# SMALL BALL PROBABILITIES FOR THE FRACTIONAL STOCHASTIC HEAT EQUATION DRIVEN
BY A COLORED NOISE

Jiaming Chen  Dept. of Mathematics
University of Rochester
Rochester, NY 14627<EMAIL_ADDRESS>
###### Abstract.

We consider the fractional stochastic heat equation on the $d$-dimensional
torus $\mathbb{T}^{d}:=[-1,1]^{d}$, $d\geq 1$, with periodic boundary
conditions:

$\partial_{t}u(t,x)=-(-\Delta)^{\alpha/2}u(t,x)+\sigma(t,x,u)\dot{F}(t,x)\quad
x\in\mathbb{T}^{d},t\in\mathbb{R}^{+},$

where $\alpha\in(1,2]$ and $\dot{F}(t,x)$ is a white in time and colored in
space noise. We assume that $\sigma$ is Lipschitz in $u$ and uniformly
bounded. We provide small ball probabilities for the solution $u$ when
$u(0,x)\equiv 0$.

###### Key words and phrases:

fractional stochastic heat equation, colored noise, small ball probability.

###### 2020 Mathematics Subject Classification:

Primary, 60H15; Secondary, 60G60.

##  1\. Introduction

In this paper we consider small ball probabilities for solutions to the
fractional stochastic heat equation of the type:

(1.1) $\partial_{t}u(t,{\bf x})=-(-\Delta)^{\alpha/2}u(t,{\bf
x})+\sigma(t,{\bf x},u)\dot{F}(t,{\bf x})\quad{\bf
x}\in\mathbb{T}^{d},t\in\mathbb{R}^{+},$

with given initial profile $u(0,\cdot)=u_{0}:\mathbb{T}^{d}\to\mathbb{R}$
where $\mathbb{T}^{d}:=[-1,1]^{d}$ is a $d$-dimensional torus. The operator
$-(-\Delta)^{\alpha/2}$, where $1<\alpha\leq 2$, is the fractional power
Laplacian on $\mathbb{T}^{d}$. The centered Gaussian noise $\dot{F}$ is white
in time and colored in space, i.e.,

$\mathbb{E}\left(\dot{F}(t,{\bf x}),\dot{F}(s,{\bf
y})\right)=\delta_{0}(t-s)\Lambda({\bf x-y}),$

where $\delta_{0}$ is the Dirac delta generalized function and
$\Lambda:\mathbb{T}^{d}\to\mathbb{R}_{+}$ is a nonnegative generalized
function whose Fourier series is given by

(1.2) $\Lambda({\bf x})=\sum_{{\bf n}\in\mathbb{Z}^{d}}\lambda({\bf
n})\exp(\pi i{\bf n\cdot x})$

where ${\bf n}\cdot{\bf x}$ represents the dot product of two $d$-dimensional
vectors. We will need the following two assumptions on the function
$\sigma:\mathbb{R}_{+}\times\mathbb{T}^{d}\times\mathbb{R}\to\mathbb{R}$.

######  Hypothesis 1.1.

There exists a constant $\mathcal{D}>0$ such that for all $t\geq 0$, ${\bf
x}\in\mathbb{T}^{d}$, $u,v\in\mathbb{R}$,

(1.3) $|\sigma(t,{\bf x},u)-\sigma(t,{\bf x},v)|\leq\mathcal{D}|u-v|.$

######  Hypothesis 1.2.

There exist constants $\mathcal{C}_{1}$, $\mathcal{C}_{2}>0$ such that for all
$t\geq 0$, ${\bf x}\in\mathbb{T}^{d}$, $u\in\mathbb{R}$,

(1.4) $\mathcal{C}_{1}\leq\sigma(t,{\bf x},u)\leq\mathcal{C}_{2}.$

In fact, (1.1) is not well-posed since the solution $u$ is not differentiable
and $\dot{F}$ exists as a generalized function. However, under the assumptions
(1.3) and (1.4), we define the mild solution $u(t,{\bf x})$ to (1.1) in the
sense of Walsh [Wal86] satisfying

(1.5) $u(t,{\bf x})=\int_{\mathbb{T}^{d}}\bar{p}(t,{\bf x-y})u_{0}({\bf
y})d{\bf y}+\int_{[0,t]\times\mathbb{T}^{d}}\bar{p}(t-s,{\bf
x-y})\sigma(s,{\bf y},u(s,{\bf y}))F(dsd{\bf y}),$

where $\bar{p}:\mathbb{R}_{+}\times\mathbb{T}^{d}\to\mathbb{R}_{+}$ is the
fundamental solution of the fractional heat equation on $\mathbb{T}^{d}$

(1.6) $\begin{split}\partial_{t}&\bar{p}(t,{\bf
x})=-(-\Delta)^{\alpha/2}\bar{p}(t,{\bf x})\\\ &\bar{p}(0,{\bf
x})=\delta_{0}({\bf x}).\end{split}$

Following [Dal99], it is well known (see also [DKM+09]) that if $\lambda({\bf
n})$, the Fourier coefficients of $\Lambda({\bf x})$, satisfy

(1.7) $\sum_{{\bf n}\in\mathbb{Z}^{d}}\frac{\lambda({\bf n})}{1+|{\bf
n}|^{\alpha}}<\infty,$

where $|\cdot|$ is the Euclidean norm, then there exists a unique random field
solution $u(t,{\bf x})$ to equation (1.5). Examples of spatial correlation
satisfying (1.7) are:

  1. 1.

The Riesz kernel $\Lambda({\bf x})=|{\bf x}|^{-\beta}$, $0<\beta<d$. In this
case, there exist positive constants $c_{1}$, $c_{2}$ such that for all ${\bf
n}\in\mathbb{Z}^{d}$,

(1.8) $c_{1}|{\bf n}|^{-(d-\beta)}\leq\lambda({\bf n})\leq c_{2}|{\bf
n}|^{-(d-\beta)},$

and it is easy to check that condition (1.7) holds whenever $\beta<\alpha$.

  2. 2.

The space-time white noise case $\Lambda({\bf x})=\delta_{0}({\bf x})$. In
this case, $\lambda({\bf n})$ is a constant and (1.7) is only satisfied when
$\alpha>d$, that is, $d=1$ and $1<\alpha\leq 2$.

Small ball probability problems have a long history, and one can see [LS01]
for more surveys. In short, we are interested in the probability that a
stochastic process $X_{t}$ starting at 0 stays in a small ball for a long time
period, i.e.,

$P\left(\sup_{0\leq t\leq T}|X_{t}|<\varepsilon\right)$

where $\varepsilon>0$ is small. A recent paper [AJM21] has studied this
problem when $X_{t}$ is the solution of the stochastic heat equation with
$d=1$, $\alpha=2$ and $\Lambda=\delta_{0}$. The objective of this paper is to
generalize their results with the Riesz kernel.

##  2\. Main Result

######  Theorem 2.1.

Under the assumptions (1.3) and (1.4), if $u(t,{\bf x})$ is the solution to
(1.1) with $u_{0}({\bf x})\equiv 0$, then there are positive constants
$\textbf{C}_{0},\textbf{C}_{1},\textbf{C}_{2},\textbf{C}_{3},\mathcal{D}_{0}$
depending only on $\mathcal{C}_{1},\mathcal{C}_{2}$, $\alpha$, $\beta$ and
$d$, such that for all
$\mathcal{D}<\mathcal{D}_{0},\varepsilon_{0}>\varepsilon>0$, $T>1$, we have

  1. (a)

when $d=1$ and $\alpha\geq 2\beta$,

$\textbf{C}_{0}\exp\left(-\frac{\textbf{C}_{1}T}{\varepsilon^{\frac{2(2\alpha-\beta)}{\beta}}}\right)<P\left(\sup_{\begin{subarray}{c}0\leq
t\leq T\\\ {\bf x}\in\mathbb{T}^{d}\end{subarray}}|u(t,{\bf
x})|\leq\varepsilon\right)<\textbf{C}_{2}\exp\left(-\frac{\textbf{C}_{3}T}{\varepsilon^{\frac{2(\alpha+\beta)}{\alpha-\beta}}}\right),$

  2. (b)

or in other cases,

$0\leq P\left(\sup_{\begin{subarray}{c}0\leq t\leq T\\\ {\bf
x}\in\mathbb{T}^{d}\end{subarray}}|u(t,{\bf x})|\leq\varepsilon\right)<{\bf
C_{2}}\exp\left(-\frac{{\bf
C_{3}}T}{\varepsilon^{\frac{2\alpha}{\alpha-\beta}\left(\left(1+\frac{\beta}{\alpha
d}\right)\wedge\left(\frac{2\alpha-\beta}{\alpha}\right)\right)}}\right).$

Here we make a couple of remarks. These could be of independent interests.

######  Remark 2.1.

  1. (a)

The lack of lower bound for small ball probability in part $(b)$ is due to an
exponential growth number of grids in space.

  2. (b)

When $d\geq 2$ and $\Lambda({\bf x})=\delta_{0}({\bf x})$, the solution exists
as a distribution. Is there a way to estimate the small probability for some
norm of this solution?

  3. (c)

The small ball probability estimation has a close relation with the Chung’s
type Law of the Iterated Logarithm (see [LS01] for more details). Can we
follow the idea from [LX21] to get a similar result for non-Gaussian random
fields/strings?

Here is the organization of this paper. In Section 3 we state the key
proposition and how this proposition relates to the main result. In Section 4
we give some useful estimations. In Section 5 we prove the key proposition.

Throughout the entire paper, $C$ and $C^{\prime}$ denote positive constants
whose values may vary from line to line. The dependence of constants on
parameters will be denoted by mentioning the parameters in parenthesis.

##  3\. Key Proposition

We decompose $[-1,1]$ into intervals of length $\varepsilon^{2}$ on each
dimension and divide $[0,T]$ into intervals of length $c_{0}\varepsilon^{4}$
where $c_{0}$ satisfies

(3.1) $0<c_{0}<\min\left\\{1,\left(\frac{C_{6}}{36\mathcal{C}_{2}^{2}\ln
C_{5}}\right)^{\frac{\alpha}{\alpha-\beta}}\right\\}$

where $C_{5},C_{6}$ are constants specified in Lemma 4.5. Moreover, for
$\forall\varepsilon>0$ and $\mathcal{C}$ is specified in Lemma 5.2, we require

(3.2) $0<c_{0}<\mathcal{C}\varepsilon^{\frac{2\alpha d-4\beta}{\beta}}.$

######  Remark 3.1.

Unlike the white noise case in [AJM21], $c_{0}$ needs to be selected depending
on $\varepsilon$ in this paper. Indeed, $c_{0}$ does not appear in the bounds
for small ball probability.

Define $t_{i}=ic_{0}\varepsilon^{4},x_{j}=j\varepsilon^{2}$ and

$n_{1}:=\min\\{n\in\mathbb{Z}:n\varepsilon^{2}>1\\},$

where $i\in\mathbb{N}$ and $j\in\mathbb{Z}$. Consider a sequence of sets
$R_{i,j}\subset\mathbb{R}\times\mathbb{R}^{d}$ as

(3.3) $R_{i,j}=\left\\{(t_{i},x_{j_{1}},x_{j_{2}},...x_{j_{d}})|-n_{1}+1\leq
j_{k}\leq j,k=1,2,...,d\right\\}.$

By symmetry, $(x_{j_{1}},x_{j_{2}},...x_{j_{d}})$ lies in $[-1,1]^{d}$ when

(3.4) $-n_{1}+1\leq j_{k}\leq n_{1}-1\text{~{}for $k=1,2,...,d$}$

For $n\geq 0$, we define a sequence of events that we can use for the upper
bound in Theorem 2.1,

(3.5) $F_{n}=\left\\{|u(t,{\bf x})|\leq
t_{1}^{\frac{\alpha-\beta}{2\alpha}}\text{~{}for all $(t,{\bf x})\in
R_{n,n_{1}-1}$}\right\\}.$

In addition, let $E_{-1}=\Omega$ and for $n\geq 0$, we define a sequence of
events that we can use for the lower bound in Theorem 2.1,

(3.6) $E_{n}=\left\\{|u(t_{n+1},{\bf
x})|\leq\frac{1}{3}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\text{~{}and~{}}|u(t,{\bf
x})|\leq\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\text{~{}for
all~{}}t\in[t_{n},t_{n+1}],{\bf x}\in[-1,1]^{d}\right\\}.$

The following proposition along with the Markov property will lead to Theorem
2.1.

######  Proposition 3.1.

Consider the solution to (1.1) with $u_{0}({\bf x})\equiv 0$. Then, there
exist $\varepsilon_{1}>0$ and ${\bf
C_{4},C_{5},C_{6},C_{7}},\mathcal{D}_{0}>0$ depending only on
$\mathcal{C}_{1}$, $\mathcal{C}_{2}$, $\alpha$, $\beta$, and $d$ such that for
any $0<\varepsilon<\varepsilon_{1}$ and $\mathcal{D}<\mathcal{D}_{0}$,

  1. (a)

$P\left(F_{n}\bigg{|}\bigcap_{k=0}^{n-1}F_{k}\right)\leq{\bf
C_{4}}\exp\left(-\frac{{\bf
C_{5}}}{\varepsilon^{2}+\mathcal{D}^{2}t_{1}^{\frac{\alpha-\beta}{\alpha}}}\right),$

  2. (b)

and when $d=1$ and $\alpha\geq 2\beta$,

$P\left(E_{n}\bigg{|}\bigcap_{k=-1}^{n-1}E_{k}\right)\geq{\bf
C_{6}}\exp\left(-\frac{{\bf
C_{7}}}{\varepsilon^{\frac{4(\alpha-\beta)^{2}}{\alpha\beta}}}\right).$

Next we show how Theorem 2.1 follows from Proposition 3.1.

Proof of Theorem 2.1: The event $F_{n}$ deals with $u(t,{\bf x})$ at the time
$t_{n}$, so putting these events together indicates

$F:=\bigcap_{n=0}^{\left\lfloor\frac{T}{t_{1}}\right\rfloor}F_{n}\supset\left\\{|u(t,{\bf
x})|\leq t_{1}^{\frac{\alpha-\beta}{2\alpha}},t\in[0,T],{\bf
x}\in[-1,1]^{d}\right\\},$

and

$P(F)=P\left(\bigcap_{n=0}^{\left\lfloor\frac{T}{t_{1}}\right\rfloor}F_{n}\right)=P(F_{0})\prod_{n=1}^{\left\lfloor\frac{T}{t_{1}}\right\rfloor}P\left(F_{n}\bigg{|}\bigcap_{k=0}^{n-1}F_{k}\right).$

With $u_{0}({\bf x})\equiv 0$, Proposition 3.1 immediately yields

$\displaystyle P(F)$ $\displaystyle\leq\left[{\bf C_{4}}\exp\left(-\frac{{\bf
C_{5}}}{\varepsilon^{2}+\mathcal{D}^{2}t_{1}^{\frac{\alpha-\beta}{\alpha}}}\right)\right]^{\left\lfloor\frac{T}{t_{1}}\right\rfloor}\leq{\bf
C_{4}^{\prime}}\exp\left(-\frac{{\bf
C_{5}^{\prime}}T}{\varepsilon^{2}t_{1}+\mathcal{D}^{2}t_{1}^{\frac{2\alpha-\beta}{\alpha}}}\right)$
$\displaystyle\leq{\bf C_{2}}\exp\left(-\frac{{\bf
C_{3}}T}{t_{1}^{\left(1+\frac{\beta}{\alpha
d}\right)\wedge\left(\frac{2\alpha-\beta}{\alpha}\right)}}\right).$

The last inequality follows from the inequality of $c_{0}$ in (3.2) and
$\mathcal{D}<\mathcal{D}_{0}$. Therefore we have

$P\left(\left\\{|u(t,{\bf x})|\leq
t_{1}^{\frac{\alpha-\beta}{2\alpha}},t\in[0,T],{\bf
x}\in[-1,1]^{d}\right\\}\right)<{\bf C_{2}}\exp\left(-\frac{{\bf
C_{3}}T}{t_{1}^{\left(1+\frac{\beta}{\alpha
d}\right)\wedge\left(\frac{2\alpha-\beta}{\alpha}\right)}}\right),$

then replacing $t_{1}^{\frac{\alpha-\beta}{2\alpha}}$ with $\varepsilon$ and
adjusting $\varepsilon_{1}$ to $\varepsilon_{0}$ give the upper bound in
Theorem 2.1.

For the lower bound, the event $E_{n}$ deals with $u(t,{\bf x})$ in the time
interval $[t_{n},t_{n+1}]$, so putting these events together indicates

$E:=\bigcap_{n=-1}^{\left\lfloor\frac{T}{t_{1}}\right\rfloor-1}E_{n}\subset\left\\{|u(t,{\bf
x})|\leq\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}},t\in[0,T],{\bf
x}\in[-1,1]^{d}\right\\},$

and

$P(E)=P\left(\bigcap_{n=-1}^{\left\lfloor\frac{T}{t_{1}}\right\rfloor-1}E_{n}\right)=P(E_{-1})\prod_{n=0}^{\left\lfloor\frac{T}{t_{1}}\right\rfloor-1}P\left(E_{n}\bigg{|}\bigcap_{k=-1}^{n-1}E_{k}\right).$

With $u_{0}({\bf x})\equiv 0$, Proposition 3.1 immediately yields

$P(E)\geq\left[{\bf C_{6}}\exp\left(-\frac{{\bf
C_{7}}}{\varepsilon^{\frac{4(\alpha-\beta)^{2}}{\alpha\beta}}}\right)\right]^{\frac{T}{t_{1}}}\geq{\bf
C_{0}}\exp\left(-\frac{{\bf
C_{1}}T}{t_{1}\varepsilon^{\frac{4(\alpha-\beta)^{2}}{\alpha\beta}}}\right).$

Therefore, from the inequality of $c_{0}$ in (5.27), we have

$P\left(\left\\{|u(t,{\bf
x})|\leq\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}},t\in[0,T],{\bf
x}\in[-1,1]^{d}\right\\}\right)>{\bf C_{0}}\exp\left(-\frac{{\bf
C_{1}}T}{\varepsilon^{\frac{4(\alpha-\beta)(2\alpha-\beta)}{\alpha\beta}}}\right),$

then replacing $\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}$ with
$\varepsilon$ and adjusting $\varepsilon_{1}$ to $\varepsilon_{0}$ give the
lower bound in Theorem 2.1.

##  4\. Preliminary

In this section, we provide some preliminary results that are used to prove
the key proposition 3.1.

###  4.1. Heat Kernel Estimates

For ${\bf x}\in\mathbb{R}^{d}$, $p(t,{\bf x})$ is the smooth function
determined by its Fourier transform in ${\bf x}$

$\hat{p}(t,\nu):=\int_{\mathbb{R}^{d}}p(t,{\bf x})\exp(2\pi i\nu\cdot{\bf
x})d{\bf x}=\exp(-t(2\pi|\nu|)^{\alpha}),\quad\nu\in\mathbb{R}^{d}.$

For ${\bf x}\in\mathbb{T}^{d}$, from the standard Fourier decomposition we
have

(4.1) $\bar{p}(t,{\bf x})=2^{-d}\sum_{{\bf
n}\in\mathbb{Z}^{d}}\exp\left(-\pi^{\alpha}|{\bf n}|^{\alpha}t\right)\exp(\pi
i{\bf n}\cdot{\bf x}).$

The following lemma gives an estimation on heat kernel $\bar{p}(t,{\bf x})$,
which is similar to Lemma 2.1 and Lemma 2.2 in [Li17].

######  Lemma 4.1.

For all $t\geq s>0$ and ${\bf x,y}\in\mathbb{T}^{d}$, there exist constants
$C,C^{\prime}>0$ depending only on $\alpha,d$ such that

(4.2) $\int_{\mathbb{T}^{d}}|\bar{p}(t,{\bf y-x})-\bar{p}(t,{\bf y})|d{\bf
y}\leq C\left(\frac{|{\bf x}|}{t^{1/\alpha}}\wedge 1\right),$ (4.3)
$\int_{\mathbb{T}^{d}}|\bar{p}(t,{\bf x})-\bar{p}(s,{\bf x})|d{\bf x}\leq
C^{\prime}\left(\log\left(\frac{t}{s}\right)\wedge 1\right).$

###### Proof.

We begin with inequality (4.2),

(4.4) $\begin{split}\int_{\mathbb{T}^{d}}|\bar{p}(t,{{\bf
y-x}})-\bar{p}(t,{\bf y})|d{\bf y}&=\int_{\mathbb{T}^{d}}\left|\sum_{{\bf
n}\in\mathbb{Z}^{d}}\left[p(t,{\bf y-x}+2{\bf n})-p(t,{\bf y}+2{\bf
n})\right]\right|d{\bf y}\\\ &\leq\int_{\mathbb{T}^{d}}\sum_{{\bf
n}\in\mathbb{Z}^{d}}\left|p(t,{\bf y-x}+2{\bf n})-p(t,{\bf y}+2{\bf
n})\right|d{\bf y}\\\ &=\int_{\mathbb{R}^{d}}|p(t,{\bf y-x})-p(t,{\bf
y})|d{\bf y}\\\ &\leq\int_{\mathbb{R}^{d}}|{\bf
x}|\cdot\sup_{c_{0}\in[0,1]}\left|\nabla_{{\bf z}}p(t,{\bf y}-c_{0}{\bf
x})\right|d{\bf y}.\end{split}$

By Lemma 5 in [BJ07] and $(2.3)$ of [JS16], we have

(4.5) $\left|\nabla_{{\bf z}}p(t,{\bf z})\right|\leq C(d,\alpha)|{\bf
z}|\left(\frac{t}{|{\bf z}|^{d+2+\alpha}}\wedge t^{-(d+2)/\alpha}\right)\leq
C(d,\alpha)\frac{t|{\bf z}|}{(t^{1/\alpha}+|{\bf z}|)^{d+2+\alpha}}.$

We put (4.5) into (4.4) to get

$\begin{split}\int_{\mathbb{T}^{d}}|\bar{p}(t,{{\bf y-x}})-\bar{p}(t,{\bf
y})|d{\bf y}&\leq C(d,\alpha)|{\bf x}|\int_{\mathbb{R}^{d}}\frac{t|{\bf
y}|}{(t^{1/\alpha}+|{\bf y}|)^{d+2+\alpha}}d{\bf y}\\\ &\leq C(d,\alpha)|{\bf
x}|\int_{0}^{\infty}\frac{tx}{(t^{1/\alpha}+x)^{d+2+\alpha}}x^{d-1}dx\\\
&=\frac{C(d,\alpha)|{\bf
x}|}{t^{1/\alpha}}\int_{0}^{\infty}\frac{w^{d}}{(1+w)^{d+2+\alpha}}dw\\\
&\leq\frac{C(d,\alpha)|{\bf x}|}{t^{1/\alpha}}.\end{split}$

Clearly, $\int_{\mathbb{T}^{d}}|\bar{p}(t,{{\bf y-x}})-\bar{p}(t,{\bf
y})|d{\bf y}\leq 2$, so that (4.2) follows. For inequality (4.3), we have

(4.6) $\begin{split}\int_{\mathbb{T}^{d}}|\bar{p}(t,{{\bf x}})-\bar{p}(s,{\bf
x})|d{\bf x}&=\int_{\mathbb{T}^{d}}\left|\sum_{{\bf
n}\in\mathbb{Z}^{d}}\left[p(t,{\bf x}+2{\bf n})-p(s,{\bf x}+2{\bf
n})\right]\right|d{\bf x}\\\ &\leq\int_{\mathbb{T}^{d}}\sum_{{\bf
n}\in\mathbb{Z}^{d}}\left|p(t,{\bf x}+2{\bf n})-p(s,{\bf x}+2{\bf
n})\right|d{\bf x}\\\ &=\int_{\mathbb{R}^{d}}|p(t,{\bf x})-p(s,{\bf x})|d{\bf
x}\\\ &\leq\int_{\mathbb{R}^{d}}\int_{s}^{t}|\partial_{r}p(r,{\bf x})|drd{\bf
x}.\end{split}$

Proposition 2.1 in [VdPQR17] shows

(4.7) $|(-\Delta)^{\alpha/2}p(r,{\bf
x})|\leq\frac{C(d,\alpha)}{(r^{2/\alpha}+|{\bf
x}|^{2})^{\frac{d+\alpha}{2}}}.$

Applying (1.6), (4.7) and Fubini’s theorem to (4.6) yields that

$\displaystyle\int_{\mathbb{T}^{d}}|\bar{p}(t,{{\bf x}})-\bar{p}(s,{\bf
x})|d{\bf x}$ $\displaystyle\leq
C(d,\alpha)\int_{\mathbb{R}^{d}}\int_{s}^{t}\frac{1}{(r^{2/\alpha}+|{\bf
x}|^{2})^{\frac{d+\alpha}{2}}}drd{\bf x}$
$\displaystyle=C(d,\alpha)\int_{s}^{t}\int_{0}^{\infty}\frac{x^{d-1}}{(r^{2/\alpha}+x^{2})^{\frac{d+\alpha}{2}}}dxdr$
$\displaystyle=C(d,\alpha)\int_{s}^{t}\frac{dr}{r}\int_{0}^{\infty}\frac{w^{d-1}}{(1+w^{2})^{\frac{d+\alpha}{2}}}dw$
$\displaystyle\leq C(d,\alpha)\left(\log(t)-\log(s)\right).$

Similarly, $\int_{\mathbb{T}^{d}}|\bar{p}(t,{{\bf x}})-\bar{p}(s,{\bf
x})|d{\bf x}\leq 2$, so that (4.3) follows. ∎

###  4.2. Noise Term Estimates

We denote the second integral of (1.5), i.e. noise term, by

(4.8) $N(t,{\bf x}):=\int_{[0,t]\times\mathbb{T}^{d}}\bar{p}(t-s,{\bf
x-y})\sigma(s,{\bf y},u(s,{\bf y}))F(dsd{\bf y}),$

We will now estimate the regularity of $N(t,{\bf x})$ in the following two
lemmas.

######  Lemma 4.2.

There exists a constant $C>0$ depending only on $\alpha,\beta,d$ and
$\mathcal{C}_{2}$ in (1.4) such that for any
$\xi\in\left(0,\frac{1}{\alpha}\wedge\frac{\alpha-\beta}{\alpha}\right)$,
$t\in[0,1]$ and ${\bf x,y}\in\mathbb{T}^{d}$, we have

$\mathbb{E}\left[(N(t,{\bf x})-N(t,{\bf y}))^{2}\right]\leq
C\mathcal{C}_{2}^{2}|{\bf x-y}|^{\alpha\xi}.$

###### Proof.

To simplify our computation, fix $t,s,{\bf x},{\bf y}$ and we denote

$K({\bf z}):=\bar{p}(t-s,{\bf x-z})-\bar{p}(t-s,{\bf y-z}).$

Using Fubini’s theorem, (1.4) and the triangle inequality, we have

(4.9) $\begin{split}&\mathbb{E}\left[(N(t,{\bf x})-N(t,{\bf y}))^{2}\right]\\\
&=\int_{0}^{t}\int_{\mathbb{T}^{d}}\int_{\mathbb{T}^{d}}K({\bf z})K({\bf
w})\Lambda({\bf w-z})\mathbb{E}[\sigma(s,{\bf z},u(s,{\bf z}))\sigma(s,{\bf
w},u(s,{\bf w}))]d{\bf w}d{\bf z}ds\\\ &\leq\sup_{r,{\bf
u}}\mathbb{E}\left[\sigma(r,{\bf u},u(r,{\bf
u}))^{2}\right]\int_{0}^{t}\int_{\mathbb{T}^{d}}\int_{\mathbb{T}^{d}}|K({\bf
z})||K({\bf w})|\Lambda({\bf w-z})d{\bf w}d{\bf z}ds\\\
&\leq\mathcal{C}_{2}^{2}\int_{0}^{t}\int_{\mathbb{T}^{d}}\int_{\mathbb{T}^{d}}|K({\bf
z})|[\bar{p}(t-s,{\bf x-w})+\bar{p}(t-s,{\bf y-w})]\Lambda({\bf w-z})d{\bf
w}d{\bf z}ds.\end{split}$

Then we use the standard Fourier decomposition (4.1) to estimate the spatial
convolution,

(4.10) $\begin{split}\int_{\mathbb{T}^{d}}\bar{p}(t-s,{\bf x-w})\Lambda({\bf
w-z})d{\bf w}&=C(d)\sum_{{\bf n}\in\mathbb{Z}^{d}}\lambda({\bf
n})\exp(-\pi^{\alpha}|{\bf n}|^{\alpha}(t-s))\exp(\pi i{\bf n}\cdot({\bf
x-z}))\\\ &\leq C(d)\sum_{{\bf n}\in\mathbb{Z}^{d}}\lambda({\bf
n})\exp(-\pi^{\alpha}|{\bf n}|^{\alpha}(t-s))\\\ &\leq C(d)\sum_{{\bf
n}\in\mathbb{Z}^{d}}|{\bf n}|^{-d+\beta}\exp(-\pi^{\alpha}|{\bf
n}|^{\alpha}(t-s))\\\ &\leq
C(d)\int_{0}^{\infty}x^{-d+\beta}\exp(-\pi^{\alpha}x^{\alpha}(t-s))x^{d-1}dx\\\
&=C(d)(t-s)^{-\beta/\alpha}\int_{0}^{\infty}x^{\frac{\beta-\alpha}{\alpha}}\exp(-x)dx\\\
&=C(\alpha,\beta,d)(t-s)^{-\beta/\alpha}.\end{split}$

We can get a similar result for $\int_{\mathbb{T}^{d}}\bar{p}(t-s,{\bf
y-w})\Lambda({\bf w-z})d{\bf w}$. Applying (4.10), Lemma 4.1 to (4.9) and
since $1\wedge x<x^{\alpha\xi}$ for all $x>0,\xi\in(0,1/\alpha)$, we get

$\begin{split}\mathbb{E}\left[(N(t,{\bf x})-N(t,{\bf y}))^{2}\right]&\leq
C(\alpha,\beta,d)\mathcal{C}_{2}^{2}\int_{0}^{t}\int_{\mathbb{T}^{d}}|K({\bf
z})|(t-s)^{-\beta/\alpha}d{\bf z}ds\\\ &\leq
C(\alpha,\beta,d)\mathcal{C}_{2}^{2}\int_{0}^{t}(t-s)^{-\beta/\alpha}\left(\frac{|{\bf
x-y}|}{(t-s)^{1/\alpha}}\wedge 1\right)ds\\\ &\leq
C(\alpha,\beta,d)\mathcal{C}_{2}^{2}|{\bf
x-y}|^{\alpha\xi}\int_{0}^{t}(t-s)^{-\xi-\beta/\alpha}ds\\\ &\leq
C(\alpha,\beta,d)\mathcal{C}_{2}^{2}|{\bf x-y}|^{\alpha\xi}.\end{split}$

Note that the integral $\int_{0}^{t}(t-s)^{-\xi-\beta/\alpha}ds$ converges
provided
$\xi\in\left(0,\frac{1}{\alpha}\wedge\frac{\alpha-\beta}{\alpha}\right)$. ∎

######  Lemma 4.3.

There exists a constant $C>0$ depending only on $\alpha,\beta,d$ and
$\mathcal{C}_{2}$ in (1.4) such that for any
$\zeta\in\left(0,\frac{\alpha-\beta}{\alpha}\right)$, $1\geq t\geq s>0$ and
${\bf x,y}\in\mathbb{T}^{d}$, we have

$\mathbb{E}\left[(N(t,{\bf x})-N(s,{\bf x}))^{2}\right]\leq
C\mathcal{C}_{2}^{2}|t-s|^{\zeta}.$

###### Proof.

Using Fubini’s theorem, (1.4) and the triangle inequality, we have

(4.11) $\begin{split}\mathbb{E}&[(N(t,{\bf x})-N(s,{\bf x}))^{2}]\\\
&=\mathbb{E}\left[\left(\int_{0}^{s}\int_{\mathbb{T}^{d}}\left[\bar{p}(t-r,{\bf
x-z})-\bar{p}(s-r,{\bf x-z})\right]\sigma(r,{\bf z},u(r,{\bf z}))F(d{\bf
z}dr)\right.\right.\\\ &\hskip
142.26378pt\left.\left.+\int_{s}^{t}\int_{\mathbb{T}^{d}}\bar{p}(t-r,{\bf
x-z})\sigma(r,{\bf z},u(r,{\bf z}))F(d{\bf z}dr)\right)^{2}\right]\\\
&\leq\sup_{r,{\bf u}}\mathbb{E}\left[\sigma(r,{\bf u},u(r,{\bf
u}))^{2}\right]\left(\int_{0}^{s}\int_{\mathbb{T}^{d}}\int_{\mathbb{T}^{d}}|\bar{p}(t-r,{\bf
x-z})-\bar{p}(s-r,{\bf x-z})|\cdot\right.\\\ &\hskip
113.81102pt[\bar{p}(t-r,{\bf x-w})+\bar{p}(s-r,{\bf x-w})]\Lambda({\bf
w-z})d{\bf w}d{\bf z}dr\\\ &\hskip
56.9055pt\left.+\int_{s}^{t}\int_{\mathbb{T}^{d}}\int_{\mathbb{T}^{d}}\bar{p}(t-r,{\bf
x-z})\bar{p}(t-r,{\bf x-w})\Lambda({\bf w-z})d{\bf w}d{\bf z}dr\right)\\\
&=:\mathcal{C}_{2}^{2}\left(I_{1}+I_{2}\right).\end{split}$

Applying (4.10) and Lemma 4.1 to $I_{1}$ and since
$1\wedge\log(1+x)<x^{\zeta}$ for all $x>0,\zeta\in(0,1)$, we get

(4.12) $\begin{split}I_{1}&\leq
C(\alpha,\beta,d)\int_{0}^{s}\left(\log\left(\frac{t-r}{s-r}\right)\wedge
1\right)\cdot[(t-r)^{-\beta/\alpha}+(s-r)^{-\beta/\alpha}]dr\\\ &\leq
C(\alpha,\beta,d)\int_{0}^{s}\left(\log\left(\frac{t-s+x}{x}\right)\wedge
1\right)x^{-\beta/\alpha}dx\\\ &\leq
C(\alpha,\beta,d)(t-s)^{\zeta}\int_{0}^{s}x^{-\beta/\alpha-\zeta}dx\\\ &\leq
C(\alpha,\beta,d)(t-s)^{\zeta}.\end{split}$

Note that the integral $\int_{0}^{s}x^{-\beta/\alpha-\zeta}ds$ converges
provided $\zeta\in\left(0,\frac{\alpha-\beta}{\alpha}\right)$. In order to
estimate $I_{2}$, we use the standard Fourier decomposition (4.1) to bound the
spatial convolution,

$\begin{split}\int_{\mathbb{T}^{d}}\int_{\mathbb{T}^{d}}\bar{p}(t-r,{\bf
x-z})\bar{p}(t-r,{\bf x-w})&\Lambda({\bf w-z})d{\bf w}=C(d)\sum_{{\bf
n}\in\mathbb{Z}^{d}}\lambda({\bf n})\exp(-2\pi^{\alpha}|{\bf
n}|^{\alpha}(t-r))\\\ &\leq C(d)\sum_{{\bf n}\in\mathbb{Z}^{d}}|{\bf
n}|^{-d+\beta}\exp(-2\pi^{\alpha}|{\bf n}|^{\alpha}(t-r))\\\ &\leq
C(d)\int_{0}^{\infty}x^{-d+\beta}\exp(-2\pi^{\alpha}x^{\alpha}(t-r))x^{d-1}dx\\\
&=C(d)(t-r)^{-\beta/\alpha}\int_{0}^{\infty}x^{\frac{\beta-\alpha}{\alpha}}\exp(-x)dx\\\
&=C(\alpha,\beta,d)(t-r)^{-\beta/\alpha}.\end{split}$

Then for $I_{2}$ in (4.11), we have

(4.13) $I_{2}\leq
C(\alpha,\beta,d)\int_{s}^{t}(t-r)^{-\beta/\alpha}dr=C(\alpha,\beta,d)(t-s)^{\frac{\alpha-\beta}{\alpha}}.$

By (4.11), (4.12) and (4.13), we conclude

$\mathbb{E}\left[(N(t,{\bf x})-N(s,{\bf x}))^{2}\right]\leq
C\mathcal{C}_{2}^{2}(t-s)^{\zeta}.$

∎

######  Lemma 4.4.

There exist constants $C_{1},C_{2},C_{3},C_{4}>0$ depending only on
$\alpha,\beta,d$ and $\mathcal{C}_{2}$ in (1.4) such that for all $0\leq
s<t\leq 1$, ${\bf x,y}\in\mathbb{T}^{d}$,
$\xi\in\left(0,\frac{1}{\alpha}\wedge\frac{\alpha-\beta}{\alpha}\right),\zeta\in\left(0,\frac{\alpha-\beta}{\alpha}\right)$,
and $\kappa>0$,

(4.14) $P(|N(t,{\bf x})-N(t,{\bf y})|>\kappa)\leq
C_{1}\exp\left(-\frac{C_{2}\kappa^{2}}{\mathcal{C}_{2}^{2}|{\bf
x-y}|^{\alpha\xi}}\right),$ (4.15) $P(|N(t,{\bf x})-N(s,{\bf x})|>\kappa)\leq
C_{3}\exp\left(-\frac{C_{4}\kappa^{2}}{\mathcal{C}_{2}^{2}|t-s|^{\zeta}}\right).$

###### Proof.

For a fixed $t$, define

$N_{t}(s,{\bf x}):=\int_{[0,s]\times\mathbb{T}^{d}}\bar{p}(t-r,{\bf
x-y})\sigma(r,{\bf y},u(r,{\bf y}))F(drd{\bf y}).$

Note that $N_{t}(t,{\bf x})=N(t,{\bf x})$ and $N_{t}(s,{\bf x})$ is a
continuous $\mathcal{F}_{s}^{F}$ adapted martingale in $s\leq t$ since the
integrand does not depend on $s$. For fixed $t,{\bf x}$ and ${\bf y}$, let

$M_{s}:=N_{t}(s,{\bf x})-N_{t}(s,{\bf
y})=\int_{[0,s]\times\mathbb{T}^{d}}[\bar{p}(t-r,{\bf x-z})-\bar{p}(t-r,{\bf
y-z})]\sigma(r,{\bf z},u(r,{\bf z}))F(drd{\bf z}),$

and it is easy to check that $M_{t}=N(t,{\bf x})-N(t,{\bf y})$. As $M_{s}$ is
a continuous $\mathcal{F}_{s}^{F}$ adapted martingale with $M_{0}=0$, it is a
time changed Brownian motion. In particular, we have

$M_{t}=B_{\langle M\rangle_{t}},$

and Lemma 4.2 gives a uniform bound on the time change as

$\langle M\rangle_{t}\leq C\mathcal{C}_{2}^{2}|{\bf x-y}|^{\alpha\xi}.$

Therefore, by the reflection principle for the Brownian motion $B_{\langle
M\rangle_{t}}$,

$\displaystyle P(N(t,{\bf x})-N(t,{\bf y})>\kappa)$
$\displaystyle=P(M_{t}>\kappa)=P(B_{\langle M\rangle_{t}}>\kappa)$
$\displaystyle\leq P\left(\sup_{s\leq C\mathcal{C}_{2}^{2}|{\bf
x-y}|^{\alpha\xi}}B_{s}>\kappa\right)=2P\left(B_{C\mathcal{C}_{2}^{2}|{\bf
x-y}|^{\alpha\xi}}>\kappa\right)$ $\displaystyle\leq
C_{1}\exp\left(-\frac{C_{2}\kappa^{2}}{\mathcal{C}_{2}^{2}|{\bf
x-y}|^{\alpha\xi}}\right).$

Switching ${\bf x}$ and ${\bf y}$ gives

$\displaystyle P(-N(t,{\bf x})+N(t,{\bf y})>\kappa)$
$\displaystyle=P(M_{t}<-\kappa)\leq 2P\left(B_{C\mathcal{C}_{2}^{2}|{\bf
x-y}|^{\alpha\xi}}<-\kappa\right)$ $\displaystyle\leq
C_{1}\exp\left(-\frac{C_{2}\kappa^{2}}{\mathcal{C}_{2}^{2}|{\bf
x-y}|^{\alpha\xi}}\right).$

Consequently, for
$\forall\xi\in\left(0,\frac{1}{\alpha}\wedge\frac{\alpha-\beta}{\alpha}\right)$,

$P(|N(t,{\bf x})-N(t,{\bf y})|>\kappa)\leq
C_{1}\exp\left(-\frac{C_{2}\kappa^{2}}{\mathcal{C}_{2}^{2}|{\bf
x-y}|^{\alpha\xi}}\right),$

which completes the proof of (4.14). For a fixed ${\bf x}$, we define

$U_{q_{1}}=\int_{[0,{q_{1}}]\times\mathbb{T}^{d}}[\bar{p}(t-r,{\bf
x-y})-\bar{p}(s-r,{\bf x-y})]\sigma(r,{\bf y},u(r,{\bf y}))F(drd{\bf y})$

where $0\leq q_{1}\leq s$. Note $U_{q_{1}}$ is a continuous
$\mathcal{F}_{q_{1}}^{F}$ adapted martingale with $U_{0}=0$. Also define

$V_{q_{2}}=\int_{[s,s+{q_{2}}]\times\mathbb{T}^{d}}\bar{p}(t-r,{\bf
x-y})\sigma(r,{\bf y},u(r,{\bf y}))F(drd{\bf y})$

where $0\leq q_{2}\leq t-s$. Note $V_{q_{2}}$ is a continuous
$\mathcal{F}_{q_{2}}^{F}$ adapted martingale with $V_{0}=0$. Thus, both
$U_{q_{1}}$ and $V_{q_{2}}$ are time changed Brownian motions, i.e.,

$U_{t}=B_{\langle U\rangle_{t}}\text{~{}and~{}}V_{t-s}=B^{\prime}_{\langle
V\rangle_{t-s}}$

where $B$, $B^{\prime}$ are two different Brownian motions. Note that
$N(t,{\bf x})-N(s,{\bf x})=U_{t}+V_{t-s}$, then

$P(N(t,{\bf x})-N(s,{\bf x})>\kappa)\leq
P(U_{t}>\kappa/2)+P(V_{t-s}>\kappa/2).$

Lemma 4.3 provides a uniform bound on the time changes as

$\langle U\rangle_{t}\leq
C\mathcal{C}_{2}^{2}(t-s)^{\zeta}\text{~{}and~{}}\langle V\rangle_{t-s}\leq
C\mathcal{C}_{2}^{2}(t-s)^{\zeta}.$

By the reflection principle for the Brownian motions $B_{\langle
U\rangle_{t}}$ and $B^{\prime}_{\langle V\rangle_{t-s}}$,

$\displaystyle P(N(t,{\bf x})-N(s,{\bf x})>\kappa)$ $\displaystyle\leq
P\left(B_{\langle U\rangle_{t}}>\kappa/2\right)+P\left(B^{\prime}_{\langle
V\rangle_{t-s}}>\kappa/2\right)$ $\displaystyle\leq 2P\left(\sup_{r\leq
C\mathcal{C}_{2}^{2}|t-s|^{\zeta}}B_{r}>\frac{\kappa}{2}\right)=4P\left(B_{C\mathcal{C}_{2}^{2}|t-s|^{\zeta}}>\frac{\kappa}{2}\right)$
$\displaystyle\leq
C_{3}\exp\left(-\frac{C_{4}\kappa^{2}}{\mathcal{C}_{2}^{2}|t-s|^{\zeta}}\right).$

In addition,

$\displaystyle P(-N(t,{\bf x})+N(s,{\bf x})>\kappa)$ $\displaystyle\leq
P(U_{t}<-\kappa/2)+P(V_{t-s}<-\kappa/2)$ $\displaystyle\leq
4P\left(B_{C\mathcal{C}_{2}^{2}|t-s|^{\zeta}}<-\frac{\kappa}{2}\right)$
$\displaystyle\leq
C_{3}\exp\left(-\frac{C_{4}\kappa^{2}}{\mathcal{C}_{2}^{2}|t-s|^{\zeta}}\right).$

Consequently, for $\forall\zeta\in\left(0,\frac{\alpha-\beta}{\alpha}\right)$,

$P(|N(t,{\bf x})-N(s,{\bf x})|>\kappa)\leq
C_{3}\exp\left(-\frac{C_{4}\kappa^{2}}{\mathcal{C}_{2}^{2}|t-s|^{\zeta}}\right),$

which completes the proof of (4.15). ∎

######  Definition 4.1.

Given a grid

$\mathbb{G}_{n}=\left\\{\left(\frac{j}{2^{2n}},\frac{k_{1}}{2^{n}},...,\frac{k_{d}}{2^{n}}\right):0\leq
j\leq 2^{2n},0\leq k_{1},...,k_{d}\leq
2^{n},j,k_{1},...,k_{d}\in\mathbb{Z}\right\\},$

we write

$\left(t_{j}^{(n)},x_{k_{1}}^{(n)},...,x_{k_{d}}^{(n)}\right)=\left(\frac{j}{2^{2n}},\frac{k_{1}}{2^{n}},...,\frac{k_{d}}{2^{n}}\right).$

Two points
$\left(t_{j}^{(n)},x_{k_{1}}^{(n)},...,x_{k_{d}}^{(n)}\right),\left(t_{j^{\prime}}^{(n)},x_{k^{\prime}_{1}}^{(n)},...,x_{k^{\prime}_{d}}^{(n)}\right)$
are called nearest neighbors if either

  1. 1.

$j=j^{\prime},|k_{i}-k^{\prime}_{i}|=1\text{~{}for only
one~{}}i\text{~{}and~{}}k_{l}=k^{\prime}_{l}\text{~{}for the other
indices~{}}l,or$

  2. 2.

$|j-j^{\prime}|=1~{}\text{and}~{}k_{i}=k^{\prime}_{i}~{}\forall i.$

The following lemma generalizes the Lemma 3.4 in [AJM21], which plays a key
role in estimating the small ball probability.

######  Lemma 4.5.

There exist constants $C_{5},C_{6}>0$ depending on $\alpha,\beta,d$ and
$\mathcal{C}_{2}$ in (1.4) such that for all $\gamma,\kappa,\varepsilon>0$ and
$\gamma\varepsilon^{4}\leq 1$, we have

$P\left(\sup_{\begin{subarray}{c}0\leq t\leq\gamma\varepsilon^{4}\\\ {\bf
x}\in[0,\varepsilon^{2}]^{d}\end{subarray}}|N(t,{\bf
x})|>\kappa\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right)\leq\frac{C_{5}}{1\wedge\sqrt{\gamma^{d}}}\exp\left(-\frac{C_{6}\kappa^{2}}{\mathcal{C}_{2}^{2}\gamma^{\frac{\alpha-\beta}{\alpha}}}\right).$

###### Proof.

Fix $\gamma\geq 1$, and consider the grid

$\mathbb{G}_{n}=\left\\{\left(\frac{j}{2^{2n}},\frac{k_{1}}{2^{n}},...,\frac{k_{d}}{2^{n}}\right):0\leq
j\leq\gamma\varepsilon^{4}2^{2n},0\leq
k_{1},...,k_{d}\leq\varepsilon^{2}2^{n},j,k_{1},...,k_{d}\in\mathbb{Z}\right\\}.$

Let

(4.16)
$n_{0}=\left\lceil\log_{2}\left(\gamma^{-1/2}\varepsilon^{-2}\right)\right\rceil,$

and for $n<n_{0}$, $\mathbb{G}_{n}$ contains only the origin. For $n\geq
n_{0}$, the grid $\mathbb{G}_{n}$ has at most
$\left(\gamma\varepsilon^{4}2^{2n}+1\right)\cdot\left(\varepsilon^{2}2^{n}+1\right)^{d}\leq
2^{d+1+(2+d)n}\varepsilon^{2d+4}\gamma\leq 2^{2d+3}2^{(2+d)(n-n_{0})}$ many
points. We will choose two parameters
$0<\delta_{1}(\alpha,\beta)<\delta_{0}(\alpha,\beta)<\frac{\alpha-\beta}{\alpha}$
satisfying the following constraint

(4.17)
$2\zeta\wedge\alpha\xi=\frac{2(\alpha-\beta)}{\alpha}+2\delta_{1}-2\delta_{0},$

where
$\xi\in\left(0,\frac{1}{\alpha}\wedge\frac{\alpha-\beta}{\alpha}\right),\zeta\in\left(0,\frac{\alpha-\beta}{\alpha}\right)$.
Fix the constant

$\mathcal{M}=\frac{1-2^{-\delta_{1}}}{(3+d)2^{(\delta_{0}-\delta_{1})n_{0}}},$

and consider the event

$A(n,\kappa)=\left\\{|N(p)-N(q)|\leq\kappa\mathcal{M}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}2^{-\delta_{1}n}2^{\delta_{0}n_{0}}\text{
for all $p,q\in\mathbb{G}_{n}$ nearest neighbors}\right\\}.$

If $p,q\in\mathbb{G}_{n}$ are the case 1 nearest neighbors in the Definition
4.1, (4.14) implies

$P\left(|N(p)-N(q)|>\kappa\mathcal{M}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}2^{-\delta_{1}n}2^{\delta_{0}n_{0}}\right)\leq
C_{1}\exp\left(-\frac{C_{2}\kappa^{2}\mathcal{M}^{2}\varepsilon^{\frac{4(\alpha-\beta)}{\alpha}}}{2^{-n\alpha\xi}\mathcal{C}_{2}^{2}}2^{-2\delta_{1}n}2^{2\delta_{0}n_{0}}\right).$

If $p,q\in\mathbb{G}_{n}$ are the case 2 nearest neighbors in the Definition
4.1, (4.15) implies

$P\left(|N(p)-N(q)|>\kappa\mathcal{M}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}2^{-\delta_{1}n}2^{\delta_{0}n_{0}}\right)\leq
C_{3}\exp\left(-\frac{C_{4}\kappa^{2}\mathcal{M}^{2}\varepsilon^{\frac{4(\alpha-\beta)}{\alpha}}}{2^{-2n\zeta}\mathcal{C}_{2}^{2}}2^{-2\delta_{1}n}2^{2\delta_{0}n_{0}}\right).$

Therefore, a union bound gives

$\displaystyle
P(A^{c}(n,\kappa))\leq\sum_{\begin{subarray}{c}p,q\in\mathbb{G}_{n}\\\
\text{nearest
neighbors}\end{subarray}}P\left(|N(p)-N(q)|>\kappa\mathcal{M}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}2^{-\delta_{1}n}2^{\delta_{0}n_{0}}\right)$
$\displaystyle\leq
C2^{(2+d)(n-n_{0})}\exp\left(-\frac{C^{\prime}\kappa^{2}\mathcal{M}^{2}\varepsilon^{\frac{4(\alpha-\beta)}{\alpha}}}{\mathcal{C}_{2}^{2}}2^{n(2\zeta\wedge\alpha\xi)}2^{-2\delta_{1}n}2^{2\delta_{0}n_{0}}\right)$
$\displaystyle=C2^{(2+d)(n-n_{0})}\exp\left(-\frac{C^{\prime}\kappa^{2}\mathcal{M}^{2}}{\mathcal{C}_{2}^{2}}\left(\varepsilon^{\frac{4(\alpha-\beta)}{\alpha}}2^{\frac{2n_{0}(\alpha-\beta)}{\alpha}}\right)2^{n(2\zeta\wedge\alpha\xi)}2^{-2\delta_{1}n}2^{2\delta_{0}n_{0}}\right)$
$\displaystyle\leq
C2^{(2+d)(n-n_{0})}\exp\left(-\frac{C^{\prime}\kappa^{2}\mathcal{M}^{2}}{\mathcal{C}_{2}^{2}\gamma^{\frac{(\alpha-\beta)}{\alpha}}}2^{(2\zeta\wedge\alpha\xi-2\delta_{1})(n-n_{0})}\right),$

where $C,C^{\prime}$ are positive constants depending only on
$\alpha,\beta,T,d$. The last inequality follows from that
$\varepsilon^{\frac{4(\alpha-\beta)}{\alpha}}2^{\frac{2n_{0}(\alpha-\beta)}{\alpha}}\geq\gamma^{-\frac{(\alpha-\beta)}{\alpha}}$
by the definition of $n_{0}$ in (4.16), and our choice of
$\delta_{0},\delta_{1}$ in (4.17). Let $A(\kappa)=\bigcap\limits_{n\geq
n_{0}}A(n,\kappa)$ and we can bound $P(A^{c}(\kappa))$ by summing all
$P(A^{c}(n,\kappa))$,

$\displaystyle P\left(A^{c}(\kappa)\right)\leq\sum_{n\geq
n_{0}}P\left(A^{c}(n,\kappa)\right)$ $\displaystyle\leq\sum_{n\geq
n_{0}}C2^{(2+d)(n-n_{0})}\exp\left(-\frac{C^{\prime}\kappa^{2}\mathcal{M}^{2}}{\mathcal{C}_{2}^{2}\gamma^{\frac{(\alpha-\beta)}{\alpha}}}2^{(2\zeta\wedge\alpha\xi-2\delta_{1})(n-n_{0})}\right)$
$\displaystyle\leq
C_{5}\exp\left(-\frac{C_{6}\kappa^{2}\mathcal{M}^{2}}{\mathcal{C}_{2}^{2}\gamma^{\frac{(\alpha-\beta)}{\alpha}}}\right).$

Now we consider a point $(t,x)$, which is in a grid $\mathbb{G}_{n}$ for some
$n\geq n_{0}$. From arguments similar to page 128 of [DKM+09], we can find a
sequence of points from the origin to $(t,{\bf x})$ as $(0,{\bf
0})=p_{0},p_{1},...,p_{k}=(t,{\bf x})$ such that each pair is the nearest
neighbor in some grid $\mathbb{G}_{m},n_{0}\leq m\leq n$, and at most $(3+d)$
such pairs are nearest neighbors in any given grid. On the event $A(\kappa)$,
we have

$\displaystyle|N(t,{\bf
x})|\leq\sum_{j=0}^{k-1}|N(p_{j})-N(p_{j+1})|\leq(3+d)\sum_{n\geq
n_{0}}\kappa\mathcal{M}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}2^{-\delta_{1}n}2^{\delta_{0}n_{0}}\leq\kappa\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}.$

Points in $\mathbb{G}_{n}$ are dense in
$[0,\gamma\varepsilon^{4}]\times[0,\varepsilon^{2}]$, and we may extend
$N(t,{\bf x})$ to a continuous version. Therefore, for $\gamma\geq 1$,

$P\left(\sup_{\begin{subarray}{c}0\leq t\leq\gamma\varepsilon^{4}\\\ {\bf
x}\in[0,\varepsilon^{2}]^{d}\end{subarray}}|N(t,{\bf
x})|>\kappa\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right)\leq
C_{5}\exp\left(-\frac{C_{6}\kappa^{2}}{\mathcal{C}_{2}^{2}\gamma^{\frac{\alpha-\beta}{\alpha}}}\right).$

For $0<\gamma<1$, a union bound and stationarity in ${\bf x}$ imply that

$\sqrt{\gamma^{d}}P\left(\sup_{\begin{subarray}{c}0\leq
t\leq\gamma\varepsilon^{4}\\\ {\bf
x}\in[0,\varepsilon^{2}]^{d}\end{subarray}}|N(t,{\bf
x})|>\kappa\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right)\leq
P\left(\sup_{\begin{subarray}{c}0\leq t\leq\gamma\varepsilon^{4}\\\ {\bf
x}\in[0,\sqrt{\gamma}\varepsilon^{2}]^{d}\end{subarray}}|N(t,{\bf
x})|>\kappa\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right)$
$=P\left(\sup_{\begin{subarray}{c}0\leq
t\leq(\sqrt{\gamma}\varepsilon^{2})^{2}\\\ {\bf
x}\in[0,\sqrt{\gamma}\varepsilon^{2}]^{d}\end{subarray}}|N(t,{\bf
x})|>\frac{\kappa}{\gamma^{\frac{\alpha-\beta}{2\alpha}}}\left(\gamma^{1/4}\varepsilon\right)^{\frac{2(\alpha-\beta)}{\alpha}}\right)\leq
C_{5}\exp\left(-\frac{C_{6}\kappa^{2}}{\mathcal{C}_{2}^{2}\gamma^{\frac{\alpha-\beta}{\alpha}}}\right).$

As a result,

$P\left(\sup_{\begin{subarray}{c}0\leq t\leq\gamma\varepsilon^{4}\\\ {\bf
x}\in[0,\varepsilon^{2}]^{d}\end{subarray}}|N(t,{\bf
x})|>\kappa\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right)\leq\frac{C_{5}}{1\wedge\sqrt{\gamma^{d}}}\exp\left(-\frac{C_{6}\kappa^{2}}{\mathcal{C}_{2}^{2}\gamma^{\frac{\alpha-\beta}{\alpha}}}\right).$

∎

######  Remark 4.1.

If we suppose $\sigma$ in (4.8) satisfies $|\sigma(s,{\bf y},u(s,{\bf
y}))|\leq C\left(\gamma\varepsilon^{4}\right)^{\frac{\alpha-\beta}{2\alpha}}$,
then the probability in Lemma 4.5 is bounded above by

$\frac{C_{5}}{1\wedge\sqrt{\gamma^{d}}}\exp\left(-\frac{C_{6}\kappa^{2}}{C^{2}(\gamma\varepsilon^{2})^{\frac{2(\alpha-\beta)}{\alpha}}}\right),$

which can be proved similarly to the above lemma.

##  5\. Proof of Proposition 3.1

The following lemma gives a lower bound for variance of the noise term
$N(t_{1},{\bf x})$ and an upper bound on the decay of covariance between two
random variables $N(t_{1},{\bf x})$, $N(t_{1},{\bf y})$ as $|{\bf x}-{\bf y}|$
increases.

######  Lemma 5.1.

Consider noise terms $N(t_{1},{\bf x})$, $N(t_{1},{\bf y})$ with a
deterministic $\sigma(t,{\bf x},u)=\sigma(t,{\bf x})$, then there exist
constants $C_{7},C_{8}>0$ depending only on $\mathcal{C}_{1}$,
$\mathcal{C}_{2}$, $d$, $\alpha$, and $\beta$ such that

$C_{7}t_{1}^{\frac{\alpha-\beta}{\alpha}}\leq{\rm Var}[N(t_{1},{\bf x})],$
${\rm Cov}[N(t_{1},{\bf x}),N(t_{1},{\bf y})]\leq C_{8}t_{1}\left|{\bf x}-{\bf
y}\right|^{-\beta}.$

###### Proof.

We use the Fubini’s theorem, (1.8) the expression (1.2) and (4.1) to show that

$\displaystyle\textrm{Var}[N(t_{1},{\bf x})]$
$\displaystyle=\int_{0}^{t_{1}}\int_{\mathbb{T}^{d}}\int_{\mathbb{T}^{d}}\bar{p}(t_{1}-s,{\bf
x}-{\bf y})\bar{p}(t_{1}-s,{\bf x}-{\bf z})\sigma(s,{\bf y})\sigma(s,{\bf
z})\Lambda({\bf y}-{\bf z})d{\bf y}d{\bf z}ds$ $\displaystyle\geq
C(d)\mathcal{C}_{1}^{2}\left(\sum_{{\bf n}\in\mathbb{Z}^{d}}\lambda({\bf
n})\int_{0}^{t_{1}}e^{-2\pi^{\alpha}|{\bf n}|^{\alpha}(t_{1}-s)}ds\right)$
$\displaystyle=C(d)\mathcal{C}_{1}^{2}\left(\lambda({\bf 0})t_{1}+\sum_{{\bf
n}\in\mathbb{Z}^{d},{\bf n}\neq{\bf 0}}\lambda({\bf
n})\frac{1-e^{-2\pi^{\alpha}|{\bf n}|^{\alpha}t_{1}}}{2\pi^{\alpha}|{\bf
n}|^{\alpha}}\right)$ $\displaystyle\geq
C(d,\mathcal{C}_{1})\int_{1}^{\infty}\frac{1-e^{-2\pi^{\alpha}x^{\alpha}t_{1}}}{2\pi^{\alpha}x^{d+\alpha-\beta}}x^{d-1}dx.$

The last inequality follows from that $\int_{0}^{t_{1}}e^{-2\pi^{\alpha}|{\bf
n}|^{\alpha}(t_{1}-s)}ds$ decreases as $|{\bf n}|$ increases. Changing
variable to $w=2\pi^{\alpha}x^{\alpha}t_{1}$ yields

$\int_{1}^{\infty}\frac{1-e^{-2\pi^{\alpha}x^{\alpha}t_{1}}}{2\pi^{\alpha}x^{\alpha-\beta+1}}dx=C(\alpha,\beta,d)t_{1}^{\frac{\alpha-\beta}{\alpha}}\int_{2\pi^{\alpha}t_{1}}^{\infty}\frac{1-e^{-w}}{w^{2-\beta/\alpha}}dw\geq
C(\alpha,\beta,d)t_{1}^{\frac{\alpha-\beta}{\alpha}}\int_{2\pi^{\alpha}}^{\infty}\frac{1-e^{-w}}{w^{2-\beta/\alpha}}dw.$

The last integral converges with $0<\beta<\alpha\wedge d$, which completes the
proof of the first part. In addition, we use the definition of $\Lambda(x)$ in
(1.2) and the fact $1-e^{-x}\leq x$ to derive the upper bound of covariance
between $N(t_{1},{\bf x})$ and $N(t_{1},{\bf y})$ when ${\bf x}\neq{\bf y}$,

$\displaystyle\textrm{Cov}[N(t_{1},{\bf x}),N(t_{1},{\bf
y})]=\mathbb{E}[N(t_{1},{\bf x})N(t_{1},{\bf y})]$ $\displaystyle\leq
C(d)\mathcal{C}_{2}^{2}\left(\sum_{{\bf n}\in\mathbb{Z}^{d}}\lambda({\bf
n})\exp(\pi i{\bf n}\cdot({\bf x}-{\bf
y}))\int_{0}^{t_{1}}e^{-2\pi^{\alpha}|{\bf n}|^{\alpha}(t_{1}-s)}ds\right)$
$\displaystyle\leq C_{8}t_{1}\sum_{{\bf n}\in\mathbb{Z}^{d}}\lambda({\bf
n})\exp(\pi i{\bf n}\cdot({\bf x}-{\bf y}))=C_{8}t_{1}\left|{\bf x}-{\bf
y}\right|^{-\beta}.$

∎

Proof of Proposition 3.1(a) The Markov property of $u(t,\cdot)$ (see page 247
in [DPZ14]) implies

$P\left(F_{j}|\sigma\\{u(t_{i},\cdot)\\}_{0\leq
i<j}\right)=P\left(F_{j}|u(t_{j-1},\cdot)\right).$

If we can prove that $P\left(F_{j}|u(t_{j-1},\cdot)\right)$ has a uniform
bound ${\bf C_{4}}\exp\left(-\frac{{\bf
C_{5}}}{\varepsilon^{2}+\mathcal{D}^{2}t_{1}^{\frac{\alpha-\beta}{\alpha}}}\right)$,
then it is still a bound for the conditional probability
$P\left(F_{j}|\bigcap_{k=0}^{j-1}F_{k}\right)$, which is conditioned on a
realization of $u(t_{k},\cdot),0\leq k<j$. Thus, it is enough to show that

$P\left(F_{1}\right)\leq{\bf C_{4}}\exp\left(-\frac{{\bf
C_{5}}}{\varepsilon^{2}+\mathcal{D}^{2}t_{1}^{\frac{\alpha-\beta}{\alpha}}}\right),$

where ${\bf C_{4}}$, ${\bf C_{5}}$ do not depend on $u_{0}$. Consider the
truncated function

$f_{\varepsilon}({\bf x})=\begin{cases}{\bf x}&|{\bf x}|\leq
t_{1}^{\frac{\alpha-\beta}{2\alpha}}\\\ \frac{{\bf x}}{|{\bf x}|}\cdot
t_{1}^{\frac{\alpha-\beta}{2\alpha}}&|{\bf
x}|>t_{1}^{\frac{\alpha-\beta}{2\alpha}}\end{cases},$

and, particularly, we have $|f_{\varepsilon}({\bf x})|\leq
t_{1}^{\frac{\alpha-\beta}{2\alpha}}$. Consider the following two equations

$\partial_{t}v(t,{\bf x})=-(-\Delta)^{\alpha/2}v(t,{\bf x})+\sigma(t,{\bf
x},f_{\varepsilon}(v(t,{\bf x})))\dot{F}(t,{\bf x}),$

and

$\partial_{t}v_{g}(t,{\bf x})=-(-\Delta)^{\alpha/2}v_{g}(t,{\bf
x})+\sigma(t,{\bf x},f_{\varepsilon}(u_{0}({\bf x})))\dot{F}(t,{\bf x})$

with the same initial $u_{0}({\bf x})$. We can decompose $v(t,{\bf x})$ by

$v(t,{\bf x})=v_{g}(t,{\bf x})+D(t,{\bf x})$

with

$D(t,{\bf x})=\int_{[0,t]\times\mathbb{T}^{d}}\bar{p}(t-s,{\bf x}-{\bf
y})[\sigma(s,{\bf y},f_{\varepsilon}(v(s,{\bf y})))-\sigma(s,{\bf
y},f_{\varepsilon}(u_{0}({\bf y})))]F(dsd{\bf y}).$

The Lipschitz property on the third variable of $\sigma(t,{\bf x},u)$ in (1.3)
gives

(5.1) $\begin{split}|\sigma(s,{\bf y},f_{\varepsilon}(v(s,{\bf
y})))-\sigma(s,{\bf y},f_{\varepsilon}(u_{0}({\bf
y})))|&\leq\mathcal{D}|f_{\varepsilon}(v(s,{\bf
y}))-f_{\varepsilon}(u_{0}({\bf y}))|\\\ &\leq
2\mathcal{D}t_{1}^{\frac{\alpha-\beta}{2\alpha}}.\end{split}$

Recall that $R_{i,j}$ in (3.3) and define a new sequence of events,

$H_{j}=\left\\{|v(t,{\bf x})|\leq
t_{1}^{\frac{\alpha-\beta}{2\alpha}},\forall(t,{\bf x})\in R_{1,j}\setminus
R_{1,j-1}\right\\}.$

Clearly, the property of $f_{\varepsilon}({\bf x})$ and (3.5) imply

$F_{1}=\bigcap_{j=-n_{1}+1}^{n_{1}-1}H_{j}.$

Also, define another two sequences of events

$A_{j}=\left\\{|v_{g}(t,{\bf x})|\leq
2t_{1}^{\frac{\alpha-\beta}{2\alpha}},\forall(t,{\bf x})\in R_{1,j}\setminus
R_{1,j-1}\right\\},$ $B_{j}=\left\\{|D(t,{\bf
x})|>t_{1}^{\frac{\alpha-\beta}{2\alpha}},\exists(t,{\bf x})\in
R_{1,j}\setminus R_{1,j-1}\right\\}.$

It is straightforward to check that

$H_{j}^{c}\supset A_{j}^{c}\cap B_{j}^{c},$

which implies

(5.2)
$\begin{split}P(F_{1})&=P\left(\bigcap_{j=-n_{1}+1}^{n_{1}-1}H_{j}\right)\leq
P\left(\bigcap_{j=-n_{1}+1}^{n_{1}-1}[A_{j}\cup B_{j}]\right)\\\ &\leq
P\left(\left(\bigcap_{j=-n_{1}+1}^{n_{1}-1}A_{j}\right)\bigcup\left(\bigcup_{j=-n_{1}+1}^{n_{1}-1}B_{j}\right)\right)\\\
&\leq
P\left(\bigcap_{j=-n_{1}+1}^{n_{1}-1}A_{j}\right)+P\left(\bigcup_{j=-n_{1}+1}^{n_{1}-1}B_{j}\right)\\\
&\leq
P\left(\bigcap_{j=-n_{1}+1}^{n_{1}-1}A_{j}\right)+\sum_{j=-n_{1}+1}^{n_{1}-1}P(B_{j}).\end{split}$

The second inequality can be showed by using induction. Moreover, for
$j=-n_{1}+1$,

(5.3) $B_{j}\subseteq\left\\{\sup_{\begin{subarray}{c}0\leq s\leq
c_{0}\varepsilon^{4}\\\ {\bf
y}\in[(-n_{1}+1)\varepsilon^{2},(-n_{1}+2)\varepsilon^{2}]^{d}\end{subarray}}|D(s,{\bf
y})|>t_{1}^{\frac{\alpha-\beta}{2\alpha}}\right\\},$

and for $j>-n_{1}+1$,

(5.4) $\begin{split}B_{j}&\subseteq\left\\{\sup_{(t,{\bf x})\in
R_{1,j}\setminus R_{1,j-1}}|D(t,{\bf
x})|>t_{1}^{\frac{\alpha-\beta}{2\alpha}}\right\\}\\\
&\subseteq\bigcup_{(t,{\bf x})\in R_{1,j-1}\setminus
R_{1,j-2}}\left\\{\sup_{\begin{subarray}{c}0\leq s\leq c_{0}\varepsilon^{4}\\\
{\bf y}\in{\bf x}+[0,\varepsilon^{2}]^{d}\end{subarray}}|D(s,{\bf
y})|>t_{1}^{\frac{\alpha-\beta}{2\alpha}}\right\\}.\\\ \end{split}$

From (5.1) and Remark 4.1, we get

(5.5) $P\left(\sup_{\begin{subarray}{c}0\leq s\leq c_{0}\varepsilon^{4}\\\
{\bf y}\in{\bf x}+[0,\varepsilon^{2}]^{d}\end{subarray}}|D(s,{\bf
y})|>t_{1}^{\frac{\alpha-\beta}{2\alpha}}\right)\leq\frac{C_{5}}{1\wedge\sqrt{{c_{0}}^{d}}}\exp\left(-\frac{C_{6}}{4\mathcal{D}^{2}t_{1}^{\frac{\alpha-\beta}{\alpha}}}\right),$

where the proof does not rely on ${\bf x}$ since $u_{0}({\bf x})\equiv 0$.
Therefore, (5.3) implies

(5.6) $P(B_{-n_{1}+1})\leq P\left(\sup_{\begin{subarray}{c}0\leq s\leq
c_{0}\varepsilon^{4}\\\ {\bf
y}\in[0,\varepsilon^{2}]^{d}\end{subarray}}|D(s,{\bf
y})|>t_{1}^{\frac{\alpha-\beta}{2\alpha}}\right),$

and (5.4) implies, for $j>-n_{1}+1$,

(5.7)
$P(B_{j})\leq\left[(j+n_{1}-1)^{d}-(j+n_{1}-2)^{d}\right]P\left(\sup_{\begin{subarray}{c}0\leq
s\leq c_{0}\varepsilon^{4}\\\ {\bf
y}\in[0,\varepsilon^{2}]^{d}\end{subarray}}|D(s,{\bf
y})|>t_{1}^{\frac{\alpha-\beta}{2\alpha}}\right).$

Hence, using (5.5), (5.6) and (5.7), we conclude that

(5.8)
$\sum_{j=-n_{1}+1}^{n_{1}-1}P(B_{j})\leq\left((2n_{1}-2)^{d}+1\right)\leq\frac{C(d)}{\varepsilon^{2d}}\cdot\frac{C_{5}}{1\wedge\sqrt{{c_{0}}^{d}}}\exp\left(-\frac{C_{6}}{4\mathcal{D}^{2}t_{1}^{\frac{\alpha-\beta}{\alpha}}}\right).$

To compute the upper bound for
$P\left(\bigcap_{j=-n_{1}+1}^{n_{1}-1}A_{j}\right)$, we define a sequence of
events involving $v_{g}$,

$I_{j}=\left\\{|v_{g}(t,{\bf x})|\leq
2t_{1}^{\frac{\alpha-\beta}{2\alpha}},\forall(t,{\bf x})\in
R_{1,j}\right\\}\quad\text{and}\quad I_{-n_{1}}=\Omega.$

Then we can write $P\left(\bigcap_{j=-n_{1}+1}^{n_{1}-1}A_{j}\right)$ in terms
of conditional probability as

(5.9)
$P\left(\bigcap_{j=-n_{1}+1}^{n_{1}-1}A_{j}\right)=P(I_{n_{1}-1})=P(I_{-n_{1}})\prod_{j=-n_{1}+1}^{n_{1}-1}\frac{P(I_{j})}{P(I_{j-1})}=\prod_{j=-n_{1}+1}^{n_{1}-1}P(I_{j}|I_{j-1}).$

Let $\mathcal{G}_{j}$ be the $\sigma-$algebra generated by

$N_{\varepsilon}(t,{\bf x})=\int_{0}^{t}\int_{\mathbb{T}^{d}}p(t-s,{\bf
x}-{\bf y})\sigma(s,{\bf y},f_{\varepsilon}(u_{0}({\bf y})))F(d{\bf
y}ds),~{}~{}(t,{\bf x})\in R_{1,j}.$

If we can show that there is a uniform bound for
$P\left(I_{j}|\mathcal{G}_{j-1}\right)$, then it is still a bound for the
conditional probability $P\left(I_{j}|I_{j-1}\right)$. Notice that
$\sigma(s,{\bf y},f_{\varepsilon}(u_{0}({\bf y})))$ is deterministic and
uniformly bounded, then by Lemma 5.1, we have

(5.10) $\textrm{Var}[N_{\varepsilon}(t_{1},{\bf x})]\geq
C_{7}t_{1}^{\frac{\alpha-\beta}{\alpha}},$

and for $(t,{\bf x})\in R_{1,j}\setminus R_{1,j-1}$, one can decompose

(5.11) $v_{g}(t,{\bf x})=\int_{\mathbb{T}^{d}}p(t,{\bf x}-{\bf y})u_{0}({\bf
y})d{\bf y}+X+Y,$

where $X=\mathbb{E}[N_{\varepsilon}(t,{\bf x})|\mathcal{G}_{j-1}]$ is a
Gaussian random variable, which can be written as

(5.12) $X=\sum_{(t,{\bf x})\in R_{1,j-1}}\eta^{(j)}(t,{\bf
x})N_{\varepsilon}(t,{\bf x}),$

for some coefficients $\left(\eta^{(j)}(t,{\bf x})\right)_{(t,{\bf x})\in
R_{1,j-1}}$. Then the conditional variance equals

$\displaystyle\textrm{Var}(Y|\mathcal{G}_{j-1})=\mathbb{E}[(N_{\varepsilon}(t,{\bf
x})-X|\mathcal{G}_{j-1})^{2}]-(\mathbb{E}[N_{\varepsilon}(t,{\bf
x})-X|\mathcal{G}_{j-1}])^{2}$
$\displaystyle=\mathbb{E}[(N_{\varepsilon}(t,{\bf
x})-\mathbb{E}[N_{\varepsilon}(t,{\bf
x})|\mathcal{G}_{j-1}]|\mathcal{G}_{j-1})^{2}]=\textrm{Var}[N_{\varepsilon}(t,{\bf
x})|\mathcal{G}_{j-1}].$

Since $Y=N_{\varepsilon}(t,{\bf x})-X$ is independent of $\mathcal{G}_{j-1}$,
we write $\textrm{Var}(Y)$ as

$\textrm{Var}(Y)=\textrm{Var}(Y|\mathcal{G}_{j-1})=\textrm{Var}[N_{\varepsilon}(t,{\bf
x})|\mathcal{G}_{j-1}].$

In fact, for a Gaussian random variable $Z\sim N(\mu,\sigma^{2})$ and any
$a>0$, the probability $P(|Z|\leq a)$ is maximized when $\mu=0$, thus

$\begin{split}P\left(I_{j}|\mathcal{G}_{j-1}\right)&\leq P\left(|v_{g}(t,{\bf
x})|\leq 2t_{1}^{\frac{\alpha-\beta}{2\alpha}},(t,{\bf x})\in R_{1,j}\setminus
R_{1,j-1}\bigg{|}\mathcal{G}_{j-1}\right)\\\ &\leq
P\left(|Z^{\prime}|\leq\frac{2t_{1}^{\frac{\alpha-\beta}{2\alpha}}}{\sqrt{\textrm{Var}[N_{\varepsilon}(t,{\bf
x})|\mathcal{G}_{j-1}]}}\right)\end{split}$

where $Z^{\prime}\sim N(0,1)$. Let’s use the notation SD to denote the
standard deviation of a random variable. By the Minkowski inequality,

$\textrm{SD}(X)\leq\sum_{(t,{\bf x})\in R_{1,j-1}}\left|\eta^{(j)}(t,{\bf
x})\right|\cdot\textrm{SD}[N_{\varepsilon}(t,{\bf x})],$

and

$\textrm{SD}[N_{\varepsilon}(t,{\bf x})]\leq\textrm{SD}(X)+\textrm{SD}(Y).$

If we can control coefficients by restricting

$\sum_{(t,{\bf x})\in R_{1,j-1}}\left|\eta^{(j)}(t,{\bf
x})\right|\leq\frac{1}{2},$

then the standard deviation of $X$ is less than one half the standard
deviation of $N_{\varepsilon}(t,{\bf x})$,

$\textrm{SD}[N_{\varepsilon}(t,{\bf
x})]\leq\textrm{SD}(X)+\textrm{SD}(Y)\leq\frac{1}{2}\textrm{SD}[N_{\varepsilon}(t,{\bf
x})]+\textrm{SD}(Y).$

From (5.10), $\textrm{Var}(Y)$ is bounded below by
$C_{7}t_{1}^{\frac{\alpha-\beta}{\alpha}}$, so that we can derive the uniform
upper bound of $P(I_{j}|\mathcal{G}_{j-1})$,

$\begin{split}P(I_{j}|\mathcal{G}_{j-1})&\leq
P\left(|Z^{\prime}|\leq\frac{2t_{1}^{\frac{\alpha-\beta}{2\alpha}}}{\sqrt{\textrm{Var}[N_{\varepsilon}(t,{\bf
x})|\mathcal{G}_{j-1}]}}\right)\\\ &\leq
P\left(|Z^{\prime}|\leq\frac{2t_{1}^{\frac{\alpha-\beta}{2\alpha}}}{\sqrt{C_{7}t_{1}^{\frac{\alpha-\beta}{\alpha}}}}\right)\\\
&=P(|Z^{\prime}|\leq C^{\prime})<1,\end{split}$

where $C^{\prime}$ depends only on $\mathcal{C}_{1}$, $d$, $\alpha$, and
$\beta$. A bound (3.4) on $j$ and (5.9) together yield

(5.13) $P\left(\bigcap_{j=-n_{1}+1}^{n_{1}-1}A_{j}\right)\leq
C^{2\varepsilon^{-2}}=C\exp\left(-\frac{C^{\prime}}{\varepsilon^{2}}\right),$

where $C,C^{\prime}$ depends only on $\mathcal{C}_{1}$, $d$, $\alpha$, and
$\beta$. The following lemma shows how to select $c_{0}$ to make
$\sum\limits_{(t,{\bf x})\in R_{1,j-1}}|\eta^{(j)}(t,{\bf
x})|\leq\frac{1}{2}$, which completes the proof.

######  Lemma 5.2.

For a given $\varepsilon>0$, we may choose $c_{0}>0$ in (3.2) such that

$\sum_{(t,{\bf x})\in R_{1,j-1}}|\eta^{(j)}(t,{\bf x})|\leq\frac{1}{2}.$

###### Proof.

Let $X$ and $Y$ be random variables defined in (5.11) and (5.12). Since $Y$
and $\mathcal{G}_{j-1}$ are independent, for $\forall(t,{\bf x})\in
R_{1,j-1}$,

$\textrm{Cov}[Y,N_{\varepsilon}(t,{\bf x})]=0$

and for $(t,{\bf y})\in R_{1,j}\setminus R_{1,j-1}$, we have

(5.14) $\textrm{Cov}[N_{\varepsilon}(t,{\bf x}),N_{\varepsilon}(t,{\bf
y})]=\textrm{Cov}[N_{\varepsilon}(t,{\bf x}),X]=\sum_{(t,{\bf x^{\prime}})\in
R_{1,j-1}}\eta^{(j)}(t,{\bf x^{\prime}})\textrm{Cov}[N_{\varepsilon}(t,{\bf
x}),N_{\varepsilon}(t,{\bf x^{\prime}})].$

We write the equation (5.14) in a matrix form as

(5.15) ${\bf X}={\bf\Sigma}{\bf\eta},$

where the vector ${\bf\eta}=\left(\eta^{(j)}(t,{\bf x})\right)_{(t,{\bf x})\in
R_{1,j-1}}^{T}$, the vector ${\bf
X}=\left\\{\textrm{Cov}[N_{\varepsilon}(t,{\bf x}),N_{\varepsilon}(t,{\bf
y})]\right\\}_{(t,{\bf x})\in R_{1,j-1}}^{T},$ and ${\bf\Sigma}$ is the
covariance matrix of $\left(N_{\varepsilon}(t,{\bf x})\right)_{(t,{\bf x})\in
R_{1,j-1}}$. Let $||\cdot||_{1,1}$ be the matrix norm induced by the
$||\cdot||_{l_{1}}$ norm, that is for a matrix ${\bf A}$,

$||{\bf A}||_{1,1}:=\sup_{{\bf x\neq 0}}\frac{||{\bf Ax}||_{l_{1}}}{||{\bf
x}||_{l_{1}}}.$

It can be shown that $||{\bf
A}||_{1,1}=\max\limits_{j}\sum\limits_{i=1}^{n}|a_{ij}|$ (see page 259 of
[RB00]). Therefore, we have

$||{\bf\eta}||_{l_{1}}=||{\bf\Sigma}^{-1}{\bf
X}||_{l_{1}}\leq||{\bf\Sigma}^{-1}||_{1,1}||{\bf X}||_{l_{1}}.$

We rewrite ${\bf\Sigma}={\bf D}{\bf T}{\bf D}$, where ${\bf D}$ is a diagonal
matrix with diagonal entries $\sqrt{\textrm{Var}[N_{\varepsilon}(t,{\bf
x})]}$, and ${\bf T}$ is the correlation matrix with entries

$e_{{\bf xx^{\prime}}}=\frac{\textrm{Cov}[N_{\varepsilon}(t,{\bf
x}),N_{\varepsilon}(t,{\bf
x^{\prime}})]}{\sqrt{\textrm{Var}[N_{\varepsilon}(t,{\bf
x})]}\cdot\sqrt{\textrm{Var}[N_{\varepsilon}(t,{\bf x^{\prime}})]}}.$

Thanks to Lemma 5.1, for ${\bf x}\neq{\bf x^{\prime}}$, $|e_{{\bf
xx^{\prime}}}|$ can be bounded above by

$|e_{{\bf xx^{\prime}}}|\leq\frac{C_{8}t_{1}\left|{\bf x}-{\bf
x^{\prime}}\right|^{-\beta}}{C_{7}t_{1}^{1-\beta/\alpha}}.$

Define ${\bf A}={\bf I-T}$. Because ${\bf A}$ has zero diagonal entries, we
can bound $||{\bf A}||_{1,1}$ by

$\displaystyle||{\bf A}||_{1,1}$ $\displaystyle=\max_{{\bf x}}\sum_{{\bf
x}\neq{\bf x^{\prime}}}|t_{{\bf xx^{\prime}}}|\leq\sum_{(t,{\bf x})\in
R_{1,n_{1}-1}}|e_{{\bf
0x}}|=\frac{C_{8}t_{1}^{\beta/\alpha}}{C_{7}}\sum_{(t,{\bf x})\in
R_{1,n_{1}-1}}|{\bf x}|^{-\beta}$
$\displaystyle\leq\frac{C(d)C_{8}t_{1}^{\beta/\alpha}}{C_{7}\varepsilon^{2\beta}}\int_{0}^{\sqrt{d}\varepsilon^{-2}}r^{d-\beta-1}dr=\frac{C(d,\beta)C_{8}}{C_{7}}\cdot\frac{(c_{0}\varepsilon^{4})^{\beta/\alpha}}{\varepsilon^{2d}}.$

For any $\varepsilon>0$, we denote
$\mathcal{C}=\left(\frac{C_{7}}{3C(d,\beta)C_{8}}\right)^{\alpha/\beta}$ and
choose $c_{0}<\mathcal{C}\varepsilon^{\frac{2\alpha d-4\beta}{\beta}}$ in
(3.2), which makes $||{\bf A}||_{1,1}<\frac{1}{3}.$ Therefore, summing the
geometric series gives that

$||{\bf T}^{-1}||_{1,1}=||{\bf(I-A)}^{-1}||_{1,1}\leq\frac{1}{1-||{\bf
A}||_{1,1}}\leq\frac{3}{2},$

and $||{\bf\Sigma}^{-1}||_{1,1}\leq||{\bf D}^{-1}||_{1,1}\cdot||{\bf
T}^{-1}||_{1,1}\cdot||{\bf
D}^{-1}||_{1,1}\leq\frac{3}{2}C_{7}^{-1}t_{1}^{-\frac{\alpha-\beta}{\alpha}}$.
Substituting the bounds into (5.15) and choosing $c_{0}$ as in (3.2), we
obtain

$||{\bf\eta}||_{l_{1}}\leq\frac{3}{2}C_{7}^{-1}t_{1}^{-\frac{\alpha-\beta}{\alpha}}||{\bf
X}||_{l_{1}}<\frac{3}{2}\cdot\frac{1}{3}=\frac{1}{2}.$

∎

Combining (5.2), (5.8) and (5.13) yields

$\begin{split}P(F_{1})&\leq\frac{C(d)C_{5}}{(1\wedge\sqrt{{c_{0}}^{d}})\varepsilon^{2d}}\exp\left(-\frac{C_{6}}{4\mathcal{D}^{2}t_{1}^{\frac{\alpha-\beta}{\alpha}}}\right)+C\exp\left(-\frac{C^{\prime}}{\varepsilon^{2}}\right)\\\
&\leq C_{5}^{\prime}\exp\left(-\frac{d}{2}\ln
t_{1}-\frac{C_{6}}{4\mathcal{D}^{2}t_{1}^{\frac{\alpha-\beta}{\alpha}}}\right)+C\exp\left(-\frac{C^{\prime}}{\varepsilon^{2}}\right)\end{split}$

We choose a $\mathcal{D}_{0}$ depending only on $\alpha,\beta$ and $d$ such
that for any $\mathcal{D}<\mathcal{D}_{0}$,

$\displaystyle P(F_{1})$ $\displaystyle\leq
C_{5}^{\prime}\exp\left(-\frac{C_{6}^{\prime}}{\mathcal{D}^{2}t_{1}^{\frac{\alpha-\beta}{\alpha}}}\right)+C\exp\left(-\frac{C^{\prime}}{\varepsilon^{2}}\right)$
$\displaystyle\leq{\bf C_{4}}\exp\left(-\frac{{\bf
C_{5}}}{\varepsilon^{2}+\mathcal{D}^{2}t_{1}^{\frac{\alpha-\beta}{\alpha}}}\right).$

which completes the proof of Proposition 3.1 (a).

Proof of Proposition 3.1 (b) We first state the Gaussian correlation
inequality, which is crucial to proof Proposition 3.1 (b).

######  Lemma 5.3.

For any convex symmetric sets $K,L$ in $\mathbb{R}^{d}$ and any centered
Gaussian measure $\mu$ on $\mathbb{R}^{d}$, we have

$\mu(K\cap L)\geq\mu(K)\mu(L).$

###### Proof.

See in paper [Roy14], [LM17]. ∎

By the Markov property of $u(t,\cdot)$, the behavior of $u(t,\cdot)$ in the
interval $[t_{n},t_{n+1}]$ depends only on $u(t_{n},\cdot)$ and
$\dot{F}(t,{\bf x})$ on $[t_{n},t]\times[-1,1]^{d}$. Therefore, it is enough
to show that

$P\left(E_{0}\right)\geq{\bf C_{6}}\exp\left(-\frac{{\bf
C_{7}}}{\varepsilon^{\frac{2d(\alpha d-\beta)}{\beta}}}\right),$

where ${\bf C_{6}}$, ${\bf C_{7}}$ do not depend on $u_{0}$ and
$|u_{0}(x)|\leq\frac{1}{3}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}$. Now
we are ready to compute the lower bound for the small ball probability with a
smooth and deterministic $\sigma(s,y,u)=\sigma(s,y)$, which is a Gaussian
case. For $n\geq 0$, define a sequence of events

(5.16) $D_{n}=\left\\{|u(t_{n+1},{\bf
x})|\leq\frac{1}{6}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}},\text{and}~{}|u(t,{\bf
x})|\leq\frac{2}{3}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}},~{}\forall
t\in[t_{n},t_{n+1}],{\bf x}\in[-1,1]^{d}\right\\}.$

Denote

$\bar{p}_{t}(u_{0})({\bf x})=\bar{p}(t,\cdot)*u_{0}({\bf
x})=\int_{\mathbb{T}^{d}}\bar{p}(t,{\bf x-y})u_{0}({\bf y})d{\bf y},$

and we have

(5.17) $\bar{p}_{t}(u_{0})({\bf x})\leq\sup_{{\bf x}}|u_{0}({\bf
x})|\leq\frac{1}{3}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}.$

We consider the measure $Q$ given by

$\frac{dQ}{dP}=\exp\left(Z_{t_{1}}-\frac{1}{2}\langle Z\rangle_{t_{1}}\right)$

where

$Z_{t_{1}}=-\int_{[0,t_{1}]\times\mathbb{T}^{d}}f(s,{\bf y})F(dsd{\bf y}).$

If $Z_{t_{1}}$ satisfies Novikov’s condition in [All98], then

$\widetilde{F}(t,{\bf x}):=F(t,{\bf x})-\langle F(\cdot,{\bf x}),Z\rangle_{t}$

is a centered spatially homogeneous Wiener process under the measure $Q$ (see
[All98] for more details). Therefore, for ${\bf x}\in[-1,1]^{d}$, Fubini’s
Theorem with the covariance structure of $\dot{F}(t,{\bf x})$ gives

$\displaystyle\dot{\widetilde{F}}(t,{\bf x})=\dot{F}(t,{\bf
x})+\int_{\mathbb{T}^{d}}f(t,{\bf y})\Lambda({\bf x-y})d{\bf y},$

which is a colored noise under measure $Q$. Since
$\frac{\bar{p}_{t}(u_{0})({\bf x})}{t_{1}\sigma(t,{\bf x})}$ is smooth and
bounded function, and $\Lambda({\bf x})$ is the Riesz kernel on
$\mathbb{T}^{d}$, [RS16] shows that there is a continuous formula for the
fractional Laplacian of $\frac{\bar{p}_{t}(u_{0})({\bf x})}{t_{1}\sigma(t,{\bf
x})}$ on $\mathbb{T}^{d}$, so that one may assume that there is a function
$f(t,{\bf y})$ such that,

$\int_{\mathbb{T}^{d}}f(t,{\bf y})\Lambda({\bf x-y})d{\bf
y}=\frac{\bar{p}_{t}(u_{0})({\bf x})}{t_{1}\sigma(t,{\bf x})}.$

Moreover, $\Lambda({\bf x})\geq d^{-\beta/2}$ and (5.17) imply

(5.18)
$\begin{split}&\mathbb{E}\left[(Z_{t_{1}})^{2}\right]=\mathbb{E}\left[\int_{[0,t_{1}]\times\mathbb{T}^{d}}\int_{[0,t_{1}]\times\mathbb{T}^{d}}f(s,{\bf
y})f(t,{\bf z})F(dsd{\bf y})F(dtd{\bf z})\right]\\\
&=\int_{0}^{t_{1}}\int_{\mathbb{T}^{d}}\int_{\mathbb{T}^{d}}f(s,{\bf
y})f(s,{\bf z})\Lambda({\bf y-z})d{\bf y}d{\bf
z}ds=\int_{0}^{t_{1}}\int_{\mathbb{T}^{d}}f(s,{\bf
y})\frac{\bar{p}_{s}(u_{0})({\bf y})}{t_{1}\sigma(s,{\bf y})}d{\bf y}ds\\\
&\leq\int_{0}^{t_{1}}\int_{\mathbb{T}^{d}}f(s,{\bf
y})\left(\frac{1}{3}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right)\cdot\frac{1}{t_{1}\mathcal{C}_{1}}d{\bf
y}ds\leq\frac{C(d,\beta)\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}}{\mathcal{C}_{1}t_{1}}\int_{0}^{t_{1}}\int_{\mathbb{T}^{d}}f(s,{\bf
y})\Lambda({\bf 1-y})d{\bf y}ds\\\
&=\frac{C(d,\beta)\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}}{\mathcal{C}_{1}t_{1}}\int_{0}^{t_{1}}\frac{\bar{p}_{s}(u_{0})({\bf
1})}{t_{1}\sigma(s,{\bf 1})}ds\leq
C(d,\beta,\mathcal{C}_{1})\frac{\varepsilon^{\frac{4(\alpha-\beta)}{\alpha}}}{t_{1}}<\infty,\end{split}$

which satisfies Novikov’s condition with a deterministic $f$. Thus, we can
rewrite equation $(1.1)$ with deterministic $\sigma$ as

$\displaystyle u(t,{\bf x})$ $\displaystyle=\bar{p}_{t}(u_{0})({\bf
x})+\int_{[0,t]\times\mathbb{T}^{d}}\bar{p}(t-s,{\bf x-y})\sigma(s,{\bf
y})\left[\widetilde{F}(dsd{\bf y})-\frac{\bar{p}_{s}(u_{0})({\bf
y})}{t_{1}\sigma(s,{\bf y})}dsd{\bf y}\right]$
$\displaystyle=\bar{p}_{t}(u_{0})({\bf x})-\frac{t\bar{p}_{t}(u_{0})({\bf
x})}{t_{1}}+\int_{[0,t]\times\mathbb{T}^{d}}\bar{p}(t-s,{\bf
x-y})\sigma(s,{\bf y})\widetilde{F}(dsd{\bf y})$
$\displaystyle=\left(1-\frac{t}{t_{1}}\right)\bar{p}_{t}(u_{0})({\bf
x})+\int_{[0,t]\times\mathbb{T}^{d}}\bar{p}(t-s,{\bf x-y})\sigma(s,{\bf
y})\widetilde{F}(dsd{\bf y}).$

The first term is $0$ at $t_{1}$, and $|u_{0}({\bf
x})|\leq\frac{1}{3}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}$, we have

(5.19) $\left|\left(1-\frac{t}{t_{1}}\right)\bar{p}_{t}(u_{0})({\bf
x})\right|\leq\frac{1}{3}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}},{\bf
x}\in[-1,1]^{d},t<t_{1}.$

Define

$\widetilde{N}(t,{\bf x})=\int_{[0,t]\times\mathbb{T}^{d}}\bar{p}(t-s,{\bf
x-y})\sigma(s,{\bf y})\widetilde{F}(d{\bf y}ds),$

and suppose $c_{0}<1$, then applying Lemma 4.5 to $\widetilde{F}$ gives

$Q\left(\sup_{\begin{subarray}{c}0\leq t\leq c_{0}\varepsilon^{4}\\\ {\bf
x}\in[0,c_{0}\varepsilon^{2}]^{d}\end{subarray}}|\widetilde{N}(t,{\bf
x})|>\frac{1}{6}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right)\leq
C_{5}\exp\left(-\frac{C_{6}}{36\mathcal{C}_{2}^{2}c_{0}^{\frac{\alpha-\beta}{\alpha}}}\right),$

where $\gamma=c_{0}^{-1}>1$ and
$\kappa=\left(6c_{0}^{\frac{\alpha-\beta}{\alpha}}\right)^{-1}$. To make sure
that the right hand side is strictly less than 1, we require

$c_{0}<\min\left\\{1,\left(\frac{C_{6}}{36\mathcal{C}_{2}^{2}\ln
C_{5}}\right)^{\frac{\alpha}{\alpha-\beta}}\right\\},$

which is mentioned in (3.1). By the Gaussian correlation inequality in Lemma
5.3, we obtain

$\displaystyle Q\left(\sup_{\begin{subarray}{c}0\leq t\leq t_{1}\\\ {\bf
x}\in[-1,1]^{d}\end{subarray}}|\widetilde{N}(t,{\bf
x})|\leq\frac{1}{6}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right)$
$\displaystyle\geq Q\left(\sup_{\begin{subarray}{c}0\leq t\leq t_{1}\\\ {\bf
x}\in[0,c_{0}\varepsilon^{2}]^{d}\end{subarray}}|\widetilde{N}(t,{\bf
x})|\leq\frac{1}{6}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right)^{\left(\frac{2}{c_{0}\varepsilon^{2}}\right)^{d}}$
$\displaystyle\geq\left[1-C_{5}\exp\left(-\frac{C_{6}}{36\mathcal{C}_{2}^{2}c_{0}^{\frac{\alpha-\beta}{\alpha}}}\right)\right]^{\left(\frac{2}{c_{0}\varepsilon^{2}}\right)^{d}}.$

From (5.16) and (5.19), we get

$Q(D_{0})\geq Q\left(\sup_{\begin{subarray}{c}0\leq t\leq t_{1}\\\ {\bf
x}\in[-1,1]^{d}\end{subarray}}|\widetilde{N}(t,{\bf
x})|\leq\frac{1}{6}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right),$

and if we replace $f(s,y)$ with $2f(s,y)$ for $Z_{t_{1}}$,

(5.20)
$1=\mathbb{E}\left[\frac{dQ}{dP}\right]=\mathbb{E}\left[\exp\left(Z_{t_{1}}-\frac{1}{2}\langle
Z\rangle_{t_{1}}\right)\right]=\mathbb{E}[\exp\left(2Z_{t_{1}}-2\langle
Z\rangle_{t_{1}}\right)].$

Because $f(s,y)$ is deterministic, we may estimate the Radon-Nikodym
derivative,

$\displaystyle\mathbb{E}\left[\left(\frac{dQ}{dP}\right)^{2}\right]$
$\displaystyle=\mathbb{E}[\exp\left(2Z_{t_{1}}-\langle
Z\rangle_{t_{1}}\right)]=\mathbb{E}[\exp\left(2Z_{t_{1}}-2\langle
Z\rangle_{t_{1}}\right)\cdot\exp(\langle Z\rangle_{t_{1}})]$
$\displaystyle\leq\exp\left(C(d,\beta,\mathcal{C}_{1})\frac{\varepsilon^{\frac{4(\alpha-\beta)}{\alpha}}}{t_{1}}\right).$

The last inequality follows from (5.18) and (5.20). The Cauchy-Schwarz
inequality implies

$Q(D_{0})\leq\sqrt{\mathbb{E}\left[\left(\frac{dQ}{dP}\right)^{2}\right]}\cdot\sqrt{P(D_{0})},$

and as a consequence, we get

(5.21)
$P(D_{0})\geq\exp\left(-\frac{C\varepsilon^{\frac{4(\alpha-\beta)}{\alpha}}}{t_{1}}\right)\exp\left(\frac{C^{\prime}}{c_{0}^{d}\varepsilon^{2d}}\ln\left[1-C_{5}\exp\left(-\frac{C_{6}}{36\mathcal{C}_{2}^{2}c_{0}^{\frac{\alpha-\beta}{\alpha}}}\right)\right]\right),$

where $C,C^{\prime}$ depend only on $d,\beta,\mathcal{C}_{1}$. For the lower
bound with a non-deterministic $\sigma(t,{\bf x},u)$, we write

$u(t,{\bf x})=u_{g}(t,{\bf x})+D(t,{\bf x})$

where $u_{g}(t,{\bf x})$ satisfies the equation

$\partial_{t}u_{g}(t,{\bf x})=-(-\Delta)^{\alpha/2}u_{g}(t,{\bf
x})+\sigma(t,{\bf x},u_{0}({\bf x}))\dot{F}(t,{\bf x})$

and

$D(t,{\bf x})=\int_{[0,t]\times\mathbb{T}^{d}}\bar{p}(t-s,{\bf
x-y})[\sigma(s,{\bf y},u(s,{\bf y}))-\sigma(s,{\bf y},u_{0}({\bf
y}))]F(dsd{\bf y})$

with an initial profile $u_{0}$. Since $u_{g}$ is Gaussian, for an event
defined as

$\widetilde{D}_{0}=\left\\{|u_{g}(t_{1},{\bf
x})|\leq\frac{1}{6}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}},\text{and}~{}|u_{g}(t,{\bf
x})|\leq\frac{2}{3}\varepsilon^{\frac{2(\alpha-\beta)}{4}}~{}\forall
t\in[0,t_{1}],{\bf x}\in[-1,1]^{d}\right\\},$

we can apply (5.21) to it and get

(5.22)
$P(\widetilde{D}_{0})\geq\exp\left(-\frac{C\varepsilon^{\frac{4(\alpha-\beta)}{\alpha}}}{t_{1}}\right)\exp\left(\frac{C^{\prime}}{c_{0}^{d}\varepsilon^{2d}}\ln\left[1-C_{5}\exp\left(-\frac{C_{6}}{36\mathcal{C}_{2}^{2}c_{0}^{\frac{\alpha-\beta}{\alpha}}}\right)\right]\right).$

Define the stopping time

$\tau=\inf\left\\{t:|u(t,{\bf x})-u_{0}({\bf
x})|>2\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\text{~{}for some ${\bf
x}\in[-1,1]^{d}$}\right\\},$

and clearly we have $\tau>t_{1}$ on the event $E_{0}$ in (3.6) since
$|u_{0}(x)|\leq\frac{1}{3}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}$, and
$|u(t,x)|\leq\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}$ for $\forall
t\in[0,t_{1}]$ on the event $E_{0}$. We make another definition

$\widetilde{D}(t,x)=\int_{[0,t]\times\mathbb{T}^{d}}\bar{p}(t-s,{\bf
x-y})[\sigma(s,{\bf y},u(s\wedge\tau,{\bf y}))-\sigma(s,{\bf y},u_{0}({\bf
y}))]F(dsd{\bf y}),$

and $D(t,{\bf x})=\widetilde{D}(t,{\bf x})$ for $t\leq t_{1}$ on the event
$\\{\tau>t_{1}\\}$. Moreover, from (3.6), we have

(5.23) $\begin{split}P(E_{0})&\geq
P\left(\widetilde{D}_{0}\bigcap\left\\{\sup_{\begin{subarray}{c}0\leq t\leq
t_{1}\\\ {\bf x}\in[-1,1]^{d}\end{subarray}}|D(t,{\bf
x})|\leq\frac{1}{6}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right\\}\right)\\\
&=P\left(\left(\widetilde{D}_{0}\bigcap\left\\{\sup_{\begin{subarray}{c}0\leq
t\leq t_{1}\\\ {\bf x}\in[-1,1]^{d}\end{subarray}}|D(t,{\bf
x})|\leq\frac{1}{6}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right\\}\bigcap\\{\tau>t_{1}\\}\right)\right.\\\
&\hskip
85.35826pt\bigcup\left.\left(\widetilde{D}_{0}\bigcap\left\\{\sup_{\begin{subarray}{c}0\leq
t\leq t_{1}\\\ {\bf x}\in[-1,1]^{d}\end{subarray}}|D(t,{\bf
x})|\leq\frac{1}{6}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right\\}\bigcap\\{\tau\leq
t_{1}\\}\right)\right).\end{split}$

On the event $\\{\tau>t_{1}\\}$, we have

$\sup_{\begin{subarray}{c}0\leq t\leq t_{1}\\\ {\bf
x}\in[-1,1]^{d}\end{subarray}}|D(t,{\bf x})|=\sup_{\begin{subarray}{c}0\leq
t\leq t_{1}\\\ {\bf x}\in[-1,1]^{d}\end{subarray}}|\widetilde{D}(t,{\bf x})|,$

and on the event $\widetilde{D}_{0}\cap\\{\tau\leq t_{1}\\}$, we have, for
some ${\bf x}$,

$|u_{g}(\tau,{\bf
x})|\leq\frac{2}{3}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}},~{}|u(\tau,{\bf
x})-u_{0}({\bf
x})|>2\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}~{}\text{and}~{}|u_{0}({\bf
x})|\leq\frac{1}{3}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}.$

The above leads to

$\displaystyle\sup_{\begin{subarray}{c}{\bf x}\end{subarray}}|D(\tau,{\bf
x})|$ $\displaystyle=\sup_{\begin{subarray}{c}{\bf
x}\end{subarray}}|u(\tau,{\bf x})-u_{g}(\tau,{\bf
x})|\geq\sup_{\begin{subarray}{c}{\bf x}\end{subarray}}(|u(\tau,{\bf
x})|-|u_{g}(\tau,{\bf x})|)$ $\displaystyle\geq\sup_{\begin{subarray}{c}{\bf
x}\end{subarray}}|u(\tau,{\bf
x})|-\frac{2}{3}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\geq
2\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}-\frac{1}{3}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}-\frac{2}{3}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}$
$\displaystyle\geq\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}>\frac{1}{6}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}},$

which implies

$\widetilde{D}_{0}\cap\left\\{\sup_{\begin{subarray}{c}0\leq t\leq t_{1}\\\
{\bf x}\in[-1,1]^{d}\end{subarray}}|D(t,{\bf
x})|\leq\frac{1}{6}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right\\}\cap\\{\tau\leq
t_{1}\\}=\phi.$

Combining the above with (5.23) yields

(5.24) $\begin{split}P(E_{0})&\geq
P\left(\widetilde{D}_{0}\bigcap\left\\{\sup_{\begin{subarray}{c}0\leq t\leq
t_{1}\\\ {\bf x}\in[-1,1]^{d}\end{subarray}}|\widetilde{D}(t,{\bf
x})|\leq\frac{1}{6}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right\\}\bigcap\\{\tau>t_{1}\\}\right)\\\
&\geq P(\widetilde{D}_{0})-P\left(\sup_{\begin{subarray}{c}0\leq t\leq
t_{1}\\\ {\bf x}\in[-1,1]^{d}\end{subarray}}|\widetilde{D}(t,{\bf
x})|>\frac{1}{6}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right),\end{split}$

and $|u(t,{\bf x})-u_{0}({\bf x})|\leq
2\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}$ for all $t\in[0,t_{1}]$ and
${\bf x}\in[-1,1]^{d}$. We apply the Lipschitz property on the third variable
of $\sigma(t,{\bf x},u)$ in (1.3) to Remark 4.1 and use a union bound from
(5.8) to get

(5.25) $P\left(\sup_{\begin{subarray}{c}0\leq t\leq t_{1}\\\ {\bf
x}\in[-1,1]^{d}\end{subarray}}|\widetilde{D}(t,{\bf
x})|>\frac{1}{6}\varepsilon^{\frac{2(\alpha-\beta)}{\alpha}}\right)\leq\frac{C_{5}}{(c_{0}\varepsilon^{4})^{d/2}}\exp\left(-\frac{C_{6}}{144\mathcal{D}^{2}(c_{0}\varepsilon^{4})^{\frac{\alpha-\beta}{\alpha}}}\right).$

Consequently, from (5.22), (5.24) and (5.25), we conclude that,

(5.26)
$\begin{split}P(E_{0})&\geq\exp\left(-\frac{C\varepsilon^{\frac{4(\alpha-\beta)}{\alpha}}}{t_{1}}\right)\exp\left(\frac{C^{\prime}}{c_{0}^{d}\varepsilon^{2d}}\ln\left[1-C_{5}\exp\left(-\frac{C_{6}}{36\mathcal{C}_{2}^{2}c_{0}^{\frac{\alpha-\beta}{\alpha}}}\right)\right]\right)\\\
&\hskip
85.35826pt-\frac{C_{5}}{(c_{0}\varepsilon^{4})^{d/2}}\exp\left(-\frac{C_{6}}{144\mathcal{D}^{2}(c_{0}\varepsilon^{4})^{\frac{\alpha-\beta}{\alpha}}}\right)\\\
&=\exp\left(-\frac{C\varepsilon^{\frac{4(\alpha-\beta)}{\alpha}}}{t_{1}}+\frac{C^{\prime}}{c_{0}^{d}\varepsilon^{2d}}\ln\left[1-C_{5}\exp\left(-\frac{C_{6}}{36\mathcal{C}_{2}^{2}c_{0}^{\frac{\alpha-\beta}{\alpha}}}\right)\right]\right)\\\
&\hskip 85.35826pt-C_{5}\exp\left(-\frac{d}{2}\ln
t_{1}-\frac{C_{6}}{144\mathcal{D}^{2}t_{1}^{\frac{\alpha-\beta}{\alpha}}}\right).\end{split}$

When $d=1$ and $\alpha\geq 2\beta$, we may choose $c_{0}$ in (3.2) satisfying

(5.27)
$\mathcal{C}^{\prime}\varepsilon^{\frac{4\alpha-8\beta}{\beta}}<c_{0}<\mathcal{C}\varepsilon^{\frac{2\alpha-4\beta}{\beta}},$

where $0<\mathcal{C}^{\prime}<\mathcal{C}$ and $\varepsilon$ is small enough.
Then choose a $\mathcal{D}_{0}$ depending only on $\alpha,\beta$ and $d$ such
that for any $\mathcal{D}<\mathcal{D}_{0}$ , we have

$\displaystyle P(E_{0})$
$\displaystyle\geq\exp\left(-C\varepsilon^{\frac{-4(\alpha-\beta)^{2}}{\alpha\beta}}\right)-C_{5}^{\prime}\exp\left(-\frac{C_{6}^{\prime}}{\mathcal{D}^{2}t_{1}^{\frac{\alpha-\beta}{\alpha}}}\right)$
$\displaystyle\geq
C\exp\left(-C^{\prime}\varepsilon^{\frac{-4(\alpha-\beta)^{2}}{\alpha\beta}}\right)-C_{5}^{\prime}\exp\left(-\frac{C_{6}^{\prime}}{\mathcal{D}^{2}\varepsilon^{\frac{4(\alpha-\beta)^{2}}{\alpha\beta}}}\right)$
$\displaystyle\geq{\bf C_{6}}\exp\left(-\frac{{\bf
C_{7}}}{\varepsilon^{\frac{4(\alpha-\beta)^{2}}{\alpha\beta}}}\right).$

When $d=1$ and $\alpha<2\beta$, we may choose $c_{0}$ in (3.1). However, the
second term could exceed the first term in (5.26) for small enough
$\varepsilon$ and we may not achieve a lower bound for small probability for
any $0<\varepsilon<\varepsilon_{0}$. Similarly, for $d\geq 2$, the first term
decays exponentially from (5.22) and the second term grows exponentially from
(5.25), hence we cannot achieve a lower bound for small probability for any
$0<\varepsilon<\varepsilon_{0}$, which completes the proof.

##  6\. Acknowledgment

The author would like to thank his advisor Carl Mueller for useful discussions
and productive comments.

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