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 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 (3.1) and H denotes the Hausdorff metric;
(H3) is an -Carathéodory multivalued map;
(H4) there exist functions and a nondecreasing function such that
for all ;
(H5) there exists a number such that
where , , , are given by (3.1), (3.2) and (3.3) respectively, and .
Then problem (1.1) has a solution on .
Proof To transform problem (1.1) to a fixed point problem, let us introduce an operator as
We shall show that the operators and satisfy all the conditions of Theorem 2.5 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 [50, 51]). Then and a.e. on .
It follows that and
Taking the supremum over the interval , we obtain
Combining the previous inequality with the corresponding one obtained by interchanging the roles of x and y, we find that
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 a completely continuous operator having a closed graph 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 and for some . Then we have
and 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 . Let with . Then 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 2.7 that is a closed graph operator. Further, we have . Since , therefore, we have
for some .
Hence has a closed graph (and therefore has closed values). In consequence, is compact valued.
Therefore the operators and satisfy all the conditions of Theorem 2.5. So the conclusion of Theorem 2.5 applies and either condition (i) or condition (ii) holds. We show that 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, by using the computations from Step 1 and Step 2, Claim I, we have
Now, if condition (ii) of Theorem 2.5 holds, then there exist and such that . This implies that x is a solution with and consequently, inequality (3.5) yields
which contradicts (3.4). Hence, has a fixed point in by Theorem 2.5, which in fact is a solution of problem (1.1). This completes the proof. □
Example 3.3 Consider a nonlocal integral boundary value problem of fractional integro-differential equations given by
where , , , , , , , .
We found , , , , , , , , .
Then we have
with . 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 problem (3.6).