Abstract
We study a new family of measures of noncompactness in the locally Sobolev space , equipped with a suitable topology. As an application of the technique associated with this family of measures of noncompactness, we study the existence of solutions for a class of nonlinear Volterra integrodifferential equations. Further, we give an illustrative example to verify the effectiveness and applicability of our results.
1. Introduction
A locally integrable function is a function which is integrable on every compact subset of its domain of definition. The importance of such functions lies in the fact that their function space is similar to spaces, but its members are not required to satisfy any growth restriction on their behavior at infinity: in other words, locally integrable functions can grow arbitrarily fast at infinity but are still manageable in a way similar to ordinary integrable functions.
Sobolev spaces [1], that is, the class of functions with derivatives in , play an important role in the analysis. In the recent times, there have been considerable efforts to study these spaces. The significant impact of them arises from the fact that solutions of partial differential equations are naturally found in Sobolev spaces. They also can be applied to approximation theory, calculus of variations, differential geometry, spectral theory, and so on.
The degree of noncompactness of a set is measured by means of functions called measures of noncompactness. Measures of noncompactness are very useful tools in Banach spaces. They are widely used in fixed point theory, differential equations, functional equations, integral and integrodifferential equations, optimization, and so on [2–8].
In this paper, we define two topological structures on the locally Sobolev space : the Fréchet metric topology given by a sequence of seminorms and the topology generated by the family of projections,where the spaces are furnished with weak topologies. Next, we introduce a weak measure of noncompactness in . Thereafter, we define a family of measures of noncompactness in with the topology and we investigate the basic properties of .
As an application, we study the existence of solutions for a class of nonlinear Volterra integrodifferential equationsFinally, we give an illustrative example to verify the effectiveness and applicability of our results.
2. Preliminaries
Let denote the Lebesgue measure of a Lebesgue measurable subset . For subset of a fixed -cell , we write .
Suppose that . We recall that the Sobolev space () is defined to consist of those real measurable functions which, together with all their distributional derivatives of order , belong to . Here is a multi-index; that is, each is a nonnegative integer, , and .
If , then the symbol stands for the operator of restriction where is the restriction of the function to the -cell .
The locally Sobolev space (shortly ) is defined to consist of those real measurable functions which, together with all their distributional derivatives of order , are locally integrable on ; that is, for each , whereFurthermore, we denote by the operator of restrictionWe consider the space with the Fréchet metric topology .
Definition 1 (see [9]). The metrizable topology induced in by the family of seminorms (4), that is, the topology defined by the distance or equivalently will be called Fréchet metric topology in and it will be denoted by .
The convergence and compactness in the topology are characterized in the following proposition.
Proposition 2. A sequence is convergent to in the topology if and only if for each . A set is relatively compact in the topology if and only if is relatively compact in the Banach space for .
Proof. The proof is easy.
We conclude this section by recalling Dunford-Pettis Theorem (see [10]) which will be used in proving our results in next section.
Theorem 3. Let be a bounded set in . Then has compact closure in the weak topology if and only if is equi-integrable; that is, for all measurable sets with and all
3. Main Results
In this section, we introduce a weak measure of noncompactness in . For this purpose, we need the following result.
Proposition 4. Assume that is a bounded set in . Then has compact closure with the weak topology if and only if for all measurable sets with and all
Proof. Take and let be a Banach space with norm Define by Obviously is one-to-one, continuous linear operator and is closed in . Thus, and are isometrically isomorphic. The proof is completed by Dunford-Pettis Theorem.
Consider the subfamily of consisting of relatively compact sets in the topology and the family of all nonempty and bounded subsets (in norm) of . Now, we introduce a weak measure of noncompactness in .
Let be a bounded subset of the space . For and let us denote The essential properties of this measure are contained in the following lemma.
Lemma 5. The weak measure of noncompactness satisfies the following conditions:(1).(2)If , then .(3).(4).(5) where is the weak closure of a set and is the closed convex hull (with respect to the norm topology) of .(6)If is a sequence of nonempty, bounded (in norm), weakly closed subsets of and with , then the intersection set is nonempty and weakly compact in
Proof. Take such that . Then . Therefore, for any there exists such that for all with and all . So, using Proposition 4 is relatively compact in the weak topology and . In order to verify , suppose that Applying Proposition 4 for any there exists such that for all and such that
Now, for all we have It enforces that and thus we get It implies that , and so . We conclude that (1) holds.
The fact that satisfies conditions (2)–(4) is obvious.
Now, we check that condition (5) holds. For this purpose, suppose that and . Thus, there is a sequence such that weakly in . On the other hand, by the definition of we have for any and with
Letting and using Fatou’s Lemma we yield for any with
Hence This implies that Moreover, we obtain by condition (2) that Particularly, By [11, Theorem ] we have that It remains to prove (6); suppose that is a sequence of the weakly closed and nonempty sets of such that for , and Now, for any , take We claim that is a weakly compact set in . To prove this assertion, we need to check that Let be fixed. Since , then there exists such that .
Hence, we can find small enough that Thus, for all , we haveAs we know that the set is weakly compact. By Proposition 4, there exists such that for all and such that ThereforeUsing (22) and (24) we obtain for . It means that .
Thus, there is a subsequence such that weakly in . Since , , and is weakly closed for all , we get which completes the proof of (6).
4. The Topology and the Family of Measures of Noncompactness
In this section, we will define the topology in . Furthermore, we will introduce the family of measures of noncompactness in with the topology and we will describe its properties.
Definition 6. The topology is the weakest topology on such that all operators defined in (5) are continuous.
A base for is the collection of all sets of the form , where and is an arbitrary open subset of with the weak topology. Also, there is the natural topological homeomorphism from onto , given by formula
In this note, we say that a subset is bounded, if is bounded for every seminorm Furthermore, we denote by the family of all nonempty and bounded subsets of and by its subfamily consisting of all relatively compact sets in the topology
Now, we are in a position to introduce a family of measures of noncompactness in with the topology
Definition 7. The family of mappings , defined by formulais said to be the family of measures of noncompactness in with topology Indeed, where is defined in (12).
Theorem 8. The family of measures of noncompactness , defined by (27), fulfills the following conditions: The family for is nonempty and If , then for for . for . for where and are the closure of the set in the topology and the closed convex hull of the set with respect to the topology , respectively.If is a sequence of nonempty, bounded, weakly closed subsets of and with for each then the intersection set is nonempty and compact in
Proof. First we show that holds. To do this, take . Therefore for each . Now, by using Lemma 5, is weakly compact in for . Applying Tychonoff’s Theorem we conclude that is weakly compact in .
On the other hand, it is easy to see that . It in turns implies that Now, since is a homeomorphism, it follows that is compact in and .
Conversely, suppose that . Thus, is compact in , which implies that is weakly compact in .
By Lemma 5 we have that is, .
Parts – are easy to verify. To prove , suppose that . Using again Lemma 5 and inclusion , we get Hence, and .
We know that for ; therefore we have This means that .
It remains to prove . Suppose that is a sequence of nonempty and weakly closed sets of such that for and for By applying Lemma 5, we have Therefore, by Lemma 5, we have that is nonempty and weakly compact in for It implies that is nonempty and weakly compact.
It is routine to see that Now, suppose that is a mapping defined by By [12, Proposition ], the family of all sets , where is a neighborhood of in the space , is a base for the subspace at the point Also, we know that if is a continuous mapping between two topological spaces, thenfor all subsets contained in the domain of
Take and . Then from (35) we have which implies thatConversely, let Then for each Hence, for each
Suppose that is a neighborhood of in the space Then This implies that and Since this is true for each , then HenceTherefore using (37) and (40) we get and so is nonempty and weakly compact. Since is a homeomorphism, we get hence, is nonempty and compact in
As an immediate consequence of Theorem 8, we have the following corollary.
Corollary 9. A subset is relatively compact in the topology if and only if is relatively weakly compact in for each .
We recall that a subset is a.e. equicontinuous on , if, for each , there exists a closed subset such that and the restriction of all functions from to the set form an equicontinuous family of functions. Furthermore, is said to be countable a.e. equicontinuous on if every sequence has a subsequence , which is a.e. equicontinuous on .
Using some ideas from [8] we have the following results.
Lemma 10. Suppose that a subset is bounded in norm. Then is relatively compact if and only if is relatively weakly compact and countable a.e. equicontinuous on .
Proof. We first suppose that is a relatively compact subset of the Banach space . Since the set is also relatively weakly compact, then it suffices to show that is countable a.e. equicontinuous. Regarding the proof of Corollary 9 of [13], is totally bounded in if and only if and are totally bounded in . Then and are relatively compact in . From [8, Lemma ], it follows that and are countable a.e. equicontinuous on with respect to . It can be easily shown that is countable a.e. equicontinuous on with respect to . Conversely, suppose that is relatively weakly compact in and countable a.e. equicontinuous on . Obviously, and are countable a.e. equicontinuous on with respect to . Using Proposition 4 and Theorem 3, it yields that and are relatively weakly compact in . Applying again [8, Lemma ], it obtains that and are relatively compact in and so is relatively compact in .
Lemma 11. Suppose that a subset is bounded. Then is relatively compact in the topology if and only if is relatively compact in the topology and is countable a.e. equicontinuous on -cells of .
Proof. Suppose that is relatively compact in the topology . According to Proposition 2, is relatively compact in for each . Applying Lemma 10, we obtain is relatively weakly compact for all and is countable a.e. equicontinuous on -cells of . From Corollary 9, we conclude that is relatively compact in the topology .
Conversely, suppose that is relatively compact in the topology and is countable a.e. equicontinuous on -cells of . From Corollary 9, we deduce that is relatively weakly compact for each . Hence from Lemma 10 and Proposition 2, we find that is relatively compact in the topology .
Lemma 12 (see [5]). Assume that is a complete metric space. Let , be nonempty subsets of such that are relatively compact for and , where denotes the Hausdorff distance between sets. Then the set is also relatively compact.
5. Example of Applications of the Family of Measures of Noncompactness
In this section, we study the applicability of the family of measures of noncompactness in the locally Sobolev space with the topology . For this purpose, we investigate the existence solutions to a special nonlinear Volterra integrodifferential equation of the form We start with some preliminary facts which will be needed further on.
Suppose that is a given measurable subset and is a metric space. We say that a function satisfies Carathéodory conditions if it is measurable in for any and is continuous in for almost all .
Definition 13. Assume that is a given set and is a function satisfying the Carathéodory conditions. Then, to every positive real-valued function which is measurable on , we may assign the function , . The operator defined in such a way is said to be the superposition (or Nemytskii) operator generated by the function . It is well known that the function is also measurable on .
Definition 14. We say that a function is locally absolutely continuous if for each nonempty compact interval , the restriction of to is absolutely continuous.
Now, let . Recall that the space consists of all real-valued measurable functions on such that, for every line parallel to one of the -coordinate axes , is locally absolutely continuous on .
Given a space of measurable functions over , we denote by the space of all measurable functions such that there exists with almost everywhere in . In particular, for we denote by the derivative where and almost everywhere in .
Following Marcus and Mizel [14] we call a function on a locally absolutely continuous Carathéodory function if belongs to for every , and is continuous and locally absolutely continuous for almost all .
By using some facts from [15] we can prove the following theorem.
Theorem 15. Let be a natural number and let be a locally absolutely continuous Carathéodory function. Then the superposition operator generated by maps the space into itself and it is bounded and continuous with the Fréchet topology if and only if the partial derivative generates superposition operator from into , and the partial derivative generates a superposition operator from into ; that is, for each and , we have where , , and .
Proof. See [15, Theorem ] and comment after it (page 185) and apply it for the special case .
Finally, we proceed by establishing some facts concerning the linear Volterra integral operator in .
We denote by the area and assume is a given function which is measurable with respect to both variables. For an arbitrary function put where The operator is defined in such a way which is called the linear Volterra integral operator. The following lemma gives us a fundamental property of the Volterra integral operator (see [16]).
Lemma 16. If the linear Volterra integral operator maps into itself, then it is continuous in the Fréchet metric topology .
Now, let us assume that ; that is, is the restriction of the operator to the Banach space . The operator is continuous (cf. [16, 17]) and the norm of the linear operator we denote by for .
Equation (42) will be discussed later under the following assumptions:(1) is a natural number and is a continuous function such that, for each .(2) is a locally absolutely continuous Carathéodory function and moreover, for each , , and , we have where , with and .(3) is a continuous operator and for each , and (4) satisfies Carathéodory conditions; that is, the function is measurable on and the function is continuous a.e. on .(5) is an operator given by the formula where and it satisfies in the following conditions: where , is in satisfies Carathéodory conditions, and (6)For all
We study the existence result in the following theorem.
Theorem 17. Under assumptions (1)–(6), (42) has at least one solution which belongs to the space .
Proof. Let us consider the operator defined by the right side of (42). Then we can assume that , where is the superposition operator generated by Fix arbitrary For , we have Hence, On the other hand, by (45), (49), and condition (3) we have where .
Applying (46), (50), and condition (3) we haveThus, combining (53), (54), and (55) it can be deduced that In addition, from (45) and using the same techniques as above, one can show that Now (56) and (57) imply thatwhere and .
Next, we define the nonnegative, nondecreasing function and the subset of as follows: The set is bounded and convex in . Inequality (58) ensures that maps into itself; that is,Furthermore, the set is convex and closed in the Banach space , so it is weakly closed in for . Sincethen is closed in the topology ; that is,Regarding condition (5) and applying majorant principle (cf. [16, 17]) we infer that the operator is continuous. In aid to this fact, together with Proposition 2, we conclude thatis continuous in the topology .
Further, take a nonempty subset of the set and arbitrary number . Fix a number and a nonempty subset such that is measurable and . Then, for arbitrarily fixed we haveTaking into account absolute continuity of the integrals of the functions and and letting , we getPut , , . In view of (60) and (63), we have for all , by which together with (66) we get It follows then from condition (6) that Due to condition in Theorem 8 we deduce that the set is nonempty, convex, and compact in This means for each .
In the sequel, let us consider the sequence of the operators , defined by formula It is readily verified thatWe now apply Lemma 12 to prove is relatively compact in with the topology For the sake of convenience, we fulfill it in 2 steps.
Step 1. We are going to show that is relatively compact in the topology To do this, we apply Lemma 11. First we show that ; that is, Let us fix a number and let us put In view of (70) and condition (5) we derive that Taking into account absolute continuity of the integral of the function , it yields thatOn the other hand, using again (70) and condition (5) we have that Regarding (74) and (75) we obtain Hence , for . It enforces that the bounded set is relatively compact in the topology It remains to prove that is countably a.e. equicontinuous on -cells. Let us fix and According to Luzin’s Theorem, there is a closed subset such that , and the functions and are continuous.
Let us take arbitrary -tuples such that, for each Then, for , by condition (5) we have where denotes the modulus of continuity of the function on the set and From uniform continuity of the function on the compact set , absolute continuity of the integral of the function , and , we yieldis arbitrarily small provided the number is small enough. This proves that is countably a.e. equicontinuous on -cells.
Note that if are arbitrary, then it is enough to break down the integration intervals into suitable subintervals and applying the same reasoning as above, to see that is countably a.e. equicontinuous on -cells. We then conclude by applying Lemma 11 that is relatively compact in
Step 2. We are going to prove that , where denotes the Hausdorff distance in the Fréchet metric space . Since all functions of the set are uniformly bounded in , Using our assumptions, for arbitrarily fixed , we haveBy absolute continuity of integral of function and applying (68) and (80) we derive that This equality together with inclusion implies thatRegarding (83), Step 1, and Lemma 12 we derive that is also relatively compact in Taking into account the continuity of (see Theorem 15), we get that is also relatively compact in Let us take We observe that is compact and convex subset of the space with the topology . From continuity of and in view of Theorem 15 the operator transforms continuously the set into itself. So, applying the classical Tychonoff fixed point principle we conclude that there is a solution of (42). This completes the proof of our theorem.
Now, we present an example which verifies the effectiveness and applicability of Theorem 17.
Example 18. Consider the following Volterra integrodifferential equation: Equation (84) is a special case of (42) with It is easy to see that satisfies in condition (1) of Theorem 17. Also, satisfies Carathéodory conditions and if we define , , and , then condition (2) of Theorem 17 holds. Moreover, let us take Then we have It can be easily shown that condition (5) of Theorem 17 holds, too. On the other hand, .
Consequently, all the conditions of Theorem 17 are satisfied. It implies that the integrodifferential equation (84) has at least one solution in the space
Conflicts of Interest
The authors declare that there are no conflicts of interest regarding the publication of this paper.