text:
\section{Introduction}

\subsection{Main result}
We consider the stochastic heat equation (SHE) on the torus $\mathbb{T}^d$:
\begin{equation}\label{e.she}
\partial_t {\mathcal U}=\tfrac12\Delta {\mathcal U}+{\mathcal U}\xi,
\end{equation}
where $\xi$ is a generalized Gaussian random field over $\mathbb{R}\times\mathbb{T}^d$ with the covariance function
\[
\mathbf{E}[\xi(t,x)\xi(s,y)]=\delta(t-s)R(x-y).
\]
Here $\mathbb{T}^d$ is the unit torus.
We assume $\xi$ is built on the probability space $(\Omega,\mathcal{F},\mathbf{P})$,
and consider two cases in the paper: (i) $R:\mathbb{T}^d\to\mathbb{R}_{\geq0}$ is a
smooth bounded function and $d\geq 1$; (ii) $R(\cdot)=\delta(\cdot)$
is the Dirac function and $d=1$, i.e.,  $\xi$ is a $1+1$ spacetime
white noise. The product between ${\mathcal U}$ and $\xi$ is interpreted in the
It\^o's sense. Without loss of generality, we assume
\begin{equation}
\label{hr0}
\int_{\mathbb{T}^d} R(x)dx=1
\end{equation}
 in case (i).  We assume that the initial condition is an arbitrary probability measure $\nu$ on $\mathbb{T}^d$.

Here is the main result of the paper:
\begin{theorem}\label{t.mainth}
There exists $\gamma,\sigma>0$, which only depends on $R,d$ and is defined in \eqref{e.defgamma} and \eqref{e.defsigma} below, such that for any $x\in\mathbb{T}^d$,
\[
\frac{\log \mathcal{U}(t,x)+\gamma t}{\sqrt{t}}\Rightarrow N(0,\sigma^2), \quad\quad \mbox{ as } t\to\infty.
\]
\end{theorem}

\subsection{Context}

Define $h(t,x)=\log \mathcal{U}(t,x)$, then $h$ formally satisfies the nonlinear SPDE
\begin{equation}\label{e.kpz}
\partial_t h=\tfrac12\Delta h+\tfrac12|\nabla h|^2+\xi,
\end{equation}
which is the KPZ equation and a default model for random surface growth \cite{kardar1986dynamic}. The study of the equation in $d=1$ with a spacetime white noise has witnessed important progress in recent years, including making sense of the equation \cite{gubinelli2015paracontrolled,gubinelli2017kpz,hairer2013solving,hairer2014theory,kupiainen2016renormalization}, studying universal behaviors  \cite{amir2011probability,balazs2011fluctuation,borodin,borodin1,duncan,kpzfix,quastelkpz,spohn,spohn1,virag}, etc. For a detailed review of singular SPDEs and the KPZ universality class, we refer to \cite{corwin2012kardar,haoshen,quastelintroduction,quastel2015one} and the references therein. For the equation \eqref{e.kpz} and other relevant  models in statistical physics including directed polymers and exclusion processes, the study of the large scale  behaviors is mostly focused on an infinite line and the limiting object is the so-called KPZ fixed point. It is expected that the   result on the torus is different and Gaussian fluctuations prevail in large time. Indeed, Theorem~\ref{t.mainth} shows that as $t\to\infty$,
\begin{equation}\label{e.conkpz}
\frac{h(t,x)+\gamma t}{\sqrt{t}}\Rightarrow N(0,\sigma^2).
\end{equation}
The result also holds in $d\geq 2$ with a smooth noise.

The recent work on periodic  totally asymmetric simple exclusion process \cite{zhipeng1,zhipeng2,zhipeng} considered the case where the size of the torus is scaled with time, and studied the fluctuations of the integrated particle current for several special initial data. A transition from the Tracy-Widom distribution in the small time limit to the Gaussian distribution in the large time limit was derived in \cite[Theorem 1.5]{zhipeng}. In our problem, the size of the torus is fixed, so it corresponds to their super-relaxation time scale

There are recent results deriving the Edwards-Wilkinson limit of the KPZ equation in $d\geq2$ in a  weak disorder regime, with the equation posed on the whole space and driven by a smooth noise, see \cite{CRS18,chatterjee2018constructing,cometskpz,cosco,kpz1,kpz2,nikos,magnen2017diffusive}. Although the limit is also of Gaussian distribution, the mixing mechanism is rather different from our setting, and it is ultimately the result of the weak disorder and is related to the diffusive behaviors of directed polymers in high dimensions with high temperatures.

Our proof of \eqref{e.conkpz} goes through the Hopf-Cole transformation so that we directly study $\mathcal{U}=\exp(h)$   and avoid dealing with the small scale singularity appearing in \eqref{e.kpz}. The recent work \cite{perkowski} studied the generators of the stochastic Burgers equation, the singular SPDE satisfied by $\nabla h(t,x)=\tfrac{\nabla \mathcal{U}(t,x)}{\mathcal{U}(t,x)}$, and their result implies the geometric ergodicity of $\nabla h$. Here we choose to work on a different quantity, the endpoint distribution of the directed polymer in random environment:
\[
u(t,x)=\frac{\mathcal{U}(t,x)}{\int_{\mathbb{T}^d} \mathcal{U}(t,x')dx'}.
\]
One of the main ingredients in our proof is the geometric ergodicity of the Markov process $\{u(t,\cdot)\}_{t\geq0}$, see Theorem~\ref{t.geomeEr}. The proof is inspired by the classical work of Sinai \cite{sinai}, which is actually on the Burgers equation.  Another key quantity is the intersection time of two independent polymer paths, similar to the so-called \emph{replica overlap}, which is used to distinguished different temperature regimes and  plays a crucial role in the study of thermodynamic limits of directed polymers, and in our case  corresponds to \[
\int_0^t \int_{\mathbb{T}^{2d}} R(x-y)u(s,x)u(s,y)dxdyds.
\]
It is an additive functional of $\{u(t,\cdot)\}_{t\geq0}$, and the idea is to utilize the aforementioned geometric ergodicity to construct a corrector and extract a martingale from the above term. The geometric ergodicity is then used to drive a martingale central limit theorem which eventually leads to \eqref{e.conkpz}. Another closely related work is \cite{rosati}, where a one force - one solution principle (modulo constants) for the KPZ equation was established.

For a more detailed discussion on directed polymers in random environment, we refer to the book \cite{comets2017directed} and the references therein.  For the endpoint distribution we are interested in, our approach is more related to that of  \cite{bakhtin1,bates2016endpoint,mukherjee}. For the polymer lying on the torus, we refer to the physics literature \cite{brunet,brunet1,brunet2} for some relevant discussions.

\subsection{Organization of the paper}

The   paper is organized as follows. In Section~\ref{s.2}, we sketch
the main ingredients used in the proof, including  the geometric
ergodicity of $\{u(t,\cdot)\}_{t\geq0}$ and  a martingale
decomposition for the free energy of the directed polymer. In
Section~\ref{s.3}, we derive a nonlocal SPDE satisfied by
$u$. Section~\ref{s.polymer} is devoted to proving the geometric
ergodicity of $\{u(t,\cdot)\}_{t\geq0}$ by following the approach of
\cite{sinai}. In Section~\ref{s.5}, we prove the main theorem through
constructing the corrector and applying the martingale central limit
theorem. In the appendix we recall some facts concerning the
Fortet-Mourier metric on the space of probability measures on polish
metric spaces, see Section \ref{s.FM} and recall some fairly standard facts
concerning  the stochastic heat equation,
see Section \ref{s.SHE}.

\subsection*{Acknowledgements}
Y.G. was partially supported by the NSF through DMS-1907928
and the Center for Nonlinear Analysis of CMU. T.K. acknowledges the support of NCN grant 2020/37/B/ST1/00426.

\section{Ideas of the proof}

\label{s.2}

\subsection{Endpoint distribution of directed polymer}

To study the SHE \eqref{e.she}, a closely related object is the directed polymer in random environment. We introduce some notations. For any $y\in \mathbb{T}^d$, Let $\mathcal{U}(t,x;y)$ be  the solution to
\begin{equation}\label{e.defGreen}
\begin{aligned}
&\partial_t \mathcal{U}(t,x;y)=\tfrac12\Delta_x \mathcal{U}(t,x;y)+\mathcal{U}(t,x;y)\xi(t,x),\quad\quad t>0,\\
&\mathcal{U}(0,x;y)=\delta(x-y).
\end{aligned}
\end{equation}
In other words, $\mathcal{U}$ is the Green's function of the SHE.
For any probability measure $\nu$, denote the solution to  \eqref{e.she} with the initial data $\nu(dx)$ by $\mathcal{U}(t,x;\nu)$, then with the Green's function, we can write
\begin{equation}\label{e.UnuUy}
\mathcal{U}(t,x;\nu)=\int_{\mathbb{T}^d} \mathcal{U}(t,x;y)\nu(dy).
\end{equation}
If $\nu(dx)=f(x)dx$ for some $0\leq f\in L^1(\mathbb{T}^d)$, we   write
instead $\mathcal{U}(t,x;f)=\mathcal{U}(t,x;\nu)$. For a directed polymer of length $t$, with the starting point distributed according to $\nu$, its endpoint distribution is given by
\begin{equation}\label{e.defPo}
u(t,x;\nu)=\frac{\mathcal{U}(t,x;\nu)}{\int_{\mathbb{T}^d} \mathcal{U}(t,x';\nu)dx'}.
\end{equation}
Throughout the paper, we will omit the dependence of either $\mathcal{U}$, or
$u$ on the initial data $\nu$ when there is no danger of
confusion. Let  $\mathcal{M}_1(\mathbb{T}^d)$ be the space of probability
measures on $\mathbb{T}^d$, then for $t\geq 0$ and each realization of the
noise $\xi$, we can view $u(t)=u(t,\cdot)$ as the density of some
probability measure on $\mathbb{T}^d$ and sometimes, with some abuse of notations, we write $u(t)\in \mathcal{M}_1(\mathbb{T}^d)$.

\begin{remark}\label{r.fkpolymer}
In the case  when $R$ is  smooth, by the Feynman-Kac formula we have
\[
\mathcal{U}(t,x;y)=\mathbb{E}_x[e^{\int_0^t\xi(t-s,B_s)ds-\frac12R(0)t}\delta(B_t-y)],
\]
where $\mathbb{E}_x$ is the expectation with respect to the Wiener  measure $\mathbb P_x$  on $ C([0,+\infty);\mathbb{R}^d)$, with the paths starting at $x$,
i.e. $\mathbb P_x[B_0=x]=1$.
Through a change of variable and the time reversal
property of a Brownian bridge, we have
\[
\begin{aligned}
\mathcal{U}(t,x;y)=&\mathbb{E}_x[e^{\int_0^t\xi(s,B_{t-s})ds-\frac12R(0)t}\delta(B_t-y)]\\
=&\mathbb{E}_y[e^{\int_0^t\xi(s,B_{s})ds-\frac12R(0)t}\delta(B_t-x)],
\end{aligned}
\]
which implies
\[
\mathcal{U}(t,x;\nu)=\int_{\mathbb{T}^d} \nu(dy) \mathbb{E}_y[e^{\int_0^t \xi(s,B_s)ds-\frac12R(0)t}\delta(B_t-x)].
\]
Therefore, one can write
\[
u(t,x;\nu)=\frac{\int_{\mathbb{T}^d} \nu(dy) \mathbb{E}_y[e^{\int_0^t \xi(s,B_s)ds}\delta(B_t-x)]}{\int_{\mathbb{T}^d} \nu(dy) \mathbb{E}_y[e^{\int_0^t \xi(s,B_s)ds } ]},
\]
which is usually how the endpoint distribution of the directed polymer is defined.
\end{remark}

Sometimes to
indicate the dependence of $\xi$ on the random event we shall write
\[
\xi=\big\{\xi(t,x;\omega)\big\}_{(t,x;\omega)\in \mathbb{R}\times\mathbb{T}^d\times\Omega}.
\]
In
addition, since the noise is space-time homogeneous, we
may and shall assume that there is an additive group
$\{\theta_{t,x}\}_{(t,x)\in  \mathbb{R}\times\mathbb{T}^d}$ of ${\mathcal F}/{\mathcal F}$
measurable maps $ \theta_{t,x}:\Omega\to\Omega$ such that $\mathbf{P}\circ
\theta_{t,x}^{-1}=\mathbf{P}$ for all $(t,x)\in \mathbb{R}\times\mathbb{T}^d$ and
\begin{equation}
\label{012903-21}
\xi(t,x; \theta_{s,y}(\omega))=\xi(t+s,x+y; \omega), \quad (t,x),(s,y)\in \mathbb{R}\times\mathbb{T}^d,\,\omega\in\Omega.
\end{equation}
  As a consequence of the Markov property of $\{{\mathcal
    U}(t)\}_{t\geq0}$ and the white-in-time nature of the noise, we conclue the following.
\begin{lemma}\label{l.markov}
$\{u(t)\}_{t\geq0}$ is   a Markov process taking values in $\mathcal{M}_1(\mathbb{T}^d)$.
\end{lemma}
The proof of the lemma can be found in Section \ref{MPd} of the Appendix.

One of the main technical parts of the paper is to prove the geometric ergodicity of the Markov process $\{u(t;\nu)\}_{t\geq 0}$. To state the result, we introduce some notations.

For any $p\ge1$ we let
$$
\mathcal{M}_1^{(p)}(\mathbb{T}^d):=\Big\{\nu\in
\mathcal{M}_1(\mathbb{T}^d)
\nu(dx)=f(x)
dx,\,f\in L^p(\mathbb{T}^d)\Big\},
$$
equipped with the relative topology from $L^p(\mathbb{T}^d)$. We
shall identify $\mathcal{M}_1^{(p)}(\mathbb{T}^d)$ with the appropriate subset
of densities, i.e.
\begin{equation}
\label{012403-21}
D^p(\mathbb{T}^d)=D(\mathbb{T}^d)\cap L^p(\mathbb{T}^d)\quad \mbox{and }D^\infty(\mathbb{T}^d)=D(\mathbb{T}^d)\cap C(\mathbb{T}^d).
\end{equation}
Here $ D(\mathbb{T}^d)$ is the set of densities w.r.t. the
 Lebesgue measure on $\mathbb{T}^d$. We will consider the Fortet-Mourier
 metric on $\mathcal{M}_1(\mathbb{T}^d)$, see Section~\ref{s.FM} for a brief
 recollection of the facts concerning this metric. For any $F: \mathcal{M}_1(\mathbb{T}^d)\to \mathbb{R}$, we define
\begin{equation}
\label{FLip}
\|F\|_{{\rm Lip}}:=\|F\|_\infty+ \sup\limits_{\mu\not=\nu,\mu,\nu\in \mathcal{M}_1(\mathbb{T}^d)}\frac{|F(\mu)-F(\nu)|}{{\rm d}_{\mathrm{TV}}(\mu,\nu)}.
\end{equation}
Denote by ${\rm Lip}(\mathcal{M}_1(\mathbb{T}^d))$ the space of all
functions $F$ for which $\|F\|_{{\rm Lip}}<+\infty$. Let $\mathcal{B}(\mathcal{M}_1(\mathbb{T}^d))$ and $\mathcal{B}(D^p(\mathbb{T}^d))$ be the Borel $\sigma-$algebra of $\mathcal{M}_1(\mathbb{T}^d)$ and $D^p(\mathbb{T}^d)$ respectively.

Define the
transition
probability densities
\begin{equation}
\label{PtA}
{\mathcal P}_t(\nu,A):=\mathbf{E} 1_A(u(t;\nu)),\quad A\in {\mathcal B}({\mathcal
  M}_1 (\mathbb{T}^d)),\,\nu\in {\mathcal
  M}_1 (\mathbb{T}^d).
\end{equation}
By the property of SHE, for any $t>0$, $u(t,\cdot;\nu)$ is a continuous function almost surely,
thus, for any $t>0$, $\mathcal{P}_t(\nu,\cdot)$ is actually supported on $D^p(\mathbb{T}^d)$ for any $p\in[1,+\infty]$, and we can consider transition probabilities ${\mathcal P}_t(\nu,\cdot)$
on ${\mathcal B}(D^p(\mathbb{T}^d))$.

Define the transition probability operator by
\begin{equation}
\label{PtF}
{\mathcal P}_tF(\nu):=\int_{{\mathcal M}_1(\mathbb{T}^d)}F(v){\mathcal P}_t(\nu,dv),\quad F\in
B_b({\mathcal M}_1 (\mathbb{T}^d)),\,\nu\in  {\mathcal M}_1 (\mathbb{T}^d).
\end{equation}
We shall also consider a semigroup $\{{\mathcal P}_t\}_{t\geq0}$ on
$B_b(D_p(\mathbb{T}^d))$ for any $p\in[1,+\infty]$.

\begin{theorem}
\label{t.geomeEr}
There exist a Borel probability measure $\pi_\infty$ on ${\mathcal M}_1(\mathbb{T}^d)$ and
constants $C,\lambda>0$, which only depend on the
covariance function $R(\cdot)$ and dimension $d$  such that
\begin{itemize}
\item[(i)]
\begin{equation}
\label{032303-21}
\pi_\infty\big( D^p(\mathbb{T}^d)\big)=1,
\quad p\in[1,+\infty],
\end{equation}
\item[(ii)] for any  $ F\in
{\rm Lip}({\mathcal M}_1(\mathbb{T}^d)) $ we have
\begin{equation}
\label{011203-21}
\left\|{\mathcal
    P}_tF-\int_{{\mathcal M}_1(\mathbb{T}^d)}F(u)\pi_\infty(du)\right\|_\infty \leq
Ce^{-\lambda t}\|F\|_{{\rm Lip}},\quad  t\geq0   .
\end{equation}
The above result also holds for $D^p(\mathbb{T}^d)$ in place of ${\mathcal M}_1(\mathbb{T}^d)$ for any $p\in[1,+\infty]$.
\end{itemize}
\end{theorem}
The proof of the theorem is given in Section~\ref{s.polymer}.

\subsection{Martingale decomposition}

We will first prove Theorem~\ref{t.mainth} for the partition of the
directed polymer, defined as
\begin{equation}\label{e.defZ}
Z_t=\int_{\mathbb{T}^d} \mathcal{U}(t,x)dx.
\end{equation}
By the mild formulation of SHE, one can write
\begin{equation}\label{e.mild}
\mathcal{U}(t,x)=\int_{\mathbb{T}^d} p_t(x-y)\nu(dy)+\int_0^t\int_{\mathbb{T}^d} p_{t-s}(x-y)\mathcal{U}(s,y)\xi(s,y)dyds,
\end{equation}
where
\begin{equation}\label{e.defptx}
p_t(x)=\frac{1}{(2\pi t)^{d/2}}\sum_{n\in\mathbb{Z}^d}\exp\left\{-\frac{|x+n|^2}{2t}\right\}
\end{equation}
is the heat kernel on $\mathbb{T}^d$. Thus, recalling that $\nu$ is a probability measure, we have
\begin{equation}\label{e.4211}
Z_t= 1+\int_0^t\int_{\mathbb{T}^d}\mathcal{U}(s,y)\xi(s,y)dyds,
\end{equation}
which is a positive  local martingale, whose quadratic variation
equals
\begin{equation}\label{e.qvZ}
\langle Z\rangle_t=\int_0^t \mathcal{R}\big(\mathcal{U}(s)\big) ds.
\end{equation}
Here for $R\in L^\infty(\mathbb{T}^d)$ we let
\begin{equation}
\label{fR}
\begin{split}
&\mathcal{R}(u,v):=\int_{\mathbb{T}^{2d}}R(x-y)u(x)v(y)dxdy,\\
&
\mathcal{R}(u):=\mathcal{R}(u,u)=\int_{\mathbb{T}^{2d}}R(x-y)u(x)u(y)dxdy,\quad  u,v\in L^1(\mathbb{T}^d).
\end{split}
\end{equation}
If $R(\cdot)=\delta(\cdot)$, the above definition holds for  $u,v\in L^2(\mathbb{T}^d)$.

By Lemma~\ref{l.bdQVZ}, we know that $\mathbf{E}\langle Z\rangle_t^p \leq C(t,p)$, for some
constant $C(t,p)<+\infty$ depending only on $t$ and $p\in[1,+\infty)$.
Hence  $\big(Z_t\big)_{t\ge0}$ is   a martingale.
Define a local martingale $(M_t)_{t\geq0}$ through the SDE
\begin{equation}
dM_t=Z_t^{-1}dZ_t, \quad\quad M_0=0.
\end{equation}
By the It\^o formula
\begin{equation}\label{e.relationZM}
Z_t=\exp\big\{M_t-\tfrac12\langle M\rangle_t\big\}.
\end{equation}
We can write  $M_t$ and its quadratic variation in terms of the endpoint distribution of the directed polymer:
\begin{equation}\label{e.ExpM}
\begin{aligned}
M_t=\int_0^t\int_{\mathbb{T}^d} Z_s^{-1}\mathcal{U}(s,y)\xi(s,y)dyds=\int_0^t\int_{\mathbb{T}^d} u(s,y)\xi(s,y)dyds,
\end{aligned}
\end{equation}
and
\begin{equation}\label{e.qvM}
\langle M\rangle_t
=\int_0^t \mathcal{R}(u(s))ds.
\end{equation}
The following lemma shows that $M$ is a square-integrable martingale.
\begin{lemma}
For any $t>0$ and $p\in[1,+\infty)$ we have $\mathbf{E}[\langle M\rangle_t^p]\leq C(t,p,d,R)$.
\end{lemma}
\begin{proof}
For the case of $R$ being bounded, we have $\langle M\rangle_t\leq
t\|R\|_{L^\infty(\mathbb{T}^d)}$. For the case of $R(\cdot)=\delta(\cdot)$,
we have $\langle M\rangle_t= \int_0^t  \|u(s)\|_{L^2(\mathbb{T}^d)}^2 ds$.  Applying
the triangle inequality we further derive
\[
\mathbf{E}[\langle M\rangle_t^p]^{1/p} \leq C\int_0^t \mathbf{E}[\|u(s)\|_{L^2(\mathbb{T}^d)}^{2p}]^{1/p}ds,
\]then the lemma is a direct consequence of Lemma~\ref{l.mmbdu}.
\end{proof}

Then we can write
\begin{equation}\label{e.delogZt}
\begin{aligned}
\log Z_t=&M_t-\tfrac12\langle M\rangle_t\\
=&\int_0^t\int_{\mathbb{T}^d} u(s,y)\xi(s,y)dyds-\tfrac12\int_0^t \mathcal{R}(u(s))ds.
\end{aligned}
\end{equation}
The term $\int_0^t \mathcal{R}(u(s))ds$ is an additive functional of the
Markov process $(u(s))_{s\geq0}$, which will be the key object to
study in the paper.
Define
\begin{equation}\label{e.defgamma}
\gamma:=\tfrac12\int_{\mathcal{M}_1(\mathbb{T}^d)} \mathcal{R}(u)\pi_\infty(du),
\end{equation}
where $\pi_\infty$ is the  unique invariant measure for the process (Theorem  \ref{t.geomeEr}),
and \begin{equation}\label{e.cenR}
\tilde{\mathcal{R}}=\mathcal{R}-2\gamma.
\end{equation}
It will become clear later (see Lemma~\ref{l.nondege1} below) that $\gamma\in(0,+\infty)$.

Now we write
\begin{equation}\label{e.delogZt1}
\log Z_t+\gamma t=\int_0^t\int_{\mathbb{T}^d} u(s,y)\xi(s,y)dyds-\tfrac12\int_0^t \tilde{\mathcal{R}}(u(s))ds.
\end{equation}
The idea is to utilize Theorem~\ref{t.geomeEr} to solve the Poisson equation associated with $\tilde{\mathcal{R}}$ and extract the martingale part from $\int_0^t \tilde{\mathcal{R}}(u(s))ds$, which combines with the martingale $M_t$ to converge to the limiting Gaussian random variable, after a rescaling:
\begin{theorem}\label{t.partitionfunction}
As $t\to\infty$, we have
\[
\frac{\log Z_t+\gamma t}{\sqrt{t}}\Rightarrow N(0,\sigma^2)
\]
In addition, we have
\begin{equation}
\label{012204-21}
\sigma^2\ge\int_{\mathbb{T}^d}R(x)dx.
\end{equation}
\end{theorem}

\begin{remark}[Process-level convergence]
Combining Theorems~\ref{t.geomeEr} and \ref{t.partitionfunction}, we actually have the following picture: as $t\to\infty$,  \begin{equation}\label{e.4271}
\log u(t,x)=\log \tfrac{\mathcal{U}(t,x)}{Z_t}=h(t,x)-\log Z_t\Rightarrow \log \rho(x)
\end{equation}
in distribution in $C(\mathbb{T}^d)$, where the limit $\rho$ is sampled from the invariant measure $\pi_\infty$. Therefore, in large time, the random height function $h$ is approximately $\log \rho$ shifted by the random constant $\log Z_t$, and this random constant is roughly  Gaussian distributed after a centering and rescaling.
 We also have
\[
h(t,x)-h(t,0)\Rightarrow \log \rho(x)-\log \rho(0)
\]
in distribution in $C(\mathbb{T}^d)$.

According to the results of  \cite{funaki}, see Theorems 1.1 and
  2.1,    in the case of $1+1$ spacetime white noise, for each $t\geq0$ and in statinarity
  the law of
  $\left(h(t,x)-h(t,0)\right)_{x\in\mathbb{T}}$ over $C(\mathbb{T})$ coincides with  that of
  $(B(x))_{x\in\mathbb{T}}$ - a
 Brownian bridge  satisfying $B(0)=B(1)=0$.
The invariant measure $\pi_\infty$ in this case is given by the law of
$$
\rho(x)=\frac{e^{B(x)}}{\int_{\mathbb{T}}e^{B(x')}dx'}, \quad x\in\mathbb{T}
$$
over $D^\infty(\mathbb{T})$,
  see also \cite{bertini,gubinelli2017kpz}.
\end{remark}

\section{SPDE for endpoint distribution of directed polymer}

\label{s.3}

Define the Fourier coefficients of the spatial covariance of the noise
\begin{equation}\label{e.defrk}
 \hat r_k= \int_{\mathbb{T}^d}R(x)e^{-i 2\pi k\cdot x}dx,\quad k\in\mathbb{Z}^d.
\end{equation}
They are non-negative.

We can write formally that
\begin{equation}
\label{xtx}
\xi(t,x)=\sum_{k\in \mathbb{Z}^d} e_k(x)\dot{w}_k(t), \quad\quad e_k(x)=\sqrt{\hat{r}_k}e^{i2\pi k\cdot x},
\end{equation}
where $(w_k)_{k\in\mathbb{Z}^d}$ are   one dimensional complex-valued
zero mean Wiener processes satisfying
\begin{equation}
\label{022204-21}
w_k^\star(t)=w_{-k}(t),\quad \mathbf{E}[w_k(t)
w_\ell^\star(s)]=\delta_{k,\ell}\min(t,s),\quad k,\ell\in\mathbb{Z}^d,\,t,s\ge0.
\end{equation}
Here $\delta_{k,\ell}$ is the Kronecker symbol.  We also define
\begin{equation}\label{e.defW}
W(t,x)=\sum_{k\in\mathbb{Z}^d} e_k(x)w_k(t)
\end{equation}
and write
\[
W(t)=W(t,\cdot),\quad\quad dW(t,x)=\xi(t,x)dt.
\]
We formulate the following assumption:
\begin{assumption}\label{a.smooth}
 There exists $N>0$ such that $\hat{r}_k=0$ for $  |k|>N$.
\end{assumption}
Under the above assumption, the summations in \eqref{xtx} and
\eqref{e.defW} are finite.

The main result of this section is to derive an SPDE satisfied by the
endpoint distribution of the directed polymer. Recall that
$u(t,x;\nu)$ has been defined in \eqref{e.defPo} for any $\nu\in\mathcal{M}_1(\mathbb{T}^d)$.
In this section  the initial data is assumed to be of the form
$\nu(dx)=v(x)dx$, where $v\in D(\mathbb{T}^d)\cap C^\infty(\mathbb{T}^d)$. To simplify the notation, we
omit the initial data and write   $u(t)=u(t,\cdot)$ and $\mathcal{U}(t)=\mathcal{U}(t,\cdot)$.
\begin{proposition}\label{p.spde}
Suppose that
Assumption~\ref{a.smooth} is in force. Furthermore, assume  that  the initial data $v\in  D(\mathbb{T}^d)\cap C^\infty(\mathbb{T}^d)$ and $u\in
C([0,\infty),H^k(\mathbb{T}^d))$, for any $k\geq1$,  is a strong solution to
\begin{equation}\label{e.spde}
\begin{aligned}
du(t)=&\Big\{\tfrac12\Delta u(t) +u(t) \int_{\mathbb{T}^d} u(t,x)R\star u(t,x)dx -u(t)R\star u(t)\Big\}dt\\
&+u(t)dW(t)-u(t)\int_{\mathbb{T}^d}u(t,x)dW(t,x),\quad u(0)=v.
\end{aligned}
\end{equation}
\end{proposition}

\begin{proof}
Under Assumption~\ref{a.smooth} and by the fact that $v\in C^\infty(\mathbb{T}^d)$, we have
\begin{equation}\label{e.4131a}
\mathcal{U}(t)=v+\int_0^t \tfrac12\Delta \mathcal{U}(s)ds+\int_0^t \mathcal{U}(s)dW(s).
\end{equation}
By \eqref{e.4211}, we have (which can also be obtained by integrating the above equation in $x$)
\begin{equation}\label{e.4131b}
Z_t= \int_{\mathbb{T}^d}\mathcal{U}(t,x)dx=1+\int_0^t \int_{\mathbb{T}^d}\mathcal{U}(s,x) dW(s,x).
\end{equation}
Using the above two equations, we have
\begin{equation}\label{e.4131}
\begin{aligned}
&\langle Z\rangle_t=\int_0^t \mathcal{R}\big(\mathcal{U}(s)\big)ds,\\
&\langle \mathcal{U},Z\rangle_t=\int_0^t\mathcal{U}(s) \mathcal{R}\star \mathcal{U}(s)ds.
\end{aligned}
\end{equation}
By \eqref{e.negammZ}, we know that  $(Z_t)_{t\geq0}$ is continuous and positive almost surely. Then by It\^o's formula, we have
\[
\begin{aligned}
du(t)=&d[\mathcal{U}(t)Z_t^{-1}]\\
=&Z_t^{-1}d\mathcal{U}(t)-\mathcal{U}(t)Z_t^{-2}dZ_t+\mathcal{U}(t)Z_t^{-3}d\langle Z\rangle_t-Z_t^{-2}d\langle \mathcal{U},Z\rangle_t.
\end{aligned}
\]
Invoking \eqref{e.4131a} - \eqref{e.4131}, the proof is complete.
\end{proof}

We introduce the following operators:
\begin{equation}\label{e.defAB}
\begin{aligned}
&\mathscr{A}(f):=\tfrac12\Delta f+f \langle f,R\star f\rangle_{L^2(\mathbb{T}^d)}-fR\star f, \quad\quad f\in H^2(\mathbb{T}^d),\\
&\mathscr{B}_k(f): =fe_k-f\langle f,e_{-k}\rangle_{L^2(\mathbb{T}^d)},\quad\quad f\in L^2(\mathbb{T}^d).
\end{aligned}
\end{equation}
Note that $e_k^*=e_{-k}$, so $\langle f,e_{-k}\rangle_{L^2(\mathbb{T}^d)}=\int_{\mathbb{T}^d} f(x)e_k(x)dx$, and for real-valued $f$, we have
\[
\mathscr{B}_k(f)^*=\mathscr{B}_{-k}(f).
\] With the above notations, the equation \eqref{e.spde} can be written in a more compact form:
\[
du(t)=\mathscr{A}(u(t))dt+\sum_{k\in\mathbb{Z}^d } \mathscr{B}_k(u(t))dw_k(t),
\]
and we have the following lemma:
\begin{lemma}\label{l.bdspde}
Under Assumption~\ref{a.smooth} and further assume $v\in D(\mathbb{T}^d)\cap C^\infty(\mathbb{T}^d)$, then for any $T>0$ and $p\in[1,+\infty)$, there exists $C=C(T,p,R,d)$ such that
\[
\begin{aligned}
&\sup_{t\in[0,T]}\mathbf{E}\big[\|\mathscr{A}(u(t))\|_{L^2(\mathbb{T}^d)}^p\big] \leq C,\\
&\sum_{k\in\mathbb{Z}^d } \sup_{t\in [0,T]}\mathbf{E}\big[\|\mathscr{B}_k(u(t))\|_{L^2(\mathbb{T}^d)}^p\big] \leq C.
\end{aligned}
\]
\end{lemma}
\begin{proof}
Since $u(t)=\mathcal{U}(t)Z_t^{-1}$, the result is standard and  follows from
properties of SHE combined with \eqref{e.negammZ}. \end{proof}

\section{Geometric ergodicity of directed polymer on torus}
\label{s.polymer}

In this section, we study the endpoint distribution of the directed polymer, defined in \eqref{e.defPo}, and prove Theorem~\ref{t.geomeEr}.

\subsection{Factorization }

Recall that we consider the solution to SHE with the initial data $\nu\in \mathcal{M}_1(\mathbb{T}^d)$:
\begin{equation}\label{e.SHEf}
\partial_t \mathcal{U}=\tfrac12\Delta\mathcal{U}+\mathcal{U}\xi,\quad\quad \mathcal{U}(0,dx)=\nu(dx).
\end{equation}

For any $t>s$ and $x,y\in\mathbb{T}^d$, define $\mathcal{Z}_{t,s}^\omega(x,y)$ as the solution to
\begin{equation}\label{e.defZts}
\begin{aligned}
&\partial_t \mathcal{Z}_{t,s}^\omega(x,y)=\tfrac12\Delta_x \mathcal{Z}_{t,s}^\omega(x,y)+\mathcal{Z}_{t,s}^\omega(x,y)\xi(t,x), \quad\quad t>s,\\
&\mathcal{Z}_{s,s}^\omega(x,y)=\delta(x-y).
\end{aligned}
\end{equation}
In other words, $\mathcal{Z}_{t,s}^\omega(x,y)$ is the propagator of the SHE from $(s,y)$ to $(t,x)$. We will keep its dependence on $\omega$ throughout this section. In light of \eqref{012903-21}, we have
\begin{equation}\label{022903-21}
\mathcal{Z}_{t,s}^{\theta_{r,z}(\omega)}
(x,y)=\mathcal{Z}_{t+r,s+r}^{\omega} (x+z,y+z).
\end{equation}
For any set $I\subset \mathbb{R}$, denote the $\sigma$-algebra
\begin{equation}
\label{cF}
\mathcal{F}_{I}=\sigma\{\xi(t,\cdot): t\in I\}.
\end{equation}
We shall also write
\begin{equation}
\label{cFt}
{\mathcal F}_t:={\mathcal F}_{[0,t]},\quad t\ge0.
\end{equation}
Note that the random variable   $\mathcal{Z}_{t,s}^\omega$
is  ${\mathcal F}_{[s,t]}$-measurable.

Fix $t>1$.
We have
\begin{equation}\label{e.3111}
\mathcal{U}(t,x;\nu)=\int_{\mathbb{T}^d} \mathcal{Z}_{t,t-1}^\omega(x,y_1) \mathcal{U}(t-1,y_1;\nu) dy_1.
\end{equation}

Let
\begin{equation}\label{e.defN}
N(t):=[t]+1.
\end{equation}
Sometimes, when it leads to no confusion, for abbreviation we shall write $N=N(t)$. By iterating \eqref{e.3111}, we further derive that
\begin{equation}\label{e.3112}
\begin{aligned}
\mathcal{U}(t,x;\nu)=\int_{\mathbb{T}^{Nd}}& \mathcal{Z}_{t,t-1}^\omega (x,y_1) \left(\prod_{j=1}^{N-2} \mathcal{Z}^\omega_{t-j,t-j-1}(y_j ,y_{j+1} ) \right)\\
&\times \mathcal{Z}^\omega_{t-N+1,0}(y_{N-1} ,y_{N} )
dy_1\ldots dy_{N-1}\nu(dy_N ).
\end{aligned}
\end{equation}
With the  convention of $
y_0=x$,
we can write
\begin{equation}\label{e.31121}
\begin{aligned}
\mathcal{U}(t,x;\nu)=  \int_{\mathbb{T}^{Nd}} \prod_{j=0}^{N-1} \mathcal{Z}^\omega_{t-j, {(t-j-1)_+}}(y_j ,y_{j+1} )  dy_1\ldots dy_{N-1}\nu(dy_N ).
\end{aligned}
\end{equation}
 {Here $(x)_+:=\max\{x,0\}$}.

\subsection{Construction of a Markov chain }

\label{sec5.1}

In this section, we construct a  time reversed Markov chain $\{Y_n\}_{n=1}^N$, which
depends on the random environment, and allows
to express the endpoint density  $ u(t,x;\nu)$, see \eqref{e.defPo},
as an average with respect to $Y_1$, see \eqref{e.densityY1} below. It turns out  that, with some
positive probability, the random realizations of the transition probability
density kernel of the chain are uniformly positive,
see \eqref{021903-21} and \eqref{021903-21b}. This, in turn, allows us to
use a coupling argument that leads to the proof of the geometric
ergodicity of the  density  process $\big( u(t;\nu)\big)_{t\ge0}$.

The  {random} Markov chain takes values in $\mathbb{T}^d$ and is
constructed as follows. For a fixed $\omega\in\Omega$, we run the chain backward in time. Let
$\pi_N^\omega(y_N)$ be the  {(random)}  density of $Y_N$ and
$\pi_k^\omega(y_k\,|\,y_{k+1})$ be the  {(random)}  transition density
kernels:
\begin{equation}\label{e.trandensity}
\begin{aligned}
&\pi_1^\omega(y_1\,|\,y_2)=\frac{\mathcal{Z}_{t-1,t-2}^\omega (y_1,y_2)}{\int_{\mathbb{T}^d} \mathcal{Z}_{t-1,t-2}^\omega (z_1,y_2)dz_1}\\
&\pi_k^\omega(y_k\,|\,y_{k+1})=\mathcal{Z} ^\omega_{t-k,(t-k-1)_+}(y_k,y_{k+1})\\
&
\times \frac{\int_{\mathbb{T}^{(k-1)d}} \mathcal{Z}_{t-1,t-2}^\omega (z_1,z_2)\ldots \mathcal{Z} ^\omega_{t-k+1,t-k}(z_{k-1},y_{k}) dz_{1,k-1}}{\int_{\mathbb{T}^{kd}} \mathcal{Z} ^\omega_{t-1,t-2}(z_1,z_2)\ldots \mathcal{Z} ^\omega_{t-k,(t-k-1)_+}(z_k,y_{k+1})dz_{1,k}}
\end{aligned}
\end{equation}
for $k=1,\ldots, N-1$. Here $dz_{1,k}:=dz_1\ldots dz_k$.
Finally we let
\begin{align}
\label{piN}
&
\pi_N^\omega(dy_N):=A^\omega_{N, \nu}  \int_{\mathbb{T}^{(N-1)d}}\left(\prod_{j=1}^{N-2}
  \mathcal{Z}^\omega_{t-j,t-j-1}(z_j ,z_{j+1} ) \right)\\
&
\times \mathcal{Z}^\omega_{t-N+1, 0}(z_{N-1} ,y_N )\nu(dy_N )dz_{1,N-1},\notag
\end{align}
with the (random) constant $A^\omega_{N, \nu}$ given by
\[
A_{N,\nu}^\omega:=\left(\int_{\mathbb{T}^{Nd}}\prod_{j=1}^{N-1}
  \mathcal{Z}^\omega_{t-j,  {(t-j-1)_+}}(z_j ,z_{j+1} )  \nu(dz_N
  )dz_{1,N-1}\right)^{-1}
\]

{Sometimes, we shall write $\pi_N^\omega(dy_N;\nu)$  when we wish to highlight the dependence of
the distribution on the initial data $\nu$. Also, when $\nu(dx)=f(x)dx$ for
$f\in D(\mathbb{T}^d)$ we shall write $\pi_N^\omega(y_N;f)$ as the density of
$\pi_N^\omega(dy_N;\nu)$ w.r.t. the Lebesgue measure}.
Note that $\mathcal{Z}_{t,s}^\omega (x,y)\in \mathcal{F}_{[s,t]}$ for all $x,y\in\mathbb{T}^d$,
cf \eqref{cF}.
Let $\mathbb P_{\pi_N}^\omega$ be the path measure on $\mathbb{T}^{Nd}$
  constructed from the above transition probabilities. It
  is
  the  law of time reversed Markov chain $\{Y_n\}_{n=1}^N$, where
  $$
Y_k(y_1,\ldots,y_N):=y_k, \quad\quad  (y_1,\ldots,y_N)\in
\mathbb{T}^{Nd},\,k\in\{1,\ldots,N\}.
$$
Let $\mathbb{E}_{\pi_N}^\omega$ denote the   expectation with respect to
  this measure. Suppose that $\nu(dx) =f(x) dx$. Then,
the joint density of $(Y_1,\ldots,Y_N)$ under  $\mathbb P_{\pi_N}^\omega$ equals
\begin{align}
\label{012003-21a}
\varphi^\omega_{1,N}(y_1,\ldots,y_N)&=
\pi_1 ^\omega (y_1\,|\,y_2)\ldots \pi_{N-1}^\omega (y_{N-1}\,|\,y_N)\pi_N ^\omega (y_N;f)\\
&
=A_{N, \nu}^\omega\left(\prod_{j=1}^{N-1} \mathcal{Z}^\omega_{t-j, (t-j-1)_+}(y_j ,y_{j+1} ) \right)f(y_N ).\notag
\end{align}
Here $A_{N, \nu}^\omega$ is the (random) normalizing constant.

For a given $2\le k\le N-1$ denote by $\mathbb P_{k}^\omega$ the path
measure on $\mathbb{T}^{Nd}$, corresponding to a time reversed Markov chain
obtained from the transition probability densities
\begin{equation}
\label{042003-21}
p_j^\omega(y_j\,|\,y_{j+1})=
\left\{
\begin{array}{ll}
\pi_j^\omega(y_j\,|\,y_{j+1}),& 1\le j\le k-1,\\
&\\
1,& k\le j\le N-1.
\end{array}
\right.
\end{equation}
and $p_N^\omega(y_N)\equiv 1$. The respective expectation shall be
denoted by $\mathbb{E}_{k}^\omega$.
The joint density of $(Y_1,\ldots,Y_k)$ under  $\mathbb P_{k}^\omega$ equals
\begin{align}
\label{012003-211}
\varphi^\omega_{1,k}(y_1,\ldots,y_k)&=
p_1 ^\omega (y_1\,|\,y_2)\ldots p_{k-1}^\omega (y_{k-1}\,|\,y_{k})\\
&
=A_{k,\mathbbm{1}}^\omega\left(\prod_{j=1}^{k-1} \mathcal{Z}^\omega_{t-j, t-j-1}(y_j ,y_{j+1} ) \right), \notag
\end{align}
with the (random) constant
\begin{equation}\label{012003-212}
A_{k,\mathbbm{1}}^\omega:=\left(\int_{\mathbb{T}^{kd}}\prod_{j=1}^{k-1}
  \mathcal{Z}^\omega_{t-j,t-j-1 }(z_j ,z_{j+1} ) dz_{1,k}\right)^{-1}
\end{equation}

In other words, comparing the Markov chains sampled from the path measures $\mathbb{P}_k^\omega$ and $\mathbb{P}_{\pi_N}^\omega$, the only difference lies in the distribution of $(Y_k,\ldots,Y_N)$. For $\mathbb{P}_k^\omega$, we sample $Y_k,\ldots,Y_N$ independently from the uniform distribution on $\mathbb{T}^d$, while we follow \eqref{e.trandensity} and \eqref{piN} for $\mathbb{P}_{\pi_N}^\omega$.

By  \eqref{e.3112} and \eqref{012003-21a}, one can write
\begin{equation}\label{e.UY1}
\begin{aligned}
\mathcal{U}(t,x;\nu)=&
\big(A^\omega_{N,\nu}\big)^{-1}\int_{\mathbb{T}^{Nd}}\mathcal{Z}_{t,t-1}^\omega(x,y_1)\pi_1^\omega(y_1\,|\,y_2)\ldots \pi_{N-1}^\omega(y_{N-1}\,|\,y_N)\pi_N^\omega(dy_N)dy_{1,N-1}\\
=&\big(A^\omega_{N,\nu}\big)^{-1}\mathbb{E}_{\pi_N}^\omega[\mathcal{Z}^\omega_{t,t-1}(x,Y_1)].
\end{aligned}
\end{equation}
This, in turn implies
\begin{equation}\label{e.densityY1}
  u(t,x;\nu)=\frac{\mathcal{U}(t,x;\nu)}{\int_{\mathbb{T}^d}\mathcal{U}(t,x';\nu)dx'}=\frac{\mathbb{E}_{\pi_N}^\omega[\mathcal{Z}_{t,t-1}^\omega(x,Y_1)]}{\int_{\mathbb{T}^d}\mathbb{E}_{\pi_N}^\omega[\mathcal{Z}_{t,t-1}^\omega (x',Y_1)]dx'}.
\end{equation}
We note that in the above formula, the dependence on $\nu$ is through the distribution $\pi_N^\omega$. It is also clear that the dependence of $u(t,x;\nu)$ on the random environment $\{\xi(s,\cdot):s\leq t-1\}$ is only through the random variable $Y_1$, hence it suffices to study the mixing property of the Markov chain.

\subsection{Properties of the Markov chain}
\begin{lemma}\label{l.ulbd}
For any $p\in[1,+\infty)$, there exists $C>0$ only depending on $p,d,R$ such that
\begin{equation}\label{e.ulbd}
\mathbf{E}\Big[\big(\inf_{x,y\in\mathbb{T}^d}
\mathcal{Z}_{t,t-1}^\omega(x,y)\big)^{-p}\Big]+\mathbf{E}\Big[\big(\sup_{x,y\in\mathbb{T}^d}\mathcal{Z}_{t,t-1}^\omega(x,y)\big)^p\Big]\leq
C,\quad \quad  t>1.
\end{equation}
\end{lemma}

It is a classical result for the stochastic heat equation, and for the convenience of readers, we present its proof in Section~\ref{s.sheOther}.

For any $t>1$ and $\delta>0$, define the event
\begin{equation}\label{e.defB}
{B_t(\delta)}:=\left\{\omega\in\Omega:\,\frac{\inf_{x,y,z\in\mathbb{T}^d}\mathcal{Z}_{t+1,t}^\omega(x,y)\mathcal{Z}_{t,t-1}^\omega (y,z)}{\sup_{x,z\in \mathbb{T}^d}\int_{\mathbb{T}^d} \mathcal{Z}_{t+1,t}^\omega (x,y')\mathcal{Z}_{t,t-1}^\omega (y',z)dy'}>\delta\right\},
\end{equation
which belongs to   $\mathcal{F}_{[t-1,t+1]}$, cf \eqref{cF}, and, {thanks to
\eqref{012903-21},
\begin{equation}\label{e.defBa}
\theta^{-1}_{s,y}\Big(B_t(\delta)\Big)=B_{t+s}(\delta),\quad (s,y)\in\mathbb{R}_+\times\mathbb{T}^d.
\end{equation}}
 Lemma~\ref{l.ulbd} implies
\begin{lemma}\label{l.lowerbd}
There exists a constant $C>0$ only depending on $d,R$ such that
\begin{equation}
\mathbf{P}[B_t(\delta)]\geq 1-C\delta, \quad \mbox{for all $t>1$ and $\delta>0$}.
\end{equation}
\end{lemma}

\begin{proof}
First it is clear that by the time homogeneity of the random
environment, the probability $\mathbf{P}[B_{t }(\delta)]$ does not depend on $t$. To simplify the notation, denote
\[
W_1=\inf_{x,y,z\in\mathbb{T}^d}\mathcal{Z}_{t+1,t}^\omega(x,y)\mathcal{Z}_{t,t-1}^\omega(y,z), \quad\quad W_2=\sup_{x,z\in \mathbb{T}^d}\int_{\mathbb{T}^d} \mathcal{Z}_{t+1,t}^\omega(x,y)\mathcal{Z}_{t,t-1}^\omega(y,z)dy,
\]
then for any $\delta>0$, we have
\[
\begin{aligned}
\mathbf{P}[B_{t }(\delta)]=&\mathbf{P}[W_1W_2^{-1}>\delta]=1-\mathbf{P}[W_1^{-1}W_2\geq \delta^{-1}]\\
\geq &1-\delta\mathbf{E}[W_1^{-1}W_2]\geq 1-\delta \sqrt{\mathbf{E}[W_1^{-2}]\mathbf{E}[W_2^2]}.
\end{aligned}
\]
By Lemma~\ref{l.ulbd}, there exists a constant $C>0$ depending on $d,R$ such that
\[
\sqrt{\mathbf{E}[W_1^{-2}]\mathbf{E}[W_2^2]}\leq C.
\]
The proof is complete
\end{proof}

The following lemma holds.
\begin{lemma}\label{l.lbdtr}
We have
\begin{equation}
\label{011903-21}
\inf_{y_k,y_{k+1}\in\mathbb{T}^d}\pi_k^\omega(y_k\,|\, y_{k+1})> \delta,\quad\quad
\omega\in B_{ t-k}(\delta)
\end{equation}
for all $k=2,\ldots,N-2$
\end{lemma}
\begin{proof}
Recall that
\[
\begin{aligned}
\pi_k^\omega(y_k\,|\,y_{k+1})=&\frac{\int_{\mathbb{T}^{(k-1)d}} \mathcal{Z}_{t-1,t-2}^\omega(y_1,y_2)\ldots \mathcal{Z}_{t-k,t-k-1}^\omega(y_k,y_{k+1})dy_1\ldots dy_{k-1}}{\int_{\mathbb{T}^{kd}} \mathcal{Z}_{t-1,t-2}^\omega(y_1,y_2)\ldots \mathcal{Z}_{t-k,t-k-1}^\omega(y_k,y_{k+1})dy_1\ldots dy_{k-1}dy_k}\\
= &\frac{ \int_{\mathbb{T}^d} g(y_{k-1})\mathcal{Z}_{t-k+1,t-k}^\omega(y_{k-1},y_k)\mathcal{Z}_{t-k,t-k-1}^\omega(y_k,y_{k+1})dy_{k-1}}{\int_{\mathbb{T}^{2d}}g(y_{k-1})\mathcal{Z}_{t-k+1,t-k}^\omega(y_{k-1},y_k)\mathcal{Z}_{t-k,t-k-1}^\omega(y_k,y_{k+1})dy_{k-1} dy_k},
\end{aligned}
\]
for some non-trivial $g\geq0$. Then it is enough to use the fact that
\[
\begin{aligned}
&\inf_{y_k,y_{k+1}\in\mathbb{T}^d}\pi_k^\omega(y_k\,|\,y_{k+1})\\
&\geq  \frac{\int_{\mathbb{T}^d} g(y)dy\cdot \inf_{y_{k-1},y_k,y_{k+1}\in\mathbb{T}^d}\mathcal{Z}_{t-k+1,t-k}^\omega(y_{k-1},y_k)\mathcal{Z}_{t-k,t-k-1}^\omega(y_k,y_{k+1})}{\int_{\mathbb{T}^d} g(y)dy\cdot \sup_{y_{k-1},y_{k+1}}\int_{\mathbb{T}^d}\mathcal{Z}_{t-k+1,t-k}^\omega(y_{k-1},y_k)\mathcal{Z}_{t-k,t-k-1}^\omega(y_k,y_{k+1})dy_k }
\end{aligned}
\]
and the definition of $B_t(\delta)$ in \eqref{e.defB} to complete the proof.
\end{proof}

Combining Lemmas~\ref{l.lowerbd} and \ref{l.lbdtr}, we conclude that
there exists $\delta>0$, depending  only on $d,R$, and  events
\begin{equation}
\label{021903-21a}
B_{t-k}(\delta)\in \mathcal{F}_{[t-k-1,t-k+1]}, \quad\quad  k=2,\ldots,N-2,
\end{equation}
such that
\begin{equation}
\label{021903-21}
\mathbf{P}[B_{t-k }(\delta)]> \delta,
\end{equation}
and the transition probability of the chain $\{Y_n\}_{n=1}^N$ satisfies
\begin{equation}
\label{021903-21b}
\inf_{y_k,y_{k+1}\in\mathbb{T}^d}\pi_k^\omega(y_k\,|\,y_{k+1})>\delta,\quad\quad\quad \mbox{ if }  \omega
\in B_{t-k}(\delta),\quad k=2,\ldots,N-2
\end{equation}

\subsection{Coupling construction}

In this section, we use a
standard couplingargument to study the dependence of $u(t,x; \nu)$ on the ``remote'' environment $\{\xi(r,\cdot): r\leq s\}$ with $t-s\gg1$.

Let $t>1$. Define
\[
I_k:=[t-k-1,t-k+1],\quad\quad k=0,\ldots,[t]-1.
\]
We decompose the interval $[0,t]$ into
subintervals
\[
[0,t]=[0,t-2m_N-1]\cup \bigcup_{m\in \Lambda_N} I_{m} \cup [t-1,t].
\]
where, as we recall $N=[t]+1$ and
\begin{equation}\label{e.defLambdaN}
\Lambda_N:=\{2,4,\ldots, 2m_N\},
\end{equation} and $ 2m_N$ is the largest   number so that $t-2m_N-1\geq 0$:
\[
2m_N=\left\{\begin{array}{cc}
N-2 & \text{ if } N \text{ is even,}\\
N-3 & \text{ if } N \text{ is odd.}
\end{array}
\right.
\]
From \eqref{021903-21b} we know that there exists $\delta>0$ that only depends on $d,R$ such that  $B_{t-k }(\delta)\in\mathcal{F}_{I_k}$, and
\[
\inf_{y_k,y_{k+1}\in\mathbb{T}^d}\pi_k^\omega(y_k\,|\,y_{k+1})>\delta,\quad\quad \mbox{ if }
\omega\in B_{t-k}(\delta),\quad k\in\Lambda_N.
\]
We fix $\delta$ from now on.
Let
\begin{equation}
\label{qk}
q_k^\omega(y_k\,|\,y_{k+1}):=  \frac{\pi_k^\omega(y_k\,|\,y_{k+1})-\delta}{1-\delta},\quad \quad \mbox{ if }\omega\in B_{t-k}(\delta).
\end{equation}
Let us introduce a sequence of i.i.d. Bernoulli random variables
$\{\tau_k\}_{k\in\mathbb{Z}}$ that is defined on a probability space $\Big(\Sigma,{\mathcal
  A},\mathbf{Q}\Big)$. We shall denote by $\mathbb{E}_{\mathbf{Q}}$ the corresponding
expectation.  {We assume
  that
\[
\mathbf{Q}[\tau_k=1]=\delta, \quad\quad \mathbf{Q}[\tau_k=0]=1-\delta.
\]
}
Consider
the following random transition  kernels: for $k=2,3,\ldots, {N-2} $,
 we have
\begin{align}
&\tilde \pi_k^{\omega,0}\Big(y_k\,|\,y_{k+1}\Big)=q_k^\omega(y_k|y_{k+1}),
  \quad \mbox{ if } \omega \in B_{t-k}(\delta),
\label{ctp1}\\
&
\tilde \pi_k^{\omega,1}\Big(y_k \,|\,
y_{k+1}\Big)=1,\qquad \mbox{ if }\omega \in
B_{t-k}(\delta), \label{ctp2}\\
&
\tilde \pi_k^{\omega,\tau}\Big(  y_k \,|\,
y_{k+1}\Big)=\pi_{k}^\omega(y_k\,|\,y_{k+1}),\qquad  \mbox{ if }\omega \in B_{t-k}^c(\delta),\tau\in\{0,1\}. \label{ctp3}
\end{align}
We also let

\begin{align}
&{\tilde{\pi}_{1}^{\omega,\tau}\Big(  y_1\,|\,
y_{2}\Big)=\pi_{1}^\omega(y_{1}\,|\,y_{2}), \quad \tau\in\{0,1\}.}\\
&{\tilde{\pi}_{N-1}^{\omega,\tau}\Big(  y_{N-1}\,|\,
y_{N}\Big)=\pi_{N-1}^\omega(  y_{N-1}\,|\,
y_{N}), \quad \tau\in\{0,1\}.}\\
&\tilde \pi_{N}^{\omega,\tau}\Big( d y_{N}\Big)=\pi_{N}^\omega(dy_{N}),\quad \tau\in\{0,1\}. \label{ctpN}
\end{align}
 We shall write $\tilde \pi_{N}^{\omega,\tau}(y; \nu)$  when we
  wish to
  highlight the dependence of $\tilde \pi_{N}^{\omega,\tau}$ on the
  initial data $\nu$, with the convention of writing $\tilde
  \pi_{N}^{\omega,\tau}(y; f)$, when $\nu(dx)=f(x)dx$.
For a given realization of $\{\tau_k(\sigma)\}_{k=1}^N$, $\sigma\in \Sigma$, we
consider the path measure $ \mathbb P^{\omega,\sigma}_{\tilde{\pi}_N}$ on $\mathbb{T}^{Nd}$ corresponding to the time
reversed
Markov dynamics  obtained by using $\tilde \pi_N^{\omega,\tau_N(\sigma)}$ as
the final density and $\tilde \pi_k^{\omega,\tau_k(\sigma)}( y_k \,|\,
y_{k+1})$ as transition probabilities between times $k+1$ and $k$,
for $k=1,\ldots,N-1$. Let $\mathbb{E}^{\omega,\sigma}_{\tilde{\pi}_N}$ be the corresponding expectation, for each fixed realization of $\omega,\sigma$.
Note that
\begin{equation}
\label{032204-21}
\int_{\Sigma}\mathbb P^{\omega,\sigma}_{\tilde{\pi}_N}(A)\mathbf{Q}(d\sigma)=\mathbb{P}^\omega_{\pi_N}(A),
\quad A\in {\mathcal B}(\mathbb{T}^{Nd}).
\end{equation}
Therefore, from \eqref{e.densityY1} and \eqref{032204-21},
\begin{equation}\label{e.densityY1c}
  u(t,x;\nu)=\frac{\int_{\Sigma}\mathbf{Q}(d\sigma)\mathbb{E}_{\tilde \pi_N}^{\omega,\sigma}[\mathcal{Z}_{t,t-1}^\omega(x,Y_1)]}{\int_{\mathbb{T}^d}dx'\int_{\Sigma}\mathbf{Q}(d\sigma)\mathbb{E}_{\tilde{\pi}_N}^{\omega,\sigma}[\mathcal{Z}_{t,t-1}^\omega (x',Y_1)]}.
\end{equation}

Define the events
\begin{equation}
\label{022203-21z}
{A}_k^t=B_{t-k}(\delta)\times \{\tau_k=1\}\in {\mathcal
  F}_{[t-k-1,t-k+1]}\otimes {\mathcal A},\quad k=2,3,\ldots,N-2.
\end{equation}
If some $A_k^t$ occurs, the transition probability is given by the uniform distribution, as can be seen from \eqref{ctp2}, so the chain ``renews''. We emphasize here that $N$ depends on $t$ as well. If there is no danger of confusion we shall omit the superscript $t$
from the notation of the events.
Note that
\begin{equation}
\label{012003-21}
\mathbf{P}\otimes\mathbf{Q}[ {A}_k^t]=\mathbf{P}[B_{t-k}(\delta)]\mathbf{Q}[\tau_k=1]>\delta^2,\quad
 {k=2,3,\ldots,N-2}.
\end{equation}
It is also clear that the events $A_m^t$,  $m\in \Lambda_{N}$ are independent under $\mathbf{P}\otimes\mathbf{Q}$ (recall that $\Lambda_N$ was defined in \eqref{e.defLambdaN}).
In addition, thanks to \eqref{e.defBa},   for any
$1<s<t$ and non-negative $\ell\in\mathbb{Z}$ we have
\begin{equation}
\label{032903-21}
\theta_{t-s,y}^{-1}\Big(A_k^s\Big)=A_{k}^t,\quad k=2,3,\ldots,N(s)-2,\,y\in\mathbb{T}^d.
\end{equation}
Here we have denoted $\theta_{r,y}(\omega,\sigma):=\Big(\theta_{r,y}(\omega),\sigma\Big)$.

Note that for any $F\in B_b(\mathbb{T}^d)$ we have
\begin{equation}
\label{022003-21}
\begin{aligned}
\mathbb{E}_{\mathbf{Q}}\otimes \mathbb{E}_{\tilde \pi_N}^{\omega,\cdot}
\Big[F(Y_1)\Big|\tau_k=1\Big]=& {\tfrac{1}{\delta}\int_{\{\tau_k=1\}}\mathbf{Q}(d\sigma)\mathbb{E}_{\tilde \pi_N}^{\omega,\sigma}[F(Y_1)]}\\
=&\mathbb{E}_{k}^{\omega}
\Big[F(Y_1)\Big],\quad \quad 2\le k\le N-2,\,\omega \in B_{t-k}(\delta),
\end{aligned}
\end{equation}
where, as we recall, $\mathbb{E}_k^\omega$ is the expectation with respect to the Markov chain constructed in \eqref{042003-21} and \eqref{012003-211}.
Now it is clear
that for any $t>1$ and $2\leq k\leq N-2$,  the random variable
\begin{equation}
\label{022003-21a}
 1_{ B_{t-k}(\delta)}(\omega) \int_{\{\tau_k=1\}}\mathbf{Q}(d\sigma)\mathbb{E}_{\tilde \pi_N}^{\omega,\sigma}[\mathcal{Z}_{t,t-1}^\omega(x,Y_1)],\quad\quad  \omega\in\Omega
\end{equation}
is ${\mathcal F}_{ {[t-k-1,t]}}$-measurable.

\subsection{Approximation}
For any $\nu\in {\mathcal M}_1(\mathbb{T}^d)$, define
\begin{equation}\label{e.defXom}
X^\omega(t,x; \nu):=\mathbb{E}_{ \pi_N(\nu)}^{\omega}[\mathcal{Z}_{t,t-1}^\omega(x,Y_1)]=\mathbb{E}_{\mathbf{Q}}\mathbb{E}_{\tilde \pi_N(\nu)}^{\omega,\cdot}[\mathcal{Z}_{t,t-1}^\omega(x,Y_1)].
\end{equation}
When $\nu(dx) =f(x)dx$ for $f\in D(\mathbb{T}^d)$ we shall write $X^\omega(t,x; f):=X^\omega(t,x; \nu)$.
We omit writing the initial data $\nu$ from our notation when they are obvious from the context.

Fix $t>10$ and $2\leq k\leq  {N(t)-2}$. We shall approximate $X^\omega(t,x)$
by some random variable that is $\mathcal{F}_{[t-k,t]}-$measurable.
Define
\begin{equation}\label{e.defLambdaNk}
\Lambda_{N(t),k}=\{m: m \leq k-1\}\cap \Lambda_{N(t)},
\end{equation}
\begin{equation}\label{e.deftildeAk}
\tilde{A}_k^t=\bigcup_{m\in \Lambda_{N(t),k}}A_m^t\in {\mathcal
  F}_{[t-k,t-1]}\otimes {\mathcal A},
\end{equation}
and
\begin{equation}\label{e.defXkom}
X_k^\omega(t,x):= \int_{\Sigma}\mathbf{Q}(d\sigma)1_{\tilde{A}_k^t}(\omega,\sigma)\mathbb{E}_{\tilde \pi_{N(t)}}^{\omega,\sigma}[\mathcal{Z}_{t,t-1}^\omega(x,Y_1)],
\end{equation}
and, thanks to
 \eqref{032903-21},
\begin{equation}\label{e.deftildeAk1}
\theta_{t-s,0}^{-1}\Big(\tilde{A}_k^s\Big)=\tilde{A}_{k}^t.
\end{equation}
In particular, as can be seen from \eqref{022003-21}, the random
  field $\{X_k(t,x)\} $ does not depend on the
  initial data.
Moreover, thanks to \eqref{022903-21} and \eqref{e.deftildeAk1}, we have
\begin{equation}\label{XtXs}
X_k^{\theta_{t-s,0}(\omega)}(s,x)=X_k^\omega(t,x),\quad k\le N(s)-2.
\end{equation}
As a direct consequence of the above and   formula \eqref{022003-21},
we conclude:
\begin{lemma}
\label{prop012303-21}
Suppose that $\nu_1,\nu_2\in D(\mathbb{T}^d)$.
For any $t\ge s>k$ and $2\leq k\leq N(s)-2$,
the laws of the fields
$X_k(t,\cdot\,;\nu_1)$ and $X_k(s,\cdot\,;\nu_2)$ over the space
$L^1(\mathbb{T}^d)$ are identical.
\end{lemma}

\begin{lemma}\label{l.appX}
For any   $2\leq k\leq N(t)-2$ and $\nu\in {\mathcal M}_1(\mathbb{T}^d)$,
the field $X_k (t,\cdot;\nu)$ is
$\mathcal{F}_{[t-k,t]}-$measurable. Moreover, for any $p\in[1,+\infty)$, there exists
$C,\lambda>0$ only depending on $p,d$ and $R$ such that
\begin{equation}\label{e.3211}
\mathbf{E}\Big[\sup_{\nu\in{\mathcal M}_1(\mathbb{T}^d)}\|X^\omega (t;\nu)-X_k^\omega (t;\nu)\|^p_{L^\infty(\mathbb{T}^d)}\Big] \leq
Ce^{-\lambda k}\quad \mbox{for   $2\leq k\leq {N(t)-2}$}.
\end{equation}
\end{lemma}
\begin{proof}
First we obviously have
\begin{equation}
\mathbb{E}_{\tilde \pi_N}^{\omega,\sigma}[\mathcal{Z}_{t,t-1}^\omega(x,Y_1)]\leq \sup_{x,y\in\mathbb{T}^d} \mathcal{Z}_{t,t-1}^\omega(x,y),
\end{equation}
which implies
\[
\sup_{x\in\mathbb{T}^d}|X^\omega(t,x;\nu)-X_k^\omega(t,x;\nu)| \leq \sup_{x,y\in\mathbb{T}^d} \mathcal{Z}_{t,t-1}^\omega(x,y)\int_{\Sigma}\mathbf{Q}(d\sigma) 1_{(\tilde{A}_k^t)^c}(\omega,\sigma).
\]
Thus, by Jensen's inequality
\begin{align*}
&\mathbf{E}\Big[\sup_{\nu\in{\mathcal
  M}_1(\mathbb{T}^d)}\|X^\omega(t;\nu)-X_k^\omega(t;\nu)\|^p_{L^\infty(\mathbb{T}^d)}\Big]\\
&
 \leq \sqrt{\mathbf{E} \Big(\sup_{x,y\in\mathbb{T}^d} \mathcal{Z}_{t,t-1}^\omega(x,y)\Big)^{2p} }\sqrt{\mathbf{P}\otimes\mathbf{Q}[(\tilde{A}_k^t)^c]}.
\end{align*}
Since $(\tilde{A}_k^t)^c=\bigcap_{m\in\Lambda_{N,k}} (A_m^t)^c$ and the events
$A_m^t$, $m\in\Lambda_{N,k}$ are independent, from \eqref{012003-21} we have
\begin{equation}
\label{022203-21}
\mathbf{P}\otimes\mathbf{Q}[(\tilde{A}_k^t)^c]=\prod_{m\in \Lambda_{N,k}} \mathbf{P}\otimes\mathbf{Q}[(A_m^t)^c]<(1-\delta^2)^{\#\Lambda_{N,k}},
\end{equation}
where $\delta$ depends only on $R$ and $d$.
Combining \eqref{022203-21} with Lemma~\ref{l.ulbd}, we derive that
\[
\mathbf{E}\Big[\sup_{\nu\in{\mathcal M}_1(\mathbb{T}^d)}\|X^\omega(t;\nu)-X_k^\omega(t;\nu)\|^p_{L^\infty(\mathbb{T}^d)}\Big] \leq  C(1-\delta^2)^{\#\Lambda_{N,k}/2},
\]
where the constant $C >0$ does not depend on $t$.
This proves \eqref{e.3211}.

We prove now that $X_k^\omega(t,x;\nu)$ is $\mathcal{F}_{[t-k,t]}-$measurable.
Consider now the event $\tilde{A}_k$ given by
\eqref{e.deftildeAk}. {From this point on we shall drop $t$ from the
notation of the events and integers.} Since $N,k$ are both fixed here, to simplify the
notations, let
$$
M:=\max\{m: m\in \Lambda_{N,k}\}
$$ and for any $m\in \Lambda_{N,k}$,  define
\[
D_m=\left\{\begin{array}{ll}
A_m\cap\left(  \mathlarger{\mathlarger{\bigcup}}\limits_{m'>m,\,m'\in \Lambda_{N,k}} A_{m'}\right)^c,  & \text{ if  } m<M,\\
 A_M, & \text{ if } m=M.
 \end{array}
 \right.
 \]
Then,
\begin{equation}
\label{042203-21}
1_{\tilde{A}_k}(\omega,\sigma)=\sum_{m\in \Lambda_{N,k}}1_{D_m}(\omega,\sigma),
\end{equation}
 and it suffices to show that for any $m\in \Lambda_{N,k}$,
 \[
  \int_{\Sigma}\mathbf{Q}(d\sigma)1_{D_m}(\omega,\sigma)\mathbb{E}_{\tilde \pi_N}^{\omega,\sigma}[\mathcal{Z}_{t,t-1}^\omega(x,Y_1)]
    \]
    is $\mathcal{F}_{[t-k,t]}-$measurable. We can write
        \[
    \begin{aligned}
  &\int_{\Sigma}\mathbf{Q}(d\sigma)1_{D_m}(\omega,\sigma)\mathbb{E}_{\tilde \pi_N}^{\omega,\sigma}[\mathcal{Z}_{t,t-1}^\omega(x,Y_1)] \\
  &= \int_{\Sigma}\mathbf{Q}(d\sigma)1_{A_m}(\omega,\sigma)
\left(\prod_{m'>m,m'\in \Lambda_{N,k}}
  1_{A_{m'}^{c}}(\omega,\sigma)\right)\mathbb{E}_{\tilde
  \pi_N}^{\omega,\sigma}[\mathcal{Z}_{t,t-1}^\omega(x,Y_1)] .
    \end{aligned}
    \]
  By convention we understand a product over an empty set to be equal to $1$.  Invoking \eqref{022003-21}, the right hand side of the above equation can be written as
    \begin{align}
\label{052203-21}
&
\left(\int_{\Sigma}\mathbf{Q}(d\sigma)1_{A_m}(\omega,\sigma)
\mathbb{E}_{ m}^{\omega}[\mathcal{Z}_{t,t-1}^\omega(x,Y_1)]\right)
 \int_{\Sigma}\mathbf{Q}(d\sigma)\left(\prod_{m'>m,m'\in \Lambda_{N,k}}
  1_{A_{m'}^{c}}(\omega,\sigma)\right)\notag\\
&
   =
\mathbb{E}_{ m}^{\omega}[\mathcal{Z}_{t,t-1}^\omega(x,Y_1)] \delta 1_{B_{t-m}(\delta)}(\omega)\prod_{m'>m,m'\in
  \Lambda_{N,k}}\Big(1-\delta 1_{B_{t-m'}(\delta)}(\omega)\Big),
\end{align}
which is $\mathcal{F}_{[t-k,t]}-$measurable. This completes the proof.
\end{proof}

From the above proof, we can actually conclude that
$X_k^{\omega}(t,x)$ only depends on $t,x,k$ and the random environment
$\{\xi(s,\cdot):s\in [t-k,t]\}$. Recall that $u(t,\cdot;\nu)$ denotes
the endpoint density of the polymer path, assuming that the starting distribution is an arbitrary probability measure $\nu\in \mathcal{M}_1(\mathbb{T}^d)$. We have the following key proposition which underlies Theorem~\ref{t.geomeEr}.
\begin{proposition}\label{p.couplingCon}
For any  $2\leq k\leq N(t)-2$, there exists $  u_k(t,x)$ that is \linebreak $\mathcal{F}_{[t-k,t]}-$measurable and does not depend on the initial  distribution $\nu$, such that for any $p\in[1,+\infty)$,
\begin{equation}
\label{072203-21}
\sup_{\nu\in{\mathcal M}_1(\mathbb{T}^d)}\mathbf{E}\Big[\|  u(t;\nu)- u_k(t)\|^p_{L^\infty(\mathbb{T}^d)}\Big] \leq Ce^{-\lambda k},
\end{equation}
where the constants $C,\lambda>0$ depend only  on $p,d,R$.
\end{proposition}
\begin{proof}
By \eqref{e.densityY1c} and \eqref{e.defXom}, we have
\[
  u(t,x;\nu)=\frac{X^\omega(t,x;\nu)}{\int_{\mathbb{T}^d}X^\omega(t,x';\nu)dx'}.
\]
Let
$$
C_k(\delta):=\bigcup_{m\in\Lambda_{N,k}}B_{t-m}(\delta),
$$
where we recall that $\Lambda_{N,k}$ was defined in \eqref{e.defLambdaNk}.
 {Assuming $\omega\in B_{t-m}(\delta)$ for some $m\in \Lambda_{N,k}$, then $1_{\tilde{A}_k}(\omega,\sigma)\geq 1_{\{\tau_m=1\}}(\sigma)$ ($\tilde{A}_k$ was defined in \eqref{e.deftildeAk}), which implies
\begin{equation}
\label{062203-21}
\begin{aligned}
\inf_{x\in\mathbb{T}^d} X_k^\omega(t,x)\geq &\left(\int_{\Sigma}\mathbf{Q}(d\sigma)1_{\{\tau_m=1\}}(\sigma)\right)\inf_{x,y\in\mathbb{T}^d} \mathcal{Z}_{t,t-1}^\omega(x,y)\\
&=\delta\inf_{x,y\in\mathbb{T}^d} \mathcal{Z}_{t,t-1}^\omega(x,y),\quad \quad \mbox{ if } \omega\in C_k(\delta),
\end{aligned}
\end{equation}}
with $X_k^\omega$ defined in \eqref{e.defXkom}.
Define
\begin{equation}\label{e.defuk}
  u_k(t,x):=
  \left\{
  \begin{array}{ll}
  \frac{X^\omega_k(t,x)}{\int_{\mathbb{T}^d}X^\omega_k(t,x')dx'}, &\mbox{ if }\omega\in C_k(\delta)\\
  0, &\mbox{ if } \omega\in C_k^c(\delta).
  \end{array}
  \right.
\end{equation}
From
the definition \eqref{e.defXom},  we conclude
\[
\inf_{x,y\in\mathbb{T}^d}\mathcal{Z}^\omega_{t,t-1}(x,y)\leq\inf_{x\in\mathbb{T}^d} X^\omega(t,x)\leq \sup_{x\in\mathbb{T}^d}X^\omega(t,x) \leq \sup_{x,y\in\mathbb{T}^d} \mathcal{Z}^\omega_{t,t-1}(x,y).
\]
Concerning  $X^\omega_k(t,x)$, we conclude from \eqref{062203-21} that
\begin{align*}
\delta 1_{C_k(\delta)}(\omega)\inf_{x,y\in\mathbb{T}^d}\mathcal{Z}^\omega_{t,t-1}(x,y)&\le  \inf_{x\in\mathbb{T}^d}
  X^\omega_k(t,x)\\
&
\leq \sup_{x\in\mathbb{T}^d}X^\omega_k(t,x) \leq \sup_{x,y\in\mathbb{T}^d} \mathcal{Z}^\omega_{t,t-1}(x,y) .
\end{align*}
Thus, for $\omega\in C_k(\delta)$, we have
\begin{equation}
\label{082203-21}
\begin{split}
\sup_{x\in\mathbb{T}^d}|  u(t,x;\nu)-  u_k(t,x)|\leq& \sup_{x\in\mathbb{T}^d}
|X^\omega(t,x)-X_k^\omega(t,x)|\\
&
\times \left(\inf_{x,y\in\mathbb{T}^d}\mathcal{Z}^\omega_{t,t-1}(x,y) \right)^{-1}\left(1+\frac{1}{\delta}
    \frac{\sup_{x,y\in\mathbb{T}^d}
      \mathcal{Z}^\omega_{t,t-1}(x,y)}{\inf_{x,y\in\mathbb{T}^d}\mathcal{Z}^\omega_{t,t-1}(x,y)}\right).
\end{split}
\end{equation}
We can therefore write
\begin{equation}
\label{072203-21a}
\mathbf{E}[\sup_{x\in\mathbb{T}^d} |  u(t,x;\nu)-  u_k(t,x)|^p] =I_1+I_2,
\end{equation}
where
\begin{equation}
\label{072203-21b}
\begin{split}
&
I_1:=\mathbf{E}\Big[\sup_{x\in\mathbb{T}^d} |  u(t,x;\nu)-  u_k(t,x)|^p1_{C_k(\delta)}\Big],\\
&
I_2:=\mathbf{E}\Big[\sup_{x\in\mathbb{T}^d} |  u(t,x;\nu)|^p1_{C_k^c(\delta)}\Big].
\end{split}
\end{equation}
To estimate $I_1$, we use the bound \eqref{082203-21} together with
  Lemmas~\ref{l.ulbd} and \ref{l.appX}, and obtain that
 there exist $C,\lambda>0$ depend only  on $p,d,R$ such that $I_1\leq Ce^{-\lambda k}$.
To deal with  $I_2$, note that by the H\"older estimate for any $r>p$
we can write
\begin{equation}
\label{072203-21d}
\begin{aligned}
I_2\leq &\left\{\mathbf{E}\Big[\left(\sup_{x\in\mathbb{T}^d}
    u(t,x;\nu)\right)^r\Big]\right\}^{p/r}\Big(\mathbf{P}[C_k^c(\delta)]\Big)^{1-p/r}\\
  &= \left\{\mathbf{E}\Big[\left(\sup_{x\in\mathbb{T}^d}
    u(t,x;\nu)\right)^r\Big]\right\}^{p/r}\Big(\mathbf{P}[\bigcap_{m\in\Lambda_{N,k}}B^c_{t-m}(\delta)]\Big)^{1-p/r}.
  \end{aligned}
\end{equation}
Since $B^c_{t-m}(\delta)$ are independent and satisfy
\eqref{021903-21}, we conclude that the right hand side can be
estimated by
$$
\left\{\mathbf{E}\Big[\left(\sup_{x\in\mathbb{T}^d}
    u(t,x;\nu)\right)^r\Big]\right\}^{p/r}\Big(1-\delta\Big)^{(1-p/r)\#\Lambda_{N,k}}.
$$
Using Lemma \ref{l.ulbd} we conclude that
 there exist $C,\lambda>0$ depend only  on $p,d,R$ such that $I_2\leq Ce^{-\lambda k}$.
The proof is complete.
\end{proof}

\subsection{Proof of Theorem~\ref{t.geomeEr}}

Recall that given an initial distribution $\nu\in {\mathcal M}_1(\mathbb{T}^d)$
the polymer endpoint density $u(t;\nu)$ is given   by \eqref{e.defPo}.
Fix $ F\in
{\rm Lip}({\mathcal M}_1(\mathbb{T}^d)) $.
 By virtue
of Lemma~\ref{prop012303-21} and \eqref{e.defuk}, for $t>s>k>s/2$, we have
$$
\mathbf{E}[F(  u_k(t))]=\mathbf{E}[F(  u_k(s))].
$$ Therefore, for any $\nu_1,\nu_2\in \mathcal{M}_1(\mathbb{T}^d)$, we have
\begin{align*}
&
\Big|{\mathcal P}_tF(\nu_1)-{\mathcal P}_sF(\nu_2)\Big|=\Big|
  \mathbf{E}[F(u(t; \nu_1))]-\mathbf{E}[F(u(s; \nu_2))]\Big|\\
&
=\Big|\mathbf{E}[F(u(t; \nu_1))]-\mathbf{E}[F(  u_k(t))]-\mathbf{E}[F(u(s; \nu_2))]+\mathbf{E}[F(
  u_k(s))]\Big|\\
&
\le
\Big|\mathbf{E}[F(  u(t; \nu_1))]-\mathbf{E}[F(  u_k(t))]\Big|+\Big|\mathbf{E}[F(  u(s;
  \nu_2))]-\mathbf{E}[F(  u_k(s))]\Big|\\
&
\le
  \|F\|_{{\rm Lip}}\Big(\sup_{\mu\in {\mathcal M}_1(\mathbb{T}^d)}\mathbf{E}\|  u(t;
  \mu)-  u_k(t)\|_{L^1(\mathbb{T}^d)}+\sup_{\mu\in {\mathcal
  M}_1(\mathbb{T}^d)}\mathbf{E}\|  u(s; \mu)-  u_k(s)\|_{L^1(\mathbb{T}^d)}\Big).
\end{align*}
Using Proposition \ref{p.couplingCon}
 we can further write   that
\begin{equation}
\label{012304-21}
\Big|{\mathcal P}_tF(\nu_1)-{\mathcal P}_sF(\nu_2)\Big|\leq C\|F\|_{{\rm
    Lip}}e^{-\lambda k}\le C\|F\|_{{\rm
    Lip}}e^{-\lambda s/2},
\end{equation}
which in turn implies that there exist $C,\lambda>0$ such that
\[
\sup_{\nu_1,\nu_2\in {\mathcal M}_1(\mathbb{T}^d)}\Big\|\delta_{\nu_1}{\mathcal P}_t-\delta_{\nu_2}{\mathcal P}_s\Big\|_{\mathrm{FM}}\leq
C e^{-\lambda s/2},\quad t>s>0.
\]
For any $\Pi\in {\rm M}_1( {\mathcal M}_1(\mathbb{T}^d))$ and $F\in {\rm Lip}({\mathcal
  M}_1(\mathbb{T}^d))$, we can write
$$
\int_{{\mathcal M}_1(\mathbb{T}^d)} F(\nu)\Pi{\mathcal P}_t(d\nu)=\int_{{\mathcal M}_1(\mathbb{T}^d)}\Pi(d\nu) \left(\int_{\mathcal{M}_1(\mathbb{T}^d)}\delta_{\nu}{\mathcal P}_t(d\mu)F(\mu)\right),
$$
so we also conclude that
{\begin{equation}
\label{042303-21}
\sup_{\Pi_1,\Pi_2\in {\rm M}_1( {\mathcal M}_1(\mathbb{T}^d))}\Big\|\Pi_1{\mathcal P}_t-\Pi_2{\mathcal P}_s\Big\|_{\mathrm{FM}}\leq
C e^{-\lambda s/2},\quad t>s>0,
\end{equation}}
which implies that there
exists a unique $\pi_\infty\in {\rm M}_1( {\mathcal M}_1(\mathbb{T}^d))$ such that
\begin{equation}\label{e.452}
\lim_{t\to+\infty}\Pi{\mathcal P}_t=\pi_\infty
\end{equation}
for any $\Pi\in  {\rm M}_1( {\mathcal M}_1(\mathbb{T}^d))$, in the
Fortet-Mourier metric on ${\rm M}_1( {\mathcal
  M}_1(\mathbb{T}^d))$. Hence
$$
\pi_\infty =\pi_\infty{\mathcal P}_t  \quad\mbox{for all $t>0$}.
$$
Suppose that $p\in[1,+\infty]$. Since $\delta_{\nu}{\mathcal
  P}_t(D^p(\mathbb{T}^d))=1$ for all $\nu\in {\mathcal M}_1(\mathbb{T}^d)$ and any $t>0$,
 we have
$$
\pi_\infty\Big(D^p(\mathbb{T}^d)\Big)=\pi_\infty{\mathcal P}_t\big(D^p(\mathbb{T}^d)\Big)=\int_{{\mathcal M}_1(\mathbb{T}^d)}\pi_\infty(d\nu) \left(\delta_{\nu}{\mathcal
  P}_t\Big(D^p(\mathbb{T}^d)\Big)\right)=1,
$$
which finishes the proof of part (i).

Part (ii) is a direct consequence of \eqref{042303-21} and \eqref{e.452}.

Suppose now that   $p\in(1,\infty]$ and  $F\in {\rm
    Lip}(D^p(\mathbb{T}^d))$. Proposition \ref{p.couplingCon} also applies in
  this case with the $L^1(\mathbb{T}^d)$ norm, used above, replaced
  by the $L^p(\mathbb{T}^d)$ norm. In consequence we conclude
  \eqref{012304-21} with the Lipschitz norm of $F$ in the respective
  $L^p(\mathbb{T}^d)$ metric. Hence, we  infer the estimate
  \eqref{042303-21} with the respective Fortet-Mourier norm on the l.h.s. The remaining part of the argument stays the same and, as
  a result, we conclude the statement of the theorem in the case of the
  space $D^p(\mathbb{T}^d)$.
\qed

\subsection{Asymptotic stability of  the transition probability operator}

A straightforward consequence of Theorem~\ref{t.geomeEr} is
\begin{corollary}
\label{cor020104-21}
Suppose that $F\in C_b ({\mathcal M}_1(\mathbb{T}^d))$ and $\Pi\in {\rm
  M}_1({\mathcal M}_1(\mathbb{T}^d))$  then
\begin{equation}
\label{040104-21}
\lim_{t\to+\infty}\int_{{\mathcal M}_1(\mathbb{T}^d)}{\mathcal
    P}_tFd\Pi= \int_{{\mathcal M}_1(\mathbb{T}^d)}F(u)\pi_\infty(du).
\end{equation}
\end{corollary}

\begin{proposition}
\label{prop021203-21}
Suppose that $F\in {\rm Lip} ({\mathcal M}_1(\mathbb{T}^d))$. Then,
\begin{equation}
\label{021203-21}
\lim_{t\to+\infty}\mathbf{E}_{\Pi}\left(\frac{1}{t} \int_0^t
  F(u(s))ds-\int_{{\mathcal M}_1(\mathbb{T}^d)}F(u)\pi_\infty(du) \right)^2=0.
\end{equation}
where $\mathbf{E}_{\Pi}$ denotes the expectation w.r.t. the noise $\xi$ realization and the initial
data $u(0)$ distributed according to a Borel probability
measure $\Pi$ on ${\mathcal M}_1(\mathbb{T}^d)$. {The same result also holds for $D^p(\mathbb{T}^d)$ in place of ${\mathcal M}_1(\mathbb{T}^d)$ for any $p\in[1,+\infty]$.}
\end{proposition}
\proof
It suffices to show that as $t\to\infty$,
\begin{equation}
\label{021203-21a}
\mathbf{E}_{\Pi}\left(\frac{1}{t} \int_0^t
  F(u(s))ds\right)\to \int_{{\mathcal M}_1(\mathbb{T}^d)}F(u)\pi_\infty(du),
\end{equation}
and
\begin{equation}
\label{021203-21b}
\mathbf{E}_{\Pi}\left(\frac{1}{t} \int_0^t
  F(u(s))ds\right)^2\to\left(\int_{{\mathcal M}_1(\mathbb{T}^d)}F(u)\pi_\infty(du)\right)^2.
\end{equation}
Equality
\eqref{021203-21a} follows directly from \eqref{040104-21}. Concerning
\eqref{021203-21b},
we can write
\begin{equation}
\label{021203-21c}
\begin{split}
&\mathbf{E}_{\Pi}\left(\frac{1}{t} \int_0^t
  F(u(s))ds\right)^2=\frac{2}{t^2} \mathbf{E}_{\Pi}\left(\int_0^t
  F(u(s))ds\int_0^{s}F(u(s'))ds' \right)\\
&
 {=\frac{2}{t^2} \int_0^t
  ds \int_0^{s}ds' \left[\int_{{\mathcal M}_1(\mathbb{T}^d)} {\mathcal
      P}_{s'}\Big(F{\mathcal P}_{s-s'}F
      \Big)d\Pi \right]}\\
      &=\frac{2}{t^2} \int_0^t
  ds \int_0^{s}ds' \left[\int_{{\mathcal M}_1(\mathbb{T}^d)} {\mathcal
      P}_{s-s'}\Big(F{\mathcal P}_{s'}F
      \Big)d\Pi \right].
\end{split}
\end{equation}
Using Theorem~\ref{t.geomeEr} part (ii), we conclude that
\[
\Big\|{\mathcal
      P}_{s-s'}\Big(F{\mathcal P}_{s'}F
      \Big)-{\mathcal
      P}_{s-s'}\Big(F\int_{{\mathcal M}_1(\mathbb{T}^d)} F(u)\pi_\infty(du)\Big)
     \Big\|_\infty\le Ce^{-\lambda s'}.
      \]
      Therefore,
  \begin{equation}
\label{021203-21cc}
\begin{split}
& \left|\frac{2}{t^2}\int_0^t
  ds \int_0^{s}ds' \left[\int_{{\mathcal M}_1(\mathbb{T}^d)} {\mathcal
      P}_{s-s'}\Big(F{\mathcal P}_{s'}F
      \Big)d\Pi \right]\right.\\
      &
      \left.-\frac{2}{t^2}\left(\int_{{\mathcal M}_1(\mathbb{T}^d)} Fd\pi_\infty\right)
 \int_0^t
  ds \int_0^{s}ds' \int_{{\mathcal M}_1(\mathbb{T}^d)} {\mathcal
      P}_{s-s'}Fd\Pi\right|\\
      &
      \le\frac{C}{t^2} \int_0^t ds\int_0^s e^{-\lambda s'}ds'\to0, \quad\quad\quad \mbox{ as } t\to\infty.
      \end{split}
\end{equation}
On the other hand, upon another
application of  Theorem~\ref{t.geomeEr} part (ii), we conclude that
\begin{equation}
\label{021203-21d}
\begin{aligned}
&  \frac{2}{t^2} \int_0^t
  ds \int_0^{s}ds' \int_{{\mathcal M}_1(\mathbb{T}^d)} {\mathcal
      P}_{s-s'}Fd\Pi\\
      &=2 \int_0^1
  ds \int_0^{s}ds' \int_{{\mathcal M}_1(\mathbb{T}^d)} {\mathcal
      P}_{t(s-s')}Fd\Pi\\
      &\to2\int_{{\mathcal M}_1(\mathbb{T}^d)} F(u)\pi_\infty(du)  \int_0^1
  ds \int_0^{s}ds'=\int_{{\mathcal M}_1(\mathbb{T}^d)} F(u)\pi_\infty(du).
  \end{aligned}
\end{equation}
Combining \eqref{021203-21cc} with \eqref{021203-21d}, we conclude  \eqref{021203-21b}, thus finishing the proof of the proposition. The proof for the case of $F\in {\rm Lip} (D^p(\mathbb{T}^d))$ is the same.
\qed

\subsection{More on Theorem~\ref{t.geomeEr}}

In the statement of Theorem~\ref{t.geomeEr}, we have only considered
the Lipschitz functional on $\mathcal{M}_1(\mathbb{T}^d)$, or $D^p(\mathbb{T}^d)$.
In what follows we shall need the stability of the semigroup ${\mathcal
  P}_t$ on functionals that are only local Lipschitz.  For that
purpose we shall need  the following lemma:
\begin{lemma}\label{l.bdut1}
For any $p\in[1,+\infty)$, there exists $C=C(d,R,p)$ such that \[
\sup_{t>1,\nu\in\mathcal{M}_1(\mathbb{T}^d)}\mathbf{E}\Big[\big(\sup_{x\in\mathbb{T}^d}u(t,x;\nu)\big)^p\Big]\leq C.
\]
\end{lemma}
\begin{proof}
By \eqref{e.densityY1}, we have when $t>1$ that
\[
\sup_{x\in\mathbb{T}^d}u(t,x;\nu) \leq \frac{\sup_{x,y\in\mathbb{T}^d} \mathcal{Z}_{t,t-1}^\omega(x,y)}{\inf_{x,y\in\mathbb{T}^d} \mathcal{Z}_{t,t-1}^\omega(x,y)},
\]
then the result is a direct consequence of Lemma~\ref{l.ulbd}.
\end{proof}

Now consider the functional $\mathcal{R}: D^2(\mathbb{T}^d)\to \mathbb{R}$ defined in \eqref{fR}:
\begin{equation}
\label{022304-21}
\mathcal{R}(v)=\int_{\mathbb{T}^{2d}} R(x-y) v(x)v(y)dxdy\quad\mbox{ and  $\tilde{\mathcal{R}}=\mathcal{R}-2\gamma$},
\end{equation}
with $\gamma$ given by \eqref{e.defgamma}. The functional is globally Lipschitz
in the case when $R(\cdot)$ is bounded, but only locally Lipschitz
when $R=\delta$, as then $\mathcal{R}(v)=\|v\|_{L^2(\mathbb{T}^d)}^2$.

\begin{proposition}\label{p.tailPtR}
There exist $C,\lambda$ only depending on $R,d$ such that
\[
\sup_{t>1,\nu\in \mathcal{M}_1(\mathbb{T}^d)}|\mathcal{P}_t\tilde{\mathcal{R}}(\nu)|\leq
Ce^{-\lambda t}.
\]
\end{proposition}
\begin{proof}
By Lemma~\ref{l.bdut1} and the Markov property, it suffices to prove the result for any $v\in D^2(\mathbb{T}^d)$ and $t>1$. First we note that $\mathcal{R}(\cdot)$ is continuous from $D^2(\mathbb{T}^d)$ to $\mathbb{R}$, therefore by Theorem~\ref{t.geomeEr}, we have as $t\to\infty$ that
\[
\mathcal{R}(u(t;v))\Rightarrow \mathcal{R}(\tilde{u}) \quad\quad \mbox{ in distribution},
\]
with $\tilde{u}$ sampled from $\pi_\infty$. Lemma~\ref{l.bdut1} ensures the uniform integrability so we have
\begin{equation}\label{e.461}
\mathcal{P}_t\mathcal{R}(v)=\mathbf{E}[\mathcal{R}(u(t;v))]\to \mathbf{E}[\mathcal{R}(\tilde{u})]=2\gamma.
\end{equation}
Next, for any $t_1>t_2>k>t_2/2$, we have
\[
\begin{aligned}
|\mathcal{P}_{t_1}\mathcal{R}(v)-\mathcal{P}_{t_2}\mathcal{R}(v)
\leq \sum_{j=1}^2|\mathbf{E}[\mathcal{R}(u(t_j;v))-\mathcal{R}(u_k(t_j))]|
\end{aligned}
\]
with $u_k$ defined in \eqref{e.defuk}. By Proposition~\ref{p.couplingCon} and the fact that
\[
\begin{aligned}
&|\mathcal{R}(u(t_j;v))-\mathcal{R}(u_k(t_j))|\\
&\leq
\|u(t_j;v)-u_k(t_j)\|_{L^\infty(\mathbb{T}^d)}(\|u(t_j;v)\|_{L^1(\mathbb{T}^d)}+\|u_k(t_j)\|_{L^1(\mathbb{T}^d)}\\
&\leq  2\|u(t_j;v)-u_k(t_j)\|_{L^\infty(\mathbb{T}^d)},
\end{aligned}
\]
we derive $|\mathcal{P}_{t_1}\mathcal{R}(v)-\mathcal{P}_{t_2}\mathcal{R}(v)|\leq  Ce^{-\lambda t_2}$. Sending $t_1\to\infty$ and applying \eqref{e.461}, we complete the proof.
\end{proof}

\section{Proofs of Theorem~\ref{t.partitionfunction} and Theorem~\ref{t.mainth}}

\label{s.5}

Recall from \eqref{e.delogZt1} that
\[
\log Z_t+\gamma t=\int_0^t\int_{\mathbb{T}^d} u(s,y)\xi(s,y)dyds-\tfrac12\int_0^t \tilde{\mathcal{R}}(u(s))ds
\]
where $\tilde{\mathcal{R}}=\mathcal{R}-2\gamma$,
\begin{equation}\label{e.uU}
u(s)=u(s;\nu)=Z_s^{-1}\mathcal{U}(s;\nu),
\end{equation}
where $\nu$ is an arbitrary probability measure on $\mathbb{T}^d$. By the definition of $\gamma$ in \eqref{e.defgamma}, we have
\[
\int_{\mathcal{M}_1(\mathbb{T}^d)} \tilde{\mathcal{R}}(u) \pi_\infty(du)=0.
\]
The goal in this section is to prove Theorem~\ref{t.partitionfunction}: as $t\to\infty$,
\begin{equation}\label{e.4201}
\frac{\log Z_t+\gamma t}{\sqrt{t}}\Rightarrow N(0,\sigma^2).
\end{equation}
With the above convergence, the proof of Theorem~\ref{t.mainth} goes as follows: for any $\nu\in\mathcal{M}_1(\mathbb{T}^d)$ and $x\in\mathbb{T}^d$, we write
\begin{equation}
\frac{\log \mathcal{U}(t,x;\nu)+\gamma t}{\sqrt{t}}=\frac{\log Z_t+\gamma t}{\sqrt{t}}+\frac{\log u(t,x;\nu)}{\sqrt{t}}.
\end{equation}
For $t>1$,  by \eqref{e.densityY1} we have
\begin{equation}\label{e.utUL}
\frac{\inf_{x,y\in\mathbb{T}^d} \mathcal{Z}_{t,t-1}^\omega(x,y)}{\sup_{x,y\in\mathbb{T}^d} \mathcal{Z}_{t,t-1}^\omega(x,y)}\leq u(t,x;\nu) \leq \frac{\sup_{x,y\in\mathbb{T}^d} \mathcal{Z}_{t,t-1}^\omega(x,y)}{\inf_{x,y\in\mathbb{T}^d} \mathcal{Z}_{t,t-1}^\omega(x,y)}.
\end{equation}
Thus, in light of Lemma~\ref{l.ulbd}, we conclude thatthere exists a constant $C>0$ such that
$$
\mathbf{E}|\log u(t,x;\nu)|\leq C\quad \mbox{ for all }
t>1,x\in\mathbb{T}^d.
$$
 As a result we obtain
\[
\frac{\log u(t,x;\nu)}{\sqrt{t}}\to0,\quad\mbox{as }t\to+\infty,
\]
in probability and the proof of Theorem~\ref{t.mainth} is complete.

\subsection{Construction of the corrector}

The  purpose of the present section is to construct the solution of
the equation
$-{\mathcal L}\chi=\tilde\mathcal{R}$, where $\tilde\mathcal{R}$ is defined in
\eqref{022304-21} and ${\mathcal L}$ is a properly understood
generator of the transition probability semigroup $(\mathcal{P}_t)_{t\geq0}$, see
Remark \ref{rmk012304-21} below. The field $\chi:D^\infty(\mathbb{T}^d)\to\mathbb{R}$
is called the  corrector and is a crucial object in the
proof of the CLT for $\log Z_t$.
We start with the following lemma:
\begin{lemma}\label{l.bdRv}
For each  $p\in[1,+\infty)$ there exists a constant $C>0$ such that
\begin{equation}\label{e.bduinfinity}
\sup_{t\geq0} \mathbf{E}[\|u(t;v)\|_{L^\infty(\mathbb{T}^d)}^p]\leq C(1+\|v\|_{L^\infty(\mathbb{T}^d)}^p),
\end{equation}
and
\begin{equation}
\label{042304-21}
|{\mathcal P}_t \tilde{\mathcal{R}}(v)|\leq  C(e^{-\lambda
  t}1_{t>1}+(1+\|v\|_{L^\infty(\mathbb{T}^d)})1_{t\in[0,1]}),\quad \mbox{for any }v\in D^\infty(\mathbb{T}^d),\,t\ge0
.
\end{equation}
\end{lemma}

\begin{proof}
The case of $t>1$ is implied by Lemma~\ref{l.bdut1} and Proposition~\ref{p.tailPtR}. For $t\in[0,1]$, we have
\[
\begin{aligned}
\mathcal{P}_t\mathcal{R}(v)&=\mathbf{E} \int_{\mathbb{T}^{2d}} R(x-y)u(t,x;v)u(t,y;v)dxdy\\
& \leq\mathbf{E} \|u(t;v)\|_{L^\infty(\mathbb{T}^d)} \int_{\mathbb{T}^{2d}} R(x-y)u(t,x;v)dxdy =\mathbf{E} \|u(t;v)\|_{L^\infty(\mathbb{T}^d)},
\end{aligned}
\]
and it suffices to apply \eqref{e.she14} to conclude the proof.
\end{proof}

\begin{lemma}\label{l.conPRv}
For any $v\in D^\infty(\mathbb{T}^d)$, the function $\mathcal{P}_t\tilde{\mathcal{R}}(v)$ is continuous in $t\geq0$.
\end{lemma}

\begin{proof}
It is a direct consequence of Lemma~\ref{l.uconti} and \eqref{e.bduinfinity}.
\end{proof}

The time dependent corrector field is defined as
\begin{equation}
\label{chit}
\chi(t,v):=\int_0^t\mathcal{P}_s\tilde{\mathcal{R}}(v)ds,\quad \quad t\ge0,\,v\in D^\infty(\mathbb{T}^d).
\end{equation}

We have the following result.
\begin{proposition}
\label{prop032403-21}
The function $\chi:[0,+\infty)\times D^\infty(\mathbb{T}^d)\to\mathbb{R}$ satisfies
\begin{equation}
\label{e.pt}
\frac{d\chi(t,v)}{dt}=\mathcal{P}_t\tilde{\mathcal{R}}(v),\quad t\ge0.
\end{equation}
In addition, for any $t\ge0$ and $v\in D^\infty(\mathbb{T}^d)$, we have
\begin{equation}
\label{cellt}
\lim_{\delta\to0} \delta^{-1}[\mathcal{P}_\delta \chi(t,v)-\chi(t,v)]  ={\mathcal P}_t \tilde{\mathcal{R}}(v)-\tilde{\mathcal{R}}(v).
\end{equation}
\end{proposition}

\begin{proof}
 First, \eqref{e.pt} is a consequence of Lemma~\ref{l.conPRv}. To show \eqref{cellt}, for any $\delta>0, t\geq0$ and $v\in D^\infty(\mathbb{T}^d)$, we have
\[
\begin{aligned}
\delta^{-1}[\mathcal{P}_\delta\chi(t,v)-\chi(t,v)]=&\delta^{-1} [\int_0^t{\mathcal P}_{s+\delta} \tilde{\mathcal{R}}(v)ds-\int_0^t{\mathcal P}_s \tilde{\mathcal{R}}(v)ds]\\
=&\delta^{-1}[\int_t^{t+\delta} {\mathcal P}_{s} \tilde{\mathcal{R}}(v)ds-\int_0^\delta {\mathcal P}_s \tilde{\mathcal{R}}(v)ds].
\end{aligned}
\]
Sending $\delta\to0$ and applying Lemma~\ref{l.conPRv} again, the proof is complete.
\end{proof}

By virtue of Lemma~\ref{l.bdRv}, we can define $\chi: D^\infty(\mathbb{T}^d)\to \mathbb{R}$
\begin{equation}\label{e.defcorrector}
\chi(v):=\int_0^\infty \mathcal{P}_t\tilde{\mathcal{R}}(v) dt.
\end{equation}
 \begin{remark}
\label{rmk012304-21}
Analogously to \eqref{cellt}, we can show that
\begin{equation}
\label{032304-21}
\lim_{\delta\to0} \delta^{-1}[\mathcal{P}_\delta \chi(v)-\chi(v)]   =-\tilde{\mathcal{R}}(v),\quad\quad  v
\in D^\infty(\mathbb{T}^d).
\end{equation}
The left hand side of \eqref{032304-21} can be treated as a pointwise
definition of the generator ${\mathcal L}$  of the semigroup   $(\mathcal{P}_t)_{t\geq0}$.
\end{remark}
As a
 consequence of Lemma~\ref{l.bdRv}, the following proposition holds:
\begin{proposition}
\label{prop022503-21}
We have for any $v\in D^\infty(\mathbb{T}^d)$,
\begin{equation}\label{e.bdchi}
|\chi(v)|\leq C(1+\|v\|_{L^\infty(\mathbb{T}^d)}),
\end{equation}
\begin{equation}
\label{cellt1}
\sup_{v\in D^\infty(\mathbb{T}^d)} |\chi(v)-\chi(t,v)|\leq  \frac{C}{\lambda}e^{-\lambda
  t}, \quad\quad t\ge 1.
\end{equation}
\end{proposition}
\begin{proof}
It is a direct consequence of \eqref{042304-21}.
\end{proof}

\subsection{Frechet gradient of corrector}

Using the corrector $\chi$ defined in \eqref{e.defcorrector} we
shall be able to write $\log Z_t+\gamma t$ as a sum of a continuous square
integrable martingale and a term of order  $1$, see \eqref{ztNt} -
\eqref{e.4183} below. This fact  is  crucial in our proof of the CLT.
The quadratic variation of the martingale can be expressed in terms of
an appropriately defined   Frechet gradient of the corrector. The
present section is devoted to provide a definition of such an object. Note that
the latter  is not completely obvious, due to the fact that $\chi$ is
defined on a set $D^\infty(\mathbb{T}^d)$ that has no interior points.

For any $t\geq0$, $z\in\mathbb{T}^d$, $v\in D^\infty(\mathbb{T}^d)$ and $h\in
L^2(\mathbb{T}^d)$, we let
\begin{equation}\label{e.defUz}
U(t,x;z,v)=\frac{\mathcal{U}(t,x;z)}{\int_{\mathbb{T}^d} \mathcal{U}(t,x';v)dx'}
\end{equation}
and
\begin{equation}\label{e.defUh}
U(t,x;h,v)=\int_{\mathbb{T}^d} U(t,x;z,v)h(z)dz
=\frac{\mathcal{U}(t,x;h)}{\int_{\mathbb{T}^d} \mathcal{U}(t,x';v)dx'},\quad x\in\mathbb{T}^d.
\end{equation}
We extend the definition of $\mathcal{P}_t\mathcal{R}$ to an open subset of
$L^2(\mathbb{T}^d)$ containing $D^\infty(\mathbb{T}^d)$.
For any $t\geq0$ and $v\in L^2(\mathbb{T}^d)$ such that $\int_{\mathbb{T}^{d}}\mathcal{U}(t,x;v)dx\neq 0$,  define
\begin{equation}\label{e.defPtomega}
\mathscr{P}_t(v):=\int_{\mathbb{T}^{2d}} R(x-y) \frac{\mathcal{U}(t,x;v)\mathcal{U}(t,y;v)}{\left(\int_{\mathbb{T}^d}\mathcal{U}(t,x';v)dx'\right)^2}dxdy.
\end{equation}
Therefore,  $\mathcal{P}_t\mathcal{R}(\cdot)$ coincides with $\mathbf{E}[\mathscr{P}_t(\cdot)]$ on $D^\infty(\mathbb{T}^d)$, but the latter is defined on a much bigger set. Define
\begin{equation}\label{e.L2plus}
L_+^2(\mathbb{T}^d):= \{f\in L^2(\mathbb{T}^d): f\geq 0,\, \,\|f\|_{L^2(\mathbb{T}^d)}\neq 0\}.
\end{equation}
For any $v\in L_+^2(\mathbb{T}^d)$, we let
\begin{equation}\label{e.deftildev}
\tilde{v}(x)=\|v\|_{L^1(\mathbb{T}^d)}^{-1}v(x).
\end{equation}

We have the following lemma for the Frechet derivative of $\mathscr{P}_t(v)$.
\begin{lemma}\label{l.fredeomega}
For any $t\geq0, v\in D^\infty(\mathbb{T}^d)$ and almost every realization of
$\xi$, $\mathscr{P}_t$ is twice Frechet differentiable at $v$ and the
respective first and second order Frechet derivatives equal
\begin{equation}\label{e.fredePomega}
\begin{aligned}
\mathcal{D}\mathscr{P}_t(v)(z)=&\,2\int_{\mathbb{T}^{2d}}R(x-y)U(t,x;z,v )u(t,y;v )dxdy\\
&-2\|U(t;z,v)\|_{L^1(\mathbb{T}^d)}\int_{\mathbb{T}^{2d}}R(x-y)u(t,x;v )u(t,y;v ) dxdy,
\end{aligned}
\end{equation}
\begin{equation}\label{e.frede2}
\begin{aligned}
&\mathcal{D}^2\mathscr{P}_t(v)(z_1,z_2)\\
&=2\int_{\mathbb{T}^{2d}} R(x-y)U(t,x;z_1,v)U(t,y;z_2,v)dxdy \\
&-4\|U(t;z_2,v)\|_{L^1(\mathbb{T}^d)}\int_{\mathbb{T}^{2d}} R(x-y)U(t,x;z_1,v)u(t,y;v)dxdy\\%  \int_{\mathbb{T}^d} U(t,x';z_2,v)dx'\\
&-4\|U(t;z_1,v)\|_{L^1(\mathbb{T}^d)}\int_{\mathbb{T}^{2d}} R(x-y)U(t,x;z_2,v)u(t,y;v)dxdy \\
&+6\|U(t;z_1,v)\|_{L^1(\mathbb{T}^d)}\|U(t;z_2,v)\|_{L^1(\mathbb{T}^d)}\int_{\mathbb{T}^{2d}} R(x-y)u(t,x;v)u(t,y;v)dxdy.
\end{aligned}
\end{equation}

\end{lemma}

\begin{proof}
We only consider $t>0$ here (the case of $t=0$ is easy to analyze).  One can write
\[
\|\mathcal{U}(t;v )\|_{L^1(\mathbb{T}^d)}=\int_{\mathbb{T}^{2d}} \mathcal{U}(t,x;y )v(y)dxdy,
\] where  $\mathcal{U}(t,x;y )$ is the Green's function of SHE, which, for almost all $\xi$,  is a continuous and positive function in $(x,y)\in\mathbb{T}^{2d}$. Fix such a realization of $\xi$ and $v\in D^\infty(\mathbb{T}^d)$, we have $\|\mathcal{U}(t;v )\|_{L^1(\mathbb{T}^d)}>0$, thus, there exists $\delta>0$ (which could depend on $\xi$) such that $
\|\mathcal{U}(t;\tilde{v})\|_{L^1(\mathbb{T}^d)}>0$ for any $\tilde{v}$ with $\|\tilde{v}-v\|_{L^2(\mathbb{T}^d)}<\delta$. Then $\mathscr{P}_t(\cdot)$ is well-defined for all such $\tilde{v}$, and it is a straightforward calculation to check that for any $h\in L^2(\mathbb{T}^d)$, as $\delta\to0$
\[
\delta^{-1}[\mathscr{P}_t(v+\delta h )-\mathscr{P}_t(v)]\to \langle\mathcal{D}\mathscr{P}_t(v),h\rangle_{L^2(\mathbb{T}^d)}
\]
The proof of \eqref{e.frede2} is similar.
\end{proof}

Using the functional $\mathcal{R}$, see \eqref{fR}, we can rewrite
\eqref{e.fredePomega} in the form
\begin{equation}\label{e.4121}
\begin{aligned}
\langle \mathcal{D}\mathscr{P}_t(v),h\rangle_{L^2(\mathbb{T}^d)}
=&2 \mathcal{R}(U(t;h,v),u(t;v))\\
&-2\mathcal{R}(u(t;v))\int_{\mathbb{T}^d} U(t,x;h,v)dx,
\end{aligned}
\end{equation}
for any $v\in D^\infty(\mathbb{T}^d), h \in L^2(\mathbb{T}^d)$.
By further restricting to $h\in L^2_+(\mathbb{T}^d)$, with $\tilde{h}=h/\|h\|_{L^1(\mathbb{T}^d)}$, we can write
\begin{equation}\label{e.4122}
\begin{aligned}
&\langle \mathcal{D}\mathscr{P}_t(v),h\rangle_{L^2(\mathbb{T}^d)}\\
&=2\int_{\mathbb{T}^{3d}} R(x-y)[u(t,x;\tilde{h})-u(t,x;v)]u(t,y;v)U(t,x';h,v)dxdydx'\\
&=2\,\mathcal{R}(u(t;\tilde{h})-u(t;v),u(t;v))\|U(t;h,v)\|_{L^1(\mathbb{T}^d)}.
\end{aligned}
\end{equation}

Next, we prove several technical lemmas on the Frechet derivatives,
where we will only use the following estimate:
\begin{equation}\label{e.conditionR}
|\mathcal{R}(f,g)| \leq \hat R_*\|f\|_{L^2(\mathbb{T}^d)}\|g\|_{L^2(\mathbb{T}^d)},\quad f,g\in L^2(\mathbb{T}^d),
\end{equation}
where
\begin{equation}
\label{Rstar}
\hat R_*:=\sup_{k\in\mathbb{Z}^d}\hat{r}_k
\end{equation} and $(\hat r_k)_{k\in\mathbb{Z}^d}$ are the Fourier coefficients of $R$, cf. \eqref{e.defrk}. By our assumption of $R\geq0$ and $\int R=1$, we actually have $\hat R_*=\hat{r}_0=1$.

\begin{lemma}\label{l.bdDe1}
For any $p\in[1,+\infty)$ and $T>0$, there exists $C>0$, depending only on
  $ p$ and $T$, such that
\[
\begin{aligned}
&\sup_{t\in[0,T]}\big(\mathbf{E}\langle |\mathcal{D}\mathscr{P}_t(v)|,h\rangle_{L^2(\mathbb{T}^d)}^p\big)^{1/p} \leq C\|h\|_{L^2(\mathbb{T}^d)}\|v\|_{L^2(\mathbb{T}^d)}(1+\|v\|_{L^2(\mathbb{T}^d)})\\
&\leq  C\|h\|_{L^2(\mathbb{T}^d)}\|v\|_{L^\infty(\mathbb{T}^d)},\quad   \, v\in D^\infty(\mathbb{T}^d), \, 0\leq h\in L^2(\mathbb{T}^d).
\end{aligned}
\]
\end{lemma}

\begin{lemma}\label{l.bdDe2}
For any $p\in[1,+\infty)$ and $T>0$, there exists $C>0$, depending only on
  $  p$ and $T$, such that
\[
\begin{aligned}
\sup_{t\in[0,T]}\big(\mathbf{E}  \langle h_1,
|\mathcal{D}^2\mathscr{P}_t(v)|h_2\rangle_{L^2(\mathbb{T}^d)}^p\big)^{1/p}\leq
C\|h_1\|_{L^2(\mathbb{T}^d)}\|h_2\|_{L^2(\mathbb{T}^d)}(1+\|v\|_{L^\infty(\mathbb{T}^d)} )
\end{aligned}
\]
for all $v\in D^\infty(\mathbb{T}^d)$, $0\leq h_1,h_2\in L^2(\mathbb{T}^d)$.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{l.bdDe1}]
From \eqref{e.fredePomega},
we have
\[
\begin{aligned}
\langle |\mathcal{D}\mathscr{P}_t(v)|,h\rangle_{L^2(\mathbb{T}^d)} \leq &C\|U(t;h,v)\|_{L^2(\mathbb{T}^d)}\|u(t;v)\|_{L^2(\mathbb{T}^d)}(1+\|u(t;v)\|_{L^2(\mathbb{T}^d)})\\
\leq & C\|U(t;h,v)\|_{L^2(\mathbb{T}^d)}\|u(t;v)\|_{L^\infty(\mathbb{T}^d)}.
\end{aligned}
\]
Here we have used the fact that $\|f\|_{L^2(\mathbb{T}^d)}^2 \leq
\|f\|_{L^\infty(\mathbb{T}^d)}$ for any $f\in D(\mathbb{T}^d)$. The proof is
complete by  applying Lemma~\ref{l.lshe1} and   the H\"older inequality.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{l.bdDe2}]
From \eqref{e.frede2}, we have
\begin{equation}\label{e.4141}
\begin{aligned}
 &\langle h_1, |\mathcal{D}^2\mathscr{P}_t(v)|h_2\rangle_{L^2(\mathbb{T}^d)}\\
 &\leq C\|U(t;h_1,v)\|_{L^2(\mathbb{T}^d)}\|U(t;h_2,v)\|_{L^2(\mathbb{T}^d)}(1+\|u(t;v)\|_{L^\infty(\mathbb{T}^d)} ).

 \end{aligned}
 \end{equation}
 Similarly, the proof is complete by applying Lemma~\ref{l.lshe1} and  H\"older inequality.
\end{proof}

Now we define the gradient of the time dependent corrector: for any
$v\in D^\infty(\mathbb{T}^d)$, $t,\,T\geq0$, we let
\begin{equation}\label{e.defDe11}
\mathcal{D}\mathcal{P}_t\mathcal{R}(v):=\mathbf{E}[\mathcal{D}\mathscr{P}_t(v)],\quad\quad \mathcal{D}^2\mathcal{P}_t\mathcal{R}(v):=\mathbf{E}[\mathcal{D}^2\mathscr{P}_t(v)],
\end{equation}
where $\mathcal{D}\mathscr{P}_t$ and $\mathcal{D}^2\mathscr{P}_t$ are given by
\eqref{e.fredePomega} and \eqref{e.frede2} respectively. Let
\begin{equation}\label{e.defgraCor}
\mathcal{D}\chi(T,v):=\int_0^T\mathcal{D}\mathcal{P}_t\mathcal{R}(v)dt.
\end{equation}
The following two lemmas concern the continuity of the Frechet
derivatives, and their proofs also only rely on the property of $\mathcal{R}$
through \eqref{e.conditionR}.
\begin{lemma}\label{l.contDe1}
There exists $C=C(T)>0$ depending on $T>0$ such that
\begin{equation}
\label{032304-21z}
\begin{aligned}
&\sup_{t\in[0,T]}|\langle \mathcal{D}\mathcal{P}_t\mathcal{R}(v_1)-\mathcal{D}\mathcal{P}_t\mathcal{R}(v_2),h\rangle_{L^2(\mathbb{T}^d)}|  \\
&\leq  C\|h\|_{L^2(\mathbb{T}^d)}\|v_1-v_2\|_{L^2(\mathbb{T}^d)}P(\|v_1\|_{L^2(\mathbb{T}^d)},\|v_2\|_{L^2(\mathbb{T}^d)})
\end{aligned}
\end{equation}
for all  $v_1,v_2\in D^\infty(\mathbb{T}^d)$ and $h\in L^2(\mathbb{T}^d)$.
Here $P(\cdot,\cdot)$ is some polynomial function.

Moreover, we have also
\begin{equation}
\label{032304-21a}
\begin{aligned}
&\sup_{t\in[0,T]}|\langle h_1, \big(\mathcal{D}^2\mathcal{P}_t\mathcal{R}(v_1)-\mathcal{D}^2\mathcal{P}_t\mathcal{R}(v_2)\big)h_2\rangle_{L^2(\mathbb{T}^d)}| \\
&\leq  C \|h_1\|_{L^2(\mathbb{T}^d)}\|h_2\|_{L^2(\mathbb{T}^d)}\|v_1-v_2\|_{L^2(\mathbb{T}^d)}P(\|v_1\|_{L^2(\mathbb{T}^d)},\|v_2\|_{L^2(\mathbb{T}^d)}),
\end{aligned}
\end{equation}
for all $v_1,v_2\in D^\infty(\mathbb{T}^d)$ and $h_1,h_2\in
L^2(\mathbb{T}^d)$. Here, again,
 $P(\cdot,\cdot)$ is some polynomial.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{l.contDe1}]
We only prove \eqref{032304-21z}, as the argument for
\eqref{032304-21a} follows the same lines. From Lemma~\ref{l.lshe1} and the fact that $\mathcal{U}$ solves a linear equation, we know that for any $t\in[0,T]$, $h\in L^2(\mathbb{T}^d), v_1,v_2\in D^\infty(\mathbb{T}^d)$, we have

(i) $\big(\mathbf{E}[\|u(t;v_1)-u(t;v_2)\|_{L^2(\mathbb{T}^d)}^p]\big)^{1/p}\leq C\|v_1-v_2\|_{L^2(\mathbb{T}^d)}(1+\|v_1\|_{L^2(\mathbb{T}^d)})$,

(ii) $\big(\mathbf{E}[ \|U(t;h,v_1)-U(t;h,v_2)\|_{L^2(\mathbb{T}^d)}^p] \big)^{1/p}\leq C \|h\|_{L^2(\mathbb{T}^d)}\|v_1-v_2\|_{L^2(\mathbb{T}^d)}$.

To see   (i) note that
\begin{align*}
\|u(t;v_1)-u(t;v_2)\|_{L^2(\mathbb{T}^d)}\le \frac{
  \|\mathcal{U}(t;v_1-v_2)\|_{L^2(\mathbb{T}^d)}}{\|\mathcal{U}(t;v_2)\|_{L^1(\mathbb{T}^d)}}+\frac{\|u(t;v_1)\|_{L^2(\mathbb{T}^d)}
  \|\mathcal{U}(t;v_1-v_2)\|_{L^1(\mathbb{T}^d)}}{\|\mathcal{U}(t;v_2)\|_{L^1(\mathbb{T}^d)}},
\end{align*}
The estimate follows then directly from Lemma~\ref{l.lshe1} and
estimate \eqref{e.negammZ} (recall that
$Z_t=\|\mathcal{U}(t;v)\|_{L^1(\mathbb{T}^d)}$).
 Formula (ii) follows by similar considerations.

 Using \eqref{e.fredePomega}, the above estimates, \eqref{e.conditionR}
and the H\"older inequality, it is a straightforward calculation to derive that
\[
\begin{aligned}
&|\langle \mathbf{E}\mathcal{D}\mathscr{P}_t(v_1),h\rangle_{L^2(\mathbb{T}^d)}-\langle \mathbf{E}\mathcal{D}\mathscr{P}_t(v_2),h\rangle_{L^2(\mathbb{T}^d)}|\\
&\leq C \|h\|_{L^2(\mathbb{T}^d)}\|v_1-v_2\|_{L^2(\mathbb{T}^d)}P(\|v_1\|_{L^2(\mathbb{T}^d)},\|v_2\|_{L^2(\mathbb{T}^d)}
\end{aligned}
\]
for some polynomial function $P$. The proof is complete.
\end{proof}

 \subsubsection{Estimate on $ \mathcal{D}\mathcal{P}_t\mathcal{R}(v)$}
For the gradient of the corrector $\mathcal{D}\chi(T,v)$ to be well-defined with $T$ approaching to infinity, one needs to refine the estimates on $\mathcal{D}\mathcal{P}_t\mathcal{R}(v)$ for large $t$. Here is the main result  of this section.
  \begin{proposition}\label{p.gracor}
There exists $C,\lambda>0$ depending only on $R,d$ such that
\begin{equation}
\label{011703-21z}
\begin{aligned}
\Big|\langle {\mathcal D}\mathcal{P}_t\mathcal{R}( v),h\rangle_{L^2(\mathbb{T}^d)}\Big|\leq C\|h\|_{L^1(\mathbb{T}^d)}\big(1_{[1,+\infty)}(t)e^{-\lambda t} +1_{[0,1]}(t)\|v\|_{L^\infty(\mathbb{T}^d)}     \big)
\end{aligned}
\end{equation}
for all $t\geq 0$, $v\in D^\infty(\mathbb{T}^d) $ and $ h\in L^2(\mathbb{T}^d)$.
\end{proposition}

Directly from Proposition \ref{p.gracor} we conclude the following.
\begin{corollary}\label{cor013103-21}
For $C,\lambda>0$ as in Proposition \ref{p.gracor}  we have
\begin{equation}
\label{011703-21}
\begin{aligned}
&\|{\mathcal D}\mathcal{P}_t\mathcal{R}( v)\|_{L^\infty(\mathbb{T}^d)} \leq C\big(1_{t>1}e^{-\lambda t} +1_{t\in[0,1]}\|v\|_{L^\infty(\mathbb{T}^d)}     \big),\\
&\|\mathcal{D}\chi(t,v)\|_{L^\infty(\mathbb{T}^d)} \leq C(1+\|v\|_{L^\infty(\mathbb{T}^d)})
\end{aligned}
\end{equation}
for all $t\geq 0$, $v\in D^\infty(\mathbb{T}^d)$.
\end{corollary}

The proof of Proposition~\ref{p.gracor} relies on the following lemmas.
\begin{lemma}\label{l.471}
For any $p\in[1,+\infty)$ there exist $C,\lambda>0$ depending only on $R,d,p$ such that
\[
\mathbf{E}\|u(t;\nu_1)-u(t;\nu_2)\|_{L^\infty(\mathbb{T}^d)}^p \leq Ce^{-\lambda
  t},\quad\quad  t> 1,\,\nu_1,\nu_2\in \mathcal{M}_1(\mathbb{T}^d).
\]
\end{lemma}
\begin{proof}
By
Proposition~\ref{p.couplingCon}, for any $\tfrac{t}{2}<k<N(t)-2$, we have
\[
\begin{aligned}
&\mathbf{E}\|  u(t;\nu_1)-  u(t;\nu_2)\|_{L^\infty(\mathbb{T}^d)}^p\\
&\leq C\big( \mathbf{E}\|  u(t;\nu_1)-  u_k(t)\|_{L^\infty(\mathbb{T}^d)}^p+\mathbf{E}\|  u(t;\nu_2)-
u_k(t)\|_{L^\infty(\mathbb{T}^d)}^p\big)\leq Ce^{-\lambda k}
\end{aligned}
\]
for some $C,\lambda>0$ only depending on $R,d,p$. The proof is complete.
\end{proof}

We have the following estimate on $U$:
\begin{lemma}\label{l.bdU}
If $p\in[1,+\infty)$, then there exists $C$ depending on $R,d,p$ such that
\[
\mathbf{E}\left[\|U(t;h,v)\|_{L^1(\mathbb{T}^d)}^p\right]\leq
C\|h\|_{L^1(\mathbb{T}^d)}^p,\quad \quad v\in D(\mathbb{T}^d),\, 0\leq h\in L^1(\mathbb{T}^d),\,t\ge0.
\]
\end{lemma}
\begin{proof}
By definition, we have
\[
\int_{\mathbb{T}^d} U(t,x;h,v)dx=\frac{\int_{\mathbb{T}^d} \mathcal{U}(t,x;h)dx}{\int_{\mathbb{T}^d}\mathcal{U}(t,x;v)dx}.
\]
Through a time reversal of the noise $\xi(\cdot,\cdot)\mapsto \xi(t-\cdot,\cdot)$, we have, for each $t\geq 0$,
\[
\frac{\int_{\mathbb{T}^d} \mathcal{U}(t,x;h)dx}{\int_{\mathbb{T}^d}\mathcal{U}(t,x;v)dx}\stackrel{\rm law}{=}\frac{\int_{\mathbb{T}^d}\mathcal{U}(t,x;\mathbbm{1})h(x)dx}{\int_{\mathbb{T}^d} \mathcal{U}(t,x;\mathbbm{1})v(x)dx}=\frac{\int_{\mathbb{T}^d}u(t,x;\mathbbm{1})h(x)dx}{\int_{\mathbb{T}^d} u(t,x;\mathbbm{1})v(x)dx},
\]
where  $\mathbbm{1}(x)\equiv1$. Thus, it suffices to estimate the moments of the r.h.s. We have
\[
\begin{aligned}
\frac{\int_{\mathbb{T}^d}u(t,x;\mathbbm{1})h(x)dx}{\int_{\mathbb{T}^d} u(t,x;\mathbbm{1})v(x)dx} \leq \|h\|_{L^1(\mathbb{T}^d)} \sup_{x\in\mathbb{T}^d} u(t,x;\mathbbm{1}) (\inf_{y\in\mathbb{T}^d} u(t,x;\mathbbm{1}))^{-1}.
\end{aligned}
\]
For $t>1$, we use \eqref{e.utUL} and apply Lemma~\ref{l.ulbd} to complete the proof. For $t\leq 1$, we use the estimate
\begin{align*}
&
\sup_{x\in\mathbb{T}^d} u(t,x;\mathbbm{1}) (\inf_{x\in\mathbb{T}^d} u(t,x;\mathbbm{1}))^{-1} =
       \sup_{x\in\mathbb{T}^d}\mathcal{U}(t,x;\mathbbm{1}) \sup_{x\in\mathbb{T}^d} \mathcal{U}(t,x;\mathbbm{1})^{-1}\\
&
\le \tfrac{1}{2} \sup_{x\in\mathbb{T}^d}\mathcal{U}(t,x;\mathbbm{1})^2+ \tfrac{1}{2} \sup_{x\in\mathbb{T}^d} \mathcal{U}(t,x;\mathbbm{1})^{-2}
 \end{align*}
 and Lemma~\ref{l.U1} to complete the proof.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{p.gracor}]
Assuming first that $h\in L_+^2(\mathbb{T}^d)$, recall  from  \eqref{e.4122}, we have
\[
\begin{aligned}
\langle {\mathcal D}\mathcal{P}_t\mathcal{R}(v),h\rangle_{L^2(\mathbb{T}^d)}=2\mathbf{E}[ \mathcal{R}(u(t;\tilde{h})-u(t;v),u(t;v))\|U(t;h,v)\|_{L^1(\mathbb{T}^d)}].
\end{aligned}
\]
Here, as we recall, $\tilde{h}=\|h\|_{L^1(\mathbb{T}^d)}^{-1}h$.
For $t>1 $, we use the fact that \[
|\langle {\mathcal D}\mathcal{P}_t\mathcal{R}(v),h\rangle_{L^2(\mathbb{T}^d)}| \leq 2\mathbf{E}[\|u(t;\tilde{h})-u(t;v)\|_{L^\infty(\mathbb{T}^d)}\|u(t;v)\|_{L^1(\mathbb{T}^d)}\|U(t;h,v)\|_{L^1(\mathbb{T}^d)}]
\]
and apply Lemmas~\ref{l.471}, \ref{l.bdU}
and the H\"older inequality to derive that
\[
|\langle {\mathcal D}\mathcal{P}_t\mathcal{R}(v),h\rangle_{L^2(\mathbb{T}^d)}| \leq Ce^{-\lambda t}\|h\|_{L^1(\mathbb{T}^d)}.
\]
For $t\leq 1$,
we use a different expression: in \eqref{e.fredePomega}, we  use the fact that
\[
\int_{\mathbb{T}^{2d}}R(x-y)U(t,x;h,v)u(t,y;v)dxdy\leq\|R\|_{L^1(\mathbb{T}^d)} \|u(t;v)\|_{L^\infty(\mathbb{T}^d)}\|U(t;h,v)\|_{L^1(\mathbb{T}^d)}
\] to derive that
\[
|\langle \mathcal{D}\mathscr{P}_t(v), h\rangle_{L^2(\mathbb{T}^d)}| \leq 4 \|R\|_{L^1(\mathbb{T}^d)}\|u(t;v)\|_{L^\infty(\mathbb{T}^d)}\|U(t;h,v)\|_{L^1(\mathbb{T}^d)},
\]
Taking the expectation, applying Lemma~\ref{l.bdU}, estimate
\eqref{e.she14}   and
the H\"older inequality, we obtain
\[
\begin{aligned}
|\langle {\mathcal D}\mathcal{P}_t\mathcal{R}(v),h\rangle_{L^2(\mathbb{T}^d)}|
&\leq C \|v\|_{L^\infty(\mathbb{T}^d)}\|h\|_{L^1(\mathbb{T}^d)}.
\end{aligned}
\]

To generalize the proof to $h\in
L^2(\mathbb{T}^d)$, it suffices to write $h=h_+-h_-$ and apply the above discussion to $h_+,h_-$ separately. The proof is complete.
\end{proof}

Define \begin{equation}\label{e.defDchi}
\mathcal{D}\chi(v)=\int_0^\infty\mathcal{D}\mathcal{P}_t\mathcal{R}(v)dt, \quad\quad v\in D^\infty(\mathbb{T}^d).
\end{equation}
As a direct consequence of \eqref{e.4122} and Corollary~\ref{cor013103-21}, we have
\begin{proposition}\label{p.bddeCor}
For any $v\in D^\infty(\mathbb{T}^d)$, we have
\begin{equation}\label{e.vDv}
\langle v,\mathcal{D}\chi(v)\rangle_{L^2(\mathbb{T}^d)}=0.
\end{equation}
In addition there exists $C>0$ such that
\begin{equation}\label{e.4171}
\|\mathcal{D} \chi(v)\|_{L^\infty(\mathbb{T}^d)} \leq
C(1+\|v\|_{L^\infty(\mathbb{T}^d)}),\quad v\in D^\infty
\end{equation}
and
\begin{equation}
\label{010104-21}
\sup_{v\in D^\infty(\mathbb{T}^d)}\| {\mathcal D}\chi(t,v)-{\mathcal D}\chi
(v)\|_{L^\infty(\mathbb{T}^d)}\leq \frac{C}{\lambda}e^{-\lambda t},\quad t>1.
\end{equation}
\end{proposition}

\subsection{A semimartingale decomposition} The goal in this section is to obtain a semimartingale decomposition of the process $\{\mathcal{P}_s\mathcal{R}(u(t;v))\}_{t\geq0}$,
for any fixed $s\geq0$ and $v\in D^\infty(\mathbb{T}^d)$. Recall that, for any (fixed) $v\in D^\infty(\mathbb{T}^d)$, we have
\[
\mathcal{P}_s\mathcal{R}(v)=\mathbf{E}[\mathcal{R}(u(s;v))]=\mathbf{E}[\mathscr{P}_s(v)].
\]
From now on, assume that, in \eqref{e.defPtomega}, we use an
independent copy of noise, denoted by $\tilde{\xi}$, to generate the
random variable
$\mathscr{P}_s(v)$. To emphasize the dependence, we will write $\mathscr{P}_s(v)=\mathscr{P}_s(v;\tilde{\xi})$, so for any $v\in D^\infty(\mathbb{T}^d)$,
\[
\mathcal{P}_s\mathcal{R}(v)=\mathbf{E}_{\tilde{\xi}}[\mathscr{P}_s(v;\tilde{\xi})],
\]
and
\[
\mathcal{P}_s\mathcal{R}(u(t;v,\xi))=\mathbf{E}_{\tilde{\xi}}[\mathscr{P}_s(u(t;v,\xi);\tilde{\xi})],
\]
where $\mathbf{E}_{\tilde{\xi}}$ denotes the expectation only with respect to
$\tilde{\xi}$, and the $u(t;v,\xi)$ in the above expression is based on the original noise $\xi$.

By Lemma~\ref{l.fredeomega}, we know that for almost every realization
of $\tilde{\xi}$, $\mathscr{P}_s(\cdot;\tilde{\xi})$ is twice Frechet
differentiable at any $v\in D^\infty(\mathbb{T}^d)$. The same proof actually
shows that $\mathscr{P}_s(\cdot;\tilde{\xi})$ is infinitely Frechet
differentiable.
The idea is to fix the realization of  $\tilde{\xi}$, apply
the It\^o formula to $\mathscr{P}_s(u(t;v);\tilde{\xi})$, then take the expectation with respect to $\tilde{\xi}$ to obtain a semimartingale decompoisition of $\mathcal{P}_s\mathcal{R}(u(t;v))$.

\subsubsection{The case of smooth noise and initial data}

In this section, we assume that $v\in D(\mathbb{T}^d)\cap C^\infty(\mathbb{T}^d)$ (so it is automatically in $D^\infty(\mathbb{T}^d)$) and the noise $\xi$  satisfies Assumption~\ref{a.smooth}, i.e., it contains only finitely many Fourier modes. Since the initial data is fixed, we omit its dependence to write $u(t)=u(t;v)$. By Proposition~\ref{p.spde}, we know that $u(t)$ is a strong solution to the following SPDE
\begin{equation}\label{e.spde1}
\begin{aligned}
du(t)=\mathscr{A}(u(t))dt+\sum_{k\in\mathbb{Z}^d} \mathscr{B}_k(u(t))dw_k(t),
\end{aligned}
\end{equation}
with the operators $\mathscr{A},\mathscr{B}_k$ defined in \eqref{e.defAB}.

Fix $s\geq0$ and a realization of $\tilde{\xi}$, applying It\^o
formula, see e.g. \cite[Theorem 4.32, p. 106]{daza},

we have
\begin{equation}\label{e.itoPs}
\begin{aligned}
\mathscr{P}_s(u(t);\tilde{\xi})-\mathscr{P}_s(v;\tilde{\xi})&=\int_0^t \langle \mathcal{D}\mathscr{P}_s(u(r);\tilde{\xi}),u(r)dW(r)\rangle_{L^2(\mathbb{T}^d)} \\
&+\int_0^t \langle \mathcal{D}\mathscr{P}_s(u(r);\tilde{\xi}), \mathscr{A}(u(r))\rangle_{L^2(\mathbb{T}^d)} dr\\
&+\tfrac12\sum_{k\in\mathbb{Z}^d}\int_0^t \langle \mathscr{B}_k(u(r)), \mathcal{D}^2\mathscr{P}_s(u(r);\tilde{\xi})\mathscr{B}_{k}(u(r))\rangle_{L^2(\mathbb{T}^d)}dr.
\end{aligned}
\end{equation
Here we have used the fact that
\[
\langle \mathcal{D}\mathscr{P}_s(v;\tilde{\xi}),v\rangle_{L^2(\mathbb{T}^d)}=0, \quad\quad \mbox{ if } v\in D^\infty(\mathbb{T}^d),
\]
which comes from \eqref{e.4122}.

Freeze $(u(r;\xi))_{r\in[0,t]}$ and take the expectation with respect to $\tilde{\xi}$ in \eqref{e.itoPs}. By  Lemmas~\ref{l.bdDe1} and \ref{l.bdDe2}, we interchange the expectation and the   integral to obtain
\begin{equation}\label{e.itoPs1}
\begin{aligned}
\mathcal{P}_s\mathcal{R}(u(t) )-\mathcal{P}_s\mathcal{R}(v)=&\int_0^t \langle
\mathcal{D}\mathcal{P}_s\mathcal{R}(u(r)),u(r)dW(r)\rangle_{L^2(\mathbb{T}^d)} \\
&+\int_0^t
\mathcal{L}\mathcal{P}_s\mathcal{R}(u(r))dr,
\end{aligned}
\end{equation}
with
\begin{equation}\label{e.defL1}
\begin{aligned}
\mathcal{L}\mathcal{P}_s\mathcal{R}(v):=&\langle \mathcal{D}\mathcal{P}_s\mathcal{R}(v),\mathscr{A}(v)\rangle_{L^2(\mathbb{T}^d)}\\
&+\frac12\sum_{k\in\mathbb{Z}^d} \langle \mathscr{B}_k(v), \mathcal{D}^2\mathcal{P}_s\mathcal{R}(v )\mathscr{B}_{k}(v)\rangle_{L^2(\mathbb{T}^d)}.
\end{aligned}
\end{equation}
for any $s\geq0$ and $v\in  D(\mathbb{T}^d)\cap C^\infty(\mathbb{T}^d)$. In addition
$\mathcal{D}\mathcal{P}_s\mathcal{R}$ and $\mathcal{D}^2\mathcal{P}_s\mathcal{R}$ are defined in \eqref{e.defDe11}.

The following lemma holds:
\begin{lemma}\label{l.generator1}
For any $T>0$, $v\in D(\mathbb{T}^d)\cap C^\infty(\mathbb{T}^d)$ and $p\in[1,+\infty)$, we have
\begin{equation}\label{e.bddrift}
\sup_{s,r\in[0,T]}\mathbf{E}|\mathcal{L}\mathcal{P}_s\mathcal{R}( u(r))|^p \leq C(T,p,v).
\end{equation}
For any $s\geq0$ and $v$ as above,  we have
\begin{equation}\label{e.generatorcon}
\begin{aligned}
\mathcal{L}\mathcal{P}_s\mathcal{R}(v )=\lim_{\delta\to0}\delta^{-1}[\mathcal{P}_{s+\delta}\mathcal{R}(v)-\mathcal{P}_s\mathcal{R}(v)].
\end{aligned}
\end{equation}
  \end{lemma}
  \begin{proof}
  First, by Lemmas~\ref{l.bdDe1} and \ref{l.bdDe2}, we have
  \[
  \begin{aligned}
  |\mathcal{L}\mathcal{P}_s\mathcal{R}( u(r))| &\leq C e^{CT}\|\mathscr{A}(u(r))\|_{L^2(\mathbb{T}^d)}\|u(r)\|_{L^\infty(\mathbb{T}^d)}\\
  &+Ce^{CT}\sum_{k\in\mathbb{Z}^d}\|\mathscr{B}_k(u(r))\|_{L^2(\mathbb{T}^d)}^2(1+\|u(r)\|_{L^\infty(\mathbb{T}^d)}).
  \end{aligned}
  \]
  Then we apply Lemmas~\ref{l.bdspde} and \ref{l.bdRv} to conclude
  \eqref{e.bddrift}. To show \eqref{e.generatorcon}, fix
  $s\geq0$. Define a continuous local martingale
  \[
  \mathbf{M}_s(t):=\int_0^t \langle \mathcal{D}\mathcal{P}_s\mathcal{R}(u(r)),u(r)dW(r)\rangle_{L^2(\mathbb{T}^d)},
  \]
   By the equality \eqref{e.defDe11} and  Lemma~\ref{l.bdDe1}, its quadratic variation satisfies
  \[
  \langle \mathbf{M}_s\rangle_t=\int_0^t\mathcal{R}\Big(\mathcal{D}\mathcal{P}_s\mathcal{R}(u(r))u(r)\Big)dr\leq C\int_0^t \|u(r)\|_{L^\infty(\mathbb{T}^d)}^4dr.
  \]
 Next we apply Lemma~\ref{l.bdRv} to conclude that $\mathbf{E}[\langle
 \mathbf{M}_s\rangle_t]\leq Ct$, so $\mathbf{M}_s$ is a bona fide, square integrable,
 continuous trajectory martingale. Therefore from \eqref{e.itoPs1} we have
  \[
  \begin{aligned}
  \delta^{-1}[\mathcal{P}_{s+\delta}\mathcal{R}(v)-\mathcal{P}_s\mathcal{R}(v)]=&\delta^{-1}\big(\mathbf{E}[\mathcal{P}_s\mathcal{R}(u(\delta))]-\mathcal{P}_s\mathcal{R}(v)\big)\\
  =&\delta^{-1}\int_0^\delta\mathbf{E}[\mathcal{L}\mathcal{P}_s\mathcal{R}(u(r))]dr.
  \end{aligned}
  \]
By \eqref{e.bddrift}, Lemma~\ref{l.contDe1}, and the fact that the sample path of $u(\cdot)$ lies in $C([0,\infty),H^k(\mathbb{T}^d))$ for any $k\geq1$, we conclude the proof.
    \end{proof}

 \begin{remark}
 Formula \eqref{e.itoPs1}
yields  the semimartingale decomposition of the process
$(\mathcal{P}_s\mathcal{R}(u(t)))_{t\ge0}$ for any $s\ge0$.
\end{remark}

The drift term in \eqref{e.itoPs1} can be simplified after a time integration:
\begin{lemma}\label{l.interchangeIn}
For any $T,t\geq0$,
\[
\int_0^T \big(\int_0^t  \mathcal{L}\mathcal{P}_s\mathcal{R}(u(r))  dr\big) ds=\int_0^t \mathcal{P}_T\mathcal{R}(u(r))dr-\int_0^t\mathcal{R}(u(r))dr.
\]
\end{lemma}

\begin{proof}
First, by \eqref{e.bddrift}, we have
\[
\int_0^T \big(\int_0^t  \mathcal{L}\mathcal{P}_s\mathcal{R}(u(r))  dr\big) ds=\int_0^t \big(\int_0^T  \mathcal{L}\mathcal{P}_s\mathcal{R}(u(r))  ds\big)dr.
\]
It remains to use \eqref{e.generatorcon} to derive that for any $v\in D(\mathbb{T}^d)\cap  C^\infty(\mathbb{T}^d)$,
\[
{\mathcal P}_T \mathcal{R}(v)-\mathcal{R}(v)=\int_0^T \mathcal{L} \mathcal{P}_s\mathcal{R}(v)ds.
\]
The proof is complete.
\end{proof}

Using the semimartingale decomposition in \eqref{e.itoPs1} and Lemma~\ref{l.interchangeIn}, we have
\begin{proposition}\label{p.madefinal}
Suppose that Assumption~\ref{a.smooth} is in force and further assume
that the initial condition $v\in D(\mathbb{T}^d)\cap C^\infty(\mathbb{T}^d)$. Then,
for any $T,t\geq0$ we have
\begin{equation}\label{e.matdecom}
\begin{aligned}
\int_0^t \mathcal{R}(u(r;v))dr=&\int_0^t \langle \mathcal{D}\chi(T,u(r;v)),u(r;v)dW(r)\rangle_{L^2(\mathbb{T}^d)}\\
&-\chi(T,u(t;v))+\chi(T,v)+\int_0^t \mathcal{P}_T \mathcal{R}(u(r;v))dr.
\end{aligned}
\end{equation}
\end{proposition}

\begin{proof}
In \eqref{e.itoPs1}, we integrate $s$ from $0$ to $T$ and apply Lemma~\ref{l.interchangeIn} to derive
\[
\begin{aligned}
\chi(T,u(t))-\chi(T,v)&=\int_0^T\big(\int_0^t \langle \mathcal{D}\mathcal{P}_s\mathcal{R}(u(r)),u(r)dW(r)\rangle_{L^2(\mathbb{T}^d)}\big) ds\\
&+\int_0^t \mathcal{P}_T \mathcal{R}(u(r))dr-\int_0^t \mathcal{R}(u(r))dr.
\end{aligned}
\]
For the first term in the right hand side, we use the stochastic
Fubini theorem, see e.g. \cite[Theorem 4.33, p. 110]{daza},  and obtain
\[
\begin{aligned}
&\int_0^T\big(\int_0^t \langle \mathcal{D}\mathcal{P}_s\mathcal{R}(u(r)),u(r)dW(r)\rangle_{L^2(\mathbb{T}^d)}\big) ds\\
&=\int_0^t \langle \int_0^T  \mathcal{D}\mathcal{P}_s\mathcal{R}(u(r))ds, u(r)dW(r)\rangle_{L^2(\mathbb{T}^d)}=\int_0^t \langle \mathcal{D}\chi(T,u(r)),u(r)dW(r)\rangle_{L^2(\mathbb{T}^d)},
\end{aligned}
\]
where $\mathcal{D}\chi$ was defined in \eqref{e.defgraCor}. The proof is complete.
\end{proof}

\subsubsection{The general case}
The following is the main result of this section, which removes the smoothness assumptions on the noise and the initial data in Proposition~\ref{p.madefinal}.
\begin{proposition}\label{p.41711}
For any $T,t\geq0$ and $v\in D^\infty(\mathbb{T}^d)$, the decomposition \eqref{e.matdecom} holds.
\end{proposition}

A direct consequence of the above proposition is the decomposition of
the additive functional appearing in the formula for $\log Z_t$, see
\eqref{e.delogZt}, into a stochastic integral with respect to $dW(t)$
and   boundary terms.
\begin{corollary}
For any $t\geq0$ and $v\in D^\infty(\mathbb{T}^d)$, we have
\begin{equation}\label{e.matdecom2}
\begin{aligned}
\int_0^t \tilde{\mathcal{R}}(u(r;v))dr=\int_0^t &\langle \mathcal{D}\chi(u(r;v)),u(r;v)dW(r)\rangle_{L^2(\mathbb{T}^d)}\\
&-\chi(u(t;v))+\chi(v).
\end{aligned}
\end{equation}
\end{corollary}
\begin{proof}
First, in both sides of \eqref{e.matdecom}, we subtract $2\gamma t$ to obtain
\begin{equation}
\begin{aligned}
\int_0^t \tilde{\mathcal{R}}(u(r;v))dr=&\int_0^t \langle \mathcal{D}\chi(T,u(r;v)),u(r;v)dW(r)\rangle_{L^2(\mathbb{T}^d)}\\
&-\chi(T,u(t;v))+\chi(T,v)+\int_0^t \mathcal{P}_T\tilde{\mathcal{R}}(u(r;v))dr.
\end{aligned}
\end{equation}
Sending $T\to\infty$ in the above equation, and applying
Lemma~\ref{l.bdRv} together with estimates \eqref{cellt1} and \eqref{010104-21}, we conclude the proof.
\end{proof}

The rest of the section is devoted to proving Proposition~\ref{p.41711}. Fix $v\in D^\infty(\mathbb{T}^d)$ from now on.
Let us introduce some notations. Recall that
$(\hat r_k)_{k\in\mathbb{Z}^d}$ are the Fourier coefficients of $R(\cdot)$, see \eqref{e.defrk}.
For any $N>0$, define
\begin{equation}
R_N(x)=\sum_{|k|\leq N} \hat{r}_k e^{i2\pi k\cdot x}.
\end{equation}
Let $\xi_N$ be a finite dimensional noise that is white in time and smooth in space with the spatial covariance function  $R_N$, and
\[
dW_N(t,x)=\xi_N(t,x)dt=\sum_{|k|\leq N} \sqrt{\hat{r}_k} e^{i2\pi k\cdot x}d{w}_k(t).
\]
Given $v\in D^\infty(\mathbb{T}^d)$, we choose and fix a sequence of $v_N\in D(\mathbb{T}^d)\cap C^\infty(\mathbb{T}^d)$ such that

\begin{equation}
\label{vN}
\|v_N-v\|_{L^\infty(\mathbb{T}^d)}\to0, \quad\quad N\to\infty.
\end{equation}

Let $\mathcal{U}_N(t;v)$ be the solution to SHE with noise $\xi_N$ and initial
data $v$. Let furthermore
\[
u_N(t,x;v)=\frac{\mathcal{U}_N(t,x;v)}{\int_{\mathbb{T}^d} \mathcal{U}_N(t,x';v)dx'},
\]
\[
U_N(t,x;h,v)=\frac{\mathcal{U}_N(t,x;h)}{\int_{\mathbb{T}^d} \mathcal{U}_N(t,x';v)dx'}.
\]
Define
\[
\begin{aligned}
&\mathcal{R}_N(v_1,v_2)=\int_{\mathbb{T}^{2d}}R_N(x-y)v_1(x)v_2(y)dxdy, \quad\quad v_1,v_2\in L^2(\mathbb{T}^d),\\
&\mathcal{R}_N(v)=\mathcal{R}_N(v,v), \quad\quad v\in L^2(\mathbb{T}^d)
\end{aligned}
\]
and denote $\mathcal{P}_t^N$ the semigroup associated with $u_N$, so
\[
\mathcal{P}_t^N\mathcal{R}_N(v)=\mathbf{E}[\mathcal{R}_N(u_N(t;v))].
\]
Define
\[
\begin{aligned}
&\chi_N(T,v)=\int_0^T \mathcal{P}_t^N\mathcal{R}_N(v)dt-2\gamma T,\\
&\mathscr{P}_t^N(v):=\int_{\mathbb{T}^{2d}} R_N(x-y) \frac{\mathcal{U}_N(t,x;v)\mathcal{U}_N(t,y;v)}{\left(\int_{\mathbb{T}^d}\mathcal{U}_N(t,x';v)dx'\right)^2}dxdy,
\end{aligned}
\]
and
\[
\mathcal{D}\chi_N(T,v)=\int_0^T \mathbf{E}[\mathcal{D}\mathscr{P}_t^N(v)]dt.
\]
Then by Proposition~\ref{p.madefinal}, we have
\begin{equation}\label{e.4151}
\begin{aligned}
\int_0^t \mathcal{R}_N(u_N&(r;v_N))dr=\int_0^t \langle \mathcal{D}\chi_N(T,u_N(r;v_N)),u_N(r;v_N)dW_N(r)\rangle_{L^2(\mathbb{T}^d)}\\
&-\chi_N(T,u_N(t;v_N))+\chi_N(T,v_N)+\int_0^t \mathcal{P}_T^N \mathcal{R}_N(u_N(r;v_N))dr.
\end{aligned}
\end{equation}

The goal is to pass to the limit  in \eqref{e.4151} as
$N\to\infty$. We first state the following elementary lemmas.

\begin{lemma}\label{l.conuN}
For any $T\geq0$, $p\in[1,+\infty)$, $h\in L^2(\mathbb{T}^d)$,

  the following convergences take place
\[
\begin{aligned}
&\sup_{t\in[0,T]}\mathbf{E}\|u_N(t;v)-u(t;v)\|_{L^2(\mathbb{T}^d)}^p\to0,\\
&\sup_{t\in[0,T]}\mathbf{E}\|u_N(t;v_N)-u(t;v)\|_{L^2(\mathbb{T}^d)}^p\to0,\\
&\sup_{t\in[0,T]}\mathbf{E}\|U_N(t;h,v)-U(t;h,v)\|_{L^2(\mathbb{T}^d)}^p\to0,\quad
\mbox{as $N\to\infty$}.
\end{aligned}
\]
Furthermore, for any $t>0$, as $N\to\infty$,
\begin{equation}
\label{012604-21}
\mathbf{E} \big(\int_{\mathbb{T}^{2d}} [\mathcal{U}_N(t,x;y)-\mathcal{U}(t,x;y)]^2 dxdy\big)^p\to0.
\end{equation}
\end{lemma}

The proof is  standard and we present it in
Section~\ref{s.sheOther} of the appendix.

\begin{lemma}\label{l.conRN}
(i) For any $v_1,v_2\in L^2(\mathbb{T}^d)$, we have
\[
\begin{aligned}
&|\mathcal{R}_N(v_1,v_2)-\mathcal{R}(v_1,v_2)|\leq  \|v_1\|_{L^2(\mathbb{T}^d)}\|v_2\|_{L^2(\mathbb{T}^d)},\\
&\mathcal{R}_N(v_1,v_2)\to \mathcal{R}(v_1,v_2), \quad\quad N\to\infty.
\end{aligned}
\]
 (ii) For any $v_j\in L^2(\mathbb{T}^d),j=1,\ldots,4$, we have
\[
|\mathcal{R}_N(v_1,v_2)-\mathcal{R}_N(v_3,v_4)| \leq  \|v_1-v_3\|_{L^2(\mathbb{T}^d)}\|v_2\|_{L^2(\mathbb{T}^d)}+ \|v_2-v_4\|_{L^2(\mathbb{T}^d)}\|v_3\|_{L^2(\mathbb{T}^d)}.
\]
\end{lemma}

\begin{proof}
It suffices to write the integrals in the Fourier domain
\[
\mathcal{R}_N(v_1,v_2)=\sum_{|k|\leq N}\hat{r}_k\hat{v}_1(k)\hat{v}_2^*(k), \quad\quad \mathcal{R}(v_1,v_2)=\sum_{k\in \mathbb{Z}^d}\hat{r}_k\hat{v}_1(k)\hat{v}_2^*(k),
\]
and use the fact that $0\leq \hat{r}_k\leq \hat R_*=1$ to conclude the
proof of (i). Similarly, for (ii), by virtue of the Cauchy-Schwarz
inequality, we have
\[
\begin{aligned}
&|\mathcal{R}_N(v_1,v_2)-\mathcal{R}_N(v_3,v_4)|\leq  \sum_{k\in\mathbb{Z}^d}   \big|\hat{v}_1(k)\hat{v}_2^*(k)-\hat{v}_3(k)\hat{v}_4^*(k)\big|\\
&\leq \|v_1-v_3\|_{L^2(\mathbb{T}^d)}\|v_2\|_{L^2(\mathbb{T}^d)}+ \|v_2-v_4\|_{L^2(\mathbb{T}^d)}\|v_3\|_{L^2(\mathbb{T}^d)}.
\end{aligned}
\]
The proof is complete.
\end{proof}

With the above two lemmas, we have
\begin{proposition}\label{p.4161}
For any $t,T\geq0$,

  we have
\begin{equation}
\begin{aligned}
&\chi_N(T,v_N)\to \chi(T,v),\\
&\chi_N(T,u_N(t;v_N))\to \chi(T,u(t;v)), \\
&\int_0^t \mathcal{R}_N(u_N(r;v_N))dr\to \int_0^t \mathcal{R}(u(r;v))dr,\\
&\int_0^t \mathcal{P}_T^N \mathcal{R}_N(u_N(r;v_N))dr\to  \int_0^t \mathcal{P}_T\mathcal{R}(u(r;v))dr,\quad\mbox{as $N\to\infty$}.
\end{aligned}
\end{equation}
The convergence holds in $L^p(\Omega)$ for any $p\in[1,+\infty)$.
\end{proposition}

\begin{proof}
The proofs for all cases are similar, so we take the last one as an example. One can write
\[
\int_0^t \mathcal{P}_T^N \mathcal{R}_N(u_N(r;v_N))dr=\int_0^t \mathbf{E}[\mathcal{R}_N(u_N(r+T;v_N))|\mathcal{F}_r]dr,
\]
and
\[
\int_0^t \mathcal{P}_T\mathcal{R}(u(r;v))dr=\int_0^t \mathbf{E}[\mathcal{R}(u(r+T;v))|\mathcal{F}_r]dr.
\]
So the difference can be decomposed into two terms:
\[
\begin{aligned}
&I_1=\int_0^t \mathbf{E}[\mathcal{R}_N(u_N(r+T;v_N))|\mathcal{F}_r]dr-\int_0^t \mathbf{E}[\mathcal{R}_N(u(r+T;v))|\mathcal{F}_r]dr,\\
&I_2=\int_0^t \mathbf{E}[\mathcal{R}_N(u(r+T;v))|\mathcal{F}_r]dr-\int_0^t \mathbf{E}[\mathcal{R}(u(r+T;v))|\mathcal{F}_r]dr.
\end{aligned}
\]
We apply Lemmas~\ref{l.conuN} and \ref{l.conRN} to conclude that as $N\to\infty$, $I_1,I_2\to0$ in $L^p(\Omega)$ for any $p\in[1,+\infty)$.
\end{proof}

To deal with the martingale term in \eqref{e.4151}, we need the following lemmas:

\begin{lemma}\label{l.appFDe}
For any $f\in D^\infty(\mathbb{T}^d)$ and $t>0$, we have as $N\to\infty$,
\[
\mathcal{D}\mathcal{P}_t^N\mathcal{R}_N(f)\to \mathcal{D}\mathcal{P}_t\mathcal{R}(f) \quad  \mbox{ in } \quad  L^2(\mathbb{T}^d).
\]
\end{lemma}
\begin{proof}
To simplify the notation we shall   assume that $\hat R_*=1$. Fix $t>0$. Recall from \eqref{e.fredePomega} that
\[
\begin{aligned}
 \mathcal{D}\mathscr{P}_t^N(f)(z)=&2\int_{\mathbb{T}^{2d}}R_N(x-y)U_N(t,x;z,f)u_N(t,y;f)dxdy\\
&-2\|U_N(t;z,f)\|_{L^1(\mathbb{T}^d)}\int_{\mathbb{T}^{2d}}R_N(x-y)u_N(t,x;f)u_N(t,y;f)dxdy
\end{aligned}
\]
and $\mathcal{D}\mathcal{P}_t^N\mathcal{R}_N(f)=\mathbf{E}\mathcal{D}\mathscr{P}_t^N(f)$.

To prove the result, it suffices to show as $N\to\infty$,
\begin{equation}
\mathbf{E}\int_{\mathbb{T}^d}|\mathcal{D}\mathscr{P}_t^N(f)(z)-\mathcal{D}\mathscr{P}_t(f)(z)|^2dz\to0.
\end{equation}
One can write the error as $\mathcal{D}\mathscr{P}_t^N(f)(z)-\mathcal{D}\mathscr{P}_t(f)(z)=2(I_1^{(N)}(z)+I_2^{(N)}(z))$ with
\[
I_1^{(N)}(z)=\int_{\mathbb{T}^{2d}}R_N(x-y)U_N(t,x;z,f)u_N(t,y;f)dxdy-\int_{\mathbb{T}^{2d}}R(x-y)U(t,x;z,f)u(t,y;f)dxdy
\]
and
\[
\begin{aligned}
I_2^{(N)}(z)=&\|U(t;z,f)\|_{L^1(\mathbb{T}^d)}\int_{\mathbb{T}^{2d}}R(x-y)u(t,x;f)u(t,y;f)dxdy\\
&-\|U_N(t;z,f)\|_{L^1(\mathbb{T}^d)}\int_{\mathbb{T}^{2d}}R_N(x-y)u_N(t,x;f)u_N(t,y;f)dxdy.
\end{aligned}
\]

By Lemma~\ref{l.conRN}, we have
\[
\begin{aligned}
|I_1^{(N)}(z)|\leq&\|U_N(t;z,f)-U(t;z,f)\|_{L^2(\mathbb{T}^d)}\|u_N(t;f)\|_{L^2(\mathbb{T}^d)}\\
&+\|U(t;z,f)\|_{L^2(\mathbb{T}^d)}\|u_N(t;f)-u(t;f)\|_{L^2(\mathbb{T}^d)}\\
&+|\int_{\mathbb{T}^{2d}} (R_N(x-y)-R(x-y))U(t,x;z,f)u(t,y;f)dxdy|\\
&=:J_1^{(N)}(z)+J_2^{(N)}(z)+J_3^{(N)}(z)
\end{aligned}
\]
To deal with $J_1$, we first
note that
\begin{align*}
&
|U_N(t,x;z,f)-U(t,x;z,f)|\le
  |\mathcal{U}_N(t,x;z)-\mathcal{U}(t,x;z)|\cdot\|\mathcal{U}_N(t;f)\|_{L^1(\mathbb{T}^d)}^{-1}\\
&
+U(t,x;z,f) \|\mathcal{U}_N(t;f)\|_{L^1(\mathbb{T}^d)}^{-1}\|\mathcal{U}_N(t;f)-\mathcal{U}(t;f)\|_{L^1(\mathbb{T}^d)}.
\end{align*}
Using Lemma~\ref{l.conuN}, \eqref{e.estiGreen1} and Lemma~\ref{l.MMZ},
together with the H\"older inequality, we
conclude that for any $p\in[1,+\infty)$
\begin{align*}
\lim_{N\to+\infty}\mathbf{E}\left(\int_{\mathbb{T}^{2d}}|U_N(t,x;z,f)-U(t,x;z,f)|^2dxdz\right)^p=0.
\end{align*}
Invoking Lemma~\ref{l.conuN} and  the H\"older inequality we conclude that
\[
\mathbf{E}\|J_1^{(N)}\|_{L^2(\mathbb{T}^d)}^2=\mathbf{E}\Big[\|u_N(t;f)\|_{L^2(\mathbb{T}^d)}^2\int_{\mathbb{T}^{2d}}|U_N(t,x;z,f)-U(t,x;z,f)|^2dxdz\Big]
\to0.
\]
Similarly, we have
\[
\mathbf{E}\|J_2^{(N)}\|_{L^2(\mathbb{T}^d)}^2=\mathbf{E}\Big[\|u_N(t;f)-u(t;f)\|_{L^2(\mathbb{T}^d)}^2\int_{\mathbb{T}^{2d}} U(t,x;z,f)^2dxdz\Big]\to0.
\]
Concerning $J_3^{(N)}$, denote by $\hat{U}_k(t;z,f), \hat{u}_k(t;f)$  the
Fourier coefficients of $U(t;z,f)$ and $u(t;f)$, respectively. We have
\[
\mathbf{E}\|J_3^{(N)}\|_{L^2(\mathbb{T}^d)}^2 \leq \mathbf{E} \left[\sum_{|k|>N} \int_{\mathbb{T}^d}|\hat{U}_k(t;z,f)|^2dz \sum_{|k|>N} |\hat{u}_k(t;f)|^2\right]\to0,
\]
by virtue of the monotone convergence theorem.
This shows that $\mathbf{E}\|I_1^{(N)}\|_{L^2(\mathbb{T}^d)}^2 \to0$. The proof that
$\mathbf{E}\|I_2^{(N)}\|_{L^2(\mathbb{T}^d)}^2 \to0$ can be carried out along similar lines.
\end{proof}
\begin{lemma}\label{l.42011}
For any $T,t\geq0$, $p\in[1,+\infty)$,

 we have
\begin{equation}\label{e.41711}
\sup_{r\in[0,t],N>1}\mathbf{E}\big( \|\mathcal{D}\chi_N(T,u_N(r;v_N))\|_{L^2(\mathbb{T}^d)}^p+\|\mathcal{D}\chi(T,u(r;v))\|_{L^2(\mathbb{T}^d)}^p\big) \leq C.
\end{equation}
In addition, for any $r\in[0,t]$,
\begin{equation}\label{e.41712}
 \mathbf{E}\|\mathcal{D}\chi_N(T,u_N(r;v_N))-\mathcal{D}\chi(T,u(r;v))\|_{L^2(\mathbb{T}^d)}^p\to0, \mbox{ as } N\to\infty.
\end{equation}
\end{lemma}
\begin{proof}
To simplify the notations, we let $f_N=u_N(r;v_N)$ and $f=u(r;v)$.
Concerning \eqref{e.41711}, note that by Lemma~\ref{l.bdDe1} for a
fixed $T>0$ we can find $C>0$ such that
\[
\begin{aligned}
\|\mathcal{D}\chi_N(T,f_N)\|_{L^2(\mathbb{T}^d)}+\|\mathcal{D}\chi(T,f)\|_{L^2(\mathbb{T}^d)} \leq C\Big(\|f_N\|_{L^\infty(\mathbb{T}^d)} +\|f\|_{L^\infty(\mathbb{T}^d)}\Big).
\end{aligned}
\]
Applying Lemma~\ref{l.bdRv}, we conclude estimate \eqref{e.41711}.
For \eqref{e.41712}, we write
\[
\begin{aligned}
&\mathcal{D}\chi_N(T,f_N)-\mathcal{D}\chi(T,f)\\
&=\int_0^T\big[ \mathcal{D}\mathcal{P}_t^N\mathcal{R}_N(f_N)-\mathcal{D}\mathcal{P}_t\mathcal{R}(f)\big]dt\\
&=\int_0^T \big[\mathcal{D}\mathcal{P}_t^N\mathcal{R}_N(f_N)-\mathcal{D}\mathcal{P}_t^N\mathcal{R}_N(f)\big]dt+\int_0^T \big[\mathcal{D}\mathcal{P}_t^N\mathcal{R}_N(f)-\mathcal{D}\mathcal{P}_t\mathcal{R}(f)\big]dt.
\end{aligned}
\]

For the first term, by Lemma~\ref{l.contDe1}, we have
\[
\begin{aligned}
\|\mathcal{D}\mathcal{P}_t^N\mathcal{R}_N(f_N)-\mathcal{D}\mathcal{P}_t^N\mathcal{R}_N(f)\|_{L^2(\mathbb{T}^d)} \leq C\|f_N-f\|_{L^2(\mathbb{T}^d)}P(\|f_N\|_{L^2(\mathbb{T}^d)},\|f\|_{L^2(\mathbb{T}^d)})
\end{aligned}
\]
where $P$ is some given polynomial function. Then we apply Lemmas~\ref{l.bdRv} and \ref{l.conuN} to derive that
\[
\mathbf{E}\int_0^T \|\mathcal{D}\mathcal{P}_t^N\mathcal{R}_N(f_N)-\mathcal{D}\mathcal{P}_t^N\mathcal{R}_N(f)\|_{L^2(\mathbb{T}^d)}^pdt\to0,  \mbox{ as } N\to\infty.
\]

For the second term, we first note that, by Lemma~\ref{l.bdDe1}, we have
\[
\|\mathcal{D}\mathcal{P}_t^N\mathcal{R}_N(f)-\mathcal{D}\mathcal{P}_t\mathcal{R}(f)\|_{L^2(\mathbb{T}^d)}\leq C\|f\|_{L^\infty(\mathbb{T}^d)}.
\]
Then we apply Lemmas~\ref{l.bdRv} and \ref{l.appFDe}  to derive that
\[
\mathbf{E}\int_0^T \|\mathcal{D}\mathcal{P}_t^N\mathcal{R}_N(f)-\mathcal{D}\mathcal{P}_t\mathcal{R}(f)\|_{L^2(\mathbb{T}^d)}^pdt\to0, \mbox{ as }N\to\infty.
\]
The proof is complete.
\end{proof}

Finally, we can prove the convergence of the martingale term in \eqref{e.4151}:
\begin{proposition}\label{p.41712}
For any $t,T\geq0$,

   we have
\[
\begin{aligned}
\int_0^t& \langle \mathcal{D}\chi_N(T,u_N(r;v_N)),u_N(r;v_N)dW_N(r)\rangle_{L^2(\mathbb{T}^d)}\\
&\to \int_0^t \langle \mathcal{D}\chi(T,u(r;v)),u(r;v)dW(r)\rangle_{L^2(\mathbb{T}^d)},
\mbox{ as $N\to\infty$, in $L^2(\Omega)$.}
\end{aligned}
\]
\end{proposition}
\begin{proof}
To simplify the notation, we denote
\[
\begin{aligned}
&f_N(r,x)=\mathcal{D}\chi_N(T,u_N(r;v_N))(x)u_N(r,x;v_N),\\
&f(r,x)=\mathcal{D}\chi(T,u(r;v))(x)u(r,x;v),
\end{aligned}
\]
then the goal is to show
$\int_0^t\int_{\mathbb{T}^d}f_N(r,x)dW_N(r,x)\to\int_0^t \int_{\mathbb{T}^d}
f(r,x)dW(r,x)$ in $L^2(\Omega)$. The error can be written as a sum of  two terms:
\[
\begin{aligned}
I_{1,N}=&\int_0^t\int_{\mathbb{T}^d} [ f_N(r,x)-f(r,x)]dW_N(t,x),\\
I_{2,N}=&\int_0^t \int_{\mathbb{T}^d} f(r,x)dW_N(t,x)-\int_0^t \int_{\mathbb{T}^d}f(r,x)dW(t,x).
\end{aligned}
\]

For the first term, we have
\[
\begin{aligned}
\mathbf{E}[I_{1,N}^2]=&\mathbf{E}\int_0^t\int_{\mathbb{T}^d}[f_N(r,x)-f(r,x)][f_N(r,y)-f(r,y)]R_N(x-y)dxdy \\
&\leq \int_0^t\mathbf{E}\|f_N(r)-f(r)\|_{L^2(\mathbb{T}^d)}^2dr.
\end{aligned}
\]
We bound the $L^2$ norm by
\[
\begin{aligned}
\|f_N(r)-f(r)\|_{L^2(\mathbb{T}^d)} \leq& \|\mathcal{D}\chi_N(T,u_N(r;v_N))-\mathcal{D}\chi(T,u(r;v))\|_{L^2(\mathbb{T}^d)}\|u_N(r;v_N)\|_{L^\infty(\mathbb{T}^d)}\\
+&\|\mathcal{D}\chi(T,u(r;v))\|_{L^\infty(\mathbb{T}^d)}\|u_N(r;v_N)-u(r;v)\|_{L^2(\mathbb{T}^d)}
\end{aligned}
\]
Then it suffices to apply Lemmas~\ref{l.bdRv}, \ref{l.conuN}, \ref{l.42011} and \eqref{011703-21} to conclude that $\mathbf{E}[I_{1,N}^2]\to0$.

For the second term, with $\hat{f}_k(r)=\int_{\mathbb{T}^d} f(r,x)e^{-i2\pi k\cdot x}dx$, we have
\[
\mathbf{E}[I_{2,N}^2]=\sum_{|k|>N}\hat{r}_k\int_0^t  \mathbf{E} |\hat{f}_k(r)|^2 dr \leq \sum_{|k|>N}\int_0^t  \mathbf{E} |\hat{f}_k(r)|^2 dr,
\]
 which goes to zero as $N\to\infty$, by invoking the fact that
 \[
 \begin{aligned}
 \mathbf{E}\|f(r)\|_{L^2(\mathbb{T}^d)}^p \leq &\mathbf{E} \Big[\|\mathcal{D}\chi(T,u(r;v))\|_{L^2(\mathbb{T}^d)}^p\|u(r;v)\|_{L^\infty}^p\Big]\\
 \leq & C\mathbf{E}\Big[ (1+\|u(r;v)\|_{L^\infty(\mathbb{T}^d)})^p\|u(r;v)\|_{L^\infty}^p \Big]\leq C,
 \end{aligned}
 \]
 where we have used \eqref{011703-21} in the second inequality and
 \eqref{e.bduinfinity} in the third one. Here $C$ is a
 generic constant, independent of $N$.
\end{proof}

Now we can combine Propositions~\ref{p.4161} and \ref{p.41712} to conclude Proposition~\ref{p.41711}.

\subsection{Martingale CLT}
Define
\begin{equation}\label{e.4181}
{\frak z}_t:=\log Z_t+\gamma t=\int_0^t\int_{\mathbb{T}^d} u(s,y;\nu)\xi(s,y)dyds-\tfrac12\int_0^t \tilde{\mathcal{R}}(u(s;\nu))ds,
\end{equation}
where $\nu\in \mathcal{M}_1(\mathbb{T}^d)$. Recall that the goal is to show
\[
\frac{{\frak z}_t}{\sqrt{t}}\Rightarrow N(0,\sigma^2), \quad\quad \mbox{ as } t\to\infty.
\]
It is clear that as $t\to\infty$, the random variables
${\frak z}_2/\sqrt{t}$ tend to $0$   in probability,
thus, it suffices to consider
\[
{\frak z}_t-{\frak z}_2=\int_2^t \int_{\mathbb{T}^d} u(s,y;\nu)\xi(s,y)dyds-\tfrac12\int_2^t \tilde{\mathcal{R}}(u(s;\nu))ds.
\]
Since $u(2;\nu)$ is a continuous function almost surely,  by the Markov property, we can assume that in \eqref{e.4181}, we start from $\nu(dx)=v(x)dx$ with some fixed $v\in D^\infty(\mathbb{T}^d)$.

By \eqref{e.matdecom2}, we write
\begin{equation}
\label{ztNt}
{\frak z}_t={\mathcal N}_t +{\frak r}_t,
\end{equation}
 with the martingale term
\begin{equation}\label{e.4182}
{\mathcal N}_t
\int_0^t \left\langle u(r;v)(1-\tfrac12\mathcal{D}\chi(u(r;v))),dW(r)\right\rangle_{L^2(\mathbb{T}^d)},
\end{equation}
and the remainder term
\begin{equation}\label{e.4183}
{\frak r}_t=\tfrac12\chi(u(t;v))-\tfrac12\chi(v).
\end{equation}
The quadratic variation of $\mathcal{N}_t$ is
\[
\langle \mathcal{N}\rangle_t
\int_0^t \mathcal{R}\big(u(r;v)(1-\tfrac12\mathcal{D}\chi(u(r;v)))\big)dr,
\]
which satisfies
\[
\langle \mathcal{N}\rangle_t \leq \int_0^t\|u(r;v)\|_{L^\infty(\mathbb{T}^d)}^2(1+\tfrac12\|\mathcal{D}\chi(u(r;v))\|_{L^\infty(\mathbb{T}^d)})^2 dr.
\]
Applying Lemma~\ref{l.bdRv} and \eqref{e.4171}, we conclude that for any $p\in[1,+\infty)$,
\begin{equation}\label{e.bdNQVt}
\mathbb{E}[\langle \mathcal{N}\rangle_t^p] \leq Ct^p.
\end{equation}
Define
\begin{equation}\label{e.defsigma}
\sigma^2=\int_{\mathcal{M}_1(\mathbb{T}^d)}\mathcal{R}\big(u(1-\tfrac12\mathcal{D}\chi(u))\big)\pi_\infty(du).
\end{equation}
The following proposition  shows the convergence of the (rescaled) quadratic variation:
\begin{proposition}\label{p.conQV}
As $t\to\infty$,
\[
\mathbf{E}[|t^{-1}\langle \mathcal{N}\rangle_t-\sigma^2|^2]\to0.
\]
\end{proposition}

\begin{proof}
 For any $T>1$, define
\[
\langle \mathcal{N}^T\rangle_t=\int_0^t \mathcal{R}\big(u(r;v)(1-\tfrac12\mathcal{D}\chi(T,u(r;v)))\big)dr,
\]
and
\[
\sigma_T^2=\int_{\mathcal{M}_1(\mathbb{T}^d)}\mathcal{R}\big(u(1-\tfrac12\mathcal{D}\chi(T,u))\big)\pi_\infty(du).
\]
By Lemma~\ref{l.conRN}, we have
\[
\begin{aligned}
|\langle \mathcal{N}^T\rangle_t-\langle \mathcal{N}\rangle_t|\leq\int_0^t &\|u(r;v)\|_{L^\infty(\mathbb{T}^d)}^2\|\mathcal{D}\chi(T,u(r;v))-\mathcal{D}\chi(u(r;v))\|_{L^\infty(\mathbb{T}^d)}\\
&\times (1+\|\mathcal{D}\chi(T,u(r;v))\|_{L^\infty(\mathbb{T}^d)} +\|\mathcal{D}\chi(u(r;v))\|_{L^\infty(\mathbb{T}^d)})dr.
\end{aligned}
\]
Further applying \eqref{e.4171} and \eqref{010104-21}, we derive
\[
|\langle \mathcal{N}^T\rangle_t-\langle \mathcal{N}\rangle_t|\leq Ce^{-\lambda T}\int_0^t \|u(r;v)\|_{L^\infty(\mathbb{T}^d)}^2(1+\|u(r;v)\|_{L^\infty(\mathbb{T}^d)})dr.
\]
Similarly, we have
\[
|\sigma_T^2-\sigma^2|\leq  Ce^{-\lambda T}\int_{\mathcal{M}_1(\mathbb{T}^d)}\|u\|_{L^\infty}^2(1+\|u\|_{L^\infty(\mathbb{T}^d)})\pi_\infty(du).
\]
Thus, applying Lemma~\ref{l.bdRv}, we derive
\[
\sup_{t>0}\mathbf{E}[t^{-2}|\langle \mathcal{N}^T\rangle_t-\langle \mathcal{N}\rangle_t|^2]+|\sigma_T^2-\sigma^2| \leq Ce^{-\lambda T},
\]
so we only need to show that for each $T$,  as $t\to\infty$,
\begin{equation}\label{e.conQVT}
\mathbf{E}[|t^{-1}\langle \mathcal{N}^T\rangle_t-\sigma_T^2|^2]\to0.
\end{equation}
Fix $T>0$. By Lemmas~\ref{l.contDe1}, \ref{l.conRN} and Corollary~\ref{cor013103-21}, the functional $\mathsf{F}:D^\infty(\mathbb{T}^d)\to\mathbb{R}$
\[
\begin{aligned}
\mathsf{F}(u):=&\mathcal{R}\big(u(1-\tfrac12\mathcal{D}\chi(T,u))\big)\\
\end{aligned}
\]
satisfies
\[
|\mathsf{F}(u_1)-\mathsf{F}(u_2)|\leq \|u_1-u_2\|_{L^\infty(\mathbb{T}^d)} P(\|u_1\|_{L^\infty(\mathbb{T}^d)},\|u_2\|_{L^\infty(\mathbb{T}^d)}),
\]
where $P$ is some polynomial function. In other words, $\mathsf{F} $ is locally Lipschitz. For any $M>0$, let $\mathsf{F}_M:D^\infty(\mathbb{T}^d)\to\mathbb{R}$ be a global Lipschitz function that is a cutoff of $\mathsf{F}$ in the sense that  $\mathsf{F}_M(u)=\mathsf{F}(u)$ if $\|u\|_{L^\infty(\mathbb{T}^d)}\leq M$, and $|\mathsf{F}_M|\leq |\mathsf{F}|$. Applying Proposition~\ref{prop021203-21}, we have \[
\mathbf{E}[|t^{-1}\int_0^t \mathsf{F}_M(u(r;v))dr-\sigma_{T,M}^2|^2]\to0,
\]
where $\sigma_{T,M}^2=\int_{\mathcal{M}_1(\mathbb{T}^d)}\mathsf{F}_M(u)\pi_\infty(du)$. For the error induced by $M$, we have
\[
\begin{aligned}
&t^{-1} \int_0^t \mathbf{E}|\mathsf{F}_M(u(r;v))-\mathsf{F}(u(r;v))|^2 dr\\
&\leq 4t^{-1}\int_0^t \mathbf{E}\big[|\mathsf{F}(u(r;v))|^21_{\|u(r;v)\|_{L^\infty(\mathbb{T}^d)}>M}\big]dr \leq \frac{C}{\sqrt{M}},
\end{aligned}
\]
where for the last estimate we have applied the H\"older inequality and Lemma~\ref{l.bdRv}.
Similarly, we can show $\sigma_{T,M}^2\to \sigma_T^2$ as $M\to\infty$. The proof is complete.
\end{proof}

The following two lemmas  combine to complete the proof of Theorem~\ref{t.partitionfunction}.
\begin{lemma}\label{l.418remainder}
As $t\to\infty$, $\frac{1}{\sqrt{t}}{\frak r}_t\to0$ in probability.
\end{lemma}
\begin{lemma}\label{l.418martingale}
As $t\to\infty$, $\frac{1}{\sqrt{t}}\mathcal{N}_t\Rightarrow N(0,\sigma^2)$ in distribution.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{l.418remainder}]
By \eqref{e.bdchi}, we have
\[
|{\frak r}_t|\leq C(1+\|v\|_{L^\infty(\mathbb{T}^d)}+\|u(t;v)\|_{L^\infty(\mathbb{T}^d)}).
\]
Taking expectation and applying Lemma~\ref{l.bdRv}, we derive that $\mathbf{E}|{\frak r}_t|\leq C(1+\|v\|_{L^\infty(\mathbb{T}^d)})$. The proof is complete.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{l.418martingale}]
By the same proof for \eqref{e.bdNQVt}, we have for any $t\geq s$
\[
\mathbf{E} (\langle \mathcal{N}\rangle_t-\langle \mathcal{N}\rangle_s)^p\leq C(t-s)^p.
\]
Together with Proposition~\ref{p.conQV}, we conclude that the process $(\varepsilon^2\langle \mathcal{N}\rangle_{t/\varepsilon^2})_{t\geq0}$ converges in $C[0,\infty)$ to $(\sigma^2t)_{t\geq0}$, as $\varepsilon\to0$. Then by the martingale central limit theorem, we derive that \[
\varepsilon \mathcal{N}_{t/\varepsilon^2}\Rightarrow \sigma B_t
\]
in $C[0,\infty)$ in distribution, where $B$ is a standard Brownian motion. The proof is complete.
\end{proof}

The following lemma guarantees the non-degeneracy of $\gamma,\sigma$.
\begin{lemma}\label{l.nondege1}
We have $\gamma,\sigma\in(0,+\infty)$.
\end{lemma}

\begin{proof}
Recall that
\[
\begin{aligned}
\gamma=\tfrac12\int_{\mathcal{M}_1(\mathbb{T}^d)} \mathcal{R}(u)\pi_\infty(du),\quad\quad \sigma^2=\int_{\mathcal{M}_1(\mathbb{T}^d)}\mathcal{R}\big(u(1-\tfrac12\mathcal{D}\chi(u))\big)\pi_\infty(du).
\end{aligned}
\]
By \eqref{e.4171}, Lemma~\ref{l.bdRv}, Theorem~\ref{t.geomeEr} and the
fact that
$$
\mathcal{R}(f)\leq \|f\|_{L^2(\mathbb{T}^d)}^2 \leq \|f\|_{L^\infty(\mathbb{T}^d)}
\quad\mbox{ for any $f\in D^\infty(\mathbb{T}^d)$},
$$
 we conclude that $\gamma,\sigma^2<+\infty$. On the other hand, by the fact that for any $t>1$,
\[
\mathcal{R}(u(t))\geq \inf_{x\in\mathbb{T}^d} u(t,x)^2\geq \left(\frac{\inf_{x,y\in\mathbb{T}^d} \mathcal{Z}_{t,t-1}^\omega(x,y)}{\sup_{x,y\in\mathbb{T}^d} \mathcal{Z}_{t,t-1}^\omega(x,y)}\right)^2
\]
we have $\mathbf{E}[\mathcal{R}(u(t))]\geq c$ for some constant $c>0$, independent of
$t>1$. This implies that $\gamma>0$.

Next, for any $u\in D^\infty(\mathbb{T}^d)$, we let
$$
g=u(1-\tfrac12\mathcal{D}\chi(u))\in L^\infty(\mathbb{T}^d).
$$
Then (recall that
$\hat{r}_0 =1$, see \eqref{hr0})
\[
\mathcal{R}(g)=\sum_{k\in\mathbb{Z}^d} \hat{r}_k |\hat{g}_k|^2\geq |\hat{g}_0|^2.
\]
Since
\[
\hat{g}_0=\int_{\mathbb{T}^d} g(x)dx=1-\tfrac12\langle u,\mathcal{D}\chi(u)\rangle_{L^2(\mathbb{T}^d)}=1,
\]
where the last equality comes from \eqref{e.vDv}, we conclude that $\sigma^2\geq1$.
\end{proof}

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