数理解析研究所講究録 1445 巻 2005 年 143-153

143

Stability Properties for Linear Volterra Difference Equations
with Convolution Kernels in a Banach Space
村上 悟 (Satoru Murakami)

岡山理科大学・理学部*

Department of Applied Mathematics,
Okayama University of Science
岡山理科大学・理学部

長渕 裕 (Yutaka Nagabuchi)

Department of Applied Science,
Okayama University of Science

1. INTRODUCTION

We consider the Volterra difference equations of convolution tyPe$x(n+1)= \sum_{r=0}^{n}B(n-r)x(r)$ ,$n\in \mathbb{Z}^{+}$,$(E_{0})$and$y(n+1)= \sum_{r=-\infty}^{n}B(n-r)y(r)$,$n$$\in \mathbb{Z}$,$(E_{\infty})$, the nonnegative integers) are bounded linear operators on a Banachwhere $B(n)$ (space $X$ over the field C. The study of Volterra difference equations has actively been
done. Indeed, in the case where $X$ is of finite dimension, the equations have extensively
been treated in the book [1] and some results on stability properties and so on wereobtained; for more details we refer the reader to [1, 2, 3] and the references therein.
Also, in $[4, 5]$ , Volterra difference equations with infinite dimensional $X$ were discussed inconnection with some partial differential equations with piecewise continuous delays, andwas investigated in connection with the invertibility
uniform asymptotic stability forof the characteristic operator together with the summability of the fundamental solution,
under additional conditions such as the mutual commutativity of the operators $B(n),$ $n\in$or the exponential decay of the norm $||B(n)||$ .
In this paper, we give a nice result on the stability properties of the zero solution ofin the context above. Indeed, without the additional conditions imposed inor$[4, 5]$ , we will establish an equivalence relation among the uniform asymptotic stability of, the summability of the fundamental solution and theorthe zero solution of
invertibility of the characteristic operator outside the unit circle in the complex plane.
$n\in \mathbb{Z}^{+}$$(E_{\infty})$$\mathbb{Z}^{+}$$(E_{0})$$(E_{\infty})$$(E_{0})$$(E_{\infty})$supported by the Grant-in-Aid for Scientific Research (C), No.16540177, of the Japanese
Ministry of Education, Culture, Sports, Science and Technology$*\mathrm{P}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{l}\mathrm{y}$144

2. NOTATIONS
the space. We denote by
Let $X$ be a (complex) Banach space with the norm
is a Banach space equipped with theof all bounded linear operators on $X$ . Clearly,, which is defined byoperator norm$|\cdot$$\mathcal{L}(X)$$|$$\mathcal{L}(X)$$||\cdot$$||$$||T||= \sup\{|Tx| :.
for any
For any interval

x\in X, |x|=1\}$$T\in \mathcal{L}(X)$,
we use the same notaion meaning the discrete one. Also, for an $X$ -valued function 4 on a discretee.g. $[0, \sigma]=\{0, 1, \ldots, \sigma\}$ for cr$\sup\{|\xi(j)| : j\in J\}$ . Let
and a function
interval , its norm is denoted by$J\cap \mathbb{Z}$$J$$J\subseteq \mathbb{R}$$\in \mathbb{Z}^{+}$$J$$\sigma\in \mathbb{Z}^{+}$$||\xi||_{J}:=$

satisfying
the solution $x(n)$ of: $[0, \sigma]arrow X$ be given. We denote by$x(n)=\phi(n)$ on. Similarly, forand a function : $(-\infty, \tau]arrow X$ , we denote.
satisfying $y(n)=\psi(n)$ on
by $y(n;\tau, \psi)$ the solution $y(n)$ of$(E_{0})$$x(n;\sigma, \phi)$$\phi$$[0, \sigma]$$\tau\in \mathbb{Z}$$\psi$$(-\infty, \tau]$$(E_{\infty})$

Definition 1. The zero solution of$(E_{0})$is said to beandsuch that if(i) uniformly stable if for any $>0$ there exists a.
for allthenwith
is an initial function on(ii) uniformly asymptotically stable if it is uniformly stable, and if there exists a $\mu>0$ suchandwith the property that, if
that, for any $>0$ there exists an$n\geq\sigma+N$.
for allthenis an initial function onwith$\sigma\in \mathbb{Z}^{+}$$\delta=\delta(\epsilon)>0$$\in$$|x(n;\sigma, \phi)|<\epsilon$$||\phi||[0,\mathrm{a}]<\delta$$[0, \sigma]$$\geq\sigma$$n$$\sigma\in \mathbb{Z}^{+}$$N=N(\epsilon)\in \mathbb{Z}^{+}$$\epsilon$$|x(n;\sigma, \phi)|<\epsilon$$||\phi||[0,\mathrm{a}]<\mu$$[0, \sigma]$$\phi$

Definition 2. The zero solution of$(E_{\infty})$$\phi$

is said to beandsuch that if(i) uniformly stable if for any $>0$ there exists a$n\geq\tau$.
for allthenis an initial fun ction onwith(ii) uniformly asymptotically stable if it is uniformly stable, and if there exists a $\mu>0$
with the property that, if
such that, for any $>0$ there exists an
and is an initial function onwith $||\psi||(-\infty,\tau]<\mu$ then $|y(n;\tau, \psi)|<\epsilon$ for all$\delta$$\epsilon$$(-\infty, \tau]$$=\delta(\epsilon)>0$$\tau\in \mathbb{Z}$$|y(n;\tau, \psi)|<\epsilon$$||\psi||(-\infty,\tau]<\delta$$\tau\in \mathbb{Z}$$N=N(\epsilon)\in \mathbb{Z}^{+}$$\epsilon$$\psi$$(-\infty, \tau]$$\psi$$n\geq\tau+N$ .

The fundamental solution of$(E_{0})$is a family in$\mathcal{L}(X)$$R(n+1)= \sum_{j=0}^{n}B(n-j)R(j)$ ,satisfying the relation$n\in \mathbb{Z}^{+}$. Then, for instance, the solution $y(n;\tau, \psi)$ ofandof constant formula as follows;$R(\mathrm{O})=I$$(E_{\infty})$

is given by the variation$y(n; \tau, \psi)=R(n-\tau)\psi(\tau)+\sum_{r=\tau}^{n-1}R(n-r-1)(\sum_{s=-\infty}^{\tau-1}B(r-s)\psi(s))$

.

(1)

145

3. MAIN RERSULTS
In what follows, we assume that $B:=\{B(n)\}\subset \mathcal{L}(X)$ is summable, that is, thecondition $\sum_{n=0}^{\infty}||B(n)||<\infty$ holds, and study stability properties of the zero solution of. Here and subsequently,, together with those of the zero solution of Eq.
Eq.
for $|z|\geq 1$ .
denotes the -transform of $B$ ; that is,In [5, Theorem 2] and [4, Theorem 2], the equivalence among the uniform asymptotic, the summability of the fundamental solution
stability of the zero solution of Eq.
, and the invertibility of the characteristic operator $zI-\hat{B}(z)$$R=\{R(n)\}$ of Eq.
has been established under some restrictions such as the mutual
associated with Eq.
or the exponential decay of the norm
commutativity of the operators $B(n)$ ,$||B(n)||$ . We will show in the following theorem that [5, Theorem 2] and [4, Theorem 2]
hold true without such restrictions.
$(E_{0})$$(E_{\infty})$$\hat{B}(z)$$\hat{B}(z).--\Sigma_{n=0}^{\infty}B(n)z^{-n}$$Z$$(E_{\infty})$$(E_{0})$$(E_{0})$$n\in \mathbb{Z}^{+}$, and assume that $B(n)$ ,Theorem 1. Let, are all compact. Then the following statements are equivalent.
$B=\{B(n)\}_{n=0}^{\infty}\in l^{1}(\mathbb{Z}^{+}):=l^{1}(\mathbb{Z}^{+}; \mathcal{L}(X))$$n\in \mathbb{Z}^{+}$

is uniformly asymptotically stable.
(i) The zero solution of Eq.
is unifo rmly asymptotically stable.
(ii) The zero solution of Eq.
.
(iii)$|z|\geq
1$ , the operator $zI-\hat{B}(z)$ is invertible in(iv) For any such that$(E_{0})$$(E_{\infty})$$R=\{R(n)\}_{n=0}^{\infty}\in l^{1}(\mathbb{Z}^{+})$$z$$\mathcal{L}(X)$.

In order to prove the theorem, we need the following preparatory results.

Proposition 1. Let
is invertible for eachsuch that$\rho\in \mathbb{R}$, there$\tilde{K}(\rho):=\Sigma_{n=-\varpi}^{\infty}K(n)e^{-i\rho n}$$\tilde{K}(\rho)(I-\tilde{K}(\rho))^{-1}=\tilde{R}(\rho)$

Proof.

, and assume that
. Then there is an$K=\{K(n)\}_{n=-\infty}^{\infty}\in l^{1}(\mathbb{Z}):=f^{1}(\mathbb{Z};\mathcal{L}(X))$(1-Step) For each (small)$\epsilon$

$>0$

,

we define a1
0$R\in l^{1}(\mathbb{Z})$$\forall\rho\in \mathbb{R}$$2\pi$.

-periodic function$\tilde{\phi}_{\epsilon}$$(|t|\leq\epsilon)$$(2\epsilon \leq|t|\leq\pi)$$\tilde{\phi}_{\epsilon}(t)=\{$$(2\epsilon -t)/\epsilon$$(\epsilon<t<2\epsilon )$$(2\epsilon +t)/\epsilon$$(-2\epsilon<t<-\epsilon)$

One can easily check that Fourier coefficients of$\tilde{\phi}_{\epsilon}$$d_{l}= \frac{1}{2\pi}\int_{-\pi}^{\pi}\tilde{\phi}_{\epsilon}(t)e^{ilt}dt$.

are given by$\frac{2}{\pi\in t^{2}}\sin\frac{3\epsilon l}{2}\sin\frac{\epsilon l}{2}$$(l=\pm 1, \pm 2, \ldots)$$=\{$$\frac{3\epsilon}{2\pi}$$I-\tilde{K}(\rho)$$(l=0)$ .

by

146

Clearly, thhe sequence
we get$\phi_{\epsilon}:=\{d_{l}\}_{l=-\infty}^{\infty}$

is summable, and by the Fourier expansion theorem$\sum_{l=-\infty}^{\infty}d_{l}e^{-il8}=\tilde{\phi}_{\epsilon}(t)$Hence it follows that for any$t_{0}\in \mathbb{R}$,R.

$\forall t\in$,$\tilde{\phi}_{\epsilon}(t-t_{0})\equiv\sum_{l=-\infty}^{\infty}c_{l}e^{-itl}$,

is a sequence defined bywhere
the sequence is summable.
(2-Step) Consider the function $f(x)$ defined by$c_{l}=d_{l}e^{it_{0}l}=\phi_{\epsilon}(l)e^{it_{0}l}$$c:=\{c_{l}\}_{l\in \mathbb{Z}}$for$l\in \mathbb{Z}$. Notice that$c$’$\frac{1}{\pi x^{2}}\sin 3x\sin x$$(x\neq 0)$$f(x)=$$(x=0)$ .

$\frac{3}{\pi}$One can easily see that$f$is continuously difFerentiable. In fact, $f’(x)$ is given by’$- \frac{2}{\pi x^{3}}\sin 3x\sin x+\frac{1}{\pi x^{2}}(3\cos 3x\sin x+\sin 3x\cos x)$$(x\neq 0)$0$(x=0)$ .

$f’(x)=$

Since $\lim|x|arrow\infty(|f(x)|+|f’(x)|)=0$ , there exists a constant $H>0$ such that $\sup_{-\infty<x<\infty}(|f(x)|+$$|f’(x)|)=H$ . Moreover,$\int_{0}^{\infty}|f’(x)|dx$$=$$\int_{0}^{1}|f’(x)|dx+\int_{1}^{\infty}|f’(x)|dx$$\leq$$H+ \int_{1}^{\infty}\frac{6}{\pi x^{2}}dx$$\leq$

where $C=6/\pi$ .
(3-Step) Put$H+C$ ,

$M= \sup_{\rho\in \mathbb{R}}||(I-\tilde{K}(\rho))^{-1}||$, and take a positive integer $N$ such that$\frac{23M}{\pi}\sum_{|\tau\}\geq N+1}||K(\tau)||\leq\frac{1}{4}$Moreover, take a positive integer$k_{\mathrm{V}}0$.

, $k_{0}\geq 3$ , such that

$2NM(H+C) \pi\sum_{l\in \mathbb{Z}}||K(l)||<\frac{3k_{0}}{4}$and set $=\pi/(3k_{0})$ andrelation holds:$\epsilon$$\rho_{n}=3\mathrm{n}\mathrm{s}$, $n=0,1$ ,$\sum_{n=0}^{2k_{0}-1}\tilde{\phi}_{\epsilon}(\rho-\rho_{n})\equiv 1$$\ldots$,,. Then $\beta 2k_{0}=2\pi$ , and the following$\forall\rho\in \mathbb{R}$

,

147

where

$\tilde{\phi}_{\epsilon}$is the one introduced in 1-Step. Set $F(\rho)=\tilde{K}(\rho)(I-\tilde{K}(\rho))^{-1}$ and $F_{n}(\rho)=$R. Then$\tilde{\phi}_{\epsilon}(\rho-\rho_{n})F(\rho))\rho\in$$F( \rho)\equiv\sum_{n=0}^{2k_{0}-1}F_{n}(\rho)$.

Therefore, in order to establish the proposition it suffices only to certify that for each
. We now set
such that
there exists an$F_{n}(\rho)\equiv\tilde{R}_{n}(\rho)$$R_{n}\in l^{1}(\mathbb{Z})$$l\in \mathbb{Z}$$K_{n}(l)=[((\phi_{2\epsilon}e^{i\rho_{n}}.)*K)(l)-\phi_{2\epsilon}(l)e^{i\rho_{n}l}\tilde{K}(\rho_{n})||(I-\tilde{K}(\rho_{n}))^{-1},$

where

$*$

denotes the convolution in$\tilde{K}_{n}(\rho)$

$=$

$=$

Observe that$n$$l^{1}(\mathbb{Z})$. Then$K_{n}\in l^{1}(\mathbb{Z}).$,,and moreover$[(\phi_{2\epsilon}e^{i\rho_{n}}.)^{-}(\rho)\tilde{K}(\rho)-(\phi_{2\epsilon}e^{i\rho_{n}}.)^{\sim}(\rho)\tilde{K}(\rho_{n})\ovalbox{\tt\small REJECT}(I-\tilde{K}(\rho_{n}))^{-1}$$\phi_{2\epsilon}(\rho-\rho_{n})(\tilde{K}(\rho)-\tilde{K}(\rho_{n}))(I-\tilde{K}(\rho_{n}))^{-1}$$\tilde{\phi}_{\epsilon}(\rho-\rho_{n})\neq 0$

implies$\tilde{K}_{n}(\rho)$$\phi_{2\epsilon}(\rho-\rho_{n})=1$.

, and hence$=$$(\tilde{K}(\rho)-\tilde{K}(\rho_{n}))(I-\tilde{K}(\rho_{n}))^{-1}$$=$$(\tilde{K}(\rho)-I)(I-\tilde{K}(\rho_{n}))^{-[perp]}+I$,or$I-\tilde{K}_{n}(\rho)=$(I$-\tilde{K}(\rho)$)$(I-\tilde{K}(\rho_{n}))^{-1}$,which implies$(I-\tilde{K}(\rho))^{-1}=(I-\tilde{K}(p_{n}))^{-1}(I-\tilde{K}_{n}(\rho))^{-1}$.

This observation leads to$F_{n}(\rho)$$\equiv$$\equiv$$\tilde{\phi}_{\epsilon}(\rho-\rho_{n})\tilde{K}(\rho)(I-\tilde{K}(\rho))^{-1}$$\tilde{\phi}_{\epsilon}(\rho-\rho_{n})\tilde{K}(\rho)(I-\tilde{K}(\rho_{n}))^{-1}(I-\tilde{K}_{n}(\rho))^{-1}$.

We claim that$|K_{n}|_{1}:= \sum_{l=-\infty}^{\infty}||K_{n}(l)||<\frac{1}{2}$.

(2)

If the claim is true, then the series $\Sigma_{\tau=0}^{\infty}K_{n}^{*\tau}:=e+K_{n}+K_{n}*K_{n}+K_{n}*K_{n}*K_{n}+\cdots$ , (here,with, converges inis the unit element into get the equality
and hence we may set$l^{1}(\mathbb{Z}))$$e$$l^{1}(\mathbb{Z})$$(I-\tilde{K}_{n}(\rho))^{-1}\equiv(\Sigma_{\tau=0}^{\infty}K_{n}^{*\tau})^{\sim}(\rho)$$R_{n}=(\phi_{\epsilon}e^{i\rho_{n}}.)*K*\{(I-\tilde{K}(\rho_{n}))^{-1}\Sigma_{\tau=0}^{\infty}K_{n}^{*\tau}\}$$F_{n}=\tilde{R}_{n}$with$R_{n}\in l^{1}(\mathbb{Z})$

.

148

In what follows we will evaluate$|K_{n}|_{1}$$|K_{n}|_{1}$to establish (2). It follows that$\leq$$M \sum_{l=-\infty}^{\infty}||\sum_{\tau=-\infty}^{\infty}K(\tau)(\phi_{2\epsilon}(l-\tau)e^{i\rho_{n}(l-\tau\rangle})-\phi_{2\epsilon}(l)e^{i\rho_{n}l}\sum_{\tau=-\infty}^{\infty}K(\tau)e^{-i\rho_{n}\tau}||$$\leq$$M \sum_{1\leq|\tau[\leq N}||K(\tau)||\sum_{l=-\infty}^{\infty}|\phi_{2\epsilon}(l-\tau)-\phi_{2\epsilon}(l)|$$+M \sum_{|\tau|\geq N+1}||K(\tau)||\sum_{l=-\infty}^{\infty}|\phi_{2\epsilon}(l-\tau)-\phi_{2\epsilon}(l)|$$=$Noting$0<\epsilon$$I_{2}$$\leq$:$I_{1}+I_{2}$.

$<1/2$ , we get$2M \sum_{|\tau|\geq N+1}||K(\tau)||\cross\sum_{l=-\infty}^{\infty}|\phi_{2\epsilon}(l)|$$( \frac{3\epsilon}{\pi}+\sum_{k=1}^{\infty}\frac{2}{\pi k^{2}\epsilon}|\sin 3\epsilon k\sin\epsilon k|)$$\leq$$2M \sum_{|\tau|\geq N+1}||K(\tau)||\mathrm{x}$$( \frac{3\epsilon}{\pi}+\frac{2}{\pi\epsilon}\sum_{k=1}^{[1/\Xi]}\frac{1}{k^{2}}|\sin 3\epsilon k\sin\epsilon k|+\frac{2}{\pi\epsilon}\sum_{k=[1/\epsilon]+1}^{\infty}\frac{1}{k^{2}})$$\leq$$2M \sum_{|\tau|\geq N+1}||K(\tau)||\cross$$( \frac{3\epsilon}{\pi}+\frac{2}{\pi\epsilon}\sum_{k=1}^{[1/\epsilon]}\frac{3\epsilon^{2}k^{2}}{k^{2}}+\frac{2}{\pi\epsilon}\oint_{[1/\epsilon]}^{\infty}\frac{dx}{x^{2}})$$\leq$$2M \sum_{|\tau|\geq N+1}||K(\tau)||\cross$$( \frac{3\epsilon}{\pi}+\frac{6\epsilon}{\pi}\mathrm{x}$$\leq$$[1/ \epsilon]+\frac{2}{\pi\epsilon}\frac{1}{[1/\epsilon]})$$2M \sum_{|\tau\}\geq N+1}||K(\tau)||\mathrm{x}$$\leq$$\frac{23M}{\pi}\sum_{|\tau|\geq N+1}||K(\tau)||$$\frac{1}{4}$$\leq$,

where
denotes the largest integer which does not exceed
introduced in 2-Step we get$[1/\Xi]$$1/\epsilon$. Also, using the function$f$$I_{1}$$\leq$$M|K|_{1} \sup_{1\leq|\tau|\leq N}(\sum_{l=-\infty}^{\infty}|f((l-\tau)\epsilon)-f(l\epsilon)|\epsilon)$$\leq$$M \epsilon|K|_{1}\sup_{1\leq|\tau\}\leq N}(\sum_{m=0}^{|\tau|-1}\sum_{s=-\infty}^{\infty}|f(\{(s+1)|\tau|+m\}\epsilon)-f(\{s|\tau|+m\}\epsilon)|)$$\leq$$M_{\Xi i}|K|_{1} \sup_{1\leq|\tau|\leq N}(\sum_{m=0}^{|\tau|-1}\sum_{s=-\infty}^{\infty}\int_{\{s|\tau|+m\}\epsilon}^{\{(s+1)|\tau|+m\}\epsilon}|f’(x)|dx)$

141$\leq$$M \epsilon|K|_{1}\sup_{1\leq|\tau|\leq N}(\sum_{m=0}^{|\tau|-1}\int_{-\infty}^{\infty}|f’(x)|dx)$$\leq$$M \epsilon|K|_{1}N\oint_{-\infty}^{\infty}|f’(x)|dx$$<$

2 $(H+C)MN\pi|K|_{1}$$3k_{0}$$<$$\frac{1}{4}$.

Thus $|K_{n}|_{1}\leq I_{1}+I_{2}<1/4+1/4=1/2$, as required.
Proposition 2. Let
and that $I-\hat{B}(z)$ is invertible for eachsuch that
Then there is an$\square$, and assume that $I-B(0)$ is invertible
.
where
with

$B=\{B(n)\}_{n=0}^{\infty}\in l^{1}(\mathbb{Z}^{+})$$z\in \mathbb{C}$$\hat{B}(z):=\Sigma_{n=0}^{\infty}B(n)z^{-n}$$|z|\geq 1_{f}$$R\in l^{1}(\mathbb{Z}^{+})$$\hat{B}(z)(I-\hat{B}(z))^{-1}=\hat{R}(z)$Proof. Consider the sequence$B’\in l^{1}(\mathbb{Z})$,$\forall|z|\geq 1$.

defined by $B’(n)=B(n)$ if$n\geq 0,\mathrm{a}\mathrm{n}\mathrm{d}B’(n)=0$

if $n<0$ . Then$I- \tilde{B}’(\rho)=I-\sum_{n=0}^{\infty}B’(n)e^{-i\rho n}=I-\hat{B}(e^{i\rho})$,$\forall\rho\in$R.

, and consequently there exists a
is invertible for eachHencein
, by Proposition 1. Define an element,
such thatis bounded and continuous. The function
by $Q_{+}(n)=Q(n)$ for any
on the domain $|z|\geq 1$ , and it is analytic on $|z|>1$ . Similarly, the function
is bounded and continuous on the domain $|z|\leq 1$ , and it is analytic on $|z|<1$ . Moreover,, thenif $|z|=1$ with
$I-\tilde{B}’(\rho)$$\tilde{B}’(\rho)(I-\tilde{B}’(\rho))^{-1}=\tilde{Q}(\rho)$$Q_{+}$$\forall\rho\in \mathbb{R}$$n\in \mathbb{Z}^{+}$$l^{1}(\mathbb{Z}^{+})$$Q\in l^{1}(\mathbb{Z})$$\rho\in \mathbb{R}$$\hat{Q}_{+}(z)$$\Sigma_{n=1}^{\infty}Q(-n)z^{n}$$z=e^{i\rho}$$\hat{B}(z)(I-\hat{B}(z))^{-1}-\hat{Q}_{+}(z)$$=$$\tilde{B}’(\rho)(I-\tilde{B}’(\rho))^{-1}-\sum_{n=0}^{\infty}Q(n)e^{-i\rho n}$$=$$\tilde{Q}(\rho)-\sum_{n=0}^{\infty}Q(n)e^{-i\rho n}$$=$$\sum_{n=-\infty}^{-1}Q(n)e^{-i\rho n}$$=$$\sum_{n=1}^{\infty}Q(-n)e^{i\rho n}$$=$$\sum_{n=1}^{\infty}Q(-n)z^{n}$

.

150

Therefore, the function $G(z)$ defined by$\hat{B}(z)(I-\hat{B}(z))^{-1}-\hat{Q}_{+}(z)$$(|z|\geq 1)$$\Sigma_{n=1}^{\infty}Q(-n)z^{n}$$(|z|<1)$$G(z)=\{$

is analytic on the entire domain by Morera’s theorem. Observe that $I-\hat{B}(z)arrow I-B(0)$as $|z|arrow\infty$ . Since $I-B(0)$ is invertible by the assumption, it follows thatin$\lim_{|z|arrow\infty}||(I-\hat{B}(z))^{-1}||=||(I-B(0))^{-1}||$ , and consequently $\sup_{|z|\geq 1}||(I-\hat{B}(z))^{-1}||<$. Therefore, $G(z)$ is bounded on the entire domain, and hence $G(z)$ is a constantfunction by Liouville’s theorem. Then $G(z)\equiv G(0)=0$ , and hence it follows that$\hat{B}(z)(I-\hat{B}(z))^{-1}=\hat{Q}_{+}(z)$ for any
with $|z|\geq 1$ . Thus we may set $Q_{+}=R$ to establish$\mathcal{L}(X)$$\infty$$z$

the proposition.

$\square$

We are now 1n a position to prove the theorem.
andholds true. Also, the implications
Clearly, the implication
have already been proved in [5, Theorem 2] and [4, Theorem 2]. In what.
andfollows, we will prove the implications$[(\mathrm{i}\mathrm{i}\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{i})]$$[(\mathrm{i}\mathrm{i})\Rightarrow(\mathrm{i})]$$[(\mathrm{i}\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{v})]$$[(\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{i})]$$[(\mathrm{i}\mathrm{v})\Rightarrow(\mathrm{i}\mathrm{i}\mathrm{i})]$defined by $D(n)=$$B(n-1)$ if $n\geq 1$ , and $D(n)=0$ if $n=0$ . Clearly, $I-D(0)$ is invertible. For any
with $|z|\geq 1$ , we get

Proof of$[(\mathrm{i}\mathrm{v})\Rightarrow(\mathrm{i}\mathrm{i}\mathrm{i})]$. Let us consider the sequence$D\in f^{1}(\mathbb{Z}^{+})$$z\in \mathbb{C}$,$\hat{D}(z)=\sum_{n=0}^{\infty}D(n)z^{-n}=z^{-1}\hat{B}(z)$and hence$I- \hat{D}(z)=\frac{1}{z}(zI-\hat{B}(z))$

Thus $I-\hat{D}(z)$ is invertible for each$z\in \mathbb{C}$with $|z|\geq 1$ , and it satisfies the relation$(I-\hat{D}(z))^{-1}=z(zI-\hat{B}(z))^{-1}$

By virtue of Proposition 2, there exists a, $|z|\geq 1$ , and hence we get.

,$Q\in l^{1}(\mathbb{Z}^{+})$$|z|\geq 1$such that $D\wedge(z)(I-\hat{D}(z))^{-1}=$$\hat{Q}(z)$$I+\hat{Q}(z)$$=$$I+\hat{D}(z)(I-\hat{D}(z))^{-1}$$=$$(I-\hat{D}(z))^{-1}$$=$$z(zI-\hat{B}(z))^{-1}$for all $|z|\geq 1$ . Consider the sequence $S=\{S(n)\}_{n=0}^{\infty}$ defined by$S(n)=\{$.

$I+Q(0)$$(n=0)$$Q(n)$$(n\geq 1)$

.

151

, and $\hat{S}(z)=I+\hat{Q}(z)=z(zI-\hat{B}(z))^{-1}$ for all $|z|\geq 1$ . Notice thatThen
the fundamental solution
is bounded exponentially, that is, $\sup_{n\geq 0}e^{-n\omega}||R(n)||<\infty$of $R$ converges for. Hence the $Z$-transform
for some constant$|z|>e^{\omega}$ .
Let us consider the -transform of both sides in the equation $R(n+1)=$$\Sigma_{k=0}^{\infty}B(n-k)R(k)$ with
(0) $=I$ to get the relation $z(\hat{R}(z)-I)=\hat{B}(z)\hat{R}(z)$ , or$(zI-\hat{B}(z))\hat{R}(z)=zI$ for. Thus it follows that $\hat{R}(z)=z(zI-\hat{B}(z))^{-1}=\hat{S}(z)$ for, which. By the uniqueness of the -transform, we get $R(n)\equiv S(n)$ ,allshows the summability of , as required.
$S\in l^{1}(\mathbb{Z}^{+})$$R$$\omega$$\geq 0$$\Sigma_{n=0}^{\infty}R(n)z^{-n}$

$Z$

$R$

$|z|>e^{\omega}$$Z$$|z|>e^{\omega}$$n\in \mathbb{Z}^{+}$$R$

Proof of$[(\mathrm{i})\supset(\mathrm{i}\mathrm{i})]$.

Let$\tau\in \mathbb{Z}$, and ,$||\phi||_{(-\infty,\tau]}<\delta(\epsilon/2)$$\phi$: $(-\infty, \tau]arrow X$ be given in such a way that$\psi$

and$|| \psi||(-\infty,\tau]<\min\{\delta(1/2), \mu\}$,andare those in Definition 1. Let us take a sequence
wherethat $n_{j}arrow\infty(jarrow\infty)$ . We may assume that $\tau+n_{j}>0$ for $j=1$ , 2,
1 : $[0, \tau+n_{j}]arrow X$ by$\delta(\cdot)$$\phi^{\gamma}(n):=\phi(n-n_{j})$and$x^{g}(n)$such. Define$\{n_{j}\}\subset \mathbb{Z}^{+}$

$\mu$

,

$n\in[0, \tau+n_{j}]$$\ldots$,by$x(n+n_{j;}\tau+n_{j}, \phi^{\mathrm{J}})$$(n\geq-n_{J})$$\phi(n)$$(n <-n_{j})$$x^{j}(n):=\{$. Since $x^{j}(n)=\psi(n+n_{j})=\phi(n)$ for $n\in[-n_{j}, \tau]$ , the uniform asymptotic
for $j=1,2$ ,yieldsstability of the zero solution of$\ldots$$(E_{0})$$|x^{j}(n)|< \frac{\epsilon}{2}$(3)

for $n\geq\tau$ .

contains a convergent
be given. We now assert that the sequence
Let anysubsequence. Indeed, in case of $n\leq\tau$ , we get $x^{j}(n)=\phi(n)$ , and hence the assertion clearlyholds. Let us consider the case $\tau<n$ . It follows that$\{x^{j}(n)\}_{j}$$n\in \mathbb{Z}$$x^{j}(n)$$=$$\sum_{k=0}^{n_{j}+n-1}B(n_{j}+n-1-k)x(k;\tau+nj, \phi^{j})$$=$$\sum_{s=-n_{j}}^{n-1}B(n-1-s)x^{\mathrm{i}}(s)$$=$$\sum_{s=-\infty}^{n-1}B(n-1-s)x^{j}(s)+\sum_{s=-\infty}^{-n_{j}-1}B(n -1-s)\phi(s)$.

By virtue of the summability of $B=\{B(n)\}_{n=0}^{\infty}$ , it is easy to certify that the term
tends to 0 as $jarrow\infty$ . Moreover, since the operator $B(n-1-s)$
is compact, we see that the sequence $\{\sum_{s=-\infty}^{n-1}B(n-1-s)x^{j}(s)\}_{j}$ contains a convergent$\Sigma_{s=-\infty}^{-n_{\mathrm{j}}-1}B(n-1-s)\phi(s)$152

subsequence. This observation leads to that the sequencesubsequence, which completes the proof of the assertion.

$\{x^{j}(n)\}_{j}$contains a convergent,, denoted by the same notation
Now one can select a subsequence ofon as $jarrow\infty$ . Obviously $\tilde{y}(n)=\phi(n)$ for $n\in(-\infty, \tau]$ .
which converges to some
Moreover, it follows that $\lim_{jarrow\infty}\Sigma_{s=-n_{j}}^{n}B(n-s)x^{j}(s)=\Sigma_{s=-\infty}^{n}B(n-s)\tilde{y}(s)$ . Thus we
obtain that$x^{j}(n)$$\{x^{j}(n)\}_{\mathrm{i}}$$\tilde{y}(n)$$\tilde{y}(n+1)$$\mathbb{Z}$$= \lim_{Jarrow\infty}x^{j}(n+1)$$= \lim_{jarrow\infty}x(n +1+n_{j}; \tau+n_{j}, \phi^{j})$$= \lim_{jarrow\infty}\sum_{r=0}^{n+n_{j}}B(n+n\mathrm{i}-r)x(r;\tau+n_{\mathit{1}}, \phi^{?})$$= \lim_{jarrow\infty}\sum_{s=-n_{j}}^{n}B(n-s)x^{j}(s)$$=$which implies that$\sum_{s=-\infty}^{n}B(n-s)\tilde{y}(s)$$\tilde{y}(n)=y(n\mathrm{j}\tau, \phi)$

on Z. Letting,$jarrow\infty$$|y(n; \tau, \phi)|\leq\frac{\epsilon}{2}<\epsilon$forin (3) we get$n\geq\tau$.

(4)

Furthermore, by the same argument we see that$|y(n; \tau, \psi)|<\frac{\epsilon}{2}$(5)

for $n\geq\tau+N(\epsilon/2)$ ,is the one in Definition 1. The inequality (4), together with (5), shows that
whereis uniformly asymptotically stable.
the zero solution of$N(\cdot)$$\square$$(E_{\infty})$

Remark 1. One can see from the proof that in Theorem 1, the implications, are compact. It is anhold true without the assumption that $B(n)$ ,) hold(orinteresting problem to ask whether or not the implication (11)good without the compactness condition on $B(n)$ . But the problem is still open for theauthors.

$(\mathrm{i}\mathrm{v})\Rightarrow(\mathrm{i}\mathrm{i}\mathrm{i})\Rightarrow$$n\in \mathbb{Z}^{+}$$(\mathrm{i}\mathrm{i})\Rightarrow(\mathrm{i})$$\Rightarrow(\mathrm{i}\mathrm{i}\mathrm{i})$$(\mathrm{i}\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{v})$Remark 2, We can apply Theorem 1 to establish the existence of bounded (resp. asymPwith a bounded (resp.
totically almost periodic) solutions for forced equations ofisasymptotically almost periodic) forcing term, provided that the zero solution of
uniformly asymptotically stable. Details will be discussed in a forth-coming paper [6].
$(E_{\infty})$$(E_{0})$REFERENCES
1. S. Elaydi, An Introduction to Difference Equations, Springer, New-York, 1996.
2. S. Elaydi and S. Murakami, Asymptotic stability versus exponential stability in linear Volterra difference equations of convolution type, J. Difference Equations and APPL, 2 (1996), 401-410

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3. S. Elaydi and S. Murakami, Uniform asymptotic stability in linear Volterra difference equations, J.
Difference Equations and APPL, 3 (1998), 203-218.
4. T. Furumochi, S. Murakami and Y. Nagabuchi, Volterra difference equations on a Banach space andabstract differential equations with piecewise continuous delays, Japanese. J. Math., Vo1.30, No. 2(2004), 387-412.
5. T. Furumochi, S. Murakami and Y. Nagabuchi, A generalization of Wiener’s lemma and its application to Volterra difference equations on a Banach space, J. Difference Equations and Appl., Vol. 10,No, 13-15 (2004), 1201-1214.
6. S. Murakami and Y. Nagabuchi, Stability properties and asymptotic almost periodicity for linear
Volterra difference equations in a Banach space, Japanese J. Math., Vol. 31-2 (in press)