Denote , where . Let
For any , define
then is a Banach space with the norm (see [17]).
The ArzelaAscoli theorem fails to work in the Banach space due to the fact that the infinite interval is noncompact. The following compactness criterion will help us to resolve this problem.
Lemma 2.1 (see [17]).
Let . Then, is relatively compact in if the following conditions hold:
(a) is bounded in ;
(b) the functions belonging to and are locally equicontinuous on ;
(c) the functions from and are equiconvergent, at .
Throughout the paper we assume the following.
Suppose that , and there exist nonnegative functions with such that
, where
Denote
Lemma 2.2.
Supposing that with , then BVP
has a unique solution
where
in which , and for .
Proof.
Integrating the differential equation from to , one has
Then, integrating the above integral equation from to , noticing that and , we have
Since , it holds that
By using arguments similar to those used to prove Lemma 2.2 in [9], we conclude that (2.7) holds. This completes the proof.
Now, BVP (1.1) is equivalent to
Letting , (2.12) becomes
For , define operator by
Then,
Set
Remark 2.3.
is the Green function for the following associated homogeneous BVP on the halfline:
It is not difficult to testify that
Let us first give the following result of completely continuous operator.
Lemma 2.4.
Supposing that and hold, then is completely continuous.
Proof.

(1)
First, we show that is well defined.
For any , there exists such that . Then,
so
Similarly,
Further,
On the other hand, for any and , by Remark 2.3, we have
Hence, by , the Lebesgue dominated convergence theorem, and the continuity of , for any , we have
So, for any .
We can show that . In fact, by (2.23) and (2.24), we obtain
Hence, is well defined.

(2)
We show that is continuous.
Suppose , and . Then, as , and there exists such that . The continuity of implies that
as . Moreover, since
we have from the Lebesgue dominated convergence theorem that
Thus, is continuous.

(3)
We show that is relatively compact.
(a) Let be a bounded subset. Then, there exists such that for all . By the similar proof of (2.20) and (2.22), if , one has
which implies that is uniformly bounded.
(b) For any , if , we have
Thus, for any there exists such that if , then
Since is arbitrary, then and are locally equicontinuous on .
(c) For , from (2.27), we have
which means that and are equiconvergent at . By Lemma 2.1, is relatively compact.
Therefore, is completely continuous. The proof is complete.
Lemma 2.5 (see [28, 29]).
Let be Banach space, be a bounded open subset of , and be a completely continuous operator. Then either there exist such that , or there exists a fixed point .
Lemma 2.6 (see [28, 29]).
Let be a bounded open set in real Banach space , let be a cone of , and let be completely continuous. Suppose that
Then,