1 Introduction and preliminaries

In this paper, we consider the following nonlinear Volterra-type difference equation:

x(n)= i = 1 λ [ f i ( n , x ( n ) ) m Z a i ( n , m ) g i ( m , x ( m ) ) ] ,nZ,
(1.1)

where λ is a fixed positive integer and f i :Z×RR, a i :Z×ZR, g i :Z×RR (i=1,2,,λ) satisfy some conditions recalled in Section 3.

For the background of discrete Volterra equations, we refer the reader to the well-known monograph [1] by Agarwal. The first motivation for this paper is some recent work on asymptotical periodicity for Volterra-type difference equations in [26] by Diblík et al. In fact, asymptotical behavior for Volterra-type difference equations, including periodicity, asymptotical periodicity, etc., has been of great interest for many mathematicians. However, to the best of our knowledge, there is seldom literature available about asymptotically almost periodicity for Equation (1.1). Thus, in this paper, we will investigate this problem. In addition, it is needed to note that compared with asymptotically periodic sequences, in general, it is more difficult to obtain the compactness for a set of asymptotically almost periodic sequences.

On the other hand, in a recent work [7], by using the classical Schauder fixed point theorem, Ding et al. established an interesting existence theorem for the following functional integral equation:

y(t)=e ( t , y ( α ( t ) ) ) +g ( t , y ( β ( t ) ) ) [ h ( t ) + R k ( t , s ) f ( s , y ( γ ( s ) ) ) d s ] ,tR.
(1.2)

In fact, the existence of almost periodic type solutions has been an interesting and important topic in the study of qualitative theory of difference equations. We refer the reader to [813] and references therein for some recent developments on this topic. Equation (1.1) can be seen as a discrete analogue (but more general) of Equation (1.2). That is another main motivation for this work.

Throughout the rest of this paper, we denote by ℤ ( Z + ) the set of (nonnegative) integers, by ℕ the set of positive integers, by ℝ ( R + ) the set of (nonnegative) real numbers, by Ω a subset of ℝ, and by X a Banach space.

First, let us recall some notations and basic results of almost periodic type sequences (for more details, see [11, 14, 15]).

Definition 1.1 [14]

A function f:ZX is called almost periodic if ∀ε, N(ε)N such that among any N(ε) consecutive integers there exists an integer p with the property that

f ( k + p ) f ( k ) <ε,kZ.

Denote by AP(Z,X) the set of all such functions. Moreover, we denote AP(Z,R) by AP(Z) for convenience.

Lemma 1.2 [[14], Theorem 1.26]

A necessary and sufficient condition for the sequence f:ZR to be almost periodic is that for any integer sequence { n k }, one can extract a subsequence { n k } such that {f(n+ n k )} converges uniformly with respect to nZ.

Remark 1.3 Let f,gAP(Z). By Lemma 1.2, it is not difficult to show that ∀ε, N(ε)N such that among any N(ε) consecutive integers there exists a common integer p with the property that

| f ( k + p ) f ( k ) | <εand | g ( k + p ) g ( k ) | <ε

for all kZ.

Next, we denote by C 0 (Z,X) the space of all the functions f:ZX such that lim | n | f(n)=0.

Definition 1.4 A function f:ZX is called asymptotically almost periodic if it admits a decomposition f=g+h, where gAP(Z,X) and h C 0 (Z,X). Denote by AAP(Z,X) the set of all such functions. Moreover, we denote AAP(Z,R) by AAP(Z) for convenience.

Definition 1.5 Let ΩR and f be a function from Z×Ω to ℝ such that f(n,) is continuous for each nZ. Then f is called almost periodic in n Z uniformly for ωΩ if for every ε>0 and every compact ΣΩ, there corresponds an integer N ε (Σ)>0 such that among N ε (Σ) consecutive integers there exists an integer p with the property that

| f ( k + p , ω ) f ( k , ω ) | <ε

for all kZ and ωΣ. Denote by AP(Z×Ω) the set of all such functions.

Similarly, for each subset ΩR, we denote by C 0 (Z×Ω) the space of all the functions f:Z×ΩR such that f(n,) is continuous for each nZ, and lim | n | f(n,x)=0 uniformly for x in any compact subset of Ω.

Definition 1.6 A function f:Z×ΩR is called asymptotically almost periodic in n uniformly for xΩ if it admits a decomposition f=g+h, where gAP(Z×Ω) and h C 0 (Z×Ω). Denote by AAP(Z×Ω) the set of all such functions.

Lemma 1.7 Let E{AP(Z,X),AAP(Z,X)}. Then the following hold true:

  1. (a)

    fE implies that f is bounded.

  2. (b)

    f,gE implies that f+gE. Moreover, fgE if X=R.

  3. (c)

    E is a Banach space equipped with the supremum norm.

Proof The proof is similar to that of the continuous case (cf. [14, 15]). So, we omit the details. □

2 A compactness criterion

The following theorem is a well-known result for the continuous case (see, e.g., [[16], p.24, Theorem 2.5]). Here, we give a discrete version.

Theorem 2.1 Let f be a function fromto ℝ. Then fAAP(Z) if and only ifε, M(ε),N(ε)N such that among any N(ε) consecutive integers there exists an integer p with the property that

| f ( k + p ) f ( k ) | <ε

for all kZ with |k|M(ε) and |k+p|M(ε).

Proof We first show the ‘only if’ part. Let fAAP(Z). Then there exist gAP(Z) and h C 0 (Z,R) such that f=g+h. By gAP(Z), for each ε>0, N(ε)N such that among any N(ε) consecutive integers there exists an integer p with the property that

| g ( k + p ) g ( k ) | < ε 3 ,kZ.

In addition, since h C 0 (Z,R), for the above ε>0, there exists M(ε)N such that |h(k)|< ε 3 for all kZ with |k|M(ε). Thus, we have

| f ( k + p ) f ( k ) | | g ( k + p ) g ( k ) | + | h ( k + p ) | + | h ( k ) | <ε

for all kZ with |k|M(ε) and |k+p|M(ε).

Next, let us prove the ‘if’ part. First, let us show that f is bounded. Letting ε=1, there exists M(1),N(1)N such that among any N(1) consecutive integers there exists an integer p with the property that

| f ( k + p ) f ( k ) | <1

for all kZ with |k|M(1) and |k+p|M(1). Then, for each kZ with |k|M(1), there exists p k [M(1)k,M(1)+N(1)k]Z such that

| f ( k + p k ) f ( k ) | <1.

Noting that k+ p k [M(1),M(1)+N(1)], we get

| f ( k ) | | f ( k + p k ) | +1 max k [ M ( 1 ) , M ( 1 ) + N ( 1 ) ] | f ( k ) | +1

for all kZ with |k|M(1). Thus,

sup k Z | f ( k ) | max k [ M ( 1 ) , M ( 1 ) + N ( 1 ) ] | f ( k ) | +1<+.

Now, let us show that fAAP(Z). We divide the remaining proof into three steps.

Step 1. Since f is bounded, we can choose a sequence { s n }N such that lim n + s n =+ and lim n + f(k+ s n ) exists for each kZ. Let

g ¯ (k)= lim n + f(k+ s n ),kZ.

For each ε>0, among any N(ε) consecutive integers there exists an integer p with the property that

| f ( k + p ) f ( k ) | <ε

for all kZ with |k|M(ε) and |k+p|M(ε). Then, for each fixed kZ, we have

| f ( k + s n + p ) f ( k + s n ) | <ε

for sufficiently large n, which yields that

| g ¯ ( k + p ) g ¯ ( k ) | ε.

Thus, g ¯ AP(Z).

Step 2. Now fix ε>0. Then, for each nN, there exists t n [ s n N(ε), s n ]Z such that

| f ( k + t n ) f ( k ) | <ε
(2.1)

for all kZ with |k|M(ε) and |k+ t n |M(ε). Let r n = s n t n . Then r n {0,1,2,,N(ε)}, which means that there exist a subsequence { r n }{ r n } and r(ε){0,1,2,,N(ε)} such that

r n r(ε).

Thus, for all kZ with |k|M(ε), we have

| f ( k ) g ¯ ( k r ( ε ) ) | = | f ( k ) f ( k + t n ) | + | f ( k r ( ε ) + s n ) g ¯ ( k r ( ε ) ) | .

Combining this with (2.1), lim n + t n =+, and

g ¯ (k)= lim n + f(k+ s n ),kZ,

we conclude

| f ( k ) g ¯ ( k r ( ε ) ) | ε

for all kZ with |k|M(ε).

Step 3. By Step 2, we know that for each ε>0, there exists r(ε){0,1,2,,N(ε)} such that

| f ( k ) g ¯ ( k r ( ε ) ) | ε

for all kZ with |k|M(ε). Taking ε=1,1/2, , we get a sequence {r(1/m)}. On the other hand, it follows from Step 1 that g ¯ AP(Z). Thus, going to a subsequence, if necessary, we may assume that g ¯ (r(1/m)) is uniformly convergent on ℤ. Let

g(k)= lim m + g ¯ ( k r ( 1 / m ) ) ,kZ.

Then gAP(Z). In addition, noting that

| f ( k ) g ( k ) | | f ( k ) g ¯ ( k r ( 1 / m ) ) | + | g ¯ ( k r ( 1 / m ) ) g ( k ) | 1 m + | g ¯ ( k r ( 1 / m ) ) g ( k ) |

for all kZ with |k|M(1/m), we know that fg C 0 (Z). This completes the proof. □

Definition 2.2 FAAP(Z) is said to be equi-asymptotically almost periodic if for each ε>0, there exist M(ε),N(ε)N such that among any N(ε) consecutive integers there exists an integer p with the property that

sup f F | f ( k + p ) f ( k ) | <ε

for all kZ with |k|M(ε) and |k+p|M(ε).

Theorem 2.3 Let FAAP(Z). Then F is precompact in AAP(Z) if and only if the following two conditions hold:

  1. (i)

    for each kZ, {f(k):fF} is bounded;

  2. (ii)

    F is equi-asymptotically almost periodic.

Proof ‘only if’ part

Let FAAP(Z) be precompact. Then F is bounded in AAP(Z). So, (i) obviously holds. In addition, ε>0, there exists NN and f 1 , f 2 ,, f N F such that

F i = 1 N B( f i ,ε).
(2.2)

By Remark 1.3, we can get that { f 1 , f 2 ,, f N } is equi-asymptotically almost periodic. Combing this with (2.2), we can show that F is equi-asymptotically almost periodic, i.e., (ii) holds.

‘if part’

Let { f n }F. Since { f n (k)} is bounded for each kZ, we can assume that (if necessary going to a subsequence) { f n (k)} is convergent for each kZ. On the other hand, since F is equi-asymptotically almost periodic, for each ε>0, there exist M(ε),N(ε)N such that among any N(ε) consecutive integers there exists an integer p with the property that

sup n Z | f n ( k + p ) f n ( k ) | <ε/3
(2.3)

for all kZ with |k|M(ε) and |k+p|M(ε). For the above ε>0, there exists a positive integer K such that for all n,m>K, the following hold:

| f n ( k ) f m ( k ) | <ε/3,k [ M ( ε ) , M ( ε ) + N ( ε ) ] Z.
(2.4)

For all kZ with |k|M(ε), taking p[k+M(ε),k+M(ε)+N(ε)]Z, by (2.3) and (2.4), we get

| f n ( k ) f m ( k ) | | f n ( k ) f n ( k + p ) | + | f n ( k + p ) f m ( k + p ) | + | f m ( k + p ) f m ( k ) | < ε / 3 + ε / 3 + ε / 3 = ε , n , m > K ;

also, for all kZ with |k|<M(ε), by (2.4), we have

| f n ( k ) f m ( k ) | <ε/3<ε,n,m>K.

Thus, we get

sup k Z | f n ( k ) f m ( k ) | ε,n,m>K,

which means that { f n (k)} is uniformly convergent on ℤ, i.e., { f n } is convergent in AAP(Z). So, F is precompact in AAP(Z). □

3 Application to Volterra difference equations

In this section, we discuss the existence of an asymptotically almost periodic solution to Volterra difference equation (1.1). Throughout the rest of this paper, p,q1 are two fixed real numbers and

1 p + 1 q =1.

In addition, we denote by l p (Z) (resp. l q (Z)) the space of all the functions f:ZR satisfying

f p := ( k Z | f ( k ) | p ) 1 / p <+ ( resp.  f q : = ( k Z | f ( k ) | q ) 1 / q < + ) .

For convenience, we first list some assumptions.

(H1) For each i{1,2,,λ}, f i (,x)AAP(Z) for any fixed xR, and there exists a constant L i 0 such that

| f i ( k , x ) f i ( k , y ) | L i |xy|,kZ,x,yR.

(H2) For each i{1,2,,λ}, g i (k,) is continuous for each kZ, and for each r>0, there exists a sequence { μ i r } l p (Z) such that

| g i ( k , x ) | μ i r (k),|x|r,kZ.

(H3) For each i{1,2,,λ}, a ˜ i AAP(Z, l q (Z)), where [ a ˜ i (k)](l)= a i (k,l), k,lZ.

(H4) There exists a constant M>0 such that

i = 1 λ α i L i μ i M p <1,

where α i = sup n Z a ˜ i ( n ) q ; and

i = 1 λ [ sup n Z , | x | K | f i ( n , x ) | α i μ i M p ] <K,K>M,

Theorem 3.1 Assume that (H1)-(H4) hold. Then Equation (1.1) has an asymptotically almost periodic solution.

Proof We denote

( A i x ) ( n ) = f i ( n , x ( n ) ) , n Z , x AAP ( Z ) , i = 1 , 2 , , λ ; ( B i x ) ( n ) = m Z a i ( n , m ) g i ( m , x ( m ) ) , n Z , x AAP ( Z ) , i = 1 , 2 , , λ ;

and

(Mx)(n)= i = 1 λ ( A i x)(n)( B i x)(n),nZ,xAAP(Z).

It suffices to prove that ℳ has a fixed point in AAP(Z). We give the proof in three steps.

Step 1. A i and B i both map AAP(Z) into AAP(Z), i=1,2,,λ.

Since f i is Lipschitz, by Remark 1.3, we can first show that for each compact subset KR and each i{1,2,,λ}, { f i (,x):xK} is equi-asymptotically almost periodic. Then it is easy to show that A i xAAP(Z) for each xAAP(Z).

Since a ˜ i AAP(Z, l q (Z)), there exist b i AP(Z, l q (Z)) and c i C 0 (Z, l q (Z)) such that a ˜ i = b i + c i . For each xAAP(Z), noting that for n,pZ,

and

| m Z [ c i ( n ) ] (m) g i ( m , x ( m ) ) | c i ( n ) q μ i x p ,

we know that B i xAAP(Z).

Step 2. For each yAAP(Z) with yM, there exists a unique x y AAP(Z) such that

x y = i = 1 λ A i x y B i y.

Let

(Yx)(n)= i = 1 λ ( A i x)(n)( B i y)(n),nZ,xAAP(Z).

Then, by Step 1, Y maps AAP(Z) into AAP(Z). For all x 1 , x 2 AAP(Z) and nZ, we have

| ( Y x 1 ) ( n ) ( Y x 2 ) ( n ) | i = 1 λ | ( A i x 1 ) ( n ) ( A i x 2 ) ( n ) | | ( B i y ) ( n ) | = i = 1 λ | f i ( n , x 1 ( n ) ) f i ( n , x 2 ( n ) ) | | ( B i y ) ( n ) | i = 1 λ L i | x 1 ( n ) x 2 ( n ) | | m Z a i ( n , m ) g i ( m , y ( m ) ) | i = 1 λ L i x 1 x 2 | m Z a i ( n , m ) g i ( m , y ( m ) ) | i = 1 λ L i x 1 x 2 m Z | [ a ˜ i ( n ) ] ( m ) | μ i M ( m ) i = 1 λ L i x 1 x 2 a ˜ i ( n ) q μ i M p ( i = 1 λ α i L i μ i M p ) x 1 x 2 ,

which yields that

Y x 1 Y x 2 ( i = 1 λ α i L i μ i M p ) x 1 x 2 .

Noting that i = 1 λ α i L i μ i M p <1, Y has a unique fixed point x y in AAP(Z).

Step 3. ℳ has a fixed point in AAP(Z).

Let E={yAAP(Z):yM} and

Ny= x y ,yE,

where x y is the unique fixed point of Y (see Step 2).

We claim that N(E)E. In fact, if there exists y 0 E such that N y 0 >M, then by (H4), we have

N y 0 = x y 0 = sup n Z | i = 1 λ ( A i x y 0 ) ( n ) ( B i y 0 ) ( n ) | sup n Z ( i = 1 λ | f i ( n , x y 0 ( n ) ) | | m Z a i ( n , m ) g i ( m , y 0 ( m ) ) | ) sup n Z ( i = 1 λ | f i ( n , x y 0 ( n ) ) | α i μ i M p ) i = 1 λ [ sup n Z , | x | N y 0 | f i ( n , x ) | α i μ i M p ] < N y 0 ,

which is a contradiction.

Next, let us show that N:EE is continuous. For all y 1 , y 2 E, we have

N y 1 N y 2 = x y 1 x y 2 = i = 1 λ A i x y 1 B i y 1 i = 1 λ A i x y 2 B i y 2 i = 1 λ A i x y 1 B i y 1 A i x y 2 B i y 1 + A i x y 2 B i y 1 A i x y 2 B i y 2 ( i = 1 λ α i L i μ i M p ) x y 1 x y 2 + i = 1 λ ( M L i + sup n Z | f i ( n , 0 ) | ) B i y 1 B i y 2 ,

which gives that

N y 1 N y 2 i = 1 λ β i B i y 1 B i y 2 ,
(3.1)

where

β i := M L i + sup n Z | f i ( n , 0 ) | 1 ( i = 1 λ α i L i μ i M p ) ,i=1,2,,λ.

Letting y k y in E, by (3.1), we have

N y k N y i = 1 λ β i B i y k B i y i = 1 λ β i sup n Z ( m Z | a i ( n , m ) | | g i ( m , y k ( m ) ) g i ( m , y ( m ) ) | ) i = 1 λ β i sup n Z ( a ˜ i ( n ) q g i ( , y k ( ) ) g i ( , y ( ) ) p ) i = 1 λ α i β i g i ( , y k ( ) ) g i ( , y ( ) ) p .
(3.2)

For each i=1,2,,λ, noting that

| g i ( m , y k ( m ) ) g i ( m , y ( m ) ) | 2 μ i M (m),mZ,

g i (m,) is continuous for each mZ, and y k (m)y(m) for each mZ, we conclude that

g i ( , y k ( ) ) g i ( , y ( ) ) p 0.

Combining this with (3.2), we know that N y k Ny. So N:EE is continuous.

Now, let us show that N(E) is precompact in AAP(Z). In order to show that, we first prove each B i (E) is precompact in AAP(Z). By a direct calculation, we can get

| ( B i y ) ( n ) | α i μ i M p ,i=1,2,,λ,

for all yE and nZ. In addition, for all n 1 , n 2 Z and yE, we have

| ( B i y ) ( n 1 ) ( B i y ) ( n 2 ) | m Z | a i ( n 1 , m ) a i ( n 2 , m ) | | g i ( m , y ( m ) ) | a ˜ i ( n 1 ) a ˜ i ( n 2 ) q μ i M p ,

which yields that each B i (E) is equi-asymptotically almost periodic since . Then, by Theorem 2.3, each B i (E) is precompact in AAP(Z). Let { y k }E. Then { B i y k }, if necessary going to a subsequence, is convergent in AAP(Z) for each i{1,2,,λ}. By (3.1), we conclude that {N y k } is convergent in AAP(Z). So, N(E) is precompact in AAP(Z).

By applying Schauder’s fixed point theorem, there exists a fixed point y of N in E. Then we have

y =N y = x y = i = 1 λ A i x y B i y = i = 1 λ A i y B i y =M y ,

which means that y is a fixed point of ℳ. This completes the proof. □

Finally, we give a simple example to illustrate our result.

Example 3.2 Let λ=2, p=1, q=,

f 1 ( n , x ) = x 10 ( sin n + sin π n + 1 | n | + 1 ) , g 1 ( n , x ) = sin ( x e n 2 ) 2 ( 1 + n 2 ) , a 1 ( n , m ) = cos n + cos 2 n + 1 n 2 + 1 3 ( 1 + m 2 ) ,

and

f 2 (n,x)= cos n sin x 20 , g 2 (n,x)= arctan ( n x ) 1 + n 2 , a 2 (n,m)= 1 3 e m 2 sinn.

It is easy to see that (H1) holds with L 1 = 3 10 and L 2 = 1 20 . Also, (H2) holds with μ 1 r (n) 1 2 ( 1 + n 2 ) and μ 2 r (n) π 2 1 1 + n 2 . In addition, (H3) can be easily verified. By a direct calculation, we can get

α 1 1, α 2 1 3 ,

and

μ 1 r 1 π + 1 2 , μ 2 r 1 π 2 + π 2 ,r>0.

Letting M=1, we have

i = 1 2 α i L i μ i M 1 3 ( π + 1 ) 20 + π 2 + π 120 <1,

and

i = 1 2 [ sup n Z , | x | K | f i ( n , x ) | α i μ i M 1 ] 3 ( π + 1 ) 20 K+ π 2 + π 120 <K,K>1.

Thus, (H4) holds with M=1. Then, by using Theorem 3.1, Equation (1.1) has an asymptotically almost periodic solution.