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 .