Proof. By means of the given sub-supersolution
for problem (
P), we introduce some related mappings. The cut-off function for problem (
P), we introduce some related mappings. The cut-off function
is defined by
where
s and
r are the constants given in (1) and Hypothesis 1. Using (
14) in conjunction with
enables us to find that
with a constant
and a function
. Moreover, proceeding as in [
4], we can establish that
with positive constants
and
.
In view of (
15), the Nemytskij operator
generated by
maps continuously
to
. Therefore, the mapping
defined by
is completely continuous. This is true because the inclusion
is compact being the adjoint of the compact inclusion
(note that
owing to the assumption
in Hypothesis 1).
Hypothesis 1 and (
5) imply that the Nemytskij operator
maps continuously
to
with
. Composing the preceding Nemytskij operator with the inclusion
, which is compact because it is the adjoint operator of the compact inclusion
(note that
since
in Hypothesis 1), we obtain a completely continuous mapping
given by
for all
and
.
We also make use of the truncation operator
given by
for all
and a.e.
. It is a continuous and bounded mapping (in the sense that it maps bounded sets to bounded sets). Notice that its range lies in
, so
T can be composed with the operator
.
Now we consider for every
the operator
defined by
From Proposition 1
it is known that the operator
is bounded, while the above comments demonstrate that the operators
,
and
T are all of them bounded. Therefore from (
18) we infer that the operator
is bounded.
We claim that
is a pseudomonotone operator. In this respect, let a sequence
satisfy
in
and
The sequence
is bounded in
, while
in
by the compact embedding
, thus
The sequence
is bounded in
, while
in
by the compact embedding
, producing
Consequently, complying with (
18), we see that (
20) reduces to (
12). This, in conjunction with the weak convergence
, enables us to apply Proposition 1
ensuring that the strong convergence
in
holds.
From the strong convergence
in
it follows the strong convergence
in
. This amounts to saying that
in
since
Again, from the strong convergence
in
we infer that
as
. Taking into account the continuity of the mappings
and
, we have
and
as
, for every
. We can conclude that
is a pseudomonotone operator (see, e.g., ([
2], Definition 2.97)).
The next step in the proof is to show that the operator
is coercive provided
is large enough. Taking advantage of the fact that
whenever
, let us note by (
16), (
19) and Hypothesis 1 that
for all
. Now we estimate the last term in (
21) based on the fact that by (
5) we know that
, and so
. Using the definition of
in (
17), Hölder’s inequality and the continuous embedding in (
9) it turns out that
with a constant
. We can insert the preceding inequality in (
21) to derive
with a constant
. The Hölder’s and Young’s inequalities in conjunction with embedding (
5) imply
with constants
and
. Then (
22) entails
for all
. Recalling from (16) that
, we can choose
so large to have
. Hence due to
(see (1)), (
23) yields the coercivity of
, i.e.,
We have shown that the nonlinear operator
is bounded, pseudomonotone and coercive provided
is sufficiently large. Therefore, for such an
we can apply the main theorem of pseudomonotone operators (see, e.g., ([
2], Theorem 2.99)) ensuring that there exists a solution
to the equation
Fix an admissible
as pointed out above. We are going to prove that
resolving (
24) is a weak solution of the original problem (
P), which means that (
2) is satisfied. To this end, notice that (
19) and (
24) yield
We proceed by comparing
u with the subsolution
and supersolution
postulated in Hypothesis 1. We claim that
a.e. in
. Towards this, it can be readily checked that
, where the condition
on
in the sense of traces is essentially used. Thus, we can insert
in (
25) and (
4) which gives
and
From (
26) and (
27), by subtraction we are led to
By (
14), (
17), and the preceding inequality we get
Since the function
is positive almost everywhere in
and the mapping
on
is monotone, we arrive at
Therefore, the Lebesgue measure of the set is zero, i.e., a.e. in .
Similarly, we can prove that
a.e. in
. Specifically, relying on the condition
on
(in the sense of traces), it holds
, which allows us to test (
25) and (
3) with
. This results in
and
Arguing as before, we deduce from (
28), (
29), (
14), and (
17) the following estimate
At this point, the positivity of the function
on
and the monotonicity of the mapping
on
confirm that
from which we can readily derive that
a.e in
.
Based on the enclosure property
a.e. in
, it follows through (
17) that
and through (14) that
. As a result, (
25) takes the form of (
2), thus the proof is complete. □
Proof. Our goal is to apply Theorem 1 by constructing an appropriate sub-supersolution. In order to determine a subsolution, we use an eigenfunction
corresponding to the first eigenvalue
of problem (
13) with the properties
for a.e.
,
, and
as mentioned in
Section 2. Then we choose an
sufficiently small to verify
where
is the positive constant postulated in assumption
. Then assumption
implies
For a possibly smaller
we can suppose
with
in assumption
.
Let us fix an
for which (
30) and (
32) are fulfilled. We claim that
is a subsolution to problem (
P). Indeed, by (
13) with
in place of
u and (
31) we note that
for all
,
a.e. in
, thereby proving the claim.
Next we claim that the constant function
, with
in assumption
, is a supersolution to problem (
P). Accordingly, from assumption
we find that
for all
,
a.e. in
, which proves the claim.
It is clear from (
32) that
for a.e. in
. Assumption
ensures that the growth condition required in Hypothesis 1 of Theorem 1 holds true. Therefore, all the hypotheses of Theorem 1 are verified, which permits the conclusion that there exists a solution
of problem (
P) within the ordered interval
. Since the function
is nontrivial and nonnegative, and
, we have that
u is nontrivial and nonnegative, whereas
renders the boundedness of
u and the a priori estimate
. The proof is complete. □
We end the paper with a simple example for which Theorem 2 applies.