0\endSb
I_A(x)\,\frac
{\nu^y(A)}{\mu(A)}.$$
Then $\DDD{\varphi_k^y=\frac{d\nu^y}{d\mu}\bigg|_{\BB_k}}$ and
also $\DDD{\varphi_k=\frac{d\nu}{d\mu\otimes\kappa}
\bigg|_{\BB_k\otimes\BB_Y}}$, where the measure $\nu$
on $X\times Y$ is defined by $\nu(dx,dy)=\nu^y(dx)\,\kappa(dy)$
and $\BB_Y$ is the Borel field of $Y$. Under the measure
$\mu\otimes\kappa$, $\varphi_k$ is a
nonnegative martingale
with respect to the filtration $\{\BB_k\otimes\BB_Y\}$,
hence there is a $\mu\otimes\kappa$-a.s. limit
$\DDD{\varphi(x,y)=\lim_{k\to\infty}\varphi_k(x,y)}$.
If $\{\varphi_k\}$ were uniformly $\mu\otimes\kappa$-integrable,
then $\varphi_k\to\varphi$ would also hold in $L\sosup 1
(\mu\otimes\kappa)$, and consequently, for any $m$ and
$A\in\BB_m\otimes\BB_Y$,
$$\nu(A)=\lim_{k\to\infty}\int_A \varphi_k\,d\mu\otimes\kappa
=\int_A \varphi\,d\mu\otimes\kappa.
$$
The filtration $\{\BB_k\otimes\BB_Y\}$ generates $\BB_X\otimes
\BB_Y$, which in turn equals $\BB_{X\times Y}$ by the
2nd countability of the topologies, hence the above implies
that $\DDD{\varphi=\frac{d\nu}{d\mu\otimes\kappa}}$. In
particular, since $\nu$ and $\mu\otimes\kappa$ have a common
$Y$-marginal,
$\DDD{\varphi^y=\frac{d\nu^y}{d\mu}}$ and $\nu^y\ll\mu$
for $\kappa$-a.e. $y$.
It remains to prove the uniform integrability of
$\{\varphi_k\}$ and (). For both it suffices to show
that
$\mu\otimes\kappa(\varphi_k^p)\le
C$ for all $k$.
$$\aligned
\mu\otimes\kappa(\varphi_k^p)&=\nu(\varphi_k^{p-1})
=\sum\Sb A: A\in\BB_k\\ \mu(A)>0\endSb\iint I_A(x)\,\frac
{\nu^y(A)^{p-1}}{\mu(A)^{p-1}}\,\nu(dx,dy)\\
&=\sum\Sb A: A\in\BB_k\\ \mu(A)>0\endSb
\frac 1{\mu(A)^{p-1}}\int \nu^y(A)^p \,\kappa(dy)\le
C\,\sum_{A\in\BB_k}\mu(A)\\
&= C.\qed
\endaligned
$$
\enddemo
\vskip .4in
%\input part4
\flushpar
{\it 3.3. The local equilibrium}
\hbox{}
\flushpar
Now fix a smooth function $\phi$ on $\btd$.
We start by showing that, for any $\delta>0$,
$$\lim_{N\to\infty}
P^N\lbrakk \sup_{0\le t\le T}\biggl|
\a^N_t(\phi)-\a^N_0(\phi)-\int_0^t
\kerroin N^{-d}\sum_{x\in\bzdN}\eta_s^2(x)\,A\phi(\tfrac xN)
\,ds\biggr| \ge\delta\rbrakk =0.
\tag 25
$$
Using estimates (13) and (15) (for the short-range and
long-range models, respectively) the expression in
$|\ \ |$'s can be bounded by
$$\sup_{0\le t\le T} \biggl( |M_t| + C\,N^{-1}
\int_0^t N^{-d}\sum_x\eta_s^2(x)\,ds \biggr)
\le
\sup_{0\le t\le T} |M_t| + C\,N^{-1}
\int_0^T N^{-d}\sum_x\eta_s^2(x)\,ds
$$
for a constant $C$ that depends
on $\phi$ alone. The expectation of this quantity
is bounded by
$$E^N\lbrak \sup_{0\le t\le T} M_t^2 \rbrak^{1/2}
+O(N^{-1}),$$
which vanishes as $N\to\infty$ as was shown in (21) above.
This establishes (25).
Let $\LaNe=\{z\in\bzd:\text{$0\le z_i< N\e $
for $i=1,\ldots,d$}\}$ for $\e>0$.
Next we turn (25) into
$$\aligned
&\lim_{\e\to 0}\limsup_{N\to\infty}
P^N\lbrakk \sup_{0\le t\le T}\biggl|
\a^N_t(\phi)-\a^N_0(\phi)\\
&\qquad\qquad -\int_0^t
\kerroin N^{-d}\sum_{x\in\bzdN}
A\phi(\tfrac xN)\,\biggl(
\frac1{|\LaNe|}\sum_{y\in x+\LaNe}\eta_s^2(y)\biggr)
\,ds\biggr| \ge\delta\rbrakk =0.
\endaligned
\tag 26
$$
Of course, $y\in x+\LaNe$ is again interpreted
with periodic boundary conditions of $\bzdN$ in mind.
To prove (26), notice first that by changing the order of
summation
$$
\aligned
&N^{-d}\sum_xA\phi(\tfrac xN)\,\biggl(
\frac1{|\LaNe|}\sum_{y\in x+\LaNe}\eta_s^2(y)\biggr)
=N^{-d}\sum_x\eta_s^2(x)\,A\phi(\tfrac xN)
\\
&\qquad\qquad +O\bigl( \;\sup_{\|\xi-\xi'\|\le\sqrt d\e}
|A\phi(\xi)-A\phi(\xi')|\;\bigr)\ett N^{-d}\sum_x\eta_s^2(x).
\endaligned
$$
By the a priori bound the expectation of the
error term vanishes as $\e\to 0$, uniformly in $N$
and $t$, so (26) follows from (25).
The next proposition
establishes a weak form of
local equilibrium, sufficient for our
needs: The empirical second
moment of the sticks in the
cube $\LaNe$
is asymptotically the same as for i.i.d. exponential
random variables with expectation given by the
empirical mean:
\proclaim{Proposition 3.7}
$$\aligned
\lim_{\e\to 0}\limsup_{N\to\infty}
E^N\lbrakk\, \int_0^T N^{-d}\sum_{x\in\bzdN} \biggl|
&\frac1{|\LaNe|}\sum_{y\in x+\LaNe}\eta_t^2(y) \\
&- 2\,\biggl (\frac1{|\LaNe|}\sum_{y\in x+\LaNe}\eta_t(y)
\biggr)^2
\,\biggr| \,dt\,\rbrakk =0.
\endaligned
\tag 27
$$
\endproclaim
Proposition 3.7 will be proved via a number of intermediate
results.
Write
$S_\La(\eta)={|\La|^{-1}}\sum_{x\in\La}\eta(x)$ for any
set $\La\subset\bzdN$, and similarly $S_\La(\eta^2)$ for
the average of squares. Define the probability measure
$\mubar^{N}$ on $\OO_N$ by
$$\mubar^N=\frac1T\int_0^T N^{-d}\sum_{x\in\bzdN}
\mu^N_t\circ\theta_x\,dt,
\tag 28
$$
where $\mu^N_t$ is the distribution of the stick process
at time $t$ and $\theta_x$ are the translations defined
on $\OO_N$ by $(\theta_x\eta)_y=\eta_{x+y}$, again
modulo the cube $\bzdN$. Then $\mubar^N$ is a translation
invariant measure that continues, by Lemma 3.2, to satisfy
the a priori bounds:
$$\text{ $\mubar^N\{\eta^k(x)\}\le T\,C_k$ for all
$k$, $x$, and $N$.} \tag 29
$$
The claim (27) of the proposition now reads
$$
\lim_{\e\to 0}\limsup_{N\to\infty}
\mubar^N\lbrak \bigl| S_{\LaNe}(\eta^2)-2\,S_{\LaNe}^2(\eta)
\bigr|\rbrak=0.
\tag 30
$$
As a standing notational convention, $\nu$ always denotes
a product probability measure under which the
sticks are i.i.d. exponential random variables
with expectation $K_0$.
For example, on each $\OO_N$ $\nu=
\gamma_{K_0}^{\otimes\bzdN}$, but we shall have occasion
to consider other sets of sites too besides the $\bzdN$'s.
The choice of $K_0$ for the expectation
is technically convenient
for then we have
derivatives
$\DDD{f^N_t(\eta)=\frac{d\mu^N_t}{d\nu}}(\eta)$ on $\OO_N$
that are
bounded uniformly over both $\eta$ and $t$. This holds for
$t=0$ by Assumption 1, and for $t>0$ by
general principles: Suppose $\nu$ is invariant for a Markov
process and $f_0=d\mu/d\nu$.
If $1\le p\le\infty$ and $q$ is the dual
exponent, then for $0\le g\in L\sosup q(\nu)$
$$\mu_t(g)=\nu\{f_0\,E^\piste g(\eta_t)\}\le
\|f_0\|_{L^p(\nu)}\|E^\piste g(\eta_t)\|_{L^q(\nu)}.
$$
First with $p=1$ this shows that $\mu_t\ll\nu$. Letting
$f_t=d\mu_t/d\nu$ we see
for all $p$ that $\|f_t\|_{L^p(\nu)}$
is nonincreasing in $t$.
The relative entropy or
Kullback-Leibler information
$H(Q\,|\,P)$ of two probability measures $Q$ and
$P$ is defined by
$$H(Q\,|\,P)=\cases \DDD{P\lbrak \frac{dQ}{dP}
\log\frac{dQ}{dP}\rbrak}
&\text{if $Q\ll P$}\\
\infty &\text{otherwise.}
\endcases
$$
$H(Q\,|\,P)$ measures a certain statistical distance
between $Q$ and $P$.
$H(Q\,|\,P)\ge 0$ holds always and $H(Q\,|\,P)=0$ if and
only if $Q=P$. A straightforward computation shows
that Assumptions 1 and 3 imply that the entropy bound
$$H(\mu^N_0|\,\nu)\le C\, N^d
\tag 31
$$
holds for all $N$
with a constant $C=C(K_0,\e_0)$ that depends only on the
constants of Assumptions 1 and 3. It is for the sake
of (31) that we need to make Assumption 3.
To handle simultaneously both the short-range and the
long-range model, let $p_N(x,y)$ be the uniform
distribution on $x+V_N$ for the long-range model, and let
$\beta=2$ for the short-range model. For Borel functions
$f\ge 0$ on $\OO_N$, set
$$\sigma_N(f)=
\sum_{x,y}p_N(x,y)\,\nu\lbrak \int_0^{\eta(x)}
\bigl[f(\eta^{u,x,y})-f(\eta)\bigr]\,
\bigl[\log f(\eta^{u,x,y})-\log f(\eta)\bigr]\,du\rbrak.
\tag 32
$$
The functional $\sigma_N$ is nonnegative, convex, and
translation invariant: $\sigma_N(f)=\sigma_N(f\circ\theta_x)$.
Set $H^N_t=H(\mu^N_t|\,\nu)$.
\proclaim{Lemma 3.8} $\DDD{\frac{d}{dt}H^N_t=-\frac{N^{\beta}}2\,
\sigma_N(f^N_t).}$
\endproclaim
\demo{Proof}
First let $\e>0$ and
deduce the conclusion
for $\DDD{g^\e_t=
(1-\e)f^N_t+\e}$
by direct calculation:
$$\aligned
\frac d{dt}\nu\lbrak g^\e_t\log g^\e_t\rbrak
&=\nu\lbrak (1+\log g^\e_t)\,\LL_Ng^\e_t\rbrak \\
&=\nu\lbrak \log g^\e_t\,\LL_Ng^\e_t
+g^\e_t\,\LL_N(\log g^\e_t)\rbrak\\
&= -\frac{N^{\beta}}2\,\sigma_N(g^\e_t)
\endaligned
$$
by using reversibility,
by substituting in (1) or (3), and by applying (9) to one of
the resulting terms. Nothing is problematic here since
everything in sight is bounded. Then write, for $s0$, pick $0\le a

**0$ for some $a>0$,
then
$$\delta\le \int \rho[a,\infty)^k\,Q(d\rho)
=\mu\{\eta_{x_1}\ge ka\}\searrow 0
$$
as $k\nearrow \infty$, a contradiction. Hence there is a
finite function
$r_n(\rho)$ such that for $Q$-a.e. $\rho$,
$$ \rho[\tfrac 1n,\infty)= \exp\lbrak-\frac 1{n \,r_n(\rho)}
\rbrak.
$$
Then by (46)
$$\aligned
\int \rho[\tfrac kn,\tfrac{k+1}n)\,Q(d\rho)
&=\int \rho[\tfrac 1n,\infty)^k-\rho[\tfrac 1n,\infty)^{k+1}
\,Q(d\rho)\\
&=\int e^{- k(n\, r_n(\rho))^{-1}}-
e^{-(k+1)(n\, r_n(\rho))^{-1}} \,Q(d\rho)\\
&=\int \lbrakk \int_{k/n}^{(k+1)/n} \frac{e^{-w/r_n(\rho)}}
{r_n(\rho)}\,dw\,\rbrakk \,Q(d\rho).
\endaligned
\tag 47
$$
Setting $Q_n(B)=Q\{\rho: \gamma_{r_n(\rho)}\in B\}$
for Borel sets $B$ of probability measures defines a
sequence of measures
$Q_n$ supported by exponential distributions. (Recall
that $\gamma_r$ was defined as the exponential distribution
with expectation $r$, see (7).) Let $f$ be a bounded
uniformly continuous function on $[0,\infty)$ and
$\delta_n(f)=\sup\{|f(w)-f(w')|: |w-w'|\le 1/n\}$.
Utilizing (47) we get
$$\aligned
\int \rho(f)\,Q(d\rho)&=\sum_{k=0}^\infty
f(\tfrac kn)\int \rho[\tfrac kn,\tfrac{k+1}n)\,Q(d\rho)
+O(\delta_n(f))\\
&=\int \lbrakk \,\sum_{k=0}^\infty
f(\tfrac kn) \int_{k/n}^{(k+1)/n} \frac{e^{-w/r_n(\rho)}}
{r_n(\rho)}\,dw\rbrakk\,Q(d\rho)+O(\delta_n(f))\\
&=\int \gamma_{r_n(\rho)}(f)\,Q(d\rho)+O(\delta_n(f))\\
&=\int \gamma(f)\,Q_n(d\gamma)+O(\delta_n(f)).
\endaligned
\tag 48
$$
Taking $f(w)=w$ (truncate and pass to a limit
in (48)) gives
$$\int\gamma(f)\,Q_n(d\gamma)=\mu\{\eta(x)\}+O(\tfrac 1n)\le C$$
by the a priori bound, consequently
$$Q_n\{\gamma: \gamma(f)\ge A\}\le C/A$$
for all $A>0$ and so
$\{Q_n\}$ is tight,
because $\gamma\mapsto\gamma(f)$ is a
homeomorphism from the set of exponential distributions
onto $[0,\infty)$.
Let $\Qtil$ be a limit point of $\{Q_n\}$,
still a measure supported by exponential distributions.
Letting $n\to\infty$ in (48) along a suitable subsequence
shows that,
at least on a single site, $\mu$
behaves as a $\Qtil$-mixture of exponentials. But then for
any finite set $\Ga\subset\bzd_+$ and $x\in\Ga$,
Lemma 3.10 implies that
$$\aligned
\mu\{\eta:\eta_\Ga\ge q\}
&=\mu\lbrak\eta:\eta_x\ge \sum_{y\in\Ga}q(y)\rbrak\\
&=\int \gamma\bigl[\sum_{y\in\Ga}q(y),\infty\bigr)\,\Qtil(d\gamma)\\
&=\int \gamma^{\otimes\bzd_+}\{\eta:\eta_\Ga\ge q\}\,
\Qtil(d\gamma)
\endaligned
$$
where the last equality
used an elementary property of exponential distributions.
Since the class of sets $\{\eta:\eta_\Ga\ge q\}$ is
rich enough to determine a probability measure,
$\mu=\int \gamma^{\otimes\bzd_+}\,\Qtil(d\gamma)$ and
the corollary is proved.
\qed
\enddemo
We are in a position to finish off the proof of the
one-block estimate. Let $\nu_r=\gamma_r^{\otimes\bzd_+}$
be the i.i.d. exponential distribution on $\OO$ with expectation $r$.
Since
$$E\lbrakk \bigg|\frac1n\sum_{i=1}^nX_i-EX_1\biggl|^2
\rbrakk=\frac 1n E\lbrak \bigl| X_1-EX_1\bigr|^2\rbrak
$$
holds for any square-integrable i.i.d. random variables, and
exponential variables satisfy the formulas
$E\{(X-EX)^2\}=(EX)^2$ and $E\{(X^2-E(X^2))^2\}=20(EX)^4$,
we can estimate as follows:
$$\aligned
&\nu_r\lbrak \bigl| S_{\La_L}(\eta^2)-2\,S_{\La_L}^2(\eta)
\bigr|\rbrak \\
\le\,&\nu_r\lbrak \bigl| S_{\La_L}(\eta^2)-2 \,r^2
\bigr|\rbrak + 2\,\nu_r\lbrak \bigl| r+S_{\La_L}(\eta)
\bigr|\cdot \bigl| r-S_{\La_L}(\eta)
\bigr|\rbrak\\
\le\,&\bigl\| S_{\La_L}(\eta^2)-2 \,r^2 \bigr\|_{L^2(\nu_r)}
+ C\,r\, \bigl\| S_{\La_L}(\eta)-r \bigr\|_{L^2(\nu_r)}\\
\le\,& C\,L^{-d/2}\,r^2\\
\le\,& C\,L^{-d/2}\,\nu_r\{\eta^2(x)\}.
\endaligned
$$
Let $\mu=\int \nu_r \,Q(dr)$ be the decomposition of the
limit point $\mu$ we have been considering.
Then the above gives, together with the a priori bound,
$$\mu\lbrak \bigl| S_{\La_L}(\eta^2)-2\,S_{\La_L}^2(\eta)
\bigr|\rbrak \le C\,L^{-d/2}\,\mu\{\eta^2(x)\}\le
C\,L^{-d/2}.
$$
This estimate holds for all limit points $\mu$ with
the same constant $C$; hence it holds
uniformly over $L$ in (40). We have established (39) and
completed the one-block estimate.
\hbox{}
\flushpar
{\it 3.3.2. Two-block estimate }
\hbox{}
\flushpar
In this subsection we prove
$$\lim_{L\to\infty}\limsup_{\e\to 0}
\limsup_{N\to\infty}
K^{-2d}\sum_{1\le i,j\le d}
\mubar^N\lbrak |S_{\De_i}(\eta)-S_{\De_j}(\eta)|^2\rbrak
=0.
\tag 49
$$
We remind the reader that $K^d$ is the maximal number
of disjoint translates of $\La_L$ that fit inside
$\LaNe$, and $\De_1$, $\ldots$, $\De_{K^d}$ is a tiling
of $\La_{KL}$ with a maximal collection
of such translates.
The first task is to get rid of the $i,j$-dependence
in the integral appearing in (49). Define a two-site
Dirichlet form by
$$D_{x,y}(g)=
\nu\lbrak \int_0^{\eta(x)} \bigl[ g(\eta^{u,x,y})-g(\eta)
\bigr]^2 \,du\rbrak.
$$
Set $D_{x,y}=D_{x,y}(g^N)$
where as before $g^N=\sqrt{\fbar^N}$.
\proclaim{Lemma 3.12} There is a constant $C$ such that
$D_{x,y}\le C\,\e^2$ for all $N$, $\e$, and
$x$, $y\in\LaNe$.
\endproclaim
\demo{Proof}
We begin with the short-range model. Fix $R>0$ and
for all pairs $z$, $w\in\La_R$ pick and fix a
path $z=z_0,z_1,\ldots,z_{b(z,w)}=w$ such that
$p(z_i-z_{i-1})>0$ for $i=1,\ldots,b(z,w)$.
Set
$$A=\inf_{z,w\in\La_R}\,\inf_{1\le i\le b(z,w)}
p(z_i-z_{i-1})\qquad\text{ and }\qquad
B=\max_{z,w\in\La_R}b(z,w).
$$
Then $A>0$ and $B<\infty$. For each
pair $z$, $w\in \La_R$ and for each $N>R$ there is
a path $z=\zbar_0,\zbar_1,\ldots,\zbar_{b(z,w)}=w$
inside $\bzdN$ such that $p_N(\zbar_{i-1},\zbar_i)\ge
p(z_i-z_{i-1})\ge A$.
Given $x$, $y\in\LaNe$, construct first a path
$x=w_0,w_1,\ldots,w_\ell=y$ so that each consecutive pair
$w_{i-1}$,
$w_{i}$ is contained in
a translate of $\La_R$.
By proceeding along each coordinate axis in turn,
this can be achieved with
$$\ell\le 3\,d\,N\e/R. \tag 50$$
Now fill in between each pair $w_{i-1},w_i$ with translates
of the
paths constructed earlier. This results in the path
$x=x_0,x_1,\ldots,x_m=y$ with $m\le B\,\ell$ and
$p(x_i-x_{i-1})\ge A>0$ for each $i$.
$$
\aligned
\bigl[ g^N(\eta^{u,x,y})-g^N(\eta)
\bigr]^2&=\lbrak \sum_{i=1}^m\bigl[
g^N(\eta^{u,x_0,x_i})-g^N(\eta^{u,x_0,x_{i-1}})
\bigr]\,\rbrak^2\\
&\le m\sum_{i=1}^m \bigl[ g^N(\eta^{u,x_0,x_i})
-g^N(\eta^{u,x_0,x_{i-1}}) \bigr]^2,
\endaligned
$$
hence an application of (9) and the above development gives
$$
\aligned
&\nu\lbrak \int_0^{\eta(x)} \bigl[ g^N(\eta^{u,x,y})-g^N(\eta)
\bigr]^2 \,du\rbrak \\
\le\,&m\sum_{i=1}^m \nu\lbrak \int_0^{\eta(x_{i-1})}
\bigl[ g^N(\eta^{u,x_{i-1},x_i})-g^N(\eta)
\bigr]^2 \,du\rbrak\\
\le\,&m\sum_{i=1}^m \frac 1{p(x_i-x_{i-1})}
\sum_z p_N(x_{i-1},z)\, \nu\lbrak \int_0^{\eta(x_{i-1})}
\bigl[ g^N(\eta^{u,x_{i-1},z})-g^N(\eta)
\bigr]^2 \,du\rbrak\\
\le\,&A^{-1}m^2\, D^N_{x}(g^N)\\
\le\,&C\,\e^2
\endaligned
$$
by (36) and (50), remembering that $\beta=2$ for this model.
This proves the lemma for the short-range
model.
For the long-range model, arrange a sequence
$E_0,E_1,\ldots, E_m$ of cubes with side length
$N^\a/4$ so that $x\in E_0$, $y\in E_m$, and
$E_i\subset z+V_N$ for each $z\in E_{i-1}$ for all $i$.
This can be achieved with $m\le C\,N^{1-\a}\e$.
Consider a path
$x=x_0,x_1,\ldots,x_m=y$ with $x_i\in E_i$ for all $i$.
Reasoning as above, we get
$$D_{x,y}\le m\sum_{i=1}^m D_{x_{i-1},x_i}.$$
Sum over $x_m\in E_m$ to get
$$\aligned
D_{x,y}&\le m\sum_{i=1}^{m-1} D_{x_{i-1},x_i}
+ 4^dN^{-\a d}m\,\sum_{z\in E_m} D_{x_{m-1},z}\\
&\le m\sum_{i=1}^{m-1} D_{x_{i-1},x_i}
+ C\,m\,D_{x_{m-1}}^N(g^N).
\endaligned
$$
Now iterate: sum over $x_{m-1}\in E_{m-1}$, $\ldots$,
$x_1\in E_1$ in turn, and use (36) together with
$2\a+\beta=2$.
\qed
\enddemo
As far as this estimate and the a priori bound
goes, the relative positions of $x$ and $y$, and consequenly
of $\De_i$ and $\De_j$, are immaterial, so we can
simply think of two
disjoint cubes $\La_L$ and $\La_L'$ and a probability
measure on the sticks in the union $\La_L\cup\La_L'$.
$N$ disappears, and instead of (49) we prove
$$\lim_{L\to\infty}\lim_{\e\to 0}
\sup_{\mu\in\NN_{L,\e}} \mu\lbrak
|S_{\La_L}(\eta)-S_{\La_L'}(\eta)|^2\rbrak
=0,
\tag 51
$$
where $\NN_{L,\e}$ is the class of probability measures
$\mu$ on $\OO_{\La_L\cup\La_L'}$ that satisfy $\mu\ll\nu$,
$D_{x,y}(\sqrt f)$ $\le C\e^2$
for $\DDD{f=\frac{d\mu}{d\nu}}$,
and the a priori bound $\mu\{\eta^k(x)\}\le C_k$, for
all $x,y\in \La_L\cup\La_L'$.
For each $L$, let $\mu_L$ be a measure that
satisfies
$$\mu_L\lbrak
|S_{\La_L}(\eta)-S_{\La_L'}(\eta)|^2\rbrak
=\lim_{\e\to 0}
\sup_{\mu\in\NN_{L,\e}} \mu\lbrak
|S_{\La_L}(\eta)-S_{\La_L'}(\eta)|^2\rbrak.
$$
Existence of the $\mu_L$'s is justified as in
the discussion following (40).
Take $\Ga=\La_L\cup\La_L'$ in Lemma 3.10, and observe
that the last bound in (43) is precisely what we have
control over with Lemma 3.12. Thus the proof of Lemma 3.10
goes through again, and (41) holds for $\mu_L$. In particular, the
variables $(\eta(x): x\in \La_L\cup\La_L')$ are
exchangeable under $\mu$. (But finite
exchangeability does not imply a decomposition
into product measures, so Corollary 3.11 does not follow
yet.)
Fix $R$, and for $L>>R$ replace $\La_L$ with a disjoint union
$\Ga_1\cup\ldots\cup\Ga_{K^d}$ of translates of
$\La_R$, maximal with respect to the property of
fitting inside $\La_L$. Let $\Ga_i'$ be the translate
inside $\La_L'$ that sits relative to $\La_L'$ as
$\Ga_i$ sits relative to $\La_L$. Then (51) follows from
$$\lim_{R\to\infty}\limsup_{L\to \infty}
K^{-d}\sum_i \mu_L\lbrak
|S_{\Ga_i}(\eta)-S_{\Ga_i'}(\eta)|^2\rbrak
=0,
\tag 52
$$
which
by exchangeability is equivalent to
$$\lim_{R\to\infty}\limsup_{L\to \infty}
\mu_L\lbrak
|S_{\La_R}(\eta)-S_{\La_R'}(\eta)|^2\rbrak
=0.
\tag 53
$$
But any limit point of the $\mu_L$'s is infinitely
exchangeable, hence we may prove
$$\lim_{R\to\infty}\sup_\mu \mu\lbrak
|S_{\La_R}(\eta)-S_{\La_R'}(\eta)|^2\rbrak
=0,
\tag 54
$$
where the supremum is over infinitely
exchangeable $\mu$ that satisfy the moment
conditions $\mu\{\eta^k(x)\}\le C_k$ for all $k$. The
proof can be completed as was done for the
one-block estimate.
\hbox{}
We are ready to prove the local
equilibrium:
\demo{Proof of Proposition 3.7} Combine (30), (37), (39), and (49).
\qed
\enddemo
Before utilizing (27) we wish to change it slightly.
For $\xi,\theta\in\btd$,
let
$$\chi_{\e,\xi}(\theta)=\e^{-d}I_{\xi+[0,\e)^d}(\theta),
\tag 55
$$
i.e. the indicator function of the $\e$-cube on $\btd$ with
lower left corner at $\xi$, normalized by the volume.
Since $y\in x+\LaNe$ if and only if
$y/N\in x/N+[0,\e)^d$, (27) is equivalent to
$$\aligned
\lim_{\e\to 0}\limsup_{N\to\infty}
E^N\lbrakk\, \int_0^T N^{-d}\sum_{x\in\bzdN} \biggl|
&\frac1{|\LaNe|}\sum_{y\in x+\LaNe}\eta_t^2(y)
- 2\,\bigl[\a^N_t(\chi_{\e,x/N})\bigr]^2
\,\biggr| \,dt\,\rbrakk =0.
\endaligned
\tag 56
$$
By inserting (56) into (26) we get as conclusion of this
subsection:
$$\aligned
&\lim_{\e\to 0}\limsup_{N\to\infty}
P^N\lbrakk \sup_{0\le t\le T}\biggl|
\a^N_t(\phi)-\a^N_0(\phi)\\
&\qquad\qquad -2\,\kerroin \,\int_0^t
N^{-d}\sum_{x\in\bzdN}
A\phi(\tfrac xN)\,\bigl[\a^N_s(\chi_{\e,x/N})
\bigr]^2
\,ds\biggr| \ge\delta\rbrakk =0.
\endaligned
\tag 57
$$
\vskip .3in
%\input part5
\flushpar
{\it 3.4. Further technicalities}
\hbox{}
\flushpar
First we turn (57) into
$$\aligned
\lim_{\e\to 0}\limsup_{N\to\infty}
P^N\lbrakk \sup_{0\le t\le T}\biggl|&
\a^N_t(\phi)-\a^N_0(\phi)\\
&\quad-
2\kerroin \int_0^t\int_{\btd}
A\phi(\xi)\,\bigl[\a^N_s(\chi_{\e,\xi})
\bigr]^2\,d\xi
\,ds\biggr| \ge\delta\rbrakk =0.
\endaligned
\tag 58
$$
This step requires the a priori estimate and the
continuity of $A\phi$: For $\xi\in x/N+[0,N^{-1})^d$,
$$\aligned
&\bigl| \;A\phi(\xi)\,\bigl[\a^N_s(\chi_{\e,\xi})
\bigr]^2 -
A\phi(\tfrac xN)\,\bigl[\a^N_s(\chi_{\e,x/N})
\bigr]^2\;\bigr|\\
\le\;&2\,\|A\phi\|_\infty
\biggl(N^{-d}\sum_y\eta(y)\biggr)\,\biggl(
N^{-d}\sum_{y\in \La(x,N\xi,N\e)}\eta(y)\biggr)\\
&\qquad\qquad
+ \biggl(N^{-d}\sum_y\eta(y)\biggr)\,|A\phi(\xi)-
A\phi(\tfrac xN)|,
\endaligned
$$
where $\La(x,y,K)=(x+\La_K)\triangle(y+\La_K)$.
Integrate this bound over $\btd$ and note
that $|\La(x,N\xi,N\e)|=O((N\e)^{d-1})$. Then (58) follows from (57).
\proclaim{Lemma 3.13} For $\psi\in C(\btd)$ and fixed $\e>0$,
$\DDD{ \nu\mapsto\int_{\btd}
\psi(\xi)\,\nu(\chi_{\e,\xi})^2\,d\xi}$
is a continuous function
of $\nu\in\MM$.
\endproclaim
\demo{Proof} Suppose $\nu_n\to\nu$ in the topology
of $\MM$. Let $0\le f_k\le g_k\le \e^{-d}$
be bounded continuous
functions such that $f_k\nearrow \e^{-d}I_{(0,\e)^d}$
and $g_k\searrow \e^{-d}I_{[0,\e]^d}$ on $\btd$, and
let $f_k^\xi(\theta)=f_k(\theta-\xi)$,
similarly for $g_k^\xi$.
Then for all $k$,
$$\aligned
\int \psi(\xi)\,\nu(f_k^\xi)^2d\xi
&\le \liminf_{n\to\infty}
\int \psi(\xi)\,\nu_n(\chi_{\e,\xi})^2\,d\xi\\
&\le
\limsup_{n\to\infty}
\int \psi(\xi)\,\nu_n(\chi_{\e,\xi})^2\,d\xi\le
\int \psi(\xi)\,\nu(g_k^\xi)^2d\xi.
\endaligned
$$
Thus it suffices to show that
$\DDD{\lim_{k\to\infty} \bigl[ \nu(g^\xi_k)-\nu(f^\xi_k)\bigr]
=0}$
for Lebesgue a.e. $\xi$. This comes by a simple Fubini
argument:
$$\aligned
&\int \lim_{k\to\infty} \bigl[ \nu(g^\xi_k)-\nu(f^\xi_k)\bigr]
\,d\xi \\
=\,&\int \lbrak \int I_{[0,\e]^d}(\theta-\xi)-
I_{(0,\e)^d}(\theta-\xi)\,\nu(d\theta)\,\rbrak\,d\xi\\
=\,&\int \lbrak \int I_{[0,\e]^d}(\theta-\xi)-
I_{(0,\e)^d}(\theta-\xi)\,d\xi\, \rbrak\,
\nu(d\theta)=0.
\qed
\endaligned
$$
\enddemo
Thus in terms of the distribution $\PP^N$ of $\a^N_\piste$, the
probability in (58) equals
$\PP^N\{\oo_\piste\in\DD_\MM: \Phi_\e(\oo_\piste)\ge\delta\}$
for a certain continuous function $\Phi_\e$ on
$\DD_\MM$. Let $\PP$ be any limit point
of $\{\PP^N\}$. Since (58) holds for all $\delta>0$,
it implies that
$$\lim_{\e\to 0}\PP\{ \oo_\piste: \Phi_\e(\oo_\piste)
\ge\delta\} =0,\tag 59
$$
again for all $\delta>0$.
Now recall that according to Lemma 3.5
there exists a $\PP(d\oo_\piste)\otimes
dt\otimes d\xi$-a.e. defined jointly
measurable function $u(\oo_\piste, t,\xi)$
such that
$\oo_t(d\xi)=u(\oo_\piste, t,\xi)\,d\xi$. We suppress
the $\oo_\piste$-argument, and then (59) may be
written as
$$\aligned
&\lim_{\e\to 0}
\PP\lbrakk \sup_{0\le t\le T}\biggl|
\oo_t(\phi)-\oo_0(\phi)\\
&\qquad\qquad -2\kerroin\int_0^t \int_{\btd}
A\phi(\xi)\,\biggl( \e^{-d}\int_{\xi+[0,\e)^d}
u(s,\theta)\,d\theta \biggr)^2\,d\xi
\,ds\biggr| \ge\delta\rbrakk =0.
\endaligned
\tag 60
$$
For a.e. $\oo_\piste$ and $s$,
$$\lim_{\e\to 0} \biggl( \e^{-d}\int_{\xi+[0,\e)^d}
u(s,\theta)\,d\theta\biggr)^2 =u^2(s,\xi)
$$
for a.e. $\xi$
by Lebesgue's differentiation theorem.
To extend this convergence to the integral inside (60),
introduce the maximal function
$$M_p(s,\xi)=\sup_{\e>0} \, \e^{-d}\int_{\xi+[0,\e)^d}
u^p(s,\theta)\,d\theta.
$$
Write $\EE$ for expectation under the measure $\PP$.
\proclaim{Lemma 3.14} $\DDD{\EE\int_0^T\int_{\btd}
M_2(s,\xi)\,d\xi\,ds <\infty.}$
\endproclaim
\demo{Proof} Since $M_2^2(s,\xi)\le M_4(s,\xi)$,
the maximal theorem gives (see p. 91 in \cite{Fo}):
$$|\{\xi: M_2(s,\xi)> r\}|\le
|\{\xi: M_4(s,\xi)> r^2\}|\le C\,r^{-2}\int_{\btd}
u^4(s,\xi)\,d\xi,
$$
where $|\ett|$ denotes normalized Lebesgue measure on
$\btd$.
Hence by (22)
$$\aligned
\EE\int_0^T\int_{\btd}M_2(s,\xi)\,d\xi\,ds
&\le \EE\int_0^T\int_1^\infty |\{\xi: M_2(s,\xi)>r\}|
\,dr\,ds +T\\
&\le C\,\int_1^\infty r^{-2}\,dr\ett
\EE\int_0^T \int_{\btd} u^4(s,\xi)\,d\xi\,ds +T\\
&<\infty. \qed
\endaligned
$$
\enddemo
This lemma and the dominated convergence theorem imply
that
$$\aligned
\lim_{\e\to 0}\EE\lbrakk
\sup_{0\le t\le T} \biggl|&
\int_0^t \int_{\btd}
A\phi(\xi)\,\biggl( \e^{-d}\int_{\xi+[0,\e)^d}
u(s,\theta)\,d\theta \biggr)^2\,d\xi\,ds \\
&\qquad\qquad- \int_0^t \int_{\btd}
A\phi(\xi)\,u^2(s,\xi)\,d\xi\,ds \biggr|\,\rbrakk =0,
\endaligned
$$
and consequently (60) turns into the statement
$$
\aligned
\PP\lbrakk \sup_{0\le t\le T}\biggl|
\oo_t(\phi)-\oo_0(\phi) -2\kerroin\int_0^t \int_{\btd}
A\phi(\xi)\,u^2(s,\xi)\,d\xi
\,ds\biggr| \ge\delta\rbrakk =0.
\endaligned
\tag 61
$$
Letting $\delta\searrow 0$ shows that
$\PP$-a.e. $\oo_\piste$ is a continuous $\MM$-valued
path with a derivative for a.e. $t$, and a weak solution
of the equation
$\partial_tu=2\kerroin A(u^2)$. By the uniqueness lemma
of the next subsection
$\PP$ is supported by a single path.
This has two consequences: (i) It implies
that $\oo_t$ has a density $u(t,\xi)$ for all $t$.
For we can reprove Lemma 3.5 for a fixed
time $t$ and conclude that $\oo_t\ll d\xi$ for
$\PP$-a.e. $\oo_\piste$, in particular, for
the unique $\oo_\piste$ supporting $\PP$.
(ii) It promotes the
weak convergence $\PP^N\to\PP$ to convergence in
probability of the $\DD_\MM$-valued
random variables $\a^N_\piste$ to the path $u(\piste,\xi)\,
d\xi$ as stated in Theorem 1. This comes from a general fact:
Weak convergence to a degenerate distribution implies
convergence in probability.
For Theorems 1 and 2, it remains to prove the bound
$u(t,\xi)\le \|u_0\|_\infty$, for all $(t,\xi)\in Q_T$.
For any $0\le \phi\in C(\btd)$,
$$\aligned
\int_{\btd}\phi\,u(t)&=\lim_{N\to\infty}
E^N\{\a^N_t(\phi)\}=\lim_{N\to\infty}
N^{-d}\sum_x\phi(\tfrac xN)\,E^N\{\eta_t(x)\}\\
&\le K_0 \int_{\btd}\phi,
\endaligned
$$
where the precise constant $K_0$ of Assumption 1 comes
from the last line of the proof of the a priori
estimate Lemma 3.2.
But now note that it is perfectly possible to choose
the initial distribution so that $K_0\le \|u_0\|_\infty$.
Thus with the uniqueness lemma we have proved
Theorems 1 and 2 under Assumption 3, that the initial
density is bounded away from 0. This assumption will
be lifted after the uniqueness proof.
\vskip .3in
\flushpar
{\it 3.5. Uniqueness lemma}
\hbox{}
\flushpar
\proclaim{Lemma 3.15} Let $(a_{i,j})$ be a symmetric positive
semidefinite matrix and set
$$A\phi=\sum_{i,j}
a_{i,j}\partial_{\xi_i}\partial_{\xi_j}\phi.$$
Suppose $t\mapsto \oo(t,d\xi)$ is a
continuous $\MM$-valued path on $0\le t\le T$ such that
\roster
\item"{(i)}" there exists a jointly measurable function
$u(t,\xi)$ on $Q_T$ such that
$\oo(t,d\xi)=u(t,\xi)\,d\xi$ for a.e. $t$,
\item"{(ii)}" $\DDD{\int_0^T\int_{\btd} u^4(t,\xi)\,d\xi\,
dt<\infty\,}$, and
\item"{(iii)}" $\DDD{ \int_{\btd} \phi(\xi)\, \oo(t,d\xi)
- \int_{\btd} \phi(\xi)\, \oo(0,d\xi)
=\int_0^t\int_{\btd} A\phi(\xi)\,u^2(s,\xi)\,d\xi\,ds\,}$
for all smooth $\phi$ on $\btd$ and for all $0\le t\le T$.
\endroster
Then if $\sigma(t,d\xi)$ is another continuous
$\MM$-valued path that satisfies the
analogues of
{\rm (i)-(iii)} and $\sigma(0)=\oo(0)$, then $\sigma(t)=\oo(t)$
for all $0\le t\le T$.
\endproclaim
\demo{Proof} Let $\sigma(t,d\xi)=v(t,\xi)\,d\xi$
when the density exists.
Let $f_\e(\xi)=\e^{-d}f(\e^{-1}\xi)$ be a compactly
supported, symmetric, smooth approximation to the
identity. Set
$$u_\e(t,\xi)=[\oo(t)\ast f_\e](\xi)=\int_{\btd}
f_\e(\xi-\theta)
\,\oo(t,d\theta)$$
and similarly define $v_\e(t,\xi)$. Then $u_\e$
(and also $v_\e$) satisfies
$$\int_{\btd} \psi(\xi)\, u_\e(t,\xi)\,d\xi
- \int_{\btd} \psi(\xi)\,u_\e(0,\xi)\,d\xi
=\int_0^t\int_{\btd} A\psi(\xi)\,[u^2(s)\ast
f_\e](\xi)\,d\xi\,ds
$$
for all smooth $\psi$ and {\it all} $t$. We wrote
$u^2(s)$ for the function $u^2(s)(\theta)=u^2(s,\theta)$.
Subtracting the equation for $v_\e$ from the equation
for $u_\e$ and writing
$\phi^s_\e=u^2(s)\ast f_\e-v^2(s)\ast f_\e$ (a well-defined
smooth function on $\btd$ for a.e. $s$) gives
$$\int_{\btd} \psi\,\bigl[ u_\e(t)- v_\e(t)\bigr]
=\int_{\btd} \sum_{i,j}a_{i,j}
\bigl(\partial_{\xi_i}\partial_{\xi_j}\psi\bigr)
\,\biggl(\int_0^t
\phi^s_\e\, \,ds\biggr). \tag 62
$$
Now take $\psi=\phi^t_\e$ for those a.e. $t$ for which
this makes sense and integrate over $0\le t\le T$ to
render the exceptional set of $t$'s harmless.
After an integration by parts on
the right-hand side we have
$$\aligned
\int_0^T dt\,\int_{\btd}
\phi^t_\e\,\bigl[ u_\e(t)- v_\e(t)\bigr]
&=-\int_{\btd} \sum_{i,j}a_{i,j}\int_0^T
\bigl(\partial_{\xi_i}\phi^t_\e\bigr)\,\biggl(\int_0^t
\partial_{\xi_j}\phi^s_\e\, \,ds\biggr)\,dt\\
&=
-\int_{\btd} \sum_{i,j}a_{i,j} \int_0^T\int_0^T
\bigl(\partial_{\xi_i}\phi^t_\e\bigr)\,\bigl(
\partial_{\xi_j}\phi^s_\e\bigr)\,I_{\{t\ge s\}}\,dt\,ds.
\endaligned
$$
By the symmetry of $(a_{i,j})$
$$
2\,\int_0^T dt\,\int_{\btd} \phi^t_\e \bigl[ u(t)\ast f_\e
-v(t)\ast f_\e\bigr] =
-\int_{\btd} \sum_{i,j}a_{i,j} \biggl(\int_0^T
\partial_{\xi_i}\phi^t_\e\,dt\biggr)\,\biggl(\int_0^T
\partial_{\xi_j}\phi^t_\e\,dt\biggr).
$$
The conclusion we derive from this, by $(a_{i,j})$'s
positive semidefiniteness, is that for all $\e>0$,
$$\int_0^T dt\,\int_{\btd} \bigl[ u^2(t)\ast f_\e
-v^2(t)\ast f_\e \bigr]\, \bigl[ u(t)\ast f_\e
-v(t)\ast f_\e\bigr] \le 0.
$$
Hypothesis (ii) gives sufficient integrability to let
$\e\searrow 0$ and recover
$$\int_{Q_T} \bigl[ u^2 -v^2 \bigr]\,
\bigl[ u -v\bigr] \le 0
$$
which implies that $u(t,\xi)=v(t,\xi)$ almost everywhere.
Hence $\oo(t)=\sigma(t)$ for a.e. $t$, and by continuity
for all $t$.
\qed
\enddemo
\vskip .3in
\flushpar
{\it 3.6. Removing Assumption 3}
\hbox{}
\flushpar
Assume given an initial density $u_0$, not necessarily bounded
away fom zero, and initial distributions
$\mu^N_0$ satisfying Assumption 1.
For $\e>0$, the versions of the
theorems thus far proved apply to the initial densities
$u_0^\e=u_0+\e$ with initial
distributions $\mu^{\e,N}_0$ arranged to satisfy
$\mu^{\e,N}_0\{\eta(x)\}=\mu^N_0\{\eta(x)\}+\e$.
Fix $\e>0$ for the moment.
Let $Q^N$ be the coupling of $\mu^N_0$ and $\mu^{\e,N}_0$
given in Lemma 3.3. Let $\mmP^N$ be the distribution of
the joint process with initial distribution $Q^N$,
constructed as in the proof of Lemma 3.2 so that
$\mmP^N\{(\eta_\piste,\zeta_\piste): \eta_\piste\le\zeta_\piste\}
=1$, and the $\eta_\piste$ and $\zeta_\piste$ marginals
of $\mmP^N$ are the processes $P^N$ and $P^{\e,N}$ with
initial distributions $\mu^N_0$ and $\mu^{\e,N}_0$,
respectively. Let $\PPs^N$ be the distribution of
$$(\a^N_\piste,\a^{\e,N}_\piste)=
\biggl( N^{-d}\sum_x\eta_\piste(x)\,
\delta_{\frac xN}\,,\, N^{-d}\sum_x\zeta_\piste(x)\,
\delta_{\frac xN}\,\biggr)
$$
on the space $\DD_\MM\times\DD_\MM$.
The tightness proof did not depend on Assumption 3,
hence $\{\PPs^N\}$ has tight marginals and consequently
is itself tight. Let $\PPs$ be a limit point with
marginals $\PP$ and $\PP^\e$. We know that $\PP^\e$ is
supported by the unique path $u^\e(t,\xi)\,d\xi$
described in Theorem 1 or 2, whichever model we are
talking about. Nor did Lemma 3.5 depend on Assumption 3,
and so $\PP(d\oo_\piste)\otimes dt$-a.e. $\oo_t$
has a density $v(t,\xi)$. For all $0\le a 0$ holds for
some constant $\e_0$, uniformly over $N$ and $x\in\bzdN$.
Thus for the case $\varkappa=0$ we need to
proceed as for the stick model, by first assuming
$u_0$ bounded away from zero and then removing this
assumption in the end, but for the case $\varkappa>0$
no such assumption is needed. The quantities
$\sigma_N(g)$ and $D_x^N(g)$ are defined as in (32) and
(34) except that
the integral $\DDD{\int_0^{\eta(x)} du}$ and $u$ are replaced by
the sum $\DDD{\hiukkas N\sum_{k=1}^{\eta(x)/\shiukkas N}}$
and $k\hiukkas N$, respectively.
(36) then follows from the entropy bound as before, with
$\beta=2$.
For the local equilibrium we need to prove
$$\aligned
\lim_{\e\to 0}\limsup_{N\to\infty}
E^N\lbrakk\, \int_0^T N^{-d}\sum_{x\in\bzdN} \biggl|
&\frac1{|\LaNe|}\sum_{y\in x+\LaNe}\eta_t(y)(\hiukkas N
+\eta_t(y)) \\
&- 2\,\bigl( \varkappa S_{ x+\LaNe}(\eta_t) +
S_{ x+\LaNe}^2(\eta_t) \bigr)
\,\biggr| \,dt\,\rbrakk =0.
\endaligned
\tag 65
$$
As before $S_\Ga(\eta)=|\Ga|^{-1}\sum_{y\in\Ga} \eta(y)$.
(65) follows from showing
$$\aligned
\lim_{\e\to 0}\limsup_{N\to\infty}
E^N\lbrakk\, \int_0^T N^{-d}\sum_{x\in\bzdN} \biggl|
&S_{ x+\LaNe}(\eta_t^2)\\
&-
\bigl( \varkappa S_{ x+\LaNe}(\eta_t) +
2\,S_{ x+\LaNe}^2(\eta_t) \bigr)
\,\biggr| \,dt\,\rbrakk =0.
\endaligned
\tag 66
$$
Repeating Lemma 3.9,
our tasks are again to prove the one-block estimate
$$\lim_{L\to\infty}\limsup_{N\to\infty}
\mubar^N\lbrak \bigl| S_{\La_L}(\eta^2)-\varkappa
S_{\La_L}(\eta) -2\,S_{\La_L}^2(\eta) \bigr|\rbrak =0
\tag 67
$$
and the two-block estimate
$$\lim_{L\to\infty}\limsup_{\e\to 0}
\limsup_{N\to\infty}
K^{-2d}\sum_{1\le i,j\le d}
\mubar^N\lbrak |S_{\De_i}(\eta)-S_{\De_j}(\eta)|^2\rbrak
=0.
\tag 68
$$
The measure $\mubar^N$ was defined by (28).
\hbox{}
\flushpar
{\it 4.1. One-block estimate, case $\varkappa=0$}
\hbox{}
\flushpar
Letting $\mu_L$ denote a limit point of a
subsequence of $\{\mubar^{N}\}$ that realizes the
$\DDD{\limsup_{N\to\infty}}$ in (67), we need to prove
$$\lim_{L\to\infty}
\mu_L\lbrak \bigl| S_{\La_L}(\eta^2)-
2\,S_{\La_L}^2(\eta) \bigr|\rbrak =0
\tag 69
$$
exactly as in section 3.3.1. Thus it suffices to show
that Lemma 3.10 holds for an arbitrary limit point $\mu$
of $\{\mubar^{N}\}$. Given $q=(q(x):x\in\Ga)
\in [0,\infty)^\Ga$, define the configuration $q_N$
by $q_N(x)=[q(x)/\hiukkas N]\hiukkas N$, where
$[\ett]$ denotes integer part.
Then we get
$$\int_a^b \mubar^N\{\eta_\Ga\ge q^{u,x,y}\}\,du
=\hiukkas N\sum_{k=[a/\shiukkas N]}^{[b/\shiukkas N]-1}
\mubar^N\{\eta_\Ga\ge q_N^{k\shiukkas N,x,y}\}+O(\hiukkas N).
$$
The error is $O(\hiukkas N)$,
because for any $x$ and any fixed constants $a_0$ and
$a_1$,
$$\mubar^N\{ a_0 \le \eta(x) < a_0+a_1 \hiukkas N\}
\le K_1 \,a_1\,\sup_{k} \nu^N\{\eta(x)=k\hiukkas N\}
=O(\hiukkas N).
$$
Reasoning as in (43) yields
$$\aligned
&\biggl| \int_a^b \mubar^N\{\eta: \eta_\Ga\ge q\} -
\mubar^N\{\eta: \eta_\Ga\ge q^{u,x,y}\}\,du \biggr|\\
\le\,&C\,\biggl(\nu^N\lbrak\hiukkas N
\sum_{k=1}^{\eta(x)/\shiukkas N}\bigl[
{g^N(\eta^{k\shiukkas N,x,y})}-{g^N(\eta)} \bigr]^2
\rbrak\biggr)^{1/2}+O(\hiukkas N).
\endaligned
\tag 70
$$
Now proceed as in the proof of Lemma 3.10.
\hbox{}
\flushpar
{\it 4.2. One-block estimate, case $\varkappa>0$}
\hbox{}
\flushpar
Let $\mu$ be any limit point of $\{\mubar^N\}$.
It is a probability measure on
$\{k\varkappa: k=0,1,2,\ldots\}^{\bzd_+}$.
Let $\Ga$ be a finite subset of $\bzd_+$ and
$q=(k(x)\varkappa: x\in\Ga)$ a particle configuration
on $\Ga$, where $k(x)$ are nonnegative
integers.
\proclaim{Lemma 3.16} For any $x,y\in\Ga$ and $k\le k(x)$,
$$\mu\{\eta: \eta_\Ga=q\}=\mu\{\eta: \eta_\Ga=q^{k\varkappa,x,y}\}.
\tag 71
$$
\endproclaim
\demo{Proof}
Let $q_N=(k(x)\hiukkas N:x\in\Ga)$ be
the corresponding configuration for the $N$th process.
Proceeding as in (43), then constructing a suitable path
from $x$ to $y$ and using the bound (36) give, as in the proof
of Lemma 3.10,
$$\aligned
&\bigl|
\mubar^N\{\eta_\Ga=q_N\}-\mubar^N
\{\eta_\Ga=q_N^{k\shiukkas N,x,y}\}\bigr|\\
\le\,&C\,\hiukkas N^{-1}\,\biggl( \nu^N\lbrak
\hiukkas N\sum_{k=1}^{\eta(x)/\shiukkas N}
\bigl[ g^N(\eta^{k\shiukkas N,x,y})-g^N(\eta)\bigr]^2
\rbrak \biggr)^{1/2}\\
\le\,&C\,N^{-1}.
\endaligned
$$
This suffices for (71).
\qed
\enddemo
Corresponding to Corollary 3.11 we now have:
\proclaim{Corollary 3.17} $\mu$ is a mixture of i.i.d.
geometric distributions.
\endproclaim
\demo{Proof} Exchangeability is immediate from (71), so there
is a representation
$$\mu=\int \rho^{\otimes\bzd_+}\,
Q(d\rho)$$
of $\mu$ in terms of i.i.d. distributions.
The integration variable $\rho$ is a probability
measure on $\{k\varkappa: k=0,1,2,\ldots\}$.
Consider first a single site $x$, and
a set $\{x_1,\ldots,x_m\}$ of mutually distinct sites
distinct from $x$. By (71),
$$\aligned
\mu\{\eta(x)=m\varkappa\}&=\sum_{k_1,\ldots,k_m\ge 0}
\mu\{\eta(x)=m\varkappa, \eta(x_1)=k_1\varkappa,\ldots,
\eta(x_m)=k_m\varkappa \}\\
&=\sum_{k_1,\ldots,k_m\ge 0}
\mu\{\eta(x)=0, \eta(x_1)=(k_1+1)\varkappa,\ldots,
\eta(x_m)=(k_m+1)\varkappa \}\\
&=
\mu\{\eta(x)=0, \eta(x_1)\ge\varkappa,\ldots,
\eta(x_m)\ge\varkappa \}\\
&=\int \rho(0)(1-\rho(0))^m\,Q(d\rho),
\endaligned
$$
so the distribution $\mu\{\eta(x)\in\ett\}$ is a mixture
of geometric distributions. But then
$$\aligned
\mu\{\eta_\Ga\ge q\}&=
\mu\lbrak\eta(x)\ge \varkappa\sum_{y\in\Ga}k(y)\rbrak\\
&=\int (1-\rho(0))^{\sum_{y\in\Ga}k(y)}\,Q(d\rho)\\
&=\int\prod_{y\in\Ga} (1-\rho(0))^{k(y)}\,Q(d\rho),
\endaligned
$$
and we see that $\mu$ is indeed a mixture of i.i.d.
geometrics.
\qed
\enddemo
The one-block estimate is now completed as was done
in section 3.1.1, utilizing the fact that if $X$ is a
$\varkappa\mmZ_+$-valued geometric random variable, then
$E(X^2)=\varkappa EX+2(EX)^2$.
\hbox{}
\flushpar
{\it 4.3. Two-block estimates}
\hbox{}
\flushpar
This time we have to define the two-site Dirichlet form
separately for each $N$:
$$D^N_{x,y}=\nu^N\lbrak\hiukkas N
\sum_{k=1}^{\eta(x)/\shiukkas N}\bigl[
{g^N(\eta^{k\shiukkas N,x,y})}-{g^N(\eta)} \bigr]^2
\rbrak.
$$
But the argument of Lemma 3.12 works again to give
$$D^N_{x,y}\le C\,\e^2 $$
for all $N$, $\e$, and
$x$, $y\in\LaNe$. As in section 3.3.2,
this bound allows us to deduce (68) by proving
$$\lim_{L\to\infty}\limsup_{\e\to 0}\limsup_{N\to\infty}
\sup_{\mu} \mu\lbrak
|S_{\La_L}(\eta)-S_{\La_L'}(\eta)|^2\rbrak
=0,
\tag 72
$$
where the supremum is over the class of probability measures
$\mu$ on $\OO^{(\shiukkas N)}_{\La_L\cup\La_L'}$
that satisfy $\mu\ll\nu^N$,
$D^N_{x,y}$ $\le C\e^2$,
and $\mu\{\eta^k(x)\}\le C_k$ for
all $x,y\in \La_L\cup\La_L'$.
For each $L$, let $\mu_L$ be a measure that
satisfies
$$\mu_L\lbrak
|S_{\La_L}(\eta)-S_{\La_L'}(\eta)|^2\rbrak
= \limsup_{\e\to 0}\limsup_{N\to\infty}
\sup_{\mu}\mu\lbrak
|S_{\La_L}(\eta)-S_{\La_L'}(\eta)|^2\rbrak.
$$
For the case $\varkappa=0$
the computation done in (70) shows that
$$\aligned
\biggl| \int_a^b \mubar^N\{\eta: \eta_\Ga\ge q\} -
\mubar^N\{\eta: \eta_\Ga\ge q^{u,x,y}\}\,du \biggr|
\le\,C\,\e +O(\hiukkas N).
\endaligned
$$
Thus, as we let first $N\to\infty$ and then $\e\to 0$,
Lemma 3.10 holds for $\mu_L$ and we can complete the proof
of the two-block estimate for the case $\varkappa=0$
as was done in section
3.3.2.
The pattern is clear by now so we leave the details of
the two-block estimate for the case $\varkappa>0$ to
the reader.
\hbox{}
After establishing the local equilibrium we have the
analogue of (57) for the particle model:
$$\aligned
&\lim_{\e\to 0}\limsup_{N\to\infty}
P^N\lbrakk \sup_{0\le t\le T}\biggl|
\a^N_t(\phi)-\a^N_0(\phi)\\
&\qquad\qquad -2\,\kerroin \,\int_0^t
N^{-d}\sum_{x\in\bzdN}
A\phi(\tfrac xN)\,\bigl(\varkappa \,\a^N_s(\chi_{\e,x/N})+
\bigl[\a^N_s(\chi_{\e,x/N})
\bigr]^2\bigr)
\,ds\biggr| \ge\delta\rbrakk =0.
\endaligned
$$
The remaining technical steps follow as before, and
with this we consider Theorem 3 proved.
\vskip .4in
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\enddocument
**