We first prove Theorem 2.2 via the following Lemmas.

Lemma 3.1.

If

,

, and

, then

is a Fredholm operator of index zero. Furthermore, the linear continuous projector operator

can be defined by

and the linear operator

can be written by

Proof.

has a solution

satisfying

,

, if and only if

In fact, if (3.5) has solution

satisfying

,

, then from (3.5) we have

According to

,

, we obtain

On the other hand, if (3.6) holds, setting

where

is an arbitrary constant, then

is a solution of (3.5), and

, and from

and (3.6), we have

Then
. Hence (3.7) is valid.

For

, define

Let

, and we have

then

, thus

. Hence,

, where

, also

. So we have

, and

Thus,
is a Fredholm operator of index zero.

We define a projector
by
. Then we show that
defined in (3.2) is a generalized inverse of
.

In fact, for

, we have

and, for

, we know

In view of

,

, and

, thus

This shows that

. Also we have

then
. The proof of Lemma 3.1 is finished.

Lemma 3.2.

Under conditions (2.5) and (2.9), there are nonnegative functions

satisfying

Proof.

Without loss of generality, we suppose that

. Take

, then there exists

such that

Obviously,

, and

and from (2.5) and (3.21), we have

Hence we can take

,

, 0, and

to replace

,

,

, and

, respectively, in (2.5), and for the convenience omit the bar above

,

, and

, that is,

Lemma 3.3.

If assumptions (H1), (H2) and
,
, and
hold, then the set
for some
is a bounded subset of
.

Proof.

Suppose that

and

. Thus

and

, so that

thus by assumption (H2), there exists

, such that

. In view of

Again for

,

, then

,

thus from Lemma 3.1, we know

From (3.29) and (3.30), we have

If (2.5) holds, from (3.31), and (3.26), we obtain

Thus, from

and (3.32), we have

From

, (3.32), and (3.33), one has

From (3.35) and (3.33), there exists

, such that

Again from (2.5), (3.35), and (3.36), we have

Then we show that
is bounded.

Lemma 3.4.

If assumption (H2) holds, then the set
is bounded.

Proof.

Let

, then

and

; therefore,

From assumption (H2),
, so
, clearly
is bounded.

Lemma 3.5.

If the first part of condition (H3) of Theorem 2.2 holds, then

for all

. Let

where
is the linear isomorphism given by
, for all
,
. Then
is bounded.

Proof.

Suppose that

, then we obtain

If

, then

. Otherwise, if

, in view of (3.40), one has

which contradicts
. Then
=
and we obtain
; therefore,
is bounded.

The proof of Theorem 2.2 is now an easy consequence of the above lemmas and Theorem 2.1.

Proof of Theorem 2.2.

Let
such that
. By the Ascoli-Arzela theorem, it can be shown that
is compact; thus
is
-compact on
. Then by the above Lemmas, we have the following.

(i)
for every
.

(ii)
for every
.

(iii)Let

, with

as in Lemma 3.5. We know

, for

. Thus, by the homotopy property of degree, we get

According to definition of degree on a space which is isomorphic to

,

, and

Then by Theorem 2.1,
has at least one solution in
, so that the BVP (1.1), (1.2) has at least one solution in
. The proof is completed.

Remark 3.6.

If the second part of condition (H3) of Theorem 2.2 holds, that is,

for all

, then in Lemma 3.5, we take

and exactly as there, we can prove that

is bounded. Then in the proof of Theorem 2.2, we have

since
. The remainder of the proof is the same.

By using the same method as in the proof of Theorem 2.2 and Lemmas 3.1–3.5, we can show Lemma 3.7 and Theorem 2.3.

Lemma 3.7.

If

,

, and

, then

is a Fredholm operator of index zero. Furthermore, the linear continuous projector operator

can be defined by

and the linear operator

can be written by

Proof of Theorem 2.3.

Then, for

,

; thus

,

; hence

thus, from assumption (H4), there exists

, such that

and in view of

, we obtain

From

, there exists

, such that

. Thus, from

, one has

We let

; hence from (3.58) and (3.59), we have

thus, by using the same method as in the proof of Lemmas 3.2 and 3.3, we can prove that
is bounded too. Similar to the other proof of Lemmas 3.4–3.7 and Theorem 2.2, we can verify Theorem 2.3.

Finally, we give two examples to demonstrate our results.

Example 3.8.

Consider the following boundary value problem:

where

,

and

satisfying

,

, and

, then we can choose

,

, and

, for

; thus

and
has the same sign as
when
, we may choose
, and then the conditions (H1)–(H3) of Theorem 2.2 are satisfied. Theorem 2.2 implies that BVP (3.61) has at least one solution,
.

Example 3.9.

Consider the following boundary value problem:

where

,

and

satisfying

, and

, then we can choose

,

, and

, for

; thus

Similar to Example 3.8, we have

and
has the same sign as
when
, we may choose
, and then all conditions of Theorem 2.3 are satisfied. Theorem 2.3 implies that BVP (3.65) has at least one solution
.