On Solutions of Fractional Order Boundary Value Problems with Integral Boundary Conditions in Banach Spaces

The object of this paper is to investigate the existence of a class of solutions for some boundary value problems of fractional order with integral boundary conditions. The considered problems are very interesting and important from an application point of view. They include two, three, multipoint, and nonlocal boundary value problems as special cases. We stress on single and multivalued problems for which the nonlinear term is assumed only to be Pettis integrable and depends on the fractional derivative of an unknown function. Some investigations on fractional Pettis integrability for functions and multifunctions are also presented. An example illustrating the main result is given.


Introduction
The theory of boundary value problems is one of the most important and useful branches of mathematical analysis.Boundary value problems of various types create a significant subject of several mathematical investigations and appear often in many applications, especially in solving numerous problems in physics and engineering.For example, heat conduction, chemical engineering, underground water flow, thermoelasticity, and plasma physics can be reduced to nonlocal problems with integral boundary conditions.For boundary value problems with integral boundary conditions and comments on their importance, we refer the reader to [1][2][3] and the references therein.
The class of boundary value problems with integral boundary conditions considered below contains as special cases numerous two, three, multipoint, and nonlocal boundary value problems.Such problems are mainly investigated when considering functions satisfying some conditions expressed in terms of the strong topology of a Banach space .We will investigate the case, when functions are not strongly continuous and strongly integrable.In this situation we need to introduce more general notion of a solution.We should note that the considered case seems to be a natural case and cover many particular cases considered for both the strong and weak topologies (cf.Lemma 19).A more general notion of solutions allows us to solve the problem under very general assumptions, not so restrictive as before (see our last section).
In contrast to the classical approach for the theory of boundary value problems, the theory for fractional order BVP's is still developing one and not satisfactorily described.It is caused by the fact that it is very difficult to find convenient and handy conditions ensuring the existence of solutions of several nonlinear boundary value problems of fractional order.In the considered case of a weak topology on  our results form a relatively new branch of investigations.
In comparison with the existence results in the above list, our assumptions seem to be more natural.In contrast to earlier results, we drop the requirement that  is a real-valued function independent of the fractional derivatives and we consider the case of vector-valued Pettis, but not necessarily Bochner, integrable functions.As we mentioned above, the assumptions in the existence theorem are expressed in terms of the weak topology.Such a result does not appear in the earlier literature and so it seems to be new.We collect all interesting properties for the fractional Pettis integral.Moreover, we are able also to start some studies for multivalued fractional boundary value problems with Pettis-integral boundary conditions and fractionally Pettis integrable multifunctions.
In the paper we stress also on comparison results for Pettis integrals and fractional Pettis integrals.This is also done for the multivalued integrals and seems to be interesting by itself, independently of applicability of our results.The properties of fractional integral operators on the spaces of Pettis integrable functions as well as on some of its subspaces are also investigated.
Finally, we remark that, in the Banach spaces, the existence of solutions of some boundary value problems of fractional orders has been considered in terms of Pettis integrals, for the first time, by Salem [20].In this paper, for clarity of proofs, we restrict ourselves to the case of reflexive spaces, but it is easy to extend our results for nonreflexive spaces by putting contraction hypothesis with respect to some measure of weak noncompactness and by using appropriate fixed point theorem (cf.[22]).Nevertheless, all auxiliary results in this paper are not restricted to reflexive spaces.
The question of proving the existence of solutions to the problem (1) reduces to proving the existence of solutions of a Fredholm integral equation.Since the space of all Pettis integrable functions is not complete (in general), we restrict our attention to the case of weakly continuous solution of the Fredholm integral equation (modeled off the problem (1)); hence we are ready to find the so-called pseudo-solutions of the problem (1) (cf.[22,23]).

Preliminaries and Auxiliary Results
For the sake of the reader's convenience here we collect a few facts which will be needed further on.Let  = [0, 1].According to the custom   (), 1 ≤  ≤ ∞ will denote the Banach space of real-valued measurable functions  defined on .Let  ∞ () denote the Banach space of real-valued essentially bounded and measurable functions defined on .Through the paper,  is considered to be a Banach space with norm ‖ ⋅ ‖ and with its dual space  * .Moreover, let   = (, ) = (, (,  * )) denote the space  with its weak topology.By [, ] we will denote the Banach space of strongly continuous functions  :  →  endowed with a standard ‖‖ 0 = sup ∈ ‖()‖, while [, ] denotes the space of all -valued Pettis integrable functions in the interval  (see [24,25] for the definition).Let us also recall that a function ℎ :  →  is said to be weakly-weakly sequentially continuous if ℎ takes each weakly convergent sequence in  into weakly convergent sequence in .We point out that a bounded weakly measurable function  :  →  need not to be Pettis integrable even if  is reflexive.However, in reflexive Banach spaces, the weakly measurable function  :  →  is Pettis integrable if and only if ((⋅)) is Lebesgue integrable on  for every  ∈  * [26].
Let us recall some basic facts.The following Mazur's lemma can be found in [24,26].

Lemma 1. A convex subset of a normed space 𝐸 is closed if and only if it is weakly closed.
A simple consequence of the Hahn-Banach theorem is as follows.
Proposition 2. Let  be a normed space with  0 ̸ = 0. Then there exits  ∈  * with ‖‖ = 1 and  0 = ‖ 0 ‖.Now, we are in a position to recall a fixed point theorem being an extension of results from [27].(3) Remark 5.In a reflexive Banach space  the set H 1 0 () coincides with the space [, ].This is due to the fact that in reflexive Banach spaces, the weakly measurable function  :  →  is Pettis integrable if and only if ((⋅)) is Lebesgue integrable on  for every  ∈  * [26].In general, this is the space of Dunford integrable functions.
It is worthwhile to recall the following.Definition 7. Let  :  → .The (left-sided) fractional Pettisintegral (shortly LS-FPI) of  of order  > 0 is defined by In the above definition the sign "∫" denotes the Pettis integral.For further purpose, we define the right-sided fractional Pettis-integral (shortly RS-FPI) by We will call a function fractionally Pettis integrable provided this integral exists as an element of  (for arbitrary  < 1).
We need to clarify the relations between Pettis integrability and fractional Pettis integrability.Similar results will be proved for classes H  0 ().This will be important in our consideration, but it seems to be really interesting in itself.
Here we restrict ourselves to the case of left-sided fractional Pettis-integrals.
To make the paper more expository, we will consider fractional Pettis integrability for both cases:  < 1 and  > 1.
The last case is more important in our paper, but the first one is necessary to compare our results with some earlier theorems.

𝜑 (𝐼
As a consequence of some properties of a convolution for the Pettis integral [28, Proposition 9], for arbitrary , we have the following.
In the case  = R, it is a well-known consequence of an inequality of Young that the linear fractional integral operators [29]) (a deep result from interpolation theory implies that even ) is compact (see, e.g., [20,30]).
The following results plays a major rule in our analysis.
Proof.Only the proof in case of the LS-FPI is given since the case of the RS-FPI is very similar.
This means that for  ≥ 1 we have "uniform" estimations for all , but for 0 <  < 1 the situation is more complicated (a weakly singular case).
Our consideration as well as Theorem 4.1.1 in [29] gives us a new property.As a consequence of Lemma 10, we are able to prove the following.
The following Lemma is well known in the case  = R, but to see that it also holds in the vector-valued case, we provide a proof.Lemma 16.For 0 <  ≤  we have for every weakly continuous function  : [0, 1] → ,  0  = , and In particular, when  = , (20) means that the operator     + is defined in (,   ) and that   is the left-inverse of   + .
Proof.The first claim, that is,  1 +  = , follows from the fact that the integral of weakly continuous function is weakly continuous, then pseudo-differentiable with respect to the right endpoint of the integration interval.Let  =  +  and  −  =  +  with ,  ∈ N 0 and ,  ∈ [0, 1).Then we have, in view of  1  =  and Lemmas 9 and 10, that The following auxiliary Lemma will be needed in our techniques.
Proof.Let V ∈ [,   ] be a pseudo-solution to the problem (23) and  ∈  * .As in the proof of Lemma 9 it follows that   + V exists and the real function  is continuous for every  ∈  * ; moreover Thus (0) = 0 for every  ∈  * ; that is, (0) = 0. Further In the view of Lemma 16 we also have If otherwise is not stated, we will assume from now that  ∈ (1, 2] and  ∈ (0, 1].
To obtain the Hammerstein type integral equation modeled off the problem (23), we keep the boundary value problem (23) in mind and we formally put (cf.[11 In the view of Lemma 13, we obtain Let us present two remarks about the above assumptions. (i) For the interesting discussion about the growth conditions for Pettis integrable functions of the above type see [31].For differential equations with Caputo fractional integrals (i.e., solutions in the space   (0, 1), R  ), where  − 1 <  <  the problem of dominants for considered functions (Assumption (3)) was considered in [32] It follows that Therefore
Remark 21.We point out that if  is reflexive, it is not necessary to assume any compactness conditions on the nonlinearity of .This will be due to [33,Lemma 2] and the fact that a subset of reflexive Banach spaces is weakly compact if and only if it is weakly closed and norm bounded.

Weak Solutions of the Hammerstein Integral Equation
In this section, in the light of the Assumptions (1)-( 3) imposed on , we proceed to obtain a result which relies on the fixed point Theorem 3 to ensure the existence of weak solution to the integral equation (35).For the sake of convenience, we introduce the following.
Definition 22.By a solution to (35) we mean a function V ∈ (, ) which satisfies the integral equation (35).This is equivalent to the finding V ∈ (, ) with We need to explain why we consider continuous solutions.By the properties of the Pettis integral this should be weakly continuous function.Since  is continuous and we impose (local) boundedness hypothesis for , our solutions are strongly continuous (cf.[22]).We restrict our attention to the space (, ) and then our integral operators will be defined on this space.In contrast to the case of weakly-weakly continuous functions , we need to replace the space (,   ) endowed with its topology of weak uniform convergence by the space of (strongly) continuous functions (, ) with its weak topology.We will utilize in our proofs some characterization of its weak topology.Now, we are in the position to state and prove the first existence result.
We need to prove now that  :  →  is weaklyweakly sequentially continuous.Let us recall that the weak convergence in  ⊂ (, ) is exactly the weak pointwise convergence.Let (V  ) be a sequence in  weakly convergent to V. Then V  () → V() in   for each  ∈ [0, 1].Since  is closed, by Lemma 1 we have V ∈ .
Fix  ∈  and note, in the view of Lebesgue dominated convergence theorem for the Pettis integral (see [31,34]), that   + V  () →   + V() in   .Let us recall that the topology on (,   ) on equicontinuous subsets coincides with the topology of weak pointwise convergence.Since  satisfies Assumption (1), we have (,   − V  (), V  ()) converging weakly to (,   − V(), V()); hence again the Lebesgue dominated convergence theorem for Pettis integral yields V  () converging weakly to V() in , but  is an equicontinuous subset of (, ), and then  :  →  is weakly-weakly sequentially continuous.Applying now Theorem 3, we conclude that  has a fixed point in , which completes the proof.
Let us present a multivalued problem: Some basic results for multivalued boundary value problems with Pettis integrals are due to Maruyama [35], Azzam et al. [36], Azzam-Laouir and Boutana [37], and Satco [38].However these results are devoted to study the standard case  −  = 2 and three-point boundary conditions.Our result is an essential extension for the previous ones.
By () and () we denote the family of all nonempty convex compact and nonempty convex weakly compact subsets of , respectively.For every nonempty convex bounded set  ⊂  the support function of  is denoted by (⋅, ) and defined on  * by (, ) = sup ∈ , for each  ∈  * .Definition 24.A multifunction  :  → 2  with nonempty, closed values is weakly sequentially upper hemicontinuous if and only if for each  ∈  * (, (⋅)) :  → R is sequentially upper semicontinuous from (, ) into R.
In the remaining part of the paper a multifunction  is supposed to be Pettis integrable in the sense of Aumann.
where    denotes the set of all Pettis integrable selections of  provided that this set is not empty.
Let us note that the multivalued Pettis integral can be defined by other methods.The above definition is the best choice for our consideration.This can be deduced from the following theorem.
Theorem 26 (see [39]).Let  :  → () [()] be measurable and scalarly integrable multifunction (i.e., the support functions are real-valued integrable functions).Then the following statements are equivalent: Taking into account Theorem 8, we are able to add one more condition to the above theorem, which seems to be important in our consideration.Since  −  > 1,  is continuous, and by taking arbitrary Pettis integrable selection we obtain Pettis integrability of (, ⋅)(⋅).Let us recall that we restrict ourselves to the case of the (left-sided) fractional Pettis-integral.
Note that for multivalued mappings we will utilize Kakutani's fixed point theorem (for continuity concepts see [27]).
Theorem 28 (see [27]).If  is a nonempty weakly compact convex subset of  and  :  → 2  is sequentially weakly upper semi-continuous, then there exists a fixed point of ; that is,  ∈  with  ∈ ().
For 1 ≤  ≤ ∞, we define the class H  () to be the class of all functions  :  →  having  ∈   () for every  ∈  * .If  = ∞, the added condition

Theorem 8 .
If  :  →  is Pettis integrable, then (a)   +  is defined almost a.e. on , (b)  is fractionally Pettis integrable on , (c) if  is Pettis integrable and strongly measurable, then   +  :  →  is bounded, weakly continuous and −1 (),  < , is Pettis integrable.That is, the operator   makes sense.Further,   + is well defined.To see this, define  :  →  by () :=   + (),  ∈ [0, 1].From the definition of fractional Pettis integrals, we have for every  ∈  * that 0 () and is well defined.Proof.Only the proof in case of the LS-FPI is given since the case of the RS-FPI is very similar.Note first that, for  ∈ H  0 (), we have in the view of Proposition 6 that the function  → ( − ) page 73]), (⋅) is weakly measurable.However, in reflexive Banach spaces, weakly measurable functions  :  →  are Pettis integrable if and only if  is Dunford integrable; that is, ((⋅)) is Lebesgue integrable on  for every  ∈  * .Thus (⋅) is Pettis integrable.That is,   +  ∈ H 0 ().Remark 11.Let us discuss some properties of H  () and