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central limit theorem, linear systems,  
binary contact path process, diffusive behavior, delocalization.
---------------0805030243107
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\begin{document}
\bcenter

\large{\bf Central Limit Theorem for a Class of 
Linear Systems}\footnote{\today}

\vvs \normalsize

\noindent 
Yukio NAGAHATA and 
Nobuo YOSHIDA\footnote{
Supported in part by JSPS Grant-in-Aid for Scientific
Research, Kiban (C) 17540112}\\

\ecenter

\begin{abstract}
We consider a class of interacting particle systems with values 
in $[0,\8)^{\zd}$, of which the binary contact path process 
is an example. 
For $d \ge 3$ and under a certain square integrability condition 
on the total number of the particles, we prove a central limit theorem 
for the density of the particles, together 
with upper bounds for the density of the most populated site and 
the replica overlap. 
\end{abstract}

\small
\noindent Abbreviated Title:  CLT for Linear Systems.\\
 AMS 2000 subject classification :
Primary 60K35; secondary 60F05, 60J25. \\
Key words and phrases: central limit theorem, linear systems, 
binary contact path process, diffusive behavior, delocalization.
%%%%%%%%%%%
\tableofcontents
%%%%%

\normalsize
%%%%%%
\SSC{Introduction}
%%%%%%
We write $\N=\{0,1,2,...\}$, 
$\N^*=\{1,2,...\}$ and 
$\Z=\{ \pm x \; ; \; x \in \N \}$. For 
$x=(x_1,..,x_d) \in \rd$, $|x|$ stands for the $\ell^1$-norm: 
$|x|=\sum_{i=1}^d|x_i|$. For $\h=(\h_x)_{x \in \zd} \in \R^{\zd}$, 
$|\h |=\sum_{x \in \zd}|\h_x|$. 
Let 
$(\W, \cF, P)$ be 
a probability space.
We write $P[X]=\int X \; dP$ and 
$P[X:A]=\int_A X \; dP$ for a r.v.(random variable) $X$ and an 
event $A$. 
%%%%%%%%%
\subsection{The binary contact path process (BCPP)}
%%%%%%%%%%%%%%
We start with a motivating simple example. 
Let $\h_t =(\h_{t,x})_{x \in \zd} \in \N^{\zd}$, $t \ge 0$ be 
binary contact path process (BCPP for short) with parameter $\lm>0$.
Roughly speaking, the BCPP is an extended version of the basic 
contact process, in which not only the presence/absence of 
the particles at each site, but also their number is considered. 
The BCPP was originally introduced by D. Griffeath 
\cite{Gri83}. Here, we explain the process following 
the formulation in the book of T. Liggett \cite[Chapter IX]{Lig85}.
Let $N^z=(N^z_t)_{t \ge 0}$, ($z \in \zd$) be independent rate one Poisson 
processes. We suppose that the process $(\h_t)$ 
starts from a deterministic configuration 
$\h_0=(\h_{0,x})_{x \in \zd}\in \N^{\zd}$  
with $|\h_0|<\8$. 
At the $i$-th jump time 
$t=T^{z,i}$ of $N^z$, $\h_{t-}$ is replaced by $\h_t$ randomly 
as follows: for each $e \in \zd$ with $|e|=1$, 
$$
\h_{t,x}=\lef\{ \barray{ll}
\h_{t-,x}+\h_{t-,z} & \mbox{if $x = z+e$,}\\
\h_{t-,x} & \mbox{if otherwise} \\
\earray \ri.\; \; 
\mbox{with probability ${\lm \over 2d\lm +1}$,} 
$$
(all the particles at site $z$ are 
duplicated and added to those on the site $z=x+e$), and 
$$
\h_{t,x}=\lef\{ \barray{ll}
0 & \mbox{if $x=z$}, \\
\h_{t-,x} & 
\mbox{if $x \neq z$}
\earray \ri.\; \; 
\mbox{with probability ${1 \over 2d\lm +1}$}
$$
(all the particles at site $z$ disappear).
The replacement occurs independently for different $(z,i)$ and independently 
from the Poisson processes. A motivation to study the BCPP comes from the 
fact that the projected process
$$
\lef( \h_{t,x} \wedge 1\ri)_{x \in \zd},\; \; \; t \ge 0
$$
is the basic contact process \cite{Gri83}.

Let 
$$
\kp_p={2d\lm -1 \over 2d\lm +1}\; \; \; \mbox{and}\; \; \;  
\ov{\h}_t=(\exp(-\kp_1t)\h_{t,x})_{x \in \zd}.
$$
Then,  $(|\ov{\h}_t|)_{t \ge 0}$
is a martingale and therefore, the 
following limit exists almost surely:
$$
|\ov{\h}_\8|\st{\rm def}{=}
\lim_t|\ov{\h}_t|.
$$
Moreover, $P (|\ov{\h}_\8| >0)>0$ if 
\bdnl{|h_8|>0}
\mbox{$d \ge 3$ and $\lm > {1 \over 2d (1-2\pi_d)}$,} 
\edn
where $\pi_d$ is the return probability for the simple random 
walk on $\zd$ \cite{Gri83}. 
It is known that $\pi_d \le \pi_3=0.3405...$ for $d \ge 3$ 
\cite[page 103]{Spi76}.


%%%%%%%%%%%%%%%%%%%%%%%
%We now look at the density:
%$$
%\rh_{t,x}={\h_{t,x} \over |\h_t|}.
%$$
%%%%%%
We denote the
density of the particles by: 
\bdnl{rh}
\rh_{t,x}=\frac{\h_{t,x}}{|\h_t|} =\frac{\ov{\h}_{t,x}}{|\ov{\h}_t|}, \;
\; t >0, x \in \zd. 
\edn 
Interesting objects related to the
density would be 
\bdnl{rh^*} \rh^*_t=\max_{x \in \zd}\rh_{t,x}, \;
\; \mbox{and}\; \; \cR_t=\sum_{x \in \zd}\rh_{t,x}^{2}. 
\edn
$\rh^*_t$ is the density at the most populated site, while $\cR_t$
is the probability that a given pair of particles at time $t$ are
at the same site. We call $\cR_t$ the {\it replica overlap}, in
analogy with the spin glass theory. Clearly, $(\rh^*_t)^{2} \le \cR_t
\le \rh^*_t$. These quantities convey information on
localization/delocalization of the particles. Roughly speaking,
large values of $\rh^*_t$ or $\cR_t$ indicates that the most of
the particles are concentrated on small numbers of ``favorite
sites" ({\it localization}), whereas small values of them implies
that the particles are spread out over large number of sites ({\it
delocalization}).

As  a special case of \Cor{CLT} below, we have 
the following result, which shows the diffusive behavior 
and the delocalization of the BCPP under the 
condition (\ref{|h_8|>0}):
%%%%%%%%%%
\Theorem{BCPP}
%%%%%%%
Suppose (\ref{|h_8|>0}). Then, for any $f \in C_{\rm b}(\rd)$, 
$$
\lim_{t \ra \8} \sum_{x \in \zd}
f\lef(x/\sqrt{t}\ri)\rh_{t,x}
=\int_{\rd}fd\n\; \; \; 
\mbox{in $P(\; \cdot \; | |\ov{\h}_\8|>0)$-probability,}
$$
where 
$C_{\rm b} (\rd)$ 
stands for the set of bounded continuous functions on $\rd$, and 
$\n$ is the Gaussian measure with
$$
\int_{\rd}x_id\n (x)=0, 
\; \; \; \int_{\rd}x_ix_jd\n (x)={\lm \over 2d\lm +1}\del_{ij},\; \; \; 
i,j=1,..,d.
$$
Furthermore, 
$$
\cR_t=\cO (t^{-d/2})\; \; \; 
\mbox{as $t \nearrow \8$ in $P(\; \cdot \; | |\ov{\h}_\8|>0)$-probability}.
$$
%%%%%%%
\end{theorem}
%%%%%
\subsection{The results}
%%%%%%
We generalize \Thm{BCPP} to a certain class of linear interacting 
particle systems with values in $[0,\8)^{\zd}$\cite[Chapter IX]{Lig85}.
Recall that the particles in BCPP either die, or make binary branching. 
To describe more general ``branching mechanism", 
we introduce a random vector 
$K=(K_x)_{x \in \zd }$ which is bounded and of 
finite range in the sense that
\bdnl{K_x}
K_x \lef\{\barray{l}
\in [0,b_K] \; \; \mbox{a.s. for all $x \in \zd$,}\\
=0, \; \; \mbox{a.s. if $|x| >r_K$}
\earray \rig.\; \; 
\mbox{for some non-random $b_K,r_K \in [0,\8)$.}
\edn 
Let $N^z=(N^z_t)_{t \ge 0}$, ($z \in \zd$) be independent rate one Poisson 
processes, and let $K^{z,i}=(K_x^{z,i})_{x \in \zd }$ 
($z \in \zd$, $i \in \N^*$) be i.i.d. 
random vectors with the same distributions as $K$, 
independent of $\{N^z\}_{z \in \zd}$. 
We suppose that the process $(\h_t)_{t \ge 0}$ 
starts from a deterministic configuration 
$\h_0=(\h_{0,x})_{x \in \zd} \in [0,\8)^{\zd}$ 
with $|\h_0|<\8$.
At the $i$-th jump time 
$t=T^{z,i}$ of $N^z$, $\h_{t-}$ is replaced by $\h_t$, where 
\bdnl{h_(x,t)}
\h_{t,x}=\lef\{ \barray{ll}
K^{z,i}_0\h_{t-,z} & \mbox{if $x=z$}, \\
\h_{t-,x}+K^{z,i}_{x-z}\h_{t-,z} & 
\mbox{if $x \neq z$}.
\earray \ri.
\edn
The BCPP is a 
special case of this set-up, in which 
\bdnl{binK} 
K= \lef\{ \barray{ll}
0 & \mbox{with probability ${1 \over 2d\lm +1}$} \\
\lef(\del_{x,0}+\del_{x,e} \rig)_{x \in \zd} & 
\mbox{with probability ${\lm \over 2d\lm +1}$, for each $2d$ neighbour 
$e$ of 0.} \earray  \rig. \edn
%%%%%%%%
%As compared with BCPP where only binary branching occurs, 
%we allow more general, possibly random ``branching".
%%%%%%%
It is convenient to introduce 
the following matrix 
representation of the 
linear transformation $\h_{t-} \mapsto \h_t$:
$$
\h_{t,x}=\sum_{y \in \zd}A^{z,i}_{x,y}\h_{t-,y},
$$
where 
\bdnl{Miz}
A^{z,i}_{x,y}=\lef\{ \barray{ll}
1+\del_{x,z}(K^{z,i}_0-1) & \mbox{if $x=y$}, \\
\del_{y,z}K^{z,i}_{x-y} & 
\mbox{if $x \neq y$}.
\earray \ri.
\edn 
%%%%%%%%%%
%Letting 
%$$
%\tl{A}^{z,s}_{x,y}=\lef\{ \barray{ll}
%A^{z,i}_{x,y} & \mbox{if $s=T^{z,i}$}, \\
%\del_{x,y} & 
%\mbox{if $s \neq T^{z,i}$},
%\earray \ri.
%\; \; \; s \ge 0,
%$$
%%%%%%%
The precise definition of the process $(\h_t)_{t \ge 0}$ is then  given by 
the following stochastic differential equation:
\bdnl{sde}
\h_{t,x}=\h_{0,x}
+\sum_{z \in \zd}\int^t_0
\lef( \sum_{y \in \zd}\tl{A}^{z,s}_{x,y}\h_{s-,y}-\h_{s-,x}\ri)d N^z_s,
\edn
where $\tl{A}^{z,s}_{x,y}=A^{z,N^z_s}_{x,y}$. 
By (\ref{K_x}), it is standard to see that (\ref{sde}) defines  
a unique process $\h_t=(\h_{t,x})$, ($t \ge 0$) 
and that $(\h_t)$ is Markovian. 

We set
\bdmn
\kp_1&=&\sum_{x \in \zd}P[(K_x-\del_{x,0})^p] \; \; \; 
p=1,2,\label{kp12} \\
\ov{\h}_t &=&(\exp (-\kp_1t )\h_{t,x})_{x \in \zd}. \label{ovh_t}
\edmn
We will see that $(|\ov{\h}_t|)_{t \ge 0}$ 
is a martingale (\Lem{FK1} below) and therefore, the 
following limit exists almost surely:
\bdnl{ovh_8}
|\ov{\h}_\8|\st{\rm def}{=}
\lim_t|\ov{\h}_t|.
\edn
To state \Thm{CLT}, 
We define
\bdnl{G(x)}
G(x)=\int^\8_0P_R^0(R_t=x)dt,
\edn
where $((R_t)_{t \ge 0}, P_R^x)$ is the continuous-time random walk on 
$\zd$ starting from $x \in \zd$, with the generator 
\bdnl{L_R}
L_Rf (x)=\half \sum_{y \in \zd}\lef( P[K_{x-y}]+P[K_{y-x}] \ri)
\lef( f(y)-f(x)\rig).
\edn
As before, 
$C_{\rm b} (\rd)$ stands for the set of bounded continuous functions on $\rd$.
%%%%%%%%%%%%
\Theorem{CLT}
%%%%%%
Suppose (\ref{K_x}) and that
\bdmn
& & \mbox{the set $\{x \in \zd\; ;\; P[K_x]\neq 0\}$ contains a 
linear basis of $\rd$,} \label{K1}\\
%%%%%%%%%%%%%%%%%%%%%
%& & P[K_0K_x]=P[K_x] \; \; \mbox{for $x \in \zd \bsh \{ 0\}$,}\label{K3}\\
%%%%%%%%%%%%%%%%%%%%%%
& & P[(K_x-\del_{x,0})(K_y-\del_{y,0})]=0
\; \; \mbox{for $x,y \in \zd$ with $x \neq y$.}
\label{K4}
\edmn
Then, referring to (\ref{kp12})--(\ref{L_R}), 
the following are equivalent:
\bds
\item[(a)] ${\kp_2 \over 2}G(0)<1$,
\item[(b)] 
${\dps \sup_{t \ge 0} P[|\ov{\h}_t|^2]<\8}$,
\item[(c)]
${\dps  \lim_{t \ra \8} \sum_{x \in \zd}
f\lef((x -mt)/\sqrt{t}\ri)\ov{\h}_{t,x}=|\ov{\h}_\8|\int_{\rd}fd\n}$  
in $\bL^2 (P)$ for all $f \in C_{\rm b} (\rd)$, 
\eds
where $ m=\sum_{x \in \zd}xP[K_x] \in \rd $ 
and $\n$ is the Gaussian measure with
\bdnl{nu}
\int_{\rd}x_id\n (x)=0, 
\; \; \; \int_{\rd}x_ix_jd\n (x)
=\sum_{x \in \zd}(x_i-m_i)(x_j-m_j)P[K_x],\; \; \; 
i,j=1,..,d.
\edn
Moreover, if ${\kp_2 \over 2}G(0)<1$, then, 
there exists $C \in (0,\8)$ such that
\bdnl{RepDec}
\sum_{x, \tl{x}\in \zd}f(x-\tl{x})P[\ov{\h}_{x,t}\ov{\h}_{\tl{x},t}]  
\le Ct^{-d/2}|\h_0|^2\sum_{x \in \zd}f(x)
\edn
for all $t>0$ and $f :\zd \ra [0,\8)$ with $\sum_{x \in \zd}f(x)<\8$. 
%%%%%%%%
\end{theorem}
%%%%%%%
The proof of \Thm{CLT} will be presented in section \ref{pCLT}. 

\vvs
\noindent {\bf Remarks:} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf 1)} 
The condition (\ref{K1}) guarantees a reasonable non-degeneracy
for the transition mechanism (\ref{h_(x,t)}). 
On the other hand, (\ref{K4}) amounts to saying that 
the transition mechanism (\ref{h_(x,t)})
updates the configuration 
by ``at most one coordinate at a time". 
Here is an explanation for the relevance of the condition (\ref{K4}).
To prove \Thm{CLT}, we use a certain Markov chain 
on $\zd \times \zd$, which is introduced in \Lem{FK2} below. 
Thanks to (\ref{K4}), the Markov chain is stationary with respect 
to the counting measure on $\zd \times \zd$. 
The stationarity plays an important role in the proof of \Thm{CLT}--
see \Lem{FK3} below. 
%%%%%%%%%%%%%%%%%%%%%%

\vs
\noindent{\bf 2)}
%%%%%%%%%%%%%%%%%%%5
Because of (\ref{K1}), the random walk 
$(R_t)$ is recurrent for 
$d=1,2$ and transient for $d \ge 3$. Therefore, 
${\kp_2 \over 2}G(0)<1$ is possible only if $d \ge 3$.
As will be explained in the proof, 
${\kp_2 \over 2}G(0)<1$ is equivalent to 
$$
P_R^0\lef[\exp \lef( {\kp_2 \over 2}\int^\8_0 \del_0 (R_t)dt\ri) \ri]<\8.
$$
%%%%%%%%%%%%%%%%%%%%%%5
\noindent{\bf 3)}
If, in particular, 
\bdnl{SRW}
P[K_x] = \lef\{ \barray{ll} 
c >0 & \mbox{for $|x|=1$,} \\  0 & \mbox{for $|x| \ge 2$,}
\earray \rig.
\edn
then, $(R_t)_{t \ge 0} \st{\mbox{\scriptsize law}}{=}
(\widehat{R}_{2dct})_{t \ge 0}$, 
where $(\widehat{R}_\cdot)$ is the simple 
random walk. Therefore, the condition (a) becomes
\bdnl{Khas2}
{\kp_2 \over 4dc(1-\pi_d)}<1.
\edn
By (\ref{binK}), the BCPP satisfies 
(\ref{K1})--(\ref{K4}). Furthermore, $\kp_2=1$ and 
we have (\ref{SRW}) with $c={\lm \over 2d\lm +1}$.  
Therefore, (\ref{Khas2}) is equivalent to (\ref{|h_8|>0}).

\vs 
%%%%%%%%%%%%%%%%%%%
\noindent{\bf 4)}
%%%%%%%%%%%%%%%%%%
The proof of \Thm{CLT} is roughly divided into two steps; \\
\bds
\item[(i)] to represent the two-point function 
$P[\h_{t,x}\h_{t,\tl{x}}]$ in terms of a continuous-time 
Markov chain on $\zd \times \zd$ via the Feynman-Kac formula 
(\Lem{FK2} and \Lem{FK3} below), 
\item[(ii)] to show the central limit theorem for the ``weighted" 
Markov chain, where the weight comes from the additive functional 
due to the Feynman-Kac formula (\Lem{perturb2} below).
\eds
The above strategy was adopted earlier by one of the authors for 
branching random walk in random 
environment \cite{Yos08}. There, the Markov chain alluded to above 
is simply the product of simple random walks on $\zd$, so that the central 
limit theorem with the Feynman-Kac weight is relatively easy.
Since the Markov chain in the present paper is no longer a random walk, 
it requires more work.
However, the good news here is that the Markov chain we have to work on is 
``close" to a random walk. In fact, we get the central limit 
theorem by perturbation from that for a random walk case. 

\vs
%%%%%%%%%%%%%%%%%%%
\noindent{\bf 5)}
%%%%%%%%%%%%%%%%%%
In connection to 4) above, D. Griffeath discussed 
a Feynman-Kac formula for 
$$
\sum_{y \in \zd}P[\h_{t,x}\h_{t,\tl{x}+y}]
$$
 in the case of BCPP 
(cf. proof of Theorem 1 in \cite{Gri83}). In the present paper, we have the 
formula for each summand of the above summation, 
so that finer information is available. 

\vs
%%%%%%%%%%%%%%%%%%%
\noindent{\bf 6)}
%%%%%%%%%%%%%%%%%%
The dual process of $(\h_t)$ above (in the sense of \cite[page 432]{Lig85})
is given by replacing (\ref{h_(x,t)}) by:
\bdnl{h_(x,t)*}
\h_{t,x}=\lef\{ \barray{ll}
\sum_{y \in \zd}K^{z,i}_{y-x}\h_{t-,y} & \mbox{if $x=z$}, \\
\h_{t-,x} & \mbox{if $x \neq z$}.
\earray \ri.
\edn
This amounts to replacing the matrix $(A^{z,i}_{x,y})_{x,y}$ by its 
transpose. As can be seen from the proofs, all the results in 
this paper remain true for the dual process. 
%%%%%%%%%%%%%%%%%%%%%

\vvs

We define the density and the replica overlap in the same way 
as (\ref{rh})--(\ref{rh^*}). Then, as 
an immediate consequence of \Thm{CLT}, we have the following 
%%%%%%%%%%
\Corollary{CLT}
%%%%%%%
Suppose (\ref{K_x}), (\ref{K1})--(\ref{K4}) and that 
${\kp_2 \over 2}G(0)<1$. 
Then, $P (|\ov{\h}_\8|>0)>0$ and for all $f \in C_{\rm b} (\rd)$, 
$$
\lim_{t \ra \8} \sum_{x \in \zd}
f\lef((x -mt)/ \sqrt{t}\ri)\rh_{t,x}
=\int_{\rd}fd\n\; \; \; 
\mbox{in $P(\; \cdot \; | |\ov{\h}_\8|>0)$-probability,}
$$
where $ m=\sum_{x \in \zd}xP[K_x] \in \rd $ and 
$\n$ is the same Gaussian measure defined by (\ref{nu}).
Furthermore, 
$$
\cR_t=\cO (t^{-d/2})\; \; \; 
\mbox{as $t \nearrow \8$ in $P(\; \cdot \; | |\ov{\h}_\8|>0)$-probability}.
$$
%%%%%%%
\end{corollary}
%%%%
Proof: The first statement is immediate from \Thm{CLT}(c). Taking 
$f(x)=\del_{x,0}$ in (\ref{RepDec}), we see that
$$
P[\sum_{x \in \zd}\ov{\h}_{x,t}^2 ]\le Ct^{-d/2}|\h_0|^2\; \; \; 
\mbox{for $t >0$}.
$$
This implies the second statement. 
\hfill $\Box$
%%%%%%%%%%%%%%%%

\vvs
For $a \in \zd$, let $\h_t^a$ be the process starting from  
$\h_0 =(\del_{a,x})_{x \in \zd}$. 
As a by-product of \Thm{CLT}, we have the following formula 
for the covariance of $(|\ov{\h}_\8^a|)_{a \in \zd}$.
For BCPP, this formula was obtained by D. Griffeath 
as the main result of \cite{Gri83} (cf. Theorem 1 in that paper). 
%%%%%%%%%%%%
%We define 
%$\ov{\h}_t^a$ and $|\ov{\h}_\8^a|$ exactly in the same way as 
%(\ref{ovh_t}) and (\ref{ovh_8}) ($|\ov{\h}_\8^a|$ exists, 
%since $(|\ov{\h}_t^a|)_{t \ge 0}$ is a martingale). 
%%%%%%%%%%%%%
\Theorem{cov}
%%%%%%%%%
Suppose (\ref{K_x}), (\ref{K1})--(\ref{K4}) and that 
${\kp_2 \over 2}G(0)<1$. Then,
$$
P[|\ov{\h}_\8^a||\ov{\h}_\8^b|]=1+{\kp_2G(a-b) \over 2-\kp_2G(0)}, 
\; \; \; a,b \in \zd.
$$
%%%%%%%%%
\end{theorem}
%%%%%%%%%%%
The proof of \Thm{cov} will be presented in section \ref{pcov}. 
%%%%%%%%%%%%%%%%%
%Proof: The first statement is immediate from \Thm{CLT}(c). Taking 
%$f(x)=\del_{x,0}$ in (\ref{RepDec}), we see that
%$$
%P[\sum_{x \in \zd}\ov{\h}_{x,t}^2 ]\le Ct^{-d/2}|\h_0|^2\; \; \; 
%\mbox{for $t >0$}.
%$$
%This implies the second statement. 
%\hfill $\Box$
%%%%%%%%%%%%%%%%
%%%%%%%%%
\SSC{Lemmas}
%%%%%%%%%
\subsection{Markov chain representations for the point functions}
%%%%%%%%%%%%%
We assume (\ref{K_x}) throughout, but not  (\ref{K1})--(\ref{K4}) 
for the moment.
We start with the Feynman-Kac formula for one-point function: 
%%%%%%%%%%%%%
\Lemma{FK1}
%%%%%%%%%
\bdnl{FK1}
P[\h_{t,x}]=e^{\kp_1t}P^x[\h_{0,S_t}], \; \; (t,x)\in [0,\8) \times \zd,
\edn
where $\kp_1$ is defined by (\ref{kp12}) and 
$((S_t)_{t \ge 0}, P^x)$ is the continuous-time random walk on 
$\zd$ with the generator
$$
L_S f (x)= \sum_{y \in \zd}P[K_{x-y}]
\lef( f(y)-f(x)\rig).
$$ 
As a consequence, $(|\ov{\h}_t|)_{t \ge 0}$ (cf. (\ref{ovh_t})) 
is a martingale.
%%%%%%%%%%
\end{lemma}
%%%%%%%
Proof: It is easy to see that, 
\bds
\item[(1)] \hspace{1cm}
${\dps \gm_{x,y}  \st{\rm def}{=} 
\sum_{z \in \zd}\lef( P[A^{z,i}_{x,y}]-\del_{x,y}\ri)=
P[K_{x-y}]-\del_{x,y},}$ for $x,y\in \zd$, 
\item[(2)] \hspace{1cm}
${\dps \kp_1=\sum_{x \in \zd}P[K_x]-1=\sum_{y \in \zd}\gm_{x,y}}$, 
(cf. (\ref{kp12})).
\eds
We show that $u(t,x)\st{\rm def}{=}P[\h_{t,x}]$ solves 
the integral equation 
\bds
\item[(3)] \hspace{1cm}
${\dps u(t,x)-u(0,x)=\int^t_0( L_S +\kp_1)u(s,x)ds,}$
\eds
and that
\bds
\item[(4)] \hspace{1cm}
${\dps \sup_{t \in [0,T]}\sup_{x \in \zd}|u(t,x)|<\8}$
 for any $T \in (0,\8)$.
\eds
To show (3), we average (\ref{sde}):
\bdnn
u(t,x)-u(0,x)
&=&\sum_{z \in \zd}\int^t_0
\lef( \sum_{y \in \zd}P[A^{z,1}_{x,y}]u(s,y)-u(s,x)\ri)ds \\
&=& \int^t_0
 \sum_{y \in \zd}\sum_{z \in \zd}
\lef(P[A^{z,1}_{x,y}]-\del_{x,y}\ri)u(s,y)ds \\
&\st{\mbox{\scriptsize (1)}}{=}& \int^t_0
 \sum_{y \in \zd}\gm_{xy}u(s,y)ds \\
&\st{\mbox{\scriptsize (2)}}{=}&
\lef( \sum_{y \in \zd}\gm_{xy}(u(s,y)-u(s,x))+\kp_1 u(s,x) \ri)ds 
=\mbox{RHS of (3)}.
\ednn 
On the other hand, we have by (\ref{K_x}) and (\ref{sde}) 
that, for any $p \in \N^*$, 
there exists $C_1 \in (0,\8)$ such that
$$
P[\h_{t,x}^p] \le C_1\sum_{y:|x-y| \le r_K}\int^t_0 P[\h_{s,y}^p]ds,
 \; \; \; t \ge 0.
$$
By iteration, we see that 
there exists $C_2 \in (0,\8)$ such that
\bdnl{sde2}
P[\h_{t,x}^p] \le e^{C_2t}\sum_{y \in \zd}e^{-|x-y|}(1+\h_{0,y}^p),
 \; \; \; t \ge 0,
\edn
which implies (4). \\
The solution to (3) subject to (4) is unique, for each given $\h_0$. 
This can be seen by using Gronwall's inequality with respect to the 
norm $\| u\|=\sum_{x \in \zd}e^{-|x|}|u(x)|$. Moreover, the RHS 
of (\ref{FK1}) is a solution to (3) subject to the bound (4). 
This can be seen by adapting 
the argument in \cite[page 5,Theorem 1.1]{Szn98}. Therefore, we 
get (\ref{FK1}). Now, 
$$
P[|\ov{\h}_t|]\st{\mbox{\scriptsize (\ref{FK1})}}{=}
\sum_{x \in \zd}P^x[\h_{0,S_t}]
=\sum_{x,y \in \zd}\h_{0,y}P^x[S_t=y]
=\sum_{y \in \zd}\h_{0,y}=|\h_0|.
$$
Then, from this and the Markov 
property of $(\h_t)$, we see 
that $|\ov{\h}_t|$ is a martingale.
\hfill $\Box$
%%%%%%%%%%%

\vvs
To prove the Feynman-Kac formula for two-point function, we prepare:
%%%%%%%%%%%%%
\Lemma{Gm}
%%%%%%%%%
For $x,y,\tl{x},\tl{y} \in \zd$, 
\bdmn
\Gm_{x,\tl{x},y,\tl{y}} 
& \st{\rm def}{=}& 
\sum_{z \in \zd}\lef( P[A^{z,i}_{x,y}A^{z,i}_{\tl{x},\tl{y}}]
-\del_{x,y}\del_{\tl{x},\tl{y}}\ri) \nn \\
&=&
\lef\{ \barray{ll}
P[K_{x-y}K_{\tl{x}-\tl{y}}]\del_{y,\tl{y}} 
& \mbox{if $x \neq y$ and $\tl{x} \neq \tl{y}$,}\\
P[K_{\tl{x}-\tl{y}}]+ P[(K_0-1)K_{\tl{x}-\tl{y}}]\del_{y,\tl{y}} 
& \mbox{if $x=y$ and $\tl{x} \neq \tl{y}$,}\\
P[K_{x-y}]+ P[(K_0-1)K_{x-y}]\del_{y,\tl{y}} 
& \mbox{if $x \neq y$ and $\tl{x} =\tl{y}$,}\\
2P[K_0-1]+\del_{x,\tl{x}}P[(K_0-1)^2]
& \mbox{if $x=y$ and $\tl{x}=\tl{y}$.} \earray \rig.
\label{Gm2} \\
V(x,\tl{x}) & \st{\rm def}{=}& 
\sum_{y,\tl{y} \in \zd}\Gm_{x,\tl{x},y,\tl{y}} =2\kp_1 +
\sum_{y \in \zd }P[(K_{x-y}-\del_{x,y})(K_{\tl{x}-y}-\del_{\tl{x},y})],
\label{C_Gm2}
\edmn
%%%%%%%%
\end{lemma}
%%%%%%%%%
Proof: Direct computations.
\hfill $\Box$
%%%%%%%%%%%

\vvs
We have the Feynman-Kac formula for two-point function:
%%%%%%%%%%%%%
\Lemma{FK2}
%%%%%%%%%
Let $(X,\tl{X})=((X_t, \tl{X}_t)_{t \ge 0}, P_{X,\tl{X}}^{x,\tl{x}})$ be 
the continuous-time Markov chain on 
$\zd \times \zd$ starting from $(x,\tl{x})$, with the generator 
$$
L_{X,\tl{X}}f (x,\tl{x})=\sum_{y,\tl{y} \in \zd}
\Gm_{x,\tl{x},y,\tl{y}} \lef( f(y,\tl{y})-f(x,\tl{x})\rig),
$$
where $\Gm_{x,\tl{x},y,\tl{y}}$ is defined by (\ref{Gm2}). 
Then, for $(t,x,\tl{x})\in [0,\8) \times \zd \times \zd$, 
%%%%
%\bdnl{FK2}
%%%%
$$
P[\h_{t,x}\h_{t,\tl{x}}]=P^{x,\tl{x}}_{X,\tl{X}}
\lef[ \exp \lef( \int^t_0 V(X_s,\tl{X}_s)ds\ri)
\h_{0,X_t}\h_{0,\tl{X}_t} \ri],
$$
where $V$ is defined by (\ref{C_Gm2}).
%%%%%%%%%%
\end{lemma}
%%%%%%%
Proof: 
We show that $u(t,x,\tl{x})\st{\rm def}{=}P[\h_{t,x}\h_{t,\tl{x}}]$ 
solves the integral equation 
\bds
\item[(1)] \hspace{1cm}
${\dps u(t,x,\tl{x})-u(0,x,\tl{x})=
\int^t_0( L_{X,\tl{X}}+V(x,\tl{x}))u(s,x,\tl{x})ds.}$
\eds
Then, the lemma can be obtained in the same way as in \Lem{FK1}.
By (\ref{sde}), we have 
$$
\h_{t,x}\h_{t,\tl{x}}-\h_{0,x}\h_{0,\tl{x}}
=\sum_{z \in \zd}\int^t_0
\lef( \sum_{y, \tl{y}\in \zd}
\tl{A}^{z,s}_{x,y}\tl{A}^{z,s}_{\tl{x},\tl{y}}\h_{s-,y}\h_{s-,\tl{y}}
-\h_{s-,x}\h_{s-,\tl{x}}\ri)d N^z_s.
$$
Therefore, 
\bdnn
u(t,x,\tl{x})-u(0,x,\tl{x})
&=&
\sum_{z \in \zd}\int^t_0
\lef( \sum_{y, \tl{y}\in \zd}
P[A^{z,1}_{x,y}A^{z,1}_{\tl{x},\tl{y}}]u(s,y,\tl{y})
-u(s,x,\tl{x})\ri)ds \\
&=&
\int^t_0 \sum_{y, \tl{y}\in \zd}
\sum_{z \in \zd}\lef( P[A^{z,1}_{x,y}A^{z,1}_{\tl{x},\tl{y}}]
-\del_{x,y}\del_{\tl{x},\tl{y}}\ri) u(s,y,\tl{y})ds \\
&=&
\int^t_0 \sum_{y, \tl{y}\in \zd}\Gm_{x,\tl{x},y,\tl{y}} u(s,y,\tl{y})ds\\
&=&
\int^t_0 \lef(\sum_{y, \tl{y}\in \zd}\Gm_{x,\tl{x},y,\tl{y}} 
( u(s,y,\tl{y})-u(s,x,\tl{x}) )+V(x,\tl{x})u(s,x,\tl{x})\ri) ds \\
&=& \int^t_0( L_{X,\tl{X}}+V(x,\tl{x}))u(s,x,\tl{x})ds.
\ednn
\hfill $\Box$
%%%%%%%%%%%%%%%%%%

\vvs 
We assume (\ref{K1})--(\ref{K4}) from here on. Then, 
(\ref{Gm2})--(\ref{C_Gm2}) are reduced to:
\bdmn
\Gm_{x,\tl{x},y,\tl{y}} 
&=&
\lef\{ \barray{ll}
P[K_{x-y}^2]\del_{x,\tl{x}} \del_{y,\tl{y}} 
& \mbox{if $x \neq y$ and $\tl{x} \neq \tl{y}$,}\\
P[K_{\tl{x}-\tl{y}}]
& \mbox{if $x=y$ and $\tl{x} \neq \tl{y}$,}\\
P[K_{x-y}] 
& \mbox{if $x \neq y$ and $\tl{x} =\tl{y}$,}\\
2P[K_0-1]+\del_{x,\tl{x}}P[(K_0-1)^2]
& \mbox{if $x=y$ and $\tl{x}=\tl{y}$.} \earray \rig.
\label{Gm} \\
V (x,\tl{x})
&=&  2\kp_1 +\kp_2 \del_{x,\tl{x}}, 
\; \; \; \mbox{(cf.(\ref{kp12})).}
\label{C_Gm}
\edmn
It is easy to see from (\ref{Gm})--(\ref{C_Gm}) that
\bdnl{stat}
\sum_{y,\tl{y} \in \zd}\Gm_{x,\tl{x},y,\tl{y}}
=\sum_{y,\tl{y} \in \zd}\Gm_{y,\tl{y},x,\tl{x}}=2\kp_1 +\kp_2 \del_{x,\tl{x}},
\edn
which implies that $(X,\tl{X})$ is stationary with respect 
to the counting measure on $\zd \times \zd$. 
We denote the dual process of $(X,\tl{X})$ 
by $(Y,\tl{Y})=((Y_t,\tl{Y}_t)_{t \ge 0}, P_{Y,\tl{Y}}^{x,\tl{x}})$, 
that is, the continuous time Markov chain on 
$\zd \times \zd$ with the generator 
\bdnl{L_Y*}
L_{Y,\tl{Y}}f (x,\tl{x})=\sum_{y,\tl{y} \in \zd}
\Gm_{y,\tl{y},x,\tl{x}} \lef( f(y,\tl{y})-f(x,\tl{x})\rig).
\edn
Thanks to (\ref{stat}), $L_{X,\tl{X}}$ and 
$L_{Y,\tl{Y}}$ are dual operators on $\ell^2 (\zd \times \zd)$. 
%%%%%%%%
%where $\Gm_{x,\tl{x},y,\tl{y}}$ is defined by (\ref{Gm2}). 
%\sum_{z \in \zd}\Gm_{x,0,y+z,z}&=&2P[K_{x-y}],\; \; \;
%\mbox{if $x \neq y$.}
%\label{Gm^(-)}
%%%%%%%%%%%%
%In particular, $\Gm_{x,\tl{x},y,\tl{y}}=\Gm_{y,\tl{y},x,\tl{x}}$ 
%and hence that the Markov chain $((X_t, \tl{X}_t)_{t \ge 0}, P^{x,\tl{x}})$ 
%is symmetric in this case.
%%%%%%%%%%%

\vvs
\noindent {\bf Remark:} If we additionally 
suppose that $P[K_x^p]=P[K_{-x}^p]$ for $p=1,2$ and $x \in \zd$, 
then, $\Gm_{x,\tl{x},y,\tl{y}}=\Gm_{y,\tl{y},x,\tl{x}}$ for 
all $x,\tl{x},y,\tl{y} \in \zd$. Thus, $(X,\tl{X})$ and 
$(Y,\tl{Y})$ are the same in this case. 


\vvs
The relative motion $Y_t-\tl{Y}_t$ of the components of 
$(Y,\tl{Y})$ is nicely identified by: 
%%%%%%%%%%%%%
\Lemma{Y-Y}
%%%%%%%%%
$((Y_t-\tl{Y}_t)_{t \ge 0}, P_{Y,\tl{Y}}^{x,\tl{x}})$ and 
$((R_{2t})_{t \ge 0}, P_R^{x-\tl{x}})$ (cf. (\ref{L_R})) have the same law.
%%%%%%%%%%
\end{lemma}
%%%%%%%
Proof: 
Since $(Y,\tl{Y})$ is shift invariant, in the sense that 
$\Gm_{x+v,\tl{x}+v,y+v,\tl{y}+v} =\Gm_{x,\tl{x},y,\tl{y}} $ 
for all $v \in \zd$, $((Y_t-\tl{Y}_t)_{t \ge 0}, P_{Y,\tl{Y}}^{x,\tl{x}})$ 
is a Markov chain. Moreover, its 
jump rate is computed as follows.
For $x \neq y$, 
\bdnn
\sum_{z \in \zd}\Gm_{y+z,z,x,0}
& = & P[K_{x-y}]+P[K_{y-x}] +\del_{x,0}\sum_{z \in \zd}
P[(K_{y+z}-\del_{y,z})(K_{z}-\del_{0,z})] \\
& \st{\mbox{\scriptsize (\ref{K4})}}{=} & P[K_{x-y}]+P[K_{y-x}].
\ednn
\hfill $\Box$
%%%%%%%%%%

\vvs
To prove \Thm{CLT}, the use of \Lem{FK2} is made not in itself, but 
via the following lemma. It is the proof of this lemma, where the 
duality of $(X,\tl{X})$ and $(Y,\tl{Y})$ plays its role.
%%%%%%%%%%%%%
\Lemma{FK3}
%%%%%%%%%
For $g:\zd \times \zd \ra [0,\8)$, 
\bdmn
\lefteqn{\sum_{x,\tl{x} \in \zd}
P[\ov{\h}_{t,x}\ov{\h}_{t,\tl{x}}]g(x,\tl{x})} \nn \\
&=&\sum_{x,\tl{x} \in \zd}\h_{0,x}\h_{0,\tl{x}}
P_{Y,\tl{Y}}^{x,\tl{x}}
\lef[ \exp \lef( \kp_2\int^t_0 \del_0(Y_s-\tl{Y}_s)ds\ri)
g(Y_t,\tl{Y}_t) \ri].\label{ell_2}
\edmn
In particular, for $f:\zd  \ra [0,\8)$, 
\bdnl{FK3}
\sum_{x,\tl{x} \in \zd}P[\ov{\h}_{t,x}\ov{\h}_{t,\tl{x}}]f(x-\tl{x})
=\sum_{x,\tl{x} \in \zd}\h_{0,x}\h_{0,\tl{x}}
P_R^{x-\tl{x}}\lef[ \exp \lef( {\kp_2 \over 2}\int^{2t}_0 \del_0(R_u)du\ri) 
f(R_{2t})\ri].
\edn
%%%%%%%%%%
\end{lemma}
%%%%%%%
Proof: 
It follows from \Lem{FK2} and (\ref{C_Gm}) that
\bds
\item[(1)]
\hspace{1cm} ${\dps 
\mbox{LHS of (\ref{ell_2})}=\sum_{x,\tl{x} \in \zd}
P_{X,\tl{X}}^{x,\tl{x}}
\lef[ \exp \lef( \kp_2\int^t_0 \del_0(X_s-\tl{X}_s)ds\ri)
\h_{0,X_t}\h_{0,\tl{X}_t} \ri]g(x,\tl{x}).}$
\eds
We now observe that the operators 
\bdnn
f (x,\tl{x}) & \mapsto &
P_{X,\tl{X}}^{x,\tl{x}}
\lef[ \exp \lef( \kp_2\int^t_0 \del_0(X_s-\tl{X}_s)ds\ri)
f(X_t, \tl{X}_t) \ri], \\
f (x,\tl{x}) & \mapsto &
P_{Y,\tl{Y}}^{x,\tl{x}}
\lef[ \exp \lef( \kp_2\int^t_0 \del_0(Y_s-\tl{Y}_s)ds\ri)
f(Y_t, \tl{Y}_t) \ri]
\ednn
are dual to each other with respect to the counting measure on
 $\zd \times \zd $. Therefore, 
$$
\mbox{RHS of (1)}=\mbox{RHS of (\ref{ell_2})}.
$$
Taking $g (x,\tl{x})=f(x-\tl{x})$ in particular, we have by (\ref{ell_2}) 
and \Lem{Y-Y} that
\bdnn
\mbox{LHS of (\ref{FK3})}
&=&\sum_{x,\tl{x} \in \zd}\h_{0,x}\h_{0,\tl{x}}P^{x,\tl{x}}_{Y,\tl{Y}}
\lef[ \exp \lef( \kp_2\int^t_0 \del_0(Y_s-\tl{Y}_s)ds\ri)
f(Y_t-\tl{Y}_t)\ri] \\
&=& \sum_{x,\tl{x} \in \zd}\h_{0,x}\h_{0,\tl{x}}
P_R^{x-\tl{x}}\lef[ \exp \lef( \kp_2\int^t_0 \del_0(R_{2u})du\ri)
f(R_{2t}) \ri]
=\mbox{RHS of (\ref{FK3})}.
\ednn
\hfill $\Box$
%%%%%%%%%%%%%%%%
\subsection{Central limit theorems for Markov chains}
%%%%%%%%%%%%%%%
We prepare central limit theorems for Markov chains, which is obtained 
by perturbation of random walks. 
%%%%%%%%
\Lemma{perturb1}
%%%%%%%
Let $((Z_t)_{t \ge 0}, P^x)$ be a continuous-time random walk 
on $\zd$
with the generator 
$$
L_Z f(x)=\sum_{y \in \zd}a_{y-x}(f(y)-f(x)),
$$ 
where we assume that
$$
\sum_{x \in \zd}|x|^2a_x<\8.
$$
Then, for any $B \in \s [Z_u\; ; \; u \in [0,\8)]$, 
$x \in \zd$, and $f \in C_{\rm b}(\rd)$,
$$
\lim_{t \ra \8}P^x [f((Z_t -mt)/\sqrt{t}):B]
=P^x (B)\int_{\rd}fd\n,
$$
where $m=\sum_{x \in \zd}xa_x$ and $\n$ is the Gaussian measure with
\bdnl{nua}
\int_{\rd}x_id\n (x)=0, 
\; \; \; \int_{\rd}x_ix_jd\n (x)
=\sum_{x \in \zd}(x_i-m_i)(x_j-m_j)a_x,\; \; \; 
i,j=1,..,d.
\edn
%%%%%%
\end{lemma}
%%%%%%%
Proof: 
By subtracting a constant, we may assume that $\int_{\rd}fd\n =0$. 
We first consider the case that 
$B \in \cF_s \st{\rm def}{=}\s [Z_u\; ; \; u \in [0,s]]$ 
for some $s \in (0,\8)$. It is easy to see 
from the central limit theorem for $(Z_t)$ that for any  $x \in \zd$,
$$
\lim_{t \ra \8}P^x [f((Z_{t-s} -mt)/\sqrt{t})]=0.
$$
With this and the bounded convergence theorem, we have 
$$
P^x[f((Z_t -mt)/\sqrt{t}):B]
=P^x[P^{Z_s}[f((Z_{t-s} -mt)/\sqrt{t})]:B]
\lra 0\; \; \mbox{as $t \nearrow \8$}.
$$
Next, we take $B \in \s [Z_u\; ; \; u \in [0,\8)]$. 
For any $\e>0$, there exist $s \in (0,\8)$ and $\tl{B} \in \cF_s$  
such that $P^x[|{\bf 1}_B-{\bf 1}_{\tl{B}}|] <\e$. 
Then, by what we already have seen, 
$$
\suplim_{t \ra \8}
P^x[f((Z_t -mt)/\sqrt{t}):B] 
 \le \suplim_{t \ra \8}P^x[f((Z_t -mt)/\sqrt{t}):\tl{B}] 
+\| f \|\e  = \| f \|\e,
$$
%%%%%%
%\le (P^x (B)+\e)\int_{\rd}fd\n +\| f \|\e,
%%%%%%
where $\| f\|$ is the sup norm of $f$. Similarly, 
$$
\inflim_{t \ra \8}
P^x[f((Z_t -mt)/\sqrt{t}):B] \ge   -\| f \|\e.
$$
Since $\e>0$ is arbitrary, we are done.
\hfill $\Box$
%%%%%%%%
\Lemma{perturb2}
%%%%%%%
Let $Z=((Z_t)_{t \ge 0}, P^x)$ be as in \Lem{perturb1} and 
and $D \sub \zd$ be transient for $Z$.
On the other hand, let $\tl{Z}=((\tl{Z}_t)_{t \ge 0}, \tl{P}^x)$ 
be the continuous-time Markov chain on $\zd$
with the generator 
$$
L_{\tl{Z}} f(x)=\sum_{y \in \zd}\tl{a}_{x,y}(f(y)-f(x)),
$$ 
where we assume that $\tl{a}_{x,y}=a_{y-x}$ if $x \not\in D$ and that 
$D$ is also transient for $\tl{Z}$. 
Furthermore, we assume that 
a function $v:\zd \ra \R$ satisfies
\bdnn
& & \mbox{$v \equiv 0$ outside $D$,}\\
& & \tl{P}^z\lef[\exp \lef( \int_0^\8 |v(\tl{Z}_t)|dt\ri)\ri]<\8\; \;
\mbox{for some $z \in \zd$}.
\ednn
Then, for $f \in C_{\rm b}(\rd)$, 
$$
\lim_{t \ra \8}\tl{P}^z
\lef[\exp \lef( \int_0^t v(\tl{Z}_u)du\ri)f((\tl{Z}_t-mt)/\sqrt{t})\ri]
=\tl{P}^z\lef[\exp \lef( \int_0^\8 v(\tl{Z}_t)dt\ri)\ri]\int_{\rd}fd\n,
$$
where $\n$ is the Gaussian measure such that (\ref{nua}) holds.
%%%%%%
\end{lemma}
%%%%%%%
Proof:
Define  
\bdnn
H_D(\tl{Z})&=&\inf \{ t \ge 0\; ; \; \tl{Z}_t \in D\}, \; \; 
T_D(\tl{Z})=\sup \{ t \ge 0\; ; \; \tl{Z}_t \in D\},\\
e_t&=&\exp \lef( \int_0^t v(\tl{Z}_s)ds \ri).
\ednn
Then,
\bdmn
\lefteqn{\tl{P}^z\lef[e_tf((\tl{Z}_t-mt)/\sqrt{t})\ri]} \nn \\
&=&\tl{P}^z\lef[e_tf((\tl{Z}_t-mt)/\sqrt{t}):T_D(\tl{Z})<s\ri]+\e_{s,t}\nn\\
&=&\tl{P}^z\lef[e_sf((\tl{Z}_t-mt)/\sqrt{t}):T_D(\tl{Z})<s\ri] +\e_{s,t} \nn\\
&=& \tl{P}^z\lef[e_s{\bf 1}_{\tl{Z}_s \not\in D}
\tl{P}^{\tl{Z}_s}\lef[f((\tl{Z}_{t-s}-mt)/\sqrt{t}):H_D(\tl{Z})=\8 \ri]\ri]
+\e_{s,t},\label{est}
\edmn
where
\bdnn
|\e_{s,t}| 
& = & 
\lef| \tl{P}^z\lef[e_tf((\tl{Z}_t-mt)/\sqrt{t}):T_D(\tl{Z})\ge s\ri]\rig| \\
& \le &\| f \|\tl{P}^z\lef[
\exp \lef( \int_0^\8 |v(\tl{Z}_t)|dt\ri):T_D(\tl{Z})\ge s\ri] \ra 0\; \; 
\mbox{as $s \ra \8$.}
\ednn
We now observe that
$$
\tl{P}^x( \; \cdot \; | H_D(\tl{Z})=\8)=P^x( \; \cdot \; | H_D(Z)=\8)\; \; \; 
\mbox{for $x \not\in D$,}
$$
where $H_D(Z)$ is defined similarly as $H_D(\tl{Z})$.
Hence, for $x \not\in D$ and fixed $s>0$, we have by \Lem{perturb1} that
$$
\lim_{t \ra \8}\tl{P}^x\lef[f((\tl{Z}_{t-s}-mt)/\sqrt{t}):H_D(\tl{Z})=\8 \ri]
= \tl{P}^x[H_D(\tl{Z})=\8] \int_{\rd}fd\n.
$$
Therefore, 
\bdnn
\lefteqn{\lim_{t \ra \8}
\tl{P}^z\lef[e_s{\bf 1}_{\tl{Z}_s \not\in D}
\tl{P}^{\tl{Z}_s}\lef[f((\tl{Z}_{t-s}-mt)/\sqrt{t}):H_D(\tl{Z})=\8 \ri]\ri]}\\
&=&\tl{P}^z\lef[e_s{\bf 1}_{\tl{Z}_s \not\in D}
\tl{P}^{\tl{Z}_s}[H_D(\tl{Z})=\8] \rig]\int_{\rd}fd\n \\
&=& \tl{P}^z\lef[e_s :T_D (\tl{Z})<s \rig]\int_{\rd}fd\n. \\
\ednn
Thus, letting $t \ra \8$ first, and then $s \ra \8$, in (\ref{est}), 
we get the lemma.
\hfill $\Box$
%%%%%%%%%%%%%%%%%
\subsection{A Nash type upper bound for the Schr\"odinger semi-group}
%%%%%%%%%%%%%%%%%%%
We will use the following lemma to prove (\ref{RepDec}). The lemma 
can be generalized to symmetric Markov chains on more general graphs.
However, we restrict ourselves to random walks on $\zd$, since 
it is enough for our purpose.
%%%%%%%%%%%%%
\Lemma{Nash}
%%%%%%%%%%%%
Let $((Z_t)_{t \ge 0}, P^x)$ be continuous-time random walk 
on $\zd$
with the generator:
$$
L_Z f(x)=\sum_{y \in \zd}a_{y-x}(f(y)-f(x)),
$$ 
where we assume that
\bdnn
& &
\mbox{the set $\{x \in \zd\; ;\; a_x\neq 0\}$ is bounded and contains a 
linear basis of $\rd$,}\\
& & a_x=a_{-x}\; \; \mbox{for all $x \in \zd$},
\ednn
Let $v:\zd \ra \R$ be a function such that
$$
C_v \st{\rm def}{=}
\sup_{x \in \zd}P^x\lef[\exp \lef( \int_0^\8 |v(Z_t)|dt\ri)\ri]<\8.
$$
Then, there exists $C \in (0,\8)$ such that
\bdnl{RepDec2}
\sup_{x \in \zd} P^x\lef[\exp \lef( \int_0^t v(Z_u)du\ri)f(Z_t)\ri]
\le Ct^{-d/2}\sum_{x \in \zd}f(x)
\edn
for all $t>0$ and $f :\zd \ra [0,\8)$ with $\sum_{x \in \zd}f(x)<\8$. 
%%%%%%%%%%%%
\end{lemma}
%%%%%%%%%
Proof: We adapt the argument in \cite[Lemma 3.1.3]{CY04}.
For a bounded function $f:\zd \ra \R$, we introduce
\bdnn
(T_tf)(x)&=&P^x \lef[ 
\exp \lef( \int^t_0v(Z_u)du\rig)f(Z_t)\rig],
\; \; \; x \in \zd,\\
T_t^hf&=&\frac{1}{h}T_t [fh], \; \; 
\mbox{where ${\dps h(x)=P^x\lef[\exp \lef( \int_0^\8 v(Z_t)dt\ri)\ri]}$.}
\ednn
Then,
$(T_t)_{t \ge 0}$ extends to a symmetric,
strongly continuous semi-group on $\ell^2(\zd)$.
We now consider the measure $\sum_{x \in \zd}h(x)^2\del_x$ on $\zd$, 
and denote by $(\ell^{p,h}(\zd), \| \; \cdot\;  \|_{p,h})$ the associated 
$\bL^p$-space. Then, it is standard to see that $(T_t^h)_{t \ge 0}$
defines a symmetric strongly continuous semi-group on $\ell^{2,h}(\zd)$ 
and that for $f \in \ell^{2,h}(\zd)$,
\bdnn
\cE^h (f,f)  & \st{\rm def.}{=} & \lim_{t \searrow 0}{1 \over t}
\sum_{x\in \zd} f (x)(f-T_t^hf)(x)h(x)^2  \\
&=& \half \sum_{x,y \in \zd}a_{y-x}| f(y)-f(x)|^2h(x)h(y).
\ednn
%%%%%%%
%where the bracket stands for the inner product of $\ell^{2,h}(\zd)$.
%%%%%%%%
By the assumptions on $(a_x)$, we have the Sobolev inequality:
\bds
\item[(1)] \hspace{1cm}
${\dps \sum_{x \in \zd}| f (x)|^{\frac{2d}{d-2}} \le c_1
\lef( \half\sum_{x,y \in \zd}a_{y-x}| f(y)-f(x)|^2 \rig)^{\frac{d}{d-2}}
}$ for all $f \in \ell^2 (\zd)$, 
\eds
where $c_1\in (0,\8)$ is independent of $f$. This can be seen via 
isoperimetric inequality \cite[page 40, (4.3)]{Woe00}. 
We have on the other hand that
\bds
\item[(2)] \hspace{1cm}
$ 1/C_v\le h(x) \le C_v.$
\eds  
We see from (1) and (2) that 
$$
\sum_{x \in \zd}| f (x)|^{\frac{2d}{d-2}}h(x)^2\le c_2
\cE^h (f,f) ^{\frac{d}{d-2}}
\; \; \; \mbox{for all $f \in \ell^{2,h} (\zd)$,}
$$
where $c_2\in (0,\8)$ is independent of $f$.
This implies that
there is a constant $C$ such that
$$
\| T^h_t \|_{2 \ra \8,h} \le Ct^{-d/4}
\; \; \; \mbox{for all $t >0$},
$$
e.g.,\cite[page 75, Theorem 2.4.2]{Dav89},
where $\| \cdot \|_{p \ra q,h}$ denotes
the operator norm from $\ell^{p,h}(\zd)$ to $\ell^{q,h}(\zd)$.
Note that $\| T^h_t   \|_{1 \ra 2,h}
=\| T^h_t   \|_{2 \ra \8,h}$
by duality. We therefore have via semi-group property that
\bds
\item[(3)] \hspace{1cm}
$\| T^h_t \|_{1 \ra \8,h} \le
\| T^h_{t/2} \|_{2 \ra \8,h}^2 \le C^2t^{-d/2}$ for all $t >0$.
\eds
Since $T_tf=h T^h_t[f/h]$,
the desired bound (\ref{RepDec2})
follows from (2) and (3).
\hfill $\Box$
%%%%%%%%%
\SSC{Proof of \Thm{CLT} and \Thm{cov}}
%%%%%%%%%%%%%%%%%%
\subsection{Proof of \Thm{CLT}} \label{pCLT}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%
(a) $\LRa$ (b): 
%%%%%%%%%%%%%
Define 
$$
h (x)=P_R^x\lef[\exp \lef( {\kp_2 \over 2}\int^\8_0 \del_0 (R_t)dt\ri)\ri].
$$
Then, $\sup_{t \ge 0} P[|\ov{\h}_t|^2]=h(0)$ by (\ref{FK3}).
Therefore, it is enough to show that (a) is equivalent to $h(0)<\8$.
In fact, (a) implies that $h(0)<\8$,
 by Khas'minskii's lemma \cite[page 7]{Szn98}.
Note that $\sup_{x \in \zd}G(x)=G(0)$ to use Khas'minskii's lemma.
On the other hand, we have that
%%%%%%
%by differentiating 
%$\exp \lef( {\kp_2 \over 2}\int^t_0 \del_0 (R_s)ds\rig)$ 
%and then integrating, 
%%%%%%%%%%%%
$$
\exp \lef( {\kp_2 \over 2}\int^t_0 \del_0 (R_s)ds\rig)
=1+{\kp_2 \over 2}\int^t_0\del_{0,R_s}
\exp \lef( {\kp_2 \over 2}\int^t_s \del_0 (R_u)du\rig)ds, 
$$
and hence that $h (x)=1+{\kp_2 \over 2} h(0)G(x)$. Thus, $h(0)<\8$ 
implies (a) and that 
\bdnl{h(x)}
h(x)=1+{\kp_2G(x) \over 2-\kp_2G(0)}.
\edn
%%%%%%%%
% the condition (a) is equivalent to 
%$\sup_{t \ge 0}P[|\ov{\h}_t|^2]<\8$, which in turn 
%is equivalent to the $\bL^2 (P)$-convergence of the martingale 
%$|\ov{\h}_t|$.
%%%%%%%%%%%%%%%
(a),(b) $\Ra$ (c): 
%%%%%%%%%%%%
%By a standard approximation arguments, we may assume that 
%$f \in C_{b,u}(\rd)$, where $C_{b,u}(\rd)$ 
%denotes the set of bounded, uniformly continuous functions.
%%%%%%%%%%%%%%%%%
Since (b) implies that 
$\lim_{t \ra \8}|\ov{\h}_t|=|\ov{\h}_\8|$ in $\bL^2 (P)$,
 it is enough to prove that, 
$$
U_t\st{\rm def.}{=}\sum_{x \in \zd}\ov{\h}_{t,x}f \lef( (x-mt)/\sqrt{t}\rig)
\lra 0\; \; \; \mbox{in $\bL^2 (P)$ as $t \nearrow \8$}
$$
for $f \in C_{\rm b} (\rd)$ such that  $\int_{\rd}fd\n =0$.

We set $f_t(x, \tl{x}) =f ((x-m)/\sqrt{t})f ((\tl{x}-m)/\sqrt{t})$. 
Then, by \Lem{FK3},
$$
P[U_t^2]
 =  \sum_{x,\tl{x} \in \zd}
P[\ov{\h}_{t,x}\ov{\h}_{t,\tl{x}}]f_t(x, \tl{x})
= \sum_{x,\tl{x} \in \zd}\h_{0,x}\h_{0,\tl{x}}
P_{Y,\tl{Y}}^{x,\tl{x}}\lef[ e_tf_t(Y_t, \tl{Y}_t) \ri],
$$
where $e_t=\exp \lef( \kp_2\int^t_0 \del_0(Y_s-\tl{Y}_s)ds\ri)$.
Note that by \Lem{Y-Y} and (a), 
\bds
\item[(1)]\hspace{1cm}
${\dps P_{Y,\tl{Y}}^{x,\tl{x}}\lef[ e_\8\ri] =h(x-\tl{x}) \le h(0) <\8}$.
\eds
Since $|\h_0|<\8$, it is enough to prove that for each $x,\tl{x} \in \zd$
\bds
\item[(2)]\hspace{1cm}
${\dps
\lim_{t \ra \8}
P_{Y,\tl{Y}}^{x,\tl{x}}\lef[ e_tf_t(Y_t, \tl{Y}_t) \ri]=0}$.
\eds
We now apply \Lem{perturb2} to the Markov chain 
$\tl{Z}_t\st{\rm def.}{=}(Y_t,\tl{Y}_t)$ and the random walk 
$(Z_t)$ on $\zd \times \zd$ with the genaretor 
$$
L_Zf (x,\tl{x})
=\sum_{y,\tl{y} \in \zd}a_{x,\tl{x},y,\tl{y}} 
\lef( f(y,\tl{y})-f(x,\tl{x})\rig)\; \; \; 
\mbox{with}\; \; 
a_{x,\tl{x},y,\tl{y}}=\lef\{\barray{ll}
\Gm_{y,\tl{y},x,\tl{x}} & \mbox{if $x=\tl{x}$ or $y=\tl{y}$,} \\
0 & \mbox{if otherwise}.\earray \rig.
$$
Let $D=\{ (x,\tl{x}) \in \zd \times \zd\; ; \; x=\tl{x}\}$. 
Then, $a_{x,\tl{x},y,\tl{y}}=\Gm_{x,\tl{x},y,\tl{y}}$ if 
$(x,\tl{x}) \not\in D$. 
Moreover, $D$ is transient both for $(Z_t)$ and for $(\tl{Z}_t)$.
Finally, the Gaussian measure $\n \otimes \n $ 
is the limit law in the central limit theorem 
for the random walk $(Z_t)$.
Therefore, by (1),(2) and \Lem{perturb2},
$$
\lim_{t \ra \8}
P_{Y,\tl{Y}}^{x,\tl{x}}\lef[ e_tf_t(Y_t, \tl{Y}_t)  \ri]
=P_{Y,\tl{Y}}^{x,\tl{x}}\lef[ e_\8\ri]
\lef( \int_{\rd}fd\n \rig)^2=0.
$$
(c) $\Ra$ (b): This can be seen by taking $f \equiv 1$.\\
(\ref{RepDec}):By (\ref{FK3}),
$$
\sum_{x,\tl{x} \in \zd}P[\ov{\h}_{t,x}\ov{\h}_{t,\tl{x}}]f(x-\tl{x})
=\sum_{x,\tl{x} \in \zd}\h_{0,x}\h_{0,\tl{x}}P^{x-\tl{x}}_R\lef[ 
\exp \lef( {\kp_2 \over 2}\int^{2t}_0 \del_0(R_u)du\ri) f(R_{2t})\ri].
$$
We apply \Lem{Nash} to the right-hand-side to get  (\ref{RepDec}).
\hfill $\Box$
%%%%%%%%%%%%%%%%%%
\subsection{Proof of \Thm{cov}} \label{pcov}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
By the shift-invariance, we may assume that $b=0$. We have by \Lem{FK3} 
that
$$
P[\ov{\h}^a_{t,x}\ov{\h}^0_{t,\tl{x}}]
=P_{Y,\tl{Y}}^{a,0}
\lef[ \exp \lef( \kp_2 \int^{t}_0 \del_0(Y_u-\tl{Y}_u)du\ri) 
:(Y_t,\tl{Y}_t)=(x,\tl{x}) \ri],
$$
and hence by \Lem{Y-Y} that
$$
P[|\ov{\h}_t^a||\ov{\h}_t^0|]
=P^{a,0}_{Y,\tl{Y}}
\lef[ \exp \lef( \kp_2 \int^{2t}_0 \del_0(Y_u-\tl{Y}_u)du\ri)\ri]
=P^{a}_R\lef[ \exp \lef( {\kp_2 \over 2}\int^{2t}_0 \del_0(R_u)du\ri)\ri].
$$
By \Thm{CLT}, both $|\ov{\h}_t^a|$ and 
$|\ov{\h}_t^0|$ are convergent in $\bL^2 (P)$ if $ {\kp_2 \over 2}G(0)<1$.
Therefore, letting $t \nearrow \8$, we conclude that
$$
P[|\ov{\h}_\8^a||\ov{\h}_\8^0|]=
P^{a}_R\lef[ \exp \lef( {\kp_2 \over 2}\int^{\8}_0 \del_0(R_u)du\ri)\ri]
\st{\mbox{\scriptsize (\ref{h(x)}) }}{=}1+{\kp_2G(a) \over 2-\kp_2G(0)}.
$$
\hfill $\Box$
%%%%%%%%%%%%%%
\small

\vvs
\noindent{\bf Acknowledgements:}
The authors thank Shinzo Watanabe for the improvement of \Lem{perturb1}.
%%%%%%%%
%He also thanks Yukio Nagahata for discussions. 
%%%%%%%%%%%
\begin{thebibliography}{99}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\bibitem{CY04}
Comets, F., Yoshida, N.:
Some New Results on Brownian Directed Polymers in Random Environment, 
RIMS Kokyuroku {\bf 1386}, 50--66, avilable at authors' web pages. 
(2004).  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\bibitem{Dav89}
Davies, E. B.:
``Heat kernels and spectral theory", 
Cambridge University  Press (1989).                                         %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\bibitem{Gri83}
Griffeath, D.:
The Binary Contact Path Process, 
Ann. Probab. Volume 11, Number 3 (1983), 692-705. 
%%%%%%%%%%%%%%%%%%%%%%%%%%
\bibitem{Lig85}
Liggett, T. M. :
``Interacting Particle Systems",  
Springer Verlag, Berlin-Heidelberg-Tokyo
(1985).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
\bibitem{Spi76}
Spitzer, F.: ``Principles of Random Walks", Springer Verlag, 
New York, Heiderberg, Berlin (1976).
%%%%%%%%%%%%%%%%%%%
\bibitem{Szn98} 
Sznitman, A.-S., : 
Brownian Motion, Obstacles and Random Media, 
Springer monographs in mathematics, Springer (1998).
%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%
\bibitem{Woe00}
Woess, W.: 
``Random Walks on Infinite Graphs and Groups",
Cambridge University Press. (2000).
%%%%%%%%%%%%%%%%%%%%%%
\bibitem{Yos08} 
Yoshida, N.: Central Limit Theorem for 
Branching Random Walk in Random Environment, to appear in Ann. 
Appl. Proba., arXiv:0712.0649 (2008).
%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%
\end{thebibliography}
%%%%%


\vvs
\noindent   
Yukio Nagahata \\
Department of Mathematics, \\
Graduate School of Engineering Science \\
Osaka University,\\
Toyonaka 560-8531, Japan.\\
email: {\tt nagahata@sigmath.es.osaka-u.ac.jp}\\
URL: {\tt http://www.sigmath.osaka-u.ac.jp/}$\widetilde{}$ {\tt nagahata/}


\vvs
\noindent Nobuo Yoshida \\
Division of Mathematics \\
Graduate School of Science \\
Kyoto University,\\
Kyoto 606-8502, Japan.\\
email: {\tt nobuo@math.kyoto-u.ac.jp}\\
URL: {\tt http://www.math.kyoto-u.ac.jp/}$\widetilde{}$ {\tt nobuo/}

%%%%%%%%%%
\end{document}
%%%%%%%%
\Lemma{perturb1}
%%%%%%%
Let $((Y_t)_{t \ge 0}, P^x)$ be a continuous-time, mean-zero random walk 
on $\zd$
with the generator 
$$
L_Y f(x)=\sum_{y \in \zd}a_{y-x}(f(y)-f(x)),
$$ 
where we assume
$$
\sum_{x \in \zd}|x|^2a_x<\8.
$$
Let $H_D=\inf \{t \ge 0\; ; \; Y_t \in D\}$ for a 
transient set $D \sub \zd$. Then, $x \not\in \zd$,
$$
\lim_{t \ra \8}P^x (Y_t /\sqrt{t} \in \cdot |H_D=\8)=\n,\; \; \; \mbox{weakly,}
$$
where $\n$ is the Gaussian measure with
\bdnl{nua}
\int_{\rd}x_id\n (x)=0, 
\; \; \; \int_{\rd}x_ix_jd\n (x)=\sum_{x \in \zd}x_ix_ja_x,\; \; \; 
i,j=1,..,d.
\edn
%%%%%%
\end{lemma}
%%%%%%%
Proof: We have to show that for $f \in C_{\rm b}(\rd)$,
\bds
\item[(1)]
${\dps \lim_{t \ra \8}P^x [f(Y_t /\sqrt{t}):H_D=\8]
=P^x [H_D=\8]\int_{\rd}fd\n}$.
\eds
We may assume that $f$ is Lipschitz continuous. 
Define  
$$
T_D =\sup \{ t \ge 0\; ; \; Y_t \in D\},\; \; 
e_{t,\a}=\exp \lef( -\a\int^t_0{\bf 1}_D(Y_s)ds \ri).
$$
Since ${\dps \lim_{\a \ra \8}e_{\8,\a}={\bf 1}_{\{H_D=\8\}}}$, 
it is enough to prove that
\bds
\item[(2)]
${\dps \lim_{t \ra \8}P^x [e_{\8,\a}f(Y_t /\sqrt{t})]
=P^x[e_{\8,\a}]\int_{\rd}fd\n}$, uniformly in $\a >0$.
\eds
We write $s=t^{1/3}$ and approximate the LHS of (2) by the RHS:
\bdnn
P^x[e_{\8,\a}f(Y_t /\sqrt{t})]
&=&P^x[e_{\8,\a}f(Y_t /\sqrt{t}):T_D<s]+\e^{(1)}_{t} \\
&=&P^x[e_{s,\a}f(Y_t /\sqrt{t}):T_D<s]+\e^{(1)}_{t} \\
&=&P^x[e_{s,\a}f(Y_t /\sqrt{t})]+\e^{(1)}_{t}+\e^{(2)}_{t} \\
&=&P^x[e_{s,\a}f((Y_t-Y_s) /\sqrt{t})]+\e^{(1)}_{t}+...+\e^{(3)}_{t}\\
&=&P^x[e_{s,\a}]P^x[f((Y_t-Y_s) /\sqrt{t})]+\e^{(1)}_{t}+...+\e^{(3)}_{t} \\
&=&P^x[e_{s,\a}]P^x[f(Y_t /\sqrt{t})]+\e^{(1)}_{t}+...+\e^{(4)}_{t}\\
&=&P^x[e_{\8,\a}]P^x[f(Y_t /\sqrt{t})]+\e^{(1)}_{t}+...+\e^{(5)}_{t} \\
&=&P^x[e_{\8,\a}]\int_{\rd}fd\n+\e^{(1)}_{t}+...+\e^{(6)}_{t}.
\ednn
The error terms $\e^{(j)}_{t}$, $j=1,..,6$ are bounded as follows:
\bdnn
|\e^{(1)}_{t}|
&=&\lef| P^x[e_{\8,\a}f(Y_t /\sqrt{t}):T_D \ge s]\ri|
\le \| f \| P^x[T_D \ge s],\\
|\e^{(2)}_{t}|
&=&\lef| P^x[e_{s,\a}f(Y_t /\sqrt{t}):T_D \ge s]\ri|
\le \| f \| P^x[T_D \ge s],\\
|\e^{(3)}_{t}|
&=&\lef| P^x\lef[ 
e_{s,\a}\lef(f((Y_t-Y_s) /\sqrt{t})-f(Y_t /\sqrt{t})\ri) \ri]\ri|
\le  {\rm Lip}_f P[|Y_s|]/\sqrt{t},\\
|\e^{(4)}_{t}|
&=&P^x[e_{s,\a}]\lef| P^x\lef[ 
f((Y_t-Y_s) /\sqrt{t})-f(Y_t /\sqrt{t}) \ri]\ri|
\le  {\rm Lip}_f P[|Y_s|]/\sqrt{t},\\
|\e^{(5)}_{t}|
&=&P^x[e_{s,\a}-e_{\8,\a}]\lef| P^x[f(Y_t /\sqrt{t})]\ri|
\le \| f \| P^x[T_D \ge s],\\
|\e^{(6)}_{t}|
&=&P^x[e_{\8,\a}]
\lef| P^x[f(Y_t /\sqrt{t})]-\int_{\rd}fd\n \ri| 
\le  \lef| P^x[f(Y_t /\sqrt{t})]-\int_{\rd}fd\n \ri|,
\ednn
where $\| f\|$ and 
${\rm Lip}_f$  are the sup norm and the Lipschitz 
constant of $f$.
Since $D$ is transient, 
${\dps \lim_{t \ra \8}P^x[T_D \ge s]=0}$. 
Moreover, ${\dps \lim_{t \ra \8}P[|Y_s|]/\sqrt{t}=0}$ by the chioce of $s$. 
These, together with the central limit theorem for $(Y_t)$ without 
conditioning, imply (2).
\hfill $\Box$

$$
\ell^{p,h}(\zd)= \{ f:\zd \ra \R\; ; \; \| f \|_{p,h} <\8\},
$$
where the norm $\| f\|_{p,h}$ is defined by:
$$
\mbox{
${\dps \| f \|_{p,h} =\lef( \sum_{x \in \zd}|f(x)|^p \rig)^{1/p}}$ 
for $1 \le p <\8$ and 
${\dps \| f \|_{\8,h} = \sup_{x \in \zd}|f(x)|}$.}
$$
In particular, $\ell^{2,h}(\zd)$ is a Hilbert space. 
---------------0805030243107--
