Journal of Function Spaces
Volume 2018 (2018), Article ID 8746148, 13 pages
https://doi.org/10.1155/2018/8746148

Research Article

On the Theory of Fractional Calculus in the Pettis-Function Spaces

Hussein A. H. Salem

Department of Mathematics and Computer Science, Faculty of Science, Alexandria University, Alexandria, Egypt

Correspondence should be addressed to Hussein A. H. Salem; hssdina@alex-sci.edu.eg

Received 19 February 2018; Accepted 2 April 2018; Published 15 May 2018

Academic Editor: Gestur Ólafsson

Copyright © 2018 Hussein A. H. Salem. 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.

Abstract.

Throughout this paper, we outline some aspects of fractional calculus in Banach spaces. Some examples are demonstrated. In our investigations, the integrals and the derivatives are understood as Pettis integrals and the corresponding derivatives. Our results here extended all previous contributions in this context and therefore are new. To encompass the full scope of our paper, we show that a weakly continuous solution of a fractional order integral equation, which is modeled off some fractional order boundary value problem (where the derivatives are taken in the usual definition of the Caputo fractional weak derivative), may not solve the problem.

1. Introduction and Preliminaries

The issue of fractional calculus for the functions that take values in Banach space where the integrals and the derivatives are understood as Pettis integrals and the corresponding derivatives has been studied for the first time by the authors of [1, 2]. Following the appearance of [1], there has been a significant interest in the study of this topic (see, e.g., [3–6]; see also [7–9]). This paper is devoted to presenting general results and examples for the existence of the fractional integral (and corresponding fractional differential) operators in arbitrary Banach space where it is endowed with its weak topology. In our investigations, we show that the well-known properties of the fractional calculus for functions taking values in finite dimensional spaces also hold in infinite dimensional spaces. Our results extend all previous contributions of the same type in the Bochner integrability setting and in the Pettis integrability one.

For the readers convenience, here we present some notations and the main properties for the Pettis integrals. For further background, unexplained terminology and details pertaining to this paper can be found in Diestel et al. [10, 11] and Pettis [12].

Throughout this paper, we consider the measure space , where , denote a fixed interval of the real line, denotes the Lebesgue -algebra , and stands for the Lebesgue measure. denotes a real Banach space with a norm and is its dual. By we denotes the space when endowed with the weak topology generated by the continuous linear functionals on . We will let denote the Banach space of weakly continuous functions , with the topology of weak uniform convergence. And denotes the space of -valued Pettis integrable functions in the interval (see [10, 12] for the definition). Recall that (see, e.g., [10, 13–19]) the weakly measurable function is said to be Dunford (or Gelfand) integrable on if and only if is Lebesgue integrable on for each .

Definition 1. Let . Define to be the class of all weakly measurable functions having for every
If , the added condition

must be satisfied by each We also define the class by

Further, we define the space , by

We also define .

In the following proposition, we summarize some important facts which are the main tool in carrying out our investigations (see [10, 12, 16, 17]).

Proposition 2. Let be of conjugate exponents (that is, with the convention that ). If is weakly measurable, then

(1)if and , then the following hold ,(2), if and only if for every ,(3)if is reflexive (containing no isometric copy of ), the weakly (strongly) measurable function is Pettis integrable on if and only if is Dunford integrable on ,(4)for any , we have . If is weakly sequentially complete, this is also true for .

We remark that there is a bounded weakly measurable function which is not Pettis integrable (see, e.g., [19]).

A fundamental property of Pettis integral is contained in the following.

Proposition 3 (see [12] Corollary 2.51). If , then for any bounded subset of elements of , the integrals

are weakly equi-absolutely continuous.

Theorem 4 (mean value theorem for Pettis integral). If the function is Pettis integrable on , then

where , is the length of and is the closed convex hull of .

2. Fractional Integrals of Vector-Valued Functions

In this section, we define and study the Riemann-Liouville fractional integral operators and the corresponding fractional derivatives in Banach spaces.

Devoted by the definition of the Riemann-Liouville fractional integral of real-valued function, we introduce the following.

Definition 5. Let The Riemann-Liouville fractional Pettis integral (shortly RFPI) of of order is defined by

In the preceding definition, “” stands for the Pettis integral.

When , it is well known (see, e.g., [20, 21]) that the operator sends , continuously to if satisfy .

This seems to be a good place to put the following.

Example 6. Let be an infinite dimensional Banach space that fails cotype (see [22] and the references therein). Define the strongly measurable function by

with similar notations as in ([13], Corollary 4) where we choose , , and

to be the fat Cantor sets (that is, holds for every ).
As cited in ([13], Corollary 4), is Pettis integrable functions on whose indefinite integral is nowhere weakly differentiable on . Here we will show that has RFPI of all order and

Arguing similarly as in ([13], page 368), we have in view of that holds for every .
Also, for any and fixed , we have for some that

Therefore, for any measurable , we arrive at

Obviously, the latter series converges whenever , which allows us to interchange the integral and summation below to see that

It remains to prove that

Evidently, we have

which approaches zero as as needed for (9).

Remark 7. Observe Example 6. We remark the following:

(1)There is a reflexive Banach space for which the indefinite Pettis integral of the function defined by (7) is nowhere weakly differentiable on (see [13], the remark below Corollary 4). Meanwhile, has a RFPI of all order .(2)The function is weakly continuous on : this follows easily from the definition of the Pettis integral. In fact, we have in view of (9) that

 holds for every . Since

 then the infinite series of continuous functions in the left hand side of (15) converges uniformly in . Hence, the function is continuous on (this yields the weak continuity of on ).

In the following lemma, we gather together some simple particular sufficient conditions that ensure the existence of the Riemann-Liouville fractional integral of the functions from .

Lemma 8. Let be weakly measurable function. The RFPI of the function of order makes sense a.e. on if at least one of the following cases holds:

(a), and .(b), .(c) is strongly measurable which lies in , where provided that is weakly complete or contains no copy of .

If is reflexive, this is also true for any , and .
In all cases, holds for every .

In the assertions ( and ), we find sufficient conditions needed for the existence of in the situation in which no restriction is placed on . In the third assertion, the properties of allow us to characterize a function for which exists.

Proof. Firstly, assertion is direct consequence of Proposition 2 (part ) since holds for almost every whenever .
Secondly, to prove assertion , let and be the conjugate exponents to . Since , we have that holds for every . Thus, the assertion is now an easy consequence of Proposition 2 (part ).
Thirdly, to prove we let , and be strongly measurable. Since the strong measurability is preserved under multiplication operation on functions, the product is strongly measurable on for almost every . In view of Young’s inequality, for every , the real-valued function,

is Lebesgue integrable on , for almost every . So the assertion follows immediately from Proposition 2 (parts (3, 4)).
Similarly, when is reflexive, the result follows from part of Proposition 2.
However, in all cases, the function is Pettis integrable on for almost every . That is, for almost every , there exists an element in denoted by such that

holds for every . This completes the proof.

Remark 9. If such that does not exist for some , then it does not exist even when we “enlarge” the space into . To see this, let such that . If exists for some , then . Since assumes only values in , it follows by the mean value theorem for Pettis integral (Theorem 4) that the RFPI of should lie in .

Before we come to a deep study of the mathematical properties of the RFPI operator, let us take a look at the following miscellaneous examples.

Example 10. Let . Define the function by .
This function is weakly measurable, Pettis integrable on , and is a function of bounded variation (see, e.g., [18]). That is, . Hence, in view of Lemma 8 with , the RFPI of exists on . Further, calculations (cf. [4]) show that

Example 11. Let . Define the function from the interval into the Hilbert space as

We note that

Thus, the function is well defined. We claim that is Dunford integrable on . Once our claim is established, Lemma 8 guarantees the existence of on . It remains to prove this claim and to calculate .
To see this, let . According to the Riesz representation theorem on Hilbert spaces there exists a uniquely determined such that . A standard arguments using Beppo Levi’s Theorem yields

So is Dunford integrable on and hence Pettis integrable on since is reflexive. Consequently, in view of Lemma 8, exists on . To calculate the RFPI of fix and let . We have

Since the series is uniformly convergent on , it follows in view of the generalized linearity of the fractional integrals [23, Lemma 5], that (cf. [21, Table 9.1])

where is the Gauss hypergeometric function evaluated at and

Since

we see that . Thus,

Example 12. Let . Define the countable-valued function by

This function is strongly measurable, Pettis integrable function on (see, e.g., [10, 16]). We claim that with . Once our claim is established, Lemma 8 guarantees the existence of , on . It remains to prove this claim by showing firstly that and to calculate . To do this, let . Then there corresponds to a unique such that . By noting that

holds for any , a standard argument using Levi’s Theorem (or Lebesgue dominated convergence theorem) and Minkowski’s inequality yields

Thus, , holds for every (that is, ). Since , it follows by Proposition 2, in view of the strong measurability of , that . Owing to Lemma 8, we infer that the RFPI of of any order exists on the interval . To compute this integral, fix for some , the set of natural numbers, and let . Then

Therefore, by Beppo Levi Theorem it follows that

Consequently, we conclude that

where the nonzero coordinate started from the place. It can be easily seen that for any . Evidently, we have that

hold for any . This yields for any (this is precisely what we would expect from Definition 5).

Look at part of Lemma 8, with and . Below we give an example showing that the strongly measurability hypothesis imposed on a function is not sufficient for the existence of even when is Denjoy-Pettis integrable (cf. [16]).

Example 13. Let . Define the strongly measurable function by

where

It is immediate that (cf. [16]) is a well-defined, Denjoy-Pettis (but it is not Pettis) integrable on .
In what follows, we will show that the RFPI of does not exist on a subinterval of positive measure on .
To see this, we make use of Proposition 3 as follows: define for each interval the functionals (required by Proposition 3) to be the corresponding to the element where the nonzero coordinate is in the th place. Then and thus . Clearly, the family runs through the unit ball of the dual of . Evidently, the isometric isomorphism between and yields .
Now take and define the continuous function by

It can be easily seen that, in view of , is negative on . By standard results from (classical) calculus, it follows that is strictly decreasing on , in particular for all . Thus, for any and any we have

An explicit calculation using L’Hospital’s rule two times reveals

from which it follows that

Therefore, in view of Proposition 3, . Hence, does not exist, which is what we wished to show.

The following theorem provides a useful characterization of the space , for which the statements reveal how much the fractional integral is better than the function . Indeed, based on Lemma 8 using an inequality of Young, we can easily prove the following.

Lemma 14. For any , the following holds.

(a)If is reflexive, then for every , the operator maps into . In particular, if , then however small is.(b)If contains no copy of , then for every , the operator maps into . In particular, if , the operator maps into itself.(c)If , the operator sends to (if we define ). In particular, if , .

Proof. At the beginning, let and with and define the real-valued function by . If is reflexive, it follows, in view of Lemma 8, that exists a.e. in and for every . Young’s inequality guarantees that the convolution product lies in . Consequently for every . The reflexivity of together with Proposition 2 yields .
In particular, if , it can be easily seen that for every . That is, . By Young’s inequality, it follows that for every however small is. Now, the assertion follows because of the reflexivity of .
Next, in order to prove the assertion let and note, in view of Lemma 8, that exists on . Define by . Since for every , it follows that . Moreover, for any , we have

where

Since , owing to Proposition 2 (part ), it follows that and are Pettis integrable on and so .
A combination of these results yields for every and there exists an element such that for every . Since contains no copy of , it follows that (cf. [17, Theorem 23.]) . Consequently . This is the claim .
To prove the assertion , let and . By Lemma 8, we deduce that exists a.e. on . Now, let be such that . Since , . Therefore, as a direct consequence of Young’ inequality it follows that

holds for every . Now, we claim that in as . Once our claim is established, the definition guarantees that , which is what we wished to show. It remains to prove our claim: without loss of generality, assume that . Then there exists (as a consequence of the Hahn-Banach theorem) with such that . By Hölder’s inequality we obtain

This is equivalent with the following estimate:

Owing to we get in as and consequently, in view of , . This proves the first part of the assertion .
Finally, let and . Without loss of generality, assume . Then for any we have by Hölder inequality with , in view , that

By noting that when , it can be easily seen that

That is,

A combination of these results yields

Thus, . This completes the proof.

Corollary 15. For any , .

Proof. Let , then holds for every . By noting that the weak continuity implies a strong measurability ([24] page 73), it follows that is strongly measurable on . Hence, in view of Proposition 2, . Consequently for every . That is, . Hence, the desired result is obtained.

Example 16. Let . Define the function by We have (in view of Example 12) that

To show that is norm continuous on , let . With no loss of generality, we may assume that , for some . Since the nonzero terms of the sequence in (50) are nonincreasing, then, in view of

we have

Thus, the RFPI of is norm continuous on . Precisely, since , for any , then . This is precisely what one would expect from Lemma 14 (part ).
Analogously, an explicit calculation reveals that lies in , , where defined by Example 10.

We now consider additional mapping properties of the operator . Precisely, we will show that the RFPI enjoys the following commutative property which is folklore in case . However, the proof is completely similar to that of [8], Lemma 3.5.

Lemma 17. Let and . Then

holds for every . If is reflexive, this is also true for every .

Proof. Since , it follows that . So, by Lemmas 8 and 14, the mappings , , and belong to for every . Now repeating the same process as in ([8], Lemma 3.5) implies the claim.
When is reflexive, the result follows as a direct application of Lemma 14.

3. Fractional Derivatives of Vector-Valued Functions

After the notation of the fractional integrals of vector-valued functions, the fractional derivatives become a natural requirement. Before we come to the definitions and a detailed study of the mathematical properties of fractional differential operators, we recall the following.

Definition 18. Consider the vector-valued function :

(1)Let be differentiable on for every . The function is said to be weakly differentiable on if there exists such that for every we have

 The function is called the weak derivative of the function .(2)Let be differentiable a.e. on for every (the null set may vary with ). The function is said to be pseudo differentiable on if there exists a function such that for every there exists a null set such that

 In this case, the function is called the pseudo derivative of . If is pseudo differentiable on and the null set invariant for every , then is a.e. weakly differentiable on .

Clearly, if is a.e. weakly differentiable on , then is a.e. differentiable on . The converse holds in a weakly sequentially complete space (see [25], Theorem 7.3.3).

For more details of the derivatives of vector-valued functions we refer to [10, 12, 26].

The following results play a major role in our analysis

Proposition 19 (see [27], Theorem 5.1). The function is an indefinite Pettis integrable, if and only if is weakly absolutely continuous on and have a pseudo derivative on . In this case, is an indefinite Pettis integral of any of its pseudo derivatives.

Now we are in the position to define the fractional-type derivatives of vector-valued functions.

Definition 20. Let . For the positive integer such that , we define the Caputo fractional-pseudo (weak) derivative “shortly CFPD (CFWD)” of of order by

where the sign “” denotes the pseudo (or weak) differential operator. We use the notation and to characterize the Caputo fractional-pseudo derivatives and Caputo fractional weak derivatives, respectively.

It is well known that, although the weak derivative of a weakly differentiable function is uniquely determined, the pseudo derivative of the pseudo differentiable function is not unique. Also, although any two pseudo derivatives , of a function need not be a.e. equal (see [12, Example 9.1] and [26, p. 2]), the functions , are weakly equivalent on (that is, holds a.e. for every ). The next lemma provides a useful characterization property of the CFPD. Really, it can be easily seen that the CFPD of a Caputo fractional-pseudo differentiable function does not depend on the choice of a pseudo derivative of the function.

Lemma 21. Let be pseudo differentiable function where pseudo derivatives lie in . If exists on , then the CFPD of depends on the choice of the pseudo derivatives of .

Proof. Let be two pseudo derivatives of the pseudo differentiable function . Since and are weakly equivalent on , then for every there exists a null set which depends on such that for every . Consequently, for almost every ,

holds for every . Thus,

which is what we wished to show.

We consider the following examples.

Example 22. Define by

We claim that the CFWD of exists on . To see this, we let . Then there corresponds to a sequence such that

Since the series in (60) converges uniformly on , the formal differentiation of yields

It is not hard to justify the differentiation by noting that the series in the right hand side of (61) is uniformly and absolutely convergent on . Consequently, is differentiable on for , meaning that is weakly differentiable on (because is weakly complete (cf. [25], Theorem 7.3.3) and . Owing to Lemma 25, the CFWD of any order of exists on . To calculate the CFWD of on we observe, in view of Example 11, that

Example 23. Let be an infinite dimension Banach space that fails cotype. Define by , where given by formula (7). As cited in ([13], Corollary 4), is nowhere weakly differentiable on . Consequently the CFWD of any order of loses its meaning on .

Remark 24. As shown in Example 23, there is an infinite dimension Banach space and weakly absolutely continuous function which is nowhere weakly differentiable (hence the CFWD of does not exist). Also, even when is separable, and is Lipschitz function, the pseudo derivatives (hence the CFPD) of need not to exist [28].
However, Definition 20 of the CFPD (CFWD) has the disadvantage that it completely loses its meaning if the function fails to be pseudo (weakly) differentiable. Precisely, the CFPD (in particular the CFWD) of a function loses its meaning if is not weakly absolutely continuous.

The next lemma gives sufficient conditions that ensure the existence of the Caputo fractional derivatives of a function .

Lemma 25. Let . For the function , the following hold:

(a)If has a pseudo derivative , where , then the CFPD of of order exists on . Moreover, .(b)If is weakly complete or contains no copy of and if has a weak derivative , where , then the CFWD of of order exists on . Moreover, .

This holds for any with if is reflexive.
In all cases,    holds for every .

Proof. Since the weak (pseudo) derivative of an a.e. weakly (pseudo) differentiable function is strongly (weakly) measurable [12, 26, 28], the proof is readily available, in view of the definition of Caputo fractional derivatives and Lemma 8. Moreover, since with , it follows, in view of Lemma 14, that . This completes the proof.

Besides the Caputo fractional-pseudo (weak) derivatives, we define the Riemann-Liouville fractional-pseudo (weak) derivatives.

Definition 26. Let . For the positive integer such that , we define the Riemann-Liouville fractional-pseudo (weak) derivative “shortly RFPD (RFWD)” of of order by

where is defined as in Definition 20. We use the notation () to characterize the Riemann-Liouville fractional-pseudo (weak) derivatives.

Clearly, in infinite dimension Banach spaces, the weak absolute continuity of , is necessarily (but not sufficient) condition for the existence of RFPD (in particular RFWD) of .

Lemma 27. Let . For any with , we have

If is reflexive, this is also true for every .

Proof. Our assumption with yields, in view of Lemma 17, that

The claim now follows immediately, since (cf. Proposition 19) the indefinite integral of Pettis integrable function is weakly absolutely continuous and it is pseudo differentiable with respect to the right endpoint of the integration interval and its pseudo derivative equals the integrand at that point.

Remark 28. When we replace by , then Lemma 27 is no longer necessarily true for arbitrary even when is reflexive: evidently, in [13, remark below Corollary 4] the existence of a reflexive Banach space and a strong measurable Pettis integrable function was proved such that has nowhere weakly differentiable integral. In this case, lost its meaning. This gives a reason to believe that (65) (hence (64)) with could not happen.

However, we have the following result.

Lemma 29. Let . For any , we have

Proof. Our assumption (see the proof of Corollary 15) yields, in view of Lemma 17, that

The claim now follows immediately, since the indefinite integral of weakly contentious function is weakly absolutely continuous and it is weakly differentiable with respect to the right endpoint of the integration interval and its weak derivative equals the integrand at that point.

The following lemma is folklore in case , but to see that it also holds in the vector-valued case, we provide a proof.

Lemma 30. Let and . If the function is weakly absolutely continuous on and passes a pseudo derivative in , then

In particular, if passes a weak derivative in , then

If is reflexive, this is also true for every .

Proof. We observe that, under the assumption imposed on together with Proposition 3, the weakly absolutely continuity of is equivalent to

Hence, owing to Lemma 27, it follows that

which is what we wished to show.
The proof of (69) is very similar to that in (68); therefore, we omit the details.

Remark 31. As we remark above, the definition of the CFPD of a function loses completely its meaning if is not weakly absolutely continuous. For this reason, we are able to use Lemma 30 to define the Caputo fractional derivative in general; that is, we put

Similarly, we define

However, Lemma 30 claims that, for the weakly absolutely continuous functions having Pettis integrable pseudo [weak] derivatives, definitions (72) and (73) of the Caputo fractional-pseudo [weak] derivatives coincide with Definition 20.

4. An Application

Let be given function. Consider the boundary value problem of the fractional type

with certain constants ,  ,  . To obtain the integral equation modeled off the problem (74), we let be a weakly continuous solution to problem (74); then formally we have

that is

This reads (cf. [29])

with some (presently unknown) quantity .

Now, we solve (77) for by , and it follows that

Thus,

Assume that the function is weakly-weakly continuous function such that

(1)for any , there is a constant such that for all and ,(2)there exists a nondecreasing continuous function , and for all such that for every bounded set , where stands for De Blasi’s weak measure of noncompactness (see [30]).

Occasionally, if is weakly-weakly continuous and is reflexive, then the assumptions (1) and (2) are automatically satisfied (see, e.g., [31]).

Theorem 32 (see [6], Theorem 3.3). Let and . If is weakly-weakly continuous and satisfies the assumptions (1) and (2), then the integral equation (77) has a weakly continuous solution defined on .

Proof. We omit the proof since it is almost identical to that in the proof in ([6], Theorem 3.3) with (small) necessary changes.

In the following example we assume that solves (77) and we will show that, not only do we have that no longer necessarily solves (74) (when the Caputo fractional weak derivative is taken in the sense of Definition 20), but even worse, it could happen that the problem (74) is “meaningless” on .

Example 33. Let . Let be an indefinite dimensional reflexive Banach space that fails cotype. Define the weakly-weakly continuous function by , where is defined by formula (7). Obviously (cf. Example 6, in view of Remark 7), satisfies the assumptions of Theorem 32.
Now, consider the integral equation (77) with . Namely, we consider the integral equation

Obviously, the solution is weakly continuous on (this is an immediate consequence of Remark 7 and Lemma 17).
Since holds for every , it follows by the commutative property of the fractional integral operators calculus over the space of Lebesgue integrable functions (see [21], Section ) that

that is

Hence, . Consequently, problem (80) becomes

As showed in ([13], Corollary 4), the indefinite integral is nowhere weakly differentiable on . Thus, is nowhere weakly differentiable on ; hence the CFWD of (hence the boundary value problem) completely loses its meaning if the CFWD is taken in the sense of Definition 20. This is what we wished to show.

Now, we are in the position to state and prove the following existence theorem.

Theorem 34. If is a function such that all conditions from Theorem 32 hold, then problem (74) (where the Caputo fractional weak derivative is taken in the sense (73)) has a weak solution on .

Proof. Let be a solution to (77). Equip and I × E with weak topology and note that is continuous as a mapping from into Since is weakly-weakly continuous on , then is a composition of this mapping with and thus is weakly continuous on .
Now, operating by the operator on both sides of (77), it follows by Lemma 29 that

Now, insert definition (73) of the Caputo fractional derivative, and we get

With some further efforts, one can get the boundary condition
Therefore, satisfies problem (74) which is what we wished to show.

Data Availability

No data were used to support this study

Conflicts of Interest

The author declares no conflicts of interest.

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