Convex Sweeping Processes with Noncompact Perturbations and with Delay in Banach Spaces

Ibrahim, A. G.; Aladsani, F.

doi:https://doi.org/10.1155/2016/3853205

Abstract and Applied Analysis

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Research Article | Open Access

Volume 2016 | Article ID 3853205 | https://doi.org/10.1155/2016/3853205

Convex Sweeping Processes with Noncompact Perturbations and with Delay in Banach Spaces

A. G. Ibrahim¹and F. Aladsani¹

Academic Editor: Khalil Ezzinbi

Received13 Sept 2015

Accepted20 Sept 2015

Published23 Feb 2016

Abstract

We prove two results concerning the existence of solutions for functional differential inclusions that are governed by sweeping processes, with noncompact valued perturbations in Banach spaces. Indeed, we have two goals. The first is to give a technique that allows considering sweeping processes with noncompact valued perturbations and associated with a multivalued function depending on time. The second is to give a technique to overcome the arising problem from the nonlinearity of the normalized mappings, when we deal with sweeping processes with noncompact valued perturbations and associated with a multivalued function depending on time and state.

1. Introduction

In his leading paper, Moreau [1] proposed and studied the following differential inclusion governed by sweeping process of first order:where is a multifunction from the interval to the family of nonempty closed convex subsets of a Hilbert space and is the normal cone of the subset at the point . Problem (1) corresponds to several important mechanical problems. For more details concerning the applications of (1), we refer to [1].

Since then, important improvements have been developed by several authors. For some existence results of solutions for sweeping processes in Hilbert or Banach spaces, we refer to [2–10].

Let be a Banach space, a positive real number, the Banach space of continuous functions from to endowed with the uniform norm , a multifunction from to the family of nonempty closed convex subsets of , a multifunction defined on and with nonempty closed values in the topological dual space, , of , and a multifunction from to the family of nonempty closed convex subsets of . Let and be the normalized duality mappings and for each , and , . Let be given.

In this paper, we prove two existence results. In the first result (Theorem 15), we prove, in -uniformly convex and -uniformly smooth Banach space, the existence of solutions for the following sweeping process with noncompact valued perturbation and with delay:

In the second result (Theorem 16), we prove, in -uniformly convex and -uniformly smooth Banach spaces, the existence of solutions for the following sweeping process with a noncompact valued perturbation and with delay:

We prove the existence of solutions for (3) without imposing that the values of are contained in a fixed convex compact subset; instead, we suppose that satisfies a condition in terms of the Hausdorff measure of noncompactness.

It is important to note that when the sweeping process is associated with a multifunction depending on time and state, the nonlinearity of the normalized duality mappings causes a difficulty when we deal with sweeping process with a perturbation; see, for example, the remark which has been given by Bounkhel and Castaing [6]. For this reason we define the multifunction on

In order to explain the mathematical motivation for this work, we mention some recent results in this domain.

Aitalioubrahim [2] considered (2) when is a Hilbert space and the value of is nonempty compact and not necessarily convex.

In Theorem (4.3) in [5], Bounkhel and Al-Yusof established the existence of solutions for (2), in a separable, -uniformly convex and -uniformly smooth Banach space without delay and when , for any , for some convex compact set ; is upper semicontinuous with convex compact values and , , for some convex compact subset .

In Theorem (4.5) in [6], Bounkhel and Castaing considered (3) in a separable, -uniformly convex and -uniformly smooth Banach space, , and when and is defined from to the family of nonempty closed convex subsets of and satisfies a condition in terms of the Hausdorff measure of noncompactness.

Castaing et al. [10] considered a second-order sweeping process without delay in a separable Hilbert space in the case when is a Lipschitz multifunction defined on and taking closed -proxy regular () values in ; and is a convex weakly compact valued scalary upper semicontinuous defined on .

Ibrahim and AL-Adsani [7] considered a second order sweeping process with delay in a separable -uniformly convex and -uniformly smooth Banach space and the values of perturbation are nonempty convex weakly compact.

Noel and Thibault [8] established the existence of solutions for nonconvex sweeping processes with a moving set depending on the state in Hilbert spaces.

For other contributions on differential inclusions, see Gomaa [11].

We note that if is a Hilbert space, then is equal to the identity mapping, is -uniformly convex and -uniformly smooth Banach space. Therefore, our technique allows discussing some sweeping process problems with noncompact perturbation in Hilbert spaces or in Banach spaces, whether the moving set depends on time or on time and state.

The paper is organized as follows. Section 2 is devoted to some definitions and notations needed later. In Section 3, we prove the existence of solutions for (2). In Section 4, we give existence of solutions for (3).

2. Preliminaries and Notations

Let is nonempty closed, is nonempty closed convex, and is nonempty, convex and compact. Let and .

Definition 1 (see Def. 2.4.1 [12]). The multivalued mappings are called the normalized duality mappings.

In the following Lemma, we recall some properties of and

Lemma 2 (see [12], Prop. 2.4.5, 2.4.12, and 2.4.15). We have the following:(1)If is a Hilbert space, then , for all .(2)For each , is nonempty closed convex and bounded subset of .(3), for all and all .(4)If is strictly convex, is single-valued.(5)If is strictly convex, is one to one; that is, .(6)If is uniformly convex, then is uniformly continuous on each bounded set in ; that is, and there is a such that for and with and we have (note that if is uniformly convex, then it is strictly convex and hence is single-valued mapping).(7)If is reflexive, then is a mapping from onto , that is (8)If is reflexive strictly convex space with strictly convex conjugate space , then and are one-to-one, onto and single-valued mapping and where is the identity mapping on and is the identity mapping on .

For more properties of the duality mapping, we refer to [12–14].

Definition 3 (see [12], Def. 2.8.1). The Banach space is said to be uniformly smooth if where is the modulus of smoothness of .

Definition 4 (see [12–14]). Let be a real number. A Banach space is said to be -uniformly smooth if there exists a constant such that , .

Clearly, every -uniformly smooth Banach space is uniformly smooth.

Lemma 5 (see [12–14]). The following properties are satisfied:(1) is uniformly smooth if and only if is uniformly convex.(2) is uniformly convex if and only if is uniformly smooth.(3)If is uniformly smooth, then is reflexive.

Lemma 6 (see [12, 13]). Let :(1)If is -uniformly convex, then is -uniformly smooth where .(2)If is -uniformly smooth, then is -uniformly convex where .

Remark 7. It is known that(1)the Banach space is uniformly smooth;(2)if , then is -uniformly convex; if , then is -uniformly convex.

Now, let be two functions defined by

Observe that , , and if is a Hilbert space, then

Definition 8 (see [12]). Let be a nonempty subset of and . If there exists a point satisfying , then is called a generalized projection of onto , where .

The set of all such points is denoted by ; that is,

The following lemma summarizes some important properties of and (see [12]).

Lemma 9. Let be a reflexive Banach space with dual space and let be a nonempty closed convex subset of . The following properties hold:(1), .(2)If is uniformly convex or uniformly smooth, then (3) is strictly convex if and only if is singleton for all .(4)If is also smooth, then, for any given ,

Definition 10 (see [4, 14]). Let be a nonempty closed convex subset of and . The convex normal cone of at is defined by

The following lemma gives a closedness property of the subdifferential of the distance function associated with a set-valued mapping with closed convex values.

Lemma 11 (see [14]). Let be a nonempty, closed, and convex subset of a Banach space and . Then,(1), where is the subdifferential of the distance function;(2)if is reflexive and smooth, then

Let us recall the following lemmas that will be used in the sequel.

Lemma 12 (see [5], Lemma and Prop. 4.2). Let , , let be a -uniformly convex and -uniformly smooth Banach space, and let be a nonempty bounded subset of ; then, there exist two constants and such that Moreover, if , in addition, is closed and , then

Lemma 13 (see [6], Prop. 4.3). Let , and let be a -uniformly convex and -uniformly smooth Banach space. The normalized duality mapping is Lipschitz on bounded sets; that is, for any , there is a positive constant dependent on , such that

Lemma 14 (see [15], Lemma ). Let be a separable Banach space, let be a measurable multifunction, and let be a measurable function. Then, for any positive measurable function , there exists a measurable selection of such that for almost all

3. Existence Results of Solutions for Problem (2)

Theorem 15. Let , , be a separable, -uniformly convex and -uniformly smooth Banach space, and . Suppose that the following conditions hold:There is a convex compact subset , such that , .There are two constants and such that for any , and any , we have where There is a continuous function such that, for any and , There is a continuous function such that, for any and any ,

Then, for any fixed with , there is a continuous function such that is Lipschitz on and satisfies (2).

Proof. Since is uniformly convex, it is reflexive and strictly convex. Moreover, because is uniformly smooth, its topological conjugate is uniformly convex and hence is strictly convex. Then, by property (8) in Lemma 2, the normalized duality maps and are one-to-one, onto, single-valued maps and ,, where and are the identity maps on and , respectively. Moreover, since is reflexive and strictly convex, Lemma 9 ensures that the generalized projection is singleton for any closed convex subset of and for any . Finally, from the reflexivity and smoothness of , Lemma 11 tells us thatNow, for any fixed natural number , we consider the following partition for : , We put , , , and . Also, let be defined as , , , , , , , and . In order to make the proof easy for the reader, we divide the rest of the proof into steps.
Step 1. In this step, we show that if is a fixed natural number and is a fixed function, then there are , , and such that and are absolutely continuous on and the following properties hold:Indeed, in view of Lemma 14, there is such that , and Put andNext, we define for andObserve that, according to Lemma 9, this construction is well defined. Moreover, from the definition of the generalized projection, relation (26) gives us and this relation together with (27) implies Next, by induction, we can define for whereNow, let be such that , , , and . Note that , for . Therefore, the functions , , and satisfy properties (i)–(vi) in (24).
Step 2. In this step, we show that there are three sequences , , and such that and are absolutely continuous on and for any we haveIndeed, let be any fixed element in ; in view of Step 1, there are , , and such that are absolutely continuous on and properties (i)–(vi) in (24) are satisfied for . Now, since , then in view of Step 1, there are , , and such that are absolutely continuous on and (32) is satisfied at So, we can define inductively three sequences , and such that properties (i)–(vi) in (32) are satisfied.
Step 3. In this step, we show that there is a positive number such that for any Let be a fixed natural number. According to () and (), for any ,where . Then, for every ,Let and Note that is -uniformly convex and -uniformly smooth ( Therefore, in view of Lemma 12 and (35), there exists a positive constant such that Then, for every ,Further, since is -uniformly convex and -uniformly smooth, then, again, by Lemma 12, there is a positive constant such that So, for every , This relation together with (37) yields According to , the last inequality impliesMoreover, from the assumption , we infer that Then, is -uniformly convex and -uniformly smooth. Hence, from Lemma 13, there is a constant , depending on , such that, for all , , , and , we haveConsequently, This inequality and (41) yield Thus, where and
Therefore, for , , and , we get This proves that (33) is true.
Step 4. Our claim in this step is to show that the sequence has a convergent subsequence, still denoted by , converging uniformly to a Lipschitz function converges weakly to , and converges uniformly to .
Indeed, according to the definition of and (33), the sequence is equicontinuous. Moreover, let . Then, there are , , such that Hence, in view of (30) and (), This shows that the set is relatively compact in . Therefore, Theorem . Ch.1. in [16] implies that there is a Lipschitz function , such that the sequence has a subsequence, still denoted by , which converges uniformly to , and converges weakly to We extend the definition of on by putting , . Thus, converges uniformly to on . Since is uniformly continuous on the compact set , then the sequence is uniformly convergent to with . This proves our claim in this step.
Step 5. Let us show that , .
Let . According to (), we haveSince , therefore, ,
Step 6. In this step, we show that , for every .
Let We haveBy the continuity of , the uniform convergence of towards , and the preceding estimate, we get .
Step 7. In this step, we show that the sequence , defined by , converges almost everywhere to a function and , a.e. .
To prove this, let and let be a fixed point such that (32)(iv) and (v) are satisfied. In view of (), we have Thus, for any two natural numbers and (), we infer that Since , is continuous, and , then the right-hand side of the last inequality tends to zero when Hence, the sequence is a Cauchy sequence in Thus, there is a function such thatObserve that, by (34), , So, . It remains to show that , a.e. . Indeed, for every and almost everywhere , we have This relation and () imply Because converges to almost everywhere, is continuous, and , for every , we conclude that , a.e. .
Step 8. We proceed to show that We follow the arguments used in [10]. In view of (32)(vi), (37), and Lemma 11, for every natural number one obtainswhere . . Observe that the uppersemicontinuity of the subdifferential of Lipschtz function ([17], Prop. 1.7) imply the multivalued function is scalary uppersemicontinuous with convex and compact values. That is, let , in , , in , , and Then, for any , So, for any ,where is the support function.
Now, let be a sequence in which separates the points. Hence, from the weak convergence of the sequence to in , for any Lebesgue measurable subset , we have This relation with (55) and (57) yields Therefore, Again, since the multivalued function is measurable with convex and compact, then by ([18], III.35) it follows that As , , we get This completes the proof.

4. Existence Result for (3)

In the following theorem, we present an existence theorem of solutions for (3). We do not assume that the values of are contained in a convex compact fixed subset.

Theorem 16. Let , let , let be a separable, -uniformly convex and -uniformly smooth Banach space, let , let , and let . Suppose that and the following conditions hold:There is a positive constant such that , .There are three positive constants with such that for any , any , and any we have where and the coefficients are given in Lemma 12.For any and any bounded subset in with and any , one has where

Then, for any fixed satisfying , there is a continuous function such that is absolutely continuous on and

Proof. We divide the proof into the following steps.
Step 1. For any fixed natural number , let , , ,, and be as in the proof of Theorem 15.
By following the same lines in Steps 1 and 2 in the proof of Theorem 15, with the following modifications: ,we can show that there are , , and such that and are absolutely continuous on and the following properties hold:Step 2. In this step, we show that there is a positive number such that, for any ,Let be a fixed natural number. According to () and (), for any , we haveThen, for ,Note that is -uniformly convex and -uniformly smooth. Therefore, in view of (73), (74), and Lemma 12, there exists a positive constant such that In addition, since is -uniformly convex and -uniformly smooth, then, again, by Lemma 12, there is a positive constant such that Then,This inequality with (66), (73), and , give usMoreover, from the assumption , we infer that Then, is -uniformly convex and -uniformly smooth. Hence, from Lemma 13, there is a constant , depending on , such that, for all , , , and , we haveConsequently,This inequality and (78) yield Thus,where , , and . By repeating the same procedure as (82), -times, we infer thatNote that and Then, . Hence, by arguing as in (78) and using (79), one obtains Thus,Relations (83) and (85) and the fact that yield Therefore, for , for , and for , we get This shows that (72) is true.
Step 3. We show that there is a Lipschitz function , such that the sequence has a convergent uniform subsequence, still denoted by , to and converges weakly to
Indeed, by (72), for any , , we haveThen, the sequence is equicontinuous on . We want to show that, for any , the subset is relatively compact in .
As above, by Lemma 12, there is a constant , depending on , such thatSo, for and ,But, by (66), Then, (90) yieldsOn the other hand, in view of (88) and , for any and any , we have This means thatIn view of (92) and (94), one obtainswhere .
Now, assume by contradiction that there is such that is not relatively compact in . Since is continuous on bounded sets, the set is not relatively compact in . Then, . Observe that, by (90), for ,Therefore, in view of , Then, we can find such thatLet be a natural number such that and , . So, by using (95) and (96),Then, from the properties of , (99), and , we infer that which is a contradiction.
Therefore, Theorem 4. Ch.1. in [16] implies that there is a Lipschitz function , such that the sequence has a convergent uniform subsequence, still denoted by , to and converges weakly to We extend the definition of on by putting , . Thus, converges uniformly to on . Since is uniformly continuous on bounded sets, then the sequence is uniformly convergent to with .
Step 4. Let us show that , .
Let . According to (), we have Note that Then, , .
Step 5. Following the same lines in Steps and 6 in the proof of Theorem 15, we can show that , for every . Moreover, the sequence defined by , , converges almost everywhere to a function and , a.e. .
Step 6. We proceed to show that In view of (72) and (73) and Lemma 11, for every natural number , one obtainsBy and ([19], Prop. 1.7), the multivalued function is scalary uppersemicontinuous with convex and compact values. Then, for any ,where is the support function.
Now, let be a sequence in which separates the points. Hence, from the weak convergence of the sequence to in , for any Lebesgue measurable subset , we have This relation with (103) and (104) yields So, for almost , Again, since the multivalued function is measurable with convex and compact, then by ([18], III.35) it follows that As , we get This completes the proof.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The authors gratefully acknowledge the Deanship of Scientific Research, King Faisal University of Saudi Arabia, for their financial support for this Research Project no. 150153.

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Copyright © 2016 A. G. Ibrahim and F. Aladsani. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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