Denote

, where

. Let

For any

, define

then
is a Banach space with the norm
(see [17]).

The Arzela-Ascoli 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

Lemma 2.2.

Supposing that

with

, then BVP

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

Remark 2.3.

is the Green function for the following associated homogeneous BVP on the half-line:

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,

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