In this section, by using the techniques of approximate solutions and fixed points, we establish a result on the existence of mild solutions for the nonlocal impulsive problem (1.1) when the nonlocal item and the impulsive functions are only assumed to be continuous in and , respectively.

In practical applications, the values of for near zero often do not affect . For example, it is the case when

So, to prove our main results, we introduce the following assumptions.

(H2) is a continuous function, and there is a such that for any with , . Moreover, there exist such that for any .

(H3)There exists a such that is a continuous function, and for any with , . Moreover, there exist such that

for any , , and

for any , .

(H4)The function is continuous a.e. ; the function is strongly measurable for all . Moreover, for each , there exists a function such that for a.e. and all , and

(H5) is continuous for every , and there exist positive numbers such that for any and .

We note that, by Theorem 2.1, there exist and such that and

For simplicity, in the following we set and will substitute by below.

Theorem 3.1.

Let (H1)–(H5) hold. Then the nonlocal impulsive Cauchy problem (1.1) has at least one mild solution on , provided

To prove the theorem, we need some lemmas. Next, for , we denote by the maps defined by

In addition, we introduce the decomposition , where

for and .

Lemma 3.2.

Assume that all the conditions in Theorem 3.1 are satisfied. Then for any , the map defined by (3.7) has at least one fixed point .

Proof.

To prove the existence of a fixed point for , we will use Darbu-Sadovskii's fixed point theorem.

Firstly, we prove that the map is a contraction on . For this purpose, let . Then for each and by condition (H3), we have

Thus,

which implies that is a contraction by condition (3.6).

Secondly, we prove that , , are completely continuous operators. Let be a sequence in with

in . By the continuity of with respect to the second argument, we deduce that for each , converges to in , and we have

Then by the continuity of , , , and using the dominated convergence theorem, we get

in , which implies that are continuous on .

Next, for the compactness of we refer to the proof of [4, Theorem 3.1].

For and any bounded subset of , we have

which implies that is relatively compact in for every by the compactness of . On the other hand, for , we have

Since is relatively compact in , we conclude that

which implies that is equicontinuous on . Therefore, is a compact operator.

Now, we prove the compactness of . For this purpose, let

Note that

Thus according to Lemma 2.3, we only need to prove that

is precompact in , as the remaining cases for , , can be dealt with in the same way; here is any bounded subset in . And, we recall that , , which means that

Thus, by the compactness of , we know that is relatively compact in for every .

Next, for , we have

Thus, the set is equicontinuous due to the compactness of and the strong continuity of operator . By the Arzela-Ascoli theorem, we conclude that is precompact in . The same idea can be used to prove that is precompact for each . Therefore, is precompact in , that is, the operator is compact.

Thus, for any bounded subset , we have by Lemma 2.4,

Hence, the map is an -contraction in .

Now, in order to apply Lemma 2.5, it remains to prove that there exists a constant such that . Suppose this is not true; then for each positive integer , there are and such that . Then

Dividing on both sides by and taking the lower limit as , we obtain that

This is a contradiction with inequality (3.6). Therefore, there exists such that the mapping maps into itself. By Darbu-Sadovskii's fixed point theorem, the operator has at least one fixed point in . This completes the proof.

Lemma 3.3.

Assume that all the conditions in Theorem 3.1 are satisfied. Then the set is precompact in for all , where

and is the constant in (H2).

Proof.

The proof will be given in several steps. In the following is a number in .

Step 1.

is precompact in .

For , define by

For , let , , , , and we define by

By condition (H3), is well defined and for , we have

On the other hand, for , , we have , . So,

Now, for , we have

By the compactness of , , we get that is relatively compact in for every and is equicontinuous on , which implies that is precompact in .

By the same reasoning, is precompact in .

For , we claim that is Lipschitz continuous with constant . In fact, (H3) implies that for every and ,

that is,

Therefore, is Lipschitz continuous with constant .

Clearly, is precompact in , and so is in .

Thus, by (3.29) and Lemma 2.4, we obtain

By (3.6), , which implies . Consequently, is precompact in .

Step 2.

is precompact in .

For , let

and define by

By (H3), is well defined and for , we have

So, for , , we have

where

According to the proof of Step 1, we know that

are all precompact in and is Lipschitz continuous with constant .

Next, we will show that is precompact in . Firstly, it is easy to see that is precompact in . Thus according to Lemma 2.3, it remains to prove that

is precompact in . And, we recall that , , which means that

By Step 1, is precompact in . Without loss of generality, we may suppose that

Therefore, , as in . Thus, by the continuity of and , we get

as , which implies that is relatively compact in . And, for , by the compactness of , , is also relatively compact in . Therefore, is relatively compact in for every .

Next, for , we have

Thus, the set is equicontinuous on due to the compactness of and the strong continuity of operator , . By the Arzela-Ascoli theorem, we conclude that is precompact in . Therefore, is precompact in .

Thus, by Lemma 2.4, we obtain

By (3.6), , which implies . Consequently, is precompact in .

Step 3.

The same idea can be used to prove the compactness of in for , where . This completes the proof.

Proof of Theorem 3.1.

For , , let

where comes from the condition (H2). Then, by condition (H2), .

By Lemma 3.3, without loss of generality, we may suppose that , as . Thus, by the continuity of and , we get

as . Thus,

is precompact in . Moreover, and are both precompact in . And is Lipschitz continuous with constant . Note that

Therefore, by Lemma 2.4, we know that the set is precompact in . Without loss of generality, we may suppose that in . On the other hand, we also have

Letting in both sides, we obtain

which implies that is a mild solution of the nonlocal impulsive problem (1.1). This completes the proof.

Remark 3.4.

From Lemma 3.3 and the above proof, it is easy to see that we can also prove Theorem 3.1 by showing that is precompact in .

The following results are immediate consequences of Theorem 3.5.

Theorem 3.5.

Assume (H1), (H3)–(H5) hold. If , then the impulsive Cauchy problem (1.1) has at least one mild solution on , provided

Theorem 3.6.

Assume (H1), (H2), (H4), and (H5) hold. If , then the nonlocal impulsive problem (1.1) has at least one mild solution on , provided .

Theorem 3.7.

Assume (H1), (H4), and (H5) hold. If , then the impulsive problem (1.1) has at least one mild solution on , provided .

Remark 3.8.

Theorems 3.5-3.6 are new even for many special cases discussed before, since neither the Lipschitz continuity nor compactness assumption on the impulsive functions is required.