In order to apply Lemma 2.7 to prove our main results, we first give
,
,
as follows. Let
,
Clearly,
,
Now, we make the following hypotheses.
is continuous. There exists a nonnegative function
with
on a subinterval of
. Also there exists a nondecreasing function
such that
for any
.
There exist two positive constants
such that
. Moreover,
, and
are continuous. There exists a positive constant
such that
Let
where
.
Now, we state our main results.
Theorem 3.1.
Assume that
,
, and
are satisfied; moreover,
, where
, then the problem (1.5) has at least one solution.
Proof.
The proof will be given in several steps.
Step 1.
The operator
is completely continuous.
Let
. In fact, by
,
can be replaced by
. For any
, we have
These imply that
, where
is a positive constant, that is,
is uniformly bounded. In addition, for any
, for all
,
, we can obtain
Taking into account the uniform continuity of the function
on
, we get that
is equicontinuity on
. By the Lemma 5.4.1 in [23], we have
as relatively compact. Due to the continuity of
,
,
, it is clear that
is continuous. Hence, we complete the proof of Step 1.
Step 2.
is bounded.
From
, it follows that there exists a positive constant
, such that
Now, we verify the validity of all the conditions in Lemma 2.6 with respect to the operator
,
, and
. Let
. From
, we have
Combining with the property that
is bounded (Step 1), we have
bounded on
. Hence, we can assume that
,
is a constant.
Step 3.
is a nonlinear contraction.
Let
,
,
. By
, we obtain
, and
,
. Since
, we have
, that is,
is a nonlinear contraction
.
Step 4.
in Lemma 2.7 does not occur.
To this end, we perform the argument by contradiction. Suppose that
holds, then there exist
,
, such that
. Hence, we can obtain
and
Therefore,
. However, it contradicts with (3.8).
Hence, by using Steps 1–4, Lemmas 2.6 and 2.7,
has at least one fixed point
, which is the solution of problem (1.5).
Next, we will give some corollaries.
Corollary 3.2.
Assume that
,
, and
are satisfied; moreover,
, where
; then the problem (1.5) has at least one solution.
Assume that,
()(sublinear growth),
is continuous. There exists a nonnegative function
with
on a subinterval of
. Also there exists a constant
, such that
for any
.
Corollary 3.3.
Assume that
,
, and
are satisfied, then the problem (1.5) has at least one solution.
Assume that
is continuous. There exists a nonnegative function
with
on a subinterval of
. Also there exists a constant
such that
for any
,
there exist two positive constants
such that
. Moreover,
, and
are continuous. There exists a positive constant
, such that
Corollary 3.4.
Assume that
,
, and
are satisfied, then the problem (1.4) has at least one solution.
Assume that
is continuous. There exists a nonnegative function
with
on a subinterval of
.
for any
.
Corollary 3.5.
Assume that
,
, and
are satisfied, then the problem (1.4) has at least one solution.
Remark 3.6.
Compared with Theorem 3.2 in [16], Corollary 3.5 does not need conditions
, and
. Moreover, we only need
.
Example 3.7.
Consider the following problem:
where
. Here,
,
,
. Let
and
, then we can see that
holds. Choosing
,
, we can easily obtain that
holds. Let
, then we have that
also holds. Moreover,
,
. Hence, we get
for any given
. Therefore, By Theorem 3.1, the above problem (3.13) has at least one solution for
.