From now on we will assume that the system (1.1) has a unique solution for a given initial condition on and without loss of generality , thus .
We will make the following assumptions on (1.1).

(H1)
is continuous in the second variable for any fixed .

(H2)
System (1.1) has a bounded solution , passing through , , that is, .
For this bounded solution , there is an such that for all . So, we will have to assume that for all , and . Next, we will point out the definitions of stability for functional difference equations adapting it from the continuous case according to Hino et al. in [3].
Definition 4.1.
A bounded solution of (1.1) is said to be:

(i)
stable, if for any and any integer , there is such that implies that for all , where is any solution of (1.1);

(ii)
uniformly stable, abbreviated as "'', if for any and any integer , there is ( does not depend on ) such that implies that for all , where is any solution of (1.1);

(iii)
uniformly asymptotically stable, abbreviated as "'', if it is uniformly stable and there is such that for any , there is a positive integer such that if and , then for all , where is any solution of (1.1);

(iv)
globally uniformly asymptotically stable, abbreviated as "'', if it is uniformly stable and as , whenever is any solution of (1.1).
Remark 4.2.
It is easy to see that an equivalent definition for , being , is the following:

(iii)
is , if it is uniformly stable, and there exists such that if and , then as , where is any solution of (1.1).
4.1. The Periodic Case
Here, we will assume what follows.

(H3)
The function in (1.1) is periodic in , that is, there exists a positive integer such that for all .
Moreover, we will assume what follows.
Lemma 4.3.
Suppose that condition () holds. If is a bounded solution of (1.1) such that , then is also bounded in .
Proof.
Let us say that for all . Then by Axiom (A)(iii) and hypothesis () we have
Lemma 4.4.
Suppose that condition () holds. Let be a sequence in such that for all . Assume that as for every and , then in as for each . In particular, if as uniformly in , then in as uniformly in .
Proof.
By Axiom (A)(iii) and hypotheses we have that
In the particular case we obtain
and so as . On the other hand, since is fixed, it follows that
for each . Therefore, we have concluded the proof.
Theorem 4.5.
Suppose that condition () and (H1)–(H3) hold. If the bounded solution of (1.1) is , then is an a.a.p. sequence in , equivalently, (1.1) has an a.a.p. solution.
Proof.
By Lemma 4.3 there exists such that for all , and a bounded (or compact) set such that for all . Let be any integer sequence such that and as . For each , there exists a nonnegative integer such that . Set . Then for all . Since is a bounded set, we can assume that, taking a subsequence if necessary, for all , where . Now, set . Thus,
which implies that is a solution of the system,
through . It is clear that if is , then is also with the same pair as the one for .
Since is bounded for all and , we can use the diagonal method to get a subsequence of such that converges for each as . Thus, we can assume that the sequence converges for each as . Since , by Lemma 4.4 it follows that is also convergent for each . In particular, for any there exists a positive integer such that if ( is the constant given in Axiom A(ii), then
where is the number given by the uniform stability of . Since , it follows from Definition 4.1 and (4.7) that
and by Axiom A(ii) it follows that
This implies that for any positive integer sequence , as , there is a subsequence of for which converges uniformly on as . Thus, the conclusion of the theorem follows from Lemma 3.9(d).
Before proving our following result we remark that if is a.a.p. then there are unique sequences such that , with a.p. and as as . By Lemma 3.9(a) it follows that is bounded and thus . Hence, by Axiom (C) we must have that for all . In particular, for all .
Theorem 4.6.
Suppose that and (H1)–(H3) hold and the bounded solution of (1.1) is , then system (1.1) has an a.p. solution, which is also .
Proof.
It follows from Theorem 4.5 that is an a.a.p. Set (), where is a.p. sequence and as . For the positive integer sequence , by Lemma 3.9(b)–(d) and arguments of the previous theorem, we can choice a subsequence of such that converges uniformly in and uniformly on as and is also a.p. Then, uniformly in , and thus by Lemma 4.4 uniformly in on as and . Since
as , we have for , that is, the system (1.1) has an almost periodic solution, and so we have proved the first statement of the theorem.
In order to prove the second affirmation, notice that since . For any , let be a solution of (1.1) such that and . Again, by Lemma 4.4 as for each , so there is a positive integer such that if , then
Thus, for , we have
Then,
Therefore, there is such that if , then
and hence, for all , where is a pair for the uniform stability of . This shows that if , then
for all , which implies that for all if because is arbitrary. This proves that is .
In the case when we have an asymptotically stable solution of (1.1) we obtain the following result.
Theorem 4.7.
Suppose that and (H1)–(H3) hold and the bounded solution of (1.1) is , then the system (1.1) has a periodic solution of period for some positive integer , which is also .
Proof.
Set , . By the proof of Theorem 4.5, there is a subsequence which converges to a solution of (4.6) for each and hence by Lemma 4.4, as . Thus, there is a positive integer such that (), where is obtained from the uniformly asymptotic stability of . Let , and notice that is a solution of (1.1). Since for , that is, , we have
and hence,
because is (see also Remark 4.2). On the other hand, is a.a.p. by Theorem 4.5, then
where is a.p. and as . It follows from (4.17) and (4.18) that
which implies that for all because is a.p.
For the integer sequence , , we have . Then uniformly for all as , and again by Lemma 4.4, uniformly in as . Since , we have for , which implies that (1.1) has a periodic solution of period .
Now, we will proceed to prove that by the use of definition in Remark 4.2. Notice that since then is a solution of (1.1) with the same as the one for . Let be any solution of (1.1) such that . Set . Again, for sufficient large , we have the similar relations (4.12) and (4.14) with and . Thus,
as if , because , , and satisfy (1.1). This completes the proof.
Finally, if the particular solution is , we will prove that system (1.1) has a periodic solution.
Theorem 4.8.
Suppose that and (H1)–(H3) hold and that the bounded solution of (1.1) is , then the system (1.1) has a periodic solution of period .
Proof.
By Theorem 4.5, is a.a.p. Then ), where () is an a.p. sequence and as . Notice that is also a solution of (1.1) satisfying . Since is , we have that as , which implies that for all . Using same technique as in the proof of Theorem 4.7, we can show that is a periodic solution of (1.1).
4.2. The Almost Periodic Case
Here, we will assume that

(H4)
the function in (1.1) is almost periodic in uniformly in the second variable.
By we denote the uniform closure of , that is, . Note that by Lemma 3.12 and by Lemma 3.13.
Lemma 4.9.
Suppose that Axiom (C) is true, and that is an a.p. sequence with , then is a.p.
Proof.
We know that, given , there exists an integer such that each discrete interval of length contains a such that
By Axiom (C) we have
Lemma 4.10.
Suppose that is a fading memory space and is a.a.p. with , then is a.a.p.
Proof.
Since is a.a.p. there are unique sequences and such that is a.p. and as . Then by Lemma 4.9 it follows that is a.p., and by Lemma 2.3 it follows that as . Therefore, is a.a.p.
Theorem 4.11.
Suppose that conditions , (H1)(H2), and (H4) hold and that is a fading memory space. If the bounded solution of (1.1) is an a.a.p. sequence, then the system (1.1) has an a.p. solution.
Proof.
Since the solution is a.a.p., it follows from Lemma 3.10 that has a unique decomposition , where is a.p. and as . Notice that is bounded. By Lemma 4.3 there is a compact set in such that for all . By Lemma 3.13, there is an integer sequence , , such that as and uniformly on as . Taking a subsequence if necessary, we can also assume that uniformly on , and by Lemma 3.9(b) we have that is also an a.p. sequence. For any , there is a positive integer such that if , then . In this case, we see that uniformly for all as , and hence by Lemma 4.4 in in as . Since
and from the previous considerations the first term of the righthand side of (4.23) tends to zero as and since as , we have that for all , which implies that (1.1) has an a.p. solution passing through , where for .
We are now in a position to prove the following result.
Theorem 4.12.
Suppose that the assumptions , (H1), (H2), and (H4) hold, and that is a fading memory space. If the bounded solution of (1.1) is , then is a.a.p. Consequently, (1.1) has an a.p. solution which is .
Proof.
Let the bounded solution of (1.1) be with the triple . Let be any positive integer such that as . Set . As previously is a solution of
and is with the same triple . By Lemma A.2, for the set and any there exists such that and for some implies that for all , where is a bounded solution of
passing through and for . Since is uniformly bounded for all and , taking a subsequence if necessary, we can assume that is convergent for each and uniformly on , for some a.p. function . In this case, by Lemma 4.4 there is a positive integer such that if , then
On the other hand, for is a solution of (4.25) with , that is,
where is defined by the relation
To apply Lemma A.2 to (4.24) and its associated equation (4.27), we will point out some properties of the sequence . Since uniformly on , for the above , there is a positive integer such that if , then
which implies that for all . Applying Lemma A.2 to (4.24) and its associated equation (4.27) with the above arguments and condition (4.26), we conclude that for any positive integer sequence , as , and , there is a positive integer such that
and hence by Axiom A(ii) for all if . This implies that the bounded solution of (1.1) is a.a.p. by Lemma 3.9(d). Furthermore, (1.1) has an a.p. solution, which is by Theorem 4.11. This ends the proof.