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`Journal of Function SpacesVolume 2018, Article ID 8746148, 13 pageshttps://doi.org/10.1155/2018/8746148`
Research Article

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

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

Correspondence should be addressed to Hussein A. H. Salem; ge.ude.ics-xela@anidssh

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

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., [36]; see also [79]). 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, 1319]) 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 bywith 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 andArguing 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) thatholds 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 thatConsequently, 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 bywhereIt 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 thatTherefore, 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