence compact in .
4.2. The Convex Case: An MNC Approach
First, we gather together some material on the measure of noncompactness. For more details, we refer the reader to [36, 45] and the references therein.
Definition 4.6.
Let be a Banach space and a partially ordered set. A map is called a measure of noncompactness on , MNC for short, if
for every .
Notice that if is dense in , then and hence
Definition 4.7.
A measure of noncompactness is called

(a)
monotone if , implies .

(b)
nonsingular if for every .

(c)
invariant with respect to the union with compact sets if for every relatively compact set , and .

(d)
real if and for every bounded .

(e)
semiadditive if for every .

(f)
regular if the condition is equivalent to the relative compactness of .
As example of an MNC, one may consider the Hausdorf MNC
Recall that a bounded set has a finite net if there exits a finite subset such that where is a closed ball in .
Other examples are given by the following measures of noncompactness defined on the space of continuous functions with values in a Banach space :

(i)
the modulus of fiber noncompactness
(4.38)where is the Hausdorff MNC in and ;

(ii)
the modulus of equicontinuity
(4.39)It should be mentioned that these MNC satisfy all abovementioned properties except regularity.
Definition 4.8.
Let M be a closed subset of a Banach space and an MNC on . A multivalued map is said to be condensing if for every , the relation
implies the relative compactness of .
Some important results on fixed point theory with MNCs are recalled hereafter (see, e.g., [36] for the proofs and further details). The first one is a compactness criterion.
Lemma 4.9 (see [36, Theorem 5.1.1]).
Let be an abstract operator satisfying the following conditions:
(S_{1}) is Lipschitz: there exists such that for every
(S_{2}) is weaklystrongly sequentially continuous on compact subsets: for any compact and any sequence such that for a.e. , the weak convergence implies the strong convergence as .
Then for every semicompact sequence , the image sequence is relatively compact in .
Lemma 4.10 (see [36, Theorem 5.2.2]).
Let an operator satisfy conditions together with the following:
(S_{3}) there exits such that for every integrable bounded sequence , one has
where is the Hausdorff MNC.
Then
where is the constant in .
The next result is concerned with the nonlinear alternative for condensing u.s.c. multivalued maps.
Lemma 4.11 (see [36]).
Let be a bounded open neighborhood of zero and a condensing u.s.c. multivalued map, where is a nonsingular measure of noncompactness defined on subsets of , satisfying the boundary condition
for all and . Then .
Lemma 4.12 (see [36]).
Let be a closed subset of a Banach space and is a closed condensing multivalued map where is a monotone MNC on . If the fixed point set Fix is bounded, then it is compact.
4.2.1. Main Results
In all this part, we assume that there exists such that
Let be a Carathéodory multivalued map which satisfies Lipschitz conditions with respect to the Hausdorf MNC.
(B_{4}) There exists such that for every bounded in ,
Lemma 4.13.
Under conditions and , the operator is closed and , for every where is as defined in the proof of Theorem 4.3.
Proof.
We have the following steps.
Step 1 ( is closed).
Let , and. We will prove that . means that there exists such that for a.e.
Since , Assumption implies that is integrably bounded. In addition, the set is relatively compact for a.e. because Assumption both with the convergence of imply that
Hence the sequence is semicompact, hence weakly compact in to some limit by Lemma 2.15. Arguing as in the proof of Theorem 4.3 Part 2, and passing to the limit in (4.47), we obtain that and for each
As a consequence, , as claimed.
Step 2 ( has compact, convex values).
The convexity of follows immediately by the convexity of the values of . To prove the compactness of the values of , let for some and . Then there exists satisfying (4.47). Arguing again as in Step 1, we prove that is semicompact and converges weakly to some limit , a.e. hence passing to the limit in (4.47), tends to some limit in the closed set with satisfying (4.49). Therefore the set is sequentially compact, hence compact.
Lemma 4.14.
Under the conditions and , the operator is u.s.c.
Proof.
Using Lemmas 2.10 and 4.13, we only prove that is quasicompact. Let be a compact set in and such that . Then there exists such that
Since is compact, we may pass to a subsequence, if necessary, to get that converges to some limit in . Arguing as in the proof of Theorem 4.3 Step 1, we can prove the existence of a subsequence which converges weakly to some limit and hence converges to , where
As a consequence, is u.s.c.
We are now in position to prove our second existence result in the convex case.
Theorem 4.15.
Assume that satisfies Assumptions and . If
then the set of solutions for problem (1.1) is nonempty and compact.
Proof.
It is clear that all solutions of problem (1.1) are fixed points of the multivalued operator defined in Theorem 4.3. By Lemmas 4.13 and 4.14, and it is u.s.c. Next, we prove that is a condensing operator for a suitable MNC . Given a bounded subset , let the modulus of quasiequicontinuity of the set of functions denote
It is well known (see, e.g., [36, Example 2.1.2]) that defines an MNC in which satisfies all of the properties in Definition 4.7 except regularity. Given the Hausdorff MNC , let be the real MNC defined on bounded subsets on by
Finally, define the following MNC on bounded subsets of by
where is the collection of all countable subsets of . Then the MNC is monotone, regular and nonsingular (see [36, Example 2.1.4]).
To show that is condensing, let be a bounded set in such that
We will show that is relatively compact. Let and let where is defined by
is defined by
Then
Moreover, each element in can be represented as
Moreover (4.56) yields
From Assumption , it holds that for a.e. ,
Lemmas 4.9 and 4.10 imply that
Therefore
Since , we infer that
implies that , for a.e. . In turn, (4.62) implies that
Hence (4.60) implies that . To show that , i.e, the set is equicontinuous, we proceed as in the proof of Theorem 4.3 Step 1 Part (b). It follows that which implies, by (4.61), that . We have proved that is relatively compact. Hence is u.s.c. and condensing, where is as in the proof of Theorem 4.3. From the choice of , there is no such that for some . As a consequence of the nonlinear alternative of LeraySchauder type for condensing maps (Lemma 4.11), we deduce that has a fixed point in , which is a solution to problem (1.1). Finally, since is bounded, by Lemma 4.12, is further compact.
4.3. The Nonconvex Case
In this section, we present a second existence result for problem (1.1) when the multivalued nonlinearity is not necessarily convex. In the proof, we will make use of the nonlinear alternative of LeraySchauder type [44] combined with a selection theorem due to Bressan and Colombo [46] for lower semicontinuous multivalued maps with decomposable values. The main ingredients are presented hereafter. We first start with some definitions (see, e.g., [47]). Consider a topological space and a family of subsets of .
Definition 4.16.
is called measurable if belongs to the algebra generated by all sets of the form where is Lebesgue measurable in and is Borel measurable in .
Definition 4.17.
A subset is decomposable if for all and for every Lebesgue measurable set , we have:
where stands for the characteristic function of the set .
Let be a multivalued map with nonempty closed values. Assign to the multivalued operator defined by . The operator is called the Nemyts'kiĭ operator associated to .
Definition 4.18.
Let be a multivalued map with nonempty compact values. We say that is of lower semicontinuous type (l.s.c. type) if its associated Nemyts'kiĭ operator is lower semicontinuous and has nonempty closed and decomposable values.
Next, we state a classical selection theorem due to Bressan and Colombo.
Let be a separable metric space and let be a Banach space. Then every l.s.c. multivalued operator , with closed decomposable values has a continuous selection, that is, there exists a continuous singlevalued function such that for every .
Let us introduce the following hypothesis.
(H_{1}) is a nonempty compact valued multivalued map such that

(a)
the mapping is measurable;

(b)
the mapping is lower semicontinuous for a.e. .
The following lemma is crucial in the proof of our existence theorem.
Lemma 4.20 (see, e.g., [48]).
Let be an integrably bounded multivalued map satisfying . Then is of lower semicontinuous type.
Theorem 4.21.
Suppose that the hypotheses or and are satisfied. Then problem (1.1) has at least one solution.
Proof.
imply, by Lemma 4.20, that is of lower semicontinuous type. From Lemma 4.19, there is a continuous selection such that for all . Consider the problem
and the operator defined by
As in Theorem 4.3, we can prove that the singlevalued operator is compact and there exists such that for all possible solutions , we have . Now, we only check that is continuous. Let be a sequence such that in , as . Then
Since the function is continuous, we have
Let
From the choice of , there is no such that for in . As a consequence of the nonlinear alternative of the LeraySchauder type (Lemma 4.5), we deduce that has a fixed point which is a solution of problem (4.68), hence a solution to the problem (1.1).
4.4. A Further Result
In this part, we present a second existence result to problem (1.1) with a nonconvex valued righthand side. First, consider the Hausdorff pseudometric distance
defined by
where and . Then is a metric space and is a generalized metric space (see [49]). In particular, satisfies the triangle inequality.
Definition 4.22.
A multivalued operator is called

(a)
Lipschitz if there exists such that
(4.75) 
(b)
a contraction if it is Lipschitz with .
Notice that if is Lipschitz, then for every ,
Our proofs are based on the following classical fixed point theorem for contraction multivalued operators proved by Covitz and Nadler in 1970 [50] (see also Deimling [33, Theorem 11.1]).
Lemma 4.23.
Let be a complete metric space. If is a contraction, then .
Let us introduce the following hypotheses:
(A_{1}) ; is measurable for each ;
(A_{2}) there exists a function such that
with
Theorem 4.24.
Let Assumptions be satisfied. Then problem (1.1) has at least one solution.
Proof.
In order to transform the problem (1.1) into a fixed point problem, let the multivalued operator be as defined in Theorem 4.3. We will show that satisfies the assumptions of Lemma 4.23.

(a)
for each . Indeed, let be a sequence converge to . Then there exists a sequence such that
(4.79)
Since has compact values, let be such that . From and , we infer that for a.e.
Then the Lebesgue dominated convergence theorem implies that, as ,
with
proving that .
(b) There exists , such that
Let and . Then there exists ( is a measurable selection) such that for each
() tells us that
Hence there is such that
Then consider the mapping , given by
that is . Since are measurable, Theorem III.4.1 in [30] tells us that the closed ball is measurable. Finally the set is nonempty since it contains . Therefore the intersection multivalued operator is measurable with nonempty, closed values (see [29–31]). By Lemma 2.5, there exists a function , which is a measurable selection for . Thus and
Let us define for a.e.
Then
Thus
By an analogous relation, obtained by interchanging the roles of and , we finally arrive at
where and
is the Bieleckitype norm on . So, is a contraction and thus, by Lemma 4.23, has a fixed point , which is a mild solution to (1.1).
Arguing as in Theorem 4.3, we can also prove the following result the proof of which is omitted.
Theorem 4.25.
Let be a reflexive Banach space. Suppose that all conditions of Theorem 4.24 are satisfied and . Then the solution set of problem (1.1) is nonempty and compact.
5. Filippov's Theorem
5.1. Filippov's Theorem on a Bounded Interval
Let be a mild solution of the integral equation:
We will consider the following two assumptions.
(C_{1})The function is such that

(a)
for all is measurable,

(b)
the map is integrable.
(C_{2})There exist a function and a positive constant such that
Theorem 5.1.
Assume that the conditions . Then, for every problem (1.1) has at least one solution satisfying, for a.e. , the estimates
where .
Proof.
We construct a sequence of functions which will be shown to converge to some solution of problem (1.1) on the interval , namely, to
Let on and , that is,
Then define the multivalued map by . Since and are measurable, Theorem III.4.1 in [30] tells us that the ball is measurable. Moreover is measurable (see [29]) and is nonempty. Indeed, since is a measurable function, from Lemma 2.6, there exists a function which is a measurable selection of and such that
Then , proving our claim. We deduce that the intersection multivalued operator is measurable (see [29–31]). By Lemma 2.7 (KuratowskiRyllNardzewski selection theorem), there exists a function which is a measurable selection for . Consider
For each , we have
Hence
Using the fact that is measurable, the ball is also measurable by Theorm III.4.1 in [30]. From we have
Hence there exist such that
We consider the following multivalued map is nonempty. Therefore the intersection multivalued operator is measurable with nonempty, closed values (see [29–31]). By Lemma 2.5, there exists a function , which is a measurable selection for . Thus and
Define
Using (5.8) and (5.12), a simple integration by parts yields the following estimates, valid for every :
Let . Arguing as for , we can prove that is a measurable multivalued map with nonempty values; so there exists a measurable selection . This allows us to define
For , we have
Then
Performing an integration by parts, we obtain, since is a nondecreasing function, the following estimates:
Let . Then, arguing again as for , , , we show that is a measurable multivalued map with nonempty values and that there exists a measurable selection in . Define
For , we have
Repeating the process for , we arrive at the following bound:
By induction, suppose that (5.21) holds for some and check (5.21) for . Let . Since is a nonempty measurable set, there exists a measurable selection , which allows us to define for
Therefore, for a.e. , we have
Again, an integration by parts leads to
Consequently, (5.21) holds true for all . We infer that is a Cauchy sequence in , converging uniformly to a limit function . Moreover, from the definition of , we have
Hence, for a.e. , is also a Cauchy sequence in and then converges almost everywhere to some measurable function in . In addition, since , we have for a.e.
Hence
where
From the Lebesgue dominated convergence theorem, we deduce that converges to in . Passing to the limit in (5.22), we find that the function
is solution to problem (1.1) on . Moreover, for a.e. , we have
Passing to the limit as , we get
with
6. The Relaxed Problem
More precisely, we compare, in this section, trajectories of the following problem,
and those of the convexified Volttera integral inclusion problem
where refers to the closure of the convex hull of the set . We will need the following auxiliary results in order to prove our main relaxation theorem.
Lemma 6.1 (see [29]).
Let be a measurable, integrably bounded setvalued map and let be an integrable map. Then the integral is convex, the map is measurable and, for every , and every measurable selection of , there exists a measurable selection of such that
With being a reflexive Banach space, the following hypotheses will be assumed in this section:
The function satisfies
(a)for all the map is measurable,
(b)the map is integrable bounded (i.e., there exists such that .
Then our main contribution is the following.
Theorem 6.2.
Assume that and hold. Then problem (6.2) has at least one solution. In addition, for all and every solution of problem (6.2), has a solution defined on satisfying
In particular , where
Remark 6.3.
Notice that the multivalued map also satisfies .
Proof.
Part 1 ()
For this, we first transform problem (6.2) into a fixed point problem and then make use of Lemma 4.23. It is clear that all solutions of problem (6.2) are fixed points of the multivalued operator defined by
where
To show that satisfies the assumptions of Lemma 4.23, the proof will be given in two steps.
Step 1.
for each . Indeed, let be such that in , as . Then and there exists a sequence such that
Then is integrably bounded. Since has closed values and integrable bounded, then from Corollary 4.14 for every we have such that
Since is a convergent sequence the there exists such that
Then
that is
Since is weakly compact in the reflexive Banach space , there exists a subsequence, still denoted , which converges weakly to by the DunfordPettis theorem. By Mazur's Lemma (Lemma 2.16), there exists a second subsequence which converges strongly to in , hence almost everywhere. Then the Lebesgue dominated convergence theorem implies that, as ,
proving that .
Step 2.
There exists such that for each where the norm will be chosen conveniently. Indeed, let and . Then there exists such that for each
Since, for each ,
then there exists some such that
Consider the multimap defined by
As in the proof of Theorem 5.1, we can show that the multivalued operator is measurable and takes nonempty values. Then there exists a function , which is a measurable selection for . Thus, and
For each , let
Therefore, for each , we have
Hence
where
By an analogous relation, obtained by interchanging the roles of and , we find that
Then is a contraction and hence, by Lemma 4.23, has a fixed point , which is solution to problem (6.2).
Part 2
Let be a solution of problem (6.2). Then, there exists such that
that is, is a solution of the problem
Let and be given by the relation . From Lemma 6.1, there exists a measurable selection of such that
Let
Hence
With Assumption , we infer from Corollary 4.14 that there exists such that
Then
Since, under and , is measurable (see [29]), by the above inequality, we deduce that . From Theorem 5.1, problem (6.1) has a solution which satisfies
where
Using the definition of , we obtain the upper bound
Since is arbitrary, , showing the density relation .
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This paper was completed when the second and forth authors visited the Department of Mathematical Analysis of the University of Santiago de Compostela. The authors would like to thank the department for its hospitality and support. The authors are grateful to the referee for carefully reading the paper.
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Agarwal, R.P., Benchohra, M., Nieto, J.J. et al. Some Results for Integral Inclusions of Volterra Type in Banach Spaces. Adv Differ Equ 2010, 798067 (2010). https://doi.org/10.1155/2010/798067
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DOI: https://doi.org/10.1155/2010/798067