Before presenting the main results, we define the solutions of the boundary value problem (1.1).
Definition 3.1 A function is said to be a solution of the problem (1.1) if , and there exist functions , such that
In the sequel, we set
(H1) is an Carathéodory multivalued map;
(H2) there exists a function such that
for all and , where is given by (3.2);
(H3) is an Carathéodory multivalued map;
(H4) there exists a function with for a.e. and a nondecreasing function such that
for all ;
(H5) there exists a number such that
where , are given by (3.2) and (3.3), respectively, and .
Then the problem (1.1) has a solution on .
Proof To transform the problem (1.1) to a fixed-point problem, let us define an operator by
for , , where Q is given by (3.1).
We study the integral inclusion in the space of all continuous real valued functions on with supremum norm . Define two multivalued maps by
Observe that . We shall show that the operators and satisfy all the conditions of Theorem 2.4 on . For the sake of clarity, we split the proof into a sequence of steps and claims.
Step 1. We show that is a multivalued contraction on .
Let and . Then and
for some . Since , there exists such that . Thus the multivalued operator U is defined by , where
has nonempty values and is measurable. Let be a measurable selection for U (which exists by Kuratowski-Ryll-Nardzewski’s selection theorem [21, 22]). Then and a.e. on .
It follows that and
Taking the supremum over the interval , we obtain
Combining the inequality (3.5) with the corresponding one obtained by interchanging the roles of x and y, we get
for all . This shows that is a multivalued contraction as
Step 2. We shall show that the operator is u.s.c. and compact. It is well known [, Proposition 1.2] that if an operator is completely continuous and has a closed graph, then it is u.s.c. Therefore we will prove that is completely continuous and has a closed graph. This step involves several claims.
Claim I maps bounded sets into bounded sets in .
Let be a bounded set in .
Now for each , there exists a such that
Then for each ,
which implies that
Hence is bounded.
Claim II maps bounded sets into equicontinuous sets.
As in the proof of Claim I, let be a bounded set and for some . Then there exists such that
Then for any with we have
Obviously the right hand side of the above inequality tends to zero independently of as . Therefore it follows by the Arzelá-Ascoli theorem that is completely continuous.
Claim III Next we prove that has a closed graph.
Let , and . Then we need to show that . Associated with , there exists such that for each ,
Thus it suffices to show that there exists such that for each ,
Let us consider the linear operator given by
as . Thus, it follows by Lemma 2.6 that is a closed graph operator. Further, we have . Since , we have
for some .
Hence has a closed graph (and therefore it has closed values). In consequence, is compact valued.
Therefore the operators and satisfy all the conditions of Theorem 2.4. So the conclusion of Theorem 2.4 applies and either condition (i) or condition (ii) holds. We show that the conclusion (ii) is not possible. If for , then there exist and such that
By hypothesis (H2), for all , we have
Hence for any ,
for all . Then we have
Now, if condition (ii) of Theorem 2.4 holds, then there exist and such that . Then x is a solution of (3.6) with and consequently, the inequality (3.7) yields
which contradicts (3.4). Hence, has a fixed point in by Theorem 2.4, which in fact is a solution of the problem (1.1). This completes the proof. □
3.1 The lower semi-continuous case
This section is devoted to the study of the case that the maps in (1.1) are not necessarily convex-valued. We establish the existence result for the problem at hand by applying the nonlinear alternative of Leray-Schauder type and a selection theorem due to Bressan and Colombo  for lower semi-continuous maps with decomposable values. Before presenting this result, we revisit some basic concepts.
Let be a nonempty closed subset of a Banach space and be a multivalued operator with nonempty closed values. is lower semi-continuous (l.s.c.) if the set is open for any open set in . Let be a subset of . is measurable if belongs to the σ algebra generated by all sets of the form , where is Lebesgue measurable in and is Borel measurable in ℝ. A subset of is decomposable if for all and measurable , the function , where stands for the characteristic function of .
Definition 3.3 Let Y be a separable metric space and let be a multivalued operator. We say has a property (BC) if is lower semi-continuous (l.s.c.) and has nonempty closed and decomposable values.
Let be a multivalued map with nonempty compact values. Define a multivalued operator associated with F as
which is called the Nemytskii operator associated with F.
Definition 3.4 Let be a multivalued function with nonempty compact values. We say F is of lower semi-continuous type (l.s.c. type) if its associated Nemytskii operator ℱ is lower semi-continuous and has nonempty closed and decomposable values.
Lemma 3.5 ()
Let Y be a separable metric space and let be a multivalued operator satisfying the property (BC). Then has a continuous selection, that is, there exists a continuous function (single-valued) such that for every .
Theorem 3.6 Assume that (H2), (H4), (H5), and the following condition hold:
(H6) are nonempty compact-valued multivalued maps such that
, are measurable,
are lower semicontinuous for each ;
Then the boundary value problem (1.1) has at least one solution on .
Proof It follows from (H2), (H4), and (H6) that F and G are of l.s.c. type. Then from Lemma 3.5, there exist continuous functions such that , for all .
Consider the problem
Observe that if is a solution of (3.8), then x is a solution to the problem (1.1). Now, we define two multivalued operators by
Clearly are continuous. Also the argument in Theorem 3.2 guarantees that and satisfy all the conditions of the nonlinear alternative for contractive maps in the single-valued setting  and hence the problem (3.8) has a solution. □
Example 3.7 Consider the following fractional boundary value problem:
with . Using the given data, we find that
Clearly , and by the condition:
it is found that , where . Thus, all the assumptions of Theorem 3.2 are satisfied. Hence, the conclusion of Theorem 3.2 applies to the problem (3.9).