Consider the space endowed with the norm
Obviously is a Banach space.
We need the following result  in the sequel.
Let E be a Banach space, a completely continuous operator and
a bounded set. Then S has a fixed point in E.
Let us define the operator by
For the sake of convenience, we set
The operatoris completely continuous.
First, we show that the operator is continuous. Let be a sequence in X with and . Then we have
Since , uniformly on I. Similarly, uniformly on I for , uniformly on I for . Also, we get , , and uniformly on I for . Since
using the continuity of f, , , we get . Thus, T is continuous on X. Now, let be a bounded subset. Then there exists a positive constant such that for all and . We show that T Ω is a bounded set. We have
for all . Hence, we get
This implies that the operator T maps bounded sets of X into bounded sets. Now, we prove that the sets , , , are equicontinuous on I. For , we have
In a similar manner, one can find that
Clearly the right-hand sides of the above inequalities tend to zero as . So T is completely continuous. This completes the proof. □
Assume that there exist positive constants, (), (), (), (), (), such that
for, all, and all. In addition, assume that
whereand. Then problem (1.1)-(1.3) has at least one solution.
In view of Theorem 3.2, the operator is completely continuous. Next we show that the set is bounded. Let and . Then we have
and . Thus, we get
This implies that
and so . Thus, the set V is bounded. Hence it follows by Theorem 3.1 that the operator T has at least one fixed point, which in turn implies that problem (1.1)-(1.3) has a solution. □
Assume thatandare continuous functions and there exist constants (), (), (), (), () such that
for all, and, and also
for, , and. In addition, suppose that
Then problem (1.1)-(1.3) has a unique solution.
Set , for , and choose . We show that , where . Let . Then we have