# Moderate deviation principles for the WASEP

Linjie Zhao  School of Mathematics and Statistics & Hubei Key Laboratory of
Engineering Modeling and Scientific Computing, Huazhong University of Science
and Technology, Wuhan 430074, China<EMAIL_ADDRESS>
###### Abstract.

We study the weakly asymmetric simple exclusion process on the integer
lattice. Under suitable constraints on the strength of the weak asymmetry of
the dynamics, we prove moderate deviation principles for the fluctuation
fields when the process starts from stationary measures. As an application, we
obtain sample path moderate deviation principles for the occupation time of
the process in one dimension.

###### Key words and phrases:

Exclusion process; fluctuation fields; moderate deviations; occupation time.

Acknowledgments. The project is supported by the Fundamental Research Funds
for the Central Universities in China, and by the National Natural Science
Foundation of China with grant numbers 11971038 and 12371142.

##  1\. Introduction

The exclusion process is one of the most widely studied models in statistical
physics and in the theory of interacting particle systems. In the dynamics,
particles perform random walks on some graph subject to the exclusion rule,
that is, there is at most one particle at each site. The theory of
hydrodynamic limits studies the behaviors of large systems of interacting
particle systems at a macroscopic scale. Law of large numbers, fluctuations
and large deviation principles from hydrodynamic limits have been widely
investigated so far, see [13] and references therein. Moderate deviations
studies probabilities of rare events occurring with intermediate frequency,
and fills the gap between large deviations and fluctuations [4, Section 3.7].
As far as we know, moderate deviations from hydrodynamic limits was only
considered in [8, 23, 24] for the symmetric simple exclusion process, and in
[22] for a specific Ginzburg-Landau model.

One of the main purposes of this article is to study moderate deviations from
hydrodynamic limits for a more general class of exclusion processes, _i.e._ ,
the weakly asymmetric simple exclusion process (WASEP) on the integer lattice
$\mathbb{Z}^{d}$, $d\geq 1$, where particles jump to the left at rate $1/(2d)$
and to the right at rate $1/(2d)+\alpha n^{-\beta}/d$ in the $i$-th direction
for $1\leq i\leq d$. Here, $\alpha\geq 0$ and $\beta>0$ are two parameters,
and $n$ is the scaling parameter which will go to infinity. Notice that when
$\alpha=0$, we recover the standard symmetric simple exclusion process. Under
diffusive scaling (time sped up by $n^{2}$) and suitable constraints on
$\beta$, we prove moderate deviation principles for the density fluctuation
fields of the process when starting from the stationary measure, see Theorem
2.1.

The proof of Theorem 2.1 follows the ideas in [8]. For the upper bound, we
investigate an exponential martingale associated with the density fluctuation
fields of the process; for the lower bound, a perturbation of the WASEP is
considered. Compared with [8], due to the asymmetric part of the dynamics, we
need to prove an extra super-exponential estimate, see Proposition 3.2.

As an application of Theorem 2.1, we further investigate occupation times of
the WASEP. Stationary fluctuations for the occupation times, or more generally
the additive functionals, of the exclusion process have been widely studied,
see [1, 2, 10, 12, 14, 19, 20, 21]. We refer the readers to [15] for an
excellent review of this topic. Large deviations for the occupation times of
the exclusion process was considered in [3, 16]. Very recently, moderate
deviations for the occupation times were proved at any fixed time in the
symmetric simple exclusion process [7] and in the Kawasaki dynamics with
mixing conditions [9]. Under suitable constraints on $\beta$, we prove in
Theorem 2.2 sample path moderate deviation principles for the occupation time
of the WASEP in one dimension.

The main idea of the proof of Theorem 2.2 is to relate the occupation time to
the density fluctuation fields of the WASEP. Then by Theorem 2.1 and the
contraction principle, we obtain a variational formula as the moderate
deviations rate functional of the occupation time, which can be solved
explicitly.

The article is organized as follows. In Section 2 we state the model and our
main results. Moderate deviations upper and lower bounds from hydrodynamic
limits are proved respectively in Sections 3 and 4. Finally, Section 5 is
devoted to the proof of moderate deviations for the occupation time.

##  2\. Model and results

The state space of the weakly asymmetric simple exclusion process (WASEP) is
$\mathcal{X}_{d}:=\\{0,1\\}^{\mathbb{Z}^{d}}$, whose elements are denoted by
$\eta=\\{\eta_{x}\\}_{x\in\mathbb{Z}^{d}}$. We call a function $f$ on
${\mathcal{X}}_{d}$ is local if its value depends on $\eta$ only through a
finite number of coordinates. Fix parameters $\alpha\geq 0$ and $\beta>0$. The
infinitesimal generator of the process is $L_{n}=n^{2}(L_{s}+\alpha
n^{-\beta}L_{a})$, where for local functions
$f:\mathcal{X}_{d}\rightarrow\mathbb{R}$,

$\displaystyle L_{s}f(\eta)$
$\displaystyle=\frac{1}{2d}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}\big{[}f(\eta^{x,x+e_{i}})-f(\eta)\big{]},$
$\displaystyle L_{a}f(\eta)$
$\displaystyle=\frac{1}{d}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}\eta_{x}(1-\eta_{x+e_{i}})\big{[}f(\eta^{x,x+e_{i}})-f(\eta)\big{]}.$

Here, $n$ is the scaling parameter, $\\{e_{i}\\}_{1\leq i\leq d}$ are the
canonical basis of $\mathbb{R}^{d}$, and for $x,y\in\mathbb{Z}^{d}$,
$\eta^{x,y}$ is the configuration obtained from $\eta$ after swapping the
values of $\eta_{x}$ and $\eta_{y}$, that is, $(\eta^{x,y})_{z}=\eta_{x}$ if
$z=y$, $=\eta_{y}$ if $z=x$, and $=\eta_{z}$ otherwise.

For a constant $\rho\in[0,1]$, let $\nu_{\rho}$ be the Bernoulli product
measure on $\mathcal{X}_{d}$ with particle density $\rho$,

$\nu_{\rho}(\eta_{x}=1)=\rho,\quad\forall x\in\mathbb{Z}^{d}.$

It is well known that $\nu_{\rho}$, $\rho\in[0,1]$, are reversible for the
generator $L_{s}$, and invariant for $L_{a}$, see [17, Chapter VIII].

Throughout this article, we fix $\rho\in(0,1)$, and let $\eta^{n}(t)$ be the
process with generator $L_{n}$ and with initial distribution $\nu_{\rho}$.
Without confusion, we write $\eta(t)$ for $\eta^{n}(t)$ to make notation
short. Fix $T>0$ a time horizon. Denote by $\mathbb{P}^{n}_{\rho}$ the
probability measure on the path space ${\mathcal{D}}([0,T],{\mathcal{X}}_{d})$
associated with the law of the process $\\{\eta(t)\\}$, and by
$\mathbb{E}^{n}_{\rho}$ the corresponding expectation.

###  2.1. Density fluctuation fields

In this subsection, we state the moderate deviation principles (MDP) from
hydrodynamic limits. The rescaled density fluctuation field $\mu^{n}_{t}$ of
the process, which acts on Schwartz functions
$H\in{\mathcal{S}}:={\mathcal{S}}(\mathbb{R}^{d})$, is defined as

$\big{\langle}\mu^{n}_{t},H\big{\rangle}=\frac{1}{a_{n}}\sum_{x\in\mathbb{Z}^{d}}(\eta_{x}(t)-\rho)H(\tfrac{x-v_{n}tm}{n}),$

where

$v_{n}:=\frac{\alpha(1-2\rho)}{d}n^{2-\beta},\quad m=(1,1,\ldots,1).$

For two real numbers $a,b$, denote $a\vee b=\max\\{a,b\\},a\wedge
b=\min\\{a,b\\}$. We make the following assumptions on $\beta$ and $a_{n}$.

######  Assumption 2.1.

Assume

$n^{d/2}\sqrt{\log n}\ll a_{n}\ll n^{d}\wedge n^{d+\beta-1}.$

Assume further that

  * •

for $d=1$,

$\beta>\frac{2}{3},\quad a_{n}\gg n^{1-\tfrac{\beta}{2}};$

  * •

for $d=2$, there exists some $0<\varepsilon_{0}<1$ such that

$\beta>\frac{1}{2},\quad a_{n}\gg n^{2-\beta}(\log
n)^{\tfrac{1}{1-\varepsilon_{0}}};$

  * •

for $d\geq 3$,

$\beta>\tfrac{1}{2},\quad a_{n}\gg n^{d-\beta}.$

Next, we introduce the moderate deviations rate function. For
$H\in{\mathcal{C}}^{1,\infty}_{c}([0,T]\times\mathbb{R}^{d})$ and
$\mu\in{\mathcal{D}}([0,T],\mathcal{S}^{\prime})$, where
${\mathcal{S}}^{\prime}:=\mathcal{S}^{\prime}({\mathbb{R}}^{d})$ is the dual
of the Schwartz space ${\mathcal{S}}$, define the linear functional
$l(\cdot,\cdot)$ as

$l(\mu,H)=\left\langle\mu_{T},H_{T}\right\rangle-\left\langle\mu_{0},H_{0}\right\rangle-\int_{0}^{T}\langle\mu_{s},\left(\partial_{s}+(1/2d)\Delta\right)H_{s}\rangle
ds.$

Above, $H_{t}(u)=H(t,u)$ for $u\in\mathbb{R}^{d}$, and
$\Delta=\sum_{i=1}^{d}\partial_{u_{i}}^{2}$ is the Laplacian operator. The
rate function
$\mathcal{Q}(\mu):=\mathcal{Q}_{dyn}(\mu)+\mathcal{Q}_{0}(\mu_{0})$ is defined
as

$\displaystyle\mathcal{Q}_{dyn}(\mu)$
$\displaystyle=\sup_{H\in{\mathcal{C}}^{1,\infty}_{c}([0,T]\times\mathbb{R}^{d})}\left\\{l(\mu,H)-\frac{\chi(\rho)}{2d}\sum_{i=1}^{d}\|\partial_{u_{i}}H\|_{L^{2}([0,T]\times\mathbb{R}^{d})}^{2}\right\\},$
$\displaystyle\mathcal{Q}_{0}\left(\mu_{0}\right)$
$\displaystyle=\sup_{\phi\in{\mathcal{C}}_{c}^{\infty}\left(\mathbb{R}^{d}\right)}\left\\{\left\langle\mu_{0},\phi\right\rangle-\frac{\chi(\rho)}{2}\|\phi\|_{L^{2}(\mathbb{R}^{d})}^{2}\right\\},$

where $\chi(\rho)=\rho(1-\rho)$, and

$\|\partial_{u_{i}}H\|_{L^{2}([0,T]\times\mathbb{R}^{d})}^{2}:=\int_{0}^{T}\int_{\mathbb{R}^{d}}[\partial_{u_{i}}H_{s}(u)]^{2}\,du\,ds,\quad\|\phi\|_{L^{2}(\mathbb{R}^{d})}^{2}:=\int_{\mathbb{R}^{d}}\phi(u)^{2}du.$

We state the first main result of this article.

######  Theorem 2.1.

Under Assumption 2.1, the sequence of processes $\\{\mu^{n}_{t}:0\leq t\leq
T\\}_{n\geq 1}$ satisfies the MDP with decay rate $a_{n}^{2}/n^{d}$ and with
rate function $\mathcal{Q}(\mu)$. More precisely, for any closed set
$C\subseteq{\mathcal{D}}([0,T],\mathcal{S}^{\prime})$, and for any open set
$O\subseteq{\mathcal{D}}([0,T],\mathcal{S}^{\prime})$,

$\displaystyle\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\big{(}\\{\mu^{n}_{t}:0\leq
t\leq T\\}\in C\big{)}\leq-\inf_{\mu\in C}\mathcal{Q}(\mu),$ (2.1)
$\displaystyle\liminf_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\big{(}\\{\mu^{n}_{t}:0\leq
t\leq T\\}\in O\big{)}\geq-\inf_{\mu\in O}\mathcal{Q}(\mu).$ (2.2)

###  2.2. Additive functionals

As an application of Theorem 2.1, we obtain the MDP for the additive
functionals of the WASEP in dimension $d=1$. For any local function $f$ on
${\mathcal{X}}_{1}$, define

$\Gamma^{n}_{t}(f)=\frac{n}{a_{n}}\int_{0}^{t}\\{f(\eta(s))-\tilde{f}(\rho)\\}ds,$

where $\tilde{f}(\rho)=E_{\nu_{\rho}}[f]$. When $f(\eta)=\eta_{0}$, we simply
write $\Gamma^{n}_{t}=\Gamma^{n}_{t}(f)$ the occupation time of the WASEP.
Invariance principles for the above additive functional was proved by
Gonçalves and Jara in [10]. In particular, if $\beta>1$ or $\rho=1/2$, then
the limit is given by $\tilde{f}^{\prime}(\rho)$ multiple of a fractional
Brownian motion with Hurst parameter $3/4$.

Now, we introduce the rate function for the MDP of the additive functional
$\Gamma^{n}_{t}(f)$. Let $\\{B_{t}^{3/4}:t\geq 0\\}$ be the fractional
Brownian motion with Hurst parameter $3/4$, which has the following
representation:

$B_{t}^{3/4}=\int_{0}^{t}K(t,s)dB_{s},$

where $\\{B_{t}\\}$ is the standard one dimensional Brownian motion, and

$K(t,s)=\sqrt{\frac{3}{8\int_{0}^{1}(1-x)^{-\frac{1}{2}}x^{-\frac{3}{4}}dx}}s^{-\frac{1}{4}}\int_{s}^{t}(u-s)^{-\frac{3}{4}}u^{\frac{1}{4}}du,$

see [18, Proposition 2.5, Chapter 2]. Let ${\mathcal{H}}$ be the set of
continuous functions $\gamma:[0,T]\rightarrow\mathbb{R}$ such that there
exists $\dot{\gamma}\in L^{2}([0,T])$ satisfying that

$\gamma(t)=\int_{0}^{t}K(t,s)\dot{\gamma}(s)ds,\quad t\in[0,T].$

Finally, if $\tilde{f}^{\prime}(\rho)\neq 0$, define

$I_{\rm
path}(\gamma)=\begin{cases}\frac{1}{2}\int_{0}^{T}\dot{\gamma}(t)^{2}dt,\quad&\text{if
}\;\gamma\in{\mathcal{H}};\\\ +\infty,\quad&\text{otherwise},\end{cases}\quad
I_{f}(\gamma)=\frac{1}{\sigma^{2}\tilde{f}^{\prime}(\rho)^{2}}I_{\rm
path}(\gamma),$

where $\sigma^{2}=4\sqrt{2}\chi(\rho)/(3\sqrt{\pi})$ is the limiting variance
of the occupation time, see [12]. Notice that $I_{\rm path}$ is the large
deviations rate function of the fractional Brownian motion with Hurst
parameter $3/4$, see [5, Theorem 3.4.12] for example.

The following is the second main result of this article.

######  Theorem 2.2.

Let $d=1$ and Assumption 2.1 hold. Assume further that $\rho=1/2$ if
$2/3<\beta\leq 1$. If $\tilde{f}^{\prime}(\rho)\neq 0$, then the sequence
$\\{\Gamma^{n}_{t}(f):0\leq t\leq T\\}_{n\geq 1}$ satisfies the MDP with decay
rate $a_{n}^{2}/n$ and with rate function $I_{f}(\cdot)$. More precisely, for
any closed set $C\subseteq{\mathcal{C}}([0,T],\mathbb{R})$, and for any open
set $O\subseteq{\mathcal{C}}([0,T],\mathbb{R})$,

$\displaystyle\limsup_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\big{(}\\{\Gamma^{n}_{t}(f):0\leq
t\leq T\\}\in C\big{)}\leq-\inf_{\gamma\in C}I_{f}(\gamma),$ (2.3)
$\displaystyle\liminf_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\big{(}\\{\Gamma^{n}_{t}(f):0\leq
t\leq T\\}\in O\big{)}\geq-\inf_{\gamma\in O}I_{f}(\gamma).$ (2.4)

##  3\. Proof of Theorem 2.1: upper bound

###  3.1. An exponential martingale

In this subsection, we introduce the following exponential martingale, which
plays an important role when proving the upper bound (2.1). For
$H\in{\mathcal{C}}^{1,\infty}_{c}([0,T]\times\mathbb{R}^{d})$, by Feynman-Kac
formula [13, Appendix 1.7],

$\displaystyle{\mathcal{M}}^{n}_{t}(H)=\exp\Big{\\{}\frac{a_{n}^{2}}{n^{d}}\big{\langle}\mu^{n}_{t},H_{t}\big{\rangle}-\frac{a_{n}^{2}}{n^{d}}\big{\langle}\mu^{n}_{0},H_{0}\big{\rangle}-\int_{0}^{t}e^{-\frac{a_{n}^{2}}{n^{d}}\big{\langle}\mu^{n}_{s},H_{s}\big{\rangle}}(\partial_{s}+L_{n})e^{\frac{a_{n}^{2}}{n^{d}}\big{\langle}\mu^{n}_{s},H_{s}\big{\rangle}}ds\Big{\\}}$

is a martingale. We want to write the above martingale as a functional of the
density fluctuation field.

In the following computations, we simply write $\eta_{x}=\eta_{x}(s)$ when
there are no confusions. By Taylor’s expansion and summation by parts formula,

$\displaystyle
e^{-\frac{a_{n}^{2}}{n^{d}}\big{\langle}\mu^{n}_{s},H_{s}\big{\rangle}}n^{2}L_{s}e^{\frac{a_{n}^{2}}{n^{d}}\big{\langle}\mu^{n}_{s},H_{s}\big{\rangle}}$
(3.1)
$\displaystyle=\frac{n^{2}}{2d}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}\Big{[}e^{\frac{a_{n}}{n^{d}}(\eta_{x}-\eta_{x+e_{i}})(H_{s}(\frac{x+e_{i}-v_{n}sm}{n})-H_{s}(\frac{x-v_{n}sm}{n}))}-1\Big{]}$
$\displaystyle=\frac{a_{n}^{2}}{n^{d}}\Big{\\{}\frac{1}{2d}\big{\langle}\mu^{n}_{s},\Delta_{n}H_{s}\big{\rangle}+\frac{1}{4dn^{d}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}(\eta_{x}-\eta_{x+e_{i}})^{2}\big{(}\nabla_{n,i}H_{s}(\tfrac{x-v_{n}sm}{n})\big{)}^{2}+O_{H}\Big{(}\frac{a_{n}}{n^{d+1}}\Big{)}\Big{\\}}.$

Above, $O_{H}(a_{n}/n^{d+1})$ is bounded by $C(H)a_{n}/n^{d+1}$ which vanishes
as $n\rightarrow\infty$ since $a_{n}\ll n^{d}$, and $\nabla_{n,i}$ and
$\Delta_{n}=\sum_{i=1}^{d}\Delta_{n,i}$ are the discrete gradient and
Laplacian respectively,

$\displaystyle\nabla_{n,i}H_{s}(\tfrac{x-v_{n}sm}{n})$
$\displaystyle=n\big{[}H_{s}(\tfrac{x+e_{i}-v_{n}sm}{n})-H_{s}(\tfrac{x-v_{n}sm}{n})\big{]},$
$\displaystyle\Delta_{n,i}H_{s}(\tfrac{x-v_{n}sm}{n})$
$\displaystyle=n^{2}\big{[}H_{s}(\tfrac{x+e_{i}-v_{n}sm}{n})+H_{s}(\tfrac{x-e_{i}-v_{n}sm}{n})-2H_{s}(\tfrac{x-v_{n}sm}{n})\big{]}.$

Similarly,

$\displaystyle
e^{-\frac{a_{n}^{2}}{n^{d}}\big{\langle}\mu^{n}_{s},H_{s}\big{\rangle}}\alpha
n^{2-\beta}L_{a}e^{\frac{a_{n}^{2}}{n^{d}}\big{\langle}\mu^{n}_{s},H_{s}\big{\rangle}}$
(3.2) $\displaystyle=\frac{\alpha
n^{2-\beta}}{d}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}\eta_{x}(1-\eta_{x+e_{i}})\Big{[}e^{\frac{a_{n}}{n^{d}}(H_{s}(\frac{x+e_{i}-v_{n}sm}{n})-H_{s}(\frac{x-v_{n}sm}{n}))}-1\Big{]}$
$\displaystyle=\frac{a_{n}^{2}}{n^{d}}\Big{\\{}\frac{\alpha
n^{1-\beta}}{da_{n}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}[\eta_{x}(1-\eta_{x+e_{i}})-\chi(\rho)]\nabla_{n,i}H_{s}(\tfrac{x-v_{n}sm}{n})+O_{H}(n^{-\beta})\Big{\\}}.$

We introduced $\chi(\rho)$ in the last line since
$\sum_{x\in\mathbb{Z}^{d}}\nabla_{n,i}H_{s}(\tfrac{x-v_{n}sm}{n})=0$. Finally,

$\displaystyle
e^{-\frac{a_{n}^{2}}{n^{d}}\big{\langle}\mu^{n}_{s},H_{s}\big{\rangle}}\partial_{s}e^{\frac{a_{n}^{2}}{n^{d}}\big{\langle}\mu^{n}_{s},H_{s}\big{\rangle}}$
$\displaystyle=\frac{a_{n}^{2}}{n^{d}}\Big{\\{}\big{\langle}\mu^{n}_{s},\partial_{s}{H}_{s}\big{\rangle}-\frac{\alpha(1-2\rho)n^{1-\beta}}{da_{n}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}[\eta_{x}-\rho]\partial_{u_{i}}H_{s}(\tfrac{x-v_{n}sm}{n})\Big{\\}}.$

To replace the above discrete gradient and Laplacian by the continuous ones,
we need the following lemma.

######  Lemma 3.1.

Let $r_{n}\gg a_{n}$ be a sequence of real numbers, and let
$f:\mathcal{X}_{d}\rightarrow\mathbb{R}$ be some local function. Then, for any
$H\in{\mathcal{C}}^{1,\infty}_{c}([0,T]\times\mathbb{R}^{d})$ and any
$\delta>0$,

$\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0\leq
t\leq
T}\Big{|}\int_{0}^{t}\frac{1}{r_{n}}\sum_{x\in\mathbb{Z}^{d}}\big{(}\tau_{x}f(\eta(s))-\tilde{f}(\rho)\big{)}H(s,\tfrac{x}{n})ds\Big{|}>\delta\Big{)}=-\infty,$

where $\tilde{f}(\rho)=E_{\nu_{\rho}}[f]$.

###### Proof.

By Markov’s inequality, we only need to show that for any $A>0$,

$\lim_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\exp\Big{\\{}\int_{0}^{T}\Big{|}\frac{Aa_{n}^{2}}{n^{d}r_{n}}\sum_{x\in\mathbb{Z}^{d}}\big{(}\tau_{x}f(\eta(s))-\tilde{f}(\rho)\big{)}H(s,\tfrac{x}{n})\Big{|}ds\Big{\\}}\Big{]}=0.$
(3.3)

By Jensen’s inequality and stationary of the process, the above expectation is
bounded by

$\frac{1}{T}\int_{0}^{T}dsE_{\nu_{\rho}}\Big{[}\exp\Big{\\{}\Big{|}\frac{ATa_{n}^{2}}{n^{d}r_{n}}\sum_{x\in\mathbb{Z}^{d}}\big{(}\tau_{x}f(\eta)-\tilde{f}(\rho)\big{)}H(s,\tfrac{x}{n})\Big{|}\Big{\\}}\Big{]}.$

Using the basic inequalities $e^{|x|}\leq e^{x}+e^{-x}$ and

$\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log(x_{n}+y_{n})=\max\Big{\\{}\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log
x_{n},\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log y_{n}\Big{\\}},$
(3.4)

we can remove the absolute value inside the expectation. Since $\nu_{\rho}$ is
a product measure and $e^{x}\leq 1+x+(x^{2}/2)e^{|x|}$, for $n$ large enough,
we can bound the expectation by

$\displaystyle\prod_{x\in\mathbb{Z}^{d}}\Big{(}E_{\nu_{\rho}}\Big{[}\exp\Big{\\{}\frac{C(f)ATa_{n}^{2}}{n^{d}r_{n}}\big{(}\tau_{x}f(\eta)-\tilde{f}(\rho)\big{)}H(s,\tfrac{x}{n})\Big{\\}}\Big{]}\Big{)}^{1/C(f)}$
$\displaystyle\leq\Big{(}1+\frac{C(f,A,T,H)a_{n}^{4}}{n^{2d}r_{n}^{2}}\Big{)}^{n^{d}/C(f,H)}.$

Thus, the limit on the left side of (3.3) is bounded by

$\lim_{n\rightarrow\infty}\frac{C(f,A,T,H)a_{n}^{2}}{r_{n}^{2}}=0,$

which concludes the proof. ∎

Using the above lemma, we have

$\displaystyle\eqref{cal 4}$
$\displaystyle=\frac{a_{n}^{2}}{n^{d}}\Big{\\{}\frac{1}{2d}\big{\langle}\mu^{n}_{s},\Delta
H_{s}\big{\rangle}+\frac{\chi(\rho)}{2dn^{d}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}\big{(}\partial_{u_{i}}H_{s}(\tfrac{x-v_{n}sm}{n})\big{)}^{2}+o_{H}(n)+\varepsilon^{n}_{1,H}(s)\Big{\\}},$
$\displaystyle\eqref{cal 5}$
$\displaystyle=\frac{a_{n}^{2}}{n^{d}}\Big{\\{}\frac{\alpha
n^{1-\beta}}{da_{n}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}[\eta_{x}(1-\eta_{x+e_{i}})-\chi(\rho)]\partial_{u_{i}}H_{s}(\tfrac{x-v_{n}sm}{n})+o_{H}(n)+\varepsilon^{n}_{2,H}(s)\Big{\\}}.$

Above, $o_{H}(n)$ denotes some constant depending on $H$ and vanishes as
$n\rightarrow\infty$, and for $j=1,2$, for any $\delta>0$,

$\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0\leq
t\leq
T}\Big{|}\int_{0}^{t}\varepsilon^{n}_{j,H}(s)ds\Big{|}>\delta\Big{)}=-\infty.$
(3.5)

Since

$\eta_{x}(1-\eta_{x+e_{i}})-\chi(\rho)-(1-2\rho)(\eta_{x}-\rho)=-(\eta_{x}-\rho)(\eta_{x+e_{i}}-\rho)+\rho(\eta_{x}-\eta_{x+e_{i}}),$
(3.6)

by summation by parts formula and Lemma 3.1,

$\displaystyle\frac{\alpha
n^{1-\beta}}{da_{n}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}[\eta_{x}(1-\eta_{x+e_{i}})-\chi(\rho)-(1-2\rho)(\eta_{x}-\rho)]\partial_{u_{i}}H_{s}(\tfrac{x-v_{n}sm}{n})=-Q^{n}_{s}(H)+\varepsilon^{n}_{3,H}(s),$

where $\varepsilon^{n}_{3,H}(s)$ satisfies the estimates in (3.5), and

$Q^{n}_{s}(H)=\frac{\alpha
n^{1-\beta}}{da_{n}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}(\eta_{x}(s)-\rho)(\eta_{x+e_{i}}(s)-\rho)\partial_{u}H_{s}(\tfrac{x-v_{n}sm}{n}).$

To sum up, we have shown that

${\mathcal{M}}^{n}_{t}(H)=\exp\Big{\\{}\frac{a_{n}^{2}}{n^{d}}\Big{(}\big{\langle}\mu^{n}_{t},H_{t}\big{\rangle}-\big{\langle}\mu^{n}_{0},H_{0}\big{\rangle}-\int_{0}^{t}\big{\langle}\mu^{n}_{s},\big{(}\partial_{s}+\tfrac{1}{2d}\Delta\big{)}H_{s}\big{\rangle}ds\\\
-\frac{\chi(\rho)}{2d}\int_{0}^{t}\int_{\mathbb{R}^{d}}\sum_{i=1}^{d}\big{(}\partial_{u_{i}}H_{s}(u)\big{)}^{2}du\,ds+\int_{0}^{t}Q^{n}_{s}(H)ds+o_{H}(n)t+\int_{0}^{t}\varepsilon^{n}_{H}(s)ds\Big{)}\Big{\\}},$
(3.7)

where $\varepsilon^{n}_{H}(s)=\sum_{j=1}^{3}\varepsilon^{n}_{j,H}(s)$.

The next result states that the time integral of $Q^{n}_{s}(H)$ is
superexponentially small uniformly in time as $n\rightarrow\infty$.

######  Proposition 3.2.

Under Assumption 2.1, for any $\delta>0$,

$\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0\leq
t\leq T}\Big{|}\int_{0}^{t}Q^{n}_{s}(H)ds\Big{|}>\delta\Big{)}=-\infty.$

To prove the above result, we need to introduce more notation. For an integer
$\ell>0$, let $p_{\ell}$ be the uniform measure on
$\Lambda_{\ell}^{d}:=\\{0,1,\ldots,\ell-1\\}^{d}$, and let $q_{\ell}$ be the
convolution of $p_{\ell}$ with itself,

$p_{\ell}(x)=\ell^{-d}\mathbf{1}\\{x\in\Lambda_{\ell}^{d}\\},\quad
q_{\ell}(x)=\sum_{y\in\mathbb{Z}^{d}}p_{\ell}(x-y)p_{\ell}(y),\quad
x\in\mathbb{Z}^{d}.$

Notice that $q_{\ell}$ is supported in $\Lambda_{2\ell-1}^{d}$. Let
$\delta_{0}$ be the Dirac measure at the origin. The following result was
proved in [11, Lemma 3.2].

######  Lemma 3.3.

There exists some function $\Phi_{\ell}:\mathbb{Z}^{d}\times\\{e_{i}\\}_{1\leq
i\leq d}\rightarrow\mathbb{R}$ such that for any $z\in\mathbb{Z}^{d}$,

$\delta_{0}(z)-q_{\ell}(z)=\sum_{i=1}^{d}\\{\Phi_{\ell}(z,e_{i})-\Phi_{\ell}(z-e_{i},e_{i})\\}.$

Moreover, $\Phi_{\ell}(z,e_{i})=0$ for $z\notin\Lambda_{2\ell-1}^{d}$, and
there exists some constant $C_{0}=C_{0}(d)$ such that

$\sum_{i=1}^{d}\sum_{z\in\mathbb{Z}^{d}}\Phi_{\ell}(z,e_{i})^{2}\leq
C_{0}g_{d}(\ell),$

where

$g_{d}(\ell)=\begin{cases}\ell,\quad&\text{if }d=1,\\\ \log\ell,\quad&\text{if
}d=2,\\\ 1,\quad&\text{if }d\geq 3.\end{cases}$

For $\eta\in{\mathcal{X}}_{d}$, define $\eta^{\ell}_{x}$ as the spatial
average of $\eta_{x+\cdot}$ with respect to the weight function $q_{\ell}$,

$\eta_{x}^{\ell}=\sum_{z\in\mathbb{Z}^{d}}\eta_{x+z}q_{\ell}(z),\quad
x\in\mathbb{Z}^{d}.$

Finally, for an integer $\ell>0$, let

$Q^{n,\ell}_{s}(H)=\frac{\alpha
n^{1-\beta}}{da_{n}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}(\eta_{x}(s)-\rho)(\eta_{x+e_{i}}^{\ell}(s)-\rho)\partial_{u}H_{s}(\tfrac{x-v_{n}sm}{n}).$

To prove Proposition 3.2, we shall take $\ell=\ell(n)\leq n$ such that

$\ell^{d}g_{d}(\ell)\ll n^{2\beta},\quad\frac{n^{2d}}{a_{n}^{2}}\ll\ell^{d}.$
(3.8)

Such $\ell$ exists by Assumption 2.1. Indeed, let

$M_{n}=\begin{cases}n^{\beta}\wedge n,\quad&\text{if }d=1,\\\
n^{2\beta/d}\wedge n,\quad&\text{if }d\geq 2,\end{cases},\quad
m_{n}=\frac{n^{2}}{a_{n}^{2/d}}.$

Then, Assumption 2.1 ensures that $M_{n}\gg m_{n}$ and that for $d=2$,

$\Big{(}\frac{M_{n}}{m_{n}}\Big{)}^{1-\varepsilon_{0}}\gg\log M_{n}.$

Thus, one can take $\ell(n)=M_{n}^{\varepsilon_{0}}m_{n}^{1-\varepsilon_{0}}$.

###### Proof of Proposition 3.2.

By (3.4), we only need to prove that, for any $\delta>0$,

$\displaystyle\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0\leq
t\leq
T}\Big{|}\int_{0}^{t}[Q^{n}_{s}(H)-Q^{n,\ell}_{s}(H)]ds\Big{|}>\delta\Big{)}=-\infty,$
(3.9)
$\displaystyle\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0\leq
t\leq T}\Big{|}\int_{0}^{t}Q^{n,\ell}_{s}(H)ds\Big{|}>\delta\Big{)}=-\infty.$
(3.10)

We remark that the first constraint on $\ell$ in (3.8) is used to prove (3.9),
while the second one is used to prove (3.10).

We first prove (3.9). By Markov’s inequality, we only need to show that for
any $A>0$,

$\lim_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\exp\Big{\\{}\sup_{0\leq
t\leq
T}\Big{|}\frac{Aa_{n}^{2}}{n^{d}}\int_{0}^{t}[Q^{n}_{s}(H)-Q^{n,\ell}_{s}(H)]ds\Big{|}\Big{\\}}\Big{]}=0.$
(3.11)

Let $t_{i}=iT/n^{d+1}$ for $0\leq i\leq n^{d+1}$. Since

$|Q^{n}_{s}(H)-Q_{s}^{n,\ell}(H)|\leq\frac{C(H,d)n^{d+1-\beta}}{a_{n}},$

we can replace the supremum over time $0\leq t\leq T$ by over $\\{t_{i}\\}$ in
(3.11) with an error bounded by $C(H,A,d)/(a_{n}n^{\beta})$. Thus, we only
need to show that

$\lim_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\exp\Big{\\{}\sup_{0\leq
i\leq
n^{d+1}}\Big{|}\frac{Aa_{n}^{2}}{n^{d}}\int_{0}^{t_{i}}[Q^{n}_{s}(H)-Q^{n,\ell}_{s}(H)]ds\Big{|}\Big{\\}}\Big{]}=0.$

Since for any $K$ random variables $X_{1},\ldots,X_{K}$,

$\log E\big{[}\exp\\{\max_{1\leq i\leq
K}X_{i}\\}\big{]}\leq\log\Big{(}\sum_{i=1}^{K}E\big{[}\exp\\{X_{i}\\}\big{]}\Big{)}\leq\log
K+\max_{1\leq i\leq K}\log E\big{[}\exp\\{X_{i}\\}\big{]},$ (3.12)

and $a_{n}\gg n^{d/2}\sqrt{\log n}$, we can first put the supremum over $0\leq
i\leq n^{d+1}$ outside, then remove the absolute value, and finally only need
to prove that

$\lim_{n\rightarrow\infty}\sup_{0\leq i\leq
n^{d+1}}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\exp\Big{\\{}\frac{Aa_{n}^{2}}{n^{d}}\int_{0}^{t_{i}}[Q^{n}_{s}(H)-Q^{n,\ell}_{s}(H)]ds\Big{\\}}\Big{]}=0.$
(3.13)

By Feynman-Kac formula [13, Appendix 1.7],

$\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\exp\Big{\\{}\frac{Aa_{n}^{2}}{n^{d}}\int_{0}^{t_{i}}[Q^{n}_{s}(H)-Q^{n,\ell}_{s}(H)]ds\Big{\\}}\Big{]}\leq\frac{\alpha
An^{1-\beta}}{da_{n}}\int_{0}^{t_{i}}ds\\\ \sup_{f:\;\nu_{\rho}{\rm-
density}}\Big{\\{}\int_{{\mathcal{X}}_{d}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}(\eta_{x}-\rho)(\eta_{x+e_{i}}-\eta_{x+e_{i}}^{\ell})\partial_{u}H_{s}(\tfrac{x-v_{n}sm}{n})f(\eta)d\nu_{\rho}-\frac{dn^{d+1+\beta}}{\alpha
Aa_{n}}D(\sqrt{f})\Big{\\}},$ (3.14)

where the supremum is over local functions $f\geq 0$ on ${\mathcal{X}}_{d}$
such that $E_{\nu_{\rho}}[f]=0$, and $D(\sqrt{f})$ is the Dirichlet form of
the function $f$ with respect to the generator $L_{s}$ under $\nu_{\rho}$,

$D(\sqrt{f})=E_{\nu_{\rho}}[\sqrt{f}(-L_{s})\sqrt{f}]=\frac{1}{4d}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}E_{\nu_{\rho}}\big{[}\big{(}\sqrt{f(\eta^{x,x+e_{i}})}-\sqrt{f(\eta)}\big{)}^{2}\big{]}.$

Since

$\displaystyle\eta_{x+e_{i}}-\eta_{x+e_{i}}^{\ell}$
$\displaystyle=\sum_{z\in\mathbb{Z}^{d}}\eta_{x+e_{i}+z}[\delta_{0}(z)-q_{\ell}(z)]$
$\displaystyle=\sum_{z\in\mathbb{Z}^{d}}\sum_{j=1}^{d}\eta_{x+e_{i}+z}[\Phi_{\ell}(z,e_{j})-\Phi_{\ell}(z-e_{j},e_{j})]$
$\displaystyle=\sum_{z\in\mathbb{Z}^{d}}\sum_{j=1}^{d}[\eta_{x+e_{i}+z}-\eta_{x+e_{i}+z+e_{j}}]\Phi_{\ell}(z,e_{j}),$

by changing of variables $\eta\mapsto\eta^{x+e_{i}+z,x+e_{i}+z+e_{j}}$ and
Young’s inequality, the integral inside the supremum in (3.14) is bounded by

$\displaystyle\frac{1}{2}\sum_{x\in\mathbb{Z}^{d},z\in\Lambda_{2\ell-1}^{d}}\sum_{i,j=1}^{d}\Phi_{\ell}(z,e_{j})\partial_{u}H_{s}(\tfrac{x-v_{n}sm}{n})$
$\displaystyle\qquad\times\int_{{\mathcal{X}}_{d}}(\eta_{x}-\rho)[\eta_{x+e_{i}+z}-\eta_{x+e_{i}+z+e_{j}}][f(\eta)-f(\eta^{x+e_{i}+z,x+e_{i}+z+e_{j}})]d\nu_{\rho}$
$\displaystyle\leq\frac{1}{4B}\sum_{x\in\mathbb{Z}^{d},z\in\Lambda_{2\ell-1}^{d}}\sum_{i,j=1}^{d}\Phi_{\ell}(z,e_{j})^{2}\partial_{u}H_{s}(\tfrac{x-v_{n}sm}{n})^{2}\int_{{\mathcal{X}}_{d}}\big{[}\sqrt{f(\eta)}+\sqrt{f(\eta^{x+e_{i}+z,x+e_{i}+z+e_{j}})}\big{]}^{2}d\nu_{\rho}$
$\displaystyle\quad+\frac{B}{4}\sum_{x\in\mathbb{Z}^{d},z\in\Lambda_{2\ell-1}^{d}}\sum_{i,j=1}^{d}\int_{{\mathcal{X}}_{d}}\big{[}\sqrt{f(\eta)}-\sqrt{f(\eta^{x+e_{i}+z,x+e_{i}+z+e_{j}})}\big{]}^{2}d\nu_{\rho}$

for any $B>0$. By Lemma 3.3 and the fact that $f$ is a $\nu_{\rho}$-density,
the last expression is bounded by

$\frac{C(H,d,C_{0})}{B}n^{d}g_{d}(\ell)+Bd^{2}(2\ell)^{d}D(\sqrt{f}).$

Taking $B$ such that

$Bd^{2}(2\ell)^{d}=\frac{dn^{d+1+\beta}}{\alpha Aa_{n}},$

we bound the limit on the left side of (3.13) by

$\lim_{n\rightarrow\infty}C(H,d,C_{0},A)\frac{\ell^{d}g_{d}(\ell)}{n^{2\beta}}=0,$

thus proving (3.13).

Now, we prove (3.10). As before, we only need to prove that, for any $A>0$,

$\lim_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\exp\Big{\\{}\frac{Aa_{n}^{2}}{n^{d}}\int_{0}^{T}\big{|}Q^{n,\ell}_{s}(H)\big{|}ds\Big{\\}}\Big{]}=0.$

By Jensen’s inequality, stationary of the process and (3.4), we only need to
show that

$\lim_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\Big{(}\frac{1}{T}\int_{0}^{T}ds\\\
E_{\nu_{\rho}}\Big{[}\exp\Big{\\{}\frac{ATa_{n}\alpha
n^{1-\beta}}{dn^{d}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}(\eta_{x}-\rho)(\eta_{x+e_{i}}^{\ell}-\rho)\partial_{u}H_{s}(\tfrac{x-v_{n}sm}{n})\Big{\\}}\Big{]}\Big{)}=0.$
(3.15)

By definition,

$\displaystyle\sum_{x\in\mathbb{Z}^{d}}(\eta_{x}-\rho)(\eta_{x+e_{i}}^{\ell}-\rho)\partial_{u}H_{s}(\tfrac{x-v_{n}sm}{n})$
$\displaystyle=\sum_{x,y,z\in\mathbb{Z}^{d}}(\eta_{x}-\rho)(\eta_{x+e_{i}+z}^{\ell}-\rho)p_{\ell}(z-y)p_{\ell}(y)\partial_{u}H_{s}(\tfrac{x-v_{n}sm}{n}).$

Making the change of variables $z\mapsto z-x-e_{i},y\mapsto y-x-e_{i}$, the
last line equals

$\sum_{y\in\mathbb{Z}^{d}}\Big{(}\sum_{x\in\mathbb{Z}^{d}}(\eta_{x}-\rho)p_{\ell}(y-x-
e_{i})\partial_{u}H_{s}(\tfrac{x-v_{n}sm}{n})\Big{)}\Big{(}\sum_{z\in\mathbb{Z}^{d}}(\eta^{\ell}_{z}-\rho)p_{\ell}(z-y)\Big{)}.$

Notice that if $|y-y^{\prime}|:=\max_{1\leq i\leq
d}|y_{i}-y^{\prime}_{i}|>5\ell$, then for $\ell$ large enough,

$\Big{(}\sum_{x\in\mathbb{Z}^{d}}(\eta_{x}-\rho)p_{\ell}(y-x-
e_{i})\partial_{u}H_{s}(\tfrac{x-v_{n}sm}{n})\Big{)}\Big{(}\sum_{z\in\mathbb{Z}^{d}}(\eta^{\ell}_{z}-\rho)p_{\ell}(z-y)\Big{)}$

and

$\Big{(}\sum_{x\in\mathbb{Z}^{d}}(\eta_{x}-\rho)p_{\ell}(y^{\prime}-x-e_{i})\partial_{u}H_{s}(\tfrac{x-v_{n}sm}{n})\Big{)}\Big{(}\sum_{z\in\mathbb{Z}^{d}}(\eta^{\ell}_{z}-\rho)p_{\ell}(z-y^{\prime})\Big{)}$

are independent under $\nu_{\rho}$. By Hölder’s inequality, the expectation in
(3.15) is bounded by

$\prod_{y\in\mathbb{Z}^{d}}\prod_{i=1}^{d}\Big{(}E_{\nu_{\rho}}\Big{[}\exp\Big{\\{}\frac{ATa_{n}\alpha
n^{1-\beta}d(5\ell)^{d}}{n^{d}}\\\
\times\Big{(}\sum_{x\in\mathbb{Z}^{d}}(\eta_{x}-\rho)p_{\ell}(y-x-
e_{i})\partial_{u}H_{s}(\tfrac{x-v_{n}sm}{n})\Big{)}\Big{(}\sum_{z\in\mathbb{Z}^{d}}(\eta^{\ell}_{z}-\rho)p_{\ell}(z-y)\Big{)}\Big{\\}}\Big{]}\Big{)}^{\frac{1}{d(5\ell)^{d}}}.$

We say a random variable $X$ is sub-Gaussian of order $\sigma^{2}$ if

$\log E[e^{\theta
X}]\leq\frac{\theta^{2}\sigma^{2}}{2},\quad\forall\theta\in\mathbb{R}.$

Then, one can easily check that
$\sum_{x\in\mathbb{Z}^{d}}(\eta_{x}-\rho)p_{\ell}(y-x-
e_{i})\partial_{u}H_{s}(\tfrac{x-v_{n}sm}{n})$ and
$\sum_{z\in\mathbb{Z}^{d}}(\eta^{\ell}_{z}-\rho)p_{\ell}(z-y)$ are both sub-
Gaussian of order $C(H)\ell^{-d}$. Since by Assumption 2.1, $a_{n}\ll
n^{d+\beta-1}$, using [11, Lemma F.8], for $n$ large enough, the expectation
in (3.15) is bounded by

$C(A,T,H,\alpha,d)^{\frac{n^{d}}{\ell^{d}}}.$

Thus, the limit in (3.15) is bounded by

$C(A,T,H,\alpha,d)\lim_{n\rightarrow\infty}\frac{n^{2d}}{\ell^{d}a_{n}^{2}}=0,$

which concludes the proof. ∎

###  3.2. Proof of the upper bound

In this subsection, we prove the moderate deviations upper bound (2.1). The
strategy is as follows: we first prove the upper bound for compact sets in
${\mathcal{D}}([0,T],{\mathcal{S}}^{\prime})$, and then prove the exponential
tightness of the sequence $\\{\mu^{n}_{t}:0\leq t\leq T\\}$. Finally, from the
above two results, we conclude the proof of the upper bound for closed sets in
${\mathcal{D}}([0,T],{\mathcal{S}}^{\prime})$ immediately.

We start with the former. Let
$K\subset{\mathcal{D}}([0,T];{\mathcal{S}}^{\prime})$ be compact and
$H\in{\mathcal{C}}^{1,\infty}_{c}([0,T]\times\mathbb{R}^{d})$. By (3.7) and
Proposition 3.2, for any $\delta>0$,

$\displaystyle\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\mu^{n}\in
K\Big{)}=\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\mathcal{M}^{n}_{T}(H)\mathcal{M}^{n}_{T}(H)^{-1}\mathbf{1}_{\\{\mu^{n}\in
K,|{\mathcal{R}}^{n}_{T}(H)|\leq\delta\\}}\Big{]},$

where

${\mathcal{R}}^{n}_{t}(H)=\int_{0}^{t}Q^{n}_{s}(H)ds+o_{H}(n)t+\int_{0}^{t}\varepsilon^{n}_{H}(s)ds.$
(3.16)

On the event $\\{\mu^{n}\in K,|{\mathcal{R}}^{n}_{T}(H)|\leq\delta\\}$, for
any $\phi\in{\mathcal{C}}_{c}^{\infty}(\mathbb{R}^{d})$,
$\mathcal{M}^{n}_{T}(H)^{-1}$ is bounded from above by

$\exp\Big{\\{}\frac{a_{n}^{2}}{n^{d}}\sup_{\mu\in
K}\Big{(}-l(\mu,H)+\frac{\chi(\rho)}{2d}\sum_{i=1}^{d}\|\partial_{u_{i}}H\|_{L^{2}([0,T]\times\mathbb{R}^{d})}^{2}+\delta-\mu_{0}(\phi)\Big{)}\Big{\\}}\exp\Big{\\{}\frac{a_{n}^{2}}{n^{d}}\mu_{0}(\phi)\Big{\\}}.$

Therefore,

$\displaystyle\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\mu^{n}\in
K\Big{)}\leq-$ $\displaystyle\inf_{\mu\in
K}\Big{(}l(\mu,H)-\frac{\chi(\rho)}{2d}\sum_{i=1}^{d}\|\partial_{u_{i}}H\|_{L^{2}([0,T]\times\mathbb{R}^{d})}^{2}+\mu_{0}(\phi)\Big{)}$
$\displaystyle+\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\exp\Big{\\{}\frac{a_{n}^{2}}{n^{d}}\mu_{0}(\phi)\Big{\\}}\Big{]}+\delta.$

For the initial density fluctuation field, since $a_{n}\ll n^{d}$, by Taylor’s
expansion,

$\displaystyle\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}$
$\displaystyle\log\mathbb{E}^{n}_{\rho}\Big{[}\exp\Big{\\{}\frac{a_{n}^{2}}{n^{d}}\mu^{n}_{0}(\phi)\Big{\\}}\Big{]}$
(3.17)
$\displaystyle=\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\sum_{x\in\mathbb{Z}^{d}}\log
E_{\nu_{\rho}}\Big{[}\exp\Big{\\{}\frac{a_{n}}{n^{d}}(\eta_{x}-\rho)\phi(x/n)\Big{\\}}\Big{]}$
$\displaystyle=\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\sum_{x\in\mathbb{Z}^{d}}\log\Big{(}1+\frac{a_{n}^{2}\chi(\rho)}{2n^{2d}}\phi(x/n)^{2}+O_{\phi}(a_{n}^{3}/n^{3d})\Big{)}$
$\displaystyle=\frac{1}{2}\chi(\rho)\|\phi\|_{L^{2}(\mathbb{R}^{d})}^{2}.$

By first letting $\delta\rightarrow 0$, and then optimizing over
$H\in{\mathcal{C}}_{c}^{1,\infty}([0,T]\times\mathbb{R}^{d})$ and
$\phi\in{\mathcal{C}}_{c}^{\infty}(\mathbb{R}^{d})$,

$\displaystyle\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\mu^{n}\in
K\Big{)}$
$\displaystyle\leq-\sup_{H\in{\mathcal{C}}_{c}^{1,\infty}([0,T]\times\mathbb{R}^{d}),\atop\phi\in{\mathcal{C}}_{c}^{\infty}(\mathbb{R}^{d})}\inf_{\mu\in
K}\Big{\\{}l(\mu,H)-\frac{\chi(\rho)}{2d}\sum_{i=1}^{d}\|\partial_{u_{i}}H\|_{L^{2}([0,T]\times\mathbb{R}^{d})}^{2}+\mu_{0}(\phi)-\frac{\chi(\rho)}{2}\|\phi\|_{L^{2}(\mathbb{R}^{d})}^{2}\Big{\\}}.$

Notice that the term inside the above brace is continuous and linear in $\mu$
for fixed $H,\phi$, and continuous and concave in $H,\phi$ for fixed $\mu$.
Thus, by the Minimax theorem (see [8] for example), we can exchange the order
of the above supremum and infimum. This concludes the proof of the upper bound
(2.1) for compact sets.

It remains to prove that the sequence $\\{\mu^{n}\\}$ is exponentially tight,
which holds true once we can show the following two estimates:

  1. (i)

for any $H\in{\mathcal{S}}$,

$\limsup_{A\rightarrow\infty}\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0\leq
t\leq T}|\big{\langle}\mu^{n}_{t},H\big{\rangle}|>A\Big{)}=-\infty;$ (3.18)

  2. (ii)

for any $H\in{\mathcal{S}}$, and for any $\varepsilon>0$,

$\limsup_{\delta\rightarrow
0}\limsup_{n\rightarrow\infty}\sup_{\tau\in{\mathcal{T}}}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0<t\leq\delta}|\big{\langle}\mu^{n}_{t+\tau}-\mu^{n}_{\tau},H\big{\rangle}|>\varepsilon\Big{)}=-\infty,$
(3.19)

where ${\mathcal{T}}$ is the set of all stopping times bounded by $T$.

For any $H\in{\mathcal{S}}$, we first rewrite (3.7) as

$\displaystyle\frac{n^{d}}{a_{n}^{2}}\log{\mathcal{M}}^{n}_{t}(H)=$
$\displaystyle\big{\langle}\mu^{n}_{t},H\big{\rangle}-\big{\langle}\mu^{n}_{0},H\big{\rangle}-\int_{0}^{t}\big{\langle}\mu^{n}_{s},\tfrac{1}{2d}\Delta
H\big{\rangle}ds$ (3.20)
$\displaystyle-\frac{\chi(\rho)}{2d}\int_{0}^{t}\int_{\mathbb{R}^{d}}\sum_{i=1}^{d}\big{(}\partial_{u_{i}}H_{s}(u)\big{)}^{2}du\,ds+{\mathcal{R}}^{n}_{t}(H),$

where ${\mathcal{R}}^{n}_{t}(H)$ is introduced in (3.16). Since the
penultimate term on the right hand side is bounded by $C(H,d)t$, it satisfies
the estimates in (3.18) and (3.19) obviously. The second and the last terms on
the right hand side also satisfy the above estimates by (3.17) and Proposition
3.2. For the third one on the right hand side, we have the following lemma.

######  Lemma 3.4.

For any $H\in{\mathcal{S}}$, and for any $\varepsilon>0$,

$\displaystyle\limsup_{A\rightarrow\infty}\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0\leq
t\leq
T}\Big{|}\int_{0}^{t}\big{\langle}\mu^{n}_{s},H\big{\rangle}ds\Big{|}>A\Big{)}=-\infty,$
$\displaystyle\limsup_{\delta\rightarrow
0}\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0<t\leq\delta}\Big{|}\int_{s}^{s+t}\big{\langle}\mu^{n}_{r},H\big{\rangle}dr\Big{|}>\varepsilon\Big{)}=-\infty.$

The proof of the above lemma is the same as [8, Lemma 2.2], and thus is
omitted.

It remains to deal with the martingale term. By Markov’s inequality,

$\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0\leq t\leq
T}\frac{n^{d}}{a_{n}^{2}}|\log{\mathcal{M}}^{n}_{t}(H)|>A\Big{)}\leq-A+\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\sup_{0\leq
t\leq T}|{\mathcal{M}}^{n}_{t}(H)|\Big{]}.$

By (3.7), for any $\lambda\in\mathbb{R}$,

$\log\mathbb{E}^{n}_{\rho}\Big{[}{\mathcal{M}}^{n}_{t}(H)^{\lambda}\Big{]}\leq
C(H)|\lambda|(|\lambda|+1)ta_{n}^{2}/n^{d}.$ (3.21)

Then, by Doob’s inequality,

$\mathbb{E}^{n}_{\rho}\Big{[}\sup_{0\leq t\leq
T}|{\mathcal{M}}^{n}_{t}(H)|^{2}\Big{]}\leq
4\mathbb{E}^{n}_{\rho}\Big{[}{\mathcal{M}}^{n}_{T}(H)^{2}\Big{]}\leq
4e^{C(H)Ta_{n}^{2}/n^{d}}.$

Thus,

$\limsup_{A\rightarrow\infty}\limsup_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0\leq
t\leq
T}\frac{n^{d}}{a_{n}^{2}}|\log{\mathcal{M}}^{n}_{t}(H)|>A\Big{)}=-\infty.$

For the estimate in (3.19), for any $\varepsilon>0$ and for any $\lambda>0$,

$\displaystyle\limsup_{n\rightarrow\infty}\sup_{\tau\in{\mathcal{T}}}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0<t\leq\delta}|\log{\mathcal{M}}^{n}_{t+\tau}(H)-\log{\mathcal{M}}^{n}_{\tau}(H)|>\varepsilon
a_{n}^{2}/n^{d}\Big{)}$
$\displaystyle\leq-\lambda\varepsilon+\limsup_{n\rightarrow\infty}\sup_{\tau\in{\mathcal{T}}}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\sup_{0<t\leq\delta}\Big{(}{\mathcal{M}}^{n}_{t+\tau}(H)/{\mathcal{M}}^{n}_{\tau}(H)\Big{)}^{\lambda}\Big{]}$
$\displaystyle\leq-\lambda\varepsilon+\limsup_{n\rightarrow\infty}\sup_{\tau\in{\mathcal{T}}}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\Big{(}{\mathcal{M}}^{n}_{\delta+\tau}(H)/{\mathcal{M}}^{n}_{\tau}(H)\Big{)}^{2\lambda}\Big{]}$
$\displaystyle\leq-\lambda\varepsilon+C(H)\lambda(\lambda+1)\delta.$

Above, we used Doob’s inequality and (3.21). We conclude the proof by first
letting $\delta\rightarrow 0$ and then $\lambda\rightarrow+\infty$.

##  4\. Proof of Theorem 2.1: lower bound

To prove the lower bound from hydrodynamic limits, we consider the following
perturbation of the generator $L_{n}$. For any
$H\in{\mathcal{C}}^{1,\infty}_{c}([0,T]\times\mathbb{R}^{d})$, define
$L_{n}^{H}=n^{2}(L_{n,s}^{H}+\alpha n^{-\beta}L_{n,a}^{H})$, where for local
functions $f:\mathcal{X}\rightarrow\mathbb{R}$,

$\displaystyle L_{n,s}^{H}f(\eta)$
$\displaystyle=\frac{1}{2d}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}\exp\big{\\{}\tfrac{a_{n}}{n^{d}}\big{(}\eta_{x}-\eta_{x+e_{i}}\big{)}\big{(}H_{s}(\tfrac{x+e_{i}-v_{n}sm}{n})-H_{s}(\tfrac{x-v_{n}sm}{n})\big{)}\big{\\}}$
$\displaystyle\qquad\times\big{[}f(\eta^{x,x+e_{i}})-f(\eta)\big{]},$
$\displaystyle L_{n,a}^{H}f(\eta)$
$\displaystyle=\frac{1}{d}\sum_{x\in\mathbb{Z}^{d}}\eta_{x}(1-\eta_{x+e_{i}})\exp\big{\\{}\tfrac{a_{n}}{n^{d}}\big{(}H_{s}(\tfrac{x+e_{i}-v_{n}sm}{n})-H_{s}(\tfrac{x-v_{n}sm}{n})\big{)}\big{\\}}\big{[}f(\eta^{x,x+e_{i}})-f(\eta)\big{]}.$

Notice that we introduced a spatial translation to the function $H$ in the
above definition. For $\phi\in{\mathcal{C}}^{\infty}_{c}(\mathbb{R}^{d})$,
define $\nu_{\rho}^{n,\phi}$ as the product measure on ${\mathcal{X}}_{d}$
with marginals

$\nu_{\rho}^{n,\phi}(\eta_{x}=1)=\rho+\frac{\chi(\rho)a_{n}}{n^{d}}\phi(\tfrac{x}{n}),\quad
x\in\mathbb{Z}^{d}.$

Last, we denote by $\mathbb{P}^{n}_{H,\phi}$ the probability measure on
${\mathcal{D}}([0,T],{\mathcal{X}}_{d})$ induced by the process with generator
$L_{n}^{H}$ starting from the measure $\nu_{\rho}^{n,\phi}$, and by
$\mathbb{E}^{n}_{H,\phi}$ the corresponding expectation. By [13, Proposition
7.3, Appendix 1], we have

$\frac{d\mathbb{P}^{n}_{H,\phi}}{d\mathbb{P}^{n}_{\rho}}={\mathcal{M}}^{n}_{T}(H)\frac{d\nu_{\rho}^{n,\phi}}{d\nu_{\rho}}.$
(4.1)

The following result concerns hydrodynamic limits for the measure
$\mu^{n}=\\{\mu^{n}_{t}:0\leq t\leq T\\}$ under $\mathbb{P}^{n}_{H,\phi}$,
whose proof is presented in Subsection 4.1.

######  Theorem 4.1.

As $n\rightarrow\infty$ , the sequence of measures $\\{\mu^{n}_{t}(du):0\leq
t\leq T\\}_{n\geq 1}$ converges in $\mathbb{P}^{n}_{H,\phi}$-probability to
the absolutely continuous measure $\\{\mu(t,u)du:0\leq t\leq T\\}$, where
$\mu(t,u)$ is the unique weak solution to

$\begin{cases}\partial_{t}\mu=\frac{1}{2d}\Delta\mu-\frac{\chi(\rho)}{d}\Delta
H,\quad&\text{in}\quad(0,T]\times\mathbb{R}^{d},\\\
\mu(0,\cdot)=\chi(\rho)\phi(\cdot),\quad&\text{in}\quad\mathbb{R}^{d}.\end{cases}$
(4.2)

Here, we say $\mu$ is a weak solution to (4.2) if for any
$G\in{\mathcal{C}}_{c}^{1,2}([0,T]\times\mathbb{R}^{d})$ and for any $0<t\leq
T$,

$\int_{\mathbb{R}^{d}}\mu(t,u)G_{t}(u)du=\chi(\rho)\int_{\mathbb{R}^{d}}\phi(u)G_{0}(u)du\\\
+\int_{0}^{t}\int_{\mathbb{R}^{d}}\mu(s,u)\Big{(}\partial_{s}+\frac{1}{2d}\Delta\Big{)}G_{s}(u)\,du\,ds+\frac{\chi(\rho)}{d}\sum_{i=1}^{d}\int_{0}^{t}\int_{\mathbb{R}^{d}}\partial_{u_{i}}G_{s}(u)\partial_{u_{i}}H_{s}(u)\,du\,ds.$

Actually, $\mu$ has the following explicit expression: for any $t\geq 0$ and
for any $u\in\mathbb{R}^{d}$,

$\mu(t,u)=\chi(\rho)\int_{\mathbb{R}^{d}}p_{t/d}(u-v)\phi(v)dv-\frac{\chi(\rho)}{d}\sum_{i=1}^{d}\int_{0}^{t}\int_{\mathbb{R}^{d}}\partial_{v_{i}}p_{(t-s)/d}(u-v)\partial_{v_{i}}H(s,v)\,dv\,ds,$
(4.3)

where $p_{t}(u)$ is the transition probability of the standard $d$-dimensional
Brownian motion

$p_{t}(u)=(2\pi
t)^{-d/2}\exp\Big{\\{}-\sum_{i=1}^{d}\frac{u_{i}^{2}}{2t}\Big{\\}}.$

Before proving the lower bound, we also need to introduce some properties of
the rate function ${\mathcal{Q}}$. For
$H,G\in{\mathcal{C}}^{1,\infty}_{c}([0,T]\times\mathbb{R}^{d})$, define the
scalar product

$[H,G]=\frac{1}{d}\sum_{i=1}^{d}\int_{0}^{T}\int_{\mathbb{R}^{d}}\partial_{u_{i}}H(t,u)\partial_{u_{i}}G(t,u)\,du\,dt.$

Define further the equivalence relation $G\sim H$ if and only if
$[H-G,H-G]=0$. Last, let ${\mathcal{H}}^{1}$ be the Hilbert space defined as
the completion of
${\mathcal{C}}^{1,\infty}_{c}([0,T]\times\mathbb{R}^{d})/\sim$ with respect to
the scalar product $[\cdot,\cdot]$.

######  Lemma 4.2.

If ${\mathcal{Q}}(\mu)<+\infty$, then there exist $\phi\in
L^{2}(\mathbb{R}^{d})$ and $H\in{\mathcal{H}}^{1}$ such that

${\mathcal{Q}}_{0}(\mu_{0})=\frac{\chi(\rho)}{2}\|\phi\|_{L^{2}(\mathbb{R}^{d})}^{2},\quad{\mathcal{Q}}_{dyn}(\mu)=\frac{\chi(\rho)}{2}[H,H].$

Moreover, $\mu$ is the unique weak solution of (4.2).

The proof of the above result uses Riesz representation theorem, and is
standard, see [8, Lemma 5.1] for example. Indeed, if
${\mathcal{Q}}(\mu)<+\infty$, then there exists $\phi\in
L^{2}(\mathbb{R}^{d})$ and $H\in{\mathcal{H}}^{1}$ such that

$\big{\langle}\mu_{0},\psi\big{\rangle}=\chi(\rho)\big{\langle}\phi,\psi\big{\rangle},\;\forall\psi\in
L^{2}(\mathbb{R}^{d});\quad l(\mu,G)=\chi(\rho)[H,G],\;\forall
G\in{\mathcal{H}}^{1},$

from which the above lemma follows easily. For this reason, we omit the proof.

Now, we are ready to prove the lower bound.

###### Proof of the lower bound (2.2).

Let $O$ be an open set of ${\mathcal{D}}([0,T],{\mathcal{S}}^{\prime})$. We
only need to show that for any $\mu\in O$,

$\liminf_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}(\\{\mu^{n}_{t}:0\leq
t\leq T\\}\in O)\geq-{\mathcal{Q}}(\mu).$

Without loss of generality, we can assume ${\mathcal{Q}}(\mu)<+\infty$;
otherwise, the last inequality is trivial. Let $\phi=\phi(\mu)$ and $H=H(\mu)$
be identified as in Lemma 4.2. For $\delta>0$, choose
$\phi^{\delta}\in{\mathcal{C}}_{c}^{\infty}(\mathbb{R}^{d}),H^{\delta}\in{\mathcal{C}}^{1,\infty}_{c}([0,T]\times\mathbb{R}^{d})$,
such that

$\lim_{\delta\rightarrow
0}\|\phi^{\delta}-\phi\|_{L^{2}(\mathbb{R}^{d})}=\lim_{\delta\rightarrow
0}\|H^{\delta}-H\|_{{\mathcal{H}}^{1}}=0.$

Above, $\|H\|_{{\mathcal{H}}^{1}}:=\sqrt{[H,H]}$. Then, by (4.3) and Lemma
4.2, $\mu^{\delta}=\mu^{\delta}(\phi^{\delta},H^{\delta})$ converges, as
$\delta\rightarrow 0$, to $\mu$ in
${\mathcal{D}}([0,T],{\mathcal{S}}^{\prime})$, and $\lim_{\delta\rightarrow
0}{\mathcal{Q}}(\mu^{\delta})={\mathcal{Q}}(\mu)$. Thus, for any $\delta>0$,
there exists $\mu^{\delta}\in O$ with
$\phi^{\delta}\in{\mathcal{C}}_{c}^{\infty}(\mathbb{R}^{d}),H^{\delta}\in{\mathcal{C}}^{1,\infty}_{c}([0,T]\times\mathbb{R}^{d})$
such that

$-{\mathcal{Q}}(\mu^{\delta})\geq-{\mathcal{Q}}(\mu)-o_{\mu}(\delta),$

where $o_{\mu}(\delta)$ is some constant depending on $\mu$ and converges to
zero as $\delta\rightarrow 0$. Therefore, we only need to show that, for
$\delta$ small enough,

$\liminf_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}(\\{\mu^{n}_{t}:0\leq
t\leq T\\}\in O)\geq-{\mathcal{Q}}(\mu^{\delta}).$

Since $\mu^{\delta}\in O$ for $\delta$ small enough, to make notations short,
let us denote $\mu^{\delta}$ by $\mu$, and then
$\phi=\phi(\mu)\in{\mathcal{C}}^{\infty}_{c}(\mathbb{R}^{d})$ and
$H=H(\mu)\in{\mathcal{C}}^{1,\infty}_{c}([0,T]\times\mathbb{R}^{d})$. By
(4.1), we first write

$\mathbb{P}^{n}_{\rho}(\mu^{n}\in
O)=\mathbb{E}^{n}_{H,\phi}\Big{[}\frac{d\mathbb{P}^{n}_{\rho}}{d\mathbb{P}^{n}_{H,\phi}}\mathbf{1}\\{\mu^{n}\in
O\\}\Big{]}=\mathbb{E}^{n}_{H,\phi}\Big{[}{\mathcal{M}}^{n}_{T}(H)^{-1}\frac{d\nu^{n}_{\rho}}{d\nu^{n,\phi}_{\rho}}\Big{|}\mu^{n}\in
O\Big{]}\mathbb{P}^{n}_{H,\phi}(\mu^{n}\in O).$

Then, by Jensen’s inequality,

$\displaystyle\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}(\mu^{n}\in O)$
$\displaystyle\geq\mathbb{E}^{n}_{H,\phi}\Big{[}\frac{n^{d}}{a_{n}^{2}}\log{\mathcal{M}}^{n}_{T}(H)^{-1}\Big{|}\mu^{n}\in
O\Big{]}$
$\displaystyle+\mathbb{E}^{n}_{H,\phi}\Big{[}\frac{n^{d}}{a_{n}^{2}}\log\frac{d\nu^{n}_{\rho}}{d\nu^{n,\phi}_{\rho}}\Big{|}\mu^{n}\in
O\Big{]}+\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{H,\phi}(\mu^{n}\in
O)=:\sum_{j=1}^{3}A^{n}_{j}.$

The term $A^{n}_{3}$ is simpler: by Theorem 4.1, under
$\mathbb{P}^{n}_{H,\phi}$, $\mu^{n}\rightarrow\mu\in O$ in probability as
$n\rightarrow\infty$. Thus,

$\lim_{n\rightarrow\infty}A^{n}_{3}=0.$

For $A^{n}_{1}$, by (3.7) and (3.16),

$\displaystyle\frac{n^{d}}{a_{n}^{2}}\log{\mathcal{M}}^{n}_{T}(H)^{-1}=-\Big{(}\ell(\mu^{n},H)-\frac{\chi(\rho)}{2}[H,H]+{\mathcal{R}}^{n}_{T}(H)\Big{)}.$

By Lemma 4.3 below, ${\mathcal{R}}^{n}_{T}(H)$ converges in
$\mathbb{P}^{n}_{H,\phi}$-probability to zero as $n\rightarrow\infty$.
Together with Theorem 4.1 and Lemma 4.2, the last line converges, as
$n\rightarrow\infty$, in $\mathbb{P}^{n}_{H,\phi}$-probability to

$-\Big{(}\ell(\mu,H)-\frac{\chi(\rho)}{2}[H,H]\Big{)}=-\frac{\chi(\rho)}{2}[H,H]=-{\mathcal{Q}}_{dyn}(\mu).$

Last, by dominated convergence theorem,

$\displaystyle\lim_{n\rightarrow\infty}A^{n}_{1}=\lim_{n\rightarrow\infty}\frac{1}{\mathbb{P}^{n}_{H,\phi}(\mu^{n}\in
O)}\mathbb{E}^{n}_{H,\phi}\Big{[}\frac{n^{d}}{a_{n}^{2}}\log{\mathcal{M}}^{n}_{T}(H)^{-1}\mathbf{1}\\{\mu^{n}\in
O\\}\Big{]}=-{\mathcal{Q}}_{dyn}(\mu).$

It remains to deal with $A^{n}_{2}$. Since $\nu^{n,\phi}_{\rho}$ and
$\nu^{n}_{\rho}$ are both product measures,

$\displaystyle\frac{n^{d}}{a_{n}^{2}}\log\frac{d\nu^{n,\phi}_{\rho}}{d\nu^{n}_{\rho}}=\frac{n^{d}}{a_{n}^{2}}\sum_{x\in\mathbb{Z}^{d}}\Big{\\{}\eta_{x}\log\Big{(}1+\frac{(1-\rho)a_{n}}{n^{d}}\phi(\tfrac{x}{n})\Big{)}+(1-\eta_{x})\log\Big{(}1-\frac{\rho
a_{n}}{n^{d}}\phi(\tfrac{x}{n})\Big{)}\Big{\\}}.$

Since $|\log(1+x)-x+x^{2}/2|\leq Cx^{3}$ for $x$ small, the last line equals

$\frac{1}{a_{n}}\sum_{x\in\mathbb{Z}^{d}}(\eta_{x}-\rho)\phi(\tfrac{x}{n})-\frac{1}{2n^{d}}\sum_{x\in\mathbb{Z}^{d}}\big{(}\eta_{x}(1-\rho)^{2}+(1-\eta_{x})\rho^{2}\big{)}\phi(\tfrac{x}{n})^{2}+O_{\phi}(a_{n}/n^{d}).$

Notice that

$\eta_{x}(1-\rho)^{2}+(1-\eta_{x})\rho^{2}=(1-2\rho)(\eta_{x}-\rho)+\chi(\rho).$

Thus,

$\displaystyle\frac{n^{d}}{a_{n}^{2}}\log\frac{d\nu^{n,\phi}_{\rho}}{d\nu^{n}_{\rho}}=$
$\displaystyle\frac{1}{a_{n}}\sum_{x\in\mathbb{Z}^{d}}(\eta_{x}-\rho)\phi(\tfrac{x}{n})$
(4.4)
$\displaystyle\qquad+\frac{(2\rho-1)}{2n^{d}}\sum_{x\in\mathbb{Z}^{d}}(\eta_{x}-\rho)\phi(\tfrac{x}{n})^{2}-\frac{\chi(\rho)}{2}\|\phi\|_{L^{2}(\mathbb{R}^{d})}^{2}+o_{\phi}(n),$

where $o_{\phi}(n)$ is a constant depending on $\phi$ and vanishes as
$n\rightarrow\infty$. Under $\nu_{\rho}^{n,\phi}$, the last expression
converges in probability to $\chi(\rho)\|\phi\|_{L^{2}(\mathbb{R}^{d})}^{2}/2$
as $n\rightarrow\infty$. As we dealt with $A^{n}_{1}$, by dominated
convergence theorem,

$\lim_{n\rightarrow\infty}A^{n}_{2}=-\frac{\chi(\rho)}{2}\|\phi\|_{L^{2}(\mathbb{R}^{d})}^{2}=-{\mathcal{Q}}_{0}(\mu_{0}).$

This concludes the proof. ∎

###  4.1. Proof of Theorem 4.1

In this subsection, we prove Theorem 4.1. By Dynkin’s martingale formula,
under $\mathbb{P}^{n}_{H,\phi}$, for any
$G\in{\mathcal{C}}^{1,2}_{c}([0,T]\times\mathbb{R}^{d})$,

$\displaystyle M^{n}_{t}(G)$
$\displaystyle=\big{\langle}\mu^{n}_{t},G_{t}\big{\rangle}-\big{\langle}\mu^{n}_{0},G_{0}\big{\rangle}-\int_{0}^{t}(\partial_{s}+L^{H}_{n})\big{\langle}\mu^{n}_{s},G_{s}\big{\rangle}ds,$
(4.5) $\displaystyle M^{n}_{t}(G)^{2}$
$\displaystyle-\int_{0}^{t}\big{\\{}L^{H}_{n}\big{\langle}\mu^{n}_{s},G_{s}\big{\rangle}^{2}-2\big{\langle}\mu^{n}_{s},G_{s}\big{\rangle}L^{H}_{n}\big{\langle}\mu^{n}_{s},G_{s}\big{\rangle}\big{\\}}ds$
(4.6)

are both martingales.

We first calculate the time integral in (4.5). For the part associated to the
symmetric generator $L^{H}_{n,s}$,

$\displaystyle
n^{2}L^{H}_{n,s}\big{\langle}\mu^{n}_{s},G_{s}\big{\rangle}=\frac{n^{2}}{2da_{n}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}$
$\displaystyle\exp\Big{\\{}\frac{a_{n}}{n^{d}}(\eta_{x}-\eta_{x+e_{i}})(H_{s}(\tfrac{x+e_{i}-v_{n}sm}{n})-H_{s}(\tfrac{x-v_{n}sm}{n}))\Big{\\}}$
$\displaystyle\times(\eta_{x}-\eta_{x+e_{i}})\big{(}G_{s}(\tfrac{x+e_{i}-v_{n}sm}{n})-G_{s}(\tfrac{x-v_{n}sm}{n})\big{)}.$

Using the basic inequality $|e^{x}-1-x|\leq(x^{2}/2)e^{|x|}$, and the
summation by parts formula, the term on the right-hand side of the last
expression equals

$\frac{1}{2da_{n}}\sum_{x\in\mathbb{Z}^{d}}(\eta_{x}-\rho)\Delta_{n}G_{s}(\tfrac{x-v_{n}sm}{n})\\\
+\frac{1}{2dn^{d}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}(\eta_{x}-\eta_{x+e_{i}})^{2}\partial_{u_{i}}H_{s}(\tfrac{x-v_{n}sm}{n})\partial_{u_{i}}G_{s}(\tfrac{x-v_{n}sm}{n})+o_{H,G}(n),$
(4.7)

where $o_{H,G}(n)$ is some constant depending on $H,G$ and vanishes as
$n\rightarrow\infty$. Similarly, for the asymmetric part,

$\alpha n^{2-\beta}L^{H}_{n,a}\big{\langle}\mu^{n}_{s},G_{s}\big{\rangle}\\\
=\frac{\alpha
n^{1-\beta}}{da_{n}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}\eta_{x}(1-\eta_{x+e_{i}})\exp\Big{\\{}\frac{a_{n}}{n^{d}}(H_{s}(\tfrac{x+e_{i}-v_{n}sm}{n})-H_{s}(\tfrac{x-v_{n}sm}{n}))\Big{\\}}\nabla_{n,i}G_{s}(\tfrac{x-v_{n}sm}{n}).$

Using the inequality $|e^{x}-1|\leq|x|e^{|x|}$, the last line equals

$\frac{\alpha
n^{1-\beta}}{da_{n}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}\big{(}\eta_{x}(1-\eta_{x+e_{i}})-\chi(\rho)\big{)}\nabla_{n,i}G_{s}(\tfrac{x-v_{n}s}{n})+o_{H,G}(n).$
(4.8)

By Lemma 3.1, we can replace $\nabla_{n,i}G_{s}$ by $\partial_{u_{i}}G_{s}$ in
the last expression, producing an error term $\varepsilon^{n}_{4,G}(s)$ that
is super-exponentially small under the measure $\mathbb{P}^{n}_{\rho}$. For
the time derivative,

$\partial_{s}\big{\langle}\mu^{n}_{s},G_{s}\big{\rangle}=\big{\langle}\mu^{n}_{s},\partial_{s}G_{s}\big{\rangle}-\frac{\alpha
n^{1-\beta}}{da_{n}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}(1-2\rho)(\eta_{x}-\rho)\partial_{u_{i}}G_{s}(\tfrac{x-v_{n}sm}{n}).$

By (3.6), summation by parts formula and Lemma 3.1,

$(\partial_{s}+\alpha
n^{2-\beta}L^{H}_{n,a})\big{\langle}\mu^{n}_{s},G_{s}\big{\rangle}\\\
=\big{\langle}\mu^{n}_{s},\partial_{s}G_{s}\big{\rangle}-\frac{\alpha
n^{1-\beta}}{da_{n}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}(\eta_{x}-\rho)(\eta_{x+e_{i}}-\rho)\partial_{u_{i}}G_{s}(\tfrac{x-v_{n}sm}{n})+\varepsilon^{n}_{5,G}(s)+o_{H,G}(n),$

where $\varepsilon^{n}_{5,G}(s)$ satisfies the estimates in (3.5). To sum up,
the time integral in (4.5) equals

$\begin{split}&\int_{0}^{t}\big{\langle}\mu^{n}_{s},\Big{(}\partial_{s}+\frac{1}{2d}\Delta_{n}\Big{)}G_{s}\big{\rangle}ds\\\
&+\int_{0}^{t}\frac{1}{2dn^{d}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}\big{(}\eta_{x}(s)-\eta_{x+e_{i}}(s)\big{)}^{2}\partial_{u_{i}}H_{s}(\tfrac{x-v_{n}sm}{n})\partial_{u_{i}}G_{s}(\tfrac{x-v_{n}sm}{n})ds\\\
&-\int_{0}^{t}\frac{\alpha
n^{1-\beta}}{da_{n}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}(\eta_{x}(s)-\rho)(\eta_{x+e_{i}}(s)-\rho)\partial_{u_{i}}G_{s}(\tfrac{x-v_{n}sm}{n})ds\\\
&+o_{H,G}(n)+\int_{0}^{t}\sum_{j=4,5}\varepsilon^{n}_{j,G}(s)ds.\end{split}$
(4.9)

A simple calculation also shows that the time integral in (4.6) equals

$\int_{0}^{t}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}\Big{(}\frac{1}{2da_{n}^{2}}(\eta_{x}(s)-\eta_{x+e_{i}}(s))^{2}+\frac{\alpha}{dn^{\beta}a_{n}^{2}}\eta_{x}(s)(1-\eta_{x+e_{i}}(s))\Big{)}\\\
\times\exp\big{\\{}\tfrac{a_{n}}{n^{d}}\big{(}\eta_{x}(s)-\eta_{x+e_{i}}(s)\big{)}\big{(}H_{s}(\tfrac{x+e_{i}-v_{n}sm}{n})-H_{s}(\tfrac{x-v_{n}sm}{n})\big{)}\big{\\}}(\nabla_{n,i}G_{s}(\tfrac{x-v_{n}sm}{n}))^{2}ds.$

By Doob’s inequality and Cauchy-Schwarz inequality, since $a_{n}\gg n^{d/2}$,

$\lim_{n\rightarrow\infty}\mathbb{E}^{n}_{H,\phi}\Big{[}\sup_{0\leq t\leq
T}M^{n}_{t}(G)^{2}\Big{]}\leq\lim_{n\rightarrow\infty}\frac{C(H,G)T^{2}n^{d}}{a_{n}^{2}}=0.$
(4.10)

To deal with (4.9), we need the following lemma.

######  Lemma 4.3.

Let ${\mathcal{A}}\in{\mathcal{D}}([0,T],{\mathcal{X}}_{d})$ satisfy

$\lim_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}({\mathcal{A}})=-\infty.$

Then,

$\lim_{n\rightarrow\infty}\mathbb{P}^{n}_{H,\phi}({\mathcal{A}})=0.$

###### Proof.

For any two probability measures $\mu,\nu$ on the same space, recall that the
relative entropy of $\mu$ with respect to $\nu$ is defined as

$H(\mu|\nu)=\int\log\Big{(}\frac{d\mu}{d\nu}\Big{)}d\mu$

if $\mu$ is absolutely continuous with respect to $\nu$, and
$H(\mu|\nu)=+\infty$ otherwise.

We claim that

$H(\mathbb{P}^{n}_{H,\phi}|\mathbb{P}^{n}_{H,0})=H(\nu^{n,\phi}_{\rho}|\nu_{\rho})\leq\frac{K_{0}a_{n}^{2}}{n^{d}}$
(4.11)

for some constant $K_{0}=K_{0}(\phi)>0$. Indeed, by (4.4),

$\Big{|}\log\frac{d\nu_{\rho}^{n,\phi}}{d\nu_{\rho}}-\frac{a_{n}}{n^{d}}\sum_{x\in\mathbb{Z}^{d}}(\eta_{x}-\rho)\phi(x/n)\Big{|}\leq
C(\phi)\frac{a_{n}^{2}}{n^{d}}.$

Thus,

$\Big{|}H(\nu^{n,\phi}_{\rho}|\nu_{\rho})-\int\frac{a_{n}}{n^{d}}\sum_{x\in\mathbb{Z}^{d}}(\eta_{x}-\rho)\phi(x/n)d\nu^{n,\phi}_{\rho}\Big{|}\leq
C(\phi)\frac{a_{n}^{2}}{n^{d}}.$

We conclude the proof by noting that

$\Big{|}\int\frac{a_{n}}{n^{d}}\sum_{x\in\mathbb{Z}^{d}}(\eta_{x}-\rho)\phi(x/n)d\nu^{n,\phi}_{\rho}\Big{|}=\frac{a_{n}^{2}\chi(\rho)}{n^{2d}}\sum_{x\in\mathbb{Z}^{d}}\phi(x/n)^{2}\leq
C(\phi)\frac{a_{n}^{2}}{n^{d}}.$

By entropy inequality (see [13, Proposition 8.2, Appendix 1]),

$\mathbb{P}^{n}_{H,\phi}({\mathcal{A}})\leq\frac{\log
2+H(\mathbb{P}^{n}_{H,\phi}|\mathbb{P}^{n}_{H,0})}{\log[1+1/\mathbb{P}^{n}_{H,0}({\mathcal{A}})]}.$

Thus, we conclude the proof once we can show that

$\lim_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{H,0}({\mathcal{A}})=-\infty.$

To prove it, by (4.1), Cauchy-Schwarz inequality, and (3.21),

$\displaystyle\lim_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{P}^{n}_{H,0}({\mathcal{A}})=\lim_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\big{[}{\mathcal{M}}^{n}_{T}(H)\mathbf{1}_{{\mathcal{A}}}\big{]}$
$\displaystyle\leq\limsup_{n\rightarrow\infty}\frac{n^{d}}{2a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\big{[}{\mathcal{M}}^{n}_{T}(H)^{2}\big{]}+\lim_{n\rightarrow\infty}\frac{n^{d}}{2a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}({\mathcal{A}})=-\infty.$

∎

Let $Q^{n}_{H,\phi}$ be the probability measure on
${\mathcal{D}}([0,T],\mathcal{S}^{\prime})$ induced by the process
$\\{\mu^{n}_{t}:0\leq t\leq T\\}$ under the measure $\mathbb{P}^{n}_{H,\phi}$.
Theorem 4.1 follows directly from the following lemma and the uniqueness of
the weak solution to the PDE (4.2).

######  Lemma 4.4.

The following statements hold.

  1. $(1)$

The sequence of probability measures $\\{Q^{n}_{H,\phi}\\}_{n\geq 1}$ is
tight.

  2. $(2)$

Let $Q^{*}$ be any limit point of the probability measures
$\\{Q^{n}_{H,\phi}\\}_{n\geq 1}$ along some subsequence. Then, $Q^{*}$ is
concentrated on trajectories which are absolutely continuous with respect to
the Lebesgue measure,

$Q^{*}\big{(}\mu_{t}(du)=\mu_{t}(u)du\;\text{for some }\mu_{t}(u),\;\forall
t\in[0,T]\big{)}=1.$

  3. $(3)$

$Q^{*}$ is concentrated on trajectories whose densities are the weak solutions
to the PDE (4.2).

###### Proof.

(1) For the tightness, we only need to show that for any $G\in{\mathcal{S}}$
and for any $\varepsilon>0$,

$\displaystyle\limsup_{A\rightarrow\infty}\limsup_{n\rightarrow\infty}\mathbb{P}^{n}_{H,\phi}\Big{(}\sup_{0\leq
t\leq T}|\big{\langle}\mu^{n}_{t},G\big{\rangle}|>A\Big{)}$ $\displaystyle=0,$
$\displaystyle\limsup_{\delta\rightarrow
0}\limsup_{n\rightarrow\infty}\sup_{\tau\in{\mathcal{T}}}\mathbb{P}^{n}_{H,\phi}\Big{(}\sup_{0<t\leq\delta}|\big{\langle}\mu^{n}_{t+\tau}-\mu^{n}_{\tau},G\big{\rangle})|>\varepsilon\Big{)}$
$\displaystyle=0,$

where ${\mathcal{T}}$ is the set of all stopping times bounded by $T$. The
above two estimates follow directly from Lemma 4.3 and (3.18), (3.19).

(2) For the absolute continuity, we follow the proof of [13, Lemma 1.6 and
Remark 1.8, Chapter 5] and thus only sketch it. For any bounded and continuous
function $J:{\mathcal{S}}^{\prime}\rightarrow\mathbb{R}_{+}$, by entropy
inequality (see [13, Section 8, Appendix 1]),

$\mathbb{E}^{n}_{H,\phi}[J(\mu^{n}_{t})]\leq\frac{n^{d}}{a_{n}^{2}}H(\mathbb{P}^{n}_{H,\phi}|\mathbb{P}^{n}_{H,0})+\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{H,0}\Big{[}\exp\Big{\\{}\frac{a_{n}^{2}}{n^{d}}J(\mu^{n}_{t})\Big{\\}}\Big{]}.$

By (4.11), the first term on the right side above is bounded by $K_{0}$. By
the definition of $\mathbb{E}^{n}_{H,0}$, Cauchy-Schwarz inequality, and
(3.21),

$\displaystyle\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{H,0}\Big{[}\exp\Big{\\{}\frac{a_{n}^{2}}{n^{d}}J(\mu^{n}_{t})\Big{\\}}\Big{]}=\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}{\mathcal{M}}^{n}_{T}(H)\exp\Big{\\{}\frac{a_{n}^{2}}{n^{d}}J(\mu^{n}_{t})\Big{\\}}\Big{]}$
$\displaystyle\leq\frac{n^{d}}{2a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}{\mathcal{M}}^{n}_{T}(H)^{2}\Big{]}+\frac{n^{d}}{2a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\exp\Big{\\{}\frac{2a_{n}^{2}}{n^{d}}J(\mu^{n}_{t})\Big{\\}}\Big{]}$
$\displaystyle\leq
C(H)T+\frac{n^{d}}{2a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\exp\Big{\\{}\frac{2a_{n}^{2}}{n^{d}}J(\mu^{n}_{t})\Big{\\}}\Big{]}.$

It remains to deal with the second term on the right side in the last
expression. Since $\nu_{\rho}$ is invariant for the WASEP and is a product
measure, the measure $\mu^{n}_{t}$ satisfies the large deviations principle
with decay rate $a_{n}^{2}/n^{d}$ and with rate function

$I_{0}(\mu)=\sup_{f\in{\mathcal{S}}}\Big{\\{}\int_{\mathbb{R}^{d}}f(u)\mu(du)-\int_{\mathbb{R}^{d}}\log
M_{\rho}(f(u))du\Big{\\}},$

where, for $\theta\in\mathbb{R}$,

$M_{\rho}(\theta)=E_{\nu_{\rho}}[e^{\theta(\eta_{0}-\rho)}]=\rho
e^{\theta(1-\rho)}+(1-\rho)e^{-\theta\rho}.$

Moreover, the rate function $I_{0}$ is equal to

$I_{0}(\mu)=\begin{cases}\int_{\mathbb{R}^{d}}h(\mu(u))du,\quad&\text{if
}\mu(du)=\mu(u)du,\\\ +\infty,\quad&\text{otherwise},\end{cases}$ (4.12)

where $h$ is the Legendre transform of $M_{\rho}$,

$h(\gamma)=\sup_{\theta}\big{\\{}\theta\gamma-\log M_{\rho}(\theta)\big{\\}}.$

Thus, by Laplace-Varadhan Theorem (see [13, Theorem 3.1, Appendix 2]),

$\limsup_{n\rightarrow\infty}\frac{n^{d}}{2a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\exp\Big{\\{}\frac{2a_{n}^{2}}{n^{d}}J(\mu^{n}_{t})\Big{\\}}\Big{]}\leq\sup_{\mu\in{\mathcal{S}}^{\prime}}\big{\\{}J(\mu)-I_{0}(\mu)/2\big{\\}}.$

Formally, taking $J=I_{0}/2$, the last line is zero, and thus

$E_{Q^{*}}[I_{0}(\mu_{t})]=\lim_{n\rightarrow\infty}\frac{n^{d}}{a_{n}^{2}}\log\mathbb{E}^{n}_{H,\phi}[I_{0}(\mu^{n}_{t})]\leq
2[K_{0}+C(H)T].$

Thus, by Fubini’s Theorem,

$E_{Q^{*}}\Big{[}\int_{0}^{T}I_{0}(\mu_{t})dt\Big{]}=\int_{0}^{T}E_{Q^{*}}[I_{0}(\mu_{t})]dt\leq
2[K_{0}T+C(H)T^{2}].$

Together with (4.12), $Q^{*}$ concentrates on measures which are absolutely
continuous with respect to the Lebesgue measure for $t\in[0,T]$ almost
everywhere. We conclude the proof by changing the measure $\mu_{t}(du)$ is a
time set of measure zero if necessary.

(3) To characterize the limit point, for any
$G\in{\mathcal{C}}^{1,2}_{c}([0,T]\times\mathbb{R}^{d})$ and for any
$\varepsilon>0$, by (4.5), (4.9) and (4.10),

$\displaystyle\lim_{n\rightarrow\infty}\mathbb{P}^{n}_{H,\phi}\Big{(}$
$\displaystyle\Big{|}\big{\langle}\mu^{n}_{t},G_{t}\big{\rangle}-\big{\langle}\mu^{n}_{0},G_{0}\big{\rangle}-\int_{0}^{t}\big{\langle}\mu^{n}_{s},\Big{(}\partial_{s}+\frac{1}{2d}\Delta_{n}\Big{)}G_{s}\big{\rangle}ds$
$\displaystyle-\int_{0}^{t}\frac{1}{2dn^{d}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}\big{(}\eta_{x}(s)-\eta_{x+e_{i}}(s)\big{)}^{2}\partial_{u_{i}}H_{s}(\tfrac{x-v_{n}sm}{n})\partial_{u_{i}}G_{s}(\tfrac{x-v_{n}sm}{n})ds$
$\displaystyle+\int_{0}^{t}\frac{\alpha
n^{1-\beta}}{da_{n}}\sum_{x\in\mathbb{Z}^{d}}\sum_{i=1}^{d}(\eta_{x}(s)-\rho)(\eta_{x+e_{i}}(s)-\rho)\partial_{u_{i}}G_{s}(\tfrac{x-v_{n}sm}{n})ds\Big{|}>\varepsilon\Big{)}=0.$

By Lemmas 4.3, 3.1, and Proposition 3.2, we could replace $\Delta_{n}G_{s}$ by
$\Delta G_{s}$ in the third term inside the absolute value, replace
$\big{(}\eta_{x}(s)-\eta_{x+e_{i}}(s)\big{)}^{2}$ by $2\chi(\rho)$ in the
fourth one, and the last term inside the absolute value vanishes in
$\mathbb{P}^{n}_{H,\phi}$-probability. Thus, for any $\varepsilon>0$,

$\lim_{n\rightarrow\infty}\mathbb{P}^{n}_{H,\phi}\Big{(}\Big{|}\big{\langle}\mu^{n}_{t},G_{t}\big{\rangle}-\big{\langle}\mu^{n}_{0},G_{0}\big{\rangle}-\int_{0}^{t}\big{\langle}\mu^{n}_{s},\Big{(}\partial_{s}+\frac{1}{2d}\Delta\Big{)}G_{s}\big{\rangle}ds\\\
-\frac{\chi(\rho)}{d}\sum_{i=1}^{d}\int_{0}^{t}\int_{\mathbb{R}^{d}}\partial_{u_{i}}H_{s}(u)\partial_{u_{i}}G_{s}(u)duds\Big{|}>\varepsilon\Big{)}=0.$

This concludes the proof. ∎

##  5\. Proof of Theorem 2.2

In this section, we assume $d=1$. The following lemma states that if
$\tilde{f}^{\prime}(\rho)\neq 0$, then we only need to consider the MDP for
the occupation time.

######  Lemma 5.1.

If $\tilde{f}^{\prime}(\rho)\neq 0$, then for any $\varepsilon>0$,

$\lim_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0\leq
t\leq
T}\Big{|}\Gamma^{n}_{t}(f)-\tilde{f}^{\prime}(\rho)\Gamma^{n}_{t}\Big{|}>\varepsilon\Big{)}=-\infty.$

The above result was proved in [7, Section 5] without the supremum over time
inside the parenthesis. However, since $a_{n}\gg\sqrt{n\log n}$, by dividing
the time interval $[0,T]$ into small time intervals as in Proposition 3.2, it
is not hard to extend the result to be uniform in time. For this reason, we
omit the proof.

In the rest of this section, we prove the sample path MDP for the occupation
time. In subsection 5.1, we show the sequence $\\{\Gamma^{n}_{t}:0\leq t\leq
T\\}_{n\geq 1}$ is exponentially tight. Subsections 5.2 and 5.3 are devoted to
the proof of the finite-dimensional MDP of the occupation time. We conclude
the proof of the sample path MDP in Subsection 5.4. Finally, in Subsection
5.5, we calculate a related variational problem.

###  5.1. Exponential tightness

The exponential tightness of the sequence $\\{\Gamma^{n}_{t}:0\leq t\leq
T\\}_{n\geq 1}$ follows directly from the following result and Lemma 3.4. Let
$\varphi\in{\mathcal{C}}^{\infty}_{c}(\mathbb{R})$ have support contained in
$(0,1)$ such that $\int_{\mathbb{R}}\varphi(u)du=1$. For any $\varepsilon>0$,
define $\varphi_{\varepsilon}(u)=\varepsilon^{-1}\varphi(u/\varepsilon)$.

######  Lemma 5.2.

Assume $\beta>1$ or $\rho=1/2$. Then, for any $\delta>0$,

$\limsup_{\varepsilon\rightarrow
0}\limsup_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0\leq
t\leq
T}\Big{|}\Gamma^{n}_{t}-\int_{0}^{t}\big{\langle}\mu^{n}_{s},\varphi_{\varepsilon}\big{\rangle}ds\Big{|}>\delta\Big{)}=-\infty.$
(5.1)

###### Proof.

For any configuration $\eta\in\mathcal{X}_{1}$, let
$(\eta\ast\varphi^{n}_{\varepsilon})_{x}$ be the discrete convolution of
$\eta$ with $\varphi_{\varepsilon}$,

$(\eta\ast\varphi^{n}_{\varepsilon})_{x}=\frac{1}{n}\sum_{y}\eta_{x+y}\varphi_{\varepsilon}(y/n),\quad
x\in\mathbb{Z}.$

Then, it suffices to show that for any $\delta>0$,

$\displaystyle\limsup_{\varepsilon\rightarrow\infty}\limsup_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0\leq
t\leq
T}\Big{|}\frac{n}{a_{n}}\int_{0}^{t}[\eta_{0}(s)-(\eta(s)\ast\varphi^{n}_{\varepsilon})_{0}]ds\Big{|}>\delta\Big{)}=-\infty,$
(5.2)
$\displaystyle\limsup_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\sup_{0\leq
t\leq
T}\Big{|}\int_{0}^{t}\Big{\\{}\frac{n}{a_{n}}\big{[}(\eta(s)\ast\varphi^{n}_{\varepsilon})_{0}-\rho\big{]}-\big{\langle}\mu^{n}_{s},\varphi_{\varepsilon}\big{\rangle}\Big{\\}}ds\Big{|}>\delta\Big{)}=-\infty.$
(5.3)

We first prove (5.3). Notice that

$\frac{n}{a_{n}}\big{[}(\eta(s)\ast\varphi^{n}_{\varepsilon})_{0}-\rho\big{]}-\big{\langle}\mu^{n}_{s},\varphi_{\varepsilon}\big{\rangle}=\frac{-v_{n}s}{na_{n}}\sum_{y}(\eta_{y}(s)-\rho)\varphi_{\varepsilon}^{\prime}(\theta_{n}(y))+O_{\varphi_{\varepsilon}}(1/a_{n})$

for some $\theta_{n}(y)$ between $y/n$ and $(y-v_{n}s)/n$. Thus, (5.3) holds
if $\beta>1$ by Lemma 3.1, and also holds if $\rho=1/2$ since $v_{n}=0$ in
this case.

The proof of (5.2) is similar to (3.9), so we only sketch it. By Chebyshev’s
inequality, we only need to show that, for any $A>0$,

$\limsup_{\varepsilon\rightarrow
0}\limsup_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{E}^{n}_{\rho}\Big{[}\exp\Big{\\{}\sup_{0\leq
t\leq
T}\Big{|}a_{n}A\int_{0}^{t}[\eta_{0}(s)-(\eta(s)\ast\varphi^{n}_{\varepsilon})_{0}]ds\Big{|}\Big{\\}}\Big{]}=0.$
(5.4)

Take $t_{i}=iT/n$ for $0\leq i\leq n$. As in the proof of Proposition 3.2,
since $a_{n}\gg\sqrt{n\log n}$, by Feynman-Kac formula, the last limit is
bounded by

$\limsup_{\varepsilon\rightarrow
0}\limsup_{n\rightarrow\infty}\,T\sup_{f:\nu_{\rho}{\rm-
density}}\Big{\\{}\frac{An}{a_{n}}\int[\eta_{0}-(\eta\ast\varphi^{n}_{\varepsilon})_{0}]fd\nu_{\rho}-\frac{n^{3}}{a_{n}^{2}}D(\sqrt{f})\Big{\\}}.$
(5.5)

Notice that

$\displaystyle\eta_{0}-(\eta\ast\varphi^{n}_{\varepsilon})_{0}=\frac{1}{n}\sum_{y=0}^{\varepsilon
n}\sum_{z=0}^{y-1}(\eta_{z}-\eta_{z+1})\varphi_{\varepsilon}(y/n)+O_{\varphi_{\varepsilon}}(1/n),$

where $O_{\varphi_{\varepsilon}}(1/n)\leq C/n$ for some constant
$C=C(\varphi_{\varepsilon})$. By making change of variables
$\eta\mapsto\eta^{z,z+1}$, the first term inside the supremum in (5.5) equals

$\frac{A}{2a_{n}}\sum_{y=0}^{\varepsilon
n}\sum_{z=0}^{y-1}\int(\eta_{z}-\eta_{z+1})(f(\eta)-f(\eta^{z,z+1}))\varphi_{\varepsilon}(y/n)d\nu_{\rho}+O_{\varphi_{\varepsilon}}(A/a_{n}).$

By Young’s inequality, for any $B>0$, the first term in the last line is
bounded by

$\displaystyle\frac{A}{4Ba_{n}}\sum_{y=0}^{\varepsilon
n}\sum_{z=0}^{y-1}\int(\sqrt{f(\eta)}-\sqrt{f(\eta^{z,z+1})})^{2}d\nu_{\rho}$
$\displaystyle\qquad+\frac{BA}{4a_{n}}\sum_{y=0}^{\varepsilon
n}\sum_{z=0}^{y-1}\int(\sqrt{f(\eta)}+\sqrt{f(\eta^{z,z+1})})^{2}\varphi_{\varepsilon}(y/n)^{2}d\nu_{\rho}$
$\displaystyle\leq\frac{A\varepsilon
n}{Ba_{n}}D(\sqrt{f})+\frac{C(\varphi)BAn^{2}}{a_{n}}.$

Taking $B=A\varepsilon a_{n}/n^{2}$, we bound the expression in (5.5) by
$C(\varphi)TA^{2}\varepsilon+TO_{\varphi_{\varepsilon}}(A/a_{n})$, thus
concluding the proof. ∎

###  5.2. Finite-dimensional upper bound

In this subsection, we prove the upper bound for the MDP of the occupation
time in the sense of finite-dimensional distributions. For any integer $k>0$,
denote
$\boldsymbol{\alpha}=(\alpha_{1},\ldots,\alpha_{k})^{T}\in\mathbb{R}^{k}$. For
$t_{1}<t_{2}<\ldots<t_{k}$, let
$A=A(t_{1},\ldots,t_{k})=(a(t_{i},t_{j}))_{1\leq i,j\leq k}$ be the $k\times
k$ matrix with $a(s,t)$ being the covariance function of the fractional
Brownian motion with parameter $3/4$,

$a(s,t)=\frac{1}{2}\big{(}t^{3/2}+s^{3/2}-|t-s|^{3/2}\big{)},\quad s,t\geq 0.$

######  Lemma 5.3.

Fix integer $k>0$. For any closed set $F_{i}\subset\mathbb{R}$, $1\leq i\leq
k$,

$\limsup_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\Gamma^{n}_{t_{i}}\in
F_{i},\;1\leq i\leq
k\Big{)}\leq-\inf\Big{\\{}\frac{1}{2\sigma^{2}}\boldsymbol{\alpha}^{T}A^{-1}\boldsymbol{\alpha}:\alpha_{i}\in
F_{i},1\leq i\leq k\Big{\\}},$

where $\sigma^{2}=4\sqrt{2}\chi(\rho)/(3\sqrt{\pi})$.

###### Proof.

For any $\delta>0$ and any set $F\subset\mathbb{R}$, let
$F^{\delta}:=\\{x:|y-x|\leq\delta\text{ for some }y\in F\\}$ be the
$\delta$-dilation of the set $F$. By Lemma 5.2, the limit on the left side in
the lemma is bounded by, for any $\delta>0$,

$\limsup_{\varepsilon\rightarrow
0}\limsup_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\int_{0}^{t_{i}}\big{\langle}\mu^{n}_{s},\varphi_{\varepsilon}\big{\rangle}ds\in
F_{i}^{\delta},\;1\leq i\leq k\Big{)}.$

Since the mapping
$\mu\mapsto\int_{0}^{t}\big{\langle}\mu_{s},\varphi_{\varepsilon}\big{\rangle}ds$
is continuous, by Theorem 2.1, the last line is bounded by

$-\sup_{\delta>0}\liminf_{\varepsilon\rightarrow
0}\inf\Big{\\{}{\mathcal{Q}}(\mu):\int_{0}^{t_{i}}\big{\langle}\mu_{s},\varphi_{\varepsilon}\big{\rangle}ds\in
F_{i}^{\delta},\;1\leq i\leq k\Big{\\}}.$ (5.6)

Define

${\mathcal{A}}=\Big{\\{}\mu:\mu\;\text{given by \eqref{mu formula} with
}\phi\in\mathcal{S}(\mathbb{R})\text{ and
}H(t,\cdot),\partial_{t}H(t,\cdot)\in{\mathcal{S}}(\mathbb{R}),\forall
t\in[0,T]\Big{\\}}.$ (5.7)

We claim that (5.6) is bounded from above by

$-\inf_{\alpha_{i}\in F_{i},1\leq i\leq
k}\inf\Big{\\{}{\mathcal{Q}}(\mu):\mu\in{\mathcal{A}},\int_{0}^{t_{i}}\mu_{s}(0)ds=\alpha_{i},\;1\leq
i\leq k\Big{\\}}.$ (5.8)

We then conclude the proof of the lemma by using Proposition 5.4 below.

For the claim, it is trivial if the infimum in (5.6) is infinity, so we only
need to consider the $\mu$’s such that ${\mathcal{Q}}(\mu)<+\infty$. Then, by
Lemma 4.2, $\mu$ is the unique weak solution to (4.2). For $\varepsilon>0$,
define $\mu_{s}^{\varepsilon}=\mu_{s}\ast\bar{\varphi}_{\varepsilon}$ as the
convolution of $\mu_{s}$ with
$\bar{\varphi}_{\varepsilon}(\cdot):=\varphi_{\varepsilon}(-\cdot)$. By
definition,
$\big{\langle}\mu_{s},\varphi_{\varepsilon}\big{\rangle}=\mu_{s}^{\varepsilon}(0)$.
Since ${\mathcal{Q}}$ is convex, we have
${\mathcal{Q}}(\mu^{\varepsilon})\leq{\mathcal{Q}}(\mu)$. Thus, for any
$\varepsilon>0$,

$\inf\Big{\\{}{\mathcal{Q}}(\mu):\int_{0}^{t_{i}}\big{\langle}\mu_{s},\varphi_{\varepsilon}\big{\rangle}ds\in
F_{i}^{\delta},\;1\leq i\leq
k\Big{\\}}\geq\inf\Big{\\{}{\mathcal{Q}}(\mu):\int_{0}^{t_{i}}\mu_{s}(0)ds\in
F_{i}^{\delta},\;1\leq i\leq k\Big{\\}},$

which permits us to bound (5.6) from above by

$-\inf\Big{\\{}{\mathcal{Q}}(\mu):\int_{0}^{t_{i}}\mu_{s}(0)ds\in
F_{i},\;1\leq i\leq k\Big{\\}}.$ (5.9)

For any $\delta>0$, for any $H\in{\mathcal{H}}^{1}$ and any $\phi\in
L^{2}(\mathbb{R})$, by Lemma 4.2 and (4.3), there exists
$H^{\delta}\in{\mathcal{C}}^{1,\infty}_{c}([0,T]\times\mathbb{R})$ and
$\phi\in{\mathcal{C}}^{\infty}_{c}(\mathbb{R})$ such that

${\mathcal{Q}}(\mu)\geq{\mathcal{Q}}(\mu^{\delta})-\delta,\quad\big{|}\int_{0}^{t_{i}}\mu_{s}(0)-\mu_{s}^{\delta}(0)ds\big{|}\leq\delta,\;1\leq
i\leq k.$

Since $\mu^{\delta}\in{\mathcal{A}}$, we bound (5.9) from above by

$-\inf\Big{\\{}{\mathcal{Q}}(\mu):\mu\in{\mathcal{A}},\int_{0}^{t_{i}}\mu_{s}(0)ds\in
F_{i}^{\delta},\;1\leq i\leq k\Big{\\}}+\delta.$

Finally, we prove the claim by letting $\delta\rightarrow 0$. ∎

######  Proposition 5.4.

For any
$\boldsymbol{\alpha}=(\alpha_{1},\ldots,\alpha_{k})^{T}\in\mathbb{R}^{k}$ and
for any $0<t_{1}<t_{2}<\ldots<t_{k}\leq T$,

$\inf\Big{\\{}{\mathcal{Q}}(\mu):\mu\in{\mathcal{A}},\int_{0}^{t_{i}}\mu_{s}(0)ds=\alpha_{i},1\leq
i\leq
k\Big{\\}}=\frac{1}{2\sigma^{2}}\boldsymbol{\alpha}^{T}A^{-1}\boldsymbol{\alpha}.$
(5.10)

The proof of the above result is postponed to Subsection 5.5.

###  5.3. Finite-dimensional lower bound

The next result concerns the lower bound for the MDP of the occupation time in
the sense of finite-dimensional distributions.

######  Lemma 5.5.

Fix integer $k>0$. For any closed set $O_{i}\subset\mathbb{R}$, $1\leq i\leq
k$,

$\liminf_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\Gamma^{n}_{t_{i}}\in
O_{i},\;1\leq i\leq
k\Big{)}\geq-\inf\Big{\\{}\frac{1}{2\sigma^{2}}\boldsymbol{\alpha}^{T}A^{-1}\boldsymbol{\alpha}:\alpha_{i}\in
O_{i},1\leq i\leq k\Big{\\}}.$

###### Proof.

For any $\alpha_{i}\in O_{i}$, $1\leq i\leq k$, and for any small $\delta>0$
such that $(\alpha_{i}-\delta,\alpha_{i}+\delta)\in O_{i}$, $1\leq i\leq k$,
by Lemma 5.2,

$\displaystyle\liminf_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\Gamma^{n}_{t_{i}}\in
O_{i},\;1\leq i\leq
k\Big{)}\geq\liminf_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\Gamma^{n}_{t_{i}}\in(\alpha_{i}-\delta,\alpha_{i}+\delta),\;1\leq
i\leq k\Big{)}$
$\displaystyle\geq\liminf_{\varepsilon\rightarrow\infty}\limsup_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\int_{0}^{t_{i}}\big{\langle}\mu^{n}_{s},\varphi_{\varepsilon}\big{\rangle}ds\in(\alpha_{i}-\delta/2,\alpha_{i}+\delta/2),\;1\leq
i\leq k\Big{)}.$

By Theorem 2.1, the last expression is bounded from below by

$-\limsup_{\varepsilon\rightarrow
0}\inf\big{\\{}{\mathcal{Q}}(\mu):\int_{0}^{t_{i}}\big{\langle}\mu_{s},\varphi_{\varepsilon}\big{\rangle}ds\in(\alpha_{i}-\delta/2,\alpha_{i}+\delta/2),\;1\leq
i\leq k\big{\\}}.$

Let $\hat{\mu}$ be the minimizer of the variational problem

$\inf\Big{\\{}{\mathcal{Q}}(\mu):\mu\in{\mathcal{A}},\int_{0}^{t_{i}}\mu_{s}(0)ds=\alpha_{i},\;1\leq
i\leq k\Big{\\}}.$

It is constructed in the proof of Proposition 5.4, and satisfies that

${\mathcal{Q}}(\hat{\mu})=\frac{1}{2\sigma^{2}}\boldsymbol{\alpha}^{T}A^{-1}\boldsymbol{\alpha}.$

Since for any $1\leq i\leq k$,

$\lim_{\varepsilon\rightarrow
0}\int_{0}^{t_{i}}\big{(}\hat{\mu}_{s}(0)-\big{\langle}\hat{\mu}_{s},\varphi_{\varepsilon}\big{\rangle}\big{)}ds=0,\quad\int_{0}^{t_{i}}\hat{\mu}_{s}(0)ds=\alpha_{i},$

there exists some $\varepsilon_{0}=\varepsilon_{0}(\delta,\hat{\mu})$ such
that for any $\varepsilon<\varepsilon_{0}$ and for any $1\leq i\leq k$,

$\int_{0}^{t_{i}}\big{\langle}\hat{\mu}_{s},\varphi_{\varepsilon}\big{\rangle}ds\in(\alpha_{i}-\delta/2,\alpha_{i}+\delta/2).$

Therefore, for $\varepsilon<\varepsilon_{0}$,

$\inf\big{\\{}{\mathcal{Q}}(\mu):\int_{0}^{t_{i}}\big{\langle}\mu_{s},\varphi_{\varepsilon}\big{\rangle}ds\in(\alpha_{i}-\delta/2,\alpha_{i}+\delta/2),\;1\leq
i\leq k\big{\\}}\leq{\mathcal{Q}}(\hat{\mu}).$

This implies that for any $\alpha_{i}\in O_{i}$, $1\leq i\leq k$,

$\liminf_{n\rightarrow\infty}\frac{n}{a_{n}^{2}}\log\mathbb{P}^{n}_{\rho}\Big{(}\Gamma^{n}_{t_{i}}\in
O_{i},\;1\leq i\leq
k\Big{)}\geq-\frac{1}{2\sigma^{2}}\boldsymbol{\alpha}^{T}A^{-1}\boldsymbol{\alpha}.$

We conclude the proof by optimizing over $\alpha_{i}\in O_{i}$, $1\leq i\leq
k$. ∎

###  5.4. Concluding the proof of Theorem 2.2

In the last three subsections, we have shown the MDP for the occupation time
in the sense of finite-dimensional distributions and exponential tightness of
the process $\\{\Gamma^{n}_{t}:0\leq t\leq T\\}$. By [6, Theorem 4.28], the
process $\\{\Gamma^{n}_{t}:0\leq t\leq T\\}$ satisfies the MDP with rate
function

$\frac{1}{\sigma^{2}}\sup\Big{\\{}\frac{1}{2}\boldsymbol{\alpha}^{T}A^{-1}\boldsymbol{\alpha}:k\geq
1,0\leq t_{1}<t_{2}<\ldots<t_{k}\leq T\Big{\\}}$

for any $\alpha\in{\mathcal{C}}([0,T])$, where
$\boldsymbol{\alpha}=(\alpha(t_{1}),\ldots,\alpha(t_{k}))^{T}$ and
$A=(a(t_{i},t_{j}))_{1\leq i,j\leq k}$. The last supremum is exactly the large
deviation rate function of the fractional Brownian motion
$\\{n^{-1/2}B^{3/4}_{t}:0\leq t\leq T\\}$, and thus equals $I_{\rm path}$.
This concludes the proof.

###  5.5. Proof of Proposition 5.4

The proof is divided into several steps.

Step $1$. We first show that, for any $\alpha\in\mathbb{R}$,

$\inf\Big{\\{}{\mathcal{Q}}(\mu):\mu\in{\mathcal{A}},\int_{0}^{T}\mu_{s}(0)ds=\alpha\Big{\\}}=\frac{\alpha^{2}}{2\sigma^{2}T^{3/2}}=\frac{3\sqrt{\pi}\alpha^{2}}{8\sqrt{2}T^{3/2}\chi(\rho)}.$
(5.11)

By (4.3), the constraint $\int_{0}^{T}\mu_{s}(0)ds=\alpha$ is equivalent to

$\displaystyle\frac{\alpha}{\chi(\rho)}$
$\displaystyle=\int_{0}^{T}\int_{\mathbb{R}}p_{s}(v)\phi(v)\,dv\,ds-\int_{0}^{T}\int_{0}^{s}\int_{\mathbb{R}}p^{\prime}_{s-r}(v)\partial_{v}H(r,v)\,dv\,dr\,ds$
$\displaystyle=\int_{\mathbb{R}}\Big{(}\int_{0}^{T}p_{s}(v)ds\Big{)}\phi(v)\,dv-\int_{0}^{T}\int_{\mathbb{R}}\Big{(}\int_{0}^{T-r}p^{\prime}_{s}(v)ds\Big{)}\partial_{v}H(r,v)\,dv\,dr.$

By Cauchy-Schwarz inequality,

$\displaystyle\frac{\alpha^{2}}{\chi(\rho)^{2}}\leq\big{\\{}\|\phi\|_{L^{2}(\mathbb{R})}^{2}+[H,H]\big{\\}}\Big{\\{}\int_{\mathbb{R}}\Big{(}\int_{0}^{T}p_{s}(v)ds\Big{)}^{2}dv+\int_{0}^{T}\int_{\mathbb{R}}\Big{(}\int_{0}^{T-r}p^{\prime}_{s}(v)ds\Big{)}^{2}dv\,dr\Big{\\}}.$

Moreover, the equality holds at the point $(\phi^{T,\alpha},H^{T,\alpha})$,
where

$\phi^{T,\alpha}(v)=c_{0}\int_{0}^{T}p_{s}(v)ds,\quad\partial_{v}H^{T,\alpha}(r,v)=-c_{0}\int_{0}^{T-r}p^{\prime}_{s}(v)ds,$

with the constant $c_{0}$ determined by

$\frac{\alpha}{\chi(\rho)}=c_{0}\,\Big{\\{}\int_{\mathbb{R}}\Big{(}\int_{0}^{T}p_{s}(v)ds\Big{)}^{2}dv+\int_{0}^{T}\int_{\mathbb{R}}\Big{(}\int_{0}^{T-r}p^{\prime}_{s}(v)ds\Big{)}^{2}dv\,dr\Big{\\}}.$
(5.12)

Notice that $\phi^{T,\alpha},H^{T,\alpha}\in{\mathcal{A}}$. At the end of this
step, we show that

$\displaystyle\int_{\mathbb{R}}\Big{(}\int_{0}^{T}p_{s}(v)ds\Big{)}^{2}dv$
$\displaystyle=\frac{4(2-\sqrt{2})T^{3/2}}{3\sqrt{\pi}},$ (5.13)
$\displaystyle\int_{0}^{T}\int_{\mathbb{R}}\Big{(}\int_{0}^{T-r}p^{\prime}_{s}(v)ds\Big{)}^{2}dv\,dr$
$\displaystyle=\frac{8(\sqrt{2}-1)T^{3/2}}{3\sqrt{\pi}}.$ (5.14)

Thus,

$c_{0}=c_{0}(T,\alpha)=\frac{3\sqrt{\pi}}{4\sqrt{2}T^{3/2}}\frac{\alpha}{\chi(\rho)}.$
(5.15)

Then, we conclude the proof of (5.11) by Lemma 4.2.

It remains to prove (5.13) and (5.14). For (5.13),

$\displaystyle\int_{\mathbb{R}}\Big{(}\int_{0}^{T}p_{s}(v)ds\Big{)}^{2}dv=\int_{\mathbb{R}}dv\int_{0}^{T}ds\int_{0}^{T}dt\frac{1}{2\pi}\frac{1}{\sqrt{ts}}e^{-\tfrac{v^{2}}{2}(\tfrac{1}{t}+\tfrac{1}{s})}.$

Since

$\int_{\mathbb{R}}dv\frac{1}{\sqrt{2\pi}}e^{-\tfrac{v^{2}}{2}(\tfrac{1}{t}+\tfrac{1}{s})}=\sqrt{\frac{ts}{t+s}},$

the last expression equals

$\displaystyle\int_{0}^{T}ds\int_{0}^{T}dt\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{t+s}}=\int_{0}^{T}ds\frac{\sqrt{2}}{\sqrt{\pi}}\big{[}(T+s)^{1/2}-s^{1/2}\big{]}=\frac{4(2-\sqrt{2})T^{3/2}}{3\sqrt{\pi}}.$

For (5.14), notice that

$\int_{0}^{T}\int_{\mathbb{R}}\Big{(}\int_{0}^{T-r}p^{\prime}_{s}(v)ds\Big{)}^{2}dv\,dr=\int_{0}^{T}dr\int_{\mathbb{R}}dv\int_{0}^{T-r}ds\int_{0}^{T-r}dt\frac{1}{2\pi}\frac{v^{2}}{(ts)^{3/2}}e^{-\tfrac{v^{2}}{2}(\tfrac{1}{t}+\tfrac{1}{s})}.$

Since

$\int_{\mathbb{R}}dv\frac{v^{2}}{\sqrt{2\pi}}e^{-\tfrac{v^{2}}{2}(\tfrac{1}{t}+\tfrac{1}{s})}=\Big{(}\frac{ts}{t+s}\Big{)}^{3/2},$

the last expression equals

$\displaystyle\int_{0}^{T}dr\int_{0}^{T-r}ds\int_{0}^{T-r}dt\frac{1}{\sqrt{2\pi}}(t+s)^{-3/2}=\int_{0}^{T}dr\int_{0}^{T-r}ds\frac{\sqrt{2}}{\sqrt{\pi}}[s^{-1/2}-(T-r+s)^{-1/2}]$
$\displaystyle=\int_{0}^{T}dr\frac{4(\sqrt{2}-1)}{\sqrt{\pi}}(T-r)^{1/2}=\frac{8(\sqrt{2}-1)T^{3/2}}{3\sqrt{\pi}}.$

Step $2$. We then show that, for any $\alpha\in\mathbb{R}$ and for any $0\leq
t\leq T$,

$\inf\Big{\\{}{\mathcal{Q}}(\mu):\mu\in{\mathcal{A}},\int_{0}^{t}\mu_{s}(0)ds=\alpha\Big{\\}}=\frac{\alpha^{2}}{2\sigma^{2}t^{3/2}}=\frac{3\sqrt{\pi}\alpha^{2}}{8\sqrt{2}t^{3/2}\chi(\rho)}.$
(5.16)

By replacing $T$ by $t$ in the last step, one can prove that the left hand
side is not smaller than the right hand side. For the opposite direction,
define $(\phi^{t,\alpha},H^{t,\alpha})$ as

$\phi^{t,\alpha}(v)=c_{0}(t,\alpha)\int_{0}^{t}p_{s}(v)ds,\quad\partial_{v}H^{t,\alpha}(r,v)=\begin{cases}-c_{0}(t,\alpha)\int_{0}^{t-r}p^{\prime}_{s}(v)ds,\quad&0\leq
r\leq t,\\\ 0,&r\geq t,\end{cases}$ (5.17)

where $c_{0}(t,\alpha)$ is given by (5.15) with $T$ replaced by $t$. By the
above calculations, the right hand side in (5.16) equals
${\mathcal{Q}}(\mu^{t,\alpha})$, where $\mu^{t,\alpha}$ is defined from (4.3)
corresponding to $(\phi^{t,\alpha},H^{t,\alpha})$. Thus, we only need to show

$\int_{0}^{t}\mu^{t,\alpha}(0,s)ds=\alpha.$

It is straightforward by using (4.3) and (5.12), thus concluding the proof.

Step 3. Last, we claim that the infimum in the proposition is obtained at the
point $(\hat{\phi},\hat{H})$, where

$\hat{\phi}=\sum_{i=1}^{k}\beta_{i}\phi^{t_{i},1},\quad\hat{H}=\sum_{i=1}^{k}\beta_{i}H^{t_{i},1}.$
(5.18)

Above, the coefficients $\\{\beta_{i}\\}$ are determined by the constraints

$\int_{0}^{t_{i}}\hat{\mu}(s,0)ds=\alpha_{i},\quad 1\leq i\leq k.$

Equivalently,

$\sum_{j=1}^{k}\beta_{j}\int_{0}^{t_{i}}\mu^{t_{j},1}(s,0)ds=\alpha_{i},\quad
1\leq i\leq k.$

Indeed, for the claim, we need to show that, for any $\mu\in{\mathcal{A}}$
satisfying

$\int_{0}^{t_{i}}\mu_{s}(0)ds=0,\quad\forall i=1,2,\ldots,k,$

we have

${\mathcal{Q}}(\hat{\mu}+\mu)=\frac{\chi(\rho)}{2}\Big{(}\|\phi+\hat{\phi}\|_{L^{2}(\mathbb{R})}^{2}+\|H+\hat{H}\|_{{\mathcal{H}}^{1}}^{2}\Big{)}\geq\frac{\chi(\rho)}{2}\Big{(}\|\hat{\phi}\|_{L^{2}(\mathbb{R})}^{2}+\|\hat{H}\|_{{\mathcal{H}}^{1}}^{2}\Big{)}={\mathcal{Q}}(\hat{\mu}).$

The above result follows directly if we can show that

$\big{\langle}\phi,\hat{\phi}\big{\rangle}+[H,\hat{H}]=0.$

By (5.17) and (4.3), the left hand side of the last line equals

$\displaystyle\sum_{i=1}^{k}\beta_{i}\Big{[}\int_{\mathbb{R}}dv\int_{0}^{t_{i}}dr\partial_{v}H(r,v)\big{(}-c_{0}(t_{i},1)\int_{0}^{t_{i}-r}p_{s}^{\prime}(v)ds\big{)}+\int_{\mathbb{R}}dv\phi(v)\big{(}c_{0}(t_{i},1)\int_{0}^{t_{i}}p_{s}(v)ds\big{)}\Big{]}$
$\displaystyle=\sum_{i=1}^{k}\frac{\beta_{i}c_{0}(t_{i},1)}{\chi(\rho)}\int_{0}^{t_{i}}\mu(s,0)ds=0,$

thus concluding the proof.

Next, we show that

$\int_{0}^{t_{i}}\mu^{t_{j},\alpha_{j}}(s,0)ds=\frac{\alpha_{j}a(t_{i},t_{j})}{t_{j}^{3/2}}.$
(5.19)

By (4.3) and (5.17), the term on the left hand side equals

$\displaystyle\chi(\rho)c_{0}(t_{j},\alpha_{j})\Big{\\{}\int_{0}^{t_{i}}ds\int_{0}^{t_{j}}dr\int_{\mathbb{R}}dvp_{s}(v)p_{r}(v)+\int_{0}^{t_{i}\wedge
t_{j}}dr\int_{0}^{t_{i}-r}ds\int_{0}^{t_{j}-r}d\tau\int_{\mathbb{R}}dvp^{\prime}_{\tau}(v)p^{\prime}_{s}(v)\Big{\\}}.$

By direct calculations, the first term inside the above brace equals

$\displaystyle\int_{0}^{t_{i}}ds\int_{0}^{t_{j}}dr\frac{1}{\sqrt{2\pi}}(r+s)^{-1/2}=\frac{2\sqrt{2}}{3\sqrt{\pi}}\big{[}(t_{i}+t_{j})^{3/2}-t_{i}^{3/2}-t_{j}^{3/2}\big{]},$

and the second one equals

$\int_{0}^{t_{i}\wedge
t_{j}}dr\int_{0}^{t_{i}-r}ds\int_{0}^{t_{j}-r}d\tau\frac{1}{\sqrt{2\pi}}(\tau+s)^{-3/2}\\\
=\frac{4\sqrt{2}}{3\sqrt{\pi}}\big{[}t_{i}^{3/2}+t_{j}^{3/2}+(t_{i}+t_{j}-2t_{i}\wedge
t_{j})^{3/2}/2-(t_{i}-t_{i}\wedge t_{j})^{3/2}-(t_{j}-t_{i}\wedge
t_{j})^{3/2}-(t_{i}+t_{j})^{3/2}/2\big{]}.$

Thus,

$\int_{0}^{t_{i}}\mu^{t_{j},\alpha_{j}}(s,0)ds=\frac{4\sqrt{2}}{3\sqrt{\pi}}\chi(\rho)c_{0}(t_{j},\alpha_{j})a(t_{i},t_{j})=\frac{\alpha_{j}}{t_{j}^{3/2}}a(t_{i},t_{j}).$

Let $D={\rm Diag}(d_{j})_{1\leq j\leq k}$ be the diagonal $k\times k$ matrix
with $d_{j}=t_{j}^{-3/2}$. Then, the constraints on $\boldsymbol{\beta}$
reduces to

$AD\boldsymbol{\beta}=\boldsymbol{\alpha},$

and thus

$\boldsymbol{\beta}=D^{-1}A^{-1}\boldsymbol{\alpha},$

where $\boldsymbol{\beta}=(\beta_{1},\ldots,\beta_{k})^{T}$.

With the above construction, by Lemma 4.2, the infimum in the proposition
equals

$\displaystyle\frac{\chi(\rho)}{2}$
$\displaystyle\sum_{i,j=1}^{k}\beta_{i}\beta_{j}\big{\\{}\big{\langle}\phi^{t_{i},1},\phi^{t_{j},1}\big{\rangle}+[H^{t_{i},1},H^{t_{j},1}]\big{\\}}=\frac{\chi(\rho)}{2}\sum_{i,j=1}^{k}\beta_{i}\beta_{j}c_{0}(t_{i},1)c_{0}(t_{j},1)$
$\displaystyle\times\Big{\\{}\int_{0}^{t_{i}}ds\int_{0}^{t_{j}}dr\int_{\mathbb{R}}dvp_{s}(v)p_{r}(v)+\int_{0}^{t_{i}\wedge
t_{j}}dr\int_{0}^{t_{i}-r}ds\int_{0}^{t_{j}-r}d\tau\int_{\mathbb{R}}dvp^{\prime}_{\tau}(v)p^{\prime}_{s}(v)\Big{\\}}.$

By the above calculations, the last expression equals

$\displaystyle\frac{1}{2\sigma^{2}}\sum_{i,j=1}^{k}\beta_{i}\beta_{j}a(t_{i},t_{j})t_{i}^{-3/2}t_{j}^{-3/2}=\frac{1}{2\sigma^{2}}\boldsymbol{\beta}^{T}DAD\boldsymbol{\beta}=\frac{1}{2\sigma^{2}}\boldsymbol{\alpha}^{T}A^{-1}\boldsymbol{\alpha},$

thus concluding the proof.

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