Let

and denote by

the Green function of the problem

By [

2,

3,

20], the Green function

can be expressed as

and it is known that (see, e.g., [

3,

20])

Lemma 3.1 (see [10, Lemmas 2.1 and 2.3]).

For

and

, the inequalities

hold.

Let

and let

be a solution of the differential equation

satisfying the Lidstone boundary conditions

It follows from the definition of the Green function

that

It is easy to check that

is a solution of problem (2.8), (1.2) if and only if

, and its derivative

is a solution of a problem involving the functional differential equation

and the Lidstone boundary conditions (3.8). From (3.9) (for

), we see that

is a solution of problem (3.10), (3.8) exactly if it is a solution of the equation

in the set

. Consequently,

is a solution of problem (2.8), (1.2) if and only if it is a solution of the equation

in the set

. It means that

is a solution of problem (2.8), (1.2) if

is a fixed point of the operator

defined as

We prove the existence of a fixed point of
by the following fixed point result of cone compression type according to Guo-Krasnosel'skii (see, e.g., [18, 19]).

Lemma 3.2.

Let

be a Banach space, and let

be a cone in

. Let

be bounded open balls of

centered at the origin with

. Suppose that

is completely continuous operator such that

holds. Then,
has a fixed point in
.

We are now in the position to prove that problem (2.8), (1.2) has a solution.

Lemma 3.3.

Let (
) and (
) hold. Then, problem (2.8), (1.2) has a solution.

Proof.

Let the operator

be given in (3.13), and let

Then,
is a cone in
and since
for
by (3.4) and
satisfies (2.5), we see that
. The fact that
is a completely continuous operator follows from
, from Lebesgue dominated convergence theorem, and from the Arzelà-Ascoli theorem.

Choose

and put

for

. Then, (cf. (2.5))

Since

and

for

, the equality

holds with some

for

. We now use the equality

and have

Hence,

, and so

Next, we deduce from the relation

where

. Since

for

, we have

The last inequality together with (3.21) gives

where

is from (

). Since

is arbitrary, relations (3.18) and (3.21) imply that for all

, inequalities (3.18) and

hold. By (

), there exists

such that

Then, it follows from (3.18), (3.24), and (3.26) that

The conclusion now follows from Lemma 3.2 (for
and
).

The properties of solutions to problem (2.8), (1.2) are collected in the following lemma.

Lemma 3.4.

Let (
) and (
) be satisfied. Let
be a solution of problem (2.8), (1.2). Then, for all
, the following assertions hold:

(i)
for
,
, and
for a.e.
,

(ii)
is increasing on
, and for
,
is decreasing on
, and there is a unique
such that
,

(iii) there exists a positive constant

such that

for
,

(iv) the sequence
is bounded in
.

Proof.

Let us choose an arbitrary

. By (2.5),

and it follows from the definition of the Green function

that the equality

holds for
and
. Now, using (1.2), (3.4), (3.30), and (3.31), we see that assertion (i) is true. Hence,
is decreasing on
for
and
is increasing on this interval. Due to
for
, there exists a unique
such that
for
. Consequently, assertion (ii) holds.

Next, in view of (2.5), (3.6), and (3.31),

and (cf. (3.32) for

)

since

on

by assertion (ii). Let

Then estimate (3.29) follows from relations (3.32)–(3.37).

It remains to prove the boundedness of the sequence

in

. We use estimate (3.29), the properties of

given in (

), and the inequality

for all

. Now, from the above estimates, from (2.6) and from

for some

, which is proved in (ii), we get

Notice that

by (

). Consequently,

Since

for

, which follows from the fact that

vanishes in

by (1.2) and assertion (ii), inequality (3.45) yields

where

is from (

). Due to the condition

in (

), there exists a positive constant

such that for all

the inequality

is fulfilled. The last inequality together with estimate (3.46) gives
for
. Consequently,
for
,
, and assertion (iv) follows.

The following result gives the important property of
for applying the Vitali convergent theorem in the proof of Theorem 4.1.

Lemma 3.5.

Let (

) and (

) hold. Let

be a solution of problem (2.8), (1.2). Then, the sequence

is uniformly integrable on

, that is, for each

, there exists

such that if

and

, then

Proof.

By Lemma 3.4 (iv), there exists

such that for

, the inequality

holds. Now, we conclude from (2.5) and (2.6), from the properties of

and

given in

, and finally from (3.29) that for

and

, the estimate

is fulfilled, where

is a positive constant. Since the functions

,

, and

(

) belong to the set

by assumption (

), in order to prove that

is uniformly integrable on

, it suffices to show that the sequences

are uniformly integrable on
. Due to
and
for
by (
), this fact follows from [13, Criterion 11.10 (with
and
)].