Let

To present the main results of this paper, we need the following assumption.

(H) is such that there exists such that , for all .

Remark 2.1. (1) is a translation invariant Banach space. For every , one has too. Set , then satisfies (H), and therefore there exist a great number of functions satisfying the assumption (H). (2) Reference [5] uses an assumption similar to (H) implicitly.

Let . We have the following lemma.

Lemma 2.2.

Under the assumption (H), one has .

Proof.

By (H), there exists such that , . Let and , it is easy to verify that ,, , ,, , , and , , for all that is, , . Set , similarly we can obtain , , and , for all that is, .

Lemma 2.3.

Suppose that is a Banach space, denotes the set of bounded linear operators from to , and , then is bounded invertible and

where , is an identical operator.

The proof of Lemma 2.3 can be found in any book of functional analysis. We remark that if is a linear operator and its inverse exists, then is also a linear operator.

To get w-solutions or solutions of (1.2), we start with its corresponding difference equations.

Suppose that is an ow-solution of (1.2), then satisfies the three conditions in Definition 1.3. By a process of integrating (1.2) two times in as in [6–10], we can easily get

Similarly if is an ew-solution of (1.2), by the process of integrating (1.2) two times in , we get

These lead to the difference equations

From the analysis above, one sees that if is an ow-solution (resp., ew-solution) of (1.2), then one gets (2.5) (resp., (2.6)); if is a solution of (1.2), then one gets both (2.5) and (2.6). Conversely, we will show that the w-solutions or solutions of (1.2) are obtained via the solutions of (2.5) and (2.6). In order to get the solutions of (2.5) and (2.6), we will consider the following difference equations:

Notice that for any sequences , and , one has . Especially, In virtue of studying (2.7) and (2.8), we have the following theorem.

Theorem 2.4.

Under the assumption (H), (2.7) (resp., (2.8)) has a unique solution (resp., ).

Proof.

As the proof of [7, Theorem 9], define by , where is the Banach space consisting of all bounded sequences in with . Notice Lemmas 1.4 and 2.3, we know that (2.7) has a unique solution . By the process of proving Lemma 2.2, we have that is, where (this follows in the same way as [7, Theorem 9]). Therefore, (2.7) has a unique solution .

Similarly, (2.8) has a unique solution and , that is, where . Therefore, (2.8) has a unique solution . This completes the proof.

Remark 2.5. (i) In Theorem 2.4, since and , we can easily get

It must be stressed that (2.9) and (2.10) are important, since they can guarantee the continuity of the w-solutions or solutions of (1.2) constructed in Theorems 2.6, 2.7, and 2.8.

(ii) Let with satisfying (2.9) , and with satisfying (2.10) . Notice the fact that the solution of (2.7) (resp., (2.8)) must be a solution of (2.5) (resp., (2.6)), it is false conversely. So, suppose the assumption (H) holds, it follows from Theorem 2.4 that (2.5) (resp., (2.6)) has solution (resp., ). Moreover, such solutions may not be unique. See Example 3.1 at the end of this paper.

In the following, we focus on seeking the almost-periodic w-solutions or solutions of (1.2) via the almost-periodic sequence solutions of (2.5) and (2.6). As mentioned above, it is due to that, to get almost-periodic w-solutions or solutions of (1.2), we have to use a way quite different from [6–10]. Our main idea is to construct solutions or w-solutions of (1.2) piecewise. It seems that this is a new technique.

Without loss of generality, suppose that (resp., ) is an arbitrary solution of (2.5) (resp., (2.6)). To prove the following theorems, we need to introduce some notations firstly:

where and . It can be easily verified that

For the existence of the almost-periodic ow-solution of (1.2), we have the following.

Theorem 2.6.

Under the assumption (H), (1.2) has an ow-solution such that .

Proof.

Under the assumption (H), define as

where

From (2.9), it follows that is continuous on and , . Moreover, for , , one has ; for , , one has . By simple calculation, for , , we have

Note that , this implies that the one-sided derivatives exist at . Since , the second-order derivatives are continuous at , . Therefore, is an ow-solution of (1.2) such that , . Furthermore, it is easy to check that is almost periodic, we omit the details. The proof is complete.

For the existence of the almost-periodic ew-solution of (1.2), we have the following.

Theorem 2.7.

Under the assumption (H), (1.2) has an ew-solution such that .

Proof.

Under the assumption (H), define as

where

From (2.10), it follows that is continuous on and , . The rest of the proof is similar to that of Theorem 2.6, we omit the details.

For the existence of almost-periodic solution of (1.2), we have the following.

Theorem 2.8.

Under the assumption (H), if is the common solution of (2.5) and (2.6), then (1.2) has a solution such that , . If replaces , the conclusion is still true.

Proof.

Since and are solutions of (2.5) and (2.9) respectively, and they are also solutions of (2.6) and (2.10), respectively, it follows from Theorems 2.6 and 2.7 that, by simple calculation, the almost-periodic ow-solution constructed as the proof of Theorem 2.6 with , , is the same as the almost-periodic ew-solution constructed as the proof of Theorem 2.7 with , . This implies is an almost-periodic solution of (1.2) such that , . If replaces , the proof is similar, we omit the details.

Remark 2.9.

As mentioned above, an ow-solution of (1.2) is not equivalent to an ew-solution of (1.2), and a solution of (1.2) is an ow-solution of (1.2) as well as an ew-solution of (1.2). See the examples in Section 3.

The following theorem is usually used for judging whether or not a w-solution of (1.2) is a solution of (1.2).

Theorem 2.10.

Suppose that is a solution of (1.2), then

Proof.

If is a solution of (1.2), then must be common solution of (2.5) and (2.6). Moreover, (2.6) is equivalent to

Substituting into both the above equation and (2.5), then add the resulting equations to get the result.